Unified
d Desig
gn of Steel
S
S
Structures, 3rd Ed
d.
Lo
ouis F. Geeschwind
dner
Forrmer Vice Prresident of Engineering
E
g and Researrch, Americaan Institute of Steel Connstruction;
Prof
ofessor Emerritus of Arch
hitectural En
ngineering, T
The Pennsylvvania State U
University; aand Senior
Con
nsultant, Pro
ovidence Eng
gineering Co
orporation
Judy Liu
Prof
ofessor, Scho
ool of Civil and
a Construcction Engineeering, Oreggon State University
Ch
harles J. Carter
C
President, Amerrican Institu
ute of Steel Construction
C
State Co
ollege, Pennsyllvania
Dedication
This book is dedicated to our mentors, teachers and former students, each of whom taught us something
and, in the process, contributed to who we are today.
Copyright © first and second editions 2008, 2012, John Wiley & Sons, Inc., third edition 2017, Louis F.
Geschwindner. All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording,
scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States
Copyright Act, without prior written permission of the copyright holder.
Anyone making use of the information presented in this publication assumes all liability arising from
such use.
Cover photo: Rutgers University School of Business, Piscataway, NJ
Photo courtesy of WSP Parsons Brinckerhoff
Printed by CreateSpace, an Amazon.com Company
Available from Amazon.com, CreateSpace.com, and other retail outlets
For additional information see www.SteelStuff.com
ISBN-13: 978-1543207521
ISBN-10: 1543207529
Library of Congress Control Number: 2017906318
CreateSpace Independent Publishing Platform, North Charleston, SC
Preface
INTENDED AUDIENCE
This book presents the basics of design of steel building structures, and is based on the 2016
unified specification, ANSI/AISC 360-16 Specification for Structural Steel Buildings. It is
intended primarily as a text for a first course in steel design for civil and architectural engineers.
Such a course usually occurs in the third or fourth year of an engineering program. The book can
also be used in a second, building-oriented course in steel design, depending on the coverage in
the first course. In addition to its use as a textbook, it provides a good review for practicing
engineers looking to learn the provisions of the latest specification or to convert their practice
from any of the old specifications to the new specification. Users are expected to have a firm
knowledge of statics and strength of materials and have easy access to the AISC Steel
Construction Manual, 15th Edition.
UNIFIED ASD AND LRFD
A preferred approach to the design of steel structures had been elusive over a 20 year period from
1986 to 2005. In 1986, the American Institute of Steel Construction (AISC) issued its first Load
and Resistance Factor Design (LRFD) Specification for Structural Steel Buildings. This
specification came after almost 50 years of publication of an Allowable Stress Design (ASD)
specification. Unfortunately, LRFD was accepted by the academic community but not by the
professional engineering community. Although AISC revised the format of the ASD specification
in 1989, it had not updated its provisions for over 25 years. This use of two specifications was
seen as an undesirable situation by the professions related to the building industry, and in 2001
AISC began the development of a combined ASD and LRFD specification. In 2005, AISC
published its first unified specification, combining the provisions of both the LRFD and ASD
specifications into a single standard for the design of steel building structures. That specification,
ANSI/AISC 360-05 Specification for Structural Steel Buildings, reflected a major change in
philosophy by AISC, one that made the use of ASD and LRFD equally acceptable as approaches
for the design of steel buildings. That has been the case for the past 11 years. Now, the newest
specification, ANSI/AISC 360-16 Specification for Structural Steel Buildings continues that
philosophy of equal status for design by ASD and LRFD.
The reader familiar with past editions of the separate ASD and LRFD specifications but
not with either the 2005 or 2010 editions will undoubtedly question how these two diverse design
philosophies could be effectively combined into one specification. This is a reasonable question
to ask. The primary answer is that the 2005 specification was not a combination of the old ASD
and LRFD provisions. It was a new approach with a new ASD and a new LRFD that used the best
of both previous approaches. The first unified specification took a different approach. It was
based on the understanding that the strength of an element or structure, called the nominal
strength in the specification, can be determined independent of the design philosophy. Once that
nominal strength is determined, the available strength for ASD or LRFD is determined as a
function of that nominal strength. Thus, the available strength of the element is always based on
the same behavior and no inconsistency in behavior results from the use of ASD or LRFD. This
important aspect of the unified specification is further explained in Chapter 1.
The 2016 specification shows AISC’s commitment to maintaining both design
approaches. With only one strength equation for both methods, any updates are applied equally to
the ASD and LRFD provisions. The designer continues to be given the opportunity to apply
engineering judgment in deciding what approach to use for design, either by ASD or LRFD.
iii
iv Preface
AISC has also issued ANSI/AISC 341-16 Seismic Provisions for Structural Steel Buildings, the
standard that guides the design of steel building structures to resist seismic loads.
CHANGES IN BUILDING LOADS
In addition to the provisions for steel design issued by AISC, structural engineering has seen
many changes in the area of loads for which buildings must be designed. The American Society
of Civil Engineers (ASCE) is continually revising ASCE 7 Minimum Design Loads for Buildings
and Other Structures, its standard for building loads. The International Code Council (ICC)
continues to issue its International Building Code (IBC) on a three year cycle, and it in turn is
adopted by a local jurisdiction on the schedule chosen by that jurisdiction. The IBC adopts ASCE
7, ANSI/AISC 360, ANSI/AISC 341, and other structural standards. The major changes for the
structural engineer brought about by these continually changing standards are the inclusion of
requirements for consideration of seismic loading, which applies to the entire country, and
revised wind and snow load requirements. For the calculation of loads within this text, ASCE 716 provisions are used. For any actual design, the designer must use the loadings established by
the governing building code. ANSI/AISC 341 is discussed in Chapter 13.
UNITS
ANSI/AISC 360-16 is, as much as possible, a unitless specification. In those rare instances where
equations could not be written in a unitless form, two equations are given, one in U.S. customary
units and one in SI units. The Manual presents all of its material in U.S. customary units. The
construction industry in this country has not adopted SI units in any visible way, and it is not
clear that they will in the foreseeable future. Thus, this book uses only U.S. customary units.
TOPICAL ORGANIZATION
Chapters 1 through 3 present the general material applicable to all steel structures. This is
followed in Chapters 4 through 9 with a presentation of member design. Chapters 10 through 12
discuss connections and Chapter 13 provides an introduction to seismic design.
In Chapter 1, the text addresses the principles of limit states design upon which all steel
design is based. It shows how these principles are incorporated into both LRFD and ASD
approaches and shows how reliability varies with live-to-dead load ratio for both approaches.
Chapter 1 also provides a description of ANSI/AISC 360-16, the Steel Construction Manual and
the AISC web site. These are the major source documents for all that is presented in this book.
Chapter 2 introduces the development of load factors, resistance factors, and safety factors. It
discusses load combinations and compares the calculation of required strength for both LRFD
and ASD. The calibration of ASD and LRFD and the resulting system reliability is also
addressed. Chapter 3 discusses steel as a structural material. It describes the availability of steel in
a variety of shapes and the grades of steel available for construction.
Once the foundation for steel design is established, the various member types are
considered. Tension members are addressed in Chapter 4, compression members in Chapter 5,
and bending members in Chapter 6. Chapter 7 covers plate girders, which are simply bending
members made from individual plates. Chapter 8 treats members subjected to combined axial
load and bending as well as design of bracing. Chapter 9 deals with composite members, that is,
members composed of both steel and concrete working together to provide the available strength.
Each of these chapters begins with a discussion of that particular member type and how it is used
in buildings. This is followed by a discussion of the specification provisions and the behavior
from which those provisions have been derived. The LRFD and ASD design philosophies of the
2016 specification are used throughout. Design examples that use the specification provisions
Preface v
directly are provided along with examples using the variety of design aids available in the AISC
Steel Construction Manual. All examples that have an LRFD and ASD component are provided
for both approaches. Throughout this book, ASD examples, or portions of examples that address
the ASD approach, are presented with shaded background for ease of identification.
The member-oriented chapters are followed by chapters addressing connection design.
Chapter 10 introduces the variety of potential connection types and discusses the strength of
bolts, welds, and connecting elements. Chapter 11 addresses simple connections. This includes
simple beam shear connections, light bracing connections and direct bearing connections. Chapter
12 deals with moment-resisting connections. As with the member-oriented chapters, the basic
principles of limit states design are developed first. This is followed by the application of the
provisions to simple shear connections and beam-to-column moment connections through
extensive examples in both LRFD and ASD.
The text concludes in Chapter 13 with an introduction to steel systems for seismic force
resistance. It discusses the variety of structural framing systems available and approved for
inclusion in the seismic force resisting system.
NEW TO THIS EDITION
This third edition is based on ANSI/AISC 360-16 Specification for Structural Steel Buildings, the
2016 edition of the AISC specification. The committee responsible for developing the
specification was charged by its chair to develop a new specification that would show minimal
changes. Thus, changes in the specification upon which this book is based are limited to those
areas where the committee felt important knowledge had been gained since the 2010 edition. As a
result, the changes in this book related to the specification provisions are also somewhat limited.
Other changes have been implemented here to better explain some topics and to expand others
while revising the entire book wherever needed to reflect the latest specification. Throughout, the
use of Specification and Manual equation numbers has been implemented to assist the reader in
navigating the Specification and Manual. Also, all examples have been revised to provide the
actual equations being used prior to entering the numbers for the calculations.
Example Problems for ANSI/AISC 360-16: Twenty new examples have been added and all
other examples have been updated to the new specification and the new table of member
properties, as well as being revised to improve clarity of intent.
New Homework Problems: Fifty-five homework problems have been added for a total of 489
and a large percentage of old problems have been revised to expand the opportunity for
instructors to provide meaningful activities for their students. Problems continue to be included
that carry over from one chapter to another so that an opportunity exists to link concepts of design
to one or two specific structures.
The following addresses the changes to be found in each chapter:
Chapter 1 includes an expanded discussion of structural integrity along with a discussion of the
timing of adoption of the new provisions into the International Building Code. The integrated
project introduced in this chapter for use throughout the book has been relocated to a new city
from the 2nd edition and the framing system modified. This will provide new homework options
for those who have implemented this project. A computer model using the RAM Structural
System will be available on the book website to support inclusion of the integrated project in
courses. Finally, an expanded discussion of reliability and statistics as it applies to structural steel
design has been included.
vi Preface
Chapter 2 provides an expanded discussion of snow, wind and seismic loads and additional
calculations for these environmental loads using ASCE 7.
Chapter 3 discusses the new steels approved by the 2016 Specification and the new approach
taken by ASTM to the specification of high strength bolts.
Chapter 4 addresses tension members. The provisions have not changed, but there has been a
revision in standard hole sizes for bolts. These new sizes have been implemented in the examples
where appropriate.
Chapter 5 looks at compression members and the Specification nomenclature change of KL to Lc
has been implemented. A section and an example have been added to address gravity-only
columns and their influence on the effective length of columns in lateral load resisting systems.
The completely new approach for treatment of columns with slender elements, introduced with
the 2016 Specification, is addressed. Single angle compression members and built-up
compression members are discussed and examples provided.
Chapter 6 on flexural members includes a discussion of the shape factor and its significance. The
use of Manual Table 3-10, the beam curves, with Cb not equal to 1.0, is expanded and a new
example is included to illustrate the use of Manual Table 3-2, the economy tables, for
noncompact beams. The treatment of tees, single angles and double angle beams has been
expanded and examples included. Determination of shear strength for wide-flange members when
the reduced resistance factor or increased safety factor must be used is now illustrated.
Chapter 7 addresses plate girders as doubly symmetric I-shapes built up from plates. It now
includes a discussion of these plate girders with compact webs. The completely revised treatment
of shear in plate girders included in the 2016 Specification has been incorporated, and the
corresponding stiffener design has been expanded.
Chapter 8 has an extensively expanded discussion of stability analysis and design including
comparisons between the direct analysis method, the effective length method and the first-order
analysis method. The discussion of beam column design has been revised to reflect the changes in
Manual Part 6, and an example of interaction of tension and bending is included.
Chapter 9 now includes a discussion of partial composite action, recognizing the concern
expressed in the Specification with low levels of partial composite action and the corresponding
ductility demand. Emphasis is given to the plastic neutral axis locations in Manual Table 3-19
and the corresponding percent of composite action. Composite column discussion is expanded to
include examples for composite HSS.
Chapter 10 has added discussions of the new ASTM bolt specifications and the new high
strength Group C bolts. A discussion of bolt installation has been included. The new
interpretation of the controlling bolt limit states is discussed, and examples have been revised to
reflect this new approach. Slip-critical bolts in combined shear and tension are now addressed.
The discussion of weld strength when loaded at an angle and strength at the base metal has been
expanded.
Chapter 11 has been revised to account for the new interpretation of the controlling bolt limit
states. Weld strength in the base metal has been included in all examples where applicable and
examples have been expanded to add clarity.
Preface vii
Chapter 12 continues to address moment connections and also reflects the changes in controlling
bolt limit states as applicable. Base metal weld strength is also treated more directly and overall
discussion within the examples has been expanded for clarity.
Chapter 13 continues to outline the application of the Seismic Provisions.
Throughout the book, new figures are included to better illustrate the corresponding material.
EXAMPLES AND HOMEWORK PROBLEMS IN LRFD AND ASD
The LRFD and ASD design philosophies of the 2016 specification are used throughout the book.
It is anticipated, however, that instructors and students will concentrate on only one design
approach. This will be the most effective use of time and resources except for situations where it
will be valuable to illustrate the differences and similarities of the LRFD and ASD philosophies.
ASD is presented in this book primarily for those who are looking to compare methods in order to
better understand the concepts of the unified specification and designers who have been
practicing with ASD and wish to move their practice to LRFD. LRFD should be the primary
approach used to teach structural steel design to new students.
LRFD and ASD in Examples
Design examples that use the specification provisions directly are provided along with examples
using the variety of design aids available in the AISC Steel Construction Manual. All examples
that have an LRFD and ASD component are provided for both approaches. Throughout this book,
ASD examples, or portions of examples that address the ASD approach, are presented with
shaded background for ease of identification.
EXAMPLE 4.10b
Tension Member
Design by ASD
SOLUTION
Goal:
Select a double-angle tension member for use as a web member in a truss
and determine the maximum area reduction that would be permitted for
holes and shear lag.
Given:
The member must carry a dead load of PD = 67.5 kips and a live load of
PL = 202.5 kips. For the load combination PD + PL, the ASD required
strength is Pa = 270 kips. Use equal leg angles of A36 steel.
Step 1:
Determine the minimum required gross area based on the limit state of
yielding where Ω = 1.67:
Ag
Step 2:
min
= Pa
( Fy
Ω ) = 270 ( 36 1.67 ) = 12.5 in.2
Based on this minimum gross area, from Manual Table 1-15, select
2L6×6×9/16 with Ag = 12.9 in.2
LRFD and ASD in Homework Problems
Each chapter includes homework problems at the end of the chapter. These problems are
organized to follow the order of presentation of the material in the chapters. Several problems are
provided for each general subject. Problems are provided for both LRFD and ASD solutions.
There are also problems designed to show comparisons between ASD and LRFD solutions. These
problems show that in some instances one method might give a more economical design, whereas
in other instances the reverse is true.
viii Preface
WEBSITE
Additional resources are available from the book website at www.SteelStuff.com . The following
resources are available on the student section of the website.
●
Answers: Selected homework problem answers are available on the student
section of the website.
●
Errata: We have reviewed the text to make sure that it is as error-free as
possible. However, if any errors are discovered, they will be listed on the book
website as a reference.
If you encounter any errors as you are using the book, please send them directly to the authors
(LFG@psu.edu) so we may include them on the website, and correct these errors in future
editions.
RESOURCES FOR INSTRUCTORS
All resources for instructors are available through an Instructor link on the website at
www.SteelStuff.com .
The following resources are available to instructors who adopt the text:
• Solutions Manual: Solutions for all homework problems in the text.
• Integrated Building Project RAM Structural System: Computer model and
example output for one solution of integrated Building project
• Textbook Figures: Select figures available in PowerPoint format
Visit the Instructor link on the website at www.SteelStuff.com to register and request access to
these resources.
ACKNOWLEDGEMENTS
The previous two editions of this book were under the sole authorship of the first author. Many
individuals contributed to those editions and we want to thank them for their contributions to
those works as those contributions clearly are carried over to this new edition. This third edition
has been produced through the cooperative efforts of three individuals who have a long standing
common academic and professional relationship. For the many individuals who have motivated
and guided each of us, we acknowledge your contribution to all that we do. We thank those who
have provided input and clarity to our understanding of structural steel design through our
involvement in the development of the AISC Specification and Manual. Finally, we acknowledge
the contributions of our spouses and families through whose support we are able to accomplish
this project.
Louis F. Geschwindner
Judy Liu
Charles J. Carter
Contents
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
3
3.1
3.2
3.3
3.4
Introduction 1
Scope 1
The Specification 1
The Manual 4
AISC Web Site Resources 5
Principles of Structural Design 5
Parts of the Steel Structure 6
Types of Steel Structures 11
1.7.1
Bearing Wall Construction 11
1.7.2
Beam-and-Column Construction 12
1.7.3
Long-Span Construction 14
1.7.4
High-Rise Construction 15
1.7.5
Gable-Frame Construction 16
Design Philosophies 17
Fundamentals of Allowable Strength Design
(ASD) 18
Fundamentals of Load and Resistance Factor
Design (LRFD) 19
Inelastic Design 20
Structural Safety and Integrity 20
Limit States 26
Building Codes and Design Specifications 27
Integrated Design Project 27
Problems 29
Loads, Load Factors, and Load
Combinations 30
Introduction 30
Building Load Sources 31
2.2.1
Dead Load 31
2.2.2
Live Load 31
2.2.3
Snow Load 32
2.2.4
Wind Load 32
2.2.5
Seismic Load 33
2.2.6
Special Loads 33
Building Load Determination 34
2.3.1
Dead Load 34
2.3.2
Live Load 35
2.3.3
Snow Load 37
2.3.4
Wind Load 38
2.3.5
Seismic Load 40
Load Combinations for ASD and LRFD
Load Calculations 43
Calibration 51
Problems 53
3.5
3.6
3.7
3.8
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
5
5.1
5.2
5.3
42
Steel Building Materials 56
Introduction 56
Applicability of the AISC Specification 56
Steel for Construction 58
Structural Steel Shapes 62
3.4.1
ASTM A6 Standard Shapes 63
3.4.2
Hollow Shapes 65
3.4.3
Plates and Bars 66
5.4
5.5
5.6
5.7
5.8
3.4.4
Built-up Shapes 68
Chemical Components of Structural Steel 68
Grades of Structural Steel 70
3.6.1
Steel for Shapes 70
3.6.2
Steel for Plates and Bars 74
3.6.3
Steel for Fasteners 75
3.6.4
Steel for Welding 77
3.6.5
Steel for Headed Stud Anchors 78
Availability of Structural Steel 78
Problems 78
Tension Members 80
Introduction 80
Tension Members in Structures 80
Cross-Sectional Shapes for Tension
Members 82
Behavior and Strength of Tension
Members 84
4.4.1
Yielding 85
4.4.2
Rupture 85
Computation of Areas 87
4.5.1
Gross Area 87
4.5.2
Net Area 87
4.5.3
Influence of Hole Placement
4.5.4
Effective Net Area 96
Design of Tension Members 101
Block Shear 104
Pin-Connected Members 112
Eyebars and Rods 115
Built-Up Tension Members 116
Truss Members 116
Bracing Members 117
Problems 120
91
Compression Members 124
Compression Members in Structures 124
Cross-Sectional Shapes for Compression
Members 124
Compression Member Strength 126
5.3.1
Euler Column 126
5.3.2
Other Boundary Conditions 130
5.3.3
Combination of Bracing and End
Conditions 131
5.3.4
Real Column 134
5.3.5
AISC Provisions 137
Additional Limit States for Compression 145
Length Effects 146
5.5.1
Effective Length for Inelastic
Columns 152
5.5.2
Effective Length when Supporting
Gravity Only Columns 154
Slender Elements in Compression 157
Column Design Tables 163
Torsional Buckling and Flexural-Torsional
Buckling 169
ix
x Contents
5.9
5.10
5.11
5.12
Single-Angle Compression Members
Built-Up Members 177
Column Base Plates 181
Problems 182
6
6.1
6.2
6.3
Bending Members 188
Bending Members in Structures 188
Strength of Beams 189
Design of Compact Laterally Supported WideFlange Beams 194
Design of Compact Laterally Unsupported WideFlange Beams 201
6.4.1
Lateral-Torsional Buckling 201
6.4.2
Moment Gradient 206
Design of Noncompact Beams 216
6.5.1
Local Buckling 216
6.5.2
Flange Local Buckling 218
6.5.3
Web Local Buckling 219
Design of Beams for Weak Axis Bending 222
Design of Beams for Shear 224
Continuous Beams 227
Plastic Analysis and Design of Continuous
Beams 229
T-Shaped Members in Bending 232
6.10.1 Yielding of Tees 232
6.10.2 Lateral-Torsional Buckling of
Tees 233
6.10.3 Flange Local Buckling of Tees 234
6.10.4 Stem Local Buckling of Tees 234
Single-Angle Bending Members 237
6.11.1 Yielding 238
6.11.2 Leg Local Buckling 238
6.11.3 Lateral-Torsional Buckling 239
Double-Angle Members in Bending 243
6.12.1 Yielding of Double Angles 243
6.12.2 Lateral-Torsional Buckling of Double
Angles 243
6.12.3 Leg Local Buckling of Double
Angles 244
Members in Biaxial Bending 247
Serviceability Criteria for Beams 247
6.14.1 Deflection 248
6.14.2 Vibration 248
6.14.3 Drift 248
Concentrated Forces on Beams 250
6.15.1 Web Local Yielding 251
6.15.2 Web Local Crippling 252
Open Web Steel Joists and Joist Girders 256
Problems 259
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
7
7.1
7.2
7.3
174
Plate Girders 264
Background 264
Homogeneous Plate Girders in Bending 266
7.2.1
Noncompact Web Plate Girders 267
7.2.2
Slender Web Plate Girders 271
7.2.3
Compact Web Plate Girders 279
Homogeneous Plate Girders in Shear 281
7.3.1
Nontension Field Action 282
7.4
7.5
8
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11
8.12
8.13
8.14
9
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
7.3.2
Tension Field Action 284
Stiffeners for Plate Girders 286
7.4.1
Intermediate Stiffeners 286
7.4.2
Bearing Stiffeners 288
7.4.3
Bearing Stiffener Design 291
Problems 295
Beam-Columns and Frame Behavior 297
Introduction 297
Second-Order Effects 298
Interaction Principles 300
Interaction Equations 301
Braced Frames 305
Moment Frames 312
Specification Provisions for Stability Analysis
and Design 324
8.7.1
Direct Analysis Method 325
8.7.2
Effective Length Method 325
8.7.3
First-Order Analysis Method 326
8.7.4
Geometric Imperfections 326
8.7.5
Comparison of Methods 327
Initial Beam-Column Selection 333
Beam-Column Design Using Manual
Part 6 336
Combined Simple and Moment Frames 337
Partially Restrained Frames 349
Stability Bracing Design 358
8.12.1 Column Bracing 359
8.12.2 Beam Bracing 360
8.12.3 Frame Bracing 361
Tension Plus Bending 363
Problems 366
Composite Construction 372
Introduction 372
Advantages and Disadvantages of Composite
Beam Construction 375
Shored versus Unshored Construction 375
Effective Flange 376
Strength of Composite Beams and Slab 377
9.5.1
Fully Composite Beams 378
9.5.2
Partially Composite Beams 383
9.5.3
Composite Beam Design Tables 386
9.5.4
Negative Moment Strength 391
Shear Stud Strength 391
9.6.1
Number and Placement of Shear
Studs 392
Composite Beams with Formed Metal
Deck 394
9.7.1
Deck Ribs Perpendicular to Steel
Beam 395
9.7.2
Deck Ribs Parallel to Steel Beam 396
Fully Encased Steel Beams 403
Selecting a Section 403
Serviceability Considerations 408
9.10.1 Deflection During Construction 408
9.10.2 Vibration Under Service Loads 409
9.10.3 Live Load Deflections 409
Contents xi
9.11
9.12
9.13
Composite Columns 412
Composite Beam-Columns
Problems 426
10
10.1
10.2
10.3
10.4
10.5
Connection Elements 429
Introduction 429
Basic Connections 429
Beam-to-Column Connections 430
Fully Restrained Connections 432
Simple and Partially Restrained
Connections 433
Mechanical Fasteners 434
10.6.1 Common Bolts 434
10.6.2 High-Strength Bolts 435
10.6.3 Bolt Holes 436
Bolt Limit States 438
10.7.1 Bolt Shear 439
10.7.2 Bolt Bearing and Tearout 440
10.7.3 Strength at Bolt Holes 441
10.7.4 Bolt Tension 442
10.7.5 Slip 448
10.7.6 Combined Tension and Shear in
Bearing-Type Connections 450
10.7.7 Combined Tension and Shear in SlipCritical Connections 451
Welds 451
10.8.1 Welding Processes 452
10.8.2 Types of Welds 453
10.8.3 Weld Sizes 454
Weld Limit States 454
10.9.1 Fillet Weld Strength 455
10.9.2 Groove Weld Strength 461
Connecting Elements 461
10.10.1
Connecting Elements in
Tension 461
10.10.2
Connecting Elements in
Compression 462
10.10.3
Connecting Elements in
Flexure 462
10.10.4
Connecting Elements in Shear 462
10.10.5
Block Shear Strength 463
10.10.6
Connecting Element Rupture
Strength at Welds 466
Problems 467
10.6
10.7
10.8
10.9
10.10
10.11
11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
421
Simple Connections 471
Types of Simple Connections 471
Simple Shear Connections 471
Double-Angle Connections: BoltedBolted 473
Double-Angle Connections: WeldedBolted 492
Double-Angle Connections: BoltedWelded 496
Double-Angle Connections: WeldedWelded 498
Single-Angle Connections 498
Single-Plate Shear Connections 509
11.9
11.10
11.11
11.12
12
12.1
12.2
12.3
12.4
12.5
13
13.1
13.2
13.3
Seated Connections 517
Light Bracing Connections 524
Beam Bearing Plates and Column Base
Plates 536
Problems 542
Moment Connections 546
Types of Moment Connections 546
Limit States 548
Moment Connection Design 549
12.3.1 Direct-Welded Flange
Connection 549
12.3.2 Welded Flange Plate Connection 556
12.3.3 Bolted Flange Plate Connection 565
Column Stiffening 578
12.4.1 Flange Local Bending 578
12.4.2 Web Local Yielding 579
12.4.3 Web Local Crippling 580
12.4.4 Web Compression Buckling 580
12.4.5 Web Panel Zone Shear and
Stiffeners 581
Problems 589
13.7
13.8
Steel Systems for Seismic Resistance 591
Introduction 591
Expected Behavior 592
Moment-Frame Systems 594
13.3.1 Special Moment Frames (SMF) 595
13.3.2 Intermediate Moment Frames (IMF)
and Ordinary Moment Frames
(OMF) 597
Braced-Frame Systems 597
13.4.1 Special Concentrically Braced Frames
(SCBF) 598
13.4.2 Ordinary Concentrically Braced Frames
(OCBF) 601
13.4.3 Eccentrically Braced Frames
(EBF) 601
Other Framing Systems
603
13.5.1 Special Truss Moment Frames
(STMF) 604
13.5.2 Buckling-Restrained Braced Frames
(BRBF) 604
13.5.3 Special Plate Shear Walls
(SPSW) 605
13.5.4 Composite Systems 606
Other General Requirements 606
13.6.1 Bolted and Welded Connections 606
13.6.2 Protected Zones 607
13.6.3 Local Buckling 607
13.6.4 Column Requirements 607
13.6.5 Column Bases 608
Conclusions 608
Problems 608
Index
610
13.4
13.5
13.6
xii Contents
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Chapter
1
Introductio
on
One Worlld Trade Centeer, New York
Phooto Courtesy M
Michael Mahessh, Port Authorrity of NYNJ
1.1
SCOP
PE
A wide variety
v
of desiigns can be ch
haracterized aas structural ssteel design. This book deeals with the
design of steel structu
ures for build
dings as goveerned by the ANSI/AISC 3360-16 Specif
ification for
Structura
al Steel Build
dings, publish
hed by the Am
merican Instiitute of Steel Constructionn (AISC) in
2016, and referred to as the Speciffication in thiis book. The aareas of appliication given throughout
this book
k specifically
y focus on thee design of stteel building structures. T
The treatment of subjects
associateed with bridgees and industrrial structuress, if addressedd at all, is keppt relatively brief.
The
T book ad
ddresses the concepts andd design critteria for the two design approaches
detailed by
b the Specifi
fication: Load
d and Resistannce Factor Deesign (LRFD)) and Allowabble Strength
Design (A
ASD). Both methods
m
are discussed
d
laterr in this chapter.
In
I addition to
o the Specifica
ation, the prim
mary referencce for this boook is the 15thh edition of
the AISC
C Steel Constrruction Manu
ual. This referrence handbook contains tables of the bbasic values
needed for
f structurall steel design
n, design tabbles to simpliify actual deesign, and thhe complete
Specifica
ation. Through
hout this book
k, this is referrred to as the Manual.
1.2
THE SPECIFIC
CATION
The ANS
SI/AISC 360-1
16 Specificatiion for Structtural Steel Buuildings is thhe latest in a llong line of
standard specification
ns published by
b the Ameriican Institute of Steel Connstruction forr the design
and consstruction of structural
s
steeel buildings. The first ediition was pubblished in 1923. For the
reader in
nterested in th
he historical aspects of thhese specificaations, AISC
C has two ressources that
provide detailed
d
guid
dance on the historical struuctural steel standards. Thhe first is AIISC Design
Guide 15, AISC Reh
habilitation and Retrofit Guide: A R
Reference foor Historic SShapes and
Specifica
ations. This Design
D
Guide provides outlline comparissons of the prrovisions in thhe different
editions of the Specifi
fication. The second
s
resourrce is found oon the AISC web site, ww
ww.aisc.org,
AISC Sp
pecifications 1923–2010,
1
which
w
containns a searchabble compenddium of all of the AISC
Specifica
ations for Stru
uctural Steel Buildings
B
prooduced from 11923 through 2010.
1
2 Chapter 1
Introduction
Current design is carried out under the provisions published in the 2016 edition of the
AISC Specification. In addition to the detailed provisions, the Specification contains User Notes
and a detailed Commentary that provides insights into the source and application of the
provisions. The reader interested in additional background on the provisions discussed in this
book is encouraged to investigate the materials cited in the appropriate sections of the
Commentary. The Specification contains 14 chapters and 8 appendices. To provide a concise
guide to the use of the Specification, a brief description is given here.
Chapter A: General Provision. This chapter provides the scope of the Specification and
summarizes all referenced specifications, codes, and standards. It also provides the requirements
for materials to be used in structural steel design and the design documents necessary to
communicate that design.
Chapter B: Design Requirements. This chapter gives the general requirements for analysis and
design that are applicable throughout the entire Specification. It provides the charging language
needed for application of the subsequent chapters.
Chapter C: Design for Stability. This chapter, along with Appendix 7, addresses the requirements
for the design of structures to ensure stability. It details those factors that must be taken into
consideration in any analysis and design.
Chapter D: Design of Members for Tension. This chapter applies to the design of members
subjected to axial tension resulting from forces acting through the centroidal axis.
Chapter E: Design of Members for Compression. This chapter addresses members subjected to
axial compression resulting from forces applied at the centroidal axis.
Chapter F: Design of Members for Flexure. This chapter applies to members loaded in a plane
parallel to a principal axis that passes through the shear center or is restrained against twisting.
This is referred to as simple bending about one axis.
Chapter G: Design of Members for Shear. This chapter addresses webs of singly or doubly
symmetric members subject to shear in the plane of the web. It also addresses other shapes such
as single angles and hollow structural sections.
Chapter H: Design of Members for Combined Forces and Torsion. This chapter addresses
design of members subject to an axial force in combination with flexure about one or both axes,
with or without torsion. It also applies to members subjected to torsion only.
Chapter I: Design of Composite Members. This chapter addresses the design of members
composed of steel shapes and concrete working together as a member. It addresses compression,
flexure, and combined forces.
Chapter J: Design of Connections. This chapter addresses the design of connections, including
the connecting elements, the connectors, and the connected portions of members.
Chapter K: Additional Requirements for HSS and Box Section Connections. This chapter
addresses requirements in addition to those given in Chapter J for the design of connections to
hollow structural sections and built-up box sections of uniform thickness. Its also addresses
connections between HSS and box members.
Introduction
Chapter 1 3
Chapter L: Design for Serviceability. This chapter summarizes the performance requirements for
the design of a serviceable structure.
Chapter M: Fabrication and Erection. This chapter addresses the requirements for shop
drawings, fabrication, shop painting, and erection.
Chapter N: Quality Control and Quality Assurance. This chapter addresses the requirements for
ensuring quality of the constructed project.
Appendix 1: Design by Advanced Analysis. The body of the Specification addresses design based
on an elastic analysis. This appendix addresses design by alternative methods generally referred
to as advanced methods. It includes the classical plastic design method and design by direct
modeling of imperfections.
Appendix 2: Design for Ponding. This appendix provides methods for determining whether a
roof system has sufficient strength and stiffness to resist the influence of water collecting on the
surface and forming a pond.
Appendix 3: Fatigue. This appendix provides requirements for addressing the influence of high
cycle loading on members and connections that could lead to cracking and progressive failure.
For most building structures, fatigue is not an issue of concern.
Appendix 4: Structural Design for Fire Conditions. This appendix provides the criteria for
evaluation of structural steel subjected to fire conditions, including (1) the prescriptive approach
provided for in the model building code and most commonly used in current practice and (2) the
engineered approach.
Appendix 5: Evaluation of Existing Structures. This appendix provides guidance on the
determination of the strength and stiffness of existing structures by load tests or a combination of
tests and analysis.
Appendix 6: Member Stability Bracing. This appendix details the criteria for ensuring that
column, beam and beam-column bracing has sufficient strength and stiffness to meet the
requirements for member bracing assumed in the provisions of the Specification for design of
those members.
Appendix 7: Alternative Methods of Design for Stability. This appendix, along with Chapter C,
provides methods of designing structures to ensure stability. Two alternative methods are
provided here, including the method most commonly used in past practice.
Appendix 8: Approximate Second-Order Analysis. This appendix provides a method for
obtaining second-order effects by an amplified first-order analysis. The provisions are limited to
structures supporting load primarily through vertical columns.
Each chapter of this book will identify those chapters of the Specification that are
pertinent to that chapter. The reader is encouraged to become familiar with the organization of the
Specification.
4 Chapter 1
Introduction
1.3 THE MANUAL
The AISC Steel Construction Manual, 15th edition, is the latest in a series of manuals published
to assist the building industry in designing safe and economical steel building structures. The first
edition was published in 1928 and the ninth edition in 1989. These manuals addressed design by
the allowable stress method. In 1986 the first edition of the load and resistance factor design
method manual was published, with the third edition published in 1999. The next in this unbroken
string of manuals published in support of steel design and construction was the first manual to
unify these two design methods and was published in 2005 as the 13th edition. The current
edition of the Manual is the 15th. Students who purchase the Manual through the AISC Student
Discount Program also have an opportunity to apply for a free AISC Student Membership at the
same time. Students are encouraged to become AISC Student Members in order to take full
advantage of all free member benefits.
As is the case for the Specification, AISC has two resources to assist in addressing the
historic aspects of steel design and construction. The first is, again, AISC Design Guide 15, AISC
Rehabilitation and Retrofit Guide: A Reference for Historic Shapes and Specifications. This
Guide provides properties of beam and column sections as old as the wrought iron shapes
produced as early as 1873. The second resource is the electronic AISC Shapes Database. This
database is available through the AISC web site www.aisc.org. It is a searchable database with
properties for all shapes produced since 1873, consistent with the printed data in Design Guide
15. Access to the electronic shapes database is free to AISC members.
The Manual is presented in 17 parts as follows:
Part 1: Dimensions and Properties
Part 2: General Design Considerations
Part 3: Design of Flexural Members
Part 4: Design of Compression Members
Part 5: Design of Tension Members
Part 6: Design of Members Subject to Combined Forces
Part 7: Design Considerations for Bolts
Part 8: Design Considerations for Welds
Part 9: Design of Connecting Elements
Part 10: Design of Simple Shear Connections
Part 11: Design of Partially Restrained Moment Connections
Part 12: Design of Fully Restrained Moment Connections
Part 13: Design of Bracing Connections and Truss Connections
Part 14: Design of Beam Bearing Plates, Column Base Plates, Anchor Rods, and Column Splices
Introduction
Chapter 1 5
Part 15: Design of Hanger Connections, Bracket Plates, and Crane-Rail Connections
Part 16: Specifications and Codes
Part 17: Miscellaneous Data and Mathematical Information
Each chapter of this book identifies those parts of the Manual that will be used with the
material to be addressed. In many instances, the user will need to look in several parts of the
Manual to fully understand the topics or solve the problems presented.
1.4
AISC WEB SITE RESOURCES
Another primary resource is the AISC web site, where there is information that is free to all
visitors and additional electronic resources that are free to members only. Students will find a
great deal of useful information on the AISC publications web site, www.aisc.org/epubs. The
primary resources include electronic versions of the Specification, the Shapes Database, the Steel
Construction Manual References, and the Steel Construction Manual Design Examples. The
Specification, as described in Section 1.2 and the historic Shapes Database, as mentioned in
Section 1.3, are available free to all through the web site. The 15th edition Steel Construction
Manual Shapes Database is also available free to all. The AISC web site also includes an
extensive array of journal and proceedings papers. All of the references cited in the Commentary
and the Manual, for which AISC owns the copyright, are accessible under Steel Construction
Manual Resources; Interactive Reference List.
Probably the most valuable aspect of the AISC web site for readers of this book is the
complete set of the 15th edition Steel Construction Manual Design Examples. These examples are
presented in four sections.
Section I: Examples Based on the AISC Specification. This section contains examples
demonstrating the use of the specific provisions of the Specification, organized by Specification
chapter.
Section II: Examples Based on the AISC Steel Construction Manual. This section contains
examples of connection design using the Specification and the tables found in the Manual.
Section III: System Design Examples. This section contains examples associated with the design
of a specific building and the application of the system-wide requirements.
Section IV: Additional Resources. This section provides design tables for higher-strength steels
than published in the printed Manual.
Although the topics covered in this book are supported by calculated example problems,
the reader might find the electronic Steel Construction Manual Design Examples helpful for
further understanding of some of the specific provisions or design aids described in the book. In
addition, some of the Design Examples go beyond the coverage in this book and provide
additional useful information regarding typical design or detailing. The reader is encouraged to
investigate what the AISC web site has to offer through both free and member only publications.
1.5
PRINCIPLES OF STRUCTURAL DESIGN
From the time an owner determines a need to build a building, through the development of
conceptual and detailed plans, to completion and occupancy, a building project is a multi-faceted
6 Chapter 1
Introduction
task that involves many professionals. The owner and the financial analysis team evaluate the
basic economic criteria for the building. The architects and engineers form the design team and
prepare the initial proposals for the building, demonstrating how the users’ needs will be met.
This teamwork continues through the final planning and design stages, where the design
drawings, specifications, and contract documents are readied for the construction phase. During
this process, input may also be provided by the individuals who will transform the plans into a
real-life structure. The steel detailer, fabricator and erector all have a role in that process, and add
their respective expertise to make the design constructible. Thus, those responsible for the
construction phase of the project often help improve the design by taking into account the actual
on-site requirements for efficient construction.
Once a project is completed and turned over to the owner, the work of the design teams is
normally over. The operation and maintenance of the building, although major factors in the life
of the structure, are not usually within the scope of the designer’s responsibilities, except when
significant changes in building use are anticipated. In such cases, a design team should verify that
the proposed changes can be accommodated.
The basic goals of the design team can be summarized by the words safety, function, and
economy. The building must be safe for its occupants and all others who may come in contact
with it. It must neither fail locally nor overall, nor exhibit behavioral characteristics that test the
confidence of rational human beings. To help achieve that level of safety, building codes and
design specifications are published that outline the minimum criteria that any structure must meet.
The building must also serve its owner in the best possible way to ensure that the
functional criteria are met. Although structural safety and integrity are of paramount importance,
a building that does not serve its intended purpose will not have met the goals of the owner.
Last, but not least, the design, construction, and long-term use of the building should be
economical. The degree of financial success of any structure will depend on a wide range of
factors. Some are established prior to the work of the design team, whereas others are determined
after the building is in operation. Nevertheless, the final design should, within all reasonable
constraints, produce the lowest combined short- and long-term expenditures.
The AISC Specification follows the same principles. The mission of the AISC Committee
on Specifications is to “develop the practice-oriented specification for structural steel buildings
that provide for life safety, economical building systems, predictable behavior and response, and
efficient use.” Thus, this book emphasizes the practical orientation of this Specification.
1.6 PARTS OF THE STEEL STRUCTURE
All structures incorporate some or all of the following basic types of structural components:
1.
Tension members
2.
Compression members
3.
Bending members
4.
Combined force members
5.
Connections
The first four items represent structural members. The fifth, connections, represents the
contact regions between the structural members, which ensure that all components work together
as a structure.
Introoduction
C
Chapter 1 7
Detailed
D
evalu
uations of thee strength, beehavior, and ddesign criteria for these m
members are
presented
d in the follow
wing chapterss:
Tension
T
mem
mbers:
Chapteer 4
members:
Compression
C
Chapteer 5
Bending
B
mem
mbers:
Chapteers 6 and 7
Combined
C
forrce members:
Chapteer 8
Connections:
C
Chapteers 10, 11, andd 12
The
T strength and behavio
or of structurral frames coomposed of a combinatioon of these
elementss are covered in Chapters 8 and 13, and the special considerationss that apply too composite
(steel and
d concrete wo
orking togeth
her) constructiion are presennted in Chaptter 9. An intrroduction to
the desig
gn of steel stru
uctures for eaarthquake loaading is presennted in Chapter 13. The properties of
structural steel and th
he various sh
hapes commoonly used are discussed inn Chapter 3, and a brief
discussio
on of the types of loads and
d load combinnations is presented in Chaapter 2.
Tension
T
memb
bers are typiccally found ass web and choord members in trusses andd open-web
steel joissts; as diagon
nal members in structurall bracing sysstems; and ass hangers forr balconies,
mezzanin
ne floors, and
d pedestrian walkways.
w
Theey are also ussed as sag rodds for purlins and girts in
many bu
uilding types, as well as to
t support plaatforms for m
mechanical eqquipment andd pipelines.
Figures 1.1
1 and 1.2 illlustrate typicaal applicationns of tension m
members in acctual structures.
In
I the idealizeed case, tensiion members transmit conncentric tensille forces onlyy. In certain
structures, reversals of
o the overalll load may cchange the tennsion membeer force from
m tension to
compresssion. Some members
m
will be
b designed ffor this actionn; others will hhave been designed with
the assum
mption that th
hey will carry tension only..
The
T idealized
d tension mem
mber is analyyzed with thee assumption that its end cconnections
are pins,, which prev
vent any mom
ment or sheaar force from
m being transsmitted to thhe member.
Howeverr, in an actuall structure, th
he type of connnection norm
mally dictatess that some beending may
be introd
duced to the tension
t
memb
ber. This is aalso the case when the tennsion memberr is directly
exposed to some form
m of transversse load. Mom
ments will alsso be introducced if the eleement is not
perfectly
y straight, or if the axial loaad is not appliied along the centroidal axxis of the mem
mber.
The
T primary load
l
effect in
n the tension member is a concentric aaxial force, w
with bending
and shearr considered as
a secondary effects.
members aree also known as columns, sstruts, or postts. They are uused as web
Compression
C
and chorrd members in trusses and
a
joists andd as verticall members inn all types oof building
structures. Figure 1.3 shows a typiccal use of struuctural comprression membbers.
Figure 1.1
1
Use of Tension
T
Memb
bers in a Trusss
Photo cou
urtesy Ruby + Associates
A
8 Chapter 1
Introduction
n
Figure 1.2 Use of Tenssion Memberss as Hangers
The idealized co
ompression member
m
carriies only a cooncentric, coompressive foorce. Its
strength is heavily
h
influeenced by thee distance bettween the suppports, as weell as by the support
conditions. The
T basic collumn is thereffore defined aas an axially loaded membber with pinnned ends.
Historically,, design ruless for compresssion memberss have been bbased on the bbehavior and strength
of this idealiized compression memberr.
The basic colum
mn is practicaally nonexisttent in real sstructures. Reealistic end ssupports
rarely resem
mble perfect pins; the ax
xial load is nnormally nott concentric, due to the w
way the
surrounding
g structure traansmits its load to the meember; and beeams and sim
milar componnents are
likely to be connected to
o the column
n in such a w
way that mom
ments are intrroduced. All of these
conditions produce
p
bendiing effects in the member,, making it a combined forrce member oor beamcolumn, as distinct
d
from the
t idealized column.
The primary loaad effect in the pinned--end column is thereforee a concentrric axial
compressivee force accom
mpanied by thee secondary eeffects of bending and sheaar.
Bending memberrs are known
n as beams, girders, joistts, spandrels, purlins, linttels, and
girts. Altho
ough all of these
t
are bending membbers, each naame implies a certain sttructural
application within
w
a build
ding:
umns in a Buiilding Frame
Figure 1.3 Use of Colu
Introoduction
C
Chapter 1 9
1.
1 Beams, girders, and jo
oists form parrt of commonn floor system
ms. The beam
ms are most
often considered as thee members thhat are directlly supported bby girders, whhich in turn
are usuallly supported by
b columns. JJoists are beam
ms with fairlyy close spacinng. A girder
may geneerally be conssidered a highher-order bennding memberr compared w
with a beam
or joist. However,
H
variiations to this basic schemee are commonn.
2.
2 The bending members that form thee perimeter off a floor or rooof plan in a bbuilding are
known ass spandrels or
o spandrel b eams. Their design may be different from other
beams and
d girders becaause the load comes primaarily from onee side of the m
member.
3.
3 Bending members in roof systemss that span bbetween otheer bending m
members are
usually reeferred to as purlins.
p
4.
4 Lintels arre bending members
m
that span across tthe top of oppenings in waalls, usually
carrying the
t weight off the wall aboove the openiing as well ass any other looad brought
into that area. They ty
ypically are sseen spanninng across the openings forr doors and
windows.
5.
5 Girts are used in exterrior wall systtems. They trransfer the laateral load froom the wall
surface to
o the exterior columns. Thhey may also assist in suppporting the weight of the
wall.
Figure
F
1.4 sh
hows beams and girders in an actuall structure unnder construcction. The
idealized
d beam is sho
own in the figure
fi
as a m
member with a uniform looad supportedd on simple
supports..
The
T basic ben
nding member carries traansverse loadds that act inn a plane conntaining the
longitudiinal centroidaal axis of thee member. Thhe primary looad effects arre bending m
moment and
shear forrce. Axial forcces and torsio
on may occur as secondaryy effects.
The
T most com
mmon combin
ned force meember is know
wn as a beam
m-column, im
mplying that
this strucctural elemen
nt is simultan
neously subjeccted to bendiing and axiall compressionn. Although
the preseence of both
h bending an
nd axial tenssion represennts a potentiaal loading caase for the
combined
d force memb
ber, this case is not as criti cal or commoon as the beam
m-column loaading case.
Iddealized bendiing member
Figure 1.4
1
Building
g Structure Sh
howing Bendiing Members
100 Chapter 1
Introductio
on
R
n of Steel Fraames in Whicch the Verticaal Members A
Are
Figure 1.5 Schematic Representatio
o Axial Loadss and Bending
g Moments
Subjected to
Figu
ure 1.5a is a schematic illu
ustration of a multi-story steel frame w
where the beaams and
columns aree joined with rigid connecctions. Becausse of the geo metric configguration, the types of
connections, and the loaading pattern
n, the verticaal members aare subjectedd to axial looads and
bending mom
ments. This is a typical ap
pplication of ppractical beam
m-columns; otther examples are the
members off the gable fraame shown in
n Figure 1.5bb and the verttical componeents of a singgle-story
portal framee shown in Fig
gure 1.5c.
The beam-colum
mn may be reg
garded as the general strucctural elemennt, where axiaal forces,
shear forces, and bending
g moments acct simultaneouusly. Thus, thhe basic colum
mn may be thoought of
as a special case, represeenting a beam
m-column wiith no momennts or transveerse loads. Siimilarly,
the basic ben
nding membeer may be thought of as a bbeam-columnn with no axiaal load. Therefore, the
consideratio
ons that must be accounted
d for in the deesign of bothh columns andd beams mustt also be
applied to beeam-columnss.
Because of the generalized
g
nature
n
of the combined foorce elementt, all load efffects are
considered primary.
p
How
wever, when the ratio of aaxial load to axial strengtth in a beam-column
becomes hig
gh, column behavior
b
will overshadow other influennces. Similarlly, when the ratio of
applied mom
ment to mom
ment strength
h is high, beaam behavior will outweiggh other effects. The
beam-colum
mn is an elem
ment in which
h a variety off different forrce types inteeract. Thus, ppractical
design appro
oaches are no
ormally based
d on interactioon equations.
Con
nnections are the collectio
on of elemennts that join the memberss of a steel sstructure
together. Wh
hether they co
onnect the axially loaded m
members in a truss or the bbeams and columns
Introdduction
Figure 1.6
1
Chhapter 1 11
Building
g Connectionss
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with permissionn. All rights resserved
of a multti-story framee, connectionss must ensuree that the strucctural membeers function toogether as a
unit, con
nsistent with th
he assumption
ns made in thhe design.
The
T fastenerss used in stru
uctural steel connections today are alm
most entirelyy limited to
bolts and
d welds. The load effects that the variious elementss of the connnection must resist are a
function of the specifi
fic connection
n type being cconsidered. T
They include aall of the possible forces
and mom
ments. Figuree 1.6 illustratees a variety of connectionns. The ideallized represenntations for
connectio
ons are presen
nted in Chaptters 10, 11, annd 12.
1.7
TYPE
ES OF STEEL STRUC
CTURES
It is diffi
ficult to classiify steel strucctures into neeat categoriess, due to the wide variety of systems
availablee to the design
ner. The elem
ments of the sttructure, as deefined in Secttion 1.6, are ccombined to
form thee total structu
ure of a build
ding, which must safely and economiically carry aall imposed
loads. Th
his combinatio
on of members is usually rreferred to as the framing ssystem.
buildings com
Steel-framed
S
me in a widee variety of shapes and sizzes and in coombinations
with otheer structural materials.
m
A few
f exampless are given inn the followinng paragraphss, to set the
stage for the applicatio
on of structurral design preesented in subbsequent chappters.
1.7.1
Bea
aring Wall Constructio
C
n
Bearing wall constru
uction is prim
marily used ffor one- or ttwo-story buiildings, such as storage
warehouses, shopping
g centers, offiice buildings,, and schoolss. This system
m normally usses brick or
concrete block mason
nry walls, on which are plaaced the endss of the flexuural memberss supporting
the floor or roof. The flexural mem
mbers are usuaally hot-rolledd structural stteel shapes, allone or in
122 Chapter 1
Introductio
on
Figuree 1.7 Bearinng Wall Buildding
Photo ccourtesy Dougglas Steel Fabriicating Corporaation
n with open web
w steel joistts or cold-form
med steel shap
apes. An exam
mple of a bearring wall
combination
building is shown
s
in Figu
ure 1.7.
1.77.2
Beam--and-Colum
mn Construcction
Beam-and-ccolumn constrruction is the most commoonly used systtem for steel structures todday. It is
suitable for large-area bu
uildings such as schools annd shopping ccenters, whichh often have nno more
than two sto
ories but may have a large number of sppans. It is alsoo suitable for buildings wiith many
stories. Colu
umns are plaaced accordin
ng to a reguular, repetitiouus grid that supports thee beams,
girders, and joists, which
h are used forr the floor andd roof system
ms. The regulaarity of the floor plan
lends itself to
t economy in
i fabrication
n and erectionn, because moost of the meembers will bbe of the
same size. An
A example of this type of structure is shhown in Figuure 1.8.
For multi-story buildings, th
he use of com
mposite steell and concreete flexural m
members
affords addiitional saving
gs. Further ad
dvances can bbe expected ass designers bbecome more familiar
with the use of compositee columns and
d other elemeents of mixed construction systems.
Figure 1.8 Beam-and-C
Column Build
ding
Photo courtessy Douglas Steeel Fabricating Corporation
Introdduction
Chhapter 1 13
Figure 1.9
1 Braced Beeam-and-Colu
umn Buildingg
Photo cou
urtesy Douglas Steel Fabricatting Corporatioon
Beam-and-col
B
lumn structurres rely on eitther their connnections or a separate braccing system
to resist lateral
l
loads. A frame in which
w
all connnections are m
moment resistant provides rresistance
against th
he action of laateral loads, such
s
as wind and earthquak
akes, and overrall structural stability,
through the
t bending stiffness of thee overall fram
me.
Figure 1.10
1
Idealizeed Illustration
n of Several T
Types of Beam
m-and-Colum
mn Framed Strructures (a)
moment--resistant fram
me; (b) truss-b
braced frame;; (c) core-bracced frame; (dd) floor plan oof shear
wall and core-braced building;
b
(e) floor
f
plan of bbuilding withh a combinatioon of braced and
unbraced
d bents.
A frame with
hout memberr-end restrainnt needs a seeparate lateraal load resistiing system,
which is often afford
ded by having
g the elemennts along onee or more of the column llines act as
144 Chapter 1
on
Introductio
braced fram
mes, as seen in
n Figure 1.9.. One of the most commoon types of bbracing is the vertical
truss, which
h is designed to take the loads imposeed by wind aand seismic aaction. Other bracing
schemes inv
volve shear walls and reinfforced concrette cores. The latter type m
may also be refferred to
as a braced core
c
system and
a can be hig
ghly efficientt because of thhe rigidity of the box-shapped cross
section of th
he core. Thee core serves a dual purppose in this ccase: in addittion to providding the
bracing systtem for the bu
uilding, it serv
ves as the verrtical conduitt in the comppleted structurre for all
of the necessary services,, including eleevators, stairccases, electriccity, and otheer utilities.
Com
mbinations off these types of
o constructioon are also coommon. For eexample, fram
mes may
have been designed
d
as moment resistaant in one direection of the bbuilding and as truss braceed in the
other. Of course, such a choice
c
recogn
nizes the threee-dimensionall nature of thee structure.
Figu
ure 1.10 shows an idealiized represenntation of sevveral types oof beam-and-column
framed strucctures.
1.77.3
Long-S
Span Consttruction
This type of
o constructio
on encompassses steel-fram
med structurres with longg spans betw
ween the
vertical load
d-carrying eleements, such as covered aarenas. The loong distancess may be spaanned by
one-way tru
usses, two-way
y space trussees, or plate annd box girderrs. A long-spaan structure iis shown
in Figure 1.1
11. Arches orr cables could
d also be usedd, although theey are not connsidered here.
Lon
ng-span constrruction is also
o used in builldings that reequire large, ccolumn-free iinteriors.
In such casees the buildin
ng may be a corec
or othe rwise bracedd structure, whhere the longg span is
the distance from the exteerior wall to the
t core.
Man
ny designers would also characterize
c
ssingle-story rrigid frames aas examples of longspan constru
uction system
ms. Dependin
ng on the geoometry of thee frame, suchh structures ccan span
substantial distances,
d
ofteen with excelllent economyy.
Figure 1.11 Long-Span Structure
Photo courtessy Douglas Steeel Fabricating Corporation
Introdduction
Chhapter 1 15
Figu
ure 1.12 Higgh-Rise Buildding
Struucture
Photto courtesy Doouglas Steel Faabricating
Corpporation
1.7.4
gh-Rise Con
nstruction
Hig
High-risee construction
n refers to mu
ulti-story buiildings of signnificant heighht; an example is shown
in Figure 1.12. The large heightts and uniquue problems encountered in the desiggn of such
structures warrant treeating them in
ndependentlyy from typicaal beam-and-column consstruction. In
addition, over the passt 40 years seeveral designners have devveloped a num
mber of new concepts in
multi-sto
ory frame desiign, such as th
he super com
mposite colum
mn and the steeel plate shear wall.
Particular
P
carre must be ex
xercised in thhe choice andd design of tthe lateral loaad resisting
system in
n high-rise co
onstruction. Itt is not just a m
matter of extrrapolating froom the principples used in
the analy
ysis of lower-rrise structures, because maany effects pllay a major roole in the desiign of highrise build
dings but hav
ve significanttly less impacct on frames of smaller hheight. These effects are
crucial to
o the proper design
d
of the high-rise
h
struccture.
Some
S
of thesee effects may
y be referred tto as second--order effects,, because theyy cannot be
quantifieed through a normal,
n
linearrly elastic anaalysis of the fframe. Althouugh second-oorder effects
are preseent in all strucctures, they may
m be more significant inn high-rise sttructures. Foor example,
when a structure is displaced
d
lateerally, additioonal momentt is induced in a column due to the
eccentriccity of the column loads. When
W
added too the momentts and shears produced by gravity and
wind loaads, the resu
ulting effectss may be siggnificantly laarger than thhose computted without
consideriing the secon
nd-order effeccts. A designeer who does not incorporaate both typees of effects
will be making
m
a serio
ous and perhaaps unconservvative error.
Framing
F
systeems for high--rise buildinggs reflect the increased im
mportance of lateral load
resistance. Thus, attem
mpts at makin
ng the perimeeter of a buildding act as a uunit or tube hhave proven
quite succcessful. Thiss tube may bee in the form of a truss, ass with the John Hancock Building in
Chicago,, Illinois, show
wn in Figure 1.13a, or a frrame, as in thhe former Worrld Trade Cennter in New
York Citty, shown in Figure
F
1.13b; a solid wall ttube with cuttouts for winddows, as usedd in the Aon
Center in
n Chicago, sh
hown in Figurre 1.13c; or seeveral intercoonnected or bbundled tubess, such as in
the Williis Tower in Chicago (form
merly known aas the Sears Tower), shownn in Figure 1.13d.
166 Chapter 1
Introductio
on
(a)
(b)
(c)
(d)
Figure 1.13 High-Rise Buildings (a)) the John Haancock Centerr; (b) the Worrld Trade Cennter;
(c) the Aon Center; (d) th
he Willis Tow
wer.
Photo (b) cou
urtesy Leslie E. Robertson Asssociates, RLL
LP
1.77.5
Gable--Frame Con
nstruction
Many design
ners include the
t single-storry frame as ppart of the lonng-span constrruction categoory. The
metal buildiing industry has
h capitalizeed on the usee of this system through ffine-tuned deesigns of
frames for storage
s
wareh
houses, indusstrial buildinggs, temporaryy and permannent office buuildings,
and similar types
t
of strucctures. The gaable frames arre typically buuilt-up, web ttapered membbers that
can be optim
mized for thee forces preseent. Memberrs are conneccted in the fiield using bolted end
plate connecctions. Metall buildings, co
ommonly refferred to as prre-engineeredd metal buildiings, are
now availab
ble as custom engineered for
fo specific appplications. A
An example off a gable-fram
me metal
building is seen
s
in Figuree 1.14.
Introdduction
Chhapter 1 17
Figure 1.14
1
Gable-F
Frame Metal Building
Photo cou
urtesy Metal Bu
uilding Manufa
facturers Assocciation
1.8
DESIIGN PHILO
OSOPHIES
A successsful structuraal design resu
ults in a struucture that is safe for its ooccupants, caan carry the
design lo
oads without overstressing
g any componnents, does noot deform or vibrate excesssively, and
is econom
mical to build
d and operate for its intendded life span. Although ecoonomy may aappear to be
the primary concern of
o an owner, safety must be the primaary concern oof the engineeer. Costs of
labor an
nd materials will vary frrom one geoographic locaation to anoother, makingg it almost
impossib
ble to design a structure th
hat is equally economical iin all locationns. Because thhe foremost
task of th
he designer is
i to produce a safe and s erviceable strructure, desiggn criteria suuch as those
published
d by the Am
merican Institu
ute of Steel Constructionn are based oon technical m
models and
consideraations that prredict structurral behavior aand material rresponse. Thee use of thesee provisions
by the designer
d
will dictate the economy
e
of a particular ssolution in a particular loocation and
business climate.
To
T perform a structural deesign, it is neecessary to qquantify the ccauses and efffects of the
loads thaat will be exeerted on each element throoughout the life of the struucture. This iis generally
termed th
he load effectt or the requirred strength. It is also neccessary to acccount for the behavior of
the mateerial and the shapes that compose theese elements.. This is refeerred to as thhe nominal
strength or capacity of
o the elementt.
In
I its simplesst form, strucctural designn is the deterrmination of member sizees and their
correspon
nding connecctions, so thatt the strength of the structuure is greater than the loadd effect. The
degree to
o which this is
i accomplish
hed is often teermed the maargin of safetyy. Numerous approaches
for accom
mplishing thiss goal have beeen used overr the years.
Although
A
passt experience might seem to indicate thhat the structtural designerr knows the
exact maagnitude of th
he loads that will
w be applieed to the struucture, and thee exact strenggth of all of
the structtural elementts, this is usuaally not the caase. Design looads are provvided by manyy codes and
standardss and, althou
ugh the valuess given are sspecific, signiificant uncerttainty is assoociated with
those maagnitudes. Loaads, load factors, and load combinationns are discusseed in Chapterr 2.
As
A is the casee for loading, significant uuncertainty is associated w
with the determ
mination of
the behav
vior and stren
ngth of structtural memberrs. The true inndication of lload-carryingg capacity is
given by
y the magnitu
ude of the loaad that causess the failure of a componnent or the strructure as a
whole. Failure
F
may eiither occur ass the physicall collapse of ppart of the buuilding, or be considered
to have occurred
o
if deflections,
d
fo
or instance, eexceed certainn predeterminned values. W
Whether the
failure iss the result of
o a lack of strength
s
(colllapse) or stifffness (deflecction), these phenomena
18 Chapter 1
Introduction
reflect the limits of acceptable behavior of the structure. Based on these criteria, the structure is
said to have reached a specific limit state. A strength failure is termed an ultimate limit state,
whereas a failure to meet operational requirements, such as deflection, is termed a serviceability
limit state.
Regardless of the approach to the design problem, the goal of the designer is to ensure
that the load on the structure and its resulting load effect, such as bending moment, shear force,
and axial force, in all cases are sufficiently below each of the applicable limit states. This ensures
that the structure meets the required level of safety or reliability.
Three approaches to the design of steel structures are permitted by the AISC
Specification:
1.
Allowable strength design (ASD)
2.
Load and resistance factor design (LRFD)
3.
Design by inelastic analysis
The design approaches represent alternative ways of formulating the same problem, and all
have the same goal. All three are based on the nominal strength of the element or structure. The
nominal strength, most generally expressed as Rn, is determined in exactly the same way, from
the exact same equations, whether used in ASD or LRFD. Some formulations of design by
inelastic analysis, such as plastic design, also use these same nominal strength equations whereas
other approaches to inelastic design model in detail every aspect of the structural behavior and do
not rely on the equations provided through the Specification. The use of a single nominal strength
expression for both ASD and LRFD permits the unification of these two design approaches. It
will become clear throughout this book how this approach has simplified steel design for those
who have struggled in the past with comparing the two available philosophies. The following
sections describe these three design approaches, any one of which is an acceptable approach to
structural steel design according to the AISC Specification.
1.9 FUNDAMENTALS OF ALLOWABLE STRENGTH DESIGN (ASD)
Prior to 2005, allowable strength design was referred to as allowable stress design. It is the oldest
approach to structural design in use today and has been the foundation of AISC Specifications
since the original provisions of 1923. Allowable stress design was based on the assumption that
under actual load, stresses in all members and elements would remain elastic. To meet this
requirement, a safety factor was established for each potential stress-producing state. Although
historically ASD was thought of as a stress-based design approach, the allowable strength was
always obtained by using the proper combination of the allowable stress and the corresponding
section property, such as area or elastic section modulus.
The current allowable strength design approach is based on the concept that the required
strength of a component is not to exceed a certain permitted or allowable strength under normal
in-service conditions. The required strength is determined on the basis of specific ASD load
combinations and an elastic analysis of the structure. The allowable strength incorporates a factor
of safety, Ω, and uses the nominal strength of the element under consideration. This strength
could be presented in the form of a stress if the appropriate section property is used. As a result of
doing this, the resulting stresses will most likely again be within the elastic range, although this is
not a preset requirement of the Specification.
The magnitude of the factor of safety and the resulting allowable strength depend on the
particular governing limit state against which the design must produce a certain margin of safety.
Introduction
Chapter 1 19
Safety factors are obtained from the Specification. This requirement for ASD is provided in
Section B3.2 of the Specification as
Ra ≤
Rn
Ω
(AISC B3-2)
which can be stated as
Required Strength (ASD) ≤
Nominal Strength
= Allowable Strength
Safety Factor
The governing strength depends on the type of structural element and the limit states
being considered. Any single element can have multiple limit states that must be assessed. The
safety factor specified for each limit state is a function of material behavior and the limit state
being considered. Thus, it is possible for each limit state to have its own unique safety factor. For
example, the limit state of yielding of a tension member is given by
Pn = Fy Ag
where Fy is the steel yield strength and Ag is the gross area of the member. The safety factor is
Ω = 1.67. Thus, for steel with a yield strength of 50 ksi, the allowable strength is
Pn 50.0 Ag
=
= 30 Ag
1.67
Ω
Design by ASD requires that the allowable stress load combinations of the building code
be used. Loads and load combinations are discussed in detail in Chapter 2.
1.10 FUNDAMENTALS OF LOAD AND RESISTANCE FACTOR DESIGN (LRFD)
Load and resistance factor design explicitly incorporates the effects of the random variability of
both strength and load. Because the method includes the effects of these random variations and
formulates the safety criteria on that basis, it is expected that a more uniform level of reliability,
and thus safety, for the structure and all of its components will be attained.
LRFD is based on the concept that the required strength of a component under LRFD
load combinations is not to exceed the design strength. The required strength is obtained by
increasing the load magnitude by load factors that account for load variability and load
combinations. The design strength is obtained by reducing the nominal strength by a resistance
factor that accounts for the many variables that impact the determination of member strength.
Load factors for LRFD are obtained from the building codes for strength design and will be
discussed in Chapter 2. As for ASD safety factors, the resistance factors are obtained from the
Specification.
The basic LRFD provision is provided in Section B3.1 of the Specification as
Ru ≤ φRn
(AISC B3-1)
which can be stated as
Required Strength (LRFD) ≤ Resistance Factor × Nominal Strength=Design Strength
20 Chapter 1
Introduction
Again considering the limit state of yielding of a tension member,
Pn = Fy Ag
and the resistance factor is φ = 0.90 . For steel with a yield strength of 50 ksi, the design strength
is
φPn = 0.90(50) Ag = 45 Ag
LRFD has been a part of the AISC Specifications since it was first issued in 1986.
1.11 INELASTIC DESIGN
The Specification permits a wide variety of formulations for the inelastic analysis of steel
structures through the use of Appendix 1. Any inelastic analysis method will require that the
structure and its elements be modeled in sufficient detail to account for all types of behavior. An
analysis of this type must be able to track the structure’s behavior from the unloaded condition
through every load increment to complete structural failure. The only inelastic design approach
that will be discussed in this book is plastic design (PD).
Plastic design is an approach that has been available as an optional method for steel
design since 1961, when it was introduced as Part 2 of the then current Specification. The limiting
condition for the structure and its members is attainment of the load that would cause the
structure to collapse, usually called the ultimate strength or the plastic collapse load. For an
individual structural member this means that its plastic moment capacity has been reached. In
most cases, due to the ductility of the material and the member, the ultimate strength of the entire
structure will not have been reached at this stage. The less stressed members can take additional
load until a sufficient number of members have exhausted their individual capacities so that no
further redistribution or load sharing is possible. At the point where the structure can take no
additional load, the structure is said to have collapsed. This load magnitude is called the collapse
load and is associated with a particular collapse mechanism.
The collapse load for plastic design is the service load times a certain load factor. The
limit state for a structure that is designed according to the principles of plastic design is therefore
the attainment of a mechanism. For this to occur, all of the structural members must be able to
develop the yield stress in all fibers at the most highly loaded locations.
There is a fine line of distinction between the load factor of PD and the safety factor of
ASD. The former is the ratio between the plastic collapse load and the service or specified load
for the structure as a whole, whereas the latter is an empirically developed, experience-based term
that represents the relationship between the elastic strength of the elements of the structure and
the various limiting conditions for those components. Although numerically close, the load factor
of plastic design and the factor of safety of allowable stress design are not the same parameter.
1.12 STRUCTURAL SAFETY AND INTEGRITY
The preceding discussions of design philosophies indicate that although the basic goal of any
design process is to ensure that the end product is a safe and reliable structure, the ways in which
this is achieved may vary substantially.
In the past, the primary goal for safety was to provide an adequate margin against the
consequences of overload. Load factor design and its offshoots were developed to take these
considerations into account. In real life, however, many other factors also play a role. These
include, but are not limited to the following:
1.
Variations of material strength
Introdduction
2.
2
Variaations of crosss-sectional sizze and shape
3.
3
Accurracy of metho
od of analysiss
4.
4
Influeence of workm
manship in shhop and field
5.
5
Presence and variaation of residuual stresses
6.
6
Lack of member sttraightness
7.
7
Variaations of locattions of load aapplication pooints
Chhapter 1 21
These factors consideer only some of
o the sourcees of variationn of the strenggth of a struccture and its
componeents. An even
n greater sourrce of variatioon is the loadding, which iss further com
mplicated by
the fact that
t different types
t
of load have differennt variational characteristiccs.
Thus,
T
a metho
od of design that does nott attempt to inncorporate thhe effects of sstrength and
load varriability will be burdened
d with unacccounted-for ssources of unncertainty. T
The realistic
solution, therefore, is to deal with safety as a prrobabilistic cooncept. This iis the foundattion of load
and resisstance factor design, wheere the probaabilistic charracteristics off load and sstrength are
evaluated
d and the ressulting safety
y margins dettermined statiistically. Eacch load type is given its
own speccific factor in
n each combin
nation, and eaach material liimit state is aalso given its own factor.
This metthod recognizzes that theree is always a finite, thouggh very smalll, chance thaat structural
failure will
w actually occur.
o
Howev
ver, this methhod does not attempt to atttach a speciffic value to
this probability. No sp
pecific level of
o probabilityy of failure is ggiven or impllied by the Sppecification.
In
I ASD, the variabilities of load andd strength arre not treatedd explicitly as separate
parameteers. They are lumped togeether through the use of a single factorr of safety. Thhe factor of
safety vaaries with eaach strength limit
l
state buut does not vvary with loaad source. A
ASD can be
thought of
o as LRFD with a singlee load factor.. LRFD desiggns are generrally expectedd to have a
more uniiform level of
o reliability than ASD deesigns. That is, the probaability of failuure of each
element in an LRFD design
d
will be the same, reegardless of tthe type of looad or load coombination.
Howeverr, a detailed analysis
a
of reeliability undeer the LRFD provisions shhows that reliability still
varies un
nder different load combin
nations. In AS
SD there is noo attempt to aattain uniform
m reliability;
rather, th
he goal is to simply havee a safe struccture, althouggh some elem
ments will bee safer than
others.
For
F the development of LR
RFD, load efffect (memberr force), Q, aand resistancee (strength),
R, are assumed to each have a variability that caan be describbed by the norrmal distributtions shown
by the bell-shaped
b
curves
c
in Fig
gure 1.15. Sttructures can be considerred safe as llong as the
resistance is greater th
han the load effect,
e
R > Q. If it were apppropriate to cconcentrate soolely on the
mean vallues, Qm and Rm, it would be relativelyy easy to ensuure a structuree’s safety. Hoowever, the
full reprresentation off the data shows an area where the ttwo curves ooverlap. Thiss area
Figure 1.15
1
Probability Distributtion, R and Q
222 Chapter 1
Introductio
on
Figure 1.16 Probability
y Distribution
n, (R – Q)
represents cases where th
he load effectt exceeds the resistance annd therefore identifies occuurrences
of failure. Saafety of the sttructure is a function
f
of th e size of this region of oveerlap. The sm
maller the
region of ov
verlap is, the lower
l
the prob
bability of faiilure.
Ano
other approacch to presentiing the data iis to look at the differencce between reesistance
and load efffect. Figure 1..16 shows thee same data aas Figure 1.155 but presentss it as (R – Q)). For all
cases where (R – Q) < 0, the structure is said to havve failed, and for all cases where this diifference
is positive, the
t structure is
i considered
d safe. In this presentation of the data, tthe shaded areea to the
left of the orrigin represen
nts the probab
bility of failurre. To limit thhat probabilityy of failure, thhe mean
value, (R – Q)m, must bee maintained at an approprriate distancee from the orrigin. This disstance is
shown in Fig
gure 1.16 as βσ
β (R–Q), wheree β is the reliaability index aand σ(R–Q) is thhe standard ddeviation
of (R – Q).
A th
hird represen
ntation of thee data is shoown in Figurre 1.17. In thhis case, thee data is
presented ass ln(R/Q). Thee logarithmic form of the ddata is a welll-conditioned representatioon and is
more useful in the derivaation of the factors
fa
requireed in LRFD. If we know tthe exact disttribution
of the resisttance and loaad effect data, the probabbility of failuure can be directly related to the
reliability in
ndex β. Unforrtunately, wee know the acctual distribut
utions for relaatively few reesistance
and load efffect componeents. Thus, we
w must relyy on other chharacteristics of the data, such as
means and standard
s
deviaations.
The statistical an
nalyses requirred to establissh an approprriate level off reliability haave been
carried out by
b the approp
priate specificcation commiittees, and thee resulting looad factors, reesistance
factors, and safety factorss have been established.
e
L
Load factors aare presented in the buildinng codes
whereas resiistance factorrs and safety factors
f
for eacch limit state are given in tthe Specificattion.
Figure 1.17
7 Probability
y Distribution
n, ln(R/Q)
Introduction
Chapter 1 23
Since the load combinations and resistance and safety factors have been established, the
reliability can be determined for specific design situations. The reliability index, β, is given in the
Specification Commentary as
β=
ln ( Rm Qm )
(AISC C-B3-2)
VR2 + VQ2
where Rm is the mean resistance, Qm is the mean load effect, as discussed earlier, and VR and VQ
are the coefficients of variation of resistance and load effect respectively. Design according to
LRFD is given by
Ru ≤ φRn
(AISC B3-1)
where the required strength, Ru is another term for the load effect, Q, Rn is the nominal strength
and ϕ is the resistance factor. The reliability of design is determined when the required strength is
exactly equal to the available strength. Thus, Equation B3-1 can be rewritten as Q = φRn . The
load effect will depend on the load combination being considered. Thus, for the LRFD live load
plus dead load combination, written in terms of the live-to-dead load ratio, L/D,
Q = 1.2 D + 1.6 L = (1.2 + 1.6 ( L D ) ) D = φRn
(1)
From Ravindra and Galambos1 the mean resistance is given by
Rm = Rn M m Fm Pm
(2)
and the coefficient of variation of the resistance is given by
VR = Vm2 + VF2 + VP2
(3)
Mm is the mean of the ratio of the actual yield stress to the specified yield stress and VM is
the coefficient of variation; Fm is the mean of the ratio of the actual section property to the
Manual value and VF is the coefficient of variation; and Pm is the mean of the ratio of the test
specimen strength to the predicted strength using the Specification equations and the actual
material and geometric properties and VP is the coefficient of variation.
Solving Equation 1 for Rn and substituting into Equation 2 yields
Rm =
(1.2 + 1.6 ( L D ) ) D M
φ
m
Fm Pm
(4)
Rearranging Equation 2 yields
M m Fm Pm =
Rm
Rn
(5)
Thus, combining Equations 4 and 5 gives
1
Ravindra, M.K. and Galambos, T.V. (1978), “Load and Resistance Factor Design for Steel,” Journal of the Structural
Division, American Society of Civil Engineers, Vol. 104, No. ST9, September, pp. 1,337–1,353.
24 Chapter 1
Introduction
⎛ R ⎞
Rm = D (1.2 + 1.6 ( L D ) ) ⎜ m ⎟
⎝ φRn ⎠
(6)
From Ravindra and Galambos the mean load effect for dead load plus live load is
Qm = Dm + Lm
(7)
which can, with some manipulation, be rewritten as
Qm = Dm + Lm = ( Dm D + ( Lm L )( L D ) ) D
(8)
They also give the coefficient of variation of the load effect which can be written as a
function of the live-to-dead load ratio as
VQ =
(( D
D )VD ) + ( ( Lm L )( L D )VL )
2
m
2
Qm
(9)
If Equations 6 and 8 are substituted into Equation C-B3-2 the reliability index, β, will be
given in terms of the live-to-dead load ratio, L D , and the resistance factor, ϕ. Thus,
β=
⎡R
ln ⎢ m
VR2 + VQ2 ⎢⎣ φRn
1
⎛
⎞⎤
1.2 + 1.6 ( L D )
⎜⎜
⎟⎟ ⎥
⎝ ( Dm D ) + ( Lm L )( L D ) ⎠ ⎥⎦
(10)
For LRFD, the live plus dead load combinations are 1.4D and (1.2D + 1.6L). The
effective dead load factor as a function of the live-to-dead-load ratio can be taken as
1.4
⎡
⎤
γ LRFDi = max ⎢
⎥
⎣1.2 + 1.6 ( L D )i ⎦
(11)
and the mean load effect dead load multiplier as
Qmi = Dm D + ( Lm L )( L D )i
(12)
Thus, Equation 10 can be generalized to address other LRFD load combinations as
follows:
⎡R ⎛
⎞⎤
⎡ R ⎛γ
1.2 + 1.6 ( L D )
⎞⎤
1
1
ln ⎢ m ⎜⎜
ln ⎢ m ⎜ LRFDi ⎟ ⎥
β=
(13)
⎟⎟ ⎥ =
2
2
2
2
VR + VQ ⎢⎣ φRn ⎝ ( Dm D ) + ( Lm L )( L D ) ⎠ ⎥⎦
VR + VQi ⎣ φRn ⎝ Qmi ⎠ ⎦
where γ LRFDi is the effective LRFD load factor for the load combination under consideration, Qmi
is the mean load effect multiplier for that load combination, and VQ is the coefficient of variation
of the load effect, all as a function of the varying load ratio as indicated by the subscript i.
To convert Equation 13 for use with ASD load combinations, γ LRFDi is replaced by γ ASDi
and φRn is replaced by Rn Ω . Thus, Equation 13 becomes
i
Introduction
β=
⎡ Rm ⎛ γ ASDi
ln ⎢
⎜
⎢⎣ ( Rn Ω ) ⎝ Qmi
V +V
1
2
R
2
Qi
⎞⎤
⎟⎥
⎠ ⎥⎦
Chapter 1 25
(14)
Based on extensive studies for A992 steel (Bartlett et al.2) and the original work for the
development of the 1986 AISC Specification by Ravindra and Galambos, the following values
can be used:
Mm = 1.055; VM = 0.058
Fm = 1.00; VF = 0.05
Pm = 1.02; VP = 0.06
Thus,
Rm = Rn (1.055 )(1.00 )(1.02 ) = 1.076 Rn
VR =
( 0.058)
2
+ ( 0.05 ) + ( 0.06 ) = 0.097
2
2
Based on Galambos et al.3 (1982), the ratio of mean to code specified dead and live loads
can be taken as, for dead load
Dm/D = 1.05; VD = 0.10
and for live load
Lm/L = 1.00; VL = 0.25
The values for Qmi and VQi will be functions of the live-to-dead load ratio.
The reliability index, β, based on Equations 13 and 14, for a live-to-dead load ratio from
1.0 to 5.7 is presented in Figure 1.18 for a compact wide-flange beam under uniform moment for
both LRFD and ASD. The figure is based on the load combination of live load plus dead load and
statistical variations consistent with those used in the development of the Specification. It is seen
that the reliability of design by LRFD is somewhat more uniform for this condition than design
by ASD and that at a live to dead load ratio of approximately 3, the two approaches yield the
same reliability. The higher the reliability index is, the safer the structure. Regardless of the
numerical value of β, any structure that meets the requirements of the Specification will be
sufficiently safe. A more detailed discussion of the statistical basis of steel design is available in
Load and Resistance Factor Design of Steel Structures.4 Since the introduction of the 2005 AISC
Specification, design by ASD and LRFD have essentially been equivalent and differ only by the
effect of load combinations.
2
Bartlett, R.M., Dexter, R.J., Graeser, M.D., Jelinek, J.J., Schmidt, B.J. and Galambos, T.V. (2003), “Updating Standard
Shape Material Properties Database for Design and Reliability,” Engineering Journal, American Institute of Steel
Construction, Vol. 40, No. 1, pp. 2–14.
3
Galambos, T.V., Ellingwood, B., MacGregor, J.G. and Cornell, C.A. (1982), “Probability-Based Load Criteria: Assessment
of Current Design Practice,” Journal of the Structural Division, American Society of Civil Engineers, Vol. 108, No. ST5,
May, pp. 959–977.
4
Geschwindner, L. F., Disque, R. O., and Bjorhovde, R. Load and Resistance Factor Design of Steel Structures. Englewood
Cliffs, NJ: Prentice Hall, 1994.
26 Chapter 1
Introduction
4.0
Reliability Index, β
3.5
3.0
2.5
2.0
1.5
ASD
LRFD
1.0
0.5
0.0
0
2
4
6
Live-to-Dead Load Ratio, L/D
Figure 1.18 Reliability Index vs Live-to-Dead Load Ratio for Compact Simply Supported
Wide-Flange Beams with Uniform Moment
General structural integrity requires a continuous load path to the ground for resisting all
gravity and lateral loads that might be applied to the structure. With the introduction of the 2016
AISC Specification, provisions that address structural integrity beyond these general requirements
have been introduced. The requirements in Section B3.9 are beyond normal strength requirements
and are intended to improve the connectivity of the structure and thus the performance of the
structure under undefined extraordinary events. These requirements apply only to a small set of
structures where additional structural integrity is mandated.
1.13
LIMIT STATES
Regardless of the design approach, ASD or LRFD, or the period in history of the design’s
execution, 1923 or 2018, all design is based on the ability of a structure or its elements to resist
load. This ability is directly related to how an element carries that load and how it might be
expected to fail, which is referred to as the element’s limit state. Each structural element can have
multiple limit states, and the designer is required to determine which of these limit states will
actually limit the structure’s strength.
There are two types of limit states to be considered: strength limit states and
serviceability limit states. Strength limit states are those limiting conditions that, if exceeded, will
lead to collapse of the structure or a portion of the structure, or to such serious deformations that
the structure can no longer be expected to resist the applied load. Strength limit states are
identified by the Specification, and guidance is provided for determination of the nominal
strength, Rn, the safety factor, Ω, and the resistance factor, φ. Examples of the more common
strength limit states found in the Specification are yielding, rupture, and buckling.
Serviceability limit states are not as well defined as strength limit states. If a
serviceability limit state is exceeded, it usually means that the structure has reached some
performance level that someone would find objectionable. The Specification addresses design for
serviceability in Chapter L and defines serviceability in Section L1 as “a state in which the
function of a building, its appearance, maintainability, durability, and the comfort of its occupants
are preserved under typical usage.” Chapter L lists deflections, drift, vibration, wind-induced
motion, thermal expansion and contraction, and connection slip as items to be considered,
although no specific limitations are given for any of these limit states.
Introduction
Chapter 1 27
Strength and serviceability limit states will be addressed throughout this book as
appropriate for the elements or systems being considered.
1.14 BUILDING CODES AND DESIGN SPECIFICATIONS
The design of building structures is regulated by a number of official, legal documents that are
known commonly as building codes. These cover all aspects of the design, construction, and
operation of buildings and are not limited to just the structural design aspects.
The model code currently in use in the United States is the ICC International Building
Code. Model codes are published by private organizations and have been adopted, in whole or in
part, by state and local governments as the legal requirements for buildings within their area of
jurisdiction. In addition to the model codes, cities and other governmental entities have written
their own local building codes. Unfortunately, since the adoption of a building code is in great
part a political activity, the regulations in use across the country are not uniform. A new
International Building Code is published every 3 years but not adopted as quickly as issued. Thus,
building codes with effective dates from 2003 to 2015 are still in use. In addition, governmental
bodies will often adopt a model code with local amendments. Because of the technical nature of
the AISC Specification, local amendments normally do not affect those aspects of steel design but
they often do modify the loading definitions and thus do ultimately affect steel design.
To the structural engineer, the most important sections of a building code deal with the
loads that must be used in the design, and the requirements pertaining to the use of specific
structural materials. The load magnitudes are normally taken from Minimum Design Loads for
Buildings and Other Structures, a national standard published by the American Society of Civil
Engineers (Structural Engineering Institute) as ASCE/SEI 7. The loads presented in ASCE/SEI 7
may be altered by the model code authority or the local building authority upon adoption,
although this practice adds complexity for designers who may be called upon to design structures
in numerous locations under different political entities. Throughout this book, ASCE/SEI 7 will
be referred to simply as ASCE 7 as it is most commonly referred to in the profession.
The AISC Specification is incorporated into the model building code by reference. The
Specification, therefore, becomes part of the code, and thus part of the legal requirements of any
locality where the model code is adopted. Locally written building codes also exist and the AISC
Specification is normally adopted within those codes by reference also. Through these adoptions
the AISC Specification becomes the legally binding standard by which all structural steel
buildings must be designed. However, regardless of the Specification rules, it is always the
responsibility of the engineer to ensure that their structure can carry the intended loads safely,
without endangering the occupants.
1.15
INTEGRATED DESIGN PROJECT
This section introduces a building to be used in subsequent chapters of this book as an integrated
design project. It is a relatively open-ended design project in that only a limited set of design
parameters are set at this point. Several options will be presented in subsequent chapters so that
the project can be tailored at the desire of the instructor.
The building is a four-story office building with one story below grade. It is located in
Downers Grove, Illinois, at approximately 42°N latitude and 88°W longitude. This is a 102,000
ft2 building with approximately 25,500 ft2 per above-grade floor. For the first three floors, the
floor-to-floor height is 13 ft 6 in. For the top floor, the floor-to-roof height is 14 ft 6 in. The
below-grade floor-to-floor height is 15 ft 6 in. The façade is a lightweight metal curtain wall that
extends 2.0 ft above the roof surface, and there is a 6.0 ft screen wall around the middle bay at the
roof to conceal mechanical equipment and roof access. All steel will receive spray-applied
fireproofing as necessary.
288 Chapter 1
Introductio
on
Based on prelimiinary discussiions with the architecturall design team
m, the design w
will start
with bay sizzes of 30.0 ftt in the east-w
west directionn and 45.0 ft,, 30.0 ft, andd 45.0 ft in thhe northsouth directiion, as shown
n in Figure 1.19. Represenntative floor aand roof fram
ming plans aree shown
in Figures 1.20
1
through 1.22. To acco
ommodate a two-story atrrium on the ffirst floor, thee second
floor framin
ng plan shows an opening bounded by column liness A, C, 4, annd 5. The lateeral load
resisting sysstem consists of a pair off three-bay m
moment frames in the east--west directioon and a
pair of one-b
bay chevron braced
b
framess in the north--south directiion.
Figure 1.19
1 Schemattic Plan for In
ntegrated
Design Problem
P
F
Figure 1.20 R
Representativve Second-Flooor
F
Framing Plann
ural analysis and design i s accomplishhed through tthe use of inntegrated
Much of today’s structu
d design softtware. Figure 1.23 shows an example of a compleete three-dim
mensional
analysis and
model of thee given prelim
minary framin
ng system deeveloped using RAM Strucctural System
m. Figure
1.24 shows the results of
o the same computer
c
moodel with thee gravity-onlyy structural eelements
removed.
F
Figure 1.21 Representativ
R
e First-, Third
d-, and
F
Fourth-Floor Framing Plan
n
Figu
ure 1.22 Reprresentative Rooof Framing
Plann
Introdduction
Figure 1.2
23 Three-Dim
mensional Com
mputer Modell
of Compleete Structure from
f
RAM Sttructural
System
11.16 PROBLEMS
11. Where could
d one find information about the provisionss
oof the 1961 AIS
SC Specificatio
on?
22. What resourrce would be most likely to
o assist in thee
ddetermination of
o properties of
o a steel mem
mber found in a
bbuilding built in
n 1954?
33. Which chaapter of the AISC
A
Specifica
ation providess
innformation abo
out:
a.
b.
c.
d.
e.
geeneral requirem
ments for analy
ysis and design
n
deesign of memb
bers for flexure
deesign of connections
deesign of memb
bers for combin
ned forces and
d
teension
reequirements fo
or design of structures to
o
en
nsure stability
44. In the AISC Steel Construcction Manual, where can onee
ffind:
a. th
he AISC Speciffication
b. deesign consideraations for boltss
c. diimensions and
d properties for structurall
stteel shapes
d. deesign of comprression membeers
55. List and deffine the three basic
b
goals of a design team
m
ffor the design of
o any building
g.
66. All structurees are composed of some or all
a of five basicc
sstructural typ
pes. List theese five basic structurall
ccomponents and provide an ex
xample of each
h.
Chhapter 1 29
Figurre 1.24 Threee-Dimensionaal Computer M
Model
Show
wing Only Latteral Load Reesisting System
ms from
RAM
M Structural Syystem
mple of each oof the followinng types of
7. Prrovide an exam
consttruction. To tthe extent poossible, identiffy specific
builddings in your ow
wn locale.
a.
b.
c.
d.
e.
Bearinng wall
Beam
m-and-column
Long--span
High--rise
Gablee-frame
8. W
What type of structural system uses the combined
propeerties of two oor more differrent types of m
materials to
resistt the applied looads?
9. Liist and describbe two types of lateral loaad resisting
systeems
commoonly
used
in
beam-aand-column
consttruction.
10. IIn designing a steel structture, what muust be the
primaary concern off the design enggineer?
11. P
Provide a simpple definition oof structural dessign.
12. D
Describe the ddifference betw
ween a strengthh limit state
of a sstructure and a serviceability limit state.
13. G
Give a descripttion of both thhe LRFD and A
ASD design
approoaches. What is the fundam
mental differencce between
the m
methods?
14. P
Provide a brieff description off plastic designn (PD).
15. IIdentify three ssources of variiation in the strrength of a
struccture and its com
mponents.
16. P
Provide three eexamples of strrength limit staates.
Provide three eexamples of seerviceability lim
mit states.
17. P
C
Chapterr 2
L
Loads,, Load
d
F
Factorrs, and
d
L
Load
C
Combiination
ns
Septem
mber 11 Memorrial Pavilion, N
New York
Coopyright Buro--Happold
2.11
INTROD
DUCTION
Material deesign specificcations, like the AISC SSpecification,, do not norrmally prescrribe the
magnitudes of loads thatt are to be ussed as the baasis for designn. These loadds vary basedd on the
usage or typ
pe of occupan
ncy of the bu
uilding, and thheir magnituddes are dictatted by the appplicable
local, region
nal, or state laaws, as prescrribed through the relevant bbuilding codee.
Buillding code lo
oads are given
n as nominal values. Thesse values are to be used inn design,
even though
h it is well known
k
that th
he actual loaad magnitudee will differ from these sspecified
values. Thiss is a commo
on usage of th
he term nomiinal, the same as will be uused for the nominal
depth of a stteel member, to be discusssed later. Theese nominal vvalues are dettermined on tthe basis
of material dimensions
d
and
a densities for dead loadd, load surveyys for live loads, weather data for
rain, snow and
a wind load
ds, and geolo
ogical data forr earthquake or seismic looads. These looads are
further described in Secttion 2.2. To be
b reasonablyy certain that these loads aare not exceeded in a
given structu
ure, code loaad values hav
ve generally bbeen higher th
than the actuaal loads on a random
structure at an arbitrary point
p
in time. This somew
what higher looad level also accounts forr the fact
that all strucctural loads will
w exhibit som
me random v ariations as a function of ttime and loadd type.
To properly
p
addrress this rand
dom variationn of load, an aanalysis refleecting time annd space
interdependeence should be
b used. This is called a stoochastic analy
lysis. Many sttudies have deealt with
this highly complex
c
phen
nomenon, esp
pecially as it ppertains to livve load in buiildings. Howeever, the
use of timee-dependent loads
l
is cum
mbersome andd does not aadd significanntly to the safety or
economy off the final design.
d
For most
m
design ssituations thee building coode will speecify the
magnitude of
o the loads as
a if they werre constant o r unchangingg. Their time and space vaariations
are accounteed for throug
gh the use off the maximuum load occuurring over a certain referrence or
return period. As an exam
mple, the Am
merican live l oad criteria aare based on a reference pperiod of
50 years, wh
hereas the Can
nadian criteriia use a 30-yeear interval.
The geographicaal location off a structure pplays an impoortant role foor several loaad types,
such as tho
ose from snow
w, wind, or earthquake. These loads also are speecified based upon a
reference peeriod appropriiate for the ty
ype of load.
300
Chapter 2
2.2
Loads, Load Factors and Load Combinations 31
BUILDING LOAD SOURCES
Many types of loads may act on a building structure at one time or another, and detailed data for
each are given later. Loads of primary concern to the building designer include:
1.
2.
3.
4.
5.
6.
Dead load
Live load
Snow load
Wind load
Seismic load
Special loads
These primary load types are characterized as to their magnitude and variability by the building
code, and are described in the ensuing paragraphs.
2.2.1
Dead Load
Theoretically, the dead load of a structure remains constant throughout its lifespan, unless
modifications are made. The dead load includes the self-weight of the structure, as well as the
weight of any permanent construction materials, such as stay-in-place formwork, partitions, floor
and ceiling materials, machinery, and other equipment. The dead load may vary from the
magnitude used in the design, even in cases where actual element weights are accurately
calculated.
The weight of all dead load elements can be exactly determined only by actually
weighing and/or measuring the various pieces that compose the structure. This is almost always
an impractical solution and the designer therefore usually relies on published data of building
material properties to obtain the nominal dead loads to be used in design. These data can be found
in such publications as ASCE 7, the model building codes, and product literature. Some variation
will thus likely occur in the real structure. Similarly, differences are bound to occur between the
weights of otherwise identical structures, representing another source of dead load variability.
However, compared to other structural loads, dead load variations are relatively small and the
actual mean values are quite close to the published data.
2.2.2
Live Load
Live load is the load on the structure that results from all of the non-permanent installations. It
includes the weight of the occupants, the furniture and moveable equipment, plus anything else
that the designer could possibly anticipate might occur in the structure. The fluctuations in live
load are potentially quite substantial. They vary from being essentially zero immediately before
the occupants take possession to a maximum value at some arbitrary point in time during the life
of the structure. The magnitude of the live load to be used in a design is obtained from the
appropriate building code. The actual live load on the structure at any given time may differ
significantly from that specified by the building code. This is one reason why numerous attempts
have been made to model live load and its variation and why measurements in actual buildings
continue to be made. Although the nominal live loads found in modern building codes have not
changed much over the years, the actual use of buildings has, and load surveys continue to show
that the specified load levels are still an adequate representation of the loads the structure should
be designed to resist.
The actual live load on a structure at any given point in time is called the arbitrary pointin-time live load (Lapt). Figure 2.1 shows the variation of the live load on a structure as might be
obtained from a live load survey. The load specified by the building code is always higher than
322 Chapter 2
Loads, Loaad Factors and
d Load Comb
binations
Figure 2.1 V
Variation of L
Live
me
Load with Tim
the actual load found in
n the building survey. In addition, a portion of thhe live load remains
constant. Th
his load comees from the reelatively perm
manent fixturees and furnishhings associaated with
a particular occupancy and
a can be reeferred to as the sustainedd live load (S
SLL). The occcupants
who enter an
nd leave the space
s
form an
nother part off the live loadd, raising andd lowering thee overall
live load maagnitude with time. This vaarying live loaad is called a transient livee load (TLL)..
2.22.3
Snow Load
L
now load miight be consiidered a form
m of live loaad, unique cconditions goovern its
Although sn
magnitude and
a distributio
on and it is trreated separattely in the loaad combinatioons used in ddesign. It
is the primaary roof load
d in many geographical
g
areas and heeavily depennds on local climate,
building exp
posure, and bu
uilding geometry.
Snow load data are
a normally based on surrveys that ressult in isoline maps showinng areas
of equal deepth of grou
und snow. Using
U
this m
method, annuaal extreme ssnowfalls havve been
determined over a perio
od of many years. Thesee data have bbeen analyzeed through sttatistical
models and the expected
d lifetime max
ximum snow loads estimaated. The refeerence period is again
the 50-year anticipated life of the struccture.
A major
m
difficultty is encounttered in transslating the groound snow looad into a roof snow
load. This iss accomplisheed through a semi-empiriccal relationshhip whereby thhe ground snnow load
is multiplieed by factorss to account for propertties such as roof geomettry and the thermal
characteristiics of the roo
of. In additio
on to the unifform snow looad determinned using thee ground
snow load, drifts often lead to areass of roofs w
with significanntly higher loads which m
must be
or. Work con
ntinues to be done
d
to improove the methhod of snow lload computaation and
accounted fo
to collect sn
nowfall data.
2.22.4
Wind Load
L
By its very nature, wind
d is a highly dynamic nattural phenom
menon. For thhis reason it iis also a
complex pro
oblem from a structural peerspective. W
Wind forces fl
fluctuate signiificantly and are also
influenced by
b the geomettry of the stru
ucture, includiing the heighht, width, deptth, plan and eelevation
shape, and the
t surroundin
ng landscape. The basic b uilding code approach to wind load annalysis is
to treat wind
d as a static load
l
problem
m, using the B
Bernoulli equuation to transslate wind sppeed into
wind pressu
ure. In an app
proach similaar to that use d for snow load, a semi-eempirical equuation is
used to givee the wind lo
oad at certain
n levels as a function of a number off factors representing
effects of features such ass wind gusts, topography, and structuraal geometry.
The data used for determin
ning wind l oads are baased on meaasured wind speeds.
Meteorologiical data for 3-second win
nd speed gustts have been accumulatedd over the conntiguous
United Statees and corrected to a stand
dard height oof 33 ft. Thesse data are theen used to m
model the
long-term characteristicss for a mean recurrence iinterval of 300, 700, 17000 or 3000 yyears, as
Chapter 2
Loads, Load Factors and Load Combinations 33
required. ASCE 7 and the model codes provide maps to be used as the foundation of wind force
calculations. Because local site characteristics often dictate wind behavior, there are locations for
which special attention must be given to wind load calculation. In addition, some buildings
require (or can benefit from) special attention in determining wind load magnitude. In these cases
it might be valuable to conduct wind tunnel tests before the structural design for wind is carried
out.
2.2.5
Seismic Load
The treatment of seismic load effects is extremely complicated because of the high variability of
this natural phenomenon and the many factors that influence the impact of an earthquake on any
particular structure. In addition, because the force the building is subject to is the result of the
reaction of the mass of the building responding to the ground moving, inertia effects must be
considered.
For most buildings it is sufficient to treat seismic effects through the use of equivalent
static loads over the height of the building, provided that the magnitudes of these equivalent static
loads properly reflect the dynamic characteristics of the seismic event. Many characteristics of the
problem must be quantified in order to establish the correct magnitudes of these static loads.
These characteristics include such factors as the ground motion and response spectra for the
seismic event and the structural and site characteristics for the specific project. Meanwhile, the
extension of seismic design requirements to all areas of the country through the current model
building codes is making seismic design compulsory for many more structures. Earthquakes are
no longer a design consideration only on the West coast.
2.2.6
Special Loads
Several other loads are sometimes important. These include impact, blast, and thermal effects.
Impact: Most building loads are static or essentially so, meaning that their rate of application is
so slow that the kinetic energy associated with their motion is insignificant. For example, a
person entering a room is actually exerting a dynamic load on the structure by virtue of his or her
motion. However, because of the relatively small mass and slow movement of the individual, the
kinetic energy is essentially zero.
When loads are large and/or their rate of application is very high, the influence of the
energy brought to bear on the structure as the movement of the load is suddenly restrained must
be taken into account. This phenomenon, known as impact, occurs as the kinetic energy of the
moving mass is translated into a load on the structure. Depending on the rate of application, the
effect of the impact is that the structure experiences a load that may be as large as twice the static
value of the same mass.
Impact is of particular importance for structures where machinery processes and similar
actions occur. Cranes, elevators, and equipment such as repetitive-action industrial machinery
could all produce impact loads that would need to be considered in a design. In addition,
vibrations may be induced into a structure either by these high-magnitude impact loads or by the
normally occurring occupancy loads. Although normal live load occupancy, such as walking, is
not likely to produce increased design load magnitudes, the potential for vibration from these
activities may require consideration in the design. Additional guidance may be found in AISC
Design Guide 11, Vibrations of Steel Framed Structural Systems Due to Human Activity.
Blast: Blast effects on buildings have become a more important design consideration during the
first years of the twenty-first century. Prior to that time, blast effects were primarily considered to
be accidental. These types of blast do not occur as often as impact for normal structures, but
should be considered under certain circumstances. Many structures designed for industrial
34 Chapter 2
Loads, Load Factors and Load Combinations
installations, where products of a volatile nature are manufactured, are designed with resistance to
blast as a design consideration. When the structure is called upon to resist the effects of blast, a
great deal of effort must be devoted to determining the magnitude of the blast to be resisted.
The threat of terrorism has been increasingly recognized since the attacks on the World
Trade Center and Pentagon on September 11, 2001. In order to take that threat into account,
owners must determine the level of threat to be designed for and design engineers must establish
the extent to which a particular threat will influence the design of a particular structure. Generally
speaking, analysis and design data for blast effects are somewhat limited. Researchers continue to
work toward establishing design guidelines that help determine blast effects and member strength
in response to blast. Guidance may be found in AISC Design Guide 26, Design of Blast Resistant
Structures and AISC Facts for Steel Buildings 2, Blast and Progressive Collapse.
Thermal Effects: Steel expands or contracts under changing temperatures and in so doing may
exert considerable forces on the structure if the members are restrained from moving. For most
building structures, the thermal effects are less significant than other loads for structural strength.
Because the movement of the structure results from the total temperature change and is directly
proportional to the length of the member experiencing the change, the use of expansion joints
becomes important in long dimensions of structural framing. When expansion or contraction is
not permitted, the resulting forces must be accommodated in the members. Additional guidance
may be found in AISC Design Guide 3, Serviceability Design Considerations for Steel Buildings
and AISC Design Guide 7, Industrial Buildings: Roofs to Anchor Rods.
The AISC Specification includes guidance on the design of steel structures exposed to
fire. Appendix 4 provides criteria for the design and evaluation of structural steel components,
systems, and frames for fire conditions. In the current building design environment, design for
fire is usually accomplished by means of a prescriptive approach defined in the Specification as
design by qualification testing. If the actual thermal effects of a fire are to be addressed, the
Specification permits design by engineering analysis. Additional guidance may be found in AISC
Design Guide 19, Fire Resistance of Structural Steel Framing.
2.3 BUILDING LOAD DETERMINATION
Once the appropriate building load sources are identified, their magnitudes must be determined.
Methods to determine these magnitudes are set by the applicable building code for each load
source. The following sections provide general guidance to determine the building load
magnitudes, but for specific details, the applicable building code must be consulted.
2.3.1
Dead Load
Building dead load determination can be either quite straightforward or very complex. If the sizes
of all elements of the structural system are known before an analysis is conducted, actual material
weights may be determined and applied in the structural analysis. Selected unit weights of typical
building materials are given in Table 2.1. Manual Table 17-13 and ASCE 7 Table C3-1 provide
the weights of building materials, and product catalogs provide weights of things such as building
mechanical equipment.
If the final sizes are not known, as is normally the case in the early stages of design,
assumptions need to be made to estimate the self-weight of the structure. This necessitates an
iterative process of refinement as the design and its corresponding weight are brought together.
Chapter 2
Table 2.1
Unit Weights of Typical Building Materials
Weight (lb/ft3)
Material
Aluminum
Brick
Concrete
Reinforced, with stone aggregate
Block, 60 percent void
Steel, rolled
Wood
Fir
Plywood
2.3.2
Loads, Load Factors and Load Combinations 35
170
120
150
87
490
32–44
36
Live Load
As discussed earlier, live load magnitudes are established by the applicable building code. Table
2.2 provides values for the minimum uniformly distributed live loads for buildings for selected
occupancies.
In design, how much of the load is supported by a particular element can be determined
by multiplying the uniformly distributed load (lb/ft2) by the tributary area, AT (ft2). The tributary
area method is a simplified approach for visualizing the load on a structural element without
performing the actual equilibrium calculations. It does, however, provide the same result because
it is fundamentally based on an equilibrium analysis. Simplified tributary areas for some
structural members are given in Figure 2.2.
Although the concept of the tributary area can be used to determine the load on a
member, an equally important concept is the influence area, AI. The influence area is significant
because it reflects the area over which any applied load has an influence on the member of
interest. The member under consideration will experience no portion of the load applied outside
of the influence area. Table 2.3 provides the relationship between tributary area and influence
area, as defined by ASCE 7, for several specific structural elements, where AI = KLLAT. Several of
the values in this table are simplified relative to the actual relationship, and it is always
permissible to calculate the actual influence area.
As the influence area increases for a particular member, the likelihood of the full codespecified nominal live load actually occurring on it decreases. Because the code cannot predict
the likelihood of that full area being loaded, the magnitude of the specified load is set without
consideration of loaded area. Thus, the tabulated values are referred to as the unreduced nominal
Table 2.2
Minimum Uniformly Distributed Live Loads for Building Designa
Occupancy or use
Roof, flat
Residential dwellings, apartments, hotel rooms, school classrooms
Offices
Corridors above first floor
Auditoriums (fixed seats)
Retail stores
Bleachers
Library stacks
Heavy manufacturing and warehouses
a
Data are taken from ASCE 7.
Load (lb/ft2)
20
40
50
80
60
75–100
100
150
250
366 Chapter 2
Loads, Loaad Factors and
d Load Comb
binations
Figu
ure 2.2
Sim
mplified Tribu
utary Areas foor Some Strucctural Membeers
live load. To
T account fo
or the size of the influennce area and thereby provvide a more realistic
predictor off the actual live load on th
he structure, a live load reeduction facttor is introducced. For
2
influence arreas greater than
t
400 ft the live loadd may be redduced accordding to the liive load
reduction eq
quation:
⎛
15 ⎞
L = Lo ⎜⎜ 0.25 +
(2.1)
⎟
AI ⎟⎠
⎝
Table 2.3 Live Load Ellement Factorr, KLLa
Element
Interior colu
umns
Exterior colu
umns withoutt cantilever sllabs
Edge columns with cantillever slabs
Corner colum
mns with can
ntilever slabs
Edge beamss without canttilever slabs
Interior beam
ms
All other meembers not id
dentified abov
ve, including
Edge beeams with can
ntilever slabs
Cantilev
ver beams
One-way slabs
Two-waay slabs
Memberrs without pro
ovisions for continuous
c
sheear
tran
nsfer normal to
o their span
Note: In lieu
u of the values above, calcu
ulation of KLLL is permittedd.
a
Data are tak
ken from ASC
CE 7.
KLL
4
4
3
2
2
2
1
Loads, Load Factors and Load Combinations 37
Chapter 2
where
L = reduced live load
Lo = code-specified nominal live load
AI = influence area = KLLAT
Limitations on the use of this live load reduction are spelled out in ASCE 7. The most
important one is that L may not be taken less than 0.50Lo for members supporting one floor nor
less than 0.40Lo for members supporting more than one floor. Other limitations and exceptions
are given for loads over 100 lb/ft2, passenger vehicle garages and assembly buildings.
2.3.3
Snow Load
Roof snow load calculations start with determination of the ground snow load for the building
site. Table 2.4 provides typical ground snow load values for selected locations. The complete
picture of ground snow load is provided in the appropriate building code. In many locations,
however, the snowfall depth is a very localized phenomenon and the variability is such that it is
not appropriate to map those values. In these situations, local building officials should be
consulted to determine what the local requirements are.
The determination of roof snow load is more complex and there are many acceptable
approaches. Roof snow load on an unobstructed flat roof, as given in ASCE 7, is
pf = 0.7CeCtIspg
(2.2)
where pg is the ground snow load determined from the appropriate map, as shown in Figure 2.3.
Ground snow load by location may also be found using the website
http://snowload.atcouncil.org/.
The exposure factor, Ce, varies from a low of 0.7 for a windswept site above the tree line
to a high of 1.2 for a sheltered site. The thermal factor, Ct, accounts for the melting effect of heat
escaping through the roof. It varies from 0.85 for continuously heated greenhouses to 1.3 for
freezer buildings. The importance factor for snow, Is, varies from 0.8 for buildings that represent
a low risk to human life, risk category I, to a high of 1.2 for buildings designated as essential, risk
category IV. Buildings of normal risk fall into risk category II where the snow importance factor
is 1.0. Numerous other factors enter into the determination of roof snow load, including roof
slope, roof configuration, snowdrift, and additional load due to rain on the snow. The applicable
building code or ASCE 7 should be referred to for the complete provisions regarding snow load
determination.
Table 2.4
Typical Ground Snow Loads, pga
Location
Whittier, Alaska
Talkeetna, Alaska
Portland, Maine
Minneapolis, Minnesota
Hartford, Connecticut
Chicago, Illinois
St. Louis, Missouri
Raleigh, North Carolina
Memphis, Tennessee
Atlanta, Georgia
a
Data are taken from ASCE 7.
Load (lb/ft2)
300
120
60
50
30
25
20
15
10
5
388 Chapter 2
Loads, Loaad Factors and
d Load Comb
binations
Figurre 2.3
2.33.4
Groun
nd Snow Load
d Map from A
ASCE 7 (Usedd with permisssion from ASC
CE)
Wind Load
L
As with sno
ow load and other geograaphically linkked environm
mental loads, tthe starting ppoint for
wind load calculation is map based. In the case off wind loads, maps are proovided based upon 3second gust wind velocitties in the buiilding code (ssee Figure 2.44). Different wind speed m
maps are
provided forr the differen
nt risk categorries of buildiings. The diffferent maps take the placce of the
importance factor used in
n the determin
nation of som
me other desiggn loads such as was illustrrated for
snow. Thus there are fou
ur basic wind speed maps ffor the four rrisk categories. Table 2.5 pprovides
the wind speeed data for several
s
selectted locations with varyingg wind velociities for risk ccategory
II. Wind speeed by locatio
on may also be found usingg the website http://windsppeed.atcounciil.org/.
These data must be transform
med into wind pressures onn a given buildding to determ
mine the
appropriate design wind
d loads. This transformatioon must takee into accounnt, in additionn to the
map, such faactors as
importance of the buildiing included through seleection of the appropriate m
height abov
ve the ground
d, relative sh
heltering of tthe site, topoography, and the directionn of the
dominant wiinds.
ASCE 7 offers seeveral approacches to the d etermination of design wiind load for tthe main
wind force resisting
r
systeem (MWFRS
S). The directtional proceduure found in ASCE 7, Chhapter 27
for building
gs of all heigh
hts, where thee load must bbe applied too the windwarrd, leeward, aand side
walls converrts the mappeed data to velo
ocity pressuree at height z:
qz = 0.00256
0
K z K zzt K d K eV 2
(2.3)
where V is the wind sp
peed obtained
d from the aappropriate rrisk category map. The eexposure
coefficient, Kz, ranges from
f
0.57 at grade for thhe least criticcal exposure to 1.89 at a 500 ft
elevation forr the most criitical exposure. The topogrraphic factor, Kzt, is a calcuulated factor w
which is
intended to account for th
he wind speed
d-up in locatiions of a hill, ridge or escaarpment. For all other
locations, Kzt = 1.0. The directionality
y coefficient, Kd, for the M
MWFRS is taaken as 0.85 for both
buildings an
nd componentts and claddin
ng. For other structural typpes it may vaary up to a m
maximum
of 1.0. The ground elev
vation factor,, Ke, is intennded accountt for air denssity at the siite. It is
permitted to
o take Ke =1.0
0 for all locattions. For a siite with grounnd elevation at 6,000 ft abbove sea
level, Ke =0..80, its lowestt value.
Chapter 2
Loads, Loadd Factors and Load Combinnations 39
Figure 2.4
2
Basic Wind
W
Speed, 3--Second Gustt at 33 ft Abovve Ground foor Risk Categoory II
Exposuree Category C from ASCE 7 (Used with PPermission from
m ASCE)
Once
O
the velo
ocity pressuree is determinned through E
Equation 2.3, it must be converted to
the extern
nal design wiind pressure. For
F the MWF
FRS, this is giiven by
p = qGCp – qi(GCpi)
(2.4)
where
p
=
q
=
qi =
G =
Cp =
GC
G pi =
design
n wind pressu
ure
velocity pressure from
f
Equationn 2.3 as a funnction of wall location and height
velocity pressure from
f
Equationn 2.3 at meann roof height
gust-eeffect factor taken
t
as 0.85 for rigid buildings
extern
nal pressure coefficient
c
dettermined as a function of bbuilding geom
metry
intern
nal pressure coefficient takken as ±0.18 for enclosed buildings
The
T actual forces applied to the structuure are then ddetermined byy multiplyingg the design
wind preessures by thee tributary areeas at each leevel. Wind prressure distribbution on the main wind
force resisting system
m will vary with height accoording to the variation of tthe exposure coefficient,
Kz. The pattern of disstribution is illustrated
i
in Figure 2.5. T
To account foor the fact thaat wind may
come fro
om any directiion, ASCE 7 requires that the wind loadd be determinned for eight w
wind
5 Approxim
mate Represen
ntative Wind Velocities annd Resulting D
Dynamic Presssuresa
Table 2.5
Dynam
mic pressure (llb/ft2)
Wind velo city (mph)
Location
n
V
= 0.00256V 2
Miami, Florida
F
1770
74.0
Houston,, Texas
1330
43.3
36.9
New Yorrk, New York
k
1220
33.9
Chicago,, Illinois
1115
31.0
San Fran
ncisco, Califorrnia
1110
a
Data aree taken from ASCE
A
7 for Risk
R Categoryy II.
400 Chapter 2
Loads, Loaad Factors and
d Load Comb
binations
Figure 22.5 Wind Looad Distributiion
over Buillding Height
directions at 45° intervaals in addition
n to three quuarters of thee load along the principall axis in
conjunction with a torsio
onal moment. This results in the wind lload being appplied to the sstructure
in 12 different direction
ns and magn
nitudes withoout considerration yet off load combiinations.
Because buiilding codes have differen
nt requiremennts for wind load determ
mination, the ddesigner
ASCE 7
must review
w the provisions specified in
i the governning code. If tthere is no buuilding code, A
should be ussed.
2.33.5
Seismicc Load
Perhaps thee most rapidlly fluctuating
g area of buuilding load ddeterminationn is that for seismic
design. Alth
hough there have
h
been maany advances in the use off dynamic annalysis for earrthquake
response, co
ommon practiice is still to model the phhenomenon uusing a static load. Seismicc design
requirementts are based on
n the seismic design categgory, A througgh F, with A bbeing the leasst severe
and requirin
ng no consideeration of seissmic loads. F
For a large m
majority of buuildings in all seismic
risk categories, ASCE 7 permits the equivalent laateral force prrocedure for determinatioon of the
building basse shear throu
ugh the expresssion
V = C sW
(2.5)
where
V = seismic base shear
Cs = seismic resp
ponse coefficiient
W = effective seismic weight of
o the buildinng
The seismic response coeefficient is dettermined by
Cs =
SDS
( I)
R
(2.6a)
e
but for T ≤ TL neeed not be greeater than
Cs =
S D1
T (R / Ie )
(2.6b)
Cs =
S D1TL
T (R / Ie )
(2.6c)
and for T > TL
2
Chapter 2
Loads, Loadd Factors and Load Combinnations 41
where
SDS =
SD1 =
T =
TL =
R =
Ie =
design
n spectral response accelerration for shoort period
design
n spectral response accelerration for 1 seec period
fundaamental buildiing period
long period
p
transition period
response modificattion factor
seism
mic importancee factor whichh is 1.0 for risk categories I and II
The
T U.S. Geo
ological Surveey provides a web based toool to determ
mine site speciific data for
seismic design that precludes
p
thee need to usee the maps ffound in ASC
CE 7. It is aavailable at
http://earrthquake.usgss.gov/designm
maps/us/appliccation.php.
Once
O
the seism
mic base sheaar is determinned it must bee converted to a horizontaal load to be
applied to
t the seismiic force resissting system. For a buildding with uniiform story hheights and
uniform seismic weight distributiion over the height of thhe building, this results in a linear
variation
n of load from
m a maximum at the top to zzero at the boottom as illusttrated in Figuure 2.6.
Each
E
seismic force resistin
ng system is aassigned a response modiffication factoor, R, by the
building code, accord
ding to its ab
bility to resisst seismic forrces through a ductile ressponse. The
most ducctile steel sysstems, e.g., special
s
momeent frames, arre assigned R = 8 while less ductile
systems are assigned
d lower valuees. Higher R values leadd, through Eqquation 2.6, to a lower
required base shear and
a thus to lo
ower lateral lload on the sstructure. Thee resulting loower lateral
load, ho
owever, leadss to higher demand forr ductility aand corresponnding speciaal detailing
requirem
ments. In casees where app
propriate, thee selection of R = 3 usuually permits a building
structure to be design
ned according
g to the AISC
C Specificatioon without use of any speccial seismic
provision
ns. If a valuee of R greaterr than 3 is uused in designn of a buildinng structure, or in cases
where sp
pecifically req
quired, the design
d
must pproceed accorrding to the additional prrovisions of
ANSI/AIISC 341-16 Seismic Provvisions for Strructural Steeel Buildings. T
This is discusssed further
in Chapteer 13.
As
A with the other
o
environmental loads discussed heere, the details of load determination
for seism
mic response must be foun
nd in the apppropriate buildding code. A
ASCE 7 requiires that the
seismic load
l
be applieed to the struccture in such a way as to prroduce the moost critical load effects.
Figurre 2.6 Seism
mic Force Disstribution
over H
Height for Buuilding with U
Uniform
Story Height and S
Story Seismicc Weight
42 Chapter 2
Loads, Load Factors and Load Combinations
2.4 LOAD COMBINATIONS FOR ASD AND LRFD
In addition to specifying the load magnitudes for which building structures must be designed,
building codes specify how the individually defined loads should be combined to obtain the
maximum load effect. Care must be exercised in combining loads to determine the most critical
combination because all loads are not likely to be at their maximum magnitude at the same time.
For instance, it is unlikely that the maximum snow load and maximum wind load would occur
simultaneously. Another unlikely occurrence would be a design earthquake occurring at the same
time as the maximum design wind. Thus, building codes specify which loads are to be combined
and at what magnitude they should be considered. The designer must exercise judgment when
combining loads in situations where the normal expectations of the building code might not be
satisfied or where some particular combination would result in a greater demand than previously
identified.
The two design philosophies addressed in the AISC Specification are the direct result of
the two approaches to load combinations presented in current building codes and ASCE 7. ASD
uses the load combinations defined in ASCE 7 as being for allowable stress design and LRFD
uses the load combinations defined as being for strength design. The provisions in ASCE 7 for
allowable stress design combine loads normally at or below their nominal or serviceability levels,
the load magnitudes discussed in Section 2.3. These load combinations were historically used to
determine the load effect under elastic stress distributions, and those stresses were compared to
the allowable stresses established at some arbitrary level below failure, indicated by either the
yield stress or the ultimate stress. Including only dead, live, wind, and snow loads, the load
combinations presented in ASCE 7 Section 2.4 for ASD can be written as
1.
2.
3.
4.
5.
6.
7.
Dead
Dead + Live
Dead + Roof Live
Dead + 0.75 Live + 0.75 Roof Live
Dead + (0.6 Wind)
Dead + 0.75 Live + 0.75 (0.6 Wind) + 0.75 Roof Live
0.6 Dead + 0.6 Wind
In load combinations 3, 4, and 6, Roof Live refers to a code-specified roof live load, a snow load,
or a rain load, whichever is greater.
Used with the current AISC Specification, these load combinations are not restricted to an
elastic stress distribution as was the practice in the distant past. Even recent past ASD editions of
the AISC Specification had transitioned to use strength past the elastic level in many cases. The
current Specification is a strength-based specification, not a stress-based one, and the requirement
for elastic stress distribution is no longer applicable. This has no impact on the use of these load
combinations but may have some historical significance to those who were educated primarily
with this previously used interpretation.
The second approach available in ASCE 7 combines loads at an amplified level. These
combinations, referred to as strength load combinations, permit one to investigate the ability of
the structure to resist loads at its ultimate strength. In this approach, loads are multiplied by a load
factor that incorporates both the likelihood of the loads occurring simultaneously at their
maximum levels and the margin against which failure of the structure is measured. Again, using
only dead, live, wind, and snow loads, the load combinations presented in ASCE 7 Section 2.3, if
the live load is not greater than 100 lb/ft2 (psf), for LRFD are
Chapter 2
1.
1
2.
2
3.
3
4.
4
5.
5
Loads, Loadd Factors and Load Combinnations 43
1.4 Dead
D
1.2 Dead
D
+ 1.6 Liv
ve + 0.5 Rooff Live
1.2 Dead
D
+ 1.6 Roof Live + (0.55 Live or 0.5 Wind)
1.2 Dead
D
+ 1.0 Wiind + 0.5 Livee + 0.5 Roof L
Live
0.9 Dead
D
+ 1.0 Wiind
As was the case for ASD load combinations
c
, Roof Live is intended to be taken as a codespecified
d roof live loaad, a snow loaad, or a rain looad, whicheveer is greatest.
For
F both ASD
D and LRFD, load combinnations that innclude wind m
must be appliied with the
wind loaad acting ind
dependently in each of tthe directionns specified in ASCE 7. For those
combinattions that include seismic load, that loaad must also be applied inn the directionns specified
by ASCE
E 7. Load casses with a red
duced dead looad in combiination with w
wind or seism
mic load are
used to account
a
for siituations wheere the appliccation of deadd load might reduce the eeffect of the
lateral lo
oad. This cou
uld be importaant if the actuual dead loadd is less than anticipated. The design
method to
t be used, an
nd thus the ch
hoice of load combinationns, is at the diiscretion of thhe designer.
All curreent building codes
c
permit either ASD or LRFD, annd the AISC Specificationn provisions
address all
a limit statees for each ap
pproach. As discussed in Chapter 1, thhe resulting ddesign may
differ slig
ghtly for each
h design philo
osophy becauuse the approaach taken to eensure safety is different,
but safety
y is assured when
w
the applicable buildiing code and the AISC Specification arre followed,
regardlesss of the desig
gn approach.
2.5
Load
d Calculation
ns
To underrstand the im
mpact of thesee two approacches on analyysis, it is helppful to compuute the load
effect forr a variety off structural members
m
accorrding to bothh ASD and LR
RFD load com
mbinations.
The floorr plan of a tw
wo-story officee building is ggiven in Figuure 2.7. Load case 2 for dead plus live
load is considered
c
fo
or several beaams and coluumns. The bbuilding is ann office buildding with a
nominal live load of 50
5 psf and a calculated deead load of 700 psf. Assum
me all beams and girders
are simplly supported.
Figure 2.7
2
Floor Plaan of a Two-S
Story Office B
Building
44 Chapter 2
Loads, Load Factors and Load Combinations
Example 2.1
Girder Load
Calculation
Goal:
Determine the required moment strength for a girder from the framing plan
shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.
Given:
Girder AB on line 2-2 if the floor deck spans from line 1-1 to 2-2 to 3-3,
resulting in a uniformly distributed load on Girder AB:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area: AT = (40)(20) = 800 ft2
Influence area: AI = 2AT = 2(800) = 1600 ft2
Live load reduction:
15
15
0.25 +
= 0.25 +
= 0.625 ≥ 0.50
1600
AI
Since this member supports one floor the maximum reduction is 0.50.
For
LRFD
Step 2:
Determine the required moment strength for the LRFD load combination
LRFD loads per foot:
1.2 Dead = 1.2(70 psf)(20 ft) = 1680 lbs/ft
1.6 Live = 1.6(0.625)(50 psf)(20 ft) = 1000 lbs/ft
wu = 1.2 Dead + 1.6 Live
= 1680 + 1000 = 2680 lbs/ft = 2.68 kips/ft
Required moment strength (LRFD),
w l 2 2.68(40) 2
Mu = u =
= 536 ft-kips
8
8
For
ASD
Step 2:
Determine the required moment strength for the ASD load combination
ASD loads per foot:
Dead = (70 psf)(20 ft) = 1400 lbs/ft
Live = 0.625(50 psf)(20 ft) = 625 lbs/ft
wa = Dead + Live = 1400 + 625 = 2030 lbs/ft = 2.03 kips/ft
Required moment strength (ASD),
w l 2 2.03(40) 2
Ma = a =
= 406 ft-kips
8
8
Example 2.2
Beam Load
Calculation
Goal:
Determine the required moment strength for a beam from the framing plan
shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.
Given:
Floor beam 2-3 on line D-D if the floor deck spans from line C-C to D-D to
E-E, resulting in a uniformly distributed load on floor beam 2-3:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area: AT = (20)(30) = 600 ft2
Chapter 2
Loads, Load Factors and Load Combinations 45
Influence area: AI = 2AT = 2(600) = 1200 ft2
Live load reduction:
15
0.25 +
= 0.683 ≥ 0.50
1200
Since this member supports one floor the maximum reduction is 0.50.
For
LRFD
Step 2:
Determine the required moment strength for the LRFD load combination
LRFD loads per foot:
1.2 Dead = 1.2(70 psf)(30 ft) = 2520 lbs/ft
1.6 Live = 1.6(0.683)(50 psf)(30 ft) = 1640 lbs/ft
wu = 1.2 Dead + 1.6 Live
= 2520 + 1640 = 4160 lbs/ft = 4.16 kips/ft
Required moment strength (LRFD),
w l 2 4.16(20) 2
Mu = u =
= 208 ft-kips
8
8
For
ASD
Step 2:
Determine the required moment strength for the ASD load combination
ASD loads per foot:
Dead = (70 psf)(30 ft) = 2100 lbs/ft
Live = 0.683(50 psf)(30 ft) = 1020 lbs/ft
wa = Dead + Live = 2100 + 1020 = 3120 lbs/ft = 3.12 kips/ft
Required moment strength (ASD), M a =
wa l 2 3.12(20) 2
=
= 156 ft-kips
8
8
Example 2.3
Beam Load
Calculation
Goal:
Determine the required moment strength for a beam from the framing plan
shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.
Given:
Floor beam 2-3 on line C-C if the floor deck spans from line B-B to C-C to
D-D, resulting in a uniformly distributed load on floor beam 2-3:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area: AT = (20)(30+40)/2 = 700 ft2
Influence area: AI = 2AT = 2(700) = 1400 ft2
Live load reduction:
15
0.25 +
= 0.651 ≥ 0.50
1400
Since this member supports one floor the maximum reduction is 0.50.
For
LRFD
Step 2:
Determine the required moment strength for the LRFD load combination
LRFD loads per foot:
46 Chapter 2
Loads, Load Factors and Load Combinations
1.2 Dead = 1.2(70 psf)(30 ft + 40 ft)/2 = 2940 lbs/ft
1.6 Live = 1.6(0.651)(50 psf)(30 ft + 40 ft)/2 = 1820 lbs/ft
wu =1.2 Dead + 1.6 Live = 2940 + 1820 = 4760 lbs/ft = 4.76 kips/ft
Required moment strength (LRFD),
w l 2 4.76(20) 2
Mu = u =
= 238 ft-kips
8
8
For
ASD
Step 2:
Determine the required moment strength for the ASD load combination
ASD loads per foot:
Dead = (70 psf)(30 ft + 40 ft)/2 = 2450 lbs/ft
Live = 0.651(50 psf)(30 ft + 40 ft)/2 = 1140 lbs/ft
wa = Dead + Live = 2450 + 1140 = 3590 lbs/ft = 3.59 kips/ft
Required moment strength (ASD),
w l 2 3.59(20) 2
Ma = a =
= 180 ft-kips
8
8
Example 2.4
Column Load
Calculation
Goal:
Determine the required axial strength for a column from the framing plan
shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.
Given:
Interior column D-2 to support one floor regardless of deck span direction:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area: AT = (30)(20) = 600 ft2
Influence area: AI = 4AT = 4(600) = 2400 ft2
Live load reduction:
15
0.25 +
= 0.556 ≥ 0.50
2400
Since this member supports one floor the maximum reduction is 0.50.
For
LRFD
Step 2:
Determine the required axial strength for the LRFD load combination
LRFD load entering column at this level:
1.2 Dead = 1.2(70 psf)(600 ft2) = 50,400 lbs
1.6 Live = 1.6(0.556)(50 psf)(600 ft2) = 26,700 lbs
1.2 Dead + 1.6 Live = 50,400 + 26,700 = 77,100 lbs = 77.1 kips
For
ASD
Step 2:
Required axial strength (LRFD), Pu = 77.1 kips
Determine the required axial strength for the ASD load combination
ASD load entering column at this level:
Dead = (70 psf)(600 ft2) = 42,000 lbs
Chapter 2
Loads, Load Factors and Load Combinations 47
Live = 0.556(50 psf)(600 ft2) = 16,700 lbs
Dead + Live = 42,000 + 16,700 = 58,700 lbs = 58.7 kips
Required axial strength (ASD), Pa = 58.7 kips
Example 2.5
Column Load
Calculation
Goal:
Determine the required axial strength for a column from the framing plan
shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.
Given:
Exterior column D-4 to support one floor regardless of deck span direction:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area: AT = (30)(10) = 300 ft2
Influence area: AI = 4AT = 4(300) = 1200 ft2
Live load reduction:
15
0.25 +
= 0.683 ≥ 0.50
1200
Since this member supports one floor the maximum reduction is 0.50.
For
LRFD
Step 2:
Determine the required axial strength for the LRFD load combination
LRFD load entering column at this level:
1.2 Dead = 1.2(70 psf)(300 ft2) = 25,200 lbs
1.6 Live = 1.6(0.683)(50 psf)(300 ft2) = 16,400 lbs
1.2 Dead + 1.6 Live = 25,200 + 16,400 = 41,600 lbs = 41.6 kips
For
ASD
Step 2:
Required axial strength (LRFD), Pu = 41.6 kips
Determine the required axial strength for the ASD load combination
ASD load entering column at this level:
Dead = (70 psf)(300 ft2) = 21,000 lbs
Live = 0.683(50 psf)(300 ft2) = 10,200 lbs
Dead + Live = 21,000 + 10,200 = 31,200 lbs = 31.2 kips
Required axial strength (ASD), Pa = 31.2 kips
The floor plan of a 10-story office building is given in Figure 2.8. As with the previous
examples, load case 2 for dead plus live load is considered for several columns to illustrate the
application of live load reduction to members that support loads from two or more floors.
The building is to be designed for flexibility so the nominal live load will be 80 psf and
the calculated dead load is 100 psf.
488 Chapter 2
Loads, Loaad Factors and
d Load Comb
binations
Fiigure 2.8 Floor Plan of a
100-Story Office Building
Exxample 2.6
Coolumn Load
d
Caalculation
Goa
al:
Deterrmine the req
quired axial sttrength for a column from
m the framingg plan
shown
n in Figure 2..8 due to live load of 80 pssf and dead looad of 100 psff.
Giv
ven:
Interior column C--3 supporting one level witth 18K3 steell joists spanniing as
shown
n in all panelss:
SO
OLUTION
Step
p 1:
Deterrmine the livee load reductioon from Equaation 2.1
Trributary area pper floor: AT = (30)(24) = 720 ft2
In
nfluence area pper floor: AI = 4AT = 4(7200) = 2880 ft2
l
reduction
n:
Live load
15
0..25 +
= 0.530 ≥ 0.5 0
2880
Sincee this memberr supports onee floor the maaximum reducction is 0.50.
Forr
LRF
FD
Step
p 2:
Deterrmine the requ
uired axial strrength for thee LRFD load ccombination
LRFD
D load enterin
ng column at tthis level:
1.2 Dead = 1.2(100 psf)
f)(720 ft2) = 86,400 lbs
1.6 Live = 1.6(0.530)(880 psf)(720 ft2) = 48,800 lbbs
1.2 Dead + 1.6 Live = 886,400 + 48,8800 = 135,0000 lbs = 135 kiips
Requiired axial streength (LRFD)), Pu = 135 kiips
Forr
ASD
D
Step
p 2:
Deterrmine the requ
uired axial strrength for thee ASD load coombination
ASD load entering
g column at thhis level:
Chapter 2
Loads, Load Factors and Load Combinations 49
Dead = (100 psf)(720 ft2) = 72,000 lbs
Live = 0.530(80 psf)(720 ft2) = 30,500 lbs
Dead + Live = 72,000 + 30,500 = 103,000 lbs = 103 kips
Required axial strength (ASD), Pa = 103 kips
Example 2.7
Column Load
Calculation
Goal:
Determine the required axial strength for a column from the framing plan
shown in Figure 2.8 due to live load of 80 psf and dead load of 100 psf.
Given:
Interior column C-3 supporting five levels with 18K3 steel joists spanning
as shown in all panels:
SOLUTION
Step 1:
Determine the live load reduction from Equation 2.1
Tributary area per floor: AT = (30)(24) = 720 ft2
Tributary area for five floors: AT = 5(720) = 3600 ft2
Influence area per floor: AI = 4AT = 4(720) = 2880 ft2
Influence area for five floors: AI = 5(2880) = 14,400 ft2
Live load reduction:
0.25 +
15
= 0.375 < 0.40
14, 400
Therefore the live load reduction is limited to 0.40 since the member
supports more than one level.
