Modeling for Structural Analysis
Behavior and Basics
Graham H. Powell
Professor Emeritus of Structural Engineering
University of California at Berkeley
Computers and Structures, Inc., Berkeley, California, USA
Copyright© 2010 Computers and Structures, Inc.
All rights reserved.
No part of this publication may be reproduced or distributed in any form or
by any means, or stored in a database or retrieval system, without the prior
· ___,,
·
explicit written permission of the publisher.
Computers and Structures, Inc., 1995 University Avenue, Berkeley,
California 94704 USA web: www.csiberkeley.com
SAP2000® is a registered trademark of Computers and Structures, Inc.
Considerable time, effort and expense have gone into the development and
documentation of SAP2000®, including thorough testing and use. The user
must accept and understand that no warranty is expressed or implied by
the developers or the· distributors on the accuracy or ·the reliability of the
program. SAP2000® is a practical tool for the design/ check of structures.
The user must thoroughly read the manuals and must clearly recognize the
aspects of design that the program algorithms do not address. The user
must explicitly understand the assumptions of the programs and must
independently verify the results.
Library of Congress Cataloging-in-Publication Data
Powell, G. H. (Graham Harcourt), 1937Modeling for structural analysis : behavior and basics I Graham H.
Powell.
p.cm.
Summary: "Explains purpose and limitations of structural analysis as tool
for designing buildings, other structures. Describes linear and nonlinear
behavior of structures and structural components, and how to model this
for analysis. Uses physical explanations rather than formal theory or
mathematics. Reference for students, educators, practicing engineers at all
levels"-- Provided by publisher.
ISBN978-0-923907-88-4 (hardcover: alk. paper)
1. Structural analysis (Engineering)--Mathematical models. I. Title.
TA645.P64 2010
624.1 '71015118--dc22
2010020065
Printed in China
10 9 8 7 6 5 4 3 2 1
Preface
This book is aimed at a wide audience, and it has ambitious goals. If you are
a student, the goal is to provide you with a foundation for the classes that
you are taking in structural analysis and structural design. If you are a
young engineer, the goal is to help you understand what you are doing
when you use a computer program for structural analysis, and to help you
become a better engineer. If you are an experienced structural engineer, the
goal is to help you ·keep things in a proper perspective. If you are a
university professor who teaches structural analysis, the goal is to persuade
you to change the way that you teach the subject. In short, the goal of this
book is to change how structural analysis is perceived and taught.
At the same time, the scope of this book is rather narrow. It covers the basics
of modeling for structural analysis, but does not include many details. It
covers the Direct Stiffness Method of analysis, using physical explanations
rather than formal theory. It covers both material nonlinearity and
geometric nonlinearity in considerable depth, with emphasis on physical
understanding not on theory or mathematics. It also puts structural analysis
in its proper place, as a tool for use in structural design, not as an end in
itself. This book does not consider structural analysis theory, or how to
program structural analysis for a computer. It considers linear, nonlinear,
static and dynamic analysis, but does not explain the analysis theories in
detail. Many of the details are topics for future volumes. Throughout the
book the emphasis is on physical understanding, not on formal theory or
mathematics.
There is a reason for this approach. I have often heard it said that young
engineers use computer programs blindly, without understanding what
they are doing. This is probably true, and it is unfortunate. However, my
experience tells me that young engineers are not to blame.
~
The problem, I believe, is that engineering students are trained to see
structural analysis as some magical thing that can tell us everything we
need to know about the behavior of a structure, with a high degree of
accuracy. This is an illusion. Structural analysis is at best highly approximate, and any predictions about structural behavior that are made by a
computer program should be viewed with skepticism. Structural analysis is·
not some magical thing. It is merely a tool to help with structural design,
and a highly imperfect one.
ii
Preface
[have also heard it argued that the developers of computer programs are to
blame (not CSI, but some competitors). I disagree. A computer program for
structural analysis is a tool, and like any tool its primary goal is to enhance
productivity. The program developer's task is to produce the best possible
tool. The engineer's job is to use it with skill. It is the job of somebody else to
provide young engineers with the education and training that they need to
develop the skills. What are these skills, and who is the "somebody else"?
The following are my opinions on the required skills.
(1)
For the vast majority of engineers the skills do not include writing a
computer program to do structural analysis. This may have been a
useful skill in 1975, when structural analysis programs had limited
capabilities and often had to be augmented. It is not true with today's
computer software, which can do some amazing things. Computer
program development is now a task for specialists. For the vast
majority of engineers the challenge is to use computer programs, not
develop them.
·
It is, however, a valuable skill to write a program to process analysis
results in a specialized way, using languages such as Matlab, Mathcad
or Visual Basic. These are general purpose tools that most engineers
should be faIIliliar with and use routinely.
(2)
The skills do not include arialyzing a structure using classical "hand"
calculation methods such as Moment Distribution. I have heard it
argued that students should learn Moment Distribution in order to
develop a "feel" for structural behavior. On this point I emphatically
disagree. Moment Distribution was an excellent tool in its day, but it
is outdated and of only historical interest. I have not used Moment
Distribution in decades, and it does little to develop "feel".
Nevertheless, some hand calculation skills are definitely valuable.
Free body diagrams and equilibrium equations are extremely useful
for understanding the flow of forces and for checking that the forces
from a computer analysis satisfy equilibrium. The moment-area
method is extremely useful for checking that the deflections from a
computer analysis are reasonable. These methods, and some other
simple techniques, are essential skills. They also help to develop
"feel".
·
(3)
The skills (for most engineers) do not include a detailed understanding of "matrix methods". To begin with, there is no such thing as a
Preface
iii
-''matrix method". There is matrix notation (which is extremely useful
and should be used routinely by all engineers), and there are matrix
formulations of structural analysis methods. Most analysis methods
can be formulated with or without matrices. If a method is of only
historical interest when formulated without matrices, adding matrix
notation does not make it modem or useful.
There is, however, one analysis method that relies heavily on
matrices.. This is the Direct ·Stiffness Method. Almost all computer
programs for structural analysis are based on this method. All
engineers who do structural analysis should have a basic understanding of this method. They do not, however, need to understand the
mathematical details, and they do not need to be able to program it
for a computer. The Direct Stiffness Method is a very physical process,
and most engineers heed to understand it only in physical, not
mathematical, terms. This means understanding nodes, elements,
degrees-of-freedom, the physical meaning of a stiffness coefficient and
a stiffness matrix, how the stiffness matrices for the elements in a
structure can be assembled into a structure stiffness matrix, the need
to solve thousands of simultaneous equations, and how things can go
wrong if the analysis model is poorly conceived. The theoretical and
computational details need to be mastered only by the relatively few
engineers who work on computer program development. ·
(4)
Understanding how structural components behave is an essential
skill. For a component that is elastic (or more correctly, that can be
assumed to be elastic for analysis purposes), the key property is the
stiffness, or stiffness matrix. For beam and column components this is
usually in terms of bending stiffness (EI), axial stiffness (EA) and
possibly shear stiffness (GA'). Most textbooks on structural analysis
imply that these values are well defined and easy to calculate. In a
real structure that is often not the case. For example, how does one
calculate EI for a reinforced concrete beam that has substantial
cracking, and where· the amount of cracking varies along the beam
length? How does one calculate EI for a reinforced concrete column
where the amount of crat;king depends on the axial force? How does
· one know whether shear deformations are important or can be
ignored? Textbooks rarely address such issues.
Also, this is just for elastic analysis. In many cases, especially for
earthquake motions, a structural component can be loaded beyond
yield and become inelastic. How does the component behave? What
aspects of the behavior are important for analysis and design? What
iv
Preface
properties are needed to capture these aspects in an analysis model?
How can values for these properties be estimated? Since the properties are probably not known accurately, how does one account for the
.uncertainty? These are important issues that are rarely addressed in
structural analysis courses or textbooks.
(5)
The ability to set up an analysis model that captures the important
aspects of structural behavior is an essential skill. It is also every bit as
challenging academically as "matrix methods". Indeed, in my opinion
modeling for analysis is more challenging academically than analysis
theory.
Related to this, the ability to check computer results for consistency is
an essential skill (this is "feel" - do the results look right?). Much of
this skill develops with experience, but it can be taught. It can not,
however, be developed by learning Moment Distribution or Matrix
Methods. It can be developed much more effectively by analyzing
structures on a computer, examinirig the results critically, doing
"what-ifs" by varying. the structure properties, specifying unrealistically large stiffnesses to see what happens, and so on. For most
engineers this is much more useful than analysis theory.
(6)
Knowing how computer results are used for making design decisions
is an essential skill. In a typical structural analysis textbook the, end
result is a deflected shape and a bending moment diagram. In practice
this is only the beginning. The important thing is how the analysis
results are used to support decision making for design. Students
should understand that structural analysis is not an end in itself, but
merely a tool for use in design.
If I am correct, and these skills are the most important, why are they not
being taught? The following are my opinions on the cause of the problem.
(1)
There are three phases in structural analysis, namely "modeling" at
the beginning, "interpretation" at the end, and "computation" in the
middle. For most engineers the most important phases are modeling
and interpretation. The least important phase is computation (which ·
includes analysis theory as well as number crunching).
(2)
The computation phase is always handled by a computer program.
The program developers take care of the theory (and of many other
things, such as data management and graphics) and the computer
crunches the numbers. Most engineers can treat a computer program
Preface
v
as a "black box" that takes a model of the structure at one end and
produces "results" at the other. An engineer must have confidence
that the computations are done correctly, and must have an overall
understanding of how the computations are performed, but he or she
does not need to be concerned with the computational details. For
most engineers the most important phases are modeling and
interpretation. These phases generally require human skills and
intelligence, and generally are not handled well by computer programs.
(Automated modeling and interpretation is done to some extent, and
it is a goal of program developers. Some engineers may look forward
to the day when all three phases are automated, but be careful what
you wish for.)
(3)
This brings me to what Ibelieve is the problem. Engineering students,
in Universities around the world, are being taught almost exclusively
"computation", with little attention being paid to "modeling" or
"interpretation". The skills that students are being taught are not
useful, and the skills that are useful are not being taught. The
"somebody else" who is responsible for teaching the needed skills is
the University Professor, and he or she is often not doing a very good
job. It is relatively easy to teach computation, which is mainly theory,
and professors are usually good at theory. It is harder to teach
modeling and interpretation, and this is something that professors
often do not do so well.
It does not help that there are few, if any, textbooks that deal with modeling
and interpretation. This book does not cover all of the above skills, but it
does fill some of the gap. It covers the behavior of structural components,
the direct stiffness method, and the basic principles of modeling and
interpretation. It is planned as the first in a series, with future volumes that
consider element modeling in depth, and explain in detail the assumptions
and procedures for linear, nonlinear, static and dynamic analyses.
This iS not a textbook in the usual sense, with worked examples and
problems to be assigned. Rather, it provides background information on
behavior and modeling. In order to teach the sorts of skills that I have
referred to, a course in structural analysis would need to use exercises such
as the following.
(1)
Set up free bodies of a variety of types. Use equilibrium equations and
the virtual displacements principle to solve equilibrium problems.
Use free bodies to check the results of computer analyses, for example
ri
Preface
the forces on a beam-to-column connection. Emphasize to students
that there is no excuse for errors in free bodies and equilibrium.
2)
Sketch deflected shapes, to get a feeling for how structures deform
and how deformed elements fit together to produce the deflected
shape for a structure.
:3)
Solve simple deflection problems. Also check that the deflections
calculated by computer analysis are reasonable. I like the momentarea method, because it is physical. I also like to use simple standard
results, such as PL3 /3EI for the deflection of a cantilever beam with a
load at the tip. I do not like the virtual forces principle (the "dummy
unit load" method), because it is too much of a mathematical process
rather than a physical one.
~4)
Require students to use a computer program for structural analysis,
starting on the first day of the first analysis course. Set up linear
elastic· models for structures of a variety of types, vary the stiffnesses
of the elements, and run computer analyses. See the effects of the
changes, and explain these effects. This is, I believe, the best way to
develop a "feel" for structural behavior.
(5)
As students develop modeling and interpretation skills, add nonlinear
· analyses with material and geometric nonlinearity, always emphasizing the modeling assumptions and requiring explanations of the
behavior. Also add dynamic analyses. A first undergraduate course in
structural analysis could progress as far as simple inelastic analysis. A
second course should include dynamic analysis.
(6)
Show how analysis results are used for de8ign. Emphasize that structural
analysis is at best very approximate, and that it is not an end in itself
but merely a tool to support design. Coordinate course material in
analysis and design - all too often they are taught as independent
discipl~es.
The following are some additional points on this book:
(1)
The words "in the author's opinion" could be added in many places.
These words have been omitted to avoid excessive repetition. Much of
the book is simply the author's opinion. Disagreement is welcomed.
(2)
There is no list of references. The task of compiling one and making
the appropriate citations is simply too great. Since all analysis will be
Preface
vii
done by computer, the best initial reference is the computer program
documentation. This will lead to additional references.
(3)
This book considers design as well as analysis, and it references
design codes and standards of practice. These are mainly the U.S.
codes for steel and concrete, including ANSI/AISC 360 and ACI 318,
and also ASCE 7 and ASCE 41. These are all well known and readily
available. The notation in this book does nqt follow the notation in
any particular code.
(4)
I would like to mention one book that is a rare ex~ple of a textbook
that deals with structural behavior. This is The Elements of Structure by
W. Morgan (edited by I. Buckle), second edition, 1977, Pitman. That
book covers structural behavior, not modeling. However, successful
modeling starts with an understanding of behavior. If you can find a
copy of this book, it is well worth reading.
I would like to acknowledge Jeff Hollings for his help in reviewing the text,
Iqbal Suharwardy and G. Robert Morris for technical help, Ashraf Habibullah
for being so patient, and my wife, Lynette, for being ever so patient.
Finally, I would like to dedicate this book to the memory of Professor Tom
Paulay, the father of Capacity Design, the best teacher that ever was, and the
nicest person I ever met.
Graham Powell
Berkeley, California
graham@csiberkeley.com
January 2010
cl.,
Contents
Introduction ................................................................................. 1
1.1
Overview •...........•.........•..•........•......................•....•...........•.:........•.....•.... 1
1.2
The Phases of Structural Analysis .....•...•...........•.................•....••....2
1.2.1
Modeling ...............................................................................2
1.2.2
Interpretation ......................................................................2
1.2.3
Computation .......................................................................3
. 1.3
Relative Importance of the Three Phases ....................................4
1.4
Demand and Capacity .......................................................................5
1.4.1
Performance Assessment ................................................5
1.4.2
Direct Design ·········~········.....................................................6
1.5
Elastic vs. Inelastic Analysis ..............................................................?
1.5.1
Behavior of a Structural Component...........................7
l.5.2
Elastic vs. Inelastic Behavior:...........................................8
1.S.3
Strength-Based Design Using Elastic Analysis .......•.8
l.5.4
Strength-Based Design Using Inelastic Analysis ......9
1.5.5
Deformation-Based Design for Earthquake
Loads .................;................................................................. 11
1.5.6
Strength-Based Design for Earthquake Loads ....•. 12
1.5.7
Capacity Design Using Elastic Analysis ...•.••.....•...•... 13-
1.6
Static vs. Dynamic Analysis .........................:.................;............... 14
1.7
Small vs. Large Displacements Analysis ................................... 15
1.7.1
Overview ........;•.•;.,............................................................. 15
- 1.7.2
Equilibrium ........................................................................ 16
1.7.3 · Compatibility (Continuity) ........................................... 17
1.7.4
AnalysisTypes •..•.,,.:......................................................... 17
1.7.5
Catenary Effe.ct ................................................................. 19
1.8
Demand Analysis vs. Capacity Analysis .................................... 20
1.8.1
Overview.;:.. ~...................................................................;;. 20
1;8.2
Lateral Load at First Yield ............................................. 21
1.8.3
Earthquake Intensity at Collapse ............................... 22
1.8.4
Bending Strength of a Beam :············.......................... 22
1.8.5
Plastic Hinge Rotation Capacity .•......•.........•..••.....•..:. 23
1.8.6
Conclusion for this Section •.•••••...•~...........................,.• 26
1.9
Conclusion for this Chapter .......................................................... 26
1.1 O topics for the Following Chapters.............................................. 26
ix
Contents
Chapter2
Chapter3
What is an Analysis Model? ...................................................... 29
2.1
Actual Structure vs. Analysis Model •.•••••.......•................•...••...•.. 29
2.2
Two Types of Analysis Model ....................................................... 29
2.3
Features of Node-Element Model...................................~........... 31
2.4
Some Element Types................................_
....................................... 33
25
Connection between Nodes and Elements.•.....•..........••••..••... 34
2.5.1
Overview ............................................................................ 34
2.5.2
Connection for Bar Elements ...................................... 34
2.5.3
Rigid End Zones ............................................................... 35
2.6
Gaps and Ovedaps between Elements .................;................... 36
2.6.1
Surface Elements .............-.............................................. 36
2.6.2
Element-to-Element Contact ...................................... 37
2.7
Equilibrium between Elements ................................................... 37
2.7.1
Equilibrium at Element Boundaries .......................... 37
2.7 .2
Equilibrium at Connections ......................................... 39
2.8
Discrete Model with Finite Size Nodes and Zero Length
Elements .............................................................................................. 39
2.9
Continuum Model ............................................................................ 40
2.9.1
FrameStructure ............................................................... 40
2.9.2
Wall Structure ...:............................................................... 42
2.1 0
Elements and Components .......................................................... 43
The Direct Stiffness Method ..................~.................................. 45
3.1
Element Stiffness and Flexibility ................................................. 45
3.1.1
Overview ..............._.........................................:................. 45
Bar Element ....................................................................... 46
3.1.2
3.1.3
Rigid Body Displacements and Deformation
Modes ...•..•.•.....•...••..••.••.•..••.•••......,..................................... 48
3.1.4
Beam Element .................................................................. 49
3~1.5
Surface and Solid Elements ......................................... 51
3.1.6
Slab and Shell Elements ................................................ 54
3.2
· Stiffness and Flexibility Analysis Methods ............................... 55
3.3
The Direct Stiffness Method ......................................................... 56
3.3.1
Overview ............................................................................ 56
3.3.2
Degrees of Freedom ...................................................... 56
Addition of Stiffness ....................................................... 58
3.3.3
3.3.4
Assembly of Structure Stiffness.................................. 60
Contents
3.3.5
3.3.6
3.3.7
3.3.8
3.3.9
3.3.10 ·
3.3.11
3.3.12
3.3.13
3.3.14
Addition of Flexibility..................................................... 61
Optimal Node Numbering ...:.......................................62
Equilibrium Equations ..........................;:....................... 63
Load Matrix.....................................................;.................. 64
Equation Solving .............................................................64
Element Deformations and Forces ............................ 65
Equilibrium ChecL..................................................-...• 65
Cause of Equilibrium Unba,lance..........•.............._. .... 65
Reactions at Rigid Supports ......................................... 66
Forces Corresponding to Slaving Constraints ....... 66
3.4
Slaving Constraints - Rigid Floor Diaphragm ......................... 67
3.5
Equilibrium Errors in Linear Analysis ..........................;.............. 69
3.5.1
Overview ............................................................................ 69
3.5.2
Unstable Structure .......................................................... 69
3.5.3
Stiffness Coefficient Mismatch .........•...•..•.•................ 72
3.5.4
Programming Error•..••..•....,............................................ 75
3.5.5
Detection of Unstable Mode ....................................... 76
3.6
Element Loads ................................................................................... 79
3.6.1
Overview ............................................................................ 79
3.6.2
Element Force-Displacement Relationship ............ 79
3.6.3
Structure Equilibrium Equations ................................ 80
3.6.4
Element Initial Deformations ...................................... 81
3.6.5
Imposed Displacements at Rigid Supports ............ 83
3.7
Dynamic and Nonlinear Analysis ...................:............................ 84
3.7.1
Dynamic Loads.................................................................84
3.7.2
Material Nonlinearity .••.,................................................ 85
3.7.3
Geometric Nonlinearity................................................. 85
/
Chapter 4
xi
Substructures and Superelements ............................................. 86
Component Behavior- Uniaxial F-D Relationships •.•••••••.-. ... 89
4.1
4.2
Overview .............................................................................................89
4.1.1
Components and Elements·-·········-··:....................... 89
4.1.2
Modeling Goals for Components .............................. 90
Component Force-Deformation Relationships ...................... 91
4.2.1
A Common F-D Relationship ....................................... 91
4.2.2
F-D Relationship for Monotonic Deformation ....... 92
4.2.3
Complications for Cyclic Deformation ..................... 95.
4.2.4
Elastic and Plastic Deformations ................................ 97
4.2.5
Ductility Ratio ..............._ ................................................ 99
4.2.6
Rigid-Plastic Hinges ......................................................100
4.2.7
Other Nonlinear F-D Relationships ......................... 101
di
Contents
4.2.8
.:hapter 5 ·
Summary for this Section ........................................... 103
•
4.3
What Type of F-D Relationship is Needed? ........................... 103
4.4
Stiffness for Elastic Analysis ........................................................ 104
4.4.1
Bending Stiffness for Beams, Columns and
Walls .................................................................................. 104
4.4.2
Shear Deformation in Reinforced Concrete ......... 107
4.4.3
Connections ...............:.................................................... 108
4.4.4
Summary for this Section ........................................... 108
4.5
F-D Relationships for Inelastic Analysis .................................. 108
45.1
Overview .;........................................................................ 108
4.5.2
Amount of Inelastic Behavior.................................... 109
45.3
Practical F-D Relationship - Generic Form ........... 110
4.5.4
F-D Relationships in ASCE 41 .................................... 112
4.5.5
Backbone Relationship................................................ 114
4.6
Hysteresis loops for Inelastic Analysis .................................... 115
4.6.1
Loop Anchored to Backbone Relationship........... 115
4.6.2
Hysteresis Loop in CSI PERFORM-3D ...................... 117
4.6.3
Practical Modeling of Cyclic Degradation ............ 118
4.6.4
"In-Cycle" and "Between-Cycle" Strength Loss ... 119
4.7
Conclusion for this Chapter ........................................................ 121
Component Behavior- Multi-Axial F-D Relationships with
lnteradion •.•••..••.•..•••••.••.........•.~·············································- 123
5.1
Overview ....................................................................: ..................... 124
5.2
. Stiffnes.s Interaction....................................................................... 124
5.3
Strength Interaction ..:................................................................... 125
5.4
Inelastic Interaction : Behavior after Yield ............................. 127
5.5
Plasticity Theory for Yield of Metals ...........- ........................... 129
55.1
Overview .......................................................................... 129·
55:2
Yield of Efastic-Perfectly-Plastic Metals ................. 129
5.5.3
Strain Hardening ........................................................... 131
5.6
Interaction Surface for Friction ..................................................135
5.6.1
Bearing Component with Friction........................... 135
5.6.2
Cohesion vs. Friction .................................................... 137
5.7
Ext~nsion to P-M-M Interaction ..................,.............................. 137
5.7.1
5.7.2
5.7.3
Overview ........_ ........................._.......- .......................... 137
Steel Section.- The Analogy Works ......................... 139
Sharp Peak in Yield Surface ....................................... 141
Contents
5.7.4
5.7.5
5.7.6
5.7.7
5.7.8
xiii
RC Section - The Analogy Does Not Work
SoWell ..............................................................................142
Ductile Limit and Strength Loss ...............................145
Hysteresis Loops and Stiffness Degradation ........ 147
Cyclic Degradation .......................................................148
Other Cross Section Shapes ....................................... 148
5.8
Is Plasticity Theory Useful for P-M Interaction? .................... 148
5.8.1
Overview .............................. :~.......................................... 148
Assumptions and Approximations.......................... 148
5.8.2
5.83
Importance of Axial Deformation ,' Steel Column ..................................................................149
Is the Axial Deformation in a Plastic
5.8.4
Hinge Correct? ...............................................................152
Accumulated Axial Deformation - Concrete
5.8.5
Column .............................................................................154
Can the Axial Extension be Assumed
5.8.6
tobeZero?.......................................................................155
Axial Deformation in Static Push-Over
5.8.7
Analysis .............................................................:............... 156
Conclusion for this Section ........................................ 157
5.8.8
5.9
Axial Extension in Concrete Beams ..........................................158
5.10 Fiber Sections for P-M interaction ............................................ 159
5.10.1 Overview ...........;..............................................................159
5.10.2 Fiber Sections for Beams ............................................ 160
5.10.3 Fiber Sections for Columns ........................................160
5.10.4 Fiber Sections for Walls ............................................... 161
5.10.5 Fiber Segments ..............................................................162
5.10.6 Rigid-Plastic Fiber Hinge ............................................. 162
5.10.7 Limitations of Fiber Models ............................,.......... 163
5.11
Inelastic Shear in Beams and Columns ................................... 164
5.11.1 Overview ...................................................................;...... 164
5.11.2 M-V Interaction in Steel Beams................................. 164
5.11.3 P-M-V Interaction in Steel Columns ........................ 165
5.11.4 M-V Interaction in Reinforced Concrete
Beams ................................................................................166
5.11.5 P-M-V Interaction in Reinforced Concrete
Columns ...........................................................................167
5.11.6 P-M-V Interaction in Connections............................ 168
5.11.7 Analysis vs. Design ........................................................169
5.12 Shear in Concrete Walls ................................................................169
5.13 Multi-Axial Material Models for Plain Concrete .................... 173
5.13.1 Motivation .......................................................................173
xiv
Contents
5.13.2
5.13.3
5.13.4
Plasticity Theory ................................................_.......... 174
Compression Field Theory ....................•.............•...... 175
Simple Models Based on Uniaxial
5.13.5
5.13.6
5.13.7
5.13.8
Stress-Strain ·························'····················--········;.•.•.•.•.. 178
Possible Model with P-V lnteraction .••.................... 180
Plain Concrete Models for 3D Stress ....................... 180
Over-Reliance on Analysis ....................•....••............... 181
A Note on Demand and Capacity Analyses.......... 182
5.14 Capacity Interaction ..............•.........................•.............•............•.. 183
5.14.1 Overview ..........................................................................183
5.14.2 Effect of Axial Force on Bending Ductility .......•.... 184
5.14.3 Effect of Shear Force on Bending Ductility.....•..... 184
5.14.4 Effect of Hinge Rotation on Shear Strength •.......• 184
Chapter6
5.15
Plastic Deformation and Deformation DIC Ratios .............• 184
5.16
Summary for this Chapter ........................................................... 186
5.17
Conclusion for this Chapter ......................•......•...........•.............. 189 ·
P-A Effects, Stability and Buckling ......................................... 191
6.1
Overview ......................................................•....•......•....•••....•..........•.191
6.1.1
Causes of Geometric Nonlinearity.....•........•............ 191
6.1.2
Types of Analysis for Geometric Nonlinearity ..... 192
6.1.3
P-~ Effect.......................................................................... 192
6.1.4
Load Types and Corresponding Analyses ............ 193
6.1.5
Elastic and Inelastic Analysis ....•......•................•.....•.. 193
6.1.6
Topics for this Chapter ...........................•••...•.............. 194
6.2
P-~ and
6.2.1
6.2.2
6.2.3
6.2.4
6.3
Relative Importance of P-~ and P-o Contributions ......•...... 200
6.3.1
Cantilever Column ........................................................ 200
6.3.2
Column in an Unbraced Frame .....•....•...••..•..•.......... 200
6.4
Modeling of P-~ and P-8 Contributions......•.......•..•.......•.......• 203
6.4.1
Overview ..........................................................................203
6.4.2
Model for a Cantilever Column .......•........................ 204
6.4.3
Behavior of P-o Cable ..........•...•......•...•..•....•................. 205
6.4.4
Behavior of Elastic Cantilever Column ..•..•.........•... 207
6.4.S
Moments and Shears for Column Design ............. 208
6.4.6
Elastic Column in an Unbraced Frame •.................. 210
P-o Contributions in a Single Column..................... 197
Overview ..........................................................•............... 197
Cantilever Column ....................................•.................•. 198
Approximations in P-~ Analysis .....•....•.•.................. 198
P-~ vs. Small Displacements Effects ..•...•.•..........•... 199
Contents .
6.4.7
6.4.8
xv
Inelastic Column in an Unbraced Frame ............... 212
Element Models .............................................................213
65
Lateral Load Behavior of Frames ...............................................214
6.5.1
Overview ..........................................................................214
6.5.2
P-d Struts and P-d Columns ......................................214
6.5.3
Review of Terminology ...............................................216
65.4
Main Structure and P-d Column .............................. 216
6.5.5
Effect of P-d Column on Stiffness and
Strength............................................................................217
6.5.6
Load Increase or Strength Decrease? .....................219
6.5.7
Load on P-d Column vs. Load on Main
Structure............................................................................220
6.5.8
P-d Effect in Beams .......................................................220
6.S.9
P-d Effect in Braced Frames .......................................220
6.5.10 Summary for this Section ...........................................222
6.6
Buckling BehaviorofFrames ......................................................223
6.6.1
Overview ..........................................................................223
6.6.2
Stability of Equilibrium State .....................................223
6.6.3
Bifurcation of Equilibrium ..........................................225
6.6.4
Inelastic Behavior after Buckling ..............................226
6.6.5
Inelastic Behavior before Buckling ..........................227
6.6.6
Effect of Initial Imperfection - Elastic Case............ 228
6.6.7
Effect of Initial Imperfection - Inelastic Case ........229
6.6.8
Effect of Gravity Sway ..................................................230
6.6.9
Are Buckling loads Useful? ........................................231
6.6.10 Summary for this Section ...........................................231
6.7
P-d Columns in Multi-Story and 3D Buildings ......................232
6.7.1
P-d Column in Multistory Buildings ........................ 232
6.7.2
P-d Column in 30 Buildings ......................................233
' 6.7.3
Buildings with Multiple P-d Columns.....................235
6.8
Buckling of Multi-Story and 3D Frames ..................................236
6.9
Buckling of an Axially loaded Column ...................................238
6.9.1
Overview ..........................................................................238
6.9.2
Ideal Elastic Column .....................................................239
6.9.3
Imperfect and Inelastic Column ...............................241
6.9.4
Steel Column ..................................................................242
6.9.5
Reinforced Concrete Column ....................................243
6.9.6
Theories for Buckling Strength .................................243
6.9.7
Combined Material and Geometric
Nonlinearity._..._.............................................................244
6.9.8
Analysis Model for Buckling Strength ....................246
6.9.9
Summary for this Section ...........................................247
xvi
Contents
6.10 Simple Structure with Pin-Ended Members .......................... 248
6.10.1 Overview ........................................::................................ 248
6.10.2 Example Structure and Analysis Model ................. 248
6.10.3 Strength-Based Design Using Elastic Analysis .... 249
6.10.4 Deformation-Based Design Using
Inelastic Analysis ...........................................................251
6.10.5 ·Modeling ofDiagonal Brace Behavior ...•............... 251
6.11
Pin-Ended Elastic Column with Bending ...........•.................... 252
6.11.1 Overview ..........................................................................252
6.11.2 Elastic Column with Sinusoidal Lateral Load .....•. 253
6.11.3 Elastic Column with Other Lateral Loads .............. 254
6.11.4 Elastic Column with End Moments ......................... 254
6.11.5 Direct Calculation of Amplified Moments ............ 256
6.1 l.6 Elastic Column in a Frame ..........................................256
6.12 Beam-Column Strength ...............................................................257
6.12.1 Overview ..........................................................................257
6.12.2 Elastic-Perfectly-Plastic Behavior............................. 257
6.12.3 . Behavior with Progressive Yield ............................... 259
6.12.4 Amplification Using Tangent Modulus Theory ... 260
6.12.5 · Tangent, Reduced or Secant Modulus? ................. 263
6.13 Strength-Based Design of Beam-Columns ............................264
6.13.1 Overview ..........................................................................264
6.13.2 Steel Beam-Columns ....................................................264
6.13.3 ·Reinforced Concrete Beam-Columns ..................... 266
6.14 Deformation-Based Design of Beam-ColumQs .................... 267
6.15 Compression Membersin Braced Erames •..•......................... 268
6.15.1 Overview ..........................................................................268
6.15.2 Frame Braced by a'Wall .....................•.......•.................268
6.15.3 Strength-Based Design Using Elastic Analysis .... 269
6.15.4 Deformation-Based Design Using
Inelastic Analysis ••;........................................................270
6.15.5 ; Diagonally Braced Frame ........................................... 270
6.16 Columns in Unbraced Frames ....................................................271
6. l6.1 Overview ..........................................................................271
6.16.2 Frame and Column Buckling .....................................272
6.16.3 Strength-Based Design Using Elastic Analysis ...• 273
6.16.4 Deformation-Based Design Using
Inelastic Analysis ..'...- ................................................... 274
6.17 A Complication - Initial Drifts ,_................................................. 275
6.17.1 Overview ..........................................................................275
6.17.2 Modeling Methods .......................................................275
Contents
xvii
6.18 A Second Complication - Stiffness Reduction ..................... 277
6.18.1 Overview ..........••....•..............•.,........................................277
6.18.2 Effect on Lateral Load Analysis .................................278
6.18.3 Modeling of Stiffness Reduction .•,........................... 281
6.18.4 Effect on the P-8 Contribution ..................................282
6.18.5 Effect on Buckling Analysis ...............•......•.................282
6.18.6 Is a Reduced Stiffness Analysis Necessary?........... 283
6.19
Some Theory - Geometric Stiffness .........................................283
6.19.1 Overview ..........................................................................283
6.19.2 Geometric Stiffness Matrix for P-6 Strut........••....•. 284
6.19.3 Structure Stiffness Matrix ...........................................286
6.19.4 P-o Contribution ............................................................287
6.20
Methods for Elastic Lateral Load Analysis ..............................287
6.20.1 Overview .............................;.......................•....................287
6.20.2 Sway and Non-Sway Amplification .......•................. 288
6.20.3 Methods for Sway Amplification ..•........•.•••........•..... 288
6.20.4 Methods for Non-Sway Amplification .................... 290
6.20.5 Sway and Non-Sway Similarities .•............................293
6.20.6 Analysis Methods ..........................................................294
6.20.7 "Double-B" Analysis ......................................................295
6.20.8 "Single-B" Analysis ............................•...........................296
6.20.9 "Zero-B" Analysis ...........................................................298
6.20.10 End Moments in Zero-B Analysis ........•..•.;............... 298
6.20.11 Initial Imperfections ...........................................,......... 302
6.20.12 Stiffness Reduction and Sway Amplification .......303
6.20.13 Stiffness Reduction and Non-Sway
Amplification ..................................................................304
6.20.14 Demand/Capacity Calculation .....•....•......•.........•......305
6.20.15 Conclusion for this Section ................•.......................305
. 6.21
Direct Analysis Method for Steel Frames ...............................306
6.21.1 Overview ......................................•......•........................•...306
6.21.2 Acceptable Second-Order Analysis .........................306 ·
6.21.3 Benchmark for Sway Amplification ................•........306
6.21.4 Benchmark for Non-Sway Amplification ••.............307
6.21.5 · Acceptable Methods .....................••••~ ..........................308
6.21.6 Initial Drifts ......................................................................308
6.21.7 Basic Stiffness Reductiqn ............................................309
6.21.8 Advanced Stiffness Reduction ..................................310
6.21.9 Alternative to Advanced Stiffness Reduction ......312
6.21.10 Conclusion for this Section ....•...................................312
6.22
Inelastic Lateral Load Analysis of Frames ...............................313
6.23
Buckling Analysis ..............................................;.............................315.
xviii
Contents
6.23.1
6.23.2
6.23.3
Overview ..........................................................................315
Analysis Method with Equilibrium Bifurcation ...315
Buckling Analysis With lmperfections..•.•...............317
6.24 Some Other Structures ............................•..............•........•••..........318
6.24.1. Overview ......................•......................•......•.•...................318
6.24.2 Walls With Out-of-Plane Bending .•......••.................. 318
6.24.3 Long Span Roofs............................................................318
6.24.4 Bridge Columns .•.•..........•.....................•........•.••........~ .. 320
6.24.S Buckling of Pipe With Zero Axial Foree .•. ,...••....•.... 320
6.24.6 Buckling of Buried Pipe ...........................•...................321
6.24.7 Large Displacements ofBuried Pipe ....•..•............... 322
6.24.8 Structures With "Follower" Forces ..•.........•..............323
6.25 Lateral-Torsional Buckling ofBeams....................•.........••........ 325
6.25.1 Overview ..........................................................................325
6.25.2 Causes of Lateral-Torsional Buckling ......................325
6.25.3 Analysis Models ................................•....•.......•............... 329
6.25.4 Torsional Behavior of I-Section Beams...................329
6.25.S Compression Flange as a Column ...........,••.............331
6.25.6 Effect of Shear Force on Beam Buckling ................331
6.25.7 Practical Modeling ...............................•:...•......•............333
6.26 Bracing to Prevent Buckling ....................................................... 333
6.27 P-~ Effects in Seismic Isolators .........................•......;.........•.....•. 335
6.27.1 Overview ...........~ ............................................................335
6.27.2 Friction-Pendulum Isolator with Flat Sliding
Surface ..............................................................................335
6.27.3 Friction-Pendulum Isolator with Curved
Surface •.............•.•...•.•..-................................................... 33.7
6.27.4 Rubber-Type Isolator ................................................... 338
6.27.5 Alternative Model for Rubber-Type Isolator ........ 339
6.27.6 Which Model is Correct? ........................•...•................ 340
6.28 Some Other Types of Buckling ....................•.•.•......•......•...........342
6.29 True Large Displacements.....•...........•.............•.............•.............342
6.30 Conclusion for this Chapter .,......................................................343
Chapter7
Some Other Aspects of Behavior............................................ 345
7.1
Plastic Mechanisms........................................................................ 345
Collapse Mechanism vs. Plastic Mechanism ........ 345
Push-Over Analysis for Earthquake Loads .•..•.......347
7.1.3
Desirable and Undesirable Mechanisms ......
348
7.1.1
7.1.2
m ••••••
Contents
7.1.4
7.1.5
xix
Lateral Strength Calculation Given a Mechanism
................_ ...........................................................................349 ..
Other Causes of a Mechanism ..................................351
7.2
Mechanism Control Using Capacity Design ..........................351
7.2.1
Concept ............................................................................351
7.2.2
Examples of Mechanism Control .............................351
7.2.3
Higher Mode Effects in Tall Buildings .....................352
7.3
Static Indeterminacy and Redundancy...................................353
7.3.1
Statically Determinate Structure..............................353
7.3.2
Statically Indeterminate Structure ..........................354
7.3.3
Redundancy ..............................................,..................... 357
7.4
Nonstructural Components ........................................................357
7.5
Work and Energy ............................................................................358
7.5.1
Energy Balance in a Real Structure ..........................358
7.5.2
Energy Balance in an Elastic Analysis Model........358
7.5.3
Energy Balance in an Inelastic Analysis Model....360 .
7.5.4
Energy Balance Check During Analysis .......:..........361
7.5.5
Amount and Distribution of Dissipated
Energy .........................;..........................................•..........362
7.5.6
Dissipated Energy as a Demand-Capacity
Measure ............................................................................363
7.6
Living With Uncertainty ...............................................................363
CHAPTER
1
Introduction
Structural analysis is rarely, if ever, an end in itself. Rather, it is almost
always just a tool for use in structural design.
There are three distinct phases in structural analysis, namely ''modeling",
"computation'.' and "interpretation". This chapter argues that for most
engineers the modeling and interpretation phases are by far the most
important. The computation phase involves structural analysis theory as
well as computational methods. This phase is relatively unimportant.
This chapter also notes that there are many different reasons for doing
structural analysis and many different types of analysis, all with different
needs for modeling and interpretation.
1.1 Overview
- This chapter considers the following topics.
(1)
(2)
(3)
(4)
(5)
{6)
(7)
(8)
The three phases of structural analysis (modeling, computation and
interpretation), and their relative importance.
Demands, capacities and demand/ capacity ratios.
Performance assessment vs. direct design.
Capacity design.
Ela8tic vs. inelastic analysis.
Static load vs: dynamic load.
Small displacements vs. large displacements.
Demand analysis vs. capacity analysis.
2
Chapter 1 Introduction
These topics are considered only briefly in this chapter. They form a foundation for later chapters.
1.2 The Phases of Strudural Analysis
1.2.1
Modeling
It goes without saying that structural analysis is carried out not on actual
structures but on models of actual structures. The challenge in the modeling
phase is to set up a useful analysis model. Some key points are as follows.
(1)
An analysis model must capture the important aspects of behavior of
the real structure. A useful model does this with sufficient accuracy,
economy and detail for practical purposes. A model does not have to
be "exact", and never will be.
(2)
A model will almost alwaysbe a "node-element" model, consisting of
rigid nodes connected by deformable elements. See Chapter 2 for a
detailed description.
(3)
One major task is choosing appropriate elements, and assigning them
properties such as stiffness and strength. This is a challenging task,
requiring skill and judgment.
(4)
A second major task is choosing appropriate demand-capacity measures
for assessing performance. This also requires skill and judgment.
(5)
Just because an analysis model looks like the actual structure does not
mean that it has the same behavior. Elaborate graphical renderings of
structures are nice, and they can be useful. However, an analysis
model that looks like the real structure does not necessarily behave
like it.
1.2.2
Interpretation
The end of the modeling phase is a detailed analysis model. The end of the
computation phase is a set of "results" for the model, consisting mainly of
displacements at the nodes and forces on the elements (but also including
other things). The challenge in the interpretation phase is to use these results
to make design decisions for the actual structure. Some· key points are as
follows.
The Phases of Structural Analysis
3
(1)
It is important to consider the purpose of the analysis, and to interpret
the results in a way that supports that purpose. Analysis is not an end
in itself, and there is no one-size-fits-all procedure for interpreting
analysis results.
(2)
In most structural analysis textbooks, the results of an analysis are
likely to be a deflected shape and a beri.ding moment diagram. These
can be interesting, but they may riot be very useful for making design
decisions. For making design decisions it is necessary to think in
. terms of demands, capacities and demand/ capacity ratios.
(3)
There can often be errors in an analysis model, and possibly also in
the numerical computations. It is important to check that the results
look reasonable.
(4)
Always keep in mind that the results are for the analysis model, not
for the actual structure. An analysis model will never be an exact
representation of the actual structure, and it does not need to ~· It
must, however, be close enough for practical purposes.
·
.. 1.2.3
Computation ·
Computation, as defined in this book,· includes everything that is not
included in modeling and interpretation. Given an analysis model, the
computation phase involves everything .that is needed to get the analysis
results. This can be a· complex process, involving finite element theory,
complex logic, and extremely large numerical computations. Some key
points are as follows.
(1)
The numerical computations (given the analysis model, get the
analysis results) will almost always be done by computer. Since the
·costs of computer program development. are large, and the
development requires a great deal of specialized knowledge, a
commercial computer program will almost always be the most
economical choice. Engineers rely heavily on the skill and expertise of
computer program developers.
(2)
The many engineers who use computer programs need a basic understanding of the computation phase. This book is aimed mainly at
those engineers. One goal of the book is to provide this basic
understanding.
4
Chapter 1 Introduction
(3)
The few engineers who write computer programs need a deep
understanding of the computation phase. The development of
computer programs has become a specialized task. This book does not
consider computational details.
1.3 Relative Importance of the Three Phases
For an engineer who uses structural analysis as a practical tool, there are
two main challenges. The first challenge is to set up an analysis model that
gives useful results. The second is ·to use those results to make design
decisions. The computational details of how the results are obtained are of
secondary importance - they can be, and should be, taken care of by the
computer program (that is, by the engineers who wrote the program). For
most engineers, the modeling and interpretation phases are of primary
importance, and the computation phase is of much less importance.
For an engineer who writes computer programs (most likely a team, not a
single individual), the computation phase is of primary importance.
However, the modeling and interpretation phases are also important (in
addition to interface design, realistic graphic rendering, and many other
things). As this book shows, modeling and interpretation are complex tasks.
If a program developer does not provide the user with a practical and
efficient tool for modeling and interpretation, the program is uruikely to be
successful.
When a computer program is used as a practical tool, productivity can be
boosted if the program has automated features for modeling and
interpretation. Engineers may even look forward to the day when the
computer program creates the analysis model automatically, quickly
performs the numerical computations, and presents the results in exactly the
way that they are needed. This can already be done for relatively simple
structures, and it is a goal that program developers strive towards. The day
may come when automated modeling can be done for structures and
analyses of all types.
Demand and Capacity
5
1.4 Demand and Capacity
1.4.1
Performance Assessment
Structural analysis is primarily a tool for assessing the performance of a
structure. The design of a structure will usually proceed through a number
of phases, from preliminary to final design. In the preliminary phases the
overall proportions and dimensions of the structure are progressively
refined, and in the ·final phases the member sizes and other details are
chosen. Structural analysis is likely to be used rather informally for
preliminary design, using simplified models, and more formally in final
design, using more detailed models. In each phase, structural analysis is a
tool for assessing the performance of the structure, considering serviceability under commonly occurring loads and safety under more extreme
loads.
· Performance is almost always assessed by comparing Demands and
Capacities. Some simple examples are as follows.
(1)
If the deflection at a point is a concern for serviceability, one demand
value might be the calculated deflection at the point. The corresponding capacity is the allowable deflection.
(2)
If the bending strength of a beam is a safety concern, one demand
value might be the calculated bending moment at a cross section in
the beam. The corresponding capacity is the bending strength at that
·
section.
(3)
If yielding of a structure is allowed, a safety concern for a beam might
be whether the amount of yield in bending exceeds the ductile
capacity (the ability to yield without excessive loss of strength
through fracture, buckling, or other causes). The beam might be
modeled using plastic hinges. One demand value might be the
calculated rotation at a particular hinge. The corresponding capacity
is the allowable rotation at that hinge.
(4)
If the cost of repairing a building after an earthquake is an economic
concern, the demand might .be the calculated repair cost. The
corresponding capacity is the acceptable cost. This type of demandcapacity comparison requires complex analyses, on both the demand
and capacity sides, and at the time of writing it is a long way from
becoming standard practice.
6
Chapter 1 Introduction
When structural analysis is used for performance assessment, a simplified
view of the process is as follows.
(1)
The structure geometry and the member sizes (including reinforcement details, etc.) are known in sufficient detail for the structure to be
modeled and analyzed.
(2)
Demand-capacity measures of various types are identified (displacements, bending moments, plastic hinge rotations, etc.).
(3)
The structure is modeled and analyzed, and the demand values are
obtained from the analysis results.
(4)
Corresponding capacity values are chosen or calculated, using
judgment, building code formulas, etc.
(5)
Demand/capacity (D/C) ratios are calculated.
(6)
If no DIC ratio exceeds 1, the performance is satisfactory.
If the performance is not satisfactory, the structure must be modified,
and/ or the analysis model must be refined to provide more accurate
demand values, and/ or the capacity values must be refined.
1.4.2
Direct Design
Structural analysis also may be used for direct design. A simplified view of
the process is as follows.
(1)
The overall structural geometry is known, but not all the member
sizes. The goal is to determine the member sizes.
(2)
An analysis model is set up, using estimated member sizes.
(3)
The-structure is analyzed, and strength demands on the members are
calculated.
(4)
Member sizes are calculated to satisfy these demands. This is the
"direct design" aspect.
(5)
If the member sizes change significantly from the estimated sizes,
iteration may be needed.
Elastic vs. Inelastic Analysis
7
It is, of course, never this simple. However, there are important differences
between analyses that are used for direct design and those that are used for
performance assessment. It is important to be clear on what is expected of
the analysis and exactly how the analysis results will be used.
1.5 Elastic vs. Inelastic Analysis
1.5.1
Behavior of a Structural Component
Structural analysis makes use of relationships between forces or loads and
corresponding deformations or displacements.
In this book, the term "force-deformation relationship" is used for a structural
·component in an analysis model. The "force" can be the axial force in a bar,
the bending moment at a plastic hinge, shear force, normal stress, shear
stress, etc., depending on the component type. The corresponding "deformation" can be axial extension, plastic hinge rotation, etc.
·
The term "load-displacement relationship" is used for a complete structure
or a complete analysis model. The load-displacement relationship for a
complete structure depends on the force-deformation relationships of its
components.
The force-deformation relationship for a typical structural component has
the form shown in Figure 1.1. The load-displacement relationship for a
complete structure is likely to have a similar form.
Force (F)
Strain
Hardening
Ultimate
strength
Ductile limit
Complete failure
Hysteresis loop
Figure 1.1 Force-Deformation Relationship
for a Typical Structllral Component
8
Chapter 1 Introduction
The key parts of the relationship are as follows.
(1) Initial behavior that is essentially linear.
(2) First yield, at a point that may or may not be well defined.
(3) A region of increasing strength (strain hardening).
(4) Ultimate strength.
(5) Ductile limit, at a point that may or may not be well defined.
(6) Progressive strength loss.
·
(7) Residual strength, where the strength stabilizes.
(8) Possible complete failure.
(9) Cyclic unloading-reloading, with a hysteresis loop.
(10) Cyclic degradation, where the stiffness, strength and/ or ductility progressively deteriorate.
1.5.2
Elastic vs. Inelastic Behavior
In Figure 1.1, the component has linear (or m~arly linear) behavior up to first
yield. The behavior in this region is essentially elastic, whi~h means that
when a force is applied to the component it stores strain energy, and when
the force is removed this energy is recovered. If the force on the component
exceeds the yield force, the behavior becomes nonlinear and inelastic. For
inelastic behavior only a part of the energy is recovered. when the force is
removed.
"Elastic" is not necessarily the same as "linear". A component is elastic if all
of the work done on the component as it deforms is stored as recoverable
strain energy. Elastic components are usually linear, but can be nonlinear.
An example is a spring that has a gap. The stiffness of the component is zero
when the gap is open, and increases when the gap closes. Since the stiffness
changes this is nonlinear behavior, but the spring is elastic.
In an inelastic component, some of the work done on the component as it
deforms is dissipated, as plastic work, friction, facture energy, etc. An
inelastic component will always be nonlinear.
1.5.3
Strength-Based Design Using Elastic Analysis
The traditional approach to structural design is to make the structure strong
enough to resist the external loads with essentially elastic behavior. It is also
important to satisfy serviceability requirements, which usually means
providing enough stiffness to control deflections and vibrations,
Elastic vs. Inelastic Analysis
9
For strength-based design the structural analysis can be elastic, and its main
purpose is to calculate force demands on the structural components.
Corresponding force capacities are obtained, in most cases, from formulas in
design codes. The force capacity from a code formula is usually somewhat
lower than the actual ultimate strength.
If the force demand on a single component is close to its force capacity,
there could be significant inelastic deformation of the component. If a
substantial proportion of the components in a structure are close to their
force capacities, there could be significant inelastic deformation of the
structure as a whole. Hence, the behavior of a . structure could be
significantly inelastic under the design loads, and elastic ·analysis is not
necessarily accurate. This is especially true if the design of a structure is
optimized to reduce its weight, so that many of the components are fully
stressed. However, decades of experience have shown that elastic analysis is
accurate enough for most design purposes.
To account for uncertainty in the loading, the expected loads are multiplied
by load factors that increase the force demands. For example, the gravity
load for demand calculation might be 1.2 times the calculated dead load
plus 1.6 times the expected live load. To account for uncertainty in component strength, the estimated strength capacities are multiplied by capacity
reduction factors (or resistance factors), typically between about 0.75 and
0.9. Components that are especially important to the integrity of a structure
may be assigned smaller capacity reduction factors. In some cases the
calculated demand on a component may be multiplied by a demand
increase factor. The details can be found in design codes and are not
important for this book. The important points are that elastic analysis can be
used, and the main purpose of the analysis is to calculate force demands.
1.5.4
Strength-Based Design Using Inelastic Analysis
Strength-based design using elastic analysis considers strength at the component level. The strength of the structure as a whole is not explicitly
calculated. An alternative might be to consider strength at the structure
level, using the external load as the demand and the structure strength as
the capacity. For this alternative, the demand is known and structural
analysis is used to calculate the capacity. This is a more direct way of
ensuring sufficient strength, and it has the potential to produce more
economical designs. However, it is impractical for most structures, for the
following reasons.
1O
Chapter 1 Introduction
(1)
The calculated strength capacity of a structure can depend greatly on
the modeling assumptions. One possible approach is to model (or
attempt to. model) every mode of behavior that might contribute
significantly to collapse of the structure. This includes relatively simple
modes of behavior such as inelastic bending in beams (plastic hinge
formation), more complex ones such as column buckling, very
complex ones such as inelastic shear in reinforced concrete walls, and
extremely complex ones such as brittle fracture in welds. This approach
is not merely impractical, it is impossible. For a complex structure
even the most sophisticated analysis model can not hope to account
for all significant effects.
(2)
A more practical approach is to deliberately limit the modes of behavior
that contribute to collapse. For example, inelastic shear in reinforced ·
concrete walls can be prevented, by requiring that the shear strength
be sufficiently large that inelastic shear behavior can never occur.
. Hence, inelastic shear does not have to be considered,. and the analysis
model is simpler and more reliable. Since inelastic shear is likely to be
brittle, the performance of the structure under extreme loads is also
likely to be improved. This is an example of "capacity design".
(3)
Even when capacity design is used, the direct calculation of structure
strength requires inelastic analysis, which is much more complex and
expensive than elastic analysis. one complication is that only the
component stiffnesses are needed for an elastic analysis, whereas stiffnesses, strengths, strain hardening behavior and other properties are
needed for an inelastic analysis. An elastic model can be created based
on rough estimates of the member sizes, and can easily be updated as
the sizes are determined with greater accuracy (many computer
programs will do this automatically). This is not so easy for an
inelastic model.
(4)
There are many effects that must be accounted for in design besides
external loads. These include thermal expansion, creep (especially in
reinforced concrete) and foundation movement. Some of these affect
only serviceability, not strength. However, this has to be checked. It is
much easier to account for such effects in elastic analysis.
For most structures, elastic analysis likely is to be the standard approach for
the foreseeable future. One exception is the design of structures to resist
large earthquakes, as considered in the next section.
Elastic vs. Inelastic Analysis 11
1.5.S
Deformation-Based Design for Earthquake Loads
For earthquake resistant design there is a high probability that a small
earthquake will occur during the life of the structure, and a low probability
of a large earthquake. For a small earthquake, a structure will usually be
designed to remain essentially elastic. However, for a large earthquake it is
usually argued that it is uneconomical to design the structure to remain
elastic, and it is common practice to allow substantial,inelastic behavior.
Hence, for a large earthquake the elastic strength demand on a structure is
likely to exceed its strength capacity. This is illustrated in Figure 1.2.
LOAD
For a large earthquake, if the
1'+- structure is elastic the load
1
can exceed the yield strength;
/--I
~
But if the structure is allowed
to yield, the displacement
may be acceptable.
,•
For a small earthquake,
the load on the structure
is likely to be below yield.
DISPLACEMENT
Figure 1.2 Behavior for Earthquake Load
If the earthquake load were a static load, acting for a sustained period of
time, the structure woUld collapse. However, earthquake loads fluctuate
rapidly. A load that exceeds the structure strength may be applied several
times during an earthquake, but since the load is dynamic and acts for only
a short time, it does not necessarily cause collapse. As shown in Figure 1.2,
the maximum displacement of the structure may be acceptable, and
although some structural components become inelastic, the structure can
perform satisfactorily. For those components that become inelastic the
concern for design i.s deformation, not strength. For satisfactory performance, the deformation demand on an inelastic component must usually
be smaller than its ductile limit (as. defined in Figure 1.1).
The most rational . approach for a large earthquake is to use inelastic
analysis. This has the following advantages.
(1)
For components that become inelastic the main concern for design is
deformation (or ductility), not strength. Inelastic analysis can calculate
deformation demands directly. Elastic analysis can not.
12
Chapter 1 Introduction
(2)
As a structure becomes inelastic the forces in,. the structure can be
redistributed, and the distribution of forces in the structure can be very
different from that calculated by elastic analysis. This can have
substantial. effects on the behavior of the structure. Inelastic analysis
accounts for force redistnbution. Elastic analysis does not.
(3)
When capacity design is used, force demands must be calculated for
those components that are required to remain elastic. For example, if
inelastic shear is not allowed in a reinforced concrete wall, the shear
force demand must be calculated, and the wall must be designed with
a shear force capacity that exceeds the demand. In an inelastic analysis,
the calctilated force demands for a component can depend on the
strength of its surrounding components, or on the strength of the
component itself. For example, the shear force demand in a wall can
depend on its bending strength. This can be accounted for directly in
· an inelastic analysis, which means that capacity design can be applied
more rationally. Capacity design can still be used with elastic analysis,
as considered later, but it is a less direct process.
For a simple structure, inelastic static analysis ("static push-over" analysis)
may be sufficient. For a large or compl!:!x :;l.ructure, h1el.astic dynamic unalysw
may be necessary.
1.5.6
Strength-Based Design for Earthquake Loads
-
It is not essential to use inelastic analysis for earthquake resistant design.
For design purposes it is possible to use elastic analysis, considering inelas-
tic behavior implicitly rather than explicitly.
If a structure were designed to remam elastic in a large earthquake, loads
that correspond to this earthquake wotild be applied, and the structural
components wotild be designed for the calculated force demands using
elastic analysis. Call these the '.'elastic force demands". When inelastic
behavior is· allowed, the usual procedure is to design the components for
force demands that are ·substantially smaller than the elastic demands.
These demands are obtained by applying a "response modification .
coefficient" or "R factor" to the elastic force demands, where this coefficient
depends on the ductility of the ~omponent and its importance in the
structure.
For example, if a component is very ductile it might be designed for a force
.demand of, say, l/8th of the elastic force demand (an R factor of 8). This
means that the component is relatively weak and is likely to have substantial
Elastic vs. Inelastic Analysis 13
inelastic deformation. Conversely, if a component is very brittle it might be
designed for the full elastic force demand (an R factor of 1). Such a component
is relatively strong and is likely to have little or no inelastic deformation.
This method tends, therefore, to produce structures th.it have desirable
behavior, saving money by allowing ductile components to become inelastic,
and preserving safety by keeping brittle components elastic. In addition, it is
a practical method because it is basically strength-based design using elastic
analysis. However, elastic analysis does not account for force redistribution
as a structure becomes inelastic, and hence the use of this method does not
necessarily ensure that a structure will perform well. In some cases elastic
analysis can even give misleading information. The method can be more
reliable when it is combined with capacity design, as considered in the next
sectioi:i. The method should not be used blindly.
This emphasizes that it is not necessary for an analysis to be "exact" - it
needs only to give results that are accurate enough for design purposes.
With good engineering, structures that are designed using elastic analysis
can perform well in large earthquakes.
1.5.7 Capacity Design Using Elastic Analysis
In the basic form of strength-based design, elastic analysis is used to
calculate force demands on the structural components, and the components
are designed to have strength capacities that equal or exceed the demands.
This process must be modified when capacity design is used.
As an example, consider a reinforced concrete frame structure where inelastic bending is allowed in the beams, but where the shear behavior must
be essentially elastic. This may be done because reinforced concrete beruns
can be ductile in bending but tend to be brittle in shear.
As consi9.ered in the preceding section, when elastic analysis is used for
earthquake loads it is usual to allow implicitly for inelastic behavior, using
R factors, One-way to account for the relative ductility is to use a smaller R
factor for shear than for bending. However, there is a better and more direct
method, using capacity design.
To illustrate this, consider the frame in Figure l.3(a). The figure shows
gravity loads plus a static lateral load to represent an earthquake. For these
loads, Figure 1.3(b) shows the bending moment demands. These are the
elastic demands, With the moments for earthquake loads divided by an R
factor. For a reinforced concrete beam this factor can be as high as 8.
14
Chapter 1 Introduction
Moment diagram
/
for shear demands'"'--.,/
/,//
(a) Gravity and lateral loads
Actual strength
Moment demands
from analysis
(b) Moment diagram for beam
Figure 1.3 Capacity Design for Shear
As shown in the figure, the actual strength of the beam in bending is likely
to be significantly larger than the bending moment demand, for two main
reasons, as follows.
(1)
The reinforcement area is likely to be larger than the area needed to
provide a c~pacity that exactly equals the demand.
(2)
The actual material· strengths are likely to be larger than those
assumed for design. For designing the beam, it is usual to apply a
capacity reduction factor to the nominal, or expected, bending strength,
to account for uncertainty. For estimating the actual bending strength
capadty, the expected material strengths (or more) should be used,
with no capacity reduction factor. ·
Using capacity design, the shear. force demands are based on the dashed
moment diagram in Figure 1.3(b). Since the shear forces depend on the
bending moments, and since the bending moments are the maximum values
that can be reached regardless of the earthquake strength, if the beam is
designed for these shear forces the behavior in shear is always essentially
elastic.
An alternative is to use different R factors for bending and shear, and to
design the beam for the calculated demands. This may or may not achieve
the goal of keeping the shear behavior essentially elastic. Capacity design is
a more rational approach.
1.6
Stat~c vs.
Dynamic Analysis
The external loads on a structure can be static or dynamic. Most loads are
actually dynamic, but for analysis purposes a load can be assumed to be
Small vs. Large Displacements Analysis 15
static if it is applied slowly (relative to the period of vibration of the
structure).
For the static case, the external loads on the structure are resisted entirely by
the static forces in the structural components. For the dynamic case, the
external loads can also be resisted by inertia forces associated with the mass
of the structure. If the inertia forces and the forces in the structural
components act in the same direction, the component forces are smaller than
they would be for the same loads applied statically. If the inertia and
component forces act in opposite directions, the component forces are larger
than they would be for the same loads applied statically. In general, the
inertia and component forces act in varying directions in different parts of
the structure and at different times in the analysis, and the interactions
among the forces are complex.
Inertia forces are acceleration dependent and vary linearly with acceleration
(assuming that the mass does not change). There may also be "viscous"
forces that are velocity dependent. Experience shows that when a structure
is loaded dynamically, there is a loss of energy (energy disi;ipation), even if
the structure is elastic. It is common to assume that energy dissipation is
caused by a viscous damping mechanism, which implies the presence of
velocity-dependent forces. These forces may be linearly or nonlinearly
dependent on velocity, depending on the assumed viscous damping
mechanism.
A further complication is that the force-deformation relationship for a
structural component may depend on the deformation rate of the component,
which in tum depends on velocity. For example, a component may be
substantially stronger when loaded dynamically, with a high deformation
rate, than when loaded statically.
1.7 Small vs. Large Displacements Analysis
1.7.1
Overview
If a structure has inelastic components, its behavior will be nonlinear, as
considered earlier. This is usually referred to as material nonlinearity. There.
can also be nonlinear behavior when a structure undergoes large
displacements, even if it remains elastic. This is usually referred to as
geometric nonlinearity.
16
Chapter 1 Introduction
There are two causes of geometric nonlinearity, the first based on
equilibrium and the second on compatibility (continuity). This section gives
brief explanations. Chapter 6 goes into more detail.
1.7.2 Equilibrium
When a structure is loaded it changes shape. Strictly speaking, equilibrium
between externalloads and internal forces must be satisfied in the deformed
position of the structure. However, if the displacements are small, it can be a
reasonable approximation to consider equilibrium in the initial, undeformed
position. Since this position is fixed, the equilibrium relationships are linear.
For example, doubling the external loads exactly doubles the internal forces
(assuming no material nonlinearity).
If the displacements are not small, equilibrium must be considered in the
deformed position. In this case the equilibrium relationships are not linear
(doubling the external loads does not exactly double the internal forces).
This is illustrated in Figure 1.4
h
Force in spring
=H
(a) Undeformed Position
Force in spring
= H + PA/h
(b) Deformed Position
Figure 1.4 Equilibrium in Undeformed and Deformed Positions
Figure l.4(a) shows the undeformed position. The bending moment at the
pinned base must be zero, so by simple equilibrium the force in the spring is
equal to the horizontal load.
Figure l.4(b) shows the deformed position, assuming that the spring
compresses and the top of the bar moves horizontally by an amount ll..
Again, the bending moment at the base is zero, so to satisfy equihDriUm. the
force in the spring must be larger than the applied load, Also, the spring
Small vs. Large Displacements Analysis 17
force is not proportional to ~e load. For example, if P and H are doubled,
the force in the spring more than doubles.
1.7.3 Compatibility (Continuity)
There is a geometrical relationship between the displacements of a structure
and the deformations of its components. Figure 1.5 shows such a relation·
ship.
Bar
extension.
I
i
(a) Imposed Displacement
(b) Compatibility Relationship
Figure 15 Nonlinear Compatibility Relationship
In Figure l.S(a), the top of the bar moves horizontally. Hence, the bar must
extend to maintain continuity. Figure 1.S(b) shows the relationship between
displacement and bar extension. The bar extension is the deformed length
minus the undeformed length, h.
For a very small horizontal displacement the bar extension is close to zero
(in the limit, for a vanishingly small displacement, the bar extension is
exactly zero). For larger displacements the bar extends, with a nonlinear
relationship between displacement and extension.
1.7.4 Analysis Types
For analysis, the effects of large displacements on the equilibrium and
compaboility relationships can be treated separately. Consequently, there
are three diff~rent types of analysis that can be carried out, as follows.
(1)
Small displacements analysis. Equilibrium is .considered in the
undeformed position, and for compatibility the displacements are
assumed to be vanishingly small.
18
Chapter ·1 Introduction
(2)
True large displacements analysis. Equilibrium is considered in the
deformed position, and for compatibility the displacements are assumed
to be finite.
(3)
P-A analysis. Equilibrium is considered in the deformed position, and
for compatibility the displacements are assumed to be vanishingly
small.
There is a fourth type (deformed position for equilibrium, small displacements for compatibility), but this is never used.
Figure 1.6 illustrates the difference for a simple structure.
Moves
horizontally
I
{ Force in spring
=H
I
'
{ Force in bar
I
=P
(a) Small Displacements
(b) P-Ll
(c) Large Displacements
Figure 1.6 Different Analysis Types
For this structure, assume that the bar is stiff axially, so that it has negligible
axial deformation. Figure l.6(a) shows the loads and forces for small
displacements analysis, Figure 1.6(b) for P-A analysis, and Figure l.6(c)for
large displacements analysis. Note that in Figure 1.6(c} the spring is
assumed to remain horizontal.
The differences among the three methods depend on the relative values of
the loads P and H, and on the displacement A. Consider two example cases
as follows.
·
(1)
V =0, and Mh =0.1 (i.e., 10% drift ratio, which is a very large drift for
most structures). For all three methods the force in the spring is H and
the force in the bar is zero. The only difference is that the vertical
Small vs. Large Displacements Analysis 19
displacement is negligible for small displacements and P-A analysis,
and equal to a small value (0.005h) for large displacements analysis.
(2)
P/H =5, Mh =0.10. For the small displacements case the forces in the
spring and bar are respectively Hand P. For the P-A case the forces
are I.SH and 0.995P. For the large displacements case the forces are
1.503H and 0.995P. The vertical displacements are essentially the same
as for P=O.
These examples show that small displacements analysis can be in error
when there are substantial gravity loads and large drifts, but only in the
force in the spring (in the second example above there is an error of 50% in
the spring force). For all three analyses the axial force in the bar is very close
to P (because cos9 is very close to 1.0). When P-A and large displacement
analyses are compared there is very little difference in the spring forces. The
only significant difference is that the calculated vertical deflection is zero for
P-A analysis and a small value for large displacements analysis.
These examples roughly represent a single-story building structure (the ·
spring models the horizontal stiffness). They indicate that it can be important
to consider P-Li effects, but that it is not necessary to consider true large
displacements. This is important because P-A analysis can be much more
efficient computationally than large displacements analysis. For building
structures under gravity plus lateral loads, it is often important to consider
P-A effects, but it is rarely necessary to consider true large displacements.
1.7.5
Catenary Effect
The type of behavior described in the preceding section does not apply to all
structures. Figure 1.7 shows behavior of a different type.
p
Large displacements analysis
predicts increasing stiffness
Bars, or flexible cable
ip
....................,
......
A
......""()".::: .........
a
Small displacements and P-a
analyses predict P O
=
Figure 1.7 Catenary Effect
20
Chapter 1 Introduction
struc~e deflects it gets progressively stiffer. This is
usually referred to as the "catenary" effect. Only a large displacements
analysis accounts for this effect. The reason is that the small displacements
and P-~ analyses assume a linear compatibility relationship between the
structure displacement and the bar extension. Jn this example, the linear
compatibility relationship gives zero bar extension, even for large values of
fl. Consequently the bar force is zero, and hence P =0 in both the undeformed
and deformed positions. The large displacements analysis uses a nonlinear
compatibility relationship, as in Figure 1.5, and he.nee accounts for the
catenary effect.
Jn this example, as the
An important difference between material and geometric nonlinearity is
that geometrical nonlinearity has two well-defined causes.(equilibrium and
compatibility), both of which are governed by clear mathematical rules,
whereas material nonlinearity can have many causes and many fopns. Our
knowledge of mat~al nonlinearity depends almost entirely on what we
observe in experiments on actual structures and structural components.
1.8 Demand Analysis vs. Capacity Analysis
1.8.1
Overview
Jn most cases, structural analysis 1s used to calculate demand values for
performance as8essment or direct design. It is also possible to use structural
analysis to calculate capacity values. This is a very different process.
There are four broad types of analysis, as follows.
(1)
Strength demand. This is by far the most common type of analysis.
Analysis is used to calculate bending moments, shear forces, etc. for
structural components. Corresponding capacities are obtained from
design codes or other sources.
(2)
Deformation demand. Most inelastic analyses are of this type. Analysis
is used to calculate plastic hinge rotations, shear strains, etc.
Corresponding capacities are obtained from design codes or from
guidelines such as ASCE 41.
(3)
Strength Capacity. Analysis, rather than experiment, is used to estimate ·
the strength of a structure or structural component.
Demand Analysis vs. Capacity Analysis 21
(4)
Deformation capacity. Analysis, rather than experiment, is used to
estimate the force-deformation relationship for a structural component
(or possibly the load-displacement relationship for a structure). From
this relationship, an estimate can be made of the maximum deformation that can be imposed on the component (i.e., its deformation
capacity).
This section gives some examples of capacity analysis.
1.8.2
Lateral Load at First Yield
This is an example of a strength capacity analysis. It is an artificial example,
but it illustrates the process.
Suppose that a frame structure is required to remain elastic (no plastic hinge
formation) under combined gravity and static lateral load (representing,
say, wind load). Suppose that (a) the gravity load is known, in both
distribution and magnitude, and (b) the distribution of the lateral load is
known but not its magnitude. It is required to calculate the lateral load
magnitude at which the frame ceases to be elastic (i.e., at which the first
hinge forms). This lateral load magnitude is the elastic capacity of the frame.
The analysis is easy to carry out. The steps are as follows,
(1)
(2)
(3)
(4)
(5)
Set up an elastic model of the structure.
Identify all locations at which plastic hinges may form. Calculate the
bending moment capacities at these locations. ·
·
Analyze for the gravity load.
Add the lateral load, and progressively increase the load magnitude
until the bending moment demand at one of the potential hinge
locations exceeds the bending moment capacity. (Alternatively, since
the behavior is linear, run separate analyses for gravity load and unit
lateral load, then calculate the minimum lateral load magnitude
needed to obtain a D/C ratio equal to 1.0.)
This load magnitude is the required lateral load capacity.
Note that it is not necessary to set up an inelastic model with plastic hinges.
22
1.8.3
Chapter 1 Introduction
Earthquake Intensity at Collapse
This is a more realistic example of a strength capacity analysis. Analyses of
this type have been used to estimate collapse probabilities for actual
structures.
Consider a frame structure with known gravity load and a specified
earthquake ground motion. For the ground motion, the variation of ground
acceleration with time is known, but the intensity can vary (i.e., the ground
accelerations can be scaled up or down). Calculate the earthquake intensity
required to collapse the structure. This intensity is the collapse capacity of
the frame.
The process for calculating the collapse capacity is usually termed
"incremental dynamic analysis". The steps are as follows.
(1)
(2)
(3)
(4)
(5)
Set up an inelastic model of the structure.
Analyze for the gravity load.
Choose an intensity for the earthquake, and run an inelastic dynamic
analysis, keeping the gravity load constant. Determine whether the
structure collapses. See below for a comment on this.
Repeat the analysis for a number of earthquake intensities (usually
progressively increasing). H all goes well, at some intensity the
structure will not collapse, and at a slightly larger intensity it will
collapse.
This determines the collapse capacity, within a tolerance that depends
on the number of earthquake intensities that are considered.
The overall goal is similar to that in the preceding example, namely to
calculate a capacity. However, the details are dramatically different. It is
relatively easy to set up an elastic analysis model and to check strength DIC
ratios. It is much more difficult to set up a model that accounts for the many
types of inelastic behavior that can influence collapse and to carry out a
meaningfql dynamic analysis of such a model. Analyses of this type are
likely to be very approximate, and the analysis results can depend greatly
on the assumptions in the analysis model.
1.8.4
Bending Strength of a Beam
The preceding two examples considered complete structures. This example
considers a small part of a· complete structure. This is also a strength
capacity analysis.
Demand Analysis vs. Capacity Analysis 23
Suppose that a reinforced concrete beam has a complex cross section, so that
comm.only used formulas for calculating the bending moment capacity do
not apply. Structural analysis might be used to calculate this capacity, as
follows.
(1)
(2)
(3)
(4)
(5)
(6)
Set up an analysis model that consists of a short length of the beam,
loaded so that the beam has constant bending moment (i.e., pure
bending).
Divide the beam cross section into a number of longitudinal "fibers"
or "filaments", some representing concrete and some steel.
Model the concrete fibers using an inelastic material model that
accounts for cracking and crushing.
Model the steel fibers using an inelastic material that accounts for
yield and strain hardening.
Assume that plane sections remain plane, as is usual for beam theory.
Do an inelastic analysis, progressively increasing the bending
moment until the maximum strength is reached.
This is a relatively simple inelastic analysis. Even so, the result must be used
with caution. The reason is that the behavior in bending of an actual
reinforced concrete beam may be much more complex than is assumed in
the analysis. In particular, the analysis assumes pure bending, whereas
bending is almost always accompanied by shear. In a reinforced concrete
beam, shear force can have a substantial effect on the behavior in bending.
1.8.5
Plastic Hinge Rotation Capacity
This is an example of a deformation capacity analysis.
The inelastic behavior of beams in bending can often be modeled using plastic
hinges. This is shown in Figure 1.8.
Beam element
(a) Frame with beam element
Figure 1.8 Plastic Hinge
24
Chapter 1 Introduction
The beam is modeled as an elastic beam with plastic hinges at each end. A
plastic hinge is initially rigid, and begins to rotate at first yield. The properties of a plastic hinge include its strength in bending and its rotation
capacity, which for this example is the rotation at the ductile limit.
When an inelastic beam is modeled using plastic hinges, it is assumed that
all inelastic deformations are concentrated in the hinges, and that the rest of
the beam remains elastic. Since the inelastic behavior in an actual beam is
likely to be distributed over a significant length of the beam, a zero~length
plastic hinge is an approximation.
Figure 1.9 shows how the properties of a plastic hinge can be determined..
Yielding occurs
in a plastic zone
Most of beam stays
essentially elastic
Yielding
occurs
ofbeam=9b
(a) Actual beam with plastic zones
L
/
Hinge rotation, ah, is
measured from this line
a. 1 . - - -
Elastic·
rotation, 00
r.~
End moment
'\
on beam, Mb \.." - - ----- :.-' Mb
""'-. Total beam
Plastic hinge
rotation, 9b
rotation, ah
(c) Beam behavior
(b) Beam model with plastic hinges
Figure 1.9 Elastic and Hinge Rotations
Figure l.9(a) shows a beam with equal and opposite end moments. Figure
l.9(b) shows a model of the beam, with plastic hinges. H the relationship
between end moment and total end rotation is known, the plastic hinge
properties can be determined as shown in Figure l.9(c).
Plastic hinge properties are usually determined experimentally. However,
for this example suppose that the beam has a thin-walled steel section for
which experimental results are not available. The plastic hinge properties
Demand Analysis vs. Capacity Analysis 25
are to be estimated by analysis. The main properties to be calculated are the
bending strength and the hinge rotation capacity, which is the rotation at
the ductile limit. These can be calculated by performing an inelastic analysis
of a cantilever beam, as shown in Figure 1.10.
U2
U2
(a) Beam with end moments
~Load =M,t
Mb c-d-LJOeflection = 0b
~
(c) Finite element model
(b) Equivalent cantilever
Figure 1.10 Estimation of Hinge Properties by Analysis
Figure 1.lO(a) shows the beam. Figure l.lO(b) shows an equivalent cantilever.
Figure 1.lO(c) shows the type of finite element mesh that might be used to
analyze the cantilever.
The elements in the hinge region must account for yield of the steel, in order
to calculate the bending strength. More importantly, the ductile limit for the
beam is likely to be governed by local buckling of the steel section, so the
elements must also account for geometric nonlinearity. The buckling
behavior may be sensitive to geometrical imperfections in the beam, such as
variations in the wall thickness and local distortions caused by welding the
end connection. The stress-strain relationship for the steel may be uncertain
and may vary from point to point in the beam.
This is a capacity analysis, where analysis is used as a substitute for
experiment. Even though the structure is small, it may be difficult to set up
an accurate analysis model, especially if cyclic deformation is to be.considered.
26
1.8.6
Chapter 1 Introduction
Conclusion for this Section
Most structural analyses are demand analyses, where the goal is to estimate
strength or deformation demands. In a capacity analysis the goal is to
estimate a strength or deformation capacity.
The goal for capacity analysis is essentially "exact" simulation. This can be
difficult when there is complex inelastic behavior involving such things as
fracture and local buckling in steel or crushing, cracking, inelastic shear and
bond slip in reinforced concrete.
Capacity analyses must be used cautiously. As a general rule, capacities that
are detei:mined by experiment are likely to be more reliable.
1.9 Conclusion for this Chapter
There are different types of structural analysis, with different goals. It is
important to be clear on the purpose of the analysis, the required results,
and how the results are used. Most analyses are Demand Analyses, where
the purpose is to calculate strength and/or. deformation demands. The
corresponding capacities are obtained from other sources, such as design
codes. Some analyses may be Capacity Analyses, where the purpose is to
calculate strength and/ or deformation capaci:ties. The corresponding demands
are obtained from other sources. Capacity analysis is mor~ difficult than
demand analysis.
An analysis can never be "exact". It is literally impossible to use -analysis to
do an exact simulation of a real structure, no matter how much computing
power is available or how complex the model. Structural analysis is at best
approximate, and must always be used with a heavy dose of engineering
judgment.
1.10 Topics for the Following Chapters
This book is concerned mainly with analysis models of "finite element" or
"node-element" type, consisting of discrete nodes and elements. It is important
to understand what is meant by an "analysis model". Chapter 2 describes
the features and limitations of node-element models.
The computational part of the analysis is almost always done using the
Direct Stiffness Method. For engineers who are doing practical analysis, it is
Topics for the Following Chapters 27
not necessary to understand the details of this method, but it is important to
have an overall understanding of the process. This is covered in Chapter 3.
The most important step in setting up a node-element model is choosing the
elements and Cl$Signing them properties. In general, an element can be made
up of a number of components, with stiffness, strength and other properties.
In order to understand the features and limitations of an element, it is
important to understand the behavior of its underlying components, and ·
the assumptions that are made to capture this behavior in an analysis
model. Chapter 4 considers component behavior, with emphasis on material
nonlinearity. This chapter is limited to "uniaxial" components, where the
component has a single force (such as the axial force in a bar) and a single
corresponding deformation (such as the bar extension). Chapter 5 also
considers material nonlinearity, but for components with multi-axial forcedeformation relatiC>nships with interaction. An example is P-M-M strength
interaction in a column.
Chapters 4 and 5 consider mainly material nonlinearity. Geometric
nonlinearity is a separate issue that can be important at both the structure
level and the element level. Chapter 6 considers several aspects of geometric
nonlinearity, including the P-.1\ and P-0 contril>utions, instability, and the
buckling of columns, beams and complete structures.
Chapter. 7 considers some other aspects of structural behavior, including
plastic mechanisms, redundancy, non-structural components and energy
balance.·
CHAPTER
2
What is an Analysis Model?
When w~ analyze a structure, we analyze a model of the structure (an
"analysis model'), not the structure itself. This may seem obvious but it is
easy to forget.
For the analysis of a complete structure, a "node-element" or "finite element"
model is almost always used. This type of model consists of a finite number
of nodes and elements (often a very large number). This chapter reviews
this type of model.
For. setting up· the properties of the elements in a node-element model,
"continuum" models are often used. This chapter also provides a brief
review of this type of model.
2.1 Actual Structure vs. Analysis Model
It is important to keep in mind that an analysis model is an approximation
of the actual structure, and possibly a rather crude approximation. An analysis
model can be expected to capture only the important aspects of behavior,
omitting many details. It is unrealistic to expect an analysis to be exact.
2.2 Two Types of Analysis Model
There are two distinctly different types of model, namely (1) a "finite
element" model, consisting of a finite number of elements of finite size, and
(2) a "continuum" model, consisting of an infinite number of infinitesimally
29
30
Chapter 2 What is an Analysis Model?
small elements. In a finite element model the elements connect to each other
at a finite number of nodes. This is the main type ·of model considered in
this book, and it is referred to as a "node-element" model. In a 'continuum
model the elements connect directly to each other, without nodes.
Figure 2.1 shows a node-element model, consisting of elements of several
types. This is, of course, an artificial example. Practical models will be threedimensional and much larger
·
Nodal loads
+
Nodal
loads
Wall
Figure 2.1 Node-Element Model
Figure 2.2 shows some simple continuum models.
w
~
i l I I
+----!
I~
+
w
8
YLx
t.cry
m
V ffiv+dV
Met[]
-+lctxl+(a) Beam
Thickness = t
sid'.--Sla,
)M+dM
(b)Wall
Figure 2.2 Continuum Models
m
Features of Node-Element Model
31
In Figure 2.2(a) each element is an infinitesimally short length of beam,
subjected to bending moments and shear forces, and possibly to external
loads. In Figure 2.2(b) each element is an infinitesimally small volume of
material, subjected to normal and shear stresses, and possibly to external
loads (e.g., the self weight of the element).
The following sections describe node-element and continuum models in
more detail.
2.3
Features of Node-Element Model
The main features of a node-element model are as follows.
(1)
Nodes are rigid and are points in space. In principle; a node could
have finite dimensions, but this is usually not the case.
(2)
Elements are, and must be, deformable. Elements can be one, two or
three-dimensional, or can have zero dimensions. An element can have
zero length yet be deformable, which is impossible physically but not
mathematically.
(3)
Nodes can displace (as rigid bodies). The displacements can be
translational and/ or rotational. The displacements at most nodes are
initially unknown. The displacements at some nodes (usually support
points) can have known values, usually, but not necessarily zero.
(4)
Elements deform in a variety of ways. All deformations of the
structure originate in the elements. Elements can also displace as rigid
bodies. In· general, an element will both deform and displace. for
example, a truss bar element can extend, translate and rotate.
(5)
Forces of three types can act on a node, namely (1) external loads
(nodal loads), (2) external restraining forces (support reactions, to
impose zero or known displacements), and (3) internal forces exerted
on the node by the elementi/that connect to that node.. The term
"forces" includes moments (rotational forces).
(6)
Forces of two types can act on an element, namely (1) external loads
applied to the element (element loads, which generally can be applied
anywhere in the element),. and (2) internal forces from the nodes to
which the element connects, applied on the element "ends" (the points
where the element connects to the nodes). For a bar or beam element,
the element ends are the actual ends of the element. For a 2D surface
32
Chapter 2 What is an Analysis Model?
element or a 30 solid element the ends are the points that connect to
the adjacent nodes. There is no direct element-to-element contact or
force transfer.
(7)
Each node and each element must be in equilibrium as a free body. In
addition, the forces exerted by a node on an element must be exactly
equal and opposite to the forces exerted by the element on the node.
These are the equilibrium requirements.
(8)
There must be no gaps or overlaps at the element ends. This means
that the deformations and displacements of every element must be
such that the element exactly fits between the nodes to which it
connects. This is the compatibility (or continuity) requirement. This
requirement does not prevent gaps or overlaps along the boundaries
between 20 or 30 elements (see later for more discussion of this).
(9) . The force-displacement relationship must be known for every
element. The force-displacement relationship for an element is essentially the relationship between the forces exerted on the element ends
and the corresponding element end displacements. See later for more
details, including whether the relationship is in stiffness form or
flexibility form.
(10) If any element has element loads, a method must be available to
calculate the element end forces when the element ends are fixed (i.e.,
when the nodes to which the element connects have zero displacements). These are the element "fixed-end forces", usually expressed as
the forces acting on the element. The forces on the nodes are equal
and opposite to those on the element.
The properties of the model, including node locations, elements, forcedisplacement relationships, loads, etc., must all be specified by the engineer
doing the analysis. In some cases there may be accepted guidelines for
doing this, but often this is not the case. In some cases the properties may be
generated by a computer program, which means, in effect, that they are
specified by a computer programmer.
Given an analysis model, an analysis method is used to calculate the node
displacements, element deformations, element forces, support reactions, etc.
The analysis method must ensure that the equilibrium and compatibility
requirements are satisfied (not necessarily exactly, but with sufficient
accuracy for practical purposes). In practice, the method that is almost
Some Element Types
33
always used is the Direct Stiffness MethOd. The important features of this
method are described later.
The term "finite element model" usually (but not necessarily) refers to a
model· with 2D or 30 elements. One-dimensional line elements for trusses
and frames are usually referred to simply as "elements". The difference is
minor, and the terms "node-element model'' and "finite element model" are
largely interchangeable. The term "node-element'' model is used in this
book.
·
2.4
Some Element Types .
The number of possible element types is unlimited. Some common types are
as follows.
·
(1)
Simple bar element, which has axial stiffness but no bending stiffness.
This is a line element with two nodes.
(2)
Frame element, with axial, bending and torsional stiffness, for
modeling beams and columns. This is a line element with two nodes.
(3)
2D membrane element, in a state of plane stress. This is a surface
element, where only in-plane effects are considered. A triangular
element has at least three nodes and a quadrilateral. element at least
four nodes, at the comers. Some elements have additional nodes
along the edges, and possibly internally.
(4)
3D solid element. A tetrahedron element has at least four nodes, artd a
''brick" element has at least eight.
(5)
Shell element, with in-plane membrane stiffness· and out-of-plane
bending. stiffness. Usually this is like a membrane element with
bending stiffness added, but it might be a special case of a 3D solid
element. A triangular element has. at least three nodes, and a quadrilateral element has at least four nodes.
·
(6)
Support spring element, with up to three translational and three
rotational stiffnesses. This is usually a single-node element
(7)
Zero-length connection element, with two nodes that have identical
coordinates.
34
Chapter 2 What is an Analysis Model?
2.5 Connection betw~en Nodes and Elements
2.5.1
Overview
In a node-element model, the elements connect to each other only at the
nodes, and the element ends have the same displacements as the nodes. The
nature of the node-to-element connection warrants some elaboration.
2.5.2
Connection for Bar Elements
A simple bar element resists only IDciaI force, With no bending moments or
shear forces. One way to do this is to say that the bar element has only aXial
stiffness and zero bending stiffness. A better way, however, is to say that a
· bar element is "pin connected" or ''hinge connected" to the node, so that the
bending moment is zero at the element ends. This is shown in Figure 2.3.
·Node
Element ·
.Node
Node
Element
Node
~·~
···-·"''{a) Hinge is part of node
(b) Hinge is part of element
•I\.
Figure 2.3 Hinge Connection to Node
In Figure 2.3(a) the hinge is a part of the node. This is not a good idea
because it makes the node more complicated. The correct way is to make the
hinge a part of the element, as shown in Figure 2.3(b). This means thiit the
node is still a simple rigid body.
When hinges are used a bar element can have a non~zero bending stiffness,
so that it does not buckle in compression. ·The element ends are where the
element connects to the node, which is outside the hinge.. The end
displacements of the bar· are equal to the nOde displacements, which can
~ude both translations and rotations.
It
is usually an approximation to ~ that a bar has Zero moments at its
ends. In a real structure most ·~ar" members will have end connections that
can transmit significant bending moments.
Figure 23 shows a bar .element in 2D. In 30 there are also torsionM
moments. These can be made zero by adding a tors.ional ltjnge. In this case
Connection between Nodes and Elements
35
there must be only one torsional hinge. If there are two (one at each end),
the length of bar between the hinges can spin freely about the longitudinal
axis, which means that the bar is unstable. An alternative is to omit the
torsional hinge and assume that the torsional stiffness of the bar is zero.
2.5.3
Rigid End Zones
This book assumes that nodes are points in space, with zero dimensions.
This is not an essential assumption (see later for a mention of the "Applied
Element" method). However, there are major complications when nodes
have finite dimensions, and it is simpler to treat them as points.
This can have consequences for modeling member-to-member connections
in a structure. For example, Figure 2.4(a) shows the connection between a
beam and a column in a frame.
Often assumed
to be rigid
(i) Connection region
(b) Stiff zones are included
. in node
(c) Stiff zones are included
in elements
Figure 2.4 Connection Between Beam and Column
It is common practice to assume that bending deformations in the connection region are much smaller than in the adjacent beams and columns. Often
it is assumed that the connection region is rigid. With this assumption, it
might be tempting to assign finite dimensions to the node ·at the beam-tocolumn connection, as shown in Figure 2.4(b).
As already noted, this is not a good idea because it adds complexity if a
node can have finite dimensions. It is easier, and more rational, to incorporate the rigid connection assumption into the beam and column elements, by
adding "rigid end zones" as shown in Figure 2.4(c).
An advantage of adding rigid end zones to an element is that it is only a
minor change to allow some flexibility in the end zones, by modeling them
36 . Chapter 2 What is an Analysis Model?
not as rigid segments but as segments that are very stiff yet still deformable.
This is more realistic, as there will usually be significant deformations in the
connection region.
2.6 Gaps and Overlaps between Elements
2.6.1
Surface Elements
When surface elements are used for structural components such as walls
and slabs, the elements often have quadrilateral shapes, for example as shown
in Figure 2.S(a).
.
(a) Solid elements
(e.g. for a wall)
(b) Element theory requires
edges to remain straight
(c) Element theory allows
edges to become curved
Figure 2.5 Gaps and Overlaps. Between Elements
The elements are connected together only at the nodes. There is no elementto-element connection along the edges between nodes. In some cases the
theory used to form the element properties ensures that there is continuity
along the element edges. For example, if the element theory requires that the
element edges remain straight, when adjacent elements deform there is
continuity between adjacent elements at all points, as shown in Figure
2.S(b).
However, if the element theory allows the element edges to curve, there can
be gaps or overlaps along the edges of adjacent elements, and continuity is
generally not satisfied. This is shown in Figure 2.S(c). This is not necessarily
bad. Sometimes, greater accuracy can be obtained by relaxing the continuity
requirements.
·
Figure 2.6(a) shows rectangular elements with more nodes. When elements
of this type deform, the edges can curve and still satisfy continuity, as
shown in Figure 2.6(b).
Equilibrium between Elements
(a) Solid elements
37
(b) Element edges curve
and do not overlap
Figure 2.6 Higher Order Elements Allow Edges to Curve While Satisfying Continuity
2.6.2
Element-to-Element Contact
For elements of all types, including line elements for bars and beams, there
is nothing in a basic node-elemen~ model that forbids elements from
overlapping. The model does not prevent elements, or nodes, from occupying the same points in space.
H overlap is to be prevented, it is usually necessary to define one or more
lines or surfaces that define boundaries between parts of the structure that
. should not overlap. Usually the surfaces are defined using nodes, and the
nodes on one surface are monitored to see if they penetrate another surface.
H this happens, steps must be taken to prevent the penetration, accounting
for normal (bearing) and tangential (friction) forces at the interface. This can
be a complex computational task.
·
2.7 Equilibrium between Elements
.i
2.7.1
Equilibrium at Element Boundaries
In a node-element model, equilibrium is satisfied if each node is in
equilibrium as a free body, if each element is in equilibrium as a free body,
and if the forces. exerted on the elements by the nodes are equal and
opposite to the forees exerted on the nodes by the elements.
Figure 2.7(a) shows a steel plate, loaded in-plane. Figure 2.7(b) shows a
node-element model with surface elements. The thickness of the plate
changes at mid-height. Assume that the thickness changes suddenly,
although in practice the plates would probably be welded together. The
l
!
38
Chapter 2 What is an Analysis Model?
bending and shear stressesjust below mid-height are only one half of the
stresses just above mid~height.
_.
____..
_.
Thickness = t
=
Thickness 2t
(a) Steel plate
(b)Model
Figure 2.7. Equilibrium at Element BouQdaries
.In the actual plate the stresses change over a length roughly equal to the
thicklless of the plate (following St. Venant's principle). Exactly at midheight,·equilibpum requires that the stresses be the same in the two parts of
·
theplate.
In th~ node-el~~ent model, each element. has constant thickness and the
· plate thickness changes suddenly at mid-height. There will be bending and
shear stresses at the element interface. flowever, the stresses at the top of
lhe lower element will be different from those at the bottom of the upper
element. Hence, although the solution satisfies the equilibrium requirements
for a node-element model, stress equilibrium is not satisfied at the element
interface.
In a model with surface or solid elements, stress equilibrium will rarely be
satisfied at the boundaries between elements. The above example is
extreme, but even if the plate thickness is constant the stresses at the
elemE?J1t botindaries are generally not the same in adjacent elements.
This does not mean that analyses using surface or solid elements are
inaccurate - when used properly they can be of ample accuracy for practical
purposes. In the above example, a sudden change in stress is not greatly
different from a change that occurs over a distance roughly equal to the
plate thickness.
Discrete Model with Finite Size Nodes and Zero Length Elements
2.7 .2
39
Equilibrium at Connedions
Figure 2.8(a) shows the connection between a beam and column in a steel .
frame.
,>
:.
v
(a) Connection
(b) Model
(c) Forces
Figure 2.8 Forces at a Connection
Figure 2.8(b) shows a node and the beam and column elements. For the case
with no external loads on the node, Figure 2.8(c) shows the forces acting on
the node. Since the node is in equilibrium, the axial force in the beam
becomes the shear force in the column, and the shear force in the beam
becomes the axial force in the column.
In the node-element model, the node transfers the forces from the beam to
the column. The model does not consider how forces are transferred across
the node. In a node-element model the nodes are rigid points that can
transmit forces of any type and any magnitude. In an actual frame, the
forces are transferred· from the beam to the column through a physical
connection, and it may not be obvious how, for example, the shear stresses
in a beam becomes an ID<lal force in a column. For a physical connection it
requires understanding of connection behavior to determine how the forces
are actually transferred.
.
2.8
Discrete Model with Finite Size Nodes and Zero Length
Elements
In this book, a node-element model has nodes that are points and elements
that have finite dimensions. The opposite of this is a model where the
elements have zero dimensions (but are still deformable) and the nodes
have finite dimensions. This is the basis of the "Applied Element" model. In
this model, a frame might be modeled with rigid segments (the nodes)
40
Chapter 2 What is an Analysis Model?
connected by zero length deformable "hinges" (the elements), as shown in
Figure2.9.
This type of model has some advantages for structures that undergo such
large displacements that parts of the structure make contact with each other.
As noted earlier, to consider contact between two surfaces in a typical nodeelement model, the surfaces must be defined, and the displacements of the
nodes on a surface must be monitored to see if they penetrate another
surface. The problem is different in an applied element model. In this case,
the motions of the nodes are monitored, and contact occurs when one node
overlaps another. The nodes are rigid bodies, and there are well- established
methods for solving the interaction problem.
-
Zero length element
.Finite size node
Figure 2.9 Applied Element Model of a Frame
The disadvantage of the applied element method is that all deformations are
concentrated in zero-length elements that connect the nodes. It is difficult to
capture the behavior ofa structure using only zero length elements. In particular, a frame model, such as that in Figure 2.9i requires many more nodes
and elements than are needed for a typical node-element model.
. 2.9 Continuum Model
2.9.1
Frame Structure
A discrete model has a finite number of elements that connect to each other
at a finite number of nodes. A continuum model has an infinite number of
infinitesimally small elements that connect directly to each other, without
nodes.
This frame could be analyzed as a single continuum, but this is unlikely. It
is f:ir more likely that the frame will be analyzed using a node-element
Continuum Model
41
model, probably using three elements as shown in Figure 2.lO(b). The
properties of each element must be determined, and this is likely to be done
using a continuum model. Among other things( the fixed-end forces must be
determined for the beam element when the external load is applied and the
nodes are fixed. This might be done using a continuum model, as shown in
Figure 2.lO(c). An infinitesimal element in this model is shown in Figure
2.lO(d).
.
w per unit length
w
t l l I t
{a) Frame
{b) Node-element model
w
w
t l l l t
V ffiv+dV
-----tm
-+11+-dx
M( i-+!dxl+II !)M+dM
{c) Continuum model of beam
{d) Infinitesimal element
~---+•
~
Figure 2.1 OSimple Frame
The analysis results for a continuum model must satisfy the following
conditions.
(1)
Geometric compatibility (material continuity) must be satisfied, which
means that the element deformations must be consistent with the
beam deflected shape, with no gaps or overlaps between elements.
Also, in Figure 2.lO(c) the displacements at the ends of the beam must
be zero.
(2)
For each infinitesimal element there must be a "constitutive relationship" (essentially a force-deformation relationship), relating the forces
on the element to its deformations. For a beam, the relationship is a
moment-curvature relationship, which must be satisfied at every
point along the beam. If shear deformations are significant, the
relationship between shear force and shear deformation is also
needed.
42
Chapter 2 What is an Analysis Model?
(3)
Every infinitesimal element must be in equilibrium as a free body,
acted on·by forces from the adjacent elements plus any external loads.
The forces that adjacent elements exert on each other must be equal
and opposite.
There are a number of possible analysis methods for continuum models of
beams, including (a) setting up and solving a differential equation and (b)
the Moment-Area method. The details can be found in any book on elementary mechanics.
It may be noted that if the shear deformations in a beam are assumed to be.
zero, it is reasonable to assume that plane sections remain plane. Hence,
adjacent infinitesimal elements can fit together exactly. However, if there
are significant shear strains, plane sections do not remain plane, and
adjacent elements may not fit together exactly. Hence, the compatibility
conditions may not be satisfied.exactly. Any errors of this type are usually
small and are ignored.
2.9.2
Wall Structure
As a second example, Figure 2.ll{a) shows a simple wall.
(a) Wall
(b) Node-element model
lflJ
.
.
y
x
Wall thickness = t
(c) Finite element
(d) Infinitesimal element
Figure 2.11 Solid Finite Elements Are Continuum Models
Elements and Components
43
Such a wall could possibly be analyzed using a continuum model, but it is
far more likely that a node-eJement model will be used, with finite elements
of surface or solid type, as shown in Figure 2.ll(b). The properties for any
finite element are, however, obtained using a continuum model, as
indicated in Figures 2.ll(c) and (d).
For a finite element, the compatibility, force-deformation (i.e., stress-strain)
and equilibrium relationships must be satisfied within the element. The
details of how this is done can be found in any book on finite element theory.
2.10 Elements and Components
Section 2.4 listed several possible elements. Some elements can consist of a
single structural component. For example, a bar element might consist of a
single bar component, with aX:ial stiffness but no bending stiffness. How~ ·
ever, a bar element might also consist of three components, nam~y a· bar
component with both axial and bending stiffness, plus a hinge component at
each end, as shown earlier in Figure 2.3(b). Similarly, a beam element might
consist of a rigid end zone component at each end, with a deformable beam
component in between. A complex beam element, with end zones, semi.,
rigid end connections, a non-uniform cross section, and inelastic behavior,
inight consist of several components of a variety of types.
In a beam element with end zones and a uniform beam in between, the
uniform beam might be regarded as a continuum model with an infinite
number of infi?itesimally short components.
In all such cases the properties of the element are determined by the
properties of its components. There are many theories and mathematical
procedures that can be used to combine the component properties to give
the element properties. The details are not important for this book. The
important point is that element properties originate at the component level.
CHAPTER
3
The DirectStiffness Method
For analysis of node-element models by computer, the method of analysis
that is almost always used is the Direct Stiffness Method. To make effective
use of computer programs for analysis it is not necessary to understand the
theory and computation in depth - it is much more important to understand
the modeling procedures and assumptions. However, an engineer should
have a basic understanding of the Direct Stiffness Method. This chapter
considers the essential aspects of the method, using physical reasoning with
a minimum of mathematics.
Most structures will be large and three-dimensional. This chapter considers
small examples, mainly tWo-dimensional, to illustrate the concepts and
procedures.
3.1
3.1.1
Element Stiffness and Flexibility
Overview
The Direct Stiffness Method uses stiffness matrices, for elements and for
complete structures. The terms in these matrices are not just numbers - they
have important physical meanings. Each term in a stiffness matrix is a
stiffness coefficient that defines how much force is required to cause a unit
displacement.
This section uses single elements to illustrate the physical significance of
. stiffness coefficients. Later sections consider complete structures.
45
46
Chapter 3 The Direct Stiffness Method
A flexibility relationship is the inverse of a stiffness relationship. Each term
in a flexibility matrix is a flexibility coefficient that defines how much
displacement is caused when a unit force is applied. This section also
considers flexibility coefficients.
3.1.2
Bar Element
Figure 3.1 shows a simple 2D bar element, consisting of a single. bar
component.
Modulus= E
Area = A
\
Force = P
Extension = p
~-°,-"""'"""--~~
I·
L
E,A
Q1, q1
\
Q2, q2
~~-:----""--&~
I·
~1
Stiffness : P = (EA/L) p
_Flexibility: p = (UEA) P
[ Q1]
Q2
L
~1
=[
ENL -EA/L].[q1]
-EA/L EA/L q2
Flexibility does not exist
(a) Force vs. Deformation
+~~
Impose
Keep
qt-;1
q2i°
(b) End Force vs. End Displacement
~
=+i&·'----~
. . . +'End I
EndJ
(c) Stiffness coefficients
Ap~y
Displacements
q1 and q2 are infinite
014
Q2=0
~
~
End I
EndJ
(d) Flexibility coefficients
Figure 3.1 Stiffness and Flexibility for a Bar Element
In Figure 3.l(a) the element is supported to prevent rigid-body displacement (both translation and rotation). The element has only one mode of
deformation, namely axial extension. For a uniform elastic bar, the relationships between axial force and axial extension are as shown. The stiffness is
EA/L. Th.e flexibility is the inverse of the stiffness, or L/EA. These can be
termed the "deformation stiffness" and "deformation flexibility".
In Figure 3.l(b) the supports are changed to allow horizontal rigid body
tra.nslation. The element now has one possible rigid body displacement
(horizontal translation) and one deformation (extension). These can be
captured using the horizontal displacements at the element ends, q1 and qaThe bar extension is q2 - q1, and the rigid body displacement is 0.5 (q1 + q2).
The element stiffness is now a 2-by-2 matrix, as shown in the figure.
Element Stiffness and Flexibility
47
In this case there is no 2·by-2 flexibility matrix. The mathematical explanation is that the stiffness matrix has no inverse (it is a "singular" matrix - its
inverse has terms that are infinite). The physical explanation is as follows.
Figure 3.l(c) shows the bar with a unit displacement imposed on the
element at end I, with zero displacement at end J. The forces required to
impose these displacements are +EA/L at end I and an equal and opposite
force -EA/L at end J. This follows from physical reasoning. It also follows
mathematically from the stiffness matrix. Since qt = 1 and %= 0, the required
forces, by matrix multiplication, are Qt= EA/Land Q2 =-EA/L.
The terms in the stiffness matrix are "stiffness coefficients", which define the
element end forces when small ~t displacements are imposed. $pecifically,
if the stiffness matrix is k, the term·k..,. in this matrix is the force Q,. when a
unit displacement q. is imposed, with all other displacements zero. The first
subscript, m, in km,. is the number of the force (the row in the stiffness
matrix) and the second subscript, n, is the number of the displacement (the
column in the stiffness matrix).
The meaning of a "small unit displacement" may need explaining. If, for
example, a bar element has a unit length, it does· not mean that the imposed
"unit" displacement is equal to the length of the bar. That would be a very
large displacement, and stiffness coefficients are defined for very small
displacements. More correctly, a vanishingly small displacement, dqt, should
be imposed in Figure 3.l(c), and the end forces should be plus and minus
(EA/L).dqr Such a displacement does not change the geometry of the
element, so large displacements issues do not arise. The key point is that the
imposed displacement must be small. A "unit" displacement is used to
avoid including factors like "dqt" in the equations.
·
The flexibility counterpart is to apply a single unit fore~, with all other end
forces equal to zero, for example, Qt = 1 with Q2 = 0, as shown in Figure·
3.l(d). The corresponding displacements define the terms in the flexibility
.matrix (the flexibility coefficients). Since the bar is Unsupported, the furce
causes an infinite displacement of the bar. That is, the fleXIoility coefficients
are infinite.
In a 20 structure, a bar element can generally have any orientation, as
shown in Figure 3.2(a). In this case there are two displacements at each end
of the element, and the element stiffness matrix is 4-by-4. This allows fur
three rigid body displacements and one deformation, as shown in Figure·
3.2(b).
48
Chapter 3 The Direct Stiffness Method
,.
Q4,q4
t
y
·V-:"
~·"
~;:V
(1) X translation
/
x
/
(3)Rotation
(a) End forces and displacements
(2) Y translation
/V
\f
(4) Axial deformation
(b) Displacement and deformation modes
Figure 3.2 General 2D Bar Element
In a 3D structure there are three displacements at each end and the stiffness
matrix is 6-by-6. This allows for three rigid body translations, two rigid
body rotations and one deformation. Torsional rigid body rotation is not
included, and is assumed to be zero. If end rotations of the element are
considered as well as translations, there are six displacements at each end in
3D and the element stiffness matrix is 12-by-12. In this case a bar element is
a special case of a beam element, with only axial stiffness and no bending or
torsional stiffness.
l~by-1or2-by-2 stiffness to a 4-by4, 6-by-6 or 12-by-12 stiffness. The details are not important.
If is a simple process to transform from a
3.1.3
Rigid Body Displacement$ and Deformation Modes
A bar element is one of the simplest element types, but it shares important
features with elements that are far more complex. In particular, a bar
element has several rigid body displacement modes, all with zero stiffness,
and one deformation mode that has a finite stiffness. Almost all elements
have several rigid body modes, to allow free rigid body displacement. Most
elements have more than one deformation mode. Elements represent
members in the real structure, and there must be enough deformation
modes to capture the essential modes of behavior of the real members.
In the case of a bar element, the only important mode of deformation is (or
is assumed to be) axial extension or shortening, and the only important
force is axial tension or compression. Bending and torsional deformations,
and the corresponding bending moments, shear forces and torsional
moments, are assumed to be unimportant. The axial stiffness of a uniform
elastic bar is EA/ L. The element stiffness matrix depends only on this value.
Element Stiffness and Flexibility
49
Most other elements are more ·complex, and the element stiffness matrix
depends on properties that are more complex than EA/L. However, the
principles are similar. An element will have a finite number of deformation
modes, and these modes must be chosen to capture the behavior of real
members in sufficient detail, and with sufficient accuracy, to produce useful
results when the structure is analyzed.
The procedures for calculating element stiffness matrices can be complex,
both the0retically and numerically. However, even the most complex elements
have a relatively small number of deformation modes, which can capture
only certain types of behavior. It can be useful to consider these
deformation modes, and the corresponding types of behavior, if only in
a qualitative way. The following sections illustrate this for elements of a
variety of types.
3.1.4
Beam Element
Figure 3.3 shows a simple 2D beam element, consisting of a single beam
component.
.
Modulus=E
Section properties =A, I
. \.
P2,p2
(~-·-· ·-·-·-·-·£+.
P1, p1
1:
;r
L
P3, p3
Stiffness:
Flexibility :
l
l
[ :~ =[~~ ~:~: ~ [:~i
0
0
UEAP3
k2,1 =2El/l
~~=O
Unit rotation
k1.2 = 2El/l
[:]=[; =~ .L][=l
p3
k1,1 =4El/l
J<i.2 =4El/L
~k:2=0
Unit rotation
k1,a=O
k23 =0
(;,...------.,........,·~
~
--~
=
ka,3 EA/L
Unit extension
(a) Deformation Stiffness and
· Fexibility Matrices
(b) Deformation modes
Figure 3.3 Stiffness and Flexibility for a 2D Beam Element
Figure 3.3(a) shows a horizontal element with a uniform cross section. The
axis of the element passes. through the centroid of the beam cross section.··
50
Chapter 3 The Direct Stiffness Method
This is important because it means that when the beam is subjected to an
axial force it does not bend, and when it is bent it does not extend (or
shorten). Hence, axial and bending effects are independent, or "uncoupled".
When the element is supported to prevent rigid body displacement, as in
Figure 3.3(a), it has three deformation modes, and hence a 3-by-3 deformation stiffness matrix. The deformation modes are shown in Figure 3.3(b). A
"unit" rotation is a small rotation, not one radian, which would be extremely
large.
If shear deformations are assumed to be zero, the stiffness coefficients and
the deformation stiffness matrix are as shown. The 3-by-3 stiffness matrix
can be inverted ·to give a deformation flexibility matrix, as shown.
If a beam element beam does not have a uniform cross section, or if it consists
of several components, the stiffness matrix is different from that for a simple
uniform beam, but the stiffness coefficients that are zero in Figure 3.3
remain zero provided the centroids of all cross sections lie on the element
axis. Strictly speaking, if the centroid of any cross section does not lie on the
element axis, there is coupling between axial and bending effects. However,
such coupling is often ignored.
Jn a 2D structure, a beam element can generally have any orientation, as shown
iri Figure 3.4(a).
(a) End forces and displacements
(b) Displacement and deformation modes
Figure 3.4 General 2D Beam Element
Jn this case there are three displacements at each end of the element, and the
element stiffness matrix is 6-by-6. This allows for three rigid body displacements and three deformations, as shown in Figure 3.4(b).
The stiffness matrix is singular, and there is no correspondirig flexibility.
Element Stiffness and Flexibility
51
A 30 beam element has three translations and three rotations at each end.
The stiffness matrix is 12-by-12. This accounts for six ~d body displacements
and six deformations (two bending deformations in each principal plane,
one extension, and one torsional deformation).
Numerical values of the stiffness coefficients for the 6-by-6 and 12-by-12
stiffness matrices can be calculated quite easily. For this book the details are
not important.