LRFD
Step 2:
Determine the required axial strength for the LRFD load combination
LRFD load in the column from five levels:
1.2 Dead = 1.2(100 psf)(3600 ft2) = 432,000 lbs
1.6 Live = 1.6(0.40)(80 psf)(3600 ft2) = 184,000 lbs
1.2 Dead + 1.6 Live = 432,000 + 184,000 = 616,000 lbs = 616 kips
Required axial strength (LRFD), Pu = 616 kips
ASD
Step 2:
Determine the required axial strength for the ASD load combination
ASD load in column from five levels:
Dead = (100 psf)(3600 ft2) = 360,000 lbs
Live = 0.40(80 psf)(3600 ft2) =115,000 lbs
Dead + Live = 360,000 + 115,000 = 475,000 lbs = 475 kips
Required axial strength (ASD), Pa = 475 kips
Example 2.8
Column Load
Calculation
Goal:
Determine the required axial strength for a column from the framing plan
shown in Figure 2.8 due to live load of 80 psf and dead load of 100 psf.
Given:
Exterior column D-4 supporting seven levels with 18K3 steel joists
spanning as shown in all panels:
50 Chapter 2
SOLUTION
Loads, Load Factors and Load Combinations
Step 1:
Determine the live load reduction from Equation 2.1
Note that the exterior column supports an 8 in. (0.67 ft) portion of slab
outside the column center line. This must be included in the tributary and
influence areas.
Tributary area: AT = 7(24)(15) + 7(24)(0.67) = 2630 ft2
Influence area: AI = 4(7)(24)(15) + 2(7)(24)(0.67) =
10,300 ft2
Live load reduction:
15
0.25 +
= 0.398 < 0.40
10,300
Therefore, the live load reduction is limited to 0.40 since the member
supports more than one level.
For
LRFD
Step 2:
Determine the required axial strength for the LRFD load combination
LRFD load in column from seven levels:
1.2 Dead = 1.2(100 psf)(2630 ft2) = 316,000 lbs
1.6 Live = 1.6(0.40)(80 psf)(2630 ft2) = 135,000 lbs
1.2 Dead + 1.6 Live = 316,000 + 135,000 = 451,000 lbs = 451
kips
Required axial strength (LRFD), Pu =451 kips
For
ASD
Step 2:
Determine the required axial strength for the ASD load combination
ASD load in column from seven levels:
Dead = (100 psf)(2630 ft2) = 263,000 lbs
Live = 0.40(80 psf)(2630 ft2) = 84,200 lbs
Dead + Live = 263,000 + 84,200 = 347,000 lbs = 347 kips
Required axial strength (ASD), Pa = 347 kips
The 10-story building given in Figure 2.8 and just discussed for gravity load has uniform
story heights of 13.5 ft. The wind load is determined to be 40 psf over the upper 40 ft and the
lateral load is to be resisted by the perimeter moment frames in each direction. Use the load case
that includes dead, live and wind loads.
Example 2.9
Wind Load
Calculation
Goal:
Determine the wind force applied to the wind resisting frame for the
framing plan shown in Figure 2.8.
Given:
For each 90 ft wide frame the force applied at the roof level of each of two
frames:
SOLUTION
Step 1:
Determine the area of loading:
Chapter 2
Loads, Load Factors and Load Combinations 51
Tributary area for upper half of the top story:
AT = (145.3/2)(13.5/2)= 490 ft2
For
LRFD
Step 2:
Determine the lateral load for the LRFD load combination
LRFD wind load, load case 4:
1.0(Wind) = 1.0(40 psf) = 40 psf
Wu = (40 psf)(490 ft2) = 19,600 lbs
For
ASD
Step 2:
Determine the lateral load for the ASD load combination
ASD wind load, load case 6:
0.75(0.6Wind) = 0.75(0.6(40 psf)) = 18 psf
Wa = (18 psf)(490 ft2) = 8,820 lbs
Example 2.10
Column Load
Calculation
Goal:
Determine the wind force applied to the wind resisting frame for the
framing plan shown in Figure 2.8.
Given:
For each 145 ft wide frame the force applied at the first level below the roof
level of each of two frames:
SOLUTION
Step 1:
Determine the area of loading:
For
LRFD
Step 2:
Tributary area for the first level below the roof:
AT = (91.3/2)(13.5)= 616 ft2
Determine the lateral load for the LRFD load combination
LRFD wind load, load case 4:
1.0(Wind) = 1.0(40 psf) = 40 psf
Wu = (40 psf)(616 ft2) = 24,600 lbs
For
ASD
Step 2:
Determine the lateral load for the ASD load combination
ASD wind load, load case 6:
0.75(0.6Wind) = 0.75(0.6(40 psf)) = 18 psf
Wa = (18 psf)(616 ft2) = 11,100 lbs
2.6 CALIBRATION
The basic requirements of the ASD and LRFD design philosophies were presented in
Sections 1.9 and 1.10 and Equations B3.2 and B3.1. The required load combinations for
52 Chapter 2
Loads, Load Factors and Load Combinations
ASD and LRFD, as found in ASCE 7, were presented earlier in the chapter. This section
establishes the relationship between the resistance factor, φ, and the safety factor, Ω, as
found in the AISC Specification.
Early development of the LRFD approach to design concentrated on the
determination of resistance factors and load factors that would result in a level of
structural reliability consistent with previous practice but more uniform for different load
combinations. Because the design of steel structures before that time had no particular
safety-related concerns, the LRFD approach was calibrated to the then-current ASD
approach. This calibration was carried out for the live load plus dead load combination at
a live-to-dead-load ratio, L/D, of 3.0. This can be seen in Figure 1.18, where for a live to
dead load ratio of 3.0 the reliabilities for ASD and LRFD are essentially the same. It was
well known that for any other load combination or live-to-dead-load ratio, the two
methods could give different answers for the same design situation.
The current Specification has been developed with this same calibration, which
results in a direct relationship between the resistance factor of LRFD and the safety factor
of ASD. For the live load plus dead load combination in ASD, using Equation B3.2, and
representing the load effect simply in terms of D and L,
( D + L) ≤
Rn
Ω
This same combination in LRFD, using Equation B3.1 is
(1.2D + 1.6L) ≤ φRn
If it is assumed that the load effect is equal to the available strength and each
equation is solved for the nominal strength, the results for ASD are
Ω(D + L) = Rn
and for LRFD,
(1.2 D + 1.6 L )
= Rn
φ
With L/D taken as 3, the above equations are set equal. They are then solved for the
safety factor, which gives
Ω=
1.5
φ
(2.7)
The resistance factors in the Specification were developed through a stochastic
analysis to be consistent with the specified load factors and result in the desired reliability
for each limit state. More detail on the development of these resistance factors can be
found in Section B3.1 of the Commentary to the Specification. Once the resistance
factors were established, the corresponding safety factors were determined using
Equation 2.7. This relationship has been used throughout the Specification to set the
safety factor for each limit state.
Although the relationship is simple, there is actually no reason to use it to
determine safety factors. This has already been done in the Specification, which explicitly
defines resistance factors and safety factors for every limit state.
Chapter 2
22.7
Loads, Loadd Factors and Load Combinnations 53
PROBLEMS
11. Name and deescribe five baasic types/sourcces of building
g
looads.
22. Categorize th
he following loads as dead load,
l
live load,,
ssnow load, win
nd load, seismicc load, or speciial load.
a. Lo
oad on an office
o
floor due to filing
g
cabineets, desks, and computers.
b. Load on a ro
oof from a permanent
p
airr
handliing unit.
c. Lo
oad on stadiu
um bleachers from studentss
jumpin
ng up and dow
wn during a co
ollege footballl
game.
d. Load on a buildiing caused by an explosion.
e. Weight
W
on a steeel beam from a concrete slab
b
that it is supporting.
f. Lo
oad experienceed by an officce building in
n
Califo
ornia as it shakees during an eaarthquake.
g. Lo
oad on a skysscraper in Chiccago on a day
y
with blustery
b
condittions causing the
t building to
o
sway back
b
and forth.
11. Iff a response m
modification facctor of 3 is choosen in the
desiggn of a steel bbuilding to reesist seismic looads, what
desiggn specificationn should be connsulted?
12. W
Which designn approach coombines loads that are
norm
mally at their noominal or serviiceability levells?
13. S
Strength load combinations that are incorpporated by
the L
LRFD method ttake into accouunt what two faactors?
14. U
Using ASCE 7, determine the minimum uniformly
distriibuted live loadd for a hospitall operating rooom.
15. U
Using ASCE 7, determine the minimum uniformly
distriibuted live loadd for library staacks.
16. U
Using ASCE 7, determine the minimum uniformly
distriibuted live loadd for an apartm
ment building.
17. D
Determine thee nominal uniiformly distribbuted selfweigght of a 6 in. thiick reinforced concrete slab.
33. What is onee source you can
c consult to find the snow
w
looad data for a particular regiion as well as maps showing
g
w
wind gust data to allow you to
o calculate win
nd loads?
44. Where in th
he AISC Manual can you find
f
a table off
sselected unit weights of comm
mon building materials?
m
55. What analyssis method allo
ows the design
ner to visualizee
thhe load on a particular structural eleement withoutt
pperforming an actual
a
equilibriium calculation
n?
66. What is the rationale for th
he live load reduction factor,,
w
when can this reduction facto
or be used, and what are thee
liimitations on its
i use?
77. In determinin
ng the snow lo
oad on a structu
ure, what valuee
thhat can be obttained from thee applicable bu
uilding code iss
m
multiplied by a series of factors to obtain the roof snow
w
looad?
88. Identify and briefly describ
be three factorss that affect thee
m
magnitude of th
he snow load on
o an unobstruccted flat roof.
99. Name four factors
f
that mu
ust be taken in
nto account in
n
cconverting win
nd speed data referenced by
y the building
g
ccode into wind pressure on a given
g
building
g.
110. If the locall building codee specifies a deesign load thatt
ddiffers from what is stated in
n ASCE 7, wh
hich documentt
sshould you folllow?
P2.188
18. A building hass a column layyout as shownn in Figure
P2.188 with 30 ft baays in each dirrection. It musst support a
unifoorm dead load of 90 psf and a uniform live load of 80
psf. Determine thhe required mooment and shear strength
for bbeams and girdders, and axial strength for coolumns, as
notedd below for design by (a)) LRFD and (b) ASD.
Calcuulate with annd without llive load redduction, as
approopriate.
i. The beeam on columnn line 3 betweeen column
lines A annd B if the decck spans from line 2-2 to
3-3 to 4-4.
554 Chapter 2
Loads, Lo
oad Factors and
a Load Com
mbinations
ii. The girder on
n column line C between
n
colum
mn lines 3 and 4 if the deck spans from linee
B-B to
o C-C to D-D.
iii. The
T column at the corner on lines 4 and A
and su
upporting one floor.
f
iv. The
T column on
n the edge at th
he intersection
n
of linees C and 4 and supporting onee floor.
v. Th
he interior co
olumn at the intersection
i
off
colum
mn lines D and 3 and supportin
ng one floor.
119. If the fram
ming plan show
wn in Figure P2.19
P
were forr
thhe roof of a sttructure that carried a dead load of 55 psff
aand a roof live load of 30 psf, determine the required
d
m
moment and shear strength for beams and girders, and
d
aaxial strength for columns, as required below for (a))
ddesign by LRF
FD and (b) dessign by ASD. Do not reducee
thhe roof live loaads.
i. Th
he girder on column line A beetween column
n
lines 1 and 2 if the deck
d
spans fro
om line A-A to
o
B-B.
below
w for (a) desiggn by LRFD and (b) designn by ASD.
Calcuulate with annd without llive load redduction, as
approopriate.
i. The bbeam between column liness 2 and 3
along line D.
ii. The girder on coolumn line 3 between
column linne E and midw
way between liines D and
C.
iii. The bbeam on the lline between liines C and
D and coluumn lines 3 annd 4.
iv. The ccolumn at the corner on linees 1 and E
that suppoorts eight levelss.
v. The ccolumn on the edge at the inntersection
of lines 4 aand A that suppports eight levvels.
vi. The interior colum
mn at the interrsection of
column liine 4 and the point midwaay between
lines C annd D that suppoorts three levelss.
ii. The beam on co
olumn line 3 beetween column
n
lines B and C if the deck spans frrom line 2-2 to
o
3-3 to 4-4.
iii. The
T column at the
t corner on lines
l
1 and E.
iv. The
T column on
n the edge at th
he intersection
n
of linees 1 and B.
P2.200
21. IIntegrated Deesign Project.. For a Downners Grove,
Illinoois, office building, determiine the dead, live, snow
and w
wind loads baased on ASCE
E 7 and the coonstruction
sugg ested below.
A reasonable aassumption forr the constructtion of the
rooof is the folloowing:
v. Th
he interior co
olumn at the intersection
i
off
colum
mn lines C and 2.
2
P
P2.19
220. The framin
ng plan shown in Figure P2.2
20 is for an 18-sstory office buiilding. It must support a dead
d load of 80 psff
aand a live load of 50 psf. In all
a cases, the deecking spans in
n
a direction fro
om line A tow
ward line E. Determine
D
thee
rrequired momeent and shear strength for the
t beams and
d
ggirders, and ax
xial strength for
f the column
ns, as required
d
Roofing
Insulation
Deck
Beams
Joists
Misc.
Total
5 ppsf
2 ppsf
2 ppsf
3 ppsf
3 ppsf
5 ppsf
20 psf
Dow
wners Grove is at an elevationn of approximaately 690 ft
so a case-specific snow load callculation is noot required.
walls, snow
Becaause of the perrimeter parapett and screen w
drift loading must bbe considered.
Chapter 2
A reasonable assumption for the floor construction is
Slab and deck
60 psf
Beams and girders
8 psf
Misc.
12 psf
Total
80 psf
Since no information has been given regarding office
layout, it is prudent to use the corridor loading over the
entire floor plan. This will permit architectural flexibility.
The two-story area bounded by column lines A, C, 4, and
5 should be treated as a lobby.
For wind load, this is a rigid enclosed building
located on a site where there are no special terrain
features to consider nor is air density a concern since the
elevation is approximately 700 ft above sea level. Thus, to
determine the velocity pressure, Kd = 0.85, Kzt = Ke = 1.0
and Kz at the top of the building is 1.09. To determine the
external design wind pressure use G = 0.85 for a rigid
building, Cp = 0.8 for the windward wall and GCpi =
+0.18 for an enclosed structure.
a.
Develop a table showing the gravity loads in
all columns based on tributary area and live
load reductions. Remember that the columns
in the lateral load resisting systems will also
carry wind or seismic forces. Wind loads
will be determined in part c; seismic loads
are in the Chapter 13 Integrated Design
Project problem.
b. Develop a table showing beam loading
based on tributary area and live load
reductions. Remember that beams will see a
uniform load and girders will see
concentrated loads and some uniform load
due to their self-weight. Beam and girder
self-weight is included in the construction
loads given above. Perimeter beams and
girders will also support the lightweight
metal curtain wall at each level. A
reasonable assumption for the weight of the
curtain wall is 18 psf.
c. Using a uniform wind pressure as
determined at the top of the building and
spread over the entire windward wall,
determine the concentrated wind force at
each level for each main wind force resisting
system in each direction.
Loads, Load Factors and Load Combinations 55
C
Chapterr 3
S
Steel Buildin
B
ng Ma
aterials
wer, Cape Canaveral, FL
Exxploration Tow
Photo courteesy Thornton T
Tomasetti
3.11
INTROD
DUCTION
Steel has beeen produced in the United States sincee the 1800s. Its first use iin a bridge w
was for a
railroad brid
dge across th
he Mississippii River in Stt. Louis, builtt in 1874 by James B. Eaads. The
bridge, know
wn as the Eaads Bridge an
nd shown in Figure 3.1, iis still an insspiring steel sstructure
crossing thee river in the shadow
s
of the St. Louis G
Gateway Archh. The first skkyscraper is ggenerally
considered to
t be the Hom
me Insurance Building, dessigned by Willliam LeBaroon Jenney andd erected
at 135 South
h La Salle Strreet, Chicago
o. The buildinng, shown in Figure 3.2, w
was started onn May 1,
1884, and completed
c
in the fall of 1885.
1
It was originally a 10-story buillding but lateer had 2
stories added. The origin
nal structural design
d
calledd for wrought iron beams bbolted througgh angleiron bracketts to cast iron columns. As the framewoork reached thhe sixth floor,, the Carnegiee-Phipps
Steel Comp
pany of Pittsb
burgh, Pennssylvania, indiicated that thhey were now
w rolling “B
Bessemer
Steel” and requested
r
perrmission to su
ubstitute steeel members fo
for the wrougght iron beam
ms on all
remaining floors.
f
This was
w the firstt use of steell beams in a building. T
The Home Innsurance
Building waas demolished
d in 1929.
The first all-steell skyscraper was
w the Randd-McNally Buuilding at 1655 West Adam
ms Street
in Chicago, designed by Daniel Burnh
ham and Johnn Root. This 110-story buildding, shown inn Figure
3.3, was bu
uilt from 1888
8–1890 and was
w construccted of built-uup members made from sstandard
rolled steel bridge shapees that were riveted togetther. It markeed the beginnning of a conntinuous
evolution in
n steel buildin
ng structures that
t
is ongoinng today as nnew ideas are brought into play by
architects an
nd engineers who
w build witth steel.
Thiss evolution in steel build
dings encomppasses the m
materials usedd, the applicaations of
innovative designers,
d
and
d the specificaations that dirrect their desiigns.
3.22
APPLIC
CABILITY OF THE AISC SPECIIFICATION
N
The specificcation that gu
uides the dessign of our m
modern steel buildings waas first publiished by
AISC in 192
23. Its purpose was to prom
mote uniform practice in thhe design of ssteel buildingss. At
566
Chappter 3 Steel Building Matterials
Figure 3.1
57
Eads Bridge, St. Louis, Missouri
M
Photo courtessy Luke Gatta
Figure 3.2 Home Insuraance Building
g, Chicago, Illlinois.
Copyright © American
A
Instiitute of Steel Construction
C
Reprinted witth Permission. All Rights Resserved
Figu
ure 3.3 Raand McNally B
Building,
Chiccago, Illinois
Coppyright © Amerrican Institute of Steel
Connstruction
Reprrinted with Perrmission. All R
Rights
Reseerved
58 Chapter 3
Steel Building Materials
that time, numerous approaches were being used across the industry. Steel producers each had
their own standard for design, and the larger cities also required that their own standards be used.
This multiplicity of standards meant no standard at all. It led to a confusion of approaches in
which designers were continually called upon to change the way they designed, depending on
where their current building project was to be located.
The 1923 Specification defined “the practice adopted by the American Institute of Steel
Construction for the design, fabrication, and erection of structural steel buildings.” It went on to
provide direction on how to obtain a satisfactory structure. The following requirements were to be
fulfilled:
1.
2.
3.
4.
5.
The material used must be suitable, of uniform quality, and without defects
affecting the strength or service of the structure.
Proper loads and conditions must be assumed in the design.
The unit stresses must be suitable for the material used.
The workmanship must be good, so that defects or injuries are not produced in
the manufacture.
The computations and design must be properly performed so that the unit stresses
specified shall not be exceeded, and the structure and its details shall possess the
requisite strength and rigidity.
The Specification also provided guidance on the material to be used, stating, “Structural
steel shall conform to the Standard Specifications of the American Society for Testing Materials
for Structural Steel for Buildings, Serial Designation A9-21, as amended to date.” These
principles from 1923 are still important to steel construction almost a century later.
The 2016 AISC Specification for Structural Steel Buildings supersedes all previous AISC
Specifications. Over the years, the Specification has lost the terms fabrication and erection from
its title, nonetheless, the Specification continues to address issues pertinent to design, fabrication
and erection of steel structures. In addition, the AISC Specification has regularly been used to
guide the design of structures other than buildings. In recognition of this practice, and to ensure
that the Specification is properly applied, the scope of the Specification states, “The specification
sets forth criteria for the design fabrication and erection of structural steel buildings and other
structures, where other structures are defined as structures designed, fabricated, and erected in a
manner similar to buildings, with building-like vertical and lateral load resisting elements.”
Additionally, the Specification indicates that it “shall apply to the design, fabrication and
erection of the structural steel system or systems with structural steel acting compositely with
reinforced concrete, where the steel elements are defined in Section 2.1 of the AISC Code of
Standard Practice for Steel Buildings and Bridges.” In that document, structural steel is defined as
“the elements of the structural frame that are shown and sized in the structural design drawings,
essential to support the design loads,” and they are given here in Table 3.1. A quick review of the
table contents indicates that elements are structural steel if they are part of or attached to the
structural frame. Examples of the use of many of these elements are shown in Figure 3.4. All
elements discussed in this text will meet the above definition.
3.3 STEEL FOR CONSTRUCTION
Since the introduction of the first AISC Specification, a variety of steels have been approved for
use in steel construction. The specific steels approved at any given time has changed along with
the techniques of manufacture and steel chemistry. Steels available for use in construction have
increased in strength as manufacturing has become more refined. One important aspect of all steel
is that it generally behaves in a uniform and consistent manner. Thus, although the strength might
be different for different grades of steel, the steel can be expected to behave the same, regardless
of grade, up to its various strength limits.
Chapter 3 Steel Building Materials
59
Table 3.1 Definitions of Structural Steela
Anchor rods that will receive structural steel.
Base plates, if part of the structural steel frame.
Beams, including built-up beams, if made from standard structural shapes and/or plates.
Bearing plates, if part of the structural steel frame.
Bearings of steel for girders, trusses, or bridges.
Bracing, if permanent.
Canopy framing, if made from standard structural shapes and/or plates.
Columns, including built-up columns, if made from standard structural shapes and/or plates.
Connection materials for framing structural steel to structural steel.
Crane stops, if made from standard structural shapes and/or plates.
Door frames, if made from standard structural shapes and/or plates and if part of the structural steel
frame.
Edge angles and plates, if attached to the structural steel frame or steel (open-web) joists.
Embedded structural steel parts, other than bearing plates, that will receive structural steel.
Expansion joints, if attached to the structural steel frame.
Fasteners for connecting structural steel items: permanent shop bolts, nuts, and washers; shop bolts,
nuts, and washers for shipment; field bolts, nuts, and washers for permanent connections; and
permanent pins.
Floor-opening frames, if made from standard structural shapes and/or plates and attached to the
structural steel frame or steel (open-web) joists.
Floor plates (checkered or plain), if attached to the structural steel frame.
Girders, including built-up girders, if made from standard structural shapes and/or plates.
Girts, if made from standard structural shapes.
Grillage beams and girders.
Hangers, if made from standard structural shapes, plates, and/or rods and framing structural steel to
structural steel.
Leveling nuts and washers.
Leveling plates.
Leveling screws.
Lintels, if attached to the structural steel frame.
Machinery supports, if made from standard structural shapes and/or plates and attached to the structural
steel frame.
Marquee framing, if made from standard structural shapes and/or plates.
Monorail elements, if made from standard structural shapes and/or plates and attached to the structural
steel frame.
Posts, if part of the structural steel frame.
Purlins, if made from standard structural shapes.
Relieving angles, if attached to the structural steel frame.
Roof-opening frames, if made from standard structural shapes and/or plates and attached to the
structural steel frame or steel (open-web) joists.
Roof-screen support frames, if made from standard structural shapes.
Sag rods, if part of the structural steel frame and connecting structural steel to structural steel.
Shear stud connectors, if specified to be shop attached.
Shims, if permanent.
Struts, if permanent and part of the structural steel frame.
Tie rods, if part of the structural steel frame.
Trusses, if made from standard structural shapes and/or built-up members.
Wall-opening frames, if made from standard structural shapes and/or plates and attached to the
structural steel frame.
Wedges, if permanent.
a
From Code of Standard Practice for Steel Buildings and Bridges, AISC 2016.
Steel Elements
Reprinted from CISC Code of Standard Practice for Structural Steel, with permission, Copyright © Canadian Institute of Steel
Construction All rights reserved
Construction.
Figure 3.4
600 Chapter 3
Steel Build
ding Materialss
Chappter 3 Steel Building Matterials
61
The
T characterristics of steell that are impoortant to the sstructural enggineer can be determined
through a simple un
niaxial tensio
on test. Thi s standard ttest is conduucted accordding to the
requirem
ments of ASTM
M A370, Stan
ndard Test M
Methods and D
Definitions foor Mechanicaal Testing of
Steel Products. A sp
pecimen of specific
s
dimeensions is suubjected to a tensile forcce, and the
resulting stress and sttrain are plottted for the duuration of thee test. The sttress, f, and strain, ε, are
plotted in
n Figure 3.5 and
a defined ass follows:
f =
P
A
aand ε =
ΔL
L
where
A =
L =
P =
f =
ΔL
Δ =
ε =
cross--sectional areea at start of teest
length
h of specimen
n at start of teest
tensile force
axial tensile stress
chang
ge in length of specimen unnder stress
axial strain
The curve shown in Fig
gure 3.5 is tyypical of mildd carbon steell. Several chaaracteristics
of this sttress-strain cu
urve are worth
h noting. Firsst, the initial portion of the curve, whicch indicates
the respo
onse that wou
uld be expecteed under mosst nominal or service loadiing conditions, follows a
straight line
l
up to a point
p
called th
he proportionnal limit. For structural steeel with yieldd stresses of
65 ksi orr less, this pro
oportional lim
mit is the poinnt where the ccurve first deviates from liinear and is
called the yield point. The ratio off stress to straain in this reggion is constannt and is calleed Young’s
modulus or the modu
ulus of elasticity, E. All strructural steelss exhibit the same initial sstress-strain
behaviorr and thus hav
ve the same E value. The vvalue of E obttained throughh a large num
mber of tests
is consisstently betweeen 29,000 ksi
k and 30,0000 ksi. For aall calculatioons based onn the AISC
Specifica
ation, E = 29,000 ksi has historically bbeen used. W
Within the straaight-line porrtion of the
curve, th
he material is said to behav
ve elastically. A load can bbe applied and then removved with the
structure returning to its original co
onfiguration, showing no ppermanent deformation.
After
A
reaching
g the yield strress, the stresss-strain curvee for mild carrbon steel exhhibits a long
plateau where
w
the strress remains essentially coonstant whilee the strain increases. Thiis region is
called thee plastic region. Any struccture that is looaded into thhis region exhhibits a permaanent plastic
deformattion as show
wn by the unlloading line in Figure 3.5. The lengtth of this plaastic region
depends on the particu
ular type of stteel but typicaally is 15 to 220 times the strain at yield..
3
Figure 3.5
Typical Stress-Strain Plot for Mildd Carbon Steeel
622 Chapter 3
Steel Build
ding Materialss
At the
t end of the plastic regiion, the curvee again rises with increasing stress and strain.
This increasse is called strain
s
harden
ning and conntinues until the specimeen reaches itss tensile
strength, or ultimate stress, Fu, at the peak of the sstress-strain ccurve. Once tthe tensile strrength is
pidly sheds lo
oad and increaases strain unntil complete rrupture occurrs.
reached, the specimen rap
Yielld stress, tenssile strength, and moduluus of elasticityy are the enggineering dataa values
used throug
ghout design to fully desscribe the m
material and tto determine the strengthh of the
structural ellements. Thee ratio of thee tensile strenngth to the yield stress is also an im
mportant
characteristiic of steel. Itt is used to control
c
the baasic material behavior so that at varioous limit
states, the ex
xpected behav
vior can be en
nsured.
Figu
ure 3.6 showss the lower strain region oof the stress-strain curves for three steeels with
different yieeld stresses: 36
3 ksi, 50 ksi,, and 100 ksi . Elastic behaavior for the hhigher-strenggth steels
is the same as for the lo
ower-strength steels. As allready noted, E = 29,000 ksi for all stteel. The
differences occur
o
after th
he proportionaal limit is reaached. For steeels with a yieeld stress lesss than or
equal to 65 ksi, the plateeau defining the plastic reegion can be expected to occur. Howeever, for
steels with a yield stress greater than 65 ksi, it is expected thaat no well-deffined yield pooint will
exist and no
o well-defined
d plastic platteau will occuur, as illustraated for the steel having Fy = 100
ksi. For thesse steels it iss necessary to
o define yieldd strength by some other means. ASTM
M A370
provides forr yield streng
gth determinaation by the 0.2 percent offset methood or the 0.5 percent
elongation method.
m
In eitther case, the stress-strain ccurve must bee obtained annd the specifieed offset
or elongatio
on used to dettermine the ap
ppropriate strress value. Thhe results of tthese two appproaches
are shown in
n Figure 3.6, and the two methods
m
woulld yield slighttly different yyield strength values.
3.44
STRUCTURAL ST
TEEL SHAP
PES
Structural stteel design seerves to deterrmine the apppropriate shappe and amouunt of steel neeeded to
carry a given
n applied load. This is norrmally accom
mplished by seelecting, from
m a predeterm
mined list
of availablee shapes, th
he lightest-weeight membeer. Howeverr, options coould also innclude a
combination
n of steel elements in some
s
particuular desired form. The early days of steel
construction
n witnessed very
v
little stan
ndardization oof available sshapes. Althoough each milll would
produce its own
o shapes, the
t variety off available shaapes was limited and most structural meembers
Figure 3.6 Enlarged Ty
ypical Stress-Strain Curvess for Steels w
with Different Yield Stressees
Chappter 3 Steel Building Matterials
63
were com
mposed of theese available shapes riveteed together. O
One of AISC
C’s original ggoals was to
standardiize the shapees being prod
duced. Over tthe years, shhapes becamee standardizedd and more
shapes, designed
d
speccifically for the
t needs of building connstruction, beccame availabble. Modern
productio
on practices now
n
make a wide
w
variety of shapes avvailable to thee designer so that design
can almo
ost always be accomplished
d by selectingg one of thesee standard shaapes. In situattions where
these staandard shapess do not meett the needs oof a project, m
members com
mposed of plaate material
can be prroduced to caarry the impossed loading.
3.4.1
AST
TM A6 Stand
dard Shapes
The firstt standard shaapes to be disscussed are th
those definedd by ASTM A
A6: W-shapess, S-shapes,
HP-shapees, M-shapess, C-shapes, MC-shapes,
M
aand L-shapess. Cross-sectiions of these shapes are
shown in
n Figure 3.7, where
w
it can be
b seen that W
W-, M-, S-, aand HP-shapees all take thee form of an
I. C- and
d MC-shapess are called channels
c
and take the form
m of a C, whhile L-shapess are called
angles. Part
P 1 of the Manual
M
contaiins tables of pproperties forr all of the stanndard shapes.
W-Shapes
a wide-flangee shapes and are the mostt commonly uused shapes
W-shapes are usually referred to as
in buildin
ngs. They haave two flang
ges with essenntially paralleel inner and outer faces aand a single
web locaated midway along
a
the flan
nges. The oveerall shape of the wide flannge may vary from a
3
Figure 3.7
Structuraal Shapes
64 Chapter 3
Steel Building Materials
fairly deep and narrow section, as shown in Figure 3.7a, to an almost square section, as shown in
Figure 3.7b. These shapes have two axes of symmetry; the x-axis is the strong axis and the y-axis
is the weak axis. Wide-flange shapes can be as deep as 44 in. and as shallow as 4 in. A typical
wide-flange shape is designated as a W16×26, where the W indicates it is a W-shape, the 16
indicates it has a nominal depth of 16 in., and the 26 indicates its weight is 26 pounds per foot.
The nominal depth indicates an approximate member depth but not its actual depth. The
production of wide-flange shapes results in their grouping in families according to the size of the
rolls used to produce the shape. All shapes in a family have the same dimension between the
inner faces of the flanges. The different weights are achieved by increasing the actual depth of the
member and thus the flange thickness. Manual Table 1-1 provides the dimensions and section
properties needed for design for all W-shapes.
HP-Shapes
HP-shapes are similar to wide-flange shapes and normally used as bearing piles. These shapes
also have parallel face flanges, but unlike the W-shapes, their webs and flanges are of the same
nominal thickness and they are all close to being square, as shown in Figure 3.7c. An HP14×117
is an HP-shape with a nominal depth of 14 in. and a weight of 117 pounds per foot. Manual Table
1-4 provides the dimensions and section properties needed for design for all HP-shapes.
S-Shapes
S-shapes are American Standard beams and were previously referred to as I-beams. They were
the standard shapes used in construction prior to the development of the rolling process that
permitted the introduction of the wide-flange shapes. Although these shapes are still available,
their use today is infrequent and their availability should be confirmed prior to their specification.
These shapes have narrow flanges in relation to their depth, and the flanges have a sloping
interior face, as shown in Figure 3.7d. The Manual lists 28 S-shapes and gives their properties in
Table 1-3. As with the shapes previously discussed, the numbers in the name refer to the nominal
depth and the weight per foot. In all cases except for two S24s and two S20s, the nominal depth
and the actual depth are the same.
M-Shapes
M-shapes are miscellaneous shapes that do not fit the definitions of W-, HP-, and S-shapes. The
Manual lists 18 miscellaneous shapes. They are not particularly common and should be used in
design only after confirmation that they are economically available. A typical designation is
M12×11.8; as with the other shapes, the 12 indicates the nominal depth and the 11.8 indicates the
weight per foot. Dimensions and properties for M-shapes are found in Manual Table 1-2.
C-Shapes
C-shapes are American Standard channels and are produced using essentially the same process as
for S-shapes. They have two flanges and a single web located at the end of the flanges, as shown
in Figure 3.7e. These shapes have only one axis of symmetry and, like the W-shapes, the x-axis is
the strong axis and the y-axis is the weak axis. As with the S-shapes, the flanges have sloping
inner faces. One of the 32 C-shapes found in Manual Table 1-5 is a C8×18.75. All C-shapes have
an actual depth equal to the nominal depth.
Chapter 3 Steel Building Materials
65
MC-Shapes
MC-shapes are miscellaneous channels that cannot be classified as C-shapes. Their designations
follow the same rules as the previous shapes, with a typical shape being an MC6×18. Manual
Table 1-6 lists 40 MC-shapes, and their sizes fall into the same overall range as the C-shapes.
L-Shapes
L-shapes are angles that can have equal or unequal legs. The largest angle legs are 12 in. and the
smallest are 2 in., with the dimension taken from heel to toe of the angle. A typical angle
designation is L6×4×7/8, where the first two numbers are the dimensions of the legs and the third
is the leg thickness. Leg dimensions are actual dimensions, and the leg thickness is the same for
both legs. For unequal leg angles, the longest leg is given first. Equal leg angles have one axis of
symmetry, whereas unequal leg angles have no axis of symmetry.
All angles have two sets of two axes of interest to the designer: the geometric and
principal axes, which are illustrated in Figure 3-7(f). The geometric axes are parallel to the faces
of the legs with the x-axis, parallel to the short leg, and the y-axis, parallel to the long leg. For
equal leg angles, the x- and y-axes are similarly oriented even though one cannot distinguish
between longer and shorter legs. Alternatively, the principal axes can be used. The minor
principal axis, which for equal leg angles is perpendicular to the axis of symmetry, is the z-axis;
and the major principal axis, the axis of symmetry for equal leg angles, is the w-axis. The
principal axes are similar for unequal leg angles, except there is no axis of symmetry.
Manual Table 1-7 provides the dimensions and section properties for the design for all
angles. For some special cases, properties that may be needed for the w-axis are not available in
Table 1-7. These can be found in the 15th edition electronic shapes database.
WT-Shapes
WT-shapes are tees that have been cut from W-shapes. They are also called split tees because that
is how they are produced – by splitting a W-shape. These shapes are designated as WT5×56,
where both numbers are one-half of the corresponding numbers of the parent W-shape they were
cut from. Dimensions and properties for WT-shapes are given in Manual Table 1-8.
MT-Shapes and ST-Shapes
MT-shapes and ST-shapes are tees that have been cut from parent M- and S-shapes. The
properties and dimensions for these shapes are found in Manual Tables 1-9 and 1-10.
3.4.2
Hollow Shapes
Another group of shapes commonly used in building construction are the hollow shapes
commonly referred to as tubes or pipes. These shapes are produced by bending and welding flat
plates or by hot rolling to form a seamless section. For all hollow structural sections (HSS),
ASTM specifications set the requirements for both the materials and the sizes.
Round HSS
Round hollow structural sections are manufactured through a process called “formed from
round,” which takes a flat strip of steel and gradually bends it around its longitudinal axis and
joins the edges by welding. Once the weld has cooled, the round shape is passed through
additional shaping and sizing rolls to fix the final diameter. An example of a round HSS is
HSS5.563×0.258, where the first number is the outside diameter and the second is the nominal
666 Chapter 3
Steel Build
ding Materialss
wall thickneess. For round
d HSS, the diameter
d
and nnominal walll thickness arre always shoown as a
decimal num
mber to three places.
p
Thesee shapes are fo
found in Manuual Table 1-13.
Square and
d Rectangular HSS
Square and rectangular HSS may iniitially be form
med from rouund, with thee final sizing used to
also change the formed shape into a rectangle, orr formed from
m a flat platee through a ““formedsquare weld
d-square” proccess, in which
h the plate is gradually beent into its neaar final size. Another
process starrts with two flat
f pieces that are each bbent, with thee two half sections then jooined to
form the fin
nal shape. A typical
t
rectan
ngular HSS iss the HSS12×
×8×1/2. The ffirst number iindicates
the actual heeight of the seection, the seccond the actuual width, andd the third the nominal thickness of
the section wall.
w
For square and rectan
ngular HSS, tthe nominal w
wall thicknesss is always shhown as
a fraction. Manual
M
Tablees 1-11 and 1-12
1
provide the dimensioons and section propertiess needed
for the desig
gn of rectangu
ular and squarre HSS-shapees, respectivelly.
Steel Pipes
Steel pipes are
a another ho
ollow round section used in building construction. T
They are prodduced to
different maaterial standaards than the round HSS. Pipes are avvailable as staandard weighht (Std.),
extra strong (x-Strong), and
a double ex
xtra strong (xxx-Strong), whhich refer to tthe wall thickkness for
a given outsside diameterr. The standaard designatioon for a pipee section is inn the form Pipe 5 xStrong, indiicating that itt has to meett the pipe maaterial standarrds, have a nnominal 5 in.. outside
diameter, an
nd a thicknesss correspondin
ng to the “exttra strong” deesignation. Thhis particular pipe has
an actual ou
utside diameteer of 5.56 in. and a nominnal wall thicknness of 0.3755 in. Manual T
Table 114 provides the propertiees for steel piipes. Note thaat many rounnd HSS cross sections are made to
match the steel
s
pipe crross sections.. Although tthey are dim
mensionally innterchangeabble, it is
important to
o remember that round HSS and steeel pipes aree produced tto different material
standards.
3.44.3
Plates and
a Bars
In addition to the shapees already disscussed, steell is availablee as plates annd bars, as shhown in
Figure 3.8. These elemeents are rarely
y used alone as shapes buut are combined to form built-up
shapes or ussed alone as connecting eleements to joinn other shapess.
Figure 3.8 Plate and Baar Products
Chappter 3 Steel Building Matterials
67
Table 3.2
2 Preferred Dimensions for Plates andd Bars
Rangee of Thicknessses/Diameterrs
Product
t ≤ 3/8 in.
3/8 in. < t ≤ 1 in.
1 in. < t
Plates
1/16
1//8
1/4
Square an
nd rectangulaar bars
1/8
1//8
1/8
Circular bars
1//8
1/8
1/8
Note: Taable gives incrrements in thiickness or diaameter.
Plates
Plates arre flat rectang
gular elemen
nts that are hoot-rolled to a given thickkness and sheeared to the
appropriaate width. Att one time, plates were allso available that were rollled to a giveen width as
well as a given thicckness. These plates werre called universal mill plates. Becaause of the
manufactturing processs, these plaates had patteerns of residdual stresses that differedd from the
patterns in sheared plates and ressulted in low
wer strength. C
Current manuufacturing prractice is to
produce all plates as sheared
s
platess. By industryy definition, pplates are a m
minimum of 8 in. in width
and may vary in thick
kness from 3/1
16 in. up. Thee designation for a typical plate is PL 1//2×10×2 ft–
4 in., wh
here the first number
n
is thee thickness, thhe second thee plate width, and the thirdd the length.
Table 3.2
2 reflects the preferred
p
stan
ndard practicee for plate thiickness increm
ments.
Bars
Bars are available in
n rectangular, circular, andd hexagonal shapes, with the rectanguular bar the
most com
mmonly used
d shape in bu
uilding constrruction. The only differennce between rectangular
bars and plates is the width. Any rectangular
r
soolid element less than 8 inn. in width is technically
referred to as a bar. Because the distinction bbetween barss and plates iis not signifiicant to the
designer,, the designattion for these narrow elem
ments is the saame as for a pplate. Thus, P
PL 1/2×6×2
ft–4 in. iss a 6 in. wide bar.
3
Examp
ples of Built--up Shapes
Figure 3.9
68 Chapter 3
3.4.4
Steel Building Materials
Built-up Shapes
Other shapes are available and may be found in the Manual, but they are of limited application in
building construction. The Manual also contains tables for combinations of standard shapes that
have, over the years, been found to be useful to the designer. Figure 3.9 shows a variety of builtup shapes formed from combining plates and shapes.
3.5 CHEMICAL COMPONENTS OF STRUCTURAL STEEL
The basic mechanical properties of structural steel were presented in Section 3.3 with a limited
discussion of the types of steel available for use by the building industry. Essentially three types
of steel are used for shapes in the construction industry: carbon steel, sometimes referred to as
carbon-manganese steel, high-strength low-alloy steel (HSLA), and corrosion-resistant highstrength low-alloy steel. For plates and bars, quenched and tempered steels are also available.
The chemical composition of steel significantly influences the properties that are of
ultimate importance to the engineer. Steel is primarily made of iron but also contains such other
elements as carbon, silicon, nickel, manganese, and copper. The primary element that influences
the characteristics of steel, not counting iron, is carbon. The addition of carbon increases steel
strength but decreases ductility and weldability. Even though carbon is the most significant
component of steel (after iron), it still represents a very small percentage of the final product.
Steels generally have a carbon content of not more than 0.3 percent by weight. Steels that contain
strengthening elements in addition to carbon and manganese are referred to as HSLA steels
although there are no strict rules that apply to that designation.
Although the formula for a specific steel might be different from that of any other steel,
certain elements are required in order to meet a specific set of criteria. These specifications come
from the ASTM standards for each steel type and are discussed in Section 3.6. However, the
chemical elements that may be found in the most dominant steel for wide flange shapes are
reviewed here. The specific percentage requirements for ASTM A992 steel are given in Table
3.3.
Carbon
Carbon (C) is the most common element, excluding iron, found in all steel. It is the most
economical element used to increase strength. However, it also decreases ductility. Carbon
content usually ranges from about 0.15 to 0.30 percent. Anything lower than 0.15 percent would
produce steel with too low a strength, and anything higher than 0.30 percent would yield steel
with poor characteristics for use in construction.
Table 3.3
Chemical Requirements for A992 Steel
Element
Carbon, max
Manganese
Silicon, max
Vanadium, max
Columbium, max
Phosphorus, max
Sulfur, max
Copper, max
Nickel, max
Chromium, max
Molybdenum, max
Composition, %
0.23
0.50 to 1.60
0.40
0.15
0.05
0.035
0.045
0.60
0.45
0.35
0.15
Chapter 3 Steel Building Materials
69
Manganese
Manganese (Mn) has an effect on strength similar to that of carbon. It is a necessary component
because of the way it combines with oxygen and sulfur and its impact on the rolling process. In
addition, manganese improves the notch toughness of steel. It is added to steel to offset
reductions in notch toughness due to the presence of other elements. It has a negative effect on
material weldability.
Silicon
Silicon (Si) is an important element for removing oxygen from hot steel.
Vanadium
Vanadium (V) is another strengthening element. It refines the grain size and thus increases
strength. Its biggest advantage is that while increasing strength, it does not negatively impact
weldability or notch toughness.
Columbium
Columbium (Cb) is a strengthening element that, in small quantities, can increase the yield point
and, to a lesser extent, the tensile strength. However, it has a significant negative impact on notch
toughness.
Phosphorus
Phosphorus (P) increases strength and decreases ductility. It improves resistance to atmospheric
corrosion, particularly when used in combination with copper. It has a negative impact on
weldability that is more severe than that of manganese. It is generally an undesirable element but
is permitted in very limited quantities in all steel.
Sulfur
Sulfur (S) is also permitted in very limited quantities in all steel. It has a negative impact on
weldability comparable to that of phosphorus. Generally, steelmaking practice works to remove
as much sulfur as possible.
Copper
Copper (Cu) in limited quantities is beneficial to steel. It increases strength with only a limited
negative impact on ductility. If its content is held relatively low, it will have little effect on
weldability. It is the most significant contributing element in the production of corrosion-resistant
steel.
Nickel
Nickel (Ni) can provide a moderate improvement in strength and enhances corrosion resistance. It
can also improve resistance to corrosion for steel subjected to seawater when used in combination
with copper or phosphorus. It generally leads to a slight improvement in notch toughness.
70 Chapter 3
Steel Building Materials
Chromium
Chromium (Cr) is typically used in combination with copper to improve corrosion resistance. It
also provides some strengthening of steels containing copper and vanadium. Chromium is an
integral component of stainless steel.
Molybdenum
Molybdenum (Mo) increases strength but significantly decreases notch toughness. However, this
negative impact can be controlled by appropriate processing or balancing with other elements.
3.6 GRADES OF STRUCTURAL STEEL
3.6.1
Steel for Shapes
Many more grades of steel are produced than are approved by AISC for use in structures. A
unique ASTM number designates each type of approved steel. The steels approved for structural
shapes are grouped as carbon steel (A36, A53, A500, A501, A529, A709, A1043 and A1085),
high-strength low-alloy steel (A572, A618, A709, A913, A992, and A1065), and corrosionresistant high-strength low-alloy steel (A588, A847, and A1065). Figure 3.10 lists these approved
steels, their minimum yield and tensile stresses, and the shapes to which they are applicable. The
table also identifies which combination of shape and material specification are preferred, such as
A992 for W-shapes, and which material specifications do not apply for particular shapes.
A36 Steel
A36 steel was the most commonly available structural steel for many years. It was introduced in
the 1961 AISC Specification and until the late 1990s was the steel of choice for most steel shapes
except for HSS, pipe, and plates. It is a mild carbon steel, so it is well suited for bolted or welded
construction. Even when higher-strength steels were used for members, this steel was the usual
choice for connecting elements. It continues to be the preferred steel for M-, S-, C-, MC-, and Lshapes. It has a minimum yield stress, Fy = 36 ksi and a tensile stress, Fu = 58 to 80 ksi, although
Fu = 58 ksi is used for calculations throughout the Specification, Manual and AISC Design
Examples.
A53 Steel
A53 steel is the single standard for steel pipes approved for construction. It is available in three
types and two grades; however, they are not all approved for structural applications. These pipes
are generally intended for mechanical and pressure applications, and the only grade approved for
construction is Grade B. This grade is available as Type E, which denotes electric-resistance
welding of the seam, or Type S, which is a seamless pipe. A53 Grade B has a minimum yield
stress, Fy = 35 ksi and a minimum tensile stress, Fu = 60 ksi. A53 Grade B steel pipe are treated
as HSS in the Specification
A500 Steel
A500 steel is a carbon steel used for structural tubing in round, square, and rectangular shapes,
otherwise known as HSS. It comes in two grades approved by AISC for construction: Grade C,
which is the preferred grade, and Grade B. The standard permits either welded or seamless
manufacture. Round HSS Grade C has a minimum yield stress, Fy = 46 ksi and a minimum tensile
Chappter 3 Steel Building Matterials
Figure 3.10
3
Applicaable ASTM Specifications
S
s for Various Structural Shhapes
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with permissionn. All rights resserved.
71
72 Chapter 3
Steel Building Materials
stress, Fu = 62 ksi; whereas rectangular HSS Grade C, has a minimum yield stress, Fy = 50 ksi,
with a minimum tensile stress, Fu = 62 ksi.
A501 Steel
A501 steel is a carbon steel similar to A36 but used for round and rectangular HSS. It is approved
in Grade A, which has a minimum yield stress, Fy = 36 ksi, and a minimum tensile stress, Fu =
58 ksi, and in Grade B, which has a minimum yield stress, Fy = 50 ksi, and a minimum tensile
stress, Fu = 70 ksi.
A529 Steel
A529 steel is a carbon-manganese steel available in grades 50 and 55. It is approved for the
smaller shapes with flange thickness no greater than 1.5 in. A529 Grade 50 has a minimum yield
stress, Fy = 50 ksi and a tensile stress, Fu = 65 to 100 ksi whereas Grade 55 has a minimum
yield stress, Fy = 55 ksi, and a tensile stress, Fu = 70 to 100 ksi.
A709 Steel
A709 steel represents two types of steel, carbon steel, and high-strength low-alloy steel. Seven
grades in four yield strengths are included. This standard is developed specifically for bridges
where requirements necessary for bridge applications are added to the basic requirements for
A36, A572, A992, and A588 steels. In buildings, the corresponding grade of A709 steel may be
used in place of these four steels.
A1043 Steel
A1043 steel is a carbon steel available in two grades. Grade 36 has a yield stress, Fy = 36 to 50
ksi and a minimum tensile strength, Fu = 58 ksi. Grade 50 has a yield stress, Fy = 50 to 65 ksi and
a minimum tensile strength, Fu = 65 ksi. This steel is available when a maximum yield-to-tensile
ratio of 0.80 is required. W-shapes with a flange width of 6 in. or greater and plates up to 5 in.
thickness are included.
A1085
A1085 steel is a carbon steel for hollow structural sections. For HSS produced according to this
standard, the wall thickness is permitted to be no more than 5% under the nominal wall thickness
and the mass cannot be more than 3.5% under the nominal mass. This standard was developed so
that the design properties for HSS could be developed based on the nominal dimensions of the
section. For other HSS material specifications, the tolerance on thickness is so large that design
must be carried out using properties determined by using 0.93 times the nominal thickness, rather
than the nominal thickness. Grade A is the only grade provided for in the standard. It is available
with a yield stress, Fy = 50 to 70 ksi, and a minimum tensile stress, Fu = 65 ksi.
A572 Steel
A572 is a high-strength low-alloy steel, also referred to as columbium-vanadium structural steel,
available in five grades. It is a versatile high-strength steel with good weldability. Availability of
shapes and plates is a function of grade, generally depending on element thickness. It is available
in all shapes other than HSS and pipe. The full range of minimum yield stress is 42 to 65 ksi,
depending on grade, and the minimum tensile stress ranges from 60 to 80 ksi, again depending on
grade. A572 Grade 50 is the preferred steel for HP-shapes.
Chapter 3 Steel Building Materials
73
A618 Steel
A618 is a high-strength low-alloy steel used for HSS. Grades I, II, and III are approved for use in
structures by AISC. It is one of two high-strength low-alloy steels available in HSS. Grade II has
limited atmospheric corrosion resistance, and Grade III can be produced with increased corrosion
resistance if required. The minimum yield stress depends on the particular product and may vary
from 46 to 50 ksi. The minimum tensile stress varies from 65 to 70 ksi, again depending on grade
and product wall thickness.
A913 Steel
A913 is a high-strength low-alloy steel produced by quenching and self-tempering. It is available
in grades 50, 60, 65, and 70. The minimum yield stress ranges from 50 to 70 ksi and the
minimum tensile stress ranges from 65 to 90 ksi.
A992 Steel
A992 steel is a high-strength low-alloy steel that has become the steel of choice for wide-flange
shapes. It was first approved for use in 1998 as a replacement for steel that had come to be dualcertified as meeting both A36 and A572 Grade 50. This standard was developed partly as the
result of an improved understanding of the impact of material property variations on structural
behavior, and partly as the result of the changes in properties caused by the use of scrap as the
main resource for steel production. The chemical components for A992 steel were given in Table
3.3 and discussed in Section 3.5. It has a minimum yield stress, Fy = 50 ksi, and a minimum
tensile stress, Fu = 65 ksi. An additional requirement is that the yield-to-tensile ratio cannot
exceed 0.85.
A1065 Steel
A1065 steel is a high-strength low-alloy steel specifically for HSS. This steel is produced to the
standards for steels previously approved for use as plates. Thus, the tolerances are tighter than
those for other common HSS steels such as A500 and A501. HSS produced according to this
standard are first formed into two channels on a press brake and then welded together to form the
tube. This material is produced in Grade 50 and in Grade 50W, a corrosion resistant grade, with a
minimum yield stress, Fy = 50 ksi, and a minimum tensile stress, Fu = 60 ksi. A1065 steel
appears in the 2016 Specification for the first time. Before specifying this steel, the designer
should confirm its availability.
A588 Steel
A588 is a high-strength low-alloy corrosion-resistant steel with substantially better corrosion
resistance than carbon steel with or without copper. It is available for all shapes, except HSS and
pipe, as well as plate. For all shapes, and for plates up to 4 in., it has a minimum yield stress of 50
ksi and a minimum tensile stress of 70 ksi. Plates up to 8 in. are available at reduced stress values.
A847 Steel
A847 is a high-strength low-alloy corrosion-resistant steel used for HSS. It has the same
minimum yield and tensile stresses as A588.
744 Chapter 3
ding Materialss
Steel Build
Figure 3.11 Applicablee ASTM Speccifications forr Plates and B
Bars
Copyright © American Insttitute of Steel Construction.
C
R
Reprinted with Permission. A
All rights reservved.
3.66.2
Steel for Plates and Bars
Many of thee steels alread
dy discussed for
f shapes aree also availabble for plates aand bars. Figgure 3.11
shows the ASTM desig
gnations, the correspondiing yield andd tensile streesses, and thhe plate
Chapter 3 Steel Building Materials
75
thickness for which they apply. The only steels available for plates and bars that are not also
available for shapes are A283, A242, A514, and A1066.
A283 Steel
A283 steel is a low to intermediate strength carbon steel that is available in Grade C and Grade D.
Grade C has a minimum yield stress, Fy = 30 ksi, and a tensile stress, Fu = 55 to 75 ksi while
Grade D has a minimum yield stress, Fy = 33 ksi, and a tensile stress, Fu = 60 to 80 ksi. Although
there is no thickness limit specified, plates thicker than 2 in. may be more difficult to obtain and
the designer should check availability prior to specifying.
A242 Steel
A242 is a high-strength low-alloy corrosion-resistant steel also called weathering steel. It was one
of the first corrosion-resistant steels and has a corrosion resistance approximately four times that
of normal carbon steel. It is available in three grades but is now less common than the newer
A588. The minimum yield stress ranges from 42 to 50 ksi and the minimum tensile stress ranges
from 63 to 70 ksi.
A514 Steel
A514 is a high-yield strength-quenched and tempered alloy steel suitable for welding. It is
available as plate material up to 6 in. There are 14 different grades, which vary according to the
chemical content and maximum thickness. The minimum yield stress is either 90 or 100 ksi, and
the ultimate tensile stress ranges from 100 to 130 ksi. This is the highest-yield-stress steel
approved for use, except for fasteners, according to the AISC Specification.
A1066 Steel
A1066 steel is a high-strength low-alloy steel produced in 5 grades, Grade 50, Grade 60, Grade
65, Grade 70, and Grade 80 with a minimum yield stress corresponding to the grade designation.
Minimum tensile strength for Grade 50 is 65 ksi, Grade 60 is 75 ksi, Grade 65 is 80 ksi, Grade 70
is 85 ksi and Grade 80 is 90 ksi. Grades 50 and 60 are available up to 4 in. thickness, Grade 65 to
3 in. Grade 70 to 2 in. and Grade 80 to 1 in.
3.6.3
Steel for Fasteners
Fasteners for steel construction today include high-strength bolts, common bolts, threaded rods,
and anchor rods. In addition, nuts, washers, and direct-tension indicators must be specified. The
ASTM steels approved for these elements are listed in Figure 3.12. Many grades of steel are
appropriate for the variety of mechanical fasteners used in steel construction, but only the three
steels commonly specified for bolts are discussed here.
A307 Bolts
A307 bolts are also called common bolts or black bolts. Although the ASTM standard specifies
three grades, only Grade A is approved for use as bolts in general applications. These bolts have
an ultimate tensile strength of 60 ksi and are thus at a strength level similar to A36 steel.
Although these bolts continue to be listed by AISC, they are rarely used in steel-to-steel structural
connections.
766 Chapter 3
Steel Build
ding Materialss
Figgure 3.12 Applicable
A
AS
STM Specificcations for Vaarious Types of Structural Fasteners
Coopyright © Am
merican Institutee of Steel Consstruction. Reprrinted with Perm
mission. All riights reserved.
Chapter 3 Steel Building Materials
77
F3125 Bolts
F3125 is a standard that consolidates what had previously been given in multiple separate
standards. It provides for quenched and tempered bolts manufactured in two strength grades, two
types, and two styles. Thus, there are 6 grades specified, two of those six being metric equivalents
to two other grades.
Grade A325 Bolts
Grade A325 bolts are heavy hex head bolts that are the predominant high-strength bolts used in
construction. Two types are available: type 1, the normal medium carbon bolt; and type 3, which
is the same bolt provided in a weathering steel. Bolts are available from 1/2 in. to 1-1/2 in.
diameter and the minimum tensile strength is 120 ksi. These bolts are commonly referred to as
A325 bolts.
Grade A490 Bolts
Grade A490 also designates a heavy hex head structural bolt. These fasteners are used when a
higher tensile strength is required. As with the Grade A325 bolts, they are available as type 1 or
type 3 with the same distinction. Again, the bolts are available from 1/2 in. to 1-1/2 in. diameter
and have a minimum tensile strength of 150 ksi. These bolts are commonly referred to as A490
bolts.
Grade F1852 Bolts
Grade F1852 provides the standard specification for “twist-off” tension control bolt-nut-washer
assemblies with a tensile strength of 120 ksi. These structural fasteners are unique in that they do
not have a hex head but rather have a splined shank that permits installation through the use of a
special torque wrench. Grade F1852 connectors are essentially Grade A325 bolts but are
manufactured as a different grade because their geometric characteristics differ from those of
heavy hex head bolts. As with Grade A325, the tensile strength of these fasteners is 120 ksi for
the diameters available, from 1/2 to 1-1/4 in. These bolts are commonly referred to as A325 TC
bolts.
Grade F2280 Bolts
Grade F2280 provides the standard specification for “twist-off” tension control bolt-nut-washer
assemblies, with a tensile strength of 150 ksi. These connectors are essentially Grade A490 bolts
but, like Grade F1852 connectors, must be manufactured as a separate grade because their
geometric characteristics differ from those of heavy hex head bolts. These fasteners are available
in diameters of 1/2 to 1-1/4 in. These bolts are commonly referred to as A490 TC bolts.
3.6.4
Steel for Welding
Steel used for welding is called filler metal because it essentially fills the gap between the base
metal pieces it is joining. The most critical aspect of selecting filler metal, which actually
corresponds to the welding electrode, is matching the electrode with the base metal. In all cases,
the weld must not form the weak part of the joint. The American Welding Society provides the
specification for appropriate matching of the base metal and electrodes in Table 3.1 of their
standard ANSI/AWS D1.1 and the AISC Specification includes some information in Chapter J.
The most commonly used weld strength is 70 ksi. A discussion of welding processes and material
matching is presented in Chapter 10.
78 Chapter 3
3.6.5
Steel Building Materials
Steel for Headed Stud Anchors
Steel headed stud anchors, more commonly known simply as shear studs, are mechanical
fasteners welded to structural shapes and embedded in concrete that permit steel and concrete to
work together. This is called composite construction. Because these studs are welded to the steel
shape, their properties are specified jointly between AWS and ASTM. Shear studs are specified
as given in AWS D1.1 Clause 7, with material as required in Clause 7.2.6. Type B is usual and
the corresponding mechanical requirements are stated in AWS D1.1 Table 7.1 (Fy = 51 ksi, Fu =
65 ksi).
3.7 AVAILABILITY OF STRUCTURAL STEEL
Structural engineers normally use the list of shapes found in the AISC Manual as the basis for
design. Unfortunately, all shapes are not equally available in the marketplace and the selection of
shapes that are difficult to obtain could negatively impact the overall cost and schedule of a
project. Some shapes are available from a wide variety of producers; for instance, a W10×30
could be obtained from six different mills as of April 2016. However, the largest shapes, such as
the W44×335, are not produced by all mills. Also, several of the smaller M-shapes are not rolled
by any mill. Shape availability data are maintained by the mills on the AISC Web site at
www.aisc.org/steelavailability.
Another important source – increasingly the primary source – of steel is the steel service
center. These organizations are warehouses throughout the country that obtain steel directly from
the mills; many stock nearly the full range of shapes. Although the task of obtaining the steel
needed for any given project falls to the steel fabricator, it is always beneficial to the engineer to
have some knowledge of availability. AISC maintains a list of steel service centers at
www.aisc.org which in April 2016 identified 20 AISC Member Steel Service Centers and the
states they serve.
3.8 PROBLEMS
1. When was the first AISC Specification published and
what was its purpose?
2. In addition to buildings, what other types of structures
are included in the scope of the 2010 AISC Specification?
3. Sketch and label a typical stress-strain curve for steel
subjected to a simple uniaxial tension test.
4. What is the value of the modulus of elasticity used for
calculations according to the AISC Specification, and
what does this value represent in relation to the graph of
stress versus strain for steel?
5. What happens to a steel element when it is loaded
beyond the elastic limit and then unloaded?
6. Describe the difference between the yield stress and
ultimate stress of a steel element.
7. Sketch and label 10 different structural shape cross
sections whose properties are given in the AISC Manual.
8. What are the nominal and actual depths of a W36×135
wide-flange member? What is the weight of this member
per linear foot? (Hint: Use your AISC Manual.)
9. What are the nominal and actual depths of a W16×100
wide-flange member? What is the weight of this member
per linear foot? (Hint: Use your AISC Manual.)
10. What are the nominal and actual depths of a W18×130
wide-flange member? What is the weight of this member
per linear foot? (Hint: Use your AISC Manual.)
11. What are the nominal and actual depths of a W14×730
wide-flange member? Compare these to the nominal and
actual depths of a W14×145. (Hint: Use your AISC
Manual.)
12. What are the nominal and actual depths of a W12×336
wide-flange member? Compare these to the nominal and
actual depths of a W12×16. (Hint: Use your AISC
Manual.)
Chapter 3 Steel Building Materials
13. What are the nominal and actual depths of a W10×112
wide-flange member? Compare these to the nominal and
actual depths of a W10×12. (Hint: Use your AISC
Manual.)
14. What are the actual depth, flange width, and flange
thickness of a W14×132? (Hint: Use your AISC Manual.)
79
27. What is the outside diameter of a round
HSS10.750×0.375? What are the nominal and design wall
thicknesses? (Hint: Use your AISC Manual.)
28. What is the distinction between a round HSS and a
pipe?
15. What are the actual depth, flange width, and flange
thickness of a W27×84? (Hint: Use your AISC Manual.)
29. What are the outside diameter and nominal and design
wall thicknesses of a Pipe 4 xx-Strong? (Hint: Use your
AISC Manual.)
16. What are the actual depth and flange width of an
HP12×84? What are the thicknesses of the web and
flange? (Hint: Use your AISC Manual.)
30. What are the outside diameter and nominal and design
wall thicknesses of a Pipe 10 Std.? (Hint: Use your AISC
Manual.)
17. What are the actual depth, average flange thickness,
and web thickness of a C12×30? (Hint: Use your AISC
Manual.)
31. What is the difference between a rectangular bar
section and a plate?
18. What are the actual depth, average flange thickness,
and web thickness of a C9×15? (Hint: Use your AISC
Manual.)
32. What are the three preferred types of steel for shapes
in the construction industry?
33. What effects does the addition of carbon have on
steel?
19. What are the cross sectional area, leg dimensions and
thickness of an L8×8×5/8? (Hint: Use your AISC
Manual.)
34. Name three elements that improve the corrosion
resistance of steel.
20. What are the cross sectional area, leg dimensions and
thickness of an L6×4×1/2? (Hint: Use your AISC
Manual.)
35. What grade of steel is most commonly used today in
the production of W-shapes, and what are its yield stress
and tensile stress?
21. What are the cross sectional area and weight per linear
foot of an L5×3×1/2 member? (Hint: Use your AISC
Manual.)
36. What grade of steel is most commonly used today in
the production of round HSS? Rectangular HSS?
22. What are the actual depth, flange width, flange
thickness, and stem thickness of a WT7×45? Compare to
the properties for a W14×90. (Hint: Use your AISC
Manual.)
23. What are the actual depth, flange width, flange
thickness, and stem thickness of a WT10.5×22? Compare
to the properties for a W21×44. (Hint: Use your AISC
Manual.)
24. What are the outside dimensions of a rectangular
HSS12×10×3/8? What are the nominal and design wall
thicknesses? (Hint: Use your AISC Manual.)
25. What are the outside dimensions of a rectangular
HSS8×6×1/2? What are the nominal and design wall
thicknesses? (Hint: Use your AISC Manual.)
26. What is the outside diameter of a round
HSS6.000×0.500? What are the nominal and design wall
thicknesses? (Hint: Use your AISC Manual.)
37. What are the differences between an A500 Grade C
rectangular HSS and an A1085 rectangular HSS?
38. What grade of steel is preferred for the fabrication of
most structural shapes other than W-shapes, and what are
its yield stress and tensile stress? (Hint: See Figure 3.10.)
39. What grade of steel is typically used for high-strength
bolts in construction?
40. What resources can be consulted to determine the
availability of a particular steel structural shape?
C
Chapterr 4
T
Tension Mem
mberss
111 South M
Main, Salt Lake City, UT
Photo courteesy Hassett Enggineering
4.11
INTROD
DUCTION
The most effficient way to
o carry a forcee in steel is thhrough tensioon. Because teension forces result in
a fairly unifform stress disstribution in the
t member ccross section, all of the maaterial is able to work
to its fullest capacity. Th
he normal assu
umption that tensile forcess are applied to a member through
the centroid of the cross section mean
ns that other sstructural actiions, such as buckling or bbending,
are not norm
mally present to reduce thee material’s abbility to carryy load. Thus, tension mem
mbers are
perhaps the simplest to deesign and a good starting ppoint for studdying structuraal steel designn.
Ten
nsion memberrs are fairly common
c
elem
ments in buildding structurees, although they are
not found in
i every stru
ucture. The structural meembers considered in thiis chapter arre those
subjected to
o a concentricc tensile forcce as their prrimary force.. Secondary effects, such as load
misalignmen
nt and the in
nfluence of co
onnections, w
will be addreessed; howeveer, the interaaction of
tension and bending is saaved for treatm
ment in Chaptter 8.
Table 4.1 lists th
he sections off the Specificaation and part
rts of the Mannual discussed in this
chapter.
4.22
TENSIO
ON MEMBE
ERS IN STR
RUCTURES
A wide varriety of tensiion memberss can be fouund in buildiing structurees. Among thhe more
important arre members of trusses, braccing memberss, hangers, annd sag rods.
Ten
nsion memberrs are found in trusses ass chords, diaagonals, and verticals. Figgure 4.1
shows a typ
pical simply supported tru
uss, with the tension bottoom chord and two of thee tension
diagonals in
ndicated. Ten
nsion membeers used as bbracing for sstructures aree normally loong and
slender, as seen
s
in Figuree 4.2. Becausee these slendeer members arre relatively fflexible, they must be
carefully designed and errected, particcularly if therre is any channce of load reeversal, which would
call on them
m to carry a compression load. Even thee smallest coompressive foorce in a mem
mber that
has been deesigned as a tension-only
y member caan cause signnificant strenngth or servicceability
problems in the final stru
ucture.
800
Chapter 4
Table 4.1
Tension Mem
mbers
81
Sections of
o Specification and Parts of Manual Coovered in thiss Chapter
Specificattion
B3
B4.2
B4.3
D1
D2
D3
D4
D5
D6
J3.2
J3.5
J3.6
J4.1
J4.3
J7
M2.5
Design Basiis
Design Walll Thickness fo
for HSS
Gross and Net
N Area Deteermination
Slenderness Limitations
Tensile Stren
ngth
Effective Neet Area
Built-up Meembers
Pin-Connectted Members
Eyebars
Size and Usee of Holes
Maximum Spacing
S
and E
Edge Distancee
Tensile and Shear Strenggth of Bolts annd Threaded P
Parts
Strength of Elements
E
in T
Tension
Block Shearr Strength
Bearing Streength
Bolted Consstruction
Manuall
Part 1
Part 5
Dimensions and Propertiees
Design of Teension Membbers
Other
O
examples of tensio
on members are hangers that connect lower floors to some
support above,
a
as seen
n in Figure 4.3, and sag roods that suppoort purlins in the roof struucture or the
girts in the
t walls of a steel-frameed building. IIt is easy to see the importance of m
most tension
memberss because they normally carry
c
an obvioous, direct looad. For mem
mbers like sagg rods, their
failure caan produce unsightly
u
displacements in the walls, annd could causse stability prroblems for
the purlin
ns or girts, bu
ut they are nott likely to be found carryinng significantt direct loadinng.
Figure 4.1
4
A Simply Supported Truss
T
with Teension Membbers Indicatedd
Photo cou
urtesy Matt Meelrose/LERA Consulting
C
Struuctural Engineeers
822 Chapter 4
Tension Members
Figure 4.2 Tension Braacing Membeers
ngers
Figure 4.3 Tension Han
4.33
CROSS--SECTIONAL SHAPE
ES FOR TEN
NSION ME
EMBERS
Tension meembers can be structural steel
s
shapes, plates, or coombinations oof shapes andd plates;
eyebars and
d pin-connected plates; rod
ds and bars; or wire ropee and steel caables. Wire rrope and
steel cables are not coverred by the Speecification noor consideredd here, althouggh they are im
mportant
elements in the special strructures wherre they occur .
Eyebars are not in
i common use today but ccan be foundd in older applications, partticularly
in bridges, trrusses and sim
milar structurres, as seen inn Figure 4.4. A
Although rareely used, theyy are still
covered in the
t Specificattion. The pin--connected pllate shown inn Figure 4.5 iis actually a ppart of a
connection. This configu
uration is used in a varietyy of applicatiions, such as industrial strructures,
anchors for tension
t
memb
bers, and con
nnections in brridge girders..
Chapter 4
Figure 4.4
4
Eyebars in an Historicc Building Rooof Structure
Figure 4.5
4
A Pin-Co
onnected Mem
mber
Tension Mem
mbers
83
Photo cou
urtesy of LeJeu
une Steel Co.
Several
S
comm
mon shapes used
u
for tensiion memberss are shown iin Figure 4.66, and some
typical bu
uilt-up shapes are given in
n Figure 4.7. T
The solid rounnd bar is freqquently used, either as a
threaded rod or weldeed to other members.
m
The threaded endd provides a ssimple connection to the
gn must take into account the reductionn in cross-secttional area caaused by the
structure, but the desig
threads. Rods
R
with up
pset ends are occasionally used instead of the normaal rods; the ennlarged end
permits threading
t
witthout reducin
ng the cross-ssectional areaa below the m
main portion of the rod.
The diffeerences betweeen these two types of rodss can be seen in Figure 4.88.
Square,
S
rectan
ngular, and circular
c
HSS have becomee more comm
mon as tensioon members
over the past few yeears, largely due to their attractive apppearance annd ease of m
maintenance.
Howeverr, the end connections may
m become complicatedd and expenssive, dependding on the
particular application.. HSS are especially usefull for longer teension compoonents, when slenderness
and related serviceability consideraations may bee important.
Single
S
angless, as shown in
i Figure 4.77a, are used eextensively inn towers, succh as those
supportin
ng cellular telephone com
mmunications and high-volltage power lines. Double angles and
double ch
hannels, as sh
hown in Figu
ure 4.7b and dd, are probablly the most ppopular tensioon members
for planaar trusses due to the fact thaat gusset plattes can be connveniently plaaced in the space
844 Chapter 4
Tension Members
Figure 4.6 Common Sh
hapes for Ten
nsion Memberrs
between th
he individuall shapes. Th
he end connnections forr these mem
mbers are ttherefore
straightforw
ward to design
n and fabricatee, and allow ffor symmetryy in the verticaal plane.
Larg
ge tensile forcces may requ
uire cross secttions with an area dictatingg that the meember be
made from a wide-flangee shape, a tee,, double channnels, or builtt-up shapes, ssuch as those given in
Figure 4.7e and f. Built--up cross secctions were m
more commonn in the past,, when rolledd shapes
were compaaratively smalller in cross section
s
and thhe cost of labbor was loweer; today largeer rolled
shapes are available
a
and labor costs mean
m
it is moore economiccal to use them. Current sttructural
applications of such elem
ments are fou
und in long- span roof truusses, bridge trusses, and bracing
members in large industriial structures..
4.44
BEHAV
VIOR AND STRENGTH
H OF TENS
SION MEM
MBERS
Tension meembers are co
overed in Ch
hapter D of thhe Specificattion. The firsst requiremennt given,
Section D1, is not really a requiremen
nt at all but reccognition thaat in the past tthere had beenn a limit
on the slend
derness, the raatio of length
h to radius off gyration, L/rr, for tension members. Thhat limit
was really a recommend
dation since very slenderr members m
may be difficcult to handlee during
construction
n but it had no
o impact on th
he strength off tension mem
mbers. Thus, tthe user note ssuggests
that a limit on
o L/r of 300 would be reaasonable.
Figure 4.7
Typicaal Built-up Shapes Used as Tension Mem
mbers
Chapter 4
Figu
ure 4.8
Tension Mem
mbers
85
Thrreaded and Up
pset Rods
Two
T
possiblee limit states are defined iin Section D22 for tensionn members: yyielding and
rupture. The controllling limit staate depends oon the abilityy of the mem
mber to undeergo plastic
deformattion. Both of these failure modes repressent limit stattes of strengthh that must bee taken into
account in the desig
gn of the ten
nsion membeer. The desiggn basis for ASD and L
LRFD were
presented
d in Sections 1.9 and 1.10
0, respectivelly. Equations B3-2 and B3-1 are repeaated here to
reinforcee the relation
nship betweeen the nominnal strength, resistance faactor, and saafety factor
presented
d throughout the Specificattion.
The
T requirement for ASD is
i
Ra ≤
Rn
Ω
(A
AISC B3-2)
The
T requirement for LRFD
D is
Ru ≤ φRn
4.4.1
(A
AISC B3-1)
Yield
ding
Yielding occurs when
n the uniform
mly distributeed stress throughout the ccross section reaches the
yield streess over the length of the member. Althhough the meember will coontinue to ressist the load
that caussed yielding to
t occur, it wiill undergo exxcessive strettching, and thhis elongationn will make
the mem
mber unusable. The longer the
t member, the greater thhe elongationn. Because thee limit state
of yieldiing on the grross section of the membber is accomp
mpanied by thhis large defoormation, it
readily seerves as a waarning of any impending
i
faailure.
The
T yield limiit state is defiined as
(A
AISC D2-1)
Pn = Fy Ag
where
Pn = nominal tensile yield strength
Fy = yield streess
Ag = gross areea of the mem
mber
4.4.2
The desig
gn strength an
nd allowable strength are tto be determinned using
φt = 0.90 (LRFD)
Ωt = 1.667 (ASD)
Rupture
Holes in a member wiill cause stresss concentrati ons to develoop under the sservice load, aas shown in
Figure 4.9.
4
Elastic theory show
ws that the stress conccentration ressults in a ppeak stress
approxim
mately three times
t
the average stress. As the peakk stress reachhes the yieldd stress, the
member will continuee to strain and
d load can conntinue to incrrease. With inncreasing loadd, the strain
in the reg
gion of the ho
ole increases into the strainn hardening rregion, and thhe member ruuptures once
the stresss in this areaa exceeds thee ultimate streength. Althouugh the mateerial in the reegion of the
holes yieelds initially, it does so over
o
a very shhort length, rresulting in a small total elongation.
Thus, the material caan reach its ultimate
u
strenngth through strain hardeening, withouut excessive
on, and failuree occurs throu
ugh rupture. T
The limit statee of rupture oon the effectivve net area
elongatio
866 Chapter 4
Tension Members
Figure 4.9 Stress Conccentration Duee to Hole in M
Member
of the crosss section is accompanied
a
by small deeformations ffrom yieldingg, giving littlle or no
warning of the impendin
ng sudden failure, and off
ffering limitedd opportunitiies to take coorrective
action beforre the rupture..
The rupture limitt state is defin
ned as
Pn = Fu Ae
(AISC D2-2)
where
Pn = nominal tensile rupture sttrength
Fu = ultimate streess
Ae = effective nett area of the member
m
The design strength
s
and allowable
a
streength are to bbe determinedd using
ASD)
φt = 0.75 (LRFD)
(
Ωt = 2.00 (A
If the two lim
mit states werre to result in
n the same avaailable strengtth, using the L
LRFD formuulation,
0.75
0 Fu Ae = 0.990 Fy Ag
(4.1)
Ae Ag = 0.90 Fy 0.75 Fu
(4.2)
or
The limit staate of yieldin
ng on the grosss section govverns when thhe right-handd side of Equaation 4.1
is less than the
t left-hand side. Using Equation
E
4.2, yyielding on thhe gross section governs iff
(4.3)
Ae Ag > 0.90 Fy 0.75 Fu
and rupture on the effectiive net section
n governs if
Ae Ag < 0.90 Fy 0.75 Fu
(4.4)
Steel with a small (Fy /F
Fu) value, such
h as ASTM A
A36, with 0.99Fy/0.75Fu = 0.9(36)/(0.755(58)) =
0.74, will allow more off the cross seection to be removed in tthe form of bbolt holes beefore the
ns than will stteel with a hiigher (Fy /Fu) value, such aas ASTM A992, with
rupture limitt state govern
0.9Fy/0.75F
Fu = 0.9(50)/(0
0.75(65)) = 0.92.
The comparisonss discussed ab
bove are appllicable only ffor normal boolted connectiions and
their corresp
ponding areass. The Specifiication equatiions are not inntended to coover tension m
members
with large cutouts.
c
Thesee require speccial design coonsiderations,, and are beyyond the scope of this
book becausse they are no
ot common in most buildinng structures.
Chapter 4
Tension Members
87
Although welded connections do not normally require the removal of material from the
cross section, the placement of the welds and the type of cross section may require a reduction
from the gross area to determine the effective net area.
4.5 COMPUTATION OF AREAS
The design of tension members uses the following cross-sectional area definitions:
1.
2.
3.
gross area, Ag
net area, An
effective net area, Ae
The criteria governing the computation of the various areas required for tension member analysis
and design are given in Sections B4.3 and D3 of the Specification. They are discussed in further
detail here.
4.5.1
Gross Area
The gross area of a member might also be thought of as the full cross-sectional area. A section is
made perpendicular to the longitudinal axis of the element, along which the tensile force is acting,
and the gross area, Ag, is the area of that cross section. No holes or other area reductions are
considered when calculating the gross area.
In the case of plates, bars, and solid circular shapes, the value of Ag is found directly as
the value of width times thickness, bt, for plates and bars, and as πd 2 4 for circular shapes,
where d is the diameter. For structural steel shapes commonly used in construction, the Manual
provides values for gross areas in Part 1. However, in lieu of using the tabulated values, Ag may
be approximated as
(4.5)
Ag = Σwi ti
where wi and ti are the width and thickness, respectively, of the rectangular cross-sectional
element, i, of the shape. Equation 4.5 applies only to shapes that are composed of flat plate
components, such as wide flanges. The calculation for hollow circular shapes is similarly
straightforward. The gross area of HSS meeting the requirements of ASTM A500 is determined
using 93 percent of the nominal wall thickness of the shape, as indicated in Section B4.2. Because
HSS are consistently manufactured with a thickness at the low end of the tolerance limit, the
values provided in the Manual are all based on this reduced thickness. For channels and S-shapes,
both with sloping flanges, the inaccuracy of this approach means that the approach is rarely used.
For all hot-rolled shapes, the area is easily obtained from the tables in Part 1 of the Manual.
The procedure for angles requires a slight modification. The angle may be treated as an
equivalent flat plate, where the effective width is taken as the sum of the leg dimensions less the
thickness, and the gross area is this effective width times the angle thickness.
An angle and its equivalent flat plate are shown in Figure 4.10.
4.5.2
Net Area
The net area is obtained by subtracting the area of any holes occurring at a particular section from
the gross area at that section. Thus, holes associated with mechanical fasteners, such as bolts, and
with welds, such as plug welds and slot welds, are considered. These fastening elements are
normally used only to connect tension members to the adjacent parts of the structure, and the
reduced areas therefore normally appear at member ends. However, if holes occur at any point
along a tension member, their effect must be considered.
888 Chapter 4
Tension Members
Figu
ure 4.10 An
ngle and its Equivalent
E
Flaat Plate
Plug
g and slot welds are made through holees in a membeer and are relatively uncom
mmon in
most structu
ures in which the framin
ng members are made frrom structuraal shapes andd plates.
Normal wellded joints do
o not involvee making holles in membeers. Thus, weelds do not nnormally
reduce the cross-sectionaal area of tension members, so the net arrea is equal too the gross areea.
In the
t computattion of net area
a
for a teension membber with boltted end connnections,
determining
g the size of the holes is important. T
The criteria foor standard, oversize, andd slotted
holes are cov
vered in Speccification Secttion J3.2.
H
Standard Holes
Normal steeel constructio
on requires th
he specificati on of fasteneer size ratherr than hole siize. The
hole is then
n sized accord
ding to whatt is required to accommodate the fastener. The maanner in
which the ho
ole is fabricatted is also crittical.
In recognizing th
he needs for fabrication an
and erection ttolerances, staandard bolt hholes for
bolts up to 7/8
7 in. in diam
meter are mad
de 1/16 in. laarger in diameeter than the bolt to be insserted in
the hole. Fo
or 1 in. diameeter bolts and
d larger, the sstandard hole may be 1/8 in. larger in ddiameter
than the bolt to be inserrted in the hole.
h
This inccrease for larrger bolts is an option thhat some
fabricators may
m deem un
nnecessary; ho
owever, sincee that may noot be known bby the designeer ahead
of time, the larger hole sh
hould be used
d in calculatioons. Thus, a 3/4 in. bolt rrequires a holle with a
(3/4+1/16) = 13/16 in. diameter
d
and a 1 in. diameeter bolt requuires a 1-1/8 diameter holee. In the
case of pun
nched holes, the punching
g process maay damage soome of the m
material imm
mediately
adjacent to the
t hole. Thaat material maay not be connsidered fully effective in ttransmitting lload and
must also bee deducted fro
om the gross area along wiith the materiial that has acctually been reemoved.
Figure 4.11 shows an ang
gle leg with a hole along w
with the punchhed out piece.. As the punchh is
g with Puncheed Hole and P
Plug that was removed by P
Punching
Figure 4.11 Angle Leg
Chapter 4
Tension Members
89
applied to the material, the edges around the hole are deformed, as seen in the taper of the
punched out piece. In discounting this region, the effective hole diameter is increased by another
1/16 in., regardless of bolt diameter, according to Specification Section B4.3b. The deduction is
conservatively applied to all holes, since one generally does not know whether holes will be
punched or drilled.
The most common practice for producing holes is through punching. However, any
method that meets the hole surface quality requirements of Specification Section M2.5 is
permitted.
Because the decision to punch or drill a hole is a function of the steel fabricator’s
equipment capacity, for the design of tension members it is standard practice to deduct for holes
with a diameter 1/8 in. greater than the specified bolt size up to 7/8 in. and 3/16 for 1 in. bolts and
above.
The following examples demonstrate gross and net area calculations for several shapes.
EXAMPLE 4.1
Gross and Net
Area
Goal:
Determine the gross and net areas of a plate with a single line of holes.
Given:
A single line of standard holes for 3/4 in. bolts is placed in a 6×1/2 plate,
as shown in Figure 4.12a.
SOLUTION
Step 1:
Determine the gross area at Section 1-1.
Ag = 6 (1 2 ) = 3.0 in.2
Step 2:
Determine the effective hole size for a 3/4 in. diameter bolt.
Step 3:
Determine the net area at section 2-2.
An = ( b − d e ) t = ( 6.0 − 7 8 )(1 2 ) = 2.56 in.2
de = ( 3 4 + 1 16 + 1 16) = 7 8 in.
EXAMPLE 4.2
Gross and Net
Area
Goal:
Determine the gross and net areas of a plate with a double line of holes.
Given:
A double line of standard holes for 7/8 in. bolts are placed in a 10×3/4
plate, as shown in Figure 4.12b.
SOLUTION
Step 1:
Determine the gross area at section 1-1.
Ag = 10 ( 3 4 ) = 7.5 in.2
Step 2:
Determine the effective hole size for a 7/8 in. diameter bolt.
Step 3:
Determine the net area at section 2-2.
An = ( b − nd e ) t = (10.0 − 2 (1.0 ) ) ( 3 4 ) = 6.00 in.2
de = ( 7 8 + 1 16 + 1 16) = 1.0 in.
900 Chapter 4
Tension Members
Figure 4.12 Plates and Shapes with Holes for Us e with Exampples 4.1 throuugh 4.4
EX
XAMPLE 4.3
4
Grross and Net
Arrea
Goa
al:
Deterrmine the gross and net areeas of an anggle with a singgle line of hooles in
each leg.
l
Giv
ven:
A sin
ngle line of sttandard holess for 7/8 in. bbolts is placed on each legg of a
6×6×¾
¾ angle as sh
hown in Figurre 4.12c.
SO
OLUTION
Step
p 1:
Deterrmine the grosss area at secttion 1-1 from
m Manual Tabble 1-7.
Ag = 8.46 in.2
Step
p 2:
Deterrmine the effeective hole sizze for a 7/8 inn. diameter boolt.
Step
p 3:
Deterrmine the net area at sectioon 2-2.
An = Ag − nd e t = 8.46 − 2 (1.0 ))( 3 4 ) = 6.96 in.2
de = ( 7 8 + 1 16 + 1 16) = 1.0 in.
EX
XAMPLE 4.4
4
Grross and Net
Arrea
Goal:
Determine the gross and net are as of a channnel with muultiple lines oof
holes.
Given:
Four lines of standard
d holes for 1 in. bolts are placed in an MC12×31, aas
shown in
n Figure 4.12
2d. Two liness are in the w
web and one line is in eacch
flange.
SO
OLUTION
Step 1:
Determine the gross area
a at sectionn 1-1 from M
Manual Table 1-6.
Ag = 9.12 in.2
Step 2:
Determine the effective hole size ffor a 1.0 in. ddiameter bolt.
Chapter 4
Tension Mem
mbers
91
de = (1.0 + 1 8 + 1 16) = 1.19 in.
Step 3:
Deterrmine the nett area at sectioon 2-2.
First determine thee web and flaange thicknesses from Mannual Table 1-6.
tw = 0.3700 in. and tf = 00.700 in.
An = Ag − nw d e t w − n f d e t f
= 9.12 − 2 (1.119 )( 0.370 ) − 2 (1.19 )( 0.7000 )
= 6.57 in.2
Oversizeed and Slotteed Holes
Section J3.2
J
of the Specification
S
gives the reequired measuurements forr larger-than-standard or
oversized
d holes, as well
w as for sh
hort-slotted aand long-slottted holes. Figgure 4.13 illuustrates the
criteria that
t
apply fo
or nominal bolt
b
diameterrs of 5/8 in., 3/4 in. annd 7/8 in.; rrefer to the
Specifica
ation for data for other bollt sizes. Thesse types of hooles are used to facilitate tthe erection
of the strructure and, in
n some casess, to permit laarger rotationss or deformattions to take pplace under
loading.
Short Co
onnecting Ellements
Tension members witthin connectio
ons are usuallly short connnecting elemeents such as liinks, flange
plates, orr gusset platees. When the member is sshort, and thee net area andd gross area aare close to
equal, th
here may not be sufficient length for thhe load to sprread to a uniform distribuution on the
entire cro
oss section. In
I this case, the
t area that is the first too yield may rreach rupture at an early
stage, an
nd the rupturee limit state would
w
therefoore be reachedd prematurelyy. This is an undesirable
mode off failure, prim
marily because it is not duuctile and couuld occur sudddenly, with little or no
warning. Section J4.1 indicates thaat the effectivve net area in these short cconnecting eleements may
be limiteed due to strress distributiion as calcullated by methhods such ass the Whitmoore section,
which is discussed in Chapter 11.
4.5.3
Influ
uence of Holee Placement
The exam
mples in Sectiion 4.5.2 reprresent simple cases in which the net areea is found inn the section
that prod
duces the larg
gest reduction
n in area, usuually the sectiion with the llargest number of holes.
Howeverr, hole placem
ment does no
ot always folllow simple ppatterns wherre every secttion has the
same num
mber of holess. It is sometim
mes advantaggeous to use a pattern of staaggered holess, such as
Figure 4.13
4
Size Crriteria for 5/8 to 3/4 in. Bollts in Oversizze and Slottedd Holes
922 Chapter 4
Tension Members
Figure 4.14 Staggered Hole Patternss in Plate andd Angle
those shown
n in Figure 4.14. Figure 4.14a
4
shows aan arrangemeent of staggerred holes for a plate,
and Figure 4.14b
4
shows an example for an angle. When there are multiple holes, the ceenter-tocenter distan
nce between adjacent holes in the direection paralleel to the prim
mary applied force is
defined as th
he pitch, s. When
W
there is more than onne line of holes parallel to the line of foorce, the
center-to-cen
nter distance between adjjacent holes in the directtion perpendiicular to the primary
applied force is the gage, g.
It iss not clear fro
om Figure 4.1
14 what the ggoverning nett section wouuld be for eithher case.
For the platee, sections 1--1 and 2-2 giv
ve identical A n values, in w
which a deduuction for onee hole is
taken for eaach line. Ano
other possibillity would bee to follow a line that inccorporates tw
wo holes,
starting alon
ng line 1 an
nd ending along line 2, as shown byy the diagonnal dashed liine. The
Specification
n refers to thiis line as a “chain” becaus e it links indiividual holes. In this case, the area
of two holess would be deducted
d
from
m the gross crross section. H
However, thiis approach w
would be
no differentt than if both
h holes weree along the ssame straightt line, line 1 or line 2. IIt seems
reasonable in this situatio
on that an app
proach that woould deduct bboth holes woould deduct tooo much,
because thesse holes are staggered
s
and
d are not alonng the same straight line. The correct solution
should be so
omewhere between the ressult of deductting for one hhole and that of deductingg for two
holes.
A siimplified approach to addrress the interaaction of stagggered holes w
was adopted llong ago
by previouss AISC Speciifications. Allthough numeerous studiess have been conducted siince this
original sim
mplification was
w introducced, none haave proposedd a significaantly more accurate
approach thaat is equally easy
e
to implem
ment.
The Specification
n approach, given
g
in Sectiion B4.3b, reequires that evvery potentiaal failure
line be assesssed with the full area of eaach intersecteed hole deduccted and someething added back for
the increaseed strength prrovided by th
he diagonal ppath. For every diagonal oon a potentiaal failure
path, the quaantity s 2 4 g is added bacck into the nett width to acccount for the overestimatioon of the
required ded
duction when
n a full adjaceent hole has bbeen deductedd. Examples 4.5 and 4.6 sshow the
application of the staggeered hole criteerion in the m
middle of the member where the tensioon flows
through the member. Wh
hen holes are at the end off a member aas part of a coonnection thee flow of
the tension force
f
through
h the connecto
ors should be considered.
Chapter 4
Tension Mem
mbers
93
ure 4.15 Holle Pattern forr 18 in. Plate U
Used in Exam
mple 4.5
Figu
EXAMPLE
E 4.5
Net Width and Net
Area of Pla
ate
Goal:
G
Deetermine the net
n width of a plate with sttaggered holees. Then deterrmine the
nett area.
Given:
G
The hole pattern
n for an 18×77/8 in. plate with holes foor 3/4 in. bollts that is
on as shown in Figure 4.115. A and F rrepresent the edges of
loaaded in tensio
thee plate and
d B, C, D
D, and E represent the hole llocations.
SOLUTION
N
Step
S
1:
Ch
hain ABF (a sttraight line thhrough one hoole):
Deeduct for one hole
h
( 3 4 + 1 8) = −00.875 in.
Step
S
2:
Ch
hain ABCF:
Deeduct for two holes
h
2 ( 3 4 + 1 8) = −1.75 in.
Forr BC add
s 4 g = ( 2.0 )
2
2
Tottal deduction
Step
S
3:
= −1.50 in.
Ch
hain ACEF
Deeduct for two holes
h
Forr CE add
2 ( 3 4 + 1 8) = −1.75 in.
s 2 4 g = ( 2.5 )
( 4 (10.0 ) ) = +0.16 in.
2
= −1.59 in.
Tottal deduction
Step
S
4:
Ch
hain ABCEF
Deeduct for threee holes
3( 3 4 + 1 8) = −2.63 in.
( 4 ( 4.0 ) ) = +0.25 in.
4 g = ( 2.5 ) ( 4 (10.0 ) ) = +0.16 in.
Forr BC add
s 2 4 g = ( 2.0 )
Forr CE add
2
Tottal deduction
( 4 ( 4.0 ) ) = +0.25 in.
s2
2
= −2.22 in.
94 Chapter 4
Tension Members
Step 5:
Chain ABCDEF
Deduct for four holes
4 ( 3 4 + 1 8) = −3.50 in.
( 4 ( 4.0 ) ) = +0.25 in.
4 g = ( 4.5 ) ( 4 ( 6.0 ) ) = +0.84 in.
4 g = ( 2.0 ) ( 4 ( 4.0 ) ) = +0.25 in.
s 2 4 g = ( 2.0 )
2
For CD add
s2
2
For DF add
s2
For BC add
2
= −2.16 in.
Total deduction
Step 6:
Deduct the largest quantity to obtain the least net width. For chain ABCEF,
the net width is
bn = 18.0 − 2.22 = 15.8 in.
Step 7:
Use the least net width to determine the net area
An = bn t = 15.8 ( 7 8 ) = 13.8 in.2
EXAMPLE 4.6
Gross Area and
Net Area
Goal:
Determine the gross and net areas of an angle with staggered holes.
Given:
An L6×4×1/2 with holes for 7/8 in. bolts placed in the middle of the
member as shown in Figure 4.16.
SOLUTION
Step 1:
Determine the width of the equivalent flat plate representing the width of
the angle.
we = l1 + l2 − t = 6 + 4 − 1 2 = 9.50 in.
Step 2:
Determine the gross area of the equivalent plate.
Ag = we t = 9.50 (1 2 ) = 4.75 in.2
This is the same as found in Manual Table 1-7
Step 3:
Determine the gages for each bolt line.
The workable gages for the holes as found in Table 1-7A are shown in
Figure 4.16. The gage between the holes closest to the heel of the angle in
the two legs must be adjusted to account for the angle thickness. Thus,
( g + g1 − t ) = 2.50 + 2.25 − 0.50 = 4.25 in.
Step 4:
Determine the net area.
The governing net section will be section 2-2 or section 2-1-2. There is no
need to consider section 1-1 because bn1 will clearly be greater than bn2.
For section 2-2, the net width is
bn 2 = ( 9.5 − 2 ( 7 8 + 1 8 ) ) = 7.5 in.
and the net area is
An 2 = 7.5 ( 0.5 ) = 3.75 in.2
Chapter 4
Tension Mem
mbers
95
Fo
or section 2-1
1-2, this chainn has two staaggers of boltt holes, and bboth have
thee same pitch (s = 2.50 in)). The gages are differentt, with one att 4.25 in.
an
nd the other att 2.50 in.
Th
he net width for
f this chain is
⎛ 2.50 2 ⎞ ⎛ 2.50 2 ⎞
bn 3 = 9.5 − 3(7 8 + 1 8) + ⎜
⎟+⎜
⎟ = 7.49 in.
⎝ 4(2.5 ) ⎠ ⎝ 4(4.25)) ⎠
an
nd the net areaa is
An 3 = 7.49(0.5) = 3.75 in.
Step
S
5:
Seelect the least net area.
west net areaa controls. In this case, thee two chains yield the
The low
same net
n area,
An = An 2 = An3 = 3.75 in.2
Figure 4.16
4
Hole Paattern for L6×
×4×1/2 Used in Example 44.6
Figure 4.17
4
Concep
ptual Basis forr Shear Lag R
Reduction Facctor
96 Chapter 4
4.5.4
Tension Members
Effective Net Area
When all elements of a tension member are attached to connecting elements, the entire member
participates fully in transferring the load to the connection. However, when not all elements are
attached to connecting elements, they cannot all participate fully. Figure 4.17 shows an angle
with one leg attached to a connecting element and the other, the outstanding leg, unattached. To
account for the inability of this unattached leg to transfer load, the net area used in calculating the
rupture strength must be reduced to an effective net area, Ae.
This phenomenon occurs because the uniform stresses, occurring near the mid-length of
the member at some distance from the connection, must be transferred through the more restricted
area where the connection is located. The portion of the member area that is participating
effectively in the transfer of force is smaller than the full net area. Thus, the net area is reduced to
the effective net area.
This general behavior is called shear lag. Since its introduction into the Specification, it
has been approximated by the use of the shear lag reduction factor such that
Ae = UAn
Specification Table D3.1 provides values of the shear lag factor, U, for a wide variety of
elements. For all tension members, except plates, HSS and those with longitudinal welds, when
the tension load is transmitted to some but not all of the cross-sectional elements, the effective
length of the connection is reduced to l′ = l – x , where x is the distance from the attached face
to the member centroid and l is the length of the connection, as shown in Figure 4.17. The
reduction in net area is then taken in proportion to the reduction in effective length, l′/l. Thus, the
reduction becomes
l′ l − x
x
(4.6)
U= =
=1−
l
l
l
Figure 4.18 shows the definition of connection length, l, for both a bolted and a welded
connection. For a welded connection, the length of the connection, l, is the length of the weld. For
a bolted connection, l is the distance between the bolt holes at each end of the connection.
Table D3.1 of the Specification also provides simplified approaches to finding the shear
lag factor when certain criteria are met.
Case 7 of Table D3.1: for W-, M-, S-, and HP-shapes or for tees cut from
these shapes, the following apply:
Flange connected with three or more
fasteners per line in the direction of
loading
bf ≥ 2/3d
bf < 2/3d
Web connected with four or more
fasteners per line in the direction of
loading
U = 0.90
U = 0.85
U = 0.70
Case 8 of Table D3.1: for single and double angles:
With four or more fasteners per line in
the direction of loading
With three fasteners per line in the
direction of loading
U = 0.80
U = 0.60
Chapter 4
Figure 4.18
4
mbers
Tension Mem
97
Definitiion of Connection Length,, L, for Boltedd and Weldedd Connection
If
I U is calculaated using Caase 7 or Casee 8 as well as Case 2, the llarger value iis permitted
to be useed. For any other
o
conditio
ons, the tablee values shouuld be used. H
However, for open cross
sections (such as W-,, M-, S-, C-, HP-, WT-, aand ST-shapees, single anggles, and doubble angles),
n need to tak
ke U as less th
han the ratio of the gross area of the coonnected elem
ments to the
there is no
member gross area. This will also prove useeful when thhere is only one row of connectors
mber axis.
transversse to the mem
When
W
tension
n members arre connected with staggerred bolts andd not all elem
ments of the
member are connecteed, the effeccts of bolt sttagger, shearr lag and tennsion flow tthrough the
on must be co
onsidered.
connectio
Figure 4.19
4
Single-Angle Tensio
on Member foor Example 4.7
(a)
Figure 4.20
4
(b)
A WT Welded
W
to a Gusset
G
Plate ffor Examples 4.8 and 4.9
98 Chapter 4
Tension Members
EXAMPLE 4.7
Tensile Strength of
an Angle
Goal:
Determine the design strength (LRFD) and the allowable strength (ASD) of
an angle.
Given:
Consider an L4×4×1/2 attached through one leg to a gusset plate with 3/4
in. bolts in standard holes as shown in Figure 4.19. Use A36 steel.
SOLUTION
Step 1:
Determine the gross and net areas needed for calculations.
Ag = 3.75 in.2 (from Manual Table 1-7)
An = 3.75 − ( 3 4 + 1 8 )(1 2 ) = 3.31 in.2
Step 2:
Determine the shear lag factor and the effective net area.
First using Case 8 from Table D3.1, there are only three fasteners in a line
for this connection, U = 0.60.
The minimum U for this open shape connected as shown is
Gross area of connected leg 4.0(0.5)
U=
=
= 0.533
Gross area of angle
3.75
Based on connection length, the shear lag factor for a 6 in. connection
length and an angle with x = 1.18 in. (from Manual Table 1-7) is
1.18
x
U =1− =1−
= 0.80
6
L
Thus, use U = 0.80, the largest permitted value, and
Ae = 0.80(3.31) = 2.65 in.2
For
LRFD
Step 3:
For the limit state of yielding,
Pn = Fy Ag = 36 ( 3.75 ) = 135 kips
φPn = 0.90 (135 ) = 122 kips
Step 4:
For the limit state of rupture,
Pn = Fu Ae = 58 ( 2.65 ) = 154 kips
φPn = 0.75 (154 ) = 116 kips
Step 5:
For
ASD
Step 3:
Determine the design strength. The limit state of rupture controls and the
design strength is
φPn = 116 kips
For the limit state of yielding,
Pn = Fy Ag = 36(3.75) = 135 kips
Pn 135
=
= 80.8 kips
Ωt 1.67
Step 4:
For the limit state of rupture,
Chapter 4
Tension Members
99
Pn = Fu Ae = 58(2.65) = 154 kips
Pn 154
=
= 77.0 kips
Ωt 2.00
Determine the allowable strength. The limit state of rupture controls and the
allowable strength is
Pn
= 77.0 kips
Ωt
Note that the controlling limit state is the same for ASD and LRFD.
Step 5:
EXAMPLE 4.8
Tensile Strength of
a Tee
Goal:
Determine the design strength (LRFD) and the allowable strength (ASD) of
a WT.
Given:
Consider a WT6×32.5 attached to a gusset plate with equal length
longitudinal welds as shown in Figure 4.20a. Use A992 steel.
SOLUTION
Step 1:
Determine the gross and net areas:
Ag = 9.54 in.2 (from Manual Table 1-8)
Because the force is transferred by welds only, there are no holes and
An = Ag
Step 2:
Determine the shear lag factor and the effective net area.
The minimum U for this open shape connected as shown is
Gross area of connected flange 12.0(0.605)
U=
=
= 0.761
Gross area of WT
9.54
Based on connection length, the shear lag factor for a 12 in. connection
length with longitudinal welds only and a tee with bf = 12.0 and x = 0.985
in. (from Manual Table 1-8) is
2
3 (12.0 )
3l 2 ⎛
x⎞
⎛ 0.985 ⎞
U= 2
1
−
=
1−
⎟
⎟ = 0.688
2
2 ⎜
2 ⎜
3l + w ⎝
l ⎠ 3 (12.0 ) + 12.0 ⎝
12 ⎠
Thus, use U = 0.761 and
Ae = 0.761(9.54) = 7.26 in.2
For
LRFD
Step 3
For the limit state of yielding,
Pn = Fy Ag = 50 ( 9.54 ) = 477 kips
φPn = 0.90 ( 477 ) = 429 kips
Step 4:
For the limit state of rupture,
Pn = Fu Ae = 65 ( 7.26 ) = 472 kips
φPn = 0.75 ( 472 ) = 354 kips
100 Chapter 4
Tension Members
Step 5:
For
ASD
Step 3:
Therefore, the limit state of rupture controls and the design strength is
φPn = 354 kips
For the limit state of yielding,
Pn = 50(9.54) = 477 kips
Pn 477
=
= 286 kips
Ωt 1.67
Step 4:
For the limit state of rupture,
Pn = 65(7.26) = 472 kips
Pn 472
=
= 236 kips
Ωt 2.0
Step 5:
Therefore, the limit state of rupture controls and the allowable strength is
Pn
= 236 kips
Ωt
EXAMPLE 4.9
Tensile Strength of
a Tee
Goal:
Determine the design strength (LRFD) and the allowable strength (ASD)
of a WT.
Given:
Consider a WT6×32.5 attached to a gusset plate with unequal length
longitudinal welds as shown in Figure 4.20b. Use A992 steel.
SOLUTION
Step 1:
Determine the gross and net areas:
Ag = 9.54 in.2 (from Manual Table 1-8)
Step 2:
Because the force is transferred by welds only,
An = Ag
Determine the shear lag factor.
The minimum U for this open shape connected as shown is
Gross area of connected flange 12.0(0.605)
U=
=
= 0.761
Gross area of WT
9.54
Based on connection length,
6 + 18
l +l
l= 1 2 =
= 12 in.
2
2
the shear lag factor for a 12 in. connection length with longitudinal welds
only and a tee with bf = 12.0 and x = 0.985 in. (from Manual Table 1-8)
is
2
3 (12 )
3l 2 ⎛
x⎞
⎛ 0.985 ⎞
U= 2
1
−
=
⎜
⎟
⎜1 −
⎟ = 0.688
3l + w2 ⎝
12 ⎠
l ⎠ 3 (12 )2 + 12.0 2 ⎝
Thus, use U = 0.761
Chapter 4
Step 3:
For
LRFD
Step 4:
For
ASD
Step 4:
Tension Members
101
Determine the effective net area.
Ae = 0.761(9.54) = 7.26 in.2
The effective net area is the same as it was for Example 4.8, thus tension
rupture will again control
φPn = 354 kips
The effective net area is the same as it was for Example 4.8, thus tension
rupture will again control
Pn
= 236 kips
Ωt
4.6 DESIGN OF TENSION MEMBERS
To design a structural steel tension member, the member size must be determined and then the
appropriate limit states checked. The only additional issue to address is the slenderness of the
member. For tension members, slenderness is defined as the member length divided by the least
radius of gyration, L/r. The Specification has, in the past, placed a limit on the slenderness of
tension members. However, since slenderness of a tension member is primarily a serviceability
limit state, there is currently no specified limitation on tension member slenderness, as indicated
in Section D1. The designer should exercise caution when selecting tension members with very
high slenderness ratios, that is, those near the former limit of L/r = 300, because these members
could easily be damaged during erection and might cause other problems due to their flexibility in
the transverse direction.
Because the main task in tension member design is to determine the area of the member,
the two limit states of yielding and rupture can be used to determine minimum gross and net areas
such that
P
ΩP
Ag min = u (LRFD) or Ag min = t a (ASD)
Fy
φt Fy
and
P
ΩP
Ae min = u (LRFD) or Ae min = t a (ASD)
Fu
φt Fu
Because connection details are not normally known in the early stages of member selection, it
may not be possible to determine the actual deductions necessary to obtain the exact effective net
area of the member being designed. One approach would be to assume a fixed percentage
deduction for the effective net area. The designer would decide the magnitude of this deduction
and then confirm the adequacy of this deduction at the completion of the design.
Part 5 of the Manual provides tables for tension member design that give the strength of
tension members based on the limit states of yielding on the gross area and rupture on an
effective net area equal to 0.75Ag. This is a reasonable estimate of a value of Ae that is practical to
achieve with typical end connections, not a minimum value or a prescriptive way to calculate Ae.
If the actual effective net area differs from this assumed value, the designer can simply adjust the
strength accordingly.
102 Chapter 4
Tension Members
EXAMPLE 4.10a
Tension Member
Design by LRFD
SOLUTION
EXAMPLE 4.10b
Tension Member
Design by ASD
SOLUTION
Goal:
Select a double-angle tension member for use as a web member in a truss,
and determine the maximum area reduction that would be permitted for
holes and shear lag.
Given:
The member must carry a dead load of PD = 67.5 kips and a live load of PL
= 202.5 kips. For the load combination 1.2PD + 1.6PL, the LRFD required
strength is Pu = 405 kips. Use equal leg angles of A36 steel.
Step 1:
Determine the minimum required gross area based on the limit state of
yielding where ϕ = 0.9:
Ag min = Pu φFy = 405 ( 0.9 ( 36 ) ) = 12.5 in.2
Step 2:
Based on this minimum gross area, from Manual Table 1-15, select
2L6×6×9/16 with Ag = 12.9 in.2
Step 3:
Determine the minimum effective net area based on the limit state of
rupture where ϕ = 0.75:
Ae min = Pu φFu = 405 ( 0.75 ( 58 ) ) = 9.31 in.2
Step 4:
Thus, the combination of holes and shear lag may not reduce the area of
this pair of angles by more than
Ae Ag = 9.31 12.9 = 0.722
Goal:
Select a double-angle tension member for use as a web member in a truss
and determine the maximum area reduction that would be permitted for
holes and shear lag.
Given:
The member must carry a dead load of PD = 67.5 kips and a live load of PL
= 202.5 kips. For the load combination PD + PL, the ASD required strength
is Pa = 270 kips. Use equal leg angles of A36 steel.
Step 1:
Determine the minimum required gross area based on the limit state of
yielding where Ω = 1.67:
Ag min = Pa ( Fy Ω ) = 270 ( 36 1.67 ) = 12.5 in.2
Step 2:
Based on this minimum gross area, from Manual Table 1-15, select
2L6×6×9/16 with Ag = 12.9 in.2
Step 3:
Determine the minimum effective net area based on the limit state of
rupture where Ω = 2.00:
Ae min = Pa ( Fu Ω ) = 270 ( 58 2.00 ) = 9.31 in.2
Thus, the combination of holes and shear lag may not reduce the area of
this pair of angles by more than
Ae Ag = 9.31 12.9 = 0.722
Note that since the ratio PL/PD = 3.0, the required areas for the LRFD and ASD
solutions are the same.
Step 4:
Chapter 4
Tension Members
103
EXAMPLE 4.11a
Tension Member
Design by LRFD
Goal:
Select a WT9 for use as a tension member and determine the maximum
area reduction that would be permitted for holes and shear lag.
Given:
The member must carry an LRFD required strength of Pu = 818 kips. Use
A992 steel.
SOLUTION
Step 1:
Determine the minimum required gross area based on the limit state of
yielding where ϕ = 0.9:
Ag min = Pu φFy = 818 ( 0.9 ( 50 ) ) = 18.2 in.2
Step 2:
Based on the minimum gross area, from Manual Table 1-8, select
WT9×65 with Ag = 19.2 in.2
Step 3:
Determine the minimum effective net area needed to resist the applied force
where ϕ = 0.75:
Ae min = Pu φFu = 818 ( 0.75 ( 65 ) ) = 16.8 in.2
Step 4:
The combination of holes and shear lag may not reduce the area of this WT
by more than
Ae Ag = 16.8 19.2 = 0.875
EXAMPLE 4.11b
Tension Member
Design by ASD
Goal:
Select a WT9 for use as a tension member and determine the maximum
area reduction that would be permitted for holes and shear lag.
Given:
The member must carry an ASD required strength of Pa = 545 kips. Use
A992 steel.
SOLUTION
Step 1:
Determine the minimum required gross area based on the limit state of
yielding where Ω = 1.67:
Ag min = Pa ( Fy Ω ) = 545 ( 50 1.67 ) = 18.2 in.2
Step 2:
Based on the minimum gross area, from Manual Table 1-8, select
WT9×65 with Ag = 19.2 in.2
Step 3:
Determine the minimum effective net area needed to resist the applied force
where Ω = 2.00:
Ae min = Pa ( Fu Ω ) = 545 ( 65 2.00 ) = 16.8 in.2
Step 4:
The combination of holes and shear lag may not reduce the area of this WT
by more than
Ae Ag = 16.8 19.2 = 0.875
1004 Chapter 4
Tension Members
M
(a)
(
Figure 4.21 Example of
o Block Sheaar Failure of a Plate
((b)
Photo courtessy Robert Driv
ver
Figgure 4.22 B
Block Shear G
Geometry forr
Exaample 4.10
4.77
BLOCK
K SHEAR
mber tears ou
ut as shown iin Figure 4.221a, a combinnation of tenssion and
When a porrtion of a mem
shear resultss as shown in
n Figure 4.21
1b, and the faailure is know
wn as a blockk shear failurre. Even
though this failure
f
mode is primarily the
t result of a connection ffailure, it mayy possibly conntrol the
overall stren
ngth of a tenssion member. The resistannce to tear-ouut is provided by a combinnation of
shear on the plane paralleel to the tensio
on force and ttension on the plane perpeendicular to itt.
Rup
pture will alw
ways be the co
ontrolling mo de on the tennsion face of tthe failure bloock, due
to the relativ
vely short len
ngth of material that will bee available too yield. The coontrolling lim
mit states
on the shearr face will be either yieldin
ng or rupturee, whichever hhas the lowerr strength. Unnlike the
situation forr overall mem
mber strength
h, in which thhe yield and rrupture limit states were ffound to
have different resistance and safety faactors, block sshear uses thee same valuess for both lim
mit states.
mple comparisson of nomin
nal strengths is appropriatte for determ
mining the conntrolling
Thus, a sim
limit state. Section
S
J4.3 of
o the Specificcation gives thhe block sheaar strength as
Rn = 0.6 Fu Anv + U bs Fu Ant ≤ 00.6 Fy Agv + U bbs Fu Ant
(AIS
SC J4-5)
where
i shear
Agv = gross area in
Chapter 4
Tension Members
105
Anv = net area in shear
Ant = net area in tension
Ubs = 1.0 if the tension stress is uniform and 0.5 if the tension stress is not uniform. For
tension members, the tensile stress is assumed to be uniform. Thus, Ubs = 1.0 will be used
(Ubs = 0.5 is addressed in Chapter 10.)
The design and allowable strengths are determined using
φ = 0.75 (LRFD)
Ω = 2.0 (ASD)
The block shear strength given by Equation J4-5 can be simplified by recognizing that it simply
requires the combination of the minimum shear strength, yielding or rupture, with the tensile
rupture strength. It is also important to note that the calculation of net area for both tension and
shear is determined by deducting the area of the holes from the gross areas.
EXAMPLE 4.12
Gusset Plate
Tension Strength
Goal:
Determine whether the gusset plate has sufficient strength in block shear.
Given:
The gusset plate shown in Figure 4.22 has a plate thickness of 1/2 in. The
required strength for LRFD is Pu = 225 kips and for ASD is Pa = 150 kips.
The steel is A36 and the holes are punched for 7/8 in. bolts.
SOLUTION
Step 1:
Determine the areas needed to perform the calculations.
Ant = ( 6 − ( 7 8 + 1 8 ) ) (1 2 ) = 2.50 in.2
Agv = 2 (11)(1 2 ) = 11.0 in.2
Anv = 2 (11.0 − 3.5 ( 7 8 + 1 8 ) ) (1 2 ) = 7.50 in.2
Step 2:
Determine the nominal block shear strength:
Determine the tension rupture strength
Fu Ant = 58 ( 2.50) = 145 kips
Consider shear yield and shear rupture and select the least nominal
strength; thus,
0.6Fy Agv = 0.6 ( 36)(11.0 ) = 238 kips
0.6Fu Anv = 0.6 ( 58)( 7.50) = 261 kips
Selecting the shear yield term and combining it with the tension rupture
term gives a block shear nominal strength , using Ubs = 1.0, of
Rn = 238 + 1.0 (145 ) = 383 kips
For
LRFD
Step 3:
For LRFD, the design strength is
φRn = 0.75 ( 383) = 287 > 225 kips
Because the design strength is greater than the required strength of 225
kips, the gusset plate is adequate to resist this force based on block shear.
For
ASD
Step 3:
For ASD, the allowable strength is
1006 Chapter 4
Tension Members
M
Rn Ω = 3883 2.00 = 192 > 150 kips
Because the allo
owable streng
gth is greater than the reqquired strengtth of 150 kipps, the
gussset plate is ad
dequate to resiist this force bbased on blocck shear.
Figure 4.23 Spliced Teension Membeer for Exampple 4.13
EX
XAMPLE 4.13a
4
Teension Stren
ngth of
Sppliced Memb
bers by
LR
RFD
Go
oal:
Deteermine the deesign strengthh of a splice bbetween two W
W-shapes.
Giiven:
Two
o W14×43 A992 wide flannges are splicced by flangee plates, as shhown in
Figu
ure 4.23, witth 7/8 in. diaameter bolts arranged as shown in standard
holees. The LRFD
D available sstrength of a group of six bolts in a fllange is
227 kips. The plaates will be seelected such tthat they do nnot limit the m
member
ngth.
stren
SO
OLUTION
Sttep 1:
Deteermine the deesign strengthh for the limit state of yieldding of the W
W14×43.
m Manual Taable 1-1, Ag = 12.6 in.2
From
Thu
us, Pn = Fy Ag = 50 (12.6) = 630 kips
φPn = 0.90 ( 630) = 567 kips
Sttep 2:
Deteermine the net area of thee W14×43. T
The area to bbe deducted ffor each
flan
nge, with tf = 0.530
0
in., is
2 ( 7 8 + 1 8 )( 0.530 ) = 1.06 in.2
Thu
us, with the deeduction for tw
two flanges frrom the gross area
An = 122.6 − 2 (1.06 ) = 10.5 in.2
Sttep 3:
Deteermine the sh
hear lag factoor. The W14×
×43 is treatedd as two tee seections,
each
h a WT7×21.5. The x foor each WT iis found in M
Manual Tablee 1-8 as
Chapter 4
Tension Members
107
1.31 in. With L = 2(3.0) = 6.0 in.,
U =1−
1.31
= 0.782
6.0
Specification Table D3.1 provides that for this case, if b f ≥ 2d 3 , a value
of U = 0.90 may be used and if b f < 2 d 3 U = 0.85.
From Manual Table 1-1, bf = 8.0 in. and d = 13.7 in. thus:
b f = 8.0 < 2 (13.7 ) 3 = 9.13 in.
Therefore use U = 0.85
Step 4:
Determine the design strength for the limit state of rupture.
Ae = UAn = 0.85 (10.5 ) = 8.93 in.2
Pn = Fu Ae = 65 ( 8.93) = 580 kips
φPn = 0.75 ( 580) = 435 kips
Step 5:
Determine the design block shear strength of the flanges.
The block shear limit state must be checked for tear-out of the flanges, as
shown in Figure 4.24. The calculations will be carried out for one block as
shown in the figure and the total obtained by adding the results for all four
flange sections.
Rupture on the tension plane:
Fu Ant = 65 ( 2.0 − (1 2 )( 7 8 + 1 8 ) ) ( 0.530 ) = 51.7 kips
Yield on the shear plane:
0.6Fy Agv = 0.6 ( 50)( 8.00)( 0.530) = 127 kips
Rupture on the shear plane:
0.6 Fu Anv = 0.6 ( 65 ) ( 8.00 − 2.5 ( 7 8 + 1 8 ) ) ( 0.530 ) = 114 kips
Because shear rupture is less than shear yield, the design strength for a
single block shear element is, with Ubs = 1.0,
Rn = (114 + 1.0 ( 51.7 ) ) = 166 kips
φRn = 0.75 (166) = 125 kips
and the block shear strength of the W14×43 is
φRn = 4 (125) = 500 kips
Step 6:
Compare the design strength for each limit state.
Bolt design shear strength
(given, 2 flanges ×227 kips/(bolt group))
Yielding of member
Rupture of the member
Block shear for the member
454 kips
576 kips
435 kips
500 kips
1008 Chapter 4
Tension Members
M
Step 7:
Rup
pture of the member
m
contrrols the desiggn. Therefore,, the design sstrength
of th
he splice is
φRn = 435 kiips
Figuree 4.24 Block
k Shear Checck for Exam
mple 4.13
EX
XAMPLE 4.13b
4
Teension Stren
ngth of
Sppliced Memb
bers by
AS
SD
Go
oal:
W-shapes.
Deteermine the deesign strengthh of a splice bbetween two W
Giiven:
Two
o W14×43 A992 wide flannges are splicced by flangee plates, as shhown in
Figu
ure 4.23, witth 7/8 in. diaameter bolts arranged as shown in standard
holees. The ASD available streength of a grooup of six bollts in a flangee is 151
member
kipss. The plates will be seleccted such thaat they do noot limit the m
stren
ngth.
SO
OLUTION
Sttep 1:
W14×43.
Deteermine the deesign strengthh for the limit state of yieldding of the W
From
m Manual Taable 1-1, Ag = 12.6 in.2
Pn = Fy Ag = 50 (122.6 ) = 630 kipps
Pn Ω = 630 1.67 = 377 kips
Sttep 2:
The area to bbe deducted ffor each
Deteermine the net area of thee W14×43. T
flan
nge, with tf = 0.530
0
in., is
2 ( 7 8 + 1 8 )( 0.530 ) = 1.06 in.2
two flanges frrom the gross area
Thu
us, with the deeduction for tw
An = 122.6 − 2 (1.06 ) = 10.5 in.2
Sttep 3:
×43 is treatedd as two tee seections,
Deteermine the sh
hear lag factoor. The W14×
for each WT is found in M
Manual Tablee 1-8 as
each
h a WT7×21
1.5. The x fo
1.31
1 in. With L = 2(3.0) = 6.0 in.,
1.31
= 00.782
U =1−
6.0
Chapter 4
Tension Members
109
Specification Table D3.1 provides that for this case, if b f ≥ 2d 3 , a value
of U = 0.90 may be used and if b f < 2 d 3 U = 0.85.
From Manual Table 1-1, bf = 8.0 in. and d = 13.7 in. thus:
b f = 8.0 < 2 (13.7 ) 3 = 9.13 in.
Therefore use U = 0.85
Step 4:
Determine the allowable strength for the limit state of rupture.
Ae = UAn = 0.85 (10.5 ) = 8.93 in.2
Pn = Fu Ae = 65 ( 8.93) = 580 kips
Pn Ω = 580 2.00 = 290 kips
Step 5:
Determine the allowable block shear strength of the flanges.
The block shear limit state must be checked for tear-out of the flanges, as
shown in Figure 4.24. The calculations will be carried out for one block as
shown in the figure and the total obtained by adding the results for all four
flange sections.
Rupture on the tension plane:
Fu Ant = 65 ( 2.0 − (1 2 )( 7 8 + 1 8 ) ) ( 0.530 ) = 51.7 kips
Yield on the shear plane:
0.6Fy Agv = 0.6 ( 50)( 8.00)( 0.530) = 127 kips
Rupture on the shear plane:
0.6 Fu Anv = 0.6 ( 65 ) ( 8.00 − 2.5 ( 7 8 + 1 8 ) ) ( 0.530 ) = 114 kips
Because shear rupture is less than shear yield, the allowable strength for a
single block shear element is, with Ubs = 1.0,
Rn = (114 + 1.0 ( 51.7 ) ) = 166 kips
Rn Ω = 166 2.00 = 83.0 kips
and the block shear strength of the W14×43 is
Rn Ω = 4 ( 83.0) = 332 kips
Step 6:
Compare the design strength for each limit state.
Bolt design shear strength
(given, 2 flanges ×151 kips/(bolt group))
Yielding of member
Rupture of the member
Block shear for the member
Step 7:
302 kips
377 kips
290 kips
332 kips
Rupture of the member controls the design. Therefore, the design strength
of the splice is
110 Chapter 4
Tension Members
Rn Ω = 290 kips
EXAMPLE 4.14a
Tension Strength of
an Angle by LRFD
Goal:
Determine the design strength of one of a pair of angles in a tension
member.
Given:
The truss diagonal member in Figure 4.25 consists of a pair of angles
L4×3×3/8 that are loaded in tension. The bolts to be used are 3/4 in. and the
steel is A36. The bolt design shear strength for one angle in this connection
is 53.7 kips.
SOLUTION
Step 1:
Determine the single angle design strength for the limit state of yielding.
Ag = 2.49 in.2
Pn = Fy Ag = 36 ( 2.49) = 89.6 kips
φPn = 0.90 ( 89.6 ) = 80.6 kips
Step 2:
Determine the single angle design strength for the limit state of rupture.
An = 2.49 − ( 3 4 + 1 8 )( 3 8 ) = 2.16 in.2
The shear lag coefficient is
U =1−
x
0.775
=1−
= 0.871
6.0
l
and the effective net area is
Ae = AnU = 2.16 ( 0.871) = 1.88 in.2
The nominal strength is
Pn = Fu Ae = 58 (1.88) = 109 kips
Therefore the design strength is
φPn = 0.75 (109) = 81.8 kips
Step 3:
Determine the single angle design strength in block shear.
Rupture on the tension plane:
Fu Ant = 58 (1.5 − (1 2 )( 3 4 + 1 8 ) ) ( 3 8 ) = 23.1 kips
Yield on the shear plane:
0.6Fy Agv = 0.6 ( 36)( 7.25)( 3 8) = 58.7 kips
Rupture on the shear plane:
0.6 Fu Anv = 0.6 ( 58 ) ( 7.25 − 2.5 ( 3 4 + 1 8 ) ) ( 3 8 ) = 66.1 kips
Because shear yield is less than shear rupture, the design strength for a
single block shear element is, with Ubs = 1.0,
Rn = ( 58.7 + 1.0 ( 23.1) ) = 81.8 kips
φRn = 0.75 ( 81.8) = 61.4 kips
C
Chapter 4
Step 4:
T
Tension Mem
mbers
Compare
C
the single angle ddesign strengtth for each lim
mit state.
53.7 kipps
80.6 kipps
81.8 kipps
61.4 kipps
Bolt
B design sttrength
Yielding
Y
of th
he member
Rupture
R
of thee member
Block
B
shear for
fo the membeer
Step 5:
111
The
T bolt desig
gn strength controls the deesign. Thereffore, the desiggn strength
of
o one angle is
i
φRn = 53. 7 kips
The
T design strrength for thhe pair of anglles is
φRn = 2 ( 53.7 ) =107 kips
Figure 4.225 Truss Diiagonal Mem
mber
for Exampple 4.14
EXAMPLE
E 4.14b
Tension Sttrength of
an Angle by ASD
Goal:
Determine
D
the allowable sstrength of oone of a pair of angles inn a tension
member.
m
Given:
The
T truss diaagonal membber in Figuree 4.25 consissts of a pair of angles
L4×3×3/8
L
thaat are loaded iin tension. Thhe bolts to be used are 3/4 in. and the
steel
s
is A36. The bolt aallowable sheear strength for one anggle in this
connection
c
is 35.7 kips.
SOLUTION
N
Step 1:
Determine
D
thee single anglee allowable sttrength for thee limit state oof yielding.
Ag = 2.49 in.2
Pn = Fy Ag = 366 ( 2.49 ) = 89.66 kips
Pn Ω = 89.6 1. 67 = 53.7 kipps
Step 2:
Determine
D
thee single anglee allowable sttrength for thee limit state oof rupture.
An = 2.449 − ( 3 4 + 1 8 )( 3 8 ) = 2.166 in.2
112 Chapter 4
Tension Members
The shear lag coefficient is
U =1−
x
0.775
=1−
= 0.871
l
6.0
and the effective net area is
Ae = AnU = 2.16 ( 0.871) = 1.88 in.2
The nominal strength is
Pn = Fu Ae = 58 (1.88) = 109 kips
Therefore the allowable strength is
Pn Ω = 109 2.00 = 54.5 kips
Step 3:
Determine the single angle allowable strength in block shear.
Rupture on the tension plane:
Fu Ant = 58 (1.5 − (1 2 )( 3 4 + 1 8 ) ) ( 3 8 ) = 23.1 kips
Yield on the shear plane:
0.6Fy Agv = 0.6 ( 36)( 7.25)( 3 8) = 58.7 kips
Rupture on the shear plane:
0.6 Fu Anv = 0.6 ( 58 ) ( 7.25 − 2.5 ( 3 4 + 1 8 ) ) ( 3 8 ) = 66.1 kips
Because shear yield is less than shear rupture, the allowable strength for a
single block shear element is, with Ubs = 1.0,
Rn = ( 58.7 + 1.0 ( 23.1) ) = 81.8 kips
Rn Ω = 81.8 2.00 = 40.9 kips
Step 4:
Compare the single angle allowable strength for each limit state.
Bolt allowable strength
Yielding of the member
Rupture of the member
Block shear for the member
Step 5:
35.7 kips
53.7 kips
54.5 kips
40.9 kips
The bolt allowable strength controls the design. Therefore, the allowable
strength of one angle is
Rn Ω = 35.7 kips
The design strength for the pair of angles is
Rn Ω = 2 ( 35.7 ) =71.4 kips
4.8
PIN-CONNECTED MEMBERS
When a pin connection is to be made in a tension member, a hole is cut in both the member and
the parts to which it is to be attached. A pin is inserted in the hole and a mechanical means is
C
Chapter 4
T
Tension Mem
mbers
113
found to keep the eleements togeth
her. This typee of connectioon is the clossest to a true frictionless
F
4.26 shows
s
the endd of a pin-coonnected mem
mber and the dimensions
pin as caan be made. Figure
needed to
o determine its
i strength. These
T
membeers are not paarticularly com
mmon in builldings; they
are used mainly for special applications, such aas hangers inn suspension structures or connecting
b
structu
ures.
links in bridge
Specification
S
Section D5 identifies thee limit statess for which pin-connected members
must be designed. Th
hese are (1) teension on the effective nett area, (2) sheear on the efffective area,
ng on the pro
ojected area of the pin, andd (4) yieldingg on the gross section. The strength of
(3) bearin
the pin-cconnected tension member is taken as thhe lowest streength predicteed by each off these limit
states.
For
F tension on
n the effectivee net area, acttually a ruptuure limit state
Pn = Fu ( 2tbe )
(A
AISC D5-1)
φt = 0.75(LRFD)
Ωt = 2.000 (ASD)
where t is
i the thickness of the plate and be is thhe effective w
width of the pllate, taken as (2t + 0.63)
in inchess, but not morre than the acctual distancee from the edgge of the holee to the edge of the part,
measured
d perpendicullar to the direction of the fo
force, be ≤ b .
For
F shear on the
t effective area,
a
again a rrupture limit state
Pn = 0.6 Fu Asf
(A
AISC D5-2)
φsf = 0..75(LRFD)
Ωsf = 2.000 (ASD)
where Asf = 2t(a + d/2
2), a is the sh
hortest distancce from the eddge of the pinn hole to the edge of the
member measured parrallel to the diirection of thee force, and d is the pin diameter.
For
F bearing on
n the projecteed area of the pin, from Seection J7,
Pn = 1.8Fy Apb
((AISC J7-1)
φt = 0.75 (LRFD))
Ωt = 2.000 (ASD)
where Apb
ojected area of
o the pin.
p = td, the pro
For
F yielding in the gross seection,
Pn = Fy Ag
φt = 0.90 (LRFD))
(A
AISC D2-1)
Ωt = 1.67 (ASD)
Figure 4.226 Pin-Connnected Tensioon Member
EXAMPLE
E 4.15a
Pin-Conneected
Member Design by
LRFD
Goal:
Design
D
a pin-connected meember using L
LRFD.
Given:
A dead load of
o 30 kips annd a live loadd of 70 kips aare to be suppported. The
steel
s
is A572 Gr. 50 with a yield stress of 50 ksi andd an ultimate strength of
65
6 ksi. Use a 3/4 in. plate w
with a 4 in. piin.
114 Chapter 4
Tension Members
SOLUTION
Step 1:
Determine the required strength.
Pu = 1.2 ( 30) + 1.6 ( 70) = 148 kips
Step 2:
Determine the minimum required effective net width for the limit state of
rupture based on Equation D5-1,
P
148
= 2.02 in.
( be )min = u =
φFu 2t 0.75(65)(2)(0.750)
and the maximum be from Section D5.1 is
( be )max = 2t + 0.63 = 2 ( 0.750 ) + 0.63 = 2.13 in.
Therefore, try a 9.0 in. plate, which will give an actual distance to the edge
of the plate equal to 2.5 in. which is greater than (be)min. Thus, (be)max is
used to calculate the rupture strength of the plate.
Step 3:
Determine the design strength of this 9×3/4 plate for the limit state of
tension rupture using Equation D5-1.
φPn = φFu ( 2tbe ) = 0.75 ( 65 ) ( 2 ( 0.750 ) ) ( 2.13 ) = 156 > 148 kips
Step 4:
Determine the design strength for the limit state of shear rupture using
Equation D5-2. For a 9 in. plate and a 4 in. pin, a = b = 2.5 in.
Asf = 2t ( a + d 2 ) = 2 ( 0.750 )( 2.5 + 4.0 2 ) = 6.75 in.2
φPn = φ0.6 Fu Asf = 0.75 ( 0.6 ( 65 ) ) ( 6.75 ) = 197 > 148 kips
Step 5:
Determine the design strength for the limit state of bearing on the projected
area of the 4 in. pin using Equation J7-1.
Apb = td = 0.750 ( 4.0 ) = 3.0 in.2
φPn = φ1.8Fy Apb = 0.75 (1.8)( 50)( 3.0) = 203 > 148 kips
Step 6:
Determine the design strength for the limit state of yielding on the gross
area of the member using Equation D2-1.
φPn = φFy Ag = 0.9 ( 50)( 0.750)( 9.00 ) = 304 > 148 kips
Step 7:
Since each limit state has design strength greater than the required strength,
the 9×3/4 pin-connected member with a 4 in. pin will be sufficient to carry
the applied load.
EXAMPLE 4.15b
Pin-Connected
Member Design by
ASD
Goal:
Design a pin-connected member using ASD.
Given:
A dead load of 30 kips and a live load of 70 kips are to be supported. The
steel is A572 Gr. 50 with a yield stress of 50 ksi and an ultimate strength of
65 ksi. Use a 3/4 in. plate with a 4 in. pin.
SOLUTION
Step 1:
Determine the required strength.
Chapter 4
Tension Members
115
Pa = 30 + 70 = 100 kips
Step 2:
Determine the minimum required effective net width for the limit state of
rupture based on Equation D5-1,
100 ( 2.00 )
PΩ
= 2.05 in.
( be )min = a =
Fu 2t 65(2)(0.750)
and the maximum be from Section D5.1 is
( be )max = 2t + 0.63 = 2 ( 0.750 ) + 0.63 = 2.13 in.
Therefore, try a 9.0 in. plate, which will give an actual distance to the edge
of the plate equal to 2.5 in. which is greater than (be)min. Thus, (be)max is
used to calculate the rupture strength of the plate.
Step 3:
Determine the allowable strength of this 9×3/4 plate for the limit state of
tension rupture using Equation D5-1.
Pn Ω = Fu ( 2tbe ) Ω = 65 ( 2 ( 0.750 ) ) ( 2.13 ) 2.00 = 104 > 100 kips
Step 4:
Determine the allowable strength for the limit state of shear rupture using
Equation D5-2. For a 9 in. plate and a 4 in. pin, a = b = 2.5 in.
Asf = 2t ( a + d 2 ) = 2 ( 0.750 )( 2.5 + 4.0 2 ) = 6.75 in.2
Pn Ω = 0.6Fu Asf Ω = 0.6 ( 65)( 6.75) 2.00 = 132 > 100 kips
Step 5:
Determine the allowable strength for the limit state of bearing on the
projected area of the 4 in. pin using Equation J7-1.
Apb = td = 0.750 ( 4.0 ) = 3.0 in.2
Pn Ω = 1.8Fy Apb Ω = 1.8 ( 50)( 3.0 ) 2.00 = 135 > 100 kips
Step 6:
Determine the allowable strength for the limit state of yielding on the gross
area of the member using Equation D2-1.
Pn Ω = Fy Ag Ω = 50 ( 0.750)( 9.00) 1.67 = 202 > 100 kips
Step 7:
Since each limit state has allowable strength greater than the required
strength, the 9×3/4 pin-connected member with a 4 in. pin will be sufficient
to carry the applied load.
4.9 EYEBARS AND RODS
Eyebar tension members are not commonly used in new construction but may be found in special
applications where the objective is a design that has some historical context. Thus, the provisions
for the design of these members are still found in Section D6 of the Specification. Historically,
they were commonly used as tension members in trusses and as links forming the main tension
member in suspension bridges. Eyebars are designed only for the limit state of yielding on the
1116 Chapter 4
Tension Members
M
gross section
n because thee dimensionall requirementts preclude thhe possibility of failure at aany load
below that leevel. Figure 4.27
4
shows a schematic off an eyebar annd Figure 4.4 shows an eyeebar in a
building app
plication.
Rod
ds are commo
only used forr tension mem
mbers in situuations wheree the requiredd tensile
strength is small. Thesee tension meembers are ggenerally connsidered secoondary membbers and
include sag rods, hangerss, and tie rodss, although thhere are cases where rods pplay a significcant role
p
forcees. Rods may
y also be usedd as part of thhe lateral braacing system in walls
in carrying primary
and roofs.
hough it is po
ossible to con
nnect rods by welding to thhe structure, tthreading andd bolting
Alth
is the most common
c
meaans of connecttion. Rods caan be threadedd in two wayss. Standard roods have
threads that reduce the cross-sectiona
c
al area througgh the removval of materiaal. The upset rod has
ds, with the threads
t
reducing that area to somethingg larger than the gross areea of the
enlarged end
rod. The streength of the rod depends on
o the mannerr in which thee threads are aapplied.
For a standard th
hreaded rod, th
he nominal sttrength is given in Specificcation Sectionn J3.6 as
Fn = 0.75Fu over the areaa of the unthreeaded body oof the rod, which gives
Pn = 0.755Fu Ab
(4-7)
and for desig
gn,
φt = 0.75 (LRFD)
(
Ωt = 2.000 (ASD)
4.110
BUILT
T-UP TENSION MEMB
BERS
o the Specifiication allowss tension mem
mbers that aree fabricated frrom a combinnation of
Section D4 of
shapes and plates.
p
Their strength is determined in the same waay as the strenngth for single-shape
tension mem
mbers. Howev
ver, the desig
gner must rem
member that in bolted buiilt-up membeers, bolts
are usually placed
p
along the member length to tie the various sshapes togethher. These bollts result
in holes alon
ng the membeer length, nott just at the ennds, so that ruupture on thee effective nett section
may becomee the controllling limit state at a locatioon other than the member eend. Welded built-up
members wiill not normallly experiencee rupture on thhe effective nnet area.
Perfforated coverr plates or tie plates can be used to tie the separrate shapes ttogether.
Limitations on the spacin
ng of these elements are aalso providedd in Section D
D4, and requiirements
for the placeement of boltss can be found in Section JJ3.5.
Figure 4.277 Eyebar Geeometry
4.111
TRUSS
S MEMBER
RS
ommon tensio
on members found in buillding structurres are the tennsion web annd chord
The most co
members off trusses. Tru
usses are norrmally foundd as roof struuctures and aas transfer sttructures
within a buiilding. Depen
nding on the particular
p
loadd patterns thaat a truss mighht experiencee, a truss
member mig
ght be called
d upon to alw
ways resist teension or to resist tensionn in some caases and
C
Chapter 4
T
Tension Mem
mbers
117
compresssion in otherrs. In cases in which a member is required to carry both ttension and
compresssion, it will need
n
to be sizeed accordinglly. Because thhe compressioon strength oof a member
is normaally significan
ntly less than the tension strength of thhat same mem
mber, as willl be seen in
Chapter 5,
5 compressio
on may actuallly control thee design.
The
T typical trruss member can be compposed of either single shappes or a com
mbination of
shapes. When
W
composed of a com
mbination of shhapes, the reqquirements ddiscussed in S
Section 4.10
must be included. Ottherwise, trusss members aare designed the same way as any otther tension
member discussed in this chapter. Examples 4 .10, 4.11, andd 4.14 showeed the applicaation of the
p
to several
s
truss tension
t
membbers.
tension provisions
4.12
BRA
ACING ME
EMBERS
t
memberrs, members used
u
to providde lateral loaad resistance ffor a buildingg might also
As with truss
be called
d upon to carrry tension un
nder some coonditions and compressionn under others. They too
would neeed to be desiigned to resisst both loads. However, it is often morre economicall to provide
twice as many tension
n members an
nd to assume tthat if a tensioon member w
were called uppon to resist
d buckle and
d therefore caarry no load. This would permit all brracing to be
compresssion, it would
designed
d as tension-o
only memberrs and almosst certainly ppermit them tto have a sm
maller cross
section than
t
if they were
w
required
d to resist coompression. A
An additionall simplificatioon that this
assumptiion entails is the eliminattion of potenntial compression memberrs from the aanalysis for
member forces. This may result in the structuure being a determinate structure rathher than an
minate one, thu
us simplifying
g the analysiss.
indeterm
F igure 4.28 Brraced Frame for Example 4.16
EXAMPLE
E 4.16a
Tension Member
M
Design by LRFD
L
Goal:
Select
S
an apprropriate tensi on member fo
for use as a brrace in an X-bbraced
frame.
f
Given:
The
T member is
i a tension-oonly diagonal as shown in F
Figure 4.28 aand must
resist
r
a force resulting
r
from
m the horizonntal wind loadd, W = 120 kipps. Use a
pair
p of equal leg
l angles or a threaded rood of A36 steeel.
SOLUTION
N
Step 1:
Determine
D
thee force in the brace labeledd AC if the brrace labeled B
BD is
assumed
a
not to
t act since it would be in compression for this loadiing. The
most
m critical load
l
combinaation would bee LRFD Loadd Combinatioon 5, 0.9
Dead
D
+ 1.0 Wind.
W
Using siimilar trianglees,
⎛ 72.1 ⎞
Pu = 1.0 (120 ) ⎜
⎟ = 144 kips
⎝ 60.0 ⎠
118 Chapter 4
Tension Members
Step 2:
Determine the minimum required gross area based on the limit state of
yielding.
Ag min = 144 ( 0.90 ( 36 ) ) = 4.44 in.2
Step 3:
Based on this minimum gross area, from Manual Table 1-15, select
2L2-1/2×2-1/2×1/2 with Ag = 4.52 in.2
Step 4:
Determine the minimum effective net area based on the limit state of
rupture.
Ae min = 144 ( 0.75 ( 58 ) ) = 3.31 in.2
Step 5:
Thus, the combination of holes and shear lag may not reduce the area of
this pair of angles by more than
Ae Ag = 3.31 4.52 = 0.732
Step 6:
Determine the slenderness ratio of the selected double angles and compare
it with the recommended maximum of 300.
From Manual Table 1-15, rx = 0.735 in. and
L 72.1(12 )
=
= 1180 > 300
rx
0.735
A check of Manual Table 1-15 shows that no pair of double angles will
satisfy the slenderness limit of 300.
Step 7:
Select a threaded rod to meet the strength requirement. The maximum
slenderness limit does not apply to rods.
For yielding, the minimum gross area is the same as for the double angles.
For rupture of a threaded rod, it is reasonable to assume that the effective
net area will not be less than 0.75Ag. Thus, using the minimum effective net
area from step 3, Ae = 3.31, the corresponding minimum gross area is
3.31
Ag =
= 4.41
0.75
Thus, the limit state of yielding controls and the minimum required area is
4.44 in.2.
Therefore, select a 2½ in. diameter threaded rod with A = 4.91 in.2.
EXAMPLE 4.16b
Tension Member
Design by ASD
Goal:
Select an appropriate tension member for use as a brace in an X-braced
frame.
Given:
The member is a tension-only diagonal as shown in Figure 4.28 and must
resist a force resulting from the horizontal wind load, W = 120 kips. Use a
pair of equal leg angles or a threaded rod of A36 steel.
Chapter 4
SOLUTION
Tension Members
119
Step 1:
Determine the force in the brace labeled AC if the brace labeled BD is
assumed not to act since it would be in compression for this loading. The
most critical load combination would be ASD Load Combination 5, Dead
+ 0.6Wind. Using similar triangles,
⎛ 72.1 ⎞
Pa = 0.6 (120 ) ⎜
⎟ = 86.5 kips
⎝ 60.0 ⎠
Step 2:
Determine the minimum required gross area based on the limit state of
yielding.
Ag min = 86.5 ( 36 1.67 ) = 4.01 in.2
Step 3:
Based on this minimum gross area, from Manual Table 1-15, select
2L3-1/2×3-1/2×5/16 with Ag = 4.20 in.2
Step 4:
Determine the minimum effective net area based on the limit state of
rupture.
Ae min = 86.5 ( 58 2.00 ) = 2.98 in.2
Step 5:
Thus, the combination of holes and shear lag may not reduce the area of
this pair of angles by more than
Ae Ag = 2.98 4.20 = 0.710
Step 6:
Determine the slenderness ratio of the selected double angles and compare
it with the recommended maximum of 300.
From Manual Table 1-15, rx = 1.08 in. and
L 72.1(12 )
=
= 801 > 300
rx
1.08
A check of Manual Table 1-15 shows that no pair of double angles will
satisfy the slenderness limit of 300.
Step 7:
Select a threaded rod to meet the strength requirement. The maximum
slenderness limit does not apply to rods.
For yielding, the minimum gross area is the same as for the double angles.
For rupture of a threaded rod, it is reasonable to assume that the effective
net area will not be less than 0.75Ag. Thus, using the minimum effective net
area from step 3, Ae = 2.98, the corresponding minimum gross area is
2.98
Ag =
= 3.97
0.75
Thus, the limit state of yielding controls and the minimum required area is
4.01 in.2.
Therefore, select a 2-3/8 in. diameter threaded rod with A = 4.43 in.2.
1120 Chapterr 4
44.13
Tension
n Members
PROB
BLEMS
11. Determine th
he gross and neet areas for an 8×3/4 in. platee
w
with a single lin
ne of standard holes for 7/8 in
n. bolts.
22. Determine th
he gross and neet areas for a 10×1/2 in. platee
w
with a single lin
ne of standard holes for 3/4 in
n. bolts.
33. Determine th
he gross and net
n areas for a 6×5/8 in. platee
w
with a single lin
ne of standard holes for 1 in. bolts.
15. D
Determine the gross and net areas for a C115×40 with
four lines of standaard holes for 33/4 in. bolts. E
Each flange
will ccontain one linne of bolts and the web will ccontain two
lines of bolts.
16. D
Determine the gross and net areas for a C110×25 with
two llines of standarrd holes in the web for 7/8 inn. bolts.
17. D
Determine the net width forr a 10×1/2 in. plate with
3/4 iin. bolts placeed in three linnes as shown in Figure
P4.177.
44. Determine th
he gross and neet areas for a 10×1/2 in. platee
w
with two lines of
o standard holles for 7/8 in. bolts.
b
55. Determine th
he gross and neet areas for a 12×5/8 in. platee
w
with two lines of
o standard holles for 3/4 in. bolts.
b
66. Determine th
he gross and net
n areas for a 14×1 in. platee
w
with three liness of standard ho
oles for 3/4 in. bolts.
77. Determine the
t gross and net areas forr an L4×4×1/2
2
w
with two lines,, one in each leg, of standard
d holes for 3/4
4
inn. bolts.
88. Determine the
t gross and net areas forr an L5×5×5/8
8
w
with two lines,, one in each leg, of standard
d holes for 7/8
8
inn. bolts.
99. Determine the
t gross and net areas for an L8×4×9/16
6
w
with three liness of standard holes,
h
two in th
he 8 in. leg and
d
oone in the 4 in. leg, for 7/8 in.. bolts.
P4.177
18. D
Determine the net width forr a 12×1/2 in. plate with
3/4 iin. bolts placeed in three linnes as shown in Figure
P4.188.
110. Determine the gross and net
n areas for a WT8×20 with
h
thhree lines off standard holles for 3/4 in
n. bolts. Each
h
eelement of the WT
W will be atttached to the co
onnection.
111. Determine the gross and
d net areas fo
or a WT12×31
1
w
with three lines of standard holes for 7/8 in.
i bolts. Each
h
eelement of the WT
W will be atttached to the co
onnection.
112. Determine the gross and net
n areas for a WT7×41 with
h
thhree lines off standard holles for 3/4 in
n. bolts. Each
h
eelement of the WT
W will be atttached to the co
onnection.
P4.188
19. D
Determine the net area for thhe L6×4×5/8 w
with 7/8 in.
boltss shown in Figuure P4.19.
113. Determine the gross and net
n areas for a WT9×25 with
h
ffour lines of staandard holes fo
or 3/4 in. bolts. Each elementt
oof the WT will be attached to
o the connection (two lines off
bbolts in the stem
m).
114. Determine the gross and net areas for a C15×50 with
h
ffive lines of staandard holes for
f 7/8 in. boltts. Each flangee
w
will contain on
ne line of boltts and the weeb will contain
n
thhree lines of bo
olts.
P4.199
Chhapter 4
Teension Membbers
121
220. Determine the net area fo
or the L6×4×5/8 with 7/8 in.
bbolts shown in Figure P4.20.
P4.233
P
P4.20
221. Determine the gross and net areas for a double-anglee
tension membeer composed of
o two L4×4×
×1/2 shapes ass
sshown in Figu
ure P4.21 wiith holes for 3/4 in. boltss
sstaggered in each leg.
24. F
For a WT7×34 attached throuugh a flange to a 10×1 in.
platee with six 3/4 iin. bolts at a sppacing of 3 in.., placed in
two rows of threee bolts as shown in Figuure P4.24,
deterrmine the sheaar lag factor (consider Case 2 and Case
7) annd effective nett area of the W
WT.
P4.244
P
P4.21
222. Determine the gross and net areas for a double-anglee
tension membeer composed of
o two L6×6×
×5/8 shapes ass
sshown in Figu
ure P4.22 wiith holes for 3/4 in. boltss
sstaggered in each leg.
25. A single L6×6×
×1 is used as a tension brace in a multistoryy building. Onee leg of the anggle is attached to a gusset
platee with a single line of three 77/8 in. bolts att a spacing
of 3 in. Determinee the shear lagg factor (considder Case 2
and C
Case 8) and efffective net areaa.
26. A single L7×44×3/4 is used as a tension brace in a
multii-story buildingg. The 7 in. legg of the angle is attached
to a ggusset plate wiith a single linee of four 7/8 inn. bolts at a
spaciing of 3 in. Deetermine the sshear lag factorr (consider
Casee 2 and Case 8)) and effective net area.
P
P4.22
223. For a WT8
8×50 attached through
t
a flang
ge to a 12×3/4
4
inn. plate with eight
e
7/8 in. bo
olts at a spacin
ng of 3 in. and
d
pplaced in two rows
r
as shown
n in Figure P4
4.23, determinee
thhe shear lag factor
f
(consideer Case 2 and
d Case 7) and
d
eeffective net area of the WT.
27. A single L8×66×1/2 is used as a tension brace in a
multii-story buildingg. The 8 in. legg of the angle is attached
to a gusset plate w
with two lines of four 3/4 in. bolts at a
spaciing of 3 in. Deetermine the sshear lag factorr (consider
Casee 2 and Case 8)) and effective net area.
28. T
The WT8×50 oof Problem 23 is welded aloong the tips
of thhe flange for a length of 12 in. on eaach flange.
Deterrmine the sheaar lag factor (C
Case 2) and efffective net
area for the WT.
29. D
Determine the available strenngth of a 12×11/2 in. A36
platee connected to two 12 in. plates, as shownn in Figure
P4.299, with two lines of 3/4 in. bolts. C
Considering
1122 Chapterr 4
Tension
n Members
yyielding and neet section ruptu
ure, determine the (a) design
n
sstrength by LRF
FD and (b) allo
owable strengtth by ASD.
36. D
Determine the available strenngth of a WT9×20, A992
steel,, with the flangges welded to a 1/2 in. gussett plate with
uneqqual length lonngitudinal wellds as shown in Figure
4.20bb. Consideringg yielding annd net sectioon rupture,
deterrmine the (a) design strenngth by LRFD
D and (b)
allow
wable strength bby ASD.
37. D
Determine the available strenngth of a WT7×15, A992
steel,, with the flangges welded to a 1/2 in. gussett plate with
uneqqual length lonngitudinal weldds, a 5 in. weldd and a 11
in. w
weld. Considerring yielding and net sectioon rupture,
deterrmine the (a) design strenngth by LRFD
D and (b)
allow
wable strength bby ASD.
P
P4.29
330. Determine the availablee strength of an L6×4×3/4
4
aattached throug
gh the long leg to a gusset plaate with ten 3/4
4
inn. bolts, in two
o lines of five bolts, at a 3 in
n. spacing. Usee
A
A36 steel. Con
nsidering yield
ding and net seection rupture,,
ddetermine the (a) design strength
s
by LRFD
L
and (b))
aallowable stren
ngth by ASD.
331. Determinee the available strength of
o a C8×11.5
5
aattached throug
gh the web to a gusset plate with eight 7/8
8
inn. bolts, in two
o lines of four bolts, at a 3 in
n. spacing. Usee
A
A36 steel. Con
nsidering yield
ding and net seection rupture,,
ddetermine the (a) design strength
s
by LRFD
L
and (b))
aallowable stren
ngth by ASD.
332. Determine the
t available sttrength of a C1
12×30 attached
d
thhrough the weeb to a gusset plate
p
with ten 7/8
7 in. bolts, in
n
tw
wo lines of fiv
ve bolts, at a 3 in. spacing. Use
U A36 steel.
C
Considering yiielding and neet section ruptu
ure, determinee
thhe (a) design
n strength by LRFD and (b) allowablee
sstrength by ASD
D.
333. Determine the available strength
s
of an 8×1/2
8
in. A572
2
G
Gr. 50 plate co
onnected with three lines off 7/8 in. bolts.
C
Considering yiielding and neet section ruptu
ure, determinee
thhe (a) design
n strength by LRFD and (b) allowablee
sstrength by ASD
D.
334. Determine the available strength
s
of a WT7×15,
W
A992
2
ssteel, with the flanges
f
welded
d to a 1/2 in. gu
usset plate by a
110 in. weld along each sidee of the flangee. Considering
g
yyielding and neet section ruptu
ure, determine the (a) design
n
sstrength by LRF
FD and (b) allo
owable strengtth by ASD.
335. Determine the available strength
s
of a WT9×20,
W
A992
2
ssteel, with the flanges
f
welded
d to a 1/2 in. gu
usset plate by a
112 in. weld along each sidee of the flangee. Considering
g
yyielding and neet section ruptu
ure, determine the (a) design
n
sstrength by LRF
FD and (b) allo
owable strengtth by ASD.
38. D
Design a 12 ftt long, single-aangle tension m
member to
suppoort a live loadd of 45 kips annd a dead load of 15 kips
(L/D = 3). The meember is to bee connected thhrough one
leg. Estimate threee bolts in a sinngle line. Usee A36 steel
and llimit the slendderness ratio too 300. Considerr yield and
net seection rupture,, and design byy (a) LRFD andd (b) ASD.
39. D
Design a 12 ft llong, single-anngle tension meember as in
Probllem 38 with the same totaal service loadd, 60 kips.
Usingg a live load oof 6.7 kips andd a dead load oof 53.3 kips
(L/D = 0.126), (a)) design by L
LRFD and (b) design by
ASD
D.
40. D
Design a 12 ft llong, single-anngle tension meember as in
Probllem 38 with thhe same servicce load using a live load
of 500 kips and a dead load of 10 kkips (L/D = 5). Design by
(a) L
LRFD and (b) A
ASD.
41. D
Design a 33 fft long WT teension wind bbrace for a
multii-story buildinng to resist a wind force off 290 kips.
Use A
A992 steel annd 7/8 in. boltss connected to the flange
only.. Assume two lines of bolts with at least three bolts
per lline. Limit thee slenderness rratio to 300. Consider
yieldd and net sectioon rupture, andd design by (a) LRFD and
(b) A
ASD.
Design a W12 A992 tensionn member for a truss that
42. D
will carry a dead lload of 70 kipps and a live looad of 210
kips. The flanges w
will be bolted to the conneccting plates
with 7/8 in. bolts llocated so thatt four bolts wiill occur in
any net section. A
Assume at leaast three boltss per line.
Conssider yield andd net section ruupture, and dessign by (a)
LRFD
D and (b) ASD
D.
43. D
Design a W14 A992 tensionn member for a truss that
will ccarry a dead looad of 120 kipps and a live looad of 360
kips. The flanges w
will be bolted to the conneccting plates
with 7/8 in. bolts llocated so thatt four bolts wiill occur in
any net section. A
Assume at leaast three boltss per line.
Conssider yield andd net section ruupture, and dessign by (a)
LRFD
D and (b) ASD
D.
Chhapter 4
444. An L4×3×3
3/4 is attached
d to a gusset plate with threee
33/4 in. bolts spaaced at 3 in. with
w an end and
d edge distancee
oof 1.5 in. as shown in Fiigure P4.44. Determine
D
thee
aavailable block
k shear strength of the A36
6 angle by (a))
L
LRFD and (b) ASD.
A
Teension Membbers
123
48. A
A36 steel and 33/4 in. bolts aree used in the boolted splice
show
wn in Figure P4.48. Conssider yield, nnet section
ruptuure, and block shear rupture. Determine thee (a) design
strenngth by LRFD aand (b) the alloowable strengthh by ASD.
P
P4.44
445. A 7×3/4 in.
i A36 plate is shown in Figure P4.45,,
w
where the holles are for 3//4 in. bolts. Determine
D
thee
aavailable block
k shear strength
h by (a) LRFD and (b) ASD.
P4.488
49. A
A572 Grade 500 steel and 7/88 in. bolts are uused in the
bolteed splice shownn in Figure P44.48. Consider yield, net
sectioon rupture, annd block shear rupture. Dettermine the
(a) ddesign strengtth by LRFD and (b) the allowable
strenngth by ASD.
P
P4.45
446. Determine the available block shear sttrength for thee
A
A992 steel WT
T6×20, attach
hed through th
he flange with
h
eeight 3/4 in. bo
olts as shown in
n Figure P4.46
6, by (a) LRFD
D
aand (b) ASD.
P
P4.46
447. Determine the available block shear sttrength for thee
A
A992 steel WT
T12×34, attacched through th
he flange with
h
eeight 7/8 in. bo
olts as shown in
n Figure P4.46
6, by (a) LRFD
D
aand (b) ASD.
50. IIntegrated Dessign Project. L
Lateral load reesistance in
the nnorth-south direection is proviided by a chevvron braced
framee as shown in Figure 1.24. B
Before the forcces in these
braciing members ccan be determ
mined, the speccified wind
load must be deterrmined. At thiis stage in thee design, a
simpplified approachh to wind loadd calculation m
might yield
the fo
following loadss at each level.
Roof
Fourth floor
Third floor
Second floor
Total wind loaad
54.0 kipss
102.0 kipps
93.0 kipss
88.0 kipss
337.0 kipps
It is perm
missible to trreat the chevrron braced
framees as verticaal trusses witth pinned joiints at all
interssections. Usinng the two bbraced framess to share
equallly the givenn wind loadinng, analyze thhe vertical
truss es for the wiind in one diirection only. It will be
windward bracce is in tensioon and the
apparrent that the w
leewaard brace is inn compression. When the winnd blows in
the oother direction, the brace forces reverse.
Design alll the braces as if the tenssion forces
contrrol. Although it is usually compression fforces that
contrrol brace desiggn, in this chaapter we are foocusing on
tensioon. It would bbe useful to coonsider the usee of several
typess of members, e.g., angles, cchannels, and W-shapes,
to ressist these forcees.
.
C
Chapterr 5
C
Comprression
n
M
Members
Ruttgers Universitty School of Buusiness, Piscataaway, NJ
Photo coourtesy of WSP Parsons Brinnckerhoff
5.11
COMPR
RESSION MEMBERS
M
IN STRUC
CTURES
Compression
n members arre structural elements
e
subjjected to axiaal forces that ttend to push the ends
of the mem
mbers toward
d each other. The most ccommon com
mpression meember in a bbuilding
structure is a column. Columns are vertical
v
membbers that suppport the horiizontal elemeents of a
roof or floor system. Sev
veral columns can be seenn in Figure 55.1 as part off a building sttructure.
They are th
he primary ellements that provide
p
the vvertical spacee to form ann occupiable volume.
Other comp
pression mem
mbers are fou
und in trussess as chord annd web mem
mbers and as bracing
members in floors and walls.
w
Other naames often ussed to identify
fy compressioon members aare struts
and posts. Throughout
T
this
t
chapter the
t terms coompression m
member and ccolumn will be used
interchangeaably.
The compression
n members diiscussed in thhis chapter exxperience onlyy axial forcess. In real
structures, additional
a
load
d effects are often exertedd on a compreession member that wouldd tend to
combine ben
nding with th
he axial force. These combbined force m
members are ccalled beam-ccolumns
and are disscussed in Chapter
C
8. Th
he majority oof the provisions that appply to comp
mpression
members aree located in Chapter
C
E of th
he Specificatiion.
Table 5.1 lists th
he sections off the Specificaation and part
rts of the Mannual discussed in this
chapter.
5.22
CROSS--SECTIONAL SHAPE
ES FOR CO
OMPRESSIO
ON MEMB
BERS
n members caarry axial forcces, so the prrimary cross-ssectional propperty of intereest is the
Compression
area. Thus, the
t simple rellationship bettween force annd stress,
P
(5.1)
f =
A
is applicablle. As long as this relationship dictaates compresssion membeer strength, aall cross
sections witth the same area will perrform in the same way. IIn real structu
tures, howeveer, other
factors influ
uence the strrength of thee compressioon member, and the disttribution of tthe area
becomes imp
portant.
1224
Chapter 5
Figure 5.1
5
Comppression Mem
mbers
125
Columnss in a Multisto
ory Building
Photo cou
urtesy Greg Griieco
In
I building structures,
s
thee typical com
mpression m
member is a column and the typical
column is
i a rolled wiide-flange meember. Later discussions oof compressioon member sttrength will
show thaat the W-shap
pe does not have the mostt efficient disttribution of m
material for compression
memberss. It does, how
wever, provid
de a compressiion member tthat can easilyy be connecteed to other
memberss of the system
m such as beaams and otherr columns. Thhis feature siggnificantly innfluences its
selection
n as an approp
priate column cross sectionn.
Figure
F
5.2 sh
hows examplees of rolled aand built-up shapes that aare used as compression
memberss. Many of th
hese are the saame shapes uused for the teension members discussedd in Chapter
4. This is
i reasonablee because thee forces beinng consideredd in these tw
wo cases are both axial,
although
h they act in the
t opposite direction.
d
How
wever, other factors that iinfluence the strength of
compresssion memberss will dictate additional crriteria for the selection of tthe most efficcient shapes
for these members.
The
T tee and angle
a
shown in Figure 5.2cc and d are coommonly useed as chords aand webs of
trusses. In
I these appliications, the geometry
g
of tthe shapes helps simplify tthe connectioons between
memberss. Angles are also used in
n pairs as buiilt-up compreession membeers, with the connecting
element between
b
the two
t angles as shown in Figgure 5.2h. Thhe channel cann be found in trusses as a
single eleement or com
mbined with another
a
channnel as shown in Figure 5.22b, i, l, and m
m. Built-up
Table 5.1
Sections of
o Specification and Parts of Manual Coovered in thiss Chapter
Specificationn
B3
B4
E1
E2
E3
E4
E5
E6
E7
Design Baasis
Classificattion of Sections for Local B
Buckling
General Prrovisions
Slendernesss Limitationss and Effectivve Length
Compressiive Strength for Flexuraal Buckling of Members without
Slendeer Elements
Compressiive Strength for Torsionall and Flexuraal-Torsional B
Buckling
of Mem
mbers withou
ut Slender Eleements
Single-Angle Compresssion Memberrs
Built-up Members
M
Members with
w Slender Elements
Manual
Part 1
Part 4
Part 6
Dimension
ns and Properrties
Design of Compression
n Members
Design of Members Sub
bject to Combbined Loadinng
1226 Chapter 5
Figgure 5.2
Compresssion Memberss
Ro
olled Shapes and Built-up Shapes for Compression M
Members
columns can
n also be foun
nd using chan
nnels. The holllow structuraal sections (H
HSS) shown inn Figure
5.2e, f, and
d g are comm
monly found as columns in buildings,, particularly one-story sttructures
where the connections to
o the shape can be simpliffied by carry ing beams ovver the colum
mns. The
distribution of the materiaal in these shaapes is the moost efficient ffor columns.
5.33
COMPR
RESSION MEMBER
M
STRENGTH
S
H
f
were to
t impact thee strength of a compressioon member, thhe simple axiial stress
If no other factors
relationship given in Equation
E
5.1 could be ussed to descrribe member strength. Thhus, the
maximum fo
orce that a compression meember could rresist at yieldd would be
Py = Fy Ag
(5.2)
where Py is the yield load, sometimess called the sqquash load; Fy is the yieldd stress; and Ag is the
gross area. This is the response
r
thatt would be exxpected if a very short sspecimen, onee whose
length appro
oximates its other two dim
mensions, weere to be tessted in comprression. This type of
column test specimen, sh
hown in Figurre 5.3a, is callled a stub coolumn. Becauuse most comppression
members wiill have a len
ngth that greaatly exceeds its other dim
mensions, lenggth effects caannot be
ignored. A more
m
realistic column is sh
hown in a testt frame in Figuure 5.3b.
5.33.1
Euler Column
C
To address the
t impact off length on co
ompression m
member behavvior, a simplee model, as shown in
Figure 5.4, is
i used. The Swiss
S
mathem
matician Leonnard Euler firrst presented tthis analysis in 1759.
A number of
o assumption
ns are made in
i this colum
mn model: (1)) the column ends are fricctionless
pins, (2) thee column is peerfectly straiight, (3) the looad is appliedd along the ceentroidal axis,, and (4)
Chapter 5
(a)
Comppression Mem
mbers
127
((b)
Figure 5.3
5 Column Testing.
T
(a) Sttub column. (bb) Long colum
umn
(a) Photo courtesy Proff. Dr. Mario Fontana,
F
(b) P
Photo courtesyy Mohammed Ali Morovat aand Michael
Engelhard
dt, University of
o Texas at Austin
the materrial behaves elastically.
e
Baased on thesee assumptionss, this columnn model is usuually called
the perfeect column or the pure colu
umn.
Figure
F
5.4a sh
hows the perffect column w
with an applieed load that w
will not causee any lateral
displacem
ment or yiellding. In this arrangemeent, the loadd can be inccreased with no lateral
displacem
ment of the column.
c
How
wever, at a paarticular loadd, defined as the critical load or the
buckling load, Pcr, the column will displacce laterally as shown inn Figure 5.44b. In this
configuraation, the dasshed line reprresents the o riginal positiion of the meember and thhe solid line
representts the displacced position. Note
N
that an axis system iis presented iin the figure, with the zaxis alon
ng the membeer length and the y-axis traansverse to thhe member leength. This places the xaxis perp
pendicular to the plane of the figure. Thhe x- and y-axxes corresponnd to the centtroidal axes
of the cro
oss section.
A free body diagram
d
of thee lower portioon of the coluumn in its dispplaced positioon is shown
in Figuree 5.4c. If mom
ments are takeen about pointt C, equilibriuum requires
M z = Pcr y
From thee principles of
o mechanics and using sm
mall displacem
ment theory, the differentiial equation
relating moment
m
to cu
urvature of thee deflected m
member is giveen as
d2y
M
=− z
2
EI x
dz
Combining these two
o equations and
a rearrangiing the terms yields the differential eequation of
equilibriu
um,
d 2 y Pcr
y=0
+
dz 2 EI x
1228 Chapter 5
Compresssion Memberss
onditions for Elastic
E
Colum
mns
Figure 5.4 Stability Co
If the coeffi
ficient of the second term
m is taken as k 2 = Pcr EI x , the differenntial equationn for the
column beco
omes
d2y
+ k2 y = 0
2
dz
which is a standard seccond-order liinear ordinarry differentiaal equation. T
The solutionn to this
equation is given
g
by
y = A sin kz + B cos kz
(5.3)
where A an
nd B are con
nstants of integration. To further evalluate this equuation, the bboundary
conditions must
m be applieed. Because at
a z = 0, y = 0 and at z = L, y = 0, we find that
B=0
and
A sin kL = 0
For Equation 5.3 to havee a nontrivial solution, (sinn kL) must equal zero. Thhis requires thhat kL =
nπ, where n is any integer. Substituting
g for k and reearranging yieelds
n 2 π 2 EI x
(5.4)
Pcr =
L2
Because n can be taken as
a any integer, Equation 5.44 has a minim
mum when n = 1. This is caalled the
Euler buckliing load or th
he critical bucckling load annd is given as
π 2 EI x
(5.5)
Pcr =
L2
If values forr B and kL arre substituted into Equatioon 5.3, the shaape of the buuckled columnn can be
determined from
f
z⎞
⎛
(5.6)
y = A sin ⎜ nπ ⎟
⎝ L⎠
129
Chapter 5
Comppression Mem
mbers
Figurre 5.5
Shappe of Buckledd Columns
Because any value for
fo A will sattisfy Equatio n 5.6, a uniqque magnitudde for the diisplacement
cannot be
b determined
d; however, it is clear thaat the shape oof the buckleed column is a half sine
curve wh
hen n = 1. Th
his is shown in
i Figure 5.55a. For other vvalues of n, ddifferent buckkled shapes
will result along with the higher criitical bucklinng load. Whenn n > 1, these shapes are reeferred to as
higher mode
m
shapes. Several
S
cases are shown inn Figure 5.5b,, c, and d. In aall cases, the basic shape
is the sin
ne curve. In orrder for thesee higher modees to occur, soome type of pphysical restrraint against
buckling is required at
a the point where
w
the buckkled shape crosses the origginal, undefleected shape.
This can be accomplisshed with the addition of bbraces, which is discussed later.
We
W now havee two equation
ns to predict tthe column sttrength: Equaation 5.2, whiich does not
address length;
l
and Equation
E
5.5,, which does . These two equations arre plotted in Figure 5.6.
Because the derivation
n of the Eulerr equation waas based on ellastic behavioor and the coluumn cannot
carry mo
ore load than the
t yield load
d, there is an uupper limit too the column sstrength.
If
I the length at which thiss limit occurrs is taken ass Ly, it can bbe determinedd by setting
Equation
n 5.2 equal to Equation 5.5 and solving ffor length, givving
EI x
Ly = π
Fy Ag
To simpllify this equattion, the radiu
us of gyrationn, r, will be ussed, where
I
r=
A
Because the moment of inertia dep
pends on the axis being cconsidered, annd A is the grross area of
the sectio
on, which is independent of
o axis, r will depend on thhe buckling axxis. In the derrivation just
developeed, the axis off buckling forr the column oof Figure 5.4 was taken as the x-axis; thhus,
E
Ly = πrx
Fy
For this theoretical
t
deevelopment, a column whoose length is leess than Ly would fail by yyielding and
could bee called a sho
ort column, whereas
w
a collumn with a length greateer than Ly woould fail by
buckling and be called
d a long colum
mn.
1330 Chapter 5
Compresssion Memberss
Fiigure 5.6
C
Column Strenggth Based on Length
It iss also helpful to write Equ
uation 5.5 in tterms of stresss. Dividing bboth sides by the area
and substitu
uting again forr the radius off gyration yieelds
π2 E
(5.7)
Fcr =
2
⎛ L⎞
⎜ ⎟
⎝r⎠
In this equattion, the radiu
us of gyration
n is left unsubbscripted so tthat it can be applied to whhichever
axis is determined to be the
t critical ax
xis. A plot of sstress versus L/r would be of the same sshape as
the plot of fo
orce versus L in Figure 5.6
6.
5.33.2
Other Boundary
B
Co
onditions
Derivation of
o the bucklin
ng equations presented ass Equations 55.5 and 5.7 inncluded the bboundary
condition off frictionless pins at both ends. For perrfect columnss with other bboundary connditions,
the momentt will not be zero at the ends,
e
and thiss will result iin a nonhomoogeneous diffferential
equation. Solving the resulting diffe
ferential equaation and appplying the aappropriate bboundary
conditions will
w lead to a buckling equation
e
of a form similaar to the preevious equatiions. To
generalize th
he buckling equation
e
for other end condditions, the coolumn length,, L, is replaceed by the
column effeective length, KL, where K is the effecctive length ffactor. Thus, the general bbuckling
equations beecome
π2 EI
Pcr =
(5.8)
2
( KL )
and
π2 E
(5.9)
Fcr =
2
⎛ KL ⎞
⎜
⎟
⎝ r ⎠
Figu
ure 5.7 depiccts the origin
nal pin-endedd column wiith several examples of ccolumns
showing thee influence off different end
d conditions. All columns are shown w
with the lowerr support
fixed againsst lateral tran
nslation. Threee of the coluumns have uppper ends thaat are also reestrained
from lateral translation, and
a three others have upp er ends that aare free to traanslate. The eeffective
length can be
b visualized
d as the lengtth between innflection poinnts, where thhe curvature rreverses.
This result is similar to th
he original deerivation whenn n was takenn as some inteeger other thaan 1. It is
most easily seen in Figure 5.7b and c but
b can also bbe seen in Figgure 5.7d by vvisualizing thee
Chapter 5
Figure 5.7
5
Comppression Mem
mbers
131
Column Buckled Shap
pe for Differeent End Condditions
Figure 5.88 Extended S
Shape of
Buckled Column from F
Figure 5.7d
extended
d buckled shap
pe above the column as shhown in Figurre 5.8. In all cases, the buckled curve
is a segm
ment of the sin
ne curve. Thee most importtant thing to oobserve is thaat the columnn with fixed
ends in Figure
F
5.7b has
h an effectiive length off 0.5L, whereas the colum
mn in Figure 55.7a has an
effective length of L. Thus, the fiixed-end coluumn will havve four times the strength of the pinended co
olumn.
5.3.3
Com
mbination of Bracing
B
and End Conditiions
The influ
uence of interrmediate braccing on the efffective lengthh was touched upon in thee discussion
of the hiigher modes of
o buckling. In
I those casees, the bucklinng resulted inn equal-lengtth segments
that refleected the mod
de number. Thus,
T
a colum
mn with n = 2 had two equual segments, whereas a
Figure 5.9
5
Buckled Shape for Co
olumns with IIntermediate B
Braces
1332 Chapter 5
Compresssion Memberss
column with
h n = 3 buck
kled with three equal segm
ments. If phyysical braces are used to provide
buckling ressistance to thee column, thee effective lenngth will deppend on the loocation of thee braces.
Figure 5.9 shows three columns witth pinned endds and interm
mediate suppports. The coolumn in
Figure 5.9a is the same as the column in Figure 5.55b. The effecttive length is 0.5L, so K = 0.5. The
column in Figure
F
5.9b sh
hows lateral braces
b
in an unsymmetriccal arrangemeent with one segment
equal to L/3
3 and the otheer to 2L/3. Although
A
the eexact locationn of the inflection point w
would be
slightly into
o the longer segment,
s
norrmal practice is to take thhe longest unnbraced lengthh as the
effective len
ngth; thus KL = 2L/3, so K = 2/3. The coolumn in Figuure 5.9c is braaced at two loocations.
The longest unbraced len
ngth for this case gives an eeffective lenggth KL = 0.5L and a corressponding
K = 0.5. A general
g
rule caan be stated that,
t
when thee column endds are pinned, the longest uunbraced
length is thee effective len
ngth for bucklling in that dirrection.
Wheen other end
d conditions are
a present, tthese two influences musst be combinned. The
columns of Figure 5.10 illustrate the influence off combinationns of end supports and braacing on
the column effective leng
gth. The end conditions w
would influencce only the effective lengtth of the
end segmentt of the colum
mn. For the co
olumn in Figuure 5.10a, the lower segmeent has L = a, and that
segment would buckle with
w an effectiv
ve length KL = a. The uppper segment hhas L = b but also has
a fixed end. Thus, it wou
uld buckle witth an effectivve length KL = 0.7b, obtainned by combiining the
end conditio
ons of Figure 5.7c with th
he length, b. T
Thus, the relaationship betw
ween lengths a and b
determine which
w
end of the
t column dictates
d
the ovverall columnn effective lenngth. As an eexample,
the column in Figure 5.10b shows thaat the lowest ssegment wouuld set the collumn effectivve length
at 0.35L.
Figure 5.10 Bu
uckled Shapee for Columnss with Differeent End Condditions and Inttermediate Brraces
EX
XAMPLE 5.1
5
Th
heoretical
Coolumn Stren
ngth
Goa
al:
Deterrmine the theeoretical strenngth for a pinn-ended colum
mn and whetther it
will first
f
buckle orr yield.
Giv
ven:
A W1
10×33, A992,, column withh a length of 220 ft.
SO
OLUTION
Step
p 1:
Deterrmine the load
d that would ccause bucklinng.
With no other info
ormation, it m
must be assum
med that this ccolumn will bbuckle
aboutt its weak axiis, if it buckl es at all, because the effeective length, KL =
20 ft for both axes.
Chapter 5
Compression Members
133
From Manual Table 1-1, Iy = 36.6 in.4 and Ag = 9.71 in.2. The load that
would cause it to buckle is
2
π2 EI y π ( 29,000 )( 36.6 )
Pcr =
−
= 182 kips
2
2
( KL )
( 20 (12 ) )
Step 2:
Determine the load that would cause yielding.
Py = Fy Ag = 50 ( 9.71) = 486 kips
Step 3:
Conclusion: Because Pcr < Py, the theoretical column strength is
P = 182 kips
and the column would buckle before it could reach its yield stress.
EXAMPLE 5.2
Critical Buckling
Load
Goal:
Determine the overall column length that, if exceeded, would theoretically
cause the column to buckle elastically before yielding.
Given:
A W8×31 column with fixed supports. Use steels with (a) Fy = 40 ksi and
(b) Fy = 100 ksi.
SOLUTION
Step 1:
From Manual Table 1-1, Iy = 37.1 in.4 and Ag = 9.13 in.2.
Part a
Step 2:
Step 3:
Determine the force that would cause the column to yield when Fy = 40 ksi.
Py = Fy Ag = 40 ( 9.13 ) = 365 kips
To determine the length that would cause this same load to be the buckling
load for the pinned-pinned case, set this force equal to the buckling load
equation and determine the length from
2
π2 EI y π ( 29,000 )( 37.1)
365 kips = 2 =
L
L2
which gives
L=
π2 ( 29,000 )( 37.1)
365
= 171 in.
So the length is
L=
Step 4:
171
= 14.3 ft
12
Since a fixed-end column has an effective length equal to one-half the
actual length, buckling will not occur if the actual length is less than or
equal to:
L = 2 (14.3) = 28.6 ft for a column with Fy = 40 ksi
134 Chapter 5
Compression Members
Part b
Step 5:
Part 6:
Determine the force that would cause the column to yield when Fy = 100
ksi.
Py = Fy Ag = 100 ( 9.13 ) = 913 kips
To determine the length that would cause this same load to be the buckling
load for the pinned-pinned case, set this force equal to the buckling force
and determine the length from
2
π2 EI y π ( 29,000 )( 37.1)
913 kips = 2 =
L
L2
which gives
L=
π2 ( 29,000 )( 37.1)
913
= 108 in.
So the length is
L=
Step 7:
108
= 9.0 ft
12
Since a fixed-end column has an effective length equal to one-half the
actual length, buckling will not occur if the actual length is less than or
equal to:
L = 2 ( 9.0 ) = 18.0 ft for a column with Fy = 100 ksi
5.3.4
Real Column
Physical testing of specimens that effectively model columns found in real building structures,
like that seen in Figure 5.3b, has shown that column strength was not as great as either the
buckling load predicted by the Euler buckling equation or the squash load predicted by material
yielding. This inability of the theory to predict actual behavior was recognized early, and
numerous factors were found to be the cause. Three main factors influence column strength:
material inelasticity, column initial out-of-straightness, and end conditions. The influence of
column end conditions has already been discussed with respect to effective length determination.
Material inelasticity and initial out-of-straightness, which also significantly impact real column
strength, are discussed here.
Inelastic behavior of a column directly results from built-in or residual stresses in the
cross section. These residual stresses are, in turn, the direct result of the manufacturing process.
Steel is produced with heat, and heat is also necessary to form the steel into the shapes used in
construction. Once the shape is fully formed, it is cooled. During this cooling process residual
stresses are developed. Figure 5.11 shows a wide-flange cross section in various stages of
cooling. Initially, as shown in Figure 5.11a, the tips of the flanges with the most surface area to
give off heat begin to cool. This material contracts as it cools, eventually reaching the ambient
temperature. At this point, the fibers in this part of the section reach what is expected to be their
final length.
Chapter 5
Comppression Mem
mbers
135
Figu
ure 5.11 Diistribution of Residual Streesses
As
A adjacent fibers cool,, they too ccontract. In the processs of contraccting, these
subsequeently cooling fibers pull on
o the previoously cooled ffibers, placinng the latter uunder some
amount of
o compressiv
ve stress. Figu
ure 5.11b shoows a cross seection with addditional flangge elements
cooled. When
W
the preeviously cooled portion oof the cross section proviides enough stiffness to
restrain the
t contractio
on of the subssequently coooling materiaal, a tensile sttress is develloped in the
now-cooling materiall because it cannot cont
ntract as it w
would withouut this restraaint. When
completeely cooled, as shown in Fig
gure 5.11c, thhe tips of the fflanges and thhe middle of the web are
put into compression,
c
and the flang
ge-web junctuure is put intoo tension. Thuus, the first fibbers to cool
are in com
mpression, whereas
w
the lasst to cool are in tension.
Several
S
differrent representtations of thee residual streess distributioon have beenn suggested.
One distrribution is sh
hown in Figurre 5.11c. Thee magnitude of the maxim
mum residual stress does
not depend on the maaterial yield strength but iss a function oof material thhickness. In aaddition, the
compresssive residual stress is of critical
c
intereest in the connsideration off compressionn members.
The mag
gnitude of thiss residual streess varies from
m 10 ksi to aabout 30 ksi, depending onn the shape.
The high
her values are found in wid
de flanges witth the thickestt flange elemeents.
To
T understand
d the overall impact of thhese residual stresses on ccolumn behavvior, a stub
column will
w again be investigated. Figure 5.12 sshows the stress-strain relaation for a shhort column,
one that will not buck
kle but exhib
bits the influeence of residuual stresses. A
As the colum
mn is loaded
with an axial
a
load, thee member sho
ortens and thee correspondiing strain andd stress are deeveloped, as
if this were
w
a perfecttly elastic sp
pecimen. The response off a perfectly elastic, perfeectly plastic
column is
i shown by th
he dashed lin
ne in Figure 5 .12. When thhe applied streess is added too a member
with residual compresssive stress, th
he stub colum
mn begins to shorten at a ggreater rate ass the tips of
the flang
ge become strressed beyond
d the yield str
tress. This pooint is identifiied in Figure 5.12 as Fp,
the propo
ortional limit Thus, the strress-strain currve moves offf the straight dashed line aand follows
the curveed solid line. Continuing to add load tto the columnn results in ggreater strain for a given
stress, an
nd the column
n eventually reaches the yyield stress off the perfectly elastic matterial. Thus,
the only difference beetween the beh
havior of the actual colum
mn and the usuual test specim
men used to
ne the stress-strain relatio
onship is thatt the real coolumn behavees inelasticallly as those
determin
portions of its cross seection with co
ompressive reesidual stressees reach the m
material yield stress.
If
I a new term
m, the tangentt modulus, ET , is defined aas the slope oof a tangent too the actual
stress-strrain curve at any
a point sho
own in Figuree 5.12, an impproved predicction of colum
mn buckling
strength can be obtain
ned by modify
ying the Eulerr buckling equ
quation so thatt
π2 ET I x
Pcr =
2
( KL )
1336 Chapter 5
Compresssion Memberss
Figu
ure 5.12 Stuub Column StrressStrainn Diagrams w
with and without
Residdual Stress
Thus, as the column is loaded beyond its elastic lim
mit, ET decreases and the buuckling strenggth does
also. This paartially accou
unts for the in
nability of thee Euler bucklling equation to accuratelyy predict
column stren
ngth.
Ano
other factor to
t significanttly impact c olumn strenggth is the coolumn initiall out-ofstraightness.. Once again
n, the manufaacturing proceess for steel shapes impaacts the abilityy of the
column to caarry the prediicted load. In this case, thee problem is rrelated to the fact that no sttructural
steel membeer comes out of the producction process perfectly straaight. The AIS
SC Code of SStandard
Practice perrmits an initiaal out-of-straightness equaal to 1/1000 oof the length between poiints with
lateral support. Although
h this appearss to be a smaall variation from straighttness, it still impacts
column stren
ngth.
Figu
ure 5.13a sh
hows a perffectly elastic,, pin-ended column witth an initial out-ofstraightness,, δ. A compaarison of this column
c
diagrram with thaat used to deriive the Euler column,
Figure 5.13 Influence of
o Initial Out--of-Straightneess on Colum
mn Strength
Chapter 5
Compression Members
137
Figure 5.4, shows that the moment along the column length will be greater for this initially
crooked column in its buckled position than it would have been for an initially straight column.
Thus, the solution to the differential equation would be different. In addition, because the applied
load works at an eccentricity from the column along its length, even before buckling, a moment is
applied to the column that has not yet been accounted for. Figure 5.13b shows the load versus
lateral displacement diagram for this initially crooked column compared to that of the initially
straight column. This column not only exhibits greater lateral displacement, it also has a lower
maximum strength.
When these two factors are combined, the Euler equation cannot properly describe
column behavior on its own. Thus, the development of curves to predict column behavior has
historically been a matter of curve-fitting the test data in an attempt to present a simple
representation of column behavior.
5.3.5
AISC Provisions
The compression members discussed thus far have either yielding or overall column buckling as
the controlling limit state. Figure 5.14 plots sample column test data compared to the Euler
equation and the squash load. The Structural Stability Research Council proposed three equations
to predict column behavior. To simplify column design, AISC selected a single curve described
using two segments as their representation of column strength.
The design basis for ASD and LRFD were presented in Sections 1.9 and 1.10,
respectively. The strength equations are repeated here in order to reinforce the relationship
between the nominal strength, resistance factor, and safety factor presented throughout the
Specification.
The requirement for ASD is
Rn
Ω
(AISC B3.2)
Ru ≤ φRn
(AISC B3.1)
Ra ≤
The requirement for LRFD is
As indicated earlier, the Specification provides the relationship to determine nominal
strength and the corresponding resistance factor and safety factor for each limit state to be
considered. The provisions for compression members with nonslender elements, i.e., no local
buckling, are given in Specification Section E3. The nominal column strength for the limit state
of flexural buckling of members with nonslender elements is
Pn = Fcr Ag
and
φc = 0.9 ( LRFD )
(AISC E3-1)
Ω c = 1.67 ( ASD )
where Ag is the gross area of the section and Fcr is the flexural buckling stress. (The Euler column
derivation in Section 5.3.1 addressed the limit state of flexural buckling.)
1338 Chapter 5
Compresssion Memberss
Figure 5.14 Sample Co
olumn Test Data Comparedd to Theoreticcal Column S
Strength
The Specification
n defines Lc as
a the effectivve length and shows it equual to KL. Thiss is the
same effectiive length factor, K, discusssed earlier. T
To capture collumn behavioor when inelasstic
buckling dom
minates colum
mn strength, that
t is, wheree residual stresses become important, the
Specification
n provides thaat when
Fy
Lc / r ≤ 4.71 E / Fy or
≤ 2.225
Fe
Fy
⎡
Fcr = ⎢ 0.6658 Fe
⎢⎣
⎤
⎥ Fy
⎥⎦
(AIS
SC E3-2)
To capture behaavior when in
nelastic bucklling is not a factor and iinitial crookedness is
hat is when
dominant, th
Fy
Lc / r > 4.71 E / Fy or
> 2.225
Fe
Fcr = 00.877 Fe
SC E3-3)
(AIS
where Fe is the elastic bu
uckling stress; the Euler buuckling stresss previously ppresented as E
Equation
5.9 and restaated here is
π2 E
(AIS
SC E3-4)
Fe =
2
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
Chapter 5
Figure 5.15
5
Comppression Mem
mbers
139
Lc/r verrsus Critical Strength
S
The
T flexural buckling
b
stressses for three different steeels, A36, A9992, and A5144, versus the
slenderneess ratio, Lc/r, are shown in
i Figure 5.155. For very sllender columnns, the buckliing stress is
independ
dent of the maaterial yield. The division between elasstic and inelaastic behaviorr, Equations
E3-2 and
d E3-3, corressponds to Lc/r values of 1334, 113, and 80.2 for steells with a yieldd of 36, 50,
and 100 ksi,
k respectiv
vely.
Early
E
editionss of the LRF
FD Specificattion defined the exponentt of Equationn E3-2 in a
slightly different
d
form
m that makes the
t presentatioon a bit simpler. If a new tterm is defineed such that
2
Fy ⎛ Lc ⎞ Fy
λ c2 =
=⎜ ⎟
Fe ⎝ πr ⎠ E
then the dividing
d
poin
nt between elaastic and inelaastic behaviorr, where
Lc
E
= 44.71
r
Fy
becomes
λc =
Lc Fy 4.71
=
= 1.55
π
πr E
By substituting λ c2 = Fy Fe , the criitical flexurall buckling streess for λ c ≤ 1.5 becomes
(
2
)
Fcr = 0.658λc Fy
(5.10)
and for λ c > 1.5 ,
Fcr =
0.877
Fy
λ c2
(5.11)
A plot of
o the ratio of critical flexural
fl
buckl
kling stress too yield stresss as a functtion of the
slenderneess parameteer, λc, is giveen in Figuree 5.16. Usingg this formullation it is eevident that
regardlesss of the steell yield stress, the ratio of fflexural buckkling stress too yield stress is the same
when plo
otted against the
t slenderness parameter,, λc. Table 5.22 provides theese numericall values in a
convenieent, usable forrm.
1440 Chapter 5
Compresssion Memberss
Figure 5.16 λc versus Critical
C
Stresss
Earllier editions of
o the ASD and
a LRFD Sppecifications indicated thaat there shoulld be an
upper limit on
o the magnittude of the sleenderness rattio at Lc/r = 2000. The intennt for this limiit was to
have the eng
gineer recogn
nize that for veery slender coolumns, the fl
flexural bucklling stress waas so low
as to make the column very
v
inefficieent. This lim
mit has been rremoved in rrecent editionns of the
Specification
n because theere are many factors influeencing columnn strength thaat indicate thaat a very
slender colu
umn might acttually be acceeptable. Sectioon E2 simplyy informs the designer thatt column
slenderness should prefeerably b e kep
pt to somethinng less than 200. Table 55.3 gives the flexural
Table 5.2 Ratio of Critical Stress to Yield Stress
λc
Fcr/Fy
λc
Fcr/Fy
λc
Fcr/Fy
λc
Fcrr/Fy
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
1.000
0.999
0.996
0.991
0.983
0.974
0.963
0.950
0.935
0.919
0.901
0.881
0.860
0.838
0.815
0.790
0.765
0.739
0.712
0.95
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
0.685
0.658
0.630
0.603
0.575
0.547
0.520
0.493
0.466
0.440
0.415
0.390
0.365
0.343
0.322
0.303
0.286
0.271
0.256
1.90
1.95
2.00
2.05
2.10
2.15
2.20
2.25
2.30
2.35
2.40
2.45
2.50
2.55
2.60
2.65
2.70
2.75
2.80
0.243
0.231
0.219
0.209
0.199
0.190
0.181
0.173
0.166
0.159
0.152
0.146
0.140
0.135
0.130
0.125
0.120
0.116
0.112
2.85
2.90
2.95
3.00
3.05
3.10
3.15
3.20
3.25
3.30
3.35
3.40
3.45
3.50
3.55
3.60
3.65
3.70
3.75
0.1108
0.1104
0.1101
0.09974
0.09943
0.09913
0.08884
0.08856
0.08830
0.08805
0.07781
0.07759
0.07737
0.07716
0.06696
0.06677
0.06658
0.06641
0.06624
Chapter 5
Table 5.3
Critical Stress Values for Three Steels
Fcr (ksi)
KL/r
0
10
14
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
52
54
56
58
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
100
Compression Members
Fy = 36 ksi
36.0
35.8
35.6
35.4
35.2
35.1
34.9
34.7
34.5
34.3
34.1
33.9
33.6
33.4
33.1
32.8
32.5
32.2
31.9
31.6
31.2
30.9
30.5
30.2
29.8
29.4
29.0
28.6
28.2
27.8
27.4
27.0
26.6
26.1
25.7
25.3
24.8
24.4
23.9
23.5
23.1
22.6
22.2
21.7
21.3
Fy = 50 ksi
50.0
49.6
49.3
48.8
48.6
48.3
47.9
47.6
47.2
46.8
46.4
45.9
45.5
45.0
44.5
43.9
43.4
42.8
42.2
41.6
41.0
40.4
39.8
39.1
38.4
37.7
37.1
36.4
35.7
34.9
34.2
33.5
32.8
32.0
31.3
30.6
29.8
29.1
28.4
27.7
26.9
26.2
25.5
24.8
24.1
Fcr (ksi)
Fy = 100 ksi
100.0
98.5
97.2
95.4
94.3
93.2
91.9
90.6
89.2
87.7
86.1
84.4
82.7
81.0
79.1
77.3
75.3
73.4
71.4
69.4
67.3
65.3
63.2
61.1
59.1
57.0
54.9
52.9
50.9
48.8
46.9
44.9
43.0
41.1
39.2
37.3
35.6
33.9
32.4
31.0
29.7
28.4
27.2
26.1
25.1
KL/r
102
104
106
108
110
112
114
116
118
120
122
124
126
128
130
132
134
136
138
140
142
144
146
148
150
152
154
156
158
160
162
164
166
168
170
172
174
176
178
180
182
184
186
190
200
Fy = 36 ksi
20.8
20.4
19.9
19.5
19.0
18.6
18.2
17.7
17.3
16.9
16.4
16.0
15.6
15.2
14.8
14.4
14.0
13.6
13.2
12.8
12.4
12.1
11.8
11.5
11.2
10.9
10.6
10.3
10.1
9.81
9.56
9.33
9.11
8.89
8.69
8.48
8.29
8.10
7.92
7.75
7.58
7.41
7.26
6.95
6.28
Fy = 50 ksi
23.4
22.7
22.0
21.3
20.6
20.0
19.3
18.7
18.0
17.4
16.9
16.3
15.8
15.3
14.9
14.4
14.0
13.6
13.2
12.8
12.4
12.1
11.8
11.5
11.2
10.9
10.6
10.3
10.1
9.81
9.56
9.33
9.11
8.89
8.69
8.48
8.29
8.10
7.92
7.75
7.58
7.41
7.26
6.95
6.28
Fy = 100 ksi
24.1
23.2
22.3
21.5
20.7
20.0
19.3
18.7
18.0
17.4
16.9
16.3
15.8
15.3
14.9
14.4
14.0
13.6
13.2
12.8
12.4
12.1
11.8
11.5
11.2
10.9
10.6
10.3
10.1
9.81
9.56
9.33
9.11
8.89
8.69
8.48
8.29
8.10
7.92
7.75
7.58
7.41
7.26
6.95
6.28
141
1442 Chapter 5
Compresssion Memberss
Figure 5.17
7 Columns for
f Examples 5.3, 5.4, and 5.5.
buckling strress for valuees of Lc/r fro
om 0 to 200 for steels w
with three diffferent yield stresses.
Manual Tab
ble 4-14 prov
vides an expanded versionn of this tablee for six diffeerent yield strresses at
slenderness ratio incremeents of 1.0 ft.
Goa
al:
EX
XAMPLE 5.3
5
Coolumn Stren
ngth
byy AISC Provvisions Givven:
SO
OLUTION
Deterrmine the available columnn strength.
A W1
12×79 pin-en
nded column w
with a lengthh of 10.0 ft. aas shown in F
Figure
5.17aa. Use A992 steel.
Step
p 1:
From
m Manual Table 1-1, rx = 5..34 in., ry = 3..05 in., and A = 23.2 in.2.
Step
p 2:
Deterrmine the con
ntrolling effecctive slendernness ratio.
use the length
h is 10.0 ft aand the colum
mn has pinned ends, Lc = KL =
Becau
10.0 ft
f for both thee x-axis and y -axis. Thus,
Lcx 10.0 (12 )
=
= 22.5
5.34
rx
and
Lcy 10.0 (12 )
=
= 39.3
3.05
ry
Sincee
Lcy Lcx
>
ry
rx
Chapter 5
Compression Members
143
the y-axis controls.
Step 3:
Determine which column strength equation to use. Since
Lc
E
29,000
= 39.3 < 4.71
= 4.71
= 113
r
Fy
50
use Equation E3-2
Step 4:
Determine the Euler buckling stress.
π2 ( 29,000 )
Fe =
= 185 ksi
2
( 39.3)
Step 5:
Determine the critical stress from Equation E3-2.
⎛ Fy ⎞
⎜
⎟
⎛ 50 ⎞
⎜
⎟
Fcr = 0.658⎝ Fe ⎠ Fy = 0.658⎝ 185 ⎠ ( 50 ) = 44.7 ksi
Step 6:
For
LRFD
Step 7:
For
ASD
Step 7:
Determine the nominal strength.
Pn = 44.7 ( 23.2 ) = 1040 kips
Determine the design strength for LRFD.
φPn = 0.9 (1040 ) = 936 kips
Determine the allowable strength for ASD.
Pn 1040
=
= 623 kips
Ω 1.67
Goal:
EXAMPLE 5.4
Column Strength
by AISC Provisions Given:
SOLUTION
Determine the available column strength.
A W10×49 column with a length of 20.0 ft, one end pinned and the other
end fixed for the y-axis, and both ends pinned for the x-axis, as shown in
Figure 5.17b. Use A992 steel.
Step 1:
From Manual Table 1-1, rx = 4.35 in., ry = 2.54 in., and A = 14.4 in.2.
Step 2:
Determine the effective length factors from Figure 5.7.
Comparing the columns shown in Figure 5.17b with those shown in Figure
5.7, the effective length factors are Ky = 0.7 and Kx = 1.0.
Step 3:
Determine the x- and y-axis slenderness ratios.
Lcx K x L 1.0 ( 20.0 )(12 )
=
=
= 55.2
4.35
rx
rx
144 Chapter 5
Compression Members
Lcy K y L 0.7 ( 20.0 )(12 )
=
=
= 66.1
2.54
ry
ry
Step 4:
Using the larger slenderness ratio, determine which column strength
equation to use. Since
Lc
E
29,000
= 66.1 < 4.71
= 4.71
= 113,
r
Fy
50
use Equation E3-2
Step 5:
Determine the Euler buckling stress.
π2 ( 29,000 )
Fe =
= 65.5 ksi
2
( 66.1)
Step 6:
Determine the critical stress from Equation E3-2.
Fcr = 0.658
Step 7:
For
LRFD
Step 8:
For
ASD
Step 8:
⎛ Fy ⎞
⎜
⎟
⎝ Fe ⎠
Fy = 0.658
⎛ 50 ⎞
⎜
⎟
⎝ 65.5 ⎠
( 50 ) = 36.3 ksi
Determine the nominal strength.
Pn = 36.3 (14.4 ) = 523 kips
Determine the design strength for LRFD.
φPn = 0.9 ( 523 ) = 471 kips
Determine the allowable strength for ASD.
Pn Ω = 523 1.67 = 313 kips
Goal:
EXAMPLE 5.5
Column Strength
by AISC Provisions Given:
SOLUTION
Determine the available column strength.
A W14×53 column with a length of 40.0 ft, both ends fixed for the y-axis,
and one end pinned and one end fixed for the x-axis, as shown in Figure
5.17c. Use A992 steel.
Step 1:
From Manual Table 1-1, rx = 5.89 in., ry = 1.92 in., and A = 15.6 in.2.
Step 2:
Determine the effective length factors from Figure 5.7.
Comparing the columns shown in Figure 5.17c with those shown in Figure
5.7, the effective length factors are Ky = 0.5 and Kx = 0.7.
Step 3:
Determine the x- and y-axis slenderness ratios.
Lcx K x L 0.7 ( 40.0 )(12 )
=
=
= 57.0
5.89
rx
rx
Chapter 5
Compression Members 145
Lcy K y L 0.5 ( 40.0 )(12 )
=
=
= 125
1.92
ry
ry
Step 4:
Using the larger slenderness ratio, determine which column strength
equation to use. Since
Lc
E
29,000
= 125 > 4.71
= 4.71
= 113
r
Fy
50
use Equation E3-3
Step 5:
Determine the Euler buckling stress.
π2 ( 29,000 )
Fe =
= 18.3 ksi
2
(125)
Step 6:
Determine the critical stress from Equation E3-3.
Fcr = 0.877 Fe = 0.877 (18.3 ) = 16.0 ksi
Step 7:
Determine the nominal strength.
Pn = 16.0 (15.6 ) = 250 kips
For
LRFD
Step 8:
For
ASD
Step 8
Determine the design strength for LRFD.
φPn = 0.9 ( 250 ) = 225 kips
Determine the allowable strength for ASD.
Pn Ω = 250 1.67 = 150 kips
5.4 ADDITIONAL LIMIT STATES FOR COMPRESSION
Two limit states for compression members were discussed in Section 5.3, yielding and flexural
buckling. The strength equations provided in Specification Section E3 clearly show that the upper
limit for column strength, FyAg, is reached only for the zero-length column. Thus, the provisions
are presented in the Specification as applying to the limit state of flexural buckling only, even
though they do consider yielding.
Singly symmetric, unsymmetric, and certain doubly symmetric members may also be
limited by torsional buckling or flexural-torsional buckling. The strength provisions for these
limit states are given in Section E4 of the Specification and are discussed here in Section 5.8.
For some column profiles, another limit state may actually control overall column
strength. The individual elements of a column cross section may buckle locally at a stress below
the stress that would cause the overall column to buckle. If this is the case, the column is said to
be a column with slender elements. The impact of these slender elements on column strength is
determined through the use of an effective area which is smaller than the actual area of the
member. The additional provisions for these types of members are presented in Section 5.6.
1446 Chapter 5
Compresssion Memberss
Figure 5.18 Values of Effective
E
Len
ngth Factor, K
Copyright © American Insttitute of Steel Construction,
C
R
Reprinted with Permission. A
All rights reservved.
5.55
LENGT
TH EFFECT
TS
The effectiv
ve lengths thaat have been
n discussed w
were all relateed to fairly ssimple colum
mns with
easily defineed end condittions and braccing locationss. Once a collumn is recoggnized as beinng a part
of a real stru
ucture, determ
mining the efffective lengthh becomes m
more involved. Moreover, ffor more
complex strructures, it might
m
be sim
mpler to dete rmine the buuckling strenngth of the sstructure
through anaalysis than thrrough the usee of the effecctive length ffactor, K. Usiing that analyysis, the
elastic buck
kling stress off the individu
ual columns, Fe, can be ddetermined. T
This can then be used
directly in the
t column strength
s
equaations. Howevver, for this book, colum
mn elastic bucckling is
determined through a calculation
c
off effective leength. This approach maay incorporatte some
simplificatio
ons that wou
uld not be maade in an acttual bucklingg analysis annd, dependingg on the
approach used to determiine K, may in
nclude assumpptions of behaavior that thee actual structture may
not satisfy.
A fiirst attempt att incorporatin
ng some realiistic aspects oof structures iis shown in T
Table CA-7.1 of thee Commentary
y and here in Figure 5.18. The columnss shown in thiis figure are tthe same
as those shown in Figu
ure 5.7, and the
t same K- factors are sshown and iddentified heree as the
theoretical K-values. Wh
hat is new herre is the pressentation of rrecommendedd design values when
ideal conditiions are apprroximated. Most of these rrecommendedd values are bbased on the fact that
perfectly rig
gid connection
ns are difficu
ult to obtain. T
Thus, for exaample, a fixedd-end columnn (case a
of Figure 5.18) would have
h
a theoretical K = 0.55, but if the end connectiions were to actually
rotate, even just a small amount,
a
the efffective lengtth would increease. As the eend rotation inncreases
toward whatt would occurr for a pin-end
ded column, K would apprroach 1.0. Thuus, the recom
mmended
value of K is 0.65. A sim
milar assessm
ment of the oother cases w
with a fixed end should leaad to an
understandin
ng of the ideea behind theese recommennded values, each being a bit higher tthan the
theoretical value
v
because actual colu
umn end connditions are uunlikely to m
match the theeoretical
assumptionss.
Chapter 5
Comppression Mem
mbers
147
Figurre 5.19 Typpical Moment Frame
When
W
a colum
mn is part of a frame, as shhown in Figuure 5.19, the sstiffness of thhe members
framing into
i
the colum
mn impact thee rotation thatt could occur at the columnn ends.
As
A with the riigid supports discussed forr the columnss in Figure 5..18, these endd conditions
permit th
he column en
nd to rotate. The amountt of this rotaation is someething betweeen the zero
rotation of a fixed support
s
and the
t free rotaation of a pinn support. W
When the collumn under
consideraation is part of a frame where
w
the ennds of the coolumn are noot permitted to displace
laterally relative to eaach other, the frame is calleed a braced fr
frame, a sidessway preventeed frame, or
a sideswa
ay inhibited frame—
f
shown
n as cases a, b, and d in F
Figure 5.18. F
For a column in a braced
frame, th
he possible K-factors
rangee from 0.5 to 1.0. In framees of this typee, K is often taaken as 1.0,
a conserv
vative approx
ximation thatt simplifies ddesign. In facct, Specification Appendixx 7, Section
7.2.3(a) says that in braced
b
framees K shall be taken as 1.00 unless analysis shows thhat a lower
value is appropriate. When the co
olumn under cconsiderationn is in a fram
me in which thhe ends are
permitted
d to move latterally, the fraame is calledd a moment frrame, an unbrraced frame, a sidesway
permitted
d frame, or a sidesway uniinhibited fram
me— shown aas cases c, e, aand f in Figurre 5.18. For
the threee cases shown
n there, the lo
owest value oof K is 1.0. T
The other extrreme case, noot shown in
Figure 5.18, is a pin--ended colum
mn in an unbrraced frame. The effectivve length of tthis column
would th
heoretically bee infinite. Th
hus, K-values for columns in moment fframes range from 1.0 to
infinity.
The
T determinaation of reliab
ble effective llength factorss and thus relliable effectivve lengths is
a critical aspect of collumn design. Several apprroaches are prresented in thhe literature bbut the most
commonly used apprroach is thro
ough the aliggnment chartts presented in the Com
mmentary to
Appendix
x 7. The dev
velopment off these chartss is based onn a set of asssumptions thaat are often
violated in real strucctures; neverttheless, the aalignment chharts are usedd extensivelyy and often
modified
d in an attemp
pt to account for
f variations from these aassumptions.
These
T
assump
ptions, as giveen in the Com
mmentary to A
Appendix 7, aare:
1.
1
2.
2
3.
3
Behav
vior is purely
y elastic.
All members
m
have a constant crross section.
All jo
oints are rigid
d.
148 Chapter 5
Compression Members
4.
5.
6.
7.
8.
9.
10.
For columns in frames with sidesway inhibited, rotations at opposite ends of the
restraining beams are equal in magnitude and opposite in direction, producing
single curvature bending.
For columns in frames with sidesway uninhibited, rotations at opposite ends of
the restraining beams are equal in magnitude and direction, producing reverse
curvature bending.
The stiffness parameter L P /EI of all columns is equal.
Joint restraint is distributed to the column above and below the joint in
proportion to EI/L for the two columns.
All columns buckle simultaneously.
No significant axial compression force exists in the girders.
Shear deformations are neglected
Using these assumptions, the following equation can be obtained for columns in sidesway
inhibited frames.
GAGB
G +G ⎛
π/K ⎞ 2 tan ( π/ 2Κ )
2
−1 = 0
(AISC C-A-7-1)
( π/K ) + ⎛⎜ A B ⎞⎟ ⎜⎜1 −
⎟+
4
2
( π/Κ )
⎝
⎠ ⎝ tan ( π/Κ ) ⎠⎟
For sidesway uninhibited frames, the following equation is obtained.
GAGB ( π/Κ ) − 36
2
6 ( GA + GB )
−
( π/K ) = 0
tan ( π/K )
(AISC C-A-7-2)
In Equations C-A-7-1 and C-A-7-2, the terms GA and GB relate to the relative stiffness of the
columns and beams framing into the column at ends A and B, respectively, as given by
∑ ( EI /L )col
G=
(AISC C-A-7-3)
∑ ( EI /L ) g
If the beams and columns behave elastically, as noted in assumption 1, this reduces to
∑ ( I /L )col
G=
∑ ( I /L ) g
(5.12)
Equations C-A-7-1 and C-A-7-2 are transcendental equations that do not have a closed-form
solution. With the computer methods readily available today, iterative solutions are easily
obtained. However, that was not always the case, and a graphical solution was developed in the
early 1960s that has become a standard approach for obtaining solutions. Such graphical solutions
are called nomographs or alignment charts. Figure 5.20 shows the nomograph for sidesway
inhibited frames, and Figure 5.21 gives the chart for sidesway uninhibited frames.
Approximate solutions to Equations C-A-7-1 and C-A-7-2 have also been presented in
design rules and the literature. The French have used the following equations in their design rules
since 1966. For sidesway inhibited,
3GAGB + 1.4 ( GA + GB ) + 0.64
K=
(5.13)
3GAGB + 2 ( GA + GB ) + 1.28
For sidesway uninhibited,
K=
1.6GAGB + 4 ( G A + GB ) + 7.5
GA + GB + 7.5
(5.14)
Chapter 5
Figure 5.20
5
Alignm
ment Chart forr a
Braced Frame
F
(Sidesw
way Inhibited
d)
Copyright © American Institute of Steeel
Constructtion, Reprinted
d with Permissiion. All
rights reseerved.
Comppression Mem
mbers
149
Figure 5.211 Alignmentt Chart for ann Unbraced
Frame (Sideesway Uninhiibited)
Copyright © American Insttitute of Steel
Construction,, Reprinted witth Permission. All rights
reserved.
These ap
pproximate eq
quations are said to be aaccurate withiin 2 percent. For design this should
easily yieeld results as accurate as th
hose obtainedd by reading a value from tthe alignmentt charts.
For
F the speciial case wheere GA = GB, even simpller equations can be expressed. For
sidesway
y inhibited,
G + 0.4
(5.15)
K=
G + 0.8
For sidessway uninhibiited,
(5.16)
K = 0.8G + 1.0
Equation
ns 5.15 and 5.16 might be particularly
p
uuseful for preliminary desiggn.
EXAMPLE
E 5.6
Column Efffective
Length
Goal:
G
Deetermine the column
c
effecttive length ussing (a) the allignment charrt and (b)
Eq
quation 5.14.
Given:
G
Th
he column AB
B in a momennt frame is shhown in Figuure 5.22. Asssume that
thee column hass its web in thhe plane of thee frame. The beams also hhave their
weebs in the plaane of the fraame and thuss beams and columns aree bending
ab
bout their major, x-axis. It would be verry unusual for a beam in a moment
fraame to have the primary bending mom
ments about other than thhe x-axis.
Ho
owever, colum
mns may be ooriented for bending aboutt either princippal axis.
SOLUTION
N
Part
P
a
Step
S
1:
Deetermine mem
mber propertiees from Manuual Table 1-1.
150 Chapter 5
Compression Members
end A:
W16×36; Igx = 448 in.4
W10×88; Icx = 534 in.4
end B:
W16×77; Igx = 1110 in.4
W10×88; Icx = 534 in.4
Step 2:
Determine the stiffness ratio at each end using Equation 5.12
⎛ 534 ⎞
2⎜
⎟
14 ⎠
GA = ⎝
= 2.04
⎛ 448 ⎞
2⎜
⎟
⎝ 24 ⎠
⎛ 534 ⎞
2⎜
⎟
14 ⎠
GB = ⎝
= 0.825
⎛ 1110 ⎞
2⎜
⎟
⎝ 24 ⎠
Step 3:
Use the alignment chart shown in Figure 5.21 for a sidesway uninhibited
frame. Enter GA and GB on the appropriate scales and construct a straight
line between them, as shown in Figure 5.23. The intersection with the scale
for K gives the effective length factor, in this case,
K = 1.42
Thus,
Part b
Step 4:
Lc = KL = 1.42 (14.0 ) = 19.9 ft
Determine K using the stiffness ratios, GA and GB, determined in part (a)
Step 2 and Equation 5.14.
K=
Thus,
1.6 ( 2.04 )( 0.825) + 4 ( 2.04 + 0.825) + 7.5
2.04 + 0.825 + 7.5
= 1.45
Lc = KL = 1.45 (14.0 ) = 20.3 ft
Note that K determined graphically from the alignment chart and K
calculated with Equation 5.14 are very close, as might be expected.
EXAMPLE 5.7
Column Effective
Length
Goal:
Determine the column effective length for the column of Example 5.6 using
the alignment chart if the column is bending about its weak axis.
Given:
The column AB in a moment frame is shown in Figure 5.22. However, for
this example assume that the column has its web perpendicular to the plane
of the frame, thus it is bending about its minor or weak axis.
SOLUTION
Step 1:
Find member properties from Manual Table 1-1.
end A:
W16×36; Igx = 448 in.4
W10×88; Icy = 179 in.4
Chapter 5
en
nd B:
Comppression Mem
mbers
151
W116×77; Igx = 1110 in.4
W
W10×88; Icy = 179 in.4
Step
S
2:
Deetermine the stiffness
s
ratioo at each end uusing Equatioon 5.12.
⎛ 179 ⎞
2⎜
⎟
14 ⎠
GA = ⎝
= 0.685
⎛ 448 ⎞
2⎜
⎟
⎝ 24 ⎠
⎛ 179 ⎞
2⎜
⎟
14 ⎠
GB = ⎝
= 0.276
⎛ 1110 ⎞
2⎜
⎟
⎝ 24 ⎠
Step
S
3:
Usse the alignm
ment chart shoown in Figurre 5.21 for a sidesway unninhibited
fraame. Enter th
he values for G A and GB onn the appropriate scales annd draw a
strraight line beetween them. The line’s intersection with the scaale for K
giv
ves the effecttive length facctor—in this case,
K = 1.16
Th
hus,
Step
S
4:
5
Figure 5.22
Lc = K
KL = 1.16 (144.0 ) = 16.2 ft
No
ote that the reeduction in m
moment of inerrtia of the collumns results in the
beeams providin
ng more end rrestraint, reduucing the effecctive length ffactor for
thee column and
d thus reducinng the columnn effective lenngth.
Multi-sstory Frame fo
or Examples 55.6 and 5.7
1552 Chapter 5
Compresssion Memberss
Figure 5.23 Alignment Chart for
Example 5.66
Copyright © A
American Instiitute of Steel
Construction, Inc. Reprintedd with Permissiion.
All rights reseerved.
5.55.1
Effectiv
ve Length forr Inelastic Co
olumns
The assump
ption of elastiic behavior for
f all membeers of a fram
me is regularlly violated. W
We have
already seen
n the role thatt residual streesses play in determining column strenngth through inelastic
behavior. Th
hus, it is useeful to accom
mmodate this inelastic behhavior in the determinatioon of Kfactors. Thee assumption
n of elastic behavior is important iin the calcuulation of G as the
simplificatio
on is made to
o move from Equation C-A
A-7-3 to Equuation 5.12. R
Returning to E
Equation
C-A-7-3 an
nd assuming that all collumns framinng into a jooint have thee same moddulus of
elasticity—w
which is equaal to the tangeent modulus, ET shown inn Figure 5.12—
—and that thhe beams
behave elasttically, the deefinition of G for inelastic bbehavior becoomes
E (∑( I /L)c )
Ginnelastic = T
(5.17)
E (∑( I /L) g )
If G for elasstic behavior is
i taken as Gelastic
, then Gineelastic can be foormulated as
e
⎛E ⎞
Ginelastic = ⎜ T ⎟ Gelastic
⎝ E ⎠
(5.18)
Thus, includ
ding inelasticc column beh
havior simplyy results in a modificationn of G. The ratio of
tangent mod
dulus to elasttic modulus iss always lesss than 1, so thhe assumptioon of elastic bbehavior
for this application leadss to a conserv
vative estimaate, as can bee seen by entering the nom
mograph
with lower G-values an
nd determinin
ng the correesponding K--factor. Before a straighttforward
approach to including in
nelastic effects in the deterrmination of effective length can be prroposed,
the relationsship between the tangent modulus
m
and thhe elastic moodulus must be established.
The Commentary
y to Appendix
x 7 of the Speecification inndicates that τb = ET /E, as given in
Chapter C for
fo the direct analysis
a
meth
hod, should bee used to account for coluumn inelasticity in the
effective len
ngth method. Thus, if αPr /Pns ≤ 0.5
τb = 1.0
(AISC
C C2-2a)
Chapter 5
Compression Members 153
and if αPr /Pns > 0.5
⎛ αP
τb = 4 ⎜ r
⎝ Pns
⎞ ⎡ ⎛ αPr
⎟ ⎢1 − ⎜ P
⎠ ⎣ ⎝ ns
⎞⎤
⎟⎥
⎠⎦
(AISC C2-2b)
where α = 1.0 for LRFD and α = 1.6 for ASD. Pns is the cross section compression strength. For
members without slender elements Pns = Py. For compression members with slender elements, Pns
= FyAe which is addressed in the section 5.6. Manual Table 4-13 provides values for τb based on
the required strength, Pu/Ag and Pa/Ag. The use of Table 4-13 assumes that the column is loaded to
its full available strength. If it is not, the table provides a conservative assessment of the inelastic
stiffness reduction factor and the effective length.
EXAMPLE 5.8
Inelastic Column
Effective Length
Goal:
Determine the inelastic column effective length using the alignment chart.
Given:
Determine the inelastic effective length for the column in Example 5.6. The
column has an LRFD required strength of Pu = 950 kips and an ASD
required strength of Pa = 633 kips. Use Equation 5.14 in place of the
alignment chart. The column is A992 steel.
SOLUTION
Step 1:
From Manual Table 1-1, for a W10×88 A = 26.0 in.2, and from Example
5.6, the elastic stiffness ratios are GA = 2.04 and GB = 0.825.
For
LRFD
Step 2:
Determine the required stress based on the required strength.
Pu 950
=
= 36.5 ksi
A 26.0
Step 3:
Determine the stiffness reduction factor from Manual Table 4-13,
interpolating between 36 and 37 ksi.
τb = 0.788
Step 4:
Determine the inelastic stiffness ratios by multiplying the elastic stiffness
ratios by the stiffness reduction factor.
GiA = 0.788 ( 2.04 ) = 1.61
GiB = 0.788 ( 0.825) = 0.650
Step 5:
Determine K from Equation 5.14.
1.6 (1.61)( 0.650 ) + 4 (1.61 + 0.650 ) + 7.5
K=
= 1.37
1.61 + 0.650 + 7.5
Thus,
Lc = KL = 1.37 (14.0 ) = 19.2 ft
Note that the effective length factor and thus the effective length is less
than that determined in Example 5.6, as expected.
For
ASD
Step 2:
Determine the required stress based on the required strength.
Pa 633
=
= 24.3 ksi
A 26.0
1554 Chapter 5
Compresssion Memberss
Step
p 3:
Deterrmine the stiffness reduuction factorr from Mannual Table 4-13,
interp
polating betweeen 24 and 255 ksi.
τb = 0.691
Step
p 4:
Deterrmine the ineelastic stiffnesss ratios by m
multiplying thhe elastic stifffness
ratioss by the stiffneess reductionn factor.
GiA = 00.691( 2.04 ) = 1.41
GiB = 00.691( 0.825 ) = 0.570
Deterrmine K from Equation 5.114.
1 (1.41)( 0.5770 ) + 4 (1.41 + 0.570 ) + 7.55
1.6
K=
= 1.33
1.441 + 0.570 + 77.5
Thus,,
Lc = KL = 1.33 (14.0 ) = 18.6 ft
Step
p 5:
Note that the effeective length factor and thhus the effecctive length iis less
than that
t determineed in Examplle 5.6, as expeected.
5.55.2 Effective Length when
n Supporting
g Gravity On
nly Columnss
Another con
ndition that influences
i
co
olumn bucklinng and thus the effectivee length factoor is the
existence off columns th
hat carry only
y gravity loaad and contrribute nothingg to the lateeral load
resistance or
o stability off the structurre. Figure 5.224a illustratees a simple sstructure of tthis type
where the sttability or lateral load resiisting columnn is the flagpoole column on the left, column A,
and the gravity only collumn is the pin
p ended coolumn on thee right, colum
mn B. The looad P is
applied to co
olumn A and the load Q is applied to coolumn B.
P
Q
L
P
B
P
A
B
A
P+Q
Q
M = PΔ + Q Δ
(b))
olumn Providiing Lateral Reestraint for a Gravity Onlyy Column
Flagpole Co
(a))
P+Q
L
P
Q
Δ
Q
QΔ
L
L
A
Figgure 5.24
Δ
M = PΔ + Q Δ
(c)
Chapter 5
Compression Members 155
If Q = 0, column A behaves as if column B did not exist since it just goes along for the
ride. However, when Q is not zero, buckling of column A leads to lateral displacement, Δ, at the
top of columns A and B. Thus, equilibrium requires a lateral force be exerted at the top of column
B. This force must be resisted by column A as shown in Figure 5.24b. In the displaced position
illustrated in Figure 5.24b, equilibrium of column A requires a resisting moment at the support of
M = PΔ + QΔ . Figure 5.24c shows column A with an applied load, ( P + Q ) , which in the
displaced position produces a moment at the support of M = PΔ + QΔ . Thus, the column in
Figure 5.24c can be thought of as a representation of column A in Figure 5.24b with only slight
error.
Since these two columns are considered equal, if column A can support the load ( P + Q ) ,
it should be adequate for column A to support its load, P, and the effect of the load Q on column
B. Considering elastic buckling this can be stated as
π2 EI
P
+
Q
=
(5.19)
(
)
2
( Ko L )
Where Ko is the K-factor for column A. For this example, the theoretical K-factor for the flagpole
column is Ko = 2. Another way to approach this problem would be to continue to consider that
column A supports only the load P but use a K-factor that accounts for the influence of the load
on the gravity only column, Kn. This can be stated as
π2 EI
P=
(5.20)
2
( Kn L )
Since these two equations represent the same structure, they can be solved for π 2 EI L2 . Thus,
from Equation 5.19 π2 EI L2 = K o2 ( P + Q ) and from Equation 5.20 π 2 EI L2 = K n2 P . Setting
these equal and solving for Kn yields
P+Q
Q
Kn = Ko
= Ko 1 +
(5.21)
P
P
Thus, a column that supports a load P and also must provide stability for load Q on gravity only
columns may be designed using this modified effective length factor Kn.
EXAMPLE 5.9
Gravity Only
Columns and
Effective Length
SOLUTION
Goal:
Determine the in-plane nominal strength of the column that is required to
carry a concentrated load and provide lateral stability for gravity only
columns. Also, determine the strength of the column if there is no load on
the gravity only column.
Given:
A W14×90 column shown in Figure 5.25 as Column A is to a) carry an
applied load, P, and provide lateral restraint to a gravity only column,
Column B, carrying the load Q = 2P and b) carry an applied load, P, with
no load on the gravity only column. The W14×90 is oriented so the web is
in the plane of the frame. Use A992 steel.
Step 1:
Determine the effective length factor for the W14×90 column without
considering the gravity only column. Since this is a flagpole column, from
Figure 5.7f, the theoretical K-factor is 2.0
156 Chapter 5
Compression Members
Part a
Step 2:
Using Equation 5.21, determine the modified effective length factor to
account for the gravity only column load, Q = 2P. Thus
Q
2P
K n = K o 1 + = 2.0 1 +
= 3.46
P
P
Step 3:
From Manual Table 1-1
Step 4:
Determine which column strength equation to use.
Since
K n L 3.46 (15 (12 ) )
29,000
=
= 101 < 4.71
= 113
rx
6.14
50
A = 26.5 in.2 and rx = 6.14
use Equation E3-2
Step 5:
Determine the Euler buckling stress
π2 ( 29,000 )
= 28.1 ksi
Fe =
2
(101)
Step 6:
Determine the critical stress from Equation E3-2
⎛ Fy ⎞
⎛ 50 ⎞
⎛
⎛
⎜ ⎟⎞
⎜
⎟⎞
Fcr = ⎜ 0.658⎝ Fe ⎠ ⎟ Fy = ⎜ 0.658⎝ 28.1 ⎠ ⎟ 50 = 23.7 ksi
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
Step 7:
Determine the nominal strength
Pn = 23.7 ( 26.5 ) = 628 kips
Part b
Step 8:
With no load on the gravity only column, Kn = Ko = 2.0. Determine which
column strength equation to use.
Since
29,000
K n L 2.0 (15 (12 ) )
=
= 58.6 < 4.71
= 113
rx
6.14
50
use Equation E3-2
Step 9:
Determine the Euler buckling stress
π2 ( 29,000 )
= 83.3 ksi
Fe =
2
( 58.6)
Step 10:
Determine the critical stress from Equation E3-2
⎛ Fy ⎞
⎛ 50 ⎞
⎛
⎛
⎜ ⎟⎞
⎜
⎟⎞
Fcr = ⎜ 0.658⎝ Fe ⎠ ⎟ Fy = ⎜ 0.658⎝ 83.3 ⎠ ⎟ 50 = 38.9 ksi
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
Step 11:
Determine the nominal strength
Pn = 38.9 ( 26.5 ) = 1030 kips
Chapter 5
P
Comppression Mem
mbers
157
Q
Fiigure 5.25
Single Storyy Frame
forr Example 5.99
5.6
SLEN
NDER ELEMENTS IN
N COMPRE
ESSION
As mentioned in Section 5.4, the columns dis cussed thus ffar are controolled by overrall column
buckling. For some sh
hapes, anotheer form of bucckling may acctually controol column streength: local
buckling of the elemeents that makee up the colum
mn shape. W
Whether the shhape is rolled or built up,
it can bee thought of as
a being comp
posed of a grroup of intercconnected plaates. Dependiing on how
these plaates are suppo
orted by each other, they ccould buckle aat a stress bellow the criticcal buckling
stress off the overall column.
c
This is local buckkling, also caalled plate buuckling, and iis shown in
Figure 5.26. Local bu
uckling is desscribed througgh a plate crittical bucklingg equation sim
milar to the
Euler bucckling equatio
on for column
ns. The criticaal buckling sttress for an axxially loaded plate is
k π2 E
(5.22)
Fcr =
2
2 ⎛b⎞
112 (1 − ν ) ⎜ ⎟
⎝t⎠
where k is a plate bu
uckling coeffiicient that deepends on thee plate loadinng, edge condditions, and
length-to
o-width ratio; ν is Poisson
n’s ratio; and b/t is the rattio of the widdth perpendiccular to the
compresssion force to the thickness of the platee. The width--to-thickness ratio is calleed the plate
slenderneess and is sim
milar in functtion to the coolumn slenderrness. This crritical stress pplotted as a
function of width-to-th
hickness ratio
o is shown as the dashed cuurve in Figurre 5.27.
Figure 5.2
26
Column
n Tested to Failure
through Lo
ocal Buckling
g
Photo courttesy Perry Greeen
Figure 5.277 Plate Streength in Comppression
158 Chapter 5
Compression Members
As with overall column buckling, an inelastic transition exists between elastic buckling
and element yielding. This transition is due to the existence of residual stresses and imperfections
in the element, just as in the case of overall column buckling, and results in the inelastic portion
of the curve shown in Figure 5.27. In addition, for plates with low b/t ratios, strain hardening
plays a critical role in their behavior, and plates with large b/t ratios have significant postbuckling strength.
To ensure that local buckling will not control column strength, the critical plate buckling
stress for local buckling is limited to the critical buckling stress for overall column buckling. This
approach results in a different maximum plate slenderness value for each corresponding column
slenderness value. However, to alert the designer to the need for consideration of plate buckling,
an initial check on element slenderness is made assuming that the stress in the plate has reached
the yield stress. The development of the Specification provisions starts by finding a plate
slenderness that sets the plate buckling stress equal to the column yield stress. Equation 5.22 then
becomes
b
k π2 E
=
t
12 (1 − ν 2 ) Fy
Taking ν = 0.3, the standard value for steel, this plate slenderness becomes
b
kE
= 0.95
t
Fy
(5.23)
(5.24)
which is shown as point O in Figure 5.27. This point is well above the inelastic buckling curve. In
order to obtain a b/t that would bring the inelastic buckling stress closer to the yield stress, a
somewhat arbitrary slenderness limit is taken as 0.7 times the limit that corresponds to the column
yield stress, which gives
b
kE
= 0.665
t
Fy
This is indicated as point D in Figure 5.27.
The remaining factor to be determined is the plate buckling coefficient, k. This factor is a
function of the stress distribution, the edge support conditions and the aspect ratio of the plate.
For a plate with uniform compression on opposite ends and simply supported on all four sides,
the minimum value of k can be shown to be 4.0. For actual columns, with the variety of potential
cross section shapes available, the determination of the value of k becomes much more
complicated as the actual edge supports, stress distribution and aspect ratio vary.
The limiting width-to-thickness ratios are given in Specification Table B4.1a. These
limits may be given as
E
λ r = c3
(5.25)
Fy
where c3 is given in Table 5.4 for several elements in uniform compression taken from Table
B4.1a of the Specification. The apparent plate buckling coefficient, k, used to obtain these values
is also given in Table 5.4. For shapes with element slenderness exceeding these λr values plate
buckling must be considered. As already shown, these limits are based on the assumption that the
column is stressed to Fy. Since columns are rarely stressed to that level, it is very possible that
what appears to be a slender element compression member based on Table B4.1a may not
actually see its strength limited by the limit state of local buckling.
Compression Members 159
Chapter 5
Table 5.4
Parameters for Consideration of Compression Member Local Buckling
*
Case
λ
c3
k
c4
c5
1
b/t
0.56
0.71
0.834
0.184
b/t
0.45
0.46
0.671
0.148
d/t
h/tw
0.75
1.49
1.27
5.0
1.12
1.95
0.246
0.351
b/t
1.40
4.43
1.93
0.386
Flanges of rolled I-shaped sections,
plates projecting from rolled Ishaped sections; outstanding legs of
pairs of angles connected with
continuous contact, flanges of
channels, and flanges of tees
3
Legs of single angles, legs of double
angles with separators, and all other
unstiffened elements
4
Stems of tees
5
Webs of doubly symmetric I-shaped
sections and channels
6
Walls of rectangular HSS and boxes of
uniform thickness
* From Table B4.1a in the Specification.
For W-shapes, Case 1 in Table B4.1a, with Fy = 50 ksi, the flange slenderness limit is
λ rf = 0.56 E Fy = 13.5 , and all W-shapes have a flange slenderness less than this limit. For
webs of these W-shapes, Case 5 in Table B4.1a, λ rw = 1.49 E Fy = 35.9 , and many available
W-shapes have a web slenderness that exceeds this limit and are classified as slender.
Design of slender element compression members according to the Specification follows
the same requirements as those for compression members without slender elements, with one
modification. To account for slender element behavior, the full area of the slender element cannot
be used. Thus, a reduced effective area, Ae, is used in place of the gross area, Ag, to determine
column strength. For columns with slender elements, Section E7 indicates that column strength is
given by
Pn = Fcr Ae
(AISC E7-1)
which is to be used in place of Equation E3-1 but with the same critical stress, Fcr.
Once the designer is directed to the slender element provisions of Section E7 it is
apparent that the critical stress based on the controlling limit state must first be determined. Then,
using that stress, the actual plate element slenderness at the transition from elastic to inelastic
buckling can be determined. For the web of a rolled W-shape, Case 5 in Table B4.1a, the limiting
width-to-thickness ratio becomes
h
E
E
λ r = = c3
= 1.49
tw
Fcr
Fcr
If the width-to-thickness ratio of the web does not exceed this value, the usable width of the web
is the actual width and no change in area is required. If the width-to-thickness ratio of the web
does exceed this value, a reduced width must be determined through
⎛
F ⎞ F
be = b ⎜⎜1 − c1 el ⎟⎟ el
(AISC E7-3)
Fcr ⎠ Fcr
⎝
where the elastic plate buckling stress from Equation 5.22 is presented as
160 Chapter 5
Compression Members
2
⎛ λ ⎞
Fel = ⎜ c2 r ⎟ Fy
(AISC E7-4)
⎝ λ ⎠
The constants c1 and c2 are given in Specification Table E7-1. If Equations 5.25 and E7-4 are
substituted into Equation E7-3, the effective width becomes
E ⎛ c1c2 c3 E ⎞
be = c2 c3t
(5.26)
⎜1 −
⎟
Fcr ⎜⎝ ( b t ) Fcr ⎟⎠
Equation 5.26 may be simplified by substituting c4 = c2c3 and c5 = c1c2c3 which yields
be = c4t
E
Fcr
⎛
c5
E ⎞
⎜⎜1 −
⎟⎟
⎝ ( b t ) Fcr ⎠
(5.27)
Values for c4 and c5 are given in Table 5.4.
Continuing the consideration of W-shape webs by making the appropriate substitutions
into Equation 5.27 yields
E ⎡ 0.351 E ⎤
be = 1.95t
(5.28)
⎢1 −
⎥
Fcr ⎣ (b /t ) Fcr ⎦
Once the effective width of a slender element is obtained, the corresponding effective area of the
member can be determined. Since hot-rolled shapes have fillets at the junction of the plate
elements, the best approach for determining the effective area is to use the gross area and deduct
the appropriate ineffective element area. This will be illustrated in the example.
EXAMPLE 5.10
Strength of
Column with
Slender Elements
SOLUTION
Goal:
Determine the available strength of a compression member with a slender
web.
Given:
Use a W16×26 as a column with Lcy = 5.0 ft.
Note:
In Manual Table 1-1, this shape is identified with footnote c indicating that
it must be considered as a slender element member for compression. It is
the most slender-web W-shape available and is not normally used as a
column.
Step 1:
From Manual Table 1-1,
A = 7.68 in.2, h/tw = 56.8, tw = 0.250 in., ry = 1.12 in.
Step 2:
Determine the web slenderness limit from Specification Table B4.1a, Case
5.
E
29,000
λ rw = 1.49
= 1.49
= 35.9
Fy
50
Step 3:
Check the slenderness of the web.
h
= 56.8 > λ rw = 35.9
tw
Thus, the shape must be treated as one with a slender web. It has already
been established that all W-shapes have nonslender flanges, so that check
Chapter 5
Compression Members 161
will not be made here.
Step 4:
Determine the Euler buckling stress, Fe, for Lc = 5.0 ft.
π 2 (29,000)
Fe =
= 99.7 ksi
2
⎛ 5 (12 ) ⎞
⎜
⎟
⎝ 1.12 ⎠
Step 5:
Determine Fcr
Fe = 99.7 ksi > Fy 2.25 = 50 2.25 = 22.2 ksi
Therefore use Equation E3-2
⎛ 50 ⎞
⎜
⎟
Fcr = 0.658⎝ 99.7 ⎠ ( 50 ) = 40.5 ksi
Step 6:
Check the slenderness of the web against the new limit using Fcr in place of
Fy.
E
29,000
λ rw = 1.49
= 1.49
= 39.9
40.5
Fcr
since
h
= 56.8 > λ rw = 39.9
tw
the effective web width must be determined.
Step 7:
Determine the effective width of the web using Equation 5.28
E ⎡ 0.351 E ⎤
be = 1.95t
⎢1 −
⎥
Fcr ⎣ (b /t ) Fcr ⎦
= 1.95(0.250)
Step 8:
29,000
40.5
⎡ 0.351 29,000 ⎤
⎢1 −
⎥ = 10.9 in.
56.8
40.5 ⎦
⎣
Determine the actual web width.
The width of the web plate is given by h. However, a value of h is not
specifically available in the Manual, so with h/tw = 56.8 and tw = 0.250, h
can be determined as
h = ( h tw ) tw = 56.8 ( 0.250 ) = 14.2 in.
Step 9:
Determine the effective area.
Because be < h, use be to determine Ae. To properly account for the fillets at
the web-flange junction, the area of the ineffective web is deducted from
the gross area of the shape; thus,
Ae = Ag − ( h − be ) tw = 7.68 − (14.2 − 10.9 )( 0.250 ) = 6.86 in.2
Step 10:
Determine the nominal strength of the column.
Pn = Fcr Ae = 40.5 ( 6.86) = 278 kips
162 Chapter 5
Compression Members
For
LRFD
Step 11:
For
ASD
Step 11:
EXAMPLE 5.11
Strength of
Column with
Slender Elements
SOLUTION
Determine the design strength for this slender web column with Lc = 5.0 ft.
φPn = 0.9 ( 278) =250 kips
Determine the design strength for this slender web column with Lc = 5.0 ft.
Pn Ω = 278 1.67 =166 kips
Goal:
Determine the available strength of a compression member with a slender
web.
Given:
Use the W16×26 column from Example 5.10 but with Lcy = 15.0 ft.
Note:
This shape has already been shown to have a slender web based on Table
B4.1a
Step 1:
From Manual Table 1-1,
A = 7.68 in.2, h/tw = 56.8, tw = 0.250 in., ry = 1.12 in.
Step 2:
Determine the web slenderness limit from Specification Table B4.1a, Case
5.
E
29.000
λ rw = 1.49
= 1.49
= 35.9
Fy
50
Step 3:
Check the slenderness of the web.
h
= 56.8 > λ rw = 35.9
tw
Thus, the shape has a slender web. It has already been established that all
W-shapes have nonslender flanges.
Step 4:
Determine the Euler buckling stress, Fe, for Lc = 15.0 ft.
π 2 (29,000)
Fe =
= 11.1 ksi
2
⎛ 15 (12 ) ⎞
⎜
⎟
⎝ 1.12 ⎠
Step 5:
Determine Fcr .
Fe = 11.1 ksi < Fy 2.25 = 50 2.25 = 22.2 ksi
Therefore use Equation E3-3
Fcr = 0.877 (11.1) = 9.73 ksi
Step 6:
Check the slenderness of the web against the new limit using Fcr in place of
Fy.
Chapter 5
λ rw = 1.49
Compression Members 163
E
29,000
= 1.49
= 81.3
9.73
Fcr
since
h
= 56.8 < λ rw = 81.3
tw
the column will not be limited by local buckling of the web
Step 7:
For
LRFD
Step 8:
For
ASD
Step 8:
Determine the nominal strength of the column.
Pn = Fcr Ag = 9.73( 7.68) = 74.7 kips
Determine the design strength for this slender web column with Lc = 15.0
ft.
φPn = 0.9 ( 74.7 ) =67.2 kips
Determine the design strength for this slender web column with Lc = 15.0
ft.
Pn Ω = 74.7 1.67 =44.7 kips
Examples 5.10 and 5.11 illustrate that a W-shape that appears to be a slender element shape based
on Table B4.1a may not actually be limited in strength because of that slender element, the limit
state of local buckling. It can be shown that the W16×26 considered in these examples will have
its strength limited by local buckling for columns with effective length, Lc, up to about 10.8 ft.
Above that effective length, the slender elements will not impact overall column strength.
5.7
COLUMN DESIGN TABLES
A review of the AISC column equations, E3-2 and E3-3, shows that the only factor other than
shape geometry and material strength that influences the determination of column strength is the
slenderness ratio. Therefore, it is convenient to tabulate column strength as a function of
slenderness. Part 4 of the Manual contains tables for W-shapes, HP-shapes, and HSS and several
singly symmetric shapes. Figure 5.28 shows a sample of Manual Table 4-1a for several W14
sections with Fy = 50 ksi. As with all of the available strength tables in the Manual, both
allowable strength (ASD) and design strength (LRFD) values are given. Tables 4-1b and 4-1c are
provided for selected W-shapes that are commonly available with Fy = 65 and 70 ksi respectively.
The values in these column tables are based on the assumption that the column will
buckle about its weak axis. For all W-shapes this is the y-axis, so the values in the tables are
given in terms of the effective length with respect to the least radius of gyration, ry. Their use is
quite straightforward when the critical buckling length is about this axis. An approach that
permits the use of these tables when the strong axis controls will be addressed following the
example.
1664 Chapter 5
Compresssion Memberss
A
Compresssion
Figure 5.28 Available Strength in Axial
Copyright © American Insttitute of Steel Construction,
C
R
Reprinted with Permission. A
All rights reservved.
Chapter 5
Comppression Mem
mbers
165
Figurre 5.29 Collumns for Exaamples
5.12 and 5.13
EXAMPLE
E 5.12a
Column Deesign by
LRFD
Goal:
G
Deetermine the least-weightt section to carry the looads given uusing the
lim
mited selectio
on available inn Figure 5.28.
Given:
G
Th
he column is shown in Figgure 5.29a. Itt must resist tthe followingg loads in
thee appropriatee combinationns: PD = 56 kkips, PL = 1772 kips, and PW = 176
kip
ps. Use A992
2 steel. Assum
me the live looad comes frrom a distribuuted load
lesss than 100 psf, so that thee LRFD load factor on livee load may bee taken as
0.5
5 for load com
mbination 4.
SOLUTION
N
Step
S
1:
Deetermine thee maximum required sstrength usinng the LRF
FD load
co
ombinations from
fr
Section 22.4.
1.
1.4PD = 1.4(56) = 788.4 kips
2.
1.2PD + 1.6PL = 1.2((56) + 1.6(172) = 342 kipss
4.
1.2PD + 0.5PL + 1.0P W = 1.2(56) + 0.5(172) + 1.0(176) = 3229 kips
6.
0.9PD + 1.0PW = 0.99 (56) + 1.0(176) = 226 kipps
So
o the column must
m carry Pu = 342 kips.
Step
S
2:
Th
he column haas the same efffective lengtth about the x- and y-axes,, so enter
thee table in Fig
gure 5.28 withh Lc = 18 ft. S
Scanning acrooss the table aat Lc = 18
ft and checkin
ng the LRFD
D values, seleect the least--weight shape in this
po
ortion of the taable that can support this lload.
Seelect a W14×6
61 with a desiign compresssion strength
φPn = 456 kips
166 Chapter 5
Compression Members
EXAMPLE 5.12b
Column Design by
ASD
Goal:
Determine the least-weight section to carry the loads given using the
limited selection available in Figure 5.28.
Given:
The column is shown in Figure 5.29a. It must resist the following loads in
the appropriate combinations: PD = 56 kips, PL = 172 kips, and PW = 176
kips. Use A992 steel.
SOLUTION
Step 1:
Determine the maximum required strength using the ASD load
combinations from Section 2.4.
1. PD = 56 kips
2. PD + PL = 56 + 172 = 228 kips
5. PD + 0.60PW = 56 + 0.6(176) = 162 kips
6. PD + 0.75 PL + 0.75(0.6PW )
= 56 + 0.75(172) +0.75(0.6(176)) = 264 kips
7. 0.6 PD +0.6 PW = 0.6(56) + 0.6(176) = 139 kips
So the column must carry Pa = 264 kips.
Step 2:
The column has the same effective length about the x- and y-axes, so enter
the table in Figure 5.28 with Lc = 18 ft. Scanning across the table at Lc = 18
ft and checking the ASD values, select the least-weight shape in this
portion of the table that can support this load
Select a W14×61 with an allowable compression strength
Pn Ω = 304 kips
If the largest slenderness ratio for a particular column happens to be for x-axis buckling,
the tables may not be entered directly with the x-axis effective length because the table effective
length is intended to be used in conjunction with the least radius of gyration. However, it is
possible to determine a modified effective length that, when used in the table, will result in the
correct column strength.
When the x-axis controls column strength, the slenderness ratio used in the column
equations is Lcx/rx. To use the column tables, an effective length, (Lc)eff, must be determined that,
when combined with ry, gives the same slenderness ratio. So
( Lc )eff Lcx
=
ry
rx
Solving this equation for (Lc)eff yields
Lcx
( Lc )eff =
(rx / ry )
With this modified effective length, the tables can be entered and a suitable column selected.
There is one difficulty with this process, however. Until a column section is known, the value for
rx/ry cannot be determined. To account for this, a quick scan of the column tables should be done
to estimate rx/ry. Then, when a section is selected, the assumption can be verified and an
adjustment made if necessary.
Chapter 5
EXAMPLE 5.13
Column Design
SOLUTION
Compression Members 167
Goal:
Determine the least-weight section to carry the force given using the
limited selection available through Figure 5.28. Design by LRFD and
ASD.
Given:
The column is shown in Figure 5.29b. Use the loading from Example
5.12.
Step 1:
Determine the effective length for each axis.
Bracing of the y-axis, shown in Figure 5.29b, yields Lcy = 10.0 ft. The
unbraced x-axis has Lcx = 30.0 ft.
Step 2:
Determine (Lc)eff for the x-axis.
Select a representative rx/ry from Figure 5.28. There are two general
possibilities. Assume that the larger shapes might be needed to carry the
load, and try rx/ry = 2.44. Thus,
L
30.0
( Lc )eff = cx =
= 12.3 ft
(rx /ry ) 2.44
Step 3:
Determine the controlling effective length.
Because (Lc)eff = 12.3 ft is greater than Lcy = 10.0 ft, enter the table with Lc
= 12.3 ft and interpolate between 12 ft and 13 ft.
For
LRFD
Step 4:
From Example 5.12a the column must have a design strength greater than
Pu = 342 kips with Lc = 12.3 ft. Try a W14×43, which happens to be the
smallest column available with the limited selection available in Figure
5.28. This column has rx/ry = 3.08.
Step 5:
Determine (Lc)eff with this new rx/ry. Thus,
30.0
= 9.74 ft
( Lc )eff =
3.08
Step 6:
Determine the new controlling effective length.
Because (Lc)eff = 9.74 ft is now less than Lcy = 10.0 ft, enter the table with
10.0 ft and note that the W14×43 has a design strength of 422 kips, which
is greater than the required strength of 342 kips.
Step 7:
Therefore, use the selected
W14×43
Note: The W14×43 is identified in the table by a footnote as slender for
Fy = 50 ksi. This is not an issue for our design because the impact of any
slender element has already been taken into account in generating the
table as stated in the same footnote.
Using the full complement of tables available in the Manual results in a
smaller W12 section having the ability to carry the given load.
168 Chapter 5
Compression Members
For
ASD
Step 4:
From Example 5.12b the column must have an allowable strength greater
than Pa = 264 kips with Lc = 12.3 ft. Try a W14×48. This column has rx/ry
= 3.06.
Step 5:
Determine (Lc)eff with this new rx/ry. Thus,
30.0
= 9.80 ft
( Lc )eff =
3.06
Step 6:
Determine the new controlling effective length.
Because (Lc)eff = 9.80 ft is now less than Lcy = 10.0 ft, enter the table with
10.0 ft and see that the W14×43 has an allowable strength of 281 kips,
which is greater than the required strength of 268 kips and rx/ry = 3.08
which is greater than that for the W14×48 so Lcy = 10.0 ft will still
control.
Step 7:
Therefore, use the selected
W14×43
Note: The W14×43 is identified in the table by a footnote as slender for
Fy = 50 ksi. This is not an issue for our design because the impact of any
slender element has already been taken into account in generating the
table as stated in the same footnote.
Using the full complement of tables available in the Manual results in a
smaller W12 section not having the ability to carry the given load.
Table 4-1a in Part 4 of the Manual includes shapes from a W8×31 up to a W14×873. All
of the shapes included are considered column shapes and have reasonably similar strengths about
the x- and y-axes. That is, the shapes are close to being square and rx/ry is not extremely large,
ranging from 1.59 to 3.08. Any of the other available W-shapes may be used for columns if
desired, but it must be recognized that the relationship between the x- and y-axes is such that the
y-axis will control unless significant bracing is provided. These shapes are generally considered
beam shapes. Since beams are intended to be used to carry flexure, the relationship between the xand y-axes is not as critical. For example, for a W16×26 with a length of 24 ft braced at the ends
only for the x-axis and rx/ry = 5.59, the y-axis will control unless it is braced at least every 4.29 ft.
The W-shape column tables in Part 4 of the Manual for Fy = 50 ksi also exclude the
smallest W-shapes in an attempt to direct the design engineer toward using shapes that are more
appropriate when considering connections. That does not mean that these smaller shapes are not
acceptable for use as columns. The tables in Part 6 of the Manual, which will be discussed in
Chapter 8, can be used for the design of columns and they include all of the W-shapes.
Chapter 5
EXAMPLE 5.14
Column Design
SOLUTION
Compression Members 169
Goal:
Determine the least-weight section to carry the force given using the
small shapes provided in the column tables in Manual Part 6. Design by
LRFD and ASD.
Given:
The column has an effective length for both axes of 10 ft and must carry a
concentrated dead load of 8 kips and a concentrated live load of 24 kips.
For
LRFD
Step 1:
Step 2:
Determine the required strength for the load combination 1.2D + 1.6L.
Pu = 1.2(8.0) + 1.6(24.0) = 48.0 kips
Using Manual Table 6-2, select the lightest column to support this load.
Select the W4×13.
φPn = 60.1 kips
For
ASD
Step 1:
Step 2:
Determine the required strength for the load combination D + L.
Pa = 8.0 + 24.0 = 32.0 kips
Using Manual Table 6-2, select the lightest column to support this load.
Select the W4×13.
Pn
= 40.0 kips
Ω
5.8 TORSIONAL BUCKLING AND FLEXURAL-TORSIONAL BUCKLING
Up to this point, the discussion has addressed the limit states of flexural buckling and local
buckling. Two additional limit states for column behavior must be addressed: torsional buckling
and flexural-torsional buckling. Doubly symmetric shapes normally fail through flexural
buckling, as discussed earlier in this chapter, or through torsional buckling. Singly symmetric and
nonsymmetric shapes can fail through flexural, torsional, or flexural-torsional buckling. Because
the shapes normally used for steel members are not well suited to resist torsion, except for closed
HSS, it is usually desirable to avoid any torsional limit states through proper bracing of the
column or by avoiding torsional loading.
If either of the torsional limit states must be evaluated, the applicable Specification
provisions are found in Section E4, except for the special cases associated with single angles,
which are found in Section E5. All members addressed in Section E4 are assumed to buckle
torsionally about their shear center. For these shapes an elastic buckling stress, Fe, is determined,
which is then used in Equations E3-2 and E3-3 to determine the column critical stress, Fcr. The
equations given in Section E4 defining the elastic buckling stress for the limit states of torsional
and flexural-torsional buckling are also found in several other books with varying notation,
including Buckling Strength of Metal Structures. 1 The equations of Section E3 are used to
account for such factors as inelastic buckling, initial out-of-straightness, and residual stresses.
1
Bleich, F. Buckling Strength of Metal Structures. New York: McGraw-Hill, 1952.
1770 Chapter 5
Compresssion Memberss
F
Figure 5.30
Column for Example 5.15
The provisions for single an
ngles meetinng a specificc set of criteeria take a ddifferent
approach. By
B limiting th
he way that load
l
is applieed to the endds of a singlle-angle comppression
member, an effective slen
nderness is esstablished, whhich is then used in Equations E3-2 andd E3-3 to
determine th
he column critical stress, Fcr.
The limit state off torsional bu
uckling is not normally connsidered in thhe design of W
W-shape
columns wh
hen the y-ax
xis is the co
ontrolling axxis for flexurral buckling. Torsional bbuckling
generally do
oes not goverrn, and when it does, the ccritical load ddiffers very litttle from the strength
determined from flexuraal buckling. For other m
member typees, such as W
WT or doubble-angle
compression
n members offten used in trrusses, torsionnal limit states are quite im
mportant.
An additional factor in determ
mining strenggth based on these limit sstates is the ttorsional
effective len
ngth. The Com
mmentary reccommends thaat, conservativvely, the torsional effectivve length
be taken ass the column length and provides sevveral other possibilities iff greater accuuracy is
desired.
EX
XAMPLE 5.15
5
Str
trength of a WSh
hape Column
n
wiith Torsiona
al
Bu
uckling
Goa
al:
Deterrmine the av
vailable strenngth of a W
W-shape coluumn and connsider
torsio
onal buckling..
Giv
ven:
A W1
14×48 A992 column as shhown in Figuure 5.30 is brraced laterallly and
torsio
onally at its en
nds. At mid-hheight it is braaced to resist buckling aboout the
y-axiss, but it canno
ot resist torsioonal bucklingg based on thee bracing shoown in
Figurre 5.30c.
SO
OLUTION
Step
p 1:
From
m Manual Table 1-1,
Ag = 14.1 in.2, Ix = 484 in.4, Iy = 51.4 in.4, rx = 5.85 inn., ry = 1.91 inn.,
Cw = 2240 in..6, J = 1.45 inn.4, h/tw = 33.66
Step
p 2:
Deterrmine the web
b slenderness limit.
E
299,000
λ rw = 1.49
= 1.49
= 35.9
Fy
50
Chapter 5
Step 3:
Compression Members 171
Check slenderness of the web and flange.
h
= 33.6 < λ rw = 35.9
tw
Therefore, the web is not slender. As previously discussed, all W-shapes
have nonslender flanges.
Step 4:
Determine the critical stress for y-axis buckling.
⎛ Lc ⎞ 10 (12 )
= 62.8
⎜ ⎟ =
1.91
⎝ r ⎠y
Fe =
π2 E
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
=
2
π2 ( 29,000 )
( 62.8)
2
= 72.6 ksi >
Fy
= 22.2 ksi
2.25
Therefore, use Equation E3-2:
⎛ 50 ⎞
Fcr = ( 0.658)⎜⎝ 72.6 ⎟⎠ ( 50 ) = 37.5 ksi
Step 5:
Determine the critical stress for x-axis buckling.
20 (12 )
⎛ Lc ⎞
= 41.0
⎜ ⎟ =
5.85
⎝ r ⎠x
Fe =
π2 E
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
2
=
π2 ( 29, 000 )
( 41.0 )
2
= 170 ksi >
Fy
= 22.2 ksi
2.25
Therefore, use Equation E3-2:
⎛ 50 ⎞
Fcr = ( 0.658)⎜⎝ 171 ⎟⎠ ( 50 ) = 44.2 ksi
Step 6:
Determine the critical stress for z-axis buckling, or twisting about the shear
center, using Section E4(a) Equation E4-2.
⎡ π2 ECw
⎤ 1
Fe = ⎢
+ GJ ⎥
2
⎢⎣ ( Lcz )
⎥⎦ I x + I y
⎡ π2 ( 29,000 )( 2240 )
⎤
1
⎥
=⎢
+
11,
200
1.45
(
)
2
⎢
⎥ 484 + 51.4
( 20 (12 ) )
⎣
⎦
Fy
= 51.1 ksi >
= 22.2 ksi
2.25
Therefore use Equation E3-2:
⎛ 50 ⎞
Fcr = ( 0.658)⎜⎝ 51.1 ⎟⎠ ( 50 ) = 33.2 ksi
Step 7:
Select the lowest critical stress determined in Steps 4, 5, and 6.
Fcr = 33.2 ksi
172 Chapter 5
Compression Members
Since the controlling stress comes from Step 6, the strength of the column
is controlled by torsional buckling.
Step 8:
For
LRFD
Step 9:
For
ASD
Step 9:
Determine the nominal strength of the column
Pn = 33.2 (14.1) = 468 kips
Note: Determination of Fcr in steps 4, 5, and 6 could have been delayed
until after the controlling, smallest, value of Fe had been determined and
then Fcr determined only once.
Determine the column design strength.
φPn = 0.9 ( 468 ) = 421 kips
Determine the column allowable strength.
Pn 468
=
= 280 kips
Ω 1.67
EXAMPLE 5.16
Strength of a WTShape Column
Goal:
Determine the available strength of a WT-shape column with consideration
of flexural, torsional, and flexural-torsional buckling.
Given:
A WT7×34 A992 column is 10.0
torsionally at its ends only.
SOLUTION
Step 1:
From Manual Table 1-8,
Ag = 10.0 in.2, Ix = 32.6 in.4, Iy = 60.7 in.4, rx = 1.81 in.,
ry = 2.46 in., tf = 0.720, Cw = 3.21 in.6, J = 1.50 in.4 , d/tw = 16.9,
y = 1.29 in. , b f 2t f = 6.97
Step 2:
Determine the flange and stem slenderness limits from Table B4.1 cases 1
and 4.
E
29,000
λ rf = 0.56
= 0.56
= 13.5
Fy
50
λ rw = 0.75
Step 3:
ft long and is braced laterally and
E
29,000
= 0.75
= 18.1
Fy
50
Check slenderness of the flange and stem.
bf
= 6.97 ≤ λ rf = 13.5
2t f
d
= 16.9 < λ rw = 18.1
tw
Therefore the WT has nonslender flange and stem.
Compression Members 173
Chapter 5
Step 4:
Determine the nominal strength for flexural buckling. Since Lcx = Lcy and
the x-axis has the smallest radius of gyration, flexural buckling will be
controlled by the x-axis.
29,000
E
⎛ Lc ⎞ 10 (12 )
= 66.3 ≤ 4.71
= 4.71
= 113
⎜ ⎟ =
1.81
50
Fy
⎝ r ⎠x
Fex =
π2 E
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
2
=
π2 ( 29,000 )
( 66.3)
2
= 65.1 ksi >
Fy
= 22.2 ksi
2.25
Therefore, use Equation E3-2:
⎛ Fy ⎞
⎛ 50 ⎞
Fcr = ( 0.658 )⎜⎝ Fe . ⎟⎠ Fy = ( 0.658 )⎜⎝ 65.1 ⎟⎠ ( 50 ) = 36.3 ksi
and
Step 5:
Pn = 36.3 (10.0 ) = 363 kips
To determine flexural-torsional buckling, the elastic buckling stress for yaxis buckling is required.
⎛ Lc ⎞ 10 (12 )
= 48.8
⎜ ⎟ =
2.46
⎝ r ⎠y
Fey =
Step 6:
π2 E
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
2
=
π2 ( 29,000 )
( 48.8)
2
= 120 ksi
Determine the flexural-torsional elastic buckling stress for z-axis buckling
using Equation E4-3. The shear center of a WT-shape is at the stem-flange
intersection. Thus, the distance from the centroid to the shear center is
tf
0.720
xo = 0, yo = y − = 1.29 −
=0.930 in.
2
2
and from Equation E4-9
Ix + I y
32.6 + 60.7
ro2 = xo2 + yo2 +
= 0.0 + 0.9302 +
= 10.2
Ag
10.0
and Equation E4-8
H =1−
xo2 + yo2
0 + 0.9302
=
−
= 0.915
1
ro2
10.2
From the user note in Section E4, take Cw = 0 in Equation E4-7. Thus,
11, 200 (1.50 )
GJ
Fez =
=
= 165 ksi
2
Ag ro
10.0 (10.2 )
Step 7:
Determine the flexural-torsional elastic buckling stress for the singly
symmetric member using Equation E4-3.
174 Chapter 5
Compression Members
4 Fey Fez H ⎞
⎞ ⎛⎜
⎟
1
−
1
−
⎟⎜
2
⎠⎝
( Fey + Fez ) ⎟⎠
⎛ 120 + 165 ⎞ ⎛
4 (120 )(165 )( 0.915 ) ⎞
⎟ = 105 ksi
= ⎜⎜
⎟⎟ ⎜ 1 − 1 −
2
⎟
(120 + 165)
⎝ 2 ( 0.915 ) ⎠ ⎜⎝
⎠
⎛ Fey + Fez
Fe = ⎜
⎝ 2H
Determine the critical stress using the flexural-torsional elastic buckling
stress
Fy
Fe = 105 ksi >
= 22.2 ksi
2.25
Step 8:
Therefore, using Equation E3-2
50
Fcr = ( 0.658 )105 ( 50 ) = 41.0 ksi
Determine the nominal strength of the column for the limit state of flexuraltorsional buckling.
Pn = 41.0 (10.0 ) = 410 kips
Step 9:
For
LRFD
Step 10:
For
ASD
Step 10:
5.9
Determine the column design strength. Since the nominal strength for
flexural buckling about the x-axis is less than the flexural-torsional
buckling strength,
φPn = 0.9 ( 363) = 327 kips
Determine the column allowable strength. Since the nominal strength for
flexural buckling about the x-axis is less than the flexural-torsional
buckling strength,
Pn 363
=
= 217 kips
Ω 1.67
SINGLE-ANGLE COMPRESSION MEMBERS
Single-angle compression members would be designed for flexural-torsional buckling according
to the provisions in Specification Section E4 except for an exclusion for angles with
b t ≤ 0.71 E Fy . All hot rolled, A36 angles satisfy this exclusion limit so they need not be
checked for flexural-torsional buckling.
Studies show that the compressive strength of single angles can be reasonably predicted
using the column equations of Specification Section E3 if a modified effective length is used and
the member satisfies the following limiting criteria as found in Specification Section E5.
1.
2.
3.
4.
Members are loaded at their ends in compression through the same one leg.
Members are attached by either welding or a connection containing a minimum
of two bolts.
There are no intermediate transverse loads.
Lc r as determined in this section does not exceed 200.
Chapter 5
5.
Compression Members 175
For unequal leg angles the ratio of the long leg width to short leg width is less
than 1.7.
Two cases are given for these provisions: (1) angles that are individual members or web
members of planar trusses, and (2) angles that are web members in box or space trusses. This
distinction is intended to reflect the difference in restraint provided by the elements to which the
compression members are attached.
The first set of equations is for angles that
1.
2.
3.
Are individual members or web members of planar trusses.
Are equal-leg angles or unequal-leg angles connected through the longer leg.
Have adjacent web members attached to the same side of a gusset plate or truss
chord.
Buckling is assumed to occur about the geometric axis parallel to the attached leg. Since this may
be either the x- or y-axis, the Specification uses the subscript a and then defines ra as the radius of
gyration about the axis parallel to the attached leg.
L
If
≤ 80 ,
ra
Lc
L
(AISC E5-1)
= 72 + 0.75
r
ra
L
and if
> 80
ra
Lc
L
(AISC E5-2)
= 32 + 1.25
r
ra
These effective lengths must be modified if the unequal-leg angles are attached through the
shorter legs. The provisions of Specification Section E5 should be reviewed for these angles as
well as for similar angles in box or space trusses.
Goal:
EXAMPLE 5.17
Strength of SingleAngle Compression
Given:
Member
SOLUTION
Step 1:
Determine the available strength of a 10.0 ft single-angle compression
member using A36 steel and the provisions of Specification Section E5.
A 4×4×1/2 angle is a web member in a planar truss. It is attached by two
bolts at each end through the same leg.
From Manual Table 1-7,
A = 3.75 in.2 and rx = 1.21.
Step 2:
Determine the slenderness ratio for the axis parallel to the connected leg, ra
= rx.
L L 10.0(12)
= =
= 99.2
ra rx
1.21
Step 3:
Determine which equation will give the effective slenderness ratio.
Because
L
= 99.2 > 80
ra
use Equation E5-2.
176 Chapter 5
Compression Members
Step 4:
Determine the effective slenderness ratio from Equation E5.2.
Lc
= 32 + 1.25 ( 99.2 ) = 156 < 200
r
Step 5:
Determine which column strength equation to use.
Because
Lc
29,000
= 156 > 4.71
= 134
r
36
use Equation E3-3.
Step 6:
Determine the Euler buckling stress.
π2 E
π 2 (29,000)
Fe =
=
= 11.8 ksi
2
(156) 2
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
Step 7:
Determine the critical stress from Equation E3-3.
Fcr = 0.877 Fe = 0.877 (11.8 ) = 10.3 ksi
Step 8:
Determine the nominal strength.
Pn = Fcr A = 10.3 ( 3.75 ) = 38.6 kips
For
LRFD
Step 9:
For
ASD
Step 9:
Determine the design strength.
φPn = 0.9 ( 38.6 ) =34.7 kips
Determine the allowable strength.
Pn Ω = 38.6 1.67 =23.1 kips
For single angle compression members that do not meet the criteria set forth in Section E5 for use
of the modified slenderness ratio equations, the provisions of Sections E3 or E7 must be
followed. The provisions in Section E4 for torsional or flexural-torsional buckling do not need to
be followed for hot-rolled angles since all currently available hot-rolled A36 angles meet the leg
slenderness exclusion of b t ≤ 0.71 E Fy . Thus, the strength for flexural buckling about the
principal axes must be assessed.
Goal:
EXAMPLE 5.18
Strength of SingleAngle Compression
Given:
Member
SOLUTION
Step 1:
Determine the available strength of a 10.0 ft single-angle compression
member using A36 steel.
A 4×4×1/2 angle is a web member in a planar truss. It is attached by single
bolts at each end through the same leg.
From Manual Table 1-7,
A = 3.75 in.2 and rx = 1.21, rz = 0.776 in.2.
Chapter 5
Compression Members 177
Step 2:
Determine the slenderness ratio for the minor (weak) principal axis.
Lc 10.0(12)
=
= 155
rz
0.776
Step 3:
Determine which column strength equation to use.
Because
Lc
29,000
= 155 > 4.71
= 134
36
rz
Use Equation E3-3.
Step 4:
Determine the Euler buckling stress.
π2 E
π 2 (29,000)
Fe =
=
= 11.9 ksi
2
(155) 2
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠
Step 5:
Determine the critical stress from Equation E3-3.
Fcr = 0.877 Fe = 0.877 (11.9 ) = 10.4 ksi
Step 6:
Determine the nominal strength.
Pn = Fcr A = 10.4 ( 3.75 ) = 39.0 kips
For
LRFD
Step 7:
For
ASD
Step 7:
Determine the design strength.
φPn = 0.9 ( 39.0 ) =35.1 kips
Determine the allowable strength.
Pn Ω = 39.0 1.67 =23.4 kips
5.10 BUILT-UP MEMBERS
Members composed of more than one shape are called built-up members. Several of these were
illustrated in Figure 5.2h through n. Built-up members composed of two shapes are covered in
Specification Section E6. Compressive strength is addressed by establishing the slenderness ratio
and referring to Specification Section E3, E4, or E7 as appropriate.
If a built-up section buckles so that the fasteners between the shapes are not stressed in
shear but simply “go along for the ride,” the only requirement is that the slenderness ratio of the
shape between fasteners be no greater than 0.75 times the controlling slenderness ratio of the
built-up shape. If overall buckling would put the fasteners into shear, then the controlling
slenderness ratio will be somewhat greater than the slenderness ratio of the built-up shape. This
modified slenderness ratio is used to account for the effect of shearing deformations through the
connectors. Thus, the effective slenderness ratio for a built-up member with snug-tight connectors
will be greater than the same member with pre-tensioned or welded connectors. In addition, the
spacing of the intermediate connectors will influence the modified slenderness ratio.