The stiffness matrix for a beam element considers only forces applied at the
element ends. If i:here are loads applied along the element length, these
must be considered separately, as considered later.
3.1.5 Surface and Solid Elements
Figure 3.S(a) shows a rectangular surface element, in a state of plane stress
(a 20 membrane element).
... t.:qB
~t
.t_
Q Q6~.t/
s.q5 .·.Tl~~
•·. •. .-.;iy
.-. :· .·- .· .· .·:.· ·'. ·.•-.·.·_·-.· _.
· ..9m~ 8"• - !=•.j ·.j..dx
,, : •:,·,·:':··-·
<>y
-Ei--:•--. ·'··
y
01.cit(
Q2,q2
x··
a1,q1
tC:.q3
.
."'
dy
y
m
-+t
x
t
Q4,q4
(a) 4-Node Rectangle
Possible
Edges can be
9th node
straightor curved
(b) 4-Nooe Quadrilateral
(c) 8 or 9 Nodes
Figure 35 20 Plane Stress Elements
It is possible to support such an element so that rigid body displacements
are prevented, and hence to form a deformation stiffness matrix and a
corresponding flexibility matrix. However, this is rarely, if ever, done.
Typically, the stiffness matrix is formed directly in terms of the displacements of the nodes. For the case in Figure 3.S(a) this is an 8-by-8 matrix, and
there is no corresponding flextoility.
Figure 3.S(b) shows a similar element, but with an arbitrary quadrilateral
shape. Figure 3.S(c) shows a "higher Order" element with eight or nine nodes.
For a 4-node rectangular element the eight displacements anow for three
rigid body modes and five deformation modes, as shown in Figure 3.6.
52
Chapter 3 The Direct Stiffness Method
-tl
-+t
I Qdr--.·
_JI
," -
o••·······-,
··.
I
.:. '
..
(b) Rigid body displacements
t-+
(a) Element
--
I
ro•.·.•.·.·····.
I •
·.
I
I .... ·. ·.
o·.·•.···.·.
.·
. . .-
I
--
•.
·I . .o
•. . .·.....-,
._
..!
(c) Extension and shear
D.
'I
I
(d) Bending assumption 1..
D
I
......._
I
___ .,
\
\
/ . ·.·•·.
I
f\ro. ·. ·. ·. ·. . . .··.r\
...· ··.
\~
(e) Bending assumption 2
Figure 3.6 Rigid Body and Deformation Modes
Figures 3.6(d) and (e) show two alternative assumptions for the ''bending''.
deformations. In Figure 3.6(d) the element edges are constrained to remain
straight. This means that there will never be any gaps or overlaps between
adjacent elements, and hence that geometric compatibility is satisfied. In
Figure 3.6(e) the element edges are allowed to curve, which means that
there can be gaps or overlaps between adjacent elements, and compatibility
is generally not satisfied.
For both bending assumptions the extensional strains vary linearly in one
direction (through the "depth") and are constant in the other direction
(along the "length"). In the case where the edges are allowed to curve
(Figure 3.6(e)) these are the only strains. However, in the case where the
edges are constrained to remain straight (Figure 3.6(d)) there are also shear
strains. These strains are constant through the depth and vary linearly along
the length. It follows that there is more strain energy in the element with
straight edges than in the element with curved edges. A consequence of this
is that the stiffness is smaller for the curved edge case than for the straight
edge case. This is not necessarily a bad thing, since· arialysis .models for
surface and solid finite elements tend to overestimate the structure stiffness.
Element Stiffness and Flexibility
53
As noted earlier, stress equilibrium is generally not satisfied across element
boundaries - the stresses in the elements on either side of a boundary are ·
generally different.
Equilibrium may also not be satisfied inside the element. For the extensional
and shear modes, internal equilibrium requires zero external loads (for
example, no self weight load). The same is true for the bending modes with
curved edges. For the bending modes with straight edges the shear stresses
vary over the element in such a way that external loads are required for
equilibrium. These loads generally do not exist, so internal equilibrium is
generally not satisfied.
For both equilibrium and compatibility, the errors get smaller as the element
mesh is refined (i.e., as the elements get smaller). In the limit, when the
finite elements get infinitesimally small, the errors for a properly formulated
element should be zero.
·
The deformation modes for a rectangular element are easy to visualize, as in
Figure 3.6. It is even possible to use those modes to calculate the stiffness
matrix for a rectangular element. ff the deformation stiffness and flexibility
were to be calculated, they would be 5-by-5 matrices. As an exercise you
might like to determine the amount of coupling (i.e., which off-diagonal
stiffness coefficients are nonzero}, using physical reasoning. Note, for
example, that if Poisson's ratio i.s nonzero, the two extensional modes are
coupled (if a unit extension is imposed in the X direction, with all other
deformations-zero, there must be forces in the Y direction as well as the X
direction, because of Poisson's effect). However, the shear deformation
mode is uncoupled from the extensional modes. For the bending modes,
consider both the case with straight element edges and the case where the
edges are curved.
The deformation modes can not be visualized so easily for an element that is
not rectangular. An element that has an arbitrary quadrilateral shape, as in
Figure 3.S(b), has three rigid body modes and five deformation modes. The
deformation modes are not as simple as those in Figure 3.6, but the overall
effect is the same. The method used to calculate the stiffness matrix for an
arbitrary quadrilateral element is well defined and can be found in any
textbook on finite element theory. The method develops the 8-by-8 displacement stiffness directly. It is possible to calculate stiffnesses corresponding to
an element with edges that remain straight, or with elements that are
allowed to curve. A rectangular element is usually regarded as just a special
case of an element with an arbitrary shape.
54
Chapter 3 The Direct Stiffness Method
For an 8-node element, as in Figure 3.S(c), there are 16 displacements and
th4il element stiffness is a 16-by-16 matrix. There are still only three rigid
body modes, so there are 13 deformation modes. For a 9-node element there
are 18 displacements, and hence 15 deformation modes.
When this type of element is extended to 3D, the equivalent of a 4-node
element is an 8-node ''brick", with a node at each corner. The element has
three displacements at each node, six rigid body modes, 18 deformation
modes, and a 24-by-24 stiffness matrix. A higher order element can have
nodes at each corner, at the midpoint of each edge, at the midpoint of each
face, and at the element- center, for a total of 27 nodes and an 81-by-81
stiffness matrix.
3.1.6
Slab and Shell Elements
A slab bending element is essentially a beam element that has bending in
two directions plus torsional resistance. However, the details are complex. It
is particularly difficult to satisfy compatibility for both displacement and
rotation along the element boundaries.
A 4-node slab bending element is shown in Figure 3.7.
Bending along Y
Twist
Figure 3.7 Slab Bending Element
The element has 12 displacements, allowingfor three rigid body modes and
nine deformation modes. It is not quite this simple, but the deformation
modes have essentially the form shown in the figure.
Stiffness and Flexibility Analysis Methods
55
A shell element combines out-of-plane slab bending with in-plane membrane
action. For a plane element, the slab bending part has three displacements at
each node, as in Figure 3.7, and the membrane part has two displacements,
as in Figure 3.6. Titls omits rotation about the axis normal to the element.
Titls is usually referred to as the "drilling" displacement. With the deformation modes shown in Figures 3.6 and 3.7, this displacement has no stiffness,
and it is necessary to add a "drilling" stiffness at each node (usually some
small value). The stiffness for a 4-node element is a 24-by-24 matrix.
An additional complication for a shell element is that the element is not
necessarily.plane - it can be initially warped or even curved. A shell element
·
is the most complex type of finite element.
3.2 Stiffness and Flexibility Analysis Methods
Any method of structural analysis must satisfy three conditions, namely (1)
force equilibrium, (2) displacement compatibility (continuity), and (3) the
element force-displacement relationships (constitutive relationships).
Analysis methods can be divided into two main groups, namely Stiffness (or
Displacement) methods and Flexibility (or Force) methods.
In the stiffness method the primary unknowns of the problem are displacements. fu a node-element model these are the node displacements. The analysis
steps for a node-element model are as follows (omitting many details).
(1)
(2)
(3)
(4)
(5)
Start with geometric compatibility. Using compatibility, relate the end
displacements of each element, and hence the element deformations,
to the unknown node displacements.
Using the element force-displacement relationships in stiffness form, .·
relate the element end forces to the unknown node displacements.
Using nodal equilibrium (each node is acted on by internal element
end forces and external nodal loads), relate the external nodal loads to
the unknown node displacements. The nodal loads are known. Hence,
the result is a set of simultaneous equations (often a very large set)
relating unknown node displacements and known nodal loads. These
are often termed the "equilibrium equations".
Solve the equilibrium equations to obtain numerical values for the
node displacements.
Knowing the node displacements, again use compatibility to calculate
the element end displacements and deformations.
56
Chapter 3 The Direct Stiffness Method
(6)
Knowing the element displacements and deformations, again use the
element force-displacement relationships to calculate the element end
forces.
Knowing the element forces, again use equilibrium to check that all
nodes are in equilibrium. This is not essential, but it provides a check
on the accuracy of the calculation& If there are significant equilibrium
unbalances, something has gone wrong in the calculations.
(7)
Note that only nodal equilibrium is considered directly in the above steps
(each node is in equilibrium as a free body, acted on by external nodal loads
and internal element end forces). Element equilibrium mu:;t also be
satisfied. In particular, each element must be in equilibrium as a free body,
. acted on by the element end forces plus any external loads that may be
applied to the element. This must be taken into account in the element forcedisplacement relationships, in Step 2. That is, the theory and computations
at the element level must ensure that element equilibrium is satisfied.
A flexibility method is essentially the reverse. The primary unknowns of the
problem are forces (the "redundant" forces). The analysis starts by ensuring
that the equilibrium conditions are satisfied, uses the element forcedeformation relationships in fleXIbility form, and sets up a set of simultaneous equations using the compatibility conditions. These equations are often
called the "compatibility equations". The solution of these equations gives
the unknown forces, and then it is easy to calculate everything else. The
details can be found in most textbooks on analysis theory.
3.3
3.3.1
The Direct Stiffness Method
Overview
This section provides a more detailed description of the Direct Stiffness
Method, considering linear elastic structures. Later sections consider material
and geometric nonlinearity.
3.3.2
Degrees of Freedom
Figure 3.8(a) shows a node-element model of a simple 20 structure.
The Direct Stiffness Method
57
Axially rigid
4
6
E
2
B
c
A
3
5
(a) Nodes and elements
(b) Unknown displacements
Figure 3.8 Simple Frame Example to Illustrate DSM
The structure has six nodes and five elements. The elements are all of frame
(beam and column) type, with bending and axial stiffness. For this
discussion the dimensions, loads and element properties are not important.
Most of the nodes are free to move, with three unknown displacements per
node. However, the displacements at some nodes are '.'restrained" or
"constrained", as follows.
(1)
The supported nodes are restrained, so that some or all of their
displacements are zero.
(2)
The beam connecting nodes 4 and 6 is assumed to be rigid axially.
Hence, the horizontal displacements at these nodes are constrained to
be equal. This is an example of a "slaving" constraint, where the
dis.rlacements at one node (usually called a "slave" node) is not
independent, but depends on the displacement of another node
(usually called the "master" node). The terminology is unfortunate but
concise. In this case the horizontal displacement at node 6 (the slave
node) must be equal to the horizontal displacement at node 4 (the
master node). More examples are given later.
Taking into account restraints and constraints, there are nine unknown
displacements, as shown in Figure 3.8(b).
The unknown displacements are the "degrees of freedom" (OOFs) of the
structure. This means that the structure has only nine independent deformed
shapes, and for any load case the deformed shape must be some combination of these nine shapes. (This is not true in all cases. Some elements can be
substructured elements that have "internal" displacements in addition to the
element end displacements.. This is considered later in this chapter.)
The following are some points to note for the structure in Figure 3.8;
58
Chapter 3 The Direct Stiffness Method
(1)
Nodes 1 and 5 both have hinged supports, but they are modeled
differently. At node 1 only the translations at the node are restrained,
and the node rotation is a OOF. At node 5 the node iS fully restrained,
and there is a moment release in the column element. The model at
node 1 has the advantage that moment loads can be applied on the
node if needed, and that the rotation at the support is calculated .
directly by the analysis. The model at node 5 has the advantage that
there are fewer OOFs.
(2)
In the figures, the nodes are numbered in an arbitrary sequence, and
the OOF numbering follows the node numbering. This is only one of
many possible numbering schemes. As shown later, there are optimal
numbering schemes that minimize the computer memory requirements or the computational effort required for the analysis.
(3)
For the nodes with the slaving constraint, the lower numbered node is
chosen as the "master" node, and the other node is the "slave" node.
This is an arbitrary decision.
(4)
A 20 structure is a special case of a 30 structure, with implied
supports that restrain all out-of-plane displacements.
3.3.3 Addition of Stiffoess
When a structure is deformed, by applying forces at the nodes, the forces
that are required to deform the structure can be obtained by adding the
forces required to deform the elements. This is illustrated in Figure 3.9.
Figure 3.9(a) shows three elastic bar elements in parallel, making up a single
OOF structure. For any displacement, ll, it follows from equilibrium that the
force required to deform the structure is the sum of the forces required to
deform the bars. Hence, the structure stiffness is the sum of the bar
stiffnesses.
The situation is more complex for a multi-DOF structure, but the stiffness
coefficients for the structure can still be obtained by adding the stiffness
coefficients for the elements. Figure 3.9(b) shows the case where a unit
·horizontal displacement is imposed on node 2 of the example frame, with all
other displacements zero.
The Direct Stiffness Method
59
Ks.2=0
~..-r-------.~jl<s.2=0
(a) Simple elements in parallel
=
(b) Impose r2 = 1 with all ,bther r 0.
Forces are column 2 of structure stiffness matrix.
k-v!i.f
ks
ks.4
4,4
k
I
End J
A
4
1
End~ ~ks.1
,kz1tt..· . ~
.. 7
k1.Lft~ D
Endl
l.tki.4
k3,4 .
(c) Numbering of element
end displacements
ks1f
.4
End I
4,1
ka;i
.
(d) Forces are column 4 of
element A stiffness matrix.
(e) Forces are column 1 of
element D stiffness matrix.
Figure 3.9 Addition of Stiffness Coefficients
This is a unit value of the DOF ri- From the definition of a stiffness coefficient,
the required forces for this displacement are the stiffness coefficients in
column 2 of the structure stiffness matrix. When the displacement is imposed
on node 2, only elements A and D are deformed. The forces required to
impose the displacement on these elements are, by definition, stiffness
coefficients in the element stiffness matrices. Hence, the stiffness coefficients
in the structure stiffness matrix are obtained by adding stiffness coefficients
from the element stiffness matrices. The C<?mplication is that the numbering
of the structure DOFs is different from the numbering of the element end
displacements.
Figure 3.9(c) shows how the element end displacements might be numbered
for an element (displacements l, 2 and 3 at End I, and displacements 4, 5
and 6 at End J). For element A the structure DOF r2 corresponds to the
element end displacement q4• Hence, the forces .corresponding to a unit
value of OOF r 2 are the stiffness coefficients in column 4 of the element
stiffness matrix, as shown in Figure 3.9(d). For element D the structure DOF
r 2 corresponds to the element end displacement qr Hence, the required
forces are the coefficients in column 1 of the element stiffness matrix, as
shown in Figure 3.9(e). The structure stiffness coefficients are obtained by
adding the element stiffness coefficients, but this is an "assembly" process,
60
Chapter 3 The Direct Stiffness Method
not simple addition of the element stiffnesses. The next section considers
this process in more detail. ·
As an exercise, you might like ~o consider the case r5 = 1, which affects the
element that is assumed to be rigid axially. How do the stiffness coefficients
for this element affect the structure stiffness?
3.3.4 Assembly of Structure Stiffness
The structure stiffness matrix is assembled element-by-element. A stiffness
matrix is required for each element, where the element has the same element
orientation as in the structure and the same displacement directions. The
only difference is that the numbering of the DOFs for the structure is differ~
ent from the numbering of the end displacements for the elements.
This is shown in Figure 3.10.
Axially rigid
~-....__...,s
E
2
c
5
(b) Global (structure) numbering
(a) Nodes and elements
Local (Element) Number
Elem I J 1 2 3 4 5 6
(c) Local (element) numbering
A
1 2 0
B
3 4 0
0
0
1
0
2
3
4
5
6
7
9
c
5 6 0
0
0
5
8
D
2 4 2
3
4
5
6
7
E
4 6 5
6
7
5
8
9
(d) Location matrices (global number
for each local number)
Figure 3.10 Stiffness Coefficient Numbering
For each element, Figure 3.10 shows the relationship between the "local"
numbering for the element end displacements and the "global" numbering
for the corresponding structure OOFs. For each element the relationship can
be represented by a "location matrix", which gives the global OOF number
....
··~.
The Direct Stiffness Method
61
for each local element displacement. fu Figure 3.lO(d), each row defines the
location matfix for one element. In some cases an element end displacement
has no corresponding OOF, in which case the term in the location matrix is
zero.
· Given the stiffness matrix and location matrix for an element, the contnbu~
tion of the element to the structure stiffness iS assembled as follows.
(1)
An element stiffness coefficient k(m,n) contfi.butes to the structure
stiffness coefficient K(L(m),L(n)), where Lis the location matrix.
(2)
H either L(m) or L(n) is zero, the element stiffness coefficient does not
contnbute to the structure stiffness. It is simply discarded during the
assembly process.·
For a frame element, as in the example, the element stiffness matrix is 6-by-6
and the location matrix is 1-by-6 (or 6-by-1). The size of the location matrix
depends on the element type.
The element end displacements for the element stiffness matrix can be
ordered in any sequence. For a given OOF numbering, changing the num'."
bering of the end displacements for any element rearranges the terms in the
element stiffness matrix, but it has no effect on the structure stiffness matrix
because the location matrix also changes.
The structure OOFs can also be ordered in any sequence. H the sequence is
changed, the eleinent location matrices change. The magnitudes of the
- structure stiffness coefficients remain the same, but they change position in
the structure stiffness matrix. As noted later, this can have an effect on the
computational efficiency of the analysis.
3.3.5 Addition of Flexibility
Structure stiffnesses are obtained by adding element stiffnesses. This can be
contrasted with the case where flexibilities are added. This is illustrated in
Figure 3.11.
This figure shows two structures, each with three elastic bars. Figure 3.ll(a)
shows three bars in parallel, as considered earlier (see Figure 3.9(a)). fu this
case, all three bars have the same axial deformation, the .force on the
structure is the sum of the bar forces, and the stiffness of the structure is the·
sum of the bar stiffnesses. Figure 3.ll(b) shows three bars in series. Iri this
case, all three bars have the same axial force, the axial deformation of the
Chapter 3 The Direct Stiffness Method
62
structure is the sum of the bar deformations, and the flexibility of the
structure is the sum of the bar flexibilities.
r
1.:. 1~=(~····)A
___..
Mc
·~ k
1~
c
.
·:
r= FR
fJ
~E tE!l°t& t
fA = (UEA)A
(a) Elements in parallel
f6
i1
fc
(b) Elements in series
Figure 3.11 Addition of Stiffness and Flexibility
By analogy with the Direct Stiffness method, if there were a "Direct
FleXIbility Method" the flexibility matrix for the structure would be obtained by assembling the flexibility matrices for .the elements. However, there is
no such method (or at least, no general method that is comparable to the
Direct Stiffness Method).
3.3.6
Optimal Node Numbering
Figure 3.12 shows a building frame with two different node numbering
schemes.
2
6
3
4
2
7
5
6
3
8
7
8
4
9
9
10
(a) Numbering scheme 1
5
10
(b) Numbering scheme 2
Figure 3.12 Effect of Node Numbering on Stiffness Matrix
In Figure 3.12(a) the nodes are numbered horizontally first, then vertically.
In Figure 3.12(b) the numbering is vertical then horizontal. In each case there
are 24 DOFs, and the OOF numbering follows the node numbering.
The Direct Stiffness Method
63
Each element has a 6-by-6 stiffness matrix. Assume that each matrix is full,
with all stiffness coefficients nonzero. This can happen if axial and bending
effects are coupled.
The figure shows the form of the 24-by-24 structure stiffness matrix. Each filled
in square represents a nonzero stiffness·coefficient. In all empty squares the
stiffness coefficients are zero. For numbering scheme l, in Figure 3.12(a), the
nonzero coefficients are more tightly grouped around the stiffness matrix
diagonal than for scheme 2, in Figure 3.12(b). The reason is that the largest
node number difference between the I and J ends of any element is s,maller
in scheme I than in scheme 2. The largest node number difference in scheme
I is 2,. in the columns. The largest node number difference in scheme. 2 is 5,
in the beams.
A narrower band is better because when the equilibrium equations are solved,
the stiffness matrix is progressively modified (as shown later) and the zero
terms within the band area tend to fill in with nonzero values. The terms
outside the band always remain zero. H the band is narrow, less computer
memory is needed to store the stiffness matrix. Also, the number of numerical
operations required to solve the equations is usually smaller. The reduction
in storage requirement and computational effort for this simple example is
small. The reduction for a large structure can be very substantial.
Computer programs for structural analysis typically renumber the nodes to
optimize the structure stiffness matrix, usually by minimizing the band
width. There are a number of methods for doing this, and the details are not
important. Most schemes renumber the nodes, but some renumber the OOFs.
Some computer programs have sophisticated methods for "conditioning"
the structure stiffness matrix to reduce the time required to solve the
equilibrium equations.
3.3.7 . Equilibrium Equations
The equilibrium equations have the form
(3.1).
Kr=R
=
where K = structure stiffness matrix, r matrix of OOFs (unknown displacements) and R =matrix of known loads. The preceding sections have
considered the definition of r and the assembly of K. The next two sections
consider setting up R and solving the equations
64
3.3.8
Chapter 3 The Direct Stiffness Method
Load Matrix
If a structure has only static nodal loads, it is a straightforward process to
assemble the load matrix, R. Element loads and dynamic loads are more
complex and are considered later.
Figure 3.13 shows the frame example used earlier.
Figure 3.13(a) shows the structure DOFs. Figure 3.13(b) shows a set of nodal
loads. Figure 3.13(c) shows the load matrix. For the rigid beam slaving
constraint, the loads corresponding to r 5 are simply added.
The loads that correspond to rigid supports are ignored for the load matrix.
However, these loads are not lost. When equilibrium is checked at a
supported node, as considered later, both the loads on the node and the
support reaction are part of the equilibrium check:
0
5
-10
0
13
-20
0
-10
45
(a) Structure DOFs
(b) Nodal loads
(c) Load matrix
Figure 3.13 Load Matrix: Nodal Loads Only
H there is only one load case, the load matrix has only one column, in which
case it is often referred to as the load "vector". H there are several load cases,
the load matrix has several columns, one for each load case.
3.3.9
Equation Solving
Given K and R, an equation solver calculates r. Equation solvers are specialized
and efficient computational modules that can solve very large sets of equations quickly and reliably. The details are not important.
·
In effect, solving the equilibrium equations is equivalent to inverting the
structure stiffness matrix. However, equation solving requires far less computational effort, and far less computer memory, than matrix inversion.
The Direct Stiffness Method·
65
3.3.1 O Element Deformations and Forces
After the equations have been solved, the ncxle displacementS are known for
each load case. These displacements include the DOFs in the matrix r and
the displacements at the supports (which are usually but not necessarily
zero, as considered later). Hence, the element end displacements are known ·
for all elements, and the element deformations
be calculated. Given the
end displacements and the element stiffness matrix, ,the element end forces
can be calculated. Given the element deformations, internal element forces,
such as bending moments, can be calculated.
·
can
It may be noted that calculation of the element forces in a computer program
requires much less computation than solving the eqUihbrium equations. The ·
element force computations increase roughly linearly. with the size of the
analysis model (mainly the number of OOFs), whereas the equation solving
computations increase exponentially.
·
33.11 Equilibrium Check
To obtain an accurate solution; the numerical process for solving a large set
of equations must be carried out to a high degree of precision; In a computer
program, double precision arithmetic, using about 15 significant figures, is
used, and the solution is usually accurate. However, this may nt>t always be
the case.
The accuracy can be checked by calculating· the equilibrium error. At each
node; and for each OOF, the element forces acting on the node must balance
the external loads· on the node. Given the external loads on any node and
the end forces on the elements that connect to the node, the force unbalance
is easy to calculate. Hthere is an unbalance, this is an equilibrium error.
The unbalances Will not be exactly zero, because there are cilways small roundoff errors in the calculations. However, the unbalances should be very small
compared with the applied loads. H the unbalanced forces for all DOFs are
small, the equation solving is accurate. Computer programs will often give a
warning if there are significant unbalances (or "residuals").
3.3.12 ·cause of Equilibrium Unbalance
H there are, equilibrium ·unbalances in a linear analysis, they/are almost .
certainly caused by poor modeling of the structure, causing the: equilibriumequations to be "poorly conditioned". Even though the eqµation solving
cal~tions use high precision arithmetic, if the equations are poorly con/
66
Chapter 3 The Direct Stiffness Method
ditioned the calculated values of the DOFs, and hence of the element end
forces, can be inaccurate. Computer programs should perform equilibrium
checkS and report on any significant unbalances.
Modeling errors that can lead to equilibrium unbalances are considered in a
later section.
It is possible for equilibrium unbalances to be caused by errors in a
puter program, but that is unlikely in commercial software.
com~
3.3.13 Reactions at Rigid. Supports ·
The equilibrium check considers forces that correspond to the boFs. This
check can be used to find equilibrium errors. An equilibrium i;:heck can also
be used to calculate the reaction forces at rigid supports. If equilibrium is
checked in the direction of a known external load, the resultant force or
moment should be zero, and this determines whether there is an equilibrium
error. If the check is carried out in the. direction of an unknown support
reaction, this calculates the reaction, which is the force or moment needed to
satisfy equilibrium. The support force calculation is accurate only if there
are no equilibrium errors.
3.3.14 Forces Corresponding to Slaving Constraints
In the example frame. considered earlier (Figure 3.13), the horizontal
displacements at two of the nodes are constrained to be equal. Hence, the
axial extension is zero for the beam that connects these two nodes.
If the axial stiffness ·of the beam were finite, the axial force in this beam
would be zero (axial stiffness times zero
axial extension= zero axial force).
\
However, the slaving constraint is imposed because the axial stiffness of the
beam is assumed to be infinite. Hence, the axial force is infinity times zero,
which can be a finite number.
The axial force in the beam can be calculated in the same way that rigid
support reactions are calculated. This is illustrated in Figure 3.14.
In this figure, if horizontal equilibrium is. considered at node 4, the only
unknown is the axial force in the beam that connects nodes 4 and 6. Hence,
the axial force can be calculated. The axial force can also be calculated
considering equilibrium at node 6. If there are no equilibrium unbalances,
the axial force will be the same in both cases.
Slaving Constraints - Rigid Floor Diaphragm
·Element E is
axially rigid.
~
l
.20
10• ..,45
Node Node
4
-+
8
6
Elem
B
8
-+
· (a) Nodal loads ·
20!
Forces in element E
get axialforce from ·
horizontal equifibrium.
~~-t;.
V,
Known forces
in element D
67
Known forces
inelementB
.
(b) Equilibrium at Node 4
Figure 3:14 Load Matrix : Nodal Loads Oniy
3.4 ·Slaving Constraints - Rigid Floor Diaphragm
The example frame in the preceding section had a rigid beam constraint.
This is the 20 version of a rigid diaphragm constraint, which is often used
in models for 30 buildings.
Figure 3.15 illmtrates a simple rigid diaphragm constraint. Usually there are
many more nodes.
Figure 3.15 Rigid Diaphragm Constraint
The floor is assumed to be rigid for in-plane behavior (i.e.,· as a membrane).
It is not rigid for <>ut-of-plane bending. It follows that the diaphragm movesas a rigid body in the horizontal plane, with three degrees of freedom,
namely two ~orizontal translations and one rotation, about the vertical axis.
/
68
Chapter 3 The Direct Stiffness Method
In a rigid diaphragm constraint, the "master" node accounts for displacement of the diaphragm. This node is often located .at the centroid of the
diaphragm. All other nodes in the diaphragm are "slave" nodes, and their
in-plane displacements depend on the master node displacements. If there is
no torsional deformation of the building, the diaphragm translates without
rotating and all nodes have the same displacements. If the building has
torsional deformation, a floor diaphragm rotates about the vertical axis. In
this case, all nodes have the· same rotation but they have different
translations.
An actual floor ·diaphragm is, of course, never truly rigid. When a rigid
diaphragm constraint is used in an analysis model, the assumption is that
the diaphragm is so stiff compared with the rest of the structure that it can
be assumed to be rigid, or that the diaphragm deformations are so small
that they have no significant effects on the rest of the structure.
When a rigid floor constraint is used, all nodes in the diaphragm should, ·
strictly speaking, be in the same plane. If they are not, this could be a modeling
error. It is a little difficult to eXplain why this is the case for a floor diaphragm,
so consider a related but more obvious example, as shown in Figure 3.16..
Load
3.rr
1V 12
(a) Simple truss
t
t
(b) Deflected shape
Additional
impfied load
(c) Deflected shape
with constraint
Figure 3.16 Example of Incorrect Constraint
Figure 3.16(a) shows an analysis model for a simple 2D truss. Figure 3.16{b)
shows the deflected shape when a vertical load is applied. Only one of the
nodes displaces, as shown. Suppose, however, that the vertical displacements at nodes 3 and 4 are constrained to be equal. There is no good reason
for doing this, but it could be done. When a vertical load is applied, the
calculated deflected shape will now be as shown in Figure 3.16(c). Since
more of the elements have been extended, there must be an additional load
to satisfy equilibrium, as shown in the figure. Hence, the constraint brings
with it an implied additional external load. This is a serious modeling error.
Equilibrium Errors in Linear Analysis
69
As an exercise you might like to explain why the nodes in a rigid diaphragm
constraint should all be in the"same plane. You might also consider whether
thermal expansion loading can be specified for a rigid beam or diaphragm.
Also, in Figure 3.16(c) the vertical displacement is shown with the same
magnitude as in Figure 3.16(b). H you were to analyze the two models, would
the displacements be the same? Hnot, why not?
There are many types of constraint that might be s}>ecified. They should
always be used cautiously, considering what they mean in physical terms.
This is particularly true when simple equal displacement constraints are
used, a8 in Figure 3.16(c). In such cases, always consider whether there
might be implied external loads on the nodes.
3.5
3.5.1
Equilibrium Errors in Linear Analysis
Overview
As noted earlier, modeling errors can cause numerical problems in the
equation solving phase. There are two main errors, as follows.
(1)
Unstable structure. H the structure is unstable the equilibrium equations
have no solution. The equation solver may find a solution, but it is
meaningless, and usually involves extremely large displacements.
(2)
Poor conditioning of the structure stiffness matrix. Poor conditioning
can cause anything from small equilibrium unbalances to the same effect
as an unstable structure.
This section considers these modeling errors. The discussion applies only to
linear analysis. H the behavior is nonlinear, equilibrium unbalances can occur
for different reasons and are not necessarily errors. This is considered later.
3.5.2
Unstable Structure
A structure is unstable if it can displace as a rigid body or deform as a
mechanism. In effect, solving a set of equilibrium equations Inverts the
structure stiffness matrix. Since it requires zero load to deform a stt:ucture as
a mechanism, the ,stiffness matrix has no inverse (the flexibility of a
is infinite),
m.echanism
70
Chapter 3 The Direct Stiffness Method
The most obvious case is a structure that does not have enough supports,
allowing it to move as a rigid body. A 3D structure must have supports that
prevent rigid body translation along all three global axes, and also rigid
body rotation about all three axes.
Other mechanisms can be less obvious. Also, a mechanism can be global
(involving .deformation of the entire structure), local (involving perhaps
only a single element), or anything in between. The following are some
examples.
(1)
If a truss structure is made up entirely of pin-ended bar elements, the
element ends have no rotational stiffness. Hence, the nodes are unrestrained against rotation, and each node is a local mechanism. This
is easy for a computer program to detect, since the structure stiffness
matrix will have diagonal stiffness coefficients that are zero. In this
caSe, most computer programs will add supports to eliminate the
unrestrained rotations. Hence, although this is technically a modeling
error, most computer programs will correct it.
(2)
If a 3D truss structure has a node where all of the elements are in a
single plane, there is no stiffness normal to that plane, and there is a
..mechanism. H the normal to the plane is along a global axis, there will
be a diagonal stiffness coefficient that is zero, and a computer program
can add a support. H, however, the plane is inclined to the global axes,
there will not be a zero diagonal stiffness, and the instability will
become apparent only in the equation.solving phase. If a 3D structure
has 2D parts with no stiffness in the out-of-plane direction, this can be
a modeling error.
(3)
Figure 3.17 shows a column and some connecting beams in a steel
frame structure. The column has a base-plate that is assumed to be
non-moment-resisting. This is modeled with a hinged support, as
shown. The support is hinged for torsion as well as bending. l)le
beam-to-column connections are also assumed to be non~moment­
resisting. These are modeled using moment releases at the ends of the
beam elements, with zero rotational stiffness for both vertical and
horizontal bending. The analysis model is unstable because the
column is unrestrained in torsion and can spin as a rigid body about
its vertical axif?. The instability will become apparent. only in the
equation solving phase. The modeling error is that the support at the
column base does not restrain rotation about the vertical axis. A
column base-plate will restrain this rotation. As a general rule, always
restrain rotations about the vertical axis at a support.
Equilibrium Errors in Linear Analysis
71
Releases allow
rotation about
both bending axes.
Base is non
moment-resisting.
(a) Column
Support allows
rotation about
alUhree axes.
(b) Model
Figure 3.17 Unstable Model - Column is Unrestrained in Torsion
(4)
Figure 3.18 shows the model for some floor framing, using beam
. elements.
Figure 3.18 Unstable Model - Too Many Moment Releases
The beams all have non-moment-resisting end connections, and are
assumed to have moment releases at their ends. The releases are
assumed to have zero rotational stiffness for bending about both axes
and for torsion. The analysis model is unstable because the beams are
unrestrained in torsion. The error for the short span beam can be
detected by a computer program because a single element should not
be allowed to have torsional moment releases at both ends. The other
instability will become apparent only in the equation solving phase.
The modeling error in this case is that too many moment releases
have been used.
The following are some general rules to avoid stability problems.
72
Chapter 3 The Direct Stiffness Method
(1)
If some par~ of a 3D structure are actually 2D, make sure that there is
no out-of-plane instability.
·
(2)
If a column base is hinged, allow bending rotation but restrain torsional
rotation.
(3)
Be careful when specifying moment releases at the ends of beam and
column elements. Avoid specifying too many releases.
3.5.3
Stiffness Coefficient Mismatch
Structure stiffness matrices are usually assembled using double precision
values, with about 15 significant digits. This is usually sufficient to make the
numerical computations very robust. However, if a structure is badly modeled
and there are huge differences between the stiffness coefficients from different elements, there can be numerical problems. The following are two
examples.
·
Example 1: Frame Structure with Axially Stiff Beam
Figure 3.19(a) shows the example frame structure from earlier sections.
Extremely stiff axially
4
6
E
2
c
A
5
(a) Nodes and elements
(b) Structure DOFs
r8=1
14
N
.-----·-,
b
~1
r --··. hr
) i_
'°,,·'
""'
Force = EA/b
Force = 3El/h3
(c) Deformed shape for column 8
of structure stiffness matrix
(d) Diagonal stiffness coefficients
for elements C and E
Figure 3.19 Stiffness Coefficients for Example Frame
\,
.
··~.
Equilibrium Errors in Linear Analysis
73
As before, element E is assumed to be rigid axially, but a slaving constraint
is not used. Instead, a large value is assigned to the property EA. This can be
done by specifying a large value for the modulus E, the area A, or both.
The DOFs are numbered as show in Figure 3.19(b). There is one more OOF
than in the earlier examples, because the horizontal displacements at nodes
4 and 6 are no longer identical.
. The stiffness coefficients in column 8 of the structur~ stiffness matrix are the
forces required to deform the frame as shown in Figure 3.19(c). In particular,
the diagonal stiffness coefficient K(8,8) is the sum of the two element
stiffness coefficients shown in Figure 3.19(d). The values of these stiffness
coefficients are 3El/h3 for the column and EA/b for the beam, where h is the
column height, b is the beam span, and E, I and A have their usual meanings.
Suppose that the "large" value that is assigned to EA for the beam is
astronomically large, so that EA/b for the beam is 1(}2° uriits (axial force per
unit extension). Also, suppose that 12EI/h3 for the column is 100 units. With
these values the beam stiffness coefficient is 1018 times larger than the
column coefficient. Hence, if the structure stiffness matrix is assembled with
15 digit accuracy, when the two coefficients are added together the column
stiffness coefficient is completely lost. There is a similar loss of accuracy for
several other stiffness coefficients. This causes the structure stiffness matrix
to be poorly conditioned, and when the equilibrium equations are solved
the solution is meaningless. If the. equation solver does find a solution, an
equilibrium check will show large equilibrium unbalances.
If, a smaller (but still astronomical) value is specified for EA, so that EA/b is
1012 units, the beam stiffness coefficient is 1010 times larger than the column
coefficient. This is still a huge difference, but with 15 digit accuracy it may
be possible to get an accurate result.
If EA/b is 1015 units, the ratio of the stiffness coefficients is 1013• In this case
the equations can possibly be solved but the result is likely to be inaccurate,
with substantial equilibrium unbalances.
In this example, the modeling error is the use of astronomically large
stiffness values. In practice, no structural member can ever have such a large
stiffness. Consider, for example, a steel bar that is 100 inches square and 300
inches long. This is a block, rather than a bar, and it is extremely stiff, but its
axial stiffness, EA/b, is only about 106 kips per inch. For a similar bar With a_
2.5 m square section and 7.5 m length, EA/bis roughly 0.2x106 kN/mm.
74
Chapter 3 The Direct Stiffness Method
As a general modeling rule, always specify realistic stiffnesses, recognizing
that nothing in a real structure is ever rigid.
·
Example 2: Very Stiff Support Spring
Usually, the displacements at the supported nodes in a structure are
assumed to be zero (in reality they are not). A support displacement can be
restrained to be identically zero, in which case there is no corresponding
OOF. Alternatively, the support can be modeled by a spring element with a
large stiffness~ In this case there is a OOF at the node. However, the
displacement of the node is very small, and for practical purposes it is zero.
Figure 3.20(a) shows a beam with an inclined support.
(a) Beam with inclined support
(b) Model with support springs
(c) Model with rigid supports requires rotated coordinates
Figure 3.20 Supports Modeled using Stiff Springs
Figure 3.20(b) shows a model with springs at all three supports. Suppose
that an astronomically large stiffness is specified for each of these spring
elements, so that the axial stiffness of any spring is far larger than the
stiffness coefficients of the other elements that connect to the supported
node.
At two of the supported nodes, the spring is parallel to the vertical DOF at
the node. Hence, when the structure stiffness matrix is assembled, the stiffness coefficient from the support spring is added to a diagonal term of the
structure stiffness matrix. In Figure 3.20(b) this applies to OOFs 2 and 5.
Hence, very large values are added to the terms K(2,2) and K(5,5) in the
structitre stiffness matrix. Since only the diagonal terms are affected, even if
the support spring stiffness is so large that the stiffness coefficients of the
other elements are completely lost, this does not cause the structure stiffness
Equilibrium Errors in Linear Analysis
75
matrix to be poorly conditioned. Hall supports are of this type, the equilibrium
equations can be solved accurately.
However, the third support in Figure 320(b) is different. At this node the
support spring is inclined, and is not parallel to the OOF directions. Hence,
the support spring stiffness must be transformed to the OOF directions (the
global coordinate system) before its stiffness matrix can be assembled into
the structure stiffness. The transformed matrix affects POFs 7 and 8, and the
spring stiffness affects the off-diagonal terms K(7,8) and K(8,7) as well as the
diagonal terms K(7,7) and K(8,8). Now, if the support spring stiffness is so
large that the stiffness coefficients of the other elements are completely lost,
the structure stiffness matrix is poorly conditioned, and the equilibrium
equations will not be solved accurately. Hence, it is an error to model an
inclined support using a spring element with an astronomically large stiffness.
The supports in real structures are far from rigid. As a general rule, if you
use a spring to model a support, always estimate the actual support
stiffness, and specify a reasonable value for the spring stiffness. Usually the
support displacements in the actual structure will be small, and they will
have little effect on the behavior of the ·structure. Hence, these displacements do not have to be calculated accurately in the analysis model, and the
spring stiffness needs only to be reasonable, not "accurate". Ha spring is stiff
enough to make the support displacements small for practical purposes,
increasing the spring stiffness by a factor of lo, or 100, or 1000 will have a
negligible effect on the analysis results, unless the stiffness is made so large
that it affects the accuracy of the equation solving.
H an actual support is sufficiently flexible that the support displacements
have an effect on the structure, the stiffness of the support spring in the
analysis model must, of course, be estimated with greater accuracy.
In some computer programs, stiff inclined supports can be modeled as rigid,
with no corresponding OOF. This requires that the orientations of the DOFs
at the supported node be rotated into a local coordinate system, as shown in
Figure 3.20(c). The computer program must allow such local coordinate
systems to hf' defined,
3.5.4
Programming Error
Although it is unlikely, equilibrium unbalances can be caused by computer
programming errors. This can occur if the computations for setting up the
stiffness matrix do not exactly correspond to the computations for calculating
the element deformations and forces.
76
Chapter 3 The Direct Stiffness Method
The assembly of the structure stiffness matrix uses compatibility (element
end displacements are equal to node displacements), the element forcedisplacement relationships, and equilibrium (nodes are in equilibrium as
rigid bodies). When the node displacements· have been calculated,
compatibility is used again to get the element deformations, the element
force-displacement relationships are used to get the element end forces, and
equilibrium is used to calculate unbalances. If the equations .are solved
accurately; and if identical compatibility, force-displacement relationships
and equilibrium are used in both phases, there will be negligible unbalances.
Conversely, if the equations ate solved accurately and there are unbalances,
the two phases must· be using different compabbility conditions, different
force-displacement relationships, and/ or different equilibrium conditions.
A software developer might make this error while developing a program,
but there should be no such errors in production versions of commercial
software.
3.5.S
Detection of Unstable Mode
Equation solving is not an abstract numerical process. Rather, it has physical
significance. This is illustrated in Figure 3.21.
Figure 3.21(a) shows a 2D cantilever beam with an internal hinge. Because
of the hinge, the beam is a mechanism and is unstable.
Figure 3.2l(b) shows a 2D model with five nodes and four beam elements.
Horizontal displacements· are not considered, so that there are two DOFs
per node, namely vertical displacement and rotation. The hinge is modeled
with a bending moment release in one of the elements. The structure is
unstable, but the instability can not be seen by looking at the structure
stiffness matrix - it becomes apparent only in the equation solving phase.
a
Equation solving is a complex numerical process, but it has physical
meaning. As -the equations are solved, the structure stiffness matrix is
progressively modified, in such a way that it is always a stiffness matrix but
for a structure that progressively changes. This can be illustrated as follows.
(1) . Consider the physical significance of the stiffness coefficient K(6,6) in
the structure stiffness matrix. This is the diagonal stiffness coefficient
for the rotational OOF r6 •
Equilibrium Errors in Linear Analysis
!Load
77
Hinge
'f ,, "'''''
'*;m
r:n::(~fN> ,,,,,,,,E
(a) Beam with internal hinge
(b)Analysis model
r6 =1, all otherr=O.
(c) Deformed shape for column 6 of original structure stiffness
c ______
Modified K(6,6)
--0__-:...- -
~
r6 = 1, r5 = r7 = r8 = 0, r1 through r4 unrestrained.
(d) Shape for column 6 after reduction of equations 1, through 4
-J,!J-...----
__...o---- ,v
e-
Modified K(6,6) = 0
•
(e) Shape for column 6 after reduction of equations 1 through 5
Figure 3.21 Physical Explanation of Equation Solving
(2)
By definition, K(6,6) in the original stiffness matrix is the moment load
required to cause a unit value of r6 with all other DOFs equal to zero.
This is shown in Figure 3.21(c). Forces corresponding to DOFs r31 r41 r51
r 7 and r 8 must also be applied. By definition these forces are the offdiagonal stiffness coefficients K(3,6), K(4,6), etc. Figure 3.21(c) shows
the form of the stiffness matrix.
(3)
In the equation solving process, the stiffness matrix is progressively
modified, in effect by releasing each DOF in turn. For example, after
the first four DOFs have been released, the modified stiffness coefficient
78
Chapter 3 The Direct Stiffness Method
K(6,6) is the moment load shown in Figure 3.2l(d). This is the load
required to cause a unit value of r 6 with all other DOFs except-the first
four equal to zero. For OOFs r 1 through r 4 the corresponding nodal
loads are zero, not the displacements.
(4)
After the first four OOFs have been released the original 8-by-8
stiffness matrix, with OOFs r 1 through r8 , has been reduced to a 4-by-4
matrix with OOFs rs through r8• This is shown in Figure 3.21(d).
Those who are familiar with equation solving will recognize this as the
"forward reduction" phase. This phase continues until only a 1-by-1 matrix
remains. A similar forward reduction is performed on the load matrix. The
OOF values are then calculated in a "back substitution" phase.
At each step in the forward reduction phase there is a new stiffness matrix,
with one less row and column, and a corresponding structure with one less
OOF. For each new stiffness matrix, if the diagonal stiffuess coefficients are
all positive, the corresponding structure is stable. For example, it requires a
positive moment to cause the displacements shown in Figure 3.2l(d).
However, if any diagonal stiffness coefficient is zero, it requires no force to
displace the structure and it is a mechanism. If any diagonal stiffuess
coefficient is negative, the structure must .be supported to stop it from
collapsing. In either case the structure is unstable.
This is shown in Figure 3.2l(e). In this figure the first five OOFs have been
released1and the figure shows the displaced shape for the stiffness coefficient
K(6,6). Because of the bending moment release, it requires no moment load
to cause this displacement, and hence the modified-value of K(6,6) is zero.
This indicates that there is a mechanism. The modified stiffness coefficients
in row and column 6 are also zero. The structure stiffness matrix is a 3-by-3
matrix with OOFs r6 through r81 as shown in Figure 3.2l(e).
fa the next cycle of equation solving, the inverse of K(6,6) must be
calculated. It is unlikely that the modified value of K(6,6) will be exactly
zero, because of round-off. However, it will be very small, and possibly
negative. It is possible that the inverse of K(6,6) will be too large and cause a
numerical overflow error. It is also possible that the equations can be solved,
but the calculated displacements will be extremely large, and there will be
large equilibrium errors.
In some computer programs, the equation solver monitors the values of the
diagonal stiffness coefficients, detects when there has been an unacceptable
loss of accuracy, and issues a warning. If a mechanism forms, it is also
Element Loads
79
possible for a computer program to identify the mechanism. Essentially,
when a diagonal stiffness coefficient becomes zero, or near zero, the
corresponding DOF is given a unit value, all later DOFs are set to zero, and
the equations are back-substituted with zero load to get the remaining
DOFs. The resulting displaced shape defines the mechanism. This is the
shape shown in Figure 3.21(e).
3.6
3.6.1
Element Loads
Overview
The examples in earlier sections consider only nodal loads. This section
considers element loads. These include the following.
(1)
(2)
External loads applied on elements rather than nodes.
Element initial deformations, such as deformations caused by thermal
expansion.
This section considers only linear elastic elements.
3.6.2
Element Force-Displacement Relationship
An element stiffness matrix defines the relationship between element end
displacements and element end forces. Specifically:
Q=kq
(3.2)
where Q = matrix of element end forces, q = matrix of corresponding
element end displacements, and k = element stiffness matrix. This equation
implies that if the element end displacements are zero, then the element end
forces are also zero. This is not always the case. A more complete equation is
Q=kq+Qo
(3.3)
where Q0 = matrix of element "fixed-end forces". These are the elemE'.nt end
forces in the "fixed-end state", when q = 0.
·
Figure 3.22 illustrates this for a 2D beam element.
80
Chapter 3 The Direct Stiffness Method
(b) Fixed end forces for element load
(a) Forces and displacements for element stiffness
Figure ~.22 Fixed-End State for a Beam Element
Figure 3.22(a) shows the forces and deformations that are used for the
element stiffness matrix. Figure 3.22(b) shows the additional forces and
deformations in the fixed-end state, for a distributed transverse load on the
beam.
Figure 3.22(a) also shows the element deformed shape for each of the end
displacements. For any set of element end displacements, the element
deformed shape is (l combination of these shapes. If all end displacements
are zero, the element is undeformed. Figure 3.22(b) shows that there is an
additional deformed shape that is caused by the element loads.
The theory and procedure for calculating the fixed-end forces depends on
the element type. It is essential for each element to be in equilibrium as a
free body under the action of the external· 1oads and the fixed-end forces.
This must be considered at the element level.
3,6.3
Structure Equilibrium Equations
When there are only nodal loads, the equilibrium equations for the structure
can be written as
R=Kr
'
'~-·
(3.4)
Element Loads
81
where R = matrix of nodal loads, r =matrix of node displacements, and K =
structure stiffness matrix. When there are element fixed-end forces, this
equation must be extended to
R=Kr+Ro
(3.5)
. where Ra is the matrix of nodal loads that must be applied to satisfy equilibrium when r = 0 (i.e., when r = 0, R ·= Ra ). This is· exactly the fixed-end
state. Ra is the matrix of nodal loads that must be ~pplied to prevent any
displacements of the nodes when the element loads are applied.
For any load case, given the fixed-end forces, Oo, for all .of the loaded
elements, the matrix Ra can be assembled. The equilibrium equations that
must be solved are then
·(3.6)
The right-hand side is the effective load matrix.
After the equations have been solved, the node displacements are known.
Hence, the element end displacements, q, are known, and the element end
forces can be calculated using Equation 3.3. Nodal equilibrium is checked in
the usual way. As already noted, element equilibrium is a separate issue
that must be considered at the element level.
3.6.4
Element Initial Deformations
The most common causes of element initial deformations are (a) thermal
expansion or contraction, (b) shrinkage or swelling, and (c) tittle-dependent
creep. At the material level, expansion and shrinkage usually involve only
volume change, with no distortion. Creep usua:lly involves distortion with
no volume change. At the element level the difference may not be obvious.
For example, thermal expansion and tiine-dependent creep can have similar
effects on bar and beam elements.
The usua1 method for caku1ating initia1 deformation Joads is as follows.
Figure 3.23 shows how the loads corresponding to initial deformatio~
usually calculated.
are
··
82
Chapter 3 The Direct Stiffoess Method
Temperature increase
T2 on this side
All Q = 0. Calculate q0 .
(a) Allow unrestrained expansion.
(b) Apply forces to make end
displacements equal zero.
Figure 3.23 Fixed-End Forces for Thermal Expansion
The steps.in the calculation are as follows.
(1)
Support the element so that it is free to expand. In Figure 3.23(a) the
element is cantilevered from its 'T' end. Other support conditions could
be used.
(2)
Given the temperature change, or other information on the. initial
deformation, calculate the axial and bending deformations due to
thermal expansion, and hence the end displacements, 'lo· The element
is free to expand, and there are no external .loads, so the support
reactions are zero.
(3)
Using the element stiffness matrix, calculate the end forces, Q01 required
to cancel out the end displacements. That is
(4) · The end displacements are zero, so Q 0 is the matrix of fixed-end
forces. Note that since there are no external loads on the element, this
must be a self-equilibrating set of forces.
The rest of the computation is the same as the case with external element
loads.
Element Loads
3.6.S
83
Imposed Displacements at Rigid Supports
At a rigid support the displacement of the node is usually zero, but it can
have a nonzero value. An important point is that the displacement is
known, so it is not a DOF.
Figure 3.24(a) shows a 2D frame with support displacements.
Moves +horizontally
tMovesdown
(a) Beam with support displacements
(c) Imposed vertical displacement
r
(b) Model ~ith rigid supports
(d) Imposed horizontal displacement
----ll__
J;___
(e)Model with support springs
TMovesdown
(f) Imposed vertical displacement
Figure 3.24 Fixed-End States with Support Displacements
Figure 3.24(b) shows the analysis model. Figure 324(c) shows the deformed
shape of the structure when the vertical displacement is imposed and all of
the DOFs have zero values. This is a fixed-end state for the structure.
However, the elements that connect to the displaced support are deformed.
Figure 3.24(d) shows the fixed-end state when the horizontal displacement
·
is imposed.
H an element connects to a displaced support, it can have nonzero initial
deformations, q0• The matrix of element fixed-end forces, Q., for such an
·
element is
84
Chapter 3 The Direct Stiffness Method
(3.7)
Given the fixed:..end forces, Q0 , for the affected elements, the .matrix Ro. can
be assembled. The rest of the computation is the same as for the case with
external element loads.
Figure 3.24(e) shows an alternative model with support springs. This model
has two additional DOFs. Figure 3.24(f) shows the fixed-end state for this
model with a vertical displacement imposed. This is simpler than the state
in Figure 3.24(c), because only the support spring has an initial deformation.
When support deformations are imposed, it can be simpler to use support
springs.
When support springs are used, support displacements can also be
introduced by specifying expansion or shrinkage in the spring. For example,
in Figure 3.24(f), a contraction of the spring corresponds to a downwards
movement of the support. The initial force in the spring is tension, which
pulls the beam down.
3.7
3.7.1
Dynamic and Nonlinear Analysis
Dynamic Loads
The preceding sections have considered only static loads. For dynamic loads
there are the following differences,
(1)
The compatibility conditions are exactly the same for static and
dynamic loads. However, the element force-deformation relationships
may be different, because the material behavior can depend on the
deformation (strain) rate.
(2)
The equilibrium check must consider inertia and damping forces,
depending respectively ort acceleration and velocity, as well as the
element forces (which for a linear elastic element depend on
displacement). This' means that a mass ~trix and a damping matrix
are needed for the structure in addition to the stiffness matrix.
Analysis for dynamic loads is much more complex than analysis for static
loads.
Dynamic and Nonlinear Analysis
3.7.2
85
Material Nonlinearity
Most nonlinear analysis methods are an extension of linear methods, with
one phase that sets up a structure stiffness matrix and calculates the nodal
displacements (often called the "linearization phase") and. a second phase
that calculates the element deformations and forces (often called the "state
determination" or "state update" phase).
When there is material nonlinearity, the stiffness m'atrix for an element, in
the linearization phase, must be based on an estimate of the future behavior
of the element, which is unknown. Usually the element force-displacement
relationship is based on the current state of .the element. In the state
determination phase, however, the element deformations are known, and
hence the element forces can be (and must be) calculated for the actual
force-displacement relationship of the element, w:hich is generally nonlinear
and can change as load is applied. Hence, the force-displacement relation.:
ships in the two phases are generally different, and there are equilibrium
unbalances in the linearized solution. These unbalances can not be ignored
and must be correeted, which usually requires some kind of iteration. An
important part of a nonlinear analysis method is what kind of iteration
strategy it uses.
3.7.3
Geometric Nonlinearity
Equilibrium unbalances can also occur when there is geometric nonlinearity.
In this case, however, the unbalances result not from the element force=
displacement relationships but from the compatibility and equilibrium
conditions.