178 Chapter 5
Compression Members
For intermediate connectors that are bolted snug-tight, the modified slenderness ratio is
always greater than the slenderness ratio of the built-up member acting as a unit and is specified
as
2
⎛ Lc ⎞
⎛ Lc ⎞ ⎛ a ⎞
⎜ ⎟ = ⎜ ⎟ +⎜ ⎟
⎝ r ⎠m
⎝ r ⎠o ⎝ ri ⎠
2
(AISC E6-1)
If the intermediate connectors are welded or pre-tensioned bolted, the shearing deformation in the
connectors is significantly less than for snug-tight connectors and the modified slenderness ratio
may be equal to the slenderness ratio of the built-up member acting as a unit. For this case, the
modified slenderness ratio is specified as,
a
when ≤ 40 ,
ri
⎛ Lc ⎞
⎛ Lc ⎞
⎜ ⎟ =⎜ ⎟
r
⎝ ⎠ m ⎝ r ⎠o
and when
(AISC E6-2a)
a
> 40 ,
ri
2
⎛ Ki a ⎞
⎛ Lc ⎞
⎛ Lc ⎞
⎜ ⎟ = ⎜ ⎟ +⎜
⎟
⎝ r ⎠m
⎝ r ⎠o ⎝ ri ⎠
2
(AISC E6-2b)
where
⎛ Lc ⎞
⎜ ⎟ = column slenderness of built-up member acting as a unit
⎝ r ⎠o
K i = 0.5 for angles back-to-back
= 0.75 for channels back-to-back
= 0.86 for all other shapes
a = distance between connectors
ri = minimum radius of gyration of individual component
The remaining provisions in Specification Section E6 address dimensions and detailing
requirements. These provisions are based on judgment and experience and are provided to ensure
that the built-up member behaves in a way consistent with the strength provisions already
discussed. The ends of built-up compression members must be either welded or pre-tensioned
bolted in order to ensure that the member can work together as a unit. Even the smallest amount
of slip in the end connections could mean that the built-up member is unable to carry any more
load than the components individually. Along the length of built-up members, the longitudinal
spacing of connectors must be sufficient to provide for transfer of the required shear force in the
buckled member. The Commentary of the Specification gives guidance on how to determine the
magnitude of that requirement. A built-up compression member with connectors spaced so that
the slenderness ratio of the shape between fasteners is no greater than 0.75 times the controlling
slenderness ratio of the built-up shape will not automatically satisfy this strength requirement.
The Manual provides tables of properties for double angles, double channels, and Ishapes with cap channels in Part 1 and tables of compressive strength for double-angle
compression members in Part 4.
Chapter 5
Goal:
EXAMPLE 5.19
Strength of a Builtup Double-Angle
Given:
Compression
Member
SOLUTION
Step 1:
Compression Members 179
Determine the available strength of a 10.0 ft double-angle compression
member using A36 steel.
Two 5×3×1/4 angles, long legs back-to-back with a 3/8 in. gap are used as
a chord member in a planar truss. The angles are welded at each end to a
gusset plate and along the length at two intermediate points with a spacing
of 40 in.
From Manual Table 1-15 for double angles
A = 3.88 in.2 , rx = 1.62 in., ry = 1.19 in., ro = 2.52 in. and H = 0.638 in.
From Manual Table 1-7 for single angles
rz = 0.652 in. and J = 0.0438 in.4
Step 2:
Determine the slenderness ratio for each axis if the member works as a unit.
L 10.0(12)
=
= 74.1
rx
1.62
L 10.0(12)
=
= 101
ry
1.19
Step 3:
Determine the effective slenderness ratio for buckling about the y-axis, the
axis that will put the connectors in shear. Since the intermediate connectors
are spaced at 40 in.
a a
40
= =
= 61.3 > 40
ri rz 0.652
Therefore use Equation E6.2b
2
2
⎛ Lc ⎞
⎛ Lc ⎞ ⎛ Ki a ⎞
= 1012 + ( 0.5 ( 61.3) ) = 106
⎜ ⎟ = ⎜ ⎟ +⎜
⎟
⎝ r ⎠m
⎝ r ⎠o ⎝ ri ⎠
Step 4:
Check the maximum permitted slenderness ratio between connectors
a
= 61.3 < 0.75 (106 ) = 79.5
ri
Step 5:
Determine the elastic buckling stress for flexural buckling using the
modified slenderness ratio
π 2 ( 29,000 )
π2 E
Fey =
=
= 25.5 ksi
2
2
(106 )
⎛ Lc ⎞
⎜ ⎟
⎝ r ⎠m
Step 6:
Determine the elastic buckling stress for torsional buckling using Equation
E4-7 with Cw = 0 based on the user note.
11, 200 ( 2 ( 0.0438 ) )
GJ
Fez =
=
= 39.8 ksi
2
2
Ag ro
3.88 ( 2.52 )
Step 7:
Determine the elastic buckling stress for flexural-torsional buckling using
Equation E4-3.
180 Chapter 5
Compression Members
4 Fey Fez H ⎞
⎞ ⎛⎜
⎟
⎟ ⎜1 − 1 −
2
⎠⎝
( Fey + Fez ) ⎟⎠
⎛ 25.5 + 39.8 ⎞ ⎡
4 ( 25.5)( 39.8 )( 0.638 ) ⎤
⎢
⎥ = 19.1 ksi
= ⎜⎜
−
−
1
1
⎟⎟
2
⎥⎦
( 25.5 + 39.8)
⎝ 2 ( 0.638 ) ⎠ ⎢⎣
⎛ Fey + Fez
Fe = ⎜
⎝ 2H
Step 8:
Determine the critical stress.
Since the elastic buckling stress for flexural-torsional buckling is less than
that for flexural buckling, use that to determine the critical stress.
Fy 36
=
= 1.88 < 2.25
Fe 19.1
Therefore use Equation E3-2
Fy
⎛
Fcr = ⎜ 0.658 Fe
⎜
⎝
Step 9:
⎞
⎟ Fy = ( 0.6581.88 ) ( 36 ) = 16.4 ksi
⎟
⎠
Determine if the local buckling must be included.
For the short leg b t = 3.0 0.25 = 12.0
and for the long leg b t = 5.0 0.25 = 20.0
From Table B4.1a case 3
E
29,000
λ r = 0.45
= 0.45
= 18.9
Fcr
16.4
Since
b t = 20.0 > λ r = 18.9
the effective area for the long leg must be determined.
Step 10:
Determine the effective area using Equation 5-27 and the values of c4 =
0.671 and c5 = 0.148 from Table 5.4.
E ⎡ 0.148 E ⎤
be = 0.671t
⎢1 −
⎥
Fcr ⎣
bt
Fcr ⎦
29,000 ⎡ 0.148 29,000 ⎤
= 0.671( 0.25 )
⎢1 −
⎥ = 4.86 in.
16.4 ⎣
20
16.4 ⎦
So the effective area of the two angles is
Ae = 3.88 − 2 ( 5.0 − 4.86 )( 0.25) = 3.81 in.2
Step 11:
For
LRFD
Step 12:
Determine the nominal strength.
Pn = Fcr Ae = 16.4 ( 3.81) = 62.5 kips
Determine the design strength.
Chapter 5
Comppression Mem
mbers
181
φPn = 0.9 ( 62.5) =
=56.3 kips
For
F
ASD
A
Step
S
12:
5.11
Deetermine the allowable
a
streength.
Pn Ω = 62.5 1.677 =37.4 kips
CO
OLUMN BASE PLATES
When co
olumns are su
upported on material otheer than steel, such as conncrete or maasonry, it is
necessary
y to distributte their load over an are a significantlly larger thaan the gross area of the
column. In these situaations, a colum
mn base platee similar to thaat shown in F
Figure 5.31 is used.
Column
C
base plates may be attached to the column iin the shop, ass shown in Fiigure 5.31a,
or shippeed separately to the site and
a attached iin the field. C
Columns are normally weelded to the
plate butt may be attaached with an
ngles when llarge plates m
must be shippped separately. In either
case, the selection of the
t dimension
ns and thickneess of the plaate follows thee same rules.
Column
C
base plates are no
ormally attachhed to a footiing or pier w
with anchor roods, and the
space beetween the plate and the support is ffilled with a non-shrink ggrout. A leveling plate,
leveling nuts, or shim
ms (as shown
n in Figure 5..31b) are useed to level thhe column baase plate. In
cases wh
here the colum
mn supports an
n axial comprression only, anchor rods aare not designned to resist
a specific force. How
wever, all collumn base pllates must bee anchored w
with a minim
mum of four
anchor ro
ods according
g to the Occup
pational Safeety and Healthh Administrattion (OSHA) regulations
in OSHA
A 29 CFR 192
26 Subpart R Safety Standdards for Steeel Erection. Figure 5.32 iillustrates a
column with
w base platte in plan (Fig
gure 5.32a) annd elevation (Figure 5.32bb), including four anchor
rods.
To
T determine the area of bearing
b
that iss required, thee strength of the material uupon which
Section J8 off the Specificcation gives
the base plate is bearring must be evaluated. Foor concrete, S
provision
ns identical to
t those given in the conncrete code, A
ACI 318. Whhen the beariing plate is
covering the full area of the concreete support, thhe nominal beearing strengtth is
Figure 5.31
5
Exam
mple of a W-Shape Columnn and Base Pllate
Photos co
ourtesy Douglas Steel Fabricaating Corporatiion
1882 Chapter 5
Compresssion Memberss
Figure 5.32
Column and Base Platte Section andd Plan Includding Anchor R
Rods
Pn = Pp = 0.85 f c′A1
(AIS
SC J8-1)
where f c' is the specified
d concrete co
ompressive sstrength and A1 is the arrea of the pllate and
concrete. If the plate doess not cover alll of the concrrete, there will be an increase in strengtth due to
the spread of the load as it progresses down throughh the concretee. In this casee the nominall bearing
strength is given
g
as
(AIS
SC J8-2)
Pn = Pp = 0.85 f c′A1 A2 A1 ≤ 1.77 f c′A1
Here A2 is th
he maximum
m area of conccrete with thee same shape as the bearinng plate. The limit on
the right sid
de of the equaation imposes a maximum ratio of areass of 4:1. If thhe supporting element
is designed based on thee bearing streength of the ssoil, it will bee relatively eeasy to determ
mine the
extent to wh
hich the base plate covers the
t concrete ffoundation orr pier. In all ccases, φ = 0.665 and Ω
= 2.31.
The thickness off a column baase plate is a function of the bending strength of thhe plate.
Since bendin
ng has not yet been covereed in this bookk, this topic is deferred to Section 11.111. Those
wishing to address
a
base plate
p
design further
f
shouldd proceed to Section 11.11 and to the eexample
problems giv
ven there as well
w as AISC Design Guiddes 1, 7 and 100.
55.12
PROB
BLEMS
11. Determine the theoreticaal buckling sttrength (Eulerr
bbuckling load)) for a W8×58 A992 collumn with an
n
eeffective lengtth of 20 ft. Will
W the theorretical column
n
bbuckle or yield at this length??
4. D
Determine the theoretical bbuckling strength (Euler
buckkling load) forr an HSS 100×5×3/8 A5000 Grade C
colum
mn with an effective lenggth of 20 ft. Will the
theorretical column buckle or yieldd at this lengthh?
22. Determine the theoreticaal buckling sttrength (Eulerr
bbuckling load)) for a W16×77 A36 collumn with an
n
eeffective lengtth of 12 ft. Will
W the theorretical column
n
bbuckle or yield at this length??
5. Foor a W12×72 A992 column, determine thhe effective
lengtth at which thhe theoretical bbuckling strenngth (Euler
buckkling load) will equal the yieldd strength.
33. Determine the theoreticaal buckling sttrength (Eulerr
bbuckling load)) for a W24×370 A992 co
olumn with an
n
eeffective lengtth of 15 ft. Will
W the theorretical column
n
bbuckle or yield at this length??
6. Foor a W6×16 A
A992 column, determine the effective
lengtth at which thhe theoretical bbuckling strenngth (Euler
buckkling load) will equal the yieldd strength.
7. Foor an HP8×36 A572 Grade 550 column, dettermine the
effecctive length at which the theooretical bucklinng strength
(Euleer buckling loaad) will equal tthe yield strenggth.
Chapteer 5
88. A W14×120
0 column has an
a effective len
ngth for y-axiss
bbuckling equal to 24 ft. Deterrmine the effecctive length forr
thhe x-axis that will provide th
he same theoreetical buckling
g
sstrength (Euler buckling load)).
99. A W14×53 column has an effective len
ngth for x-axiss
bbuckling equal to 20 ft. Deterrmine the effecctive length forr
thhe y-axis that will provide th
he same theoreetical buckling
g
sstrength (Euler buckling load)).
110. An HSS12×6×1/2 column
n has an effecctive length forr
xx-axis buckling
g equal to 18 ft. Determinee the effectivee
length for the y-axis
y
that will provide the saame theoreticall
bbuckling streng
gth (Euler buck
kling load).
111. A W14×13
32 A992 colum
mn has an effecctive length off
336 ft about both axes. Determine the availablee
ccompressive sttrength for thee column. Deteermine the (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD. Is this an
n elastic or inelastic buckling condition?
112. Determine the available compressive strength for a
W
W12×152 A99
92 column witth an effectivee length aboutt
bboth axes of 40 ft. Determin
ne the (a) desig
gn strength by
y
L
LRFD and (b)) allowable sttrength by AS
SD. Is this an
n
eelastic or inelasstic buckling co
ondition?
113. A W6×15 A992
A
column has
h an effectivee length of 8 ft
aabout both ax
xes. Determinee the availablee compressivee
sstrength for thee column. Deteermine the (a) design
d
strength
h
bby LRFD and (b) allowable strength by ASD.
A
Is this an
n
eelastic or inelasstic buckling co
ondition?
Comprression Membbers
183
Deterrmine the (a)) design strenngth by LRFD
D and (b)
allow
wable strength bby ASD.
18. D
Determine the available com
mpressive strenngth for an
HSS 6×4×3/8 A5000 Grade C collumn where thhe effective
lengtth is 10 ft abouut the y-axis annd 15 ft about the x-axis.
Deterrmine the (a)) design strenngth by LRFD
D and (b)
allow
wable strength by ASD. Is thhis an elastic oor inelastic
buckkling condition??
19. A round HSS 16.000×0.3755 A500 Grade C column
has aan effective leength of 18 ft. Determine thee available
comppressive strenggth and indicatte whether this is due to
elastiic or inelasticc buckling. D
Determine the (a) design
strenngth by LRFD aand (b) allowab
able strength byy ASD.
20. A W8×40 is ussed as a 12 ft ccolumn in a braaced frame
with W16×26 beam
ms at the top and bottom ass shown in
Figurre P5.20. The ccolumns abovee and below aree also 12 ft
W8×
×40s. The beam
ms provide m
moment restraiint at each
colum
mn end. Deteermine the efffective length using the
alignnment chart annd the available compressivve strength,
and tthe (a) designn strength by L
LRFD and (b)) allowable
strenngth by ASD. Assume that tthe columns arre oriented
for ((i) buckling abbout the weakk axis and (iii) buckling
abouut the strong axxis. All steel is A992.
21. If the structuure described in Problem 20 is an
unbraaced frame, determine the effective length and
comppressive strenggth as requestedd in Problem 20.
114. Determine the available compressive strength
s
for an
n
M
M4×6 A36 collumn with an effective leng
gth about both
h
aaxes of 7 ft. Determine
D
the (a)
( design stren
ngth by LRFD
D
aand (b) allowaable strength by ASD. Is this an elastic orr
innelastic buckliing condition?
115. A W14×21
11 A992 colum
mn has an effecctive length off
440 ft about both axes. Determine the availablee
ccompressive sttrength for thee column. Deteermine the (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD. Is this an
n elastic or inelastic buckling condition?
116. Determine the available compressive strength for a
W
W12×65 A992 column when
n the effective length is 20 ft
aabout the y-axis and 40 ft abo
out the x-axis. Determine thee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD. Is thiss an elastic or inelastic bucklling condition??
D
Describe a com
mmon conditio
on where the effective length
h
is different about the differentt axes.
117. A W8×24 A992 column
n has an effecctive length off
112.5 ft about the y-axis an
nd 28 ft abo
out the x-axis.
D
Determine the available
a
comp
pressive strength and indicatee
w
whether this is due to elaastic or inelaastic buckling.
P5.200
22. A W12×136 ccolumn is show
wn in Figure P
P5.22 with
end cconditions thaat approximatee ideal conditioons. Using
the rrecommended approximate vvalues from Coommentary
1184 Chapterr 5
Compreession Membeers
T
Table C-A-7.1,, determine thee effective len
ngths for the y-aaxis and the x-axis. Which effective lengtth will controll
thhe column streength?
and a compressivee live load of 1100 kips by (a) LRFD
and ((b) ASD.
An A992 W122×53 is used as a column in a building
26. A
with an effective leength for each axis of 15 ft. Determine
whetther the columnn will carry a ccompressive deead load of
85 kkips and a com
mpressive live load of 255 kkips by (a)
LRFD
D and (b) ASD
D.
27. A
An A992 W8×
×67 is used in a structure too support a
dead load of 60 kkips and a livee load of 100 kips. The
colum
mn has an eeffective lengtth of 24 ft. Determine
whetther the colum
mn will supporrt the load by (a) LRFD
and ((b) ASD.
P
P5.22
223. A W12×87 column is sho
own in Figure P5.23
P
with end
d
cconditions thatt approximate ideal conditio
ons. Using thee
rrecommended approximate values from Commentary
y
T
Table C-A-7.1,, determine thee effective len
ngths for the y-aaxis and the x-axis. Which effective lengtth will controll
thhe column streength?
28. A
An A992 W100×54 is used as a column in a building
with an effective length of 30 ftt. Determine w
whether the
colum
mn will carry a compressive dead load of 224 kips and
a com
mpressive livee load of 72 kiips by (a) LRF
FD and (b)
ASD
D.
29. A
An A992 W16×
×77 is used as a column in a building to
suppoort a dead loaad of 130 kipss and a live looad of 200
kips. The column effective lengtth is 20 ft forr the y-axis
and 330 ft for the xx-axis. Determ
mine whether tthe column
will ssupport the loaad by (a) LRFD
D and (b) ASD.
30. A
An A992 W21×111 is used aas a column in a building
to suupport a dead lload of 120 kipps and a live load of 300
kips. The column effective lengtth is 22 ft forr the y-axis
and 333 ft for the xx-axis. Determ
mine whether tthe column
will ssupport the loaad by (a) LRFD
D and (b) ASD.
31. A
An A992 W24×131 is used aas a column in a building
to suupport a dead lload of 245 kipps and a live load of 500
kips. The column hhas an effectivve length aboutt the y-axis
of 188 ft and an efffective length aabout the x-axxis of 36 ft.
Deterrmine whetherr the column w
will support thhe load by
(a) L
LRFD and (b) A
ASD.
P
P5.23
224. A W10×60
0 column with an effective length
l
of 30 ft
ffor both axes is called upon to carry a com
mpressive dead
d
looad of 123 kip
ps and a comprressive live loaad of 170 kips.
D
Determine wheether the colum
mn will suppo
ort the load by
y
((a) LRFD and (b) ASD. Evalluate the streng
gth for (i) Fy =
550 ksi and (ii) Fy = 70 ksi.
225. A W14×25
57 A992 colum
mn in a building
g has effectivee
lengths of 16 ft for both ax
xes. Determin
ne whether thee
ccolumn will caarry a compressive dead loaad of 800 kipss
An A300 Gr. C HSS7×7×1/22 is used as a column to
32. A
suppoort a dead loaad of 175 kipss and a live looad of 100
kips. The columnn has an effeective length of 10 ft.
will support thhe load by
Deterrmine whetherr the column w
(a) L
LRFD and (b) A
ASD.
33. For the W100×88 column with bracingg and end
condditions shownn in Figure P5.33, deterrmine the
theorretical effectivve length for eaach axis and identify the
axis tthat will limit tthe column strength.
×28 column with bracingg and end
34. For the W8×
condditions shownn in Figure P5.34, deterrmine the
theorretical effectivve length for eaach axis and identify the
axis tthat will limit tthe column strength.
Chapteer 5
Comprression Membbers
185
36. A W14×176 coolumn with ennd conditions aand bracing
is shoown in Figuree P5.36. Determ
mine the least theoretical
braciing and its locaation about thee y-axis, in ordder that the
y-axiis not control thhe strength of tthe column.
37. A W12×72 coolumn is an exxterior 2nd stoory column
(griddline A) with sstrong axis bucckling in the plane of the
framee in an unbrraced multi-stoory frame. Thhe column
below
w is also a W12×72. Beam
ms and dimensiions are as
show
wn in Figure P
P5.37. Determ
mine the effective length
for tthis conditionn and the corrresponding coompressive
strenngth by (a) LRF
FD and (b) ASD
D. All steel is A992.
P
P5.33
P
P5.34
P5.366
335. A W10×100 column with
h end conditions and bracing
g
is shown in Fig
gure P5.35. Deetermine the leeast theoreticall
bbracing and its location abou
ut the y-axis, in
n order that thee
yy-axis not contrrol the strength
h of the column
n.
P5.377
38. A W12×72 coluumn is an exteerior 2nd story column (at
gridliine C) with sttrong axis buckkling in the pllane of the
framee in an unbrraced multi-stoory frame. Thhe column
below
w is also a W12×72. Beam
ms and dimensiions are as
show
wn in Figure P
P5.37. Determ
mine the effective length
for tthis conditionn and the corrresponding coompressive
strenngth by (a) LRF
FD and (b) ASD
D. All steel is A992.
P
P5.35
186 Chapter 5
Compression Members
39. A W12×72 column is an interior 2nd story column (at
gridline B) with strong axis buckling in the plane of the
frame in an unbraced multi-story frame. The column
below is also a W12×72. Beams and dimensions are as
shown in Figure P5.37. Determine the effective length
for this condition and the corresponding compressive
strength by (a) LRFD and (b) ASD. All steel is A992.
40. A W12×50 column is an interior column with strong
axis buckling in the plane of the frame in an unbraced
multi-story frame. The columns above and below are also
W12×50. The beams framing in at the top are W16×31
and those at the bottom are W16×40. The columns are 14
ft and the beam span is 25 ft. The column carries a dead
load of 75 kips and a live load of 150 kips. Determine the
inelastic effective length for this condition and the
corresponding compressive strength by (a) LRFD and (b)
ASD. All steel is A992.
41. Select the least-weight W12 A992 column to carry a
live load of 130 kips and a dead load of 100 kips with an
effective length about both axes of 14 ft by (a) LRFD and
(b) ASD.
42. Select the least-weight W14 A992 column to carry a
dead load of 200 kips and a live load of 600 kips if the
effective length about both axes is 20 ft by (a) LRFD and
(b) ASD.
43. Select the least-weight W10 A992 column to carry a
dead load of 80 kips and a live load of 280 kips with an
effective length about both axes of 15 ft by (a) LRFD and
(b) ASD.
44. Select the least-weight W8 A992 column to carry a
dead load of 20 kips and a live load of 50 kips with an
effective length about both axes of 30 ft by (a) LRFD and
(b) ASD.
45. Select the least-weight W6 A992 column to carry a
dead load of 12 kips and a live load of 36 kips with an
effective length about both axes of 8 ft by (a) LRFD and
(b) ASD.
48. If the column in Problem 44 had an effective length of
32 ft, select the lightest-weight W12 to support this load
by (a) LRFD and (b) ASD.
49. A W14 A992 column must support a dead load of 80
kips and a live load of 300 kips. The column is 22 ft long
and has end conditions that approximate the ideal
conditions of a fixed support at one end and a pin support
at the other. Select the lightest-weight W14 to support this
load by (a) LRFD and (b) ASD.
50. Select the least-weight W8 A992 column to support a
dead load of 170 kips with an effective length of 16 ft by
(a) LRFD and (b) ASD.
51. A column with an effective length of 21 ft must
support a dead load of 120 kips, a live load of 175 kips,
and a wind load of 84 kips. Select the lightest W14 A992
member to support the load by (a) LRFD and (b) ASD.
52. A W14×99 A992 column is 20 ft long, pinned at each
end, and braced at mid-height to prevent lateral
movement for buckling about the y-axis. However, the yaxis bracing is not adequate to resist torsion. Considering
flexural and torsional buckling, determine the nominal
strength of this compression member.
53. An A36 single-angle compression web member of a
truss is 10 ft long and attached to gusset plates through
the same leg at each end with a minimum of two bolts.
The member must carry a dead load of 8 kips and a live
load of 10 kips. Select the least-weight equal leg angle to
carry this load by (a) LRFD and (b) ASD.
54. If the compression web member of Problem 53 were
loaded concentrically, determine the least-weight single
angle to carry the load by (a) LRFD and (b) ASD.
55. A W16×26 A992 compression member has a slender
web when used in uniform compression. Determine the
available strength by (a) LRFD and (b) ASD when the
effective length is (i) 6 ft and (ii) 12 ft.
46. Select the least-weight W8 A992 column to carry a
dead load of 13 kips and a live load of 39 kips with an
effective length about both axes of 14 ft by (a) LRFD and
(b) ASD.
56. The W14×43 is the only A992 column shown in the
Manual column tables that has a slender web. Determine
the available strength for this column if the effective
length is 5 ft, and show whether the slender web impacts
that strength by (a) LRFD and (b) ASD.
47. A column with pin ends for both axes must be
selected to carry a compressive dead load of 95 kips and a
compressive live load of 285 kips. The column is 16 ft
long and is in a braced frame. Select the lightest-weight
W12 to support this load by (a) LRFD and (b) ASD.
57. Determine the available strength of a WT6×22.5 A992
steel compression chord of a truss with effective length Lc
= 14 ft. Consider the member braced laterally and
torsionally at its ends only. Determine by (a) LRFD and
(b) ASD.
58. Determine the available strength of a WT12×125
A992 steel column with effective length Lc = 18 ft.
Chapteer 5
C
Consider the member
m
braced
d laterally and
d torsionally att
itts ends only. Determine
D
by (aa) LRFD and (b
b) ASD.
559. Determine the
t available sttrength of a C9
9×20 A36 steell
ccompression ch
hord of a trusss with effectiv
ve length, Lc =
116 ft. Consid
der the mem
mber braced laterally and
d
toorsionally at itts ends only. Determine
D
by (a) LRFD and
d
((b) ASD.
660. Determine the availablee strength of an
a A36, 20 ft
loong, 2L6×4×5
5/8 double-angle compression
n member in a
pplanar truss. The angles are long
l
legs back--to-back with a
33/8 in. gap. Thee angles are welded
w
at each end
e to a gussett
pplate and along
g the length at two intermediaate points with
h
a spacing of 80
0 in. Determin
ne the (a) desiign strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
661. Determine the availablee strength of an
a A36, 15 ft
loong, 2L6×3-1//2×1/2 double-angle compreession memberr
inn a planar trusss. The angles are long legs back-to-back
k
w
with a 3/8 in
n. gap. The angles are co
onnected with
h
ppretensioned bo
olts at each end to a gusset plate
p
and along
g
thhe length at tw
wo intermediatee points with snug-tight boltss
aat a spacing off 60 in. Deterrmine the (a) design
d
strength
h
bby LRFD and (b)
( allowable sttrength by ASD
D.
M
Multi-Chapterr Problems
662. Using thee framing plaan shown in Figure P5.62
2
((presented earllier as Figuree 2.7), design
n the columnss
m
marked 4 and 5.
5 This is the saame structure used
u
in Section
n
22.5, where loaad calculationss with live lo
oad reductionss
w
were discussed
d. Those calcu
ulations can be reused here.
L
Load case 2 fo
or dead plus liive load is to be considered.
T
The building is an office build
ding with a nom
minal live load
d
oof 50 pounds per
p square foott (psf) and a calculated
c
dead
d
looad of 70 psf.
4: Inteerior column D-2
D regardless of
o deck span
directiion
5: Exterior column D-4
D regardless of deck span
directiion
D
Design for colu
umn length L = 14 ft and K = 1.0 using (a))
L
LRFD and (b) ASD.
A
P
P5.62
Comprression Membbers
187
63. T
The framing pplan shown in F
Figure P5.63 iis the same
as thhe one shown in Figure P2.220 for an 18-sstory office
buildding. It must suupport a dead load of 80 psff and a live
load of 50 psf. IIn all cases, tthe decking sspans in a
direcction from linne A toward line E. Deteermine the
requiired axial streength for the columns and design the
colum
mns as requireed below for ((a) design by LRFD and
(b) ddesign by ASD.. The required axial load strengths were
deterrmined in Prooblem 20 of Chapter 2. Use a story
heighht of 13.5 ft in a braced framee so that K = 1.0.
iv: The coolumn at the coorner on lines 1 and E that
supports eeight levels.
v: The collumn on the eddge at the inteersection of
lines 4 andd A that supporrts eight levelss.
vi: The interior colum
mn at the interrsection of
line 4 andd the point bettween lines C and D that
supports thhree levels.
P5.633
64. IIntegrated Dessign Project. U
Using the gravvity column
loadss determined iin Chapter 2, design the grravity-only
colum
mns. Design columns as singgle-story mem
mbers. (It is
oftenn more econoomical to usse multi-storyy columns
becauuse of construcction costs.) Seelect the final ccolumns so
that tthey are two-sstory from bellow grade to tthe second
floorr, two-story froom the second floor to the foourth floor,
and tthen use a singgle-story colum
mn to support thhe roof.
Design the columns in the braced fraame for the
graviity loads deterrmined in Chappter 2 and the wind load
deterrmined in Chaapter 4. Remem
mber that the wind load
mustt be considered to act in tw
wo directions, so use the
largeest compressionn forces from w
wind to combinne with the
graviity loads.
Using thee wind load aanalysis from Chapter 4,
desiggn all the bracces as compresssion memberss. Compare
the teension design w
with the comppression designn and select
the ap
appropriate finaal members.
C
Chapterr 6
B
Bendin
ng Memberss
Massachusetts
M
C
College of Art aand Design, Deesign and Meddia Center
Photograph
P
by Fadi Asmar/LERA Consultinng Structural E
Engineers
6.11
BENDIN
NG MEMBERS IN STRUCTURE
ES
A bending member carrries load applied normal to its longituudinal axis aand transfers it to its
support poin
nts through beending momeents and sheaars. In buildinng constructioon, the most ccommon
application of bending members
m
is to
o provide suppport for flooors or roofs. These beamss can be
either simplle span or co
ontinuous, and
d normally trransfer their load to otherr structural m
members
such as collumns, girders, and wallls. Although the terms bbeam and giirder are oftten used
interchangeaably because both are ben
nding memberrs, beam norm
mally refers tto a bending member
that directly
y supports an
n applied load
d whereas girrder usually refers to a bbending mem
mber that
supports a beam.
b
The disstinction is no
ot important fo
for design beccause the sam
me criteria appply to all
bending mem
mbers.
The most comm
monly used sh
hapes for bennding memberrs are the I-sshaped cross sections
and, of thesse, the W-shaape is domin
nant. Howeveer, there are numerous situations wheere other
shapes are used
u
as bendiing members. L-shapes arre commonly used as linteels over opennings, Tshapes are found
f
as chorrds of trusses that may be called upon to resist bendding along w
with axial
forces, and C-shapes
C
may
y coexist with
h W-shapes inn floor system
ms.
In addition
a
to thee use of the sttandard shapees, engineers often find it necessary to develop
their own sh
hapes by com
mbining shapees and/or plattes. Several eexamples of tthese built-upp shapes
are shown in Figure 6.1.
6
Although
h the use off these builtt-up shapes is permitted by the
Specification
n, they may not
n be econom
mical becausee of the laborr costs associaated with fabrication.
The compleexity that ressults from thee wide varieety of possiblle shapes is the reason sso many
separate pro
ovisions are in
ncluded in Ch
hapter F of thee Specificatioon.
The most comm
mon and econo
omical bendiing members are those thhat can reach the full
material yield strength without
w
being limited by buuckling of anny cross-sectional elements. These
members aree referred to as
a compact members
m
and aare addressedd first.
Table 6.1 lists the sections of the Specificaation and parts of the Manuual that are discussed
in this chaptter.
1888
Chapter 6
Bending Mem
mbers 189
Figure 6.1 Built-up Beaams
6.2
STRE
ENGTH OF
F BEAMS
i applied to a bending meember, resultinng in a bending moment, sstresses are developed in
As load is
the crosss section. For a load at or below
b
the noominal level, tthe load magnnitude establiished in the
building code, it is reeasonable to expect
e
the enntire beam crooss section too behave elastically. The
stresses and
a strains aree distributed as shown in F
Figure 6.2a. T
This elastic beehavior occurrs whenever
the materrial’s behavio
or correspond
ds to the initiaal straight-linne portion of tthe stress-straain curve of
Figure 3..5.
From
F
the basic principles of strength oof materials, the relationshhip between the applied
moment and resulting
g stresses is giiven by the faamiliar flexuree formula:
My
(6.1)
fy =
I
where
M = any
y applied mom
ment that stressses the sectioon in the elasttic range
y = disttance from th
he neutral axiss to the point where the strress is to be determined
I = mo
oment of inerttia
fy = resu
ulting bendin
ng stress at loccation y
Table 6.1
Sections of
o Specification and Parts of Manual Coovered in Thiis Chapter
Specifi
fication
B3
B4
F1
F2
F3
F6
F9
F10
Chapter G
H1
J10
Chapter L
Appendix
x1
Design
n Basis
Classification of Seections for Loocal Bucklingg
Generaal Provisions
Doublly Symmetricc Compact II-Shaped Meembers and C
Channels Beent about
Th
heir Major Ax
xis
Doublly Symmetricc I-Shaped M
Members withh Compact W
Webs and Nonncompact
or Slender Flan
nges Bent aboout Their Major Axis
I-Shap
ped Members and Channells Bent about Their Minor Axis
Tees and
a Double Angles
A
Loadedd in the Planee of Symmetryy
Singlee Angles
Design
n of Memberss for Shear
Doublly and Singly Symmetric M
Members Subjject to Flexurre and Axial F
Force
Flangees and Webs with
w Concenttrated Forces
Design
n for Serviceaability
Design
n by Advanceed Analysis
Mannual
Part 1
Part 3
Dimen
nsions and Pro
operties
Design
n of Flexural Members
1990 Chapter 6
Bending Members
M
Figure 6.2 Cross-Sectiional Bendin
ng Stresses aand Strains: (a) elastic; ((b) yield; (c)) partial
plastic; (d) plastic
p
Norrmally the strress at the extreme fiber, tthat is, the fiiber most distant from thee neutral
axis, is of in
nterest because the greatesst stress occuurs at this poiint. The distaance from thee neutral
axis to the extreme fiber may
m be taken
n as c and the flexure formuula becomes
M
Mc M
(6.2)
=
fb =
I
S
where
S = section modu
ulus
fb = extreme fiberr bending streess
The moment thatt causes the extreme
e
fiber to reach the yield stress, Fy, is called tthe yield
moment, My. The corresp
ponding stresss and strain diiagrams are shhown in Figuure 6.2b. If the load is
increased beeyond the yieeld moment, the
t strain in thhe extreme fiiber increasess; however, thhe stress
remains at Fy because th
hese fibers arre behaving aas depicted bby the plateauu on the stress-strain
diagram, sho
own previoussly in Figuress 3.5 and 3.6. The stress att some pointss on the crosss section
closer to thee neutral axis also reach th
he yield stresss while those even closer, from the neuutral axis
to the locatio
on dimension
ned as y in Fig
gure 6.2c, rem
main elastic.
As the
t moment continues to increase, thee portion of tthe cross secttion experienncing the
yield stress continues to increase untill the entire seection experieences the yielld stress, as sshown in
Figure 6.2d.. Note that th
here is no corrresponding sstrain diagram
m given for thhis loading condition
since the strrain would bee unlimited over
o
the entirre cross sectioon. The mom
ment that cauuses this
stress distrib
bution is calleed the plastic moment, Mp. Since the strrain is unlimiited, the crosss section
undergoes unlimited
u
rotaation and the condition is referred to aas a plastic hiinge. A crosss section
that is capab
ble of attainin
ng this stresss distribution and correspoonding momeent is referredd to as a
compact secction. At eveery stage of lo
oading, equiliibrium of thee cross sectionn requires, at all
Chapter 6
Figure 6.3
6
Bending Mem
mbers 191
Equilibriium in a Doub
bly Symmetriical Wide-Flaange Shape
times, th
hat the total in
nternal tensio
on force be eequal to the ttotal internal compressionn force. The
basic principles of strrength of matterials are adddressed in nuumerous textts, such as M
Mechanics of
Materialls.1
For
F the doubly symmetricc wide flangee shape show
wn in Figure 6.3, equilibriium for the
plastic moment
m
occurrs when the portion
p
of the shape abovee the elastic nneutral axis iss stressed to
the yield
d stress in com
mpression wh
hile the portioon below the elastic neutrral axis is streessed to the
yield streess in tension
n. For a nonssymmetric shaape, the areaa above the elastic neutrall axis is not
equal to the area below
w the elastic neutral axis. Thus, a new axis that gives equal areaas in tension
and comp
pression musst be defined. This new axxis dividing thhe section intto two equal areas is the
plastic neutral axis (P
PNA). For sym
mmetric shappes the elasticc and plastic nneutral axes ccoincide, as
was the case for the wide flange. For nonsym
mmetric shapees these neuttral axes are at different
locationss.
Because
B
equiilibrium meaans that the tension and compressionn forces are equal and
opposite,, they form a force couple.. Although m
moments can bbe taken abouut any reference point for
this case,, it is common practice to take momentts about the P
PNA. The mom
ment that corrresponds to
this fully
y yielded stresss distribution
n, Mp, is deterrmined from
(6.3)
M p = Fy ( Ac yc ) + Fy ( At yt )
where At and Ac are the
t equal tenssion and com
mpression areaas, respectiveely, and yc annd yt are the
distancess from the centroid
c
of th
he area to thhe PNA for the tension and compresssion areas,
respectiv
vely. Equation
n 6.3 may be simplified to
⎛ A⎞
M p = Fy ⎜ ⎟ ( yc + yt )
(6.4)
⎝2⎠
The
T two term
ms multiplied by the yieldd stress are fuunctions onlyy of the geom
metry of the
cross secction and aree normally co
ombined and called the pplastic sectionn modulus, Z
Z. Thus, the
plastic moment
m
is giveen as
(6.5)
M p = Fy Z
The plasttic section mo
odulus is tabu
ulated for all aavailable shappes in Part 1 oof the Manuaal.
Chapter
C
F off the Specifica
ation containns the provisiions for design of flexuraal members
subject to
o bending. Fo
or a given beaam to attain itts full plastic moment strenngth, it must be compact
and satissfy the laterall support criteeria establisheed in Sectionn F2. If these criteria are nnot met, the
strength is defined as something leess than Mp. T
The criteria too be satisfied are defined bby two limit
1
Philpot, Tim
mothy. Mechan
nics of Materia
als: An Integra
ated Learning SSystem. Wiley, 2013.
1992 Chapter 6
Bending Members
M
states in add
dition to yield
ding: local bu
uckling (a nonncompact shaape) and lateral torsional bbuckling.
These limit states and theeir impact on beam strength
th are discusseed in Sections 6.4 and 6.5..
EX
XAMPLE 6.1
6
Pllastic Momeent
Str
trength for a
Syymmetric Sh
hape
Goa
al:
Deterrmine the plaastic moment strength of a W-shape uusing the moddel of
three rectangular plates
p
and com
mpare the callculated plastiic section moodulus
to thee value determ
mined from thhe Manual.
Giv
ven:
A W2
24×192 is mo
odeled as show
wn in Figure 66.4. Assume Fy = 50 ksi.
SO
OLUTION
Step
p 1:
Deterrmine the locaation of the pllastic neutral axis.
Becau
use the shape is symmetricc, the plastic nneutral axis iss located on thhe
axis of
o symmetry.
Step
p 2:
Deterrmine the plaastic section m
modulus as tthe sum of thhe moments oof the
areas about the plaastic neutral aaxis.
A
A
⎛
⎞
Z = ( yc + yt ) = 2 ⎜ Af y f + w yw ⎟
2
2
⎠
⎝
⎡
⎛ 22.58 11.46 ⎞ 22.588 ( 0.810 ) ⎛ 22.58 ⎞ ⎤
3
+
Z = 2 ⎢13.0 (1.4
46 ) ⎜
⎟+
⎜
⎟ ⎥ = 560 inn.
2
2
2
4
⎠
⎝
⎝
⎠
⎦
⎣
Step
p 3:
Deterrmine the plasstic moment sstrength as thhe plastic secttion modulus times
the yiield stress using Equation 66.5
M p = Fy Z = 550 ( 560 ) = 288, 000 in.-kipss
or
M p = 28,,000 12 = 23330 ft-kips
Step
p 4:
Comp
pare the calcculated plasttic section m
modulus valuue with that from
Manu
ual Table 1-1..
From
m the table, Zx = 559 in.3.
This shows
s
that th
he impact of tthe simplificaation in usingg rectangular plates
and ig
gnoring the fiillets at the flaange-web junnction is smalll.
Figure
F
6.4 W24×192 M
Model for Exam
mple 6.1
Chapter 6
Figure 6.5
EXAMPLE
E 6.2
Plastic Secction
Modulus fo
or a
Nonsymmeetric
Shape
SOLUTION
N
Goal:
G
Given:
G
Step
S
1:
Bending Mem
mbers 193
T-Beam Moodel for Exam
mple 6.2
Lo
ocate the plastic neutral axxis and determ
mine the plastiic section moodulus for
a WT.
W
A WT12×51.5 modeled as ttwo plates is shown in Figgure 6.5. Asssume that
Fy = 50 ksi.
Deetermine the area
a of the T--shape.
Aflange = 9.00 ( 0.980 ) = 8.82 in.2
Astem = 0.550 (12.3 − 00.980 ) = 6.23 iin.2
Atotal = 8.82 + 6.23 = 15.1 in.2
Step
S
2:
Deetermine one--half of the arrea, because one-half of thhe area must be above
thee plastic neuttral axis and oone-half mustt be below.
Atotal 15.1
=
= 77.55 in.2
2
2
Step
S
3:
Deetermine wheether the plaastic neutral aaxis is in thee flange or tthe stem.
Beecause half of
o the area iss less than tthe area of thhe flange, thhe plastic
neeutral axis is in
i the flange aand
7.55
yp =
= 0 .839 in.
9.0
wiith the plasticc neutral axis measured froom the top of the flange.
Step
S
4:
Deetermine the plastic sectioon modulus aas the sum oof the momennts of the
areeas about the plastic neutraal axis.
⎛ 0.839 ⎞
⎛ 0.980 − 0.839 ⎞
Z = 7.55 ⎜
⎟
⎟ + ( 8.82 − 77.55 ) ⎜
2
⎝
⎠
⎝ 2 ⎠
12.3 − 0.980 ⎞
⎛
+ 6.23 ⎜ 0.9980 − 0.839 +
⎟
2
⎠
⎝
= 3.17 + 0.08895 + 36.1 = 339.4 in.3
Step
S
5:
Co
ompare these values with tthe values in M
Manual Tablee 1-8.
y p = 0.841 in. and Z = 39.2 in..3
Th
his shows thaat the impact of the simplification in ussing rectangullar plates
an
nd ignoring th
he fillets at thee flange-web junction is sm
mall.
1994 Chapter 6
Bending Members
M
Ano
other useful term is the shape factorr defined ass the ratio off the plastic section
modulus to
o the elastic section
s
modu
ulus,
shape factoor =
Z
S
Values for the shape factor
f
are no
ot given in thhe Manual bbut are readdily determinned and
their signifi
ficance will be
b discussed
d later.
6.33
DESIGN
N OF COMPACT LAT
TERALLY SUPPORTE
S
ED WIDE-F
FLANGE B
BEAMS
Section F1(b
b) of the Speecification requires that aall beams be restrained at the supportss against
twist about their longitud
dinal axis. Fo
or a compactt beam to attaain its plasticc moment streength, it
must also bee laterally sup
pported at som
me specified length along its compresssion flange. T
This may
be accompliished through
h the attachmeent of intermeediate beams or by attachinng decking att regular
intervals as shown in Fiigure 6.6. When
W
this is tthe case the beam is saidd to have fulll lateral
support. Secction 6.2 sho
owed that thee nominal strrength of a ccompact mem
mber with fulll lateral
support—that is, in whicch no bucklin
ng limit statees control—iss determined by the limit state of
yielding. For this limit staate, Specification Section F
F2 provides tthat
SC F2-1)
(AIS
M n = M p = Fy Z
Speccification Secction F1 also indicates
i
thatt for all flexurral limit statess, design strenngth
and allowab
ble strength arre to be determ
mined using
φ = 0.900 ( LRFD ) Ω = 1.67 ( AS
SD )
The design basis from Section
ns B3.1 and B
B3.2, discusseed in Chapterr 1, are repeatted here.
The requirem
ment for ASD
D is
R
(AIS
SC B3-2)
Ra ≤ n
Ω
ment for LRF
FD is
The requirem
Ru ≤ φRn
(AIS
SC B3-1)
ure 6.7 showss a portion off Manual Tabble 3-2, whichh lists W-shappes in order oof major
Figu
axis plastic section mod
dulus. Shapes are groupedd so the leastt weight mem
mber with thee largest
plastic sectio
on modulus is
i given in bo
old at the top of each grouup. This faciliitates selectioon of the
most econom
mical W-shap
pe for beams controlled byy the limit staate of yieldingg. This figuree will be
used in the following
f
exaample to aid in
n the design oof a compact,, laterally suppported beam.
Figure 6.66 Lateral Suupport of Beaam
Using Steeel Deck
Chapter 6
Figure 6.7
6
W-Shapees: Selection by Zx
Bending Mem
mbers 195
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with Permissionn. All rights resserved.
196 Chapter 6
Bending Members
EXAMPLE 6.3a
Beam Design by
LRFD
Goal:
Select the least-weight wide flange member for the conditions given.
Given:
An A992 beam, simply supported at both ends, spans 20 ft and is loaded at
midspan with a dead load of 8.0 kips and a live load of 24.0 kips, as shown
in Figure 6.8, in addition to its self-weight. Assume full lateral support and
a compact section.
SOLUTION
Step 1:
Determine the required strength using the governing LRFD load
combination from Section 2.4.
Pu = 1.2 PD + 1.6 PL = 1.2 ( 8.0 ) + 1.6 ( 24.0 ) = 48.0 kips
Mu =
Step 2:
Pu L 48 ( 20 )
=
= 240 ft-kips
4
4
Determine the required plastic section modulus. For a compact, fully braced
section,
M n = M p = Fy Z
Thus, because Specification Section B3.1 provides that the required
moment be less than or equal to the available moment, M u ≤ φM n = φFy Z ,
and
240 (12 )
M
Z req = u =
= 64.0 in.3
φFy 0.90 ( 50 )
Step 3:
Using the required plastic section modulus, select the minimum-weight Wshape from the plastic section modulus economy table shown in Figure 6.7.
Start at the bottom of the Zx column and move up until a shape in bold with
at least Zx = 64.0 in.3 is found.
Select W18×35 with Z=66.5 in.3
This is the most economical W-shape, based on the section weight, which
provides the required plastic section modulus.
Step 4:
An alternative approach, also using Figure 6.7, would be to enter the table
with the required moment, Mu = 240 ft-kips, and proceed up the φMn
column. The same section will be selected with this approach.
Step 5:
Determine the additional required strength based on the actual weight of the
chosen beam. The beam weighs 35 lb/ft, which gives an additional moment
of
⎛ 0.035 ( 20 )2 ⎞
⎟ = 1.2 (1.75 ) = 2.10 ft-kips
M u ( self -weight ) = 1.2 ⎜
⎜
⎟
8
⎝
⎠
Step 6:
Since the maximum moment for both the applied load and the self-weight
occur at the same point, combine this moment with the moment due to
superimposed load to determine the new required strength.
M u = 240 + 2.10 = 242 ft-kips
Chapter 6
Bending Members 197
Step 7:
Determine the new required plastic section modulus.
( 242 )(12 ) = 64.5 in.3
M
Z req = u =
φFy
0.90 ( 50 )
Step 8:
Make the final selection. This required plastic section modulus is less than
that provided by the W18×35 already chosen. Therefore, select the
W18×35
Note that this is a compact section.
EXAMPLE 6.3b
Beam Design By
ASD
Goal:
Select the least-weight wide flange member for the conditions given.
Given:
An A992 beam, simply supported at both ends, spans 20 ft and is loaded
at mid-span with a dead load of 8.0 kips and a live load of 24.0 kips, as
shown in Figure 6.8. Assume full lateral support and a compact section.
SOLUTION
Step 1:
Determine the required strength using the ASD load combinations from
Section 2.4.
Pa = PD + PL = ( 8.0 ) + ( 24.0 ) = 32.0 kips
Ma =
Step 2:
Determine the required plastic section modulus. For a compact, fully
braced section,
M n = M p = Fy Z
Thus, because Specification Section B3.2 provides that the required
moment be less than or equal to the available moment,
M a ≤ M n /Ω = Fy Z /Ω, and
Z req =
Step 3:
Pa L 32.0 ( 20 )
=
= 160 ft-kips
4
4
160 (12 )
160 (12 )
Ma
=
=
= 64.0 in.3
Fy /Ω ( 50 / 1.67 )
30
Using the required plastic section modulus, select the minimum-weight
W-shape from the plastic section modulus economy table shown in Figure
6.7. Start at the bottom of the Zx column and move up until a shape in
bold with at least Zx = 64.0 in.3 is found.
Select W18×35 with Z = 66.5 in.3
This is the most economical W-shape, based on section weight, which
provides the required plastic section modulus.
Step 4:
An alternative approach, again using Figure 6.7, would be to enter the
table with the required moment, Ma = 160 ft-kips, and proceed up the
Mn/Ω column. The same section will be selected with this approach.
Step 5:
Determine the additional required strength based on the actual weight of
the chosen beam. The beam weighs 35 lb/ft, which gives an additional
1998 Chapter 6
Bending Members
M
momeent of
M a ( self -weight ) =
0.035 ( 20 )
8
2
= 1.75 ft-kips
Step
p 6:
Sincee the maximum
m moment foor both the appplied load annd the self-weeight
occurr at the same point, combbine this mom
ment with thee moment duue to
superrimposed load
d to determinee the new requ
quired strengthh.
M a = 1600 + 1.75 = 1622 ft-kips
Step
p 7:
Deterrmine the new
w required plaastic section m
modulus.
162
12
(
)(
)
Ma
Z req =
= 64.8 in.3
=
Fy /Ω
30
Step
p 8:
Makee the final seleection. This reequired plastiic section moodulus is less tthan
that provided
p
by th
he W18×35 allready chosenn. Therefore, select the
W18×35
Note that this is a compact
c
sectiion
Note that since the
t live-to-deead load ratioo in Examplee 6.3 is 3, beefore inclusion of the
self-weight, the required plastic sectio
on modulus iin Examples 6.3a and 6.3bb are the sam
me. Once
the self-weig
ght is added to
t the problem
m, the live-to--dead load raatio is no longger 3 and the required
plastic sectio
on modulus iss seen to be slightly differeent between tthe LRFD andd ASD exampples.
Figure 6.8 Beam Used in Example 6.3a
6
Chapter 6
Figure 6..9
EXAMPLE
E 6.4a
Beam Desiign by
LRFD
SOLUTION
N
Bending Mem
mbers 199
Framing Plan for Exam
mple 6.4.
Goal:
G
Deesign a W-sh
hape floor beaam for the inntermediate beam marked A on the
flo
oor plan show
wn in Figure 66.9.
Given:
G
Th
he beam is loaded uniform
mly from the ffloor with a live load of 80 pounds
peer square foott (psf) and a ddead load in aaddition to thhe beam self-w
weight of
60
0 psf. The beaam will have full lateral suupport providded by the flooor deck,
an
nd a compact section will bbe selected. U
Use A992 steeel.
Step
S
1:
Deetermine the required
r
loadd and momentt.
wu = (1.2 wD + 1.6 wL ) Ltrib = (1.2 ( 600 ) + 1.6 ( 80 ) ) (10 ) = 2000 1bb/ft
wu L2 2.0 kips/ft ( 26 ft )
=
= 169 ft-kips
8
8
2
Mu =
Step
S
2:
Deetermine the required
r
plasttic section moodulus.
Fo
or a compactt, fully braceed beam, Mn = Mp = FyZ.. Section B3.1 of the
Sp
pecification reequires that
M u ≤ φM n = φFy Z
Th
herefore
(169 )(112 ) = 45.1 in..3
M
Z req = u =
φFy 0.90 ( 550 )
Step
S
3:
Ussing the plasttic section m
modulus econoomy table in Figure 6.7, sselect the
mo
ost economiccal W-shape bbased on leastt weight.
W1 4×30 with Z = 47.3 in.3
Step
S
4:
Deetermine the additional
a
reqquired strengtth based on thhe actual weigght of the
ch
hosen beam. The
T beam weiighs 30 lb/ft, which gives an additionall moment
off
200 Chapter 6
Bending Members
M u ( self -weight )
Step 5:
⎛ 0.030 ( 26 )2
= 1.2 ⎜
⎜
8
⎝
⎞
⎟ = 1.2 ( 2.54 ) = 3.04 ft-kips
⎟
⎠
Since the maximum moment for both the applied load and the self-weight
occur at the same point, combine this moment with the moment due to
superimposed load to determine the new required strength.
M u = 169 + 3.04 = 172 ft-kips
Step 6:
Determine the new required plastic section modulus.
(172 )(12 ) = 45.9 in.3
M
Z req = u =
φFy 0.90 ( 50 )
Step 7:
Make the final selection. This required plastic section modulus is less than
that provided by the W14×30 already chosen. Therefore, select the
W14×30
Step 8:
As shown in Example 6.3, an alternative approach is to use the required
moment, Mu = 172 ft-kips, and enter the φM n column to determine the
same W-shape.
EXAMPLE 6.4b
Beam Design by
ASD
Goal:
Design a W-shape floor beam for the intermediate beam marked A on the
floor plan shown in Figure 6.9.
Given:
The beam is loaded uniformly from the floor with a live load of 80 psf and
a dead load in addition to the beam self-weight of 60 psf. The beam will
have full lateral support provided by the floor deck, and a compact section
will be selected. Use A992 steel.
SOLUTION
Step 1:
Determine the required load and moment.
wa = ( wD + wL ) Ltrib = ( 60 + 80 )(10 ) = 1400 1b/ft
w L2 1.40 kips/ft ( 26 ft )
Ma = a =
= 118 ft-kips
8
8
Determine the required plastic section modulus.
2
Step 2:
For a compact, fully braced beam, Mn = Mp = FyZ. Section B3.2 of the
Specification requires that
Fy Z
M
Ma ≤ n =
Ω
Ω
Therefore,
(118 )(12 ) = 47.2 in.3
Ma
Z req =
=
30
Fy /Ω
Chapter 6
Bending Members 201
Step 3:
Using the plastic section modulus economy table in Figure 6.7, select the
most economical W-shape based on least weight.
W14×30 with Z = 47.3 in.3
Step 4:
Determine the additional required strength based on the actual weight of the
chosen beam. The beam weighs 30 lb/ft, which gives an additional moment
of
2
0.030 ( 26 )
M a ( self -weight ) =
= 2.54 ft-kips
8
Step 5:
Since the maximum moment for both the applied load and the self-weight
occur at the same point, combine this moment with the moment due to
superimposed load to determine the new required strength.
M a = 118 + 2.54 = 121 ft-kips
Step 6:
Determine the new required plastic section modulus.
(121)(12 ) = 48.4 in.3
Ma
Z req =
=
Fy /Ω
30
Step 7:
Make the final selection. This required plastic section modulus is more than
that provided by the W14×30. Therefore, select the next most economical
shape from the table
W16×31
By inspection, this additional 1 lb/ft is acceptable.
Step 8:
As shown in Example 6.3, an alternative approach is to use the required
moment, Ma = 121 ft-kips, and enter the Mn /Ω column to determine the
same W-shape.
An alternative approach for including beam self-weight is to start with a preliminary
estimate of the beam weight. That way it will already be included when the first trial shape is
selected. Then the assumed weight is compared to the weight of the selected shape. If the
assumed weight is greater than the selected weight, the beam is adequate. If there is a significant
discrepancy, either over or under, a second trial shape can be selected.
6.4 DESIGN OF COMPACT LATERALLY UNSUPPORTED WIDE-FLANGE BEAMS
6.4.1
Lateral-Torsional Buckling
The compression region of a bending member cross section has a tendency to buckle similarly to
how a pure compression member buckles. The major difference is that the bending tension region
helps to resist that buckling. The upper half of the wide flange member in bending acts as a T in
pure compression. This T is fully braced about its horizontal axis by the web so it will not buckle
in that direction, but it can be unbraced for some distance for buckling about its vertical axis.
Thus, it will have a tendency to buckle laterally. Because the tension region also tends to restrain
the lateral buckling, the shape actually buckles in a combined lateral and torsional mode. The
beam midspan deflects in the plane down and buckles laterally, causing it to twist. This is
illustrated in Figure 6.10 by a beam in a laboratory test that has failed due to lateral-torsional
2002 Chapter 6
Bending Members
M
Figure 6.10
Example of
o Lateral-Torsional Bucklling of a Beam
m in a Test Frrame
Photo courtessy of Donald W.
W White
buckling. Th
he beam appeears to have a tendency to fall over on itts weak axis. In order to reesist this
tendency, Sp
pecification Sections
S
B3.4 and F1(b) re quire that all bending mem
mbers be restrrained at
their supporrt points again
nst rotation ab
bout their longgitudinal axiss. If the beam
m has sufficiennt lateral
and/or torsio
onal support along its len
ngth, a comppact cross secction can devvelop the yielld stress
before buckling. This is the
t case that was discusseed in Sectionss 6.2 and 6.3.. If the beam tends to
buckle before the yield stress is reacched, the nom
minal momennt strength is less than thee plastic
moment.
To ensure
e
that a beam cross section
s
can deevelop its fulll plastic mom
ment strength without
lateral-torsio
onal buckling
g, Specificatio
on Section F2 .2 limits the sslenderness too
Lb
E
(6.6)
≤ 1.76
ry
Fy
where
Lb = unbraced len
ngth of the co
ompression flaange
ry = radius of gyrration for the shape about tthe y-axis
The practicaal application
n of this limiitation is to uuse the unbrraced length alone, rather than in
combination
n with the rad
dius of gyrattion in the foorm of a slennderness ratioo. This resultts in the
requirementt for attaining
g the full plaastic momennt strength giiven the Speccification as Lb ≤ L p
where
E
SC F2-5)
(AIS
L p = 1.76 ry
Fy
t maximum
m unbraced len
ngth that wouuld permit thee shape to reach its plastic moment
Thus, Lp is the
strength. Th
his value is tabulated
t
for each W-shappe and can bbe found in M
Manual Tablle 3-2, a
portion of which
w
is shown in Figure 6.7, and severaal other locatiions in the M
Manual. Since A992 is
the preferred
d specificatio
on for W-shap
pes, the tabullated values iin Table 3-2 aare based on Fy = 50
ksi.
Wheen the unbraaced length of
o a beam exxceeds Lp, itts strength is reduced duee to the
tendency off the memberr to buckle laaterally at a l oad level bellow what woould cause thee plastic
moment to be
b reached. For
F simply sup
pported doubbly symmetricc members, a closed-form solution
Chapter 6
Bending Members 203
is well established in the literature. The Guide to Stability Design Criteria for Metal Structures2
shows the development of the Specification equations based on the closed-form solution.
The elastic lateral-torsional buckling (LTB) strength of a W-shape is given in
Specification Section F2.2 as
(AISC F2-3)
M n = Fcr S x ≤ M p
where
Fcr =
Cb π2 E
⎛ Lb ⎞
⎜r ⎟
⎝ ts ⎠
2
1 + 0.078
Jc ⎛ Lb ⎞
S x ho ⎜⎝ rts ⎟⎠
2
(AISC F2-4)
which can be combined to give the nominal moment strength for elastic lateral-torsional buckling
as
Mn =
Cb π2 ES x
⎛ Lb ⎞
⎜r ⎟
⎝ ts ⎠
2
1 + 0.078
Jc ⎛ Lb ⎞
S x ho ⎜⎝ rts ⎟⎠
2
(6.7)
A beam buckles elastically if the actual stress in the member at buckling does not exceed Fy at
any point. Because all hot rolled shapes have built-in residual stresses as discussed for columns in
Section 5.3.4, there is a practical limit to the usefulness of this elastic LTB equation. The
Specification sets the level of the residual stress at 0.3Fy so that only 0.7Fy is available to resist a
bending moment from applied load while still behaving elastically. This limit results in the
maximum elastic moment, M rLTB = 0.7 Fy S x . If Equation F2-4 is set equal to 0.7Fy, a limiting
unbraced length, Lr, beyond which the member buckles elastically may be determined. This limit
as provided in Specification Section F2 is
Lr = 1.95rts
E
0.7 Fy
2
Jc
⎛ Jc ⎞
⎛ 0.7 Fy ⎞
+ ⎜
+ 6.76 ⎜
⎟
⎟
S x ho
⎝ E ⎠
⎝ S x ho ⎠
2
(AISC F2-6)
With an unbraced length between Lp and Lr, the beam behaves inelastically. In this range, the
nominal moment, Mn, is reasonably well predicted by a straight-line equation when compared to
the results of laboratory tests. The Specification equation for the nominal moment strength is
⎡
⎛ Lb − Lp ⎞ ⎤
M n = Cb ⎢ M p − ( M p − 0.7 Fy S x ) ⎜
(AISC F2-2)
⎟⎥
⎝ Lr − Lp ⎠ ⎦⎥
⎣⎢
Equation F2-2 represents interpolation between the end points of a straight line. These end points
correspond to the plastic moment, Mp given by Equation F2-1, at an unbraced length of Lp and an
elastic moment, M rLTB = 0.7 Fy S x , at an unbraced length Lr. Although the determination of Fcr and
Lr from Equations F2-4 and F2-6 may look somewhat daunting, the Manual has extensive tables
that permit their determination with little effort, and the equations also can be readily automated
using spreadsheet calculations.
The complete picture of the nominal moment strength of a beam as a function of
unbraced length is presented in Figure 6.11, where the curve segments are labeled according to
the appropriate strength equations. Curves similar to these are available in Manual Table 3-10 for
each W-shape and Table 3-11 for C- and MC-shapes. For all of these curves, the moment gradient
factor, Cb, is taken as 1.0. Examples of these curves are given in Figure 6.12.
2
Ziemian, R. D. Guide to Stability Design Criteria for Metal Structures, 6th ed. New York: John Wiley & Sons, 2010.
2004 Chapter 6
Bending Members
M
Figure 6.11 Lateral-To
orsional Buckling
EX
XAMPLE 6.5a
6
Beeam Design by
LR
RFD
Goa
al:
Select a W-shape beam
b
consideering the unbrraced length oof the compreession
flangee using Manu
ual Table 3-100.
Giv
ven:
The beam
b
spans 30.0
3
ft and haas a requiredd strength of 2282 ft-kips. IIt will
have lateral supporrt at its ends aand at midspaan. Use A9922 steel.
SO
OLUTION
Step
p 1:
Deterrmine the req
quired momeent strength aand unbracedd length. From
m the
given
n information,,
M u = 282 ft-kipps
Lb = 15.0 ft
Step
p 2:
Select the least-w
weight W-shappe from the portion of M
Manual Tablee 3-10
given
n in Figure 6.1
12.
Enterr Table 3-10 at
a a design strrength of ϕMn = 282 ft-kipps and an unbbraced
length
h of Lb = 15.0 ft. The firsst solid line tto the upper right indicatees the
least-weight shapee sufficient to carry this mooment at this unbraced lenggth.
Step
p 3:
Thereefore, select a
W18×55
EX
XAMPLE 6.5b
6
Beeam Design by
AS
SD
Goa
al:
Select a W-shape beam
b
consideering the unbrraced length oof the compreession
flangee using Manu
ual Table 3-100.
Giv
ven:
The beam
b
spans 30.0
3
ft and haas a requiredd strength of 188 ft-kips. IIt will
have lateral supporrt at its ends aand at midspaan. Use A9922 steel.
Chapter 6
Figure 6.12
6
Bending Mem
mbers 205
W-Shap
pes: Available Moment veersus Unbraceed Length
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with Permissionn. All rights resserved.
206 Chapter 6
SOLUTION
Bending Members
Step 1:
Determine the required moment strength and unbraced length. From the
given information,
M a = 188 ft-kips
Lb = 15.0 ft
Step 2:
Select the least-weight W-shape from the portion of Manual Table 3-10
given in Figure 6.12.
Enter Table 3-10 at an allowable strength of Mn/Ω = 188 ft-kips, and an
unbraced length of Lb = 15 ft. The first solid line to the upper right indicates
the least-weight shape sufficient to carry this moment at this unbraced
length.
Step 3:
Therefore, select a
W18×55
When Mn is to be determined through a calculation, an additional simplification can be
applied to the straight-line portion of the curve. From Equation F2-2, the ratio
⎛ M p − 0.7 Fy S x ⎞ ⎛ M p − M rLTB ⎞
⎜
⎟=⎜
⎟
Lr − Lp
⎝
⎠ ⎝ Lr − Lp ⎠
is a constant for each beam shape and is defined as the variable BF when written in terms of
nominal strength. This constant is tabulated in Manual Table 3-2, and shown in Figure 6.7, as
φBF and BF/Ω for LRFD and ASD, respectively. Thus, for nominal strength, with Cb again taken
as 1.0, Equation F2-2 can be rewritten as
(6.8a)
M n = M p − BF ( Lb − L p )
and for LRFD as
(6.8b)
φM n = φM p − φBF ( Lb − L p )
and for ASD as
M n M p BF
=
−
( Lb − Lp )
(6.8c)
Ω
Ω
Ω
Further inspection of Manual Table 3-2 shows that two additional moment values are given,
φb M rx and M rx Ωb . These values are the LRFD and ASD values for M rLTB = 0.7 Fy S x .
6.4.2
Moment Gradient
The nominal strength of a beam limited by lateral-torsional buckling as defined when Cb = 1.0
assumes that the moment is uniform across the entire length of the beam as shown in Figure
6.13a. For the limit state of lateral-torsional buckling this is the most severe loading case possible,
because it stresses the entire length of the beam to its maximum level, just as for a column. For
any other loading pattern, and resulting moment diagram, the compressive force in the beam
would vary with the moment diagram. Thus, the reduced stresses along the member length would
result in a reduced tendency for lateral-torsional buckling and an increase in strength. The
variation in moment over a particular unbraced segment of the beam is called the moment
gradient, which describes how the moment varies along a specific length.
Chapter 6
Figure 6.13
6
Bending Mem
mbers 207
Resistaance to the Maaximum Mom
ment under Thhree Differentt Loading Coonditions
For
F the normaal case of loaading that prooduces a mom
ment diagram
m that is not constant, the
nominal moment stren
ngth calculateed through Eqquations F2-22, F2-3, and F
F2-4 will be iincreased to
account for the mom
ment gradient when Cb is not taken ass 1.0. The laateral-torsional buckling
modificaation factor, Cb, accounts for
f nonuniform
m moment diiagrams overr the unsuppoorted length.
There haave been many
y proposed eq
quations for thhe determinat
ation of Cb in the literature.. The Guide
to Stabillity Design Criteria
C
for Metal
M
Structurres (see footnnote 2) provides a good ssummary of
these pro
oposals. The equation pressented in Speccification Secction F1 is
122.5M max
Cb =
(F1-1)
2.5M max + 3M A + 4M B + 3M C
where
Mmax = absolu
ute value of maximum
m
mom
ment in the unnbraced segm
ment
MA = absolutee value of moment at quartter point of thhe unbraced seegment
MB = absolutee value of moment at centeerline of the uunbraced segm
ment
MC = absolutee value of mo
oment at threee-quarters poinnt of the unbrraced segmennt
Cb = 1.0 for a uniform
m moment an
nd can be connservatively ttaken as 1.0 ffor other casees. In doing
so, howeever, the desig
gner may be sacrificing
s
siggnificant econnomy. Figuree 6.14 providees examples
of loadin
ng conditions,, bracing locaations, and thee correspondiing Cb valuess found in Maanual Table
3-1.
2008 Chapter 6
Bending Members
M
Figure 6.14 Values forr Cb for Simplly Supported Beams
Copyright © American Insttitute of Steel Construction.
C
R
Reprinted with Permission. A
All rights reservved.
The effect of thee moment graadient factor, Cb, is to alteer the nominnal moment–uunbraced
length relatiionship by a constant,
c
as sh
hown in Figuure 6.15. The shaded area sshows the inccrease in
moment cap
pacity as a ressult of the usee of Cb. Regaardless of how
w small the unnbraced lengtth might
be, the nom
minal moment strength of the
t member ccan never excceed the plasstic moment sstrength.
Thus, the up
pper portion of
o the curve in
n Figure 6.15 is terminatedd at Mp.
Figuree 6.15 Effecct of Moment
Gradiennt
Chapter 6
Bending Mem
mbers 209
Figure 6.116 Beam U
Used in Exampple 6.6
Goal:
G
EXAMPLE
E 6.6a
Beam Stren
ngth and
Design byy LRFD
Considerin
ng
Moment Gradient
G
SOLUTION
N
Deetermine wheether the W144×34 beam shhown in Figuure 6.16 will carry the
giv
ven load. Co
onsider the m
moment gradiient (a) Cb = 1.0 and (b) Cb from
Eq
quation F1-1, and then (c)) determine thhe least-weighht section to carry the
loaad using the correct
c
Cb .
Given:
G
Figure 6.16 shows a beam that is fixedd at one suppport and pinned at the
oth
her. The beam
m has a conceentrated deadd load of 8 kiips and a concentrated
liv
ve load of 24
4 kips at midsspan. A lateraal brace is loocated at eachh support
an
nd at the load point.
Step
S
1:
Deetermine the required
r
strenngth. For the lload combinaation of 1.2D + 1.6L,
Pu = 1.2((8.0) + 1.6(244.0) = 48.0 kipps
Step
S
2:
Deetermine thee maximum moment frrom an elasstic analysiss of the
ind
determinate beam.
b
At the fixed end thee moment is shown in Figgure 6.16
as
M u = 180 ftt-kips
Step
S
3:
Deetermine the needed valuues from Ma
Manual Table 3-2 in ordeer to use
Eq
quation 6.8b. For a W14×334,
Z = 54.6 in.3, Lp = 5.400 ft, Lr = 15.66 ft,
φMp = 2055 ft-kips, andd φBF = 7.55 kkips
210 Chapter 6
Bending Members
Part a
Step 4:
Cb = 1.0
Determine the design moment strength for lateral bracing of the
compression flange at the supports and the load, Lb = 10 ft.
Because
Lb = 10 ft > L p
< Lr
φM n = Cb ( φM p − φBF ( Lb − L p ) ) ≤ φM p
φM n = 1.0 ( 205 − 7.55(10.0 − 5.40) ) = 170 ft-kips < 205 ft-kips
As an alternative approach, Manual Table 3-10 can be entered with an
unbraced length of 10 ft and the design strength of the W14×34 determined
to be 170 ft-kips.
Therefore, since
φM n = 170 ft-kips < M u = 180 ft-kips
The W14×34 beam will not work if Cb = 1.0
Part b
Step 5:
Use the Calculated Value of Cb
Determine the correct Cb for the two unbraced segments of the beam.
For the unbraced segment BC, the first beam in Figure 6.14 can be used to
obtain Cb = 1.67 since the moment gradient for the segment from the load
to the support in Figure 6.14 is the same as the moment gradient for
segment BC of the beam being considered here. This Cb corresponds to the
maximum moment of 150 ft-kips at point B on the beam. The W14×34 can
resist this moment without consideration of Cb, as shown in part (a) Step 4
above.
For the unbraced segment AB, Figure 6.14 cannot be used and Cb must be
calculated. Using Equation F1-1 and the moment values given in Figure
6.16 at the quarter points of the segment AB,
12.5 (180 )
Cb =
= 2.24
2.5 (180 ) + 3 ( 97.5 ) + 4 (15.0 ) + 3 ( 67.5 )
Step 6:
Determine the design moment strength using the calculated value of Cb and
the design moment strength determined from part (a), with Equation 6.8b
amplified by Cb and limited to φMp.
φM n = Cb ( φM p − φBF ( Lb − L p ) ) ≤ φM p
= 2.24 ( 205 − 7.55(10.0 − 5.40) )
= 2.24(170) = 381 ft-kips > φM p = 205 ft-kips
Therefore, the limiting strength of the beam is
Chapter 6
Bending Members 211
φM n = 205 ft-kips > 180 ft-kips
So the W14×43 is adequate for bending.
Part c
Step 7:
Considering that Cb = 2.24 and 1.67 for the two segments, a smaller section
can be tried.
Assuming the beam can be treated as fully braced as a result of including
the influence of the moment gradient, Cb not equal to 1.0, select from
Manual Table 3-2 a W16×31 which is the lightest W-shape to carry the
given moment, Mu = 180 ft-kips.
Determine the values needed to evaluate the shape from Manual Table 3-2.
φM p = 203 ft-kips, L p = 4.13 ft, Lr = 11.8 ft, φBF = 10.3 kips
Step 8:
First consider unbraced segment BC. As determined in Step 5 above, Cb =
1.67. Because Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, use Equation 6.8b
with Cb.
φM n = Cb ( φM p − φBF ( Lb − Lp ) ) ≤ φM p
= 1.67 ( 203 − 10.3(10.0 − 4.13) )
= 1.67(143) = 239 ft-kips > φM p = 203 ft-kips
Note that for this unbraced segment, without the use of Cb,
φM n = 143 ft-kips < 150 ft-kips
Thus, with Cb included
φM n = 239 ft-kips > φM p = 203 ft-kips
Therefore the design strength is
φM n = 203 ft-kips > 150 ft-kips
Step 9:
For unbraced segment AB, again Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8
ft, and again use Equation 6.8b with Cb.= 2.24 as determined in Step 5.
φM n = Cb ( φM p − φBF ( Lb − Lp ) )) ≤ φM p
= 2.24 ( 203 − 10.3(10.0 − 4.13) ) = 2.24(143)
= 320 ft-kips > φM p = 203 ft-kips
where
φM n = 320 ft-kips > φM p = 203 ft-kips
Step 10:
Thus, the design strength is
φM n = 203 ft-kips > 180 ft-kips
So the W16×31 will also work.
212 Chapter 6
Bending Members
EXAMPLE 6.6b
Beam Strength and
Design by ASD
Considering
Moment Gradient
Goal:
Determine whether the W14×34 beam shown in Figure 6.16 will carry the
given load. Consider the moment gradient (a) Cb = 1.0 and (b) Cb from
Equation F1-1, and then (c) determine the least-weight section to carry the
load using the correct Cb.
Given:
Figure 6.16 shows a beam that is fixed at one support and pinned at the
other. The beam has a concentrated dead load of 8 kips and a concentrated
live load of 24 kips at midspan. A lateral brace is located at each support
and at the load point.
SOLUTION
Step 1:
Determine the required strength. For the load combination of D + L,
Pa = 8.0 + 24.0 = 32.0 kips
Step 2:
Determine the maximum moment from an elastic analysis of the
indeterminate beam. At the fixed end the moment is shown in Figure 6.16
as
M a = 120 ft-kips
Step 3:
Determine the needed values from Manual Table 3-2 in order to use
Equation 6.8b. For a W14×34,
Z = 54.6 in.3, Lp = 5.40 ft, Lr = 15.6 ft,
Mp/Ω =136 ft-kips, and BF/Ω = 5.01kips
Part a
Step 4:
Cb = 1.0
Determine the allowable moment strength for lateral bracing of the
compression flange at the supports and the load, Lb = 10 ft.
Because
Lb = 10 ft > L p
< Lr
M n Ω = ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω
= (136 − 5.01(10.0 − 5.40) ) = 113 ft-kips < 136 ft-kips
As an alternative approach, Manual Table 3-10 can be entered with an
unbraced length of 10 ft and the design strength of the W14×34 determined
to be 113 ft-kips.
Therefore, since
M n Ω = 113 ft-kips < M a = 120 ft-kips
The W14×34 beam will not work if Cb = 1.0
Part b
Step 5:
Use the Calculated Value of Cb
Determine the correct Cb for the two unbraced segments of the beam.
For the unbraced segment BC, the first beam in Figure 6.14 can be used to
obtain Cb = 1.67 since the moment gradient for the segment from the load
to the support in Figure 6.14 is the same as the moment gradient for
Chapter 6
Bending Members 213
segment BC of the beam being considered here. This Cb corresponds to the
maximum moment of 100 ft-kips at point B on the beam. The W14×34 can
resist this moment without consideration of Cb, as shown in part (a) Step 4
above.
For the unbraced segment AB, Figure 6.14 cannot be used and Cb must be
calculated. Using Equation F1-1 and the moment values given in Figure
6.16 at the quarter points of the segment AB,
Cb =
Step 6:
12.5 (120 )
2.5 (120 ) + 3 ( 65.0 ) + 4 (10.0 ) + 3 ( 45.0 )
= 2.24
Determine the allowable moment strength using the calculated value of Cb
and the allowable moment strength determined from part (a), with Equation
6.8b amplified by Cb and limited to Mp/Ω.
M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω
= 2.24 (136 − 5.01(10.0 − 5.40) )
= 2.24(113) = 253 ft-kips > M p Ω = 136 ft-kips
Therefore, the limiting strength of the beam is
M n Ω = 136 ft-kips > 120 ft-kips
So the W14×43 is adequate for bending.
Part c
Step 7:
Considering that Cb = 2.24 and 1.67 for the two segments, a smaller section
can be tried.
Assuming the beam can be treated as fully braced as a result of including
the influence of the moment gradient, Cb not equal to 1.0, select from
Manual Table 3-2 a W16×31 which is the lightest W-shape to carry the
given moment, Ma = 120 ft-kips.
Determine the values needed to evaluate the shape from Manual Table 3-2.
M p Ω = 135 ft-kips, L p = 4.13 ft, Lr = 11.8 ft, BF Ω = 6.86 kips
Step 8:
First consider unbraced segment BC. As determined in Step 5 above, Cb =
1.67. Because Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, use Equation 6.8b
with Cb.
M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω
= 1.67 (135 − 6.86 (10.0 − 4.13) ) = 1.67(94.7)
= 158 ft-kips > M p Ω = 135 ft-kips
Note that for this unbraced segment, without the use of Cb,
M n Ω = 94.7 ft-kips < 100 ft-kips
Thus, with Cb included
M n Ω = 158 ft-kips > M p Ω = 135 ft-kips
214 Chapter 6
Bending Members
Therefore the allowable strength is
M n Ω = 135 ft-kips > 100 ft-kips
Step 9:
For unbraced segment AB, again Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8
ft, and again use Equation 6.8b with Cb.= 2.24 as determined in Step 5.
M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω
= 2.24 (135 − 6.86 (10.0 − 4.13 ) ) = 2.24(94.7)
= 212 ft-kips > M p Ω = 135 ft-kips
where
M n Ω = 212 ft-kips > M p Ω = 135 ft-kips
Step 10:
Thus, the allowable strength is
M n Ω = 135 ft-kips > 120 ft-kips
So the W16×31 will also work.
Manual Table 3-10 was shown in Example 6.5 to be an efficient aid in design considering
unbraced length. When design is to consider moment gradient as well as unbraced length, the
charts in Table 3-10 can be equally useful. Since the curves in Table 3-10 are based on Cb = 1.0,
some modifications to the required moment strength must be made before entering the table.
Starting with Equation F2-2 and dividing both sides by Cb gives M n Cb equal to the nominal
strength based on Cb = 1.0. Thus, if Table 3-10 is entered with M n Cb a member can be selected.
Of course the limit on Equation F2-2 must still be addressed as will be shown in Example 6.7.
EXAMPLE 6.7a
Beam Design by
LRFD
Goal:
Select a W-shape beam, and consider the unbraced length of the
compression flange and the moment gradient, using Manual Table 3-10.
Given
The beam spans 30.0 ft and has a required strength of 282 ft-kips as in
Example 6.5a. It is loaded with a concentrated load at the midspan and will
have lateral support at its ends and at midspan. Use A992 steel.
SOLUTION
Step 1:
Determine the required moment strength and unbraced length. From the
given information,
M u = 282 ft-kips
Lb = 15.0 ft
Step 2:
Determine the appropriate Cb from Figure 6.14.
For a beam with a concentrated load at midspan and lateral supports at the
load and at the ends, Cb = 1.67.
Step 3:
Determine the moment to use for entering Manual Table 3-10.
To account for the increased strength due to the moment gradient, the
required moment is divided by Cb, and that value is used to enter the table
along with the given unbraced length.
Chapter 6
Bending Members 215
Thus, enter the table at the design moment strength
φM n 282
=
=169 ft-kips
1.67
Cb
and unbraced length Lb = 15.0 ft.
Step 4:
The first solid line above and to the right indicates the least-weight shape
sufficient to carry this moment at this unbraced length. Since this is not
found in Figure 6.12, use the actual Manual Table 3-10 and select a
W12×40. This member’s strength curve appears to be right on the
intersection of the required moment strength and given unbraced length.
Step 5:
Determine the maximum strength of the W12×40. Remember that in no
case can ϕMn be greater than ϕMp .
From Manual Table 3-2,
φM p = 214 ft-kips
Since this is less than the required strength, Mu = 282 ft-kips , this W-shape
will not be adequate even with the use of Cb.
Step 6:
Select a shape with φM p ≥ 282 ft-kips from Manual Table 3-2.
Thus , select
W18×40, φM p = 294 ft-kips
Step 7:
Confirm in Manual Table 3-10 that the line, either solid or dashed, for the
W18×40 is above and to the right of the intersection of φM n = 169 ft-kips
and Lb = 15 ft. Thus, select the
W18×40
EXAMPLE 6.7b
Beam Design by
ASD
Goal:
Select a W-shape beam, and consider the unbraced length of the
compression flange and the moment gradient, using Manual Table 3-10.
Given:
The beam spans 30.0 ft and has a required strength of 188 ft-kips as in
Example 6.5b. It is loaded with a concentrated load at the midspan and will
have lateral support at its ends and at midspan. Use A992 steel.
SOLUTION
Step 1:
Determine the required moment strength and unbraced length. From the
given information,
M a = 188 ft-kips
Lb = 15.0 ft
Step 2:
Determine the appropriate Cb from Figure 6.14.
For a beam with a concentrated load at midspan and lateral supports at the
load and at the ends, Cb = 1.67.
216 Chapter 6
Bending Members
Step 3:
Determine the moment to use for entering Manual Table 3-10.
To account for the increased strength due to the moment gradient, the
required moment is divided by Cb, and that value is used to enter the table
along with the given unbraced length.
Thus, enter the table at allowable moment strength
( M n Ω ) Cb = 188 1.67 =113 ft-kips
and unbraced length Lb = 15.0 ft.
Step 4:
The first solid line above and to the right indicates the least-weight shape
sufficient to carry this moment at this unbraced length. Since this is not
found in Figure 6.12, use the actual Manual Table 3-10 and select a
W12×40. This member’s strength curve appears to be right on the
intersection of the required moment strength and given unbraced length.
Step 5:
Determine the maximum strength of the W12×40. Remember that in no
case can M n Ω be greater than M p Ω
From Manual Table 3-2,
M p Ω = 142 ft-kips
Since this is less than the required strength, M a = 188 ft-kips , this W-shape
will not be adequate even with the use of Cb.
Step 6:
Select a shape with M p Ω ≥ 188 ft-kips from Manual Table 3-2.
Thus, select a
W18×40 with M p Ω = 196 ft-kips
Step 7:
Confirm in Manual Table 3-10 that the line, either solid or dashed, for the
W18×40 is above and to the right of the intersection of M n Ω = 113 ft-kips
and Lb = 15 ft. Thus, select the
W18×40
6.5 DESIGN OF NONCOMPACT BEAMS
6.5.1
Local Buckling
Local buckling occurs when a compression element of a cross section buckles under load before
it reaches the yield stress. Because this buckling occurs at a stress lower than the yield stress, the
shape is not capable of reaching the plastic moment. Thus, the shape is not compact and the
nominal strength of the member is something less than Mp. Buckling of the flange and web
Chapter 6
Figure 6.17
6
mbers 217
Bending Mem
Examplle of Flange Local
L
Bucklinng
Photo cou
urtesy of Donalld W. White
elementss, and lateral-torsional bucckling of the ssection, do nnot occur in issolation, so itt is difficult
to illustraate them indiv
vidually. Figu
ure 6.17 prim
marily illustrattes local buckkling of the ccompression
flange off a wide-flang
ge beam duriing loading inn an experim
mental test. Thhese failures ooccur when
the flang
ge or web aree slender, and
d they can bee predicted thhrough the uuse of the plaate buckling
equation discussed in Chapter 5. The
T projectingg flange of a w
wide-flange m
member is considered an
unstiffen
ned element because the weeb supports oonly one edgee, while the otther edge is uunsupported
and free to rotate. Th
he wide-flang
ge web is coonnected at bboth its edgess to the flangges, so it is
considereed a “stiffened” element.
Table
T
B4.1b of the Speciffication proviides the limitting width-to-thickness rattios, λp, for
the flang
ge and web to ensure that the
t full plasticc moment strrength can be reached. Whhen both the
flange an
nd web have slenderness ratios (b/tf annd h/tw, respeectively) lesss than or equal to the λp
values giiven in the taable, the shap
pes are calledd compact shhapes. If eithher element eexceeds this
value, the shape canno
ot be called compact
c
and tthe nominal sstrength must be reduced. T
These latter
shapes arre discussed here.
h
For
F the flang
ge of a W-sh
hape to be ccompact (case 10 in Table B4.1b), itts width-tothicknesss ratio must saatisfy the folllowing limit:
b
E
(6.9)
λ f = ≤ λ ppf = 0.38
t
Fy
where b /t
/ = bf/2tf. Forr the web to be
b compact (ccase 15 in Tabble B4.1b), thhe limiting rattio is
h
E
(6.10)
λ w = ≤ λ pw = 3.76
Fy
tw
Using thee common A9
992 steel with
h Fy = 50 ksi, these limits bbecome:
for a com
mpact flange
bf
≤ λ pf = 9.15
2t f
and for a compact web
b
h
≤ λ pw = 90.6
tw
A compaarison of these limits with the data giveen in Manual Table 1-1 shhows that the majority of
the W-sh
hapes have co
ompact flangees and all havee compact weebs.
2118 Chapter 6
Bending Members
M
S
frrom Specificaation Table B4.1
Figure 6.18 Definition of Element Slenderness
Copyright © American Insttitute of Steel Construction.
C
R
Reprinted with Permission. A
All rights reservved.
Figu
ure 6.18 illusstrates these dimensions
d
foor several com
mmonly usedd sections aloong with
the slendern
ness limits (w
width-to-thick
kness ratios) as found in Specificationn Table B4.1bb. Other
shapes can also
a be found in Table B4.1b.
6.55.2
Flange Local Buckliing
The full range of nominaal moment strrength, Mn, off a cross sectiion can be exxpressed as a function
of flange sleenderness, λf; this relationsship is shownn in Figure 6.19. The threee regions in thhe figure
identify threee types of beehavior. The first
f
region reppresents plasttic behavior, in which the shape is
capable of attaining
a
its full
fu plastic mo
oment strengtth. This strenngth was disccussed in Secttion 6.2.
Shapes that fall into this region
r
are callled compact.. The behavioor exhibited inn the middle rregion is
inelastic, an
nd shapes thaat fit this cateegory are callled noncomppact. Shapes that fall into the last
region exhiibit elastic buckling
b
and
d are called slender shap
apes. The prrovisions for doubly
symmetric I-shaped meembers that exhibit thesee last two fforms of behhavior are ggiven in
n Section F3.
Specification
For I-shaped secctions, the div
viding line b etween comppact and nonccompact flannges was
given in Equ
uation 6.9. Th
he division beetween noncoompact and slender flange sections is a function
of the residu
ual stresses present
p
in thee hot rolled m
member. As w
was the case with lateral-ttorsional
buckling, th
he Specificatiion assumes that elastic bbehavior cont
ntinues up to the point whhere the
elastic mom
ment, MrFLB = 0.7FySx. As with
w lateral-toorsional buckkling, this limit is a functioon of the
residual streess set at a value
v
of 0.3F
Fy. This corrresponds to a flange slenderness, as ffound in
Specification
n Table B4.1b
b, of
E
(6.11)
λ rf = 1.00
Fy
Chapter 6
Figure 6.19
6
Bending Mem
mbers 219
Flange Local Buckliing Strength.
The stren
ngth at the jun
nction of com
mpact behavioor and noncom
mpact behaviior is the plasstic moment
strength
M n = M p = Fy Z
At the ju
unction of non
ncompact and
d slender behaavior, the mom
ment is an elaastic moment defined as
M rFLB = 0.7 Fy S x
The stren
ngth for nonco
ompact shapees is representted by a straigght line betweeen these poinnts. Thus,
⎡
⎛ λ − λ pf ⎞ ⎤
M n = ⎢ M p − ( M p − M rFLB ) ⎜
(6.12)
⎟⎥
⎝ λ rf − λ pf ⎠ ⎦⎥
⎣⎢
or as giveen in the Speccification
⎡
⎛ λ − λ pf ⎞ ⎤
M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜
(A
AISC F3-1)
⎟⎥
⎝ λ rf − λ pf ⎠ ⎦⎥
⎣⎢
For A992
2 steel with Fy = 50 ksi, Equation
E
6.11 provides an upper limit too the noncom
mpact flange
of
E
29, 000
= 11.0
λ rf = 1.0
= 24.1
Fy
50
A review of Manual
M
Tablee 1-1 for bf /22tf shows thatt there are noo W-shapes w
with flanges
that exceeed this limit. Thus, all wid
de-flange shappes have either compact orr noncompactt flanges. A
further reeview of the tables showss that only 100 W-shapes hhave noncomppact flanges for Fy = 50
ksi.
6.5.3
Web
b Local Buck
kling
A compaarison of the slenderness criteria
c
for weeb local buckkling given inn Equation 6.10 with the
data avaiilable in Manual Table 1-1
1 for h/tw indiicates that all W-shapes haave compact w
webs for Fy
= 50 ksi. Thus there is no need to
o address weeb local buckkling for W-shapes. Noncoompact and
slender webs,
w
howeveer, are addressed for built--up members in Chapter 7, where platee girders are
treated.
220 Chapter 6
Bending Members
EXAMPLE 6.8
Bending Strength
of Noncompact
Beam
Goal:
Determine the nominal moment strength and then the design moment
strength (LRFD) and allowable moment strength (ASD) for a W-shape with
noncompact flange.
Given:
A simply supported W6×15 spans 10 ft. It is braced at the ends and at the
midspan (Lb = 5 ft). The steel is A992.
SOLUTION
Step 1:
Determine the nominal strength.
Check the limits for flange local buckling.
For the flange, from Manual Table 1-1,
bf
= 11.5 > λ pf = 0.38 E Fy = 9.15
2t f
Therefore, the flange is not compact. Checking for a slender flange, even
though our previous review of the Manual data indicated that no W-shapes
exceeded this requirement,
bf
= 11.5 < λ rf = 1.0 E Fy = 24.1
2t f
Because λ pf < b f /2t f < λ rf , the shape has a noncompact flange.
Step 2:
Check the limit states for web local buckling.
For the web, from Manual Table 1-1,
h
= 21.6 < λ pw = 3.76 E Fy = 90.6
tw
So the web is compact, as expected from our earlier evaluation of all Wshapes.
Step 3:
Because the shape is noncompact (flange), determine the nominal moment
strength by Equation F3-1. From Manual Table 1-1 Zx = 10.8 in.3 and Sx=
9.72 in.3 and
M p = Fy Z x = 50 (10.8 ) = 540 in.-kips
Thus, for flange local buckling
⎡
⎛ λ − λ pf ⎞ ⎤
M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜
⎟⎥
⎢⎣
⎝ λ rf − λ pf ⎠ ⎥⎦
⎡
⎛ 11.5 − 9.15 ⎞ ⎤
= ⎢540 − ( 540 − 0.7 ( 50 ) 9.72 ) ⎜
⎟ ⎥ = 509 in.-kips
⎝ 24.1 − 9.15 ⎠ ⎦
⎣
or
509 in.-kip
Mn =
= 42.4 ft-kips
12 in./ft
Step 4:
Check for the limit state of lateral-torsional buckling.
For this shape, from Manual Table 1-1 ry = 1.45 in. and
Chapter 6
Bending Members 221
L p = 1.76 ry E Fy = 1.76 (1.45 ) 29, 000 50 = 61.5 in.
Thus, Lp = 5.13 ft, which is greater than Lb = 5.0 ft, so the beam is
adequately braced to resist the plastic moment. For lateral-torsional
buckling,
M n = M p = Fy Z = 50 (10.8 ) = 540 in.-kips
or
M n = 540 /12 = 45.0 ft-kips
Step 5:
For
LRFD
Step 6:
For
ASD
Step 6:
Because the moment based on flange local buckling, 42.4 ft-kips, is less
than the moment based on lateral-torsional buckling, 45.0 ft-kips, local
buckling controls and
M n = 42.4 ft-kips
Determine the design moment for LRFD.
φMn = 0.9 ( 42.4) =38.2 ft-kips
Determine the allowable moment for ASD
M n Ω = 42.4 1.67 =25.4 ft-kips
EXAMPLE 6.9
Bending Strength
of a Noncompact
Beam
Goal:
Determine the (a) design moment strength (LRFD) and (b) allowable
moment strength (ASD) for a W-shape with noncompact flange using Table
3-2.
Given:
A simply supported W14×90 spans 15 ft and carries a uniformly
distributed load. It is braced only at the supports. The steel is A992. Note
in Table 3-2 that this W-shape is identified with a superscript f indicating
in a footnote that the flange is not compact.
SOLUTION
Step 1:
Check the unbraced length to determine if lateral-torsional buckling is a
limit state that must be considered with Lb = 15.0 ft. From Manual Table
3-2,
Lp = 15.1 > Lb = 15.0 ft
Thus, lateral-torsional buckling does not influence the beam strength.
Step 2:
Knowing that the shape is not compact, check the limits for flange local
buckling.
From Manual Table 1-1,
bf
= 10.2 > λ pf = 0.38 E Fy = 9.15
2t f
Therefore, the flange is not compact. We know that no W-shapes have
222 Chapter 6
Bending Members
slender flanges but we need that limit. Thus,
λ rf = 1.0 E Fy = 24.1
Step 3:
Determine the nominal moment strength using Equation F3-1. From
Manual Table 1-1 Zx = 157 in.3 and Sx = 143 in.3 and
M p = Fy Z x = 50 (157 ) = 7850 in.-kips
and
⎡
⎛ λ − λ pf ⎞ ⎤
M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜
⎟⎥
⎢⎣
⎝ λ rf − λ pf ⎠ ⎥⎦
⎡
⎛ 10.2 − 9.15 ⎞ ⎤
= ⎢7850 − ( 7850 − 0.7 ( 50 )(143) ) ⎜
⎟ ⎥ = 7650 in.-kips
⎝ 24.1 − 9.15 ⎠ ⎦
⎣
Thus,
Mn =
For
LRFD
Step 4:
7650
= 638 ft-kips
12
Determine the design strength and compare it to φb M px from Table 3-2
φM n = 0.9 ( 638 ) = 574 ft-kips
which is identical to the value from Table 3-2
For
ASD
Step 4:
Determine the allowable strength and compare it to M px Ωb from Table
3-2
M n 1.67 = 638 1.67 = 382 ft-kips
which is identical to the value from Table 3-2
Step 5:
Conclusion: Manual Table 3-2 includes consideration of the flange
compactness for noncompact flange W-shapes.
6.6 DESIGN OF BEAMS FOR WEAK AXIS BENDING
Up to this point, I-shaped beams have been assumed to be bending about an axis parallel to their
flanges, called the x-axis. A quick scan of the shape property tables in the Manual shows that the
section modulus and plastic section modulus about the x-axis are larger than the corresponding
values about the other orthogonal axis, the y-axis. Thus, bending about the x-axis is called strong
axis or major axis bending, whereas bending about the y-axis is called weak axis or minor axis
bending. Although beams are not normally oriented for bending about this weak axis, a situation
may arise when it is necessary to determine the strength of a beam in this orientation.
Design of I-shaped beams for weak axis bending is relatively easy. Section F6 of the
Specification applies to I-shaped members and channels bent about their minor axis. Two limit
states are identified: yielding and flange local buckling. The flange and web referred to here are
the same elements as for the case when the shape is bending about its major axis. Thus, the limits
on flange slenderness are the same as discussed earlier. For those few W-shapes with noncompact
flanges, an equation similar to that used previously for noncompact flanges is required.
Chapter 6
Bending Members 223
For the limit state of yielding,
M n = M p = Fy Z y ≤ 1.6 Fy S y
(AISC F6-1)
An I-shaped member bending about its weak axis has properties close to those of a rectangle. For
the rectangle, the ratio of the plastic moment to the elastic yield moment, called the shape factor,
equals 1.5. The addition of the web alters the elastic section modulus and plastic section modulus
so that the shape factor for these weak axis bending members exceeds 1.5. To ensure an
appropriate level of rotational capacity at the plastic limit state, the shape factor for weak axis
bending is limited to 1.6. All but five W-shapes meet this limitation.
Although I-shaped members are not often called upon to carry moment about the y-axis
as pure bending members, they are called upon to participate in combined bending as discussed in
Section 6.12 and combined with axial load as discussed in Chapter 8.
EXAMPLE 6.10
Weak Axis
Bending Strength
of Beam
Goal:
Determine the nominal moment strength and then the design moment
strength (LRFD) and the allowable moment strength (ASD) for bending
about the y-axis.
Given:
A simply supported W10×30, A992 steel, is loaded to cause bending about
its weak axis.
SOLUTION
Step 1:
Determine the nominal strength.
Check the limits for flange local buckling.
For the flange, from Manual Table 1-1,
bf
= 5.70 < λ pf = 0.38 E Fy = 9.15
2t f
Therefore, the flange is compact.
Step 2:
For the limit state of yielding, from Equation F6-1 with Zy = 8.8 in.3 and Sy
= 5.75 in.3 from Manual Table 1-1
M n = Fy Z y = 50 ( 8.84 ) = 442 in.-kips
≤ 1.6 Fy S y = 1.6 ( 50 )( 5.75) = 460 in.-kips
Thus the strength is controlled by the plastic moment and
442
Mn =
= 36.8 ft-kips
12
For
LRFD
Step 3:
For
ASD
Step 3:
For LRFD, determine the design moment.
φM n = 0.9 ( 36.8 ) =33.1 ft-kips
For ASD, determine the allowable moment.
M n Ω = 36.8 1.67 =22.0 ft-kips
224 Chapter 6
Bending Members
EXAMPLE 6.11
Weak Axis
Bending Strength
of Beam
Goal:
Determine the nominal moment strength, design moment strength (LRFD),
and allowable moment strength (ASD) for bending about the y-axis.
Given:
A simply supported W40×392, A992 steel, is loaded to cause bending
about its weak axis.
SOLUTION
Step 1:
Determine the nominal moment strength.
Check the limits for flange local buckling.
For the flange, from Manual Table 1-1,
bf
= 2.45 < λ pf = 0.38 E Fy = 9.15
2t f
Therefore, the flange is compact.
Step 2:
For the limit state of yielding, from Equation F6-1 with Zy = 212 in.3 and Sy
= 130 in.3 from Manual Table 1-1
M n = Fy Z y = 50 ( 212 ) = 10,600 in.-kips
≤ 1.6 Fy S y = 1.6 ( 50 )(130 ) = 10, 400 in.-kips
Thus the strength is controlled by the upper limit and
10, 400
Mn =
= 867 ft-kips
12
For
LRFD
Step 3:
For
ASD
Step 3:
For LRFD, determine the design moment.
φM n = 0.9 ( 867 ) = 780 ft-kips
For ASD, determine the allowable moment.
M n Ω = 867 1.67 = 519 ft-kips
6.7 DESIGN OF BEAMS FOR SHEAR
Chapter G of the Specification establishes the requirements for beam shear. Although shear
failures are uncommon with rolled sections, the strength must still be confirmed. A beam can fail
in shear by yielding or buckling. Beam webs also need to be checked for shear rupture on the net
area of the web when bolt holes are present. Shear rupture is addressed in the discussion of
connections in Chapter 10.
The nominal shear yielding strength is based on the von Mises criterion,3 which states
that for an unreinforced beam web that is stocky enough not to fail by buckling, the shear strength
can be taken as Fy 3 = 0.58Fy . The Specification rounds this stress to 0.6Fy and provides, in
Section G2, the shear strength as
(AISC G2-1)
Vn = 0.6 Fy Aw Cv1
3
Mendelson, A. Plasticity: Theory and Application, New York: Macmillan, 1968.
Chapter 6
Bending Members 225
where Aw is the area of the web, taken as the total depth times the web thickness.
The web shear coefficient, Cv1, is used to account for shear web buckling. Thus, if the
web is capable of reaching yield, Cv1 = 1.0. To ensure that the beam web is capable of reaching
yield before buckling, the Specification sets the limit on web slenderness at
h
k E
≤ 1.10 v
tw
Fy
where kv = 5.34 for unstiffened webs. All current ASTM A6 rolled I-shaped members have webs
that meet the criteria for kv = 5.34, and all A992 W-shapes meet the criteria for web yielding, Cv1
= 1.0.
Thus, the nominal shear strength of all rolled W-shapes, using Equation G2-1 with Cv1 =
1.0, can be taken as
(6.13)
Vn = 0.6 Fy Aw
Determining the shear design strength or allowable strength is complicated by a variation
in resistance and safety factors. To keep the beam shear strength provisions the same in the 2016
Specification as in the allowable stress specifications prior to 2005, the resistance and safety
factors for a particular set of rolled I-shapes was liberalized. Thus, for webs of rolled I-shapes
with h /t w ≤ 2.24 E / Fy ,
φ = 1.00
( LRFD )
Ω = 1.50
( ASD )
φ = 0.90
( LRFD )
Ω = 1.67
( ASD )
For all other shapes,
EXAMPLE 6.12
Shear Strength of
Beam
SOLUTION
Goal:
Determine the nominal shear strength, design shear strength (LRFD), and
allowable shear strength (ASD) when the beam is bending about the x-axis
and compare to the shear strength given in Manual Table 3-2.
Given:
A W16×31, A992 steel member is loaded to cause shear in its web.
Step 1:
Determine the nominal strength.
Check the limits for yielding of the web. From the user note at the end of
Section G2.1 we know that all W-shapes will yield in shear. Thus,
Cv1 = 1.0
and from Manual Table 1-1
d = 15.9 in., tw = 0.275 in., and h/tw = 51.6
Step 2:
For the limit state of yielding,
Vn = 0.6Fy AwCv1
= 0.6 ( 50)(15.9)( 0.275)(1.0) = 131 kips
Step 3:
Determine the resistance factor and the safety factor.
h
E
29, 000
= 51.6 < 2.24
= 2.24
= 53.9
tw
Fy
50
Therefore,
φv = 1.00 and Ωv = 1.50
226 Chapter 6
Bending Members
For
LRFD
Step 4:
For LRFD, determine the design shear strength.
φVn = 1.0 (131) = 131 kips
From Manual Table 3-2 φVn = 131 kips . Thus the answers are the same.
For
ASD
Step 4:
For ASD, determine the allowable shear strength.
Vn Ω = 131 1.50 = 87.3 kips
From Manual Table 3-2 Vn Ω = 87.5 kips . Thus the results are close. The
difference is due to round off.
EXAMPLE 6.13
Shear Strength of
Beam
SOLUTION
Goal:
Determine the nominal shear strength, design shear strength (LRFD), and
allowable shear strength (ASD) when the beam is bending about the x-axis
and compare to the shear strength given in Manual Table 3-2.
Given:
A W16×26, A992 steel member is loaded to cause shear in its web.
Step 1:
Determine the nominal strength.
Check the limits for yielding of the web. From the user note at the end of
Section G2.1 we know that all W-shapes will yield in shear. Thus,
Cv1 = 1.0
and from Manual Table 1-1
d = 15.7 in., tw = 0.250 in., and h/tw = 56.8
Step 2:
For the limit state of yielding,
Vn = 0.6Fy AwCv1
= 0.6 ( 50)(15.7 )( 0.250)(1.0) = 118 kips
Step 3:
For
LRFD
Step 4:
Determine the resistance factor and the safety factor.
h
E
29, 000
= 56.8 > 2.24
= 2.24
= 53.9
tw
Fy
50
Therefore,
φv = 0.90 and Ωv = 1.67
For LRFD, determine the design shear strength.
φVn = 0.90 (118 ) = 106 kips
From Manual Table 3-2 φVn = 106 kips . Thus the answers are the same
For
ASD
Step 4:
For ASD, determine the allowable shear strength.
Chapter 6
Bending Members 227
Vn Ω = 118 1.67 = 70.7 kips
From Manual Table 3-2 Vn Ω = 70.5 kips . Thus the results are close. The
difference is due to round off.
6.8 CONTINUOUS BEAMS
Beams that span over more than two supports are called continuous beams. Unlike simple beams,
continuous beams are indeterminate and must be analyzed by applying more than the three basic
equations of equilibrium. Although indeterminate analysis is not within the scope of this book, a
few aspects should be addressed, even if only briefly.
The Manual includes shears, moments, and deflections for several continuous beams with
various uniform load patterns in Manual Table 3-23. These results come from an elastic
indeterminate analysis and can be used for the design of any beams that fit the support and
loading conditions.
It has long been known that material ductility permits steel members to redistribute load.
When one section of a member becomes overloaded, it can redistribute a portion of its load to a
less loaded section. This redistribution can be accounted for through an analysis method called
plastic analysis or through a number of more modern methods capable of modeling the real
behavior of the members. These methods may collectively be called advanced analysis and may
be used in structural steel design through the provisions of Appendix 1 of the Specification. This
appendix also permits use of the simplified plastic analysis approach for continuous beams
through Appendix Section 1.3.
To allow the designer to take advantage of some of the redistribution that is accounted for
in plastic analysis, Section B3.3 gives provisions for moment redistribution in beams. Design of
beams and girders that are compact and have sufficiently braced compression flanges may take
advantage of this simplified redistribution approach. The compact criteria are those already
discussed, whereas the unbraced length criteria are different. To use the simplified redistribution,
for doubly symmetric I-shaped beams the unbraced length of the compression flange, Lb, must be
less than that given in Section F13.5 as
⎡
⎛ M ⎞⎤ ⎛ E ⎞
Lm = ⎢0.12 + 0.076 ⎜ 1 ⎟ ⎥ ⎜ ⎟ ry
(AISC F13-8)
⎝ M 2 ⎠ ⎦ ⎝ Fy ⎠
⎣
where M1 is the smaller and M2 is the larger moment at the ends of the unbraced length. The
moment ratio is positive when the moments cause reverse curvature and negative when they
cause single curvature.
When these criteria are satisfied, the beam can be proportioned for 0.9 times the negative
moments at points of support. This redistribution is permitted only for gravity-loading cases and
moments determined through an elastic analysis. When this reduction in negative moment is
used, the positive moment must be increased to maintain equilibrium. This can be accomplished
simply by adding to the maximum positive moment 0.1 times the average original negative
moments.
EXAMPLE 6.14a
Continuous Beam
Design by LRFD
Goal:
Select a compact, fully braced section for use as a continuous beam.
Given:
The beam is continuous over three spans of 30 ft each. It supports a live
load of 2.5 kip/ft and a dead load of 1.8 kip/ft. Use A992 steel.
SOLUTION
Step 1:
Determine the required strength.
228 Chapter 6
Bending Members
The design load is wu = 1.2(1.8) + 1.6(2.5) = 6.16 kip/ ft.
From the beam shear, moment, and deflection diagrams in Manual Table
3-23, case 39, the critical span is the exterior span, where the negative
moment is
2
− M BA = 0.100 wl 2 = 0.100 ( 6.16 )( 30 ) = 554 ft-kips
and the positive moment is
2
+ M BA = 0.0800 wl 2 = 0.0800 ( 6.16 )( 30 ) = 444 ft-kips
Step 2:
Consider redistribution of moments according to Specification Section
B3.3.
A design could be carried out for a maximum moment of 554 ft-kips, but
with redistribution this moment may be reduced to
M BA = 0.9 ( 554 ) = 499 ft-kips
provided that the positive moment is increased by the average negative
moment reduction. Thus,
0 + 0.1( 554 )
M AB = 444 +
= 472 ft-kips
2
Step 3:
Determine the required plastic section modulus.
Even with the increase in positive moment, the negative moment is still
the maximum moment, so for a moment of 499 ft-kips,
499 (12 )
Z req =
= 133 in.3
0.9 ( 50 )
Step 4:
.
EXAMPLE 6.14b Goal:
Continuous Beam
Given:
Design by ASD
SOLUTION
Step 1:
Select the least-weight W-shape from Manual Table 3-2.
W24×55 with Z = 134 in.3
Select a compact, fully braced section for use as a continuous beam.
The beam must be continuous over three spans of 30 ft each. It must
support a live load of 2.5 kip/ft and a dead load of 1.8 kip/ft. Use A992
steel.
Determine the required strength.
The design load is wa = (1.8) + (2.5) = 4.3 kip/ft.
From the beam shear, moment, and deflection diagrams in Manual Table 323, case 39, the critical span is the exterior span, where the negative
moment is
− M BA = 0.100 wl 2 = 0.100 (4.3)(30) 2 = 387 ft-kips
Chapter 6
Bending Members 229
and the positive moment is
+ MBA = 0.0800 wl 2 = 0.0800(4.3)(30) 2 = 310 ft-kips
Step 2:
Consider redistribution of moments according to Specification Section
B3.3.
A design could be carried out for a maximum moment of 387 ft-kips, but
with redistribution this moment may be reduced to
MBA = 0.9(387) = 348 ft-kips
provided that the positive moment is increased by the average negative
moment reduction. Thus,
0 + 0.1(387)
MAB = 310 +
= 329 ft-kips
2
Step 3:
Determine the required plastic section modulus.
Even with this increase in the positive moment, the negative moment is still
the maximum moment, so for a moment of 348 ft-kips
348 (12 )
Z req =
= 139 in.3
( 50 / 1.67 )
Step 4:
Select the least-weight W-shape from Manual Table 3-2.
W21×62 with Z = 144 in.3
Note that in the preceding example, the reduction in negative moment leads to selection of a
smaller W-shape than would have resulted from design for the original negative moment.
6.9 PLASTIC ANALYSIS AND DESIGN OF CONTINUOUS BEAMS
Up to this point, it has been assumed that the plastic moment strength of a bending member could
be compared to the maximum elastic moment on a beam to satisfy the strength requirements of
the Specification. This is accurate for determinate members in which the occurrence of the plastic
moment at the single point of maximum moment results in the development of a single plastic
hinge, which would lead to member collapse. However, for indeterminate structures, such as
continuous beams, more than one plastic hinge must form before the beam would actually
collapse, and this provides some additional capacity that an elastic analysis cannot capture. The
formation of plastic hinges in the appropriate locations causes a collapse, and the geometry of this
collapse is called a failure or collapse mechanism. This is the approach referred to as plastic
analysis, permitted by Appendix Section 1.3 of the Specification for use with LRFD only. The
Specification only addresses beam design by plastic analysis. Use beyond beams, such as for
frames, is up to the designer; proper consideration of second-order effects on stability is required.
Thus, only the determination of plastic collapse mechanisms for beams is considered here.
The formation of a beam failure mechanism may best be understood by following the
load history of a fixed-end beam with a uniformly distributed load. The beam and moment
diagrams that result from an elastic indeterminate analysis are given in Figure 6.20a. The largest
moments occur at the fixed ends and are given by wL2/12. If the load on the beam is increased,
the beam behaves elastically until the moments on the ends equal the plastic moment strength of
2330 Chapter 6
Bending Members
M
Fiigure 6.20 B
Beam and Mooment Diagraams
fo r the Developpment of a Plaastic Mechannism
the memberr, as shown in
i Figure 6.2
20b. Because the applicatiion of additional load cauuses the
member to rotate
r
at its ends
e
while maaintaining thee plastic mom
ment, these points behave as pins.
These pins are
a called pla
astic hinges. In this case, the load is designated as w1. The mem
mber can
continue to accept load beyond
b
this w1, functioninng as a simplee beam, untill a third plasttic hinge
forms at thee beam midspan. The forrmation of thhis third hingge makes the beam unstabble, thus
forming the collapse mecchanism. Thee mechanism and correspoonding momeent diagram aare given
in Figure 6.2
20c.
For the collapse mechanism just describeed, equilibriuum requires thhat the simple beam
moment, wuL2/8, equal tw
wice the plastiic moment; thhus,
M p = wu L2 /16
(6.14)
Had
d this beam been
b
designeed based on an elastic annalysis, it woould have reqquired a
moment cap
pacity greaterr than or equ
ual to wuL2/112. Using a pplastic analyssis, a smallerr plastic
moment streength, equal to
t wuL2/16, must
m
be proviided for in thhe design. Thuus, in this caase of an
indeterminatte beam, plaastic analysiss has the pootential to result in a sm
maller membeer being
required to carry
c
this sam
me load.
An additional ad
dvantage to the
t use of pllastic analysiis for indeterrminate beam
ms is the
simplicity of
o the analysiss. By observaation, regardlless of the ovverall geomettry of the conntinuous
beam, each segment betw
ween supportss can be evaluuated indepenndently of the other segmennts. This
means that any beam seg
gment, contin
nuous at eachh end and loaaded with a uuniformly disstributed
load, exhibits the same collapse
c
mech
hanism. Thus , the relation between the applied loadd and the
plastic mom
ment will be as
a given in Equation 6.14.. Plastic analyysis results ffor additional loading
s are given in
and beam configuration
c
i Figure 6.221. Additionaal examples, as well as tthe
Chapter 6
Figure 6.21
6
Bending Mem
mbers 231
Loading and Beam Configuration
C
ns Resulting ffrom Plastic A
Analysis
developm
ment of these relations thrrough applicaation of energgy principles, can be foundd in several
books inccluding, Appllied Plastic Design
D
in Stee l.4
To
T ensure thaat a given beeam cross secction can unddergo the necessary rotattion at each
plastic hiinge, the Speccification requ
uires that Fy ≤ 65 ksi , thaat the section bbe compact, aand that the
compresssion flange be braced such
h that the unbbraced lengthh in the area of the hinge is less than
Lpd given
n as Equatio
on A-1-5 in Specificationn Appendix Section 1.3. If these lim
mits are not
satisfied,, the member design must be
b based on aan elastic anallysis.
4
Disque, R. O.
O Applied Dessign in Steel. New
N York: Van
n Nostrand Reinnhold, 1971.
232 Chapter 6
Bending Members
Goal:
EXAMPLE 6.15
Beam Design using
Plastic Analysis
Given:
(LRFD only)
SOLUTION
Design a beam using plastic analysis and A992 steel. Plastic analysis is
applicable only for LRFD load combinations.
A beam is simply supported at one end and fixed at the other, similar to that
shown in Figure 6.16 for Example 6.6. It spans 20 ft and is loaded at its
midspan with a dead load of 16.0 kips and a live load of 48.0 kips. It is
assumed that the final section will be compact and adequately braced.
Step 1:
Determine the required strength.
Pu = 1.2 (16.0 ) + 1.6 ( 48.0 ) = 96.0 kips
Using the plastic analysis results from Figure 6.21c,
96.0 (10.0 )(10.0 )
Pab
φM p req =
=
= 320 ft-kips
( a + 2b ) 10.0 + 2 (10.0 )
Step 2:
Select the required W-shape from Manual Table 3-2.
W21× 44, φM p = 358 ft-kips
Step 3:
Check the initial assumptions on compactness and lateral bracing.
A check of the compact flange and web criteria shows that this shape is
compact. With continuous bracing, the beam can be designed through
plastic analysis.
Thus, select a
W 21× 44
6.10 T-SHAPED MEMBERS IN BENDING
T-shaped members are normally cut from I-shaped members by splitting the I-shape down the
longitudinal axis. Thus they are often referred to as split tees. When they are made from Wshapes they are called WT-shapes. Similarly, when made from M-shapes as MT-shapes and Sshapes as ST-shapes. Provisions for beams formed by combining a pair of angles to form a T and
for beams made from a split I-shape and loaded in the axis of symmetry are both found in Section
F9 of the Specification. These singly symmetric members may be loaded with the stem in tension
or compression. Four limit states must be considered in the design of these T-shaped members:
yielding, lateral-torsional buckling, flange local buckling, and stem local buckling. The
Specification combines the provisions for tees and double angles, and those for stem or web legs
in tension or compression. This section addresses tees while double angles will be treated after
the discussion of single angle bending members.
6.10.1
Yielding of Tees
For the limit state of yielding
Mn = M p
(AISC F9-1)
and Mp is limited, depending on the orientation of the section. For the stem in tension
(AISC F9-2)
M p = Fy Z x ≤ 1.6 M y
Chapter 6
Bending Members 233
and for the stem in compression
Mp = My
(AISC F9-4)
where M y = Fy S x .
The limit on Equation F9-2 is necessary to ensure that the member is capable of rotating
sufficiently to attain the plastic moment strength without the extreme fibers of the shape reaching
into the strain-hardening region. This is the same limit that was discussed for I-shaped members
bending about their weak axis and can be thought of as limiting the section shape factor. Limiting
the plastic moment to the elastic moment, as in Equation F9-4 when the stem is in compression, is
a conservative assumption that reflects the limited knowledge available to predict what moment
the tee in this orientation could actually attain.
6.10.2
Lateral-Torsional Buckling of Tees
Lateral-torsional buckling also must account for the orientation of the shape. For stems in tension,
the limiting unbraced length for the limit state of yielding is the same as it was for doubly
symmetric I-shapes. As given in Section F9 it is
E
Lp = 1.76ry
(AISC F9-8)
Fy
The limiting unbraced length for inelastic lateral-torsional buckling is
⎛ E ⎞ IyJ
⎛ Fy ⎞ dS
Lr = 1.95 ⎜ ⎟
2.36 ⎜ ⎟ x + 1
⎝E⎠ J
⎝ Fy ⎠ S x
(AISC F9-9)
If the unbraced length is less than Lp, the limit state of lateral-torsional buckling does not apply.
If the unbraced length is between Lp and Lr, the nominal strength, when the stem is in
tension, is given by a straight line as
⎡ Lb − Lp ⎤
Mn = M p − ( M p − M y ) ⎢
(AISC F9-6)
⎥
⎣ Lr − Lp ⎦
It should be noted that this is similar to the straight line equations used for other types of
members but when Lb = Lr , Mn is set to My.
When the unbraced length is greater than Lr and the stem is in tension, the nominal
strength is given by
Mn = Mcr
(AISC F9-7)
and
M cr =
(
1.95E
I y J B + 1 + B2
Lb
)
(AISC F9-10)
where
⎛ d ⎞ Iy
B = +2.3 ⎜ ⎟
⎝ Lb ⎠ J
(AISC F9-11)
For the stem in compression at any point along the span, whenever the unbraced length is
greater than Lp
Mn = Mcr
(AISC F9-7)
234 Chapter 6
Bending Members
and again
M cr =
(
1.95E
I y J B + 1 + B2
Lb
)
(AISC F9-10)
but the sign on B is changed to negative. Thus,
⎛ d ⎞ Iy
B = −2.3 ⎜ ⎟
⎝ Lb ⎠ J
6.10.3
(AISC F9-11)
Flange Local Buckling of Tees
The limit state of flange local buckling for tees reflects the same behavior as for the I-shapes from
which they are cut. The limiting width-to-thickness ratios are the same as discussed earlier, and
the nominal strength equation is the same, except that it is limited to 1.6My. For compact flanges,
the limit state of flange local buckling does not apply. For noncompact flanges,
λ p < λ = b f 2t f ≤ λ r ,
⎡
⎛ λ − λ pf
M n = ⎢ M p − ( M p − 0.7 Fy S xc ) ⎜
⎢⎣
⎝ λ rf − λ pf
⎞⎤
⎟ ⎥ ≤ 1.6M y
⎠ ⎥⎦
(AISC F9-14)
and for slender flanges, λr < λ,
Mn =
0.7 ES xc
⎛ bf ⎞
⎜
⎟
⎝ 2t f ⎠
(AISC F9-15)
2
Sxc is the section modulus referred to the compression flange. If the stem is in compression, this
limit state does not apply.
6.10.4
Stem Local Buckling of Tees
When the stem is in flexural compression, the nominal strength for the limit state of stem local
buckling is given by
M n = Fcr S x
(AISC F9-16)
Sx is the elastic section modulus and Fcr is the critical stress, which is dependent on the stem
slenderness, d tw . Thus, when
d
E
≤ 0.84
tw
Fy
the stem will not buckle and
Fcr = Fy
When
0.84
E
d
E
< ≤ 1.52
Fy t w
Fy
⎡
d
Fcr = ⎢1.43 − 0.515
tw
⎢⎣
Fy
E
⎤
⎥ Fy
⎥⎦
(AISC F9-18)
Chapter 6
Bending Mem
mbers 235
Figure 6.22
6
T-Beam Orientation for
f Example 66.16
and when
n
d
E
> 1.52
Fy
tw
Fcr =
1.52 E
⎛d⎞
⎜t ⎟
⎝ w⎠
2
(A
AISC F9-19)
If the stem is in tensio
on, this limit state
s
does nott apply.
EXAMPLE
E 6.16
Bending Sttrength
of WT-Shap
ape
Goal:
G
Deetermine the nominal mom
ment strengthh for the giveen WT membber if the
steem is in (a) teension (Figuree 6.22a) and ((b) compressiion (Figure 6..22b).
Given:
G
A WT9×17.5 is
i used as a beam to suppport gravity loads and haas lateral
su
upport provideed at 5 ft interrvals.
SOLUTION
N
Step
S
1:
Deetermine the section propeerties for the WT-shapes ffrom Manual Table 18.
Zx = 11
1.2 in.3, Sx = 6.21 in.3, d = 8.85 in., tw = 0.300 in.,
Iy = 7.67 inn.4, J = 0.252 in.4, ry = 1.222 in.
Part
P
a
Deetermine the nominal
n
mom
ment strength for the stem in tension. Thhe WT is
oriented as show
wn in Figure 6.22a.
Step
S
2:
Deetermine the nominal
n
mom
ment strength for the limit state of yieldiing using
Eq
quation F9-2.
M y = Fy S x = 50 ( 6.21) = 311 in.-kipps
M p = Fy Z x ≤ 1.6 M y
= 50 (11.2 ) = 5560 in.-kips
≤ 1.6 ( 311) = 4498 in.-kips
Th
hus,
Mn = 498 in.-kkips for the liimit state of yyielding
Step
S
3:
Deetermine the nominal mooment strenggth for the llimit state off lateraltorrsional buckliing. From Eqquation F9-8
Lp = 1.76ry
an
nd Equation F9-9
F
E
299,000
= 1..76 (1.22)
= 51.7 in. ⇒ 4.31 ft
Fy
50
236 Chapter 6
Bending Members
⎛ E ⎞ IyJ
Lr = 1.95 ⎜ ⎟
⎝ Fy ⎠ S x
⎛ Fy ⎞ dS
2.36 ⎜ ⎟ x + 1
⎝ E ⎠ J
⎛ 50 ⎞ 8.85 ( 6.21)
⎛ 29,000 ⎞ 7.67 ( 0.252 )
= 1.95 ⎜
+1
2.36 ⎜
⎟
⎟
6.21
⎝ 50 ⎠
⎝ 29,000 ⎠ 0.252
= 348 in. ⇒ 29.0 ft
Thus, since L p < Lb = 5.0 ft < Lr the beam will buckle inelastically and
using Equation F9-6
⎡ Lb − Lp ⎤
Mn = M p − (M p − M y ) ⎢
⎥
⎣ Lr − Lp ⎦
⎡ 5.0 − 4.31 ⎤
= 498 − ( 498 − 311) ⎢
⎥ = 493 in.-kips
⎣ 29.0 − 4.31 ⎦
Step 4:
Consider the limit state of flange local buckling.
The limit state of flange local buckling does not apply to the WT9×17.5
because the flange is compact.
Step 5:
The controlling limit state for the WT with the stem in tension is the
smaller strength given by the limit states of yielding and lateral-torsional
buckling. Thus, for the limit state of lateral-torsional buckling
M n = 493 12 = 41.1 ft-kips
Part b
Determine the nominal moment strength for the stem in compression. The
WT is oriented as shown in Figure 6.22b.
Step 6:
Determine the nominal moment strength for the limit state of yielding using
Equation F9-4
Mp is limited to My so that, from Step 2,
M n = M p = M y = 311 in.-kips
Step 7:
Determine the nominal moment strength for the limit state of lateraltorsional buckling for this orientation with Lb = 5.0 ft.
Determine B from Equation F9-11 using the negative sign since the stem is
in tension. Thus,
⎛ 8.85 ⎞ 7.67
⎛ d ⎞ Iy
B = −2.3 ⎜ ⎟
= −2.3 ⎜⎜
= −1.87
⎟⎟
5
12
0.252
(
)
⎝ Lb ⎠ J
⎝
⎠
and from Equation F9-10
M cr =
(
1.95E
I y J B + 1 + B2
Lb
)
Chapter 6
M n = M cr =
1.95 ( 29,000 )
5.0 ( 2.0 )
Bending Members 237
2
7.67 ( 0.252 ) ⎡ −1.87 + 1 + ( −1.87 ) ⎤
⎥⎦
⎣⎢
= 328 in.-kips
Step 8:
Determine the nominal moment strength for the limit state of stem local
buckling.
The stem slenderness is
d
8.85
=
= 29.5 > 0.84 E Fy = 20.2
tw 0.300
< 1.52 E Fy = 36.6
Therefore, from Equation F9-18
⎡
d Fy ⎤
Fcr = ⎢1.43 − 0.515
⎥ Fy
t w E ⎥⎦
⎢⎣
⎡
8.85
50 ⎤
= ⎢1.43 − 0.515
⎥ ( 50 ) = 40.0 ksi
0.300 29,000 ⎦
⎣
and
M n = Fcr S x = 40.0 ( 6.21) = 248 in.-kips
Step 9:
Determine the controlling limit state strength for the WT with the stem in
compression. The nominal moment strength is the smallest of the limit
states checked; thus, for the limit state of stem local buckling
M n = 248 12 = 20.7 ft-kips
Note: This example shows that using a WT-shape with the stem in
compression significantly compromises the strength of the member. Even
so, beams with this orientation are often easier to construct for such
applications as lintels in masonry walls where WTs are used in this
orientation.
6.11 SINGLE-ANGLE BENDING MEMBERS
When single angles are used as bending members, they can be bending about one of the
geometric axes, parallel to the legs, or about one of the principal axes. They are often used as
lintels over openings in masonry walls, where they are bending about the geometric axes.
Unfortunately, this most useful orientation of the single-angle bending member is also the most
complex orientation for the determination of strength. Figures 6.23a and b show a single angle
oriented for bending about the geometric axis, and Figures 6.23c and d show the angle oriented
for bending about the minor and major principal axes, respectively.
For the doubly symmetric I-shapes and singly symmetric tees discussed to this point, the
shear center is located on an axis of symmetry. Typically, loading is applied symmetrically
through the shear center. The WT-shapes considered in Section 6.10 were only discussed for
loading on the axis of symmetry. For single angles this is not the case. An angle bending about a
geometric axis and loaded through the shear center, an axis through the center of the leg
2338 Chapter 6
Bending Members
M
gle Bending about
a
Geome tric Axis andd Principal Axxis
Figure 6.23 Single-Ang
w experience single axis bending. Unffortunately itt is quite com
mmon for anglles to be
thickness, will
loaded throu
ugh a point other
o
than the shear centerr, in which caase the angle will experiennce both
bending and
d torsion. Th
he treatment of
o torsion is beyond the scope of thiss book althouugh it is
addressed in
n Specification
n Chapter H.
Speccification Secction F10 giv
ves the proviisions for sinngle-angle bennding membeers. The
limit states to
t be checked
d for these meembers are yiielding, lateraal-torsional buuckling, and lleg local
buckling.
6.111.1
Yieldin
ng
The ratio off the plastic section
s
modu
ulus to the elaastic section modulus, shaape factor, foor angles
can easily become
b
quite large. Thus, in order to be sure that the angle is not strained into the
strain–hardeening region, the nominal moment
m
for thhe limit state of yielding iss taken as
C F10-1)
(AISC
M n = 11.5 M y
where M y = Fy S , and S is taken as the
t lowest secction moduluus about the aaxis of bendinng. This
applies for either
e
bending
g about a prin
ncipal axis or a geometric aaxis.
6.111.2
Leg Lo
ocal Buckling
Legs of angles in compreession have th
he same tenddency to buckkle as other coompression ellements.
Specification
n Table B4-1b
b defines the limiting slen derness, b/t, iin case 12, as
E
λ p = 0.544
Fy
and
E
λ r = 0.911
Fy
The bending streength of a sing
mber as a funnction of leg slenderness iis shown
gle-angle mem
in Figure 6.24.
6
In the region of no
oncompact bbehavior, λ p < b t ≤ λ r , thhe nominal moment
strength is given
g
by the sttraight-line ass
⎛
⎛ b ⎞ Fy ⎞
M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟
(AISC
C F10-6)
⎟⎟
⎜
t
E
⎝
⎠
⎠
⎝
Chapter 6
Figure 6.24
6
Bending Mem
mbers 239
Strengtth of a Single Angle as a Fuunction of Leeg Slendernesss
and the nominal
n
mom
ment strength for the regionn of slender bbehavior is giiven as a com
mbination of
Equation
ns F10-7 and F10-8
F
as
0.71ESc
Mn =
2
⎛b⎞
⎜ ⎟
⎝t⎠
where Sc is the elasticc section mod
dulus to the tooe in compression, relativee to the axis of bending,
either geeometric or prrincipal axis. For bending about one off the geometrric axes for aan equal-leg
angle with no lateral-torsional resttraint, Sc musst be taken ass 0.8 times the geometric aaxis section
modulus to reflect thee effect of actu
ual bending sstresses aboutt an inclined aaxis that resullt due to the
absence of lateral-torssional restrain
nt. The largeest width-to-thhickness ratioos for each leeg width of
angles giiven in Manual Table 1-7 are shown in Figure 6.24. The angle leg with the larrgest widthto-thickn
ness ratio is b t = 5 0.25 = 20
2 . It can be sseen from thee figure that n
none of these angles will
be consid
dered slender.
6.11.3
Latteral-Torsion
nal Buckling
The limit state of lateeral-torsional buckling is a function off the axis of bbending and w
whether the
toe of the angle is at a maximum stress in tenssion or comprression. For bbending abouut the minor
principall axis and forr angles with full lateral suupport, the lim
mit state of laateral-torsionnal buckling
does not apply. Thus, provisions arre given only for bending aabout the majjor principal aaxis and for
the speciial case of ben
nding about th
he geometric axis of equall leg angles.
For
F all cases of lateral-torsional buckliing, the nomiinal moment strength is a function of
the elastiic lateral-torssional bucklin
ng moment, M cr, which iss determined for major principal axis
bending or geometric axis bending of equal leg angles such tthat
for M y M cr ≤ 1.0
⎛
My ⎞
M n = ⎜1.92 − 1.117
⎟ M y ≤ 1.5M y
⎜
M cr ⎟⎠
⎝
(A
AISC F10-2)
2440 Chapter 6
Bending Members
M
Figure 6.25
Strength of a Single Angle
A
as a Funnction of Lateeral-Torsionaal Buckling
and for M y M cr > 1.0
⎛
00.71M cr ⎞
M n = ⎜ 0.92 −
⎟ M cr
My ⎠
⎝
(AISC
C F10-3)
The nominaal bending streength divided
d by the yieldd moment forr the limit staate of lateral-ttorsional
buckling is given in Figu
ure 6.25 as a function of thhe yield mom
ment divided bby the elasticc lateraltorsional buckling momeent. It can be seen that as tthe yield mom
ment divided by the elasticc lateraltorsional buckling momeent increases, the nominal bbending strenngth of the meember divideed by the
yield momen
nt decreases.
For bending abo
out the majorr principal axxis, Figure 66.23d, the elaastic lateral-ttorsional
buckling mo
oment is
2
9 EA
Arz tCb ⎛⎜
β w rz ⎞
βw rz ⎞⎟
⎛
+
M cr =
1 + ⎜ 4.4
4.4
(AISC
C F10-4)
8Lb ⎜
Lb t ⎟⎠
Lb t ⎟
⎝
⎝
⎠
The variablees in Equation
n F10-4 are the
t same as ppreviously defined except for the new vvariable,
βw, which is a section pro
operty for sing
gle angles bennt about theirr major principal axis. It is zero for
equal leg an
ngles, positivee for angles with
w their shoort leg in com
mpression andd negative foor angles
with their lo
ong leg in com
mpression. Values
V
of βw aare given in C
Commentary Table C-F100.1. Note
that βw is in
ndependent off angle thickn
ness. All uneequal leg anggles must be evaluated forr lateraltorsional bu
uckling about the major prrincipal axis, even if the aactual appliedd load is bending the
angle about one of its geo
ometric axes. Thus, for thi s condition, th
the applied geeometric axis bending
moment must be converteed to momentts about both principal axees.
For geometric axis
a
bending the strength of equal legg angles mayy be determiined for
conditions of no lateraal-torsional reestraint or ffor lateral-torrsional restraaint at the ppoint of
maximum moment
m
only
y. For any otther conditioon of lateral--torsional resstraint, they m
must be
evaluated fo
or bending abo
out the princiipal axes.
Wheen there is no
o lateral restraaint, the yieldd moment My, is taken as 0.80FyS wheree S is the
geometric section modullus. The elasttic lateral-torrsional buckliing moment for loading tthat puts
m stress as co
ompression att the toe is
the maximum
Chapter 6
M cr =
Bending Members 241
2
⎞
0.58Eb 4tCb ⎛⎜
⎛ Lb t ⎞
⎟
1
0.88
1
+
−
⎜ 2 ⎟
⎜
⎟
L2b
b ⎠
⎝
⎝
⎠
(AISC F10-5a)
and when the maximum stress is tension at the toe
M cr =
2
⎞
0.58Eb 4tCb ⎛⎜
⎛ Lb t ⎞
⎟
1
0.88
1
+
+
⎜ 2 ⎟
⎜
⎟
L2b
b ⎠
⎝
⎝
⎠
(AISC F10-5b)
Note that when the toe is at a maximum stress in tension, the elastic lateral-torsional buckling
moment is greater than when the maximum stress is compression at the toe. This means that the
nominal moment strength for lateral-torsional buckling will be greater when the maximum stress
is tension at the toe.