·
In the linearization phase, the element stiffness matrices are U.Sually tangent
stiffnesses, which are based on vanishingly small increments of displacement, and hence use linear compatibility relationships between element end
displacements and element deformations. The state determination phase,
however, is based on finite displacement increments, using the actual
nonlinear relationships. This can cause equilibrium unbalances.
- In addition, in the linearization'phase, the assembly of the element tangent
stiffness matrices into the structure stiffness matrix is usually based on the
current geometry of the structure. In the state determination phase the node
displacements are known, and hence the new structure geometry is known.
The equilibrium check must be based on this new geometry. This can also
cause equilibrium unbalances.
86
Chapter 3 The Direct Stiffness Method
3.8 Substructures and Superelements
In the Direct Stiffness Method, the properties of an element are often
calculated using a continuum model. This is not the only way, however. In
some cases an element can consist of several components, and the element
properties are c~lculated from the component properties.
There are several ways to calculate the properties of such an element. One
way is t() treat the element as a separate node-element model, in which case
the element is a "substructure". The substructure can be processed so that it
looks like an element, in which case it becomes a "superelement". To
illustrate the process, consider the 20 frame in Figure 3.25.
3tt2 s.;-{) 9-tts 1t-f11tf4 1rt
1
4 7 • 10 13 • 16 '
_.
1 ·r .,. ·· ·· ·1
rt2
.. .. .. ..
(a) Structure DOFs
12.r.f1
B.tt5 9.rf 1tt4
__.....
10 1 4 7 13
-
(b) Substructure DOFs
.3 t2 6 t5
tl =1
9.;-t8
I!
I!
!!I
'1
1
9
When DOFs through are
released, this part fills in. -
(d) Substructure stiffness
(c) Structure DOFs with superelement
Figure 3.25 Substructure and Superelement
The frame in Figure 3.25(a) has a long span beam that is divided into four
separate elements. The reason why the beam might be subdivided is not
· important for this example. The important point is the beam can be regarded as a substructure, as shown in Figure 3.25(b). This substructure. has its
own nodes, elements and OOFs. The substructure can be reduced to a
superelement, with only two nodes, as shown in Figure 3.25(c). The superelement has a stiffness matrix and other properties, like any other beam
element.
Substructures and Superelements
87
In a real structure, a substructure would probably be much larger, such as a
span in a long span bridge, with many frame or bar· elements, or a floor
diaphragm in a building, with many slab or shell elements. The. superelement would probably have a lot more than two nodes.
In Figure 3.25(b), the internal nodes are numbered first, followed by the
external nodes. The substructure stiffness matrix thus has the form shown in
Figure 3.25(d). Since the substructure has no extemal·supports, this stiffness
matrix is singular, and it is not possible to apply loads and solve the
equilibrium equations. It is possible, howe;ver, to use equation solving to
"condense" the 15-by-15 substructure stiffness matrix to a 6-by-6 superelement stiffness matrix.
As noted earlier, the physical meaning of the forward reduction phase of the
equation solving process is that the stiffness matrix is progressively modified,
by releasing each OOF in tum, so that it is unrestrained. If the substructure
stiffness matrix is modified by releasing the first nine OOFs, aU of the
internal nodes are free to move, and the remaining 6-by-6 stiffness matrix
applies for the six external OOFs only. This is the superelement stiffness
matrix. It can be used to assemble the structure stiffness matrix for the
structure shown in Figure 3.25(c).
This process is usually termed "static condensation".
The process can also be used to get the load matrix. If there are external
loads on the substructure in Figure 3.25{b), a load matrix for the substructure. can be assembled. If equation solving is used to reduce the first
nine rows in this matrix, the remaining six rows are like fixed-end forces for
the superelement (actually they are effective loads, which are the opposite
of the fixed-end forces). These can be used just like any other fixed-end
forces. Hence, equilibrium equations for the structure in Figure 3.25(c) can
be set up and solved. The solution gives the values of OOFs 4 through 9 in
Figure 4.25(c), which are the values of OOFs 10 through 15 in Figure 4.25(b).
Given these values, back substitution can be used on the substructure
equations to calculate the valugs of· the internal OOFs. Hence the element
forces can be calculated for all of the elements in the original structure.
This may seem like a roundabout way of analyzing the original structure.
However, it can have advantages. The main advantage is that the stiffness
matrix for an analysis model with superelements can be much smaller than
the stiffness matrix for the non-substructured model. Even when the computational effort required to form the substructure and superelement
stiffnesses is considered, it can be more efficient to use substructures and
88
Chapter 3 The Direct Stiffness Method
superelements. This is particularly true if only the superelement stiffness
matrix is needed, and it· is not necessary to calculate the vfilues of the
internal OOFs in the substructures. There can also be computational savings
if there are repeated substructures, with identical properties, since only one
of the substructures needs to be processed to get the superelement stiffness.
It may be noted that static condensation is not the only way to get superelement stiffnesses and fixed'"el1d forces. By definition, the stiffness coefficients
in any column of a superelement stiffness matrix are the forces required to
impose a unit displacement for the corresponding DOF. If a substructure is
set up as a separate structure, with the external DOFs restrained, unit
support displacements can be imposed, and the required. reactions can be
calculated. For each unit displacement, the reactions define one column of
the superelement stiffness matrix. The superelement fixed-end forces can
also be calculated by applying loads and calculating the support reactions.
Finally, the condensation process is "static" condensation because it affects
the static stiffness matrix and applies only for static forces. Substructuring
can be used for dynamic analysis, but it involves inertia and damping forces
and can be a much more complex process.
CHAPTER
4
Component Behavior - Uniaxial F-D
Relationships
The behavior of a node-element model depends· on the behavior of the
.elements. An element can consist of several components, and the element
behavior depends on the behavior of its components. There can be many
different element types, with widely different behavior and properties.
However, components tend to have properties that are based on similar
concepts and theories.
This chapter considers the elastic and inelastic behavior of components. It
considers general behavior only, and does not give specific modeling
suggestions.
This chapter considers only components that have uniaxial force-deformation
relationships. Components that have multi-axial relationships, with interaction, are considered in Chapter 5.
4.1
4.1.1
Overview
Components and Elements
A node-element model consists primarily of nodes and elements. Anode is
a rigid point in space, and can not be subdivided. An element, however, can
consist of a number of components. For example, Figure 4.1 shows a beam
member in a frame and a corresponding beam element. The element is
89
90
Chapter 4 Component Behavior - Uniaxial F-D Relationships
made up of five components, as shown. The element properties depend on
the properties of these componentS.
Non moment-resisting
connection
Ductile
moment-resisting
connection
·-----·-·---·-·-·-·-
I
..
(a) Frame
.
.
~
End
zone
Moment
reiease
Elastic
beam
Plastic
hinge
End
zone
(b) Beam element
Figure 4.1 Element Composed of Several Components
It is important to keep in mind that the components only approximate the
behavior of an actual beam. For example, the non moment-resisting
connection is modeled using a moment release, which assumes that the
bending stiffness of the connection is zero. In reality, the connection is likely
to have significant bending stiffness. In some cases it might be appropriate,
or necessary, to account for this stiffness, and to model the connection using
a semi-rigid connection component. Also, the moment-resisting connection
is assiimed to be ductile and strong, so that a plastic hinge forms in the
beam. A more complex model may be needed if the connection is weaker
than the beam.
4.1.2
Modeling Goals for Components
The goal of most structural analyses is to obtain useful information for
design. An analysis model does not have to provide an exact simulation of
the behavior of the actual structure.
Since the goal is design, not exact simulation, it is not necessary to capture
every detail of the behavior of every component. An analysis model needs
to capture only those aspects of behavior that are important for design.
Component Force-Deformation Relationships 91
Unfortunately, real structures can behave in complex ways, and there may
be no accepted rules for deciding what constitutes an important aspect of
behavior. It is important, therefore, for an engineer to have a sound physical
understanding of behavior, and also a sound understanding of what is
important for design. This provides a backgro).l11d for determining what
modes of behavior must be modeled, deciding what components can be
used to capture the behavior, and choosing useful demand-capacity measures
for assessing performance.
,
This chapter considers how component properties might be specified for
practical analysis. These properties indude the following.
(1)
The component force-deformation relationship. For an elastic component
this usually requires only the· stiffness. For an inelastic component it
can require much more information.
(2)
Demand-capacity (D/C) measures for assessing performance. These
include measures for both strength and deformation.
Keep in mind that a computer program analyzes only the model, not the
actual structure. If the analysis results are to be useful, the model must be a
reasonable representation of the actual structure, and the D/C measures
must be reasonably related to actual performance.
4.2 Component Force-Deformation Relationships
4.2.1
A Common F-D Relationship
For many components the force-deformation (F-D) relationship has the form
shown in Figure 4.2.
Force (F)
Strain
Hardening
Ultimate
strength
Ductile limH
Complete failure
Hysteresis loop
Figure 4.2 A Common F-D Relationship
92
Chapter 4 Component Behavior - Uniaxial F-D Relationships
·The key parts of this relationship were identified in Chapter 1. These are
repeated below, with additional comments. The relationship for monotonic
deformation (progressively increasing deformation, with no cycling) is
considered first. The complications for cyclic deformation are considered
lat~r.
4.2.2
F-D Relationship for Monotonic Deformation
The overall shape of the F-D relationship is similar for components of a
variety of types. The details, however, can be different, and there can be
substantial uncertainty about these details. This section reviews the key
parts of the relationship (Figure 4.2), and identifies some differences in the
details (Figure 4.3).
Linear to yield, with a
well-defined yield point
F
F
LW~-- Nonlinear, with less
Hardening stiffness
can be different for
different components.
well-defined yield point
On unloading and reloading,
may be more nearly linear.
'-------+D
~-------.o
(a) Behavior before yield
F
(b) Strain hardening
F
Yield plateau may be
essentially flat, or strength
may vary continuously.
Ductile limit may or
may not be well defined.
D
D
(d) Ductile limit
(c) Yield plateau
F
F
Behavior can be
brlttie, with little
or no ductility.
Strength loss may
be gradual or rapid.
D
D
(e) Strength loss
(f) Brittle behavior
Figure 4.3 Some Variations in Behavior
Component Force-Deformation Relationships 93
The parts of the relationship in Figure 4.2 are as follows.
(1)
Initial behavior that is more or less linear
For a steel component that is essentially ela5tic until it reaches its yield
point, the initial behavior may be linear over a substantial force range. ·
For a reinforced concrete component that cracks prior to yield, the
initial behavior may be substantially nonlinear, This is illustrated in
Figure 43(a). However, once a concrete comportent has cracked, if it is
unloaded and reloaded, its effective initial behavior is closer to linear.
Some connection components, for 'example those where bolt slip can
occur, may-also be substantially nonlinear before yield.
(2)
First yield, at a point that may or may not be well defined
A steel bar with low residual stresses will tend to have a well-defined
yield force in tension (it may buckle in compression). A steel beam
with a thin flange and low residual stresses will have a fairly welldefined yield moment. However, if the flanges are thick or there are
substantial residual stresses, the yield moment is likely to be less well
defined. Reinforced concrete beams may or may not have well-defined
yield moments, depending on the distribution of reinforcement in the
cross section. Connections may or may not have well-defined yield
strengths, depending on the connection details.
(3)
A region of increasing strength (strain hardening)
If a component yields gradually, there will often be a progressive
reduction in the hardening stiffness, as in Figure 4.2. Reinforced
concrete components tend to have this type of behavior.Ha component
has a sharp yield point, it may strain harden, possibly with a roughly
constant hardening stiffness over a substantial range, as shown in
Figure 4.3(b). The amount of strain hardening can depend on several
things, and may be uncertain.
(4)
Ultimate strength
A component will almost always have an ultimate strength that can
not be exceeded. Usually this strength will be reached after a modest
amount of inelastic deformation and strain hardening. After a
component reaches its ultimate strength, it may maintain that strength
over a significant deform'ation range, so that the F-D relationship has
a "strength plateau" where the strength is essentially constant. This is
94
.Chapter 4 Component Behavior- Uniaxial F-D Relationships
illustrated in Figure 4.3(c). Some components may continue to gain
strength until the deformation is large, as suggested by Figure 4.3(b).
(5)
Ductile limit, which may or may not be well defined
Most components have limited ductility, and sooner or later they will
lose strength. The cause of the strength loss might be flange buckling
in a steel beam, concrete crushing in a concrete beam, bolt or weld
failure in a connection, etc. The point at which the amount and/ or
rate of strength loss becomes.significant is termed the "ductile limit" in
this book. There is no standard terminology. Sometimes this point is
called the "deformation capacity" of the component. In this book, a
deformation capacity is a deformation that can be compared with a
corresponding deformation demand to calculate ·a DIC ratio. This
may or may not be close to the ductile limit.
In some cases the ductile limit is well defined, with rapid strength
loss. In other cases the strength loss is gradual, with no well-defined
limit. This is illustrated in Figures 4.3(d). Sometimes the ductile limit
is defined as the deformation at which the strength reduces to 80% of
the ultimate strength. There is almost always a lot of uncertainty
about the deformation at the ductile limit.
. The length of the strength plateau can vary greatly for different
components. An extremely brittle component may reach its ductile
limit with no strain hardening range and no strength plateau, as
shown in Figure 4.3(f). A ductile component may continue to strain
harden until it reaches its ductile limit, again with no strength plateau.
(6)
Strength Loss
Strength loss is usually progressive, over a substantial deformation
range, but it may be rapid, as in Figure 4.3(e). For a brittle component
the strength loss can be can be sudden.
(7)
Residual strength
· Some components can redistribute stresses internally after losing
strength, and stabilize at a "residual" strength. For example, if a steel
beam loses strength because the compression flange buckles, the
residual bending moment might be the moment strength ignoring the
buckled part of the· flange. The residual strength might be anywhere
from about 80% of the ultimate strength to zero, depending on the
component.
Component Force-Deformation Relationships 95
(8)
Complete failure
For very large deformations, even a very ductile component may lose
essentially all of its strength. If a component reaches its residual
strength, it is often assumed that it maintains that strength until some
"ultimate" deformation is reached, then rapidly loses all strength. This
might happen when a steel component fractures, or a concrete
component cracks and crushes so badly that i~,has negligible strength.
If the goal of an analysis is to predict when a structure will collapse, it is
probably necessary to include all aspects of component behavior in the
analysis model, including complete strength loss. It is important to recognize
that any such model is likely to be very approximate for large deformations.
The more deformation that is imposed on a component, the less certain its
behavior becomes. This is especially true when there is cyclic loading. For
any real component, such as a steel or concrete beam, column or wall, the
behavior beyond the ductile limit will be known only very crudely. There is
a danger that an analysis which attempts to predict complete collapse can be
of such uncertain accuracy that it may be little more than an academic
exercise.
This does not mean that it is never useful to consider collapse as a limit
state. It does, however, mean that it must be done with great care and a lot
of skepticism.
4.2.3
Complications for Cyclic Deformation
When a component is subjected to cyclic loading (or, more precisely, cyclic
deformation), the F-D relationship for an inelastic component follows a
hysteresis loop. Figure 4.2 shows two aspects of this, as follows.
(1)
Cyclic unloading-reloading, with a hysteresis loop
Figure 4.4(a) shows a "full" loop, where the shape of the loop is the
same as the shape of the basic F-D relationship. Figure 4.4(b) shows a
loop that has "stiffness degradation". In this loop the unloadingreloading stiffness is smaller than the stiffness in the basic F~D
relationship. The area of the loop· is a measure of· the amount of
inelastic energy that is dissipated \inder cyclic deformation. The loop
in Figure 4.4(b) has a smaller area than that in Figure 4.4(a), so
stiffness degradation also leads to energy degradation. Figure 4.4(c)
shows a "pinched" loop.
96
Chapter 4 Component Behavior - Uniaxial F-0 Relationships
F.·@·.,wF··. ·
.
D
.
I
D
/,
D
'
~
(a) Full loop
(b) Stiffness degraded
In the first cycle there may
be littte or no degradation.
D
(d) Cyclic degradation
·.r::_
F
Deformation reverses
after strength loss.
~~~yormay
not be reduced.
(f) Strength loss interaction
(c) Pinched
With cyclic deformation,
the strength may increase.
D
(e) Cyclic strength increase
F
D
· After cyclic loading, effective '
relationship may have smaller
strength and smaller ductility.
(g) Ductilify degradation
Figure 4.4 Some Complications for Cyclic Deformation ·
(2)
Cyclic degradation, where the stiffness, strength and/or ductility
progressively deteriorate
There can be strength degradation as well as stiffness degradation:, as
shown in Figure 4.2. Degradation of b.oth types may develop progressively, with the amount of degradation increasing as the number of
deformation cycles increases, as shown in Figure 4.4(d). Some
components increase in strength under cyclic deformation, as shown
in Figure 4.4(e).
Degrad~tion may be affected by strength loss, as indicated in Figure
4.4{f). H a component is deformed beyond its ductile limit in one
direction, and the deformation is then reversed, the strength in the
opposite direction may or may not be reduced.
·
Component Force-Deformation Relationships 97
For example, suppose that a reinforced concrete beam loses strength
because of crushing in compression at the bottom of a beam, with
cracking at the top. When the bending deformation is reversed, the
cracks close at the top of the beam, the concrete goes into compression, but it does not necessarily crush. Hence, crushing in one direction
may have little effect on the bending strength in the other direction.
(There may be some effect, because the concrete has been cracked and
may not be as strong in compression when the cracks close. Also, the
top reinforcement has probably been yielded in tension, and it may
buckle when subjected to compression. This is just one indication of
the complex behavior that can occur under cyclic deformation.) On
the other hand, if a concrete wall is deformed inelastically in shear
and loses strength in one direction, it may well have lost strength in
the opposite direction.
The effect of cyclic deformation is often as shown in Figure 4.4(g). This
figure shows the F-D relationship for a component that is loaded only
monotonically, and also the effective relationship after the component has
been subjected to several deformation cycles. Cyclic deformation can reduce
the strength of the component and also its ductility. The effect will usually
be progressive, with the amount of degradation increasing as the number of
cycles increases
In this book, degradation of any kind (stiffness, energy, strength, ductility,
etc.) that is caused by cyclic deformation is termed "cyclic degradation".
4.2.4
Elastic and Plastic Deformations
The F-D relationship for a component will often have an initial linear region
where the behavior is elastic. When a component is loaded beyond the
elastic range, its deformation is usually regarded as partly elastic and partly
"plastic", as shown in Figure 4.5.
The F-D relationship in Figure 4.5 is initially linear, and has a fairly welldefined yield point. The elastic deformation is the current force divided by
the initial stiffness. The plastic deformation is the total deformation minus
the elastic deformation. The figure also shows the "post-yield" deformation,
which is somewhat larger than the plastic deformation.
The term "plastic" is correct when applied to ductile yielding in metals.
Other causes of nonlinear behavior, such as cracking and friction, are not
strictly plastic. However, the term "plastic" is convenient, and if is commonly
used regardless of the cause of the nonlinear behavior.
98
Chapter 4 Component Behavior - Uniaxial F-D Relationships
F
I
Plastic deformation
Initial stiffness
Post-yield deformation
.._--1-+---1---.
Yield deformation
!
J
D
l.. Total deformation
Elastic deformation
Figure 4.5 ·Elastic and Plastic Deformations
Figure 4.5 shows the deformations for monotonically increasing deformation. Figure 4.6 shows an example with cyclic deformation..
~
F
Ao,
E
I
C1
I
/
D
r-.. .r--·I
Max. positive plastic
deformation= A-A 0
A
c
Accumulated positive
plastic deformation
(A- A)o + (C - Ch + (E - Eh
=
Accumulated negative
plastic deformation =
(B - B~ + (D - D~
I
'D
0
Max. negative plastic
• deformation= D ~ D0
Figure 4.6 Maximum and Accumulated Plastic Deformations
In this example the deformation is cycled along the path 0-A-B-C-D-E. The
maximum positive and negative plastic deformations are measured from
the original elastic line, as shown. The accumulated plastic deformations are
"··!-..""'•• .
Component Force-Deformation Relationships 99
calculated for each plastic excursion. The positive and negative accumulated
deformations might be added to obtain a total accumulated deformati<?n.
To assess the performance of a component that is allowed to yield, plastic
deformation is an obvious demand-capacity measure. There are, however,
some complications, as follows.
(1)
H the F·D relationship strain hardens after yield, there is a difference
between the plastic deformation and the post-yield deformation, as
shown in Figure 4.5. There can be confusion over which should be
used. Fortunately, it often does not make much difference. For
example, consider the case where the strength at first yield is 75% of
the strength after strain hardening, and where the plastic deformation
is 3 times the elastic deformatiOJ\. In this case the post-yield
. deformation is 3.25 times larger than the elastic deformation, or 8%
larger than the plastic def()rmation. Given the many uncertainties and
approximations, an 8% difference is not large. Nevertheless, if is
important to be consistent, The plastic deformation should be used,
not the post-yield deformation.
(2)
The F-D relationship may be significantly nonlinear even for small F
values, and hence may not have a well-defined elastic range or a welldefined initial stiffness. This is often the case for reinforced concrete
components. For elastic analysis it is necessary to choose an effective
initial stiffness, and it may not be obvious how to do this.
(3)
H plastic deformation is used as a demand-capacity measure, this
measure could be the maximum plastic deformation, the accumulated
plastic deformation, or some combination of the two. The accumulated deformation is important because cyclic deformation is usually
more damaging than monotonically increasing_ deformation. For
simplicity, however, a common practice is to use the maximum
deformation for the deformation demand, and to account for cycling
indirectly by accounting for it in the deformation capacity (if more
cycling is expected, the deformation capacity is reduced). This
procedure is used in ASCE 41.
4.2.5
Ductility Ratio
The amount of plastic deformation is often expressed as a "ductility ratio",µ.
This has the advantage of being dimensionless. The definition of µ , and its ·
relationship to the plastic deformation, are shown in Figure 4.7.
Chapter 4 Component Behavior - Uniaxial F-D Relationships
100
F
Current force
/
In temlS of plastic deformation :
-+--~,.,._..
Yield force
Ductility ratio, µ
.
Plastic deformation, D P
=.E.. = 1 +
De
Ee
De
Dp= D0 (µ-1)
Post-yield deformation, D PY
~-+-i----<--+
Yield deformation, D Y
..J
!
D
In terms of post-yield deformation :
l.. Total deformation, D
Ductility ratio, µ = .E_ = 1 + EeY
Dy .
Dy
Elastic deformaiion, D e
Dpy= Dy(µ -1)
Figure 4.7 Ductility Ratio
Figure 4.7 also shows an alternative definition, in terms of post-yield
deformation. As noted earlier, the difference between plastic and post-yield
deformation may be small. However, this difference can have a substantial
effect on the ductility ratio.
For example, consider again the case where the strength at first yield is 75%
of the current strength, and where the plastic deformation is 3 times the
elastic deformation. In this case the ductility ratio expressed as (total
deformation)/(elastic deformation) is (1+3)/(1) =4, and expressed as (total
deformation)/(deformation at first yield) it is (1 + 3)/(0.75) = 5.33. This is.
33% larger. It is important, therefore, to be consistent in defining ductility
ratios. The ratio (total deformation)/(elastic deformation) should be used.
4.2.6
Rigid-Plastic Hinges .
Some components are assmii.ed to be initially rigid, with no deformation
before yield. The F-D relationship for such a component is shown in Figure
4.8(a).
(~
First yield
lnitiallyrlgid
i-;:=:!
(a) F-0 relationship
(b) Moment and shear hinges
Figure 4.8 Rigid-Plastic Hinges
Component Force-Deformation Relationships 101
Such components have zero length, and. are referred to as "rigid-plastic
hinges" or simply "plastic hinges". The most common example is a moment
hinge in a beam, with a rigid-plastic moment-rotation relationship. Shear,
axial and torsional ''hinges" can also be used.
Since the component is initially rigid there is no elastic deformation, and for
monotonically increasing deformation all of the deformati0n is plastic. Also,
there is no difference between plastic and post-yield d¢ormatiort.
Figure 4.8(b) shows plastic hinges for bending moment and for shear. In an
actual structural member, plastic deformation must be diStnbuted over a
finite length, often referred to as a "plastic zone". For bending, the plastic
zones are often short, and it can be physically reasonable. to lump the plastic
deformation into a zero-length hinge. Plastic shear deformations are likely
to be distributed over longer lengths, and it may be less reasonable to lump
the plastic deformation into a zero length hinge, with a sudden shear
displacement.
4.2.7
Other Nonlinear F-D Relationships
The F-D relationship considered in this section applies to components of a
variety of types. Some components, however, can have very different F-D
relationships. Some of these are as follows .. · . ·
(1)
The steel material used for concrete reinforcing bars may have a
stress-strain relationship as shown in Figure 4.9(a). For the first half
cycle of inelastic behavior, this relationship has· a horizontal "yield
plateau", followed by strain hardening. For cyclic strain, however,
there is no yield plateau.
(2)
Concrete crushes in compression and cracks in ·terlsiort; The stress-strain relationship for plain concrete in a state of uniaxial stress is
essentially as shown in Figure 4.9(b). A major complication is that
concrete has complex behavior 'under multi-axial stress. One aspect of
this is that when the concrete in. a beam or column is confined by
transverse hoops or ties, the longitudinal strength and ductility of the
concrete can be dramatically increased compared with tinconfined
concrete. A Second aspect.is that when·a reinforced concrete beam,
column or wall has combined longitudinal and shear stresses (or combined axial forces, bending moments and shear forces), its behavior
can be extraordinaiil.y complex. ·
·
/
102
Chapter 4 Component Behavior - Uniaxial F-D Relationships
Stress
tfk
(a) Reinforcing steel
+
j
ColJ1lf"SSion
Stress
~load-reload
1
S~n
(b) Concrete
Bond Stress·
)
+
+
I
+
(c) Buckling strut
(d) Bond slip
Figure 4.9 Some Other F-D Relationships
(3)
Slender members that can resist little or no compression are som~
times used as bracing members in steel structures. These can often be
modeled as "tension only" components, with a relationship between
axial force and axial deformation that is roughly the reverse of
concrete in compression.
(4)
A compression strut or bracing member that is not so slender may
resist substantial compression forces. A member of this type can have
a relationship between axial force and axial deformation as shown in
Figure 4.9(c). The figure shows the state of the member at different
points on the hysteresis loop. A solid dot indicates a plastic hinge. For
this F-D relationship, nonlinear behavior. is caused by geometric
nonlinearity as well as material nonlinearity.
(5)
In reinforced concrete, bond slip between concrete and steel can have
a substantial effect on the behavior of components such as beam-tocolumn connections. The relationship between bond stress (or
longitudinal shear force per unit length of a reinforcing bar) and the
corresponding bond slip has roughly the form shown in Figure 4.9(d).
Other examples of special F-D relationships include seismic isolation
devices, components made of "shape memory" alloys, and components for
What Type of F-D Relationship is Needed? 103
modeling contact forces when there is impact between two parts of a
structure.
4.2.8 Summary for this Section
This section emphasizes that structural components often have complex
behavior and uncertain properties. ''Exact" modeling of real structural
components is literally impossible.
,
However, as this book continually emphasizes, , the goal is design, not
analysis. Analysis is used to get useful· information for design, not for
"exact" Simulation. This can be done using approximate analysis models,
considering only the most important modes of behavior.
The next sections consider models for practical analysis. Elastic analysis is
considered fust, followed by inelastic analysis.
4.3 What Type of F-D Relationship is Needed?
The F-D relationship for a component can have many properties. It is not
n&essary to specify all of these properties for all analyses.
There are three broad categories of analysis, requiring progressively more
complex (and more uncertain) component properties. These are as follows.
(1)
Linear elastic.~ assumes (and requires) a constant stiffness. This is
the most common type of analysis~ It is used to calculate strength
demands, for comparison with strength-capacities.
(2)
Nonlinear ,non-cyclic (or nonlinear monotonic), _which is usually
inelastic but could be nonlinear elastic. This requires a nonlinear F-D
relationship but does not require a hysteresis loop. Within th1s
category there are a number of possible applications, including (a)
strength capacity calculation for a structure or structural assemblage,
(b) static push-over analysis for performance evaluation under
earthqUake loads, and (c) "progressive" or "disproportionate" collapse
analysis of structures damaged by explosions. This last type can use
either static· or dynamic analysis. In a dynamic analysis the F-D
relationship mtist account for unloading and re-loading, but usually
, there is no inelastic cyclic deformation.
104
(3)
Chapter4 Component Behavior - Uniaxial F-D Rel_ationships
Nonlinear cyclic. This is usually inelastic, and requires hysteresis
loops. The most common application is dynamic analysis for earth. quake loads.
For inelastic analyses, nonlinear F-D relationships are needed for the
components. However, it is not necessary to include all of the features
considered earlier. In particular, if the maximum deformation for any
component is expected to be below the ductile limit, the F-D relationship
does not have to include strength loss, which is a major simplification.
This is also the case if the deformation capacity of a component is smaller
than the ductile limit, as it usually will be. The F-D relationship can ignore
strength loss, and the deformation DIC ratio can be checked. If this ratio
exceeds 1.0, the performance requirement is not met and the design must be
revised. This is similar to the use of elastic analysis to check if a structure
yields. In that case a strength capacity is specified, and the structure yields if
this capacity is exceeded. It is not necessary to account for inelastic behavior
in the analysis model.
When specifying the F-D relationship for a component, always consider the
purpose of the analysis. As this book constantly emphasizes, analysis is
simply a tool - the goal is design, not analysis. The best analysis model is
the simplest one that will provide the needed Wormation for design.
4.4 Stiffness for Elastic Analysis
4.4.1
Bending Stiffness for Beams, Columns and Walls
If the force on a component does not exceed its yield force, the usual
assumption for analysis is that the behavior is essentially linear and elastic.
The only property that is needed for a component is its stiffness.
Beams, columns an!! walls have axial, bending and shear stiffnesses. For
beams and columns the bending stiffness is usually the most important. For
walls, and als{) for deep beams, the shear stiffness can be important.
Consider bending stiffness first.
The bending stiffness is EI, where E and I have their usual meanings. If the
bending behavior is linear before yield, it is usually not difficult to estimate
a reasonable value for the stiffness. This is usually assumed to be the case
for steel columns and for non-composite steel beams. The cross section I
.,
··~·
Stiffness for Elastic Analysis 105
value in such cases is well defined. It is common practice to assume that E is
the elastic modulus for steel (although see Chapter 6 for exceptions).
If the behavior is substantially nonlinear before yield, or if a component
does not have a well-defined yield point, it can be more difficult to estimate
an effective EI. This can be the case for reinforced concrete members, where
cracking of the concrete can substantially affect the bending stiffness. For
bending of reinforced concrete beams, a common practice is to calculate I for
the gross concrete section, ignoring the reinforcement, and to use an
effective E that is about 35% of the compression modulus for the concrete.
Columns tend to be relatively stiffer than beams, because compression in a
column suppresses cracking. A common practice is to use an effective E that
is 70% of the concrete modulus. Since concrete may creep under long-term
loads, it is also common to modify E to account for load duration.
If I for a beam is based only on the gross concrete section, and if E is the
concrete modulus, then EI does not depend on the amount of steel reinforcement in the beam. This is not correct for a reinforced concrete beam, as
shown in Figure 4.10.
-t~ti=Tension force
Moment
,(secant El
LJ atyield
Stiffness changes
as concrete cracks
Concrete cracks
to neutral axis
(a) Short length of beam
Curvature
(b) Moment-curvature relationship
Figure 4.10 Bending Stiffness for RC Beam
Figure 4.lO(a) shows a short length of beam in bending. The bending
moment is resisted by a tension-compression couple. Figure 4.lO(b) shows
likely moment-curvature relationship. When the tension reinforcement yields
there is a well-defined yield moment. The effective (or secant) EI at this
point might be a reasonable choice for the EI value in an elastic analysis.
a
_The yield moment is the yield force in the tension steel multiplied by the
lever arm of the tension-compression couple. The yield curvature is the
yield strain of the steel divided by the distance to the neutral axis, which is
somewhat smaller than the tension-compression lever arm. It follows that
the yield moment divided by the yield curvature is essentially proportional
106
Chapter 4 Component Behavior - Uniaxial F-D Relationships
to the yield force in the tension steel, and hence to the tension steel area.
Hence, the bending stiffness is not independent of the amount of reinforcement.
There are other complications and uncertainties, some of which are as
follows.
(1)
(2)
In a typical concrete beam only a part of the beam length will be fully
stressed. Most of the beam will have less cracking and be relatively
stiffer.
A reinforced concrete column is more complicated, because of the
axial force. There may also be significant axial forces in beams.
(3)
The effective El can also be difficult to estimate for a steel beam with a
composite concrete slab. In particular,_the effective El is larger when
the slab is in compression than when it is in tension.
(4)
Even for a steei beam there can be nonlinear behavior before the yield
moment is reached, because of residual stresses from the manufacturing process.
Hence, for elastic analysis the values that are assumed for EI are likely to be
very approximate. Some consequences of this are as follows.
(1)
To ensure that different engineers make similar assumptions, and
hence get consistent (if not necessarily accurate) results, guidelines are
needed. Design codes usually provide some guidelines.
(2)
If a structure is statically determinate, changes in the element stiffnesses do not change the calculated element forces (at least for a small
displacements analysis, ignoring P-~ effects). Changes in stiffness do,
however, affect the calculated deflections. This can have an effect on
performance assessment for serviceability. It is important not to
overestimate the stiffness.
(3)
If a structure is statically indeterminate, changes in the element
stiffnesses can change the calculated element forces. If the stiffness~
change in the same proportion for all elements, the calculated element
forces d~ not change, and only the deflections are affected. However,
if the relative element stiffnesses change (for example if the assumed
beam stiffnesses in a frame are reduced and the assumed column
stiffnesses stay the same), there are changes in the element forces.
Stiffness for Elastic Analysis 107
These changes may be small or they may be substantial, depending on
the type of structure.
(4)
Changes in stiffness change the P-~ effect, since this effect depends on
the deflections. This is considered in detail in Chapter 6,
(5)
Changes in stiffness can change the calculated periods of vibration for
a structure, which in turn can affect the earthquake loads for design of
the structure.
4.4.2
Shear Deformation in Reinforced Concrete
In a typical beam, shear deformations have a relativeiy small effect on the
beam stiffness, so even substantial errors in estimating the. shear stiffness
have only a small effect on the total stiffness. In a deep beam or a shear wall,
however, shear deformations can be substantial, and approxiIDations in
shear stiffness are more important.
For a linear elastic material, the shear modulus, G, is given by the well-known
equation G = O.SE/(1-tl~, where E = Young's modulus and v = Poisson's
ratio. This equation is often used for reinforced concrete, even though
concrete is not a linear elastic material. As with bending, the shear stiffness
of reinforced concrete is affected by cracking. When cracking occurs, the
effective shear modulus of concrete is far smaller than that given by the
above equation.
In a reinforced concrete wall, the shear is resisted partly by the concrete and
partly by the shear reinforcement (which is usually vertical for a beam and
horizontal for a wall). The yield force in shear is reached when the shear
reinforcement yields. It can be shown (using Mohr's circle for strain) that
when the shear reinforcement yields, the shear strain is roughly equal to the
yield strain of the reinforcement, or about 0.002. The effective shear modulus
at yield is thus the shear strength divided by about 0.002. This is a much
smaller value than 0.SEI (1 + J1.
It follows that the usual equation for the effective· shear modulus greatly
overestimates the shear stiffness. This is not important for a slender beam,
where shear deformations are small. It can be important, however, for a
wall, where shear deformations can be substantial.
108
4.4.3
Chapter 4 Component Behavior- Uniaxial F-D Relationships
C:onnections
Some structural connections are flexible, and for elastic analysis they can be
assumed to have zero stiffness. An example is a non moment-resisting
beam-to-column connection, which can often be modeled using a simple
hinge or a bending moment release. At the opposite extreme, some connections
are very stiff, and their deformations can be ignored. An example is a
welded moment-resisting beam-to-column connection.
In between, a connection may have a finite stiffness, and it may be
important to estimate this stiffness and account for it in the analysis model.
A particular case is the "panel zone" of a beam-to-column connection in a
steel frame. There can be substantial shear stresses in the panel zone, with
corresponding shear strains. Shear deformation in the panel zones can have
a significant effect on the deflections of a frame.
4.4.4 Summary for this Section
To set up a model for elastic analysis, the only property that is needed for a
component is its stiffness. In textbooks on structural analysis and structural
mechanics, it is usually implied that stiffness properties such as EA for axial
deformation, EI for bending and GA' for shear can be determined with
precision, and hence that elastic analysis is accurate. Often, however, the
stiffness properties are highly uncertam, and for this reason alone elastic
analysis may be far from accurate.
Once again, this emphasizes that structural analysis is merely a tool, and
often it is imprecise. Engineers have to live with the uncertainty.
4.5 F-D Relationships for Inelastic Analysis
4.5.1
Overview
If the force on a component exceeds the yield force, the behavior is inelastic,
and hence nonlinear. In earthquake resistant design, inelastic behavior is
often permitted for large, infrequent earthquakes. For other types of load,
inelastic behavior may not be explicitly permitted, but it is likely to be
present if the loads on the structure exceed the design loads.
It is not essential to consider inelastic behavior directly in an analysis model,
even for earthquake resistant design. The downside of inelastic analysis is
F-D Relationships for Inelastic Analysis 109
that an inelastic analysis model requires not just component stiffnesses but
also properties such as strength, strain hardening stiffness, ductile limit, and
hysteresis loop shape. Also, the analysis is nonlinear and requires much
more computer time than a linear elastic analysis. The upside is that
inelastic analysis can give better information for design than elastic analysis.
Elastic analysis may even give misleading information.
As has been emphasized in earlier sections, inelastic behavior can be
complex and uncertain, especially for large deformations and cyclic loads.
Setting up an inelastic analysis model_ does not involve simply taking an
elastic model and replacing the elastic material properties with inelastic
ones. It is almost always a much more difficult process. Setting up a useful
inelastic model requires an understanding of inelastic behavior _and how to
model it, and also an understanding- of· the design process and what
information can realistically be expected from an analysis.
This section considers, in general terms, the practical aspectS of modeling
inelastic components.
4.5.2 Amount of Inelastic Behavior
The amount of inelastic behavior that is allowed in· a structure can depend
-on the required level of performance; For earthquake resistailt design, three
commonly used performance levels are as follows.
(1)
Immediate Occupancy (IO). Only a small amount of inelastic behavior ·
is allowed, as indicated in Figure 4.11. Elastic analysis should be
sufficient.
F
LS· CP
---------•D
Figure 4.11 Allowable Inelastic. Deformation for Different Performance_ levels
(2)
Life Safety (IS). A larger amount of inelastic behavior is allowed.
Elastic analysis
be sufficient, but inelastic analysis can be justified.
An inelastic model should consider the yield and ultimate strengths of·
the structural components, and make reasonable assumptions about
may
-----~---------------------
--------···------------·--------------------------------------~--------
11 O Chapter 4 Component Behavior - Uniaxial F-D Relationships
strain hardening. The hysteresis loops can be expected to be fairly
simple at this level of deformation, with modest degradation in
stiffness and strength. It is not necessary to include strength loss in the
model, although the deformation capacities for inelastic components will
.depend on their ductile limits.
(3)
Collapse Prevention (CP). Substantial inelastic behavior is allowed,
close to the ductile limit, and possibly exceeding it for some components.
Inelastic analysis may be needed. The inelastic model should consider
yield and ultimate strengths, make reasonable assumptions about
strain hardening, include the ductile limit, and possibly account for
strength loss in some components. The hysteresis loops may be
complex atthis level of deformation, with substantial degradation. It
is a more difficult task to set up an inelastic model.
Figure 4.11 also shows the level of deformation that might be associated
with actual collapse. Although it is not standard practice, the probability of
collapse can be estimated, using "incremental dynamic analysis" (see Section
1.8.3). Collapse will usually involve inelastic deformations that are much
larger than those permitted for the Collapse Prevention performance level,
and inelastic analysis is definitely needed. Inelastic modeling becomes
increasingly complex as larger deformations must be considered, and the
analysis results become increasingly unreliable.
4.5.3
Practical F-D Relationship - Generic Form
The F-D relationships for most components are so complex that it is literally
impossible to capture all aspects of their behavior. The -best that can be
expected, and all that is needed, is to capture the aspects that are important
for design;
Ignore cyclic deformation for the time being. Figure 4.12 shows the important
features of a typical F-D relationship.
Figure 4.12(a) essentially repeats Figure 4.2. The key parts of this relationship were described in Section 4.2.2. Figure 4.12(b) divides the relationship
into a number of linear segments. This relationship has the following
properties.
\
-~'
F-D Relationships for Inelastic Analysis 111
F
Ullimale
strength
Ductile limit
Initially
Hnear
Complete failure
(a) Main features
ifll
DY
OU
fu.
Dl
DR DX
D
D
(b) Multi-linear approximation
Figure4.12 Typical F;-D Relationship and Multiclinear Approximation
Initial linear segment
This requires an initial stiffness. This is usually the stiffness that would be
used for elastic analysis. For a component 8uch as a rigid-plastic hinge, the
initial stiffness is infinite.
·
·:;",.._
Yield point
It is usually possible to make a reasonable estimate of the yield force, even if
the F-D relationship for the actual component does not have a well=<lefined
yield point. The yield deformation is the yield force divided by the initial
stiffness. For a rigid-plastic hinge the yielddeformation is zero.
Ultimate strength
The ultimate strength is usually calculated using the strength formulas in
design. codes. These forinulas may be based on "nominal" material
strengths, and the resultirig nominal component· strength may· be reduced
by •a capacity' reduction factor for strength-based design. For inelastic
analysis it is more usual to use "expected" material strengths, and capacity
reduction factors may not be applied.
Strain hardening stiffness
It may b.e possible to estimate the strain hardening stiffness, iri which case
the deformation at the start of the strength plateau follows from the yield
/
112
Chapter 4 Component Behavior - Uniaxial F-D Relationships
and ultimate strengths and the strain hardening stiffness. Alternatively, it
may be possible to estimate the deformation when the ultimate strength is
reached, in which case the strain hardening stiffness follows. If the yield and
ultimate strengths are the same, this is the special case of elastic-perfectlyplastic behavior.
Ductile limit
The ductile limit may be uncertain but a reasonable estimate is needed, to
model strength loss if the ductile limit. is exceeded, and also to establish
deformation capacities that are based on ductility.
Strength loss and residual strength
If deformations beyond the ductile limit are allowed, the rate and amount of
strength loss must be estimated. Some structural components are brittle and
lose strength rapidly. For such components it may be unwise to allow
deformation beyond the ductile limit. Other components may lose strength
gradually, and it may be reasonable to allow some deformation beyond the
ductile limit.
Complete failure
A maximum deformation, at which the component strength reduces to zero,
more-or-less suddenly, may be specified. This is likely to be little more than
a guess.
A multi-linear relationship is only an approximation of the actual relationship. In particular, in the· multi-linear relationship the stiffness changes
suddenly, whereas in the curvilinear relationship it changes continuously.
However, a multi-linear relationship is probably the best that can be
expected for practical inelastic analysis. This relationship captures the
important aspects of inelastic behavior, its shape can be varied to account
for different amounts of strain hardening and ductility, and it has only a few
key points that need to be estimated. Compared .to a curvilinear relationship, a multi-linear relationship can also be more efficient computationally,
since it has only a few discrete stiffnesses rather than a continuously varying
stiffness.
4.5.4
F-D Relationships in ASCE 41
ASCE 41 is a standard for the seismic rehabilitation (retrofit) of existing
buildings. It may also be used in the design of new buildings. It was pub-
'-.-:.......
-'\.,
F-D Relationships for Inelastic Analysis 113
lished in 2006, following earlier publications and "pre-standards", including
ATC 40 (19%), FEMA 273 (1997) and FEMA 356 (2000). Future refinements
and modifications can be expected.
Among many other things, ASCE 41 provides· modeling guidelines for
inelastic analysis and performance assessment. A major strength of ASCE 41
is thatit recommends deformation capacities for a wide range of components,
far the Immediate Occupancy, Life Safety and Collap,;e Prevention performance levels. (The term "deformation capacity" is used in this book. In ASCE
41 the term.is "acceptance criterion for nonlinear procedures".) ASCE 41 also
recommends "modeling parameters", which essentially define the shape of
the F-D relationship. ASCE 41 is not so strong in this area. The F-D relationShip is rather crude, and ·there are no guidelines on the modeling of
hysteresis loops.
ASCE 41 allows the F-D relationship for a structural component to be
obtained from experimental data. H this is not feasible, the relationship
shown in Figure 4.13 can be used. This is a special case ofthe mwti-linear
relationship in Figure 4.12.
F
A--~~~~~~--~~~._~_.
D
FEMA 356 assumption
for strength loss
Figure 4.13 ASCE 41 F-D Relationship
.· .
.
-
As shown in the figure, ASCE 41 uses the term: "plastic deformation" for the
deformation that occurs after yield: Strictly speaking, this is the post-yield
deformation, as noted earlier in this chapter, not the true plastic
deformation. The ASCE 41 plastic deformation is somewhat larger than the
true plastic deformation.
In FEMA 356, when Point C is reached in the FD relationship the strength is
assumed to drop suddenly to the residual strength. Most components lose
114
Chapter 4 Component Behavior - Uniaxial F-D Relationships
strength more or less gradually, so this assumption is generally not correct.
It probably originated with the assumption that nonlinear analysis would be
carried out by running a linear analysis, identifying the point at which a
component reached its "C" point, reducing the strength of that component,
running a new linear analysis, etc. This is not a practical analysis method,
and is probably never used. ASCE 41 softens this assumption but does not
specify how to calculate the rate of strength loss. This is one issue that can
be expected to be addressed in future modifications to ASCE 41.
As with any standard, there is a danger that engineers will apply ASCE 41
too slavishly. It provides excellent guidance, but should not be regarded as
a rigid specification. It is a first attempt at solving a very difficult series of
problems, and will almost certainly evolve over time.
4.5.5
Backbone Relationship
If experimental results are available, ASCE .41 allows them to be used to
construct F-D relationships. The concept is shown in Figure 4.14..
Backbone
D
Figure 4.14 Backbone Relationship Based On Experimental Hysteresis Loops
The details of how the relationship is constructed have been questioned by
researchers and are not considered here. The concept, however, is very
useful for practical analysis.
The resulting F-D relationship is tenned a ''backbone" relationship. It can be
uses to construct a multi-linear approximation, as in Figure 4.12 or 4.13. As
shown in the next section, it can also be used to "anchor" hysteresis loops.
...
'
·~.
Hysteresis Loops for Inelastic Analysis 115
A key issue is that many components degrade progressively as the number
of deformation cycles is increased, and hence the backbone relationship
depends on the amount of cycling. In particular, a backbone relationship
that accounts for cyclic deformation can be substantially different from the
monotonic relationship with no cycling. This is considered in the next
section.
4.6 Hysteresis Loops for Inelastic Analysis
4.6.1
Loop Anchored to Backbone Relationship
Over the decades, researchers have proposed dozens, and possibly hundreds,
of different models for hysteresis loops. Some are relatively simple~ and
some follow complex geometric rules.
In the author's opinion the most practical method (indeed, the only practical
method) is to begin with a ''backbone" force-deformation relationship, and
to anchor the hysteresis loop to that relationship. This is illustrated in Figure
4.15.
Backbone relationship
D
(a) Hysteresis loops are anchored to backbone relationship
(b) Different types of stiffness degradation
Figure 4.1 S Hysteresis Loop Anchored to a Backbone Relationship.
116
Chapter4 Component Behavior-Uniaxial F-D Relationships
Given a backbone relationship, rules ~re needed to define the shape of the
hysteresis loop. For now, ignore cyclic degradation in strength and ductility
(see the next section), and consider only the shape of the loop~
Rules are needed to define the shape of the loop, accounting for the
following.
(1)
The amount of stiffness degradation. The amount of degradation will
vary with the type of component. For any given component, the
amount of stiffness degradation is likely to increase as the maximum
deformation increases. For cyclic deformation after strength loss has
occurred, there is likely to be a large amount of stiffness degradation,
as shown in Figure 4.15(a).
(2)
The type of stiffness degradation. Figure 4.lS(b) shows three different
types of stiffness degradation, all with the same amount of energy
degradation (the same ratio of the loop area to the area of a nondegraded loop). In the first case the stiffness degrades immediately on
unloading, whereas in the second case the stiffness does not degrade
immediately. In both of these cases the stiffness reduces progressively. The third case shows a pinched loop where the stiffness decreases
and subsequently increases.
(3)
Strength loss interaction. If there is strength loss in one direction, the
strength may or may not be affected in the opposite direction.
(4)
Partial loops. Most experiments with cyclic deformation consider
progressively increasing deformation amplitudes, with complete
hysteresis loops; Actual dynamic loads are likely to impose deformations with more random amplitudes, and partial hysteresis loops.
Rules must be defined for partial loops. These rules must ensure that
energy is never gained.
(5)
Unsymmetrical F-D relationships. Many components have different
properties for. positive and negative deformations. For example, a
reinforced concrete beam usually has different positive and negative
bending strengths. The shape of the hysteresis loop may be affected,
depending on the reason for the difference in the positive and
negative prop_erties.
(6)
Special cases. A bar that yields in tension and buckles in compression ·
requires special treatment. Some other special cases were shown earlier,
in Figure 4.9.
Hysteresis Loops for Inelastic Analysis 117
4.6.2
Hysteresis Loop in CSI PERFORM-30
The CSI computer program PERFORM-30 uses, for most components, a
hysteresis loop model that is anchored to a backbone relationship and uses a
relatively simple and flexible method to account for stiffness degradation.
· The main feature of the model is that stiffness degradation is not specified
directly. Instead, the amount of energy degradatio~ is specified, and the
stiffness is degraded to provide the required ratio. between the area of the
degraded and non-degraded loops. This has the advantage that it is relatively easy, given a series of experimental hysteresis loops for a component, to
make a reasonable estimate of the amount of energy degradation. It is more
difficult to estimate stiffness degradation directly.
Some details of the model are as follows.
(1)
The backbo~e F-D relationship is a multi-linear approximation, as
shown earlier in Figure 4.12{b).
(2)
To account for energy degradation, a relationship is specified between
the amount of degradation and the maximum deformation that has
been imposed on the component, as shown in Figure 4.16{a). Usually
the amount of energy degradation will increase as the maximum
deformation increases.
F
Partial cycle
F
D
Energy ratio
(b) Maximum unloading stiffness,
minimum range
F
Maximum deformation
(a) Variation of energy ratio
(ratio of degraded loop
area to non-degraded area)
{c) Minimum unloading stiffness,
maximum rang~
Figure 4.16 Hysteresis Model in the Computer Program CSI PERFORM-30
118
Chapter 4 Component Behavior - Uniaxial F-D Relationships
(3)
The type of stiffness degradation can be varied between the two
extremes shown in Figures 4.16(b) and (c). At one extreme the
unloading stiffness is equal to the original stiffness, and in order to
obtain the required energy degradation this stiffness applies over a
small force range. At the other extreme the unloading stiffness is a
minimum, and this stiffness applies over a large force range.
(4)
If the maximum deformation in any direction exceeds the ductile
limit, so that there is strength loss, the effect on the strength in the
opposite direction can be varied between no effect and an equal effect.
(5)
A partial loop is sho\.vn in Figure 4.16(b}. For this loop there is some
inelastic energy dissipation. For a deformation range that is smaller
than the range of the unloading line, unloading and reloading follow
the unloading line and there is no energy dissipation. For a curvilinear F-D relationship, and for some hysteresis loop models based on
multi-linear relationships, energy can be di.Ssipated for such small
cycles. The amount of energy dissipation can .affect the response of a
structure. under dynamic load, and is usually referred to as
~·damping". This is a complex topic, and it is not considered in this
book.
In this model, if a constant amplitude cyclic deformation is imposed, the
stiffness does not progressively degrade as the number of cycles increases.
That is, there is no cyclic degradation. The model could be extended, based
on the procedure outlined in the next section.
4.6.3
Practical Me>deling of Cyclic Degradation
As defined in this. book, cyclic degradation causes the stiffness and/or
strength and/ or ductile limit to degrade as the number of cycles increases.
There may also be degradation in the strain hardening stiffness, the rate of
strength loss, the residual strength, and the deformation at complete failure.
For any component that has cyclic degradation, there will always be a great .
deal of uncertainty about the behavior. Do not expect accuracy in an analysis
model.
The most practical method for modeling cyclic degradation is to use a
backbone relationship, and to degrade that relationship. There are two
alternative approaches, as follows.
Hysteresis Loops for Inelastic Analysis 119
(1)
Consider cyclic degradation indirectly. Assume a certain number of
deformation cycles, ·and choose a single backbone relationship that
accounts for these cycles. This is the procedure adopted in ASCE 41.
(2)
Consider cyclic degradatio1t directly. One way to do this is to define a
"monotonic" or "virgin" relationship that applies when there is no
cycling, and a "fully degraded" or "saturated" relationship that applies
after a lot of cycling. This is illustrated in Figure 4.17. The strength,
ductility and· residual . strength may all degrade. It is necessary to
define an interpolation rule thafspecifies how the backbone relation. ship changes as the amount of cycling increases, including the effects
of partial cycles.
Monotonic backbone
F
Fully degraded
backbone
/'
I
I
/
,
/
I/
I/
1,
//
l/
__
If-/--/
,_
x./// .
I
I
D
I
. Hysteresis loop for degraded backbone
may have more stiffness degradation
than loop for monotonic backbone.
Figure 4.17 Monotonic and Fully Degraded Backbones
The indirect approach may seem crude, but given the complexity of the
problem and the many uncertainties, it can be a reasonable approach for
design. The direct approach is attractive in principle but may be too
complex for practical application.
4.6.4
"In-Cycle" and "Between-Cycle" Strength Loss
For the analysis of large structures under dynamic earthquake loads, it is
usual to use a single backbone relationship, as in Figure 4.15, because it is
simpler than a variable backbone, as in Figure 4.17. However, research
studies have been carried out for simple structures, usually with a single
OOF, comparing the analysis results for these two approaches. These
analyses show that the two approaches can give substantially different
120
Chapter 4 Component Behavior - Uniaxial F-D Relationships
results.. The reason is. the difference in behavior for "in-cycle" and "betweencycle" strength loss. This is illustrated in Figure 4.18.
F
Original relationship
Effective relationship after cycling at DA
Loops and strength loss
from cycling at DA
Loops and strength loss
from cycling at DB '-.
~
_,
YJ
D
DB
Backbone relationship
Figure 4.18 "Actual" and "Model" Behavior
Figure 4.18 shows an original F-D relationship. Consider the effect of cyclic
deformation on this relationship. In Figure 4.18, if the deformation is
increased to DA and cycled a few times, there can be cyclic degradation,
with loss of strength and ductility. This r~duces the effective F-D relationship, as shown. This is ''between-cycle" strength loss. The deformation DA is
smaller than the ductile limit for the effective F-D relationship, so there is no
strength loss caused by deformations that exceed the ductile limit. Such
strength loss would be "in-cycle" strength loss.
If the deformation is then increased to DB and cycled a few more times,
there can be more cyclic degradation, with a larger strength loss and a
greater reduction in the effective F-D relationship. Again, the deformation is
always smaller than the ductile limit, so there is only between-cycle strength
loss. This is the "actual" behavior, which is often observed when actual
components are tested.
Figure 4.18 also shows a backbone F-D relationship that might be
constructed from the test results. This relationship accounts indirectly for
cyclic degradation. If hysteresis loops are· anchored to this backbone, the
loops for cycling at deformations DA and DB might be as shown in the
figure. This is the "model" behavior.
Conclusion for this Chapter 121
It can be expected that the hysteresis loops for the model behavior will be
similar to the final loops for the actual behavior in each set of cycles.ibis
does riot mean, however, that the model behavior is an accurate representation of the actual behavior.
The reason is that in the model behavior the deformation DB is larger than
the ductile limit for the backbone relationship. Hence, when the deformation DB is first applied the ductile limit is exceed\'!d, and there is in-cycle
strength loss.
Research studies of simple structures for earthquake loads show that the
calculated displacements are substantially larger when the "model" behavior
is assumed than when the "actual" behavior is considered. That is, the
studies indicate that in-cycle strength loss is substantially more damaging
than between-cycle strength loss. The reason is that in the actual behavior,
with only between-cycle strength loss, the stiffness of the structure is always
positive, whereas in the model behavior, with in-cycle strength loss, the
stiffness can be negative. Strength loss caused by a negative stiffness tends
to be more serious than strength loss that occurs between cycles.
From this it can be-concluded that an analysis model that uses a single
backbone relationship, and hence accounts for cyclic degradation indirectly,
tends to be more conservative than a model that uses a variable backbone,
. and accounts for cyclic degradation more directly.
4.7 Conclusion for this Chapter
The behavior of structural components can be complex and uncertain,
especially if there is inelastic behavior and cyclic loading. To create a useful
analysis model it is important to have an understanding of component
behavior, how 1l: can be modeled, and what results are needed for design. A
sound feeling for structural behavior and the needs of the design process
can be more important than detailed theoretical knowledge.
Estimating the inelastic properties for a real component is not a simple task.
It may be tempting to argue that since the inelastic properties are so
uncertain, there is little point in using inelastic analysis, and elastic analysis
should be sufficient. This fails to recognize that if there is substantial
inelastic behavior in an achial structure, the results "of an elastic analysis
may be of uncertain value for making design decisions, and may even be
misleading. As a tool for obtaining information for design, even a crude
inelastic model Cat1 be more useful than an elaborate elastic model.
122
Chapter 4 Component Behavior - Uniaxial F-D Relationships
It is also important to have realistic expectations of what can be achieved
with aruilysis. Keep in mind that the goal is to get useful information for
design, not to calculate "exact" response. The former is doable and useful.
The latter is impossible and can be a waste of time.
An additional complication is that the F~D relationships considered in this
section are all uniaxial, with a single force and a single corresponding ·
deformation. For many components (possibly most) the F~D relationship is
multi-axial, with two or more forces and deformations that interact with
each other. Procedures for considering interaction are considered in the next
chapter.
CHAPTER
5
Component Behavior - Multi-Axial
F-D Relationships with lnteracJion
The preceding chapter considered components with uniaxial F-D
relationships. Many components have multi-axial relationships with two or
more forces and two or more corresponding displacements. A major
complication with multi-axial relationships is that there can be. complex
interactions among the forces. An example is a column with P-M
interaction, where, among other things, the strength in bending depends on
the magnitude of the axial force.
This chapter considers procedures for modeling such components, with
particular emphasis on interaction. Because of the interaction, there are
fundamental differences between a uniaxial relationship and a multi-axial
one. Multi-axial relationships are more complex, and modeling them
involves more uncertainty.
This chapter considers several aspects of behavior and several possible
models, with emphasis on physical explanations rather than mathematical
theory.
Interaction can also affect other component properties, not just the F-D
relationship. This chapter identifies four types of interaction, namely
stiffness, strength, inelastic and capacity interaction.
123
124
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
s.1
Overview
The preceding chapter considered uniaxial F-D relationships, with a single F
and a single corresponding D. Many components have multi-axial relation. ships, With two or more forces and deformations. If there is no interaction
(no coupling) between the forces, the F-D relationships can be separated
(uncoupled) into a number of independent uniaxial relationships. Often,
however, there is interaction, and the relationships can not be separated.
The most common example is P-M interaction in a column, where
interaction between the axial force and bending moment affects the strength
and other properties. Multi-axial F-D relationships can be much more
complex than uniaxial relationships.
This chapter considers a number of different models for components that
have multi-axial F-D relationships. The emphasis is on physical explanations of the behavior, not on the theory or computation~
As in the preceding chapter, the type of analysis is important because it
determines how much detail is needed in the F-D relationship.
5.2 Stiffness Interaction
H the F-D relationship for elastic analysis is multi-axial, this means that the
relationship is defined by a stiffness matrix, not just a single stiffness (or
flexibility). If the matrix is diagonal, there is no stiffness interaction, and.the
multi-axial relationship can be separated into a number of uncoupled uniaxial
relationships. If the matrix has any off-diagonal terms, there is interaction
and the multi-axial behavior is fundamentally different from uniaxial
behavior.
Figure 5.1 shows an example With interaction. The figure shows a reinforced
concrete wall. Usually the reference axis for setting up the elastic stiffness is
the centroidal axis of the uncracked section. In this example this is at the
center of the section.
If there is no cracking, the axial and bending effects are uncoupled and the
stiffness matrix is diagonal. If the concrete cracks, however, the neutral axis
of the cross section changes position. Hence, bending of the wall causes axial
extension at the reference axis, and there is stiffness interaction between the
axial and bending effects.
Strength Interaction
(a) Wall section
10
125
::::01
/r,M
(b) Uncracked
f;d -~--J
Reference axis and
neutral axis are at
uncracked centroid.
(c) Cracked
~
r~---_:a_ - i J
7
Reference axis
does not change.
.
~
t
Neutral axis
shifts.
· Figure 5.1 Wall Section With Cracking
In addition, as the concrete cracks, the axial and bending stiffness both get
smaller. Change in stiffness is a difficult issue for elastic analysis, and stiffness
interaction makes it even more difficult. There is no simple answer. If a wall
section cracks, elastic analysis may fail to capture important aspects of the
behavior.
There are similar effects in concrete beams and columns. However, because
the cross sections are more compact, the shift in the neutral axis is smaller
and the stiffness interaction has a smaller effect than in a wall.
5~3
Strength Interaction
The preceding section considered stiffness interaction (or stiffness coupling).
This section considers the different issue of strength interaction.
·
The F-D relationship for a structural component has two main strength
values, namely yield and ultimate. For a uniaxial relationship each of these
strengths is the value· of a single force (or moment). For a multi-axial
relationship the yield and ultimate strengths are affected by the values of
two or more forces (or moments). These strengths are usually represented as
interaction surfaces. The most common example is the P-M strength
interaction surface for the cross section of a column.
126
Chapter 5 Component Behavior - Multi-Axial. F-D Relationships
Since there are two main strengths there are two main interaction surfaces.
The ultimate strength surface is usually used to define the strength capacity
for strength-based design. Essentially, if the point representing the P-M
demand for a column lies inside this surface, the strength demand/capacity
ratio is less than 1.0. The initial yield surface is smaller than the ultimate
strength surface, and defines the limit of elastic behavior. If the point
representing the P-M demand lies inside this surface, the behavior is
essentially elastic. The yield and ultimate surfaces may have different
shapes. In between the two, in the strain hardening range, there are
transitional surfaces. For the discussion in this section, consider only the
ultimate strength surface. The initial yield surface, and strain hardening
behavior, are considered later.
Figure 5.2 shows two-dimensional P-M interaction surfaces for the ultimate
strengths of cross sections in steel and concrete columns.
~
II
II
WM
(a) Steel Column
w
~
p
.
.
Concrete stress
and steel force
M
rf\iJ
(b) Concrete Column
Figure 5.2 P-M Interaction Surfaces
Figure 5(a) shows the interaction surface for a steel section. For two points
on the surface this figure shows the corresponding stresses on the section.
Figure 5(b) shows the same for a reinforced concrete section.
For the three-dimensional case, with P-M-M interaction, the interaction
surface is a 3D solid. Most of the explanations in this chapter use twodimensional surfaces.
Inelastic Interaction : Behavior after Yield
127
5.4 Inelastic Interaction : Behavior after Yield
A P-M interaction surface such as that in Figure 5.2 defines the strength of a
cross section, but says nothing about what happens after the surface is
reached and yield occurs.
For simplicity assume elastic-perfectly-plastic behavior, so that the yield and·
ultimate interaction surfaces are the same. In this cas~, if the combination of
P and M on a cross section lies inside the surface, the behavior is elastic. If
the P-M combination is on the surface, the cross section has reached its
ultimate strength. For a steel column the cross section is yielding, with
stresses as shown in Figure 5.2(a). P-M combinations outside the surface are
not permitted.
·
A key point for a multi-axial F-D relationship is that while the interaction
surface defines when yield occurs, it does not define what happens after
yield. This is illustrated in Figure 5.3.
.:c
p
Yield
D
(a) E-p-;p behavior
Force path~.Yield
i.
I . ..
F is constant
-------··
o
-Fv
Fv
after yield.
..
F
(c) Multi-axial interaction surface
(b) Uniaxial interaction "surface"
Figure 5.3 Uniaxial vs. Multi-Axial Behavior
-
.
'
.
.
Figure 5.3(a) shows the F-D relationship for elastic-perfectly-plastic (e-p-p).
behavior. For a. uniaxial F-D relationship, Figure 5.3(b) shows the implied
interaction "surlace". This· is just a pair of points, one each for positive and
negative yield. After yield occurs the force re!Jlains. constant (until
unloading occurs). For a multi-axial relationship, with P and M, Figure
5.3(c) shows the P-M interaction surface and a possible P-M path. These .
figures show that there is a fundamental difference between the uniaxial
and multi-axial cases. Jn the uniaxial case the force does not change after
128
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
yield, but in the multi-axial case the forces can change, provided the P-M
path remains on the interaction surface.
A related difference is that after yield occurs in the uniaxial case, all of the
subsequent deformation is plastic deformation. In the multi-axial case the
forces can change, which means that some of the deformation can be elastic.
Hence, for e-p-p behavior, the behavior after yield is entirely plastic in the
uniaxial case, but partially elastic and partially plastic (i.e., elastic-plastic) in
the multi-axial case.
An interaction surface by itself says nothing about .how much of the'
deformation is plastic and how much is elastic, and hence says nothing
about how the forces can change after yield. Some additional theory is
needed to explain this behavior.
One such theory is Plasticity Theory. This theory accounts for "inelastic
interaction". It provides a mathematically and physically consistent framework for describing the multi-axial behavior after yield has occurred. The
disadvantage of plasticity theory is that it assumes '.'plastic" behavior after
yield, where the material "flows", rather like a viscous fluid. This tends to be
the case for ductile metals, but many structural components are not plastic.
For example, concrete cracks and crushes, which is not plastic behavior.
Nevertheless, plasticity theory can be useful, even for the analysis of
concrete components, and it is often used. The important aspects of
plasticity theory are described in the next section.
An alternative assumption for multi-axial e-p-p behavior might be to
assume that the forces do not change after the interaction surface is reached,
as in the uniaxial case. This assumption, however, does not match observed
behavior and could lead to substantially incorrect results.
The preceding discussion assumes e-p-p behavior, with no strain hardening.
For pure bending of a steel section. with thin flanges and no residual
stresses, the moment at first yield is close to the ultimate moment, and there
can be essentially e-p-p behavior. However, for steel sections with substantial
axial force, and also for reinforced concrete sections, there is often a
substantial -difference between the P-M combinations at first yield and at . ·
ultimate strength. This means that the interaction surfaces for yield and
ultimate strength are different, and that the multi-axial F-D relationship
strain hardens. Plasticity theory can take this into account, as described later.
Plasticity Theory for Yield of Metals
129
5.5 Plasticity Theory for Yield of Metals
5.5.1
Overview
Plasticity theory was originally developed to explain the inelastic behavior
of ductile metals, particularly steel, with multi-axial stresses. It can be
extended to other types of interaction, notably P-M interaction in columns.
This section reviews the application of plasticity theory to a ductile metal in
a state of biaxial stress. Later sections show how it can be extended to multiaxial relationships of other kinds.
Plasticity theory is not the only way to account for interaction. Some other
methods are considered later in this chapter.
5.5.2
Yield of ElastiC-Perfectly-Plastic Metals
Figure 5.4(a) shows a small piece of steel plate subjected to biaxial stress.
Assume that the behavior is elastic-perfectly-plastic (e-p-p), and that the
yield stress in simple (uniaxial) tension is cry. The uniaxial stress-strain
relationship is shown in Figure 5.4(b).
(a) Biaxial stress
(b) Uniaxial behavior
(c) Yield surface
Figure 5.4 Steel With Biaxial Stress
In the well-known von Mises theory, the interaction surface for yield of
metals is an ellipse, as shown in Figure 5.4(c). In plasticity theory this is
usually called a "yield surface" rather than an "interaction surface".
Some key points for e-p-p behavior are as follows.
130
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
(1)
If the current stres~ point is inside the yield surface, the material is
elastic. If the stress point is on the yield surface, the material is
· yielding. Stress points outside the yield surface are not allowed.
(2)
If the material is yielding, the stress point does not have to remain at
one point on the· yield surface - it can move around the surface.
Yielding continues as long as the stress point stays on the surface.
(3)
When a stress point moves around the yield surface, the stresses
change, and the behavior of the material is elastic-plastic (i.e., partly
elastic and partly plastic).
(4)
If the stress point moves back inside the yield surface, the material
unloads· to an elastic state. The stress points at yield and at subsequent
unloading can be very different.
'
Plasticity theory describes the behavior of the material after it reaches the
yield surface; The main ingredients of the theory are as follows.
(1)
If the stress point is inside the yield surface, the material is elastic.
(2)
If the stress point is on the yield surface the material is yielding. The
stresses can change after yield, even though the material is e-p-p,
which means that the stress point can move around the yield surface.'
Figure 5.5 shows a yielded state, point A, defined by stresses crlA and
cr2A. Suppose that strain increments ~A and .&2A are imposed, causing .
the stresses to change to cr18 and cr28 at point B. Plasticity theory says
that some of the 'strain increment is an elastic increment and the
remainder is plastic flow. The elastic part of the strain ·causes the
change in stress. The plastic part causes no change in stress.
(3)
Plasticity theory defines the direction of the plastic flow. That is, it
defines the ratio between the 1-axis and 2-axis components of the
plastic part of the strain. Essentially, the theory states that the
direction of plastic flow is normal to the yield surface. ·
·. For example, consider stress along the 1-axis. only, with zero stress
along the 2-axis. This is the stress path OC in Figure 5.5. As the stress
is increased, yield occurs at point C. Since the stress path is defined,
the stress stays constant after yield, and hence all subsequent strain is
plastic. The normal to the yield surface at point Chas 1-axis and 2axis components in the ratio 2:1, as shown in the figure. Hence, the .
Plasticity Theory for Yield of Metals
131
plastic strains are in this ratio, and Poisson's ratio is 0.5. This agrees
with experimental results for purely plastic deformation.
1~ 0"1
2
Normal to yield
surface at C
Path OC is
uniaxial stress
Figure 55 Some Features of the Yield Surface
A purist might note, correctly, that the normal to the yield surface is in
stress space, whereas plastic flow is in strain space. For e-p-p behavior,
plasticity theory actually states that if the stress point moves around the
yield surface, the work done by the stress increment moving through the
plastic strain increment must be zero. For the simple stress path OC in the
example, the stress change is zero after yield, so the work is zero regardless
of the direction of plastic flow. In general, however, the stress point can
move around the yield surface, so a small stress increment is always tangent
to the yield surface. For the product of this stress increment and the plastic
strain increment to be zero, the plastic strain increment must be orthogonal,
in a work sense, to the stress increment. In geometric terms, this is the same
as saying that the stress increment and the plastic strain increment are at
right angles. Since the stress increment is tangential to the yield surface, the
plastic strain increment is normal. This is just a convenient way of v1sualizing plastic flow.
5.5.3
Strain Hardening
Plasticity theory can be extended from the e'-p-p case to the case with strain
hardening.
132
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
Whereas the yield surface in the e-p~p case is fixed, when there is strain
hardening, the yield surface moves and/ or changes size and/ or changes
shape. Jn a real metal it probably does all three. There are many strain
hardening theories. Jn most of these theories the yield surface can move
and/ or change size, but is assumed not to change shape.
If the yield surface moves but does not change size (or shape), this is
"kinematic" hardening. If the yield surface changes size but does not move,
this is "isotropic" hardening. The difference is illustrated in Figure 5.6.
<r2
<rz
<ru
<l'y
0
<rz
(a) Surfaces for first yield
and ultimate stress
Figure 5.6
<r1
(b) Stress path
(c) Kinematic hardening
(d) Stress path
(e) Isotropic hardening
<rv
Kin~~tic and Isotropic Hardening
Figure 5.6(a) shows the surfaces for first yield (the ''Y" surface) and at the
ultimate stress (the 'U" surface). If the stress point is inside the Y surface the
material is elastic, between the Y and U surfaces it is strain hardening, and
on the U surface there is zero strain hardening.
Figure 5.6(b) shows the stress-strain relatioQShip for stress cr2 only (er, = 0).
~For the stress path 0-A-B-C-D, Figure 5.6{~{shows how the position of the Y
surface changes for the kinematic hardfmng assumption. The rate of move-
Plasticity Theory for Yield of Metals
133
ment in the strain hardening phase depends on the strain hardening
modulus. For unloading, along C-D, the material unloads to an elastic state
at point C, and re-yields at point D. The stress range between C and D is
2crv.
Figures 5.6(d) and (e) show the behavior for the same stress path, but with
the isotropic hardening assumption. In this case the Y swface increases in
size.· For the path 0-A-B-C the behavior is the ~e as for kinematic
hardening. On unloading, however, the stress range between points C and
D is 2cru. Also, }'Vhereas for kinematic hardening the material strain hardens
after re-yielding at point D, for isotropic hardening there is no strain
hardening.
Since the kinematic hardening assumption gives a smaller Y surface for
cyclic deformation, it is more conservative than the isotropic assumption,
and it is used more often. Neither assumptiqn is necessarily accurate.
Figure 5.6 assumes linear strain hardening, a ·stress path along one axis only,
and no stiffness degradation. Plasticity theory is more general than this. It
applies for nonlinear strain hardening and for arbitrary stress paths. Also,
the unloading modulus can be reduced if there is stiffness degradation. ·
An important point for both hardening assumptions is that the Y surface
must change in such a way that when the stre8s point reaches the U surface
the two surfaces do not overlap (which would cause theoretical and
computational difficulties). This is automatic for isotropic hardening. For·
kinematic hardening and an arbitrary stress path, the Y surface must move
so that when it reaches the U ·surface the tangents to both surfaces are the
same, and hence there is no overlap. A convenient rule to ensure that this
happens (the Mroz rule) is shown in Figure 5.7.
Corresponding point
on U surface
Current position
ofYsurface
Figure 5.7 Y Surface Motion with Mroz Rule
134
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
The figure shows the current stress point, on the Y surface, and a
corr~sponding point, with the same tangent (and hence the same flow
direction), on the U surface. For kinematic hardening, the Y surface moves
towards the corresponding point on the U surface. This ensures that when
the Y surface reaches the U surface, the two surfaces do not overlap.
Since the plastic strain is normal to the yield surface, and since it might be
expected that the strain hardening depends only · on the plastic strain, a
purist might wonder why the Y surface in Figure 5.6(c) does not move along
the normal at the current stress point. The main goal of the Mroz rule is to
ensure that the Y and U surfaces never overlap.. Although it may not be
obvious, in Figure 5.6(c) the Mroz rule does move the Y surface along the
direction of plastic flow. The reason is that the von Mises ellipse is just a 2D
section through a 3D yield surface. In 3D principal stress space, the yield
surface is a circular cylinder, as shown in Figure 5.8.
Figure 5.8 Von Mises Yield Surface in 30
The yield s:urface in the cr1-cr2 plane is the intersection of that plane with the
cylinder, which is an ellipse. When the material yields the flow direction is
normal to the cylinder, in the radial direction. If there is strain hardening,
the yield surface moves in that direction. For uniaxial stress; as in Figure
5.6(b), this corresponds to movement of the 2D yield surface as shown in
Figure 5.6(c).
Figure 5.8 illustrates an important point about the yield of ductile metals.
At a point on the axis of the cylindrical yield surface, cr1 = cr2 = crr This is a
state of "hydrostatic" stress, which causes only volume change in the
material. Since the cylindrical yield surface extends to infinity along its axis,
the theory assumes that the material always remains elastic under
'',~.
Interaction Surface for Friction
135
hydrostatic stress. This is obviously not true to infinity, but it is essentially
correct up to very large stresses. For a stress point that is not on the axis of
the cylinder, the stresses can be divided into a hydrostatic component that is
along the axis of the cylinder, and a "distortional" or "deviatoric" component
that is normal to the axis. Distortional stress is essentially shear stress. It
causes change of shape (shear strain) with no· change in volume. ·Yield
occurs when the distortional stress component equals the radius of the
,
cylinder.
5.6 Interaction Surface for Friction
5.6.1 · Bearing Component with Friction
Figure 5.9(a) shows a bearing pad that has a vertical bearing force, P, and a
horizontal shear force, V.
p
~
r
. .--.,,r
Friction
coefficient
=µ
'
Slip (flow)
direction
~ associated
Slip surface for
_ __...._ _,...
flow
v
{a) Bearing with friction
{b) Slip (yield) surface ·
Figure 5.9 Slip Surface for a Bearing with Friction
Assuming a constant friction coefficient (a questionable assumption in
practice), the horizontal shear force required to cause slip depends linearly
on the vertical bearing force. A "slip" surface defines when slip occurs, and
is essentially the same as a yield surface. This surface is shown in Figure
5.9(b). H the P-V point is inside the surface, there is no slip. H the P-V point
is on the surface, the bearing pad is slipping.
A key point is that the slip direction is horizontal, which is not normal to the
slip surface.· In plasticity theory for the yield of metals, the plastic flow
direction is normal to the yield surface. This is "associated" flow. In Figure·
5.9 l::he flow is "non-associated".
136
Chapter 5 Component Behavior - MulticAJdal F-D Relationships
A problem with non-associated flow is that after slip (or yield) occurs the
stiffness matrix is non-symmetrical, whereas for associated flow it is
symmetrical. It is not essential, but most computer programs for structural
analysis assume that the structure stiffness matrix is symmetrical, since this
reduces the memory requirements (only one half of the matrix needs to be
stored) and reduces the time required to solve the equilibnum equations (by
one half). It is possible to assume symmetry for setting up and solving the
equilibrium equations, and to consider the non-symmetry in the state
determination phase, but this leads to equilibrium unbalances and other
possible problems.
The stiffness matrix can be made symmetrical by assuming that the slip
surface is normal to the slip direction, as shown in Figure 5.9(b). Since this is
different from the true slip surface, the P-V point can move away from the
true slip surface. This can be taken into account in a computer program, and
an analysis that is accurate enough for practical use can be carried out.
However, reliability can be a ~oncem.
In particular, consider a structure with several bearings, each modeled with
a bearing element. When a load step is applied in an analysis, it is possible
that there will be a decrease in the bearing force for one bearing element, so
that its P-V point moves outside the slip surface, indicating continuing slip,
and an increase in the bearing force for another element, so that its P-V
point moves inside the slip surface, indicating return to a non-slipping state.
In the next load step, because the stiffness of the structure has changed, the
same thing can happen to another pair of elements, possibly the same pair.
This can lead·to a flip-flop condition with a large number of closely spaced
nonlinear events, and the analysis method may be unable to converge to a
solution.
A symmetrical stiffness matrix could also be obtained by assuming
associated flow. For the slip surface in Figure 5.9 this would mean that
when slip occurs there is both horizontal slip and vertical displacement in ·
the tension direction. This also causes problems.
If the sliding surface is flat, and if the friction coefficient is constant, the slip
behavior is elastic-perfectly-plastic. If the sliding surface is curved (or is
spherical in three dimensions), the slip behavior has strain hardening. It is
an interesting exercise to consi9.er why this is the case.
Extension to P-M-M Interaction
5~6.2
137
. Cohesion vs. Friction
a
Figure 5.10 shows two yield surfaces for bearing pad with lateral load,
considering 30 behavior with axial force and biaxial shear.
(a) Friction
(b) Cohesion
Figure 5.10 30 Yield Surfaces for Friction and Cohesion
Figure 5.10(a) shows a friction slip surtace, as in the preceding section. fu
this case the shear strength depends on the bearing force. Figure 5.lO(b)
shows a "cohesive" yield surface, where the shear strength is independent of
the bearing force. This is not typical behavior for a bearing, but the purpose
of the figure is to illustrate the difference between frictional and cohesive
behavior. Among other things, the cohesion case has associated plastic flow.
The yield surface in Figure 5.lO(b) is similar, conceptually, to the 30 von
Mises yield surface in Figure 5.8. In that figure, stresses along the axis of the
cylinder are similar to a bearing force~ and radial stresses are similar
a
shear force. A von Mises material is purely cohesive. As shown later, some
materials combine frictional and cohesive behavior.
to
5.7 Extension to P-M-M Interaction
5.7.1
Overview
In a piece of steel under biaxial stress, the <J1 and 0 2 stresses· interact with
each other. Plasticity. theory models this interaction. By analogy, plasticity
theory can be eXtended to other types of interaction, in particular P-M
interaction for a column cross section.
138
Chapter 5 Component Behavior - Multi-Axial F-0 Relationships
The two-dimensional and three-dimensional P-M interaction surfaces for
. steel and concrete columns have roughly the forms shown in Figure 5.11.
Axial force
I~
Moment about
strong axis
.
Figure 5.11 Yield Surfaces for Steel and Concrete Columns
This ~gure shows only the U surface. There can also be a. Y surface. By
analogy with the von Mises theory, yield occurs when the P-M point reaches
the Y surface, and strain hardening occurs until the point reaches the U
surface. For kinematic hardening the motion of the Y surface can follow the
Mroz rule, which applies in 3D as well as 2D.
This gives a mathematically consistent theory for the inelastic behavior of a
column section. The theory can be applied to yield surfaces of essentially
any shape, and it can be extended to account for stiffness degradation and
strength loss. However, it requires two key assumptions, as follows.
(1)
When there is strain hardening, it is usual to assume that the Y surface
(for first yield) and the U surface (for the ultimate strength) have the
same shape. n is then a straightforward process to consider kinematic
or isotropic hardening, or a combination of the two.
\
For an actual column the Y and U surfaces probably are not the same
shape. For example, for a steel I-section, the ratio Py/P0 between the
yield and ultimate axial forces is probably not the same as the ratio
MJ Mu between the yield and ultimate bending moments, Also, the
ratio MJ Mu for bending about the strong axis of the cross section is
probably different from the ratio for bending about the weak axis. It is
' ··~.
Extension to P-M-M Interaction
139
not essential to assume that the Y and U surfaces have the same
shape, but if they have diffetent shapes the strain hardening rules are
more complex. It is still necessary, for example, to prevent the Y and
U surfaces from overlapping. This usually requires that the Y surface
not only moves or expands but also changes shape.
(2)
Plasticity theory assumes plastic behavior, as in a ductile metal. A
steel cross section yields in much the same way that a ductile metal
yields. However, in a reinforced concrete section the concrete cracks
and crushes, and only the reinforcement yields like a metal. Hence,
plasticity theory may not apply to reinforced concrete columns.
The following sections examine whether it is reasonable to apply plasticity
theory to P-M interaction in steel and reinforced concrete.
5.7.2
Steel Section - The Analogy Works
Consider a short length of column with a cross section consisting of two
steel fibers (in effect, an I-section with one fiber for each flange and a web
that can be ignored). This is shown in Figure 5.12(a).
p
Fiber area =A
Yield stress = cry
- - Yield force
= 2Acry
Fiber 1 Fiber 2
t
.rK.
Mf T \
m
(a) Column section and loads
M
Yield moment
=Ad cry
(b) Yield surface
Figure 5.12 Simplified Steel Column
Each fiber is elastic-perfectly-plastic with area A and yield stress cry.
The column is loaded with an axial force, P, and a bending moment, M, as
shown. The axial force is applied at the reference axis for the column, which
is the axis througb the cross section centroid. This is important because it
means that when the column is elastic there is no interaction between P and
M. With the reference axis at the centroid, P alone causes axial strain but no
curvature, and M alone causes curvature but no axial strain (where axial
140
Chapter 5 Component Behavior -
Multi~Axial
F-D Relationships
strain is measured at the reference axis). If the reference axis is not at the
centroid, P and M interact even before yield.
Each fiber has only uniaxial stress, but the column has P-M interaction. It is
easy to show that the P-M interaction surface is as shown in Figure 5.12(b).
This is the yield surface for plasticit}r theory. It is a simplified version of the
2D surface in Figure 5.11 for a steel column.
To see whether plasticity theory correctly predicts the yield behavior,
consider a case where the column is subjected to axial and bending effects.
The loading and behavior are shown in Figure 5.13.
M
P=O
c
P=0.5Py
A
"'
(b) Moment-curvature relationship .
M
Unit
length
(a) Load path
(c) Plastic strain and curvature
Figure 5.13 Behavior of Simplified Steel Column
First, apply an axial compression force equal to one half the yield force. The
load path is 0-A in Figure 5.13(a). Then hold this force constant and
increase the bending moment. The load path is A-B. At Point B, Fiber 1
yields in compression while Fiber 2 ·remains elastic. The yield moment has
now been reached, and the moment-curvature relationship is e-p-p, as
shown in Figure 5.13 (b). However, when one fiber yields the neutral axis
suddenly shifts from the center of the section to the unyielded fiber. Hence,
any subsequent change in curvature is accompanied by a change in axial
strain (always measured at the reference axis). This is shown in Figure 5.13
(c).
The strains after yield are all plastic. That is, there is plastic bending of the
cross section and plastic axial deformation. When the axial force is compres-
Extension to P-M-M Interaction
141
sion, the plastic axial strain is compression and the column shortens as it
yields l'h bending. If the column were in tension, it would extend as it yields _
in bending.
Figure 5.13 (c) shows the changes in curvature( A'lf, and axial strain, A&, after
yield. The change in axial strain is A&= O.SdA'lf. This is the ratio that
plasticity theory predicts, based on the normal to the yield surface. In this
case; therefore, plasticity theory is correct.
(Another acknowledgement to purists may be appropriate here. Since axial
force ·and bending moment have different units, the normal to the yield
surface is not a simple geometric entity. However, plasticity theory accounts
for the mathematics correctly. As before, the normal to the yield surface is
useful because it shows the general direction of plastic flow.)
If the bending moment is reversed, keeping the axial force COfi!ltant, Fiber 1
immediately unloads, and the cross section returns to an elastic state, with
the neutral axis at the center of the section. When the moment is fully
reversed, Fiber 2 yields in compression while Fiber l remains elastic. This
behavior is correctly predicted by plasticity theory. Hence, for this column
the theory is also correct for cyclic load.
After yield under the reversed moment, the plastic axial strain is again
compression. Hence, as the column is cycled plastically in bending it
progressively shortens. After a number of cycles, the amount of shortening
can be substantial.
This example is for a very simple cross section and for elastic-perfectlyplastic behavior. However, it indicates that plasticity theory can correctly
account for P-M interaction in steel columns.
·
5.7.3
Sharp Peak in Yield Surface
The simplified interaction surface in Figure 5.13 has four sharp "comers" or
"peaks", along the moment and force axes. The more realistic surface in
Figure 5.11 does not have peaks along the moment axis, but does have them
along the force axis. These peaks have the following meaning.
Consider the peak on the force axis in Figure 5.13. ff the moment has a very
small positive value, the P-M point is just to the right of the peak
Physically, one of the flanges is yielding, and the other is elastic, with a
stress very close to the yield stress, and the neutral axis for the cross section
is at the unyielded flange. If, however, the moment has a very small
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
142
negative value, the neutral axis moves to the opposite flange. If the moment
is exactly zero, both flanges are yielding. There is thus a discontinuity in the
behavior, with an infinitesimally small change in moment causing a
substantial change in behavior. The situation is similar for the yield surface
in Figure 5.11, which theoretically has a sharp peak on the force axis. For the
simplified yield surface in Figure 5.13, there is similar behavior for the peak
on the moment axis.
This behavior is unlikely to occur in a real column, so real column sections
probably do not have sharp peaks. A sharp peak can also cause numerical
convergence problems in an analysis, because of the discontinuity (very
small numerical changes can cause the behavior to "flip-flop"). For
computational reasons, yield surfaces in analysis models ·usually do not
have sharp peaks.
5.7 .4
RC Section - The Analogy Does Not Work So Well
Next, consider a simple reinforced concrete section consisting of two concrete fibers and two steel fibers, as shown in Figure 5.14(a).
d
Steel fibers :
Fiber area =A,.
Yield stress = <>vs
i+-=-1
Concrete fibers :
Fiber area
Compression strength
=2 A,, <>ye+ 2 A,. cry8
=I\
Mrf\
m
Yield stress= <>ye
....._. P at balance point =A,, <>ye
_ __,_+.-..,,._-+M
Yield moment at zero axial force = A,. d cry5
{a) Column section and loads
(b) Yield surface
Figure 5.14 Simplified Concrete Column
The steel fibers are elastic-perfectly-plastic. The concrete fibers are elasticin compression and are assumed to have zero strength in
tension. The P-M strength interaction surface is shown in Figure 5.14(b). For
plasticity theory this is also the yield surface. This is a simplified version of
the concrete section interaction surface in Figure 5.11 .
pe~ecUy-plastic
....
··~.
Extension to P-M-M Interaction
143
To assess whether plasticity theory can be applied to a reinforced concrete
section, consider the behavior for cyclic bending. As for the steel section, .the
deformations are curvature and axial strain at the center of the cross section.
Consider the behavior when a very small axial compre8sion force is applied,
and then cyclic bending is imposed. The reason for applying the axial force
is considered later. The behavior is as follows.
.
I
.
(1)
In the uncracked state the neutral axis is at the center of the section.
The bending and axial stiffnesses are based on the steel plus the
uncracked concrete. When a bending moment is applied, the concrete
fiber on the tension side cracks almost immediately (if the axial force
were zero, it would crack immediately). Hence, the neutral axis shifts
towards the compression side. The P-M point is inside the yield
surface, so the behavior is elastic. However, the bending and axial
stiffnesses are smaller than for the uncracked case, and unlike the
uncracked case, there is coupling between axial and bending effects.
This issue was considered earlier, in Section 51. This behavior is not
captured by plasticity theory.
(2)
When the moment reaches the yield moment the steel fiber on the
tension side yields. The bending stiffness reduces to zero and the
neutral axis shiftS to the compression fiber.
(3)
The moment remains constant as the curvature increases. The axial
strain is tension. The relationship between axial strain and curvature
is M: = 0.5d.i\'lf. Plasticity theory captures this behavior and gives the
correct relationship.
Hence, plasticity theory correctly predicts the behavior at Steps 2 and
3, after the yield surface is reached. However, the cracking behavior is
not modeled correctly in the elastic range.
Continue by cycling the bending deformation from positive to negative,
still with a very small axial compression force. The behavior is as ·
follows.
(4)
When the bending deformation .is reduced, the steel tension fiber
immediately unloads and becomes elastic. Plasticity theory correctly
·
predicts unloading.
(5)
The section is now elastic arid cracked, as in Step 1. As the ciuvature
decreases there is axial shortenjng at the center of the section, whi.cJt is
144
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
• opposite to the axial extension in Step 1. As in Step l, this behavior is
not captured correctly.
(6)
Shortly after the moment reaches zero, the second concrete fiber
cracks (if there is no axial compression force, the fiber cracks as soon
as the bending moment changes sign). Both concrete fibers are now
cracked, and both steel fibers are elastic. The neutral axis moves to the
center of the section, and the bending and axial stiffnesses are based
on the steel only. As before, constant stiffness is usually assumed in
the elastic range, and this behavior is not captured.
(7)
When the moment reaches the yield moment, the steel yields. Because
of the small axial force, the steel fiber on the compression side yields
first, and the steel fiber on the tension side remains elastic. Hence, the
neutral axis moves to the tension fiber. As the curvature increases, the
axial strain is compression. Plasticity theory assumes that the neutral
axis moves to the compression fiber, and that the. axial strain is
tension. Hence, plasticity theory is not correct.
(8)
The steel fiber that previously yielded in tension is now yielding in
compression. Wheri the total strain in this fiber becomes zero, the
·.crack closes in the concrete fiber: The section gains stiffness, and the
neutral axis moves towards the compression fiber.
(9)
Almost immediately, the tension fiber yields, and the neutral axis
moves to the compression fiber.
(10) The moment remains constant as the curvature increases, as in Step 3.
The axial strain is tension. Plasticity theory does capture this behavior,
but it does not correct for previous inaccuracies.
(11) · H the moment is cycled a number of times, plasticity theory predicts
plastic strain in tension in each cycle, for both positive and negative
bending. Hence, the theory· predicts that there is -a progressive
increase in length. In the actual section, the cracks in the concrete
fibers close in each cycle (see Step 8), and there is no accumulated
axial extension.
If this exercise is repeated for a larger axial compression force, say one half
of the axial force at the 1balance point, cracking occurs later in the elastic
range, and the bending strength is larger. Otherwise the behavior is similar
to that with a small axial compression force. If the bending deformation is
cycled, plasticity theory incorrectly predicts progressively increasing axial
tension strain. For an axial compression force above the balance point, the
Extension to P-M-M Interaction
145
concrete does not crack, and the fibers crush when bending deformations
are applied.
If the exercise is repeated with a small axial tension force, the concrete is
cracked before the bending deformation is applied, and the cracks never
close. The section behaves like the simplified steel section of the earlier
example, except that since the axial force is tension there is progressively
increasing axial tension strain, and progressively wi.dening cracks in the
concrete fibers. Plasticity theory is more accurate in this case.
This discussion of behavior is, of course, over-simplified. However, it is
qualitatively correct, and it emphasizes that the behavior of reinforced
concrete sections can be complex. In particular:
(1)
The stiffness matrix can change in the elastic range, which complicates
the task of choosing stiffnesses for elastic analysis.
(2)
For an axial compression force below the balance point, plasticity
theory incorrectly predicts progressively increasing axial tension
strain when the bending deformation is cycled.
In summary, this example suggests that plasticity theory does a mediocre
job of modeling P-M interaction in reinforced concrete for monotonically
increasing loads, and a poor job for cyclic loads. This does not necessarily
mean that plasticity theory is inappropriate for practical modeling of
inelastic reip.forced concrete columns. Obviously, however, it raises
questions. A later section suggests answers to some of these questions.
At the risk of raising even more questions, the following sections consider
some further complications, including strength loss, stiffness degradation,
and cyclic degradation.
5.7 .5 Dudile Limit and Strength Loss
If the F-D relationship has a ductile limit, with subsequent strength loss, this
adds the following complications.
(1)
A column cross section that has P•M interaction will generally have
ductile limits for both axial strain and bending curvature. Also, for a
section with P-M-M interaction there will generally be different
ductile limits for bending curvature about the two principal axes of
the cross section.
146
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
(2)
The ductile limits can be affected by interaction. For example, as the
axial force increases in compression, the ductile limit in bending
usually decreases. That is, a column with a large axial compression
force is likely. to be less ductile than the same column with a smaller
axial force. This interaction is not considered by plasticity theory.
(3)
The rate and amount of strength loss can be affected by interaction.
For example, a column with a large axial force (or a large shear force)
is likely to lose strength more rapidly than a column with a smaller
'
axial (or shear) force.
(4)
The residual strength, as a proportion of the ultimate strength, may be
different for axial force than for bending.
This last point can be important for modeling strength loss. When strength
loss occurs, the simplest assumption is .that the yield surface reduces
uniformly in size. However, a more realistic assumption is that the axial and
bending strengths reduce by different amounts, as shown in Figure 5.15.
Yield surface before strength loss
p
Surface after strength loss, with
equal losses for bending and axial
I
M
Surface after strength loss, with .
smaller loss for axial than for bending
Similar surfaces
for concrete ·
M
Figure 5.15 Yield Surface After Strength Loss
This is not difficult to do in an analysis model, provided the reduced Y and
U surfaces still have the same shape. The assumption is that plasticity
theory still applies, but with a different shape and size for the yield surface.
\ ..~
~
...
'
Extension to P-M-M Interaction
147
Figure 5.15 assumes that if strength loss occurs for, say, positive bending,
the same amount of. loss occurs for negative bending. Uris may not always
be the Case. For a tWo-dimensional P-M yield surface it is possible to reduce
the yield surface size asymmetrically, with smaller moment strength loss
for, say, negative bending than for positive bending. It is less clear how a 3D
surface with P-M-M interaction might be modified.
5.7.6
Hysteresis Loops and Stiffness Degradation
For uniaxial F-D relationships, stiffness degradation and hysteresis loops
were considered in Section 4.6.
Hysteresis 10ops are inherently more complex for multi-axial F-D relationships. As an example, consider a column cross section with a biaxial F-D
relationship and two-dimensional P-M interaction. If P is held constant and
M is cycled, the hysteresis loops will be similar to those for a uniaxial .
moment-curvature relationship, with the complication that different P values
give loops of different sizes and shapes. The loops are more complex if P
and M both vary. They are more complex still for a section with P-M-M
interaction, where P and the two .M values can all vary cyclically and
independently.
A major consideration for the· shape of a hysteresis loop is the amount of
stiffness (and eri.ergy) degradation. For a uniaxial F-D relationship the
amount of degradation usually increases as the deformation increa8es, and
it is not too difficult to create reasonable rules for stiffness degradation. For
a multi-axial relationship it is more complex because of interaction. If there
were no interaction, separate rules could be used for P and M, based on the
axial and bending deformations. However, separate rules may be unrealistic
in. an actual column. For example, if there is P-M-M interaction, the
degradation ir\ stiffness about one bending axis is likely to depend not only
on the bending deformation . about that axis but also on the bending
deformation about the other axis, and possibly on the axial deformation.
Plasticity theory can account for stiffness degradation. The theory can
consider different stiffnesses along the P axis (based on EA) and the Maxes
(based on El), and these stiffnesses can degrade. The challenge is to establish
rules that determine how the stiffnesses degrade when there is interaction.
There are many hysteresis loop models.that have been proposed for uniaxial
F-D relationships. There are very few for multi-axial· relationships with
interaction. ·
·
,
148
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
5.7.7 Cyclic Degradation ·
As noted earlier, cyclic degradation poses modeling challenges for uniaxial
F-D relationships. The challenges are even greater for multi-axial relationships.
·
·
For · uniaxial components, two alternative modeling methods were
considered in Section 4.6.3,. namely (a) a direct method where the F-D
relationship progressively degrades under cyclic deformation and (b) an
indirect method where a single ''backbone" relationship is used, based on
an assumed amount of cyclic deformation. The direct method is complex to
apply for components with uniaxial F-D relationships. This method is even
more complex for components with multi-axial relationships, since the
entire interaction surface must progressively degrade. For practical analysis
the indirect method, with the interaction surface based on a single backbone
relationship, is likely to be used.
5.7.8 Other Cross Section Shapes
The P-M-M interaction surfaces for steel I-sections and rectangular or
circular concrete sections have fairly simple forms (see Figure 5.11).
Interaction surfaces can be constructed for more complex sections, and
plasticity theory can be applied to these surfaces. In most cases this is a
relatively minor complication. However, some complex surfaces, for
example those with re-entrant angles or surfaces that are concave outwards,
can not be considered using plasticity theory.
5.8 Is Plasticity Theory Useful for P-M lnteradion?
5.8.1
Overview
The preceding sections have considered how plasticity theory can be used to
model P-M interaction in columns. The theory has obvious limitations. This
section assesses the practical importance of these limitations.
5.8.2 Assumptions and Approximations
The application of plasticity theory to P-M interaction involves many
assumptions and approximations. Some of these apply only to reinforced
concrete cross sections, and some · apply to both steel and reinforced
concrete.
Is Plasticity Theory Useful for P-M Interaction?
149
Some issues that affect only reinfo;ced concrete sections are as follows.
(1)
Even in the elastic range (inside the yield surface), the behavior of a
real cross section is more complex than is assumed in plasticity
theory. It is usual to assume constant values for EA and EI, even
though these values change when the concrete cracks. It is also usual
to ignore the fact that cracking causes coupling between axial and
bending effects. This is not just an issue for plasticity theory - it is a
problem for any elastic analysis of reinforced concrete.
(2)
Reinforced concrete is not a plastic material. In particular, plasticity
theory does not account correctly for the effects of cracking.
(3)
A key aspect of plasticity theory is that it predicts the ratio of aXial
deformation to bending deformation when the behavior is inelastic.
For a reinforced concrete section the ratio predicted by plasticity
theory is generally not correct, mainly because of cracking.
"'
Some issues that affect both steel and reinforced concrete sections are as
follows.
(1)
H separate Y and U surfaces are used, there is likely to be uncertainty
about the amount and type of strain hardening. Plasticity theory must
usually assume that the Y and U surfaces have the same shape, and
that the shape does not change as strain hardening occurs. There is
also likely to be uncertainty about whether hardening follows the
kinematic or i~otropic assumption,
(2)
It is difficult to model hysteresis loops accurately when there is
interaction. It is even more difficult to account accurately for stiffness
degradation, strength loss, and progressive cyclic degradation in
strength and ductility.
S.8.3
Importance of Axial Deformation - Steel Column
·A key aspect of plasticity theory is that it predicts coupling between the
bending and axial deformations, and it may not do this accurately. For a
steel column in compression, plasticity theory predicts progressiVe
shortening under cyclic bending, which is qualitatively correct. For a
reinforced concrete column in compression, with an axial force that is below
the balance point, plasticity theory predicts progressive extension under
cyclic bending, which is not correct.
150
Chapter 5 ·Component Behavior - Multi-Axial F-D Relationships
A key question is whether the axial deformation has a significant effect on
the behavior of a single column or a complete rrame. This section attempts
to · answer that question for steel columns. A later section considers
reinforced concrete columns.
If a colunm is allowed to yield, its behavior in bending is likely to be the
most important consideration. This does not mean that its axial strength is
unimportant, but a ·column is unlikely to fail under axial load until its
strength has been degraded by cyclic bending deformation. If this is true,
the important aspects of behavior for a single column are (a) P-M strength
interaction and (b)the effects ofcyclic bending deformation on stiffness,
ductility and strength loss. Also, when a column is allowed to deform
inelastically, the demand-capacity measure for assessing performance is
likely to be based on bending deformation, not axial deformation.
Hence, when a column is considered· separately from the rest of the
structure, the axial deformations are likely to be. relativelt unimportant.
However, when a complete frame is· considered, axial deformations of the
columns cause vertical displacements, and these displacements may have
significant effects on the adjacent beams and connections.
To obtain a rough estimate of the amount of axial deformation, consider a
steel colunm in a 2D frame, as shown in Figure 5.16(a).
This column has P-M plastic hinges at its ends, with elastic behavior over
the rest of the column length. Assume the following._
(1)
The column has the simplifit!d steel cross section and P-M interaction
surface that were considered earlier (Figure 5.12). The hinges have
·rigid-perfectly-plastic behavior.
(2)
Apply gravity load to the frame, putting the column in compression.
Then add lateral load, causing a story drift ratio of 0.025. Plastic
hinges form at the top.and bottom of the column, as shown in Figure
5.16(b). The beams in the frame remain elastic. It may not be wise to
have hinging only in the columns; since this can lead to an
undesirable "story" collapse mechanism. However, this is a separate
issue (see Chapter 7).
·
Is Plasticity Theory Useful for P-M Interaction?
.1
T
....
Apply gravity
loadfirst
.
0.0f5h
~.1·~~1~
Add lateraI
load
(
0.005h
at yiekl
151
h~hl
.
\
I
(a) Frame and coiumn element
B •... .
I ...
I
.
. .
(b) Column deformation·
p
A~~--+--+--+
M
DriltRatiO
(c) Load path for column hinge
(d) Hinge moment and rotation ··
Figure 5.16 Frame with CycliC Loading
(3)
Assume that there is little or no change in aXial force when the lateral
load is applied. This is reasonable for an interior column (but not for
an exterior column). Also assume that initial hinge formation occurs
at a story drift ratio of 0.005. Hence, the load path for a hinge is shown
in Figure 5.16(c). The relationship between hinge moment and story
drift is shown in Figure 5.16(d). The plastic drift ratio is 0.025 - 0.005 =
0.020. It follows that the rotation of each plastic hinge iS 0.020 radians:
(4)
The behavior of a short length of column was considered earlier
- (Figures 5.12 and 5~13).
bending, the F-D relationship w~ a
relationship between bending moment and curvature,. and the flow
rule related plastic curvature and plastic aXial strain~ In the current
example the component is a zero-length plastic hinge. for bending,
the F-D relationship is a relationship between bending moment and
hinge rotation, and the flow rule relates hinge rotation and hinge aXial
deformation. Assume that plasticity theory still applies (see later for
some additional discussion). Hence, from Figure 5.13, if the section
depth is.d, when the hinge plastic rotation is 0.020 radians the hinge
axial deformation is 0.020(d /2). If the story height is h, the plastic aXial
For
152
Chapter 5 ·Component Behavior - Multi-Axial F-D Relationships
shortening of the column, as a proportion of the story height,
considering two hinges, is 0.020(d/h).
(5)
Assume that d/h is 0.1 (roughly a W12 column and a 120-inch, or 3meter, clear story height). Hence, when the drift ratio is 0.025 the
plastic axial shortening of the column is (0.020)(0.1) =0.0020 times the
story height. This is effectively an axial strain in the column. It is
roughly double the yield strain for steel. For a 120-inch story height
the plastic axial deformation is 0.24 inches (6 mm). This is a small
axial deformation and probably would not have a significant effect.
(6)
If the lateral displacement is cycled, with a drift ratio from +0.025 to
-0.025, the range of plastic hinge rotation is 0.040 radians in each half
cycle. Hence, the plastic axial shortening of the column increases by
0.004 times the story height in each half cycle, or 0.008 times the story
height in each full cycle. If the lateral displacement is cycled 5 times,
the total ax!al shortening is 0.040 times the story height. For a 120-inch
story height {3 meters), the plastic axial deformation is 4.8 inches (120
mm). This is a large axial deformation that could have substantial
effects.
This example may be extreme, because it assumes a lot of column hinging
and uses a simplified interaction surface. Nevertheless, it suggests that
accuinulated axial shortening of steel coltimns may be a concern for earthquake loads. For inelastic dynamic analysis of steel frames, if there is
column yielding, the axial deformations of the columns should be
examined. If these deformations are substantial, their effect on the behavior
of the frame should be given some thought.
5.8.4
Is the Axial Deformation in a Plastic Hinge Correct?
For a short length of a steel column, the example in Section 5.7.2 indicates
that plasticity theory correctly models axial extension when a cross section
yields. It does not necessarily follow that the axial extension of a complete
column element is predicted correctly when plastic hinges are used. This is
worth looking at in some detail.
Consider the case of a column with a constant axial force and a linear
variation of bending moment over the column height. If the momentcurvature relationship for the column cross section is elastic-perfectlyplastic, with zero stiffness after yield, all of the yield occurs in zero length
hinges at the column ends. In this case a model with plastic hinges is "exact".
However, yield in a zero length hinge means infinite curvature, and hence
.,
··~.
~--~
f:
~;
~'
~~·
~·
~·
~·
Is Plasticity Theory Useful for P-M Interaction?
153
infinite strain, which is physically impossible. In an actual column there will
be some strain hardening, and yield will occtir over finite "plastic zones" at
the column ends. This is illustrated in Figure 5.17.
Moment
~0
Yield
i
/
curvature
Curvature at
Curvaturil
•
column end
(b) Moment-curvature relationship
. Axial Force
Af column end
(a) Column moments and plastic zones
(c) Yield surfaces
Figure 5.17 Plastic Flow in Inelastic Region of a Column
Figure 5.17(a) shows the forces on the column and the bending moment
diagram. Figure 5.17(b) shows the moment-curvature relationship; with
strain hardening. At the top of the column the bending moment at yield is
A1i1 and the moment at the column end is a somewhat larger value, M2• The
column yields over a plastic zone.
· Figure 5.17(c) shows the P-M yield surfaces at the two ends of the plastic
zone. This figure assumes kinematic strain hardening where the yield
surface translates along the norinal (flow) direction. The P-M points at the
two locations are as shown. The axial forces are the same, but the bending
moments are different. The flow directions are not exactly the same at the
tWo P-M points, but they are similar. The yield surfaces for other cross
sections in the plastic zone lie between these two surfaces, and hence have
similar flow directions.
At any cross section in the plastic zone there is a plastic eurvature and a
plastic axial strain. The integral of the plastic curvature over the length of.
the plastic zone gives the plastic rotation, and the integral of the plastic a?Cial
strain gives the plastic axial extension or shortening. When a plastic hinge is
/
154
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
used there are concentrated rotations and axial deformations in the hinge.
The hinge is a close approximation of the plastic zone if the plastic rotation
in the hinge is close to the integral of the plastic curvature, and if the plastic
axial deformation is close to the integral of the plastic axial strain.
One question is whether a plastic hinge model has the same ratio of axial
deformation to rotation as a plastic zone. Figure 5.17 indicates that the ratio
between plastic axial strain and plastic curvature is similar for all cross ·
sections in a plastic zone. This ratio is also similar to the ratio between axial
deformation and rotation in a plastic hinge. Hence, it is reasonable to
assume that a plastic hinge gives the correct ratio between axial deformation
and rotation.
A second question is whether the concentrated rotation in a plastic hinge is
the same as the distributed rotation over the length of a plastic zone. In a
plastic zone, the plastic curvature is zero at one end and a maximum at the
other. If the plastic curvature varies linearly over the length of the plastic
zone, the location of an equivalent concentrated rotation is at the third point
of the plastic zone (this follows from geometric compatibility, for example
using the Moment Area method). Given an estimate of the plastic zone
length, a plastic hinge could be placed at this location. (The hinge location
must be fixed - a plastic zone can change progressively in length, and it is
generally not practical to change the plastic hinge location.) Usually,·
however, the plastic hinges in a column element are located at the element
ends (or at the top and bottom of the adjacent beams for a column element
with stiff end zones). The effect of this is that the hinge rotation is rather
smaller than the plastic rotation in the plastic zone (it is a useful exercise to
show why this is the case). The difference increases as the length of the
plastic zone increases. This difference is usually not large.
From ·this, it can be concluded that although a plastic hinge model is
approximate, it is accurate enough for most columns (and also most beams).
5.8.5
Accumulated Axial Deformation - Concrete Column
For a steel column, any accumulated axial deformations that are predicted
by plasticity theory are likely to be a real effect, and are not just a consequence of the modeling assumptions. For a reinforced concrete column,
plasticity theory does not capture the behavior accurately, especially for
cyclic deformation, so when accumulated axial deformations are calculated
they are less likely to be a real effect.
Is Plasticity Theory Useful forP-M Interaction?