For the case of lateral-torsional restraint at the point of maximum moment, Mcr is taken as
1.25 times the value determined from Equations F10-5 and My is taken as 1.0FyS.
EXAMPLE 6.17
Bending Strength
of a Single Angle
Goal:
Determine the available bending strength of an equal-leg angle loaded
about the geometric axis through the shear center.
Given:
A 6×6×5/16 A36 angle is required to span 8.0 ft on a simple span with
lateral supports at the ends only. The vertical leg is up so that it is stressed
in compression due to the uniform gravity load.
SOLUTION
Step 1:
Determine the required properties from Manual Table 1-7
S x = S y = 2.95 in.3 b = 6.0 in. t = 0.3125 in.
Step 2:
Determine the nominal moment strength for the limit state of yielding. Use
Equation F10.1.
M n = 1.5 Fy S x = 1.5 ( 36 ( 2.95 ) ) = 159 in.-kips
Step 3:
Determine the nominal moment strength for the limit state of lateraltorsional buckling, first determine the elastic lateral-torsional buckling
moment. For an equal-leg angle with maximum compression in the toe (leg
up and top in compression), use Equation F10-5a.
2
⎞
0.58 Eb 4tCb ⎛⎜
⎛ Lb t ⎞
⎟
M cr =
1
+
0.88
−
1
⎜
⎟
⎜
⎟
L2b
b2 ⎠
⎝
⎝
⎠
2
⎛
⎞
4
⎛ 8.0 (12 )( 0.3125 ) ⎞
0.58 ( 29,000 )( 6.0 ) ( 0.3125 )(1.0 ) ⎜
⎟ − 1⎟⎟
=
2
2
⎜ 1 + 0.88 ⎜⎜
⎟
( 6.0 )
(8.0 (12 ) )
⎜
⎟
⎝
⎠
⎝
⎠
= 199 in.-kips
Step 4:
Determine the yield moment to be used when there is no lateral-torsional
restraint. This accounts for the presence of some torsion by using 0.8Sx.
Thus,
M y = Fy ( 0.8 S x ) = 36 ( 0.8 ( 2.95 ) ) = 85.0 in.-kips
242 Chapter 6
Bending Members
Step 5:
Since M y M cr = 85.0 199 = 0.427 ≤ 1.0 , use Equation F10-2 to determine
the nominal moment strength for the limit state of lateral-torsional
buckling.
⎛
My ⎞
M n = ⎜ 1.92 − 1.17
⎟⎟ M y ≤ 1.5M y
⎜
M
cr
⎝
⎠
⎛
85.0 ⎞
= ⎜⎜ 1.92 − 1.17
⎟ 85.0
199 ⎟⎠
⎝
= 98.2 in.-kips ≤ 1.5M y = 1.5 ( 85.0 ) = 128 in.-kips
Thus,
M n = 98.2 in.-kips
Step 6:
Check the leg slenderness to determine if the strength limit state of leg local
buckling must be determined. For the 6×6×5/16 angle,
b t = 6.0 0.3125 = 19.2 . From Specification Table B4.1b, the slenderness
limits are
E
29, 000
λ p = 0.54
= 0.54
= 15.3
Fy
36
λ r = 0.91
E
29, 000
= 0.91
= 25.8
Fy
36
Therefore, the angle is noncompact so the strength for the limit state of leg
local buckling must be determined.
Step 7:
As with lateral-torsional buckling, the reduced section modulus must be
used when bending is about the geometric axis. Thus, using Equation F10-6
⎛
⎛ b ⎞ Fy ⎞
M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟
⎟
⎜
⎝ t ⎠ E ⎟⎠
⎝
⎛
36 ⎞
⎛ 6.0 ⎞
= 36 ( 0.8 ( 2.95) ) ⎜⎜ 2.43 − 1.72 ⎜
⎟ = 108 in.-kips
⎟
⎝ 0.3125 ⎠ 29,000 ⎟⎠
⎝
Step 8:
For
LRFD
Step 9:
For
ASD
Step 9:
The nominal bending strength is the lowest value based on the three limit
states checked. Thus, lateral-torsional buckling controls and
M n = 98.2 in.-kips
Determine the design strength
φM n = 0.9 ( 98.2 ) =88.4 in.-kips
Determine the allowable strength
M n 98.2
=
=58.8 in.-kips
Ω 1.67
Chapter 6
Bending Members 243
6.12 DOUBLE ANGLE MEMBERS IN BENDING
T-shaped bending members were discussed in Section 6.10 and it was pointed out that tees and
double angles are both treated in Specification Section F9. The same four limit states considered
for tees must be considered for double angle members: yielding, lateral-torsional buckling, flange
local buckling, and stem local buckling. As with tees, double angles must be loaded in the plane
of symmetry. Again, these limit states are treated differently if the web legs are in compression or
tension. The following describes the application of Specification Section F9 to double angle
bending members.
6.12.1
Yielding of Double Angles
For the limit state of yielding
Mn = M p
(AISC F9-1)
and Mp is limited, depending on the orientation of the section. For the web legs in tension
(AISC F9-2)
M p = Fy Z x ≤ 1.6 M y
and for web legs in compression
M p = 1.5 M y
(AISC F9-5)
where
M y = Fy S x
The limit on Equation F9-2 is required for double angles just as it was for tees, to ensure that the
member is capable of rotating sufficiently to attain the plastic moment strength without the
extreme fibers of the shape reaching into the strain-hardening region. Limiting the plastic moment
to 1.5 M y is slightly more restrictive than the limit in Equation F9-2 but is significantly greater
than it was for tees.
6.12.2
Lateral-Torsional Buckling of Double Angles
Lateral-torsional buckling also must account for the orientation of the shape. For web legs in
tension, the provisions are the same as they were for tees. The limiting unbraced lengths are
E
Lp = 1.76ry
(AISC F9-8)
Fy
and
⎛ E ⎞ IyJ
Lr = 1.95 ⎜ ⎟
⎝ Fy ⎠ S x
⎛ Fy ⎞ dS
2.36 ⎜ ⎟ x + 1
⎝E⎠ J
(AISC F9-9)
As usual, if the unbraced length is less than Lp, the limit state of lateral-torsional buckling does
not apply.
If the unbraced length is between Lp and Lr, the nominal strength, when the web legs are
in tension, is given by the same straight-line that was used for tees,
244 Chapter 6
Bending Members
⎡ Lb − Lp ⎤
Mn = M p − ( M p − M y ) ⎢
⎥
⎣ Lr − Lp ⎦
(AISC F9-6)
When the unbraced length is greater than Lr and the stem is in tension, the nominal
strength is given by
Mn = Mcr
(AISC F9-7)
and
M cr =
(
1.95E
I y J B + 1 + B2
Lb
)
(AISC F9-10)
where
⎛ d ⎞ Iy
B = +2.3 ⎜ ⎟
⎝ Lb ⎠ J
(AISC F9-11)
For web legs in compression at any point along the span, whenever the unbraced length is
greater than Lp, the nominal moment strength is determined as it was for single angles, through
Equations F10-2 and F10-3.
For M y M cr ≤ 1.0
⎛
My ⎞
M n = ⎜1.92 − 1.17
⎟ M y ≤ 1.5M y
⎜
M cr ⎟⎠
⎝
(AISC F10-2)
and for M y M cr > 1.0
⎛
0.71M cr ⎞
M n = ⎜ 0.92 −
⎟ M cr
My ⎠
⎝
The elastic lateral-torsional buckling moment is determined as
1.95E
M cr =
I y J B + 1 + B2
Lb
(
(AISC F10-3)
)
(AISC F9-10)
with
⎛ d ⎞ Iy
B = −2.3 ⎜ ⎟
⎝ Lb ⎠ J
and
M y = Fy S x
6.12.3
Leg Local Buckling of Double Angles
Leg local buckling of double angles is treated the same as leg local buckling of single angles by
reference from Specification Section F9 to Section F10. As discussed for single angles, the
limiting slenderness, b/t, in case 12, is
E
λ p = 0.54
Fy
and
Chapter 6
λ r = 0.91
Bending Members 245
E
Fy
For noncompact angles, λ p < b t ≤ λ r , the nominal moment strength is given by
⎛
⎛ b ⎞ Fy ⎞
M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟
⎟
⎜
⎝ t ⎠ E ⎟⎠
⎝
(AISC F10-6)
For slender angles the nominal moment strength is given as a combination of Equations F10-7
and F10-8 as
0.71ESc
Mn =
2
⎛b⎞
⎜ ⎟
⎝t⎠
For both, Sc is the elastic section modulus of the double angle member for the toe in compression.
EXAMPLE 6.18
Bending Strength
of a Double Angle
Goal:
Determine the available bending strength of a double equal-leg angle
loaded in the plane of symmetry.
Given:
Two 6×6×5/16 A36 angles are required to span 8.0 ft on a simple span with
lateral supports at the ends only. The vertical legs are in contact and are
oriented up so that they are stressed in compression due to the uniform
gravity load.
SOLUTION
Step 1:
Determine the required properties from Manual Table 1-7 for a single angle
S x = 2.95 in.3 I x = 13.0 in.4 b = 6.0 in. t = 0.3125 in.
From Manual Table 1-15 for double angles,
ry = 2.47 in. and 2S x = 2 ( 2.95) = 5.90 in.3
Step 2:
Determine the nominal moment strength for the limit state of yielding. Use
Equation F9.5.
M n = 1.5 Fy S x = 1.5 ( 36 ( 5.90 ) ) = 319 in.-kips
Step 3:
Determine the nominal moment strength for the limit state of lateraltorsional buckling. First determine the unbraced length beyond which
lateral-torsional buckling must be considered, Lp, based on Equation F9-8.
E
29,000
Lp = 1.76ry
= 1.76 ( 2.47 )
= 123 in.
Fy
36
Since Lb = 8.0 ft = 96.0 in.< 123 in. lateral torsional buckling does not need
to be checked.
Step 6:
Check the leg slenderness to determine if the strength limit state of leg local
buckling must be determined using Section F10.3 by reference from
Section F9.4. For two 6×6×5/16 angles with the back-to-back legs in
flexural compression, b t = 6.0 0.3125 = 19.2 . From Specification Table
B4.1b, the slenderness limits are
2446 Chapter 6
Bending Members
M
λ p = 0.54
E
299, 000
= 15.3
= 0.54
Fy
36
λ r = 0.91
E
299, 000
= 25.8
= 0.91
Fy
36
Thereefore, the an
ngles are nonncompact juust as for the single anggle of
Exam
mple 6.17 so the
t strength ffor the limit sstate of leg loocal bucklingg must
be dettermined.
Step
p 7:
Unlik
ke for the sing
gle angle, thee reduced secttion moduluss is not used. Thus,
using Equation F10-6
⎛
⎛ b ⎞ Fy ⎞
M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟
⎟
⎜
⎝ t ⎠ E ⎟⎠
⎝
⎛
36 ⎞
⎛ 6.0 ⎞
= 36 ( 5.990 ) ⎜⎜ 2.43 − 1.772 ⎜
⎟ = 2269 in.-kips
⎟
⎝ 0.3125 ⎠ 29,000 ⎟⎠
⎝
Step
p 8:
Forr
LRF
FD
Step
p 9:
Forr
ASD
D
Step
p 9:
The nominal
n
bend
ding strength is the lowestt value based on the limit states
check
ked. Thus, leg
g local bucklinng controls annd
M n = 269 in.-kiips
Deterrmine the desiign strength
φM n = 00.9 ( 269 ) =2422 in.-kips
Deterrmine the allo
owable strengtth
M n Ω = 2269 1.67 =16 1 in.-kips
h of the doublle angle beam
m in Example 6.18 is more than twice thhat of the singgle angle
The strength
beam in Exaample 6.17, ev
ven with the same limit staate, leg local buckling, conntrolling the sstrength.
Because of symmetry about the planee of loading, tthe reduced ssection moduulus that was used for
a
is not used in determining the local bucklinng strength ffor the double angle
the single angle
member. In addition, lateeral-torsional buckling doees not even nneed to be chhecked for thee double
angle beam where it poteentially could
d have controllled for the siingle angle beeam. Clearly some of
the modes of failure for a single angle beam and a ddouble angle beam can be quite differennt.
F
Figure 6.26 Biaxial Bennding of IS
Shaped Beam
m
Chapter 6
Bending Mem
mbers 247
Figure 66.27 Simple Linear Interaaction
Diagram for Biaxial B
Bending
6.13
MEM
MBERS IN
N BIAXIAL BENDING
Bending members arre often callled upon to resist forcess that result in bending about two
orthogon
nal axes. Exam
mples of thiss member typpe are crane ggirders and rroof purlins in industrial
buildingss. Regardless of the actual orientation of an appliedd moment, it is possible tto break the
moment into compon
nents about th
he two princ ipal axes, as shown in Fiigure 6.26. O
Once this is
accompliished, the ab
bility of the section to reesist the com
mbined momeents can be determined
through the
t interaction
n equation.
Chapter H addresses thee interaction of forces. F
Specification
S
For the com
mbination of
momentss, a simple lin
near interactio
on equation iss used, as shoown in Figure 6.27. This is taken from
the equattion provided
d in Specifica
ation Section H1 for combbined axial looad and mom
ment. When
the axial load is zero, Equation H1-1b reduces too
M rx M ry
+
≤ 1.0
Mcx M cy
where th
he moment terrms relate to the x- and yy-axes, the nuumerator is thhe required sttrength, and
the denominator is thee available sttrength, deterrmined as thoough the mem
mber were bennding about
only onee axis at a tim
me. Thus, if th
he required x--axis momennt is 79 percennt of the x-axxis strength,
only 21 percent
p
of thee y-axis streng
gth is availabble to resist m
moment. Moree attention is ggiven to the
use of intteraction equaations when axial
a
load is ccombined withh the bendingg moment in C
Chapter 8.
6.14
SER
RVICEABIL
LITY CRIT
TERIA FOR
R BEAMS
There are several serv
viceability co
onsiderations that the desiigner must adddress. A genneral set of
provision
ns are found in
i Specificatio
on Chapter L
L. Although faailure to satisfy these criteria may not
impact th
he strength off the memberr or overall sttructure, it m
may lead to thee first signs oof difficulty
for succeessful compleetion of a projject. The speccific criteria sshould be disscussed in dettail with the
designer’’s client so th
hat the quality
y of the final product is coonsistent withh the expectaations of the
owner. Although
A
exceessive floor deflection
d
mayy not be a saafety issue, it may lead to cracking of
finishes and an unaccceptable appeearance. It m
may also be aan indicator that other seerviceability
issues arre likely, such
h as vibration
n. Experiencee may indicaate that an annnoying amouunt of floor
vibration
n may be pressent at first, but
b occupantss become useed to it with time. The cliient may be
unwilling
g to deal witth this period
d of dissatisffaction and innsist that thee system be ddesigned to
prevent vibration
v
com
mplaints. Thiss must be knnown at the bbeginning off a project, not after the
occupantts move in and
a find the floor responnse objectionaable. The enngineer must be sure to
248 Chapter 6
Bending Members
identify these considerations for the owner so that the decisions made are appropriate to achieve
the expected outcome.
Beams generally have two serviceability issues that must be addressed directly, deflection
and vibration. With today’s high strength steels, deflection criteria may often control designs for
typical building spans and loads. A third issue is overall building drift, where the beams in
moment frames play a significant role in the response.
6.14.1
Deflection
Deflection is the normal response of a beam to its imposed load. It is impossible to erect a beam
with zero deflection under load, but the designer will be able to limit that deflection with proper
attention to this limit state. Deflections must be addressed for a variety of loading cases.
Deflection under dead load is critical because it impacts the construction process, including the
amount of concrete fill needed to form a flat and level floor. Live load deflection is critical
because it impacts the finishes of elements attached to the floor, such as ceilings and walls, and
may be visible to the occupants. Experience has demonstrated that live load deflection is not a
problem if it is limited to 1/360 of the span. Dead load deflection limitations are a function of the
particular structural element and loading. AISC Design Guide 3: Serviceability Design
Considerations for Steel Buildings covers deflection and other serviceability design criteria.
6.14.2
Vibration
Although vibration of floor systems is not a safety consideration, it can be a very annoying
response and very difficult to correct after the building is erected. The most common problem
occurs with wide-open spaces with very little damping, such as the jewelry department in a
department store. To reduce the risk of annoyance, a general rule is to space the beams or joists
sufficiently far apart so that the slab thickness is large enough to provide the needed stiffness and
damping. AISC Design Guide 11: Vibration of Steel Framed Structural Systems Due to Human
Activity covers the design of steel-framed floor systems for human comfort.
6.14.3
Drift
Under lateral loading, a building will sway sideways. This lateral displacement is called drift. As
with deflection and vibration, drift is usually not a safety consideration, but it can be annoying
and have a negative impact on nonstructural elements, causing cracks in finishes. Beams and
girders are important in reducing the drift, and their final size might actually be determined by
drift considerations. However, the impact of drift considerations on beams cannot be determined
without also evaluating the other parts of the lateral load–resisting system. This serviceability
limit state is treated in Chapter 8. Drift is also discussed in Design Guide 3.
Because beam deflection is a serviceability consideration, calculations are carried out using the
specific loads under which the serviceability considerations are to be checked. This can be live
load, deal load, or some combination of loads, but normally does not include any load factors.
Thus, regardless of whether a design is completed using LRFD or ASD, serviceability
considerations are checked for the same loads. Numerous elastic analysis techniques are available
to determine the maximum deflection of a given beam and loading. Some common loading
conditions with their corresponding maximum deflections are shown in Figure 6.28. These and
many others are given in Manual Table 3-23.
Chapter 6
Figure 6.28
6
Deflectio
ons
Bending Mem
mbers 249
Some Common
C
Loading Conditioons with Theiir Correspondding Maximum
m
EXAMPLE
E 6.19
Live Load
Deflection
Goal:
G
Ch
heck the live load deflectioon of a previoously designedd beam.
Given:
G
Usse the inform
mation from Example 6.33, where a W
W18×35 was selected.
Liimit the live lo
oad deflectionn for an accepptable designn to 1/360 of thhe span.
SOLUTION
N
Step
S
1:
Co
ollect the requ
uired informaation from Exxample 6.3.
Fo
or the W18×3
35, I = 510 inn.4. The live looad is 24 kips applied at thhe center
off a 20 ft span.
Step
S
2:
Deetermine the live load defflection. Usinng the deflecttion equation found in
Figure 6.28 for case (b),
24 ( 20.0) (12)
PL3
= 0.4667 in.
=
48EI 48 ( 29,000))( 510)
3
Δ=
Step
S
3:
3
Co
ompare the caalculated defllection to the given limit.
Th
he deflection limit is
20.0(12)
= 0.667in .
360
Beecause the callculated defleection is less tthan the defleection limit,
Δ = 0.4467 in. < Δ mmax = 0.667 in.
Δ maax =
Th
he deflection satisfies the sset criteria
250 Chapter 6
Bending Members
EXAMPLE 6.20
Beam Design
through Deflection
Limit
Goal:
Select a W-shape to satisfy a live load deflection limit.
Given:
Use the data from Example 6.19, except that the deflection limit is set to a
more severe level of 1/1000 of the span. If the selected member does not
meet the established criteria, select a W-shape that satisfies the limitation.
SOLUTION
Step 1:
Check the new deflection limit.
Δmax =
20.0 (12)
1000
= 0.240 in.
From Example 6.19 we know already that the given beam deflects too
much.
Step 2
Determine the minimum acceptable moment of inertia, Imin, necessary to
ensure that the deflection does not exceed the given limit.
Rearranging the maximum deflection for a concentrated load at mid-span
equation to solve for Imin,
24 ( 20.0) (12)
PL3
=
=
= 993 in.4
48E Δmax 48 ( 29,000)( 0.240)
3
I min
Step 3:
3
Select a beam with I ≥ 993 in.4 and, from Example 6.3, one that satisfies the
strength limit, Z ≥ 64.0 in.3 .
From the moment of inertia tables, Manual Table 3-3, select a
W21×55, I = 1140 in.4 and Z = 126 in.3
This is the lightest-weight W-shape that will satisfy the required moment of
inertia.
6.15
CONCENTRATED FORCES ON BEAMS
Before a beam can be called upon to carry a given load, that load must be transferred to the beam
through some type of connection. In a similar manner, the beam reactions must be carried to the
beam’s supporting structure through some type of connection. Although the majority of beams
are loaded through connections to their webs, some may be loaded by applying a concentrated
force to the top flange, and some will have their reactions resisted by bearing on a supporting
element. In these cases, a check must be made to establish that the beam web has sufficient
strength to resist the applied forces.
Four limit states determine the load carrying strength of the web to resist these
concentrated forces: web local yielding, web local crippling, web sidesway buckling, and flange
local bending. These limit states are all described in Section J10 of the Specification. Although it
is possible to select a beam with a web sufficiently thick that these limit states do not control, it is
normally more economical to add bearing stiffeners under the concentrated loads to provide the
necessary strength. For the limit states of web local yielding and web local crippling, it may be
easiest to increase the length of bearing, which would eliminate the requirement for bearing
stiffeners. However, if the web sidesway buckling limit state is exceeded, stiffeners are required.
In applications where W-shapes are used, this limit state is particularly critical for continuous
Chapter 6
Bending Mem
mbers 251
Figure 6.29
6
Single Concentrated
C
Force Applieed to a Beam
beams in
n the negativee moment region where tthe bottom flange is the ccompression flange. The
limit statte of flange lo
ocal bending is
i a concern oonly when tennsion loads arre applied to beams. The
design of stiffeners iss covered in Section 7.4 oof this book uunder the discussion of pllate girders,
because they are mucch more comm
monly foundd in that appliication. The llimit states of web local
yielding and web locaal crippling will
w be discusssed here withh an eye towaard using beaaring length
as the controlling factor in resisting
g these limit sstates.
6.15.1 Web Local Yieldiing
When a single conceentrated forcee is applied tto a beam ass shown in F
Figure 6.29, tthe force is
assumed to be deliverred to the beaam over a lenggth of bearingg, lb. It is thenn distributed through the
flange an
nd into the web.
w
The narro
owest portionn of the web is the criticaal section. This occurs at
the toe of
o the web-to--flange fillet, dimension k in Figure 6.229a. There arre two variablles listed in
Manual Table 1-1 forr this dimenssion, kdes and kdet. In this aapplication kdes is to be used since it
representts the lower bound
b
on the actual k dim
mension foundd in the produuction of W-sshapes. The
distributiion of the forrce along thee web takes pplace at a sloope of 1:2.5. Thus, when the critical
section iss reached, thee force has beeen distribute d over a lenggth of lb plus 22.5k in each ddirection. If
the conceentrated forcee is applied so that the forrce is distribuuted along thee web in bothh directions,
this distrribution increeases the bearring length bby 5k as show
wn in Figure 6.29b. If thee bearing is
close to the end of the member, distributionn takes placee only in onne direction, toward the
midspan.. The Specificcation definess “close to thee member ennd” as being w
within the meember depth
from the end. Thus, th
he available leength of the w
web is (lb + 2..5k), as shownn in Figure 6.29c.
The
T nominal strength of the beam weeb when the concentratedd force to bee resisted is
applied at
a a distance from
f
the mem
mber end that iis greater thann the depth off the memberr, d, is
(A
AISC J10-2)
Rn = Fyw t w ( 5k + lb )
When the concentrateed force to bee resisted is aapplied at a ddistance from
m the memberr end that is
less than or equal to th
he depth of th
he member, d,, the nominal strength is
(A
AISC J10-3)
Rn = Fyw t w ( 2.5k + lb )
252 Chapter 6
Bending Members
where
Fyw
lb
k
tw
= yield stress of the web
= length of bearing
= kdes= distance from the outer face of the flange to the web toe of the fillet weld
= web thickness
For web local yielding
φ = 1.0 (LRFD)
Ω = 1.50 (ASD)
6.15.2 Web local crippling
The criteria for the limit state of web local crippling also depend on the location of the force with
respect to the end of the member and the length of bearing, lb. The Specification equations apply
to both I-shaped members and HSS. When applied to I-shaped members, Qf = 1.0.
When the concentrated compressive force is applied at a distance from the member end
that is greater than or equal to d/2,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
2
Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
(AISC J10-4)
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
When the force is applied at a distance less than d/2 and lb/d ≤ 0.2,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
2
Rn = 0.40tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
(AISC J10-5a)
and when lb/d > 0.2,
1.5
⎡ ⎛ 4l
⎞ ⎛ tw ⎞ ⎤ EFywt f
b
Rn = 0.40t ⎢1 + ⎜
− 0.2 ⎟ ⎜ ⎟ ⎥
Qf
tw
⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣ ⎝ d
2
w
(AISC J10-5b)
For web local crippling
φ = 0.75 (LRFD)
Ω = 2.0 (ASD)
When the beam reaction is transferred to the supporting material, it is often necessary to
use a beam bearing plate to spread the load over a sufficiently large area to avoid failure of the
supporting material. These beam bearing plates are addressed in Section 11.11.
EXAMPLE 6.21a
Beam Bearing
Strength by LRFD
SOLUTION
Goal:
Determine the required length of bearing for a W-shape to support (a) a
midspan concentrated load and (b) concentrated reactions at the end of a
simple span.
Given:
A W18×35 A992 beam was selected in Example 6.3a as a simply supported
member spanning 20 ft and carrying a midspan dead load of 8.0 kips and
live load of 24.0 kips.
Step 1:
Determine section properties from Manual Table 1-1
d = 17.7 in., tw = 0.300 in., tf = 0.425 in., kdes = 0.827 in.
Step 2:
Determine the required bearing strength at the load point and at the
reactions using the LRFD load combinations from Section 2.4.
Chapter 6
Bending Members 253
Pu = 1.2 PD + 1.6 PL = 1.2 ( 8.0 ) + 1.6 ( 24.0 ) = 48.0 kips
Ru =
Part a
Step 3:
Pu 48
=
= 24 kips
2
2
At the load
For the limit state of yielding, determine the required length within the web
to resist the applied load. The required length, based on Equation J10-2
where N = ( 5k + lb ) , is then
( 5k + lb ) =
Step 4:
Pn
P φ
48 1.0
= u =
= 3.20 in.
Fywtw Fywtw 50 ( 0.300 )
Determine the minimum length of bearing, lb. Since this force is located so
that distribution takes place in both directions,
lb = 3.20 − 5 ( 0.827 ) = −0.935 in.
A negative bearing length in this calculation means that the distribution, 5k,
is more than enough to carry the required load. Thus, the minimum bearing
length for normal practice of 3.0 in. is adequate.
Step 5:
For the limit state of web local crippling, determine the nominal strength if
the bearing length is taken as a practical minimum of 3.0 in. From Equation
J10-4,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
2
Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
1.5
2 ⎡
⎛ 3.00 ⎞⎛ 0.300 ⎞ ⎤ 29,000(50)(0.425)
= 0.80 ( 0.300 ) ⎢1 + 3 ⎜
(1.0 )
⎟⎜
⎟ ⎥
0.300
⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎦⎥
⎣⎢
= 134 kips
Step 6:
Determine the design strength for web local crippling,
φRn = 0.75(134) = 101 kips
Step 7:
For the final selection at mid-span, since
Ru = 48 kips < φRn = 101 kips
a 3.0 in. bearing length will be sufficient.
Part b
Step 8:
At the end reaction
Since the end reactions are less than the concentrated load at mid-span, the
3.0 in. minimum bearing length will likely be sufficient for the end
reactions also. Thus, check a 3.0 in. bearing.
Rn = Fywtw ( 2.5k + lb ) = 50 ( 0.300 ) ( 2.5 ( 0.827 ) + 3.0 ) = 76.0 kips
φRn = 1.0 ( 76.0 ) = 76.0 kips
Step 9:
Check the limit state of web local crippling. Since the reaction is at the end
of the member at a distance less than d 2 and lb d = 3.0 17.7 = 0.17 ≤ 0.2
254 Chapter 6
Bending Members
use Equation J10-5a.
1.5
⎡
⎛ l ⎞ ⎛ t ⎞ ⎤ EFywt f
Rn = 0.40tw2 ⎢1 + 3 ⎜ b ⎟ ⎜ w ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
1.5
⎡
⎛ 3.0 ⎞⎛ 0.300 ⎞ ⎤ 29,000 ( 50 )( 0.425)
= 0.40 ( 0.300 ) ⎢1 + 3 ⎜
(1.0)
⎟⎜
⎟ ⎥
0.300
⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎦⎥
⎣⎢
2
= 67.2 kips
So,
φRn = 0.75 ( 67.2 ) = 50.4 kips
Step 10:
Determine the strength for the end reaction. Since web local crippling
controls,
φRn = 50.4 ≥ 24.0 kips
Step 11:
Final selection.
Use a minimum bearing plate length along the beam web of 3.0 in. at the
concentrated load and at the end reactions.
EXAMPLE 6.21b
Beam Bearing
Strength by ASD
SOLUTION
Goal:
Determine the required length of bearing for a W-shape to support (a) a
midspan concentrated load and (b) concentrated reactions at the end of a
simple span.
Given:
A W18×35 A992 beam was selected in Example 6.3b, as a simply
supported member spanning 20 ft and carrying a midspan dead load of 8.0
kips and live load of 24.0 kips.
Step 1:
Determine section properties from Manual Table 1-1
d = 17.7 in., tw = 0.300 in., tf = 0.425 in., kdes = 0.827 in.
Step 2:
Determine the required bearing strength at the load point and at the
reactions using the ASD load combinations from Section 2.4.
Pa = PD + PL = 8.0 + 24.0 = 32.0 kips
P 32
= 16 kips
Ra = a =
2
2
Part a
Step 3:
At the load
For the limit state of yielding, determine the required length within the web
to resist the applied load. The required length, based on Equation J10-2
where N = ( 5k + lb ) , is then
( 5k + lb ) =
1.5 ( 32 )
Pn
ΩPa
=
=
= 3.20 in.
Fywt w Fy t w 50 ( 0.300 )
Chapter 6
Step 4:
Bending Members 255
Determine the minimum length of bearing, lb. Since this force is located so
that distribution takes place in both directions,
lb = 3.20 − 5 ( 0.827 ) = −0.935 in.
A negative bearing length in this calculation means that the distribution, 5k,
is more than enough to carry the required load. Thus, the minimum bearing
length for normal practice of 3.0 in. is adequate.
Step 5:
For the limit state of web local crippling, determine the nominal strength if
the bearing length is taken as a practical minimum of 3.0 in. From Equation
J10-4,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFy t f
2
Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
1.5
⎡
⎛ 3.00 ⎞⎛ 0.300 ⎞ ⎤ 29,000(50)(0.425)
= 0.80 ( 0.300 ) ⎢1 + 3 ⎜
(1.0 )
⎟⎜
⎟ ⎥
0.300
⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎥⎦
⎢⎣
= 134 kips
Determine the allowable strength for web local crippling,
Rn 134
=
= 67.0 kips
Ω 2.00
2
Step 6:
Step 7:
For the final selection at mid-span, since
Ra = 32 kips < Rn Ω = 67 kips
a 3.0 in. bearing length will be sufficient.
Part b
Step 8:
Step 9:
At the end reaction
Since the end reactions are less than the concentrated load at mid-span, the
3.0 in. minimum bearing length will likely be sufficient for the end
reactions also. Thus, check a 3.0 in. bearing.
Rn = Fywtw ( 2.5k + lb ) = 50 ( 0.300 ) ( 2.5 ( 0.827 ) + 3.0 ) = 76.0 kips
Rn Ω = 76.0 1.5 = 50.7 kips
Check the limit state of web local crippling. Since the reaction is at the end
of the member at a distance less than d 2 and lb d = 3.0 17.7 = 0.17 ≤ 0.2
use Equation J10-5a.
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
Rn = 0.40t ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
2
w
1.5
2 ⎡
⎛ 3.0 ⎞⎛ 0.300 ⎞ ⎤ 29,000 ( 50 )( 0.425)
= 0.40 ( 0.300 ) ⎢1 + 3 ⎜
(1.0)
⎟⎜
⎟ ⎥
0.300
⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎥⎦
⎢⎣
= 67.2 kips
So,
Rn Ω = 67.2 2.0 = 33.6 kips
Step 10:
Determine the strength for the end reaction. Since web local crippling
2556 Chapter 6
Bending Members
M
contro
ols,
Rn Ω = 33.6 >16.00 kips
Step
p 11:
Final selection
Use a minimum bearing
b
plate length along the beam weeb of 3.0 in. at the
conceentrated load and at the endd reactions.
6.116
OPEN
N WEB STE
EEL JOISTS
S AND JOIS
ST GIRDER
RS
The term op
pen web steel joist
j
refers to
o a building pproduct made according too the design sttandards
of the Steel Joist Institutte (SJI). They
y are manufacctured trussess regularly ussed for buildiing floor
and roof sysstems. An ex
xample of their use in a buuilding is shoown in Figuree 6.30. As a bbuilding
product, thee structural en
ngineer is nott involved inn actually dessigning the truuss but ratherr selects
the product from a table,, much like seelecting a parrticular W-shhape. Althouggh these mem
mbers are
made from steel, they are
a not desig
gned accordinng to any A
AISC standarrds but ratherr to the
standards fo
ound in the SJJI publication
n Standard Sppecifications,, Load Tabless, and Weighht Tables
for Steel Joiists and Joist Girders (2010).
Som
me advantagess of open weeb steel joistss include the fact that theyy are lightweight and
can easily span
s
long disstances, theirr open webs can easily aaccommodatee passing mechanical
systems thro
ough the stru
ucture, and in
n some appli cations they are more ecoonomical thaan rolled
steel shapess. Disadvantaages include a lower loaad carrying ccapacity thann rolled shappes, thus
requiring mu
uch closer sp
pacing; the in
nability to eassily accommoodate concenttrated loads aat points
other than trruss panel points; and poteential vibratioon issues wheen they are ussed for floor ssystems.
In all appliccations, the economic
e
adv
vantages or ddisadvantages of using an open web stteel joist
must be asseessed for the specific
s
cond
ditions in quesstion.
Fou
ur types of op
pen web steell joists are ddefined in SJII standards: K
K-series, KCS
S-series,
LH-series an
nd DLH-seriees.
K-seeries joists arre perhaps th
he most comm
monly used jjoists for flooor and roof ssystems.
They are available in dep
pths from 10 in. to 30 in., and design taables are available coverinng spans
up to 60 ft.
ft The standard designatiion for K-seeries joists iss 16K6, wheere the first number
represents th
he depth, 16 in.; the letterr indicates thhe series desiignation, K; aand the third number
represents place
p
within th
he series, 6. The
T series deesignation ideentifies the deetails of mannufacture
of the trusss, including the
t sizes of the
t elements that make uup the truss. It is here thhat each
manufactureer has the opp
portunity to deesign its own specific joistt. For a 16K joist there are seven
n of Open Web
b Steel Joists
Figure 6.30 Application
Photo courtessy Douglas Steeel Fabricating Corporation
Chapter 6
Bending Mem
mbers 257
Figure 6.31
6
LRFD Economy
E
Tab
ble for K-Seri es Open Webb Steel Joists
Standard Specifications,, Load Tables, and Weight Ta
Tables for Steel Joists and Joisst Girders (20110),
© Steel Jo
oist Institute. Reprinted
R
with Permission. A
All rights reservved.
258 Chapter 6
Bending Members
designations within the series: 2, 3, 4, 5, 6, 7, and 9. Each increase in designation indicates an
increase in weight. For example, the seven designations of the 16K indicate joists that weigh
approximately 5.5, 6.3, 7.0, 7.5, 8.1, 8.6, and 10.0 lb/ft. Weight is given as an approximate value
since each manufacturer has the opportunity to produce joists with different cross section
elements. The one constant, however, is that whatever joist is produced, it must have the strength
indicated in the SJI standard. Figure 6.31 shows an example of the selection tables published by
SJI. K-series joists are designed to carry a uniformly distributed load. Thus, the bending strength
can vary along the span from a maximum at midspan to zero at the support, approximating a
parabolic moment diagram, and the shear strength can vary from zero at midspan to a maximum
at the support, approximating a triangular shear diagram. Design tables like those in Figure 6.31
give two numbers for each joist and span combination. Joist strength in terms of load per foot of
span is given as the upper number, and load per foot of span to cause a deflection of 1 360 of the
span is given as the lower number.
KCS-series joists are identified in the same fashion as the K-series joists. Thus, a
10KCS3 is a 10 in. deep KCS-series joist with a series designation of 3. The difference between
K- and KCS-series joists is directly related to the moment and shear diagrams. KCS-series joists
are designed for a diagram for uniform moment over all interior panel points and a constant shear.
These joists are particularly useful for supporting loads that combine uniformly distributed and
concentrated forces. Unlike the K-series joists, the design tables for the KCS-series give only the
moment and shear capacities.
LH-series joists are long-span joists. Their strengths are tabulated for spans from 25 ft to
96 ft. LH-series joists have depths from 18 in. to 48 in. A typical designation would be 28LH05.
LH-series joists are used for floor and roof systems, and, like K-series joists, they are designed
for a uniformly distributed load.
DLH-series joists are deep long-span joists intended primarily to support roof decks.
DLH-series joists start at a depth of 52 in. and go up to 120 in. The design tables indicate spans
from 62 ft up to 240 ft. The designation system is the same as for other joists; a 68DLH17 is a 68
in. deep DLH-series joist with a series designation of 17.
Another product that is designed according to the SJI standards is the joist girder. Joist
girders are essentially pre-engineered trusses that are intended to support concentrated loads at
the panel points. They are used to support open web steel joists that are evenly spaced and
introduce the same load at each panel point. A possible designation for a joist girder is
44G8N12K. The first number indicates a 44 in. deep member and the G indicates a joist girder.
The 8N indicates that there are eight joist spaces, which is the same as saying there are seven
concentrated loads on the girder. The final number indicates the load in kips. If the load is for an
LRFD design the last letter would be F, indicating a factored load, and the load magnitude would
be the LRFD required strength.
EXAMPLE 6.22
Open Web Steel
Joist Selection by
LRFD
Goal:
Select a K-series open web steel joist from the limited selection available in
Figure 6.31 to satisfy strength and deflection criteria.
Given:
Joists supporting a roof deck span 30 ft and are spaced 6.0 ft on center. The
load is a uniformly distributed dead load of 20 psf and live load of 30 psf.
Select the lightest 18 in. K-series open web steel joist available in Figure
6.31 to support this load. Limit deflection to 1/360 of the span.
SOLUTION
Step 1:
Determine the required load in pounds per foot.
Step 2:
Select a K-series joist for a span of 30 ft.
wu = (1.2wD + 1.6wL ) Ltrib = (1.2 ( 20) + 1.6 ( 30) ) ( 6) = 432 1b/ft
Chapter 6
Bending Members 259
An 18K6 will support a factored load of 451 lb/ft. This is the upper number
in the table. The other number in the table is 175 lb/ft. This is the
serviceability load that will produce a deflection of 1/360 of the span.
Step 3:
Check the deflection of the 18K6 for the serviceability live load of 30 psf.
For the 6.0 ft joist spacing, the joist carries a serviceability live load of
6(30)=180 lb/ft. Since this exceeds the load given in the table, the next
larger joist should be selected if this deflection limit is to be satisfied.
Step 4:
Check the 18K7.
For strength the capacity is
502 lb/ft > 432 lb/ft.
For deflection the capacity is
194 lb/ft > 180 lb/ft.
Step 5:
Check to be sure the joist can carry its self-weight.
The joist weighs approximately 9.0 lb/ft. The factored dead load due to
self-weight is
wself = 1.2(9.0) = 10.8 lb/ft
Thus, the total load the joist must carry is
wu = 432 + 10.8 = 443 lb/ft < 502 lb/ft
Step 6:
Final selection:
18K7
6.17
PROBLEMS
1. Determine the elastic section modulus and the plastic
section modulus for a W40×199 modeled as three
rectangles forming the flanges and the web. Compare the
calculated values to those given in the Manual.
5. Determine the elastic section modulus and the plastic
section modulus for a W18×35 modeled as three
rectangles forming the flanges and the web. Compare the
calculated values to those given in the Manual.
2. Determine the elastic section modulus and the plastic
section modulus for a W36×330 modeled as three
rectangles forming the flanges and the web. Compare the
calculated values to those given in the Manual.
6. Determine the plastic section modulus for a W44×230
modeled as three rectangles forming the flanges and the
web. Compare the calculated value to that given in the
Manual.
3. Determine the elastic section modulus and the plastic
section modulus for a W33×130 modeled as three
rectangles forming the flanges and the web. Compare the
calculated values to those given in the Manual.
7. Determine the plastic section modulus for a W27×84
modeled as three rectangles forming the flanges and the
web. Compare the calculated value to that given in the
Manual.
4. Determine the elastic section modulus and the plastic
section modulus for a W24×55 modeled as three
rectangles forming the flanges and the web. Compare the
calculated values to those given in the Manual.
8. Determine the plastic section modulus for a W16×31
modeled as three rectangles forming the flanges and the
web. Compare the calculated value to that given in the
Manual.
260 Chapter 6
Bending Members
9. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a WT15×45 modeled as two rectangles forming the
flange and the stem. Compare the calculated values to
those given in the Manual.
18. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a MC13×50 modeled as three rectangles forming the
flanges and the web. Compare the calculated values to
those given in the Manual.
10. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a WT8×50 modeled as two rectangles forming the
flange and the stem. Compare the calculated values to
those given in the Manual.
19. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus,
all about the geometric axis, for a L4×4×1/2 modeled as
two rectangles. Compare the calculated values to those
given in the Manual.
11. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a WT5×22.5 modeled as two rectangles forming the
flange and the stem. Compare the calculated values to
those given in the Manual.
20. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus,
all about the geometric axis, for a L8×8×7/8 modeled as
two rectangles. Compare the calculated values to those
given in the Manual.
12. Determine the plastic section modulus for an
HSS8×6×1/2 modeled as four rectangles forming the
flanges and webs. Remember to use the design wall
thickness for the plate thickness and ignore the corner
radius. Compare the calculated value to that given in the
Manual.
21. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus,
all about the geometric axis, for a L3×3×1/4 modeled as
two rectangles. Compare the calculated values to those
given in the Manual.
13. Determine the plastic section modulus for an
HSS12×8×5/8 modeled as four rectangles forming the
flanges and webs. Remember to use the design wall
thickness for the plate thickness and ignore the corner
radius. Compare the calculated value to that given in the
Manual.
22. A simply supported beam spans 25 ft and carries a
uniformly distributed dead load of 0.6 kip/ft, including the
beam self-weight, and a live load of 2.1 kip/ft. Determine
the minimum required plastic section modulus and select
the lightest-weight W-shape to carry the moment.
Consider only the limit state of yielding and use A992
steel. Design by (a) LRFD and (b) ASD.
14. Determine the plastic section modulus for an
HSS8×8×1/2 modeled as four rectangles forming the
flanges and webs. Remember to use the design wall
thickness for the plate thickness and ignore the corner
radius. Compare the calculated value to that given in the
Manual.
23. For flexure only, determine the lightest-weight Wshape to carry a uniform dead load of 1.2 kip/ft, including
the beam self-weight, and a live load of 3.2 kip/ft on a
simple span of 20 ft. Consider only the limit state of
yielding and use A992 steel. Design by (a) LRFD and (b)
ASD.
15. Determine the plastic section modulus for a round
HSS10×0.500. Remember to use the design wall
thickness in your calculations. Compare the calculated
value to that given in the Manual.
24. A beam is required to carry a uniform dead load of
0.80 kip/ft, including its self-weight, and a concentrated
live load of 14 kips at the center of a 30 ft span. For
bending only, determine the least-weight W-shape to
carry the load. Consider only the limit state of yielding
and use A992 steel. Design by (a) LRFD and (b) ASD.
16. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a C15×40 modeled as three rectangles forming the
flanges and the web. Compare the calculated values to
those given in the Manual.
17. Determine the elastic neutral axis, elastic section
modulus, plastic neutral axis, and plastic section modulus
for a C8×11.5 modeled as three rectangles forming the
flanges and the web. Compare the calculated values to
those given in the Manual.
25. A simply supported beam spans 32 ft and carries a
uniformly distributed dead load of 1.8 kip/ft plus the
beam self-weight and a live load of 5.4 kip/ft. Determine
the minimum required plastic section modulus and select
the lightest-weight W-shape to carry the moment.
Consider only the limit state of yielding and use A992
steel. Design by (a) LRFD and (b) ASD.
Chapter 6
26. For bending only, determine the lightest-weight Wshape to carry a uniform dead load of 4.5 kip/ft plus the
beam self-weight and a live load of 3.5 kip/ft on a simple
span of 24 ft. Consider only the limit state of yielding and
use A992 steel. Design by (a) LRFD and (b) ASD.
Bending Members 261
and at the load point. Use A992 steel and Cb = 1.0. Design
for flexure by (a) LRFD and (b) ASD.
27. A beam is required to carry a uniform dead load of 3.4
kip/ft plus its self-weight, and a concentrated dead load of
12 kips and a concentrated live load of 20 kips, both at the
center of a 40 ft span. For bending only, determine the
least-weight W-shape to carry the load. Consider only the
limit state of yielding and use A992 steel. Design by (a)
LRFD and (b) ASD.
34. A 36 ft simply supported beam is loaded with
concentrated loads 16 ft in from each support. On one
end, the dead load is 8.0 kips and the live load is 18.0
kips. At the other end, the dead load is 4.0 kips and the
live load is 9.0 kips. Include the self-weight of the beam
in the design. Lateral supports are provided at the
supports and the load points. Considering only bending,
determine the least-weight W-shape to carry the load. Use
A992 steel and Cb = 1.0. Design by (a) LRFD and (b)
ASD.
28. Considering only bending, determine the lightestweight W-shape to carry the following loads: a uniform
dead load of 0.6 kip/ft plus self-weight, a concentrated
dead load of 2.1 kips, and a concentrated live load of 6.4
kips, located at the center of a 16-ft span. Assume full
lateral support and A992 steel. Design by (a) LRFD and
(b) ASD.
35. An 18 ft simple span beam is loaded with a uniform
dead load of 1.4 kip/ft, including the beam self-weight,
and a uniform live load of 2.3 kip/ft. The lateral supports
are located at the supports and at 6.0 ft intervals.
Considering only bending, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0.
Design by (a) LRFD and (b) ASD.
29. Considering only bending, determine the lightest Wshape to carry a uniform dead load of 4.0 kip/ft plus the
self-weight and a uniform live load of 2.3 kip/ft on a
simple span of 10.0 ft. Assume full lateral support and
A992 steel. Design by (a) LRFD and (b) ASD.
36. A 40 ft simple span beam is loaded with a uniform
dead load of 2.2 kip/ft plus the beam self-weight and a
uniform live load of 3.6 kip/ft. The lateral supports are
located at the supports and at the midpoint of the span.
Considering only bending, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0.
Design by (a) LRFD and (b) ASD.
30. A 24 ft simple span laterally supported beam is
required to carry a total uniformly distributed service load
of 8.0 k/ft. Determine the lightest, A992, W-shape to
carry this load, considering only flexure and the limit state
of yielding, if the load is broken down as follows. Use
LRFD.
a. Live load = 1.0 k/ft; dead load = 7.0 kips/ft
b. Live load = 3.0 k/ft; dead load = 5.0 kips/ft
c. Live load = 5.0 k/ft; dead load = 3.0 kips/ft
d. Live load = 7.0 k/ft; dead load = 1.0 kips/ft.
31. Repeat the designs specified in Problem 30 using
ASD.
32. A 30 ft simply supported beam is loaded at the third
points of the span with concentrated dead loads of 10.0
kips and live loads of 15.0 kips. Lateral supports are
provided at the supports and at the load points. The selfweight of the beam will be ignored for this problem.
Considering only flexure, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0.
Design by (a) LRFD and (b) ASD.
33. Determine the least-weight W-shape to support a
concentrated dead load of 24 kips plus the beam selfweight and a concentrated live load of 15 kips on a 30 ft
span. The concentrated loads are located at the midpoint
of the span. Lateral supports are provided at the supports
37. An A992 W18×60 is used on a 36 ft simple span to
carry a uniformly distributed load. Using Cb = 1.0,
determine the locations of lateral supports in order to
provide just enough strength to carry (a) a design moment
of 435 ft-kips and (b) an allowable moment of 290 ft-kips.
38. An A992 W16×31 is used on an 18 ft simple span to
carry a uniformly distributed dead load of 0.95 kips/ft
plus self-weight and a live load of 1.9 kips/ft. Using Cb =
1.0, determine the locations of lateral supports in order to
provide just enough strength to carry this loading by (a)
LRFD and (b) ASD.
39. Redesign the beam of Problem 33 using the correct
value of Cb.
40. Redesign the beam of Problem 34 using the correct
value of Cb.
41. Redesign the beam of Problem 35 using the correct
value of Cb.
42. There are no tables in the Manual for the design of
beams with Fy other than 50 ksi. Construct the moment
strength vs. unbraced length curve, similar to those in
Manual Table 3-10, for a W21×68 A572 Grade 65 steel.
Show the curve up to an unbraced length of 40 ft.
262 Chapter 6
Bending Members
43. Since the chart developed for Problem 42 was similar
to those in Manual Table 3-10, it used Cb = 1.0. Redo
Problem 42 with values of Cb = 1.0, 1.14, 1.32, and 1.67.
44. Plot the nominal moment strength vs. unbraced
length for a W27×84 using A992 and A913 Grade 65
steel. Compare the nominal moment strengths for an
unbraced length of 18 ft.
45. Plot the nominal moment strength vs. unbraced
length for a W16×31 using A36 and A992 steel. Compare
the nominal moment strengths for an unbraced length of
7.0 ft.
46. Determine the available shear strength of a W24×62
A992 steel member by (a) LRFD and (b) ASD.
47. Determine the available shear strength of a W30×90
A992 steel member by (a) LRFD and (b) ASD.
48. A girder that carries a uniformly distributed dead load
of 1.7 k/ft plus its self-weight and three concentrated live
loads of 15 kips at the quarter points of the 36 ft span is to
be sized. Using A992 steel, determine the lightest Wshape to carry the load with lateral supports provided at
the supports and load points. Use the correct Cb, check
shear, and limit deflection to 1/360 of span. Design by (a)
LRFD and (b) ASD.
49. A 32 ft simple span beam carries a uniform dead load
of 2.4 k/ft plus its self-weight and a uniform live load of
3.0 k/ft. The beam is laterally supported at the supports
only. Determine the minimum-weight W-shape to carry
the load using A992 steel. Use the correct Cb, check
shear, and limit live load deflection to 1/360 of span.
Design by (a) LRFD and (b) ASD.
50. A 36 ft simple span beam carries a uniformly
distributed dead load of 3.4 kip/ft plus its self-weight and
a uniformly distributed live load of 2.4 kip/ft. Determine
the least-weight W-shape to carry the load. Use the
correct Cb, check shear, and limit live load deflection to
1/360 of the span. Use A992 steel and assume full lateral
support. Design by (a) LRFD and (b) ASD.
51. A simple span beam with a uniformly distributed dead
load of 1.1 k/ft, including the self-weight, and
concentrated dead loads of 3.4 kips and live loads of 6.0
kips at the third points of a 24-ft span, is to be designed
with lateral supports at the third points and live load
deflection limited to 1/360 of the span. Use the correct Cb,
and be sure to check shear. Determine the least-weight Wshape to carry the loads. Use A992 steel. Design by (a)
LRFD and (b) ASD.
52. A fixed-end beam on a 28 ft span is required to carry a
total ultimate uniformly distributed load of 32.0 kips.
Using plastic analysis and A992 steel, determine the
design moment and select the lightest W-shape. Assume
(a) full lateral support and (b) lateral support at the ends
and center line.
53. A beam is fixed at one support and simply supported
at the other. A concentrated ultimate load of 32.0 kips is
applied at the center of the 40 ft span. Using plastic
analysis and A992 steel, determine the lightest W-shape
to carry the load when the nominal depth of the beam is
limited to 18 in. Assume (a) full lateral support and (b)
lateral supports at the ends and the load.
54. A fixed-end beam on a 40 ft span is required to carry a
total ultimate uniformly distributed load of 72.5 kips.
Using plastic analysis and A992 steel, determine the
lightest-weight W-shape to carry the load. Assume full
lateral support.
55. A three-span continuous beam is to be selected to
carry a uniformly distributed dead load of 4.7 kip/ft,
including its self-weight, and a uniformly distributed live
load of 10.5 kip/ft. Be sure to check the shear strength of
this beam. Use A992 steel and assume full lateral support.
Design using moment redistribution by (a) LRFD and (b)
ASD.
56. Determine the available minor axis bending strength
of a W18×130 of A992 steel. Determine by (a) LRFD and
(b) ASD.
57. Determine the available minor axis bending strength
of a W10×12 of A992 steel. Determine by (a) LRFD and
(b) ASD.
58. Determine the available minor axis bending strength
of a W40×392 of A992 steel. Determine by (a) LRFD and
(b) ASD.
59. Determine the available bending strength of a
WT8×25, A992 steel, if the stem is in compression and
the unbraced length is 10 ft. Determine by (a) LRFD and
(b) ASD.
60. Determine the available bending strength of a
WT12×52, A992 steel, if the stem is in tension and the
unbraced length is 8 ft. Determine by (a) LRFD and (b)
ASD.
61. For the beam designed in Problem 34, determine the
minimum required lengths of bearing (at the concentrated
loads and at the supports) considering the limit state of
web yielding and the limit state of web local crippling.
Determine by (a) LRFD and (b) ASD.
62. For the girder designed in Problem 48, determine the
minimum required lengths of bearing (at the concentrated
C
Chapter 6
looads and at th
he supports) considering thee limit state off
w
web yielding and
a the limit state
s
of web lo
ocal crippling.
D
Determine by (a) LRFD and (b)
( ASD.
663. Obtain via the Internet a set of open web
w steel joistt
ddesign tables from any manufacturerr. Select thee
sshallowest K-series joist to span 35 ft and support a rooff
ssystem that carries a dead loaad of 25 psf and
d a live load off
335 psf. The joiists will be spaaced (i) 6.0 ft on center, (ii))
88.0 ft on centerr, and (iii) 10.0
0 ft on center. Limit
L
live load
d
ddeflection to sp
pan/360. Design
n by (a) LRFD
D and (b) ASD.
M
Multi-Chapterr Questions
664. Using thee framing plaan shown in Figure P6.64
4
((presented earliier as Figure 2..7), design the beams marked
d
11, 2, and 3. Thiis is the same structure
s
used in Section 2.5,,
w
where load calculations with
h live load reeductions weree
ddiscussed. Tho
ose calculation
ns can be reussed here. Load
d
ccase 2 for deaad plus live load is to be co
onsidered. Thee
bbuilding is an office
o
building
g with a nomin
nal live load off
550 pounds per square
s
foot (pssf) and a calcullated dead load
d
oof 70 psf.
1: Girderr AB on line 2-2 if the flo
oor deck spanss
from
m line 1-1 to 2-2
2 to 3-3:
2: Floor beam 2-3 on line D-D if the
t floor deck
k
span
ns from line C-C to D-D to E--E:
3: Floor beam 2-3 on line C-C if the
t floor deck
k
span
ns from line B-B to C-C to D--D:
D
Design by (a) LRFD
L
and (b) ASD.
A
P
P6.64
665. The framin
ng plan shown in Figure P6.6
65 is the samee
aas that shown
n in Figure P2
2.20 for an 18-story
1
officee
bbuilding. It mu
ust support a deead load of 80
0 psf and a livee
looad of 50 psff. In all casess, the decking
g spans in thee
ddirection from
m line A toward line E. Determine
D
thee
rrequired momeent and shear strength for the
t beams and
d
ddesign the beaams as requireed below for (a) design by
y
L
LRFD and (b) design by ASD
D. The requireed moment and
d
sshear strength were
w
determined in Problem 20 of Chapterr
22.
Bending Membbers 263
i. The beam b etween columnn lines 2 and 3 along line
D.
iii. The girder oon column line 3 between coluumn line E
and midwayy between liness D and C.
iiii. The beam oon the line bettween lines C and D and
column liness 3 and 4.
P6.655
66. IIntegrated Deesign Project. Design the beeams in the
graviity-only system
m for the loadds determined in Chapter
2. Bee sure to indiicate all assum
mptions regardding lateral
suppoort.
Reconsideer the proposedd framing planns given in
Figurres 1.20 and 1.21. For the panel bounded by column
lines A, C, 6, and 77, span the beaams in the 30 fft direction
and rredesign. Com
mpare the resullts with the deesigns done
abovve.
For open w
web steel joists, manufactureer literature
incluudes load tablees. These may bbe found on thhe Internet.
Selecct open web stteel joists for thhe roof using tthe loading
deterrmined in Chappter 2. Use spacing differentt from that
sugg ested in Figuure 1.22 if thhat appears too be more
approopriate.
C
Chapterr 7
P
Plate Girder
G
rs
N
New York Univversity Kimm
mel Center
Photograph
P
by Fadi Asmar/LERA Consultinng Structural E
Engineers
7.11
BACKG
GROUND
A plate gird
der is a bendiing member composed
c
of individual stteel plates. A
Although platee girders
are normally
y the memberr of choice forr situations inn which the avvailable rolledd shapes are nnot deep
or large enough to carry the intended load, there iss no requirem
ment that they will always bbe at the
deep or larg
ge end of the spectrum
s
of member
m
sizess. Beams fabrricated from individual steel plates
to meet a specific requireement are gen
nerally identiffied in the fielld as plate girrders.
Platte girders are usually used
d in building structures foor special situuations such as those
involving veery long span
ns or very laarge loads. Peerhaps their m
most commoon applicationn is as a
transfer gird
der, which iss a bending member thatt supports a structure above and perm
mits the
column spaccing to be chaanged below. They are alsso very comm
mon in industrrial structuress for use
as crane gird
ders and as support for larrge pieces off equipment. IIn commerciaal buildings, they are
often used to
o span large open
o
areas to meet particuular architectuural requiremeents; becausee of their
normally greater depth and
a resulting stiffness, theyy tend to defflect less thann other potenttial long
span solutions. An examp
ple of a buildiing applicatioon of the platee girder is shoown in Figuree 7.1.
The cross section
n of a typical plate girder is shown in F
Figure 7.2 annd a large platte girder
being transp
ported on a truck
t
is show
wn in Figure 7.3. Althouggh it is possible to combiine steel
plates into numerous
n
geo
ometries, the plate girderss addressed hhere are thosse formed froom three
plates, one for
f the web and
a two for th
he flanges. B ecause the w
web and flangges of the platte girder
are fabricateed from indiv
vidual plates, they can be designed witth the web annd flange plattes from
the same grade of steel (a
( homogeneo
ous plate girdder) or from ddifferent graddes of steel (aa hybrid
plate girder)). For hybrid
d girders, the flanges are uusually fabriccated with a hhigher grade of steel
than that useed in the web
b. This takes advantage
a
of tthe higher strresses that cann be developeed in the
flanges, whiich are locateed a greater distance
d
from
m the neutral axis than thee web, resultting in a
higher mom
ment strengtth contributiion. Hybrid girders aree relatively common inn bridge
construction
n, though theey are rarely used in builldings. In thee past, hybriid girders haave been
included in the AISC Speecification, bu
ut they are noot specificallyy addressed iin the currentt edition.
y, they are nott discussed in
n this book.
Accordingly
2664
Chapterr 7
7
Figure 7.1
Applicattion of a Platee Girder
Figure 7.2
7
Typical Plate
P
Girder Definitions
D
Figure 7.3
7
Plate Girrder Being Trransported
Photo cou
urtesy LeJeune Steel Co.
Plate Giirders 265
2666 Chapter 7
Plate Gird
ders
Fig
gure 7.4
Siingly Symmettric Plate Girdder
Ano
other type of plate girder is the singly ssymmetric girrder, one withh flanges thatt are not
of the same size, as seen in Figure 7.4
4. Although ssingly symmeetric plate girrders are addrressed in
the AISC Sp
pecification, they are not particularly common in bbuildings andd are not speecifically
addressed heere. Howeverr, the principlles for all of tthese plate girrders are the same, and thee careful
application of
o the Specifiication provissions will leadd to an econoomical and saafe design forr each of
them.
Buillt-up I-shapeed members with compacct webs are designed acccording to thhe same
provisions as
a rolled I-shaped memb
bers presenteed in Specifiication Sectioons F2 and F3 and
discussed in
n Chapter 6 off this book. The discussionn of plate girdders in this chhapter addressses these
built-up I-sh
hapes with noncompact or slender webss.
Table 7.1 lists th
he sections off the Specificaation and part
rts of the Mannual discussed in this
chapter.
7.22
HOMOG
GENEOUS PLATE GIIRDERS IN
N BENDING
G
The behavio
or of plate girrders will be addressed herre by consideering flexure and shear sepparately.
In flexure, a plate girder is considered
d in this bookk as either nooncompact or slender accoording to
the proportiions of the web.
w
Flanges can be com
mpact, noncom
mpact, or slender, and thee flange
slenderness is treated thee same as thaat discussed inn Chapter 6. Thus, it is possible, for eexample,
for a noncom
mpact web plate girder to have
h
a slendeer flange, poteentially contrrolling the cappacity of
the memberr. The design
n rules for each type of girder, noncompact web or slender w
web, are
considered separately.
s
With
h our discusssion limited to
o doubly sym
mmetric plate girders, the limit states thhat must
be considered are compression flang
ge yielding, ccompression flange local buckling, web local
buckling, an
nd lateral-torssional bucklin
ng. These are the same lim
mit states considered for thhe rolled
I-shapes in Chapter
C
6 witth the addition
n of web locaal buckling. R
Remember thaat all W-shapees have
Table 7.1 Sections of Specification
S
and
a Parts of M
Manual Coveered in This C
Chapter
Specificatioon
B4
F4
Classification of Sections for Lo
ocal Bucklingg
Otherr I-Shaped Meembers with Compact
C
or N
Noncompact W
Webs, Bent abbout Their
Major
M
Axis
F5
Doub
bly Symmetricc and Singly Symmetric
S
I- Shaped Mem
mbers with Sleender Webs
Bent
B about Their Major Axis
G
Desig
gn of Memberrs for Shear
J4.4 Streng
gth of Elemen
nts in Compreession
J10 Flang
ges and Webs with Concen
ntrated Forcess
Chapterr 7
Figuree 7.5 Platte
Flexurral Strength
Plate Giirders 267
Girder
Nominal
mit state wass not considdered for thoose bending
compact webs so the web local buckling lim
onal limit statte of tension fflange yieldinng found in Sp
Specification S
Sections F4
memberss. The additio
and F5 can be ignored
d, because thee compressionn flange alwaays controls oover the tensioon flange in
ymmetric hom
mogeneous members.
m
Platte girders witth noncompacct webs are aaddressed in
doubly sy
Specifica
ation Section F4, and thosse with slendder webs in S
Section F5. T
The nominal strength of
plate gird
ders, for all limit
l
states, can
c be describbed as shownn in Figure 7.5. As with tthe rolled Ishaped members
m
disccussed in Chaapter 6, the bbehavior is pllastic, inelastiic, or elastic. Figure 7.5
shows th
hat the plastiic behavior corresponds
c
to an area oof the figuree described aas compact.
Inelastic behavior co
orresponds to
o the area iidentified as noncompact, and elastiic behavior
correspon
nds to the areea identified as
a slender.
Applying
A
the same figuree to the lateraal-torsional bbuckling limiit state, the ffully braced
region co
orresponds to
o plastic behaavior, whereass the partiallyy braced region corresponnds to either
inelastic or elastic bucckling.
The
T flexural design
d
strengtth (LRFD) annd allowable sstrength (ASD
D) are determ
mined just as
they were for the flexu
ural memberss discussed inn Chapter 6 w
with
φ = 0.9
0 (LRFD)
Ω = 1.67 (ASD)
The requ
uirement for ASD
A
is
Ra ≤
The requ
uirement for LRFD
L
is
7.2.1
Rn
Ω
Ru ≤ φRn
(A
AISC B3-2)
(A
AISC B3-1)
Nonccompact Weeb Plate Gird
ders
uence of web slenderness on
o the strengtth of plate girrders is not trreated as a separate limit
The influ
state to be
b assessed th
hrough its ow
wn set of requuirements. Ratther, web slennderness is trreated as an
effect on
n the flange yielding
y
or flaange local buuckling strenggth and the laateral-torsionnal buckling
strength as provided in
n Specificatio
on Section F44.
The
T slenderneess parameterr of the web iis defined as λw = hc/tw. Foor the doublyy symmetric
plate gird
der, case 15 in
i Specification Table B4..1b, this can bbe simplifiedd to λw = h/tw, where h is
the clear distance betw
ween the flan
nges. Throughhout this chap
apter, hc in alll Specification equations
can be reeplaced by h.. Similarly Sxc
can
be
repl
aced
by
S
,
and Myc can bbe replaced byy My. For a
x
x a
plate gird
der to be nonccompact, the following reqquirements arre set:
2668 Chapter 7
Plate Gird
ders
λ pw < λ w ≤ λ rw
where, accorrding to Table B4.1b of the Specificatioon,
λ pw = 3.766
E
Fy
λ rw = 5.700
E
Fy
and
The influence off the noncom
mpact web iss characterizeed through thhe web plastification
factor, Rpc. This
T factor is used to assess the abilityy of the sectioon to reach itss full plastic capacity
and modifiees the flange yielding
y
or lo
ocal buckling and lateral toorsional buckkling limit staates. The
web plastificcation factor for a noncom
mpact web is ggiven in the Sp
Specification aas
⎡ Mp ⎛ Mp
⎞⎛ λ − λ pw ⎞ ⎤ M p
−⎜
− 1 ⎟⎜
R pc = ⎢
C F4-9b)
(AISC
⎟⎥ ≤
⎢⎣ M yc ⎝ M yc
⎠⎝ λ rw − λ pw ⎠ ⎦⎥ M yc
where
M p = Fy Z x ≤ 1.6 Fy S x
and
M yc = Fy S xc
Equ
uation F4-9b is
i shown in Figure
F
7.6 foor two values of Mp/Myc, oone with a m
maximum
value of 1.6
6 to account for
f the upper limit on Mp, and one arbittrarily taken aat 1.2 as an eexample.
The ratio Mp/Myc is the shape factor that was disscussed in Chhapter 6. As was the casee in that
discussion, it
i must be lim
mited to 1.6 in order to ennsure that the necessary rootation can takke place
before strain
n hardening occurs
o
as the section underrgoes plastic deformation. The minimuum Rpc is
seen to be 1.0,
1 regardless of Mp/Myc. Thus, a consservative apprroach would be to take Rppc = 1.0.
This is what the user note in Secttion F4 is im
mplying wheen it says thhat Section F5 may
conservativeely be used fo
or shapes thatt actually fall under the reqquirements off Section F4. Because
a plate gird
der with a weeb that is on
nly slightly nnoncompact w
would have ssignificant addditional
strength refl
flected throug
gh the use off Rpc, and beccause the calcculation of Rpc is not parrticularly
difficult, th
here is no ad
dvantage to this simplifiication. Thuss, Equation F4-9b will bbe used
throughout this
t chapter, as
a appropriatee.
Figure 7.6
Factor
Web Plastificcation
Chapterr 7
Figure 7.7
7
Plate Giirders 269
Nominall Flexural Streength Based oon Flange Loocal Buckling
Flange Local
L
Bucklin
ng
The requ
uirements for compression
n flange local buckling aree given in Section F4.3. T
The strength
of a nonccompact web plate girder as
a a function of flange slennderness is shhown in Figure 7.7.For a
compact flange girderr, the impact of the nonco mpact web iss to modify thhe strength byy the factor
Rpc, as giiven in Equatiion F4-1.
(A
AISC F4-1)
M n = R pc M yc
At
A the lower limit for a noncompact web, when the web is jjust barely nnoncompact,
λ w = λ pww , then
R pc =
Mp
M yc
And the nominal
n
stren
ngth for a com
mpact flange ggirder is
Mp
Mn =
M yc = M p
M yyc
(7.1)
(7.2)
At
A the upper limit for a no
oncompact w
web, that is thhe web is veryy close to beiing slender,
nd the nominaal strength forr a compact fl
flange girder bbecomes
λ w = λ pr , Rpc = 1.0 an
M n = 1. 0 M yc = Fy S xcc
(7.3)
For
F a noncom
mpact flange girder, at thhe juncture bbetween the nnoncompact aand slender
flange, λ f = λ rf , the nominal
n
mom
ment strength iis given as
M n = FL S xc = 0.7 Fy S xc
(7.4)
where the use of FL = 0.7 Fy , for double symmeetric memberss, accounts foor the stress abbove which
the mem
mber inelastic buckling app
plies. This is the same ass the residual stress assum
med to have
occurred in the hot rollled shapes, even
e
though thhis is a weldeed shape.
The
T moment strength for a noncompacct flange platte girder is ffound throughh the linear
interpolaation between the end points, as shown iin Figure 7.7,, and is givenn by Equation F4-13.
⎛ λ − λ pf ⎞
M n = R pc M yc − ( R pc M yc − FL S xxc ) ⎜
(A
AISC F4-13)
⎟
⎝ λ rf − λ pf ⎠
2770 Chapter 7
Plate Gird
ders
The influence off the noncom
mpact web is diminished as the flangee becomes m
more and
more nonco
ompact. Oncee the flange becomes
b
slendder, Rpc is noo longer needded, and flannge local
buckling con
ntrols the streength of the girder.
For a slender flaange girder, the
t behavior is an elastic buckling phenomena deppicted in
Figure 7.7 and given as
0.99 Ekc S xc
(AISC
C F4-14)
Mn =
λ2
where λ = b f 2t f and th
he plate buckliing factor is
kc =
4
(7-5)
h / tw
This plate buckling
b
facttor must be taken in the range from 0.35 to 0.766 for the purrpose of
calculation, even though it may actually be outside that range.
kling
Lateral-Torrsional Buck
Lateral-torsiional bucklin
ng for noncom
mpact web ggirders is adddressed in Seection F4.2. Lateraltorsional bu
uckling behav
vior for noncompact web plate girderss is, in princiiple, the sam
me as for
rolled beam
ms. However, the equations in the Sppecification aare slightly aaltered, and tthe web
plastification
n factor mustt be included. As when connsidering the noncompact flange, the innfluence
of the noncompact web is diminished
d as the laterral-torsional bbuckling respponse becom
mes more
dominant. Figure
F
7.8 sh
hows the stren
ngth of a nooncompact weeb plate girdder when connsidering
lateral-torsio
onal buckling
g.
For a plate girdeer with lateraal supports att a spacing no greater thaan Lp, Equatioon F4-1,
which inclu
udes the influ
uence of Rpc, again definnes the girderr strength. T
This girder w
would be
considered to
t have full laateral supportt. The definittion of this limiting unbraaced length is slightly
different thaan it was for a compact web
b member. Thhis differencee is slight, butt has been useed in the
Specification
n, because itt gives more accurate resuults when ussed for singlyy symmetric girders.
Thus, for no
oncompact weeb girders
E
L p = 1.11rt
SC F4-7)
(AIS
Fy
For I-shapess with a rectaangular comp
pression flangge, the effecttive radius off gyration forr lateraltorsional bu
uckling, rt, iss the radius of
o gyration off the compresssion flange plus one thirrd of the
compression
n portion of th
he web, given
n as
Figure 7.8 Nominal Fleexural Streng
gth Based on U
Unbraced Lenngth
Chapter 7
rt =
Plate Girders 271
b fc
(AISC F4-11)
⎛ 1 ⎞
12 ⎜ 1 + aw ⎟
⎝ 6 ⎠
where
aw =
hc t w
b fc t fc
(AISC F4-12)
The strength of a section undergoing elastic lateral-torsional buckling, Lb > Lr, can be
obtained through plate buckling theory. The Specification uses a combination of Equations F4-3
and F4-5 to determine the nominal moment strength as
Mn =
Cb π2 ES xc
⎛ Lb ⎞
⎜r ⎟
⎝ t ⎠
2
2
1 + 0.078
J ⎛ Lb ⎞
≤ R pc M yc
S xc ho ⎜⎝ rt ⎟⎠
(7.6)
This is essentially the same equation as was used for compact web girders as given through
Equations F2-3 and F2-4 but with rt replacing rts.
Because built-in stresses occur in plate girders just as residual stresses do for rolled
shapes, elastic buckling cannot occur if the applied stress pushes the actual stress on the shape
beyond the yield stress. With the built-in stress taken as 0.3Fy, the available elastic stress is again
taken as 0.7Fy. Thus, Equation 7.4 again gives the limiting strength, this time for elastic lateraltorsional buckling. Using this strength with Equation 7.6, the unbraced length that defines the
limit of elastic lateral-torsional buckling is obtained as
Lr = 1.95rt
E
FL
2
J
⎛ J ⎞
⎛F ⎞
+ ⎜
+ 6.76 ⎜ L ⎟
⎟
S xc ho
⎝ E⎠
⎝ S xc ho ⎠
2
(AISC F4-8)
If Equation F4-8 is modified to reflect only the doubly symmetric girders considered in this
chapter, and FL is taken as 0.7Fy, it becomes
Lr = 1.95rt
E
0.7 Fy
2
J
⎛ J ⎞
⎛ 0.7 Fy ⎞
+ ⎜
+ 6.76 ⎜
⎟
⎟
S x ho
⎝ E ⎠
⎝ S x ho ⎠
2
(7.7)
The lateral-torsional buckling strength when the member has an unbraced length between
Lp and Lr is given by the same straight-line equation used previously. This time, however, it
accounts for the noncompact web by including Rpc at the upper limit; thus, with FL = 0.7 Fy and
S xc = S x , Equation F4-2 becomes
⎡
⎛ Lb − Lp ⎞ ⎤
M n = Cb ⎢ R pc M yc – ( R pc M yc – 0.7Fy S x ) ⎜
⎟ ⎥ ≤ R pc M yc
⎝ Lr − L p ⎠ ⎥⎦
⎣⎢
7.2.2
(7.8)
Slender Web Plate Girders
Slender web plate girders are covered in Specification Section F5. They are those built-up
members with web slenderness, λ w = hc tw , exceeding the limit
λ rw = 5.70
E
Fy
2772 Chapter 7
Plate Gird
ders
Figgure 7.9 Beending Strenggth
Redduction Factoor
T
B4.1b, case 15.
as given in Table
As was
w the case for noncomp
pact web mem
mbers, slendeer web membbers are evaluuated by
assessing ho
ow the web sllenderness influences the oother applicaable limit statees. The impacct of the
slender web
b on member strength is characterized through the bbending strenngth reductionn factor,
Rpg. The ben
nding strength
h reduction faactor is given by Equation F5-6 and is shown in Figuure 7.9.
⎛ hc
aw
E ⎞
R pg = 1 –
(AIS
SC F5-6)
⎜⎜ – 5.7
⎟ ≤ 1.0
1200 + 300aw ⎝ tw
Fy ⎟⎠
where aw is as defined by
y Equation F4-12
F
but in tthis applicatioon is limited to a value noo greater
than 10.
Speccification Secction F13.2 places
p
limitss on the propportions of m
members thatt can be
designed acccording to itss provisions. The
T web-to-fflange area raatio, aw, is lim
mited to 10 to prevent
the designerr from using these
t
provisio
ons for membbers that are essentially w
webs with flannges that
are little mo
ore than smalll stiffeners. In
n addition, thee web slenderness is limitted so that h/ttw ≤ 260.
This ensuress that the gird
der is not so slender that tthe stated proovisions do noot properly reeflect its
behavior.
The bending streength reductio
on factor reduuces the strenngth of the giirder uniform
mly in all
ranges of flaange local bu
uckling and laateral-torsionaal buckling. T
Thus, it couldd simply be hheld as a
final reductiion, as shown
n in the Specification, or ccould be usedd within the primary equaations as
shown below
w. Because itt is uniformly
y applied to aall of the limiit states, the influence of a slender
web on flexu
ural strength is easily visuaalized.
Flange Local Buckling
The strength
h of a plate girder
g
as a fu
unction of flaange slendernness is shownn in Figure 7..10. The
slenderness parameters for
fo the flange are defined iin Table B4.11b and are the same as thoose used
c
flangge plate girdeer, λ f ≤ λ pf ,
in Chapter 6 and Section 7.2.1. For a compact
M n = R pg Fy S xc
(AIS
SC F5-1)
Because Rpgg will not exceed 1.0, thee maximum bbending strenngth of the sslender web ggirder is
limited to th
he yield momeent.
At the
t juncture between the noncompact and slender flange, λ f = λ rf , the strrength is
limited to ellastic behavio
or, after accou
unting for buillt-in stresses. Thus
(7.9)
M n = R pg ( 0.7 Fy S xc )
Chapterr 7
7
Figure 7.10
Plate Giirders 273
Nominaal Flexural Sttrength Basedd on Flange Slenderness
minal moment strength for the
t slender w
web–noncomppact flange plaate girder is ggiven by the
The nom
linear traansition which
h is a combinaation of Equaations F5-7 annd F5-8 as
⎡
⎛ λ – λ pf ⎞ ⎤
(7.10)
M n = R pg S xc ⎢ Fy – (0.3Fy ) ⎜
⎟⎥
⎢⎣
⎝ λ rf – λ pf ⎠ ⎦⎥
For
F the slend
der web–slend
der flange meember, the sttrength is thee same as it w
was for the
noncomp
pact web–slen
nder flange member
m
and ggiven in Equaation F4-14, eexcept for thee use of Rpg.
Thus, thee combination
n of Equation
ns F5-7 and F55-9 gives
0. 9Ekc S xc R pg
Mn =
(7.11)
λ2
The platee buckling facctor, kc, is as previously
p
deefined.
Lateral-Torsional Bu
uckling
Lateral-to
orsional buck
kling for thee slender weeb girder apppears quite similar to thhat for the
noncomp
pact web gird
der and is sho
own in Figuree 7.11. However, some diffferences musst be noted.
For a meember to be considered
c
ass having full llateral support, its unbraced length is llimited to a
spacing not
n greater thaan Lp, where
Figure 7.11
7
Nominaal Flexural Sttrength Basedd on Unbracedd Length
2774 Chapter 7
Plate Gird
ders
L p = 1.1rt
E
Fy
(AIS
SC F4-7)
This is the same limit used
u
for non
ncompact webb girders, buut is differennt than that uused for
compact web members.
The elastic lateraal-torsional buckling
b
strenngth of the sllender web ggirder is giveen in the
Specification
n as a combin
nation of Equations F5-1 aand F5-4 as
Cb π 2 ESS xc R pg
(7.12)
Mn =
≤ R pg Fy S xc
2
⎛ Lb ⎞
⎜r ⎟
⎝ t ⎠
Wheen this streng
gth is set equaal to the correesponding streength limit giiven by Equattion 7.9,
the limiting unbraced len
ngth, Lr, becom
mes
E
Lr = πrt
(AIS
SC F5-5)
0.7Fy
This limit for
fo elastic lateeral-torsional buckling is nnot the samee as was usedd for the nonccompact
web girder. Thus, Lr is different fo
or each of thhe three typees of plate ggirder: compaact web,
noncompactt web, and sleender web.
For the inelasticc lateral-torsiional bucklinng region, thhe strength iis given by a linear
equation sim
milar to thosse used prev
viously, with the additionn of the Rpg multiplier; tthus the
combination
n of Equationss F5-2 and F5
5-3 becomes
⎡
⎛ Lb – L p ⎞ ⎤
(7.13)
M n = Cb R pg S xc ⎢ Fy – (0.33Fy ) ⎜
x
⎟ ⎥ ≤ R pg Fy S xc
⎢⎣
⎝ Lr – Lp ⎠ ⎦⎥
Figuree 7.12 Platee Girder for E
Example 7.1
Chapter 7
EXAMPLE 7.1
Plate Girder
Flexural Strength
SOLUTION
Plate Girders 275
Goal:
Determine the available flexural strength for two different plate girder
designs, a web thickness of (a) 3/8 in. (noncompact) and (b) 1/4 in.
(slender).
Given:
The cross section of a homogeneous A36 plate girder is shown in Figure
7.12. The span is 120 ft and the unbraced length of the compression
flange is 20 ft. Assume Cb = 1.0.
Step 1:
Determine the section properties for both plate girders.
(a) 3/8 in. web
(b) 1/4 in. web
tw = 0.375 in.
tw = 0.25 in.
A = 63.5 in.2
A = 57.5 in.2
4
Ix = 30,600 in.
Ix = 29,500 in.4
4
Iy = 2560 in.4
Iy = 2560 in.
3
Sx = 1230 in.
Sx = 1190 in.3
Zx = 1330 in.3
Zx = 1260 in.3
ry = 6.35 in.
ry = 6.67 in.
Part (a)
Step 2:
For the plate girder with a 3/8 in. web plate
Check the web slenderness in order to determine which sections of the
Specification must be followed.
Thus, this is a noncompact web girder, and the provisions of Section F4
must be followed. The web plastification factor must be determined.
Step 3:
Determine the shape factor and calculate Rpc.
M p Z x 1330
=
=
= 1.08 < 1.6
M y c S x 1230
Therefore, use 1.08 in the calculation of Rpc, in Equation F4-9b.
⎡ Mp ⎛ Mp
⎞⎛ λ − λ pw ⎞ ⎤ M p
Rpc = ⎢
−⎜
− 1⎟⎜
⎟⎥ ≤
⎢⎣ M yc ⎝ M yc
⎠⎝ λ rw − λ pw ⎠ ⎥⎦ M yc
⎡
⎛ 128 – 107 ⎞ ⎤
= ⎢1.08 – (1.08 – 1.0 ) ⎜
⎟ ⎥ = 1.05 ≤ 1.08
⎝ 162 – 107 ⎠ ⎦
⎣
Step 4:
Determine the nominal bending strength for the limit state of compression
flange yielding.
M n = R pc Fy S xc
= 1.05 ( 36 )(1230 ) = 46,500in.-kips
=
Step 5:
46,500
= 3880ft-kips
12
Check the unbraced length for lateral torsional buckling, with Lb = 20 ft or
240 in. First determine the effective radius of gyration.
276 Chapter 7
Plate Girders
aw =
48 ( 0.375 )
hc t w
=
= 0.791
b fc t fc 26.0 ( 0.875 )
and from Equation F4-11
b fc
26.0
rt =
=
= 7.05 in.
⎛ aw ⎞
⎛ 0.791 ⎞
12 ⎜ 1 + ⎟
12 ⎜1 +
⎟
6 ⎠
6 ⎠
⎝
⎝
From Equation F4-7
L p = 1.1rt
E
Fy
29,000
= 220 in.
36
= 1.1(7.05)
=
220
= 18.3ft
12
And from Equation F4-8
2
J
⎛ J ⎞
⎛ E⎞
⎛F ⎞
+ ⎜
+ 6.76 ⎜ L ⎟
Lr = 1.95rt ⎜ ⎟
⎟
⎝ E⎠
⎝ FL ⎠ S xc ho
⎝ S xc ho ⎠
2
⎛ 29,000 ⎞
⎛ 0.7 ( 36 ) ⎞
⎛ 12.5
⎞
12.5
= 1.95(7.05) ⎜⎜
+ ⎜
⎟⎟
⎟
⎟ + 6.76 ⎜
⎝ 1230(48.9) ⎠
⎝ 29,000 ⎠
⎝ 0.7 ( 36 ) ⎠ 1230(48.9)
787
= 787 in. =
= 65.6 ft
12
2
where
⎛ 26(0.875)3 ⎞ 48 ( 0.375 )
1
J = ∑ bt 3 = 2 ⎜
= 12.5 in.4
⎟+
3
3
3
⎝
⎠
3
Step 6:
Determine the nominal strength based on the limit state of lateraltorsional buckling.
Because the unbraced length is between Lp and Lr, the straight line
equation, Equation 7.8, is used; thus
⎡
⎛ Lb − Lp ⎞ ⎤
M n = Cb ⎢ R pc M yc – ( R pc M yc – 0.7Fy S x ) ⎜
⎟ ⎥ ≤ R pc M yc
⎢⎣
⎝ Lr − Lp ⎠ ⎥⎦
⎡
0.7 ( 36 )(1230 ) ⎞ ⎛ 20.0 – 18.3 ⎞ ⎤
⎛
= 1.0 ⎢3880 – ⎜ 3880 –
⎟⎜
⎟ ⎥ = 3830 ft-kips
12
⎝
⎠ ⎝ 65.6 – 18.3 ⎠ ⎦⎥
⎣⎢
Step 7:
Check for the limit state of compression flange local buckling.
bf
26
=
= 14.9
λf =
2t f
2 ( 0.875 )
and the limits are
2
Chapter 7
λ pf = 0.38
and, with kc =
E
29,000
= 0.38
= 10.8
Fy
36
4
= 0.354 > 0.35 ,
48 / 0.375
λ rf = 0.95
Step 8:
Plate Girders 277
0.354 ( 29,000 )
kc E
= 0.95
= 19.2
0.7 ( 36 )
FL
Determine the nominal moment strength for the limit state of flange local
buckling.
Because λpf < λf < λrf, the shape has a noncompact flange, so that, from
Equation F4-13,
⎛ λ − λ pf ⎞
M n = R pc M yc − ( R pc M yc − FL S xc ) ⎜
⎟
⎝ λ rf − λ pf ⎠
0.7(36)(1230) ⎞⎛ 14.9 – 10.8 ⎞
⎛
= 3880 – ⎜ 3880 –
⎟⎜
⎟ = 3250 ft-kips
12
⎝
⎠⎝ 19.2 – 10.8 ⎠
Step 9:
Determine the lowest nominal moment for the limit states checked.
For compression flange local buckling,
M n = 3250 ft-kips
For
LRFD
Step 10:
φM n = 0.9 ( 3250) = 2930 ft-kips
For
ASD
Step 10:
M n 3250
=
= 1950 ft-kips
Ω
1.67
Part (b)
Step 11:
For the plate girder with a 1/4 in. web plate
Check the web slenderness in order to determine which sections of the
Specification must be followed. From step 2, λ rw = 162 and
h
48
λw = c =
= 192 > λ rw = 162
tw 0.25
Thus, this is a slender web plate girder and the provisions of Section F5
must be followed.
Step 12:
Determine the bending strength reduction factor, Equation F5-6.
48.0 ( 0.25 )
ht
aw = c w =
= 0.527<10
b fc t fc 26.0 ( 0.875 )
278 Chapter 7
Plate Girders
R pg = 1 –
=1–
⎛ hc
aw
E ⎞
⎜⎜ – 5.7
⎟ ≤ 1.0
1200 + 300aw ⎝ tw
Fy ⎟⎠
⎛ 48.0
0.527
29,000 ⎞
– 5.7
⎜⎜
⎟ = 0.988 ≤ 1.0
1200 + 300(0.527) ⎝ 0.250
36 ⎟⎠
Step 13:
Determine the nominal moment strength for the limit state of yielding.
From Equation F5-1
R p g Fy S x 0.988(36)(1190)
Mn =
=
= 3530 ft-kips
12
12
Step 14:
Check the unbraced length for the limit state of lateral torsional buckling
with Lb = 20 ft. The effective radius of gyration is
b fc
26.0
rt =
=
= 7.20 in.
⎛ aw ⎞
⎛ 0.527 ⎞
12 ⎜ 1 + ⎟
12 ⎜ 1 +
⎟
6 ⎠
6 ⎠
⎝
⎝
and from Equation F4-7
Lp = 1.1rt
E
Fy
29,000
= 225in.
36
= 1.1(7.20)
=
225
= 18.8 ft
12
From Equation F5-5
Lr = πrt
E
0.7 Fy
= π(7.20)
=
Step 15:
29,000
= 767 in.
0.7(36)
767
= 63.9 ft
12
Determine the nominal moment strength for the limit state of lateral
torsional buckling.
Because the unbraced length is between Lp and Lr, the nominal moment
strength for lateral-torsional buckling is given by Equation 7.13. Thus,
⎡
⎛ Lb – Lp ⎞ ⎤
M n = Cb Rpg S xc ⎢ Fy – (0.3Fy ) ⎜
⎟ ⎥ ≤ Rpg Fy S xc
⎝ Lr – Lp ⎠ ⎦⎥
⎣⎢
⎡
⎛ 20.0 − 18.8 ⎞ ⎤ ⎛ 1 ⎞
= 1.0 ( 0.988) (1190) ⎢36 − 0.3(36) ⎜
⎟ ⎥ ⎜ ⎟ = 3500 ft-kips
⎝ 63.9 − 18.8 ⎠ ⎦ ⎝ 12 ⎠
⎣
Step 16:
Check compression flange local buckling.
Chapter 7
Plate Girders 279
For compression flange local buckling, the flange slenderness is the same
as it was in Part (a) of this problem, λf = 14.9, and the compact flange
limit is also the same, λpf = 10.8. However, the limiting flange slenderness
for the noncompact flange is different, because it is influenced by the web
thickness through kc. For the 1/4 in. web plate,
4
kc =
h tw
=
4
48 / 0.250
= 0.289 < 0.35
therefore,
kc = 0.350
and
λ rf = 0.95
Step 17:
0.350 ( 29,000 )
kc E
= 0.95
= 19.1
0.7 Fy
0.7 ( 36 )
Determine the nominal moment strength for the limit state of flange local
buckling, Equation 7.10.
⎡
⎛ λ – λ pf ⎞ ⎤
M n = R pg S xc ⎢ Fy – (0.3Fy ) ⎜
⎟⎥
⎢⎣
⎝ λ rf – λ pf ⎠ ⎥⎦
⎡
⎛ 14.9 – 10.8 ⎞ ⎤ ⎛ 1 ⎞
= 0.988(1190) ⎢36 – 0.3(36) ⎜
⎟ ⎥ ⎜ ⎟ = 3000 ft-kips
⎝ 19.1 – 10.8 ⎠ ⎦ ⎝ 12 ⎠
⎣
Step 18:
Determine the lowest nominal moment for the limit states checked.
For compression flange local buckling
M n = 3000 ft-kips
For
LRFD
Step 19:
For
ASD
Step 19:
7.2.3
φM n = 0.9 ( 3000) = 2700 ft-kips
M n 3000
=
= 1800 ft-kips
Ω
1.67
Compact Web Plate Girders
Doubly symmetric compact web flexural members are addressed in AISC Specification Section
F3 and discussed in Chapter 6. In that presentation, the limit state of flange local buckling was
limited to a discussion of compact and noncompact flanges. Discussion of slender flanges was
delayed to this chapter since there are no hot rolled members with slender flanges.
The limiting flange width-to-thickness ratio is the same as for all other plate girders, case
11 in Table B4.1b. The nominal moment strength for elastic flange local buckling is given by
2880 Chapter 7
Plate Gird
ders
Mn =
0.9 E
Ekc S x
λ2
SC F3-2)
(AIS
All variablees in this equation are as defined
d
in eaarlier sectionss. The limit sstates of yieldding and
lateral-torsio
onal buckling
g are treated as they were ppresented in C
Chapter 6.
Figu
ure 7.13 Plaate Girder forr Example 7.22
EX
XAMPLE 7.2
7
Pllate Girder
Fllexural Stren
ngth
Goa
al:
Deterrmine the avaailable momeent strength for an A5722 Gr. 50 I-shhaped
built-u
up member.
Giv
ven:
The girder is shown in Figuure 7.13. It has lateral supports foor the
compression flange at 8 ft intervvals. Sx = 94.44 in.3 ry = 2.881 in.
SO
OLUTION
Step
p 1:
Check
k the web sllenderness inn order to deetermine whicch sections oof the
Speciffication mustt be followed.. From Table B4.1b case 15,
h
20
E
29,000
λw = c =
= 53.3 < λ pw = 3.76
= 3.76
=90.6
tw 0.. 375
Fy
50
Thus the girder haas a compact web and shoould be desiggned in accordance
with Section
S
F3. These
T
are the pprovisions thaat were discusssed in Chaptter 6.
Step
p 2:
Check
k the unbraceed length limitts for lateral-ttorsional buckkling with Lb = 8.0
ft.
From Equation F2--5
Lp = 1.76ry
=
E
29,000
= 1.76(2.81)
= 1199 in.
Fy
50
119
= 9.92 fft > Lb = 8 ftt
12
Thus lateral torsion
nal buckling iis not a factorr.
Step
p 3:
Check
k the slendern
ness limits forr flange locall buckling.
b bf
14.0
λf = =
= 28.0
=
t 2t f 2 ( 0.2550 )
Chapter 7
Plate Girders 281
The limiting slenderness from Table B4.1b case 11 is
E
29,000
= 0.38
= 9.15
λ pf = 0.38
Fy
50
and with
kc =
λ rf = 0.95
Step 4:
4
4
=
= 0.548
h tw
20.0 / 0.375
0.548 ( 29,000 )
kc E
= 0.95
= 20.2
0.7 Fy
0.7 ( 50 )
Determine the nominal strength for the limit state of flange local buckling.
Since λf = 28.0 is greater than λrf = 20.2, the flange is slender. Thus, for
flange local buckling, using Equation F3-2,
0.9Ekc S x
Mn =
λ 2f
=
Step 5:
0.9 (29,000)(0.548)(94.4)
= 144 ft-kip
(28.0) 2 (12)
Determine the lowest nominal moment for the limit states checked.
Since web local buckling and lateral torsional buckling are not factors for
this member, the only limit state considered is flange local buckling. This is
the controlling limit state. Thus
M n = 144 ft-kips
For
LRFD
Step 6:
For
ASD
Step 6:
φM n = 0.9 (144) = 130 ft-kips
M n 144
=
= 86.2 ft-kips
Ω 1.67
7.3 HOMOGENEOUS PLATE GIRDERS IN SHEAR
Shear is an important factor in the behavior and design of plate girders because the webs have the
potential to be relatively thin. Two design procedures are available for shear design of plate
girders. One accounts for the postbuckling strength available through tension field action,
whereas the other includes postbuckling buckling strength of the web, without relying on tension
field action. Transverse stiffeners can be used to increase web shear strength but are not required
unless the shear strength of the web is less than the required strength. Sometimes, a thicker web
may be required even if stiffeners are used.
The limit states for shear are web yielding and web buckling. If tension field action is not
considered, postbuckling strength is determined using a model that accounts for web stress
redistribution in a member with or without transverse stiffeners. These provisions are covered in
Specification Section G2.1. Under certain circumstances it is possible to take advantage of the
postbuckling strength of the girder web through an approach called tension field action. Research
2882 Chapter 7
Plate Gird
ders
Figure 7.14 Web Buck
kling Showing
g Tension Fieeld Action
Photo courtessy Donald Whiite
has demonsttrated that a plate
p
girder with
w transversse stiffeners aand a thin webb can act sim
milar to a
Pratt truss once
o
the web buckles, thu
us providing additional poostbuckling sttrength. The buckled
web of a plaate girder is sh
hown in Figu
ure 7.14, and tthe truss moddel behavior is illustrated inn Figure
7.15, where the buckled panel of thee girder simullates the tenssion diagonall of the truss and the
stiffener rep
presents the vertical web member. Thhe designer m
must decide whether to use this
tension field
d action or to design witho
out tension fieeld action. Onne anomaly w
with the provissions for
shear occurs when the tension
t
field action approoach actuallyy produces leess strength tthan the
approach wiithout tension
n field action
n. This is the result of the Specificationn using two ddifferent
theoretical models
m
to prredict shear strength.
s
It iss an acceptabble situation that simply must be
checked wheen using tension field actio
on if the maxiimum shear sstrength is dessired.
It will
w be seen th
hat web yieldiing controls tthe maximum
m strength of tthe girder weeb. If the
size of the girder
g
web peermits web yielding, there will be no addvantage to considering sttiffeners,
with or with
hout tension fiield action.
Avaailable shear strength
s
for webs
w
of platee girders, regaardless of weeb width-to-thhickness
ratio, is deteermined using
g φ = 0.9 and Ω = 1.67.
7.33.1
Nontension Field Acction
The provisions for sheaar strength off a plate girdder without ttension field action are ggiven in
Specification
n Section G2
2.1(b). The no
ominal shear strength is a function of thhe slendernesss of the
web defined
d as λwv = h/tw. The single limit used to ddefine the rannges of behavior is
kE
λ wvp = 1.100 v
Fy
Chapterr 7
Plate Giirders 283
Figure
F
7.15 Plate Girderr Showing Tennsion Field A
Action
Figure
F
7.16 Limitations on Plate Girdder to Permit Tension Fieldd Action
b plate buckling coefficien
nt, kv, for unsstiffened webs of I-shapedd members thhat meet the
The web
proportio
oning criteria of the Speciffication—thatt is, λwv < 2600—is taken ass kv = 5.34. For stiffened
webs, wh
here stiffenerss are spaced at
a a distance aa, as shown inn Figure 7.16,
5
kv = 5 +
( a /h ) 2
but is tak
ken as 5.34 wh
hen a/h > 3.0
0.
The
T nominal shear
s
strength
h of a girder w
without tensioon field actionn is given by
AISC G2-1)
(A
Vn = 00.6 Fy AwCv1
where
Aw = the overaall depth timees the web thiickness
Cv1, the web shear
s
coefficieent, is a functtion of web shhear slendernness
For
F λ wv ≤ 1.100 kv E /Fy ,
Cv1 = 1.0
For
F λ wv > 1.100 kv E /Fy ,
(A
AISC G2-3)
2884 Chapter 7
Plate Gird
ders
Figure 7.17 Web Sheaar Coefficientt as a Functioon of Web Shear Slenderneess
C v1 =
1..10 kv E / Fy
h tw
(AISC G2-4)
The web shear co
oefficient is sh
hown in Figuure 7.17 for tw
wo cases of thhe web plate bbuckling
coefficient, kv = 5.34 and
d kv = 10.0. For
F a web witth Cv1 = 1.0 tthe web reachhes its yield sstrength.
w shear sleenderness succh that Cv1 is less than 1.0,, the web bucckles. Compaaring the
For a web with
two curves in
i Figure 7.17 shows the impact
i
of addding stiffenerrs to a girder without tensiion field
action; a girrder with no stiffeners,
s
kv = 5.34; and a girder with sstiffeners spacced so that thhe panels
are square, a/h
a = 1.0 and kv = 10.0.
7.33.2
Tension
n Field Action
Although th
he Specification does not require that tension field action be coonsidered, a ddesigner
may take ad
dvantage of tension field acction when stiiffeners are prresent. The im
mpact of tensiion field
action is gen
nerally to incrrease the web
b shear strenggth, with the nnominal shearr strength dettermined
as a combin
nation of web buckling streength and webb postbucklinng strength. B
Both of these strength
components are functionss of stiffener spacing.
Ten
nsion field acttion is addresssed in Speciffication Sectiion G2.2. To include tensiion field
action in th
he strength caalculation, th
he plate girdeer must meett certain lim
mitations. Figuure 7.16
illustrates th
hese four limittations as desscribed here.
No Tension
n Field Action
n in End Pan
nels
Figure 7.16aa shows a plate girder wiith a potentiaally buckled w
web. The diaagonal tensionn that is
developed in
i the web brings two orthogonal ccomponents of force to the flange-stiffener
intersection.. The verticaal stiffener resists the veertical compoonent, and thhe flange ressists the
horizontal component,
c
ju
ust as for thee Pratt truss. The end pannel has no addjacent panell to help
resist the ho
orizontal comp
ponent, so thiis last panel m
must resist thhe shear forcee through beam
m shear,
without consideration of tension field action. Everyy stiffened plaate girder hass end panels thhat must
be designed
d without tenssion field acttion. This usuually results iin narrower ppanels at the ends of
tension field
d girders.
Chapter 7
Plate Girders 285
Proportions of Panels
The Specification limits the proportions of stiffened panels such that tension field action may not
be considered if
a
>3
h
Figure 7.16b shows a portion of a stiffened plate girder with stiffeners placed at the limit
of a/h = 3. The panel is quite elongated, and its effectiveness at resisting vertical forces is
significantly reduced in comparison to that of a panel with a smaller aspect ratio, such as that
shown in Figure 7.16a.
Proportions of Web to Flange
Section G2.2(b) provides two equations for shear strength with tension field action, depending on
the relative proportions of the web and flange. The first, when applied to doubly symmetric plate
girders, is that the ratio of web area to flange area not exceeding 2.5 and web height to flange
width not exceeding 6. If these limits are exceeded, the flanges are not sufficient to fully resist the
developed diagonal tension forces. Figures 7.16c and d show plate girders that meet these limits.
In this case, the nominal shear strength is give as
⎛
⎞
1 – Cv 2
⎜
⎟
Vn = 0.6Fy Aw Cv 2 +
(AISC G2-7)
2 ⎟
⎜
1.15
1
a
/
h
+
(
)
⎝
⎠
If these limits are not met, the nominal shear strength is given as
⎛
⎜
1 – Cv 2
Vn = 0.6Fy Aw ⎜ Cv 2 +
2
⎜
1.15 a h + 1 + ( a /h )
⎝
(
)
⎞
⎟
⎟
⎟
⎠
(AISC G2-8)
The web shear buckling coefficient, Cv2, is a function of the web width-to-thickness ratio. It is
based on a model of shear buckling that takes no postbuckling strength into account so it is
somewhat different from the coefficient, Cv1, given in Specification Section G2.1 when tension
field action is not considered.
When h tw ≤ λ wvp = 1.10 kv E Fy
Cv 2 = 1.0
(AISC G2-9)
when λ wvp = 1.10 kv E Fy < h tw ≤ λ wvr = 1.37 kv E Fy
Cv 2 =
1.10 kv E Fy
h tw
(AISC G2-10)
and when h tw > λ wvr = 1.37 kv E Fy
Cv 2 =
1.51kv E
( h tw )
2
Fy
(AISC G2-11)
Figure 7.18 shows the web shear strength in terms of Vn/(0.6FyAw) for a girder with and without
tension field action. These girders meet the more stringent web to flange proportion limits that
permit the use of Equation G2-7. Strength is given as a function of web shear slenderness for
three panel sizes. Equation G2-7 can be rewritten to show that the strength due to tension
2886 Chapter 7
Plate Gird
ders
Figure 7.18 Web Shearr Strength forr Girder with and without T
Tension Fieldd Action
n of the prebuuckling strenngth and the ppostbuckling strength
field action is simply thee combination
as
⎤
⎡
1 – Cv 2
⎥
Vn = 0.6Fy AwCv 2 + 0.6Fy Aw ⎢
⎢1.15 1 + a /h 2 ⎥
( ) ⎦
⎣
The prebuckling strength can be seen in Fiigure 7.18 as the strength of the girder without
tension field
d action. The addition of th
he postbucklinng strength ussing the tension field actioon model
shifts the cu
urves for each particular a/h
h shown in thhe figure.
The end panel in
n a tension field
f
plate girrder must bee especially rrigid in orderr for the
remainder of the web to properly
p
funcction as a Praatt truss; thus the stiffener spacing for thhe panel
next to the support musst be less thaan that withinn the span, aand shear in the end pannel must
t rules for a girder witho
out tension fieeld action.
conform to the
7.44
STIFFE
ENERS FOR
R PLATE GIRDERS
G
When stiffeeners are required for a plate
p
girder, they can be either interm
mediate stiffe
feners or
bearing stiff
ffeners. The purpose
p
of in
ntermediate sstiffeners is tto increase ggirder shear sstrength,
either by controlling
c
th
he buckling strength of the girder w
web or by ppermitting addditional
postbuckling
g strength to be achieved. These stiffenners are distriibuted along the girder lenngth and
result in pan
nel sizes with
h aspect ratio
os, a/h, that im
mpact girder shear strengtth. Bearing stiffeners
usually occu
ur at the locations of con
ncentrated looads or reactiions. They permit the traansfer of
concentrated
d forces that could
c
not alreeady be transfferred throughh direct bearinng on the girdder web.
7.44.1
Intermeediate Stiffen
ners
The Specificcation requireements for inttermediate stiiffeners are prrescriptive inn nature. Therre are no
forces for which
w
these stiffeners mu
ust be sized;; they are siimply sized tto meet the specific
limitations provided
p
in Section
S
G2.3.. There are tw
wo requiremeents for stiffeener sizing. F
First is a
width-to-thickness limit and
a second is a minimum m
moment of innertia.
For transverse stiiffeners, the width-to-thick
w
kness ratio, ( b t ) st , is limitted such that
E
⎛b⎞
⎜ ⎟ ≤ 0.566
Fyst
⎝ t ⎠ st
C G2-12)
(AISC
Chapterr 7
Plate Giirders 287
where Fyst
d strength of the
t stiffener. Since interm
mediate stiffenners are not ddesigned for
y is the yield
strength, it is not un
ncommon for the stiffenerr and the webb to have diffferent strenggths when a
higher strrength materiial is used forr the web.
Transverse
T
sttiffeners used
d to developp the available web shearr strength m
must have a
moment of inertia, Istt, about an axis
a
in the w
web center forr stiffener paairs or about the face in
w the web plate for sing
gle stiffeners as shown inn Figure 7.19,, such that the minimum
contact with
moment of inertia bassed on consideeration of no postbucklingg strength is
⎡ 2.5
⎤
I stt 2 = ⎢
− 2 ⎥ bp tw3 ≥ 0.55bp tw3
(AIISC G2-15)
2
⎢⎣ ( a h )
⎥⎦
mum of a and h.
h
where bp is the minim
The
T minimum
m moment of inertia requirred when the full postbuckkling strengthh or tension
field actiion strength iss included is defined
d
in thee Specificationn as
1.5
⎛ Fyw ⎞
h 4ρ1.3
st
I st1 =
⎜
⎟
440 ⎝ E ⎠
(AIISC G2-14)
where
ρst = the larger of Fyw/Fyst and 11.0
o the web
Fyw = yield stress of
quired shear strength
s
is less than the ful l available shhear strength, a linear interrpolation, as
If the req
indicated
d by Equation
n G2-13, may
y be used to ddetermine thee actual requiired stiffener moment of
inertia. Thus,
T
(AIISC G2-13)
I st ≥ I st 2 + ( I st1 − I st 2 ) ρ w
which acccounts for the
t amount of
o postbucklinng strength, either with oor without teension field
action, in
ncluded in the shear streng
gth calculatio n, where
Figure 7.19
7
Web Sttiffener Minim
mum Momentt of Inertia
2888 Chapter 7
Plate Gird
ders
Figure 7.20 Detailing Requirement
R
for Intermediiate Stiffenerss
Ist
Ist1
Ist2
Vr
Vc1
Vc2
ρw
= moment
m
of inerrtia of the tran
nsverse stiffe ner
= minimum
m
mom
ment of inertiaa of the transvverse stiffenerr required forr the developm
ment of
thee full shear po
ostbuckling sttrength, Vc1
= minimum
m
mom
ment of inertiaa of the transvverse stiffenerr required to ddevelop buckkling
streength without considering any postbuckkling strengthh, Vc2
= reequired shear strength in th
he panel beingg considered
= av
vailable postb
buckling shearr strength as ddefined in Section G2.1 orr G2.2
= av
vailable shearr strength wheen consideringg no postbuckkling strengthh, using Cv2 inn
Eq
quation G2-1 in
i place of Cv1
v .
= th
he maximum shear
s
ratio wiithin the paneels on each sidde of the transsverse stiffenner,
⎡ Vr − Vc 2 ⎤
⎢V − V ⎥ ≥ 0
⎣ c1 c 2 ⎦
provides detaailing requireements for inntermediate sttiffeners.
In addition,
a
the Specification
S
They can be stopped sh
hort of the teension flange and, when uused in pairss, do not neeed to be
T weld byy which they are attachedd to the web is to be
attached to the compresssion flange. The
b
fourr and six tim
mes the web tthickness from
m the near tooe to the webb-flange
terminated between
weld, as sho
own in Figuree 7.20, but theere is no speccific requirem
ment for sizingg that weld. N
Normally
it would be sized based on the platee thickness. W
When single stiffeners aree used, they must be
attached to the compresssion flange, if it consistts of a rectaangular plate,, to resist anny uplift
ue to torsion
n in the flang
ge. Because intermediatee stiffeners pprovide a connvenient
tendency du
mechanism to transfer brracing forces to the girder,, these stiffenners also musst be connecteed to the
n flange and must
m be capab
ble of transmiitting 1 percennt of the totall flange force.
compression
7.44.2
Bearing
g Stiffeners
quired when the strength oof the girder web is not ssufficient to rresist the
Bearing stifffeners are req
concentrated
d forces exertted on it. Alth
hough bearingg stiffeners ccan be requireed for rolled II-shaped
members an
nd were discu
ussed in Section 6.14, theyy are much m
more likely to be required ffor plate
girders, partticularly at th
he girder sup
pports. Speciffication Sectiion J10 addreesses the apppropriate
limit states. Normally th
he forces to be
b resisted arre compressivve in nature. For those caases, the
w local criippling, and web sideswaay buckling m
must be
limit states of web locaal yielding, web
checked. When the appliied load is tensile, web loocal yielding and flange loocal bending must be
considered. If the strengtth of the web is insufficiennt to resist thhe applied forrce, bearing stiffeners
can be used..
Chapterr 7
Figure 7.21
7
Plate Giirders 289
Single Concentrated
C
d Force Applieed to Beam
The
T relationsh
hip between available str ength and noominal strenggth varies forr each limit
state asssociated with web strengtth. Thus, eithher design strengths or aallowable strrengths, not
nominal strengths, mu
ust be comparred to determiine the minim
mum web strength. The apppropriate
with the follow
wing discussiion of the corrresponding
resistance factors and safety factorrs are given w
limit stattes.
Web Loccal Yielding
When a single concentrated forcee, tension or compression,, is applied tto a girder, aas shown in
t be delivereed to the girdder over a lenngth of bearinng, lb, and is
Figure 7..21, the forcee is assumed to
then disttributed throu
ugh the flangee and into thee web. For a plate girder,, the web just below the
web-flan
nge weld, dim
mensioned as k in Figure 7..21a, is the crritical locatioon. For plate ggirders, this
dimensio
on, k, is taken
n as the thick
kness of the fl
flange plus thhe dimension of the weld. As was the
case for the W-shape,, the distributtion takes plaace along a liine with a sloope of 1:2.5. T
Thus, when
the criticcal section is reached, the force has beeen distributedd over a lengtth of lb plus 22.5k in each
direction
n. If the conceentrated forcee is applied sso that the forrce distributees along the w
web in both
direction
ns, this distrib
bution increasses the bearinng length by 5k, as shownn in Figure 77.21b. If the
bearing is
i close to thee end of the member,
m
distr
tribution takes place only in one directtion, toward
the midspan. The Speecification deefines “close to the membber end” as being within tthe member
depth fro
om the end. Thus, the availlable length oof the web is ((lb + 2.5k), as shown in Figgure 7.21c.
The
T nominal strength of the
t girder weeb when the concentratedd force to bee resisted is
applied at
a a distance from
f
the mem
mber end that iis greater thann the depth off the memberr, d, is
(A
AISC J10-2)
Rn = Fyyw t w ( 5k + lb )
When
W
the con
ncentrated forrce to be resissted is applieed at a distancce from the m
member end
that is lesss than or equ
ual to the deptth of the mem
mber, d, the noominal strenggth is
AISC J10-3)
(A
Rn = Fywwt w ( 2.5k + lb )
where
Fyw =
yield stress of the web
w
lb =
length
h of bearing
k =
distan
nce from the outer
o
face of tthe flange to the web toe oof the fillet weeld
tw =
web thickness
t
290 Chapter 7
Plate Girders
For web local yielding,
φ = 1.0 (LRFD)
Ω = 1.50 (ASD)
Web local crippling
The criteria for the limit state of web local crippling of a plate girder is the same as it was for
rolled W-shapes. It again depends on the location of the force with respect to the end of the girder
and Qf =1.0 for non HSS members.
When the concentrated compressive force is applied at a distance from the member end
that is greater than or equal to d/2,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
2
Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
(AISC J10-4)
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
When the force is applied at a distance less than d/2 and lb/d ≤ 0.2,
1.5
⎡
⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f
2
Rn = 0.40tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥
Qf
tw
⎝ d ⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣
and when lb/d > 0.2,
1.5
⎡ ⎛ 4l
⎞ ⎛ tw ⎞ ⎤ EFywt f
b
2
− 0.2 ⎟ ⎜ ⎟ ⎥
Rn = 0.40tw ⎢1 + ⎜
Qf
tw
⎠ ⎝ t f ⎠ ⎥⎦
⎢⎣ ⎝ d
For web local crippling,
φ = 0.75 (LRFD)
Ω = 2.0 (ASD)
(AISC J10-5a)
AISC J10-5b)
Web Sidesway Buckling
The web of a plate girder is generally a slender element, as has already been discussed. If the
tension and compression flanges of the girder are not prevented from displacing laterally with
respect to each other at the point of a compressive load, web sidesway buckling must be assessed.
Two provisions are given for web sidesway buckling: (1) if the compression flange is restrained
against rotation, such as when it is attached to a slab, and (2) if it is not.
When the compression flange is restrained against rotation and the ratio of web
slenderness to lateral buckling slenderness, ( h tw ) ( Lb b f ) ≤ 2.3 , the nominal strength is given as
3
⎛ h /t w ⎞ ⎤
Cr tw3 t f ⎡
⎢1 + 0.4 ⎜
Rn =
⎟ ⎥
h2 ⎢
Lb /b f ⎠ ⎥
⎝
⎣
⎦
(AISC J10-6)
If ( h tw ) ( Lb b f ) > 2.3 , the limit state of web sidesway buckling does not apply.
When the compression flange is not restrained against rotation and the ratio of web
slenderness to lateral buckling slenderness, ( h tw ) ( Lb b f ) ≤ 1.7 , the nominal strength is given as
3
Cr tw3 t f ⎡ ⎛ h /tw ⎞ ⎤
⎢0.4 ⎜
Rn =
(AISC J10-7)
⎟ ⎥
h 2 ⎢ ⎝ Lb /b f ⎠ ⎥
⎣
⎦
and if ( h tw ) ( Lb b f ) > 1.7 , the limit state of web sidesway buckling does not apply.
In the above equations,
=
largest laterally unbraced length along either flange at the point of load
Lb
Chapterr 7
Plate Giirders 291
Figure 7.222 Flange Loocal
Bending foor an Applied Tension
Load
Cr
=
960,000 kssi when, at thhe location off the force, Mu < My or 1.5M
Ma < My;
480,000 ksi when, at thhe location off the force, Mu ≥ My or 1.5M
Ma ≥ My
For
F web sidessway buckling
g,
φ = 0.85
0
(LRFD)
Ω = 1.766 (ASD)
Flange Local
L
Bendin
ng
This limiit state appliees when a sin
ngle tensile c oncentrated fforce is appliied to the flannge and the
length off loading acro
oss the memb
ber flange is ggreater than oor equal to 0.15bf, as show
wn in Figure
7.22. Thee nominal streength is
Rn = 6.25 Fyf t 2f
(A
AISC J10-1)
If
I the force is applied at a distance less than 10tf from
m the membeer end, the strrength must
be reduceed by 50 perccent.
For
F flange loccal bending,
Ω= 1.67 (ASD)
φ = 0.9 (LRFD)
7.4.3
Bearring Stiffenerr Design
a checked, a decision is made regarding the need for bearing
Once thee appropriate limit states are
stiffenerss. Although itt is possible to
t select a weeb plate that would not reequire bearingg stiffeners,
this is no
ot usually the most econom
mical approacch, even thouggh the additioon of stiffenerrs is a highlabor, an
nd thus high-ccost, activity.. When bearinng stiffeners are to be sizzed, Section JJ10.8 of the
Specifica
ation requiress that they be
b sized accoording to the provisions ffor tension m
members or
compresssion memberss, as appropriiate.
Stiffeners
S
dessigned to resisst tensile forc es must be deesigned accorrding to the reequirements
of Sectio
on J4.1 which
h refer to Chaapter D for thee difference bbetween the rrequired strenngth and the
minimum
m available lim
mit state stren
ngth. The stifffener must bbe welded to the flange annd web, and
these welds must be siized to resist the force beinng transferredd to the stiffenners.
Stiffeners
S
req
quired to ressist compresssive forces m
must be designed accordding to the
provision
ns of Chapterr E, except fo
or stiffeners w
with KL/r ≤ 25, which mayy be designedd with Fcr =
Fy, accorrding to Sectiion J4.4, for the
t differencee between thee required strrength and the minimum
availablee limit state strength.
s
Theese stiffeners must either bear on or bbe welded to the loaded
flange an
nd welded to the web, and
d these welds must be sized to resist thee force being transferred
to the stiffeners.
2992 Chapter 7
Plate Gird
ders
Fig
gure 7.23 Pllate Girder foor Example 7.3
EX
XAMPLE 7.3
7
Pllate Girder
Fllexural and
Sh
hear Strengtth
Goa
al:
Deterrmine the av
vailable mom
ment and sheaar strength uusing tensionn field
action
n.
Giv
ven:
A bu
uilt-up membeer is shown in Figure 7..23. The beaam is continuuously
lateraally braced. Use
U A572 Gr.. 50 for the m
member platees and A36 fo
for the
stiffen
ners. Sx = 464
4 in.3.
SO
OLUTION
Step
p 1:
Check
k the web sleenderness to ddetermine whhich section oof the Specificcation
must be used. Usin
ng Table B4.11b case 15
h
40
29,000
λw = =
= 160 > λ rw = 5.7
= 137
t w 0.25
50
Thus,, this is a slender web plaate girder, annd the provisiions of Sectioon F5
must be applied.
Step
p 2:
Deterrmine the ben
nding strengthh reduction factor.
The ratio of web arrea to flange area is given as
40.0(0.250)
aw =
= 1.00
10.0(1.00)
f
Equation
n F5-6
And from
Rpg = 1 –
Step
p 3:
⎛ 40.0
1.00
29,000 ⎞
– 5.7
⎜⎜
⎟ = 0.985
50 ⎟⎠
1200 + 300(1..00) ⎝ 0.250
Deterrmine the nom
minal momentt strength for the limit state of yielding..
0.985(50)(4664)
M n = R pg Fy S xc =
= 1900 ft-kips
12
Chapter 7
Step 4:
Plate Girders 293
Determine the nominal moment strength for the limit state of lateraltorsional buckling.
The compression flange is fully braced, so this limit state does not apply.
Step 5:
Determine the nominal moment strength for the limit state of flange local
buckling.
The flange slenderness is, from Table B4.1b case 11
b bf
10.0
29,000
=
= 5.0 < λ pf = 0.38
= 9.15
λf = =
t 2t f 2 (1.0 )
50
Therefore, the flange is compact, and there is no reduction in strength.
Step 6:
Determine the lowest nominal moment strength for the limit states
considered, yielding, lateral-torsional buckling and flange local buckling.
For the limit state of yielding,
M n = 1900 ft-kips
For
LRFD
Step 7:
Determine the design moment strength.
φM n = 0.9 (1900) =1710 ft-kips
For
ASD
Step 7:
Determine the allowable moment strength.
Step 8:
Determine the shear strength with tension field action. First check the
girder proportions against the prescriptive requirements of Section G2.2.
M n Ω = 1900 1.67 = 1140 ft-kips
Proportions of panel:
a 60.0
=
= 1.50 < 3
h 40.0
Proportions of web to flange:
Aw 42.0(0.250)
=
= 1.05 < 2. 5
10.0(1.00)
Af
Proportions of web height to flange width:
h 40.0
=
= 4.00 < 6.0
b f 10.0
Because the criteria have been satisfied, the tension field action provisions
of Section G2.2(b)(1) may be used for all but the end panel.
Step 9:
Check the web shear slenderness limits for determination of the shear
strength coefficient. The web shear buckling coefficient from Equation G25 is
294 Chapter 7
Plate Girders
kv = 5 +
5
5
=5+
= 7.22
2
( a /h )
(60.0/40.0) 2
Thus the limits for determining Cv2 are
kE
7.22(29,000)
λ wvp = 1.1 v = 1.1
= 71.2
50
Fy
and
λ wvr = 1.37
kv E
7.22 (29,000)
= 1.37
= 88.7
50
Fy
Because λ wv = h tw = 40.0 0.250 = 160 is greater than λwvp and λwvr, the web
plate will buckle elastically.
Step 10:
Determine the shear strength coefficient from Equation G2-11.
1.51(7.22)(29,000)
Cv 2 =
= 0.247
(160) 2 (50)
The nominal shear strength with tension field action from Equation G2-7 is
⎛
⎡ 1 – 0.247
⎤⎞
Vn = 0.6(50)(42.0)(0.250) ⎜ (0.247 + ⎢
⎥ ⎟ = 192 kips
2 ⎟
⎜
⎢
⎥⎦ ⎠
1.15
1
(1.50)
+
⎣
⎝
For
LRFD
Step 11:
Determine the design shear strength.
φVn = 0.9 (192) = 173 kips
For
ASD
Step 11:
Determine the allowable shear strength.
Step 12:
Check the intermediate stiffener size for meeting the criteria.
Vn Ω = 192 1.67 = 115 kips
Check the width-to-thickness ratio of the stiffener from Equation G2-12
4.5
29,000
⎛b⎞
= 12.0 ≤ 0.56
= 15.9
⎜ ⎟ =
36
⎝ t ⎠ st 0.375
So the stiffener satisfies this requirement
For the single plate stiffener,
b h3 0.375(4.50)3
= 11.4 in.4
I st = st =
3
3
Step 13:
Check the minimum stiffener moment of inertia when tension field action is
considered.
Since the required shear strength is not given, the stiffeners will be checked
to see that they are capable of providing the full tension field strength of the
web using Equation G2-14.
Chapter 7
1.5
h 4 ρ1.3
st ⎛ Fyw ⎞
40 ⎜⎝ E ⎟⎠
1.3
50
1.5
( 40.0 ) ⎛⎜ ⎞⎟
⎝ 36 ⎠ ⎛ 50 ⎞ = 7.02 in.4
=
⎜ 29,000 ⎟
40
⎝
⎠
4
I st1 =
Plate Girders 295
and
I st1 = 7.02 < 11.4 in.4
Thus, these stiffeners are adequate to permit full use of the available shear
strength with tension field action, as given in Step 11.
7.5 PROBLEMS
1. Determine the available moment strength of an A36
plate girder with a 50×1/2 in. web plate and equal flange
plates of 12×1 in. Assume there is full lateral support.
Determine by (a) LRFD and (b) ASD.
2. Determine the available moment strength of an A572
Gr. 50 plate girder with an 80×1/2 in. web plate and equal
flange plates of 16×1/2 in. Assume there is full lateral
support. Determine by (a) LRFD and (b) ASD.
3. Determine the available moment strength of an A572
Gr. 50 plate girder with a web plate of 40×1/2 in. and
equal flange plates of 10×1 in. Assume there is full lateral
support. Determine by (a) LRFD and (b) ASD.
4. Determine the available moment strength of an A36
plate girder with a web plate of 75×3/8 in. and equal
flange plates of 14×1-1/4 in. Assume there is full lateral
support. Determine by (a) LRFD and (b) ASD.
5. For a plate girder spanning 80 ft with lateral supports at
the supports and midspan, determine the nominal moment
strength. The girder is an A36 member with a web plate
of 45×3/4 in. and flange plates of 9×3/4 in.
6. For the plate girder of Problem 1, determine the
available moment strength if the girder spans 120 ft and
has lateral supports at the ends and the third points of the
span. Determine by (a) LRFD and (b) ASD.
7. For the plate girder of Problem 2, determine the
available moment strength if the girder spans 90 ft and
has lateral supports at the ends and the quarter points of
the span. Determine by (a) LRFD and (b) ASD.
8. For the plate girder of Problem 3, determine the
available moment strength if the girder spans 70 ft and
has lateral supports at the ends only. Determine by (a)
LRFD and (b) ASD.
9. For the plate girder of Problem 4, determine the
available moment strength if the girder spans 160 ft and
has lateral supports at the ends and the third points of the
span. Determine by (a) LRFD and (b) ASD.
10. Determine the available moment strength of an A36
plate girder with a 50×1/2 in. web plate and equal flange
plates of 12×1/2 in. Assume there is full lateral support.
Determine by (a) LRFD and (b) ASD.
11. Determine the available moment strength of an A572
Gr. 50 plate girder with a web plate of 40×1/2 in. and
equal flange plates of 12×1/2 in. Assume there is full
lateral support. Determine by (a) LRFD and (b) ASD.
12. Determine the available moment strength of an A572
Gr. 50 plate girder with a web plate of 45×3/4 in. and
flange plates of 14×1/2 in. Assume there is full lateral
support. Determine by (a) LRFD and (b) ASD.
13. Determine the available shear strength of an A572 Gr.
50 plate girder without transverse stiffeners. The web
plate is 100×3/4 in. and the flange plates are 15×1-1/2 in.
Determine by (a) LRFD and (b) ASD.
14. Determine the available shear strength of the girder in
Problem 1 if there are no intermediate stiffeners.
Determine by (a) LRFD and (b) ASD.
15. Determine the available shear strength of an A572 Gr.
50 plate girder with transverse stiffeners spaced every 100
in. The web plate is 100×3/4 in. and the flange plates are
15×1-1/2 in. Determine by (a) LRFD and (b) ASD.
16. Determine the available moment and available shear
strength of an A36 plate girder on a 100 ft span with
stiffeners at the ends and 20 in. from the ends, along with
intermediate stiffeners spaced at 40 in. on center within
the span. Assume there is full lateral support. The girder
has a web plate of 40×1/4 in. and flange plates of 8×3/4
in. Determine by (a) LRFD and (b) ASD.
296 Chapter 7
Plate Girders
17. Consider the plate girder from Problem 16 and the
intermediate stiffener between a 20 in. end panel and an
adjacent 40 in. panel. Does a pair of 3×1/4 in. stiffener
plates satisfy all intermediate transverse stiffener
requirements for this location? Evaluate by (a) LRFD and
(b) ASD.
18. Consider the plate girder from Problem 16 and the
intermediate stiffener between a 20 in. end panel and an
adjacent 40 in. panel.
Does a single 3-1/2×3/8 in.
stiffener plate satisfy all intermediate transverse stiffener
requirements for this location? Evaluate by (a) LRFD and
(b) ASD.
19. A girder is to be designed to span 125 ft and carry a
uniformly distributed live load of 2.0 kip/ft and a
uniformly distributed dead load of 1.9 kip/ft including the
girder weight. Assume that the compression flange has
full lateral support, and determine the web and flange
plates necessary if the web plate is limited to a depth of
50 in. Determine any stiffener requirements for a girder
without tension field action. Use A572 Gr. 50 steel.
Design by (a) LRFD and (b) ASD.
20. Design a girder for the conditions of Problem 19 if the
compression flange is braced at 1/3 points.
Chapter
8
Beam
m-Colu
umns
and Frame
F
e
Behav
vior
KFC Yuum! Center, Loouisville, KY
photo © Bob Perzel
8.1
INTR
RODUCTIO
ON
Beam-co
olumns are meembers subjected to axial forces and beending momeents simultaneeously; thus
their beh
havior falls somewhere betw
ween that of aan axially loaaded column aand that of a bbeam under
pure ben
nding. It is thu
us possible to
o consider thee beam or axxially loaded m
member as sppecial cases
of the beeam-column. Practical applications off the beam-coolumn are nuumerous. Theey occur as
chord meembers in tru
usses, as elem
ments of rigidlly connected frameworks, and as membbers of pinconnecteed structures with transverrse or eccenttric loads. It is not alwayys possible too look at a
member and determin
ne whether it is a beam-co lumn or not; some knowleedge of the acctual forces
being caarried by the member is required
r
to ccategorize it as a beam-column. However, many
structural members are
a subjected to these coombined forcees, and the bbeam-columnn is a very
common element in building structtures.
The
T manner in
i which the combined looads are trannsferred to a particular beeam-column
significan
ntly impacts the ability of
o the membeer to resist thhose loads. S
Starting with the axially
loaded column, bendiing moments can occur fr
from various sources. Lateeral load can be applied
directly to
t the membeer, as is the caase for a trusss top chord orr a column suupporting the lateral load
from a wall.
w
Alternativ
vely, the axiaal force can bee applied at ssome eccentriicity from the centroid of
the colum
mn as a resu
ult of the speecific connecctions. In adddition, the m
member can rreceive end
momentss from its con
nnection to other memberrs of the struccture, such ass in a rigid frrame. In all
cases, th
he relation off the beam-co
olumn to thee other elemeents of the sstructure is im
mportant in
determin
ning both the applied
a
forcess and the strenngth of the m
member.
To
T understan
nd the behavior of beam--columns, it is common practice to llook at the
response predicted by
y an interactio
on equation. T
The response of a beam-coolumn to an aaxial load P,
major ax
xis moment Mx, and minor axis momentt My is presennted on the thrree-dimensionnal diagram
shown in
n Figure 8.1. Each axis in this diagram
m represents thhe capacity oof the memberr when it is
subjected
d to loading of
o one type on
nly, whereas the curves reepresent the combination oof two types
of loadin
ng. The surfacce formed by
y connecting tthe three curvves representss the interactiion of axial
load and biaxial bendiing. This interraction surfacce is of intereest to the desiggner.
297
2998
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figuree 8.1 Ultimaate Interaction Surface forr a
mn.
Stockyy Beam-Colum
The end points of
o the curves shown in Fiigure 8.1 deppend on the sstrength of thhe beammembers (Chaapter 6).
columns as described forr compression
n members (C
Chapter 5) annd bending m
The shape of
o the curvess between theese end pointts depends oon the properrties of the particular
member as well
w as the pro
operties of otther memberss of the structuure.
rts of the Mannual discussed in this
Table 8.1 lists th
he sections off the Specificaation and part
chapter.
8.22
SECON
ND-ORDER EFFECTS
The single most
m-column is w
what are
m
compliccating factor in the analy sis and desiggn of a beam
effects. Seco
known as second-order
s
ond-order eff
ffects are thee changes in member forrces and
mmonly usedd elastic
moments ass the direct result
r
of stru
uctural deform
mations. Beccause the com
methods of structural anaalysis assumee that all defoormations aree small, and bbecause the eqquations
of equilibriu
um are written
n using the un
ndeformed coonfiguration oof the structurre, these methhods are
not able to capture the additional second-order
s
effects that occur in reaal structures without
adjustment. The results of
o that type of
o analysis arre called firstt-order effectss—that is, firrst-order
forces, first--order momeents, and firstt-order displaacements. To account for the influencce of the
deformation
ns, an addition
nal analysis must
m be perforrmed. The ressults of this aadditional anaalysis are
referred to as
a the second--order effects..
Several approach
hes are availlable for inclluding secondd-order effeccts in an anaalysis. A
nelastic analyssis would takke into accounnt the actual ddeformation oof the
complete seccond-order in
Table 8.1 Sections of Specification
S
and
a Parts of M
Manual Coveered in This C
Chapter
Speecification
B3
Design Basis
C
Design for Stability
S
H
Combined Forrces and Torsion
Design of Members
M
for C
Appendix 6
Stability Braacing for Coluumns and Beaams
Appendix 7
Design for Staability
Alternative Methods
M
of D
Appendix 8
Approximatee Second-Ordder Analysis
Manual
M
Part 1
Dimensions and Propertiees
Part 3
Design of Fllexural Membbers
Part 4
Members
Design of Co
ompression M
Part 6
Design of Members
M
Subjeect to Combinned Loading
Beam
m-Columns annd Frame Behhavior
Chappter 8
299
structure and the resu
ulting forces,, as well as tthe sequencee of loading and the behaavior of the
structure after any of its compon
nents are streessed beyondd the elastic limit. This aapproach to
analysis is generally more
m
complex
x than is neceessary for noormal design. A similar appproach that
includes the actual defformations bu
ut that does nnot include ineelastic behaviior is usually sufficient.
An
A approach that is consistent with noormal design office practiice and with how beamcolumns have been handled
h
for many
m
years usses a first-ordder elastic annalysis and am
mplification
factors to
o approximatee the second-order effects.. This approaach applies theese amplificaation factors
as multip
pliers to the reesults of the first-order
f
anaalysis to obtaiin the second--order effectss.
Two
T
differentt deflection components
c
tthat could occcur in a beaam-column innfluence the
momentss in that beam
m-column. Th
he first, illusstrated in Figgure 8.2a, is tthe deflectionn along the
length off the memberr that results from the mom
ment along thhe member. IIn this case, tthe member
ends musst remain in th
heir original position
p
relatiive to each otther; thus, no sway is conssidered. The
moment created by the
t load, P, acting at ann eccentricityy δ2 from thee deformed m
member, is
superimp
posed on the moment
m
resullting from thee applied endd moments. Because the m
magnitude of
this additional momen
nt depends on
n the propertiies of the collumn itself, thhis is called tthe member
effect.
When
W
the beaam-column is part of a struucture that is permitted to sway, the dissplacements
of the ov
verall structure also influeence the mom
ments in the member. Forr a beam-coluumn that is
permitted
d to sway an amount
a
Δ2, ass shown in Fiigure 8.2b, thhe additional m
moment is givven by PΔ2.
Because the lateral diisplacement of
o a given meember is a fuunction of thee properties oof all of the
memberss in a given sttory, this is caalled the struccture effect.
To
T understand
d the magnitu
ude of the pootential increease in momeents on a coluumn due to
second-o
order effects, two simple calculations will be carriied out. The first is for a 20 ft long
column similar
s
to thatt shown in Fig
gure 8.2a. A W
W12×96 meember is used to carry an axxial load Pu
= 400 kip
ps and equal end
e momentss of Mu = 200 ft-kips bendiing the membber in single ccurvature. A
first-ordeer analysis yieelds an axial force
f
in the c olumn of 4000 kips and a bbending momeent at every
point alo
ong the colum
mn length of 200 ft-kips. The maximuum deflectionn of the mem
mber at mid
height du
ue to the mom
ment is
M u L2 200 ( 200 ) (1728 )
=
= 00.715 in.
8 EI
8 ( 29,0000 )( 833)
2
δ=
With
h the occurren
nce of this defflection, the appplied load of 400 kips is now at an ecccentricity
Figu
ure 8.2 Coluumn
Dispplacements forr SecondOrdeer Effects.
300
Chapter 8
Beam-Columns and Frame Behavior
from the member in its displaced position. Thus, an additional moment is induced into the
member equal to
400 ( 0.715 )
M additional =
= 23.8 ft-kips
12
The addition of this additional moment to the original internal moment of 200 ft-kips yields the
second-order moment,
M 2 nd = 200 + 23.8 = 224 ft-kips
Thus, there is an amplification of the moment by 224/200 = 1.12. If this were the final case,
second-order analysis would be fairly simple. Unfortunately, the additional moment just
determined also causes additional deflection, which, in turn, causes additional moment. This
process continues until equilibrium is reached. The process is an iterative one, and is nonlinear.
A second example is a column similar to that shown in Figure 8.2b. The same W12×96
member is used, and the axial force is again Pu = 400 kips. In this case, the column is a cantilever
with a moment of Mu = 200 ft-kips applied at the top. This moment will cause a horizontal
deflection at the top of the column of
M L2 200 ( 20 ) (1728 )
Δ= u =
= 2.86 in.
2 EI
2 ( 29,000 )( 833)
2
In this displaced position, the 400 kip load is now at an eccentricity from the fixed support, which
induces an additional moment
400 ( 2.86 )
M additional =
= 95.3 ft-kips
12
The addition of this additional moment to the original support moment of 200 ft-kips yields the
second-order moment
M 2 nd = 200 + 95.3 = 295 ft-kips
which is an increase of 1.48 times the first-order moment. Again, this is not the end of the
required calculations; this additional moment causes additional deflections and additional
moments.
Both of these second-order effects are significant in real structures and must be accounted
for in the design of beam-columns according to Section C1 of the Specification. Procedures for
incorporating these effects will be addressed once an overall approach to beam-column design is
established.
8.3
INTERACTION PRINCIPLES
The interaction of axial load and bending within the elastic response range of a beam-column can
be investigated through the straightforward techniques of superposition. This is the approach
normally considered in elementary strength of materials in which the normal stress due to an axial
force is added to the normal stress due to a bending moment.
Although the superposition of individual stress effects is both simple and correct for
elastic stresses, there are significant limitations when applying this approach to the limit states of
real structures. These include:
1.
Superposition of stress is correct only for behavior within the elastic range, and
only for similar stress types.
Beam-Columns and Frame Behavior
Chapter 8
301
Superposition of strain can be extended only into the inelastic range when
deformations are small.
3.
Superposition cannot account for member deformations or stability effects such
as local buckling.
4.
Superposition cannot account for structural deflections and system stability.
With these limitations in mind, it is desirable to develop interaction equations that will
reflect the true limit states behavior of beam-columns. Any limit state interaction equation must
reflect the following characteristics:
2.
Axial Load
1.
Maximum column strength
2.
Individual column slenderness
Bending Moment
1.
Lateral support conditions
2.
Sidesway conditions
3.
Member second-order effects
4.
Structure second-order effects
5.
Moment variation along the member
The resulting equations must also provide a close correlation with test results and
theoretical analyses for beam-columns, including the two limiting cases of pure bending and pure
compression.
Application of the resulting interaction equations can be regarded as a process of
determining available axial strength in the presence of a given bending moment or determining
the available moment strength in the presence of a given axial load. An applied bending moment
consumes a portion of the column strength, leaving a reduced axial load strength. When the two
actions are added together, the resulting total load must not exceed the total column strength.
Conversely, the axial load can be regarded as consuming a fraction of the moment strength. This
fraction, plus the applied moments, must not exceed the maximum beam strength.
8.4 INTERACTION EQUATIONS
A simple form of the three-dimensional interaction equation is
Pr M rx M ry
+
+
≤ 1.0
Pc M cx M cy
(8.1)
where the terms with the subscript r represent the required strength and those with the subscript c
represent the available strength.
This interaction equation is plotted in Figure 8.3. The figure shows that this results in a
straight line representation of the interaction between any two of the load components. The
3002
Chapter 8
Beam-C
Columns and Frame Behav
vior
Figure 8.3
Simpliffied Interactioon Surface.
Figuree 8.4a
Interacction Diagram
m for
Stub W
W14×82 Colum
mn.
Figurre 8.4b
Norm
malized Interacction
Diagraam for Stub
W14×
×82 Column.
Beam-Columns and Frame Behavior
Chapter 8
303
horizontal plane of Figure 8.3 represents the interaction of moments in the two principal axis
directions, called biaxial bending, whereas the vertical planes represent the interaction of axial
compression plus either major or minor axis bending. It should also be apparent that the threedimensional aspect is represented by a plane with intercepts given by the straight lines on the
three coordinate planes.
The interaction equations in Chapter H of the Specification result from fitting interaction
equations that are similar to the form of Equation 8.1 to a set of data developed from an analysis
of forces and moments for various plastic stress distributions on a stub column. Figure 8.4a shows
the actual analysis results for a W14×82 stub column. Figure 8.4b shows the same data plotted as
functions of the normalized axial strength Py and flexural strength Mp. In both cases, the influence
of length on the axial or flexural strength is not included. Using curves of this type, developed for
a wide variety of steel beam-column shapes, two equations were developed that are conservative
and accurate for x-axis bending. When applied to y-axis bending, they are significantly more
conservative; however, simplicity of design and the infrequent use of weak axis bending justify
this extra level of conservatism.
An additional modification to these equations is required to account for length effects.
Rather than normalizing the curves on the yield load and the plastic moment as was done in
Figure 8.4b, the equations were developed around the nominal strength of the column and the
nominal strength of the beam. The resulting equations are Equations H1-1a and H1-1b in the
Specification and are plotted in Figure 8.5.
The equations shown here consider bending about both principal axes, whereas the plot in
Figure 8.5 is for single-axis bending.
P
For r ≥ 0.2 ,
Pc
For
Pr 8 ⎛ M rx M ry ⎞
+ ⎜
+
⎟ ≤ 1.0
Pc 9 ⎝ M cx M cy ⎠
(AISC H1-1a)
Pr ⎛ M rx M ry ⎞
+⎜
+
⎟ ≤ 1.0
2 Pc ⎝ M cx M cy ⎠
(AISC H1-1b)
Pr
< 0.2 ,
Pc
where
Pr = required compressive strength, kips
Pc = available compressive strength, kips
Mr = required flexural strength, ft-kips
Mc = available flexural strength, ft-kips
x = subscript relating symbol to strong axis bending
y = subscript relating symbol to weak axis bending
It is important to note that
1. The available column strength, Pc, is based on the axis of the column with the
largest slenderness ratio. This is not necessarily the axis about which bending takes
place.
2. The available bending strength, Mc, is based on the bending strength of the beam
without axial load, including the influence of all the beam limit states.
3. The required compressive strength, Pr, is the force in the member, including secondorder effects.
3004
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figgure 8.5 Intteraction Equuations
H11-1a and H1-11b.
4.
The required flexural strength, Mr, is the bending m
moment in thhe member, inncluding
second-ordeer effects.
Second-ordeer forces and moments can
n be determineed through a second-orderr analysis or bby a
modification
n of the resu
ults of a first-order analyysis using am
mplification ffactors as meentioned
earlier. Thesse amplificatiion factors wiill be discusseed as they rellate to bracedd frames (Secttion 8.5)
and momentt frames (Secttion 8.6).
Add
ditional provissions are avaiilable for casees where the axial strengthh limit state iss out-ofplane buckling and the flexural
f
stren
ngth limit staate is lateral-ttorsional bucckling for bennding in
plane. Equaations H1-1a and b are co
onservative foor this situatiion, but an addditional appproach is
available. Sp
pecification Section
S
H1.3 provides thaat (1) for the limit state off in-plane insstability,
Equations H1-1a
H
and H1
1-1b should be
b used wherre the comprressive strenggth is determ
mined for
buckling in the plane of bending an
nd Mcx = Mp , and (2) forr the limit sttates of out-of-plane
buckling and
d lateral-torsiional buckling
g
Pr ⎛
Pr
⎜1..5 − 0.5
Pcy ⎝
Pcy
2
⎞ ⎛ M rx ⎞
⎟+⎜
⎟ ≤ 1.0
⎠ ⎝ Cb M cx ⎠
(AISC H1-3)
where
Pcy = available co
ompressive sttrength out off the plane of bending
Mcx = available laateral-torsional buckling sttrength for strrong axis bennding with Cb = 1.0
g factor discuussed in Chappter 6
Cb = lateral-torsiional buckling
If there is significant
s
biaxial bending
g, meaning th
that the requiired-moment--to-available--moment
ratio for y-ax
xis bending iss greater than
n or equal to 00.05, then thiss option is noot available. A
Although
this optionall approach caan provide a more
m
economiical solution in some casess, it is not useed in the
examples orr problems in this book.
Beam
m-Columns annd Frame Behhavior
Figgure 8.6
Chappter 8
305
Braaced Frame.
Figgure 8.7 Ann Axially Loaaded
Coolumn with Equal and Oppposite End
Mooments.
8.5
BRAC
CED FRAM
MES
A frame is considered
d to be braced
d if a positivee system—thaat is, an actuaal system suchh as a shear
wall (maasonry, concrrete, steel, orr other materrial) or diagoonal steel meember—as illlustrated in
Figure 8.6, serves to resist the latteral loads, sttabilize the fframe under ggravity loadss, and resist
lateral diisplacements. In these casees, columns arre consideredd braced againnst lateral trannslation and
the in-pllane K-factorr can be takeen as 1.0, acccording to A
Appendix Seection 7.2.3(aa), unless a
rational analysis
a
indiccates that a lo
ower value iss appropriate.. This is the ttype of colum
mn that was
306
Chapter 8
Beam-Columns and Frame Behavior
discussed in Chapter 6. Later in this chapter the requirements for bracing to ensure that a structure
can be considered a braced frame, as found in Appendix 6, are discussed.
If the column in a braced frame is rigidly connected to a girder, bending moments result
from the application of the gravity loads to the girder. These moments can be determined through
a first-order elastic analysis. The additional second-order moments resulting from the
displacement along the column length can be determined through the application of an
amplification factor.
The full derivation of the amplification factor has been presented by various authors.1,2
Although this derivation is quite complex, a somewhat simplified derivation is presented here to
help establish the background. An axially loaded column with equal and opposite end moments is
shown in Figure 8.7a. This is the same column that was discussed in Section 8.2. The resulting
moment diagram is shown in Figure 8.7b where the moments from both the end moments and the
secondary effects are given.
The maximum moment occurring at the mid-height of the column, Mr, is shown to be
M r = M 1 + Pδ 2
The amplification factor is defined as
AF =
M r M 1 + Pδ 2
=
M1
M1
Rearranging terms yields
AF =
1
Pδ 2
1–
M 1 + Pδ 2
Two simplifying assumptions will be made. The first is based on the assumption that δ is
sufficiently small that
δ2
δ
≈ 1
M 1 + Pδ 2 M 1
and the second, using the beam deflection, δ1 = M1 L2/8EI, assumes that
M1 8EI π 2 EI
= 2 ≈ 2 = Pe
L
L
δ1
Because these simplifying assumptions are in error in opposite directions, they tend to be
offsetting. This results in a fairly accurate prediction of the amplification. Thus,
1
(8.2)
AF =
1 – P /Pe
A comparison between the actual amplification and that given by Equation 8.2 is shown in Figure
8.8.
The discussion so far has assumed that the moments at each end of the column are equal
and opposite, and that the resulting moment diagram is uniform. This is the most severe loading
case for a beam-column braced against translation. If the moment is not uniformly distributed, the
1
Galambos, T. V., Structural Members and Frames. Englewood Cliffs, NJ: Prentice Hall, Inc., 1968.
Johnson, B. G., Ed., Guide to Stability Design Criteria for Metal Structures, 3rd ed., SSRC, New York: Wiley, 1976.
2
Beam
m-Columns annd Frame Behhavior
Chappter 8
307
F
Figure 8.8 A
Amplified Mooment:
E
Exact and Approximate.
displacem
ment along th
he member is less than ppreviously considered andd the resultingg amplified
moment is less than in
ndicated. It haas been custoomary in desiggn practice too use the casee of uniform
moment as a base an
nd to provid
de for other moment disttributions by converting tthem to an
equivalen
nt uniform mo
oment throug
gh the use of aan additional factor, Cm.
Numerous
N
stu
udies have sh
hown that a reasonably aaccurate correection resultss for beamcolumns braced again
nst translation and not subjeect to transveerse loading bbetween their supports, if
the mom
ment is reduced
d through its multiplication
m
n by Cm, wheere
Cm = 0.6 − 0.4 ( M 1 M 2 )
(A
AISC A-8-4)
t smaller to
o larger momeents at the ennds of the mem
mber unbraceed length in
M1/M2 is the ratio of the
the planee of bending
g. M1/M2 is positive
p
wheen the membber is bent inn reverse currvature and
negative when bent in
n single curvaature.
For
F beam-colu
umns in braced frames whhere the mem
mber is subjectted to transveerse loading
between supports, Cm may be taken
n from Comm
mentary Tablee C-A-8.1, orr conservativeely taken as
1.0.
The
T combinattion of the am
mplification ffactor, AF, annd the equivaalent momentt factor, Cm,
accounts for the total member seco
ondary effectss. This combiined factor is given as B1 inn Appendix
8 of the Specification
S
as
Cm
B1 =
≥ 1.0
(A
AISC A-8-3)
α Pr
1–
Pe1
where
α = 1.6 forr ASD and 1.0
1 for LRFD
D to accounnt for the nonlinear behavvior of the
structurre at its ultim
mate strength
Pr = required strength, which
w
may be taken as the ffirst-order reqquired strengtth, Pnt + Plt,
when used
u
in moment frames
Pe1 = Euler buckling
b
load for the colum
mn in the planne of bendingg with an effecctive length
factor, K = 1.0
Thus, thee value of Mr in Equations H1-1a and H
H1-1b is takenn as
M r = B1M nt
nt on the beaam-column. T
The subscript nt indicates tthat for this
where Mnt is the maxiimum momen
case, the column doess not undergo
o any lateral translation oof its ends. It is possible fo
for Cm to be
less than
n 1.0 and for Equation A-8
8-3 to give ann amplificatioon factor less than 1.0. Thhis indicates
3008
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figure 8.9 Three-Dimeensional Braced Frame forr a Single-Story Structure.
that the com
mbination of the
t Pδ effectss and the nonuuniform mom
ment results inn a moment lless than
the maximu
um moment on the beam
m-column frrom a first-oorder analysiis. In this case, the
amplification factor B1 = 1.0.
EX
XAMPLE 8.1a
8
Brraced Framee
Coolumn Desig
gn for
Coombined Ax
xial
an
nd Bending by
b
LR
RFD
Goa
al:
Desig
gn column A1
A in Figure 8.9 for the given loadss using the L
LRFD
provisions and thee second-ordeer amplificatioon factor provvided in Apppendix
8 of th
he Specificatiion.
Giv
ven:
The three-dimensional braced fframe for a siingle-story sttructure is givven in
Figurre 8.9. Rigid connections are providedd at the roof level for collumns
A1, B1,
B A4, and B4.
B All other column connnections are ppinned. Dead Load
= 50 psf,
p Snow Lo
oad = 20 psf, Roof Live Looad = 10 psf,, and Wind L
Load =
20 psf horizontal. Use
U A992 steeel. Assume thhat the X-braacing is sufficciently
stifferr than the rigiid frames to rresist all lateraal load.
Beam-Columns and Frame Behavior Chapter 8
SOLUTION
Step 1:
309
Determine the appropriate load combinations. From ASCE 7, Section 2.3,
the following two combinations are considered.
ASCE 7 load combination 3
1.2D + 1.6(Lr or S or R) + (0.5L or 0.5W)
ASCE 7 load combination 4
1.2D + 1.0W + 0.5L + 0.5(Lr or S or R)
Step 2:
Determine the factored roof gravity loads for each load combination. For
load combination 3
1.2(50) + 1.6(20) = 92 psf
and for load combination 4
1.2(50) + 0.5(20) = 70 psf
Because column A1 does not participate in the lateral load resistance, the
worst case loading will use the uniformly distributed roof load of 92 psf.
Step 3:
Carry out a preliminary first-order analysis. Because the structure is
indeterminate, a number of approaches can be taken. If an arbitrary 6:1
ratio of moment of inertia for beams to columns is assumed, a moment
distribution analysis yields the moment and force given in Figure 8.9b.
Thus, the column will be designed to carry
Pu = 29.1 kips and Mu = 37.7 ft-kips
Step 4:
Select a trial size for column A1 and determine its compressive strength
and bending strength.
Try W10×33. (Section 8.8 addresses trial section selection.)
From Manual Table 1-1
A = 9.71 in.2, rx = 4.19 in., ry = 1.94 in., Ix = 171 in.4, rx/ry = 2.16
The column is oriented so that bending is about the x-axis of the column. It
is braced against sidesway by the diagonal braces in panel A2–A3 and is
pinned at the bottom and rigidly connected at the top in the plane of
bending. The column is also braced out of the plane of bending by the brace
in panel A1–B1. Because this column is part of a braced frame, K = 1.0 can
be used. Although the Specification permits the use of a lower K-factor if
justified by analysis, this is not recommended because it would likely
require significantly more stiffness in the braced panel.
From Manual Table 4-1a, for y-axis buckling
φPn = 214 kips for Lc = 16.0 ft
From Manual Table 3-10
φMn = 113 ft-kips for Lb = 16.0 ft
310
Chapter 8
Beam-Columns and Frame Behavior
Step 5:
Check the W10×33 for combined axial load and bending in-plane.
For an unbraced length of 16 ft, the Euler load is
π 2 EI π 2 (29,000)(171)
Pe1 = 2 =
= 1330 kips
Lc1
(16.0(12)) 2
The column is bent in single curvature between bracing points, the end
points, and the moment at the base is zero, so M1/M2 = 0.0. Thus
Cm = 0.6 – 0.4(0.0) = 0.6
Therefore, the amplification factor, with α = 1.0, becomes
Cm
0.6
B1 =
=
= 0.613 ≤ 1.0
αP
1.0 ( 29.1)
1− r 1–
Pe1
1330
The Specification requires that B1 not be less than 1.0. Therefore, taking B1
= 1.0,
Mrx = B1(Mx) = 1.0(37.7) = 37.7 ft-kips
To determine which equation to use, calculate
Pu
29.1
=
= 0.136 < 0.2
φPn 214
Therefore, use Equation H1-1b
Pu
Mu
+
≤ 1.0
2φPn φM n
37.7
0.5(0.136) +
= 0.402 < 1.0
113
Thus, the W10×33 will easily carry the given loads.
The solution to Equation H1-1b indicates that there is a fairly wide extra
margin of safety. It would be appropriate to consider a smaller column for a
more economical design.
EXAMPLE 8.1b
Braced Frame
Column Design for
Combined Axial
and Bending by
ASD
Goal:
Design column A1 in Figure 8.9 for the given loads using the ASD
provisions and the second-order amplification factor provided in Appendix
8 of the Specification.
Given:
The three-dimensional braced frame for a single-story structure is given in
Figure 8.9. Rigid connections are provided at the roof level for columns
A1, B1, A4, and B4. All other column connections are pinned. Dead Load
= 50 psf, Snow Load = 20 psf, Roof Live Load = 10 psf, and Wind Load =
20 psf horizontal. Use A992 steel. Assume that the X-bracing is sufficiently
stiffer than the rigid frames to resist all lateral load.
Beam-Columns and Frame Behavior Chapter 8
SOLUTION
Step 1:
311
Determine the appropriate load combinations. From ASCE 7, Section 2.3,
the following two combinations are considered.
ASCE 7 load combination 3
D + (Lr or S or R)
ASCE 7 load combination 6
D + 0.75(0.6W) + 0.75(Lr or S or R)
Step 2:
Determine the factored roof gravity loads for each load combination. For
load combination 3
50 + 20 = 70 psf
and for load combination 6
50 + 0.75(20) = 65 psf
Because column A1 does not participate in the lateral load resistance, the
worst case loading will use the uniformly distributed roof load of 70 psf.
Step 3:
Carry out a preliminary first-order analysis. Because the structure is
indeterminate, a number of approaches can be taken. If an arbitrary 6:1
ratio of moment of inertia for beams to columns is assumed, a moment
distribution analysis yields the moment and force given in Figure 8.9c.
Thus, the column will be designed to carry
Pa = 22.1 kips and Ma = 28.7 ft-kips
Step 4:
Select a trial size for column A1 and determine its compressive strength
and bending strength.
Try W10×33. (Section 8.8 addresses trial section selection.)
From Manual Table 1-1
A = 9.71 in.2, rx = 4.19 in., ry = 1.94 in., Ix = 171 in.4, rx/ry = 2.16
The column is oriented so that bending is about the x-axis of the column. It
is braced against sidesway by the diagonal braces in panel A2–A3 and is
pinned at the bottom and rigidly connected at the top in the plane of
bending. The column is also braced out of the plane of bending by the brace
in panel A1–B1. Because this column is part of a braced frame, K = 1.0 can
be used. Although the Specification permits the use of a lower K-factor if
justified by analysis, this is not recommended because it would likely
require significantly more stiffness in the braced panel.
From Manual Table 4-1a for y-axis buckling
Pn Ω = 142 kips for Lc = 16.0 ft
From Manual Table 3-10
M n Ω = 74.9 ft-kips for Lb = 16.0 ft
312
Chapter 8
Beam-Columns and Frame Behavior
Step 5:
Check the W10×33 for combined axial load and bending in-plane.
For an unbraced length of 16 ft, the Euler load is
π 2 EI π 2 (29,000)(171)
Pe1 = 2 =
= 1330 kips
Lc1
(16.0(12)) 2
The column is bent in single curvature between bracing points, the end
points, and the moment at the base is zero, so M1/M2 = 0.0. Thus
Cm = 0.6 – 0.4(0.0) = 0.6
Therefore, the amplification factor, with α = 1.6. becomes
Cm
0.6
B1 =
=
= 0.616 ≤ 1.0
αP
1.6 ( 22.1)
1− r 1–
Pe1
1330
The Specification requires that B1 not be less than 1.0. Therefore, taking B1
= 1.0,
Mrx = B1(Mx) = 1.0(28.7) = 28.7 ft-kips
To determine which equation to use, calculate
Pu
22.1
=
= 0.156 < 0.2
Pn Ω 142
Therefore, use Equation H1-1b
Pa
Ma
+
≤ 1.0
2 Pn Ω M n Ω
28.7
0.5(0.156) +
= 0.461 < 1.0
74.9
Thus, the W10×33 will easily carry the given loads.
The solution to Equation H1-1b indicates that there is a fairly wide extra
margin of safety. It would be appropriate to consider a smaller column for a
more economical design.
8.6
Moment Frames
A moment frame depends on the stiffness of the beams and columns that make up the frame for
stability under gravity loads and under combined gravity and lateral loads. Unlike braced frames,
there is no external structure to lean against for stability. Columns in moment frames are
subjected to both axial load and moment and experience lateral translation.
The same interaction equations, Equations H1-1a and H1-1b, are used to design beamcolumns in moment frames as were previously used for braced frames. However, in addition to
the member second-order effects discussed in Section 8.5, there is the additional second-order
effect that results from the sway or lateral displacement of the frame.
Beam
m-Columns annd Frame Behhavior
8
Figure 8.10
Chappter 8
313
Structu
ure Second-Orrder Effect: S
Sway.
Figure
F
8.10 sh
hows a cantileever or flag ppole column uunder the actioon of an axial load and a
lateral lo
oad. Figure 8.10a is the column as viewed for a first-orderr elastic anallysis where
equilibriu
um requires a moment at the bottom, M lt = HL. Thhe deflection that results aat the top of
the colum
mn, Δ1, is the elastic deflecction of a canttilever, so
H
HL3
Δ1 =
(8.3)
3EI
A second-ord
der analysis yields
y
the forcces and displlacements as shown in Figure 8.10b.
The displacement, Δ2, is the total displacement
d
t, including seecond-order eeffects, and thhe moment,
including
g second-ordeer effects, is
(8.4)
B2 M lt = HL + P Δ 2
An
A equivalen
nt lateral load
d can be dettermined thatt results in thhe same mom
ment at the
bottom of
o the column
n as in the seecond-order aanalysis. Thiss load is H + PΔ2/L and iis shown in
Figure 8..10c.
It
I may be assu
umed, with only slight erroor, that the diisplacements at the top of the column
for the caases in Figurees 8.10b and c are the samee. Thus, usingg the equivaleent lateral loaad
( H + PΔ2 /L) L3 HL3 ⎛ PΔ2 ⎞
⎛ PΔ2 ⎞
=
Δ2 =
(8.5)
⎜1 +
⎟ = Δ1 ⎜1 +
⎟
HL ⎠
HL ⎠
3EII
3EI ⎝
⎝
Equation
n 8.5 can now
w be solved forr Δ2, where
Δ2 =
Δ1
P Δ1
1−
HL
(8.6)
and the result
r
substitu
uted into Equ
uation 8.4. Soolving the ressulting equatiion for the am
mplification
factor, B2, and simpliffying yields
314
Chapter 8
Beam-Columns and Frame Behavior
B2 =
Δ2
1
=
Δ1 1 – PΔ1
HL
(8.7)
Considering that the typical beam-column will be part of some larger structure, this
equation must be modified to include the effect of the multistory and multibay characteristics of
the actual structure. This is easily accomplished by summing the total gravity load on the columns
in the story and the total lateral load in the story. Thus, Equation 8.7 becomes
1
(8.8)
B2 =
ΣPΔ1
1–
ΣHL
This amplification factor is essentially that given in Appendix 8 of the Specification as
Equation A-8-6, when combined with Equation A-8-7
1
1
(AISC A-8-6)
B2 =
=
≥ 1.0
αPstory
αPstory Δ H
1–
1–
Pe story
RM HL
where
Pstory = total gravity load on the story
Pe story =
measure of lateral strength of the structure = RM
HL
ΔH
(AISC A-8-7)
ΔH = story drift from a first-order analysis due to the lateral load, H
α = 1.0 for LRFD and 1.6 for ASD to account for the nonlinear behavior of
the structure at its ultimate strength
⎛ Pmf ⎞
RM = 1 − 0.15 ⎜
(AISC A-8-8)
⎟
⎝ Pstory ⎠
Pmf = the total vertical load in columns that are part of the lateral load resisting system
The variable RM accounts for the influence of the member effect on the sidesway displacement
that could not be accounted for in the simplified derivation above. If all the columns are moment
frame columns, Pmf/Pstory = 1.0 and RM = 0.85. For braced frames, Pmf = 0 and RM = 1.0. For frames
with a combination of columns resisting lateral load through bending and gravity only or leaning
columns, the value of RM will be between these limits.
It is often desirable to limit the lateral displacement, or drift, of a structure during the
design phase. ASCE 7 Appendix C Commentary provides some general guidance. This limit can
be defined using a drift index, which is the story drift divided by the story height, ΔH/L. The
design then proceeds by selecting members so that the final structure performs as desired. This is
similar to beam design, where deflection is the serviceability criterion. Because a limit on the
drift index can be established without knowing member sizes, it can be used in Equation A-8-6;
thus an analysis with assumed member sizes is unnecessary.
With this amplification for sidesway, the moment, Mr, to be used in Equations AISC H11a and AISC H1-1b, can be evaluated. Mr must include both the member and structure secondorder effects. Thus, a first-order analysis without sidesway is carried out, yielding moments, Mnt,
that is without translation, to be amplified by B1. Next, a first-order analysis including lateral
loads and permitting translation is carried out. This yields moments, Mlt, with translation, to be
amplified by B2. The resulting second-order moment is
M r = B1M nt + B2 M lt
(AISC A8-1)
Beam-Columns and Frame Behavior
Chapter 8
315
where
B1 is given by Equation A-8-3
B2 is given by Equation A-8-6
Mnt = first-order moments when the structure is not permitted to translate laterally
Mlt = first-order moments that result from just the lateral translation
Mlt could include moments that result from unsymmetrical frame properties or loading as well as
from lateral loads. In most real structures, however, moments resulting from this lack of
symmetry are usually small and are thus often ignored.
The second-order force is
Pr = Pnt + B2 Plt
(AISC A-8-2)
The sum of Pnt and Plt for the entire structure will equal the total gravity load on the
structure, since the sum of Plt will be zero. For the individual column, however, it is important to
amplify the portion of the individual column force that comes from the lateral load.
For situations where there is no lateral load on the structure, it may be necessary to
incorporate a minimum lateral load in order to capture the second-order effects of the gravity
loads. This is covered in Section 8.7 where the three methods provided in the Specification for
treating stability analysis and design are discussed.
EXAMPLE 8.2a
Moment Frame
Strength Check for
Combined
Compression and
Bending by LRFD
Goal:
Using the LRFD provisions, determine whether the W14×90, A992 column
shown in Figure 8.11 is adequate to carry the imposed loading.
Given:
An exterior column from an intermediate level of a multi-story moment
frame is shown in Figure 8.11. The column is part of a braced frame out of
the plane of the figure. Figure 8.11a shows the elevation of the frame with
the member to be checked labeled AB. The same column section will be
used for the level above and below the column AB. A first-order analysis of
the frame for gravity loads plus the minimum lateral load (the minimum
lateral load will be discussed in Section 8.7) results in the forces shown in
Figure 8.11b, whereas the results for gravity plus wind are shown in Figure
8.11c. Assume that the frame drift under service loads is limited to
height/300 for a story shear, H = 148 kips.
SOLUTION
Step 1:
Determine the column effective length factor in the plane of bending.
Using the effective length alignment chart introduced in Chapter 5 and
given in Commentary Figure C-A-7.2, determine the effective length for
buckling in the plane of the moment frame. At each joint there are two
columns and one beam framing in. Thus,
⎛ 999 ⎞
2⎜
⎟
Σ ( I L )c
12.5 ⎠
G A = GB =
= ⎝
= 2.28
Σ ( I L )g
⎛ 2100 ⎞
⎜
⎟
⎝ 30.0 ⎠
Thus, from Figure 5.20, K = 1.66.
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
W24×76
7
A
B
8 at 12.5 ft
4
4
Ix = 21
100 in.
Ix = 999 in.
W14×90
316
A
B
7
W24×76
4
Ix = 2100 in.
4 at 300.0 ft
(a)
Figure 8.1
11 Exterior Column From
m an Intermed
diate Level off a Multistoryy Rigid Framee (Example 8.2).
Beam-Columns and Frame Behavior
Step 2:
Chapter 8
317
Determine the controlling effective length.
With rx/ry = 1.66 for the W14 × 90,
( Lcx )eff = ( KL )eff
=
( KL ) x
rx ry
=
1.66 (12.5 )
1.66
= 12.5 ft
Lcy = KLy = 1.0(12.5) = 12.5 ft
Step 3:
Since the effective length about each axis is 12.5 ft, determine the column
design axial strength using Lc = 12.5.
From the column tables, Manual Table 4-1a, for Lc = 12.5 ft,
φPn = 1060 kips
Step 4:
Determine the first-order moments and forces for the loading combination
that includes wind, 1.2D +0.5L + 1.0W.
The column end moments given in Figure 8.11c are a combination of
moments resulting from a nonsway gravity load analysis and a wind
analysis:
Moment for end A:
Moment for end B:
Mnt = 96.7 ft-kips
Mlt = 154 ft-kips
Mnt = 48.3 ft-kips
Mlt = 154 ft-kips
Compression:
Pnt = 354 kips
Plt = 99.0 kips
Step 5:
Determine the second-order moments by amplifying the first-order
moments.
No-translation amplification: The no-translation moments must be
amplified by B1. From Figure 8.11c it is seen that the end moments bend
the column in reverse curvature:
M 1 48.3
=
= 0.50
M 2 96.7
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4
Pe1 =
π2 E π 2 (29,000)(999)
=
= 12,700 kips
L2c1
(1.0(12.5)(12)) 2
Thus, with α = 1.0 for LRFD and Pr = 354 + 99 = 453 kips, Equation A-8-3
yields
318
Chapter 8
Beam-Columns and Frame Behavior
B1 =
Cm
0.4
=
= 0.415 < 1.0
(1.0)(453)
αPr
1−
1–
Pe1
12,700
Therefore, B1 = 1.0.
Translation amplification: The translation forces and moments must be
amplified by B2. The design drift limit of height/300 and Equation A-8-6
are used to determine B2.
The total service lateral load on this story is given as
H = 148 kips
Additional given information is that the total gravity load for this load
combination in Figure 8.11c is
Pstory = 2110 kips
The drift limit under the service lateral load of 148 kips is
ΔH = L/300 = 12.5(12)/300 = 0.50 in.
Remember that in the calculation of B2, H can be taken as any convenient
magnitude, as long as ΔH is the corresponding displacement. This is
because it is the ratio of H to ΔH that is used in the determination of Pe
story.
Thus, with α = 1.0 for LRFD and RM = 0.85 assuming all columns are
moment frame columns, Equation A-8-7 gives
Pe story =
RM HL 0.85(148)(12.5)(12)
=
= 37,700 kips
ΔH
0.50
and Equation A-8-6 gives
1
1
B2 =
=
= 1.06>1.0
αPstory
(1.0)2110 ⎞
⎛
1−
1– ⎜
⎟
Pe story
⎝ 37,700 ⎠
Thus, the second-order compressive force and moment are
Pr = Pnt + B2(Plt) = 354 + 1.06(99) = 459 kips
Mr = B1(Mnt) + B2(Mlt) = 1.0(96.7) + 1.06(154) = 260 ft-kips
These represent the required strength for this load combination.
Step 6:
Determine whether the W14×90 will provide the required strength based on
the appropriate interaction equation.
The unbraced length of the compression flange for pure bending is 12.5 ft,
which is less than Lp = 15.1 ft for this section, taking into account that its
flange is noncompact. Thus, from Manual Table 3-2, the design moment
strength of the section is
Beam-Columns and Frame Behavior
Chapter 8
319
φMn = 574 ft-kips
Determine the appropriate interaction equation. From Step 3, φPn = 1060
kips;
Pu
459
=
= 0.433 > 0.2
φPn 1060
so use Equation H1-1a, which yields
Pu 8 ⎛ M u
+ ⎜
φPn 9 ⎝ φM n
⎞
⎟ ≤ 1.0
⎠
8 ⎛ 260 ⎞
0.433 + ⎜
⎟ = 0.836 < 1.0
9 ⎝ 574 ⎠
Thus,
the W14×90 is adequate for this load combination.
Step 7:
Check the section for the gravity-only load combination, 1.2D + 1.6L.
Because this is a gravity-only load combination, Specification Appendix
Section 7.2.2, by reference to Section C2.2b, requires that the analysis
include a minimum lateral load of 0.002 times the gravity load. This will be
further discussed in Section 8.7. For this load combination, the total story
gravity load must also be known and is given in Figure 8.11b as Pstory =
2430 kips. Thus, for this frame the minimum lateral load is 0.002Pstory =
0.002(2430) = 4.86 kips at this level.
The forces and moments given in Figure 8.11b include the effects of this
minimum lateral load. The magnitude of the lateral translation effect is
small in this case. Since both the moment due to the minimum lateral load
and the amplification factor, B2, are expected to be small, the forces and
moments used for this check will be assumed to come from a no-translation
case, with little error. If the minimum lateral load would produce large
moments or the amplification factor, B2, calculated in Step 5, were large,
this would not be a good assumption. Therefore, at end A, Mnt = 142 ftkips, at end B Mnt = 71.0 ft-kips, and Pnt = 522 kips.
A quick review of the determination of B1 from the first part of this solution
shows that the only change is in the magnitude of the axial force and the
member end moments; thus
M 1 71.0
=
= 0.50
M 2 142
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4
Pe1 =
π2 EI π 2 (29,000)(999)
=
= 12,700 kips
L2c1
(1.0(12.5)(12)) 2
Thus, with α = 1.0 for LRFD,
Cm
0.4
B1 =
=
= 0.417 < 1.0
(1.0)(522)
αP
1− r 1 –
Pe1
12,700
320
Chapter 8
Beam-Columns and Frame Behavior
Note that B1 is again 1.0.
With the assumption that there is no lateral translation,
Mlt = 0.0 and B2 is unnecessary,thus
Pr = 522 kips, M r = 1.0(142) = 142 ft-kips
Again using Equation H1-1a,
Pu 8 ⎛ M u ⎞
+ ⎜
⎟ ≤ 1.0
φPn 8 ⎝ φM n ⎠
522 8 ⎛ 142 ⎞
+ ⎜
⎟ = 0.712 < 1.0
1060 9 ⎝ 574 ⎠
Thus,
the W14×90 is adequate for both load combinations.
EXAMPLE 8.2b
Moment Frame
Strength Check for
Combined
Compression and
Bending by ASD
Goal:
Using the ASD provisions, determine whether the W14×90, A992 column
shown in Figure 8.11 is adequate to carry the imposed loading.
Given:
An exterior column from an intermediate level of a multi-story moment
frame is shown in Figure 8.11. The column is part of a braced frame out of
the plane of the figure. Figure 8.11a shows the elevation of the frame with
the member to be checked labeled AB. The same column section will be
used for the level above and below the column AB. A first-order analysis of
the frame for gravity loads plus the minimum lateral load (the minimum
lateral load will be discussed in Section 8.7) results in the forces shown in
Figure 8.11d, whereas the results for gravity plus wind are shown in Figure
8.11e. Assume that the frame drift under service loads is limited to
height/300 for a story shear, H = 148 kips.
SOLUTION
Step 1:
Determine the column effective length factor in the plane of bending.
Using the effective length alignment chart introduced in Chapter 5 and
given in Commentary Figure C-A-7.2, determine the effective length for
buckling in the plane of the moment frame. At each joint there are two
columns and one beam framing in. Thus,
⎛ 999 ⎞
2⎜
⎟
Σ ( I L )c
12.5 ⎠
G A = GB =
= ⎝
= 2.28
Σ ( I L )g
⎛ 2100 ⎞
⎜
⎟
⎝ 30.0 ⎠
Thus, from Figure 5.20, K = 1.66.
Step 2:
Determine the controlling effective length.
With rx/ry = 1.66 for the W14 × 90,
( Lcx )eff = ( KL )eff
=
( KL ) x
rx ry
=
1.66 (12.5 )
1.66
= 12.5 ft
Beam-Columns and Frame Behavior
Chapter 8
321
Lcy = KLy = 1.0(12.5) = 12.5 ft
Step 3:
Since the effective length about each axis is 12.5 ft, determine the column
allowable axial strength using Lc = 12.5.
From the column tables, Manual Table 4-1a, for Lc = 12.5 ft,
Pn Ω = 703 kips
Step 4:
Determine the first-order moments and forces for the loading combination
that includes wind, D +0.75L + 0.75(0.6W).
The column end moments given in Figure 8.11e are a combination of
moments resulting from a nonsway gravity load analysis and a wind
analysis:
Moment for end A:
Moment for end B:
Mnt = 78.0 ft-kips
Mlt = 96.0 ft-kips
Mnt = 39.0 ft-kips
Mlt = 96.0 ft-kips
Compression:
Pnt = 280 kips
Plt = 62.0 kips
Step 5:
Determine the second-order moments by amplifying the first-order
moments.
No-translation amplification: The no-translation moments must be
amplified by B1. From Figure 8.11e it is seen that the end moments bend
the column in reverse curvature:
M 1 39.0
=
= 0.50
M 2 78.0
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4
Pe1 =
π2 E π 2 (29,000)(999)
=
= 12,700 kips
L2c1
(1.0(12.5)(12)) 2
Thus, with α = 1.6 for ASD and Pr = 280 + 62 = 342 kips, Equation A-8-3
yields
Cm
0.4
B1 =
=
= 0.418 < 1.0
(1.6)(342)
αP
1− r 1 –
Pe1
12,700
Therefore, B1 = 1.0.
Translation amplification: The translation forces and moments must be
amplified by B2. The design drift limit of height/300 and Equation A-8-6
are used to determine B2.
322
Chapter 8
Beam-Columns and Frame Behavior
The total service lateral load on this story is given as
H = 148 kips
Additional given information is that the total gravity load for this load
combination in Figure 8.11e is
Pstory = 1670 kips
The drift limit under the service lateral load of 148 kips is
ΔH = L/300 = 12.5(12)/300 = 0.50 in.
Remember that in the calculation of B2, H can be taken as any convenient
magnitude, as long as ΔH is the corresponding displacement. This is
because it is the ratio of H to ΔH that is used in the determination of Pe story.
Thus, with α = 1.6 for ASD and RM = 0.85 assuming all columns are
moment frame columns, Equation A-8-7 gives
Pe story =
RM HL 0.85(148)(12.5)(12)
=
= 37,700 kips
ΔH
0.50
and Equation A-8-6 gives
1
1
B2 =
=
= 1.08>1.0
αPstory
⎛ (1.6)1670 ⎞
1−
1– ⎜
⎟
Pe story
⎝ 37,700 ⎠
Thus, the second-order compressive force and moment are
Pr = Pnt + B2(Plt) = 280 + 1.08(62.0) = 347 kips
Mr = B1(Mnt) + B2(Mlt) = 1.0(78.0) + 1.08(96.0) = 182 ft-kips
These represent the required strength for this load combination.
Step 6:
Determine whether this shape will provide the required strength based on
the appropriate interaction equation.
The unbraced length of the compression flange for pure bending is 12.5 ft,
which is less than Lp = 15.1 ft for this section, taking into account that its
flange is noncompact. Thus, from Manual Table 3-2, the allowable moment
strength of the section is
M n Ω = 382 ft-kips
Determine the appropriate interaction equation. From Step 3, φPn = 1060
kips;
Pa
347
=
= 0.494 > 0.2
Pn Ω 703
so use Equation H1-1a, which yields
Beam-Columns and Frame Behavior
Chapter 8
323
Pa
8 ⎛ Ma ⎞
+ ⎜
⎟ ≤ 1.0
Pn Ω 9 ⎝ M n Ω ⎠
8 ⎛ 182 ⎞
0.494 + ⎜
⎟ = 0.918 < 1.0
9 ⎝ 382 ⎠
Thus,
the W14×90 is adequate for this load combination.
Step 7:
Check the section for the gravity-only load combination, D + L.
Because this is a gravity-only load combination, Specification Appendix
Section 7.2.2, by reference to Section C2.2b, requires that the analysis
include a minimum lateral load of 0.002 times the gravity load. This will be
further discussed in Section 8.7. For this load combination, the total story
gravity load must also be known and is given in Figure 8.11d as Pstory =
1690 kips. Thus, for this frame the minimum lateral load is 0.002Pstory =
0.002(1690) = 3.38 kips at this level.
The forces and moments given in Figure 8.11d include the effects of this
minimum lateral load. The magnitude of the lateral translation effect is
small in this case. Since both the moment due to the minimum lateral load
and the amplification factor, B2, are expected to be small, the forces and
moments used for this check will be assumed to come from a no-translation
case, with little error. If the minimum lateral load would produce large
moments or the amplification factor, B2, calculated in Step 5, were large,
this would not be a good assumption. Therefore, at end A, Mnt = 95.0 ftkips, at end B Mnt = 47.0 ft-kips, and Pnt = 348 kips.
A quick review of the determination of B1 from the first part of this solution
shows that the only change is in the magnitude of the axial force and the
member end moments; thus
M 1 47.0
=
= 0.50
M 2 95.0
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4
Pe1 =
π2 EI π 2 (29,000)(999)
=
= 12,700 kips
L2c1
(1.0(12.5)(12)) 2
Thus, with α = 1.6 for ASD,
Cm
0.4
B1 =
=
= 0.418 < 1.0
αPr
(1.6)(348)
1−
1–
Pe1
12,700
Note that B1 is again 1.0.
With the assumption that there is no lateral translation,
Mlt = 0.0 and B2 is unnecessary,
thus
Pr = 348 kips, M r = 1.0(95.0) = 95.0 ft-kips
Again using Equation H1-1a,
3224
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Pa
8 ⎛ Ma ⎞
+ ⎜
⎟ ≤ 1.0
Pn Ω 8 ⎝ M n Ω ⎠
348 8 ⎛ 95.0 ⎞
+ ⎜
⎟ = 0.716 < 1.0
703 9 ⎝ 382 ⎠
Thus,,
the W1
14×90 is adeqquate for bothh load combinnations.
The moments in the beams an
nd the beam-ccolumn conneections must also be ampllified for
the critical case
c
to accou
unt for the seccond-order efffects. This iss done by connsidering equuilibrium
of the beam
m-column join
nt. The ampliffied momentss in the colum
mn above andd below the jjoint are
added togeth
her and this sum distributeed to the beam
ms which fram
me into the jooint accordingg to their
stiffnesses. These
T
momen
nts then establish the conneection designn moments.
8.77
SPECIF
FICATION PROVISIO
ONS FOR ST
TABILITY
Y ANALYSIIS AND DES
SIGN
Up to this po
oint, the discu
ussion of the interaction off compressionn and bendingg has concenttrated on
the develop
pment of the interaction equations annd one approoach to incorrporate seconnd-order
effects. The Specification
n actually provides three appproaches to deal with theese two closelly linked
issues. The most
m direct ap
pproach is to use a generaal second-ordeer analysis in conjunction with the
Direct Analy
ysis Method described
d
in Chapter
C
C.
A general
g
second
d-order analy
ysis yields forrces and mom
ments that caan be used dirrectly in
the interaction equations of Chapter H without thee need to resoort to amplifiication factorrs as just
described. The
T disadvan
ntage to this approach is that, since tthe extremelyy useful prinnciple of
superpositio
on cannot be used (since the
t structurall response is nonlinear), a complete nnonlinear
analysis mu
ust be carried
d out for each
h load combiination. A ddiscussion of general, or rrigorous,
methods of second-order analysis is beyond the scoope of this boook. Thus, in the remaindeer of this
book, if seccond-order efffects have no
ot already beeen included in the analysis results given, the
amplified first-order anaalysis approacch will be ussed to obtain the requiredd second-ordeer forces
and momentts.
Figure 8.12
8
Comp
parison of thee
Effectivee Length Metthod and the
Direct Analysis
A
Meth
hod
Beam-Columns and Frame Behavior
Chapter 8
325
8.7.1 Direct Analysis Method
The advantage of the direct analysis method of Specification Chapter C is that for the design of
compression members, the effective length factor is taken as 1.0. Since for braced frames, K can
always be taken as 1.0 based on Section 7.2.3(a), the direct analysis method is particularly useful
for moment frames. The Specification requires that the stiffness of all elements contributing to the
lateral load resistance of the structure be reduced. Thus, rather than using EA for the axial
stiffness of the members, the modified stiffness, EA* = 0.8EA, is used. Similarly for flexural
stiffness, the modified stiffness EI* = 0.8τbEI, where τb accounts for the influence of residual
stresses on second-order effects, is used. It should be remembered that the influence of residual
stresses on the strength of compression and bending members was already discussed in Chapters
5 and 6. The use of τb in this instance is to capture the influence of those same residual stresses on
displacements and thus on second-order effects. This is the same τb used with the alignment
charts in the determination of the effective length in Chapter 5.
Figure 8.12 shows a comparison between the effective length method already presented
and the direct analysis method for a simple structure. Equations H1-1a and H1-1b are plotted for
the effective length method and labeled with K = 2.66. This indicates that the compressive
strength of the member has been determined using K = 2.66. The nonlinear load-moment curve is
identified with EI and EA to indicate that the nominal stiffnesses are used to determine this
behavior. The intersection of these two curves indicates that this load and moment combination
satisfy the interaction equation. Equations H1-1a and H1-1b are plotted for the direct analysis
method and identified with K = 1.0. This indicates that the compressive strength of the member
has been determined with K = 1.0. Note that regardless of which approach is selected to
determine the compressive strength, the flexural strength is the same for both methods. The
nonlinear load-moment curve is identified with EI* and EA* to indicate that the reduced
stiffnesses were used to determine this behavior. The intersection of these two curves indicates
that this load and moment combination satisfy the interaction equation. Next, note that the axial
load magnitude for both of these intersections is nearly the same. Thus, the load that satisfies the
interaction equation is the same regardless of which method is used. Since the direct analysis
method did not require the determination of K, it is a significantly simpler method than the
effective length method.
Another consideration that has only briefly been mentioned to this point is the
requirement in Section C1 that the influence of geometric imperfections be considered. As with
residual stresses, the influence of geometric imperfection on the strength of compression
members has already been addressed through the Specification column strength equations. The
requirement here is to consider the influence of out-of-plumbness on the stability of the structure.
This may be accomplished by modeling the structure in its out-of-plumb condition or through the
use of notional loads to simulate the out-of-plumbness. These notional loads will be discussed
later in this section. It should be noted that this is not a requirement of the direct analysis method
alone but a general requirement for determining required strength.
In addition to the direct analysis method, two other design methods are given in the
Specification. They are found in Appendix 7. The limitations on the application of these methods
are based on the direct analysis method.
8.7.2 Effective Length Method
Appendix 7.2 provides the requirements for the effective length method. This is the approach
already described earlier in this chapter for braced and moment frames. It is valid so long as the
ratio of second-order deflection to first-order deflection, Δ2/Δ1, is equal to or less than 1.5.
Another way to state this requirement is to remember that Δ2/Δ1 = B2, so the effective length
method is valid as long as B2 ≤ 1.5. Although this check was not made in Example 8.2, it can now
be seen that it was acceptable to use the effective length method in that example, since for both
326
Chapter 8
Beam-Columns and Frame Behavior
LRFD and ASD, B2 ≤ 1.5. A special case occurs when B2 ≤ 1.1. In this case, columns in moment
frames can be designed using K = 1.0. The effective length method is essentially the same method
used in past practice with the addition of the requirement of a minimum lateral load to be applied
in gravity-only load combinations. This is the notional load discussed above to account for initial
out-of-plumbness. It is the same as the minimum lateral load used in Example 8.2 and will be
discussed later in this section.
8.7.3 First-Order Analysis Method
A third method is given in the Appendix 7.3, the first-order analysis method. This approach
permits design without direct consideration of second-order effects except through the application
of additional notional lateral loads that account for structure out-of-plumbness and second-order
effects. This is possible because of the limits placed on the implementation of this method. As
with the effective length method, the structure must support gravity loads primarily through
vertical columns, walls or frames and the ratio of the second-order drift to first-order drift must be
less than or equal to 1.5. Additionally, compression members that participate in lateral load
resistance must behave elastically according to
αPr ≤ 0.5Pns
(AISC A-7-1)
With the foregoing limitations and the application of the notional load given by
N i = 2.1α ( Δ L ) Yi ≥ 0.0042Yi
(AISC A-7-2)
compression members may be designed using K = 1.0. Story gravity load, Yi, is defined in the
following.
8.7.4 Geometric Imperfections
Now, consider in more detail the requirement to consider geometric imperfections. Specification
Section C1 requires that geometric imperfections be considered in the analysis and design of
structures. There are two types of geometric imperfections that must be considered: initial out-ofstraightness and initial out-of-plumbness. The strength equations of Specification Chapter E
already account for initial out-of-straightness. To consider out-of-plumbness, the structure may
be modeled in the out-of-plumb position or notional loads, as given in Specification Section
C2.2b, may be used. The AISC Code of Standard Practice permits columns to be built with an
out-of-plumbness tolerance of height/500.
Figure 8.13a shows the upper two stories of a column that is out-of-plumb, L/500 =
0.002L. For level 1, with a load P1 applied, a horizontal force of 0.002P1 is required for
equilibrium as shown. The next story down, level 2, is also out-of-plumb by the same amount and
the load from above is added to the load introduced at that level so that the column must carry P1
+ P2. Since this load is also applied at the eccentricity of 0.002L, equilibrium requires that the
column be restrained by a force of 0002(P1 + P2) as shown for level 2. When these two columns
are put together, it can be seen in Figure 8.13b that the horizontal force at the intersection
becomes 0.002 P2. Thus, Figure 8.13b shows for this two story out-of-plumb column, the
horizontal forces that are required to keep it in equilibrium. The same effect could be
accomplished if the column was modeled plumb and the restraining forces from Figure 8.13b
were applied as loads as shown in Figure 8.13c.
Thus, the out-of-plumbness can be modeled with a lateral load equal to 0.2 percent of the
gravity load introduced at each level of the structure. This lateral load is called a notional load
and is taken as Ni = 0.002αYi where Yi is the total gravity load on story i and α is 1.0 for LRFD
and 1.6 for ASD, as discussed earlier. The Specification includes α as described here, but when
the amplified first-order analysis method is used to obtain second-order effects, this is not
Beam
m-Columns annd Frame Behhavior
Lev
vel 1, Top
Chappter 8
327
Level 2, Nexxt level down
Figure
F
8.13 Notional Lo
oad Model forr Geometric Im
mperfections.
necessary
y, since α is included
i
in th
he B1 and B2 calculations. This notionaal load is the ssame as the
“minimu
um lateral load
d” used in Ex
xample 8.2.
8.7.5 Comparison of Meethods
The three methods of frame stabiility analysis just describeed will be coompared usinng a simple
determin
nate structure.. Figure 8.14 shows a onee-bay unbraceed frame withh an LRFD ggravity plus
lateral lo
oad combinattion. Column A is a flag pole columnn and providees all of the lateral load
resistance while colum
mn B is a graavity only coolumn. Gravitty only colum
mns will be ddiscussed in
pth in Section
n 8.10. Colum
mn A is a W1 4×90 bendingg about its strrong axis, collumn B can
more dep
be any siize sufficientt to support th
he gravity loaad since it ddoes not contrribute to the lateral load
resistance, and the beaam is assumed to be a rigidd element.
3228
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
One-bay Unnbraced Fram
Figgure 8.14
me for
Coomparison of A
Analysis Metthods
Effective Leength Method
d: First the efffective lengthh method of A
Appendix Secction 7.2 will be used
along with the
t B1 – B2 amplification
a
for second-oorder effects. If the structuure is preventted from
swaying, thee nt analysis produces, forr column A, P nt = 200 kipps and Mnt = 0 ft-kips. Thhe lateral
translation analysis
a
produ
uces, for colu
umn A, Plt = 0 kips and Mnt = 300 ft-kips. Since theere is no
moment in the
t nt analysis, there are no
o P-δ effects (member effe
fects) and no nneed to determ
mine B1.
To assess th
he P-Δ effects (sway effectts), B2 will be determined. The 20 kip laateral load prooduces a
drift calculated as for a caantilevered beeam,
20 (15 (112 ) )
HL3
= 1.344 in.
Δ=
=
3EII 3 ( 29,000 )( 999 )
3
The total graavity load on the structure is 400 kips. Half of this lload is on thee lateral load rresisting
column A an
nd half is on the
t gravity on
nly column. T
Thus,
⎛ Pmf ⎞
⎛ 200 ⎞
RM = 1 − 0.1
15 ⎜
⎟ = 1 − 0.15 ⎜
⎟ = 0.925
⎝ 400 ⎠
⎝ Ptory ⎠
and
200 (15 (12 ) )
H
HL
= 22490 kips
Pe story = RM
= 0.925
ΔH
1.34
Thus,
1
1
=
B2 =
= 11.19
αPstory
1.0 ( 400 )
1−
1−
Pe story
2490
xt, consider the
t limitation
ns on use off the effectivve length meethod. This sstructure
Nex
supports graavity loads th
hrough verticaal columns soo it meets thee first limitattion in Sectioon 7.2.1.
The second limitation reequires that th
he second-ordder amplificaation, B2, be less than or equal to
1.5. Since B2 = 1.19 ≤ 1.5 the effective length methood may be useed for this fraame.
The required streength, includiing second-o rder effects iis found throuugh Equationns A-8-1
and A-8-2.
M r = B1M nt + B2 M lt = 0 + 11.19 ( 300 ) = 357 ft-kips
and
Pr = Pnt + B2 Plt = 200 + 1.19 ( 0 ) = 2000 kips
n the effectiv
ve length meethod is deterrmination off the effectivee length
The next step in
factor. The effective len
ngth factor for
fo the flag ppole column alone is Kx = 2.0. How
wever, as
Beam-Columns and Frame Behavior
Chapter 8
329
discussed in Chapter 5, the inclusion of the gravity only column with load will increase the
effective length of column A. Using the approach presented in Chapter 5, with the load on the
moment frame column, Pmf = 200 kips and the load on the gravity only column, Pgrav only = 200
kips, the effective length factor is
K x* = K x 1 + Pgrav only Pmf = 2.0 1 + 200 200 = 2.83
Assuming that the frame is braced out of the plane of the frame, Ky = 1.0.
The available strength of the W14×90 column can be determined from Table 6-2 for an
unbraced length of the compression flange Lb = 15 ft < Lp = 15.1 ft, φM n = 574 ft-kips . The
controlling effective length is for x-axis buckling, thus
( Lcx )eff
= 2.83(15) 1.66 = 25.6 ft and
φPn = 720 kips. With the required strength and available strength determined, the interaction
equation can be checked.
First determine which interaction equation should be used. Since
Pr φPn = 200 720 = 0.278 > 0.2 use Equation H1-1a, thus
Pr 8 ⎛ M r ⎞ 200 8 ⎛ 357 ⎞
+
=
+ ⎜
⎟ = 0.278 + 0.553 = 0.831 < 1.0
Pc 9 ⎜⎝ M c ⎟⎠ 720 9 ⎝ 574 ⎠
So the W14×90 is shown to be adequate by the effective length method.
First-Order Analysis Method: The first-order analysis method of Appendix Section 7.3 may be
used for those structures that meet the limitations of Section 7.3.1. These limitations are the same
as for the effective length method with the addition of the requirement that the columns behave
elastically such that
αPr ≤ 0.5Pns
(AISC A-7-1)
Since the W14×90 column does not have slender elements for compression,
Pns = Py = Fy Ag = 50 ( 26.5 ) = 1330 kips and for the frame of Figure 8.14 αPr = 200 kips . Thus
200 ≤ 0.5 (1330 ) = 666 kips and the first-order analysis method may be used.
The required strength for the first-order analysis method is determined from a first-order
analysis that includes a notional load defined by Equation A-7-2 added to the lateral load in all
load combinations. This notional load accounts for both the initial out-of-plumbness of the
structure and second-order effects. Thus,
N i = 2.1α ( Δ L ) Yi ≥ 0.0042Yi
(AISC A-7-2)
For our structure, Δ = 1.34 in. as before and Yi = 400 kips so the notional load for this load
combination is
(
)
Ni = 2.1(1.0) 1.34 (15 (12) ) ( 400) = 6.25 kips ≥ 0.0042 ( 400) = 1.68 kips
Thus, the lateral load in the analysis will be increased from 20 kips to 26.3 kips. The results of the
first-order analysis for the determinate structure are Pu = 200 kips and Mu =26.3(15) = 395 ftkips. Although this is called the first-order analysis method, it does require that the moment be
amplified by B1 found using
Cm
B1 =
≥ 1.0
(AISC A-8-3)
1 − αPr Pe1
330
Chapter 8
Beam-Columns and Frame Behavior
This amplification addresses the member effect and is influenced by the buckling strength of the
column as a pin ended column in a no sway condition, Pe1, and the equivalent uniform moment
factor, Cm. Thus,
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 − 0.4 ( 0 395 ) = 0.6
and with EI* = EI,
π2 EI * π2 (29,000)(999)
Pe1 = 2 =
= 8830 kips
2
Lc1
(15 (12) )
which gives
B1 =
0.6
= 0.614 < 1.0
1 − 200 8830
Therefore there is no amplification needed so Pr = Pu = 200 kips and M r = Mu = 395 ft-kips .
The available moment strength of the W14×90 column determined previously from Table
6-2 is unchanged, thus φM n = 574 ft-kips . The controlling effective length is for y-axis buckling,
thus Lcy = 15.0 ft and φPn = 1000 kips . With the required strength and available strength
determined, the interaction equation can be checked.
First determine which interaction equation should be used. Since Pr φPn = 200 1000 = 0.20 ≤ 0.2
use Equation H1-1a, thus
Pr 8 ⎛ M r ⎞ 200 8 ⎛ 395 ⎞
+
=
+ ⎜
⎟ = 0.200 + 0.612 = 0.812 < 1.0
Pc 9 ⎜⎝ M c ⎟⎠ 1000 9 ⎝ 574 ⎠
So the W14×90 is shown to be adequate by the first order analysis method.
Direct Analysis Method: The third method to be considered is the direct analysis method of
Chapter C. There are no limitations on the use of the direct analysis method like there are on the
effective length or first-order analysis methods and second-order effects and initial out-ofplumbness must be accounted for as they were for the effective length method. The only new
requirement is that the stiffness of all members that contribute to the lateral load resistance be
*
reduced in the analysis to EI * = 0.8τb EI and EA = 0.8EA . It is this stiffness reduction that
permits the use of an effective length factor equal to one when using the direct analysis method.
From the discussion of the effective length method it was seen that B2 was less than 1.5 when
using the unreduced stiffness thus the notional load to account for out-of-plumbness does not
need to be added to the lateral load. Thus, from a first order analysis of the determinate structure,
Pu = 200 kips and M u = 20.0 (15 ) = 300 ft-kips . As for the effective length method, τb = 1.0 so
*
that the flexural stiffness of column A will be taken as EI = 0.8EI . Thus the 20 kip lateral load
produces a drift calculated as for a cantilevered beam,
20 (15 (12 ) )
HL3
Δ=
=
= 1.68 in.
*
3EI
3( 0.8)( 29,000)( 999 )
3
The total gravity load on the structure is 400 kips. Half of this load is on the lateral load resisting
column A and half is on the gravity only column. Thus again,
⎛ Pmf ⎞
⎛ 200 ⎞
RM = 1 − 0.15 ⎜
⎟ = 1 − 0.15 ⎜
⎟ = 0.925
⎝ 400 ⎠
⎝ Ptory ⎠
and
Beam-Columns and Frame Behavior
Pe story = RM
Chapter 8
331
20 (15 (12 ) )
HL
= 0.925
= 1980 kips
1.68
ΔH
Thus,
1
1
=
= 1.25
αPstory
1.0 ( 400 )
1−
1−
Pe story
1980
Note that the drift increased from what was calculated for the effective length method and
therefore the second-order amplification increased.
The required strength, including second-order effects is found through Equations A-8-1
and A-8-2.
M r = B1M nt + B2 M lt = 0 + 1.25 ( 300 ) = 375 ft-kips
and
Pr = Pnt + B2 Plt = 200 + 1.25 ( 0 ) = 200 kips
B2 =
The available moment strength of the W14×90 column determined previously from Table
6-2 is unchanged, thus φM n = 574 ft-kips . The controlling effective length is for y-axis buckling,
thus Lcy = 15.0 ft and φPn = 1000 kips . With the required strength and available strength
determined, the interaction equation can be checked.
First
determine
which
interaction
equation
should
be
used.
Since
Pr φPn = 200 1000 = 0.20 ≤ 0.2 use Equation H1-1a, thus
Pr 8 ⎛ M r ⎞ 200 8 ⎛ 375 ⎞
+
=
+ ⎜
⎟ = 0.200 + 0.581 = 0.781 < 1.0
Pc 9 ⎜⎝ M c ⎟⎠ 1000 9 ⎝ 574 ⎠
So the W14×90 is adequate by the direct analysis method. Note that based on the results of the
interaction equation, this approach is less conservative than the other two methods. Since the only
new requirement of the direct analysis method is to use a reduced stiffness in calculating secondorder effects and this permits the use of an effective length factor in the lateral load resisting
direction of one, this is clearly the simplest and most direct method available.
Three methods of analysis are available and all three have their place in design. It is up to
the user to determine when to use each approach most efficiently.
EXAMPLE 8.3a
Direct Analysis
Method for
Column Design by
LRFD
Goal:
Using the LRFD provisions and the results from a second-order direct
analysis, determine if a W14×132, A992 member is adequate to carry the
given loads and moments.
Given:
The column has a length of 16 ft and is braced at the ends only. The results
of the second-order direct analysis are Pu = 800 kips, Mux = 300 ft kips, and
Muy = 76 ft kips.
SOLUTION
Step 1:
Determine the required strength.
Since the given results are from a second-order analysis, there is no need to
amplify forces and moments; thus
Pr = Pu = 800 kips, M rx = M ux = 300 ft kips,
M ry = M uy = 76 ft kips
332
Chapter 8
Beam-Columns and Frame Behavior
Step 2:
Determine the available compressive strength of the column.
Since the given results are from a direct analysis, K = 1.0; thus, from
Manual Table 4-1 with Lc = 16.0 ft,
φPn = 1440 kips
Step 3:
Determine the available strength for bending about the x-axis.
With an unbraced length Lb = 16 ft, from Manual Table 3-2,
φM p = 878 ft kips, L p = 13.3 ft, φBF = 7.74 kips
and
Step 4:
φM nx = φM p − φBF ( Lb − Lp ) = 878 − 7.74 (16.0 − 13.3) = 857 ft kips
Determine the available strength for bending about the y-axis.
From Manual Table 3-4,
φM ny = 424 ft kips
Step 5:
Check the W14×132 for combined axial load and bending. To determine
which equation to use, check
Pu
800
=
= 0.556 ≥ 0.2
φPn 1440
Therefore, use Equation H1-1a.
M ry
Pr
M rx
+
+
≤ 1.0
φPn φM nx φM ny
0.556 +
300 76
+
= 1.09 > 1.0
857 424
Thus,
the W14×132 will not carry the given load.
EXAMPLE 8.3a Goal:
Direct Analysis
Method for
Column Design
Given:
by ASD
SOLUTION
Step 1:
Using the ASD provisions and the results from a second-order direct
analysis, determine if a W14×132, A992 member is adequate to carry the
given loads and moments.
The column has a length of 16 ft and is braced at the ends only. The
results of the second-order direct analysis are Pa = 530 kips, Max = 200 ft
kips, and May = 52 ft kips.
Determine the required strength.
Since the given results are from a second-order analysis, there is no need
to amplify forces and moments. Thus,
Pr = Pa = 530 kips, M rx = M ax = 200 ft kips,
M ry = M ay = 52 ft kips
Beam-Columns and Frame Behavior
Step 2:
Chapter 8
333
Determine the available compressive strength of the column.
Since the given results are from a direct analysis, K = 1.0. Thus, from
Manual Table 4-1 with Lc = 16.0 ft,
Pn
= 960 kips
Ω
Step 3:
Determine the available strength for bending about the x-axis.
With an unbraced length Lb = 16 ft, from Manual Table 3-2,
Mp
BF
= 584 ft kips, Lp = 13.3 ft,
= 5.15 kips
Ω
Ω
And
M nx M p BF
=
−
( Lb − Lp ) = 584 − 5.15 (16.0 − 13.3) = 570 ft kips
Ω
Ω
Ω
Step 4:
Determine the available strength for bending about the y-axis.
From Manual Table 3-4
M ny
= 282 ft kips
Ω
Step 5:
Check the W14×132 for combined axial load and bending. To determine
which equation to use, check
Pu 530
=
= 0.552 ≥ 0.2
φPn 960
Therefore, use Equation H1-1a.
M ry
Pr
M rx
+
+
≤ 1.0
Pn Ω M nx Ω M ny Ω
0.552 +
200 52
+
= 1.09 > 1.0
570 282
Thus,
the W14×132 will not carry the given load.
8.8 INITIAL BEAM-COLUMN SELECTION
Beam-column design is a trial-and-error process that requires that the beam-column section be
known before any of the critical parameters can be determined for use in the appropriate
interaction equations. There are numerous approaches to determining a preliminary beam-column
size. Each incorporates its own level of sophistication and results in its own level of accuracy.
Regardless of the approach used to select the trial section, one factor remains—the trial section
must ultimately satisfy the appropriate interaction equation.
To establish a simple, yet useful, approach to selecting a trial section, Equation H1-1a is
modified by multiplying each term by Pc which yields
8 M rx Pc 8 M ry Pc
Pr +
+
≤ Pc
(8.9)
9 M cx
9 M cy
Then multiplying the third term by Mcx/Mcx, letting
334
Chapter 8
Beam-Columns and Frame Behavior
m=
8Pc
9M cx
and
U=
M cx
M cy
and substituting into Equation 8.9 yields
Pr + mM rx + mUM ry ≤ Pc
(8.10)
Because Equation 8.10 calls for the comparison of the left side of the equation to the
column strength, Pc, Equation 8.10 can be thought of as an effective axial load; thus
Peff = Pr + mM rx + mUM ry ≤ Pc
(8.11)
The accuracy used in the evaluation of m and U dictates the accuracy with which
Equation 8.11 represents the strength of the column being selected. Because at this point in a
design the actual column section is not known, exact values of m and U cannot be determined.
Past editions of the AISC Manual have presented numerous approaches to the evaluation
of these multipliers. A simpler approach however, is more useful for preliminary design. If the
influence of the length—that is, all buckling influence on Pc and Mcx—is neglected, the ratio,
Pc/Mcx, becomes A/Zx, and m = 8A/9Zx. Evaluation of this m for all W6 to W14 shapes with the
inclusion of a units correction factor of 12 results in the average m values given in Table 8.2. If
the relationship between the area, A, and the plastic section modulus, Zx, is established using an
approximate internal moment arm of 0.89d, where d is the nominal depth of the member in
inches, then m = 24/d. This value is also presented in Table 8.2. This new m is close enough to
the average m that it may be readily used for preliminary design.
When bending occurs about the y-axis, U must be evaluated. A review of the same W6 to
W14 shapes results in the average U values given in Table 8.2. However, an in-depth review of
the U values for these sections shows that only the smallest sections for each nominal depth have
U values appreciably larger than 3. Thus, a reasonable value of U = 3.0 can be used for the first
trial.
More accurate evaluations of these multipliers, including length effects, have been
conducted, but there does not appear to be a need for this additional accuracy in a preliminary
design. Once the initial section is selected, however, the actual Specification provisions must be
satisfied.
Table 8.2
Shape
W6
W8
W10
W12
W14
EXAMPLE 8.4a
Initial Trial
Section Selection
by LRFD
SOLUTION
Simplified Bending Factors
mavg
m = 24/d
4.41
4.00
3.25
3.00
2.62
2.40
2.08
2.00
1.71
1.71
Uavg
3.01
3.11
3.62
3.47
2.81
Goal:
Determine the initial trial section for a column.
Given:
The loadings of Figure 8.11c are to be used. Assume the column is a W14
and use A992 steel. Also, use the simplified values of Table 8.2, m = 24/d.
Step 1:
Obtain the required strength from Figure 8.11c. Use the first-order analysis
Beam-Columns and Frame Behavior
Chapter 8
335
results.
Pu = 453 kips
M u = 251 ft-kips
Step 2:
Determine the effective load by combining the axial force and the bending
moment.
For a W14, m = 1.71, so
Peff = 453 + 1.71( 251) = 882 kips
Step 3:
Select a trial column size to carry the required force, Peff.
Using an effective length Lc = 12.5 ft, from Manual Table 4-1, the lightest
W14 to carry this load is
W14 × 90 with φPn = 1060 kips
Example 8.2a showed that this column adequately carries the imposed load.
Because the approach used here is expected to be conservative, it would be
appropriate to consider the next smaller selection, a W14×82, and check it
against the appropriate interaction equations.
EXAMPLE 8.4b
Initial Trial
Section Selection
by ASD
SOLUTION
Goal:
Determine the initial trial section for a column.
Given:
The loadings of Figure 8.11e are to be used. Assume the column is a W14,
and use A992 steel. Also, use the simplified values of Table 8.2, m = 24/d.
Step 1:
Obtain the required strength from Figure 8.11e. Use the first-order analysis
results.
Pa = 342 kips
M a = 174 ft-kips
Step 2:
Determine the effective load by combining the axial force and the bending
moment.
For a W14, m = 1.71; thus
Peff = 342 + 1.71(174 ) = 640 kips
Step 3:
Select a trial column size to carry the required force, Peff.
Using an effective length Lc = 12.5 ft, from Manual Table 4-1, the lightest
W14 to carry this load is
W14 × 90 with Pn Ω = 703 kips
Example 8.2b showed that this column adequately carries the imposed load.
Because the approach used here is expected to be conservative, it would be
appropriate to consider the next smaller selection, a W14×82, and check it
against the appropriate interaction equations.
336
Chapter 8
Beam-Columns and Frame Behavior
Every column section selected must be checked through the appropriate interaction equations for
the second-order forces and moments. Thus, the process for the initial selection should be quick
and reasonable. The experienced designer will rapidly learn to rely on that experience rather than
these simplified approaches.
8.9
BEAM-COLUMN DESIGN USING MANUAL PART 6
Manual Part 6, Design of Members Subject to Combined Loading contains Table 6-2 which
includes the axial and flexural strength for all W-shapes. Although these tables are presented here
as they relate to combined loading, they can also be used for compression only, bending only,
tension only and shear. There is no information found in Table 6-2 that is not already included in
other Parts of the Manual already discussed. The advantage for combined loading is that all of the
available strength values needed are found in one location.
Figure 8.15 is a portion of Manual Table 6-2. It shows that the compressive strength for a
given section is a function of the effective length about the weak axis of the member. The
effective length is tabulated in the center of the table with the compressive strengths shown on the
left portion of the table. This portion of the table is used in exactly the same way as the column
tables in Part 4 of the Manual. The strong axis bending strength is a function of the unbraced
length of the compression flange of the beam. Previously, this information was available only
through the beam curves in Part 3 of the Manual. In Table 6-2 it is tabulated on the right portion
of the table with the same column of lengths now defined as the unbraced length of the
compression flange. Weak axis bending is not a function of length, so only one value is given for
each shape. Although not used for beam-columns, when tension is combined with bending, the
table also provides tension yield and rupture strength.
Goal:
EXAMPLE 8.5a
Combined Strength
Check Using
Manual Part 6 and Given:
LRFD
Check the strength of a beam-column using Manual Part 6 and compare to
the results of Example 8.2a.
Step 1:
Determine the values needed from Manual Table 6-2 (Figure 8.15). The
column is required to carry a compressive force with an effective length
about the y-axis of 12.5 ft and an x-axis moment with an unbraced length of
12.5 ft. Thus, from Figure 8.15,
φPn = 1060 kips
SOLUTION
It has already been shown that the W14×90 column of Example 8.2a is
adequate by LRFD. Use the required strength values given in Example 8.2a
and recheck this shape using the values found in Figure 8.15 or Manual
Table 6-2.
φM n = 574 ft-kips
Step 2:
Determine which interaction equation to use.
Pr
459
=
= 0.433 > 0.2
φPn 1060
Therefore, use Equation H1-1a.
200 8 ⎛ 260 ⎞
+ ⎜
⎟ = 0.433 + 0.403 = 0.836 < 1.0
1060 9 ⎝ 574 ⎠
Therefore, as previously determined in Example 8.2a, the shape is adequate
for this column and this load combination. The results from Manual Tables
Beam-Columns and Frame Behavior
Chapter 8
337
6-2 are exactly the same as those determined from Table 4-1 for
compression and Table 3-2 for bending.
EXAMPLE 8.5b
Combined
Strength Check
Using Part 6 and
ASD
Goal:
Check the strength of a beam-column using Manual Part 6 and compare to
the results of Example 8.2b.
Given:
It has already been shown that the W14×90 column of Example 8.2b is
adequate by ASD. Use the required strength values given in Example 8.2b
and recheck this shape using the values found in Figure 8.15 or Manual
Table 6-2.
SOLUTION
Step 1:
Determine the values needed from Manual Table 6-2 (Figure 8.15). The
column is required to carry a compressive force with an effective length
about the y-axis of 12.5 ft and an x-axis moment with an unbraced length of
12.5 ft. Thus, from Figure 8.15,
Pn Ω = 703 kips
M n Ω = 382 ft-kips
Step 2:
Determine which interaction equation to use.
Pr
347
=
= 0.494 > 0.2
Pn Ω 703
Thus, use Equation H1-1a.
347 8 ⎛ 182 ⎞
+ ⎜
⎟ = 0.494 + 0.424 = 0.918 < 1.0
703 9 ⎝ 382 ⎠
Thus, as previously determined in Example 8.2b, the shape is adequate for
this column and this load combination. The results from Manual Tables 6-2
are exactly the same as those determined from Table 4-1 for compression
and Table 3-2 for bending.
8.10 COMBINED SIMPLE AND MOMENT FRAMES
The practical design of steel structures often results in frames that combine segments of rigidly
connected elements with segments that are pin connected as was the case in the example frame
used in Section 8.7.5. If these structures rely on the moment frame to resist lateral load and to
provide the overall stability of the structure, the rigidly connected columns are called upon to
carry more load than appears to be directly applied to them. In these combined simple and
moment frames, the simple columns “lean” on the moment frames in order to maintain their
stability and thus are often called leaning columns. They are also called gravity columns which is
a more appropriate term since they participate only in carrying gravity loads. These columns can
be designed with an effective length factor, K = 1.0, regardless of the approach to analysis that
has been taken. Because these gravity only columns have no lateral stability of their own, the
moment frame columns must be designed to provide the lateral stability for the full frame.
Although this combination of framing types makes design of a structure more complicated, it can
also be economically advantageous, because the combination can reduce the number of moment
connections for the full structure and thereby reduce overall cost.
3338
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figure 8.15 Comb
bined Axial an
nd Bending Strength
S
for W
W-Shapes.
Copyrig
ght © Americcan Institute of
o Steel Consttruction, Inc. Reprinted wiith Permissionn. All rights
reserved.
Beam
m-Columns annd Frame Behhavior
Chappter 8
339
Fig
gure 8.16 Piinned Base U
Unbraced Fram
me.
Numerous
N
deesign approacches have beeen proposed ffor consideraation of the ggravity only
column and
a associateed moment frrame design.33,4 Yura propposes to desiggn columns thhat provide
lateral sttability for th
he total load on the fram
me at the storry in questionn, whereas L
LeMessurier
presents a modified effective
e
lengtth factor thatt accounts forr the full fram
me stability. P
Perhaps the
a
is th
hat presented by Yura, as ffollows.
most straaightforward approach
The
T two-colu
umn frame sh
hown in Figuure 8.16a is a moment fframe with ppinned base
columns and a rigidly
y connected beam.
b
The coolumn sizes are selected so that, undeer the loads
shown, they buckle simultaneou
usly in a siidesway modde, because their load is directly
proportio
onal to the sttiffness of thee members. E
Equilibrium iin the displacced position iis shown in
Figure 8.16b. The latteral displacement of the frame, Δ, ressults in a mooment at the ttop of each
column equal
e
to the lo
oad applied on
o the columnn times the diisplacement, as shown. Thhese are the
second-o
order effects discussed
d
in Section 8.6. The total loaad on the fraame is 700 kiips, and the
total PΔ moment is 70
00Δ, divided between
b
the ttwo columns based on the load that eachh carries.
If
I the load on
n the right-han
nd column iss reduced to 5500 kips, the column doess not buckle
sidewayss, because thee moment at the
t top is now
w less than 6000Δ. To reachh the bucklingg condition,
a horizon
ntal force musst be applied at the top of tthe column, aas shown in Figure 8.17b. T
This force
Fiigure 8.17 C
Columns from
m Unbraced F
Frame
wiith Revised L
Loading.
3
Yura, J. A., “The
“
Effectivee Length of Collumns in Unbraced Frames,” Engineering JJournal, AISC,, Vol. 8, No. 2,, 1971, pp.
37–42.
4
LeMessurierr, W. J., “A Praactical Method of Second Ord
der Analysis,” Engineering JJournal, AISC, Vol. 14, No. 22, 1977, pp.
49–67.
3440
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
can result on
nly from action on the lefft column thatt is transmitteed through thhe beam. Equuilibrium
of the left co
olumn, shown
n in Figure 8..17a, requiress that an addiitional columnn load of 1000 kips be
applied to th
hat column in
n order for th
he load on thhe frame to bbe in equilibriium in this ddisplaced
position. Th
he total framee capacity iss still 700 kipps and the tootal second-oorder momennt is still
700Δ.
oad that an in
ndividual collumn can resiist is limited to that perm
mitted for
The maximum lo
the column in
i a braced frrame for whicch K = 1.0. Inn this examplee case, the lefft column couuld resist
400 kips and
d the right co
olumn 2400 kips.
k
This is aan increase off four times tthe load origiinally on
the column, because the effective
e
leng
gth factor for eeach column would be redduced from 2.0 to 1.0.
nal capacity of
o the left column is only with respectt to the bendiing axis. The column
The addition
would have the same cap
pacity about th
he other axis as it did prioor to reducingg the load on tthe right
column.
ne column to carry increassed load wheen another coolumn in the frame is
The ability of on
called upon to carry lesss than its critiical load for lateral buckliing is an impportant characcteristic.
d column to lean on a m
moment framee column, provided that tthe total
This allows a pin-ended
d on the framee can be carrieed by the rigidd frame.
gravity load
Fiigure 8.18 Frame
F
Used iin Example 88.6.
Figure 8.19 Nomiinal Wind Loaad, Snow Loaad, and Dead Load (Exampple 8.6).
Beam-Columns and Frame Behavior
EXAMPLE 8.6a
Moment Frame
Strength and
Stability by LRFD
Chapter 8
341
Goal:
Determine whether the structure shown in Figure 8.18 has sufficient
strength and stability to carry the imposed loads.
Given:
The frame shown in Figures 8.18 and 8.19 is similar to that in Example 8.1,
except that the in-plane stability and lateral load resistance is provided by
the moment frame action at the four corners. The exterior columns are
W8×40, and the roof girder is assumed to be rigid. Out-of-plane stability
and lateral load resistance is provided by X-bracing along column lines 1
and 4.
The loading is the same as that for Example 8.1: Dead Load = 50 psf, Snow
Load = 20 psf, Roof Live Load = 10 psf, and Wind Load = 20 psf
horizontal. Use A992 steel.
SOLUTION
Step 1:
The analysis of the frame for gravity loads as given for Example 8.1 will be
used. Because different load combinations may be critical, however, the
analysis results for nominal Snow and nominal Dead Load are given in
Figure 8.19b. The analysis results for nominal Wind Load acting to the left
are given in Figure 8.19c.
Step 2:
Determine the first-order forces and moments for the column on lines A-1.
For ASCE 7 load combination 3:
Pu = 1.2(15.8) + 1.6(6.33) + 0.5(0.710) = 29.1 + 0.355 = 29.5 kips
Mu = 1.2(20.5) + 1.6(8.20) + 0.5(32.0) = 37.7 + 16.0 = 53.7 ft-kips
For ASCE 7 load combination 4:
Pu = 1.2(15.8) + 0.5(6.33) + 1.0(0.710) = 22.1 + 0.710 = 22.8 kips
Mu = 1.2(20.5) + 0.5(8.20) + 1.0(32.0) = 28.7 + 32.0 = 60.7 ft-kips
Step 3:
Determine the total story gravity load acting on one frame.
Dead = 0.05 ksf (90 ft)(50 ft)/2 frames = 113 kips
Snow = 0.02 ksf (90 ft)(50 ft)/2 frames = 45.0 kips
Step 4:
Determine the second-order forces and moments for load combination 3.
Gravity loads will be assumed to yield the no-translation effects, and wind
load to yield the lateral translation effects.
From Step 2,
Pnt = 29.1 kips, Plt = 0.355 kips, Mnt = 37.7 ft-kips, Mlt = 16.0 ft-kips
For the W8×40,
A = 11.7 in.2, Ix = 146 in.4, rx = 3.53 in., rx/ry = 1.73
In the plane of the frame,
⎛ 0 ⎞
Cm = 0.6 = 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜
⎟ = 0.6
⎝ 37.7 ⎠
π2 EI
π2 (29,000)(146)
Pe1 = 2 x =
= 1130 kips
(16.0(12)) 2
Lc1
342
Chapter 8
Beam-Columns and Frame Behavior
and with Pr = Pnt + Plt = 29.1 + 0.355 = 29.5
Cm
0.6
B1 =
=
= 0.616 < 1.0
29.5
αPr
1−
1–
Pe1
1130
Therefore, use B1 = 1.0.
To determine the sway amplification, the total gravity load on the frame for
this load combination from Step 3 is
Pstory = 1.2(113) + 1.6(45.0) = 208 kips
A serviceability drift index of 0.003 is maintained under the actual wind
loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine the
sway amplification factor. If this limit is not met at the completion of the
design, the second-order effects must be recalculated.
The sway amplification is given by
B2 =
1
⎛ αP
1 − ⎜ story
⎝ Pe story
⎞
⎟
⎠
> 1.0
(AISC A-8-6)
and
Pe story = RM
HL
ΔH
(AISC A-8-7)
Since one third of the load is on the moment frame corner columns,
Equation A-8-8 gives
⎛1⎞
RM = 1 − 0.15 ⎜ ⎟ = 0.95
⎝ 3⎠
Thus, with α = 1.0 for LRFD, Equation A-8-6 becomes
1
1
B2 =
=
= 1.20
1.0 ( 208)
⎛ αPstory ⎛ Δ ⎞ ⎞
1− ⎜
(0.003)
⎜ ⎟⎟ 1−
0.95 ( 4.0 )
⎝ RM H ⎝ L ⎠ ⎠
Thus, the second-order force and moment are
Mr = 1.0(37.7) + 1.20(16.0) = 56.9 ft-kips
Pr = 29.1 + 1.20(0.355) = 29.5 kips
Step 5:
Determine whether the column satisfies the interaction equation.
Because the roof beam is assumed to be rigid in this example, use the
recommended design value of K = 2.0 from Figure 5.17 case f in the plane
of the frame, Lcx = 2.0(16.0) = 32.0 ft. Out of the plane of the frame, this is
a braced frame where K = 1.0; thus, Lcy = 16.0 ft.
Determine the critical buckling axis.
Beam-Columns and Frame Behavior
( Lcx )eff
=
Chapter 8
343
Lcx
32.0
=
= 18.5 ft > Lcy = 16.0 ft
rx / ry 1.73
Thus, from Manual Table 6-2, using Lc =( Lcx) eff = 18.5 ft,
φPn = 222 kips
and from Manual Table 6-2 with an unbraced length of Lb = 16 ft
φMnx = 128 ft-kips
Determine the appropriate interaction equation to use.
Pr
29.5
=
= 0.133 < 0.2
φPn 222
Therefore, use Equation H1-1b.
Pu
M
+ u ≤ 1.0
2φPn φM n
29.5
56.9
+
= 0.511 < 1.0
2(222) 128
Thus, the column is adequate for this load combination.
Step 6:
Determine the first-order forces and moments for load combination 4 with
the same assumption as to translation and no-translation effects. From Step
2.
Pnt = 22.1 kips, Plt = 0.710 kips, Mnt = 28.7 ft-kips, Mlt = 32.0 ft-kips
Step 7:
Determine the second-order forces and moments.
In the plane of the frame, as in Step 4,
⎛ 0 ⎞
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜
⎟ = 0.6
⎝ 28.7 ⎠
π2 EI
π2 (29, 000)(146)
Pe1 = 2 x =
= 1130 kips
(16(12)) 2
Lc1
and
B1 =
Cm
0.6
=
= 0.612 < 1.0
22.8
αP
1− r 1–
Pe1
1130
Therefore, use B1 = 1.0.
To determine the sway amplification, the total gravity load on the frame is
Pu = 1.2(113) + 0.5(45.0) = 158 kips
Again, a serviceability drift index of 0.003 is maintained under the actual
wind loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine
the sway amplification factor. As before, RM = 0.95 so
344
Chapter 8
Beam-Columns and Frame Behavior
B2 =
1
1
=
= 1.14
1.0 (158)
⎛ αPstory ⎛ Δ ⎞ ⎞
1– ⎜
(0.003)
⎜ ⎟⎟ 1–
0.95 ( 4.0 )
⎝ RM H ⎝ L ⎠ ⎠
Thus, the second-order force and moment are
Mr = 1.0(28.7) + 1.14(32.0) = 65.2 ft-kips
Pr = 22.1 + 1.14(0.710) = 22.9 kips
Step 8:
Determine whether the column satisfies the interaction equation.
Using the same strength values found in Step 5, determine the appropriate
interaction equation.
Pr
22.9
=
= 0.103 < 0.2
φPn 222
Therefore, use Equation H1-1b.
Pu
M
+ u ≤ 1.0
2φPn φM n
22.9
65.2
+
= 0.561 < 1.0
2(222) 128
Thus, the column is adequate for this load combination also.
Step 9:
The W8×40 is shown to be adequate for gravity and wind loads in
combination. Now, check to see that these columns have sufficient capacity
to brace the interior pinned columns for load combination 3, which will put
the greatest load on the gravity only columns.
Step 10:
For stability in the plane of the frame, using the Yura approach discussed in
Section 8.10, the total load on the structure is to be resisted by the four
corner columns; thus
Dead Load = 0.05 ksf (50 ft)(90 ft)/4 columns = 56.3 kips/column
Snow Load = 0.02 ksf (50 ft)(90 ft)/4 columns = 22.5 kips/column
Thus, for load combination 3
Pu = 1.2(56.3) + 1.6(22.5) +0.5(0.710) = 104 kips
Mu = 1.2(20.5) + 1.6(8.20) + 0.5(32.0) = 53.7 ft-kips
Step 11:
Determine the second-order amplification.
As before, for the length Lx = 16.0 ft, Pe1 = 1130 kips, and Cm = 0.6, the
second-order amplification for member effect is
Cm
0.6
B1 =
=
= 0.66 < 1.0
104
αPr
1−
1–
Pe1
1130
Therefore, use B1 = 1.0 and Pr = Pu = 104 kips.
Sway amplification will be the same as determined in step 4, since the
gravity load is the same; thus B2 = 1.20. Therefore
Mr = Mu = 1.0(37.7) + 1.20(16.0) = 56.9 ft-kips.
Beam-Columns and Frame Behavior
Step 12:
Chapter 8
345
Check the corner columns for interaction under these forces and moments.
As determined in Step 5 for in-plane buckling,
φPnx = 222 kips
φMnx = 128 ft-kips
Checking for the appropriate interaction equation,
Pu 104
=
= 0.468 > 0.2
φPn 222
Thus, use Equation H1-1a.
Pu 8 ⎛ M u x ⎞
+ ⎜
⎟ ≤ 1.0
φPn 9 ⎝ φM n x ⎠
104 8 ⎛ 56.9 ⎞
+ ⎜
⎟ = 0.864 < 1.0
222 9 ⎝ 128 ⎠
Thus, the W8×40 is adequate for both strength under combined load and
stability for supporting the gravity only columns.
EXAMPLE 8.6b
Moment Frame
Strength and
Stability by ASD
Goal:
Determine whether the structure shown in Figure 8.18 has sufficient
strength and stability to carry the imposed loads.
Given:
The frame shown in Figures 8.18 and 8.19 is similar to that in Example 8.1
except that the in-plane stability and lateral load resistance is provided by
the rigid frame action at the four corners. The exterior columns are W8×40,
and the roof girder is assumed to be rigid. Out-of-plane stability and lateral
load resistance is provided by X-bracing along column lines 1 and 4.
The loading is the same as that for Example 8.1: Dead Load = 50 psf, Snow
Load = 20 psf, Roof Live Load = 10 psf, and Wind Load = 20 psf
horizontal. Use A992 steel.
SOLUTION
Step 1:
The analysis of the frame for gravity loads as given for Example 8.1 will be
used. Because different load combinations may be critical, however, the
analysis results for nominal Snow and nominal Dead Load are given in
Figure 8.19b. The analysis results for nominal Wind Load acting to the left
are given in Figure 8.19c.
Step 2:
Determine the first-order forces and moments for the column on lines A-1.
For ASCE 7 load combination 3:
Pa = (15.8) + (6.33) = 22.1 kips
Ma = (20.5) + (8.20) = 28.7 ft-kips
For ASCE 7 load combination 6:
Pa = (15.8) + 0.75(6.33) + 0.75(0.6(0.710)) = 20.9 kips
Ma = (20.5) + 0.75(8.20) + 0.75(0.6(32.0)) = 41.1 ft-kips
Step 3:
Determine the total story gravity load acting on one frame.
Dead = 0.05 ksf (90 ft)(50 ft)/2 frames = 113 kips
Snow = 0.02 ksf (90 ft)(50 ft)/2 frames = 45.0 kips
346
Chapter 8
Beam-Columns and Frame Behavior
Step 4:
Determine the second-order forces and moments for load combination 3.
Gravity loads will be assumed to yield the no-translation effects. With no
wind load, there will be no lateral translation effects; thus
From Step 2:
Pnt = 22.1 kips, Plt = 0 kips, Mnt = 28.7 ft-kips, Mlt = 0 ft-kips
For the W8×40:
A = 11.7 in.2, Ix = 146 in.4, rx = 3.53 in., rx/ry = 1.73
In the plane of the frame:
⎛ 0 ⎞
Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜
⎟ = 0.6
⎝ 28.7 ⎠
π2 EI
π2 (29,000)(146)
Pe1 = 2 x =
= 1130 kips
(16.0(12)) 2
Lc1
and
Cm
0.6
=
= 0.619 < 1.0
1.6(22.1)
αPr
1−
1–
Pe1
1130
Therefore, use B1 = 1.0.
B1 =
To determine the sway amplification, even though there are no lateral
translation forces or moments for this combination, the total gravity load on
the frame for this load combination from Step 3 is
Pstory = (113) + (45.0) = 158 kips
A serviceability drift index of 0.003 is maintained under the actual wind
loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine the
sway amplification factor. If this limit is not met at the completion of the
design, the second-order effects must be recalculated.
The sway amplification is given by
B2 =
1
⎛ αP
1 − ⎜ story
⎝ Pe story
⎞
⎟
⎠
> 1.0
(AISC A-8-6)
and
Pe story = RM
HL
ΔH
(AISC A-8-7)
Since one third of the load is on the moment frame corner columns,
Equation A-8-8 gives
⎛1⎞
RM = 1 − 0.15 ⎜ ⎟ = 0.95
⎝ 3⎠
Thus, with α = 1.6 for ASD, Equation A-8-6 becomes
Beam-Columns and Frame Behavior
B2 =
Chapter 8
347
1
1
=
= 1.25
⎛ αPstory ⎛ Δ ⎞ ⎞ 1 – 1.6(158) (0.003)
1– ⎜
⎜ ⎟⎟
0.95 ( 4.0 )
⎝ RM H ⎝ L ⎠ ⎠
Thus, the second-order force and moment are
Mr = 1.0(28.7) + 1.25(0) = 28.7 ft-kips
Pr = 22.1 + 1.25(0) = 22.1 kips
Step 5:
Determine whether the column satisfies the interaction equation.
Because the roof beam is assumed to be rigid in this example, use the
recommended design value of K = 2.0 from Figure 5.17 case f in the plane
of the frame, Lcx = 2(16.0) = 32.0 ft. Out of the plane of the frame, this is a
braced frame where K = 1.0; thus, Lcy = 16.0 ft.
Determining the critical buckling axis.
L
32
= 18.5 ft > Lcy = 16.0 ft
( Lcx )eff = cx =
rx /ry 1.73
Thus, from Manual Table 6-2, using Lc = (Lcx)eff = 18.5 ft,
Pn/Ω = 148 kips
and from Manual Table 6-2, with an unbraced length of Lb = 16 ft,
Mnx/Ω = 84.9 ft-kips
Determine the appropriate interaction equation to use.
Pr
22.1
=
= 0.149 < 0.2
Pn /Ω 148
Therefore, use Equation H1-1b.
Pa
Ma
+
≤ 1.0
2 ( Pn Ω ) ( M n Ω )
22.1
28.7
+
= 0.413 < 1.0
2(148) 84.9
Thus, the column is adequate for this load combination.
Step 6:
Determine the first-order forces and moments for load combination 6.
Gravity loads will be assumed to yield the no-translation effects, and wind
load will yield the lateral translation effects. From Step 2.
Pnt = 20.5 kips, Plt = 0.320 kips, Mnt = 26.7 ft-kips, Mlt = 14.4 ft-kips
Step 7:
Determine the second-order forces and moments.
In the plane of the frame, as in Step 4,
348
Chapter 8
Beam-Columns and Frame Behavior
⎛ 0 ⎞
Cm = 0.6 − 0.4 ( M 1 M 2 ) 0.6 – 0.4 ⎜
⎟ = 0.6
⎝ 26.7 ⎠
π2 EI
π2 (29,000)(146)
= 1130 kips
Pe1 = 2 x =
Lc1
(16(12)) 2
and
Cm
0.6
=
= 0.618 < 1.0
1.6(20.8)
αP
1− r 1–
Pe1
1130
Therefore, use B1 = 1.0.
B1 =
To determine the sway amplification, the total gravity load on the frame is
Pstory = (113) + 0.75(45.0) = 147 kips
Again, a serviceability drift index of 0.003 is maintained under the actual
wind loads. Therefore, H = 4.0 kips and Δ/L = 0.003 is used to determine
the sway amplification factor. As before, RM = 0.95 so
1
1
B2 =
=
= 1.23
⎛ αPstory ⎛ Δ ⎞ ⎞ 1 – 1.6(147) (0.003)
1– ⎜
⎜ ⎟⎟
0.95 ( 4.0 )
⎝ RM H ⎝ L ⎠ ⎠
Thus, the second-order force and moment are
Mr = 1.0(26.7) + 1.23(14.4) = 44.4 ft-kips
and, adding in the lateral load effect amplified by B2,
Pr = 20.5 + 1.23(0.320) = 20.9 kips
Step 8:
Determine whether the column satisfies the interaction equation.
Using the same values found in Step 5, determine the appropriate
interaction equation.
Pr
20.9
=
= 0.141 < 0.2
Pn /Ω 148
Therefore, use Equation H1-1b.
Pa
Ma
+
≤ 1.0
2 ( Pn Ω ) ( M n Ω )
20.9
44.4
+
= 0.594 < 1.0
2(148) 84.9
Thus, the column is adequate for this load combination also.
Step 9:
The W8×40 is shown to be adequate for gravity and wind loads in
combination. Now, check to see that these columns have sufficient capacity
to brace the interior pinned columns for gravity load only. This load
combination puts the greatest load in the gravity only columns.
Beam-Columns and Frame Behavior
Step 10:
Chapter 8
349
For stability in the plane of the frame, using the Yura approach discussed in
Section 8.10, the total load on the structure is to be resisted by the four
corner columns; thus
Dead Load = 0.05 ksf (50 ft)(90 ft)/4 columns = 56.3 kips
Snow Load = 0.02 ksf (50 ft)(90 ft)/4 columns = 22.5 kips
Thus, for load combination 3,
Pa = (56.3) + (22.5) = 78.8 kips
Ma = (20.5) + (8.20) = 28.7 ft-kips
Step 11:
Determine the second-order amplification
As before, for the length Lx = 16.0 ft, Pe1 = 1130 kips and Cm = 0.6, the
second-order amplification for member effect is
Cm
0.6
B1 =
=
= 0.675 < 1.0
1.6(78.8)
αP
1− r 1–
P1
1130
Therefore, use B1 = 1.0 and Pr = Pa = 78.8 kips
Sway amplification will be the same as determined in step 4, since the
gravity load is the same; thus, B2 =1.25. Therefore
M r = 1.0 ( 28.7 ) + 1.25 ( 0 ) = 28.7 ft-kips
Step 12:
Check the corner columns for interaction under this force and moment.
As determined in Step 5 for in-plane buckling,
Pn/Ω = 148 kips
Mnx/Ω = 84.9 ft-kips
Checking for the appropriate interaction equation,
Pu
78.8
=
= 0.532 > 0.2
Pn /Ω 148
Thus, use Equation H1-1a.
Pa
8 ⎛ Max ⎞
+ ⎜
≤ .0
Pn /Ω 9 ⎝ M n x /Ω ⎟⎠
78.8 8 ⎛ 28.7 ⎞
+ ⎜
⎟ = 0.833 < 1.0
148 9 ⎝ 84.9 ⎠
Thus, the W8×40 is adequate for both strength under combined load and
stability for supporting the gravity only columns.
8.11 PARTIALLY RESTRAINED FRAMES
The beams and columns in the frames considered up to this point have all been connected with
moment-resisting fully restrained (FR) connections or simple pinned connections. These latter
simple connections are defined in Specification Section B3.4a. Partially restrained connections,
3550
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figure 8.20 Moment Rotation
R
Curves for Uniform
mly Loaded B
Beam and Tyypical Connecctions.
defined in Specification
Sp
Section
S
B3.4b
b along with F
FR connectioons, have histoorically been referred
to as semirig
gid connectio
ons. When theese PR conneections are inccluded as the connecting eelements
in a structural frame, they
y influence bo
oth the strenggth and stabiliity of the struccture.
fore considerin
ng the partiallly restrained frame, it willl be helpful too look at the ppartially
Befo
restrained beam.
b
The rellationship between the ennd moment aand end rotatiion for a sym
mmetric,
uniformly lo
oaded prismattic beam can be obtained ffrom the welll-known sloppe deflection eequation
as
E
EI θ WL
+
M = –2
(8.12)
L
12
on is plotted in
n Figure 8.20
0a and labeledd as the beam line.
This equatio
F
Figure 8.21 Beam Line aand
C
Connection Cuurves.
Beam
m-Columns annd Frame Behhavior
Chappter 8
351
Fiigure 8.22 IInfluence of tthe PR
Coonnection on the Maximum
m Positive
annd Negative M
Moments of a Beam.
All
A PR connections exhibitt some rotatioon as a result of an appliedd moment. Thhe momentrotation characteristic
c
s of these con
nnections are the key to deetermining thee type of connnection and
thus the behavior off the structuree. Moment-rootation curvees for three ggeneric connnections are
shown in
n Figure 8.20
0b and are laabeled rigid, ssimple, and P
PR. Numerouus research sttudies have
been con
nducted in an
n effort to id
dentify the m
moment-rotatioon curves forr real connecctions. Two
compilations of these curves have been
b
publisheed.5,6
The
T relationsh
hip between the
t moment-rrotation charaacteristics of a connection and a beam
can be seeen by plottin
ng the beam line and connnection curvve together, aas shown in F
Figure 8.21.
Equilibriium is obtained when th
he beam linne and the cconnection ccurve intersect. Normal
engineeriing practice treats
t
connecctions capablee of resistingg at least 90 ppercent of the fixed-end
moment (FEM) as rig
gid and those capable of reesisting no m
more than 20 percent of thhe fixed-end
moment as simple. All
A connection
ns that exhibitt an ability too resist momeent between tthese limits
must be treated as partially restraained connecctions, accounnting for theeir true moment-rotation
characterristics.
The
T influencee of the PR co
onnection onn the maximum
m positive annd negative m
moments on
the beam
m is seen in Figure
F
8.22. Here,
H
the ratioo of positive or negative m
moment to thhe fixed-end
moment (FEM) is plotted against th
he ratio of beeam stiffness, EI/L, to a linnear connectioon stiffness,
M/θ. Thee moment forr which the beam
b
must bbe designed raanges from 00.75 times the fixed-end
moment to 1.5 times the
t fixed-end moment, deppending on the stiffness off the connectioon.
When
W
PR con
nnections aree used to connnect beams aand columns to form PR frames, the
analysis becomes mu
uch more com
mplex. The reesults of num
merous studiess dealing withh this issue
have beeen reported. Although
A
som
me practical deesigns have bbeen carried oout, widespreaad practical
design of PR frames is not comm
mon. In addittion to the problems assoociated with modeling a
particular connection,, the question
n of loading ssequence arisses. Because real, partiallyy restrained
connectio
ons behave nonlinearly, th
he sequence oof applied loaads influencess the structuraal response.
The apprroach to load application may
m have morre significancce than the acccuracy of the connection
model ussed in the anaalysis.
5
Goverdhan, A.
A V., A Collecction of Experiimental Momen
nt Rotation Cuurves and Evaluuation of Preddiction Equationns for SemiRigid Connecctions, Master of
o Science Theesis, Vanderbilt University, N
Nashville, TN, 1983.
6
Kishi, N., an
nd Chen, W. F,. Data Base off Steel Beam-to
o-Column Connnections, CE-S
STR-86-26, Weest Lafayette, IIN: Purdue
University, School of Engin
neering, 1986.
3552
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Alth
hough a comp
plete, theoretiical analysis oof a partially restrained fraame may currrently be
beyond the scope of norrmal engineerring practice,, a simplifiedd approach exxists that is nnot only
well within the scope of practice,
p
but also commonnly carried ouut in everydayy design and hhas been
for over half a century. This
T approach
h can be referrred to as Fleexible Momennt Connectionns. It has
historically been called Type 2 with Wind. The F
Flexible Mom
ment Connecction approacch relies
heavily on the nonlinearr moment-rottation behaviior of the PR
R connectionn although thhe actual
curve is not used. In addiition, it relies on a phenom
menon called sshake-down, which showss that the
connection, although ex
xhibiting nonlinear behaviior initially, behaves lineearly after a limited
number of applications
a
of lateral load.7
The moment-rotaation curve for
fo a typical P
PR connectioon is shown inn Figure 8.233a along
with the beaam line for a uniformly lo
oaded beam. T
The point labbeled 0 represents equilibrrium for
the applied gravity loads. The applicaation of wind load producees moments aat the beam eends that
add to the gravity
g
momeent at the leew
ward end of tthe beam andd subtract from
m the windw
ward end.
Because mo
oment at the windward
w
end
d is being rem
moved, the connnection behaaves elasticallly with a
stiffness clo
ose to the oriiginal connecction stiffnesss, whereas att the leewardd end, the connnection
continues to
o move along the nonlineaar connection curve. Pointss labeled 1 annd 1′ in Figure 8.23b
represent eq
quilibrium und
der the first ap
pplication of wind to the fr
frame.
Wheen the wind load
l
is remov
ved, the connnection movess from pointss 1 and 1′ to points 2
and 2′, as sh
hown in Figu
ure 8.23c. Th
he next applic ation of a winnd load that iss larger than the first
Figure 8.233 Moment-R
Rotation
Curves Show
wing Shake-D
Down
7
G
Geschwindner, L.
L F., and Disq
que, R. O., “Fleexible Momentt Connections ffor Unbraced F
Frames Subject to Lateral Forrces—A
Reeturn to Simpliccity.” Engineerring Journal, AISC,
A
Vol. 42, No. 2, 2005, ppp 99–112.
Beam-Columns and Frame Behavior
Chapter 8
353
and in the opposite direction will see the connection behavior move to points 3 and 3′. Note that
on the windward side, the magnitude of this applied wind moment dictates whether the
connection behaves linearly or follows the nonlinear curve, as shown in Figure 8.23d. Removal of
this wind load causes the connection on one end to unload and on the other end to load, both
linearly as shown in Figure 8.23e. Any further application of wind load, less than the maximum
already applied, will see the connection behave linearly. In addition, the maximum moment on
the connection is still close to that applied originally from the gravity load. Thus, the condition
described in Figure 8.23f shows that shake-down has taken place and the connection now behaves
linearly for both loading and unloading.
The design procedure used to account for this shake-down is straight forward. All beams
are designed as simple beams using the appropriate load combinations. This assures that the
beams are adequate, regardless of the actual connection stiffness, as was seen in Figure 8.22.
Wind load moments are determined through a modified portal analysis where the leeward column
is assumed not to participate in the lateral load resistance. Connections are sized to resist the
resulting moments, again for the appropriate load combinations. In addition, it is particularly
important to provide connections that have sufficient ductility to accommodate the large rotations
that will occur, without overloading the bolts or welds under combined gravity and wind.
Columns must be designed to provide frame stability under gravity loads as well as
gravity plus wind. The columns may be designed using the approach that was presented for
columns in moment frames, but with two essential differences from the conventional rigid frame
design:
1.
Because the gravity load is likely to load the connection to its plastic moment
capacity, the column can be restrained only by a girder on one side and this
girder will act as if it is pinned at its far end. Therefore, in computing the girder
stiffness rotation factor, Ig/Lg, for use in the effective length alignment chart, the
girder length should be doubled.
2.
One of the external columns, the leeward column for the wind loading case,
cannot participate in frame stability, because it will be attached to a connection
that is at its plastic moment capacity. The stability of the frame may be assured,
however, by designing the remaining columns to support the total frame load.
For the exterior column, the moment in the beam to column joint is equal to the capacity
of the connection. It is sufficiently accurate to assume that this moment is distributed one-half to
the upper column and one-half to the lower column. For interior columns, the greatest realistically
possible difference in moments resulting from the girders framing into the column should be
distributed equally to the columns above and below the joint.
EXAMPLE 8.7a
Column Design
with Flexible Wind
Connections by
LRFD
Goal:
Select girders and columns for a building with flexible wind connections.
Given:
An intermediate story of a three-story building is given in Figure 8.24.
Story height is 12 ft. The frame is braced in the direction normal to that
shown. Use the LRFD provisions and A992 steel.
3554
Chapter 8
Beam-C
Columns and Frame
F
Behav
vior
Figure 8.24 Intermediaate Story of a Three-Story Building (Exxample 8.7).
SO
OLUTION
Step
p 1:
Deterrmine the requ
uired forces aand moments for the load ccombination
1.2D
D + 0.5L + 1..0W
The loads shown in
i Figure 8.244 are the codde-specified nnominal loadss. The
requirred forces aree calculated uusing tributaryy areas as folllows.
Graviity loads on exterior colum
mns.
1.2D = 1.2
2(25 kips + 0..75 kips/ft (155 ft)) = 43.5 kkips
0.5L = 0.5
5 (75 kips + 2 .25 kips/ft (155 ft)) = 54.4 kkips
Total =97.9 kkips
Graviity loads on in
nterior colum
mns
1.2D = 1.2
2 (50 kips + 00.75 kips/ft (30 ft)) = 87.0 kkips
0.5L = 0.5
5 (150 kips + 22.25 kips/ft (330 ft)) = 109 kips
Total= 196 kips
Step
p 2:
Graviity load on girrders for the w
worst case 1.22D + 1.6L:
1.2D = 1.2
2 (0.75 kips/ftt (30 ft)) = 277.0 kips
1.6L = 1.6
6 (2.25 kips/ftt (30 ft)) = 1008 kips
Total = 1335 kips
Desig
gn the girder for
f the simplee beam momeent assuming full lateral suupport
using Manual Table 3-2 or 6-2.
M u = WL 8 = 135 ( 30.0 ) 8 = 506 ft-kipss
Beam-Columns and Frame Behavior
Therefore use
Step 3:
Chapter 8
355
W21×62 (φMn = 540 ft-kips, Ix = 1330 in.4)
Design the columns for the gravity load on the interior column using
Manual Table 4-1.
For buckling out of the plane in a braced frame, K = 1.0 and Lcy = 12.0
Thus, with Pu = 196 kips ,try
W14×43, (φPn = 371 kips, Ix = 428 in.4, rx/ry = 3.08)
Step 4:
To check the column for stability in the plane, determine the effective
length factor from the alignment chart with
⎛ 428 ⎞
2⎜
⎟
Σ ( I L )c
12.0 ⎠
Gtop = Gbottom =
= ⎝
= 3.21
Σ ( I L ) g ⎛ 1330 ⎞
⎜
⎟
⎝ 2(30.0) ⎠
Note that only one beam is capable of restraining the column and that the
beam is pinned at its far end; thus the effective beam length is taken as
twice its actual length.
Considering the stress in the column under load, the stiffness reduction
factor can be determined.
Pu A = 196 12.6 = 15.6 ksi
Thus, since Pu/A = 15.6 < 0.5Fy, the stiffness reduction factor from Manual
Table 4-13 is τb = 1.00. The stiffness ratio then remains
Gtop = Gbottom = 3.21
which yields, from the alignment chart, Figure 5.20
K = 1.87
Step 5:
Determine the effective length in the plane of bending.
L
1.87(12.0)
= 7.29 ft
( Lcx )eff = cx =
rx /ry
3.08
Step 6
Determine the column compressive strength from Manual Table 4-1 or 6-2
with Lc = 7.29 ft.
φPn = 484 kips
Step 7:
Determine the second-order moment.
The applied wind moment is Mu = 1.0(6.0)(12.0) = 72.0 ft-kips and the
applied force is Pu = 196 kips.
Considering all the moment as a translation moment and using
Commentary equation C-A-8-1
356
Chapter 8
Beam-Columns and Frame Behavior
Pe story =
π2 EI
( K2 L )
2
=
π2 (29,000)(428)
= 1690 kips
(1.87(12.0)(12)) 2
Therefore, for all three columns,
1
1
B2 =
=
= 1.13
1.0 ( 3(196) )
⎛ αPstory ⎞
1− ⎜
⎟ 1 – 3(1690)
⎝ Pe story ⎠
and
Mr = 1.13 (72.0) = 81.4 ft-kips
Step 8:
Determine whether the column satisfies the interaction equation
Pu 196
=
= 0.405 > 0.2
φPn 484
Therefore, use Equation H1-1a, φMn = 222, from Manual Table 3-10 or 62, which results in
Pu 8 ⎛ M u ⎞
+ ⎜
⎟ ≤ 1.0
φPn 9 ⎝ φM n ⎠
8 ⎛ 81.4 ⎞
0.405 + ⎜
⎟ = 0.731 < 1.0
9 ⎝ 222 ⎠
This indicates that the W14×43 is adequate for stability. The members can
then be used as a starting point in a more rigorous analysis.
EXAMPLE 8.7b
Column Design
with Flexible Wind
Connections by
ASD
Goal:
Select girders and columns for a building with flexible wind connections.
Given:
An intermediate story of a three-story building is given in Figure 8.24.
Story height is 12 ft. The frame is braced in the direction normal to that
shown. Use the ASD provisions and A992 steel.
Step 1:
Determine the required forces and moments for the load combination
D + 0.75L +0.75(0.6W)
The loads shown in Figure 8.24 are the code-specified nominal loads. The
required forces are calculated using tributary areas as follows.
Gravity loads on exterior columns
D = (25 k + 0.75 k/ft (15 ft)) = 36.3 kips
0.75L = 0.75 (75 k + 2.25 k/ft (15 ft)) = 81.6 kips
Total = 118 kips
Gravity loads on interior columns
D = (50 k + 0.75 k/ft (30 ft)) = 72.5 kips
0.75L = 0.75 (150 k + 2.25 k/ft (30 ft)) = 163 kips
Total = 236 kips
Gravity load on girders for the worst case, D + L
D = (0.75 k/ft (30 ft)) = 22.5 kips
L = (2.25 k/ft (30 ft)) = 67.5 kips
Total = 90.0 kips
Beam-Columns and Frame Behavior
Step 2:
357
Design the girder for the simple beam moment assuming full lateral support
using Manual Table 3-2 or 6-2.
M a = WL 8 = 90.0 ( 30.0 ) 8 = 338 ft-kips
Therefore use
Step 3:
Chapter 8
W21×62 (Mn/Ω = 359 ft-kips, Ix = 1330 in.4)
Design the columns for gravity load on the interior column using Manual
Table 4-1 or 6-2.
For buckling out of the plane in a braced frame, K = 1.0 and Lcy = 12.0
Thus, with Pa = 236 kips
try W14×43 (Pn/Ω = 247 kips, Ix = 428 in.4, rx/ry = 3.08)
Step 4:
To check the column for stability in the plane, determine the effective
length factor from the alignment chart with
⎛ 428 ⎞
2⎜
⎟
Σ ( I L )c
12.0 ⎠
= ⎝
= 3.21
Gtop = Gbottom =
Σ ( I L ) g ⎛ 1330 ⎞
⎜
⎟
⎝ 2(30.0) ⎠
Note that only one beam is capable of restraining the column and that that
beam is pinned at its far end; thus the effective beam length is taken as
twice its actual length.
Considering the stress in the column under load, the stiffness reduction
factor can be determined.
Pa 236
=
= 18.7 ksi
A 12.6
Thus, from the Manual Table 4-13, the stiffness reduction factor τb = 0.960.
The inelastic stiffness ratio then becomes
Gtop = Gbottom = 0.960(3.21) = 3.08
which yields, from the alignment chart, Figure 5.20
K = 1.84
Step 5:
Determine the effective length in the plane of bending.
L
1.84(12.0)
= 7.17 ft
( Lcx )eff = cx =
rx /ry
3.08
Step 6:
Determine the column compressive strength from Manual Table 4-1 or 6-2
with Lc = 7.17 ft.
Pn/Ω = 324 kips
358
Chapter 8
Beam-Columns and Frame Behavior
Step 7:
Determine the second-order moment.
The applied wind moment is Ma = 0.75(0.6)(6.0(12.0)) = 32.4 ft-kips and
the applied force is Pa = 236 kips.
Considering all the moment as a translation moment and using
Commentary equation C-A-8-1
π2 EI
π2 (29,000)(428)
Pe story =
=
= 1740 kips
2
2
( K 2 L ) (1.84(12)(12))
αPa = 1.6(236) = 378 kips
Therefore, for all three columns,
1
1
B2 =
=
= 1.28
3(378)
αPstory
1–
1−
3(1740)
Pe story
and
Mr = 1.28(32.4) = 41.5 ft-kips
Step 8:
Determine whether the column satisfies the interaction equation
Pr
236
=
= 0.728 < 0.2
Pn /Ω 324
Therefore, use Equation H1-1a, Mn/Ω = 148, from Manual Table 3-10 or 62, which results in
Pa
8 ⎛ Ma ⎞
+ ⎜⎜
⎟ ≤ 1.0
( Pn Ω ) 9 ⎝ ( M n Ω ) ⎟⎠
8 ⎛ 41.5 ⎞
0.728 + ⎜
⎟ = 0.98 < 1.0
9 ⎝ 148 ⎠
This indicates that the W14×43 is adequate for stability. These members
can then be used as a starting point in a more rigorous analysis.
After an acceptable column is selected, the lateral displacement of the structure must be
checked. Coverage of drift in wind moment frames is beyond the treatment intended here, but is
covered in the Geschwindner and Disque paper already referenced.
8.12 STABILITY BRACING DESIGN
Braces in steel structures are used to reduce the effective length of columns, reduce the unbraced
length of beams, and provide overall structural stability. The discussion of columns in Chapter 5
showed how braces could be effective in reducing effective length and thereby increasing column
strength. Chapter 6 demonstrated how the unbraced length of a beam influenced its strength, and
earlier in this chapter the influence of sway on the stability of a structure was discussed. Every
case assumed that the given bracing requirements were satisfied, but nothing was said about the
Beam
m-Columns annd Frame Behhavior
Chappter 8
359
Figure 8.25
8
Definitiions of Bracin
ng Types. Coppyright © Ameerican Institutee of Steel Consstruction,
Inc. Repriinted with Perm
mission. All rig
ghts reserved.
strength or stiffness of the required
d braces. For cases when bbraces are nott specifically included in
the secon
nd-order anaalysis, design
n of braces w
will follow thhe provisionss of Appendiix 6 of the
Specifica
ation. Append
dix 6 treats bracing for coolumns and beams similarlly, although tthe specific
requirem
ments are diffeerent. Two typ
pes of braces are defined: ppoint bracingg and panel brracing
Point
P
bracing controls the movement oof a point on the member without interraction with
any adjacent braced points.
p
Thesee braces woulld be attachedd to the mem
mber and thenn to a fixed
support, such as the ab
butment show
wn in Figure 88.25b.
Panel
P
bracing
g relies on oth
her braced poiints of the struucture to provvide support. A diagonal
brace wiithin a frame would be a panel brace, as shown in Figure 8.25aa. In this casse, the axial
deformattion of the diiagonal bracee is a functioon of the dispplacement att each end off the brace.
Because the horizonttal strut is ussually a part of a very sttiff floor systtem that has significant
strength in its plane, th
he strength an
nd stiffness oof the diagonaal element usuually controlss the overall
behaviorr of this braced system.
The
T brace req
quirements of the Specificcation are inttended to enaable the mem
mbers being
designed
d to reach theeir maximum
m strength bassed on the leength betweenn bracing points and an
effective length factorr, K = 1.0. A brace
b
has twoo requirementts: strength annd stiffness. A brace that
is inadeq
quate in eitheer of these reespects is nott sufficient too enable the member it iss bracing to
perform as it was desiigned.
8.12.1
Collumn Bracin
ng
For colum
mn panel braccing, the requ
uired shear strrength of the bbracing systeem is
Vbr = 0.005Pr
and the required
r
shearr stiffness is
1 ⎛ 2P
βbr = ⎜ r
φ ⎝ Lbrb
⎞
⎟ (LRFD)
⎠
φ = 0.75
5 (LRFD)
⎛ 2P ⎞
βbr = Ω ⎜ r ⎟ (ASD)
⎝ Lbr ⎠
Ω = 2..00 (ASD)
AISC A-6-1)
(A
(A
AISC A-6-2)
360
Chapter 8
Beam-Columns and Frame Behavior
where
Lbr = unbraced length of the panel under consideration
Pr = required strength of the column within the panel under consideration for ASD or
LRFD as appropriate for the design method being used.
For a column point brace, the required brace strength is
Pbr = 0.01 Pr
and the required brace stiffness is
1 ⎛ 8P
βbr = ⎜ r
φ ⎝ Lbr
⎞
⎟ (LRFD)
⎠
φ = 0.75 (LRFD)
(AISC A-6-3)
⎛ 8P ⎞
βbr = Ω ⎜ r ⎟ (ASD)
⎝ Lbr ⎠
Ω = 2.00 (ASD)
(AISC A-6-4)
where
Lbr = laterally unbraced length adjacent to the point brace
Pr = required strength for ASD or LRFD as appropriate for the design method being used.
It should be noted that the requirements for point braces are significantly greater than those for
panel braces. Thus, if a panel bracing system can be developed, it has the potential to be the more
economical approach.
8.12.2
Beam Bracing
For a beam panel brace, the required shear strength of the bracing system is
⎛M C ⎞
Vbr = 0.01⎜ r d ⎟
⎝ ho ⎠
and the required panel brace stiffness is
1 ⎛ 4M r Cd ⎞
βbr = ⎜
(LRFD)
φ ⎝ Lbr ho ⎟⎠
φ = 0.75 (LRFD)
⎛ 4M r Cd
βbr = Ω ⎜
⎝ Lbr ho
Ω = 2.00 (ASD)
(AISC A-6-5)
⎞
⎟ (ASD)
⎠
(AISC A-6-6)
where
ho = distance between flange centroids
Cd = 1.0 for single curvature bending and 2.0 for the brace closest to the inflection point
for double curvature bending
Lbr = laterally unbraced length within the panel under consideration
Mr = the largest required flexural strength of the beam within the unbraced lengths
adjacent to the point being braced
For a beam point brace, the required strength of the brace is
Pbr = 0.02MrCd/ho
and the required brace stiffness is
1 ⎛ 10M r Cd ⎞
βbr = ⎜
(LRFD)
φ ⎝ Lbr ho ⎟⎠
φ = 0.75 (LRFD)
⎛ 10M r Cd
βbr = Ω ⎜
⎝ Lbr ho
Ω = 2.00 (ASD)
(AISC A-6-7)
⎞
⎟ (ASD)
⎠
where
ho = distance between flange centroids
Cd = 1.0 for single curvature and 2.0 for double curvature as above
(AISC A-6-8)
Beam-Columns and Frame Behavior
Chapter 8
361
Lbr = laterally unbraced length adjacent to the point brace
Mr = the largest required flexural strength of the beam within the unbraced lengths
adjacent to the point being braced
As with column bracing, the requirements for point braces are greater than those for panel braces.
8.12.3
Frame Bracing
Frame bracing and column bracing are accomplished by the same panel and point braces and may
be designed using the same stiffness and strength equations. However, the most direct approach
to bracing design for frames is to include the braces in the model when a second-order analysis is
carried out. When that is the case, the provisions of Appendix 6 do not need to be checked.
EXAMPLE 8.8a
Bracing Design by
LRFD
Goal:
Determine the required bracing for a braced frame to provide stability for
the gravity load.
Given:
Using the LRFD requirements, select a rod to provide the point bracing
shown in the center panel of the three-bay frame of Figure 8.9a to provide
stability for a total gravity dead load of 113 kips and live load of 45 kips.
SOLUTION
Step 1:
Determine the required brace stiffness for gravity load.
For the gravity load, the required brace stiffness is based on 1.2D + 1.6L.
Pr = 1.2 (113) + 1.6 ( 45.0 ) = 208 kips
and from Equation A-6-4
1 ⎛ 8P
βbr = ⎜ r
φ ⎝ Lbr
Step 2:
1 ⎛ 8(208) ⎞
⎞
⎟ = 0.75 ⎜ 16.0 ⎟ = 139 kips/ft
⎝
⎠
⎠
Determine the required brace area based on required stiffness and
accounting for the angle of the brace.
Based on the geometry of the brace from Figure 8.9, where θ is the angle of
the brace with the horizontal and Lr = 34.0 ft is the length of the brace,
A E
βbr = br cos 2 θ = 139 kips/ft
Lr
This results in a required brace area
β L
139(34.0)
= 0.209 in.2
Abr = br 2r =
2
E cos θ
30
⎛ ⎞
29,000 ⎜ ⎟
⎝ 34 ⎠
Step 3:
Determine the required brace force for gravity load. The required horizontal
brace force for a point brace given by Equation A-6-3 is
Pbr = 0.01Pr = 0.01( 208) = 2.08 kips
which gives a force in the member of
Pbr ( angle ) = 2.08 ( 34 30 ) = 2.36 kips
362
Chapter 8
Beam-Columns and Frame Behavior
and a required area, assuming Fy = 36 ksi for a rod, of
Pbr ( angle)
2.36
Abr =
=
= 0.0728 in.2
φFy
0.9(36)
Step 4:
For the dead plus live load case,
Amin = 0.209 in.2
Step 5:
Select a rod to meet the required area for the controlling case of stiffness
for the dead plus live load case where Amin = 0.209 in.2
use a 5/8-in. rod with A = 0.307 in.2
EXAMPLE 8.8b
Bracing Design by
ASD
Goal:
Determine the required bracing for a braced frame to provide stability for
the gravity load.
Given:
Using the ASD requirements, select a rod to provide the point bracing
shown in the center panel of the three-bay frame of Figure 8.9a to provide
stability for a total gravity dead load of 113 kips and live load of 45 kips.
SOLUTION
Step 1:
Determine the required brace stiffness for gravity load.
For the gravity load, the required brace stiffness is based on D + L.
Pr = 113 + 45.0 = 158 kips
and from Equation A-6-4
⎛ 8P
βbr = Ω ⎜ r
⎝ Lbr
Step 2:
⎞
⎛ 8(158) ⎞
⎟ = 2.00 ⎜ 16.0 ⎟ = 158 kips/ft
⎝
⎠
⎠
Determine the required brace area based on required stiffness and
accounting for the angle of the brace.
Based on the geometry of the brace from Figure 8.9, where θ is the angle of
the brace with the horizontal and Lr =34.0 ft is the length of the brace.
A E
βbr = br cos 2 θ = 158 kips/ft
Lr
This results in a required brace area
β L
158(34.0)
Abr = br 2r =
= 0.238 in.2
2
E cos θ
⎛ 30 ⎞
29,000 ⎜ ⎟
⎝ 34 ⎠
Step 3:
Determine the required brace force for gravity load.
The required horizontal brace force for a point brace given by Equation A6-3 is
Pbr = 0.01Pr = 0.01(158 ) = 1.58 kips
Beam-Columns and Frame Behavior
Step 4:
Step 5:
Chapter 8
363
which gives a force in the member of
Pbr ( angle ) = 1.58 ( 34 30 ) = 1.79 kips
and a required area, assuming Fy = 36 ksi for a rod, of
Pbr ( angle )
1.79
Ab r =
=
= 0.0830 in.2
Fy Ω
(36 1.67)
For the dead plus live load case,
Amin = 0.238 in.2
Select a rod to meet the required area for the controlling case of stiffness
for the dead plus live load case, Amin = 0.238 in.2.
use a 5/8-in. rod with A = 0.307 in.2
8.13 TENSION PLUS BENDING
Throughout this chapter, the case of combined compression plus bending has been treated. That is
the most common case of combined loading in typical building structures. However, the
Specification also has provisions, in Section H1.2, for combining flexure and tension. The
addition of a tension force to a member already undergoing bending may be beneficial.
The interaction equations for combined tension and flexure are the same as those already
discussed and given as Equations H1-1a and H1-1b. However, if the flexural strength is
controlled by the limit state of lateral-torsional buckling, the addition of a tension force can
increase bending strength. This is accounted for in the Specification by the introduction of a
modification factor to be applied to Cb. Thus, for doubly symmetric members, Cb in Chapter F
can be multiplied by 1 + αPr Pey for axial tension that acts concurrently with flexure, where
Pey = π2 EI y L2b and α = 1.0 for LRFD and 1.6 for ASD, as before. The limit that Mn cannot
exceed Mp still must be satisfied as it was for beam design discussed in Chapter 6.
EXAMPLE 8.9a
Combined Tension
and Bending by
LRFD
Goal:
Check the given W-shape beam for combined tension and bending
Given:
A W16×77 beam spans 25 ft and carries a uniform dead load of 0.92 kips/ft
and a uniform live load of 2.79 kips/ft. It also carries a tension live load of
62.5 kips. The member is braced at the ends only for lateral-torsional
buckling. Use A992 steel.
SOLUTION
Step 1:
Determine the required moment strength
wu = 1.2 ( 0.92 ) + 1.6 ( 2.79 ) = 5.57 kips/ft
Mu =
Step 2:
Step 3:
5.57 ( 25 )
2
= 435 ft-kips
8
Determine the required tension strength
Tu = 1.6 ( 62.5 ) = 100 kips
Determine the available moment strength. With Lb = 25 ft and Cb = 1.14,
from Manual Table 6-2
φM n = 1.14 ( 382 ) = 435 ft-kips < φM p = 563 ft-kips
364
Chapter 8
Beam-Columns and Frame Behavior
Step 4:
Determine the available tension strength for the limit state of yielding.
Connections at the end of the member are at a location of zero moment so
tension rupture will not be a factor for interaction with bending. From
Table 6-2
φTn = 1020 kips
Step 5:
Determine the increase to be applied to Cb when tension is applied in
conjunction with moment strength determined for the lateral-torsional
buckling limit state.
2
π2 EI π ( 29,000 )(138 )
Pey = 2 =
= 439 kips
2
Lb
( 25 (12) )
1+
1.0 (100 )
αPr
= 1+
= 1.11
Pey
439
Step 6:
Moment strength when considered in combination with tension
Step 7:
Determine the interaction equation to use
Pr 100
=
= 0.098 < 0.2
Pc 1020
Step 8:
Use Equation H1-1b
Pr
M
0.098 435
+ r =
+
= 0.049 + 0.901 = 0.95 < 1.0
2 Pc M c
2
483
φM n = 1.11( 435 ) = 483 ft-kips < φM p = 563 ft-kips
So the beam is adequate to carry the bending moment and tension force.
Step 9:
Check the beam for bending alone, in case the tension force were not there.
Use the available moment strength from Step 3.
M r 435
=
= 1.0 ≤ 1.0
M c 435
So the beam would just be adequate. In cases where the application of the
tension force increases interaction strength and that force may not actually
occur, it is important to check the member for flexure alone.
The W16×77 is adequate to carry the applied loads.
EXAMPLE 8.9b
Combined Tension
and Bending by
ASD
Goal:
Check the given W-shape beam for combined tension and bending
Given:
A W16×77 beam spans 25 ft and carries a uniform dead load of 0.92 kips/ft
and a uniform live load of 2.79 kips/ft. It also carries a tension live load of
62.5 kips. The member is braced at the ends only for lateral-torsional
buckling. Use A992 steel.
SOLUTION
Step 1:
Determine the required moment strength
Beam-Columns and Frame Behavior
Chapter 8
365
wa = 0.92 + 2.79 = 3.71 kips/ft
Ma =
Step 2:
3.71( 25 )
2
8
Determine the required tension strength
= 290 ft-kips
Ta = 62.5 kips
Step 3:
Determine the available moment strength. With Lb = 25 ft and Cb = 1.14,
from Manual Table 6-2
Mp
Mn
= 1.14 ( 254 ) = 290 ft-kips <
= 374 ft-kips
Ω
Ω
Step 4:
Determine the available tension strength for the limit state of yielding.
Connections at the end of the member are at a location of zero moment so
tension rupture will not be a factor for interaction with bending. From
Table 6-2
Tn
= 677 kips
Ω
Step 5:
Determine the increase to be applied to Cb when tension is applied in
conjunction with moment strength determined for the lateral-torsional
buckling limit state.
Pey =
2
π2 EI π ( 29,000 )(138 )
=
= 439 kips
2
L2b
( 25 (12) )
1+
1.6 ( 62.5 )
αPr
= 1+
= 1.11
439
Pey
Step 6:
Moment strength when considered in combination with tension
Mp
Mn
= 1.11( 290 ) = 322 ft-kips <
= 374 ft-kips
Ω
Ω
Step 7:
Determine the interaction equation to use
Pr 62.5
=
= 0.092 < 0.2
Pc 677
Step 8:
Use Equation H1-1b
Pr
M
0.092 290
+ r =
+
= 0.046 + 0.901 = 0.95 < 1.0
2 Pc M c
2
322
So the beam is adequate to carry the bending moment and tension force.
Step 9:
Check the beam for bending alone, in case the tension force were not there.
Use the available moment strength from Step 3.
M r 290
=
= 1.0 ≤ 1.0
M c 290
So the beam would just be adequate. In cases where the application of the
366
Chapter 8
Beam-Columns and Frame Behavior
tension force increases interaction strength and that force may not actually
occur, it is important to check the member for flexure alone.