155
To gain some insight, consider the simplified column section that was
considered earlier (Figure 5.14). For this cross section, if the axial force is
below the balance point there is a constant ratio. between the axial· and
bending deformations. This ratio is similar to that for a steel section, but the
axial deformation is extension rather than shortening. For gravity load plus
the same lateral cycling as in the earlier steel column example (see Section
5.8.3), plasticity theory predicts an accumulated axial deformation of about
4% of the column length after 5 cycles to a drift ,ratio of 0.025, but with
extension rather than shortening. This example exaggerates the extension,
since the interaction surface for a more realistic column is not a simple
diamond. In a realistic column, as the axial force approaches the balance
point the ratio between axial and bending deformation gets smaller. An
important point, however, is that the calculated extension is a consequence
of the modeling assumptions, and is not a real effect. 'For small amounts of hinging in concrete columns, axial deformation effects
are probably not significant. Even.though plasticity theory does not predict
axial deformation correctly, it still accounts for P-M strength interaction and
for inelf1Stic bending, and is likely to be accurate enough for many practical
purposes. lf there is substantial hinging, however, plasticity theory may be
substantially in error. In such cases the analysis results need to be looked at
closely. It may be necessary to use a different type of model, in particular a
"fiber" model, as considered later.
5.8.6
Can the Axial Extension be Assumed to be Zero?
Since a major concern about plasticity theory for reinforced concrete is that
it may not predict axial deformation correctly, why not modify the theory to
a5sume that plastic axiru deformations are zero?
For a P-M-M yield surface, it is possible to assume non-associated flow,
where the plastic flow direction is normal to the M-M yield surface but has a
zero component along the axial direction. For a two-dimensional P-M
surface this is shown in Figure 5.18.
Figure 5.18 shows a point on a· P-M yield surface. The flow direction for
associated flow is as shoWn. H the flow direction is assumed to have no axial
component, and if the stiffness matrix is made symmetrical, this is the same
as assuming a yield surface that has associated flow and is diff~t from
the actual surface, as shown in the figure. During a load step in an analysis,
the P-M. point can move along this surface. lf this happens, the P-M point
can move outside the actual surface, which is not allowed, or can move
inside the actual surface, which indicates unloading to an elastic state.
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
156
p
Yield surface
Flow direction with
no axial comfl<?nent
is non-associated.
~M
11
I
11
'JfM
Associated flow direction has
axial and bending components.
Surface for associated flow
with no axial component
Figure 5.18 Non-Associated Flow
A similar situation exists if a structure has bearings that have friction, as
considered earlier (see Section 5.6.1). In both situations a numerical flip-flop
condition can develop in the analysis, with a large ntimber of closely spaced
nonlinear events, and the analysis may not converge. One difference is that
a structure will usually have relatively few friction-bearing elements,. and. a
flip-flop condition usually can be avoided. However, a structure may have
many column elements, and these elements are inherently more sensitive
numerically than friction-bearing elements. It is more likely that a flip"flop
condition will develop if the approximation in Figure 5.18 is used for
column elements. This is not necessarily fatal, but since it affects reliability it
is a concern.
S.8.7
Axial Deformation in Static Push-Over Analysis
If dynamic analysis is used for earthquake loads, cyclic behavior is modeled
directly. However, if static push-over analysis is used, cyclic behavior is
considered only indirectly.
In a steel column there can be progressively increasing axial deformation
under cyclic load. Static push-over analysis does not capture this behavior..
Somewhat paradoxically, for a reinforced concrete column a dynamic
analysis can overestimate the axial deformation under cyclic load, and pushover analysis may be more accurate.
Is Plasticity Theory Useful for P-M Interaction?
157
S.8.8 Conclusion for this Section
In the modeling of columns, the goal is to calculate strength and/ or
·deformation demands that can be compared with corresponding capacities,
with sufficient accuracy for design purposes.
The aspects of behavior that may be important are as follows.
(1)
(2)
(3)
(4)
(5)
(6)
P-M strength interaction.
Inelastic behavior, mainly in bending.
Coupling between bending and axial deformations.
Stiffness and energy degradation.
Strength loss.
Cyclic.degradation.·
When plasticity theory is used to model P-M interaction in columns, the key
issue is not whether the theory is accurate (it really isn't), but whether it
captures the yield strength, the post-yield behavioi; and the cyclic behavior
with sufficient accuracy for practical purposes.
A major advantage of plasticity theory is that it uses familiar strength
interaction surfaces that are similar to those used for conventional strengthbased design. The surfaces that are used for elastic. and inelastic analysis
may not be identical, because of differences in the expected vs. nominal
material strengths, and possibly the use. of different capacity reduction
factors. However, they are similar.
A major disadvantage is that plasticity theory can predict incorrect axial
deformations, especially for reinforced concrete columns.
The following conclusions can be reached from the discussion in this
section.
(1)
·
Plasticity theory is almost certainly reasonable for steel columns, even
in cases where there is substantial yield. In a steel column, progressively increasing axial deformation under cyclic inelastic deformation
is probably real ~ffect.
a
(2)
Plasticity theory is proqably reasonable for reinforced concrete
columns provided there is only a modest amount of inelastic behavior.
In this case plasticity theory may predict axial deformation that is not
a real effect, but the amount of this deformation is likely to be too
~mall to have a significant effect on the adjacent beams and
connections.
158
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
(3)
Plasticity theory is probably. not reasonable for reinforced concrete
columns if there is substantial inelastic cycling. One reason is that the ..
theory can overesfunate the amount of axial extension.
(4)
Allowing substantial inelastic deformation in the columns of a reinforced concrete building can lead to undesirable performance. This
may not be the case, however, for bridge ·columns, which can be
designed to accommodate substantial inelastic behavior. A fiber model,
as considered later, is likely to be better for a bridge column than a
model based on plasticity theory.
(5)
For both steel and concrete columns, plasticity theory is reasonable
only if the ductile limit is not exceeded. H the ductile limit is exceeded,
there is likely to be substantial cyclic degradation, with a great deal of
uncertainty in the behavior. Plasticity theory {or any other theory) is
unlikely to be accurate.
(6)
If substantial axial deformations can accumulate under cyclic load,
static push-over analysis may be inaccurate.
5.9 Axial Extension in Concrete Beams
The preceding section considered axial deformations in columns. Axial
deformations can also occur in beams, and can have significant effects on
the structural behavior.
·
It is usual to assume that the axial force in a beam is small, and to ignore PM interaction. For both steel and concrete beams, inelastic bending is
usually modeled using plastic hinges with uniaxial moment-rotation
relationships and no axial deformation.
In a steel beam, the normal to the P-M interaction surface for zero axial force
is usually along the M axis, with no P component. Since the axial forces in
steel beams are likely to be small, it is usually reasonable to use a uniaxial
moment hinge. In a reinforced concrete beam, however, the normal to the
P-M interaction surface for zero axial force has a substantial P component,
corresponding to .axial extension,. because the neutral axis ·of the cross
section moves to the compression side as the reinforcement yields. A plastic
hinge with. a uniaxial moment-rotation relationship ignores this axial
extension.
Fiber Sections for P-M interaction
159
If a reinforced concrete beam is unrestrained axially, it may not matter if the
axial extension is ignored. However, if there is axial restraint, from adjacent
columns or walls or from a floor slab, a substantial axial compression force
could develop in the beam. This force suppresses cracking, and hence
increases the bending strength of the beam. If the beam is stronger in
bending, it can develop larger shear forces, and it may yield in shear before
it yields in bending. The beam can also exert horizontal forces on the
adjacent columns or walls, in addition to larger than expected bending
moments. If the axial restraint is provided by a floor slab, and if the slab is
not strong enough in tension, it could crack, especially at the comer of a
building. When a uniaxial moment hinge is used, these effects are all ·
ignored.
As with a column, axial deformation after yield can be considered using a PM hinge, rather than a simple moment hinge. A P-M hinge for a reinforced
concrete beam predicts axiaj extension when the axial force is zero, and also
predicts an increase in bending strength if the axial extension is restrained,
causing a compressive axial force. However, the axial e~tension in a
reinforced concrete beam may not be predicted accurately by a P-M hinge,
especially for cyclic deformation. If it is necessary to account for axial
deformation effects with greater accuracy, it may be practical to use a fiber
model, as considered in the next section.
5.1 O Fiber Sections for P-M interaction
5.1 0.1 Overview
In the simplified examples for steel and concrete columns (Figures 5.12 and
5.14), the cross sections were modeled using uniaxial fibers. These are fiber
models, which account for P-M interaction. For reinforced concrete, a fiber
model has the· added advantage that it can account for cracking of the
concrete in the elastic range, before the steel yields.
Since a fiber model avoids the complications and approximations of plasticity
theory, why not ilse fiber models? The answer depends mainly on the type
of structure.
This section considers fiber sections for beams, columns and shear· walls.
Only general modeling procedures are considered, omitting details.
160
Chapter 5 Component Behavior- Multi-Axial F-0 Relationships
5.10.2 Fiber Sections for Beams
Figure 5.19 shows the type of fiber model that is likely to be used for a beam
cross section.
I
•••••
For bending about
horizontal axis, use fiber section.
= Steel fiber
II
.._..
I
----+---- -'I
I
I
••I··•
Concrete fiber.
Lump properties
at center of fiber.
•
I
r
!
For bending about vertical
axis, assume elastic.
Figure 5.19 Fiber Section fora Beam
A common assumption for a beam is that there is inelastic bending in only
one direction, usually in the vertical direction (bending about the horizontal
axis). For horizontal bending the behavior is assumed to be elastic. Often
there is little or no horizontal bending, ·because the beam is· braced, for
example, by a floor slab.
To model bending behavior in the vertical direction, fibers are needed only
through the depth of the beam, as indicated in the figure. For ·horizontal
bending an elastic bending stiffness is specified {i.e., an· EI value). For
vertical bending the fiber model determines EI. This model accounts for P-M
interaction, and hence for axial deformation caused by bending. For
horizontal bending the model assumes that there is no P-M interaction. It
also assumes that there is no coupling between vertical and lateral bending.
5.10.3 · Fiber Sections for Columns
A fiber model for a column must usually account for biaxial bending.
Hence, fibers are needed, as indicated in Figure 5.20. This type of model
accounts for P-M-Minteraction. ·
Fiber Sections for P-M interaction
•
•
•
•
161
-~·
-1'?¢
•
•
•
•
•
•
•
•
•
• •
•
•
;
Figure 5.20 Fiber Section for a Column
For both beams and columns, the behavior in torsion is usually assumed to
be elastic, and also to be uncoupled from the axial and bending behavior.
Shear may be important, as considered later.
S.1 OA Fiber Sections for Walls
A shear wall has bending in two directions, namely in-plane and out-ofplane. Often it is accurate enough to consider inelastic behavior only for inplane bending (membrane behavior), and to assume that the behavior is
elastic for out-of-plane bending (plate bending behavior). In this case the
fiber model can be similar to that for a beam, with fibers only for membrane
behavior, as shown in Figure 5.21.
Steel
fibers
Concrete
fibers
Figure 5.21 Fiber Section for Membrane Behavior of a Wall
As with a beam, an effective EI is specified for out-of-plane bending,· and
there is no coupling between membrane and plate bending effects. If
inelastic plate bending is to be considered, there must also be fibers through
the wall thickness. In this case the fiber model is similar to that for a column.
Figure 5.21 shows the cross section for a plane wall. More complex cross
sections can be divided into a number of plane walls, as shovm in Figure
5.22
162
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
(a) Wall cross section
(b) Model with plane walls
Figure 5.22 Wall Section Modeled as Several Plane Walls
The cross section in Figure 5.22(a) could be treated as a single section, rather
like a column. However, this is likely to be inaccurate because it does not
allow for warping of the cross section. For a fiber section it is usual to
assume that plane sections remain plane. This can be reasonable for a plane
wall, even if it is quite wide, but it can be incorrect for an open, thin-walled
section. It is more accurate to divide the section into plane parts, as in Figure
.
5.22(b).
s~10.s
FiberSegments
Figures 5.19 through 5.22 show only cross sections. In a fiber model of a
beam, column or wall, it is necessary to consider fiber segments, which are
finite member lengths with fiber cross sections. A complete beam, column or
wall element may be made up of several fiber segments.
5.10.6 Rigid-Plastic Fiber Hinge
It is possible to model a zero-length rigid-plastic hinge, using rigid-plastic
fibers. Yield of such a hinge does not occur when the first fiber reaches its
yield stress. Instead, the hinge remains rigid until it forms a mechanism. For
a beam-type hinge this means either (a) that there is yield in all except one
of the fibers and the hinge rotates about that fiber, or (b) that all fibers yield
and the hinge has independent bending and axial deformations. For a hinge
in a 3D column, a mechanism forms when all fibers yield, or all fibers yield
except one, or all fibers yield except two.
Rigid-plastic fiber hinges, with zero length, are rarely, if ever, used.
However, short fiber segments can be used to model such components as
beam-to-column connections. If a fiber segment is short, it may be initially
very stiff but it is not rigid.
Fiber Sections for P-M interaction
163
5.10.7 Limitations of Fiber Models
Fiber models can capture (a) the cracking of reinforced ·concrete cross
sections in the elastic range, (b) P-M strength interaction, and (c) both axial
and bending deformations after yield. However, fiber models can not necessarily predict ductile limits and subsequent strength loss. These depend on
complex aspects of behavior that are not necessarily included in fiber
models. Some of these are as follows.
(1)
The ductile limit for a reinforced concrete section may be reached
when the concrete crushes and loses strength. In an actual cross
section, crushing· starts at the extreme edge or comer of the cross
section and progresses continuously into the section. In a fiber model,
crushing occurs fiber-by-fiber and progresses discontinuously into the
section as more fibers yield. If relatively few fibers are used to model
a cross section, the fibers are large, crushing starts later in the model
than in the actual section; and the crushed part of the cross section
changes in relatively large jumps. As more fibers are used the model
becomes more accurate, but the computational cost increases.
(2)
The strength of concrete in compression, and also its ductility, depend
on the amount of confinement. In a column, some concrete will be
within the confinement and some will be in an unconfined outer shell.
This may need to be accounted for in the fiber model. Also, in any
given column the effectiveness of the confinement, and hence the
strength and ductility of the concrete, may be uncertain. The same is
true in a wall.
(3)
Under cyclic loading the reinforcement can yield in tension in one half
cycle and compression in the nexthalf cycle. Also, cracks that open in
OJ¥ cycle may not close completely. The ductile limit of a cross section
may be governed by buckling of the reinforcement as it yields in
compression. In a fiber model, the stress-strain relationship for steel
fibers can, in principle, account for buckling, but the buckling
behavior is uncertain and difficult to model.
(4)
The largest inelastic deformations in a column are likely to occur at
the column ends, either at a beam-to-column connection or at the
foundation level of the structure. At these locations there can be large
bond stresses, with significant penetration of bond slip into the
connection region or the foundation. This can have a substantial effect
on the column stiffness, artd possibly on its strength. Bond slip is not
considered in a basic fiber model. Bond slip can be modeled, but the
164
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
process is usually .too complex to be included in a model of· a
complete structure.
(5)
The strength ofa column may be controlled by shear, or by P, M and
V (shear force) acting in combination. The basic fiber model considers
only P-M irtteraction. As noted later, it is ·much more difficult to
model P-M-V interaction.
In summary, a fiber model can be useful, but it is not a complete solution. A
fiber model may not be accurate for large cyclic deformations, it does not
account for bond slip or shear force effects, and it probably can not predict
the ductile limit and the amount of strength loss. Fiber models can certainly
be better than models based on plasticity theory, but they still have major
limitations.
·
5.11 Inelastic Shear in Beams and Columns
5.11.1 Overview
The preceding sections have considered P-M interaction. This section adds
shear, using yield surfaces and plasticity theory, for beam and column
components. A later section considers shear in walls.
In principle it is possible to consider inelastic shear behavior in beams and
columns using P-M-V yield surfaces and plasticity theory. In .practice,
however, it is extremely difficult, especially for cyclic loading, . and
especially for reinforced concrete. There have been few, if any, practical
applications. As a consequence, much of this section is based on speculation,
not experience.
5.11.2 M-V Interaction in Steel Beams
In a steel I-beam that is bent about its strong axis, bending is carried mainly
by the flanges, and shear is carried mainly by the web. Hence, there is only
weak strength interaction between bending moment and shear force. This is
illustrated in Figure 5.23.
·
Inelastic Shear in Beams and Columns
165
Typical slender beam
Moment
Shear link in
an eccentrically
braced frame
Strong interaction
Shear
Figure 5.23 Moment-Shear Interaction
For most steel beams the strength is governed by bending, and shear has a
negligible effect within the body of the beam (but not necessarily at the end
connections). The inelastic behavior in such a beam can be modeled using
uniaxial moment hinges. The opposite is the case for a shear-controlled link
in an eccentrically braced frame. The inelastic behavior in a shear.link can be
modeled using a uniaxial shear hinge. These two cases are shown in the
figure.
The figure also shows the case with yield in both bending and shear. In
principle it is possible.to model the inelastic behavior for this case using a
biaxial hinge with M-V interaction. In practice, however, it is unlikely that
the bending and shear strengths will be known with sufficient accuracy, and
it will· be uncertain whether the beam yields in bending, shear, or some
combiriation of the two. If this situation can not be avoided, the best
approach is to bound the behavior by analyzing two models,- one with yield
in bending only and the second with yield in shear only. An analysis that
attempts to account for M-V interaction is unlikely to be reliable enough for
making design decisions.
·
The situation is· more complex for a steel I-beam bent about its weak axis,
because there is stronger interaction between M and V. It is unlikely, however, that such a beam will have a substantial shear force. In the general case
there could be biaxial bending and biaxial shear, with M-M-V-V interaction.
Inpractice this isunlikely.
5.11.3 P-M-V Interaction in Steel Columns
P-M strength interaction in columns can be modeled using P-M-M hinges.
In principle, P-M-V interaction can be considered using P-M-M-V-V hinges.
In practice it is unlikely that such a hinge component would be useful. Cases
166
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
where the strength of a steel column is clearly governed by shear are likely
to be rare, and in those cases it is probably accurate enough to use a shear
hinge with no P-V or M-V interaction. As with a beam, if it is uncertain
whether a column yields in bending, shear, or some combination of the two,
a hinge with P-M-V interaction is unlikely to be accurate, and the best
approach is to bound the behavior using simpler hinges.
5.11.4 M-V lnteradion in Reinforced Concrete Beams
The shear strength of a reinforced concrete beam can be separated into two
parts, namely (a) the contribution of the concrete and (b) the contn"bution of
the shear reinforcement. The concrete contribution is affected most by
interaction.
M-V strength interaction may be considered in strength-based design using
elastic analysis. For example, ACI 318 includes an equation for the concrete
contnbution to the shear strength at a cross section.. The form of this
equation is:
shear strength = a + b(VIM)
where the parameter a depends on the concrete strength, b depends on the
amount of longitudinal .reinforcement, and V and M are the shear force. and
bending moment demands at the cross section.
For inelastic analysis it is presumably possible to define a surface for M-V
strength interaction (although it is not obvious how to do so using the above
equation), and to use this as the yield surface for a hinge component with a
multi-axial F-D relationship. However, there are complications that make it
difficult to develop such a relationship. Two of these are as follows.
(1)
Inelastic shear deformation in concrete is not plastic, so plasticity
theory may not model the behavior correctly.
(2)
The behavior in shear is likely to be less ductile than in bending. Also,
for cyclic loading there is likely to be more rapid degradation in shear
than in bending. When there is strength loss and degradation, the size
and shape of the yield surface changes. The rules that govern these
changes are likely to be complex.
In practice, since shear in reinforced concrete tends to be brittle, it is usually
wise to avoid inelastic shear behavior by using capacity design. If inelastic
behavior in shear can occur, it may be more practical to use a uniaxial shear
Inelastic Shear in Beams and Columns
167
hinge, with a shear strength based on the above equation and a reasonable
• estimate for the ratio V/M.
5.11.5
P~M-V Interaction
in Reinforced Concrete Columns
The shear strength of a reinforced concrete column can be affected by both
axial force and bending moment. First consider P-V interaction. This is
illustrated in Figure 5.24.
,-
=-:;4~i
contribution
Compression
.".
Tension
(a) Short length of column
(b) Effect of axial force on shear strength
(c) P-Vyield surface
(d) Possible V1-V2 yield surface
Figure 5.24 Interaction Between Axial Force and Shear Strength
Figure 5.24(a) shows a short length of column, with longitudinal and shear
reinforcement. As in a beam, the shear strength can -be separated into
contributions from the concrete and the shear reinforcement. The contribution of the concrete is affected by the axial force in the column. The
contribution of the shear reinforcement is usually- assumed to be
independent of the axial force. In ACI 318, the relationship between shear
strength and axial force has the form shown in Figure 5.24(b). The details of
this relationship are not important for the current discussion, but it may be
noted that the relationship is based on experimental results, and that these
results have a lot of scatter.
The relationship in Figure 5.24(b) is used for strength-based design using
elastic analysis. It could also be used to set up a P-V yield surface for
inelastic analysis. For the two-dimensional case the P-V surface is shown in
168
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
Figure 5.24(c). In three dimensions the P-V-V surface has a funnel shape,
•
with a cross section defined by a Vl-V2 surface.
The Vl-V2 interaction surface for a particular value of P might be as shown
in Figure 5.24(d). For a circular column this surface is a circle. For a square
or rectangular column the surface may be a circle or ellipse. However, the
· shear strength along the diagonal of a square column is not necessarily the
same as the strength along a principal axis of the section, so the surface for a
square column is not necessarily a circle. The shape of the Vl-V2 surface
may also change as P changes.
The shape of the yield surface indicates that the behavior is partly cohesive
(because the shear strength is nonzero when P = 0) and partly frictional
(because the shear strength depends on P).
It is possible to set up a P-V-V hinge based on this yield surface. Some
complications for such a hinge are as follows.
(1)
Shear deformations in reinforced concrete are not plastic.
(2)
If P is constant and V is increased, the plastic "flow" drrection after
yield is essentially shear slip only (there may be some axial deformation,
but the amount is uncertain). Hence, for much of the yield surface the
flow is non-associated.
(3)
It may be necessary to account for strain hardening, the ductile limit,
cyclic deformation and cyclic degradation.
In a column element, a shear hinge of this type could be used in series with
a fiber segment, so that the fiber segment accounts for P-M-M interaction
and the shear hinge accounts for P-v~v interaction. Such a model would not
consider M-V interaction. In principle, P-M-V interaction in a column could
be modeled using a P-M-M-V-V yield surface and a corresponding hinge
component. Interaction with torsional moment might even be considered. In
practice such a hinge is almost certainly impractical.
. 5.11.6 P-M-V Interaction in Connections
··-.......
It has been said that anybody can design a beam, but it takes an engineer to
design a connection. This is also true for· setting up the analysis model for a
connection. A beam-to-column connection can be much more difficult to
model than the body of a beam or column, especially when there is P-M-V
interaction and the behavior is inelastic.
Shear in Concrete Walls
169
S.11.7 Analysis vs. Design
Because shear in beams and columns can be difficult to model, it is worth
emphasizing again that the goal is design, not analysis. Accurate modeling
of inelastic shear for analysis is impossible, especially in reinforced concrete.
This does not, of course, mean that it is impossible to design for shear.
As a general rule, the behavior of reinforced concrete in shear is brittle, with
substantial degradation under cyclic deformations/(there are exceptions,
notably coupling beams in coupled shear walls, which can be designed to be
ductile in shear). It is usually wise, therefore, to design reinforced concrete
beams and columns to remain essentially elastic in shear.
If inelastic shear must be considered, it is not necessarily more accurate to
use an elaborate analysis model. There will inevitably be major uncertainties, on both the demand side (modeling the behavior and calculating shear
deformation demands) and the capacity side (estimating shear deformation
capacities). For practical purposes it may be accurate enough to use a shear
hinge that ignores the effects of axial force and bending moment (a uniaxial
hinge in a beam, and a biaxial hinge with V-V interaction in a column). A
more elaborate model may merely give the illusion of greater accuracy.
5.12 Shear in Concrete Walls
The preceding sections have considered components, including plastic
hinges, fiber sections and fiber segments. To extend the discussion to shear
in walls, it is necessary to go beyond components and consider elements.
A beam or column element is usually a line element connecting two nodes.
Such an element can be made up from a number of components, in series
along the element length. For example, a beam element might consist of.
seven components, as shown in Figure 5.25.
This element iricludes short fiber segments to model inelastic bending with
P-M interaction (a P-M plastic hinge could also be used), and a uniaxial ·
shear hinge to model inelastic shear. These are separate components, and
there is no P-M-V strength interaction (statics requires that the maximum
possible shear force in the hinge depends on the strength in bending, but
that is not strength interaction).
170
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
End zone
End zone
Short fiber segment,
similar to a P-M hinge
Short fiber segment
Elastic beam segment
Elastic beam segment
Shear hinge
Figure 5.25 Beam Element with Components in Series
A wall element is different. Although it is possible to model some wall
structures using line elements, it is usually better to use surface finite
elements. One possible element is shown in Figure 5.26.
D
D
Vertical fibers
Horizontal fibers
Shear
(a) Wall element
(b) Element layers
Figure 5.26 A Possible Wall Element, With Layers
In this case the wall is essentially a vertical cantilever. Forthe purposes of
this discussion, consider a rectangular element and assume that the
reinforcement in the wall is vertical and horizontal. A.s noted earlier, the
behavior for out-of-plan~ bending can often be assumed to be elastic, and to
be uncoupled from the in-plane membrane behavior. For membrane
behavior there are three modes of action, as follows.
(1)
Inelastic behavior with P-M interaction in the vertical direction. This
can be modeled. using a fiber section. Some of the fibers model the
concrete and others model the steel reinforcement.
... ;";>,.
""'···
.,
Shear in Concrete Walls
(2)
(3)
171
Elastic behavior for P and M in the horizontal direction. This aspect of
behavior is secondary for a tall shear wall. It can be modeled using
elastic horizontal fibers, as in the figure, or alternatively by specifying
· EA and EI values.
Shear, considered as follows.
As shown in Figure 5.26, the element can be regarded as having a separate
layer for each mode of action. Each layer is a component, and these
components act in parallel. Conceptually this is similar to using a fiber
section component in a beam element to account for P-M interaction, and a
shear hinge component to account for shear deformation, . with the
difference that the components are in series along the beam length, not in
parallel.
·
The simplest assumption for the shear layer is that it is elastic. This requires
only the shear stiffness, which depends on the shear modulus and the
thickness of the wall.
H inelastic shear behavior must be considered, the simplest assumption is
that the material in the shear layer has a uniaxial relationship between shear
stress and shear strain, with a strength that is independent of P and M. This
· is equivalent to using a uniaxial shear hinge in a beam. Since the strength of
a wall in shear is not independent of P and M, this may not be an accurate
assumption. As noted for beams and columns, however, the goal of the
analysis is to calculate a reasonable demand-capacity ratio for making
design decisions, using inelastic shear deformation as the demand-capacity
measure. No matter how elaborate the analysis model, there will always be
uncertainty in the calculated demand. Just as importantly, there will always
be uncertainty in the capacity. Given all of the uncertainties, a relatively
simple uniaxial shear model may be the most appropriate model for
practical purposes.
This element in Figure 5.26 has the disadvantage that it does not consider
P-V or M-V interaction. It also depends on the assumption that a wall is
essentially a vertical cantilever, with axial force, bending moment and shear.
This type of element would not be appropriate for a wall with an arbitrary
shape and complex loading. A second possible element, which is more
general, is shown in Figure 5.27.
·
172
Chapter 5 Component Behavior- Multi-Axial F-D Relationships
D
D
D
r"---lr-
Integration
point
Vertical steel
(a) Wall element
(b) Element layers
Figure 5.27. A Second Possible Wall Element
This is a different type of element from that in Figure 5.26. The details are
not important here, but the element in Figure 5.26 is a "macro" or
"engineering" element, with fiber sections, whereas the element in Figure
5.27 is a more typical finite element, where the material behavior is
monitored at a number of integration points (the figure· shows 2-by-2
integration). The element in Figure 5.27 is rectangular, but in general it can
have <l!l arbitrary shape. The figure assumes two layers for the steel
reinforcement, with vertical and horizontal bars, but in general there could
be several layers, with any orientation. This type of element could be used
for walls of arbitrary shape.
Two major differences from the element in Figure 5.26 are as follows.
(1)
Jn Figure 5.26 the layers all have uniaxial stress-strain relationships. In
Figure 5.27 there is a concrete layer with a multi-axial relationship and
a 3-by-3 constitutive matrix. This is a much more complex concrete
model.
·
(2)
Jn Figure 5.26 the element extends across the width of the wall, and
there can be many fibers in the wall cross section. The element in
Figure 5.27 has, in effect, only two fibers across the element width
(one at each integration point), so a number of elements are needed
across the wall width to .provide enough· total fibers. More fibers
could be obtained within a single element by increasing the number of
;_
Multi-Axial Material Models for Plain Concrete
173
integration points, but it is usual to use 2-by-2 integration for this type
of element.
The steel layers are relatively easy to model, using uniaxial hoers. The main
· concern here is the concrete layer, where a multi-axial stress-strain relation~
ship is required for the concrete. This is considered in the next 8ection
/
5.13 Multi-Axial Material Models for Plain Concrete
5.13.1 Motivation
·
The discussion of plasticity theory in this chapter began at the material level
for steel, and progressed, by analogy, to the cross section level with P-M
interaction and P-M-V interaction. A major problem with this analogy is
that it does not work well for reinforced concrete cross sections;
As indicated in the preceding section, one way to model reinforced concrete
. is to go to the material level and model the concrete behavior directly. A
fiber model of a cross section uses this approach, but it is limited because it
assumes that the fibers have uniaxial stress-strain relationships. This
accounts only for P-M interaction. To account for P-M-V interaction it is
necessary to consider a multi-axial stress-strain relationship'. A particular
challenge is to model the behavior of concrete under multi-axial stress,
considering cracking, crushing, shear behavior, strength loss, etc.
Finite elements of the type shown in Figure 5.27 are often used to model
walls for elastic analysis. fu this case the material is elastic, and the 3-by-3
material constitutive matrix is easy to determine. An attractive approach for
inelastic analysis would be to use the same elements as in an elastic analysis
model, and simply replace the elastic stress-strain relationships with
inelastic ones. It would then be possible to use an elastic model to study the
linear behavior of a structure, and to study the nonlinear behavior by
changing the material properties from elastic to inelastic.
Unfortunately it is not so simple. For an elastic element the multi-axial
stress-strain relationship is indeed siinple, depending only on Young's
modulus and Poisson's ratio (although for reinforced concrete it is. not
obvious what their values should be). For an inelastic element the multiaxial stress-strain relationship iS far from simple, especially for cyclic,
loading. This section considers some possible models:
/
174
Chapter 5 Component Behavior- Multi-Axial F-0 Relationships
5.13.2 Plasticity Theory
One type of multi-axial model for plain concrete is a plasticity modeL This
model is conceptually similar to that used fot yield of metals and for P-M
interaction, but.is more complex because concrete is not a cohesive, plastic
material. A number.of yield surfaces, flow rules (often non-associated) and
strain hardening rules have been proposed by different researmers. All are
complicated and few, if any, are accurate, espeGiallyfor cyclic loading.
.
·.
For plane stress, the yield surface typically has the shape shown mFigure
5.28.
a 2 (compression)
(a) Biaxial stress
(b) Yield surface
Figure 5.28 20 Interaction Surface for Plain Concrete
This figure considers principal stresses, whim corresponds to a 2-by-2
constitutive relationship. This makes it easier to draw the interaction surface. At an integration point in a wall element there are three stresses, with a
3-by-3 relationship. The amount of information is the same, corresponding
essentially to two principal stresses plus an angle to the principal stress
axes.
hi the compression-compression quadrant in Figure 5.28, the concrete
crushes, whim can be treated approximately as plastic behavior provided
unloading does not occur. In the other quadrants, the concrete cracks in
tension, whim is more difficult to treat as plastic behavior, especially when
cracks open and dose.·
In 3D principal stress space, the yield surface has the shape shown in Figure
5.29. .
Multi-Axial Material Models for Plain Concrete
175
0'1
Figure 5.29 30 Interaction Surface for Plain Concrete
.
.
This is analogous to the von Mises cylinder shown earlier, in Figure 5.R
However, coocrete has both cohesive and frictional behavior. An interesting
aspect iS that a cross section through the surface at constant hydrostatic
stress is a: curvilinear triangle, not a Qi"cle.
·A material model based on this type .of interaction surface is extremely
complex, especially for cyclic loading. This type of model has been applied
successfully for monotonically increasing deformation, but it is too complex
for the analysis of most reinforced concrete structures with cyclic loading
and degradation and is not ne,cessarily accurate.
S.13.3 Compression Field Theory
The preceding section considered plasticity theory. This section considers
the main features of a very different approach, which is usually referred to
as "compression field theory", Figure ~.30 illustrates the concept.
f
+.-
i
(Je2
___. 'te12·
t-+CJe1
+-.
~
(a) WaH. Finite Element,
Stresses at a Point
(b) Element Axes
(c) Material Axes
Figure 530 Material Model for Compression Field Theory
176 · Chapter 5 Component Behavior - Multi-Axial F-D Relationships
To simplify the explanation, assume that the concrete has zero strength in
tension. The theory can be extended to include nonzero tension strength.
Consider an analysis in which the loads are applied step-by-step, in a series
of small increments. The procedure at each integration point in a wall
element is essentially as follows,
(1)
Cracking and crushing of the concrete are considered along a pair of
material axes, which are parallel to the principal stresses. Initially the
material is elastic, and the multi-axial elastic stress-strain relationship
is used. Load increments are applied, and at each increment the
principal stresses are calculated. H either principal stress is tension, or
is compression and exceeds the yield (crushing) strength, this establishes the material axes.
(2)
The multi-axial tangent stress-strain relationship is set up assuming
uniaxial behavior along the material axes, using a zero modulus along
an axis if the concrete is in tension, and using the strain hardening
modulus if it is yielded in compression. In addition, the concrete is
assumed to have a shear stiffness in the material axes. If the concrete
is cracked, this means that shear can be transmitted across a crack.
The theory specifies how the shear modulus .is calculated. The details
are not important for this discussion.
(3)
Since the shear modulus is nonzero, shear stress increments may be
calculated .when the next load increment is applied. Hence, ·the
principal stress directions can rotate. Since the material axes are
always parallel to the principal stres~, the 'material axes can also
rotate.
(4)
In each load 'step, strain increments are calculated~ The stress
increments follow from the tangent stress-strain relationship for the
step. New principal stresses are calculated, and hence new material
· axes. The strain increments along these axes are used to update the
uniaxial material states. If the material axes have rotated significantly,
or if the uniaxial modulus along either axis changes, the multi-axial
stress-strain relationship is recalculated for the next foad step.
The preceding is a fairly straightforward procedure, although it is a logical
and computational challenge to account for rotation of the material axes.
This procedure was applied by Darwin and Pecknold in 1974. However, it
ignores an important aspect of behavior that was observed, in about 1980,
by Vecchio and Collins in experiments on the shear behavior of reinforced
concrete. In the experiments, the specimens cracked along one material axis,
Multi-Axial Material Models for Plain Concrete
177
and the shear force was essentially carried by compression along the other
axis. The experiments showed that as the tension strain along the cracked
axis increased, the c<:>mpressive strength along the compression axis
decreased. The amount ·01 the.clecrease can be substantial, and has a major
effect on the shear strength.
The relationship between tension strain and compression strength has the
form shown in Figure 5..Jl. Again, the details are .not important for this
discussion.
,,
Reduction
factor, r
Strength = rfc'
~
Strain, sm2
Figure 5.31 Compressive Strength Reduction Factor
This behavior is important and must be considered in the analysis. At the
end of each load step, if the uniaxial material along one material axis is in
tension and the other is in compre~sion, the strength of the compression
material may ne~d to be reduced, based on the magnitude of the tension
strain. Figure 5.32 shows how the uniaxial material properties might be
modified.
Stress, crm1
Initial cr-&
Reduced cr-e
Strain, em1
Figure 5.32 Reduced Stress-Strain Relationship
178 · Chapter 5 Component Behavior - Multi-Axial F-D Relationships
The above theory is reasonable provided the material axes rotate through
small angles, which is often the case for non-cyclic l<Jads. If there is cyclic
deformation, however, there can be large changes in the directions of the
material axes, causing both computational and theoretical difficulties. The
theory is usually applied only for non-cyclic loads.
Jn.this model, ·shear is explicitly resisted by diagonal compression in the
concrete. There is no concept of a yield surface, plastic flow, or cohesive
versus frictional behavior, and hence this model is very different conceptually from a plasticity model.
Both models, however, depend on interaction with the reinforcement layers.
For example, for diagonal compression to be present,. the compression force
in a diagonal strut must be resisted, often by truss-type action with steel
reinforcement providing the tension resistance. At the integration points in
an element it is usual to assume that there is perfect bond between the
concrete and steel layers. This is, of course, a questio~ble assumption.
5.13.4 Simple Models Based on Uniaxial Stress-Strain
The plasticity model and the compression field model are both complex. A
much simpler model for plain concrete can be obtained by assuming that
the 3-by-3 constitutive matrix is diagonal, so that there are three uncoupled,
uniaxial materials, one each for vertical, horizontal and shear stress. With
this assumption the single concrete layer in Figure 5.27 can be replaced by
three sub-layers, one each for the vertical, horizontal and_ shear stresses in
the concrete. With two layers for the reinforcement, this makes a total of five
layers.
Figure 5.33(a) shows the three sub-layers for .plain concrete, one with
· uniaxful vertical fibers, one with uniaxial horizontal fibers, and the third .
with uniaxial shear (i.e., shear that is independent of the other stresses). This
is dearly an over-simplification of concrete behavior. It might be a reasonable approximation for the analysis of walls with well-defined vertical,
horizontal and shear directions, but it is not suitable for walls of arbitrary
shape. One reason is .that the 3-by-3 elastic constitutive matrix for concrete
(with no cracking or crushing) is diagonal for axes that are vertical and
horizontal, but changes and becomes coupled if the axes are rotated. In a
consistent theory, the elastic constitutive matrix should not change when the
·
axes are rotated.
An alternative model that partially overcomes this problem is shown in
figure 5.33(b). This model replaces the shear layer with two layers that have
Multi-Axial Material Models for Plain Concrete
179
diagonal fibers. This model has four layers, all with uniaxial fibers of
concrete type. This is essentially the same as a "lattice" model, with uniaxial
struts as shown in Figure 5.33(c). H the concrete is assumed to have no
strength in tension, shear is resisted by diagonal compression.
Shear layer
replaced by
two 45° layers
(a) Model with 3 layers
(b) Model with 4 layers
(c) Lattice
Figure 5.33 Layered .and Lattice Models for Plain Concrete
When a material model is applied to a wall of arbitrary shape, the model
should give the same results regardless of how the material axes are
oriented (for inelastic as well as elastic behavior). The four-layer "lattice"
model does not fully satisfy this condition, but it is better in this respect
than the three-layer model. In particular, there is no difference in the
behavior when the material axes are rotated 45 degrees. The behavior does
change for a rotation of, say, 22.5 degrees, but the approximation may be
acceptable.
The 3-by-3 constitutive matrix for a lattice model changes as cracking and
crushing occurs in the individual layers. An advantage of this model is that
it is easy to extend to cyclic deformation, since each layer has uniaxial fibers.
The behavior of the model can be adjusted by varying the layer thicknesses
relative to the actual thickness of the wall, and by varying the compressive
strength and other properties of the uniaxial material. The behavior can not,
however, be adjusted to agree closely with the observed behavior of actual
concrete for multi-axial stress and strain. Simple models such as this are
inevitably very approximate.
It may be noted that lattice models have been used successfully to model the
behavior of concrete at a microscopic level, considering the interaction
between the cement paste and the aggregate. These models are· complex, with 3D lattices, and are not suitable for the practical analysis of large
structures.
180
Chapter 5 Component Behavior- Multi-Axial F-D Relationships
5.13.5 Possible Model with P-V Interaction
e·
Figure 5.24 showed a P-V interaction surface that could be used for a
concrete column. This interaction surface could possibly be used at the
material level.
The element in Figure 5.26 has three layers, one for vertical P and M, one for
horizontal P and M, and one for shear. There is no strength interaction
between layers. It may be possible to modify the shear layer to use the P-V
interaction surface, so that the strength of the shear layer depends on the P
value in the vertical P-M layer. With this model, the shear layer would
effectively have a biaxial stress-strain relationship.
It may also be possible to eliminate the shear layer and use a biaxial stressstrain relationship for each concrete fiber in the vertical P-M layer. Each
fiber would have normal and shear stresses, with a P-V interaction surface.
Since each fiber in a fiber section can have a different normal stress, this
would account.for P-M-V interaction.
A possible problem with this model is that when a concrete fiber cracks in
tension, the tension stress in the fiber is zero. Hence, there is no tension P
value in the concrete, but the P-V interaction surface requires such a value
(for P-V interaction in a column, the .tension force is the force in the
reinforcement). fu this case, the concrete shear strength presumably
depends on the tension strain, not the tension stress. This is different from
basic plasticity theory. It is just one of many complications that can arise in
modeling plain concrete.
5.13.6 Plain Concrete Models for 3D Stress
The preceding sections have. considered 20. stress-strain relationships for
modeling membrane behavior in walls, assuming that the plate bending
part of the behavior is elastic. If the plate bending behavior is not elastic, a
wall element can still be regarded as a surface element, by dividing it into a
number of membrane layers through the thickness in much the same way
that the cross section of a beam element can be divided into a number of
uniaxial fibers.
If it is necessary to model a true 3D solid, using solid elements, a 30 stressstrain relationship is needed, with a 6-by-6 constitutive matrix. Of the plain
concrete models considered in this section, the plasticity theory model
applies to both 20 and 3D stress, and the lattice model bars can also be
....
-~.
Multi-Axial Material Models for Plain Concrete
181
extended to the 3D case. The layered models and the compression field ·
model apply only to the 2D case.
It is rare to do inelastic analyses of reinforced concrete using 3D solid
elements, and it is a task requiring a great deal of skill and judgment. The
vast majority of analyses use line elements for beams and columns, or
surface elements for walls.
5.13.7 Over-Reliance on Analysis
As this section shows, there are many possible models for plain concrete, ·
with different assumptions and different degrees of complexity. There is no
model that is sufficiently accurate, reliable and simple to implement for
"exact" analysis of inelastic concrete shear walls. Some methods are simple
to implement but do not accurately model the multi-axial behavior of
concrete. Others are more accurate for monotonically increasing deformation but can be complex. All existing models are of questionable accuracy
for cyclic deformation.
Even if an accurate model were available for plain concrete, it is still
necessary to model the steel reinforcement and the interaction between the
concrete and the steel. This introduces still more complications, some of
which are as follows.
(1)
(2)
(3)
(4)
(5)
(6)
Anchorage and bond slip.
"Tension stiffening", where the concrete between cracks behaves as if
it has significant tensile strength.
Crack opening and re-closing.
,
Dowel action in the reinforcement across cracks.
Sliding on construction joints, with "shear friction". This is illustrated
in Figure 5.34.
Sliding on rough cracks.
Shear force
_........_..,..__....,_~......,,.,.....·
J
Shear sliding on
construction joint ..
As reinforcement deforms
it develops a tension force.
This compresses the joint and
increases the shear strength.
Figure 534 Shear Friction
182
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
These are all difficult to model. In most practical analyses they are ignored,
for example by assuming perfect bond, or are considered indirectly by
making empirical adjustments to the model.
The end result of this complexity is that it is very difficult to set up an
accurate inelastic model of a complex concrete structure. In particular, it is
not realistic to set up an elastic finite element model, then expect that the
inelastic behavior can be predicted simply by replacing the elastic material
·properties with inelastic ones. It requires much more thought than this.
·.
This is especially true for earthquake loads, where there is likely to be
inelastic cyclic behavior. The performance of a structure for earthquake
loads may be assessed using inelastic static push-over analysis rather than
inelastic dynamic analysis. There is no cyclic deformation in static pushover analysis, so the analysis must account indirectly for cyclic loading. This
must be done by adjusting the F-D relationship, which usually means
reducing the strength and ductility below that expected for monotonically
increasing deformation. If a multi-axial material model is used for pushover analysis, and if the stress-strain relationship for the material model
applies only for monotonically increasing deformation, the analysis results
can be substantially incorrect. When analysis models of the type discussed
in this section are used, there is a danger that this mistake will be made.
Since it is not possible to account reliably for inelastic shear in an analysis
model, what can be done? Ideally,. reinforced concrete structures will be
designed to remain essentially elastic in shear, using capacity design. In this
case only P-M interaction needs to be considered in the inelastic analysis
model. If this is not possible, simple models should be considered first, since
they may be able to give sufficient information for design. These models
include uniaxial shear hinges in beams, Vl-V2 hinges in columns, and fiber
models with uniaxial fibers and separate shear layers in walls (see Figure
5.26). If a more complex model is needed, with P-V hinges or fibers with
multi-axial material models, such a model should be used with a great deal
of care and caution.
5.13.8 A Note on Demand and Capacity Analyses
The difference between a demand analysis and a· capacity analysis was
emphasized earlier (see Section 1.8). This difference applies to the types of
analysis considered in this section.
Consider the relatively simple case of P-M interaction in reinforced concrete
with two alternative models, namely (a) a fiber cross section with uniaxial
Capacity Interaction
183
fibers and (b) a P-M hinge with plasticity theory. For the P-M hinge, the P
and M strengths are based on experimental results (this includes strengths
calculated by methods or formulas that have been calibrated against
experiment). The shape of the yield surface, the ductile limits, and other
properties such as cyclic degradation are also based on experiment. In
contrast, for a fiber section the P and M strengths are not based directly on
experiment, but are calculated from the strengths of the fibers. The fibers
strengths are, of course, based on experiment, buf it does not necessarily
follow that a fiber model gives the correct P and M strengths. It also does
not necessarily follow that a fiber model gives the correct shape for the P-M
interaction surface, or accurately predicts ductile limits and cyclic degradation.
To a substantial extent, an analysis that uses P-M fiber models is a capacity
analysis, where the analysis calculates the P-M strength capacity. The
analysis is also a demand analysis, since it. calculates such things as
deformation demands. In contrast, an analysis that uses P-M hinges is only
a demand analysis, since the strength and deformation capacities are
specified directly.
·
The case of P-M-V interaction in reinforced concrete is more complex. P-MV hinge models, with plasticity theory, are likely ·to be too complex for
practical use. Fiber models that account for shear by using multi-axial stressstrain relationships for the fibers are also complex, and given the difficulty
of modeling concrete under multi-axial stress conditions, they are not
necessarily accurate. Capacity analyses are inherently more complex than
demand analyses, and unless they are thoroughly calibrated . against
experiment, there is a risk that they will give misleading results.
5.14 Capacity Interaction
5.14.1 Overview
Earlier sections in this chapter have considered stiffness, strength and
inelastic interaction. This section considers a fourth type, namely "capacity
interaction". This is where a strength or deformation capacity (i.e., the
capacity value used to calculate a demand/ capacity ratio)· depends on some
other force or deformation.
One example is P-M strength interaction, where the bending strength
depends on the axial force. H the bending strength defined by a P-M
· 184
Chapter 5 Component Behavior - Multi-Axial F-D Relationships
mteraction surface is used, as the strength capacity, this is capacity interaction. It is also an example of strength interaction.
This section give$ some other examples.. ·
5.14.2 Effect of Axial Force on Bending Ductility
In a column, the ductility is usually considered in terms of · bendlng ·
deformation only, ignoring axial deformation. Experiments on columns, for
both steel and reinforced concrete, show that as the axial compression force
on a column increases, the column becomes less ductile. That is, there is
interaction between the axial compression in the column and its ductile limit
in~~g
.
ff inelastic behavior in a column is modeled using a plastic hinge that has
P-M interaction, the rotation capacity of the hinge is usually based on the
hinge rotation at the ductile limit. As the axial force demand iricreases, the
rotation capacity decreases..
5.14.3 Effect of Shear Force on Bending Ductility
In a.reinforced concrete beam the ductility in ~~g is ·affected by the
magnitude of the shear force. As the shear force demand on a plastic hinge
increases, the rotation capacity of the hinge decreases.
5.14.4 Effect of Hinge Rotation on Shear Strength .
· In a reinforced concrete beam, when a plastic hinge forms, the strength of
the beam in shear is reduced. As the hinge rotation demand increases, the
shear strength capacity decreases.
5.15 Plastic Deformation and Deformation D/C Ratios
For a uniaxial F-D relationShip, elastic, plastic and post-yield deformations'
were defined earlier, in Section 4.2.4. That section noted that it is more
consistent to use plastic deformation than post-yield deformation. Plastic
deformations are often used for ductile limits and deformation capacities.
For a uniaxial F-D relationShip it is a straightforward process to calculate
maximum and accumulated plastic deformations, and to calculate· Separate
positive and negative values if they are needed. It is more complicated for
multi-axial F~D relationships with interaction.
\..
~~.
;,
1
i
I
I
Plastic Deformation and Deformation DIC Ratios
185
The basic concept is the same, namely that any increment of deformation for
a component can be divided into an elastic part and a f>lastic part. For the
uniaxial case the elastic deformation increment is the force increment
divided by the elastic stiffness. For the multi-axial case this becomes the
force increment multiplied by the inverse of the elastic stiffness matrix. In
the uniaxial case the force increment is the deformation increment
multiplied by the hardening stiffness. Hence, the force increment is zero for
an elastic-perfectly-plastic F-D relationship. In the mhlti-axial case there can
be a force increment as the force point moves around the yield surface, so
the force increment is not necessarily zero for an e-p-p relationship.
These are minor differences, however. The more important difference is that
the plastic deformation: for a multi-axial F-D relationship has a number of
parts. For example, for a relationship with P-M-M interaction the plastic
deformation will have axial and two bending parts. For each part it is
possible to calculate maximum and accumulated plastic deformations, and
to calculate positive and negative values, but these individual values may
not be useful. because they may interact.· For calculating deformation
demand-capacity ratios, the parts usually have to be combined in some way.
· Consider a dynamic analysis of a frame with inelastic columns. For a point
in a column, the axial force and the bending moments vary with time, and
also the plastic axial and bending deformations. For. assessing ·the
performance of the .column, it is usual to use only the plastic bending
deformation as the demand-capacity measure. One way to calculate the
bending deformation demand is to calculate the maximum. ·deformation
along each principal axis of the column- cross section, then calculate the
deformation demand as the resultant of these maxima. This is conservative
sin(.'.'.e the resultant of the maxima is an upper bound on the maximum of the
resultants. This may be accurate enough for design, but it is probably better
to calculate the resultant bending deformation at each time step and keep
the maximum.
As noted earlier, axial compression force affects bending ductility (see
Section 5.14.2). Hence, a given plastic deformation with a small axial
compression foJ,"ce is less damaging than the same deformation with a large
compression force. It follows that the important quantity is not just the
magnitude of the plastic deformation but the ratio of plastic deformation
demand to plastic deformation capacity. Since the axial compression force
can vary With time, the deformation capacity can also vary with time;c and
the deformation':demand-capacity ratio should be calculated at each tim~
step•
.
.
'
186
Chapter S Component Behavior - Multi-Axial F-0 Relationships
In general, for multi-axial F-D relationships, it is not enough to calculate just
the plastic deformations. It is important to process them appropriately to-get
. the demand-capacity ratios.
5.16 Summary for this Chapter
When the behavior of a component is affected by two or more forces, With
corresponding deformations, the forces can interact. This chapter has
identified four types of interaction, as follows.
(1)
Stiffness interaction (or stiffness coupling). This is present when the
· stiffness matrix for a component has non-zero off-diagonal terms. This
can often happen, and it usually does not cause any difficulty for
modeling. An exception is modeling for elastic analysis of a reinforced
concrete wall (see Section 5.2). ht that case, stiffness interaction is.
caused by shift of the neutral axis of the wall cross section as the
concrete cracks. This can cause modeling difficulties since the
behavior is not linear;
(2) Strength interaction. The most common example is P-M interaction in a
· · column, where the strength in bending depends on the axial force,·
artd vice versa. The interaction .can usually be represented by an
interaction surface. Strength interaction surfaces are used in strengthbased design using elastic aruilysis. In this case, strength interaction
affects only the strength demand/ capacity ratios, and it has no effect _
on the analysis model. Strength interaction surfac~ are also used for
inelastic analysis. ht this case, strength interaction defines when
elastic behavior erids and inelastic behavior begins. It does not say
anything about what happens in the inelastic range.
(3)
~-
Inelastic interaction. This is a broad term that covers the many types of
interaction that can occur after a component becomes inelastic from
yielding, crushing, etc. This type of interaction has a major effect on
modeling for · inelastic analysis. One example is the flow rule in
plasticity theory. A second example is the effect observed by Vecchio
and Collins, where cracking in concrete along one direction reduces
the strength in compression along the direction at right angles. A
· third example is the effect of axial force in a reinforced concrete
column on the shape of the hysteresis loop for bending deformation,
including degradation in stiffness, strength and ductility.
Summary for this Chapter
(4)
187
Capacity interaction. This is a broad term that covers cases where the
strength or deformation capacity for one force or deformation is
affected by the strength or deformation demand for a different force
or deformation. One example is the effect of axial force demand on the
bending strength capacity of a column (which is also strength
interaction). A very different example is the effect of axial force
demand on the bending deformation capacity of a column (i.e., large
axial compression reduces the bending ductility, and hence reduces
the deformation capacity for demand/capacity calculations). For
other examples see Section 5.14. This type of interaction affects only
the demand-capacity ratios, not the analysis model.
The most complex aspect of interaction is its effect on modeling for inelastic
analysis. This is mainly inelastic interaction. Some specific points are as
follows.
(1)
Inelastic interaction is complex, and there is no single, simple model
. that can be used to· account for it. Analyses that account for inelastic
interaction will inevitably be approximate.
(2)
Plasticity theory is often used to account for inelastic interaction. It
can be applied at the stress level (e.g., for yield of steel according to
the von Mises theory) or at the stress resultant level {e.g., for P-M
interaction in a column). Although plasticity theory has limitations, it
is a valuable tool. It is almost certainly accurate enough for modeling
steel components. It is also useful for reinforced concrete components,
but must be used more carefully because reinforced concrete is not a
plastic material.
(3)
Plastic hinge components based on plasticity theory can be used to ·
account for inelastic P-M interaction. This type of model is almost
certainly accurate enough for steel columns. It must be used more
cautiously for reinforced concrete columns, mainly because reinforced
concrete is not a plastic material.
{4)
A fiber model of a cross section, with uniaxial fibers, can be used to
account for P-M interaction. This type of model has the advantage
that it uses only uniaxial F-D relationships, and hence does not rely on
any assumptions about inelastic interaction. However, fiber models
can be expensive computationally.
(5)
Inelastic shear in reinforced concrete beams and columns can be an
important concern for design. If possible, inelastic shear in reinforced
188
Chapter 5 Component Behavior- Multi-Axial F-D Relationships
concrete should be avoided, because it tends to be brittle and also
because it is difficult to model (whicft is not a trivial reason). If
inelastic shear must be considered, the simplest and most practical
model for a beam is a uniaxial shear hinge with no interaction. For a
3D column, the most practical mod~l is a hinge with only Vl-V2
interaction, using plasticity theory. If M-V, P-V or P-M-V interaction
must be accounted for, there is no simple solution, and possibly no
reasonable solution at all. Plasticity theory does not work well, for the
reasons considered earlier in this chapter. The most practical
approach may be to use a shear hinge with no interaction and apply
design judgment when interpreting the results.