The W16×77 is adequate to carry the applied loads.
8.14
PROBLEMS
Unless noted otherwise, all columns should be considered
pinned in a braced frame out of the plane being
considered in the problem with bending about the strong
axis.
1. Determine whether a W14×90, A992 column with a
length of 12.5 ft is adequate in a braced frame to carry the
following loads from a first-order analysis: a compressive
dead load of 100 kips and live load of 300 kips, a dead
load moment of 30 ft-kips and live load moment of 70 ftkips at one end, and a dead load moment of 15 ft-kips and
a live load moment of 35 ft-kips at the other. The member
is bending in reverse curvature about the strong axis.
Determine by (a) LRFD and (b) ASD.
2. An A992 W12×58 is used as a 14 ft column in a
braced frame to carry a compressive dead load of 60 kips
and live load of 120 kips. Will this column be adequate to
carry a dead load moment of 30 ft-kips and live load
moment of 60 ft-kips at each end, bending the column in
single curvature about the strong axis? The analysis
results are from a first-order analysis. Determine by (a)
LRFD and (b) ASD.
3. Determine whether a W12×190, A992 column with a
length of 22 ft is adequate in a braced frame to carry the
following loads from a first-order analysis: a compressive
dead load of 300 kips and live load of 500 kips, a dead
load moment of 50 ft-kips and live load moment of 100
ft-kips at one end, and a dead load moment of 25 ft-kips
and a live load moment of 50 ft-kips at the other. The
member is bending in reverse curvature about the strong
axis. Determine by (a) LRFD and (b) ASD.
4. An A992 W10×60 is used as a 13 ft column in a
braced frame to carry a compressive dead load of 74 kips
and live load of 120 kips. Will this column be adequate to
carry a dead load moment of 30 ft-kips and live load
moment of 45 ft-kips at each end, bending the column in
single curvature about the strong axis? The analysis
results are from a first-order analysis. Determine by (a)
LRFD and (b) ASD.
5. Given a W14×500, A992 42 ft column in a braced
frame with a compressive dead load of 90 kips and live
load of 270 kips. Maintaining a live load to dead load
ratio of 3, determine the maximum live and dead load
second-order moments that can be applied about the
strong axis on the upper end when the lower end is pinned
by (a) LRFD and (b) ASD.
6. Given a W14×132, A992 15 ft column in a braced
frame with a compressive dead load of 350 kips and live
load of 350 kips, and maintaining a live load to dead load
ratio of 1, determine the maximum live and dead load
second-order moments that can be applied about the
strong axis on the upper end when the lower end is pinned
by (a) LRFD and (b) ASD.
7. Reconsider the column and loadings in Problem 1 if
that column were bent in single curvature by (a) LRFD
and (b) ASD.
8. Reconsider the column and loadings in Problem 2 if
that column were bent in reverse curvature by (a) LRFD
and (b) ASD.
9. Reconsider the column and loadings in Problem 3 if
that column were bent in single curvature by (a) LRFD
and (b) ASD.
10. Reconsider the column and loadings in Problem 4 if
that column were bent in reverse curvature by (a) LRFD
and (b) ASD.
11. A 14 ft pin-ended column in a braced frame must
carry a compressive dead load of 85 kips and live load of
280 kips, along with a uniformly distributed transverse
dead load of 0.4 kips/ft and live load of 1.3 kips/ft. Will a
W14×68, A992 member be adequate if the transverse load
is applied to put bending about the strong axis? Determine
by (a) LRFD and (b) ASD.
12. A pin-ended chord of a truss is treated as a member
in a braced frame. Its length is 12 ft. It must carry a
compressive dead load of 90 kips and live load of 170
kips, along with a uniformly distributed transverse dead
load of 1.1 kips/ft and live load of 2.3 kips/ft. Will a
W8×58, A992 member be adequate if the transverse load
is applied to put bending about the strong axis? Determine
by (a) LRFD and (b) ASD.
13. A moment frame is designed so that under a service
lateral load H = 150 kips, the frame drifts no more than
L/400. There are a total of 15 columns in this frame, so
Pstory is 15 times the load on this column. A 13 ft,
Beam-C
Columns and Frame Behavvior
W
W14×120, A9
992 column iss to be checcked. Analysiss
rresults are fro
om a first-order analysis. The
T column iss
ccalled upon to carry a comprressive dead load of 100 kipss
aand live load of 300 kips. This load willl be taken ass
ccoming from a no-translatio
on analysis. The
T top of thee
ccolumn is loadeed with no-tran
nslation dead lo
oad moment off
225 ft-kips and a no-translation live load mo
oment of 80 ft-kkips. The translation momentts applied to th
hat column end
d
aare a dead loaad moment off 35 ft-kips an
nd a live load
d
m
moment of 100
0 ft-kips. The lo
ower end of thee column feels
hhalf of these moments.
m
The column
c
is bend
ding in reversee
ccurvature aboutt the strong ax
xis. Will the W14×120,
W
A992
2
m
member be ad
dequate to caarry this load
ding? Analysiss
sshows that thee effective len
ngth factor in
n the plane off
bbending is 1.66
6. Determine by
y (a) LRFD and (b) ASD.
Chapteer 8
367
17. A two-story single bay fraame is shownn in Figure
P8.177. The uniforrm live and ddead loads aree indicated
alongg with the windd load. A first--order elastic aanalysis has
yieldded the results shown in the ffigure for the ggiven loads
and the appropriatte notional looads. Assuminng that the
storyy drift is limiteed to height/3000 under the ggiven wind
loadss, determine whether the first- and seecond-story
colum
mns are adequuate. The gravitty loads will pproduce the
no-trranslation resullts and the winnd load will pproduce the
transslation results. The members are shown aand are all
A9922 steel. Determ
mine by (a) LRF
FD and (b) AS
SD.
114. A W14×1
193, A992 mem
mber is propossed for use as a
112.5 ft colum
mn in an mom
ment frame. The frame iss
ddesigned so thaat under a serviice lateral load
d H = 120 kips,,
thhe frame driftss no more than
n L/500. The to
otal story load,,
Pstory, is 20 tim
mes the indiviidual column load. Analysiss
rresults are from
m a first-orderr analysis. Willl this memberr
bbe adequate to
o carry a no-trranslation com
mpressive dead
d
looad of 160 kip
ps and live load
d of 490 kips? The top of thee
ccolumn is loaded with a no-trranslation dead
d load momentt
oof 15 ft-kips an
nd a no-translaation live load moment of 30
0
fft-kips. The traanslation mom
ments applied to
t that column
n
eend are a dead load moment of 80 ft-kips and
a a live load
d
m
moment of 250
0 ft-kips. The column
c
is ben
nding about thee
sstrong axis and
d, the lower end
d of the column
n is considered
d
ppinned, and th
he effective length factor is taken as 1.5.
D
Determine by (a) LRFD and (b)
( ASD.
115. Will an A992W14×48
A
be
b adequate as a 14 ft column
n
inn a moment frame
fr
with a compressive
c
deead load of 35
5
kkips and livee load of 80
0 kips? One half of thiss
ccompressive load is taken as a no-translation load and onee
hhalf as a transslation load. The
T top and bottom
b
of thee
ccolumn are lo
oaded with a no-translatio
on dead load
d
m
moment of 20
0 ft-kips and a no-translattion live load
d
m
moment of 55 ft-kips.
f
The traanslation mom
ments applied to
o
thhe column end
ds are a dead lo
oad moment off 10 ft-kips and
d
a live load mo
oment of 50 ft-kips. Analy
ysis results aree
ffrom a first-ord
der analysis. The
T frame is deesigned so thatt
uunder a servicee lateral load H = 50 kips, th
he frame driftss
nno more than L/300.
L
The tottal story load, Pstory, is eightt
tiimes the indiv
vidual column load. The colu
umn is bent in
n
rreverse curvatu
ure about the strong axis, and
a Kx = 1.3.
D
Determine by (a) LRFD and (b)
( ASD.
116. Determine whether a 10 ft braced frrame W14×43,,
A
A992 column can carry a co
ompressive deead load of 35
5
kkips and live load of 80 kiips along with
h a dead load
d
m
moment of 20 ft-kips and live load momen
nt of 40 ft-kips.
O
One half of theese moments are
a applied at the other end,,
bbending it in sin
ngle curvature.
P8.177
18. Determine w
whether the coolumns of thhe two-bay
unbraaced frame shhown in Figurre P8.18 are aadequate to
suppoort the givenn loading. Results for the first-order
analyysis are providded. The gravitty loads will pproduce the
3368
Chapter 8
Beam--Columns and
d Frame Behaavior
nno-translation results
r
and thee wind load wiill produce thee
trranslation resu
ults. Assume th
hat the lateral drift under thee
ggiven wind loaad will be limited to a maxim
mum of 0.5 in.
A
All members are
a A992 steel and the sizes are as shown.
D
Determine by (a) LRFD and (b)
( ASD.
20. A 14 ft coluumn in a mom
ment frame muust carry a
comppressive load of 540 kips aand a momentt about the
stronng axis of 1355 ft-kips from an LRFD second-order
direcct analysis. W
Will a W14×
×74, A992 m
member be
adequuate?
119. A nonsy
ymmetric two
o-bay unbracced frame iss
rrequired to supp
port the live an
nd dead loads given
g
in Figuree
P
P8.19. Using the results from
f
the firstt-order elasticc
aanalysis provid
ded, assuming the axial forcees provided aree
ffrom the no-traanslation analyssis and the lateeral drift due to
o
a 5 kip force iss limited to 0.4
4 in., determinee whether each
h
ccolumn will be adequate. All members are A992
A
steel and
d
thhe sizes are as shown. Deteermine by (a) LRFD and (b))
A
ASD.
21. A 14 ft coluumn in a mom
ment frame muust carry a
comppressive load of 360 kips aand a momentt about the
stronng axis of 90 ftt-kips from an ASD second-oorder direct
analyysis. Will a W114×74, A992 m
member be adequate?
22. A 28 ft coluumn in a mom
ment frame muust carry a
comppressive load of 110 kips and moment about the
stronng axis of 1000 ft-kips from an LRFD second-order
direcct analysis. W
Will a W10×
×60, A992 m
member be
adequuate?
23. A 28 ft coluumn in a mom
ment frame muust carry a
comppressive load oof 73 kips and moment aboutt the strong
axis of 67 ft-kipss from an LR
RFD second-orrder direct
analyysis. Will a W110×60, A992 m
member be adequate?
24a. Select a W-sshape for a collumn with a leength of 15
ft. Thhe results of a second-orderr direct analyssis indicate
that the member m
must carry a fforce of 700 kkips and a
stronng axis momennt of 350 ft-kipss. Design by L
LRFD.
24b. Select a W-shape for a column with a leength of 15
ft. Thhe results of a second-orderr direct analyssis indicate
that the member m
must carry a fforce of 467 kkips and a
stronng axis momennt of 230 ft-kipss. Design by A
ASD.
25a. Select a W-sshape for a collumn with a leength of 28
ft. Thhe results of a second-orderr direct analyssis indicate
that tthe member m
must carry a fo
force of 1100 kips and a
stronng axis momennt of 170 ft-kipss. Design by LRFD.
25b. Select a W-shape for a column with a leength of 28
ft. Thhe results of a second-orderr direct analyssis indicate
that the member m
must carry a fforce of 730 kkips and a
stronng axis momennt of 110 ft-kipss. Design by A
ASD.
26a. Select a W-sshape for a collumn with a leength of 14
ft. T
The results of a second-ordeer direct analyssis indicate
that the member m
must carry a fforce of 350 kkips and a
stronng axis momennt of 470 ft-kipss. Design by LRFD.
26b. Select a W-shape for a column with a leength of 14
ft. T
The results of a second-ordeer direct analyssis indicate
that the member m
must carry a fforce of 230 kkips and a
ASD.
stronng axis momennt of 320 ft-kipss. Design by A
P
P8.18
27a. Select a W-sshape for a collumn with a leength of 16
ft. T
The results of a second-ordeer direct analyssis indicate
that tthe member m
must carry a fo
force of 1250 kips and a
stronng axis momennt of 450 ft-kipss. Design by L
LRFD.
Beam-C
Columns and Frame Behavvior
Chapteer 8
369
P8.19
227b. Select a W-shape for a column with a length of 16
6
fft. The resultss of a second-o
order direct an
nalysis indicatee
thhat the memb
ber must carry
y a force of 830 kips and a
sstrong axis mom
ment of 300 ft--kips. Design by
b ASD.
228. The two-b
bay moment frame
fr
shown in
n Figure P8.28
8
ccontains a sing
gle leaning collumn. The results of a first-oorder elastic an
nalysis for eacch load are giv
ven. Determinee
w
whether the ex
xterior column
ns are adequaate to providee
sstability for thee frame under dead and livee load. All W-sshapes, orienteed for bending
g about the sttrong axis, aree
ggiven and the steel
s
is A992. Determine by (a) LRFD and
d
((b) ASD.
Select an A
A36 rod to prrovide the poiint bracing
31.
wn in the centerr panel of the tthree-bay frame of Figure
show
8.9a to provide staability for a tootal gravity deead load of
150 kkips and live load of 60 kipps. Design by (a) LRFD
and ((b) ASD.
Select an A
32.
A36 rod to prrovide the poiint bracing
show
wn in the centerr panel of the tthree-bay frame of Figure
8.9a to provide staability for a tootal gravity deead load of
180 kkips and live load of 95 kipps. Design by (a) LRFD
and ((b) ASD.
229. The two-story frame sh
hown in Figurre P8.29 reliess
oon the left-hand
d columns ben
nding about thee strong axis to
o
pprovide stabiliity. Using thee first-order an
nalysis resultss
sshown, determiine whether th
he given structu
ure is adequatee
if the steel is A992.
A
Determin
ne by (a) LRFD
D and (b) ASD..
33. A 30 ft simpply supported W
W24×68 beam
m has point
braciing at third pooints. The beaam supports a uniformly
distriibuted dead lload of 1.2 kkips/ft and a uniformly
distriibuted live looad of 2.0 kkips/ft. Deteermine the
requiired point bracce strength andd the required ppoint brace
stiffnness by (a) LRF
FD and (b) AS
SD.
330. The two--bay, two-storry frame shown in Figuree
P
P8.30 is to be designed.
d
Usin
ng the Live, Deead, Snow, and
d
W
Wind Loads giiven in the fig
gure, design thee columns and
d
bbeams to provid
de the required
d strength and stability by (a))
L
LRFD and (b) ASD.
A
34. A 30 ft simpply supported W
W24×68 beam
m has panel
braciing in three eqqual panels ovver its length. The beam
suppoorts a uniform
mly distributed dead load of 1.2 kips/ft
and a uniformly distributed livve load of 22.0 kips/ft.
Deterrmine the reqquired shear strength of thhe bracing
3370
Chapter 8
Beam--Columns and
d Frame Behaavior
ssystem and the required panell brace stiffnesss by (a) LRFD
D
aand (b) ASD.
P
P8.28
P8.299
335. A simply supported W1
18×86 beam spans
s
30 ft and
d
ccarries a uniforrmly distributed
d dead load of 0.8 kips/ft and
d
a uniformly diistributed live load of 2.4 kips/ft.
k
It also
o
ccarries a tension live load of 18 kips. The member
m
is fully
y
bbraced for latteral-torsional buckling. Use A992 steel.
D
Determine if th
he W18×86 iss adequate for the combined
d
tension and ben
nding by (a) LR
RFD and (b) ASD.
A
36. A simply suppported W27×884 beam spans 30 ft and
carriees a uniformlyy distributed deead load of 1.2 kips/ft and
a uniiformly distribbuted live loadd of 2.5 kips//ft. It also
carriees a tension llive load of 330 kips. The m
member is
braceed at third poiints for laterall-torsional bucckling. Use
A9922 steel. Determ
mine if the W227×84 is adequuate for the
combbined tension aand bending byy (a) LRFD andd (b) ASD.
Beam-C
Columns and Frame Behavvior
Chapteer 8
371
Desiggn the colum
mns and beamss for the resuulting load
effeccts and redo thee analysis to chheck the strenggth of these
new members and the drift of thhe structure. Coonfirm that
the e ffective lengthh method may bbe used.
38. In
ntegrated Dessign Project – Direct Analyssis
Meth
hod
Laterral load resisstance in thee east-west diirection is
proviided by two pperimeter mom
ment frames aas seen in
Figurre 1.24. Beforre the forces inn these membbers can be
deterrmined, the speecified wind looad must be ddetermined.
At thhis stage in thee design, a sim
mplified approaach to wind
load calculation sim
milar to that uused in Chapter 4 might
yieldd the followingg loads at each llevel:
Roof
4th Floorr
3rd Floorr
2nd Floorr
Total Winnd Load
P
P8.30
337. Integrated Design Project – Effective Length
M
Method
L
Lateral load resistance
r
in the east-west direction iss
pprovided by tw
wo perimeter moment fram
mes as seen in
n
F
Figure 1.24. Beefore the forces in these meembers can bee
ddetermined the specified wind
d load must be determined.
A
At this stage in
n the design, a simplified app
proach to wind
d
looad calculation
n similar to th
hat used in Ch
hapter 4 mightt
yyield the follow
wing loads at eaach level:
Rooff
4th Floor
3rd Floor
F
2nd Floor
F
Totall Wind Load
32.0
0 kips
59.0
0 kips
54.0
0 kips
50.0
0 kips
195.0
0 kips
T
The moment frrames will shaare equally in carrying thesee
looads. They will be designed using the efffective length
h
m
method of App
pendix 7.2, and
d second-order effects will bee
inncorporated using
u
the am
mplified first-o
order analysiss
m
method of Appendix 8; thus, superposition may
m be used.
B
Before an analysis may be
b carried ou
ut, preliminary
y
m
member sizes must
m
be obtain
ned. Using thee gravity loadss
ccalculated in Chapter
C
2, select preliminarry column and
d
bbeam sizes with
hout concern for
f the frame behavior
b
of thee
sstructure.
W
With these mem
mber sizes, thee analysis is to be carried outt
ffor dead load
d, live load, roof load, an
nd wind load.
M
Members are to
o be selected for
fo the gravity plus
p wind load
d
ccombination, so
s there shou
uld be no need to includee
nnotional loads.
32.0 kips
59.0 kips
54.0 kips
50.0 kips
195.0 kips
The moment framees will share eequally in carrrying these
loadss. They will bbe designed ussing the Direcct Analysis
Methhod from Chappter C.
Befoore an analysis may be ccarried out, ppreliminary
mem
mber sizes musst be obtained. Using the graavity loads
calcuulated in Chappter 2, select preliminary coolumn and
beam
m sizes withoutt concern for tthe frame behaavior of the
struccture.
Withh these memberr sizes, the anaalysis is to be carried out
for ddead load, livve load, rooff load, and w
wind load,
follow
wing the general analysis requirements of Section
C2. Members are tto be selected for the gravityy plus wind
load combination, so there shoulld be no need to include
notioonal loads.
Desiggn the colum
mns and beamss for the resuulting load
effeccts and redo thee analysis to chheck the strenggth of these
new m
members and tthe drift of the structure.
C
Chapterr 9
C
Compo
osite Constr
C
ruction
n
7 Brryant Park, Neew York
Photo courteesy Thornton T
Tomasetti
9.11
INTROD
DUCTION
Any structurral member in
n which two or
o more mateerials having ddifferent stresss-strain relattionships
are combineed and called
d upon to work
w
as a sinngle member may be connsidered a coomposite
member.
Man
ny different types
t
of mem
mbers have beeen used thatt could be callled composite. Such
members, ass shown in Figure 9.1, incclude (a) a re inforced conccrete beam, ((b) a precast cconcrete
beam with a cast-in-placce slab, (c) a “flitch” girdder combiningg wood side members andd a steel
plate, (d) a stressed skin panel where plywood is ccombined witth solid woodd members, aand (e) a
steel shape combined
c
witth concrete.
The last type off member, an
nd similar meembers, are what are norrmally thougght of as
composite beams
b
in build
ding applicatiions. Specificaation Chapterr I provides rrules for desiggn of the
composite members
m
illusstrated in Figu
ure 9.2. Thesee members arre (a) steel beeams fully enncased in
concrete, (b) steel beamss with flat soffit concrete slabs, (c) steel beams com
mbined with cconcrete
slabs on forrmed steel deeck, (d) steell columns fuully encased in concrete, and (e) holloow steel
shapes filled
d with concrette.
Encased beams and filled columns,
c
shoown in Figurre 9.2a and e, may not require
mechanical anchorage between the steel
s
and conncrete, other than the natuural bond thaat exists
between thee two materiaals; however,, the flexurall and compreession membbers shown inn Figure
9.2b, c, and d always requ
uire some forrm of mechannical shear connnection.
Reg
gardless of thee type of mecchanical shearr device provided, it must connect the ssteel and
concrete to form a unit and permit the
t two materrials to workk together to resist the loaad. This
considerably
y increases th
he strength off the bare steeel shape. Coomposite beam
ms were firstt used in
bridge desig
gn in the United States arou
und 1935. Unntil the inventtion of the shear stud, the cconcrete
floor slab was connected to the stringeer beams by m
means of wiree spirals or chhannels weldeed to the
top flange of the beam, ass shown in Fiigure 9.3.
3772
Chaptter 9
Compposite Construuction 373
Figure 9.1 Composite Members
M
S
Beams and
a Columns
Figure 9.2 Composite Steel
In
I the l940s the
t Nelson Sttud Companyy invented thee shear stud, a headed rodd welded to
the steel beam by meaans of a speciial device or ggun, as shownn in Figure 9.4. The comppany did not
enforce its
i patent butt instead enco
ouraged nonpproprietary usse of the system, assuminng correctly
that the company
c
wou
uld get its shaare of businesss if it becam
me popular. Inn a very short time, studs
replaced spirals and channels,
c
and
d today studss are used alm
most exclusivvely in compposite beam
constructtion.
In
I 1952 AISC
C adopted com
mposite desiggn rules for eencased beam
ms in its speciification for
building design, and in l956 it ex
xtended them
m to beams w
with flat soffiits. Although the design
procedurre was based on the ultimaate strength oof the compossite section, tthe rules weree written in
terms off an allowable stress form
mulation, as w
was commonn for the timee. Since thenn, allowable
stress design for com
mposite beamss has often beeen criticizedd as being connvoluted and difficult to
understan
nd.
Figure 9.3
9
Composite Beam Usin
ng a Spiral Shhear Connecttor
3774 Chapter 9
Composiite Construction
Figure 9.4 Installation of a Shear Sttud with a Stuud Gun
Photo Courtesy W. Samuel Easterling
In the
t current Specification,
Sp
whether forr ASD or LR
RFD, the rulees for the deesign of
composite beams
b
are straightforward and surprisinngly simple. The ultimatee flexural streength of
the composiite member is
i based on plastic
p
stress distribution, with the ducctile shear coonnector
transferring shear betweeen the steel section and thee concrete slabb.
Speccification Secction I1.3 giv
ves limitationns on materiaal properties for use in coomposite
concrete meembers. Concrrete is limited
d to fc′ betweeen 3 ksi and 10 ksi for norrmal-weight cconcrete
and between
n 3 ksi and 6 ksi
k for lightw
weight concrette. The speciffied minimum
m yield strengtth of the
structural stteel used in calculating
c
th
he strength off composite m
members is nnot to exceedd 75 ksi.
Similarly, fo
or reinforcing
g steel the lim
mit for calculattions is 80 ksii.
Thiss chapter disscusses the design
d
of botth compositee beams and composite ccolumns.
Table 9.1 lissts the section
ns of the Speccification and parts of the M
Manual discuussed in this chhapter.
Table 9.1 Sections of Specification
S
and
a Parts of M
Manual Coveered in This C
Chapter
Specificationn
I1
I2
I3
I5
I6
I8
Generral Provisionss
Axiall Force
Flexu
ure
Comb
bined Axial Force
F
and Flexxure
Load Transfer
Steel Anchors
Manual
Part 3
Part 4
Desig
gn of Flexurall Members
Desig
gn of Compression Membeers
Chapter 9
Composite Construction 375
9.2 ADVANTAGES AND DISADVANTAGES OF COMPOSITE BEAM CONSTRUCTION
One feature of composite construction makes it particularly advantageous for use in building
structures. The typical building floor system is composed of two main parts: a floor structure that
carries load to supporting members, usually a concrete slab or a slab on a metal deck; and the
supporting members that span the space between girders, usually steel beams or joists. The
advantage of a composite floor system stems from the concrete slab doing “double duty”. The
concrete slab, with or without metal deck, is first designed to span between beams. Then, since
the concrete is there for that purpose in one direction, it can be used to advantage in the other
direction, that of the beam span. All other factors that could be identified as advantages of this
type of construction can be traced back to this single feature. A composite beam takes the existing
concrete slab and enables it to work with the steel beam to carry the load to the girders. Thus, the
resulting system has a greater strength than would be available from the steel beam alone. The
composite beam is stronger and stiffer than the noncomposite beam.
The advantage of this factor lies in the reduced weight and/or shallower member depth
necessary to carry a given load compared to the bare steel beam. Because the concrete slab is in
compression and the majority of the steel is in tension, the two materials are working to their best
advantage. In addition, the effective beam depth has been increased from just the depth of the
steel to the total distance from the top of the slab to the bottom of the steel, increasing the overall
efficiency of the member.
With regard to stiffness, the composite section has an increased elastic moment of inertia
compared to the bare steel beam. Although the actual calculation of the stiffness of the composite
section may be approximate in many cases, the impact of the increased stiffness profoundly
affects the static deflection.
The only disadvantage of composite construction is the added cost of the required shear
connectors, referred to as steel headed stud anchors in the Specification. Because the increased
strength, or reduction in required steel weight, is normally sufficient to offset the added cost of
providing and installing the shear connectors, this increased cost is usually not a true
disadvantage.
9.3
SHORED VERSUS UNSHORED CONSTRUCTION
Two methods of construction are available for composite beams: shored and unshored
construction. Each has advantages and disadvantages, which will be discussed briefly. The
difference between these two approaches to the construction of a composite beam is how the selfweight of the wet concrete is carried.
When the steel shape alone is called upon to carry the concrete weight, the beam is
considered to be unshored. In this case, the steel is stressed and it deflects. This is the simplest
approach to constructing the composite beam because the formwork and/or decking is supported
directly on the steel beam. Unshored construction may, however, lead to a deflection problem
during the construction phase because as the wet concrete is placed, the steel beam deflects. To
obtain a level slab, more concrete is placed where the beam deflection is greatest. This means that
the contractor must place more concrete than required based on the specified slab thickness, and
the designer needs to provide more strength than would be needed if the slab had remained of
uniform thickness.
For shored construction, temporary supports called shores are placed under the steel
beam to carry the wet concrete weight. In this case, the composite section carries the entire load
after the shores are removed. No load is carried by the bare steel beam alone, and thus, no
deflection occurs during concrete placement. Two factors must be considered in the selection of
shored construction: (1) the additional cost, in terms of both time and money, of placing and
3776 Chapter 9
Composiite Construction
Figure 9.5 Effective Fllange Width
removing th
he temporary shoring; and (2) the potenntial increase in long-term,, dead load deeflection
due to creep
p in the concrrete, which no
ow must partiicipate in carrrying the perm
manent weighht of the
slab.
hough elasticc stress distriibution and deflection unnder service load conditiions are
Alth
influenced by
b whether th
he compositee beam is shoored or unshoored, researchh has shown that the
ultimate streength of the composite section is indepeendent of the shoring situaation. Thus, thhe use of
shoring is en
ntirely a serv
viceability and
d constructabbility questionn that must be considered by both
the designer and constrructor. Wheth
her ASD or LRFD provvisions are uused, the shoored and
unshored sy
ystems have the same nominal
n
streength, as dettermined thrrough applicaation of
Specification
n Section I3.
9.44
EFFECT
TIVE FLAN
NGE
A cross secttion through a series of typ
pical compossite beams is shown in Figgure 9.5. Because the
concrete slaab is normallly part of the transverse spanning flooor system, iits thickness and the
spacing of th
he steel beam
ms are usually established pprior to the deesign of the coomposite beam
ms.
Because the abillity of the slaab to particippate in load ccarrying decreeases as the distance
from the beaam centerline increases, so
ome limit musst be establishhed to determ
mine the portioon of the
slab that can
n be used in the calculations to determ
mine the strenngth of the ccomposite beaam. The
Specification
n provides tw
wo criteria forr determiningg the effectivee width of thee concrete slaab for an
interior beam
m and an add
ditional criterion for an eddge beam in S
Section I3.1a.. As shown inn Figure
9.5, the effeective width, beff, is the su
um of b′ valuues on each sside of the centerline of tthe steel
section. For an interior beeam, b′ is the lesser of
spaan
b′ ≤
8
1
b′ ≤ distance
d
to thee adjacent beaam
2
For an edge beam, the ad
dditional criterrion is
b′ ≤ disstance to the edge of the sllab
The entire th
hickness of th
he concrete slab is availabble to carry a compressivee force. Howeever, the
depth of thee concrete used in calcullations is onnly that requiired to provide sufficient area in
compression
n to balance the
t force transsferred by thee shear conneectors to the ssteel shape. T
This may
be significan
ntly less than
n the entire slaab thickness. It should be noted that thhe slab thickness does
not influence the effective width of thee slab as it dooes for reinforrced concretee T-beam desiign.
Chaptter 9
9.5
Compposite Construuction 377
STRE
ENGTH OF
F COMPOS
SITE BEAM
MS AND SL AB
Flat soffi
fit composite beams (Figurre 9.2b) are cconstructed ussing formworrk that is set at the same
elevation
n as the top of
o the steel section. The cooncrete slab iss placed direcctly on the stteel section,
resulting in a flat surfface at the lev
vel of the top of the steel. C
Composite beeams with a fformed steel
deck (Fig
gure 9.2c) aree constructed
d with the steeel deck restinng on top of tthe steel beam
m or girder.
The conccrete is placed on top of th
he deck so thhat the concreete ribs and vvoids alternatte. Provided
that the portion
p
of con
ncrete requireed to balance tthe tension foorce in the steeel is availablle above the
tops of the
t ribs, the ultimate streength is deterrmined similaarly for the ttwo types off composite
beams. Although
A
thee steel memb
ber may be either shoreed or unshored, the strenngth of the
compositte member is independentt of the use oof shores, andd the design rrules are indeependent of
the method of constru
uction.
The
T flexural strength
s
of a composite bbeam under ppositive momeent where cooncrete is in
compresssion is preseented in Secction I3.2a oof the Speciffication. In tthis section, strength is
developeed for flat sofffit beams. Th
he required m
modifications to account fo
for the use of metal deck
are pressented in th
he next secttion. For stteel shapes with a webb slendernesss ratio of
which is the case
c
for all rollled W-, S-, aand HP-shapees), the nominnal moment,
h / t w ≤ 3 .76 E / Fy (w
Mn, is determined from
m the plastic distribution
d
oof stress on thhe composite ssection, and
φb = 0.90 (LRFD)) Ωb = 1.677 (ASD)
wn in Figuree 9.6 with threee possible plastic stress
A composite beam cross section
s
is show
distributiions. Regardlless of the strress distributiion considereed, equilibrium
m requires thhat the total
tension force
f
equals the
t total com
mpression forcce, T = C. Inn Figure 9.6a,, the plastic nneutral axis
(PNA) iss located at th
he top of the steel shape. T
The compresssion force deeveloped usinng all of the
concrete is exactly eq
qual to the ten
nsion force deeveloped usinng all of the ssteel. For the distribution
of Figuree 9.6b, the PN
NA is located within the steeel shape. In this case, all of the concreete is taking
compresssion, but this is not sufficiient to balancce the tensionn force that thhe full steel sshape could
provide. Thus, some of the steel shape
s
is in coompression inn order to saatisfy T = C. The plastic
stress disstribution sho
own in Figure 9.6c is what occurs whenn less than thee full amount of concrete
is needed
d to balance th
he tensile forcce developedd in the steel sshape. Here thhe PNA is loccated within
the concrrete and that portion of th
he concrete beelow the PNA
A is not usedd because it w
would be in
tension and
a concrete is
i not effectiv
ve in resisting tension.
In
I all three caases, equilibrrium of the crross section rrequires that the shear connnectors be
capable of
o transferring
g the force caarried by the cconcrete into the steel. Forr the cases in Figure 9.6a
and b, th
his is the full strength of th
he concrete. F
For the case in Figure 9.66c, this is the strength of
the steel shape. Becau
use the shear connectors
c
arre carrying thee full amountt of shear forcce required
Figure 9.6
9
Plastic Stress
S
Distribu
ution
378 Chapter 9
Composite Construction
to provide equilibrium using the maximum capacity of one of the elements, this is called a fully
composite beam. It is also possible to design a composite beam when the shear force that can be
transferred by the shear connectors is less than this amount. In this case, the beam is called a
partially composite beam. Although it has less strength than the fully composite member, it is
often the most economical solution.
The Specification indicates that the plastic stress distribution in the concrete shall be
taken as a uniform stress at a magnitude of 0.85fc′. This is the same distribution specified by ACI
318 Building Code Requirements for Reinforced Concrete. In addition, the distribution of stress
in the steel is taken as a uniform Fy, as was the case for determining the plastic moment strength
of a steel shape.
The Specification also provides for the use of a strain compatibility method and an elastic
stress distribution method for determining the strength of a composite section. Strain
compatibility should be considered when a section is of unusual geometry and elastic stress
distribution should be used when the web is not compact.
9.5.1
Fully Composite Beams
Establishing which stress distribution is in effect for a particular combination of steel and
concrete requires calculating the minimum compressive force as controlled by the three
components of the composite beam: concrete, steel, and shear connectors.
If all of the concrete were working in compression,
(AISC I3-1a)
Vc′ = V ′ = 0.85 f c′beff t
If all of the steel shape were working in tension,
Vs′ = V ′ = Fy As
(AISC I3-1b)
If the shear studs were carrying their full capacity, each resisting Qn,
Vq′ = V ′ = ∑ Qn
(AISC I3-1c)
Because fully composite action is being considered first, Vq′ will not control and is not considered
further. If Vs′ ≤ Vc′, the steel is fully stressed and all or only a portion of the concrete is stressed.
This is the distribution given in either Figure 9.6a or c. If Vc′ < Vs′, the concrete is fully stressed
and the steel is called upon to carry both tension and compression to ensure equilibrium. This
results in the distribution shown in Figure 9.6b. Once the proper stress distribution is known, the
corresponding forces can be determined and their point of application found. With this
information, the nominal moment, Mn, can be determined by taking moments about some
reference point. Because the internal forces are equivalent to a force couple, any point of
reference can be used for taking moments; however, it is convenient to use a consistent reference
point. These calculations use the top of the steel as the point about which moments are taken.
Determination of the PNA for the cases in Figure 9.6a and c is quite straightforward. In
both cases the steel is fully stressed in tension, so it is said that the steel controls and it is known
that the concrete must carry a compressive force equal to Vs′. Only that portion of the concrete
required to resist this force will be used, so that force is defined as Cc = 0.85 fc′beff a, where a
defines the depth of the concrete stressed to its ultimate. Setting Vs′ = Cc and solving for a yields
Fy As
Vs′
=
a=
(9.1)
0.85 f c′beff 0.85 f c′beff
Chapter 9
Composite Construction 379
For the special case where Vs′ is exactly equal to Vc′, the value of a thus obtained is equal
to the actual slab thickness, t. This is the case shown in Figure 9.6a. For all other values of a, the
distribution of Figure 9.6c results. The nominal flexural strength can then be obtained by taking
moments about the top of the steel so that
M n = Ts ( d 2 ) + Cc ( t − a 2 )
(9.2)
where Ts = Fy As and Cc = Ts because the system must be in equilibrium.
When the concrete controls, Vc′ < Vs′, the determination of the PNA is a bit more
complex. It is best to consider this case as two separate subcases: (1) the PNA occurring within
the steel flange and (2) the PNA occurring within the web. Once it is determined that Vc′ controls,
and thus Cc = Vc′, the next step is to determine the force in the steel flange and web respectively
from
(9.3)
T f = Fy b f t f
Tw = Ts − 2T f
(9.4)
A comparison between the force in the concrete and the force in the bottom flange plus
the web shows whether the PNA is in the top flange or web. If Cc > Tw + Tf, more tension is
needed for equilibrium and the PNA must be in the top flange. If Cc < Tw + Tf, less tension is
needed for equilibrium and the PNA is in the web. In either case, the difference between the
concrete force, Cc, and the available steel force, Ts, must be divided evenly between tension and
compression in order to obtain equilibrium. This allows determination of the PNA location and
the nominal moment strength. Thus, with
As–c = area of steel in compression
and
As = total area of steel
equilibrium is given by
Cc + Fy As − c = Ts − Fy As − c
Solving for the area of steel in compression yields
T − Cc
As −c = s
2Fy
(9.5)
(9.6)
For the case where the PNA is in the flange, the distance from the top of the flange to the
PNA is given by x, where
A
x = s −c
(9.7)
bf
and for the case where the PNA is in the web,
As − c − b f t f
x=
+ tf
tw
(9.8)
Equation 9.8 can be more easily understood if it is related to the areas being considered.
The area of the web in compression is the area of steel in compression less the flange area. This
web compression area is divided by the web thickness, and the result is the location of the PNA
380 Chapter 9
Composite Construction
measured from the underside of the flange. Thus, x is simply the thickness of the flange plus the
depth of the web in compression.
EXAMPLE 9.1
Fully Composite
Beam Strength
SOLUTION
Goal:
Determine the nominal moment strength for the interior composite beam
shown as Beam A in Figure 9.7. Also determine the design moment and the
allowable moment.
Given:
The section is a W21×44 and supports a 4.5 in. concrete slab. The
dimensions are as shown. Fy = 50 ksi and fc′ = 4 ksi. Assume full composite
action.
Step 1:
Determine the effective flange width, the minimum of
beff = 30.0(12 in./ft)/4 = 90.0 in.
and
beff = (10.0 + 10.0)(12 in./ft)/2 = 120 in.
Therefore use
beff = 90.0 in.
Step 2:
Determine the controlling compression force using Equations I3-1a and I31b.
Vc′ = 0.85 f c' Ac = 0.85 ( 4.0 )( 90.0 )( 4.5 ) = 1380 kips
Vs′ = Fy As = 50.0 (13.0 ) = 650 kips
Assuming full composite action, the shear connectors must carry the
smaller of Vc′ and Vs′; thus,
Vq′ = 650 kips
Because Vs′ is less than Vc′, the PNA is in the concrete.
Step 3:
Determine the depth of concrete in compression using Equation 9.1.
Fy As
650
a=
=
= 2.12 in.
0.85 f c′beff 0.85(4)(90.0)
The resulting plastic stress distribution is shown in Figure 9.7c.
Step 4:
Determine the nominal moment strength using Equation 9.2,
M n = Ts ( d 2 ) + Cc ( t − a 2 )
2.12 ⎞
⎛ 20.7 ⎞
⎛
= 650 ⎜
⎟ + 650 ⎜ 4.50 −
⎟ = 9000 in.-kips
2 ⎠
⎝ 2 ⎠
⎝
⎛ 9000 ⎞
Mn = ⎜
⎟ = 750 ft-kips
⎝ 12 ⎠
For
LRFD
Step 5:
The design moment is
φM n = 0.9 ( 750 ) = 675 ft-kips
Chaptter 9
For
F
ASD
A
Step
S
5:
Compposite Construuction 381
Th
he allowable moment
m
is
M n Ω = 750 1.67 = 449 ft-kips
Figure 9.7 Interior Com
mposite Beam
m (Examples 9.1
9 and 9.2)
EXAMPLE
E 9.2
Fully Comp
posite
Beam Stren
ngth
SOLUTION
N
Goal:
G
Deetermine the nominal mom
ment strengthh for the inteerior compossite beam
sh
hown as Beam
m A in Figure 9.7 using a llarger W-shappe. Also deterrmine the
wable momennt.
deesign momentt and the allow
Given:
G
Usse a W21×111 as shown inn Figure 9.8 for the steel m
member and the same
me full composite action.
maaterials as in Example 9.1.. Again, assum
Step
S
1:
Deetermine the effective
e
flannge width.
Th
will remain thhe same; thus,
he effective fllange width w
beff = 90.00 in.
Step
S
2:
Deetermine the controlling coompression fo
force using Eqquations I3-1a and I31b
b.
Vc′ = 0.85 f c' Ac = 0.85 ( 4.0 )( 990.0 )( 4.5 ) = 11380 kips
.
Vs′ = Fy As = 500.0 ( 32.6 ) = 16630 kips
Asssuming full composite
c
acttion,
Vq′ = 1380 kkips
Beecause Vc′ is less
l than Vs′, tthe PNA is inn the steel.
Step
S
3:
Deetermine wheether the PNA
A is in the steeel flange or w
web.
T f = Fy b f t f = 50 (12.3)( 0.875 ) = 538 kips
Tw = Fy As − 2T f = 1630 − 2 ( 538 ) = 5544 kips
Th
hus,
Cc = Vc′ = 13800 > T f + Tw = 5538 + 554 = 10090 kips
382 Chapter 9
Composite Construction
Because additional tension is required to balance the compression in the
concrete, the PNA is in the flange.
Step 4:
Determine the area of steel in compression.
Use Equation 9.6.
As −c =
Step 5:
Ts − Cc 1630 − 1380
=
= 2.50 in.2
2Fy
2(50)
Determine the location of the PNA in the flange.
The PNA is located down from the top of the steel by a distance x as given
by Equation 9.7.
A
2.50
x = s −c =
= 0.203 in.
bf
12.3
The stress distribution for this PNA location is shown in Figure 9.8b.
Step 6:
Determine the nominal moment strength of the composite beam.
Moments could be taken about any point to determine the nominal moment;
however, a simplified mathematical model shown in Figure 9.8c makes the
analysis quicker. In this case, the full area of steel is shown in tension, and
the portion in compression is first removed (130 kips on the compression
side) and then added in compression (another 130 kips on the compression
side), shown as 2(130) = 260 kips. This results in only three forces and
moment arms entering the moment equation. Thus
⎛d ⎞
⎛t⎞
⎛ x⎞
M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
⎛ 21.5 ⎞
⎛ 4.5 ⎞
⎛ 0.203 ⎞
= 1630 ⎜
⎟ + 1380 ⎜
⎟ − 2(2.50) ( 50 ) ⎜
⎟ = 20,600 in.-kips
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
20,600
Mn =
= 1720 ft-kips
12
Step 7:
For LRFD, the design moment is
φM n = 0.9 (1720 ) = 1550 ft-kips
Step 7:
For ASD, the allowable moment is
M n 1720
=
= 1030 ft-kips
Ω
1.67
Chaptter 9
Compposite Construuction 383
Figure 9.8 Interior Com
mposite Beam
m (Example 9.2)
9.5.2
PAR
RTIALLY COMPOSIT
C
TE BEAMS
S
The com
mposite members considereed thus far havve been fully composite. T
This means thhat the shear
connecto
ors have been
n assumed to
o be capable of transferriing whatever force was rrequired for
equilibriu
um when eitther the conccrete or steel were fully sstressed. Theere are manyy conditions
where th
he required strength
s
of th
he compositee beam is lesss than what would resullt from full
compositte action. In particular,
p
theere are cases where the sizze of the steeel member is dictated by
factors other
o
than thee strength off the composiite section. B
Because shearr connectors compose a
significan
nt part of th
he cost of a composite
c
beeam, econom
mies can resuult if the low
wer required
flexural strength
s
can be
b translated to
t a reduced nnumber of shhear connectors when the ssteel section
and conccrete geometry
y are already given.
If
I the composite section is
i viewed unnder elastic sttress distribuutions, partiall composite
action caan be more eaasily understo
ood. Figure 9..9 shows elasstic stress disttributions for three cases
of combiined steel an
nd concrete. The
T first casee, Figure 9.9aa, is what reesults when thhe concrete
simply rests on the steel
s
with no shear transffer between tthe two mateerials. The reesult is two
independ
dent memberss that slip paast each otheer at the inteerface. If the two materials are fully
connecteed, the elasticc stress distriibution is as shown in Fiigure 9.9c annd the materiials are not
permitted
d to slip at all.
a If some limited amounnt of slip is permitted beetween the stteel and the
concrete,, the resulting
g elastic stresss distributionn is similar tto that shownn in Figure 9.9b. This is
how the partially
p
com
mposite beam would
w
behavee in the elastic region.
The
T plastic moment
m
streng
gth for a parrtially compoosite member is the result of a stress
distributiion similar to that shown in
i Figure 9.100. The PNA w
will be in the steel and thee magnitude
of the compression
c
force in thee concrete w
will be controlled by thee strength off the shear
connecto
ors.
9
Figure 9.9
Levels of
o Composite Action for Ellastic Behavioor
3884 Chapter 9
Composiite Construction
Figure 9.10 Plastic Stress Distrribution for Partially Compposite Action
Reg
gardless of the final location of the PN
NA, the forcee in the concrrete is limitedd by the
strength of the
t shear stud
ds. Thus, an approach
a
com
mbining thosee taken for thhe three cases of fully
composite sections
s
is ussed for the partially
p
compposite membber. By the ddefinition of ppartially
composite members,
m
Cc = Vq′ = ΣQn
and the deptth of the conccrete acting in
n compressionn is given by
∑ Qn
a=
0.85 f c′beff
(9.9)
Equations 9.3 through 9.8 can then bee used to deteermine the loccation of the PNA within tthe steel
and the nom
minal moment can be obtain
ned as before .
Parttial compositee action is geenerally desccribed in term
ms of a perceent of full coomposite
action. Thuss, a beam witth shear stud strength equaal to the tensiile strength off the W-shape is said
to be 100% composite or
o fully comp
posite. A beam
m with shearr stud strengthh equal to 3//4 of the
tensile stren
ngth of the W-shape
W
is saaid to be 75%
% compositee or partially composite. T
The slip
mposite actioon requires a certain
between thee steel and th
he concrete that
t
allows fo
for partial com
amount of ductile
d
behaviior or deform
mation capacitty of the sheaar studs. Although there arre no set
requirementts in the Speccification for determining
d
a minimum leevel of ductillity, the Com
mmentary
provides gu
uidelines to in
nsure that thee partially coomposite beaams will behaave as designned. For
configuratio
ons of compo
osite beams in normal ddesign practice, three speecific guideliines are
provided thaat ensure suffficient ductility. Thus, if ppartially compposite beams meet one or more of
the followin
ng condition
ns, shear con
nnectors aree not subject to failure due to insuufficient
deformation
n capacity:
1. Beams with
w span not exceeding 30
0 ft;
2. Beams with
w a degree of compositee action of at least 50%; orr
3. Beams with
w an averaage nominal shear connectoor capacity of at least 16 kkips per ft aloong their
span. Th
his correspond
ds to a single 3/4 in. shear stud placed aat 12 in. spaciing on averagge.
Beams that do not meet these criteriaa may still peerform adequuately but theey should be checked
through a deetailed analysis that captures the actual deformation ccharacteristiccs of the shearr studs.
Chaptter 9
Figure 9.11
9
Compposite Construuction 385
Stress Distribution
D
and
a Forces Ussed in Exampple 9.3
EXAMPLE
E 9.3
Partially
Composite Beam
Strength
Goal:
G
Deetermine the nominal mooment strengtth of a partiaally composiite beam.
Allso determinee the design m
moment and thhe allowable m
moment.
Given:
G
Co
onsider the co
oncrete and ssteel beam giiven in Exam
mple 9.1 and sshown in
Figure 9.7. In this case, hoowever, assuume that the shear conneectors are
caapable of transsferring only Vq′ = 500 kipps.
SOLUTION
N
Step
S
1:
Deetermine the effective
e
flannge width.
Th
his is the same as determinned for Exampple 9.1.
beff = 90.00 in.
Step
S
2:
Deetermine the controlling
c
coompression foorce.
Frrom Example 9.1,
Frrom the given
n data
Vc′ = 1380 kips
Vs′ = 650 kkips
Vq′ = 500 kips
Beecause the low
west value off the compresssive force is ggiven by Vq′, tthis is a
paartially compo
osite memberr and Cc = Vq′ = 500 kips.
Since Vq′ Vs′ = 500
nd should
5 650 = 0.777 ≥ 0.5 , the bbeam is 77% composite an
no
ot be subject to
t insufficientt connector duuctility.
Step
S
3:
Deetermine the depth
d
of the cconcrete workking in comprression from Equation
9.9
9.
ΣQn
5000
a=
=
= 1.633 in.
′
0.85 f c beff 0.85(44)(90.0)
Step
S
4:
Deetermine the area
a of steel iin compressioon from Equaation 9.6.
T − Cc 650 − 500
As − c = s
=
= 1.50 in.2
2 Fy
2(550)
Beecause this is less than thee area of the fflange, 6.50(00.450) = 2.933 in.2, the
PN
NA is in the flange.
fl
386 Chapter 9
Composite Construction
Step 5:
Determine the location of the PNA from Equation 9.7.
A
1.50
x = s −c =
= 0.231 in.
bf
6.50
Step 6:
Determine the nominal moment strength by taking moments about the top
of the steel shape using the three forces shown in Figure 9.11.
⎛d ⎞
⎛t⎞
⎛x⎞
M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
1.63 ⎞
⎛ 20.7 ⎞
⎛
⎛ 0.231 ⎞
= 650 ⎜
⎟ + 500 ⎜ 4.50 −
⎟ − 2(1.50) ( 50 ) ⎜
⎟ = 8550 in.-kips
2 ⎠
⎝ 2 ⎠
⎝
⎝ 2 ⎠
8550
Mn =
= 713 ft-kips
12
Step 7:
For LRFD, the design moment is
φM n = 0.9 ( 713) = 642 ft-kips
Step 7:
For ASD, the allowable moment is
M n 713
=
= 427 ft-kips
Ω 1.67
The nominal moment strength decreased from 750 ft-kips for the full composite action of
Example 9.1 to 713 ft-kips for the level of partial composite action given in Example 9.3. This is
approximately a 5 percent reduction in strength, corresponding to more than a 23 percent
reduction in shear connector strength. In both cases, the strength of the composite beam is
significantly greater than that of the bare steel beam, where the plastic moment strength of the
bare steel beam is Mp = 398 ft-kips. It is acceptable to make comparisons at the nominal strength
level because for the bare steel beam and the composite beam, the resistance factors and safety
factors are the same.
9.5.3
Composite Beam Design Tables
The force transferred between the steel and concrete governs the strength of the composite beams.
The shear studs transfer that force to the concrete, so design can be linked to the total shear force,
ΣQn. Design tables have been developed that use the shear stud strength in combination with an
infinite selection of concrete areas and strengths to determine the flexural strength of the
composite beam. These are given in Manual Table 3-19, an example of which is shown here as
Figure 9.12.
The variables used in Manual Table 3-19 are defined in Figure 9.13. The beam is divided
into seven PNA locations; five are in the flange and two are in the web. When the PNA is at the
top of the flange, in position 1, the entire steel section is in tension. This is a fully composite
beam. When the PNA is in the web at location 7, 25 percent of the potential steel section force is
transferred to the concrete through the studs. As shown in Figure 9.13, the flange has five PNA
locations and the stud strength for location 6 is one-half the difference between that at locations 5
and 7. These seven PNA locations establish corresponding stud strengths, ΣQn, which are also
given in the tables. Designs using PNA location 7 are clearly of concern regarding the
deformation capacity of the shear studs since this location represents only 25% composite action.
Chaptter 9
Compposite Construuction 387
Figu
ure 9.12 Co
omposite W-S
Shapes: Availlable Strengthh in Flexure
Copy
yright © Amerrican Institute of
o Steel Constrruction. Reprinnted with Perm
mission. All righhts reserved.
3888 Chapter 9
Composiite Construction
Figure 9.12 (Conttinued)
Copyrig
ght © American
n Institute of Steel
S
Constructiion. Reprinted with Permissioon. All rights rreserved.
Chaptter 9
Figure 9.13
9
Compposite Construuction 389
Definitiion of Variab
bles for Use w
with Compositte Beam Desiign Tables
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with Permissionn. All rights resserved.
The location where
w
50% composite actiion is used w
will vary, som
metimes beingg in the web
and someetimes in the flange, and sh
hould be checcked as part oof the design pprocess.
Finding
F
the contribution of
o the concrette to the beam
m strength reqquires knowlledge of the
location of the concrrete compresssive force. A
As already disscussed, the force in the concrete is
equal to the force in the studs, ΣQ
Qn. The mom
ment arm for tthat force is defined as Y
Y2 in Figure
9.13. It is a function
n of the conccrete strengthh and concreete geometry.. These tablees are quite
flexible and accomm
modate any permitted cooncrete strenngth and effe
fective slab w
width. The
thicknesss of the slab is limited only
y by the maxiimum momennt arm given iin the table.
Although
A
thesse tables are of
o most valuee in the desiggn of a compoosite beam, thhey can also
be used to
t check a parrticular comb
bination. Sele ction of a com
mposite beam
m is illustratedd in Section
9.9.
390 Chapter 9
Composite Construction
EXAMPLE 9.4
Composite Beam
Strength Using
Tables
Goal:
Determine the design flexural strength and allowable flexural strength for
the fully composite W16×26.
Given
The W16×26 beam is used with the metal deck and slab shown in Figure
9.14. The effective flange width is given, beff = 60.0 in. Fy = 50 ksi and fc′ =
4 ksi. Use the portion of Table 3-19 from Figure 9.12.
SOLUTION
Step 1:
Determine the controlling compression force using Equations I3-1a and I31b.
Vc′ = 0.85 f c'tbeff = 0.85 ( 4.0 )( 3.0 )( 60.0 ) = 612 kips
Vs′ = Fy As = 50.0 ( 7.68 ) = 384 kips
Because Vs′ < Vc′, the steel controls and PNA is at or above the top of the
flange. In Figure 9.12 this PNA location confirms that the ΣQn = 384 kips.
Step 2:
Determine the depth of the concrete acting in compression that is needed to
balance this stud force using Equation 9.9.
ΣQn
384
=
= 1.88 in.
a=
0.85 f c′beff 0.85(4)(60.0)
Step 3:
Determine the moment arm of the compressive force from the top of the
steel.
a
1.88
Y 2 = tc − = 6.0 −
= 5.06 in.
2
2
For
LRFD
Step 4:
Determine the design moment strength from Figure 9.12
Enter the table with Y2 = 5.0, which will be slightly conservative for an
actual Y2 = 5.06, and ΣQn = 384.
For LRFD, the design moment is
φM n = 370 ft-kips
For
ASD
Step 4:
Determine the allowable moment strength from Figure 9.12 (ASD).
Enter the table with Y2 = 5.0, which will be slightly conservative for an
actual Y2 = 5.06, and ΣQn = 384.
For ASD the allowable moment is
M n Ω = 246 ft-kips
Chaptter 9
Compposite Construuction 391
Figu
ure 9.14 Co
omposite Beam for Examp
ple 9.4
9.5.4
Neg
gative Mom
ment Strengtth
Accordin
ng to Specification Section
n I3.2b, the neegative flexurral strength caan be taken eiither as that
for the bare
b
steel beaam according
g to the provvisions of Chhapter F or, iif the composite section
satisfies the criteria, as
a a compositte section usiing a plastic sstress distribuution. For thee composite
section, the
t concrete in tension is ignored andd reinforcing steel is placeed in the tenssion region.
The resisstance factor and safety factor
f
are thee same as forr the positivee moment caase, and the
nominal flexural stren
ngth is calculaated assumingg a plastic strress distribution similar to that for the
positive moment
m
case. The limitatio
ons of the Speecification staate that
1.
1
2.
2
3.
3
9.6
The steel
s
beam mu
ust be compacct and adequaately braced aaccording to C
Chapter F.
Shearr connectors (steel
(
headed stud or steel channel anchhors) are provvided in the
negattive moment region.
r
The required
r
longitudinal reinfforcing bars aare placed wiithin the effeective width
of thee slab and are properly devveloped.
SHEA
AR STUD STRENGTH
S
H
uction industtry as shear cconnectors aree called steell anchors in
What aree recognized in the constru
the Specification. Tw
wo different ty
ypes of shearr connectors: steel studs, called steel hheaded stud
anchors, and channelss, called steel channel anchhors are recoggnized by the Specificationn. The shear
connecto
or used almosst exclusively
y in building construction is the steel hheaded stud aanchor. The
diameter of the studs must be 3/4
4 in. diameterr or less exceept when useed in a flat sooffit slab in
which caase they may be up to 7/8 in. diameterr. In addition the stud diam
meter may bee no greater
than 2.5 times the flan
nge thicknesss of the beam
m to which thhey are welded if they are not located
directly over
o
the beam
m web.
The
T nominal strength of a single steell stud, Qn, ussed in a com
mposite beam is given in
Specifica
ation Section I8.2a as
Qn = 0.5
0 Asa f c′ Ec ≤ Rg Rp Asa Fu
((AISC I8-1)
392 Chapter 9
Table 9.2
Composite Construction
Nominal Shear Strength, Qn, for One Stud with Rp = Rg = 1.0 (kips)
Stud diameter (in.)
3/8
1/2
5/8
3/4
Normal-weight concrete
Lightweight concrete
wc = 145 pcf
wc = 110 pcf
fc′ = 3 ksi
5.26
9.35
14.6
21.0
fc′ = 4 ksi
6.53
11.6
18.1
26.1
fc′ = 3 ksi
fc′ = 4 ksi
4.28
7.60
11.9
17.1
5.31
9.43
14.7
21.2
Based on Fu = 65 ksi
7.18
12.8
19.9
28.7
where
Asa =
f c′ =
cross-sectional area of the shank of the stud, in.2
specified compressive strength of the concrete, ksi
Fu =
minimum specified tensile strength of the stud, ksi
Ec =
Rg , Rp =
modulus of elasticity of the concrete, wc1.5 fc′ , ksi, where wc is the unit
weight of the concrete in pounds per cubic foot and fc′ is in ksi. Although
this is somewhat different from the equation used by ACI 318, it provides
sufficiently accurate results in this instance.
reduction factors used to bring the predicted stud strength into agreement
with test results. When used in a flat soffit slab, Rg = 1.0 and Rp = 0.75.
The nominal strength of typical 3/4 in. shear studs is given, along with other results, in
Table 9.2. Values are given for normal and lightweight concrete with f c′ = 3 ksi and 4 ksi. Values
are also given for the stud strength based on the tensile strength of the stud material with Rg = Rp
= 1.0. The reductions to be applied when these studs are incorporated into a slab on a metal deck
are addressed later.
Although not normally used in practice today, channel shear connectors are also
permitted by the Specification. The nominal strength of a channel shear connector, Qn, is given as
Qn = 0.3(t f + 0.5tw )la f c′ Ec
(AISC I8-2)
where
tf = thickness of channel flange, in.
tw = thickness of channel web, in.
la = length of channel, in.
The strength of the channel shear connector must be developed by welding the channel to
the beam flange for the force Qn with appropriate consideration of the eccentricity of the force on
the connector.
9.6.1
Number and Placement of Shear Studs
Although a shear stud serves to transfer load between the steel beam and the concrete slab, it is
not necessary to place the studs in accordance with the shear diagram of the loaded beam. Tests
have demonstrated that the studs have sufficient ductility to redistribute the shear load under the
ultimate load condition. Therefore, in design, it is assumed that the studs share the load equally
and the total shear force determined according to Section 9.5 can be transferred over the distance
Chaptter 9
9
Figure 9.15
Compposite Construuction 393
Stud Pllacement for Concentrated
C
d Load
between maximum moment
m
and zero
z
moment using studs ddistributed allong this disttance. For a
uniform load this ressults in Vq′/Q
Qn connectorss on each sidde of the maaximum mom
ment at the
centerline of the beam
m span. In th
he case of cooncentrated looads placed aat the third pooints of the
beam, thee same number of studs wo
ould be used on each side of the beam bbetween the lload and the
support, and a minimu
um number of
o studs woul d be requiredd between thee loads. This is shown in
Figure 9..15.
Shear
S
studs must
m be placed
d so that they have a minim
mum of 1 in. oof lateral concrete cover,
unless ussed in ribs of formed steel deck. In the direction of tthe shear forcce the minimuum distance
from the center of thee stud to the free
fr edge is 8 in. for normaal weight conncrete and 10 in. for light
weight concrete. For studs
s
not placced in steel ddeck ribs, the minimum ceenter-to-centerr spacing is
six stud diameters alo
ong the mem
mber and fourr stud diametters transverse to the mem
mber. When
placed in
n metal deck ribs, the spaccing is to be no less than four diameteers in any dirrection. The
maximum
m center-to-ceenter stud spaacing is eight times the totaal slab thicknness or 36 in.
EXAMPLE
E 9.5
Shear Stud
d
Determinattion
Goal:
G
Deetermine the number of 33/4 in. shear studs required over the complete
beeam span.
Given:
G
SOLUTION
N
Step
S
1:
Usse the fully composite bbeam of Exam
mple 9.1. Assume normaal-weight
co
oncrete and th
he values of E
Example 9.1.
Deetermine the strength
s
of a ssingle shear sstud.
Frrom Table 9.2
2, based on thhe concrete,
Qn = 0.5 Asa fc′ Ec = 26.1 kips
an
nd based on th
he stud withouut the reduction factors
Qn = Asa Fu = 228.7 kips
Fo
or the slab in Example 9.1 , the stud is w
welded directtly to the top flange of
thee beam in a fllat soffit slab,, so
Rg = 1.00
3994 Chapter 9
Composiite Construction
and
Rp = 0.75
Qn = Rg R p Asa Fu = 1.0 ( 0.75 )( 228.7 ) = 21.5 kkips
Use th
he lowest Qn, so
Step
p 2:
Q n = 21.5 kipss
Deterrmine the num
mber of studs required.
From
m Example 9.1
Vq′ = ΣQn = 650 kkips
Thus,, the number required
r
is
650//21.5 = 30.2 sstuds
Step
p 3:
Deterrmine the totaal number of sstuds requiredd for the beam
m.
Place thirty-one 3//4 in. shear sstuds on eachh side of the beam betweeen the
maxim
mum momentt and the zeroo moment. Thhus,
Use 62 studss for the entirre beam span
LRFD
Note that these caalculations arre independennt of the use of ASD or L
n the shear fo
orce to be trransferred is the same, ass in this exaample,
when
becau
use the calculaations are carrried out at thee nominal strrength level.
9.77
COMPO
OSITE BEA
AMS WITH
H FORMED METAL D
DECK
The combin
nation of form
med steel deck
k and compossite design is considered tooday to be onne of the
most econom
mical method
ds of floor co
onstruction. T
The steel deckk is a stay-inn-place formw
work for
the concretee slab. Cells, which
w
can be formed by ennclosing the sspace below tthe deck and bbetween
the ribs, caan be used to distributee the electriccal and elecctronic system
ms of the bbuilding,
contributing
g greatly to the overall econ
nomy of the ssystem.
S
Deck Prrofiles
Figure 9.16 Common Steel
Chaptter 9
Figure 9.17
9
Compposite Construuction 395
Beam with
w Formed Metal
M
Deck
The
T Specifica
ation providess rules in Secttion I3.2c forr steel decks w
with nominall rib heights
of up to 3 in. and averrage rib width
hs of 2 in. or more. For a deck that hass ribs narroweer at the top
than at th
he interface with
w the beam
m, the width oof the rib usedd in calculatiions must be taken as no
more thaan the width at the narrow
w portion. A deck section with this proofile is show
wn in Figure
9.16 alon
ng with other common decck profiles. Sttuds must exttend at least 11-1/2 in. abovve the top of
the steel deck. The co
oncrete slab th
hickness musst be specifiedd to provide 1/2 in. of covver over the
top of the installed stu
ud. The slab thickness
t
aboove the metal deck must bee at least 2.0 in., and the
deck mu
ust be anchorred to the sup
pporting beam
am with a coombination off puddle welds or other
fastenerss and studs at a spacing nott to exceed 188 in.
9.7.1
Deck
k Ribs Perpeendicular to Steel
S
Beam
For beam
ms supporting
g the steel deck the ribs ruun perpendicuular to the beeam, as show
wn in Figure
9.17. Thee space below
w the top of th
he rib containns concrete onnly in the alteernating spacces, so there
is no opp
portunity to transfer
t
force at this level.. Thus, the onnly concrete available for calculating
the full concrete
c
forcee is above the top of the deeck.
Table 9.3
3 Shear Stu
ud Strength Adjustment Faactors
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with Permissionn. All rights
reserved.
Condition
Rg
Rp
No deckiing
Decking oriented paraallel to the steeel shape
1.0
0.75
1.0
0.75
0.85**
0.75
wr
≥ 1.5
hr
wr
< 1.5
hr
Decking oriented perp
pendicular to the steel shappe
Number of studs occu
upying the sam
me decking riib
1
1.0
0.6†
2
0.6†
0.85
3 or more
0.7
m
0.6†
hr = nom
minal rib heigh
ht, in.
wr = averrage width off concrete rib or haunch (ass defined in S
Section I3.2c), in.
**
For a siingle stud.
†
This vallue may be in
ncreased to 0.7
75 when emid-hht ≥ 2 in.
3996 Chapter 9
Composiite Construction
d Weak Shearr Stud Locatioons
Figure 9.18 Strong and
The right-hand siide of the ineequality of Eqquation I8-1, without Rp orr Rg, accounts for the
tensile stren
ngth of the stu
ud material. This
T strength must be reduuced to accouunt for the loccation of
the stud within the concrrete rib, becau
use there is a difference inn strength forr a stud placeed closer
to the rib wall
w in the dirrection of forrce versus onne placed closser to the ribb wall away ffrom the
force. The stud
s
strength value
v
specifieed on the righht-hand side of the inequaality of Equattion I8-1
includes two
o multipliers, Rg and Rp. A simplified ttable of the vvalues for these adjustmennt factors
is given in Table
T
9.3, and
d the strength
h of the stud ccontrolled by Fu, with Rp = Rg = 1.0, is given in
Table 9.2.
Rg is used to acccount for the number of sttuds in a giveen concrete riib. If a rib coontains a
single stud, Rg = 1.0; if the
t rib contain
ns two studs,, Rg = 0.85; aand if a rib coontains three or more
studs, Rg = 0.7.
0
Rp is used to acco
ount for the lo
ocation of thee stud in the rrib in either thhe strong or tthe weak
position. Fig
gure 9.18 sho
ows the stron
ng and weakk locations off a stud in reelation to the applied
force. Becau
use it is diffficult to ensu
ure that the sstuds are loccated in the sstrong positioon, it is
recommendeed that Rp = 0.6 be used unless the nneed for the sslight strengtth increase iss critical
enough to warrant
w
the extra
e
effort in
i the field tto ensure thaat studs are placed in thee strong
position. Wh
hen studs aree placed in th
he strong posiition, which is at least 2.00 in. from thee loaded
side of the rib
r edge at miid-height of th
he rib, as sho wn in Figure 9.18, Rp can be increasedd to 0.75,
which is the same value that
t was used for the flat sooffit slab.
The maximum stud
s
spacing is specified aas 36 in., whhich is conveenient becausse many
in
multiples
decks have a rib spacing
of
o 6 in.
9.77.2
Deck Ribs
R Parallel to
t Steel Beam
m
For girders supporting beams
b
that caarry steel deckk, the ribs ruun parallel too the steel secction, as
shown in Figure
F
9.19. Concrete bellow the top of the deckk can be useed in calculaating the
composite section
s
propeerties and must
m
be usedd in shear sttud calculatioons. For callculation
purposes, co
oncrete below
w the top of th
he steel deck can be negleccted unless itt is needed to balance
the shear stu
ud strength. The
T design pro
ocedure descrribed for flat ssoffit beams aapplies here aas well,
Figure 9.19 Girder with
h Formed Meetal Deck
Chaptter 9
Figure 9.20
9
Compposite Construuction 397
Examplle 9.6, case (aa)
provided
d sufficient co
oncrete is av
vailable abovve the top off the steel deeck as determ
mined using
Equation
n 9.1 or 9.9. If
I the concrette below the top of the steeel deck musst be used to balance the
stud or steel
s
beam strrength, the on
nly differencee in the deterrmination of sstrength is reelated to the
changed geometry fro
om inclusion of
o a portion oof the ribs.
When
W
the dep
pth of the steeel deck is 1 -1/2 in. or ggreater, the avverage width, wr, of the
haunch or
o rib is not to
t be less thaan 2 in. for tthe first stud in the transvverse row pluus four stud
diameters for each ad
dditional stud
d. If the deck rib is too naarrow, the deeck can be spplit over the
beam and
d spaced in su
uch a way as to allow for tthe necessary rib width witthout adverseely affecting
member strength.
For
F this deck orientation, Rg is used too account for the width-to--height ratio of the deck
rib. When wr /hr ≥ 1.5, Rg = 1.0. Wh
hen wr /hr < 11.5, Rg = 0.85..
Rp is taken as 0.75 in all caases where thhe deck ribs aare parallel to the supportinng member,
just as fo
or the flat sofffit slab.
EXAMPLE
E 9.6
Composite Beam
Design
Goal:
G
Caalculate the design momennt and allowabble moment aand determinee the stud
req
quirements for
fo a compossite section. C
Carry out thee calculations for the
following threee cases.
(a) Full composite action (F igure 9.20).
(b) Partial com
mposite actionn with ΣQn = 3387 kips , whiich results in the PNA
at the centerr of the top flaange of the stteel beam (Figgure 9.21).
(c)) Partial comp
posite action with ΣQn = 2260 kips , whiich results in the PNA
at the bottom
m of the top fflange of the ssteel beam (Figure 9.22).
SOLUTION
N
Given:
G
Usse a W18×35
5 with a 6-in. slab on a 3 iin. metal deckk perpendicular to the
m spacing is 12 ft and
beeam with the profile
p
shownn in Figure 9.16c. The beam
thee beam span is 40 ft. fc′ = 4 ksi and Fy = 50 ksi. Froom Table 1-1 d = 17.7
in. and tf = 0.42
25 in.
Step
S
1:
Deetermine the effective
e
flannge width.
beff = 440.0/4 = 10.00 ft (governs)
beff = beam spacinng = 12.0 ft
398 Chapter 9
Composite Construction
Step 2:
Determine the compression force using the full concrete and full steel areas
using Equations I3-1a and I3-1b.
Vc′ = 0.85 f c′tbeff = 0.85 ( 4 )( 3.0 )(120 ) = 1220 kips
Vs′ = Fy As = 50 (10.3) = 515 kips
Part a
Step 3:
Full Composite Action (Figure 9.20)
Determine the controlling concrete force.
For full composite action, Cc is the smaller of Vc′ and Vs′; thus,
Cc = 515 kips
Step 4:
Calculate the effective depth of the concrete using Equation 9.1.
Cc
515
a=
=
= 1.26 in.
′
0.85 f c beff 0.85(4)(120)
Because a is less than the 3 in. available in the concrete above the deck, the
procedures for a flat soffit beam can be used.
Step 5:
Determine the nominal moment strength.
⎛d ⎞
⎛ a⎞
M n = Ts ⎜ ⎟ + Cc ⎜ t − ⎟
⎝2⎠
⎝ 2⎠
1.26 ⎞
⎛ 17.7 ⎞
⎛
= 515 ⎜
⎟ + 515 ⎜ 6.0 −
⎟ = 7320 in.-kips
2 ⎠
⎝ 2 ⎠
⎝
7320
=
= 610 ft-kips
12
For
LRFD
Step 6:
The design moment is
For
ASD
Step 6:
The allowable moment is
Step 7:
Determine the strength of a single stud.
φM n = 0.9 ( 610 ) = 549 ft-kips
M n 610
=
= 365 ft-kips
Ω 1.67
From Table 9.2, the value of a single 3/4 in. stud with normal-weight
concrete, fc′ = 4 ksi, is 26.1 kips based on the concrete.
However, because the studs are used in conjunction with the metal deck, a
check for any required reduction must be made. For the deck profile given,
the ribs are spaced at 12.0 in. on center which would permit 39 studs per
beam if only one stud were placed in each rib. Thus, assuming two studs
per rib with the studs placed in the weakest location, from Tables 9.2 and
9.3,
Rg = 0.85
Rp = 0.6
Asc Fu = 28.7 kips
Chapter 9
Composite Construction 399
Rg R p Asa Fu = 0.85 ( 0.6 )( 28.7 ) = 14.6 ≤ 26.1 kips
The stud strength is the lower value, based on the stud placement in the
metal deck.
Step 8:
Determine the number of studs required on each side of the maximum
moment.
The shear that is to be transferred is 515 kips. Therefore,
Number of studs = 515/14.6 = 35.3
Step 9:
Determine the total number of studs required for the beam.
Use 36 studs on each half span or 72 studs for the full beam span
With this deck profile, studs can be placed in pairs every 12 in. This will
nicely accommodate the 72 studs on the 40 ft span with two studs placed in
each rib.
Part b
Step 10:
Partial Composite Action (Figure 9.21)
Determine the controlling concrete force.
Because the value of ΣQn = 387 kips given is less than Vc′ and Vs′ as
determined in part a, ΣQn controls and this is a partially composite beam.
With ΣQn Vs′ = 387 515 = 0.75 stud ductility will not be a concern.
Step 11:
Calculate the effective depth of the concrete using Equation 9.9 with
Cc = ΣQn = 387 kips .
Cc
387
a=
=
= 0.949 in.
′
0.85 f c beff 0.85(4)(120)
Because a < 3.0 in., sufficient concrete is available above the metal deck as
in part (a).
Step 12:
Determine the area of steel in compression.
V ′ − Cc 515 − 387
As −c = s
=
= 1.28 in.2
2 Fy
2(50)
Step 13:
Determine the location of the PNA in the steel.
Assume the PNA is in the flange.
A
1.28
= 0.213 in. < t f = 0.425 in.
x = s −c =
bf
6.00
Therefore, the PNA is in the flange.
Step 14:
Determine the nominal moment strength.
400 Chapter 9
Composite Construction
⎛d ⎞
⎛t⎞
⎛ x⎞
M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
0.949 ⎞
⎛ 17.7 ⎞
⎛
⎛ 0.213 ⎞
= 515 ⎜
⎟ + 387 ⎜ 6.0 −
⎟ − 2 (1.28)( 50 ) ⎜
⎟ = 6680 in.-kips
2 ⎠
⎝ 2 ⎠
⎝
⎝ 2 ⎠
6680
=
= 557 ft-kips
12
For
LRFD
Step 15:
The design moment is
For
ASD
Step 15:
The allowable moment is
M n ⎛ 557 ⎞
=⎜
⎟ = 334 ft-kips
Ω ⎝ 1.67 ⎠
Step 16:
Determine the stud requirements.
φM n = 0.9 ( 557 ) = 501 ft-kips
The shear to be transferred is 387 kips.
The value of stud strength previously determined in part (a) for two studs
per rib is
Qn = 14.6 kips
Step 17:
Determine the required number of shear studs.
387/14.6 = 26.5 studs
Step 18:
Determine the total number of studs required for the beam.
Use 27 studs in each half span or 54 studs for the full beam span.
With 54 studs symmetrically placed in 39 available ribs, 12 ribs will remain
without studs. It is possible to recalculate the stud requirements to see if the
higher strength when only one stud per rib is used would work for this
beam.
Since Rg = 1.0 for a single stud per rib, from part (a)
Rg R p Asa Fu = 1.0 ( 0.6 )( 28.7 ) = 17.2 ≤ 26.1 kips
and the number of studs required for the half span is
387 17.2 = 22.5
Thus, 46 studs are required and this would still require two studs per rib for
8 ribs. For constructability reasons, 54 studs placed symmetrically, two
studs per rib will be selected.
Part c
Step 19:
Partial Composite Action (Figure 9.22)
Determine the controlling concrete force.
Because the value of ΣQn = 260 kips given is less than Vc′ and Vs′, ΣQn
Chapter 9
Composite Construction 401
controls and this is a partially composite beam. With
ΣQn Vs′ = 260 515 = 0.50 stud ductility will not be a concern.
Step 20:
Calculate the effective depth of concrete using Equation 9.9.
Cc
260
a=
=
= 0.637 in.
′
0.85 f c beff 0.85(4)(120)
Step 21:
Determine the area of steel in compression.
V ′ − Cc 515 − 260
As −c = s
=
= 2.55 in.2
2 Fy
2(50)
Step 22:
Determine the location of the PNA in the steel.
Assume the PNA is in the flange.
A
2.55
x = s −c =
= 0.425 in.
bf
6.00
which is the flange thickness, as expected.
Step 23:
Determine the nominal moment strength.
⎛d ⎞
⎛t⎞
⎛ x⎞
M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟
⎝2⎠
⎝2⎠
⎝2⎠
0.637 ⎞
⎛ 17.7 ⎞
⎛
⎛ 0.425 ⎞
= 515 ⎜
⎟ + 260 ⎜ 6.0 −
⎟ − 2 ( 2.55 )( 50 ) ⎜
⎟ = 5980 in.-kips
2 ⎠
⎝ 2 ⎠
⎝
⎝ 2 ⎠
5980
=
= 498 ft-kips
12
For
LRFD
Step 24:
For
ASD
Step 24:
Step 25:
The design moment is
φM n = 0.9 ( 498 ) = 448 ft-kips
The allowable moment is
M n ⎛ 498 ⎞
=⎜
⎟ = 298 ft kips
Ω ⎝ 1.67 ⎠
Determine the stud requirements.
The shear to be transferred is 260 kips.
As before, the single stud strength is
Qn = 14.6 kips
Step 26:
Determine the required number of shear studs.
260/14.6 = 17.8 studs.
Step 27:
Determine the total number of studs required for the beam.
4002 Chapter 9
Composiite Construction
Use 18
1 studs in eacch half span oor 36 studs foor the full beaam span.
Howeever, this requ
uires only a ssingle stud inn each rib, so the Rg = 0.855 does
not neeed to be applied. Thereforre, as from paart (b)
Qn = 1.0 ( 0.6 )( 28.7 ) = 17.22 < 26.1 kips
and
260//17.2 = 15.1 sstuds
Thus,,
use a total of 32 sstuds
which
h will also be accommodatted with one sstud per rib.
Figure 9.21 Exam
mple 9.6, case (b)
Figure 9.22 Exam
mple 9.6, case (c)
A co
omparison off the results frrom Examplee 9.6 shows thhat the magniitude of the reeduction
in moment strength
s
that results
r
from a reduction inn the shear foorce transferdd by the shearr studs is
less than th
he magnitudee of shear strrength reducttion. For parrt (b) of Exaample 9.6, thhe shear
strength wass reduced app
proximately 25
2 percent buut it producedd a moment strength reduuction of
only approx
ximately 9 perrcent. Similarrly, for part (cc), the 50 perrcent reduction in stud streength led
to an 18 perrcent reductio
on in moment strength. Thiis reinforces the idea that a partially coomposite
Chaptter 9
Compposite Construuction 403
beam maay be more economical
e
th
han a fully ccomposite beaam and sugggests why theey are quite
common in building design.
d
9.8
FULL
LY ENCAS
SED STEEL
L BEAMS
Steel beaams fully enccased in conccrete that conntributes to tthe strength oof the final m
member are
called en
ncased beamss. Such beam
ms can be deesigned throuugh one of tw
wo procedurees given in
Specifica
ation Section I3.3. The fllexural strenggth can be ccalculated froom the superrposition of
elastic sttresses, considering the efffects of shorring. Or, wheen shear connnectors are prrovided, the
flexural strength can be based on the plastic sttress distributtion or strainn compatibilitty approach
for the composite
c
secction. Alternaatively, the sstrength can bbe calculatedd from the pllastic stress
distributiion on the steeel section allone. In all ccases, φb = 00.9 and Ωb = 1.67. Althouugh encased
beams arre not particu
ularly common in buildinng construction today, theese provisionns will also
apply wh
hen determining the flexuraal strength off composite beeam-columnss.
9.9
SELE
ECTING A SECTION
The desiign of a com
mposite beam is somewhatt of a trial-annd-error proccedure, as aree numerous
other dessign situation
ns. The materrial presentedd thus far in tthis chapter hhas been direccted toward
the deterrmination of section
s
streng
gth when the cross section and concretee dimensions are known.
This section addressees the prelim
minary selectiion of the steeel shape to go along w
with a given
concrete slab. This procedure
p
is followed by a discussionn of the design tables foound in the
Manual.
With
W an estim
mate of beam depth, the w
weight of the bbeam can be estimated. Thhis is based
on the asssumption thaat the PNA is within the cooncrete so thaat the full steeel section is aat yield. The
resulting dimensions are
a given in Figure
F
9.23.
The
T moment arm between the tension fforce in the stteel and the ccompression force in the
concrete is given by
d ⎛
a⎞
(9.10)
moment arm = + ⎜ t − ⎟
2 ⎝
2⎠
If the nominal momen
nt strength is divided by thhe moment aarm, the requiired tension fforce can be
determin
ned. If that fo
orce is divideed by the steeel yield streess, the requiired area is ddetermined.
Multiplying the requirred area by weight
w
of steell, 3.4 lb/ft forr each squaree inch, yields an estimate
of the beam weight. Thus,
T
3.4 M n
(9.11)
beam
m weight =
d
a⎞
⎛
⎜ + t − ⎟ Fy
2⎠
⎝2
9
Figure 9.23
Momen
nt Arm for Preeliminary Weeight Determiination
404 Chapter 9
Composite Construction
To determine the beam weight by this approach, the depth of the beam must be estimated.
Several approaches have been suggested for this. One simple approach is to take the span in feet
and divide it by 2 to get the depth, d, in inches. Another approach to determine the depth of the
total composite section (d + t) is to take the span in feet and divide by 4/3 to obtain the total depth
in inches. Any reasonable approach gives a starting point. Because the thickness of the slab is
determined from the design in the transverse direction, only the effective depth of the concrete is
left to be determined. It is generally sufficient to assume that the effective depth of the concrete is
1 in.; therefore, a/2 = 0.5 in.
Although this approach to finding a starting point for composite beam design might be
helpful, the design tables in the Manual make this much effort unnecessary. Manual Table 3-19
(Figure 9.12) has already been discussed in the context of determining the strength of a given
combination of steel and concrete. It will now be approached from the perspective of selecting a
section through the use of an example.
EXAMPLE 9.7a
Composite Beam
Design by LRFD
Goal:
Select the most economical W-shape to be used as a composite beam.
Given:
The composite beam spans 30.0 ft and is spaced at 10.0 ft from adjacent
beams. It supports a 5 in. slab on a 2 in. formed steel deck with a profile
similar to that shown in Figure 9.16a. The beam must carry a dead load
moment of 50.0 ft-kips and a live load moment of 150 ft-kips. In addition,
the bare steel beam must be checked for the dead load plus a construction
live load moment of 40.0 ft-kips. Fy = 50 ksi and fc′ = 3 ksi.
SOLUTION
Step 1:
Determine the required moment for the composite beam.
Mu = 1.2(50.0) + 1.6(150) = 300 ft-kips
Step 2:
Determine the starting moment arm for the concrete from the top of the
steel.
Manual Table 3-19 is most effective when entered with a value for Y2. In
order to start the design process, the moment arm for the compressive force
in the concrete must be estimated. It is almost always adequate to assume,
as a starting point, that a = 1.0 in. Thus
1.0
Y2 = 5 –
= 4.5 in.
2
Step 3:
Select potential W-shapes from Figure 9.12 and Manual Table 3.19. Since
the beam span is 30 ft it will be acceptable to use partial composite action
lower than 50%.
Enter the column in Figure 9.12 with Y2 = 4.5 and proceed down to identify
the potential section that will carry the design moment of 300 ft-kips.
W16×26 with ϕMn = 306 ft kips and Σ Qn = 242 kips, PNA location 4
Using portions of Table 3.19 from the Manual yields additional
possibilities:
W16×31 with ϕMn = 319 ft-kips and ΣQn = 164 kips, PNA location 6
W14×34 with ϕMn = 309 ft-kips and ΣQn = 159 kips, PNA location 6
W14×30 with ϕMn = 311 ft-kips and ΣQn = 248 kips, PNA location 4
Chapter 9
Composite Construction 405
W14×26 with ϕMn = 312 ft-kips and ΣQn = 332 kips, PNA location 2
Step 4:
Determine the effective flange width.
30.0(12)
beff ≤
= 90.0 in.
4
beff ≤ 10.0(12) = 120 in.
Therefore
beff = 90.0 in.
Step 5:
Using the W16×31, determine the depth of concrete needed to balance the
force in the shear studs.
For the beam span and spacing given using Cc = ΣQn = 164 kips and the
effective flange width already determined.
Cc
164
a=
=
= 0.715 in.
0.85 f c′beff 0.85(3)(90.0)
Because this is less than the value a = 1.0 in. that was assumed to start the
problem, the assumption was conservative. Design could continue with the
determination of a more accurate required stud strength, or this
conservative solution could be used. The required number of studs would
be determined as before, accounting for the presence of any formed steel
deck and its influence on the individual stud strength.
Step 6:
Determine the required strength of the bare steel beam under dead load plus
construction live load.
Mu = 1.2(50.0) + 1.6 (40.0) = 124 ft-kips
Step 7:
Check to verify that the bare steel beam will support the required strength.
Since this beam has the steel deck attached to its compression flange, it will
be considered as fully braced for lateral-torsional buckling under the weight
of the construction loads.
From Manual Table 3-19, the W16×31 has a design strength of
φMp = 203 ft-kips > 124 ft-kips
Therefore, the W16×31 is an acceptable selection for strength.
Step 8:
To show what happens when the assumption for a is not quite as good, the
W14×26 is considered. Again using the ΣQn determined from the table and
the effective flange width,
Cc
332
a=
=
= 1.45 in.
0.85 f c′beff 0.85(3)(90.0)
This is significantly greater than the assumed value. To consider this
section further, determine a new Y2 such that
1.45
Y2 = 5 −
= 4.28 in.
2
406 Chapter 9
Composite Construction
Entering Manual Table 3-19 with Y2 = 4.0 as a conservative number, Mu =
300 ft-kips is determined and it corresponds to the same required shear stud
strength. Thus, this section also meets the strength requirements, and the
design can proceed with stud selection.
EXAMPLE 9.7b
Composite Beam
Design by ASD
Goal:
Select the most economical W-shape to be used as a composite beam.
Given:
The composite beam spans 30.0 ft and is spaced at 10.0 ft from adjacent
beams. It supports a 5 in. slab on a 2 in. formed steel deck with a profile
similar to that shown in Figure 9.16a. The beam must carry a dead load
moment of 50.0 ft-kips and a live load moment of 150 ft-kips. In addition,
the bare steel beam must be checked for the dead load plus a construction
live load moment of 40.0 ft-kips. Fy = 50 ksi and fc′ = 3 ksi.
SOLUTION
Step 1:
Determine the required moment for the composite beam.
Ma = 50.0 + 150 = 200 ft-kips
Step 2:
Determine the starting moment arm for the concrete from the top of the
steel.
Manual Table 3-19 is most effective when entered with a value for Y2. In
order to start the design process, the moment arm for the compressive force
in the concrete must be estimated. It is almost always adequate to assume,
as a starting point, that a = 1.0 in. Thus,
1.0
Y2 = 5−
= 4.5 in.
2
Step 3:
Select potential W-shapes from Figure 9.12 and Manual Table 3.19. Since
the beam span is 30 ft it will be acceptable to use partial composite action
lower than 50%.
Enter the column in Figure 9.12 with Y2 = 4.5 and proceed down to identify
the potential section that will carry the design moment of 200 ft-kips.
W16×26 with Mn/Ω = 204 ft kips and ΣQn = 242 kips, PNA location 4
Using portions of Table 3-19 from the Manual yields additional
possibilities:
W16×31 with Mn/Ω = 212 ft-kips and ΣQn = 164 kips, PNA location 6
W14×34 with Mn/Ω = 205 ft-kips and ΣQn = 159 kips, PNA location 6
W14×30 with Mn/Ω = 207 ft-kips and ΣQn = 248 kips, PNA location 4
W14×26 with Mn/Ω = 208 ft-kips and ΣQn = 332 kips, PNA location 2
Step 4:
Determine the effective flange width.
30.0(12)
beff ≤
= 90.0 in.
4
beff ≤ 10.0(12) = 120 in.
Chapter 9
Composite Construction 407
Therefore,
beff = 90.0 in.
Step 5:
Using the W16×31, determine the depth of concrete needed to balance the
force in the shear studs.
For the beam span and spacing given using Cc = ΣQn = 164 kips and the
effective flange width already determined.
Cc
164
a=
=
= 0.715 in.
0.85 f c′beff 0.85(3)(90.0)
Because this is less than the value a = 1.0 in. that was assumed to start the
problem, the assumption was conservative. Design could continue with the
determination of a more accurate required stud strength, or this
conservative solution could be used. The required number of studs would
be determined as before, accounting for the presence of any formed steel
deck and its influence on the individual stud strength.
Step 6:
Determine the required strength of the bare steel beam under dead load plus
construction live load.
Ma = 50.0 + 40.0 = 90.0 ft-kips
Step 7:
Check to verify that the bare steel beam will support the required strength.
Since this beam has the steel deck attached to its compression flange, it will
be considered as fully braced for lateral-torsional buckling under the weight
of the construction loads.
From Manual Table 3-19, the W16×31 has an allowable strength of
M p Ω = 135 ft-kips > 90.0 ft-kips
Therefore, the W16×31 is an acceptable selection for strength.
Step 8:
To show what happens when the assumption for a is not quite as good, the
W14×26 is considered. Again using the ΣQn determined from the table and
the effective flange width,
Cc
332
a=
=
= 1.45 in.
0.85 f c′beff 0.85(3)(90.0)
This is significantly greater than the assumed value. To consider this
section further, determine a new Y2 such that
1.45
Y2 = 5−
= 4.28 in.
2
Entering Manual Table 3-19 with Y2 = 4.0 as a conservative number, Ma =
210 ft-kips is determined and it corresponds to an increased required shear
stud strength.
Thus, this section meets the strength requirements and the design can
proceed with stud selection.
408 Chapter 9
Table 9.4
Shape
W16×31
W16×26
W14×34
W14×30
W14×26
Composite Construction
Shapes Selected for Example 9.7a (LRFD)
Percent
composite
Weight of
action
steel
Σ Qn
Mu
319
164
35.9
930
306
242
63.0
780
309
159
31.8
1020
311
248
56.0
900
312
332
86.2
780
Number of
studs
16
24
16
24
32
Equivalent
weight of
studs
160
240
160
240
320
Equivalent
total weight
of beam
1090
1020
1180
1140
1100
Which of the many possible sections should be chosen as the final design depends on the
overall economics of the situation. One way to compare several choices is to look at the total
weight of the steel sections combined with the total quantity of studs required. To make this
comparison it is often effective to assume that an installed single shear stud has the equivalent
cost of 10 pounds of steel. To make this type of comparison, the five potential sections found
initially in Example 9.7a are presented in Table 9.4. Here, it was assumed that Qn = 21.0 kips.
This means that no consideration was taken for metal deck reduction. In addition, no check was
made for the assumed versus actual a dimension. This table is simply to help determine which of
the potential shapes should be considered further. Based on this table, it could be said that the
W16×26 with an equivalent weight of 1020 lbs should be investigated further.
9.10 SERVICEABILITY CONSIDERATIONS
Three important serviceability considerations are associated with the design of composite floor
systems: deflection during construction, vibration under service loads, and live load deflection
under service loads.
9.10.1
Deflection During Construction
As discussed in Section 9.3, the Specification permits either shored or unshored construction.
With unshored construction, the Specification requires that the steel section alone have adequate
strength to support all loads applied prior to the concrete’s attaining 75 percent of its specified
strength. The bare steel beam, under the weight of these loads, deflects as an elastic member.
Because of this deflection of the beam under the weight of the wet concrete, cambering of the
steel beam is often specified. Cambering is the imposition of a permanent upward deflection of
the beam in its unloaded state so that, under load, the downward deflection results in a beam
without excessive final deflection from the original horizontal position. Predicting the necessary
camber is difficult because of the varying methods and sequences of concrete placement used by
different contractors as well as such factors as the end restraint provided by the beam connection.
Even with camber it is often prudent for the designer to add a little extra concrete load into the
design dead load and for the contractor to allow for a little extra concrete in the quantity estimate.
In the case of shored construction, deflection during construction is usually not a concern,
because the shores are not removed until the concrete has achieved some strength and composite
action can be counted upon. The deflection under the wet concrete for shored construction is at a
minimum. On the other hand, long-term deflection due to creep of the concrete may have to be
investigated because the concrete is stressed under the self-weight as a permanent load along with
the sustained service loads.
Chapter 9
9.10.2
Composite Construction 409
Vibration Under Service Loads
Composite construction usually is shallower than comparable noncomposite construction and,
therefore, may be more susceptible to perceived vibrations. Because vibration calculations
assume that the beam behaves compositely, even when it is not a composite beam, the additional
stiffness of a composite beam does not improve its vibration characteristics. If problems occur,
they usually occur in applications with long spans and little damping. For instance, a large area of
a department store containing only a light jewelry display and with no partition walls connecting
one floor to the next might exhibit vibrations that would be perceptible to some customers. On the
other hand, an office building constructed with the same floor system could contain full or partial
height partitions that would provide sufficient damping to obviate any perceived vibration.
Because of wide differences in human perception of vibration and many other factors, vibration
problems do not lend themselves to simple solutions. AISC Design Guide 11, Vibrations of Steel
Framed Structural Systems Due to Human Activity, provides more information and gives the
designer an approach to vibration acceptance criteria, damping, and rational design techniques.
9.10.3
Live Load Deflections
Live load deflections can be a critical design consideration for many applications. Excessive
deflection could cause problems with the proper fit of partitions, doors, and equipment and may
also result in an unacceptable appearance, including cracking of finishes and other visible
evidence of distress. Therefore, a live load deflection calculation should be carried out for most
situations. As discussed in Chapter 6, this calculation is made with the service loads for which
deflection is of interest, usually the nominal live loads.
Because beam deflections are a function of the stiffness of the beam, the modulus of
elasticity and moment of inertia of the composite section must be determined. The true moment
of inertia at the service load level for which deflections are to be calculated is not easily
determined. In addition, the modulus of elasticity of the composite section must account for the
interaction of steel and concrete. The normal approach is to transform the concrete into a material
that behaves like steel, with the same modulus of elasticity. The moment of inertia of the new
transformed section can then be determined. Transformation of the concrete into steel is
accomplished by dividing the concrete area by the modular ratio, n = Es /Ec, and using the same
thickness. Although this seems like a fairly straightforward process, the problem is determining
the thickness of the concrete that is actually participating in resisting the deflection.
One approach is to assume that the only concrete participating in deflection resistance is
also that which is providing strength. Thus, whether it is a fully composite or partially composite
member, a moment of inertia can be determined using the known value of a from the strength
calculations. Because the nominal strength is calculated at the ultimate load level, the amount of
concrete actually participating for service loads could be significantly more than that used in the
strength calculations. Thus, a moment of inertia determined by this approach is less than what
might actually be available and is called a lower bound moment of inertia, ILB. Figure 9.24 is a
sample from Manual Table 3-20, which gives the lower bound moment of inertia in a format that
parallels the strength tables already discussed. Use of these ILB values results in a conservative
estimate of service load deflections.
4110 Chapter 9
Composiite Construction
Figure 9.24 Lower-Bound Elasstic Moment of
o Inertia, ILB, for Plastic C
Composite Secctions
Copyrig
ght © American
n Institute of Steel
S
Constructiion. Reprinted with Permissioon. All rights rreserved.
Chapter 9
EXAMPLE 9.8
Deflection
SOLUTION
Composite Construction 411
Goal:
Determine the construction load deflection of the bare steel beam and the
service load deflection for the composite beam of Example 9.7
Given:
Consider the W16×26 addressed in Example 9.7. Check the beam for a
dead load of 0.45 kips/ft, a construction live load of 0.36 kips/ft, and an inservice live load of 1.35 kips/ft. Compare the construction load deflection
to span/360. For the live load deflection, use the lower bound moment of
inertia from Figure 9.24 and compare the calculated deflection to span/360
as a design limit.
Step 1:
Determine the total construction load for deflection calculations.
w = wD + wL–const. = 0.45 + 0.36 = 0.81 kips/ft
Step 2:
Determine the moment of inertia of the W16×26 from Manual Table 1-1 or
Table 3-20.
Ix = 301 in.4
Step 3:
Calculate the construction load deflection and compare to span/360.
5(0.81)(30.0) 4 (1728)
30(12)
Δ LL =
= 1.69 in. >
= 1.0 in.
384(29,000)(301)
360
Because the deflection exceeds our limit, cambering of the beam or shoring
during construction would be required. Shoring has a significant impact on
cost as well as on scheduling; therefore, it is likely that the beam would be
cambered or a larger section used. Selecting camber as the appropriate
solution, if we must continue to use this W16×26 , the beam should be
cambered to approximately 75 percent of the dead load deflection.
Thus, 0.75(1.69) = 1.27 in., so specify a camber of 1.25 in.
Step 4:
Assuming that the construction load deflection issue is resolved, determine
the live load deflection under the in-service live load.
From Example 9.7, the W16×26 was selected using Y2 = 4.5 and the
resulting shear stud force was ΣQn = 242 kips at PNA location 4 for both
LRFD and ASD.
Step 5:
Determine the lower bound moment of inertia.
Using Manual Table 3-20 with the values given in step 4, select
ILB = 754 in.4
Step 6:
Determine the live load deflection.
5(1.35)(30.0) 4 (1728)
Δ LL =
= 1.13 in.
384(29,000)(754)
412 Chapter 9
Composite Construction
Step 7:
Compare the calculated deflection with the given limit.
span 30(12)
Δ LL = 1.13 >
=
= 1.0 in.
360
360
Because the calculated deflection is greater than the limiting value, the live
load deflection is not acceptable based on the given criteria. This result,
combined with the construction load deflection issue, would likely lead the
designer to select a larger section for this situation, as was actually done in
Example 9.7.
Deflection calculations are carried out under service loads and are
independent of design by LRFD or ASD.
9.11 COMPOSITE COLUMNS
Composite columns in building construction have been much slower to gain acceptance than
composite beams. Specification provisions were first provided in the 1986 LRFD Specification
and for ASD in the 2005 Specification. Although the use of composite columns in buildings is
still quite limited, attention to hardening of structures against blast forces will likely bring them
more to the forefront.
Specification Section I2 provides for two types of composite columns: open shapes
encased by concrete and hollow shapes filled with concrete. Composite columns exist at the
interface between specification provisions for steel and those for concrete. For a member to
qualify as a composite column under the Specification, it must meet the following limitations:
1.
2.
3.
4.
5.
6.
The cross-sectional area of the steel member must constitute at least 1 percent of
the gross area.
For an encased member the concrete encasement must be reinforced with
continuous longitudinal steel as well as lateral ties or spirals. The longitudinal
steel area must be at least 0.004 times the gross area, and the tie area must be at
least equivalent to No. 3 bars at 12 in. spacing. Tie spacing must not exceed onehalf the smallest dimension of the column.
For filled hollow sections, strength is a function of local buckling of the walls of
the hollow section. To be classified as a compact section, HSS must have a
minimum wall thickness such that b t ≤ 2.26 E Fy for rectangular HSS and
D t ≤ 0.15 ( E Fy ) for round HSS. Noncompact and slender wall hollow sections
may be used for filled composite columns provided the appropriate Specification
provisions are met. For these shapes, the compressive strength will be limited by
the limit state of local buckling.
The concrete strength, fc′, must be between 3 ksi and 10 ksi for normal-weight
concrete and 3 ksi and 6 ksi for lightweight concrete as for beams.
The maximum value of Fy for structural steel shapes to be used in calculating
strength is 75 ksi as for beams.
The maximum value of Fy for reinforcing bars to be used in calculating strength
is 80 ksi as for beams.
Although these requirements are usually readily satisfied, for situations where they are not, ACI
318 should also be consulted.
To account for the effects of column slenderness on the nominal strength of a composite
column, the equations found in Chapter E for steel columns are used with slight modification.
Chapter 9
Composite Construction 413
Because of the combination of two dissimilar materials and the general uncertainties of composite
column behavior, the resistance and safety factors are taken as
φc = 0.75 (LRFD) Ωc = 2.00 (ASD)
To convert the column equations, Equations E3-2 and E3-3, for use with a composite
column, multiply each occurrence of stress; the critical stress, Fcr, the yield stress, Fy, and the
elastic buckling stress, Fe, by the area of the column and make the following substitutions;
Fcr A = Pn , Fy A = Pno and Fe A = Pe . This is presented in Specification Section I2.1 for encased
composite columns where
P
for no ≤ 2.25
Pe
⎛ Pno ⎞
⎡
⎜
⎟⎤
Pn = Pno ⎢ 0.658⎝ Pe ⎠ ⎥
⎢⎣
⎥⎦
and for
(AISC I2-2)
Pno
> 2.25
Pe
Pn = 0.877 Pe
(AISC I2-3)
Where Pno for the bare steel column was the yield strength of the shape, for a composite column
this strength must also include the contribution of the reinforcing steel and the concrete stressed
to their ultimate strength. Thus, to account for the reinforcing steel, FysrAsr is added and to
account for the concrete, 0.85f’cAc is added, Thus,
(AISC I2-4)
Pno = Fy As + Fysr Asr + 0.85 f c′Ac
Where Pe for the bare steel column was the elastic buckling strength based on the stiffness, EI, of
the steel shape, for the composite column it must be based on an effective stiffness, EIeff, which
incorporates the contribution of the reinforcing steel and the concrete. Thus,
π 2 EI eff
(AISC I2-5)
Pe =
( Lc )2
and the effective stiffness is a direct summation of the contribution of the steel shape, the
reinforcing steel and an appropriate portion of the concrete, thus
(AISC I2-6)
EI eff = Es I s + E s I sr + C1 Ec I c
Recent research has shown that the contribution to stiffness from the concrete is a function of the
ratio of the total steel area to the gross area of the composite member, up to 70% of the stiffness
of the uncracked concrete section, given by
⎛ A + Asr ⎞
C1 = 0.25 + 3⎜ s
(AISC I2-7)
⎟ ≤ 0.7
⎝ Ag ⎠
In these equations, the s subscript refers to the steel section, the sr subscript refers to the
longitudinal reinforcing steel, and the c subscript refers to the concrete. With these variables
defined, Equations I2-2 and I2-3 are then used to determine the nominal strength of the encased
composite column.
For filled columns with compact HSS or box sections, the equivalent yield strength must
account for the behavior that these columns exhibit. Thus, as for encased composite columns, the
steel section is at yield and the concrete is at 0.85f’c for rectangular sections and 0.95f’c for round
sections. This increase to 0.95f’c reflects the increase in strength of the concrete due to the
414 Chapter 9
Composite Construction
confinement of the circular steel section just as for spiral columns designed according to ACI
318. The reinforcing steel is transformed by the modular ratio, Es Ec , into a material that
behaves like concrete. Thus
⎛
⎛ E ⎞⎞
Pno = Pp = Fy As + C2 f c' ⎜ Ac + Asr ⎜ s ⎟ ⎟
(AISC I2-9a,b)
⎝ Ec ⎠ ⎠
⎝
where
C2 = 0.85 for rectangular sections
= 0.95 for round sections
The effective stiffness of the filled hollow section also accounts for the confinement provided by
the rectangular and round steel shape. In this case, research has shown that the contribution to the
effective stiffness of the concrete is up to 90% of the stiffness of the uncracked concrete section.
Thus,
(AISC I2-12)
EI eff = Es I s + Es I sr + C3 Ec I c
and
⎛ A + Asr ⎞
C3 = 0.45 + 3⎜ s
(AISC I2-13)
⎟ ≤ 0.9
A
g
⎝
⎠
Equations I2-2 and I2-3 are then used to determine the nominal strength for the filled compact
wall column.
For what is called a noncompact wall composite column, the concrete contribution to Pno
is reduced from 0.85f’c for rectangular sections and 0.95f’c for round sections to 0.7f’c for both as
the wall slenderness increases from λp to λr. For slender wall filled composite members, the
contribution of the concrete to Pno remains unchanged but the contribution of the steel shape
becomes a function of the critical buckling stress of the wall. These requirements are given in
Specification Section I2.2b(b) and I2.2b(c). There are four square ASTM A500 Grade C HSS that
are noncompact and one that is slender; six round HSS are noncompact and none are slender.
EXAMPLE 9.9
Composite Column
Strength
Goal:
Determine the nominal strength of an encased composite column. Then
determine the design strength and the allowable strength.
Given:
The column is composed of a W14×53 encased in 18 in.×22 in. of concrete
as shown in Figure 9.25. Additional given information is as follows:
Column effective length = 15 ft.
Steel shape: Fy = 50 ksi
Reinforcing: four #9 bars, Gr. 60, Fy = 60 ksi. Area of one bar = 1.0 in.2
Concrete strength: fc′ = 5 ksi
Ec = 1451.5 5.0 = 3900 ksi
SOLUTION
Step 1:
Determine the areas of the components.
As = 15.6 in.2 from Manual Table 1-1
Asr = 4 (1.0 ) = 4.0 in.2
Ac = bh − As − Asr
= 18.0 ( 22.0 ) − 15.6 − 4.0 = 376 in.2
Ag = bh = 18.0 ( 22.0 ) = 396 in.2
Chapter 9
Step 2:
Composite Construction 415
Check the minimum steel ratios.
A
15.6
ρs = s =
= 0.0394 > 0.01
Ag 22.0(18.0)
ρsr =
Asr
4.0
=
= 0.0101 > 0.004
Ag 22.0(18.0)
So the specified minimums are satisfied.
Step 3:
Determine Pno and Pe. From Equation I2-7
⎛ 15.6 + 4.0 ⎞
C1 = 0.25 + 3 ⎜
⎟ = 0.398 < 0.7
⎝ 396 ⎠
By inspection, the y-axis will be the critical buckling axis since it is the
weak axis for both the steel shape and the concrete encasement.
I s = I y = 57.7 in.4 from Manual Table 1-1
I sr = ΣAd 2 = 4 (1.0 )( 6.625 ) = 176 in.4
2
hb3
− I s − I sr
12
22.0(18.0)3
=
− 57.7 − 176 = 10,500 in.4
12
Ic =
From Equation I2-4
Pno = Fy As + Fysr Asr + 0.85 f c′Ac
= 50 (15.6 ) + 60 ( 4.0 ) + 0.85 ( 5 )( 376 ) = 2620 kips
and from Equation I2-6
EI eff = Es I s + Es I sr + C3 Ec I c
= 29,000 ( 57.7 ) + 29,000 (176 ) + 0.398 ( 3900 )(10,500 )
= 23.1 × 106 kip-in.2
and from Equation I2-5
π 2 EI eff π 2 (23.1 × 106 )
Pe =
=
= 7040 kips
( Lc ) 2
(15.0(12)) 2
Step 4:
Determine the controlling column strength equation.
Pno 2620
=
= 0.372 < 2.25
Pe 7040
Therefore, use Equation I2-2.
Step 5:
For
LRFD
Determine the nominal compressive strength.
Pn = Pno (0.658) Pno Pe = 2620(0.658) 0.372 = 2240 kips
The design compressive strength is
4116 Chapter 9
Composiite Construction
Step
p 6:
φPn = 0.775 ( 2240 ) = 16680 kips
Forr
ASD
D
Step
p 6:
The allowable
a
com
mpressing streength is
Pn Ω = 22240 2.00 = 1 120 kips
Figure 9.25
Composite Column (Example 9.9)
The Commentary
y indicates th
hat the tablees found in A
AISC Designn Guide 6, Looad and
Resistance Factor Desig
gn of W-Sha
apes Encasedd in Concrete, give connservative ressults for
umns by folllowing earlier specificatioons. The desiign guide also includes suuggested
encased colu
details for im
mplementing encased com
mposite colum
mns in a project. Because ffilled HSS reepresents
a limited sett of possible geometries,
g
sttrength tabless have been ddeveloped forr these shapess and are
available on
n the AISC weeb page at ww
ww.aisc.org/m
manualresourcces. The tablees are used inn exactly
the same waay as the column tables for
fo the unfilleed HSS colum
mn previouslyy discussed. In cases
where the filled HSS wou
uld have stren
ngth less thann the unfilled H
HSS, the com
mposite colum
mn tables
nd the design should revertt to that for ann unfilled HSS column.
do not proviide values, an
EX
XAMPLE 9.10
9
HS
HSS Composiite
Coolumn Stren
ngth
Goa
al:
Deterrmine the avaailable strenggth of a com
mposite filled HSS columnn and
comp
pare it to the strength of an unfilled HSS
S.
Giv
ven:
The column
c
is co
omposed of aan A500 Graade C HSS7×
×4×1/2 filledd with
normaal weight con
ncrete. Additioonal given innformation is aas follows:
Lc = 15 ft
Fy = 50 ksi
fc′ = 5 ksi
SO
OLUTION
Step
p 1:
Deterrmine the walll slendernesss of the HSS ffor the most sslender wall. From
Manu
ual Table 1-11
1 and the limiit from Tablee I1.1a,
h t = 12.1 < 2.26 E Fy = 2.26 29,000 50 = 54.4
so thee column is co
ompact
Chapter 9
Step 2:
Composite Construction 417
Determine the needed properties from Manual Table 1-11.
As = 8.81 in.2 , I y = 20.7 in.4 , ry = 1.53 in.
and
Asr = 0 since there is no reinforcing steel
Ac = 6.0 ( 3.0 ) = 18 in.2 without considering the reduction in concrete
area due to the rounded corners.
Confirm that there is sufficient steel to be considered a composite column
A
8.81
ρs = s =
= 0.315 > 0.01
Ag 7.0(4.0)
Step 3:
Determine the equivalent yield strength of the column using Equations I29a and I2-9b. For the rectangular HSS, C2 = 0.85.
E ⎞
⎛
Pno = Pp = Fy As + C2 f c′ ⎜ Ac + Asr s ⎟
Ec ⎠
⎝
= 50 ( 8.81) + 0.85 ( 5 )(18 + 0 )
= 517 kips
Step 4:
Determine the effective stiffness. First determine C3 from Equation I2-13.
⎛ A + Asr ⎞
C3 = 0.45 + 3 ⎜ s
⎟
⎝ Ag ⎠
⎛ 8.81 + 0 ⎞
= 0.45 + 3 ⎜⎜
⎟⎟
⎝ 7 ( 4) ⎠
= 1.39 > 0.9
Therefore C3 = 0.9 and determine Ec
Ec = w1.5 f c′ = 1451.5 5 = 3900 ksi
Thus, from Equation I2-12
EI eff
⎛ 6.0 ( 3.0 )3 ⎞
⎟
= 29,000 ( 20.7 ) + 0.9 ( 3900 ) ⎜
⎜
⎟
12
⎝
⎠
2
= 648,000 kip-in.
Step 5:
Determine the elastic buckling strength from Equation I2-5
2
π 2 EI eff π ( 648,000 )
Pe =
=
= 197 kips
2
L2c
(12 (15 ) )
Step 6:
Determine which compressive strength equation to use
Pno 517
=
= 2.62 > 2.25
Pe 197
Therefore use Equation I2-3
418 Chapter 9
Composite Construction
Step 7:
Determine the nominal compressive strength
Pn = 0.877 Pe = 0.877 (197 ) = 173 kips
For
LRFD
Step 8:
Determine the design strength of the composite column
φPn = 0.75 (173) = 130 kips
For
ASD
Step 8:
Determine the allowable strength of the composite column
Pn Ω = 173 2.00 = 86.5 kips
Step 9:
To determine the strength of the unfilled HSS column, first determine the
elastic buckling stress from Equation E3-4
π2 ( 29,000 )
π2 E
Fe =
=
= 20.7 ksi
2
2
⎛ Lc ⎞
⎛ 12 (15 ) ⎞
⎜ ⎟
⎜
⎟
⎝ ry ⎠
⎝ 1.53 ⎠
Step 10:
Determine which critical stress equation to use
Fy
50
=
= 2.42 > 2.25
Fe 20.7
Therefore use Equation E3-3
Step 11:
Determine the critical stress
Fcr = 0.877 Fe = 0.877 ( 20.7 ) = 18.2 ksi
Step 12:
Determine the nominal strength
Pn = 18.2 ( 8.81) = 160 kips
For
LRFD
Step 13:
Determine the design strength
φPn = 0.9 (160 ) = 144 kips
For
ASD
Step 13:
Determine the allowable strength
Pn Ω = 160 1.67 = 95.8 kips
For
LRFD
Step 14:
For
ASD
Step 14:
Compare the available strength of the composite column and the unfilled
HSS.
Since the unfilled HSS has a design strength of 144 kips which is greater
than 130 kips for the filled HSS, Use the design strength of the unfilled
HSS. Thus,
φPn = 144 kips
Since the unfilled HSS has an allowable strength of 95.8 kips which is
greater than 86.5 kips for the filled HSS, Use the allowable strength of the
unfilled HSS. Thus,
Pn Ω = 95.8 kips
Chapter 9
Composite Construction 419
EXAMPLE 9.11
HSS Composite
Column Strength
Goal:
Determine the available strength of a composite filled HSS column and
compare it to the strength of an unfilled HSS.
Given:
The column is composed of an A500 Grade C HSS16×8×5/8 filled with
normal weight concrete. Additional given information is as follows:
Column effective length,
Lc = 21 ft
Fy = 50 ksi
fc′ = 5 ksi
SOLUTION
Step 1:
Determine the wall slenderness of the HSS for the most slender wall. From
Manual Table 1-11 and the limit in Table I1.1a,
h t = 16.0 0.625 = 24.5 < 2.26 E Fy = 2.26 29,000 50 = 54.4
so the column is compact
Step 2:
Determine the needed properties from Manual Table 1-11.
As = 25.7 in.2 , I y = 274 in.4 , ry = 3.27 in.
and
Asr = 0 since there is no reinforcing steel
Ac = 14.75 ( 6.75 ) = 99.6 in.2 without considering the reduction in
concrete area due to the rounded corners.
Confirm that there is sufficient steel to be considered a composite column
A
25.7
ρs = s =
= 0.201 > 0.01
Ag 16.0(8.0)
Step 3:
Determine the equivalent yield strength of the column using Equations I29a and I2-9b. For the rectangular HSS, C2 = 0.85.
E ⎞
⎛
Pno = Pp = Fy As + C2 f c′ ⎜ Ac + Asr s ⎟
Ec ⎠
⎝
= 50 ( 25.7 ) + 0.85 ( 5 )( 99.6 + 0 )
= 1710 kips
Step 4:
Determine the effective stiffness. First determine C3 using Equation I2-13.
⎛ A + Asr ⎞
C3 = 0.45 + 3 ⎜ s
⎟
⎝ Ag ⎠
⎛ 25.7 + 0 ⎞
= 0.45 + 3 ⎜⎜
⎟⎟
⎝ 16 ( 8 ) ⎠
= 1.05 > 0.9
Therefore C3 = 0.9 and determine Ec
Ec = w1.5 f c′ = 1451.5 5 = 3900 ksi
Thus, the effective stiffness from Equation I2-12 is
420 Chapter 9
Composite Construction
EI eff
⎛ 14.75 ( 6.75 )3 ⎞
⎟
= 29,000 ( 274 ) + 0.9 ( 3900 ) ⎜
⎜
⎟
12
⎝
⎠
6
2
= 9.27 × 10 kip-in.
Step 5:
Determine the elastic buckling strength from Equation I2-5
2
6
π 2 EI eff π ( 9.27 × 10 )
Pe =
=
= 1440 kips
2
L2c
(12 ( 21) )
Step 6:
Determine which compressive strength equation to use
Pno 1710
=
= 1.19 < 2.25
Pe 1440
Therefore use Equation I2-2
Step 7:
Determine the nominal compressive strength
Pn = 0.6581.19 (1710 ) = 1040 kips
For
LRFD
Step 8:
Determine the design strength of the composite column
φPn = 0.75 (1040 ) = 780 kips
For
ASD
Step 8:
Determine the allowable strength of the composite column
Pn Ω = 1040 2.00 = 520 kips
Step 9:
To determine the strength of the unfilled HSS column, first determine the
elastic buckling stress using Equation E3-4
π2 ( 29,000 )
π2 E
Fe =
=
= 48.2 ksi
2
2
⎛ Lc ⎞
⎛ 12 ( 21) ⎞
⎜ ⎟
⎜
⎟
⎝ ry ⎠
⎝ 3.27 ⎠
Step 10:
Determine which critical stress equation to use
Fy
50
=
= 1.04 < 2.25
Fe 48.2
Therefore use Equation E3-3
Step 11:
Determine the critical stress
Fcr = 0.6581.04 ( 50 ) = 32.4 ksi
Step 12:
Determine the nominal strength
Pn = 32.4 ( 25.7 ) = 833 kips
For
LRFD
Step 13:
Determine the design strength
φPn = 0.9 ( 833) = 750 kips
Chapter 9
For
ASD
Step 13:
Composite Construction 421
Determine the allowable strength
Pn Ω = 833 1.67 = 499 kips
Compare the available strength of the composite column and the unfilled
HSS.
For
LRFD
Step 14:
For
ASD
Step 14:
Since the unfilled HSS has a design strength of 750 kips which is less than
780 kips for the filled HSS, Use the design strength of the filled HSS. Thus,
φPn = 780 kips
Since the unfilled HSS has an allowable strength of 499 kips which is less
than 520 kips for the filled HSS, Use the allowable strength of the filled
HSS. Thus,
Pn Ω = 520 kips
Examples 9.10 and 9.11 illustrate that filling an HSS is not a guarantee that the column
strength will increase. Even in the case where the strength does increase, as in Example 9.11, the
increase may not be significant enough warrant the extra cost of adding the concrete. Both of
these situations result because of the increased variability of a composite column when compared
to an unfilled column and the resulting reduction in resistance factor and increase in safety factor.
9.12 COMPOSITE BEAM-COLUMNS
Composite beam-columns have the same potential to occur as bare steel beam-columns. Any
application where bending moment and axial force are applied simultaneously needs to be
addressed according to the provisions of Specification Section I5.
For doubly symmetric composite beam-columns, the most common composite beamcolumns found in building construction, the interaction equations of Specification Chapter H can
be used conservatively. For a more accurate approach to determining the available strength, the
interaction surface can be developed based on plastic stress distributions and the length
modification from Specification Section I2 made, as discussed in Section 9.11.
Figure 9.26 provides several potentially useful interaction diagrams for a composite
beam-column. Curve 1 is the interaction curve based on a strain compatibility approach similar to
that used for developing similar diagrams for reinforced concrete columns, without consideration
of length effects. Curve 2 represents a segmented straight-line approximation based on plastic
stress distributions, again without incorporating any length effects. Curve 3 is a simplification of
curve 2, incorporating length effects and using only one intermediate point between pure axial
strength and pure bending strength. Curve 4 results from the application of resistance or safety
factors to curve 3. Curve 5 is the result of applying the interaction equations of Chapter H.
Only curves 4 and 5 in Figure 9.26 account for resistance or safety factors and the effects
of length on beam-column strength. The conservatism of the Chapter H approach may not be that
great compared to the curve 4 approach.
Part 6 of the Manual includes tables for use in determining the needed points for curve 2
of the interaction diagram in Figure 9.26. The available equations address encased W-shapes for
4222 Chapter 9
Composiite Construction
Figure 9.226 Composite Column
Interactionn Diagrams
bending abo
out the majorr and minor axes, compaact rectangulaar HSS bendding about eiither the
major or miinor axis, and
d compact rou
und HSS. Figgure 9.27 is M
Manual Tablee 6-3a for an encased
W-shape beending aboutt the major axis,
a
and Figgure 9.28 is Manual Tabble 6-4 for ccompact
rectangular HSS bending
g about eitheer axis. If thee curve 4 innteraction of Figure 9.26 is to be
uation identifi
fied in Figure s 9.27 and 9.28 for pointss A, B and C need be
constructed,, only the equ
evaluated. Then
T
the lengtth effects must be applied to points A aand C, and thhe resistance oor safety
factors appliied to all threee points. Thiis process ressults in the daata for curve 4 which cann then be
used to checck the strength
h of a compossite beam-collumn.
EX
XAMPLE 9.12a
9
HS
HSS Composiite
Coolumn Stren
ngth
un
nder Combin
ned
Looading by LR
RFD
Goa
al:
Check
k the interacttion diagram of Chapter H for a givenn rectangularr HSS
underr combined ax
xial and flexuural loading.
Giv
ven:
The column
c
is com
mposed of ann HSS8×8×1/2 filled withh concrete. Itt must
carry an axial load
d of Pu = 2000 kips and a bbending mom
ment of Mu = 84 ftkips. It has an effeective length of 15 ft for bboth axes. The HSS has Fy = 46
a
fc′ = 5 ksi. Use tthe filled H
HSS column tables founnd at
ksi and
www.aisc.org/man
nualresources .
SO
OLUTION
Step
p 1:
Deterrmine the desiign axial strenngth.
From
m the on-line Table,
T
with K
KLy =15 ft,
φPn = 480 kip s
Step
p 2:
Deterrmine the flex
xural strength .
ble, the availlable flexural strength caan be found aat the
In thee on-line Tab
bottom
m of the tablee. Note that itt is not a funcction of unbraaced length.
φM n = 140 ft-kiips
Step
p 3:
Step
p 4:
Deterrmine which interaction
i
eqquation to usee, H1-1a or H1-1b.
Pu
200
=
= 0.4177 > 0.2
φPn 480
uation H1-1a..
Thereefore, use Equ
Check
k the interactiion equation.
Chapter 9
Composite Construction 423
8 ⎛ 84.0 ⎞
0.417 + ⎜
⎟ = 0.95 < 1.0
9 ⎝ 140 ⎠
Since the interaction equation is less than 1.0, this column will carry the
imposed load.
EXAMPLE 9.12b
HSS Composite
Column Strength
under Combined
Loading by ASD
Goal:
Check the interaction diagram of Chapter H for a given rectangular HSS
under combined axial and flexural loading.
Given:
The column is composed of an HSS8×8×½ filled with concrete. It must
carry an axial load of Pu = 133 kips and a bending moment of Mu = 56 ftkips. It has an effective length of 15 ft for both axes. The HSS has Fy = 46
ksi and fc′ = 5 ksi. Use the filled HSS column tables found at
www.aisc.org/manualresources.
SOLUTION
Step 1:
Determine the design axial strength.
From the on-line Table, with KLy =15 ft,
Pn Ω = 320 kips
Step 2:
Determine the flexural strength.
In the on-line Table, the available flexural strength can be found at the
bottom of the table. Note that it is not a function of unbraced length.
M n Ω = 93.0 ft-kips
Step 3:
Determine which interaction equation to use, H1-1a or H1-1b.
Pa ( Pn Ω ) = 133 320 = 0.416 > 0.2
Therefore, use Equation H1-1a.
Step 4:
Check the interaction equation.
8 ⎛ 56.0 ⎞
0.416 + ⎜
⎟ = 0.95 < 1.0
9 ⎝ 93.0 ⎠
Since the interaction equation is less than 1.0, this column will carry the
imposed load.
4224 Chapter 9
Composiite Construction
Figure 9.27
7
Composiite Column In
nteraction Diaagram Points:: Encased W--Shape
Copyright © American Institute of Steel Construction.
C
R
Reprinted with Permission. A
All rights reservved.
Chaptter 9
Figure 9.28
9
Compposite Construuction 425
Comp
posite Column
n Interaction Diagram Points: Filled HS
SS
Copyrightt © American Institute
I
of Steeel Constructioon. Reprinted w
with Permissionn. All rights resserved.
426 Chapter 9
9.13
Composite Construction
PROBLEMS
1. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W16×26 spanning 24 ft
and spaced 8 ft on center, supporting a 6 in. concrete slab.
Use fc′ = 4 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
2. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W16×45 spanning 20 ft
and spaced 8 ft on center, supporting a 5 in. concrete slab.
Use fc′ = 5 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
3. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W18×65 spanning 20 ft
and spaced 6 ft on center, supporting a 5 in. concrete slab.
Use fc′ = 5 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
4. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W18×71 spanning 18 ft
and spaced 5 ft on center, supporting a 4 in. concrete slab.
Use fc′ = 4 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
5. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W14×48 spanning 20 ft
and spaced 5 ft on center, supporting a 4 in. concrete slab.
Use fc′ = 3 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
6. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W14×68 spanning 24 ft
and spaced 6 ft on center, supporting a 4 in. concrete slab.
Use fc′ = 3 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
7. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W36×330 spanning 50 ft
and spaced 20 ft on center, supporting a 12 in. concrete
slab. Use fc′ = 6 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
8. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W27×102 spanning 30 ft
and spaced 14 ft on center, supporting a 6 in. concrete
slab. Use fc′ = 6 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
9. Determine the location of the plastic neutral axis and
the available moment strength for a flat soffit, fully
composite beam composed of a W10×12 spanning 12 ft
and spaced 4 ft on center, supporting a 5 in. concrete slab.
Use fc′ = 4 ksi and A992 steel. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
10. Repeat Problem 1 with the shear stud capacity limited
to Vq′ = 250 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
11. Repeat Problem 2 with the shear stud capacity limited
to Vq′ = 500 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
12. Repeat Problem 3 with the shear stud capacity limited
to Vq′ = 500 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
13. Repeat Problem 4 with the shear stud capacity limited
to Vq′ = 400 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
14. Repeat Problem 5 with the shear stud capacity limited
to Vq′ = 300 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
15. Repeat Problem 6 with the shear stud capacity limited
to Vq′ = 600 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
16. Repeat Problem 7 with the shear stud capacity limited
to Vq′ = 2000 kips. Determine (a) design strength by
LRFD and (b) allowable strength by ASD.
17. Repeat Problem 8 with the shear stud capacity limited
to Vq′ = 900 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
18. Repeat Problem 9 with the shear stud capacity limited
to Vq′ = 100 kips. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
19. Repeat Problems 5 and 14 with Vq′ = 450 kips and
plot the results of all three problems as a function of Vq′.
Determine (a) design strength by LRFD and (b) allowable
strength by ASD.
20. Repeat Problems 6 and 15 with Vq′ = 400 kips and
plot the results of all three problems as a function of Vq′.
Chapter 9
Determine (a) design strength by LRFD and (b) allowable
strength by ASD.
21. A W12 composite beam spaced every 8 ft is used to
support a uniform dead load of 1.0 k/ft and live load of
0.9 k/ft on a 20 ft span. Using a 4 in. flat soffit slab with
fc′ = 4 ksi, 3/4 in. shear studs, and A992 steel, determine
the least-weight shape and the required number of shear
connectors to support the load. Design by (a) LRFD and
(b) ASD.
Composite Construction 427
28. Determine the available moment strength for a
W18×46 A992 member used as a partially composite
beam to support 3 in. of concrete on a 3 in. metal deck for
a total slab thickness of 6 in. The metal deck is
perpendicular to the beam. The beam spans 30 ft and is
spaced 11 ft from adjacent beams. Shear stud strength is
Vq′ = 400 kips, fc′ = 5 ksi. Determine (a) design strength
by LRFD and (b) allowable strength by ASD.
22. A W14 composite beam is to support a uniform dead
load of 1.0 k/ft and live load of 1.4 k/ft. The beam spans
24 ft and is spaced 8 ft from adjacent beams. Using a 5 in.
flat soffit slab and 3/4 in. shear studs, determine the leastweight shape and the required number of shear connectors
to support the load if fc′ = 4 ksi and A992 steel is used.
Design by (a) LRFD and (b) ASD.
29. A composite beam is to span 20 ft and support a 4 in.
slab including a 1-1/2 in. metal deck. The deck span is 10
ft. The beam must accommodate a uniformly distributed
dead load of 75 psf including the slab weight and live
load of 100 psf. The deck has 2 in. ribs spaced 6 in. on
center, with deck ribs perpendicular to the beam.
Determine the required A992 W-shape and number of 3/4
in. shear studs. Use fc′ = 3 ksi. Design by (a) LRFD and
(b) ASD.
23. A W16 composite beam spaced every 12 ft is used to
support a uniform dead load of 1.2 k/ft and live load of
1.9 k/ft on a 20 ft span. Using a 4 in. flat soffit slab with
fc′ = 5 ksi, 3/4 in. shear studs, and A992 steel, determine
the least-weight shape and the required number of shear
connectors to support the load. Design by (a) LRFD and
(b) ASD.
30. Determine the required W-shape and number of 3/4
in. shear studs for a composite girder that spans 30 ft and
supports two concentrated dead loads of 12 kips and live
loads of 20 kips at the third points. The 1-1/2 in. metal
deck with 2 in. ribs spaced at 6 in. on center is parallel to
the girder and supports a total slab of 5 in. Use fc′ = 4 ksi
and A992 steel. Design by (a) LRFD and (b) ASD.
24. A W16 composite beam is to support a uniform dead
load of 1.8 k/ft and live load of 2.2 k/ft. The beam spans
26 ft and is spaced 10 ft from adjacent beams. Using a 5
in. flat soffit slab and 3/4 in. shear studs, determine the
least-weight shape and the required number of shear
connectors to support the load if fc′ = 4 ksi and A992 steel
is used. Design by (a) LRFD and (b) ASD.
31. Determine the live load deflection for a W24×76
A992 composite beam with a 6 in. total thickness slab on
a 3 in. metal deck. The beam spans 28 ft and is spaced at
10 ft intervals. The beam is to carry a live load of 3.4 k/ft.
Assume Y2 = 5.5 in. and ΣQn=394 kips.
25. Compare the least-weight A992 W16 and W14
members required to support a uniform dead load of 2.4
k/ft and live load of 3.2 k/ft. The beams span 18 ft and are
spaced 12 ft on center. They support a 6 in. concrete slab
with fc′ = 4 ksi. Design by (a) LRFD and (b) ASD.
26. W16×31 A992 composite beams are spaced at 10 ft
intervals and span 24 ft. The beams support a 2-1/2 in.
metal deck perpendicular to the beam with a slab whose
total thickness is 5 in. Assuming fully composite action,
determine the available moment strength and the number
of 3/4 in. shear studs required. The deck has 6 in. wide
ribs spaced at 12 in. Use fc′ = 4 ksi. Determine by (a)
LRFD and (b) ASD.
27. Determine the available moment strength of a
W18×35 A992 composite beam supporting a slab with a
total thickness of 5 in. on a 3 in. metal deck perpendicular
to the beam. The beam spans 28 ft and is spaced 12 ft
from adjacent beams. Use fc′ = 5 ksi. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
32. Determine the live load deflection for a W16×26
A992 composite beam supporting a 6 in. slab on a 2-1/2
in. metal deck. The beam spans 24 ft and is spaced at 8 ft
on center. The live load is 2.1 k/ft. Assume Y2 = 5.5 in.
and ΣQn= 384 kips.
33. Determine the live load deflection for the beam of
Problem 21.
34. Determine the live load deflection for the beam of
Problem 22.
35. Determine the available compressive strength of a 20
ft effective length 18×18 in. composite column encasing
an A992 W10×68 and eight #8, Gr. 60 reinforcing bars, fc′
= 5 ksi. Each face has three bars with their centers located
2.5 in. from the face of the concrete. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
36. Determine the available compressive strength of a
22×22 in. composite column with an effective length of
16 ft. The concrete encases an A992 W12×120 and eight
#9, Gr. 60 bars, fc′ = 5 ksi. Each face has three bars with
their centers located 2.5 in. from the face of the concrete.
428 Chapter 9
Composite Construction
Determine (a) design strength by LRFD and (b) allowable
strength by ASD.
37. Determine the available compressive strength of a
24×24 in. composite column with an effective length of
12 ft. The concrete encases an A992 W14×132 and eight
#10, Gr. 60 bars, fc′ = 5 ksi. Each face has three bars with
their centers located 2.5 in. from the face of the concrete.
Determine (a) design strength by LRFD and (b) allowable
strength by ASD.
38. Determine the available compressive strength of an
HSS10×5×3/8 composite column with an effective length
of 12 ft. The concrete has fc′ = 5 ksi. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
39. Determine the available compressive strength of an
HSS12×10×1/2 composite column with an effective
length of 16 ft. about the strong axis and 8 ft for the weak
axis. The concrete has fc′ = 5 ksi. Determine (a) design
strength by LRFD and (b) allowable strength by ASD.
40. Using interaction equations H1-1a and H1-1b,
determine if an HSS10×6×3/16 composite column with an
effective length of 20 ft will carry an axial compressive
live load of 25 kips and dead load of 18 kips, and a live
load moment of 10 ft-kips and a dead load moment of 8
ft-kips, both bending the member about its strong axis.
Use f’c = 5 ksi. Determine (a) design strength by LRFD
and (b) allowable strength by ASD.
41. Integrated Design Project. The framing plans shown
in Figures 1.20 and 1.21 were used to design beams and
girders in Chapter 6. These same gravity-only beams and
girders will now be designed as composite members.
Manufacturers’ data on composite metal decks may be
found on the Internet. For the designs carried out here,
consider a 3 in. deck with a profile similar to that shown
in Figure 9.16c. It will be used with 3 in. of concrete with
f c′ = 4 ksi for a total slab thickness of 6.0 in.
Chapter
10
Conneection
Elements
Marlins Parkk, Miami, FL
Photo: © Chhristy Radecic (www.christyrradecic.com)
10.1
INT
TRODUCTIION
A steel building
b
structure is essenttially a collecction of indiviidual memberrs attached too each other
to form a stable and serviceable wh
hole, called thhe frame. Thee assumed beehavior of the connection
between any two mem
mbers determ
mines how thee structure is analyzed to rresist gravityy and lateral
loads. Th
moments, shhears, and axxial loads forr which the
his analysis, in turn, deteermines the m
beams, columns,
c
and
d other memb
bers are desiggned. It is, ttherefore, esssential that thhe designer
understan
nd the basic behavior
b
of co
onnections.