(6)
Inelastic shear in reinforced concrete walls can be a concern, and
again it is difficult to model. An obvious approach is to use a detailed
finite element model, with steel and concrete modeled separately.
Unfortunately, the inelastic behavior of plain concrete is especially
difficult to model, and there are also difficulties in modeling the
interaction between steel and concrete. If inelastic shear in concrete is
regarded as an analysis problem, at the time of writing, there do not
appear to be any good solutions. Ultimately, however, it is a design
problem, not an analysis problem. Given the current state of the art,
design solutions must be found that do not rely heavily on analysis.
(7)
Inelastic modeling becomes inherently more uncertain as the inelastic
deformations get larger. The uncertainty is even greater when the
behavior is affected by interaction. Also, when there are cyclic loads,
inelastic interaction can affect the degradation of stiffness, strength
and ductility. For uniaxial F-D relationships, Chapter 4 considered
two methods that can be used to account for cyclic degradation, a
relatively simple one based on a single backbone relationship, and a
more elaborate one based on a degrading relationship. When there is
inelastic interaction, even the method based on a single backbone
relationship can be difficult to implement.
(8)
For deformation-based design, the demands calculated from analysis
are compared with corresponding capacities, and the performance is
assessed using demand-capacity ratios. Inelastic interaction can have
substantial effects on the calculated demands. Inelastic interaction can
also make it more dif{tcult to choose demand-capacity measures that
correlate closely with structural performance. For example, w}lerf
there is biaxial bending the demand-capacity measure for. bending
deformation should presumably be some combination of the bending ·
deformations about the two axes, allowing for inelastic interaction,
Conclusion for this Chapter
189
not just the separate rotations about the two axes. To add to the
complexity, the capacity value may be affected by the axial force
(which is capacity interaction).
5.17 Conclusion for this Chapter
To model a structure for inelastic analysis, and als6 to calculate demand/
capacity ratios for performance assessment, it is important to understand
inelastic behavior, to be aware of the possible modeling options, and to be
clear on what results are required from the analysis. Different computer
programs will provide different models, with different assumptions for
interaction. To make effective use of a computer program, it is important to
understand the available models and the assumptions on which they are
based .
.Interaction adds complications and uncertainties to an already complex and
uncertain modeling process. This chapter re-emphasizes an important
theme of this book, namely that accurate simulation of inelastic behavior is
extremely difficult, and probably impossible, particularly for cyclic loading.
Whenever analysis is used, there is a danger that it can give misleading
results. Getting useful results from an analysis requires an understanding of
structural behavior, how to capture the important aspects of this behavior,
how to use the analysis results for design, and how to design when there is
uncertainty. This does not, of course, mean that analysis is not useful. It
does, however, mean that analysis must be used intelligently, not blindly.
Always keep in mind that the goal is design, not analysis. Accurate analysis
may be impossibly complex, but sound design is always possible. When
there is inelastic behavior it is always a good idea to consider capacity
design, to make the behavior more predictable and to make it ea~ier to set
up a useful analysis model.
Other things being equal, the best design is the one that requires the least
analysis, and the best analysis is the simplest one that gives the required
information. Sound engineering judgment and an understanding of structural
behavior can be more important than analytical skill.·
This chapter and the preceding one have considered material nonlinearity.
The next chapter considers geometric nonlinearity.
CHAPTER
6
P-Li Effects, Stability and Buckling
The two preceding chapters considered material nonlinearity. This chapter
considers geometric nonlinearity.
There is a fundamental difference between geometric nonlinearity and
material nonlinearity. Material nonlinearity has a wide range of causes,
many of which are poorly understood, and it is not governed by any single
theory. Geometrical nonlinearity, on the other hand, has clear causes and is
governed by a well-defined mathematical theory. Material nonlinearity is
subject to judgment and interpretation. Geometrical nonlinearity is not. This
does not mean, however, that geometrical nonlinearity is easy to account for
in an analysis. Its effects can be complex and subtle, and they can be difficult
to capture in an analysis model.
This chapter is concerned with the practical, rather than the theoretical,
aspects of geometric nonlinearity. Two main goals of the chapter are (a) to
explain geometric nonlinearity in physical terms, with a minimum of
theory, and (b) to make recommendations for modeling and analysis. This is
not easy to do, and this is a long and complex chapter.
6.1 Overview
6.1.1
Causes ofGeometric Nonlinearity
Geometric nonlinearity oceurs when the displacements of a structure are
large enough to affect one or both of the following.
191
192
Chapter 6 P-d Effects, Stability and Buckling
(1)
The equilibrium relationships. Equilibrium in the deformed position
of the structure may be significantly different from that in the
undeformed position.
(2)
The compatibility relationships. The relationships between element
deformations and element end displacements may be significantly
·
nonlinear.
6.1.2 Types of Analysis for Geometric Nonlinearity
There are three types of analysis that can be carried out, as follows.
(1)
Small displacements analysis. This is one extreme. Equilibrium is
considered in. the undeformed position, and the compatibility relationships are assumed to be linear. In this case geometric nonlinearity is
ignored.
(2)
Large displacements analysis. This is the other extreme. Equilibrium is
considered in the deformed position, and the compatibility relationships are nonlinear. In this case, geometric nonlinearity is considered
with no approximations.
(3)
P-J analysis. This is in the middle. Equilibrium is considered in the
deformed position (with some minor approximations), and the
compabbility relationships are assumed to be linear. In this case,
geometric nonlinearity is considered approximately.
If geometric nonlinearity must be considered, it is almost always accurate
enough to use P-A analysis. Only for very flexible structures, such as cable
structures, is it necessary to use large displacements analysis. P-A analysis is
more efficient computationally than large displacements analysis. For most
structures, it is a waste of computer time to account for true large displacements.
6.1.3
P-a Effect
In this book, "P-A effects" or "the P-A effect" refers to the effects of geometric
nonlinearity for those cases where P-A analysis is sufficiently accurate. For
cases where true large displaci:;ments analysis .is needed, the term "large
displacement effects" is used.
Overview
193
P-L\ effects are most commonly associated with axial forces in colqrnns.
However, P-L\ effects can also be caused by bending moments and shear
forces, and can be present in beams, walls and other components. The early
sections in this chapter consider only axial forces. Later sections consider
bending moments and shears.
6.1.4 Load Types and Corresponding Analyses/
Most structures are subjected to gravity and lateral loads (there are, of
course, other types of load). For a building the most important load case is
usually combined gravity and lateral load. However, the performance of a
structure is usually checked for both gravity load alone and for combined
gravity and lateral load. The analysis methods for these two load types can
be different. This chapter refers to them as "Gravity Load Analysis" and
"Lateral Load Analysis".
In most cases, especially for combined gravity and lateral load, the strength
of a structure is governed by progressive yielding. In some cases, however,
especially for gravity load alone, the strength may be governed by sudden
buckling. This can require special analysis methods, and is referred to in this
chapter as "Buckling Analysis".
6.1.5 Elastic and Inelastic Analysis
Strength-based design uses elastic analysis, which is usually linear. Asnoted earlier, however, "elastic" does not always mean linear. One reason is
that a structural component can have nonlinear elastic behavior. A second
reason, which is important for this chapter, is that geometric nonlinearity
can make elastic behavior nonlinear. As before, this chapter uses the term
"elastic analysis" rather than "linear analysis".
Deformation-based design uses inelastic analysis, which is inevitably nonlinear. This chapter uses the term "inelastic analysis" rather than "nonlinear
analysis".
Elastic and inelastic analyses have different goals, as follows.
(1)
Elastic analysis is used to calculate strength demands, which are then
used to calculate strength DIC ratios. For a frame structure, the
strength demands .are usually the bending moments, shear forces and
axial forces in the.frame members. These are referred to in this chapter
as "member f()rCe demands". It can be important to account for the P-L\
194
Chapter 6
P-~
Effects, Stability and Buckling
effect, because it tends to increase, or "amplify", the member force
demands. That is, if the member forces are calculated using a P-A
analysis, these forces can be substantially larger than those calculated
using a small displacements analysis. The member force capacities are
usually not affected by the method of analysis, and are the same for
P-A and small displacements analysis. Hence, the strength D/C ratios
from a P-A analysis are usually larger than those from a small.
displacements analysis.
(2)
Inelastic analysis is used to calculate deformation demands on
members that are allowed to become substantially inelastic, for
calculating deformation D/C ratios. It can be important to account for
P-A effects because these effects tend to amplify the deformation
demands. Inelastic cinalysis may also be used to estimate the collapse
strength of a structure (in which case the analysis is a Capacity
Analysis rather than a Demand Analysis). P-A effects can substantially
reduce the calculated collapse stren~.
An important goal in elastic analysis is to calctifate the amount of amplification with sufficient accuracy for design purposes. As this chapter shows,
this is not a simple task.
·
An important goal in inelastic analysis is to account for demand amplification in a demand analysis, or capacity reduction in a capacity analysis,
accounting for both material and geometric nonlinearity. This is also not a
simple task.
6.1.6
Topics for this Chapter
The· main concern of this book is modeling for structural analysis. The main
concern for this chapter is how to account for the P-A effect in an analysis
model.
Ideally it would be possible to use the same analysis model for both small
displacements and P~A analysis, with the same nodes, elements, loads, etc.,
and to tum P-A effects on or off as desired. Unfortunately it is not always
this easy, since some aspects of P-A behavior are difficult to model. To
account rationally for the P-A effect in design, an engineer must understand
how it affects structural behavior, how this behavior can (and can not) be
accounted for in analysis, and how the analysis results are used to assess
performance. This chapter attempts to explain the process, using physical
Overview
195
explanations rather than mathematical ones, keeping in mind that the goal
is design, not analysis.
This is a long chapter, covering many topics. A list of these topics is as
follows. If you get lost as you read this chapter, it might be useful to return
to this list to re-orient yourself.
(1)
P-LI and P-0contributions. The P-A effect has twq distinct parts, referred
to in this chapter as the P-A and P-o contrib~tions. This part of the
chapter distinguishes these contributions, assesses their relative
importance, and describes methods for modeling them. The P-A
contribution affects the overall behavior of a structure, and is
relatively easy to model, using "P-A struts" or ''P-A columns". The P-0
contribution has a more local effect, and is more difficult to model,
lising "P-li cables".
(2)
P-LI effect on frame behavior. This part of the chapter describes the P-A
effect in physical terms, using simple frame examples. For these
examples the P-o contribution is small, and only the P-A contribution
is considered. The examples give a broad overview of the P-A
contribution, for elastic and inelastic frames and for lateral load and
buckling analysis.
(3)
Modeling the P-LI contribution in three-dimensional and tall buildings. The
preceding examples consider only 20 single story frames, using a P-A
column to account for the P-A contribution. P-A columns can also be
used for 3D and multi-story buildings. This part of the chapter shows
how this can be done. P-A columns have some advantages over P-A
struts.
(4)
Buckling of a pin-ended column. Up to this point only overall structural
behavior has been considered, ignoring the possibility that an
individual column could buckle in its own length. For overall
structural behavior it is usually sufficient to -consider only the P-A
contribution. For colunin buckling it is necessary to consider the
P-o contribution. A pin-ended column is the simplest case of column
buckling. This part of the chapter shows that while the buckling
behavior of an ideal elastic column is easy to calculate, it is more
complicated for an actual column with inelastic . material and
geometric imperfections. It is also more complicated if post-buckling
behavior must be modeled.
196
Chapter 6 P-L'i Effects, Stability and Buckling
(5)
Performance assessment for a structure with pin-ended members. A
building frame that has only pin-ended members is academic.
However~ it is useful to consider the steps for performance
assessment, since both the P-A contribution (for overall P-A effects)
and the P-8 contribution (for buckling of individual members) can be
·important. This part of the chapter outlines the steps' for analysis and
performance assessment; The emphasis is on strength-based design
using elastic analysis, with some discussion of deformation-based
design using inelastic analysis.
(6)
Behavior and modeling of beam-columns. Most columns have bending as
well as axial force. An important consideration for design is that
because of the P-8 contribution, the axial force in a column can
substantially amplify the bending moments. This part of the chapter
considers the causes and effects of moment amplification in beamcolumns.
(7)
Performance assessment for braced frames with beam-columns. This is a
more practical type 'Of structure than a frame with only pin-ended
members. The analysis and performance assessment must account for
moment amplification. ThiS part of the chapter considers the modeling options and analysis procedures.
(8)
Performance assessment for unbraced frames. In. a braced frame, the
moment amplification in a column is mainly from the P-8 contribution. In an unbraced frame, the moment amplificatisin is mainly from
the P-Ll contribution. This part Qf the chapter considers the modeling
options and analysis procedures for unbraced frames.
(9)
Geometric stiffness. Up to this point, the chapter has covered behavior,
modeling and performance assessment, but has said little about P-Ll
theory or the details of P-A analysis. Although this book is concerned
with modeling rather than theory, it is useful to consider the concept
of a geometric stiffness. The general theory for geometric stiffness is
mathematically complex and physically obscure. This part of the
chapter develops the geometric stiffness matrix for a bar ·element,
using a physical process rather than a mathematical one. ThiS can
help to explain what a geometric stiffness does.
(10) Second-order elastic analysis of building frames. A small displacements
analysis is a "first order" analysis. An analysis that accounts for the
P-A effect is a "second order" analysis. This part of the chapter
identifies three different ways to carry out a second order analysis.
.
.... .;,.,.
-o1. .....
'
P-Ll and P-1> Contributions in a Single Column
197
(11) Direct Analysis Method for steel building frames. Up to this point, the
chapter has considered general approaches rather than specific
procedures. The Direct Analysis Method is a specific procedure for
performing second-order analysis of steel building frames. This part
of the chapter describes and critiques the Direct Analysis Method.
(12) Buckling Analysis. Most of the emphasis to this point has been on
lateral load analysis and gravity load analysis. Jhis part of the chapter
considers buckling analysis in more detail.
(13) P-LI effects in some other structures. Up to this point, the chapter has
considered only building structures. This part of the chapter considers
P-L\ analysis for some other types of structure.
(14) Lateral-torsional buckling. Column buckling in flexure is the most
common type of buckling, but not the only type. The physical causes
of column buckling, and procedures for accounting for it in an
analysis model, have been considered earlier in the chapter. This part
of the chapter extends the concepts to lateral-torsional buckling of
beams, with emphasis on a physical explanation of how bending
moments and shear forces cause buckling.
(15) P-LI effects in seismic isolators. This part ofthe chapter considers another
type of P-L\ effect, for the case of friction-pendulum and rubber-type
seismic isolators. It is included because it illustrates some different
aspects of geometric nonlinearity.
(16) True large displacements analysis. For most structures with geometric
nonlinearity it is accurate enough to use P-L\ analysis. In some cases,
however, it is necessary to account for true large displacements. This
part of the chapter identifies some such cases.
6.2 P-Aand P-o Contributions in a Single Column
6.2.1
Overview
There are two contributions to the P-L\ effect in a column, identified in this
chapter as the "P-L\ contribution" (P-big-delta) and the "P-o contribution" (Psmall-delta). This section explains the difference between these two
contributions. Later sections consider their relative importance and how
they can be accounted for in an analysis model.
198
Chapter 6 P-8 Effects, Stability and Buckling
6.2.2
Cantilever Column
The difference between the P-A and P-15 contributions can be illustrated
using a simple cantilever column, as shown in Figure 6.l(a).
Very stiff beam
Small
displacements
contribution
P.:\
contribution
(a) Column
(b) Bending Moments
Deflected shape for combined vertical
and lateral load. Also buckling mode
shape for vertical load only.
(c) Frame with Cantilever Columns
Figure 6.1 P-8 Effect in Cantilever Column
Figure 6.l(b) shows the bending moment diagram for combined gravity and
lateral load. This diagram has three parts, as follows.
(1)
(2)
(3)
Small displacements contribution. This is the bending moment diagram
ignoring P-!l effects.
P-LI contribution. This accounts for the deflection, fl, at the top of the
column.
P-<5 contribution. This accounts for the departure of the column from a
straight line.
The simple frame shown in Figure 6.l(c) has cantilever columns. In this case
_ the deflection A is the story drift.
6.2.3
Approximations in P-Li Analysis
P-A analysis assumes that the top of the column deflects horizontally when
th~ force H is applied, rather than moving around an arc. This is because the
analysis assumes linear compatibility relationships.
P-~ and
P-1> Contributions in a Single Column
199
P-A analysis also assumes that the axial force in the column is the same at all
points.. This is not· strictly correct from an equilibrium viewpoint, but it is a
minor approximation.
Both assumptions are reasonable up to quite large values of the drift ratio,
Mh (for a building frame, up to about Mh =0.1).
6.2.4
P-..1 vs. Small Displacements Effects
In Figure 6.l(b) most of the bending moment is caused by the lateral load,
and the P-A effect amplifies this moment. This type of behavior can be
expected in a lateral load analysis. In thrligure, the P-A effect amplifies the
bending moment (and hence also the lateral deflection) by roughly one
third. This is a large amount of amplification for a lateral load analysis most structures will have smaller amounts.
As shown later in this chapter, th_ere are a number of different ways ·to
account for the P-A effect. A major concern for lateral load analysis is how
accurately this effect must be modeled. Consider a case where the moments
are amplified by 25%. Suppose that the P-A effect is not modeled accurately,
and this effect is underestimated by 20%. In this case the "exact"
amplification factor is 1.25 (i.e., the amplified moment is 1.25 times the small
displacements moment), whereas the calculated factor is 120 (i.e., 1.0 + 0.8 x
0.25). Hence, the error in the total moment is only 4% (1.20 vs. 1.25). This
indicates that it may not be necessary to model the P-A effect accurately for
lateral load analysis.
The situatio~ is different for a buckling analysis. If only the vertical loads
are considered for the frame in Figure 6.l(c), and if these loads are
progressively increased, collapse might occur by axial yield or crushing of
the columns, but it is more likely to occur by flexural buckling. The buckling
mode is a "side-sway" mode, as shown in Figure 6.l(c).
If the coluinns in the frame are initially perfectly straight and perfectly
vertical, and if the loads are perfectly vertical and act along the axes of the
columns, the bending moments in the frame are zero before buckling
occurs. After buckling there can be large bending moments. These moments
are caused entirely by the P-A effect, so the amount of moment ampljfication
is infinite.
It is impossible for a column to be perfectly straight and vertical, and the
loads will never be perfectly synunetrical or perfectly vertical, so there must
200
Chapter 6 P-A Effects, Stability and Buckling
be some bending moments in a small displacements analysis. Even so, as the
buckling load is approached the amount of moment amplification becomes
large, and when the buckling load is reached almost all of the bending is
caused by the P-A effect.
It follows that the P-A effect must be modeled more accurately for buckling
analysis, otherwise the calculated buckling load can be inaccurate.
6.3 Relative Importance of P-L1 and P-o Contributions
6.3.1
Cantilever Column
For the cantilever column in Figure 6.1, the relative importance of the P-A
and P-<i contributions can be calculated. Since the deflected shape depends
on ·the bending moment diagram, and since the P-<i contribution to the
bending moment diagram depends on the deflected shape, there is a feedback effect, and an exact calculation is not trivial. However, it can be shown
that the P-<i contribution accounts for roughly 18% of the total P-A effect,
with the P-A contribution accounting for the remaining 82%. ·
Although the P-A contribution is dominant, the P-<i contribution is
significant. From.this,.it might be concluded that both contributions should
be considered in a lateral. load analysis. However, the cantilever column
example exaggerates the importance of the P-<i contribution, as shown
below.
6.3.2
Column in an Unbraced Frame
Figure 6.1 considers a cantilever column with a fixed base. Consider the
more_ realistic case of a column in an unbraced frame. This is shown in
Figure6.2.
Figure 6.2(a) shows a column in a frame with rigid, or very stiff, beams. One
· half of a typical column behaves like a cantilever column with a fixed base.
Figure 6.2(b) shows a more realistic frame, with deformable beams. In this
frame, one half of a typical column behaves like a cantilever column with
rotation at the base. Figure 6.2(c) shows that when the base of the column·
rotates, the P-<i contribution becomes smaller relative to the P-A contribution.
....
·~,
Relative Importance of P-il and P-o Contributions
I
I
I
I
I
I
--,
I
I
I
I
I
I
I
I
I
I
I
Cantilever column
with fixed base
(a) Rigid beams
I
I
I
I
I
I
I
-,
I
I
-1
I
I
I
I
I
I
I
I
I
I
I
201
iI
I
Cantilever column
with rotation at base
A from base
A from column
-~k~~
I
I
I
I
I
I
I
I
I
~
For same A,
column has
· less bending
Base
rotates
-~
(c) Effect of rotation at base
(b) Deformable beams
Figure 6.2 Cantilever Column in a Frame
In an unbraced frame, some of the story drift originates in bending of the
beams and some in bending of the columns. In a typical frame, bending of
the beams .can be expected to cont:Ilbute roughly one half to two thirds of
the total story drift, with bending of the columns contributing the remainder.
If the beams contribute one half of the total drift, for a given total drift the
column bending is reduced by a factor of 2. Hence, the P-o contribution is
reduced by a factor of 2 compared with a fixed-base column, while the P-A
contribution is unchanged. As a result/ the P-A and P-o contributions are
about 90% and 10%, respectively, of the total P-A effect (relative total P-A
effect= 82% + 18%/2 = 0.82 + 0.09 = 0.91; P-0 contribution= 0.09/0.91 =
0.10). If the beams are more flexible and contribute 67% of the total drift, the
P-A and P-o contributiorui are about 93% and 7%, respectively.
The P-o contribution is also affected by the following.
(1)
Gravity framing. Building frames often have moment-resisting exterior
frames that resist lateral load and interior frames that are non-
202
Chapter 6 P-A Effects, Stability and Buckling
moment-resisting and support gravity load only. Figure 6.3 illustrates
this in a simple form.
Most of load is on
gravity frames.
I
Moment-resisting frames
usually have less load.
I
i'
The moment-resisting frames
resist the entire P-11 effect.
The P-11 contiibution depends
on the total gravity load.
The P-0 contribution depends
only on the loads on the
moment-resisting frames.
Gravity frames
Moment-resisting frames
Figure 63 P-A Effect in Building with Gravity Framing
In this example the gravity columns have little or no bending
deformation (they "lean" rather than bend), and. hence their P-o
contributions are essentially zero. They do, however, have P-11
contnbutions. The columns in the moment frames bend, and hence
have both P-11 and P-o contributions. However, the column axial
forces (P values) for these contnbutions depend mainly on the
tributary gravity loads on the exterior columns, which account for a
relatively small part of the total gravity load. For the structure as a
whole, all of the columns have P-11 contnbutions but only the column5
in the moment frames have P-o contributions. The P-o contributions
have a very small effect on the overall behavior of the building.
(2)
Plastic hinging. H there is inelastic behavior under combined gravity
and lateral loads, a plastic hinge may form at the end of a column.
This reduces the amount of column bending, as shown on the left of
Figure 6.4. The effect is similar if the columns remain elastic and a
hinge forms in the beam, as shown on the right of Figure 6.4. In both
cases the P-o contribution is reduced relative to the P-11 contribution.
(3) · Panel zone deformations. Deformations in the panel zones of the beamto-column connection increase /1 without increasing o. This reduces
the relative P-o contribution.
(4)
Shear defonnation. Shear deformations in a column increase /1 without
increasing o. This reduces the relative P-o contribution.
"
-~.
Relative Importance of P-Ll and P-li Contributions
203
To simplify the
diagram, assume
the beam is rigid.
Deformed shape
with plastic hinge
Elastic deformed shape,
with no plastic hinge
Figure 6.4 Effect of Plastic Hinges on P-li Contribution
Jn general, the P-o contribution is reduced by anything that increases the
story drift without changing the bending deformations in the columns, or
that keeps the same story drift while reducing bending in the columns.
For lateral load analysis of almost any practical unbraced frame building,
the P-o contnbution has only a small effect on the overall behavior. This is
particularly true for an inelastic analysis after hinges form in the beams or
·
columns.
This does not mean that the P-o contribution can always be ignored. As
emphasized later· in this chapter,· there are two distinct types of
amplification, caused respectively by the P-A and P-o contributions. For the ·
lateral load analysis of unbraced frames, P-A (or "sway") amplification
dominates and P-o (or "non-sway") amplification can usually be ignored.
This is not the case for lateral load analysis of braced frames. It is also not
the case for buckling analysis.
6.4 Modeling of P-A and P-o Contributions
6.4.1
Overview
This section considers how a small-displacements analysis model can be
extended to account for P-A effects, including both the P-A and P-o contributions.
Chapter 6 P-d Effects, Stability and Buckling
204
The P-L\ and P-o contributions can, and should, be modeled separately. The
P-L\ contribution is straightforward and easy to model. The P-o contribution
is more subtle and more difficult to model.
This section considers only the model, and not the analysis method. Later
sections consider how the model might be analyzed.
Model for a Cantilever Column
6.4.2
A cantilever column with a fixed base is not representative of a column in
an unbraced frame, since it overestimates the P-o contribution. However, it
is convenient to develop an analysis model using a cantilever column as an
example.
Figure 6.5 shows an analysis model for a cantilever column.
H
~
P-0 cable has same deflected
shape as small displs column.
Small displs _ __
column.
Forces must act on P-li
cable to satisfy equilibrium.
Equal and opposite forces
act on small displs column.
P-A
strut
P-0 ---+------..i
cable
(a) Components of model
(b) Component loads and deflections
Figure 6.5 · Model of a Cantilever Column
This model has three parts, as follows.
(1)
A "small displacements column". This has the same end conditions as
the actual column (in this case, fixed at the base and free at the top). It
has both vertical and horizontal loads, as shown. The vertical load is
the load, P, on the actual column. The horizontal load consists of the
Modeling of P-Li and P I> Contributions
205
external load, H, on the column plus loads from the P-A and P-o
contributions, as explained later.
(2)
A "P-A strut", to model the P-A contribution. This consists of a rigid,
pin-ended strut that connects horizontally to the column at its ends. In
Figure 6.S(a) there is a horizontal link at the top, and the bottom is
fixed. The P-A strut is loaded by the vertical load, P. As the column
deflects, the P-A strut rotates as shown in Figure 6.5(b). To keep the
strut in equilibrium, there must be a horizontal force, PMh, applied at
the top. This force acts on the small displacements column, increasing
the bending moment and shear force.
(3)
A "P-o cable", to model the P-o contribution. This is the complicated
part of the model. The behavior of this cable is considered in the next
'
section.
6.4.3
Behavior of
P-o Cable
An ideal cable has zero bending stiffness, but it can support load if it is in
tension. (A real cable, even a piece of string, will always have some bending
stiffness.)
The P-o cable is an ideal cable. Its shape is the same as the deflected shape of
the small displacements column, so it is a shallow cable with only a small
amount of sag. Figures 6.6(a) and (b) show the forces on a shallow cable in
tension.
The key points are as follows.
(1)
In the column model the P-o cable has the same deflected shape as the
column, and for P-A analysis the displacements are small (small
enough for linear compatibility to be a reasonable assumption). For
practical purposes, (a) the axialforce in the cable is constant over its
length, (b) the transverse loads on a horizontal cable are vertical
(horizontal for a vertical cable), and (c) the curvature of the cable can
be calculated as for a beam (curvature= second derivative of transverse
displacement).
206
Chapter 6 P-A Effects, Stability and Buckling
~-J.............:0
T
HJP··
t_.T
;;;;;;;;>"
+ * *** * * *+
(a) Uniform curvature (equal
end rotations)
(b) Linear curvature (essentially the
case for a cantilever cohimn) ·
I
(c) Forces on P-li cable for
a cantilever column
Figure 6.6 Forces on a Shallow.Cable
(2)
For equilibrium of a short length of cable, the transverse load per unit
length at any point must be _the axial force in the cable multiplied by
. the cur\rature of the cable. For a given deflected shape, the curvature
can be calculated, and hence cilso th,~. transverse loads. Figures 6.6(a)
and (b) show the transverse load distributions for two deflected
, shapes. The second of these is essentially the shape for. a cantilever
column with P and H loads. The curvature varies linearly when P is
small. As P increases, the curvature changes, but remains nearly linear
until the axial force approaches the buckling strength of the column.
(3)
From equilibrium, the vertical reactions at the cable ends are the same
as for a simple-span beam with the same transverse load. The
transverse forces on the cable, consisting of the transverse loads and
the transverse end reactions, constitute an equilibrium force· system
with a zero· resultant. The axial forces at the ends are equal and
opposite, so they are also self-equilibrating. _
Figures 6,6(b) show the loads for a horizontal cable in tension with linearly
varying curvature. Figure 6.6(c) shows the corresponding case for a P-1>
cable in a vertical cantilever column. Since the displacements are small, the
forces on the cable can be assumed to be horizontal (for small displacements
there is no difference between horizontal forces and forces normal to the
cable). The reverse of these forces act on the small displacements column.
.....
·~.
Modeling of P-d and P oContributions
207
An isolated cable can not, of course, resist compression (since an ideal cable
has no bending stiffness, its buckling strength· is zero). However, the P-0
cable in a column model is connected to the small displacements column (in
effect, it is imbedded in the column). This column provides bending
stiffness, and hence provides buckling resistance.
6.4.4 Behavior of Elastic Cantilever Column
In the column model, the P-5 cable exerts distributed transverse forces over
the length of the small displacements column, and point forces at its ends.
These forces depend on the axial force in the coluinn and the distribution of
curvature over the column length. In.contrast, a.P-L\ strut exerts transverse
forces only at the column ends. These forces depend only on the axial force
in the columTI and the L\ displacement.
The P-L\ strut and P-o'cable both exert forces on the small displacements
column. These forces are different in the following respects.
(1)
The for~es from the P-L\ strut are simply PL\/h, in opposite directions
at the top and bottom of the column. These forces affect the shear
force and bending moment in the column, and they also exert an
overall rotational couple of magnitude PL\.
(2)
The forces from the P-o cable depend on the distribution of o over the
column height. Since a column can have complex deflected shapes
(depending, in part~ on the forces from the P-o cable), these forces can
have many different distributions and magnitudes. However, the
translational and rotational resultants of these forces are always zero.
As noted earlier, the deflections of a column depend on the forces on the
small displacements column, and these forces depend on the column
deflections. Hence there is a "feedback" effect, and it is not a routine task to
calculate the deflected shape. The details of the calculation are not
important for the present discussion.. However, consider some qualitative
aspects of the behavior.
If the external loads, P and H, are applied to the small ·displacements .
column, ignoring geometric nonlinearity, the bending moment diagram and
deflected shape are easy to calculate. If the P-L\ strut is added~ it exerts a
horizontal force, PL\/h, at the top of the column, which initially is based on
the axial force and the value of L\ from the small displacements analysis.
This force increases L\, which in turn increases PL\/h, which increases L\, etc.
208
Chapter 6 P-8 Effects, Stability and Buckling
The value of /:J. will usually stabilize, and it can be calculated quite easily (an
iterative method is one option, but there are more direct methods). If Pis
large enough, the value of /:J. may not stabilize and becomes theoretically
infinite. The value of P at which this just occurs is the buckling load, or
"critical load", for a model with a small displacements column and a P-A
strut.
Assume that the value of A stabilizeswhen only the P-/:J. strut is considered.
Then add the P-o cable. This exerts additional forces on the small displacements column, consisting initially of a linearly varying distributed load (for
a column with a uniform cross section) plus a force at the top of the column.
These forces increase the column displacements, which in turn increase the
forces from both the P-o cable and the P-/:J. strut, which in turn increase the
displacements, etc. The final bending moment diagram is not exactly a
straight line, because of the distributed load from the P-o cable. Hence, the
final distributed load from the P-o cable does not vary exactly linearly along
the column length.
Again, the value of A may not stabilize, which indicates that the buckling
load for the model with both a P-A strut and a P-o cable is smaller than for a
model with only a P-A strut (it is about 18% smaller for a cantilever column
with a fixed base).
6.4.5
Moments and Shears for Column Design
For a cantilever column with a gravity load P and a lateral load H, from a
small displacements· analysis the shear force in the column is H and the
bending momentat the base is Hh, where h is the column height. These are
the forces for which the column must be designed, assuming that a small
displacements analysis is accurate enough for design purposes. The forces
are larger when P-A effects are considered.
Consider a cantilever column 1hat is modeled using a small displacements
column and a P-A strut, but no P-o cable. For this model the bending
moments and shear forces in the small displacements column are as shown
in Figure 6.7(a).
' ..... .
........
Modeling of P-~ and P1l Contributions
_.
H
rrr
-
Greater than
H+'JiA
[]
i
I
I
I
I
I
I
I
Moment
(a) P-A strut only
209
Shear
(b) P-A strut and P~ cable
Figure 6.7 Moments and Shears in a Cantilever Column
The bending moment at the base of the small displacements column is Hh +
PLi, and the shear force is H + PLi/h. The bending moment at the base is
larger than the small displacements moment for an obvious reason - the
extra moment is the gravity load, P, acting through the deflection, Li. The
reason .for the increase in the shear force may be less obvious. From
horizontal equilibrium the shear force on a cross section through the base of
the column must equal the horizontal load, H. However, the column must
be designed for a shear force that is larger than H. This is worth examining
in more detail.
The total base shear is H, which ·follows from horizontal equilibrium.
However, a cross section through the base cuts both the small displacements
column and the P-Li strut. The total shear force on this section is H, but the
horizontal shear in the P-Li strut is negative PMh. Hence, the shear force in
the small displacements 'Column is H + PLi/h. This is the shear force for
which the column must be designed.
Also, the bending moment gradient along the column length is (Hh + PLi)/h.
This is the shear force, as required by beam theory (shear force = bending
moment gradient).
This applies for the model in Figure 6.7(a). However, it is not exact for the
actual column. For the actual column the bending moment diagram is not a
straight line, so the bending moment gradient is not constant, and hence the
shear force must vary along the column height. This is accounted for when a
P-o cable is added to the model. This is shown in Figure 6.7(1J). The shear
210
Chapter 6 P-.1Hffects, Stability and Buckling
force diagram in this figure is not exactly a straight line, but it is close, until
P approaches the colwmi bucking load.
.
..
.
.
.
When a P-o cable is added, the forces that act on the small displacements
column are more complex, with ~buted.forces along the length as well
as a point force at the top. The P:A strut exerts only a point force at the top,
and this force depends only on P and the drift ratio Mh. The Ni cable exerts
both .end forces and distributed forces along· the length, both of which
depend on the deflected shape of the column.
H the forces from the Ni cable are calculated exactly, the model gives
accurate values for the bending moments and shear forces along the column
(consistent with the approximations of P-A theory). These are the bending
moments and shear forces for which the column should be designed.
. It is not easy to calculate these forces, because of the complex interactions
betWeen the deflected shape of the column and the distribution of the forces.
For most analyses the best that caµ be obtained is a. reasonable approximation-of the effects of these forces. One approximation that can be reasonable
in some cases is to ignore the P-o contribution entirely. As shown later, this
is usually reasonable for column8 in an unbraCed frame; where the P-A
contP.bution dominates. This is not reasonable, however, for a column in a
braced frame, where the P-0 contribution can dominate.
This iS addressed in more d~tail later in this chapter, when methods for
calculating sway and non-sway amplification are considered.
6.4.6 Elastic Column in an Unbraced Frame
A cantilever column is useful for .developing concepts, but is not very
realistic. Consider. the more realistic case of a colunin in an unbraced frame;
For this discussion assume that the column is elastic. Figure 6.8(a) shows a
frame and a typical column~
Figure 6.8(b) shows the deformations of the column. The beams above and
below provide rotational stiffness at the column ends.
.
...
The column can have stiff or rigid end zones, and the inflection poiD.t is not
necessarily at the column mid-height. The figure shows only the 20 case. In
the 30 case there can be bending about both axes. Figure 6.8(c) shows the forces on the small displ~cements column for the
case with no end zones. These forces include the external loads and forces
....
-
-~
.
Modeling of P-~ and Po Contributions
211
from the P-A strut and the P-8 cable. Figure 6.S(d) shows the forces for the
case with rigid end zones. •
-+"
H
r
Beams provide
rotational restraint
(a) Typical column in a frame
(b) Column deformations
From P-0 Cable
(c) Forces on small displacemenis column
(d) Forces when there are end zones to
account for beam depth
Figure 6.8 Column in an Unbraced Frame
The end zones affect the deflected shape of the column, and hence change
the forces from the P-8 cable, but they do not affect the P-A strut. This is one
reason why it is easier to model the P-A contribution than the P-8
contribution.
If the axial force is much smaller than the buckling strength, as it usually
will be,· the bending moment diagram is essentially linear (if the P-8
contribution is ignored it is exactly linear). Hence, the curvature variation
for a col~ with a uniform cross section is also linear, and for the case
with no end zones the distributed load from the P-8 cable varies linearly
over the column length. For the case with rigid end zones, the cable
curvature in the end zones is zero, and hence the distributed load is zero.
212
Chapter 6 ·P-~ Effects, Stability and Buckling
In both cases there are forces from the P-o cable at the column ends as well
as distributed forces along the column length. As noted earlier, the forces
from the P-0 cable are always self-equilibrating.
6.4~7
Inelastic Column in an Unbraced Frame
Figure 6.8 shows the forces for an elastic column.. These forces can be
substantially different when there is inelastic behavior.
H plastic hinges form only in the beams, and if the hinge behavior is elasticperfectly-plastic, ·when the hinges rotate the P-A contribution continues to
increase but the P-0 contribution stays constant. Hence, the forces from the
P-A strut continue to increase, the forces from the P-0 cable remain the same,
and the P-o contribution becomes progressively less important. This is
illustrated in Figure 6.9.
Hinges form at Ay;
Forces from P-li Cable
A continues to increase.
(a) Hinges fonn in beams.
(b)Forces after hinges fonn.
Figure 6.9 Forces When Hinges Form in Beams
The situation is more complicated if plastic hinges form in the column, as
shown in Figure 6.10.
Since the curvature at a hinge· is infinite, there must be· a point load· on the
cable· at each hinge, with a magnitude equal to the hinge rotation
multiplied by the axial force, P. These forces are shown in the figure.
P-o
Modeling of P-L'l and PIi Contributions
Hinges form at i\ Y;
• A continues to increase.
Forces from P-A Strut
continue to increase.
H+1Pih1
f
213
Forces from P-5 Cable
change as hinges rotate.
~~-=-----,.,
~
Point forces at
hinges increase as
hinges rotate.
~
(a) Hinges fonn in column.·
Endforces
also increase.
(b) Forces after hinges fonn.
Figure 6.1 O Forces When Hinges Form in Column
6.4.8
Element Models
As ·described in this section, a model for a frame element has three
components, namely a small displacements element, a P-d strut, and a P~o
cable. Two of these components (the small displacements element and the
P-ii cable) can have complex behavior, and all three can interact in complex
ways. This section has shown how these interactions can explain the
behavior ~f column members, but only in terms of broad concepts.· It says
nothing about how a frame element might be implemented in a computer
program.
Analysis by the Direct Stiffness Method requires an element stiffness matrix
that can be assembled into the structure stiffness matrix. It also requires a
procedure to determine the element deformations and· forces, given the
node displacements. There are many different ways in which this can be
done, and a mathematical formulation may not explicitly consider a model
with the above three components. Different computer programs are likely to
use different procedures, with different assumptions and computational
techniques. Different types of analysis may require different mathematical
formulations (for example, different formulations might be used for
buckling analysis and lateral load analysis, or for elastic and inelastic
analysis, or for static and dynamic loads). A single structural member, such
as a column, might be modeled as a single long element or as a number of
shorter elements. These details all depend on how the element is
implemented in a computer program.
214
Chapter 6
P-~
Effects, Stability and Buckling
Hence, the model considered in this section is only a concept. It is, however,
useful for explainittg the behavior.
6.5 Lateral Load Behavior of Frames
6.5.1
Overview
The preceding section considers only individual colUrrins, and says little
about how the P-A effect influences the behavior of complete frames. Before
going more deeply into modeling, analysis, and performance assessment, it
is useful to consider the behavior of frame structures in greater depth. This
is done in this section and the following one, for lateral load analysis and for
buckling analysis.
This section considers the P-A effect for combined gravity and lateral load,
using simple unbraced frames as examples. Since the P-o contribution is
small for lateral load analysis of unbraced frames, this section considers
only the P-A contribution. The P-o contribution is considered in more detail
later. Elastic and inelastic behavior are both considered. For inelastic
behavior, plastic hinges can form in the beams and/ or the columns.
Buckling of individual columns. is assumed not to occur. This aspect of
behavior is considered later.
In addition to discussing behavior, this section also shows how the P-A
contribution can be accounted for using a "P-A column". The example frames
in this section are two-dimensional. This is extended to three-dimensional
buildings in a later section.
. 6.5.2
P-A Struts and P-A Columns
For a single column in a frame, the P-A contribution can be modeled using a _
P-A strut. This can be extended to a structure that has many columns, with a
P-A strut for each colµmn. However, it can be convenient to combine all of
the P-A struts in a story into a single "P-A column". This is shown in Figure
6.11.
.
.
Lateral Load Behavior of Frames
Load= P3
Load= P3
P3
Load= P2
Load=P2
Load= P1 Load= P1
tIIIt tIIIt
·.·
....
c
..
(a) Each column has a P-a strut
Pa
++Ill +I Ill
+Illl ++Ill
tIIIt till}
215
t.
P2
P2
tIIIt tIIIt
P1
P1
,
""""'.
tIIIt
(b} P-A struts are combined into a P-a column
Figure 6.11 Single P-A Column vs. Several P-A Struts
In a 2D frame the P-L\ struts can be combined because the following two
conditions are usually satisfied.
(1)
For lateral load analysis it is usual to apply the gravity load first, then
to add lateral load (either static or dynamic), keeping the gravity load
constant. As the lateral load is applied, the axial forces on any column
in a story can change, but the sum of the axial loads in all colunms in
a story stays constant.
(2)
Axial deformatioris in the beams are small, so all columns in a story
have the same drift.
The P-L\ strut for a single column exerts forces of magnitude PMh on the
structure at the top and bottom of the column, where P is the axial force in
the column. Since L\ and h are the same for all columns in a story, the total P
force is ('LP)A/h. Since 'LP does not change when lateral load is applied, the
single P-8 column in any story exerts the same total forces as the several P-L\
struts.
While several P-8 struts can be combined into a single P-L\ column, P-o
cables are different. If the P-0 contribution is considered, a separate P-o cable
is required for each column, and these cables can not be combined.
216
Chapter 6 P-11 Effects, Stability and Buckling
6.5.3
Review ofTerminology
Before continuing, it may be useful to review the terminology for this
chapter. The teiminology is as follows.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
"Geometric nonlinearity" is an overall term that applies for both P-A
analysis and large displacements analysis.
The "P-A effect" is the effect of geometric nonlinearity in a P-A
analysis.
The "P-A contribution" and the "P-o contribution" are the two distinct
parts of the P-A effect.
A "P-A strut" can be used to model the P-A contribution and a "P-o
cable" can be used to model the P-o contribution.
A "P-A column" approximates the effect of several P-A struts.
In a "gravity load analysis" only the gravity load is applied. There can
be significant P-A effects.
In a "buckling analysis" the gravity load is .increased until the
structure becomes unstable and buckles. The behavior tends to be
dominated by the P-A effect.
In a '1ateral load analysis" the gravity load is applied, then lateral load
is added. There can be significant P-A effects, but the behavior tends
to be dominated by the lateral load.
This section considers the behavior that can be expected in a lateral load
analysis, considering P-A effects, ignoring the P-o contribution, and using
P-~ columns.
6.5.4
Main Structure and P-a Column
Figure 6.12(a) shows an analysis model for a 20 single story frame with
gravity and lateral loads.
This model accounts for P-A effects using a P-A column. In the analysis
model this is a pin-ended elastic bar that is very stiff axially. The load on
this column is the total gravity load on the structure.
The rest of the model is the "main structure", which is equivalent to the
"small displacements column" in the model for a single column. The-main
structure consists of the beam and column elements that make up the·
building frame. It is loaded by the external gravity and lateral loads, and by
forces from the P-A column. It .may be elastic or inelastic. There are no P-A
effects in this part of the model.
Lateral Load Behavior of Frames
Total load = P
p+
l I I I l
~lI
P-A column.
Include P-A effects.
I
l
Main structure.
Ignore P-A effects.
(a) Model with P-A column
H
t
217
Total load = P
H+PMl
l I I I l
·~- r7
'
(b) Effective load on main structure
Figure 6.12 Simple Frame with P-d Effect
The two parts of the model are connected by a stiff horizontal bar, i;o that
they have equal horizontal deflections. Alternatively, there could be a
.
horizontal slaving constraint.
Figure 6.12(b) shows the deflected structure, with a horizontal dispiacement
ll.. The external lateral· 1oad on the eritiie structUre is H, and the effective
lateral load on the main structure is H + PMh.
It follows that the sum of the shear forces in the columns of the main
structure is also H + P/1/h. This is the effective, or amplified, story shear.
The columns must be designed for this shear. The story shear for the entire
structure, including the P-ll. column, is equal to H, but the story shear in the
P-ll. column is negative, and the main structure must provide lateral suppo~
for this column.
6.5.5 Effed of P-A Column on Stiffness and Strength
Figure 6.13 shows a possible relationship between lateral load, H, and lateral
displacement, ll., for the frame.
Figure 6.13(a) shows the main stnicture, with a constant gravity load, a
varying lateral load, and no P-ll. effects. For the main structure alone, the
lateral load is HMS. As this load is progressively increased, plastic hinges
form in the beams and/or columns, resulting in a load-deflectio~·relation­
ship, as shown in Figure 6.13(c). The dots indicate hinge formation. H the
hinges have no strain hardening, the main structUre becomes a mechanism ·
when four hinges have. formed, and after that point the strength remains
constant.
Chapter 6 P-a Effects, Stability and Buckling
218
H.
Constant load, P
HMs
l l l l l
n
a
(a) Main structure
·
Hpa
t
"l~
Structure strength
a
(b) P-a column
(c) H-a Relationships
Figure 6.13 H-a Relationship
Figure 6.13(b) shows the P-L\ column, with a constant gravity load and
including P-L\ effects. The lateral load on this column is. Hp&· This load is
negative and depends linearly on L\, as shown in Figure 6.13(c).
The lateral strength of the structure is the sum of HMS and Hp&, as shown by
the dashed line in Figure 6.13(c). The following are some key points.
(1)
The P-L\ effect reduces the initial elastic stiffness of the structure. The
. · effect is unlikely to be large in a practical structure - roughly a 5% or
10% reduction for a steel frame building, and less for a concrete frame
or shear wall. In Figure 6.13(c) the reduction in stiffness is about 5%.
(2)
The strength at first yield is reduced by the same amount as the initial
stiffness. In Figure 6.13(c) this reduction is about 5%. However, there
is a larger reduction in the ultimate strength. In Figure 6.13(c) this
reduction is about 20%.
(3)
Beyond the point where the maximum strength is reached, the
stiffness of the structure is negative, and the structure is unstable. If
the structure is displaced beyond that point, and if the lateral load is
static and sustained, the structure will collapse. If the lateral load is
dynamic, however, it may act for only a short time, and even if the
structure is displaced beyond the point of instability, collapse does
not necessarily occur. Dynamic load is alsomore complex because it is
resisted .by inertia and damping ·forces as well as the structure
strength.
(4)
In Figure 6.13, the P-L\ effect causes the stiffness to be negative before
the last plastic hinge forms in the frame. That is, the frame becomes
'"'.~
..........
219
Lateral Load Behavior of frames
unstable before it forms a complete mechanism. This may not always
happen, but it is possible.
(5)
Figure 6.13 assumes no strain hardening in the hinges. If there is
strain hardening, instability will occur at a larger displacement, if at
all. However, Figure 6.13 also assumes unlimited ductility in the
hinges. If one or more hinges reach their ductile limits and lose
strength, instability is likely to occur. This ,is illustrated in Figure
6.14(a).
If there is sufficient strain
hardening, mechanism is stable.
H
,.,,--·--·--r--.
I
I
H
I
I
I
A
But as hinges reach ductile limit
and lose strength, becomes unstabl~.
----
Strength for
increasingA
I
,,/
Strength for
decreasingA
./
-------M----~-
(a) Strain Hardening and Strength Loss
(b) Load Reversal
Figure 6.14 Hardening, Strength Loss and Load Reversal
(6)
6.5.6
During a strong earthquake, a structure might be- deformed beyond
the point of maximum strength. Since the load is dynamic this does
not necessarily cause collapse. However, there is a greater likelihood
of collapse. One reason is shown in Figure 6.14(b). If the structure is
deformed beyond the point of maximum strength, and if the load is
then reversed, the strength of the structure in the reverse direction is
increased by the P-L\ effect, not decreased. Hence, a larger load is
required to reduce the displacement than to increase it. This can cause
a "ratcheting" effect and makes collapse more likely.
Load Increase or Strength Decrease?
In Section 6.5.4 it is noted that the P-L\ effect increases the effective load on
the structure. In Section 6.5.5 the P-L\ effect decreases the structure strength.
These are just opposite sides of the same coin.
·
220
Chapter 6 P-t. Effects, Stability and Buckling
If the strength of an existing frame is being calculated, this strength is
reduced by the P-A effect. If a new frame is being designed for a specified
lateral load, this load is increased by the P-A effect,
6.5.7
Load on P-L\ Column vs. Load on Main Structure
Buildings often have perimeter frames that resist lateral load, combined
with interior framing that supports gravity load but has little or no lateral
stiffness or strength. When a P-A column is used, the gravity load on the
entire structure must be applied to the P-A column, including the load on
the gravity columns. The interior framing may be omitted, to simplify the
analysis model. If this is done, the load on the P-A column will be different
from the gravity load on the main structure (the part of the building that is
included in the analysis model).
6.5.8
P-L\ Effect in Beams
In an unbraced frame, P-A effects are usually important only in the columns,
and can be ignored in the beams.
The reason is that a vertical P-A column (or a P-A strut) exerts horizontal
forces on the main structure that depend on the column axial force and the
story drift. In contrast, a P-A strut in a horizontal beam exerts vertical forces
that depend on the beam axial force and the "tilt" of the beam from the
horizontal. Beam tilt is usually much smaller than story drift, and beam
axial forces are usually much smaller than column axial forces. Hence, the
forces from a P-A strut in a beam are usually very small. Also, the forces
from a vertical P-A column or strut add to the story shears in a frame,
whereas the forces from a horizontal P-A strut add to the axial forces in the ·
columns. This has a much smaller effect on the behavior of the frame.
6.5.9 · P-L\ Effect in Braced Frames
The examples in this section have considered unbraced frames. Figure 6.15
shows models for two simple braced frames.
Figure 6.15(a) shows a diagonally braced frame. This is a two-dimensional
truss, since the connections are pinned. Figure 6.15(b) shows a frame that is
braced by a shear wall. Both models have P~A columns, although P-A struts
could be used.
Lateral Load Behavior of Frames
p+
Total load= P
!
I
.
1J><t
· P-~ column
Main structure
(a) Diagonally braced frame
P
221
Total load= P
+ rn
CE}
l'.Ir~ l
P-~ column
Main structure
(b) Frame braced by a wall
Figure 6.15 Simple Braced Frames
Braced frames have lateral load-displacement relationships that are similar
to those for unbraced frames. Some differences are as follows.
(1)
A braced frame is likely to have a larger lateral stiffness than an
unbraced frame, at least in the elastic range before inelastic behavior
occurs; Hence, P-8 effects in 'a braced frame are likely to be smaller in
the elastic range.
.
(2)
In an unbraced frame, the lateral stiffness decreases as plastic hinges
form. In a diagonally braced frame, the lateral stiffness decreases as
tension braces yield and/ or compression braces buckle. In a frame
that is braced by a wall, the lateral stiffness decreases as the wall
·
cracks or yields in bending, or becomes inelastic in shear. ·
In a diagonally braced frame, the beams, columns and braces can all have
substantial axial forces. H P-Li struts are used, the vertical struts in the
columns exert horizontal forces on the structure, the horizontal struts in the
beams exert vertical forces, and the inclined struts in the braces exert forces
with both horizontal and vertical components. As with an unbraced frame,
the vertical forces have only a small effect on the behavior of the structure.
Since the horizontal forces from the P-8 struts are dominant, it is accurate
enough in most cases to use a P-Li column. If P-Li struts are used, they must
be included in the. columns and braces, and probably should also be included in the beams.
222
Chapter 6
P-~ Effects, Stability and Buckling
6.5.1 O Summary for this Section
.
This section has considered the influence of the P-A contribution on the
behavior of building fram~ under combined gravity and lateral load. Very
simple frames have been used as examples, but the behavior is qualitatively
similar for larger structures. The following are some general observations.
(1)
The P-A contribution can be modeled using P-A struts. For many if not
most building structures, the P-A struts in each story can be combined
into a single P-A column.
(2)
The P-A contribution reduces the stiffness and strength of a building.
The reduction in the elastic stiffness and the strength at first yield are
usually modest (roughly 5 to 10%). The reduction in the ultimate
strength can be much larger.
(3) . As the structure yields, there is a progressive reduction in the lateral
stiffness of the structure. Jf this stiffness becomes negative, the
structure becomes unstable laterally, and under a sustained static load
it will collapse. Collapse does not necessarily occur under dynamic
load, such as an earthquake, but it becomes more likely. Instability
can occur before a complete mechanism forms. Strain hiirdening is
benefic.ial.
(4)
ff the ductile limit is exceeded for a substantial proportion of the
components in a structure, the lateral stiffness can become substan. tially negative, with an increased possibility uf collapse.
Earlier sections have noted the difference between "demand" and "capacity"
analyses. Lateral load analysis is used mainly to calculate demands. Elastic
lateral load analysis is used to calculate strength demands for strengthbased design, and inelastic lateral load analysis is used to calculate deformation demands for deformation-based design.
However, lateral load analysis may also be used to calculate capacities.
Spme possible capacity analyses are as follows.
(1)
Inelastic 'static push-over analysis might be used to calculate the
strength capacity of a building frame, considering ·yielding ·of the
members and also P-A.effects (i.e., considering both material and geometric nonlinearity). The calculated strength capacity is the maximum
lateral strength predicted by the analysis.
Buckling Behavior of Frames
(2)
223
Inelastic dynamic analysis might be used to calculate the earthquake
intensity that causes a building frame to collapse. This is a measure of
·
strength capacity.
The calculated demands and capacities can both be sensitive to . the
modeling assumptions. It is more difficult to calculate reliable values for
capacities than demands.
6.6 Buckling Behavior of Frames
Overview
6.6.1
The preceding section considered structures wi.th co'inbined. gravity and
lateral loads. As the lateral load is increased, the structure remains stable
until substantial yield occurs. The structure becomes unstable
the
. effective lateral stiffness ~omes.negative. ·
when
This section considers instability of a different type, where the structure
becomes unstable under gravity load alone, with no lateral load. As in the
preceding section, the concepts and behavior are illustrated using simple
unbraced frame examples. For these examples the· P-S contnbution is·
ignored, and the P-L\ contnbution is modeled using a P-L\ column. ·
6.6.2
Stability of Equilibrium State
Figure 6.16(a) ·shows the same unbraced frame that was used for earlier
examples, but with gravity load only.