Members
M
are attached to each other thhrough a variiety of conneecting elemennts, such as
plates, an
ngles, and oth
her shapes, ussing mechaniccal fasteners or welds. Thee characteristtics of these
connectin
ng elements and fastenerss must be unnderstood to assess the reesponse of thhe complete
connectio
on. With each
h connection
n, the load traansfer mechannism must bee understood so that the
applicablle limit states of the joint can
c be evaluatted.
Table
T
10.1 lissts the section
ns of the Speccification and parts of the M
Manual discuussed in this
chapter.
10.2
BAS
SIC CONNE
ECTIONS
The wide variety of potential mem
mber geomettries and arraangements avvailable for cconstruction
makes a listing of the correspondin
ng connectionns a very dem
manding exerccise. Every joint between
memberss must be analyzed and dessigned accordding to the unnique aspects oof that joint.
Figure
F
10.1 shows
s
severall examples o f tension connnections. Thhe connectionns shown in
Figure 10
0.1a, b, and c illustrate waays that a tenssion member ccan be spliced. In each casse, the bolts
are subjeected to a sh
hear force. Th
he butt jointt (Figure 10.11a) and the llap joint (Figgure 10.1b)
provide a connection between two
o members, w
whereas the jooint shown inn Figure 10.1cc shows the
connectio
on of a singlee member to a pair of mem
mbers. This tyype of joint can also be coonsidered as
a portion
n of the butt joint
j
shown in
i Figure 10..1a. The jointt shown in Figure 10.1d rrepresents a
hanger co
onnected to th
he lower flang
ge of a beam;; in this case tthe connectioon is accomplished with
429
4330 Chapter 10
1
Connecttion Elementss
Table 10.1 Sections off Specification
n and Parts off Manual Covvered in This Chapter
Specification
S
B3
Chapter D
Chapter E
G2
J1
J2
J3
J4
Design
n Basis
Design
n of Memberss for Tension
Design
n of Memberss for Compresssion
Memb
bers with Unsttiffened or Sttiffened Webss
Generaal Provisions
Welds
Bolts and
a Threaded
d Parts
Affected Elements of
o Members aand Connecting Elements
Manual
Part 7
Part 8
Part 9
Design
n Consideratio
ons for Bolts
Design
n Consideratio
ons for Weldss
Design
n of Connecting Elements
a WT-shapee, and the boltts are subjectted to a tensille load. The cconnection off a tension meember to
a gusset platte is shown in
n Figure 10.1ee. Here againn, the bolts aree subjected too a shear force. All of
these examp
ples illustratee bolted con
nnections. Sim
milar connecctions can bee accomplishhed with
welds.
The connections illustrated in Figure 10.2 aare bracket coonnections. T
The connectionn shown
in Figure 10
0.2a is a brack
ket attached to
o the flange oof a column. IIn this case, thhe bolts are suubjected
to shear and
d moment in the
t plane of th
he connectionn when loaded as shown. T
The bracket sshown in
Figure 10.2
2b, when loaaded as show
wn, subjects the bolt group to shear in the planee of the
connection and
a a momen
nt out of the pllane that resuults in a tensile force in thee top bolts.
100.3 BEAM
M-TO-COLU
UMN CONN
NECTIONS
S
Design of beam--to-column connections
c
iss addressed iin Chapters 111 and 12. T
They are
discussed briefly
b
here since
s
each connection
c
iss composed of a combinnation of connnection
elements. Th
he connection
n of a beam to
t a column ccan also be acccomplished in a variety oof ways.
Figure 10.3
3 illustrates several conn
nections of W-shape beeams to W-sshape columnns. The
classification
n of these con
nnections is a function of tthe forces being transferreed between thhe
Figure 10.1 Tension Connectio
ons
Chappter 10
Figure 10 .2
Connection Elem
ments 431
Bracket C
Connections
memberss. The connecctions shown
n in Figure 100.3a through d are usuallyy called simpple or shear
connectio
ons and are covered in Chapter 11, whereas thhose in Figurre 10.3e throough h are
generally
y referred to as
a fixed or mo
oment connecttions and covvered in Chapter 12.
In
I normal prractice beam
m-to-column connections are classifieed as simplee or fixed,
however; these connections actuallly exhibit a w
wide range of behaviors, ass discussed inn Chapter 8.
havior can be
b described through a pplot of the m
moment-rotatiion characterristics of a
This beh
particular connection
n. Typical moment-rotaation relatioonships for three beam
m-to-column
connectio
ons are preseented in Figurre 10.4. Whenn a connectioon is very stifff, it deformss very little,
even wheen subjected to
t large mom
ments. This typpe of connectiion is represeented by curvee a in
Figure 10.3 Beam-to-C
Column Connections
4332 Chapter 10
1
Connecttion Elementss
Figure 10.4 Beam-to-Column MomeentRotation Currves
Figure 10.4.. At the otherr extreme, wh
hen a connecttion is quite fflexible, it will rotate conssiderably
but will not develop a sig
gnificant mom
ment, as show
wn by curve c in Figure 100.4. Curve b inn Figure
10.4 is reprresentative off any connecction whose moment rotaation behavioor occurs som
mewhere
between currves a and c.
c Despite som
me appreciabble stiffness, these conneections still eexhibit a
degree of flexibility; th
hus, significant rotation will occur along with significant moment
resistance.
For the purposess of design, co
onnections haave usually bbeen assumedd to behave acccording
to the simpllified behavio
ors representeed by the vert
rtical axis of F
Figure 10.4 aas a fixed connnection
and the horrizontal axis of Figure 10.4 as a sim
mple connectiion. Because connectionss do not
actually beh
have in this way,
w
those thaat follow curvves a and c aand exhibit a behavior closse to the
idealized connection arre called fix
xed connecttions and siimple conneections, respeectively.
Specification
n Section B3
3.4 divides connections
c
into two cattegories: sim
mple connections and
moment con
nnections. Th
he moment co
onnection cat
ategory includdes fully resttrained (FR) moment
connections and partially
y restrained (P
PR) moment connections.. FR connectiions transfer moment
with a negliigible rotation
n between thee connected m
members, as sshown in curvve a. PR connnections
transfer mom
ment between the membeers but the rrotation is noot negligible, as demonstrrated by
curve b.
It iss the designeer’s responsib
bility to mattch connectioon behavior with the apppropriate
analysis mo
odel and to complete
c
the connection ddesign so thaat the actual connection bbehavior
matches thatt used in the analysis.
a
Ofteen, this requirres experiencee and judgmeent; the state oof the art
is such that it is usually not
n possible to
o accurately ppredict the M
M-θ curve for aanything but tthe most
basic of con
nnections.
100.4
FULLY
Y RESTRA
AINED CON
NNECTION
NS
The basic assumption fo
or frames with
h FR connecctions is that the beams annd columns m
maintain
their originaal geometric relationship over the enttire loading hhistory. This is normally called a
rigid or fixeed connection
n. The diagraams in Figuree 10.3e throuugh h show examples of bbeam-tocolumn connections that are usually treated
t
as FR
R connectionss. Although tthey may shoow some
relative rotation between members, theey have sufficcient stiffnesss to justify ignnoring this rootation.
Figu
ure 10.3e shows a connecttion with a weeb plate shopp-welded to thhe column flaange and
field-bolted to the beam
m web. The beeam flanges have been beveled in thee shop and are field-
Chapter 10
Connection Elements 433
welded to the column. Although the beam web is not continuously connected to the column, it has
been repeatedly demonstrated that this connection can adequately transfer the full plastic moment
of the beam to the column. The moment strength is derived mostly from the flange connections
and is equal to the flange force times beam depth. The small amount of moment in the web
connection and local strain hardening in the flanges add to the connection’s ability to reach the
full plastic moment of the beam. This connection is generally known as the pre-Northridge
connection because it was the de-facto standard connection for seismic applications prior to the
1994 Northridge, California, earthquake. Because its performance under the seismic load of that
event was below expectations, it is no longer used in seismic resisting frames. However, it is still
used to resist moments due to gravity and wind loads.
Figure 10.3f is similar to Figure 10.3e except that the beam frames into the web of the
column. To ensure that this connection has adequate ductility, it is important to extend the flange
connecting plates beyond the column flange and to design these plates to be a little thicker than
the beam flange. Extending the connecting plate reduces the possibility of a tri-axial stress
condition near the column flange tips. Thickening the plate reduces the average tension stress in
the plate. It also facilitates welding to the beam flange.
The connection illustrated in Figure 10.3g is a flange-plate connection. As with the
connection shown in Figure 10.3e, the web is connected to transfer the beam shear force only.
The flange forces are first transferred to the top and bottom plates and then to the column flange.
This is shown as a bolted connection, but it is also possible to fabricate this as a welded
connection. For fully welded connections, special care should be taken to address support during
erection prior to field welding; this usually means there will be some bolts used even in a fully
welded connection.
Figure 10.3h is an extended end-plate connection. For this connection, a plate is shopwelded to the end of the beam and then bolted to the column flange. Although this connection is
very popular with some fabricators, others tend to avoid it. It must be fabricated with special care
so that the end plates are parallel to each other. Also, it is not a very forgiving connection and it
can make erection difficult and expensive.
Fully restrained connections are covered in Chapter 12.
10.5 SIMPLE AND PARTIALLY RESTRAINED CONNECTIONS
Analysis of a frame with PR connections must account for the actual moment-rotation
characteristics of the connection. These connections are now referred to as partially restrained
connections but have historically been called semi-rigid connections. It is typically not possible to
determine whether a connection should be classified as PR just by looking at it. Several
connections that appear to be simple actually have the potential to resist significant moment. In
the simple connection case, the analysis assumes that the connections are pinned and free to
rotate. The rotation capacity of the connection must be sufficient to accommodate the simple
beam rotation of the beam to which it is connected.
There are basically two ways in which this type of frame can be designed to resist lateral
loads and to provide stability for gravity loads. In one case, a positive bracing system is provided,
such as diagonal steel bracing or a shear wall. In the second case, lateral stability is provided by
the limited restraint offered by the connections and members themselves. This type of connection
is called a flexible moment connection. Flexible moment connections are designed for a limited
amount of moment resistance accompanied by a significant amount of rotation. The connections
are flexible enough to rotate under gravity loads so that no gravity moments are transferred to the
columns. At the same time they are assumed to have sufficient strength and stiffness to resist the
lateral loads and to provide frame stability. This approach to frame design is a simplification of
convenience based upon experience, judgement and past usage. It was addressed in Section 8.11.
434 Chapter 10
Connection Elements
Design of these flexible moment connections will follow the same approach as that for other
connections to be discussed later.
The design of PR connections is more involved than the flexible moment connection
approach because it requires that the frame be analyzed considering the true semi-rigid behavior
of the connections. In this case, the actual M-θ curve of the connection must be known. The
resulting analysis tends to be rather complex because of the nonlinear behavior of the connection.
Although there are currently no commercially available computer programs for analysis of frames
with PR connections, there are simplified approaches that will aid in the use of these connections.
Figure 10.3 shows examples of simple and PR connections. As mentioned earlier, it is not
normally possible to tell by visual inspection whether a connection should be treated as a PR
connection. Figure 10.3a shows a double-angle connection, also referred to as a clip angle
connection. This connection has been used extensively over the years. In fact, it is usually the
standard to which other simple connections are compared. Even though it is readily accepted as a
simple connection, it has been shown that under certain circumstances it can be relied upon to
resist some moment from lateral load.
Figure 10.3b shows a single-plate framing connection that is often referred to as a shear
tab. Care must be taken when these connections are designed as simple connections to ensure that
the elements have sufficient flexibility to accommodate the simple beam rotation.
Figure 10.3c shows a seated connection and Figure 10.3d a stiffened seated connection.
Either can be bolted or welded, and they are usually used to frame a beam into the web of a Wshaped column section. Although they may appear to be stiffer than the standard double-angle
connection, they are designed to rotate sufficiently without transferring a moment to the column,
so they can be treated as simple connections.
Simple shear connections are treated in Chapter 11.
10.6
MECHANICAL FASTENERS
The mechanical fasteners most commonly used today are bolts. The Specification provides for the
use of common bolts and high-strength bolts. It also provides some direction for cases where
bolts are to be used in conjunction with rivets in new work on historic structures. There are no
provisions for rivets in new construction, however, because these connectors are no longer used
in new construction of buildings.
10.6.1
Common Bolts
Common bolts are manufactured according to the ASTM A307 specification, as discussed in
Section 3.6.3. When used, they are usually found in simple connections for such elements as girts,
purlins, light floor beams, bracing, and other applications where the loads are relatively small.
Although permitted by the Specification, they are not recommended for normal steel-to-steel
connections and should not be used where the loads are cyclic or vibratory, or where fatigue may
be a factor.
Common bolts are also called machine, unfinished, or rough bolts. They have square or
hexagonal heads and nuts and are identified by the grade designation 307A or 307B on the heads.
They are available in diameters from 1/4 in. to 4 in.
These bolts are usually installed using a spud wrench. No specified pre-tension is
required. Because no clamping force is assumed, it is necessary only to tighten the nut
sufficiently to prevent it from backing off of the bolt. The design shear and tensile strength are
given in Specification Section J3, Table J3.2.
Chappter 10
10.6.2
Connection Elem
ments 435
Hig
gh-Strength Bolts
B
High-streength bolts arre presented in Specificattion Section JJ3 in three grroups accordding to their
material strength. Fou
ur Group A high-strength
h
bolts are currrently permittted in steel sttructures as
specified
d by ASTM F3125
F
Grades A325, A325M
M and F18522 and ASTM A354, Gradee BC. Grade
A325M is
i the metric equivalent
e
off A325. Thesee bolts all havve the same m
minimum tensiile strength,
Fu = 120
0 ksi. Four Group
G
B boltts are permittted as speciffied by ASTM
M F3125 Grades A490,
A490M and
a F2280 an
nd ASTM A354, Grade BD
D. Grade A4990M is the meetric equivaleent of A490.
These bo
olts also all have
h
the samee minimum teensile strength
th, Fu = 150 kksi. Two Grooup C bolts,
ASTM F3043
F
and F3111 with min
nimum tensilee strength of 200 ksi, are ppermitted. Usse of Group
C bolts iss restricted by
y Section J3.1. Details of tthe material aand other properties of theese bolts are
described
d in Section 3.6.3.
3
Bolt strrength is baseed on the tenssile strength oof the bolt maaterial. The
nominal strength of Group
G
B boltts is 25 perceent greater thhan that of G
Group A boltss. Bolts are
generally
y referred to by
b their gradee designationn without repeeating the AS
STM F3125 ddesignation.
Thus, in this book thaat model will be followed.. All Group A and B boltss can be used for simple,
FR, or PR
P connection
ns and for bo
oth static andd dynamic looading. Bolts have alwayss been very
popular for field installation. Th
heir use in the shop haas increased considerablyy with the
introducttion of autom
mated equipmeent and tensioon control bollts (F1852 andd F2280).
A325,
A
A325M
M and F1852 bolts are avaailable in twoo types. Type 1, manufactuured from a
medium--carbon steel,, is the most commonly uused. It is avvailable in sizzes ranging fr
from 1/2 in.
through 1-1/2 in. in diameter. Ty
ype 3 is a weeathering steeel bolt with ccorrosion chaaracteristics
similar to
o those of AS
STM A242, A588,
A
and A8 47 steels. Typ
ype 3 bolts aree also availabble from 1/2
through 1-1/2
1
in. in diiameter. They
y all have a m
minimum tensiile strength Fu = 120 ksi.
A490,
A
A490M
M and F2280
0 bolts are allso available as Type 1 aand Type 3 aand in sizes
ranging from
f
1/2 in. to
o 1-1/2 in. in diameter. Alll have a minim
mum tensile sstrength Fu = 150 ksi.
A325
A
Type 1 bolts are iden
ntified by thee mark “A3255” on the bolt head. Type 3 bolts have
the desig
gnation “A325” underlined
d. A490 boltss are similar,, carry the m
marking “A490” with the
identificaation underlin
ned for Type 3. Example bbolt markings are shown iin Figure 10.5. All bolts
should allso be marked
d with a symb
bol designatinng the manufaacturer as shoown in this caase by SL in
the figuree.
A354
A
bolts arre manufacturred as quenchhed and tempeered alloy steeel in two graddes, BC and
BD. Any
y steel that meets
m
the mecchanical and cchemical prooperties in thee standard maay be used.
Grade BC
C bolts have a tensile stren
ngth of 125 orr 115 ksi, bassed on diametter, but are inccluded with
Group A bolts where a tensile stren
ngth of 120 kksi is used. G
Grade BD boltts have a tenssile strength
of 150 orr 140 ksi, based on diameteer, but are inccluded with G
Group B boltss where a tenssile strength
of 150 kssi is used. Th
hese bolts are available in sizes rangingg from 1/4 in. to 4.0 in. diaameter with
diameters over 2-1/2 in. having th
he reduced strrength. A354 bolts are maarked with thee grade and
may be marked
m
with 6 radial lines, 60° apart forr grade BD.
Figure 10.5
1
Examplle Bolt Identiification Markkings from Stt. Louis Screw
w and Bolt Coompany
4336 Chapter 10
1
Connecttion Elementss
Figure 10.6 Bolt Definitions
D
Group C bolts aree currently on
nly available as a proprietaary product annd will not bee used in
this book altthough the baasic concepts of design disccussed will eqqually apply tto them.
Figu
ure 10.6a sho
ows the principal parts aand dimensionns of a highh-strength bollt: head,
shank, bolt length,
l
and th
hread length. Figure 10.6bb shows the pprincipal partss of a tensionn control
bolt.
Both
h A325 and A490
A
bolts caan be installeed with a spud wrench, shhown in Figurre 10.7a,
or in cases where
w
a clamp
ping force is necessary, ann impact wrennch shown inn Figure 10.7bb. F1852
and F2280 bolts
b
are instaalled with a mechanical
m
ddevice that sim
multaneously holds the boolt shank
and nut and rotates them relative to eaach other. Thee end of the bbolt twists offf when the prescribed
tensile forcee is reached,, ensuring the required p re-tension. F
Figure 10.7c shows a wreench for
installation of these ten
nsion control bolts. It shhould be noteed that placeement of boolts in a
connection must
m provide sufficient cleearance for wrrench access.
100.6.3
Bolt Holes
H
w
bolts aree inserted willl impact the strength of thhe bolts in plaace, it is
Because thee holes into which
important to
o address the hole requirem
ments at this point. The SSpecification defines four types of
bolt holes th
hat are permiitted in steel construction:: standard, ovversized, shorrt slot, and loong slot.
Table 10.2 shows
s
the nom
minal hole diimensions forr each of thesse types and ffor bolts from
m 1/2 in.
diameter up. Figure 10.8 shows the fou
ur hole sizes for a 3/4 in. bbolt.
Stan
ndard holes orr short-slotted
d holes transvverse to the diirection of loaad are the staandard to
be used unlless one of th
he other typees is permitteed by the deesigner. This is because thhe other
arrangementts can reduce the final bolt strength andd allow largerr deformationns from moveement of
bolts in holees under loadiing. A standarrd hole has a maximum diameter that iss 1/16 in. greaater than
(a) Sp
pud Wrenchess
(b) Impact Wrrench
Figure 10.7
7 Wrenches used
u
to installl Bolts
Photos courteesy Tone Co., Ltd.
L
(c) Tension C
Control Wrennch
Chappter 10
Connection Elem
ments 437
Figure 10.8
1
Hole Siizes for a 3/4 in. Diameter Bolt
the bolt diameter up to 7/8 in. bo
olts and 1/8 in. for 1.0 iin. bolts and larger to accommodate
placemen
nt of the bollt. Short-slottted holes havve this same dimension iin one directiion but are
elongated
d in the otherr direction to assist in fit-uup of the connnection partss. Any slot loonger than a
short slott should be cllassified as a long slot, eveen if it is not the full lengtth of a long slot as given
in Table 10.2.
Oversized
O
holles and long--slotted holes are specifiedd when increaased tolerancce is needed
to accom
mplish the actual
a
connecction. If a ddesign includdes other thhan standard holes, the
requirem
ments of Speciification Secttion J3.2 for w
washers com
me into play. F
For the exam
mples in this
book, on
nly standard holes will be used.
u
In
I addition to
o prescribing the sizes of bolt holes, thhe Specificattion gives minnimum and
maximum
m hole spacing and edg
ge distances. Figure 10.99 shows a pplate with hoole spacing
dimensio
oned with thee standard vaariable namees used in thhe Specificatiion. The minnimum hole
spacing, s, for standaard, oversized
d, or slotted holes must nnot be less thhan 2-2/3 tim
mes the bolt
diameter. A spacing of
o 3 diameterss, 3d, is preferrred. It will be shown laterr that even at a minimum
spacing of
o 3d, bolt strrength may be
b less than w
what it could bbe if the spaccing were jusst a little bit
greater. The
T maximum
m spacing of bolts
b
in a connnection is 244 times the thiickness of thee connected
part or 12 in. This maaximum is no
ot a strength rrequirement bbut rather onee that is intendded to keep
the conneection plies in
n contact and prevent any ppotential moiisture build-up
up between thee elements.
The
T minimum
m edge distan
nces, le, speciffied are intennded to facilittate constructtion and are
not stren
ngth related. Table 10.3 shows
s
the miinimums from
m Specificatiion Table J3..4. Because
these dim
mensions willl be shown to
t directly im
mpact bolt str
trength, it is critical to prrovide edge
distancess that are com
mpatible with
h the requiredd strength of the connection. The maxximum edge
distance is 12 times the thickness of the connnected part oor 6 in. for elements nott subject to
corrosion
n, for the sam
me reasons as for
f the limits on bolt spaciing.
Tab
ble 10.2 Nom
minal Hole Dimensions,
D
in
n.
Hoole dimensionns
Bolt
diameter
1/2
5/8
3/4
7/8
1
≥1-1/8
Standaard
(dia.))
9/16
11/16
6
13/16
6
15/16
6
1-1/8
8
d + 1/8
Oversize
O
(dia.)
5/8
13/16
15/16
1-1/16
1-1/4
d + 5/16
Shortt slot
(width × length)
9/16 × 11/16
11/16 × 7/8
13/166 × 1
15/16 × 1-1/8
1-1/8 × 1-5/16
(d + 1/8) × (d + 3/8)
Long slot
(width × llength)
9/16 × 1-1/4
11/16 × 1-9/16
13/16 × 1-7/8
15/16 × 22-3/16
1-1/8 × 2-1/2
(d + 1/8) × (2.5 × d)
4338 Chapter 10
1
Connecttion Elementss
Figgure 10.9 H
Hole Spacing, Gage, and
Edgge Distances
100.7
BOLT LIMIT STA
ATES
Three basic limit states govern the response
r
of bbolts in bolteed connectionns: shear throough the
shank or thrreads of the bolt,
b
bearing or tearout onn the elementts being connnected, and teension in
the bolt. Cases where loaad reversals are
a expected, or fatigue iss a factor, havve an additionnal limit
state to preevent slip in the connectiion. This lim
mit state appllies only to connections that are
classified ass slip-criticall connectionss. For bolts iin shear connnections, the nominal connnection
strength is th
he sum of thee lowest stren
ngth for the lim
mit states of bbolt shear, beearing, and teaarout for
each bolt in the connectio
on.
Boltts may be insstalled to a sn
nug-tight conndition or preetensioned. S
Snug-tight insstallation
has no requ
uired specific level of tenssion. This is commonly aattained after a few impaccts of an
impact wren
nch or the fulll effort of an ironworker w
with an ordinaary spud wrennch. The plies should
be in firm contact, a con
ndition that means
m
the pliees are solidly seated againsst each other,, but not
necessarily in
i continuouss contact. Snu
ug-tight bolts are permittedd in connectioons that are classified
as bearing-ty
ype connectio
ons.
Prettensioned bollts are installeed to a tensioon specified iin Table J3.1 in the Specif
ification.
They must be
b used in con
nnections sub
bjected to vibbratory loads where bolt looosening is a concern
and in end connections of built-up members.
m
Preetensioned bolts are requuired for connections
designed as slip-critical connections.
c
Table 10.3 Minimum Edge
E
Distanceea from Centter of Standarrd
b
Hole to Edg
ge of Connectted Part
Bolt diameteer (in.)
Minimum eddge distance (in.)
1/2
3/4
5/8
7/8
3/4
1
7/8
1-1/8
1
1-1/4
1-1/8
1-1/2
11/4
1-5/8
>1-1/4
1--1/4 × d
a
If necessary, lesser edge
e
distancces are perm
mitted providded the
appropriate provisions from
fr
Sectionss J3.10 and JJ4 are satisfiied, but
edge distancces less than one bolt diaameter are noot permitted w
without
approval fro
om the engineeer of record.
b
For oversizzed or slotted holes, see Speecification Taable J3.5.
Chappter 10
Connection Elem
ments 439
Figure 10.10
1
Bolt Failure
F
Modess
10.7.1
Bollt Shear
pplication of bolts
b
in connnections is to resist shear. Shear throughh the shank
The most common ap
of the bo
olt is the mean
ns whereby th
he load, P, inn Figure 10.100a is transferrred from one plate to the
other. In this case, thee bolt is shearred along onee plane. Thus,, it is said to bbe a bolt in siingle shear.
The arran
ngement in Figure 10.10b shows two siide plates connnected to a ccentral plate. IIn this case,
the load, P, is transferrred from the center plate tto the side plaates and the bbolt is thereforre loaded in
double sh
hear. A bolt in
n double sheaar has twice thhe shear strenngth as a bolt in single sheaar.
For
F the limit state of bolt shear, the noominal strenggth is based oon the tensile strength of
the bolt and
a the locatiion of the sheear plane withh respect to thhe bolt threadds. Section J33.6 provides
that
Rn = Fn Ab
((AISC J3-1)
and
Ω = 2.000 (ASD)
φ = 0.75 (LRFD)
where
Fn = shear streess, Fnv, from
m Specificationn Table J3.2
Ab = area of th
he bolt shank
Tab
ble 10.4 Nom
minal Stress of
o Fasteners and
a Threadedd Parts, ksi
Description of faasteners
A30
07 bolts
Group A (e.g., A325) bolts, when
w
threads are
a not
excluded from
m shear planees (N)
Group A (e.g., A325) bolts, when
w
threads are
a excluded
from shear pllanes (X)
Group B (e.g., A4
490) bolts, when threads are
a not
excluded from
m shear planees (N)
Group B (e.g., A4
490) bolts, when threads are
a excluded
from shear pllanes (X)
Threeaded parts meeting
m
the requirements of Section A3..4,
when threadss are not exclu
uded from sheear planes (N
N)
Threeaded parts meeting
m
the requirements of Section A3..4,
when threadss are excluded
d from shear planes
p
(X)
Nominnal tensile
stresss, Fnt, ksi
Nominal sshear stress
in beariing-type
connectionns, Fnv, ksi
45
90
227
554
90
668
113
668
113
884
0.75Fu
0.4550Fu
0.75Fu
0.5663Fu
440 Chapter 10
Connection Elements
The information in Table 10.4 is taken from Specification Table J3.2. Each high-strength fastener
has two descriptions. The first is for cases where the threads are not excluded from the shear
plane, and the second is for when the threads are excluded from the shear plane. Because, in
every case, the area of the bolt shank is used to determine the nominal strength, the reduction in
area when the shear plane passes through the threads is accounted for by reducing the nominal
shear stress. This is done as a convenience in design so the area can be calculated using the
known basic bolt diameter without having to calculate the reduction for threading. When threads
are excluded from the shear plane, the bolts are called either A325-X or A490-X bolts. In these
cases, Fnv = 0.563Fu. When threads are not excluded from the shear plane, the bolts are referred to
as either A325-N or A490-N bolts. In these cases, Fnv = 0.45Fu. Only one value is provided for
A307 bolts, and that value is based on the assumption that the threads are included in the shear
plane.
Unless the designer can be sure that the final connection will result in the bolt threads
being excluded from the shear plane, it is usually best to design the connection for the worst case
of threads included in the shear plane.
10.7.2
Bolt Bearing and Tearout
The available strength for the limit state of bearing and tearout at bolt holes is specified in Section
J3.10. Because the material strength of a bolt is greater than that of the material it is bearing on,
the only bearing check is for bearing on the material of the connected parts. The Specification
provision considers two limit states for bearing strength at bolt holes: the limit state based on
shear in the material being connected, as shown in Figure 10.10c, and the limit state of material
crushing, as shown in Figure 10.10d.
When the clear distance from the edge of the hole to the edge of the part or next hole is
less than twice the bolt diameter, the limit state of shear in the plate material, also referred to as
tearout, will control. In this case, failure occurs by a piece of material tearing out of the end of the
connection as shown in Figure 10.10c or by tearing between holes in the direction of force. The
nominal strength for this failure mode, Rn, is provided by shear along the two planes. From
statics,
Rn = (shear strength)(2 planes)(clear distance)(material thickness)
Rn = 0.6 Fu ( 2lc ) t = 1.2lc tFu
(AISC J3-6c)
where
0.6Fu = ultimate shear strength of the connected material, ksi
t = thickness of the material, in.
lc = clear edge distance, measured from the edge of the hole to the edge of the
material or the next hole; unlike net shear and net tension area calculations,
no deduction is made at the hole edge for damage in bearing calculations
If the clear distance exceeds 2d, bearing on the connected material will be the controlling limit
state, as shown in Figure 10.10d. In this case, the limit state is that of hole distortion and the
calculated bolt strength will be
Rn = 2.4dtFu
(AISC J3-6a)
where
= bolt diameter
d
= connected part thickness
t
Fu = tensile strength of the connected part
If deformation at the bolt hole is not a design consideration at service loads, both of these limit
states may be increased by one-quarter, so that for tearout
Chappter 10
Connection Elem
ments 441
Rn = 1.5lc tFu
(A
AISC J3-6d)
Rn = 3.0dtFu
(A
AISC J3-6b)
and for bearing
b
olts are used in a connectio
on with long sslots and the fforce is perpeendicular to thhe slot, bolt
When bo
strength is reduced such that for teaarout
Rn = 1.0lc tFu
(A
AISC J3-6f)
b
and for bearing
Rn = 2.0dtFu
(A
AISC J3-6e)
As
A was the caase for bolt sh
hear, the resisstance and saafety factors ffor the limit sstate of bolt
bearing are
a
φ = 0.75
0
(LRFD)
Ω = 2.000 (ASD)
10.7.3
Strength at Boltt Holes
Normal use
u of bolts iss to hold two or more piecces of steel toogether. This of course reqquires holes
in each piece
p
to be co
onnected. Wh
hen bolts are pplaced througgh holes and a shear forcee is applied,
their avaailable strength will be th
he lowest vallue obtained according too the limit staates of bolt
shear, beearing on each
h connected element
e
and teearout at the bolt hole on each connectted element.
The nom
minal strength of a connectiion in shear iss the sum of tthe nominal sstrength of thee individual
bolts.
For
F a bolt in standard, oveersized, or loong slotted hooles perpendiicular to the ddirection of
loading, nominal stren
ngth will be the
t lowest vallue accordingg to Equationns J3-1, J3-6a,, and J3-6c.
It has beeen shown thaat for bolts insstalled in holees with a cleaar distance of 2d or more, bbearing will
control over
o
tearout. It
I is also possiible to determ
mine a minimuum plate thicckness to insuure that bolt
shear willl control over both bearing
g and tearout..
Figu
ure 10.11a Minimum
M
Plaate Thickness so Bolt Sheaar Controls forr A325-N bollts and Fy = 336 ksi
4442 Chapter 10
1
Connecttion Elementss
Figure 10.11b Min
nimum Plate Thickness so Bolt Shear C
Controls for A
A325-N bolts and Fy = 50 kksi
Figu
ure 10.11 sho
ows the minim
mum plate thi ckness for A3325-N bolts iin plates withh Fy = 36
ksi and Fy = 50 ksi as a function of clear
c
end disttance dividedd by bolt diam
meter, lc d ffor bolts
with diametters from 5/8 in. to 1.5 in. diameter. T
The curves staart at the minnimum clear distance
given in Tab
ble J3.4 divided by bolt diaameter and ennd at a clear ddistance of 2 bbolt diameterrs, above
which no inccrease in stren
ngth can be achieved.
a
100.7.4
Bolt Tension
T
olt tension, sttrength is direectly based oon the tensilee strength of the bolt
For the limiit state of bo
material. Section J3.6 pro
ovides that
Rn = Fn Ab
(AIS
SC J3-1)
with
Ω = 2.00 ((ASD)
φ = 0.75 (L
LRFD)
and
Fn
Ab
= tensile
t
stress, Fnt, from Speecification Taable J3.2
= area
a of the bo
olt shank
Table 10.4 shows the no
ominal tensilee stress, Fnt, for bolts takeen from Speccification Tabble J3.2.
Note that th
here is no disstinction for the
t location oof the shear plane, becauuse the bolt iss loaded
axially and the
t limiting sttress occurs over
o
the net teensile area. To ease calculaations, the areea of the
bolt shank iss again used and
a the nomin
nal tensile streess appropriaately reduced and given as 0.75Fu.
EX
XAMPLE 10.1
1
Boolt Shear
Str
trength
Goa
al:
Deterrmine the available bolt sheear strength.
Giv
ven:
(a) A single 3/4 in.. A325-N bollt.
(b) A single 7/8 in. A490-X bollt.
SO
OLUTION
Parrt (a)
Step
p 1:
For a single 3/4 in.. A325-N bollt, determine tthe bolt shankk area.
Chapter 10
Ab =
Step 2:
Connection Elements 443
πd 2 π(0.75)2
=
= 0.442 in.2
4
4
Determine the nominal shear stress.
For an A325 bolt
Fu = 120 ksi
and for the threads included (N)
Fnv = 0.45Fu = 0.45(120) = 54 ksi
For
LRFD
Step 3:
For shear, φ = 0.75 and the design strength is
φrn = 0.75(54)(0.442) = 17.9 kips
For
ASD
Step 3:
For shear, Ω = 2.00 and the allowable strength is
rn Ω = 54 ( 0.442 ) 2.00 = 11.9 kips
Part (b)
Step 4:
Step 5:
For a single 7/8 in. A490-X bolt, determine the bolt shank area.
πd 2 π(0.875)2
Ab =
=
= 0.601 in.2
4
4
Determine the nominal shear stress.
For an A490 bolt
Fu = 150 ksi
and for the threads excluded (X)
Fnv = 0.563Fu = 0.563(150) = 84 ksi
For
LRFD
Step 6:
For
ASD
Step 6:
For shear, φ = 0.75 and the design strength is
φrn = 0.75(84)(0.601) = 37.9 kips
For shear, Ω = 2.00 and the allowable strength is
rn (84)(0.601)
=
= 25.2 kips
2.00
Ω
Manual Table 7-1 provides single-bolt available shear strength values for a
wide range of bolt sizes and strengths.
4444 Chapter 10
1
Connecttion Elementss
Figure 10.12 Lap Jointt for Examplee 10.2
EX
XAMPLE 10.2
1
Laap Splice
Coonnection
Str
trength
Goa
al:
Deterrmine the available strength
th for a four-bbolt connectioon.
Giv
ven:
A lap
p joint using 1/2
1 in. A36 pllates is given in Figure 10.12. Use (a) 77/8 in.
A325
5-X bolts and (b) 7/8 in. A
A325-N bolts. The plates are labeled pplate 1
and plate
p
2 and thee bolt pairs arre labeled bollts A and boltts B. The layoout of
the co
onnection is such
s
that eacch bolt will hhave the samee limiting streength.
Thus,, the strength of the joint iss 4 times the sstrength of a single bolt.
SO
OLUTION
Parrt a
Step
p 1:
Deterrmine the nom
minal shear strrength for 7/88 in. A325-X bolts.
Fnv = 0.563Fu = 0.563 (1220 ) = 68 ksi
Ab = 0.601 in..2
rn = Fnv Ab = 68 ( 0.601) = 40.9 kips
Step
p 2:
Deterrmine the nom
minal bearing strength.
Unlesss noted otheerwise assum
me deformatioon at servicee load is a ddesign
consid
deration, thuss use Equationn J3-6a
rn = 2.4dtFu = 2.44 ( 7 8)( 0.50 ))( 58) = 60.9 kips
k
Step
p 3:
Deterrmine the tearrout strength tto the plate eddge using lc = lcA1 = lcB2.
Find the
t clear distaance from thee bolt hole to the end of thee member,
⎛1⎞
lc = 1.5 − ⎜ ⎟ ( 7/8 + 1//16 ) = 1.03 < 2d = 2 ( 7/8 ) = 1.75 in.
⎝2⎠
Sincee the clear disstance is less than 2 bolt diameters, tearout strengthh will
be lesss than bearin
ng strength. U
Using Equationn J3-6c
Chapter 10
Connection Elements 445
rn = 1.2lc tFu = 1.2 (1.03)( 0.50 )( 58) = 35.8 kips
Step 4:
Determine the tearout strength between bolt holes using lc = lcA2 = lcB1.
Find the clear distance from one bolt hole to the other,
lc = 2.5 − ( 7/8 + 1/16 ) = 1.56 < 2d = 2 ( 7/8) = 1.75 in.
Since the clear distance is less than 2 bolt diameters, tearout strength will
be less than bearing strength. Using Equation J3-6c
rn = 1.2lc tFu = 1.2 (1.56 )( 0.50 )( 58) = 54.3 kips
Step 5:
Determine the nominal strength of bolts A
Bolt shear
40.9 kips
Bearing
60.9 kips
Tearout on plate 1
35.8 kips
Tearout on plate 2
54.3 kips
Thus, these bolts have a nominal strength per bolt
rn = 35.8 kips
Determine the nominal strength of bolts B
Bolt shear
40.9 kips
Bearing
60.9 kips
Tearout on plate 1
54.3 kips
Tearout on plate 2
35.8 kips
Thus, these bolts have a nominal strength per bolt
rn = 35.8 kips
Therefore, all four bolts have the same strength.
Step 6:
Determine the nominal strength of the 4 bolt connection.
Since all 4 bolts have the same strength, the controlling limit state is tearout
to the edge of the plate, thus
Rn = 4rn = 4 ( 35.8) = 143 kips
For
LRFD
Step 7:
For
ASD
Step 7:
Part b
Step 8:
The design strength of the 4 bolt connection is
φRn = 0.75 (143) = 107 kips
The allowable strength for the 4 bolt connection is
Rn Ω = 143 2.00 = 71.5 kips
Determine the nominal shear strength for 7/8 in. A325-N bolts.
4446 Chapter 10
1
Connecttion Elementss
Fnv = 0.45Fu = 0.45 (1200 ) = 54 ksi
Ab = 0.601 in..2
rn = Fnv Ab = 54 ( 0.601) = 32.5 ksi
Step
p 9:
Deterrmine the nom
minal strengthh of the 4 boltt connection.
Sincee bearing and tearout are noot influenced by the locatioon of the sheaar
plane through the threads,
t
the nnominal strenggths determinned in Steps 22, 3,
and 4 are unchangeed. Thus, the controlling liimit state for each bolt in tthis
conneection is bolt shear. Thus,
Rn = 4rn = 4 ( 32.5) = 1130 kips
Forr
LRF
FD
Step
p 10:
Forr
ASD
D
Step
p 10:
The design
d
strengtth of the 4 bollt connection is
φRn = 00.75 (130 ) = 977.5 kips
The allowable
a
strength for the 4 bolt connecttion is
Rn Ω = 130 2.00 = 6 5.0 kips
Figure 10.13 Butt Jointt for Examplee 10.3
EX
XAMPLE 10.3
1
Bu
utt Splice
Coonnection
Str
trength
SO
OLUTION
Goa
al:
Deterrmine the available strength
th of the boltss in the butt spplice connectiion
shown
n in Figure 10
0.13.
Giv
ven:
Use 3/4
3 in. A490-N
N bolts and A
A36 plates.
Step
p 1:
Deterrmine bolt nom
minal shear sttrength.
Fnv = 0.563Fu = 0.536 (1550 ) = 68 ksi
Ab = 0.442 in..2
rn = Fnv Ab = 68 ( 0.442 ) = 30.1 kips
The bolts
b
are in do
ouble shear, thhus
rn = 2 ( 30.1) = 60.22 kips
Chapter 10
Connection Elements 447
Step 2:
Determine the nominal strength for bearing on the 1/2 in. middle plate
using Equation J3-6a,
rn = 2.4dtFu = 2.4 ( 3 4 )(1 2 )( 58) = 52.2 kips
Step 3:
Determine the nominal strength for bearing on the 3/8 in. outer plate using
Equation J3-6a,
rn = 2.4dtFu = 2.4 ( 3 4 )( 3 8)( 58) = 39.2 kips
Since there are 2 outer plates, the nominal bearing strength is
rn = 2 ( 39.2 ) = 78.4 kips
Step 4:
Determine the tearout strength in the 1/2 in. middle plate
For the bolts closest to the end of the plate
⎛1⎞
lc = 1.25 − ⎜ ⎟ ( 3/4 + 1/16 ) = 0.844<2d = 2 ( 3 4 ) = 1.5 in.
⎝2⎠
Since the clear distance is less than 2 bolt diameters, tearout strength will
be less than bearing strength. Using Equation J3-6c
rn = 1.2lc tFu = 1.2 ( 0.844 )( 0.50 )( 58) = 29.4 kips
For the interior bolts, the clear distance between bolts is
lc = 3.0 − ( 3 4 + 1 16 ) = 2.19 > 2d = 2 ( 3 4 ) = 1.5 in.
Since the clear distance is greater than 2 bolt diameters, tearout strength
will be greater than bearing strength. To confirm this, using Equation J3-6c
rn = 1.2lc tFu = 1.2 ( 2.19 )( 0.50 )( 58) = 76.2 kips
Step 5:
Determine the tearout strength in the 3/8 in outer plate
For the bolts closest to the end of the plate
⎛1⎞
lc = 1.25 − ⎜ ⎟ ( 3/4 + 1/16 ) = 0.844 < 2d = 2 ( 3/4 ) = 1.50 in.
⎝2⎠
Since the clear distance is less than 2 bolt diameters, tearout strength will
be less than bearing strength. Using Equation J3-6c
rn = 1.2lc tFu = 1.2 ( 0.844 )( 3 8)( 58) = 22.0 kips
For the two plates,
rn = 2 ( 22.0 ) = 44.0 kips
For the interior bolts, the clear distance between bolts is
lc = 3.0 − ( 3 4 + 1 16 ) = 2.19 > 2d = 2 ( 3 4 ) = 1.5 in.
Since the clear distance is greater than 2 bolt diameters, tearout strength
448 Chapter 10
Connection Elements
will be greater than bearing strength. To confirm this, using Equation J3-6c
rn = 1.2lc tFu = 1.2 ( 2.19 )( 3 8)( 58) = 57.2 kips
For the two plates,
Step 6:
rn = 2 ( 57.2 ) = 114 kips
Determine the controlling limit state and the nominal strength of each bolt.
For the bolts closest to the end of the outer plate, the limit states and
strengths are
Bolt shear
60.2 kips
Bearing on 1/2 in. plate
52.2 kips
Bearing on 2 - 3/8 in. plates 78.4 kips
Tearout on 1/2 in. plate
76.2 kips
Tearout on 2 - 3/8 in. plate 44.0 kips
Thus, these bolts have a nominal strength per bolt
rn = 44.0 kips
For the bolts closest to the end of the inner plate, the limit states and
strengths are
Bolt shear
60.2 kips
Bearing on 1/2 in. plate
52.2 kips
Bearing on 2 - 3/8 in. plates 78.4 kips
Tearout on 1/2 in. plate
29.4 kips
Tearout on 2 - 3/8 in. plate 114.0 kips
Thus, these bolts have a nominal strength per bolt
rn = 29.4 kips
Step 7:
For
LRFD
Step 8:
For
ASD
Step 8:
10.7.5
Determine the final connection nominal strength
Rn = 2 ( 44.0 ) + 2 ( 29.4 ) = 147 kips
The design strength for the connection is
φRn = 0.75 (147 ) = 110 kips
The allowable strength for the connection is
Rn Ω = 147 2.00 = 73.5 kips
Slip
The limit state of slip is associated with connections that are referred to as slip-critical. Slipcritical connections are designed to prevent slip at the required strength. They should be used
only when required, such as when the connection is subjected to fatigue or the connection has
oversized holes or slots parallel to the direction of load. Often, connections are specified to be
designed as slip-critical when these conditions do not apply. Significant connection economies
Chapter 10
Connection Elements 449
are lost, for no gain, when this is the case. Any slip-critical connection design must also include a
check for strength as a bearing-type connection by the methods discussed in the previous
sections. Bolts in slip-critical connections would be designated as A325-SC or A490-SC. The
nominal strength of a single bolt in a slip-critical connection is given in Specification Section J3.8
as
(AISC J3-4)
Rn = μDu h f Tb ns
where
μ
Du
hf
ns
Tb
= mean slip coefficient
= 0.30 for Class A surfaces
= 0.50 for Class B surfaces
= 1.13, a multiplier that reflects the ratio of mean installed bolt
pretension to specified bolt pretension.
= factor for fillers
= 1.0 for one filler between connected parts
= 0.85 for more than one filler.
If multiple fillers are directly connected independently to transfer load,
hf also may be taken as 1.0
= number of slip planes
= minimum bolt pretension specified in Table J3.1
For connections with standard size and short-slotted holes perpendicular to the direction of load,
φ = 1.00 (LRFD) Ω = 1.50 (ASD)
For oversized and short-slotted holes parallel to the direction of the load,
φ = 0.85 (LRFD) Ω = 1.76 (ASD)
For long-slotted holes
φ = 0.70 (LRFD)
Ω = 2.14 (ASD)
EXAMPLE 10.4
Slip-Critical
Connection
Strength
Goal:
Determine the available slip-critical connection strength for the connection
shown in Figure 10.13. This connection was treated as a bearing-type
connection in Example 10.3.
Given:
Use 3/4 in. A490-SC bolts in standard holes and A36 plates with a Class A
surface. There are no fillers.
SOLUTION
Step 1;
Determine bolt nominal slip strength.
From Specification Table J 3.1, the minimum bolt tension, Tb = 35 kips.
From the given information, μ = 0.30, Du = 1.13, hf = 1.0, ns = 2. Therefore,
Rn = μDh h f Tb ns
Rn = 0.30 (1.13 )(1.0 )( 35 )( 2 ) = 23.7 kips
For
LRFD
Step 2:
Determine the design strength.
From Example 10.3, the design bearing strength of the connection was
found to be
φRn = 110 kips
For the slip-critical connection,
450 Chapter 10
Connection Elements
φRn = 1.00 ( 4 )( 23.7 ) = 94.8 kips
Thus, the design strength of this slip-critical connection is
φRn = 94.8 kips
For
ASD
Step 2:
Determine the allowable strength.
From Example 10.3, the allowable bearing strength of the connection was
found to be
Rn Ω = 73.5 kips
For the slip-critical connection,
Rn Ω = 4 ( 23.7 1.5) = 63.2 kips
Thus, the allowable strength of this slip-critical connection is
Rn Ω = 63.2 kips
It can be seen through Example 10.4 that a significant amount of connection strength is lost when
a connection that otherwise could be designed as a bearing-type connection is forced to be
designed as a slip-critical connection. A situation where the connection of Example 10.4 would
need to be considered slip-critical is in a fatigue loading. By considering the resistance factor and
safety factor for other hole types, it is also seen that significant strength reductions result for those
slip-critical connections.
10.7.6
Combined Tension and Shear in Bearing-Type Connections
When bolts are subjected to simultaneous shear and tension, the available strength in each is
diminished by the presence of the other. Experimental results indicate that the interaction between
shear and tension is best represented by an elliptical relationship as shown as the solid curve in
Figure 10.14. This elliptical relationship can conveniently be replaced, with little loss of
accuracy, by three straight lines as shown by the dashed lines in Figure 10.14.
Specification Section J3.7 provides a nominal tensile stress modified to include the
effects of shearing stress, Fnt′ , to be used in determining the nominal bolt tensile strength such
that
Rn = Fnt′ Ab
(AISC J3-2)
and
φ = 0.75 (LRFD) Ω = 2.00 (ASD)
As can be seen in Figure 10.14, when the required stress in either shear or tension is less than or
equal to 30 percent of the corresponding available stress, the effects of the combined stresses can
be ignored. If both required stresses exceed this 30 percent limit, the modified tensile stress is
given by Equations J3-3a and J3-3b as
F
Fnt′ = 1.3 Fnt − nt f rv ≤ Fnt
Frv
where
Fnt = nominal tensile stress when only tension occurs
Frv = available shear stress, φFnv for LRFD or Fnv Ω for ASD
Fnv = nominal shear stress when only shear occurs
Chappter 10
Connection Elem
ments 451
frv = required shear stress, either
e
for LRF
FD or ASD
An exam
mple will be giiven in Sectio
on 11.10.
10.7.7
Com
mbined Tenssion and Shear in Slip-Crritical Conneections
When bo
olts in slip-ccritical conneections are suubjected to ssimultaneous shear and ttension, the
availablee slip strengtth is diminish
hed as the aapplied tensille force reduuces the clam
mping force
provided
d by the bolt pretension.
p
The strength oof the connecttion, thereforre, is a linear function of
the forcee compressing
g the plies. Th
his force is thhe initial pre-ttension, Tb, m
minus the appllied load, T.
The speccified slip-critical shear vaalue is, thereffore, reducedd by the factoor (1 − T Tb ) . The actual
reduction
n factor, ksc, is
i provided in
n Specificatioon Equation JJ3-5, again wiith one for AS
SD and one
for LRFD
D, as
Tu
RFD)
≥ 0 (LR
Du Tb nb
1..5Ta
ksc = 1 −
ASD)
≥ 0 (A
Du Tb nb
ksc = 1 −
(A
AISC J3-5a)
(A
AISC J3-5b)
where
Ta = required
r
tensiion force usinng ASD load ccombinationss
Tu = required
r
tensiion force usinng LRFD loadd combinationns
Tb = minimum
m
boltt pretension
Du = 1.13
nb = number
n
of bollts carrying thhe applied tennsion
Figure 10.144 Interactioon of Bolt
Shear and Teension
10.8
WELDS
Welding is the proceess of joining
g steel by m
melting and fuusing additional metal intto the joint
between the two piecees to be joineed. The ease w
with which vaarious types oof steel can bbe joined by
welding, without exh
hibiting cracks and other fflaws, is calleed weldability
ty. Most strucctural steels
used today accept weelding withou
ut the occurreence of unwannted defects. The Americaan Welding
Society (AWS)
(
defin
nes weldabilitty as “the cappacity of a m
metal to be w
welded under fabrication
4552 Chapter 10
1
Connecttion Elementss
conditions im
mposed, into a specific, su
uitably designned structure aand to perform
m satisfactoriily in the
intended serrvice.”
Welldability depeends primarily
y on the chem
mical composiition of the stteel and the thhickness
of the materrial. The impaact on weldab
bility of the vaarious chemiccal elements iin the composition of
steel was disscussed in Ch
hapter 3.
100.8.1
Weldin
ng Processess
For structuraal steel, the four
fo most pop
pular welding processes as designated bby AWS are: shielded
metal arc, su
ubmerged arcc, gas shielded
d metal arc annd flux cored arc.
M
Arc Wellding
Shielded Metal
Shielded meetal arc weld
ding (SMAW) is one of thhe oldest wellding processses. It is often called
manual or stick
s
welding. Figure 10.15
5a is a schem
matic represenntation of thiss welding proocess. A
high voltagee is induced between an electrode andd the metal ppieces that aare to be joinned. The
electrode is the source of
o the metal introduced
i
innto the joint tto make the w
weld. It is caalled the
consumable electrode. When
W
the wellding operatorr strikes an aarc between tthe electrode and the
base metal, the resulting flow of curreent melts the eelectrode andd the base mettal adjacent too it. The
electrode is coated with a special ceram
mic material called flux. T
This flux proteects the molteen metal
from absorb
bing hydrogen
n and other im
mpurities duriing the weldinng process. W
When the metal cools,
a permanentt bond exists between the electrode maaterial and thhe base materrial. Because the flux
cools at a diifferent rate th
han the metall, it separatess from the weeld and is easiily removed ffrom the
joint.
Submerged
d Arc Weldin
ng
Submerged arc welding (SAW) is an automatic or semi-automaatic process tthat is used pprimarily
when long pieces
p
of platee are to be joined. It is show
wn schematiccally in Figurre 10.15b. SA
AW
Figure 10.15 (a) Shield
ded Metal Arcc Welding annd (b) Submerrged Arc Wellding
Chappter 10
Connection Elem
ments 453
Figure 10.16 Fillet and
d Groove Wellds in Section
n
welds must
m
be madee in the nearr flat or horiizontal positiion. The fluxx is a granullar material
introduceed through a flexible tube on top of the electric aarc. It is an economical process for
applicatio
ons in which
h the use of reepetitive and automated fa
fabrication proocedures lendd efficiency
to the wo
ork.
Gas Shieelded Metal Arc
A Welding
g
Gas shieelded metal arrc welding (G
GMAW) is a pprocess in whhich a continuuous wire is ffed into the
joint to be
b welded. The
T molten metal
m
is proteccted from thee atmosphere by gas surroounding the
wire. In the field, it iss necessary to
o ensure that wind does nnot blow the ggas away from
m the joint.
This is th
he method oftten referred to
o as MIG weldding for its usse of inert gasses.
Flux Corred Arc Welding
Flux corred arc weld
ding (FCAW)) is also a ccontinuous-wiire process, except that tthe wire is
essentiallly a thin holllow tube filled with flux thhat protects tthe metal as tthe wire meltts. It can be
arranged as a semi-automatic proceess, and excepptionally highh production rrates can be aattained.
10.8.2
Typ
pes of Welds
Four bassic types of welds
w
are used
d in steel connstruction: filllet welds, grooove welds, pplug welds,
and slot welds.
w
Fillet and groove welds
w
are show
wn in Figure 10.16. Plug aand slot weldds fill a hole
or slot with weld mateerial to attach
h one piece to another.
Figure
F
10.16aa shows a filllet weld. Thee leg of the w
weld is measured along thhe interface
between the weld mettal and the baase metal. Thhe throat of thhe weld is thee shortest dim
mension and
the shearr plane of the weld. Becau
use most fillett welds are syymmetrical, w
with a 45-degree surface,
the throaat is 0.707 tim
mes the leg diimension as sshown. The siize of a fillett weld is giveen by its leg
dimensio
on, in incremeents of 1/16 in
n.
A groove welld can be either a completee joint penetraation (CJP) ggroove weld, aas shown in
Figure 10
0.16b, or a paartial joint peenetration (PJJP) groove weeld, as shownn in Figure 100.16c. Both
types of groove weldss have been prequalified
p
bby AWS. Thiis prequalificcation means that certain
weld con
nfigurations—
—including thee root openinng, R, the anggle of preparattion, α, and thhe effective
thicknesss, S—are deeemed practicaal to build aand will carryy the intendeed load. AW
WS specifies
provision
ns for prequaalifying any weld
w
configurration if circuumstances inddicate that it iis practical.
These prrequalified co
omplete and partial joint penetration ggroove weldss are shown in detail in
Manual Table
T
8-2. Th
he configurations shown inn Figure 10.166 are schemattic representattions.
4554 Chapter 10
1
Connecttion Elementss
Figgure 10.17 Terminology
y for Fillet and Groove Weeld Positions
Both
h fillet weldss and groove welds can b e laid down in a variety oof different ppositions
depending on
o the orientaation of the pieces
p
to be jjoined. The tterminology ffor these possitions is
given in Fig
gure 10.17. Flaat and horizon
ntal welds aree preferred ovver vertical annd overhead w
welds.
Wellds other than
n flat and horrizontal requiire a higher ddeposition ratee and thus result in a
higher cost final
f
connectiion.
100.8.3
Weld Sizes
S
Specification
n Section J2 addresses efffective areass and sizes ffor welds. Thhe effective aarea and
other limitattions for groo
ove welds aree given in Sppecification S
Section J2.1 aand Tables J22.1, J2.2
and J2.3. The
T effective areas of filllet welds arre given in Specification Section J2.2a. The
minimum sizes for fillet welds
w
in jointts are based oon the thinnerr of the parts bbeing joined as given
in Specificattion Table J2..4. The maxim
mum size of ffillet welds foor material lesss than 1/4 in.. thick is
the thicknesss of the mateerial, whereass for materiall 1/4 in. thickk or greater, the maximum
m size is
the material thickness lesss 1/16 in.
The minimum leength of filleet welds that are designedd on the basiis of strengthh is four
times the weeld leg size. For
F welds loaaded longituddinally, the m
maximum weldd length is 1000 times
the weld leg
g size for full effectivenesss. Welds long er than this m
must use a redduced effectivve length
as given in Section
S
J2.2b.
100.9
WELD
D LIMIT ST
TATES
The only lim
mit state to bee considered for
f a weld is rupture. Yielding of the w
weld metal will occur,
but it occurrs over such a short distaance that it iis not a factoor in connecttion behaviorr. Strain
hardening occcurs and rup
pture takes plaace without exxcessive yieldding deformaation.
The ultimate tenssile strength of
o an electrodde may vary from 60 to 120 ksi, depennding on
the specified
d composition
n. AWS classsifies electroddes according to the tensilee strength of tthe weld
metal and in
ndicates electrrode strength
h as FEXX. In thhis notation, the E represeents the electrrode and
the XX represents the ten
nsile strength
h. Thus, a typpical electrodde used to weeld A992 steeel would
have a strength of 70 ksi and be design
nated as an E770 electrode.
Chapter 10
Connection Elements 455
AWS and AISC specify that for a particular grade of structural steel, as indicated by yield
strength, there is a matching electrode. The user note table found in Section J2.6 summarizes the
AWS provisions for matching filler metals. Both organizations further specify that the steel can
be joined by welding only with the matching electrode or one that is no more than one grade
higher. This is to encourage yielding in the base metal before it occurs in the weld.
10.9.1
Fillet Weld Strength
For a fillet weld as shown in Figure 10.16a, load is transferred by shear through the throat of the
weld and the weld rupture strength is a function of the properties of the electrode. Shear strength
provisions for welds are found in Specification Section J2.4 and Table J2.5, where
Rn = Fnw Awe
(AISC J2-3)
φ = 0.75 (LRFD) Ω = 2.00 (ASD)
and
Fnw = nominal strength of the weld metal per unit area
Awe = effective area of the weld
FEXX = filler metal classification strength, the weld strength
For a longitudinally loaded fillet weld,
Fnw = 0.6 FEXX
Because the limit state of all fillet welds is one of shear rupture through the throat, the effective
area of a symmetrical fillet weld is the width of the weld at the throat, 0.707w, times the length of
the weld, l, so that
Awe = 0.707wl
The resulting nominal weld strength is
Rn = 0.60 FEXX ( 0.707 wl )
For the most commonly used weld electrode, FEXX = 70 ksi, the design strength for LRFD can be
determined as
φRn = 0.75 ( 0.6 ( 70 ) ) ( 0.707 wl ) = 22.27 wl
and the allowable strength for ASD can be determined as
Rn Ω = (0.6 (70))( 0.707wl ) 2.00 = 14.85wl
It is convenient in design to use the fillet weld strength for a fillet weld with a 1/16 in.
leg, which gives
Design strength for LRFD
φRn = 22.27wl = 22.27 (1 16 )(1.0 ) = 1.392 kips per 1 16 in. of weld per in. of length
and
Allowable strength for ASD
Rn Ω = 14.85wl = 14.85 (1 16 )(1.0 ) = 0.928 kips per 1 16 in. of weld per in. of length
Therefore, a 1/4 in. fillet weld has a design strength of 1.392×4 (sixteenths) = 5.57 kips per inch
of length and an allowable strength of 0.928×4 (sixteenths) = 3.71 kips per inch of length.
4556 Chapter 10
Connecttion Elementss
Figure 10.18
Normalizzed Strength vs
v Normalizeed Deformatioon for Welds Loaded at ann Angle
Copyright © American Insttitute of Steel Construction.
C
R
Reprinted with Permission. A
All rights reservved.
Research has sho
own that when
n load is appllied to a fillett weld at an aangle other thaan along
the length of
o the weld, more
m
strength
h is availablee than when tthe load is appplied longituudinally.
Figure 10.18 shows thatt as the anglee of load inccreases the ulltimate strenggth increasess. It also
shows a co
orresponding reduction off deformationn capacity. T
The vertical lline in Figurre 10.18
locates the maximum
m
streength of a weeld loaded lonngitudinally. Thus, welds loaded at an angle to
the weld length are seen to exhibit an increase of up to 50% in strengtth. The Specification
provides thee nominal streess, based on
n the angle off the load to the longitudiinal axis of thhe weld.
Thus,
(AIS
SC J2-5)
Fnw = 0.6 FEXX (1.00 + 0.5sin1.5 θ )
where
θ = angle of load
ding measured
d from the weeld longitudinnal axis
Thiss strength equ
uation is inten
nded to be ussed for weldss or weld grouups with unifform leg
size in whicch all elementts are in line or parallel annd loaded thrrough the cennter of gravityy. When
welds with different orieentations are combined in the same joiint, deformatiion of these ddifferent
welds, as illlustrated in Fiigure 10.18, must
m be accouunted for. Sppecification Seection J2 provvides an
approach fo
or concentriccally loaded fillet weld ggroups consiisting of elem
ments that aare both
longitudinall and transverrse to the direection of the applied loadd. For this case, the nominnal weld
strength is taken as thee larger of th
he simple suum of the w
weld strength without connsidering
orientation, given by
Rn = Rnwl + Rnwt
(AISC
C J2-6a)
or
Rn = 0.855Rnwl + 1.5Rnwwt
(AISC
C J2-6b)
where
Rnwl = totaal nominal streength of the llongitudinallyy loaded weldd without connsidering
the angle of
o load
Rnwt = totaal nominal strrength of the transversely loaded weldd without connsidering
the angle of
o load
Chappter 10
Connection Elem
ments 457
1
Weldss for Examplee 10.5
Figure 10.19
Equation
n J2-6a provid
des the streng
gth of the welld group as a function of tthe sum of thee lengths of
the welds times the sttrength per unit
u length. Im
mplementationn of Equationn J2-6b requiires that the
longitudiinally loaded weld strength
h be discounteed by 0.85 in order for thee transverse w
welds to be
increased
d by 1.5. Wh
hen welds loaaded at diffeerent angles aare combinedd, it is criticaal that their
deformattions are com
mpatible. From
m Figure 10.118 it can be seeen that a weeld loaded at 990° reaches
its ultimaate strength at
a a normalizeed deformatioon significanttly less than thhe deformatioon at which
a weld lo
oaded at 0° reeaches its ultiimate strengtth. The weld loaded at 0° reaches only about 85%
of its ultiimate strength
h when its deformation is tthe same as thhe weld loadeed at 90° wheen it is at its
ultimate strength. Thee approach of
o Equation J22-6b will be most beneficcial when thee transverse
welds aree longer than the longitudinal welds.
The
T forgoing discussion addressed
a
onlly the strength
th of the weldd material. T
The material
that the weld
w is attach
hed to is referrred to as the base materiall. This materiial must also be checked
for streng
gth according
g to the provissions of Secti on J4 which w
will be discusssed in Sectioon 10.10.
EXAMPLE
E 10.5
Weld Stren
ngth and
Load Anglee
Goal:
G
Deetermine the available
a
strength of the thhree welds givven in Figure 10.19.
Given:
G
Th
he welds are 3/4
3 in. welds,, 8.0 in. long,, and loaded ((a) along the length of
thee weld, (b) trransversely too the weld, aand (c) at a 445-degree anggle to the
weeld. Use E70 electrodes.
SOLUTION
N
Part
P
a
Step
S
1:
Weld
W Loaded Along
A
Its Len gth
Deetermine the number
n
of 1/116 units for thhe given weldd using Equattion J2-5.
A 3/4 in. weld is twelve 1/1 6 in. units accross the leg.
Step
S
2:
Deetermine the strength
s
of thhe weld when loaded alongg its length.
Th
he strength values
v
alreadyy discussed can be used because thee weld is
loaaded longitud
dinally.
For
F
LRFD
L
Step
S
3:
Th
he design streength is
φRn = 8.0 (12 )(1.392 ) = 134 kipss
For
F
ASD
A
Step
S
3:
Th
he allowable strength
s
is
Rn Ω = 8.0(12)(0.9228) = 89.1 kips
Part
P
b
Step
S
1:
Weld
W Loaded at
a 90 Degrees to the Weld L
Length
Deetermine the nominal weldd strength usiing the equattion to accounnt for the
an
ngle of load.
4558 Chapter 10
1
Connecttion Elementss
Fw = ( 0.60 FEXX ) (1.0 + 0.5sin1.5 θ )
= ( 0.60 FEXX ) (1.0 + 0.5sin1.5 ( 90 ) )
= ( 0.60 FEXX )(1.5 )
Thereefore, the streength of the w
weld is increaased by 1.5 ovver what it is when
loaded longitudinaally.
Forr
LRF
FD
Step
p 2:
Forr
ASD
D
Step
p 2:
Parrt c
Step
p 1:
d
strengtth is
The design
φRn = 1.5 (134 ) = 2001 kips
The allowable
a
strength is
Rn Ω = 1.5(89.1) = 134 kips
Weld
d Loaded at 45
5 Degrees to tthe Weld Lenngth
Deterrmine the nom
Equation J2-55 to account ffor the
minal weld strrength using E
angle of load.
Fw = ( 0.60 FEXX ) (1.0 + 0.55sin1.5 θ )
= ( 0.60 FEXX ) (1.0 + 0.55sin1.5 ( 45 ) )
= ( 0.60 FEXX )(1.30 )
Thereefore, the strength of the w
weld is increassed by 1.30 ovver what it is when
loaded longitudinaally.
Forr
LRF
FD
Step
p 2:
Forr
ASD
D
Step
p 2:
d
strengtth is
The design
φRn = 1.3 (134 ) = 1774 kips
The allowable
a
strength is
Rn Ω = 1.3(89.1) = 116 kips
Figure 10.20
C-Shaped W
Weld for Exam
mple 10.6.
Chapter 10
Connection Elements 459
EXAMPLE 10.6a
Weld Strength and
Load Angle by
LRFD
Goal:
Determine the design strength for C-shaped welds.
Given:
A C-shaped weld group is shown in Figure 10.20 to attach a tension plate to
a gusset. Use E70 electrodes and a 7/8 in. weld.
SOLUTION
Step 1:
Determine the design strength for the two 4.0 in. welds parallel to the load.
φRnwl = 2 ( 4.0 )(14 )(1.392 ) = 156 kips
Step 2:
Determine the design strength for the 6.0 in. weld transverse to the load.
φRnwt = 6 (14 )(1.392 ) = 117 kips
Step 3:
Determine the connection design strength by adding the strength based on
length of the welds.
φRn = Rnwl + Rnwt = 156 + 117 = 273 kips
Step 4:
Determine the design strength considering the added contribution of the
transverse welds while reducing the contribution of the longitudinal welds
so that
φRn = 0.85Rnwl + 1.5Rnwt = 0.85 (156 ) + 1.5 (117 ) = 308 kips
Step 5:
Determine the weld strength by selecting the larger of the results from steps
3 and 4.
Thus, accounting for the difference between the longitudinal and transverse
welds provides more strength.
φRn = 308 kips
EXAMPLE 10.6b
Weld Strength and
Load Angle by
ASD
Goal:
Determine the design strength for C-shaped welds.
Given:
A C-shaped weld group is shown in Figure 10.20 to attach a tension plate to
a gusset. Use E70 electrodes and a 7/8 in. weld.
SOLUTION
Step 1:
Determine the allowable strength for the two 4.0 in. welds parallel to the
load.
Rnwl Ω = 2(4.0)(14)(0.928) = 104 kips
Step 2:
Determine the allowable strength for the 6.0-in. weld transverse to the load.
Rnwt Ω = 6.0(14)(0.928) = 78.0 kips
Step 3:
Determine the connection allowable strength by adding the strength based
on length of the welds.
Rn Ω = 104 + 78.0 = 182 kips
Step 4:
Determine the allowable strength considering the added contribution of the
transverse welds while reducing the contribution of the longitudinal welds
so that
460 Chapter 10
Connection Elements
Rn Ω = 0.85(104) + 1.5(78.0) = 205 kips
Step 5:
Determine the weld strength by selecting the larger of the results from steps
3 and 4. Thus, accounting for the difference between the longitudinal and
transverse welds provides more strength.
Rn Ω = 205 kips
EXAMPLE 10.7a
Weld Strength and
Load Angle by
LRFD
Goal:
Determine the design strength for C-shaped welds.
Given:
The C-shaped weld group shown in Figure 10.20 is to attach a tension plate
to a gusset. However, the welds parallel to the load will be increased to 10
in. in length. Use E70 electrodes and a 7/8 in. weld.
SOLUTION
Step 1:
Determine the design strength for the two 10.0 in. welds parallel to the
load.
φRnwl = 2 (10.0 )(14 )(1.392 ) = 390 kips
Step 2:
Determine the design strength for the 6.0 in. weld transverse to the load.
φRnwt = 6 (14 )(1.392 ) = 117 kips
Step 3:
Determine the connection design strength by adding the strength based on
length of the welds.
φRn = Rnwl + Rnwt = 390 + 117 = 507 kips
Step 4:
Determine the design strength considering the added contribution of the
transverse welds while reducing the contribution of the longitudinal welds
so that
φRn = 0.85Rnwl + 1.5Rnwt = 0.85 ( 390 ) + 1.5 (117 ) = 507 kips
Step 5:
Determine the weld strength by selecting the larger of the results from steps
3 and 4. Note that both approaches give the same design strength. Thus,
φRn = 507 kips
If the welds parallel to the load were any longer, considering only weld
length would provide the most strength.
EXAMPLE 10.7b
Weld Strength and
Load Angle by
ASD
Goal:
Determine the design strength for C-shaped welds.
Given:
The C-shaped weld group shown in Figure 10.20 is to attach a tension plate
to a gusset. However, the welds parallel to the load will be increased to 10
in. in length. Use E70 electrodes and a 7/8 in. weld.
SOLUTION
Step 1:
Determine the allowable strength for the two 10.0 in. welds parallel to the
load.
Rnwl Ω = 2(10.0)(14)(0.928) = 260 kips
Chapter 10
Connection Elements 461
Step 2:
Determine the allowable strength for the 6.0 in. weld transverse to the load.
Rnwt Ω = 6.0(14)(0.928) = 78.0 kips
Step 3:
Determine the connection allowable strength by adding the strength based
on length of the welds.
Rn Ω = Rnwl + Rnwt = 260 + 78.0 = 338 kips
Step 4:
Determine the allowable strength considering the added contribution of the
transverse welds while reducing the contribution of the longitudinal welds
so that
Rn Ω = 0.85Rnwl + 1.5 Rnwt = 0.85(260) + 1.5(78.0) = 338 kips
Step 5:
Determine the weld strength by selecting the larger of the results from steps
3 and 4. Note that both approaches give the same allowable strength. Thus,
Rn Ω = 338 kips
If the welds parallel to the load were any longer, considering only weld
length would provide the most strength.
10.9.2
Groove Weld Strength
A groove weld can be either a complete or partial joint penetration weld as shown in Figure
10.16b and c. The complete joint penetration (CJP) groove weld is not designed in the usual sense
because the weld metal is always stronger than the base metal when properly matching electrodes
are used. Therefore, the strength of the base metal controls the design.
In the case of a CJP groove weld, the nominal strength of the tension joint is the product
of the yield strength of the base material and the cross-sectional area of the smallest piece joined.
The nominal strength of a partial joint penetration (PJP) groove weld in a tension joint is similar
except that the full cross-sectional area of the joined pieces is not effective. In this case, AWS
defines an effective throat dimension, S, which is a function of the configuration of the bevel as
shown in Figure 10.16c and Manual Table 8-2.
10.10
CONNECTING ELEMENTS
The plates, angles, and other elements that go into making up a connection are called connecting
elements. They, along with the region of the members actually involved in the connection, are
treated in Section J4. There are provisions for tension, compression, flexure, shear, and block
shear.
10.10.1
Connecting Elements in Tension
Although the Specification addresses tension in connecting elements in Section J4.1, it does not
alter the basic tension provisions found in Specification Chapter D. This means that two limit
states are to be considered, the limit state of yielding and the limit state of rupture. Again, for
tension, the resistance and safety factors are different for the two limit states, so any comparison
of strength must be made at the design or allowable strength level. The design strength is given
by φRn and the allowable strength by Rn/Ω, as is the case throughout the Specification. For the
limit state of yielding of connecting elements
(AISC J4-1)
Rn = Fy Ag
462 Chapter 10
Connection Elements
φ = 0.90 (LRFD)
Ω = 1.67 (ASD)
For the limit state of rupture of connecting elements
Rn = Fu Ae
φ = 0.75 (LRFD)
(AISC J4-2)
Ω = 2.00 (ASD)
The areas are determined as for the tension members previously considered.
10.10.2
Connecting Elements in Compression
Most connecting elements in compression are relatively short and have a fairly low slenderness
ratio. In addition, determination of the appropriate effective length factor requires application of
significant engineering judgment, usually amounting to making an educated guess as to an
appropriate factor. With this in mind, and to simplify connection design somewhat, the
Specification provides a simple relation in Section J4.4 for the compressive strength of
connecting elements if the slenderness ratio, lc/r, is less than or equal to 25. For this case,
(AISC J4-6)
Pn = Fy Ag
and the resistance and safety factors are the same as for other compression members,
φ = 0.90 (LRFD) Ω = 1.67 (ASD)
If the slenderness ratio of the compression element is greater than 25, the element must be
designed according to the compression member provisions of Specification Chapter E.
10.10.3
Connecting Elements in Flexure
Connecting elements that undergo flexure are to be designed with consideration of flexural
yielding, local buckling, flexural lateral-torsional buckling, and flexural rupture. These are the
same limit states to be considered for flexural members as presented in Chapter 6.
10.10.4
Connecting Elements in Shear
Member design for shear, according to Section G2, requires the consideration of the limit states
of shear yielding and shear buckling. Connecting elements and the portion of members affected
by the connection must be checked for the limit states of shear yielding and shear rupture
according to Section J4.2. Shear yielding occurs on the gross area of the element whereas shear
rupture occurs on a section containing holes. Thus, for shear yielding of the element
(AISC J4-3)
Rn = 0.6 Fy Agv
φ = 1.00 (LRFD)
Ω = 1.50 (ASD)
where Agv = gross area subject to shear. The resistance and safety factors for this case are the
same as those for the special case of rolled I-shaped members given in Specification Section
G2.1(a).
For the limit state of shear rupture
Rn = 0.6 Fu Anv
(AISC J4-4)
φ = 0.75 (LRFD) Ω = 2.00 (ASD)
where Anv = net area subjected to shear. As was the case for tension rupture, the net area is
determined by removing the area of holes from the gross area.
Chappter 10
(a))
Figu
ure 10.21
Connection Elem
ments 463
(b)
Block
B
Shear Failure:
F
(a) welded tensionn connection; (b) bolted beam end conneection
Phottos courtesy Ro
obert Driver
10.10.5
Bllock Shear Sttrength
The limiit state of bllock shear ru
upture can occcur on the cconnecting eelements or tthe affected
memberss as shown in
i Figure 10..21. Figure 110.21a showss a block shhear failure inn a welded
connectio
on of a tension member to
t the web off a W-shape.. Figure 10.21b shows a bblock shear
failure in
n a bolted end
d connection to
t a coped W--shape.
Block
B
shear combines
c
shear and tensioon failures intto a single, coomplex modee of failure.
Block sh
hear was discu
ussed in Sectiion 4.7 as it ppertained to teension membbers because iit is a major
factor in determining tension mem
mber strength. It can also be a factor in determining tthe strength
of a beam
m end reactio
on, dependin
ng on the connnection geom
metry. Thus, it is repeatedd here. The
nominal strength for the
t limit state of block sheaar rupture is
((AISC J4-5)
Rn = 0.6
6 Fu Anv + U bs Fu Ant ≤ 0.6 Fy Agv + U bs Fu Ant
where
Agv
Ant
Anv
Ubs
= gross shear area
= net teension area
= net sh
hear area
= 1.0 fo
or uniform ten
nsion stress diistribution
= 0.5 fo
or nonuniform
m tension stresss
The resisstance and saffety factors fo
or the limit staate of block sshear rupture are again
φ = 0.75 (LRFD)) Ω = 2.00 (ASD)
Figure 10
0.22 shows a single-anglee tension mem
mber attachedd to a gusset plate and a ccoped beam
end with
h the holes loccated in a sing
gle line. The tension area and shear areea are identifiied for each
and the area
a that woulld tear out is shaded.
s
4664 Chapter 10
Connecttion Elementss
Figure 10.22 Example Block Shear Failure
e
show
ws that the eexpected failuure mode willl always
A reeview of the block shear equation
include tenssion rupture, whereas the shear failure mode will be the smallerr of the shearr rupture
and the sheaar yield. The tension
t
stresss distribution factor, Ubs, iss a function oof the variatioon of the
tension stresss over the teension area. Figure
F
10.23 shows severaal elements annd the corressponding
assumed ten
nsile stress disstribution. Th
he only case iddentified by tthe Commenttary where the tensile
stress distrib
bution is not uniform is that of a copeed beam withh two rows oof bolts, as shhown in
Figure 10.23
3g.
hear Tensile Stress Distribuution
Figure 10.23 Block Sh
Chappter 10
Connection Elem
ments 465
Figgure 10.24 C
Coped Beam End for
Exaample 10.8
EXAMPLE
E 10.8
Block Shea
ar
Strength
Goal:
G
Deetermine the block shear design strenngth and alloowable strenggth for a
co
oped beam.
Given:
G
A coped W16×
×40 A992 beaam end is shown in Figurre 10.24. Asssume that
thee beam has sttandard holes for 5/8 in. boolts.
SOLUTION
N
Step
S
1:
Deetermine the gross
g
and nett shear areas aand net tensioon area for thee beam.
Reemember from
m the discusssion of tensiion members that for net area, an
ad
dditional 1/16
6 in. must be added to thee hole size to account for any hole
daamage from th
he punching ooperation. Froom Manual Table 1-1 tw = 0.305 in.
Agv = 11.0 ( 0.305 ) = 3.36 in.2
Anv = (11.0 − 33.5 ( 5 8 + 1 8 ) ) ( 0.305 ) = 2.25 in.2
Ant = ( 4.25 − 11.5 ( 5 8 + 1 8 ) ) ( 0.305 ) = 0.953 in.2
Step
S
2:
Deetermine the shear yield and rupture strength andd the tensionn rupture
strrength.
Fo
or this geomeetry, the tensiile stress disttribution is nnonuniform; ttherefore,
Ubs
b = 0.5.
Sh
hear yield
Sh
hear rupture
Teensile rupture
0.6 Fy Agvv = 0.6 ( 50 )( 33.36 ) = 101 kipps
0.6 Fu Anv = 0.6 ( 65)( 2.55) = 99.5 kiips
U bs Fu Ant = 0.5 ( 65)( 0.9953) = 31.0 kkips
466 Chapter 10
Connection Elements
Step 3:
Determine the nominal block shear strength.
Because shear rupture is less than shear yield, combine the shear rupture
with the tensile rupture. Thus,
Rn = 99.5 + 31.0 = 131 kips
For
LRFD
Step 4:
For
ASD
Step 4:
10.10.6
The design strength is
φRn = 0.75 (131) =98.3 kips
The allowable strength is
Rn Ω = 131 2.00 = 65.5 kips
Connecting Element Rupture Strength at Welds
The strength of connection elements at welds cannot always be determined directly. In those
cases, it is often convenient to determine the minimum thickness of the connecting element
required to match the rupture strength of the weld with the rupture strength of the base metal. The
rupture strength of a fillet weld was given in Section 10.9.1 as
Rn = 0.60 FEXX ( 0.707 wl )
If this is set equal to the rupture strength of the base metal, given as
Rn = 0.6 Fu tl
and the thickness is taken as tmin, the minimum required thickness for the base metal to match the
weld, with E70 electrodes and the weld size taken as D 16 is
0.60 ( 70 ) ( 0.707 ( D 16 ) l ) = 0.6 Fu tmin l
Solving for tmin yields
tmin = 3.09 D Fu
(Manual 9-2)
For a case where there are welds on both sides of the base metal, such as on the web of a beam
with angles on both sides, the minimum web thickness will be twice that for a weld on one side.
Thus,
tmin = 6.19 D Fu
(Manual 9-3)
The application of these minimum thickness calculations will be illustrated in Chapters 11 and 12
as appropriate for the connections being considered.
10.11
PROBLEMS
1. Develop a table showing the nominal shear strength
for A325-N bolts for the following sizes: 5/8, 3/4, 7/8,
and 1 in.
3. Develop a table showing the nominal shear strength
for A490-N bolts for the following sizes: 5/8, 3/4, 7/8,
and 1 in.
2. Develop a table showing the nominal shear strength
for A325-X bolts for the following sizes: 5/8, 3/4, 7/8,
and 1 in.
4. Develop a table showing the nominal shear strength
for A490-X bolts for the following sizes: 5/8, 3/4, 7/8,
and 1 in.
Chaptter 10
55. Develop a table showing the design sheear strength forr
A
A325-N, A325
5-X, A490-N, and A490-X bolts for thee
ffollowing sizess: 5/8, 3/4, 7/8, and 1 in.
66. Develop a table showing
g the allowable shear strength
h
ffor A325-N, A325-X,
A
A490-N
N, and A490-X
X bolts for thee
ffollowing sizess: 5/8, 3/4, 7/8, and 1 in.
Connnection Elements 467
with two 1/2 in. A336 plates. Deteermine (a) desiggn strength
by LR
RFD and (b) aallowable strenggth by ASD.
15. Determine thhe available sstrength of thhe 7/8 in.
A4900-N bolts in thhe lap splice shown in Figuure P10.15
with two 1/2 in. A
A572 Grade 550 plates. Dettermine (a)
desiggn strength byy LRFD and (bb) allowable sstrength by
ASD
D.
77. Develop a table showing the design sheear strength forr
F
F1852-N, F185
52-X, F2280-N
N, and F2280-X
X bolts for thee
ffollowing sizess: 5/8, 3/4, 7/8, and 1 in.
88. Develop a table showing
g the allowable shear strength
h
ffor F1852-N, F1852-X,
F
F228
80-N, and F2280-X bolts forr
thhe following sizes: 5/8, 3/4, 7/8,
7 and 1 in.
99. Determine the available strength of thee 3/4 in. A325-N bolts in the lap
l splice show
wn in Figure P10.9
P
with two
o
11/2 in. A36 plates.
p
Determ
mine (a) desig
gn strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
P10.115
16. Determine thhe available sstrength of thhe 7/8 in.
A4900-X bolts in thhe lap splice shown in Figuure P10.15
with two 1/2 in. A
A572 Grade 550 plates. Dettermine (a)
desiggn strength byy LRFD and (bb) allowable sstrength by
ASD
D.
17. Determine thhe available sstrength of thhe 7/8 in.
A3255-N bolts in thhe butt splice shown in Figuure P10.17
with two 1/2 in. siide plates andd a 1 in. main plate. Use
A36 plates. Determ
mine (a) designn strength by LRFD and
(b) alllowable strenggth by ASD.
P
P10.9
110. Determine the availab
ble strength of
o the 3/4 in.
A
A325-X bolts in
n the lap splicee shown in Fig
gure P10.9 with
h
tw
wo 1/2 in. A36 plates. Deterrmine (a) design strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
111. Determine the availab
ble strength of
o the 7/8 in.
A
A325-N bolts in
n the lap splicee shown in Fig
gure P10.9 with
h
tw
wo 1/2 in. A36 plates. Deterrmine (a) design strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
112. Determine the availablee strength of th
he 1 in. F2280-N bolts in the lap
l splice show
wn in Figure P10.9
P
with two
o
11/2 in. A36 plates.
p
Determ
mine (a) desig
gn strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
113. Determine the availablee strength of th
he 1 in. F2280-X bolts in the lap
l splice show
wn in Figure P10.9
P
with two
o
11/2 in. A36 plates.
p
Determ
mine (a) desig
gn strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
114. Determine the availablle strength off the 1-1/8 in.
F
F2280-N bolts in the lap sp
plice shown in
n Figure P10.9
9
P10.117
18. Determine thhe available sttrength of thee 1-1/8 in.
A4900-X bolts in thhe butt splice shown in Figuure P10.17
with two 1/2 in. siide plates andd a 1 in. main plate. Use
A36 plates. Determ
mine (a) designn strength by LRFD and
(b) alllowable strenggth by ASD.
19. Recreate Figgures 10.11a aand 10.11b foor A325-X
boltss.
20. Recreate Figgures 10.11a aand 10.11b foor A490-N
boltss.
4468 Chapterr 10
Connection Elemen
nts
221. Recreate Figures 10.11
1a and 10.11b
b for A490-X
X
bbolts.
26. Determine thhe available strrength of the six 3/4 in.
A3255-N bolts in a 7×1/2 in. A366 plate when thhe bolts are
placeed as shown inn Figure P10.226. Determine (a) design
strenngth by LRFD aand (b) allowab
able strength byy ASD.
222. Determine the availab
ble strength of
o the 7/8 in.
A
A490-SC boltss in the slip-ccritical lap splice shown in
n
F
Figure P10.15 with two 1//2 in. A572 Gr.
G 50 plates,,
sstandard holes,, Class B surfface, and no fillers.
f
(This iss
P
Problem 15 as a slip-criticall connection.) Determine (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD.
223. Determine the availab
ble strength of
o the 7/8 in.
A
A325-SC bolts in the butt sp
plice shown in Figure P10.17
7
w
with two 1/2 in
n. side plates and a 1 in. main
m
plate. Usee
A
A36 plates, ov
versized holess, Class A su
urface, and no
o
ffillers. (This iss Problem 17 as
a a slip-criticaal connection.))
D
Determine (a) design
d
strength
h by LRFD and
d (b) allowablee
sstrength by ASD
D.
224. Determine the availablee strength of th
he four 3/4 in.
A
A325-N bolts in
i the single L3×3×1/2 A36 when
w
the boltss
aare placed as shown in Fig
gure P10.24. Determine (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD.
P10.226
27. Determine thhe available strrength of the six 7/8 in.
A3255-N bolts in a 7×3/4 in. A366 plate when thhe bolts are
placeed as shown inn Figure P10.226. Determine (a) design
strenngth by LRFD aand (b) allowab
able strength byy ASD.
28. Determine thee available streength of the eiight 3/4 in.
A4900-N bolts in ann A992 WT6×
×17.5 when thhe bolts are
placeed as shown inn Figure P10.228. Determine (a) design
strenngth by LRFD aand (b) allowab
able strength byy ASD.
P
P10.24
225. Determine the availablee strength of th
he three 3/4 in.
A
A325-N bolts in
i the single L4
4×3×3/8 A36 when
w
the boltss
aare placed as shown in Fig
gure P10.25. Determine (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD.
P
P10.25
P10.228
29. Determine thee available streength of two 33/8 in. fillet
weldds that are loadded parallel too their length, are 10 in.
long,, and are madde from E70 eelectrodes. Dettermine (a)
desiggn strength byy LRFD and (bb) allowable sstrength by
ASD
D.
30. Determine thee available streength of two 1/2 in. fillet
weldds that are loadded parallel too their length, are 12 in.
long,, and are madde from E70 eelectrodes. Dettermine (a)
desiggn strength byy LRFD and (bb) allowable sstrength by
ASD
D.
Chaptter 10
Connnection Elements 469
331. Determine the availablee strength of tw
wo 1/4 in. fillett
w
welds that are loaded paralllel to their len
ngth, are 8 in.
loong, and are made
m
from E7
70 electrodes. Determine (a))
ddesign strength
h by LRFD an
nd (b) allowab
ble strength by
y
A
ASD.
332. If the weelds of Probleem 29 were lo
oaded at theirr
ccentroid and att 90 degrees to
o the weld len
ngth, determinee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD.
333. If the weelds of Probleem 29 were lo
oaded at theirr
ccentroid and att 45 degrees to
o the weld len
ngth, determinee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD.
334. If the weelds of Probleem 31 were lo
oaded at theirr
ccentroid and att 80 degrees to
o the weld len
ngth, determinee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD.
P10.440
41. Determine thhe available bllock shear streength for a
A992 beam witth holes for 3//4 in. bolts
copedd W21×182 A
as shhown in Figuree P10.41. Deterrmine (a) desiggn strength
by LR
RFD and (b) aallowable strenggth by ASD.
335. If the weelds of Probleem 31 were lo
oaded at theirr
ccentroid and att 35 degrees to
o the weld len
ngth, determinee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD.
336. Plot a curvee of the strengtth versus anglee of loading forr
a 12 in. long 5//16 in. weld maade from E70 electrodes.
e
Thee
pplot should incllude angles fro
om 0 degrees to
o 90 degrees.
337. Three 5/1
16 in. welds arre grouped to form a C and
d
aare loaded at th
heir centroid. Determine
D
the available weld
d
sstrength if the single transverrse weld is 9 in. and the two
o
loongitudinal welds
w
are each
h 3 in. Use E70
E
electrodes.
D
Determine (a) design
d
strength
h by LRFD and
d (b) allowablee
sstrength by ASD
D.
338. Repeat Prroblem 37 with
h the transversse weld at 3 in.
aand the two lo
ongitudinal welds at 9 in. eaach. Determinee
((a) design stren
ngth by LRFD
D and (b) allow
wable strength
h
bby ASD.
P10.441
42. Determine thhe available bllock shear streength for a
copedd W24×146 A
A992 beam witth holes for 7//8 in. bolts
as shhown in Figuree P10.42. Deterrmine (a) desiggn strength
by LR
RFD and (b) aallowable strenggth by ASD.
339. Three 3/8
8 in. welds are grouped to forrm a C and aree
looaded at theirr centroid. Deetermine the available
a
weld
d
sstrength if the single transverrse weld is 8 in. and the two
o
loongitudinal welds
w
are each
h 8 in. Use E70
E
electrodes.
D
Determine (a) design
d
strength
h by LRFD and
d (b) allowablee
sstrength by ASD
D.
440. Determine the availablee block shear strength for a
ccoped W16×26
6 A992 beam with
w holes for 3/4
3 in. bolts ass
sshown in Figurre P10.40. Deteermine (a) desiign strength by
y
L
LRFD and (b) allowable
a
stren
ngth by ASD.
P10.442
4470 Chapterr 10
Connection Elemen
nts
443. Determine the availablee block shear strength for a
ccoped W27×19
94 A992 beam
m with holes fo
or 7/8 in. boltss
aas shown in Fig
gure P10.43. Determine
D
(a) design
d
strength
h
bby LRFD and (b)
( allowable sttrength by ASD
D.
45. Determine thhe available bllock shear streength for a
A992 beam witth holes for 3//4 in. bolts
copedd W30×261 A
as shhown in Figuree P10.45. Deterrmine (a) desiggn strength
by LR
RFD and (b) aallowable strenggth by ASD.
P
P10.43
444. Determine the availablee block shear strength for a
ccoped W30×26
61 A992 beam
m with holes fo
or 7/8 in. boltss
aas shown in Fig
gure P10.44. Determine
D
(a) design
d
strength
h
bby LRFD and (b)
( allowable sttrength by ASD
D.
P
P10.44
P10.445
46. Determine thhe available bllock shear streength for a
copedd W44×230 A
A992 beam w
with holes for 12 - 1 in.
boltss spaced at 3.00 in., with horrizontal edge ddistance of
2.0 in. and locatted 1-1/2 in. down from the cope.
Deterrmine (a) desiggn strength by LRFD and (b)) allowable
strenngth by ASD.
Chapter
11
Simple Con
nnectiions
Stanford University, Bing Concert Hall
Courtesy of Degenkoolb Engineers
11.1
TYP
PES OF SIM
MPLE CON
NNECTION
NS
This chaapter addressees two types of simple c onnections, bbeam shear cconnections aand bracing
connectio
ons. Both aree commonly designed as ppinned jointss – simple, inn the terminoology of the
Specifica
ation. The connecting elem
ments and thee connectors required for these connections have
already been
b
discusseed in Chapter 10. The limitt states that ccontrol the coonnection havve also been
discussed
d individually
y, although th
heir link to coonnection dessign may not yet be complletely clear.
Connectiion design is a combinatio
on of elemennt and connecctor selectionn with a checcking of all
appropriaate limit stattes. The goal is to obtainn a connectiion with suff
fficient strenggth and the
appropriaate stiffness to carry the lo
oad in a mannner consistentt with the moddel used in thhe structural
analysis. In addition to these sim
mple shear connnections, beeam bearing plates and coolumn base
plates wiill be discusseed.
The
T limit stattes to be con
nsidered for a particular cconnection deepend on the connection
elementss, the connecttion geometry
y, and the lo ad path. Theey will be ideentified in thee following
sections as each conn
nection type is consideredd. A summary
ry of the poteential limit sttates at this
time, how
wever, may prove useful. For bolts, tthe limit stattes of tensilee rupture, sheear rupture,
bearing and
a tearout, as well as sllip, will be cconsidered. F
For welds, thhe only limit state to be
considereed is shear rupture,
r
altho
ough weld grroup geometrry will add ssome complexxity to that
consideraation. For connecting ellements, the limit states are tensionn yielding annd rupture,
compresssion buckling
g, shear yieldiing and rupturre, and the fuull range of fleexural limit sttates.
Table
T
11.1 lissts the section
ns of the Speccification and parts of the M
Manual discuussed in this
chapter.
11.2
SIM
MPLE SHEA
AR CONNE
ECTIONS
A signifiicant number of potential connection
c
geeometries aree associated w
with the varioous types of
memberss to be conn
nected. Five of the mostt commonly used simplee shear connnections are
described
d in the follow
wing sectionss with design examples folllowing. These connectionss are shown
471
472 Chapter 11
Simple Connections
Table 11.1 Sections of Specification and Parts of Manual Covered in This Chapter
Specification
B4.3
J2
J3
J4
J10
Gross and Net Area Determination
Welds
Bolts and Threaded Parts
Affected Elements of Members and Connecting Elements
Flanges and Webs with Concentrated Forces
Manual
Part 7
Part 8
Part 9
Part 10
Part 14
Part 15
Design Considerations for Bolts
Design Considerations for Welds
Design of Connecting Elements
Design of Simple Shear Connections
Design of Beam Bearing Plates, Column Base Plates, Anchor Rods, and
Column Splices
Design of Hanger Connections, Bracket Plates, and Crane-Rail Connections
in Figure 11.1a through e as double-angle, single-angle, single-plate (commonly called a shear
tab), unstiffened seated, and stiffened seated connections. Part 10 of the Manual includes many
tables that can simplify connection design; however, the examples presented here show the
required calculations when necessary to improve understanding. Once a calculation has been
sufficiently demonstrated, the Manual tables are used.
Several design considerations apply to all of the shear connections to be discussed, and in
some cases to other types of connections. It is helpful to address these before dealing with the
specific connection. The first issue to consider is the location of the hinge within the connecting
elements. It is critical that this hinge can actually occur in the real connection, because the
analytical model of the connection assumes it behaves as a hinge or pin. The location of the hinge
determines what forces and moments, if any, the individual elements must be designed for. In all
cases, the hinge is located at the most flexible point within the connection. This may be at the
face of the supporting member or at some other point within the connection. Several general
design guidelines help to ensure that the connection behaves as desired. In most cases, this means
that the hinge is located at the face of the supporting member.
For double-angle connections, angle thickness should be limited to a maximum of 5/8 in.
The bolts in the outstanding legs, those connecting to the supporting member, should be spaced at
as wide a gage as possible; for welded outstanding legs, the vertical welds should be spaced as far
apart as possible. These characteristics will ensure that the connection behaves as a simple
connection through bending of the outstanding legs.
For simple beam connections, the permitted tolerance for beam length must be
considered. Although this tolerance is not normally a consideration for member design, it
becomes important when the details of connecting members are considered. Beam length
tolerance is ±1/4 in. To accommodate fit-up, beams are held back 1/2 in. from the face of the
supporting member. Then, when considering the edge distance from a bolt hole to the end of the
member, the distance used in calculations should be taken as 1/4 in. less than that actually
detailed in cases where an under run in beam length might yield a lower strength.
For welded connections, the small loss of weld area due to stopping the weld short of the
end of the joint by one to two weld sizes is not typically considered in the calculation of weld
strength.
It is also helpful to remember the considerations for hole sizes. First, standard holes are
sized 1/16 in. larger than the bolt to be inserted for bolts up to 7/8 in. and 1/8 in. larger for 1.0 in.
and larger bolts. Then, when considering net sections for the limit states of tension rupture or
shear rupture, Specification Section B4.3 requires that an additional 1/16 in. be deducted to
Chhapter 11
Figu
ure 11.1
Siimple Connecctions 473
Sim
mple Shear Connections
C
account for any mateerial damage resulting from
m the processs of making the hole. W
When a clear
distance is calculated for the limit state
s
of tearouut, the actual hole size is uused.
These
T
design consideration
ns are includeed in the exam
mples to follow
w.
11.3
DOU
UBLE-ANG
GLE CONN
NECTIONS:: BOLTED--BOLTED
A doublee-angle shear connection, as shown in F
Figure 11.1a,, is perhaps thhe most comm
mon simple
shear con
nnection used
d in steel con
nstruction. It iis a fairly sim
mple connectiion to fabricaate and also
provides for fairly eassy erection. When
W
double- angle connecctions are to bbe installed baack to back,
there maay be some prroblems, partticularly whenn the supportting member is a column web. In the
case of attachment
a
to a column weeb, the safetyy requirementts of OSHA ccall for speciaal attention.
One solu
ution is to stag
gger the doub
ble angles. Thhis connectioon can easily accommodatee variations
in beam length
l
within
n acceptable to
olerances.
The
T double-aangle shear connection
c
m
must be checcked for the following llimit states,
grouped according to
o the elementts that make up the conneection: bolts, beam web, angles, and
supportin
ng member:
1.
1
Bolts
a.
Shear ruptture
2.
2
Beam
m
4774 Chapter 11
1
Simple Connections
C
a.
Bolt
B bearing and
a tearout onn beam web
Figure 11.2 Connection
n Geometry for
f Example 111.1
3.
4.
b.
Shear
S
yielding
g of the web
c.
Block
B
shear on
o coped beam
m web
d.
Coped
C
beam flexural
f
strenngth
Angles
a.
Bolt
B bearing and
a tearout onn angles
b.
Shear
S
rupture
c.
Shear
S
yielding
g
d.
Block
B
shear
Supportin
ng member
a.
Bolt
B bearing
h of these lim
mit states hass been addresssed in this bbook. In the eexamples thatt follow,
Each
these limit state checks arre combined into
i
a compleete connectionn design.
EX
XAMPLE 11.1a
1
Boolted-Bolted
d
Doouble-Anglee
Sh
hear Connecction
byy LRFD
Goa
al:
Desig
gn a bolted-bo
olted double-aangle shear coonnection forr a W18×50 bbeam.
Giv
ven:
The W18×50
W
beaam is conneected to the web of a W
W24×207 annd the
conneection must provide
p
a requuired strengthh Ru = 83.0 kkips. The beam
m and
colum
mn are A992
2 and the 3- 1/2×3-1/2×3//8 angles aree A36. The beam
flangee is coped 2 in.
i Use 7/8 inn. A325-N boolts in standarrd holes in thhe legs
on thee beam web and
a short slotts on the outsstanding legs. The basic sttarting
geom
metry is given in Figure 11. 2.
SO
OLUTION
Boltt Shear
Deterrmine the num
mber of bolts required baseed on the sheear rupture strrength
Step
p 1:
of thee bolts.
From
m Manual Table 7-1, the de sign shear strrength per bollt is
φrn = 24.3 kipps
Becau
use the bolts are in doublee shear, ns = 2 and the dessign shear strrength
of a single bolt is
ns φrn = 2 ( 24.3) = 488.6 kips
Chapter 11
Simple Connections 475
Values for double shear are also found in Manual Table 7-1. Because of
rounding, the number in the table is slightly greater at 48.7 kips. The total
number of bolts required is
Ru
83.0
N=
=
= 1.71
ns ( φrn ) 48.6
Therefore, based only on the limit state of bolt shear, try two bolts.
Bolt Strength at Holes in Beam Web
Determine bolt bearing and tearout strength on the beam web.
Step 2:
For the W18×50 beam, tw = 0.355 in.
For the two-bolt connection, the top bolt is 1.25 in. from the edge of the
beam cope and the second bolt is spaced 3.0 in. from the first.
For the top bolt, the clear distance is
1
1
lc = lcv − d h = 1.25 − (7/8 + 1/16) = 0.781 < 2(7/8) = 1.75
2
2
Since the clear distance is less than 2db, tearout controls over bearing. The
nominal bolt strength for tearout is
Rn = 1.2lc tFu = 1.2 ( 0.781)( 0.355)( 65) = 21.6 kips
and the design strength is
φRn = 0.75 ( 21.6) = 16.2 kips
Since this is less than the design shear strength of the bolt in double shear it
will control the strength of this bolt.
For the second bolt
lc = s − dh = 3.0 − ( 7 8 + 1 16) = 2.06 > 2 ( 7 8) = 1.75 in.
Since the clear distance is greater than 2db, bearing controls over tearout.
Thus, the nominal bolt bearing strength is
Rn = 2.4dtFu = 2.4 ( 7 8)( 0.355)( 65) = 48.5 kips
and the design strength for bearing on the beam web is
φRn = 0.75 ( 48.5) = 36.4 kips
Since this is less than the shear strength of the bolt in double shear, it will
control the strength of this bolt.
Therefore, the connection design strength, if only considering the beam
web, is based on one bolt limited by tearout and one bolt limited by
bearing. Thus,
φRn = (16.2 + 36.4) = 52.6 < 83.0 kips
476 Chapter 11
Simple Connections
and the two-bolt connection cannot support the load.
Step 3:
Determine the number of bolts required considering bearing and tearout on
the web.
Adding a third bolt spaced at 3.0 in., which means that bearing controls
over tearout, gives a connection design strength for bolt bearing and tearout
of
φRn = (16.2 + 2 ( 36.4 ) ) = 89.0 > 83.0 kips
Therefore a three bolt connection will be considered further.
Bolt Strength at Holes in Angle Leg
Determine bolt bearing and tearout strength on the A36 angle. The angle
Step 4:
legs on the beam and on the supporting member are identical and each
angle takes one half of the load.
For the bottom edge bolt, the clear distance is
1
1
lc = lev − d h = 1.25 − ( 7 8 + 1 16 ) = 0.781 < 2 ( 7 8 ) = 1.75 in.
2
2
Since the clear distance is less than 2db, tearout controls over bearing. The
nominal bolt strength for tearout is
Rn = 1.2lc tFu = 1.2 ( 0.781)( 3 8 )( 58 ) = 20.4 kips
and the design strength is
φRn = 0.75 ( 20.4 ) = 15.3 kips
Since this is less than the design shear strength of the bolt in single shear, it
will control the strength of this bolt
For the second and third bolts, the clear distance is
lc = s − d h = 3.0 − ( 7 8 + 1 16 ) = 2.06 > 2 ( 7 8 ) = 1.75 in.
Since the clear distance is greater than 2db, bearing controls over tearout.
The nominal bolt strength for bearing is
Rn = 2.4 dtFu = 2.4 ( 7 8 )( 3 8 )( 58 ) = 45.7 kips
and the design strength for bearing is
φRn = 0.75 ( 45.7 ) = 34.3 kips
Since this is greater than the design shear strength of the bolt in single
shear, bolt shear will control the strength of these two bolts.
Connection Strength at Beam
Determine the controlling strength for each bolt at the beam web
Step 5:
Top bolt
Bolt double shear
48.6 kips
Chapter 11
Web tearout
Angle bearing ×2
Middle bolt
Bolt double shear
Web bearing
Angle bearing ×2
Bottom bolt
Bolt double shear
Web bearing
Angle tearout ×2
Simple Connections 477
16.2 kips
68.6 kips
48.6 kips
36.4 kips
68.6 kips
48.6 kips
36.4 kips
30.6 kips
Therefore, the strength of the connection at the beam web is
φRn = 16.2 + 36.4 + 30.6 = 83.2 > 83.0 kips
Bolt Strength at Holes in Supporting Member
Consider the outstanding legs of the angles attached to the supporting
Step 6:
member.
In this case, the bolts are in single shear but there are twice as many bolts,
so the load per bolt is half of the load for each of the bolts in the beam web.
If the supporting member thickness is at least one-half of the beam web
thickness and the strengths are the same, the bolts in the supporting
member will be satisfactory. For the W24×207, clear distances result in all
three bolts being controlled by bearing over tearout. Thus, with tw = 0.870
in.,
Rn = 2.4dtFu = 2.4 ( 7 8)( 0.870)( 65) = 119 kips
and the design strength is
φRn = 0.75 (119) = 89.3 kips
Since this is greater than the single shear strength of the bolt,
φrn = 24.3 kips , connection strength on the supporting member will be
controlled by bolt shear.
For six bolts in single shear
φRn = 6 ( 24.3) = 146 kips
which is greater than the required strength of 83.0 kips so the bolts in the
supporting member are adequate
Connection Strength at Supporting Member
Determine the controlling strength for each bolt at the supporting member
Step 7:
Top bolt
Bolt single shear
24.3 kips
Web bearing
89.3 kips
Angle tearout
15.3 kips
Middle bolt
Bolt single shear
24.3 kips
Web bearing
89.3 kips
Angle bearing
34.3 kips
478 Chapter 11
Simple Connections
Bottom bolt
Bolt single shear
Web bearing
Angle bearing
24.3 kips
89.3 kips
34.3 kips
Therefore, the strength of the connection at the beam web is
φRn = 15.3 + 24.3 + 24.3 = 63.9 > 83.0 2 =41.5 kips
Step 8:
Evaluate the minimum depth of the connection.
The beam web connection should be at least half the depth of the beam
web, measured as the distance between the fillets, T, given in Manual Table
1-1. This requirement is to prevent twisting of the simple supports. For
W18×50, T = 15-1/2 in., so the minimum angle depth should be 7-3/4 in.
Thus, the 8-1/2 in. long angle will provide an acceptable connection depth
and will also fit within the 15.5 in. available flat length.
Remaining Beam Limit States
Check shear yield of the beam web.
Step 9:
This is a check that should be carried out during the beam design process.
At the point of connection design it is too late to find out that the beam will
not be adequate. From Manual Table 3-2,
φVn = 192 kips > 83.0 kips
Step 10:
Check shear yield of the beam web at the cope
This check is carried out according to Section J4.2 where the full area of
the remaining web is stressed to Fy and ϕ=1.0. Thus
φVn = φ0.6 Fy htw
= 1.0 ( 0.6 ( 50 ) ) (18.0 − 2.0 )( 0.355 )
= 170 kips > 83.0 kips
Therefore shear yield strength at the cope is adequate
Step 11:
Check shear rupture of the beam web at the cope
This check is carried out according to Section J4.2 where the effective area,
Ae, the full area of the remaining web less the bolt holes, is stressed to Fu
and ϕ=0.75. Thus
φVn = φ0.6 Fu Ae
⎛
⎛ 7 1 ⎞⎞
= 0.75 ( 0.6 ( 65 ) ) ⎜ 18.0 − 2.0 − 3 ⎜ + ⎟ ⎟ ( 0.355 )
⎝ 8 8 ⎠⎠
⎝
= 135 kips > 83.0 kips
Therefore shear rupture at the cope is adequate
Step 12:
Check the beam web for block shear.
Chapter 11
Simple Connections 479
The equations for block shear rupture are found in Specification Section
J4.3 and were presented in Section 10.10.5.
First calculate the areas, remembering to account for the 1/4 in. beam
length under run tolerance in the tension area calculation, so that leh = 2.0 –
0.25 = 1.75 in.:
1
⎛
⎞
Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw
2
⎝
⎠
1
⎛
⎞
= ⎜ 1.75 − ( 7 8 + 1 8 ) ⎟ (0.355) = 0.444 in.2
2
⎝
⎠
Agv = ltw
= 7.25(0.355) = 2.57 in.2
Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw
= ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 1.69 in.2
Determine the tension rupture strength
Fu Ant = 65 ( 0.444 ) = 28.9 kips
Consider shear yield and shear rupture and select the least nominal
strength; thus,
0.6 Fy Agv = 0.6 ( 50 )( 2.57 ) = 77.1 kips
0.6 Fu Anv = 0.6 ( 65 )(1.69 ) = 65.9 kips
Selecting the shear rupture term and combining it with the tension rupture
term gives the connection block shear design strength, and recalling that
Ubs = 1.0 for the case of uniform tensile stress distribution, we have
φRn = 0.75 ( 65.9 + 1.0 ( 28.9 ) ) = 0.75 ( 94.8 ) = 71.1 < 83.0 kips
Thus, the given three-bolt connection is not adequate, with block shear
being the critical limit state to this point in our calculations.
Step 13:
Revise the connection to meet the block shear strength requirements.
Consideration could be given to increasing the number of bolts and thereby
increasing the length of the connection. However, because bolt shear
required only two bolts, this would not be a particularly economical
solution. If the connection were to be lowered on the beam end so that the
distance from the center of the top bolt to the edge of the cope were 2.5 in.,
the connection would have more block shear strength. It should also be
noted that the clear end distance for the top bolt in the beam web would
increase and thus increase the tearout strength of that top bolt. There is no
need to consider that increase since at best it could contribute to reducing
the number of bolts and this would have a negative impact on the block
shear calculations being addressed in this step.
Thus, the new shear areas become
480 Chapter 11
Simple Connections
Agv = ltw
= 8.50(0.355) = 3.02 in.2
Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw
= ( 8.50 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 2.13 in.2
and the nominal shear yield and rupture strengths become
0.6 Fy Agv = 0.6 ( 50 )( 3.02 ) = 90.6 kips
0.6 Fu Anv = 0.6 ( 65 )( 2.13 ) = 83.1 kips
Selecting the shear rupture term and combining it with the tension rupture
term gives the resulting block shear design strength as
φRn = 0.75 ( 83.1 + 1.0 ( 28.9 ) ) = 0.75 (112 ) = 84.0 > 83.0 kips
Step 14:
Check the flexural strength of the coped beam.
It is a good idea to check this limit state during the initial design of the
beam. It should be anticipated during the beam design stage that a coped
connection will be required, and it is at that stage that a change in beam
section can most readily be accommodated.
Flexural strength of the coped beam is not addressed in the Specification
directly but is covered in Part 9 of the Manual. The moment in the coped
beam is taken as the shear force times the eccentricity from the face of the
support to the edge of the cope, taken as cope + setback = 7.0 + 0.50 = 7.5
in. in this example. Thus,
M u = Ru e = 83.0 ( 7.5) = 623 in.-kips
(Manual 9-5)
To determine the flexural strength of the coped beam, the web slenderness
must be checked. This requires determination of the plate buckling
coefficient, k, and the buckling adjustment factor, f. The web slenderness of
this coped beam is
ho t w = 16.0 0.355 = 45.1
The limiting web slenderness is
λ p = 0.46
k1 E
Fy
(Manual 9-12)
where
k1 = fk
(Manual 9-10)
The buckling adjustment factor, f, is a function of the ratio of the cope
length, c, to the beam depth, d, and the plate buckling coefficient, k, is a
function of the ratio of cope length, c, to the reduced beam depth, ho. For
Chapter 11
Simple Connections 481
this example, with c d = 7.0 18.0 = 0.389 ≤ 1.0
2c 2(7.0)
f =
=
= 0.778
d
18.0
(Manual 9-14a)
and with c ho = 7.0 16.0 = 0.438 ≤ 1.0
1.65
⎛h ⎞
k = 2.2 ⎜ o ⎟
⎝ c ⎠
Thus,
1.65
⎛ 16.0 ⎞
= 2.2 ⎜
⎟
⎝ 7.0 ⎠
= 8.61 (Manual 9-13a)
k1 = fk = 0.778 ( 8.61) = 6.70 ≥ 1.61
(Manual 9-10)
and
λ p = 0.46
6.70 ( 29,000 )
k1 E
= 0.46
= 28.7
Fy
50
Since λ is between λp and 2λp the net elastic and net plastic section modulus
will be needed. The net elastic section modulus is taken from Manual Table
9-2. With the depth of the cope, dc = 2.0 in.,
Snet = 23.4 in.3
The net plastic section modulus is not provided in the Manual so it must be
determined. Following the procedure discussed in Chapter 6, the net plastic
section modulus for the tee shaped beam that remains after the cope is
found to be
Z net = 42.4 in.3
and from
⎛ λ
⎞
M n = M p − (M p − M y )⎜
− 1⎟
⎝ λp
⎠
with
(Manual 9-7)
M p = Fy Z net = 50 ( 42.4 ) = 2120 in.-kips
M y = Fy S net = 50 ( 23.4 ) = 1170 in-kips
The nominal moment strength at the cope is
⎛ 45.1 ⎞
M n = 2120 − ( 2120 − 1170 ) ⎜
− 1 ⎟ = 1580 in.-kips
⎝ 28.7
⎠
and the design strength is
φM n = 0.9 (1580) = 1420 in.-kips > 623 in.-kips
So the coped beam has sufficient flexural strength.
482 Chapter 11
Simple Connections
Angles
Step 15:
Check the angles for shear rupture using Equation J4-4.
The net area of the angle on the vertical shear plane is
Anv = ( l − n ( d h + 1 16 ) ) ta = ( 8.5 − 3 ( 7 8 + 1 8 ) ) ( 3 8 ) = 2.06 in.2
and the design strength for two angles is
φVn = 2 ( φ0.6 Fu Agv ) = 2 0.75 ( 0.6 ( 58 )( 2.06 ) ) = 108 kips > 83.0 kips
(
)
So the angles are adequate for shear rupture.
Step 16:
Check the angles for shear yield using Equation J4-3.
The gross area of the angle on the vertical shear plane is
Agv = 8.5 ( 3 8 ) = 3.19 in.2
and the design strength for two angles is
φVn = 2 ( φ0.6 Fy Agv ) = 2 1.0 ( 0.6 ( 36 )( 3.19 ) ) = 138 kips > 83.0 kips
(
)
So the angles are also adequate for shear yield.
Step 17
Check the angles for block shear.
The equations for block shear in the angle are the same as those for the web
and as presented in Section 10.10.5.
First calculate the areas,
1
⎛
⎞
Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw
2
⎝
⎠
1
⎛
⎞
= ⎜ 1.0 − ( 7 8 + 1 8 ) ⎟ ( 3 8 ) = 0.188 in.2
2
⎝
⎠
Agv = ltw
= 7.25 ( 3 8 ) = 2.72 in.2
Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw
= ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) ( 3 8 ) = 1.78 in.2
Determine the tension rupture strength
Fu Ant = 58 ( 0.188) = 10.9 kips
Consider shear yield and shear rupture, and select the least nominal
strength; thus,
0.6 Fy Agv = 0.6 ( 36 )( 2.72 ) = 58.8 kips
0.6 Fu Anv = 0.6 ( 58 )(1.78 ) = 61.9 kips
Selecting the shear yield term and combining it with the tension rupture
Chhapter 11
Siimple Connecctions 483
terrm gives a co
onnection bloock shear desiign strength oof the doublee angle—
ag
gain, Ubs = 1.0
0 for this casee of uniform ttensile stress ddistribution—
—of
φRn = 2 ( 0.75 ( 558.8 + 10.9 ) ) = 105 kips > 83.0
8
kips
Step
S
18:
Prresent the finaal connection design.
Th
he three-bolt connection, rrevised as shoown in Figurre 11.3, is adeequate to
caarry the impossed load of 833.0 kips.
Figure 11.3
1
Final Connection
C
Deesign for Exam
mple 11.1
EXAMPLE
E 11.1b
Bolted-Bollted
Double-An
ngle
Shear Connection
by ASD
Goal:
G
Deesign a bolted
d-bolted doubble-angle sheaar connection for a W18×550 beam.
Given:
G
Th
he W18×50 beam is connnected to tthe web of a W24×207 and the
co
onnection musst provide a rrequired strenngth Ra = 55.0 kips. The bbeam and
co
olumn are A9
992 and the 3-1/2×3-1/22×3/8 angles are A36. T
The beam
flaange is coped
d 2 in. Use 7/88 in. A325-N
N bolts in stanndard holes inn the legs
on
n the beam weeb and short slots on the ooutstanding leegs. The basicc starting
geeometry is giv
ven in Figure 11.2.
SOLUTION
N
Step
S
1:
Bo
olt Shear
Deetermine the number
n
of boolts required bbased on the shear rupturee strength
off the bolts.
Frrom Manual Table
T
7-1, thee design shearr strength per bolt is
rn Ω = 16.22 kips
Beecause the bo
olts are in douuble shear, ns = 2 and the design shearr strength
off a single bolt is
ns rn Ω = 2 (16.2) = 32.4 kips
Vaalues for dou
uble shear aree also found in Manual T
Table 7-1. Beecause of
rou
unding, the number
n
in thee table is slighhtly greater aat 32.5 kips. The total
nu
umber of boltss required is
484 Chapter 11
Simple Connections
Ra
55.0
=
= 1.70
ns ( rn Ω ) 32.4
Therefore, based only on the limit state of bolt shear, try two bolts.
N=
Bolt Strength at Holes in Beam Web
Determine bolt bearing and tearout strength on the beam web.
Step 2:
For the W18×50 beam, tw = 0.355 in.
For the two-bolt connection, the top bolt is 1.25 in. from the edge of the
beam cope and the second bolt is spaced 3.0 in. from the first.
For the top bolt, the clear distance is
1
1
lc = lcv − d h = 1.25 − (7/8 + 1/16) = 0.781 < 2(7/8) = 1.75
2
2
Since the clear distance is less than 2db, tearout controls over bearing. The
nominal bolt strength for tearout is
Rn = 1.2lc tFu = 1.2 ( 0.781)( 0.355)( 65) = 21.6 kips
and the allowable strength is
Rn Ω = 21.6 2.00 = 10.8 kips
Since this is less than the allowable shear strength of the bolt in double
shear it will control the strength of this bolt.
For the second bolt
lc = s − dh = 3.0 − ( 7 8 + 1 16) = 2.06 > 2 ( 7 8) = 1.75 in.
Since the clear distance is greater than 2db, bearing controls over tearout.
Thus, the nominal bolt bearing strength is
Rn = 2.4dtFu = 2.4 ( 7 8)( 0.355)( 65) = 48.5 kips
and the allowable strength for bearing on the beam web is
Rn Ω = 48.5 2.00 = 24.3 kips
Since this is less than the shear strength of the bolt in double shear, it will
control the strength of this bolt.
Therefore, the connection allowable strength, if only considering the beam
web, is based on one bolt limited by tearout and one bolt limited by
bearing. Thus,
Rn Ω = (10.8 + 24.3) = 35.1 < 55.0 kips
and the two-bolt connection cannot support the load.
Step 3:
Determine the number of bolts required considering bearing and tearout on
the web.
Chapter 11
Simple Connections 485
Adding a third bolt spaced at 3.0 in., which means that bearing controls
over tearout, gives a connection allowable strength for bolt bearing and
tearout of
Rn Ω = (10.8 + 2 ( 24.3) ) = 59.4 > 55.0 kips
Therefore a three bolt connection will be considered further.
Bolt Strength at Holes in Angle Leg
Determine bolt bearing and tearout strength on the A36 angle. The angle
Step 4:
legs on the beam and on the supporting member are identical and each
angle takes one half of the load.
For the bottom edge bolt, the clear distance is
1
1
lc = lev − d h = 1.25 − ( 7 8 + 1 16 ) = 0.781 < 2 ( 7 8 ) = 1.75 in.
2
2
Since the clear distance is less than 2db, tearout controls over bearing. The
nominal bolt strength for tearout is
Rn = 1.2lc tFu = 1.2 ( 0.781)( 3 8 )( 58 ) = 20.4 kips
and the allowable strength is
Rn Ω = 20.4 2.00 = 10.2 kips
Since this is less than the allowable shear strength of the bolt in single
shear, it will control the strength of this bolt
For the second and third bolts, the clear distance is
lc = s − d h = 3.0 − ( 7 8 + 1 16 ) = 2.06 > 2 ( 7 8 ) = 1.75 in.
Since the clear distance is greater than 2db, bearing controls over tearout.
The nominal bolt strength for bearing is
Rn = 2.4 dtFu = 2.4 ( 7 8 )( 3 8 )( 58 ) = 45.7 kips
and the allowable strength for bearing is
Rn Ω = 45.7 2.00 = 22.9 kips
Since this is greater than the allowable shear strength of the bolt in single
shear, bolt shear will control the strength of these two bolts
Connection Strength at Beam
Determine the controlling strength for each bolt at the beam web
Step 5:
Top bolt
Bolt double shear
Web tearout
Angle bearing ×2
Middle bolt
Bolt double shear
32.4 kips
10.8 kips
45.8 kips
32.4 kips
486 Chapter 11
Simple Connections
Web bearing
Angle bearing ×2
Bottom bolt
Bolt double shear
Web bearing
Angle tearout ×2
24.3 kips
45.8 kips
32.4 kips
24.3 kips
20.4 kips
Therefore, the strength of the connection at the beam web is
φRn = 10.8 + 24.3 + 20.4 = 55.5 > 55.0 kips
Bolt Strength at Holes in Supporting Member
Consider the outstanding legs of the angles attached to the supporting
Step 6:
member.
In this case, the bolts are in single shear but there are twice as many bolts,
so the load per bolt is half of the load for each of the bolts in the beam web.
If the supporting member thickness is at least one-half of the beam web
thickness and the strengths are the same, the bolts in the supporting
member will be satisfactory. For the W24×207, clear distances result in all
three bolts being controlled by bearing over tearout. Thus, with tw = 0.870
in.,
Rn = 2.4dtFu = 2.4 ( 7 8)( 0.870)( 65) = 119 kips
and the allowable strength is
Rn Ω = 119 2.00 = 59.5 kips
Since this is greater than the single shear strength of the bolt,
rn Ω = 16.2 kips , connection strength on the supporting member will be
controlled by bolt shear.
For six bolts in single shear
Rn Ω = 6 (16.2) = 97.2 kips
which is greater than the required strength of 55.0 kips so the bolts in the
supporting member are adequate
Connection Strength at Supporting Member
Determine the controlling strength for each bolt at the supporting member
Step 7:
Top bolt
Bolt single shear
16.2 kips
Web bearing
59.5 kips
Angle tearout
10.2 kips
Middle bolt
Bolt single shear
16.2 kips
Web bearing
59.5 kips
Angle bearing
22.9 kips
Bottom bolt
Bolt single shear
16.2 kips
Web bearing
59.5 kips
Angle bearing
22.9 kips
Chapter 11
Simple Connections 487
Therefore, the strength of the connection at the beam web is
Rn Ω = 10.2 + 16.2 + 16.2 = 42.6 > 55.0 2 =27.5 kips
Step 8:
Evaluate the minimum depth of the connection.
The beam web connection should be at least half the depth of the beam
web, measured as the distance between the fillets, T, given in Manual Table
1-1. This requirement is to prevent twisting of the simple supports. For
W18×50, T = 15-1/2 in., so the minimum angle depth should be 7-3/4 in.
Thus, the 8-1/2 in. long angle will provide an acceptable connection depth
and will also fit within the 15.5 in. available flat length.
Remaining Beam Limit States
Check shear yield of the beam web.
Step 9:
This is a check that should be carried out during the beam design process.
At the point of connection design it is too late to find out that the beam will
not be adequate. From Manual Table 3-2,
Vn Ω = 128 kips > 55.0 kips
Step 10:
Check shear yield of the beam web at the cope
This check is carried out according to Section J4.2 where the full area of
the remaining web is stressed to Fy and Ω = 1.5. Thus
Vn Ω = 0.6 Fy ht w Ω
= ( 0.6 ( 50 ) ) (18.0 − 2.0 )( 0.355 ) 1.50
= 114 kips > 55.0 kips
Therefore shear yield strength at the cope is adequate
Step 11:
Check shear rupture of the beam web at the cope
This check is carried out according to Section J4.2 where the effective area,
Ae, the full area of the remaining web less the bolt holes, is stressed to Fu
and Ω = 2.0. Thus
Vn Ω = 0.6 Fu Ae Ω
⎛
⎛ 7 1 ⎞⎞
= ( 0.6 ( 65 ) ) ⎜ 18.0 − 2.0 − 3 ⎜ + ⎟ ⎟ ( 0.355 ) 2.00
⎝ 8 8 ⎠⎠
⎝
= 90.0 kips > 55.0 kips
Therefore shear rupture at the cope is adequate
Step 12:
Check the beam web for block shear.
The equations for block shear rupture are found in Specification Section
J4.3 and were presented in Section 10.10.5.
First calculate the areas, remembering to account for the 1/4 in. beam
488 Chapter 11
Simple Connections
length under run tolerance in the tension area calculation, so that leh = 2.0 –
0.25 = 1.75 in.:
1
⎛
⎞
Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw
2
⎝
⎠
1
⎛
⎞
= ⎜ 1.75 − ( 7 8 + 1 8 ) ⎟ (0.355) = 0.444 in.2
2
⎝
⎠
Agv = ltw
= 7.25(0.355) = 2.57 in.2
Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw
= ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 1.69 in.2
Determine the tension rupture strength
Fu Ant = 65 ( 0.444 ) = 28.9 kips
Consider shear yield and shear rupture and select the least nominal
strength; thus,
0.6 Fy Agv = 0.6 ( 50 )( 2.57 ) = 77.1 kips
0.6 Fu Anv = 0.6 ( 65 )(1.69 ) = 65.9 kips
Selecting the shear rupture term and combining it with the tension rupture
term gives the connection block shear allowable strength, and recalling that
Ubs = 1.0 for the case of uniform tensile stress distribution, we have
Rn Ω = ( 65.9 + 1.0 ( 28.9 ) ) 2.00 = 47.4 < 55.0 kips
Thus, the given three-bolt connection is not adequate, with block shear
being the critical limit state to this point in our calculations.
Step 13:
Revise the connection to meet the block shear strength requirements.
Consideration could be given to increasing the number of bolts and thereby
increasing the length of the connection. However, because bolt shear
required only two bolts, this would not be a particularly economical
solution. If the connection were to be lowered on the beam end so that the
distance from the center of the top bolt to the edge of the cope were 2.5 in.,
the connection would have more block shear strength. It should also be
noted that the clear end distance for the top bolt in the beam web would
increase and thus increase the tearout strength of that top bolt. There is no
need to consider that increase since at best it could contribute to reducing
the number of bolts and this would have a negative impact on the block
shear calculations being addressed in this step.
Thus, the new shear areas become
Agv = ltw
= 8.50(0.355) = 3.02 in.2
Chapter 11
Simple Connections 489
Anv = ( l − ( n − 0.5)( d h + 1 16 ) ) tw
= ( 8.50 − 2.5 ( 7 8 + 1 8) ) (0.355) = 2.13 in.2
and the nominal shear yield and rupture strengths become
0.6 Fy Agv = 0.6 ( 50 )( 3.02 ) = 90.6 kips
0.6 Fu Anv = 0.6 ( 65 )( 2.13 ) = 83.1 kips
Selecting the shear rupture term and combining it with the tension rupture
term gives the resulting block shear allowable strength as
Rn Ω = ( 83.1 + 1.0 ( 28.9 ) ) 2.00 = 112 2.00 = 56.0 > 55.0 kips
Step 14:
Check the flexural strength of the coped beam.
It is a good idea to check this limit state during the initial design of the
beam. It should be anticipated during the beam design stage that a coped
connection will be required, and it is at that stage that a change in beam
section can most readily be accommodated.
Flexural strength of the coped beam is not addressed in the Specification
directly but is covered in Part 9 of the Manual. The moment in the coped
beam is taken as the shear force times the eccentricity from the face of the
support to the edge of the cope, taken as cope + setback = 7.0 + 0.50 = 7.5
in. in this example. Thus,
M a = Ra e = 55.0 ( 7.5) = 413 in.-kips
(Manual 9-5)
To determine the flexural strength of the coped beam, the web slenderness
must be checked. This requires determination of the plate buckling
coefficient, k, and the buckling adjustment factor, f. The web slenderness of
this coped beam is
ho t w = 16.0 0.355 = 45.1
The limiting web slenderness is
λ p = 0.46
k1 E
Fy
(Manual 9-12)
where
k1 = fk
(Manual 9-10)
The buckling adjustment factor, f, is a function of the ratio of the cope
length, c, to the beam depth, d, and the plate buckling coefficient, k, is a
function of the ratio of cope length, c, to the reduced beam depth, ho. For
this example, with c d = 7.0 18.0 = 0.389 ≤ 1.0
f =
2c 2(7.0)
=
= 0.778
18.0
d
(Manual 9-14a)
and with c ho = 7.0 16.0 = 0.438 ≤ 1.0
1.65
⎛h ⎞
k = 2.2 ⎜ o ⎟
⎝ c ⎠
Thus,
1.65
⎛ 16.0 ⎞
= 2.2 ⎜
⎟
⎝ 7.0 ⎠
= 8.61 (Manual 9-13a)
490 Chapter 11
Simple Connections
k1 = fk = 0.778 ( 8.61) = 6.70 ≥ 1.61
(Manual 9-10)
and
λ p = 0.46
6.70 ( 29,000 )
k1 E
= 0.46
= 28.7
Fy
50
Since λ is between λp and 2λp the net elastic and net plastic section modulus
will be needed. The net elastic section modulus is taken from Manual Table
9-2. With the depth of the cope, dc = 2.0 in.,
Snet = 23.4 in.3
The net plastic section modulus is not provided in the Manual so it must be
determined. Following the procedure discussed in Chapter 6, the net plastic
section modulus for the tee shaped beam that remains after the cope is
found to be
Z net = 42.4 in.3
and from
⎛ λ
⎞
M n = M p − (M p − M y )⎜
− 1⎟
⎝ λp
⎠
with
(Manual 9-7)
M p = Fy Z net = 50 ( 42.4 ) = 2120 in.-kips
M y = Fy S net = 50 ( 23.4 ) = 1170 in-kips
The nominal moment strength at the cope is
⎛ 45.1 ⎞
M n = 2120 − ( 2120 − 1170 ) ⎜
− 1 ⎟ = 1580 in.-kips
⎝ 28.7
⎠
and the design strength is
M n Ω = 1580 1.67 = 946 in.-kips > 413 in.-kips
So the coped beam has sufficient flexural strength.
Angles
Step 15:
Check the angles for shear rupture using Equation J4-4.
The net area of the angle on the vertical shear plane is
Anv = ( l − n ( d h + 1 16 ) ) ta = ( 8.5 − 3 ( 7 8 + 1 8 ) ) ( 3 8 ) = 2.06 in.2
and the design strength for two angles is
Vn Ω = 2 ( 0.6 Fu Agv Ω ) = 2 ( 0.6 ( 58 )( 2.06 ) ) 2.00 = 71.7 kips > 55.0 kips
(
So the angles are adequate for shear rupture.
)
Simple Connections 491
Chapter 11
Step 16:
Check the angles for shear yield using Equation J4-3.
The gross area of the angle on the vertical shear plane is
Agv = 8.5 ( 3 8 ) = 3.19 in.2
and the design strength for two angles is
Vn Ω = 2 ( 0.6 Fy Agv Ω ) = 2 ( 0.6 ( 36 )( 3.19 ) ) 1.5 = 91.9 kips > 55.0 kips
(
)
So the angles are also adequate for shear yield.
Step 17
Check the angles for block shear.
The equations for block shear in the angle are the same as those for the web
and as presented in Section 10.10.5.
First calculate the areas,
1
⎛
⎞
Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw
2
⎝
⎠
1
⎛
⎞
= ⎜ 1.0 − ( 7 8 + 1 8 ) ⎟ ( 3 8 ) = 0.188 in.2
2
⎝
⎠
Agv = ltw
= 7.25 ( 3 8 ) = 2.72 in.2
Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw
= ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) ( 3 8 ) = 1.78 in.2
Determine the tension rupture strength
Fu Ant = 58 ( 0.188) = 10.9 kips
Consider shear yield and shear rupture, and select the least nominal
strength; thus,
0.6 Fy Agv = 0.6 ( 36 )( 2.72 ) = 58.8 kips
0.6 Fu Anv = 0.6 ( 58 )(1.78 ) = 61.9 kips
Selecting the shear yield term and combining it with the tension rupture
term gives a connection block shear design strength of the double angle—
again, Ubs = 1.0 for this case of uniform tensile stress distribution—of
Rn Ω = 2 ( ( 58.8 + 10.9 ) 2.00 ) = 69.7 kips > 55.0 kips
Step 18:
Present the final connection design.
The three-bolt connection, revised as shown in Figure 11.3, is adequate to
carry the imposed load of 55.0 kips.
4992 Chapter 11
1
Simple Connections
C
Figure 11.4 Welded-Bo
olted Double--Angle Conneection
111.4
DOUBLE-ANGLE
E CONNEC
CTIONS: WELDED-BO
W
OLTED
The double--angle shear connection
c
caan also be connstructed by ccombining weelding and boolting. In
this case thee angles are welded
w
to the beam
b
web, ass shown in Figgure 11.4.
The limit states to
o be considerred are
1.
Bolts
a.
Shear
S
rupture
2.
Weld
a.
Rupture
R
3.
Beam
a.
Shear
S
yielding
g of the web
b.
Block
B
shear on
o coped beam
m web
c.
Coped
C
beam flexural
f
strenngth
d.
Web
W strength at the weld
Angles
4.
a.
Bolt
B bearing and
a tearout onn angles
b.
Shear
S
rupture
c.
S