·
d~&
P
Total load= P
l
! I l l l
!
I
dH
flJ l
P-A column.
Main structure.
Include P-A effects. Ignore P-A effects.
(a) Apply P, then add vanishingly
small lateral load, dH.
(b) dH-d.1 Relationships
Figure 6.16 Buckling of Elastic Frame Under Gravity Load
224
Chapter 6 P-Li Effects, Stability and Buckling
Consider an elastic {fame and the following load sequence.
(1)
(2)
(3)
Apply gravity load, to both the main structure and the P-~ column.
For this discussion, assume that the structure remains elastic no
matter how much gravity load is applied. Since the structure and the
gravity load are both symmetrical, gravity load does not cause any
lateral deflection (i.e., no sway).
Impose a vanishingly small horizontal displacement, d~, on the
structure, and calculate the required horizontal load, dH, considering
P-~ effects. Hence calculate the lateral stiffness, dH/ ~.
Repeat for progressively increasing values of the gravity load, P.
Let the lateral stiffness of the main structure be K. Since the structure is
elastic, K is constant. The stiffness of the P-~ column is negative and equal to
P/h. The lateral stiffness of the complete structure is K - P/h. This is shown in
Figure 6.16(b).
There are three ranges of behavior, as follows.
(1)
If P < Kh, the structure lateral stiffness is positive. If a lateral displacement, ~' is imposed on the structure and then released, the structure
returns to its original position.
(2)
If P > Kh, the structure lateral stiffness is negative, and the structure is
unstable. H a lateral displacement is imposed on the structure and
then released, the structure collapses.
(3)
If P = Kh, the structure lateral stiffness is zero. If a lateral displacement is imposed on the structure and then released, the structure
stays in its displaced position. The lateral displacement can have any
value, and is not necessarily small.
Provided the structure remains exactly vertical it can be in equilibrium even
if it is unstable. However, if P > Kh even a minute disturbance, such as a
breath of wind or even the impact of a single molecule, can cause the
structure to buckle laterally.
A structure such as this can have three stability states, as follows.
(lr
Stable equilibrium, when the total lateral stiffness is positive. H the
structure is displaced a small amount and released, it returns to its
original position.
\..
·!.~'
Buckling Behavior of Frames
225
(2)
Unstable equilibrium, when the stiffness is negative. H the structure is
displaced a small amount and released, the displacements increase
and the structure buckles.
(3)
Neutral equilibrium, when the stiffness is zero. If. the structure is
displaced a small amount and released, it stays in the displaced
position.
' slightest disturbance
A structure may be balanced in an unstable state, but the
will cause it to buckle. The buckling will be dynamic. Equilibrium will be
satisfied as the structure buckles, but it will be dynamic equilibrium with
inertia and damping forces.
6.6.3
Bifurcation of Equilibrium
It is useful to draw the relationship between gravity load and lateral
displacement, to show load paths .for stable, unstable and neutral
equilibrium. Figure 6.17 shows this relationship.
Pf
t
Total load= P
+I I
I l
flJ l
!
P-~
column.
Stiffness = -P/h.
Gravity load, P
I
Main structure.
Stiffness = K.
(a) Lateral Stiffnesses
Unstable path
Neutral path
r---+
P=Kh
Bifurcation of
equilibrium paths
Stable path
Lateral displacement, .!\
(b) Relationship between gravity
load and lateral deflection
Figure 6.17 Bifurcation of Equilibrium
Figure 6.17(a) shows the structure, with gravity load only. Figure 6.17(b)
shows the relationship between gravitY load and lateral displacement.
If the gravity load, P, is smaller than Kh, where K is the lateral stiffness of
the main structure, the structure is in a state of stable equilibrium, and the
lateral displacement is zero. If the gravity load is larger than Kh, the
structure can be in equilibrium provided the lateral displacement is exactly
zero. This is the unstable equilibrium path, and the slightest lateral
226
Chapter 6
P-~ Effects, Stability and Buckling
disturbance will cause the structure to buckle. If Pis exactly equal to Kh, the
structure is in. a state of neutral equilibrium. The load P = Kh is the elastic
buckling, or critical, load for the structure.
When Pis equal to the buckling load, P-A theory says that the equilibrium is
neutral regardless of the magnitude of the buckling displacement. This is
not the case in reality, and would not be the case if a large displacements
analysis were used. For the structure in Figure 6.17(a), as the lateral
deflection increases the top of the P-A column moves in an arc, and the
height, h, progressively decreases. Hence the stiffness of the P-A column,
~p /h, gets numerically larger, and if the structure stiffness, K, is constant,
the load P must progressively decrease. In Figure 6.17(b), the neutral
equilibrium path is not a horizontal line - it actually curves downwards.
6.6.4 Inelastic Behavior after Buckling
Even if a structure remains elastic until it buckles, it is ur\.likely that it will
remain elastic after buckling occurs. For the frame in Figur~ 6.17, the likely
post-buckling behavior is shown in Figure 6.18.
Load
Elastic
buckling
load
(a) Buckling deflection and
plastic hinges
(b) Load that can be supported
Figure 6.18 Mechanism Forms After Buckling
The main structure is elastic when buckling first occurs. Plastic hinges form
as the structure deflects laterally. Four plastic hinges will ultimately form, as
shown in Figure 6.18(a).
Hinges can form after a small amount of post-buckling deformation or after
a larger amount, depending on the lateral strength of the structure and the
magnitude of the buckling load. Before the first hinge forms, the lateral
stiffness of the structure is zero (the negative stiffness of the P-A column
balances the positive elastic stiffness of the main structure). When a hinge
Buckling Behavior of Frames
227
forms, the lateral stiffness becomes negative and the structure is unstable.
Figure 6.18(b) shows the type of relationship to be expected between the •
lateral deflection and the amount of gravity load that can be supported.
This is an academic example, since it assumes elastic buckling. It is unlikely
that a real structure will remain elastic until it buckles, as considered in the
next section.
6.6.5
Inelastic Behavior before Buckling
The elastic buckling load for a building frame is usually very large, and it is
unlikely that a frame will remain elastic until it buckles.
Consider the ca8e where the frame yields before the elastic buckling load is
reached. This is illustrated in Figure 6.19.
Hinge A
unloads
Hinge B
rotates
p
Equilibrium is stable if K >
(a) Load when hinges form
f
(b) Load on main structure and P-d load
Figure 6.19 Stability When Frame Yields
The gravity load on the structure is progressively increased until buckling
occurs. When the gravity load reaches a value Py, which is smaller than the
elastic buckling load, two plastic hinges form in the beam, as shown in
Figure 6.19(a). This reduces the lateral stiffness of the main structure, and
hence reducesthe buckling load. H the reduced buckling load is larger than
Py; buckling will not occur. H the reduced buckling load is smaller than Py,
the structure suddenly becomes unstable.
To determine whether the structure is stable when two hinges form, test for
stability by imposing a small displacement, dA, and checking the corresponding load. This is shown in Figure 6.19(b).
228
Chapter 6 P-Ll Effects, Stability and Buckling
The load required to deflect the main structure is Kd~,. where K is the
stiffness with plastic hinges. The lateral force from the P-~ column is P,P~/h.
ff K is smaller than Py/h, the structure becomes suddenly unstable when the
plastic hinges form, and. the slightest lateral disturbance causes it to buckle.
If K exceeds PJh the structure is stable, and the gravity load can be
increased.
There is, however, a subtle complication in applying the stability test. In
Figure 6.19(a) the gravity load causes two hinges to form. This suggests that
the lateral stiffness, K, should be the stiffness for the main structure with
two hinges. In Figure 6.19(b), however, when a lateral displacement is
imposed on the frame, one of the hinges rotates in a direction that causes it
to unload, while the second hinge continues to rotate. It can be argued,
therefore, that K should be the stiffness with only one hinge. With this
stiffness the buckling load is larger. As shown later, it can also be argued
that the correct stiffness is the one with two hinges.
This example shows that inelastic buckling can be complex. Again, however,
the example is academic, and the buckling of real structures is even more
complex. The reason is that the example does not account for the effects of
initial imperfections. This is considered in the following sections.
6.6.6
Effect of Initial Imperfection - Elastic Case
In the buckling examples considered so far, it has been assumed that the
structure stays exactly vertical until buckling occurs. With this assumption
there is a bifurcation of the equilibrium path.
In an actual structure there will always be some geometrical imperfections,
and the structure will never be exactly vertical. This changes the behavior,
and there is no longer a bifurcation.
Consider the elastic case first. Figure 6.20(a) shows a frame in which the
columns are initially slightly tilted.
The frame is modeled using a P-.!l column, and the initial imperfection is
applied to this column only. The main structure is assumed to be exactly
vertical (this is really not important - a slight tilt would have only a small
effect on the behavior of the main structure).
Buckling Behavior of Frames
229
!1= (-f l1o)/ (K--f)
(a) Initial imperfection
(b) Equilibrium when gravity load is applied
§·
..
Gravity load, P
Elasticbuckling
load, Kh
-
----------
Smaller !10
Larger f1o
Lateral deflection, !1
(c) Relationship between gravity load and lateral deflection
Figure 6.20 Elastic Behavior with Initial Imperfection
When the gravity load is applied, the P-d column immediately exerts a
horizontal force on the main structure, as shown in Figure 6.20(b). As the
gravity load is increased, this force increases, and hence the lateral
deflection of the main structure increases-. If the elastic lateral stiffness of the
main structure is K, the displacement of the main structure, d, is given by
the equation in Figure 6.20(b).
The relationship between gravity load and lateral deflection is shown in
Figure 6.20(c). If the initial imperfection is zero, the structure buckles
suddenly at the buckling load, P = K.h. For a nonzero imperfection the
structure buckles progressively. As long as the structure stays elastic, the
load path is stable and there is no equilibrium bifurcation. ·
6.6.7
Effect of Initial Imperfection - Inelastic Case
If the structure has initial imperfections and is inelastic, its behavior follows
the equilibrium path for an elastic structure until hinges form. If the
imperfections are very small, the horizontal force from the P-d column will
be small, and hinge formation will be governed by the gravity load, as in
230
Chapter 6
P"~ Effects, Stability and Buckling
Section 6.6.5. If the imperfections are larger, the force from the P-.i\ colwnn
will be larger,· and hinge formation can be affected by this force. Jn either
case the equilibrium path is likely to be as shown in Figure 6.21.
Gravity load
Elastic
buckling --. ----------load
Hinges start to form.
~--- Enough hinges form to
make structure unstable •
.Lateral deflection
Figure 6.21 Inelastic Behavior With Initial Imperfection
Suppose that buckling occurs when two hinges form, aff before (see Figure
6.19). In the case with no imperfections, two plastic hinges form under
gravity load, and one of them unloads as the structUre buckles laterally. In
the case with imperfections, as the load that causes two hinges to form is
approached, the vertical load and lateral deflection are both increasing, and
these,continue to increase as the hinges form. It can be argued that the hinge
rotations can increase· progressively as the frame buckles, and that unloading of one of the hinges does not occur (at least not immediately - as the
frame buckles and the lateral displacements increase, Hinge A in Figure 6.19
must unload). If this argument is correct, the stiffness for checking stability,
and calculating the buckling load, is the stiffness with two hinges, not the
(larger) stiffness with one hinge.
6.6.8
Effed of Gravity Sway
If a structure is unsymmetrical, or if the load is unsymmetrical, gravity load
will cause a structUre to sway (deflect laterally). This is essentially the same
as an initial imperfection. The sway is amplified by the P-.i\ effect, in much
the same way that initial imperfections are amplified.
·-~- ..
In a real structure, neither the structure nor the loading will ever be perfectly
symmetrical.
·
·
Buckling Behavior of Frames
6.6.9
231
Are Buckling Loads Useful?
It is not difficult to calculate the elastic buckling load for a building frame
with gravity loads. However, just because something can be done does not
mean that it should be done. For a building frame the elastic buckling load
is likely to be a very large number. It may be of academic interest, but is
unlikely to have much practical value.
The inelastic buckling load is usually much smaller than the elastic buckling
load. Even this load, however, is likely to be much larger than the actual
gravity load (or the gravity load for which the structure is designed). It is
unlikely that buckling under gravity load alone will be an important design
consideration for a building structure. This is fortunate because the inelastic
buckling load is affected by many things, including initial imperfections, the
distribution of the gravity loads, the effect of yielding on the structure
stiffness, and the buckling behavior of individual columns (a topic which
has not yet been considered in this chapter). Because of this, the inelastic
buckling load is difficult to calculate with any degree of accuracy.
This does not, however, apply for structures of all types. For structures such
as long span roofs, the behavior under gravity load alone can be important,
and the strength may be governed by buckling. Buckling can also be
important for individual columns in a frame. These cases are considered in
later sections.
6.6.10 Summary for this Section
This section has considered several· aspects of buckling behavior, using a
simple frame example. Some conclusions and observations are as follows.
(1)
Multi-story and. 3D frames are obviously more complex than the
simple examples used in this section. However,- their behavior is
qualitatively similar.
(2)
The examples in this section have used a single P-A column to account
for P-A effects, rather than a separate P-A strut for each column. For
these examples a single P-A column has the same effect as a number of
P-A struts. This is not always the case, as considered later.
(3)
This section has not considered the P-o contribution. This contribution
can be significant for some structures, notably frames where individual columns can buckle within their own length. This is considered
later in this chapter.
232
Chapter 6 P-A Effects, Stability and Buckling
(4)
The elastic buckling load for a building frame can be calculated
. accurately (the details of the calrulation :ire not important). This
buckling load is usually of only academic interest.
(5)
The inelastic buckling load is more difficult to calculate accurately.
This is a strength capacity analysis, which is inherently difficult.
Fortunately, this buckling load is also of only academic interest for
most building frames. The important load case for a building frame is
usually combined lateral and gravity load, not gravity load alone.
6.7 P-L\ Columns in Multi-Story and 30 Buildings
P-a Column in Multistory Buildings
6.7.1
P-A columns can be used in multi-story buildings, as shown in Figure 6.22(a).
Load= P3
Ht
Load= P2
t
P2
'
'
Ht
!
P1
Ht'
."'
P31
i l l l 1
P3+
h3
I I I l
Load= P1
h2
I I I l
Fl
-
d3
/J.2
H2 - P3 h3 + (P2+P3) h2
H1 - (P2+P3)
h1
,[.!;J
63
~t ~P3h3
13
/J.2
Axial force = P1 +P2+P3
Main structure story shear =
lateral loads above this story=
P-A column.
Include P·A effects.
Main structure.
Ignore P-A effects.
(a) Model with P-A column
di
hz + (P1+P2+P3) h 1
di
(H1+H2+H3)+(P1+P2+P3) h 1
(b) Forces on P-A column, and
effective loads on main structure
Figure 6..22 P-A Column in a Multi-Story Frame
As with the preceding examples, P-A effects are considered using a P-A
column. Material nonlinearity may be considered in the main structure.
As with a single story frame, the main structure provides lateral support for
the P-'A column, and the P-~ column exerts lateral forces on the main
structure. For a given set of story drifts, the forces on the P-A column are
'
'-.~~'
~:
P-~ Columns in Multi-Story and
30 Buildings
233
shown in Figure 6.22(b). The forces on the right of this figure become
effective loads on the main structure. The effective load at any floor is the
external lateral load at the floor plus a "P-A load".
For the bottom story, the figure shows the effective story shear in the main
structure. This is the sum of the external lateral loads and the P-A loads
above the story. The effective stQry shears can be calculated· for the other.
two stories in the same way. The story shear from the P-A loads is the ;'P-A
story shear".
·It follows from Figure 6.22(b) that the P~A shear in any story depends only
on (a) the axial force in the P-A column at that story (equal to the gravity
load on all floors above the story) and (b) the drift in that story. The drifts in
the other stories, and the axial forces in the P-A columns in the other stories,
do not matter.
The reason is that the P-A loads from all P-A columns above any story cancel
each other out and cause no story shear. It may be noted, however, that the
P-A loads also cause overturniiig moments, and the overturning moment on
a story does depend on the drifts in the higher stories. As an exercise you
might like to calculate the P-A overturning moment at the base of the
structure in Figure 6.22.
The P-A shears have a larger effect on a structure than the P-A overturning
moments. However, the overturning moments can be significant An
analysis model with a P-A column accounts for both of them.
6.7.2
P-A Column in 30 Buildings
The earlier examples are all for 2D structures~ To be ~ for practiccil
analysis, P-A columns must be applicable to 3D structures. The 3D case is
more complex than the 2D case, and a model with a P-A column has some
- potentially serious limitations. It is useful to consider the· 3D case in some
depth because it can give extra insight into P-A effects.
Figure 6.23 shows a 3D frame building with a P-A column.
.
234
Chapter 6 P-8 Effects, Stability and Buckling
Plan
p-&column
LJ
i-,.
I
I
I
l
Elevation
Figure 6.23 P-8 Column in a 30 Frame
This type of model can be useful, but it has limitations. Some of these are as_
follows.
(1)
The floors must usually be assumed to act as rigid diaphragms for in;iplane forces, using rigid diaphragm constraints. These constraints
. connect the P~A column to the frames at each floor.
(2)
The P-A column is usually located at the centroid of the gravity loads.
If a single P-A column is to be used, as in the figure, the centroid of the
gravity loads must be at the same location in all fioors.
(3)
The P-A column exerts loads on the main structure. These loads act at
the location of the P-A column, and are based on the story-to-story
drifts at that location. Jn a 3D frame, the actual P-A loads are applied
by the P-A struts at each column, based on the column axial force and
the· drifts_ at the column location. If there is torsional· deformation of
the structure, the columns in a story can all have different drifts,
· which are not the same as the drift of the P-A column. Hence, a P-A
column does not account accurately for torsional p..,.£\ effects.
·•
Note that a P-A column can exert torsional loads on the main·
structure, depending on the location of the column relative to the
center of stiffness of the main structure. Hence, the column location
can not be chosen arbitrarily.
P-il Columns in Multi-Story and 30 Buildings
(4)
235
When the structure is loaded laterally, the axial forces in the actual
columns progressively change, some forces increasing and some
decreasing. Hence, as lateral load is applied there is a progressive
change in the distribution of the P-A loads applied by the P-A struts.
In effect, the centroid of the gravity loads progressively changes, and
there is a change in the location in plan of the resultant P-A load at
any floor. The effect of this change can be to add torsional loads on
the structure. A P-A column in a fixed location does not account for
this.
Hence, the P~A loads in a 3D structure can affect the torsional moments in
the structure as well as thestory shears. A single P-A column tends to
capture only the story shear effect (and also story overturning moments).
This does not mean that an analysis model with a· single P-A column is
inaccurate for practical analysis. Torsional effects in buildings are complex,
especially for dynamic earthquake loads and for inelastic behavior. For
earthquake loads there are torsional effects when the center of mass in .a
floor is eccentric· to the center of stiffness of the structure. For inelastic
behavior the frames on one side of the building may be weaker than those
on the other side, even in a nominally· symmetrical building. The weaker
frames can yield first, cause the center of stiffness to move, and hence
develop torsional effects. Such torsional effects are likely to be larger than
those due to P-A loads.
·
This emphasizes once again that structural analysis is at best approximate. It
is important to capture the important ·modes of behavior in an analysis
model, but it can be unproductive to worry about the fine details.
6.7.3
Buildings with Multiple P-Li Columns
In Figure 6.23, the centroid of the gravity loads has the same location at all
floors. Figure 6.24. shows a building with a setback, where the centroid of
the gravity loads suddenly changes.
One option in this case might be to use a discontinuous P-A column, as
shown in Figure 6.24(a). The total gravity load at each floor must be applied
at the top of the P-A column below that floor, and there must be a vertical . ·
support at the base of the upper P-A column. Note that P-A columns must be
vertical. It would not be correct to use a single P-A column with an inclined
column in the story below the setback.
236
Chapter 6 P-8 Effects, Stability and Buckling
(a) Discontinuous P-8 column
(b) Multiple P-A columns
Figure 6.24 Building With Setback
A better option is to use two (or more) P-A columns, as shown in Figure
6.24(b). At any floor, the load on a P-A column is the gravity load that is
tributary to that column.
Multiple P-A columns can also be used in regular buildings. For example,
the floors in a rectangular building might be divided into four quadrants,
with a P-A column in each quadrant. With this model, if the building twists,
the P-A columns have different drifts and can exert torsional moments. In
the limit, a separate P-A strut can be used for each actual column in the
building.
·
6.8 Buckling of Multi-Story and 30 Frames
In earlier sections, buckling has been considered only for single story frames.
The buckling behavior of multi-story and 3D frames is more complex because
·there can be a number of different buckling modes.
Figure 6.25(a) shows a multi-story frame with only gravity load. Assume
that the structure remains elastic, and that there are no imperfections and no
sway under gravity load. If the gravity load is progressively increased, the
structure will become unstable at the elastic buckling load.
The behavior is more complex than the single story case because a multistory structure has many possible deflected shapes. The test for stability is to
impose a vanishingly small deflected shape on the structure. For any given
deflected shape there are P-A loads, as shown in Figure 6.25(b). These loads
act on the main structure and cause it to deflect. If the deflected shapes of
Buckling of Multi-Story and 3D Frames
237
the P-A column and the main structure are exactly the same, the deflected
shape is a buckling mode shape. If the deflection magnitudes are also the
same, the structure is in a state of neutral equilibrium, and the gravity load
is the buckling load.
P3
Load= P3
Ll l l 1
Load=P2
P2
+.
{
I I I
h3
~
'
[I;
c !;!
H
.·
dA3
l!~h3
P2
dA2
H
dA3
i:IA2
I ~ ~ h3 + (P2+P3) h2
P1
Load= P1
I I I
dA3
~
-
h2
dA2
dA1
(P2+P3)hZ + (P1+P2+P3) ti1
If these loads deflect the main
structure by dA1, dA2 and dA3,
this is a buckling mode shape.
h1
rrltl
(a) Frame with P-A column
(b) Forces on P-A column,.and
effective loads on main structure
Figure 6.25 P-A loads for a Possible Buckling Mode Shape
For a three story building there are. three possible buckling mode shapes.
These shapes are roughly as shown in Figure 6.26.
Mode 1
Mode2
Mode3
Figure 6.26 Possible Buc~ling Modes for a Three Story Frame
(Beam and column curvatures are not shown.)
The buckling loads are higher for the higher modes. Since buckling will
occur in the lowest mode, the buckling loads for the higher modes are of
only academic interest.
238
Chapter 6 P,.i Effects, Stability and Buckling
If there are initial imperfections, the lateral displacements will progressively
increase as the gravity load is increased. The shape of the initial imperfection may be very different from the buckling mode shape, but as the gravity
load increases, the lateral deflected shape will change, getting progressively
closer to a buckling mode shape. ill theory, if the initial imperfections
exactly correspond to, say, the second buckling mode shape, the shape will
not change as the gravity load is increased, and the gravity load can increase
beyond the first buckling load. If thls happens, the equilibrium is unstable
and the slightest disturbance will cause the structure to buckle, in the lowest
mode.
As with the single story case, if the structure yields under gravity load
before buckling occurs, the buckling load is reduced. The inelastic buckling
· mode shape may be very different from the elastic shape.
There is added complexity for a 3D structure. In a model with a P-A column,
buckling will ~cur when the P-A loads and the structure lateral displac~
ments are in balance. However, the P-Li loads from a single P-A column may
not cause torsional deflections of the structure, whereas if multiple P-Li
struts are used, the P-Li loads often will. cause torsional deflections. It is
possible that the lowest buckling mode involves torsional deflections.
Hence~' if the buckling load is calculated using an analysis model with a
singl_e P-A column, the result could be too large.
As noted earlier, buckling loads are usually of little practical value for
building frames, so this is likely to be an academic issue.
6.9 Buckling of an Axially Loaded Column
6.9.1
Overview
The examples up to this point have considered overall collapse and
buckling of building frames. The examples consider plastic hinge formation,
but ignore the possibility that a column may buckle within its own length.
This is unlikely to happen in a building frame, but it is possible.
The examples do not account for column buckling because they consider
only the P-A contribution. Buckling of a column within its own length
depends on the P-o contnbution.
Buckling of an Axially Loaded Column
239
This section considers the buckling behavior of a single column. This
behavior is similar to that for buckling of a frame under gravity load alone,
but the details are different. The following questions. can be important.
(1)
(2)
What properties of a column affect its buckling strength?
Can the buckling strength of a colurrin be calculated reliably usmg an
analysis model?
In an analysis model for a complete structufe, can the columns be
modeled in such a way that column buckling is accounted for directly
in the analysis?
(3)
This section addresses these questions using a simple pin-ended column.
More complex cases are considered in later sections.
6.9.2
Ideal Elastic Column .
.
.
The simplest example for column buckling is a perfectly straight elastic
column, with frictionless pins at its ends, a symmetrical cross section, and
an axial load applied exactly along the column axis. Figure 6.27(a) shows a
model for such a column.
p+
El
L
~~
P-A · Small displs
column
cable
(a) Model
p+
p+
ti
\
it
\
do i--;
do!-1
I
I
I
I
I
I
I
I
I
I
I
I
I
I
\
I
I
I
I
I
flI
I
I
I
I
I
I
I
I
I
I
I
I
doi+;+--
'
I
I
I
I
I
I
J;
.b-
&
- All are sine curves
/et
p ce- L2
-
(b) Exact solution
p = 9.6EI
ce
-
t2
(c) P'-li cable has
constant curvature
(d) P'-li cable exerts
point force
Figure 6.27 Elastic Buckling of a Pin-Ended Column
This column has only o displacements, since the ends do not displace
horizontally. The model consists of a small displacements column and a P-15
cable.
240
Chapter 6
pc,1 Effects, Stability and
Buckling
When an axial force is applied, the test for stability is to impose a small
deflected shape on the column. For this deflected shape, the lorces on the
P-o cable follow from the cable curvature and the axial force. These forces
act on the small displacements column and cause it to deflect. If these
deflections are exactly the same as the imposed deflections, this is a state of
neutral equilibrium and the axial force is the buckling load for the column.
The "correct" deflected shape is a sine citrve, as shown in Figure 6.27(b). For
this shape the buckling load is P = rt2EI/L2, where E, I and L have the usual
meanings. This can be expressed as
(6.1)
where A is the area of the cross section, r is its radius of gyration, and L/r is
the slenderness ratio for the column.
The buckling load can be estimated using other deflected shapes. For
example, for an imposed deflected shape that has uniform curvature, the
load on the P-11 cable is uniform, as shown in Figure 6.27(c). When the midheight deflection of the small displacements column for a uniform load
matches the mid-height deflection of the P-o cable, it is easy to show that the
required axial force is P = 9.6EI/L2. This is 3% smaller than the exact
buckling load. In this case equilibrium is satisfied in the analysis model,
since the loads on the P-0 cable and the small displacements column are
equal and. opposite. However, compatibility is not satisfied because the
deflected shapes of the P-o cable and the small displacements column are
not identical (the curvature in the small displacements column is not
constant).
A more approximate calculation is shown in Figure 6.27(d). In this case the
deflected shape for the P-o cable is two straight lines, and the force on the
cable is a point load at mid-height. This force is 4Pdo/L. The force required
to cause a mid-height deflection do in the small displacements column is
48Eldo/L3. Hence the mid-height deflections are equal when P = 12EI/L2.
This is 22% larger than the exact buckling load. In this case the deflected ·
shapes of the P-3 cable and the small displacements column are very
different.
·
These approximate methods. are similar to the Rayleigh method for
estimating buckling loads, where a buckling mode shape is assumed and
the buckling load corresponding to this shape is calculated using energy
''-;~
'
Buckling of an Axially Loaded Column
241
principles. The Rayleigh method always gives an upper bound on the
buckling load. An interesting academic question is why the calculation
based on Figure 6.27(c) underestimates the buckling load. ·
Calculations such as these are interesting, but they are of little value for
estimating the strength of a real column, which has imperfections and
inelastic behavior.
/
6.9.3
Imperfect and Inelastic Column
A real column can never be perfectly straight, and hence there will always
be some curvature of the P-<> cable. Even for a small axial load the P-<> cable·
will exert forces on the small displacements column. As the. axial load is
increased there is a progressive increase in the forces exerted by the P-0
cable, and hence a progressive increase in the lateral deflection of the
column. This iS illustrated in Figure 6.28.
p
Elastic buckllng ·
load
---*--Enough yield occurs to
make column unstable
Ii
(a) Column with initial
imperfection
(b) Elastic column
behavior
(c) Inelastic column
behavior
Figure 6.28. Behavior with Imperfections and Yield
Figure 6.28(a) shows a column that is initially not perfectly straight (i.e., it
has an initial imperfection). Figure 6.28(b) shows the behavior assuming
elastic behavior. Figure 6.28(c) shows the likely inelastic behavior. Figure
6.28 is essentially the same as Figure 6.20 in Section 6.6.6 and Figure 6.21 in
Section 6.6.7, for a frame structure.
·
Even if a column is perfectly straight, it may yield progressively as the axial
force increases. Wheri yield Occurs the effective bending stiffness reduces,
and hence the buckling load becomes smaller. In effect, the modulus, E, in
Equation 6.1 decreases, and hence the buckling load decreases. This iS
similar to the behavior of a frame that forms plastic hinges under gravity
foad, as in Section 6.6.5.
242
Chapter 6
6.9.4
Steel Column
P-~
Effects, Stability and Buckling
From Equation 6.1, the buckling strength of an ideal elastic column depends
on its slenderness ratio. Thisis true also for a column with initial imperfections and/ or inelastic behavior.
Consider a steel column first. Figure 6.29 shows how the column strength
might vary with slenderness ratio.
Buckling
Strength
Buckling
Strength
i""' Elastic buckHng
\
_:{_
1
Axial yield
-
Aou .--""""'...:".-\
~
~------
Slenderness Ratio
(a) E-P-P material. Effect of geometrical
imperfection.
c
Stress ·-
Strain
Slenderness Ratio
(b) Perfectly straight column. Effect of
stress-strain curve.
Figure 6.29 Column Buckling Strength
Figure 629(a) shows the relationship between buckling load and slenderness ratio for a column that has initial imperfections and an elastic-perfectlyplastic material. Figure 6.29(b) shows the relationship for a column that is
perfectly straight and has a strain hardening material. The following are
some key points.
(1)
The simplest case is a perfectly straight column with an elasticperfectly-plastic material. For this case the strength is governed by
either elastic buckling or axial yield, whichever is smaller. This defines
an upper bound on the col~ strength, as shown by the dashed lines
in the figure. In Figure 6.29(a) this is the case for 00 = 0, and in Figure
6.29(b) for CJY = CJu.
(2)
For a very short column the strength is the "squash" strength for the
column cross section, Acr,,, where A is the section area and cr. is the
material strength.
(3)
For a column that is not perfectly straight and is slender (with a
slenderness ratio larger than about 120 for a steel column), the actual
Buckling of an Axially Loaded Column
243
strength is somewhat below the elastic bucl<lmg strength. This is
shown in Figure 6.29(a).
(4)
For a column that is slender and initially straight, the buckling load
can be reached before the material yields, and hence the buckling .is
elastic. This is shown in Figure 6.29(b).
(5)
For· a column with an intermediate slendell,less ratio, the strength
depends on the initial imperfection and/ or.· the material stress-strain
curve. This is shown in Figures 6.29(a) and (b).
6.9.5
Reinforced Concrete Column
The effective bending stiffness of a reinforced concrete column can be
influenced by many things, including concrete cracking, steel yield and
time-dependent creep.
Given· an effective EI, the buckling strength can be calculated using
Equation 6.1. Design codes, such as ACI 318, provide guidelines for
calculating effective EI values. Any buckling strength calculation is likely to
be very approximate.
6.9.6
Theories for Buckling Strength
There are three main theories that account for imperfections and inelastic
behavior in column buckling, as follows.
(1)
Initial lack of straightness. The column is assumed have a small initial
lateral deflection, usually assumed to be a proportion of the column
length (of the order of 0.1% of the column length). As axial load is
applied the lateral deflection progressively increases, the column
progressively yields, and at some point it buckles. This point defines
the column buckling strength. This is the type of behavior shown in
Figure 6.28(c) and Figure 629(a).
(2)
Eccentricity in the axial load. The column is initially straight, but the
axial load is not applied exactly along the column axis. As the axial
load is· increased the column bends. If the eccentricity is assumed to
be a proportion of the column length, the behavior is very similar to
that for a column with Wtial lack of straightness. If the eccentricity is
a proportion of the cross section dimension, the behavior is a little
different.
244
(3)
Chapter 6 P-Ll Effects, Stability and Buckling
Loss of stiffness through yield (tangent modulus theory). This theory is
used mainly for steel columns. The column is initially straight, and
has substantial residual stresses (as much as 50% of the steel yield
stress). As the axial load increases the column starts to yi~ld, and the
effective stress-strain relationship is as shown in Figure 6.29(b). The
column buckles at an average stress given by Equatiori 6.1, using the
tangent modulus of the material. Theoretically, the strength for a
slender column, which buckles before yield, is the ideal elastic
buckling strength. In practice this strength is usually specified to be
somewhat smaller than the ideal elastic strength. This recognizes that
a column can never be perfectly straight.
The three theories all give qualitatively similar.results. To obtain buckling
strengths for practical use, all three must be calibrated against experimental
results for real columns. Many properties can differ from column to column,
and the ~uckling strength .can be sensitive to changes in these properties.
Hence, there can be a lot of scatter in the experimental results, and substantial
uncertaint}r about the actual strength of any particular column. For example,
in the tangent modulus theory for a steel column, the effective tangent
modulus depends on the magnitude and distnbution of the residual stresses.
One consequence is that the shape of the buckling strength curve can be
different for strong axis and weak axis buckling.
6.9.7
Combined Material and Geometric Nonlinearity
Figure 6.29 shows the effects of imperfections and yield on the buckling
strength, but does not say much about the underlying causes. This is
important for modeling because a rational analysis model should capture
the underlying causes.
For both of the columns in Figure 6.29, the behavior depends on a combination of geometric and material nonlinearity. This is illustrated in Figure
6.30.
Figure 6.30(a) shows the column strength curve for the case with initial outof-straightness. Assume that the column buckles when a P-M plastic hinge
forms at the column mid-height. Also assume that this is. a rigid-plastic
hinge which forms suddenly (that is; assume elastic-perfectly-plastic
behavior). These assumptions greatly oversimplify the actual behavior, and
are made only to explain the main features of the behavior. The figure
shows the P-M interaction surface for three different slenderness ratios, as
follows.
Buckling of an Axially Loaded Column
245
Buckling
Strength
P=Aa0
Slenderness Ratio
(a) With geometric imperfection
Slenderness Ratio
(b) Perfectly straight rolumn ·
Figure 6.30 Points on P-M Interaction Surface at Buckling
(1)
For a short column, With a small slenderness ratio, there is little or no
bending in the column. Hence, when the hinge forms the axial force is
close to the axial yield force. This is much smaller than the elastic
buckling strength. In this case the effects of geometric.nonlinearity are
small, and material nonlinearity dominates.
(2)
For a large slenderness ratio there is a larger amount of bending, for,
two reasons. First, the column is longer, and hence the initial
deformation is larger, which increases the bending. Second, the axial
force is closer to the elastic buckling load, which means that any
bending moments are greatly amplified. If the axial force at failure is
P, the small displacements bending moment is Po0• This moment is
amplified by the geometric nonlinearity, and when the column fails
the moment is much larger than Po0 • In this case the geometric
nonlinearity tends to dominate.
·
(3)
For a column with an intermediate slenderness ratio, the amount of
moment amplification is smaller. Both geometric and material nonlinearity are important.
Figure 6.30(b) shows the P-M interaction surface for the case with a perfectly
straight column. In this figure the behavior is not elastic-perfectly-plastic.·
The P-M interaction surface in the figure is the ultimate surface. The surface
for initial yield is not shown.
246
Chapter 6
P-~
Effects, Stability and Buckling
Since the column is perfectly straight, there is no bending moment and the
column exhibits equilibrium bifurcation, not progressive buckling. Equilib.rium bifurcation occurs when the axial force reaches the buckling load
based on Equation 6.1, with E equal to the tangent modulus for the material.
The P-M surface iS reached only after buckling occurs.
For a buckling theory that is based orfinitial lack of straightness (or load
eccentricity), the analysis must account for inelastic behavior and moment
amplification. For a theory that is based on the tangent modulus, the analysis
must account for inelastic behavior and equilibrium bifurcation.
6.9.8
Analysis Model for Buckling Strength
As noted earlier (see Section 6.4.8) there are many possible ways to develop
a frame element that accounts for P-~ effects, and the details are not considered in this book: It is a relatively easy task to develop an element that .
accounts for ideal elastic buckling, and such an element is likely to be
available in most comp\!ter programs. It is much more. difficult to account
for progressive spread of yield, initial imperfections, and other effects that
influence the actual buckling strength.
For a single column it is possible, in many computer programs, to set up a
model that can account for these aspects and can give accurate buckling
strengths (provided. that the model is calibrated against experimental
results). Such a model is likely to be complex, requiring several elements to
model a single column, and possibly a fiber representation of the column
cross section. Analyses of such models might he used to calculate buckling
strengths for complex or unusual columns, where experimental results are
not directly applicable. These would be capacity analyses.
For a complete structure with a large number of columns it is not practical,
in most cases, to use such a model for every column, and fortunately it is
usually not necessary. Consider strength-based design first.
For strength-based design of a complete structure, elastic analysis is used
and the purpose.of the analysis is to calculate member force· demands. The
analysj.s model must account for amplification of member forces, but this
can be done using a relatively simple elastic element. The actual buckling
strength for a column, accounting for imperfections, yield, etc., is needed
only on the capacity side, and this strength can be obtained from a column
strength formula in a design code (or possibly from a detailed analysis of a
single column).
Buckling of an Axially Loaded Column
247
Deformation~based design, using inelastic analysis, is usually not much
different. The consequences of column buckling can be so severe that
buckling is likely to be avoided, using capacity design. If this is the case, one
purpose of the analysis is to calculate force demands on the columns, which
can be modeled as elastic elements, as in an elastic analysis. Again, the
actual buckling strength is needed only on the capacity side.
The modeling is more difficult if buckling is allowed to occur. In this case
the analysis model must give reasonably accurate buckling strengths, and
must account for the post-buckling behavior. It may also be necessary to
account for cyclic loading. This is considered later in this chapter.
6.9.9
Summary for this Section
This. section has considered the buckling behavior of a pin-ended column
with axial load. Both the buckling strength and the post-buckling behavior
can be important. Some points are as follows.
(1)
The buckling behavior of a column can be sensitive to the yield
strength of the material, residual stresses, initial imperfections, and
other things. These are difficult to account for in an analysis model.
(2)
For a very slender column, the actual buckling strength may be only
moderately smaller than the elastic buckling strength (roughly 12%
smaller for a steel column). For less slender columns, as in most
building frames, the actual buckling strength is much smaller than the
elastic buckling strength.
(3)
A pin-ended column can be modeled as a combination of a small
displacements column and a P-5 cable. The small displacements
column accounts for the elastic stiffness and the yielding behavior of.
the column. The P-<> cable accounts for geometric nonlinearity and
initial imperfections.
(4)
For a uniform elastic column it is fairly easy to consider the
interaction between the small displacements column and the P-<>
cable, and the elastic buckling behavior can be calculated accurately.
When there is inelastic behavior, however, the small displacements
column and the P-<> cable can interact in complex ways, and the
behavior can be sensitive to the modeling assumptions. The buckling
behavior of an inelastic column must usually be determined by
experiment.
i48
Chapter 6 P-~ Effects, Stability and Buckling
This section has considered the relatively simple case of a column with only
axial load. In most columns there will also be bending, and in a threedimensional colurnri there is biaxial bending. This is considered later.
6.1 o Simple Structure with Pin-Ended Members
6.10.1 Overview
The preceding section· considered the behavior of a single pin-ended
column. This section considers the behavior of a braced frame structure that
consists of several pin-ended columns.
This is an academic structure, with simpler behavior than a real braced
frame. However, it is useful for illustrating a number of points about
modeling and analysis. More realistic structures are considered later.
6.10.2 Example Structure and Analysis Model_
Figure 6.3l(a) shows a braced frame (a 2D truss) with pinned connections.
P-z\ column
(a) Braced frame
Main structure
(b) Analysis model
Figure 6.31 Simple Braced Frame
·~~
This is an academic example, since a real structure will rarely have pinned
connections. However, it illustrates several aspects of the modeling ru:id
analysis process.
Consider the performance assessment for combined gravity and lateral load,
using lateral load analysis (apply gravity load, then add static lateral load).
Simple Structure with Pin-Ended Members
249
Since the connections are pinned, and since there are only nodal loads, the
elements have only axial force, with no end moments and no transverse
loads along the element lengths. Assume that there are three concerns for
design, as follows.
(1)
(2)
(3)
Element yield in tension. The example structure is statically
determinate, so if one element yields the structure collapses.
Element buckling in compression. If one element buckles, the
structure collapses.
Elastic buckling of the structure as a whole. The structure might
become unstable, because of P-Li effects, before any element yields or
buckles. This is unlikely, but it could happen for a structure that is tall
and flexible.
For this example assume that the material is steel, and that only strength is
of concern, not serviceability.
Figure 6.31(b) shows an analysis model for the structure. This model has a
small displacements main structure and a P-Li column. A model with a P-A
strut for each element could also be used, and would have similar behavior.
There are no P-o cables, for the following reasons.
(1)
If the elements are assumed to be perfectly straight, there is no P-o
contribution, and including P-o cables has no effect. Straight elements
are usually assumed for an analysis model.
(2)
If the elements were specified to have initial imperfections, there
would be a P-0 contribution, and including P-o cables would have an
effect. These cables might be added to the model in an attempt to
account for column or brace buckling directly in .the analysis. As
already noted, however, it is not that simple. As shown below,
buckling is considered as a design problem, on the capacity side, not
an analysis problem on the demand side~
6.10.3 Strength-Based Design Using Elastic Analysis
Consider strength-based design using elastic struchtral analysis. The details
of the analysis are not important for this discussion. The main steps are as
follows.
·
(1)
Analysis model. Calculate the element axial stiffnesses, and hence set
up the analysis model. The usual assumption is that the axial stiffuess
is EA/L where Eis the elastic modulus, A is the element area, and Lis
~
250
Chapter 6 P-t. Effects, Stability and Buckling
the element length between the centers of the connections. It is also
usual to assume that the connections are frictionless pins.
(2)
Strength Capacities. Calculate the tension and compression strength
capacities for the structural components (in this case, the pin-ended
bar elements). The tension strength capacity may be governed by
yield in the body of the element or by the strength of the end
connection. The compression strength capacity is likely to be governed by buckling, not yield. The buckling strength must account for
initial imperfections, inelastic behavior, etc.
(3)
Gravity load analysis. Apply the factored gravity loads for each gravityonly load case (e.g., 1.2 Dead Load + 1.6 Live Load). There will usually be more than one load case. Analyze the structure, allowing for the
P-d contribution (since there may be lateral sway under the gravity
load). Calculate the strength (axial force) demands.
(4)
Strength DIC ratios for gravity load. For each gravity load case calculate
the strength DIC ratios for all elements.If any DIC ratio exceeds 1.0,
the performance requirements are not met. If this is the case the
structure must be redesigned, or the loads must be adjusted, or the
strength capacities must be calculated more accurately.
(5)
Buckling analysis. Unless the structure is slender, this step is probably
not needed. Check that the structure as a whole does not buckle under
the gravity load. Since the element strengths have been checked in
Step 4, if overall buckling occurs, it must do so elastically, with no
buckling or yield of individual elements. The demand is the gravity
load. Calculate the buckling load capacity, and hence calculate the
DIC ratio. If the structure buckles, the performance requirements are
not met.
(6)
Lateral load analysis. Analyze for q>mbined gravity and lateral loads,
allowing for the P-d contribution, and calculate tb.e axial force
demands. There will be several load cases, and the gravity loads are
usually smaller than those for the gravity-only load case (e.g., 1.2
Dead Load + 0.5 Live Load + 1.3 Wind Load). The force demands may
be significantly different from those that would be calculated using a
small displacements analysis.
(7)
Strength DIC ratios for combined gravity and lateral load. Calculate the
strength DIC ratios for all elements. If any DIC ratio exceeds 1.0, the
performance requirements are not met.
'
·~.
Simple Structure with Pin-Ended Members
251
The steps are essentially the same for a more realistic structure but the
·
details are more complicated, as considered later.
6.10.4 Deformation-Based Design Using Inelastic Analysis
The steps in the preceding section are for strength-based design using elastic
analysis. For deformation-based design using inel~tic analysis, Steps 1
through 5 are likely to be the same. However, the 'lateral load analysis in
Step 6 is different. Some differences are as follows. ·
(1)
Members in tension may be allowed to yield. The demand-capacity
measure is likely to be element extension. The actual members must
be able to yield in tension without fracturing, and deformation capacities
must be estimated.
(2)
Members in compression may be allowed to buckle. If this is the case
the buckling strength and the post-buckling behavior are both important,
and it may also be necessary to consider cyclic deformation. The
demand-capacity measure is likely to be related to the element axial
deformation, and deformation capacities must be estimated.
(3)
Alternatively, members in compression may be designed not to buckle.
Because buckling can lead to catastrophic collapse, and also because
buckling and post-buckling behavior are difficult to model, the
compression members may be protected against buckling by ensuring
that the buckling strength capacity always exceeds the axial force
demand. In this case these members can be modeled using elastic
elements and the demand-capacity measure is axial force.
An inelastic analysis is more complex than an elastic analysis. The advantage is that inelastic behavior is explicitly considered, and structural analysis
is used to ensure, with reasonable certainty, that undesirable inelastic
behavior does not occur. For strength-based design, if the loads exceed the
design loads and cause yield or buckling, the subsequent behavior can be
uncertain.
6.10.5 Modeling of Diagonal Brace Behavior
If members in cm~pression are allowed to buckle, this must be. accounted
for directly in the analysis model. As noted earlier (Section 6.9.8), an·
analysis model that accurately captures the buckling strength and the postbuckling behavior is likely to be complex, even for a pin-ended column.
252
Chapter 6 P-Ll Effects, Stability and Buckling
In a braced frame it is unlikely that the vertical columns will be permitted to
buckle., However, diagonal braces may be permitted to have inelastic behavior,
with both buckling in. compression and yielding in tension. It can often be
reasonable to assume that braces act as columns with only axial force.
Figure 6.32 shows the type of behavior that can be expected when an axially
loaded column or brace buckles and is subjected to cyclic load.
Axial
Deformation
Figure 6.32. Force-Deformation Relationship for a Buckling Strut
Since it is difficult to model both the geometric and material nonlinearity
directly in an analysis model, for practical purposes it is usually better to
model a diagonal brace as an element that has only material nonlinearity,
with a hysteresis loop of the type shown in the figure. That is, the geometric
nonlinearity is modeled as equivalent material nonlinearity.
6.1 l Pin-Ended Elastic Column with Bending
6.11.1 Overview .
The preceding section considered a frame with pin-ended members that
resist only axial force. In most frame structures the columns have bending
moments as well as axial forces. This section, and the sections that follow,
consider the behavior and modeling of such columns and their use in braced
and unbraced frames. This section starts with the simplest case, namely an
elastic, pin-ended beam-column.
Pin-Ended Elastic Column with Bending
253
6.11.2 Elastic Column with Sinusoidal Lateral Load
A compression member may be a beam-column, with lateral as well as axial
load. The simplest case is an elastic column that is initially perfectly straight,
is loaded axially, and also has a lateral load that varies sinusoidally. This is
shown in Figure 6.33.
Lateral load, deflected shape
and bending moment diagram
are all sine curves
Bending moments
with no axial load
Amplified moments
with axial load
Figure 6.33. Simplest Case of a Beam-Column
If the lateral load is applied first, with zero axial load, the bending moment
at the column mid-height is the small displacements moment, M 0 • When the
axial force is added, this moment is amplified. The "exact" solution for the
amplified moment is
M
1
-=---Mo (1-P !Pee)
(6.2)
where M = amplified moment, P = axial force, and Pce = elastic buckling·
(critical) load for the column under axial load alone. This relationship has
the form shown in Figure 6.34.
This figure shows that the amount of moment amplification can be large,
and approaches infinity as the axial force approaches the .elastic buckling
load, P..,.
The amount of amplification for any column depends on its slenderness
ratio. For a short column Pce is large, the ratio PIPce is likely to be small, and
there is little amplification. For a slender column Pce is smaller, the ratio
PIP"' may be large, and there can be a lot of amplification.
254
Chapter 6 P-8 Effects, Stability and Buckling
M/M0
St+-~~~~~~~~-.
infinite
O'--~~~~--...._~~-+~~....,.~•
0
0.5
0.8
1.0 . P/Pce
Figure 6.34. Bending Moment Amplification
6.11.3 Elastic Column with Other Lateral Loads
Equation 6.2 is for a sinusoidally varying lateral load. The amount of
moment amplification is very similar for a column with a uniform lateral
load, and roughly similar for a column• with a lateral point. Joad at midheight. It is common practice to use Equation 6.2 for a pin-ended beamcol~ with any distribution of lateral load.
6.11.4 Elastic Column with End Moments
A column can also have applied end moments, as shown in Figure 6.35.
M2~
M,l
(a) Singe curvature, Cm> 0.6.
Amplification is possible.
M2
f
(b) Double curvature, Cm< 0.6.
Amplification is unlikely.
Figure 6.35 Column with End Moments
Pin-Ended Elastic Column with Bending
255
Figure 6.35(a) shows a colurrm with erid moments that cause single curvature.
When axial load is applied, the b~ding moment diagram can be amplified
as shown. Since the end moments are applied loads they do not change, and
the maximum moment occurs within the column length. The amplification
can be approximated by the following equation:
Mmax
M2
(0.6+0.4MifM2) _
(1-P/Pc,)
Cm
>l
(1-P/Pce) i
(6.3)
where Mm"' is the maximum moment, M2 is the larger end moment, ~ is the
smaller end moment, and Cm= 0.6+0.4MJ M,. This equation is approximate,
but it agrees quite closely with the results of exact analysis for a uniform
elastic column. FOr a uniform moment, with M, = M2, Cm = 1 and Equation
6.3 is the same as Equation 6.2.
Figure 6.35(b) shows a column with double curvature. In this case Cm is
smaller than 0.6. When Equation 6.3 is used, it is common to impose a lower
limit of 0.4 on cm.
.
End moments of the type in Figure 6.35(b) are most likely to occur in an
unbraced frame, where the lateral load is resisted by column bending. A
colurrm in such a frame is unlikely to be slender for in-plane buckling, and
hence is likely. to have a large elastic buckling load, P"' and a small ratio
PIPce· For a triangular bending moment diagram with M, = 0, Equation 6.3
gives cm = 0.6. In this case, if PIP"' < 0.4 there is no moment amplification.
For the case where M, = 0.5My Cm = 0.4,_and there is no amplification if PIPce
<0.6.
It is likely, therefore, that if hinges form in the column of an unbraced
frame, these hinges will be at the top and/ or bottom of the· column, not
within its length. Note, however, that this may not apply for a 3D column,
which might buckle out-of-plane.
For a column in a braced frame, a bending moment diagram as shown in
Figure 6.35{a) is possible, and a hinge might form within the column length.
For a building frarile it is unlikely that the columns will have significant
lateral loads, and very likely that they will have end moments. Hence, for
most structures the loading in Figure 6.35 is more ~ommon than that in
Figure 6.33.
.
256
Chapter 6
P-~
Effects, Stability and Buckling
6.11.5 Direct Calculation of Amplified Moments
As an alternative to .Equation 6.3, the amplified moments might be
calculated directly, by accounting for the P-o contribution directly in the
analysis model. This alternative is considered later in this chapter.
6.11.6 Elastic Column in a Frame
End moments can also be caused by an eccentrically applied end load, and
by interaction between the beams and columns in a frame. This is shown in
Figure 6.36~
Pe
. Moment diagram ignoring
geometric nonlinearity
17-1___
v(a) Column with eccentric load
~.,';,':netric nonlinearity
(b) Column in .a braced frame
Figure 6.36 Other Causes of End Moments
In Figure 6.36(a), the end moment is caused by an applied load, and the
behavior is similar to that in Figure 6.35.
In Figure 6.36(b) the end moment is caused by interaction between the beam
and column. In this case the bending moment at· the top of the column
depends on the relative bending stiffnesses of the beam and column. As the
loads increase, the axial force in the column increases, and hence its effective
bending stiffness decreases (because of the P-o contribution). Hence, the
bending moment at the top of the column is smaller than would be calculated in a small displacements analysis. As the axial force in the column
approaches the buckling load, the bending moment at the top of the column
reduces to zero. This is not the same as a sustained moment that is applied
directly.
Another possible coniplication is as follows. If a tall unbraced frame is subjected to dynamic earthquake load, and if there are substantial higher mode
effects, some columns could, at least temporarily, have bending moment
diagrams with single curvature, as in Figure 6.35(a), rather than with double
....
·~.
Beam-Column Strength
257
curvature, as in Figure 6.35(b ). This might make such columns more vulnerable to buckling.•
6.12 Beam-Column Strength
6.12.1 Overview
The preceding section considered elastic beam-columns, where the primary
concern is moment amplification. When elastic analysis is used for strengthbased design,· the calculated bending moment demands must account for
moment amplification. The strength DIC ratio must then be calculated using
the axial force on the· column and the amplifled moments, accounting in
some way for P-M interaction
For the simple frame in Section 6.10 it was necessary to consider only axial
forces on the members. For strength-based design the axial force demands
were calculated by elastic analysis, and the corresponding capacities were
obtained from a column strength formula. The procedure is the same for
deformation-based design, provided the columns are not allowed to buckle.
If the columns are allowed to buckle, the analysis is more complex because
the buckling strength and post-buckling behavior must be accounted for
directly in the analysis model.
The process is similar for frames where the members have both axial force
and bending, but the details are more complicated, mainly because the
str-ength capacity must account for P-M interaction (i.e.; for material
nonlinearity). This section considers beam-column behavior when there is
both material and geometric nonlinearity.
6.12.2 Elastic-Perfectly-Plastic Behavior
Consider a beam-column that has elastic-perfectly-plastic behavior. Such a
column remains elastic until the combination of the axial force and the
amplified bending moment reaches the P-M interaction surface for the
column cross section. At that point the column buckles. Figure 6.37 shows
the behavior.
Figure 6.37(a) shows the column, for the case with equal end moments. The
formula that is commonly used for moment amplification is shown in the.
figure. The maximum amplified moment is at the column mid-height. Assume
258
Chapter 6
P-~
Effects, Stability and Buckling
that the column buckles when a P-M plastic hinge forms at this location (in a
real column it will occur befo~ this point, as shown later).
p
Pu
Small displacements
/ /Short column (large Pce• > Pul
Strength is reached
when a P-M plastic
hinge forms.
'/
__• Intermediate column
M
1
----Mo
1-P I Pee
long column
(small P.,,,< Pu)
M
M
(a) Column and loads
(b) P-M paths
Figure 6.37 P-M Paths and Beam-Column Strength
Consider the case where the axial force and the end moments are increased
in the same proportion. For a small displacements analysis the axial force, P,
and the bending momerit at mid-height, M, increase proportionally, and the
P-M path is a straight line. This is the "small displacements" line in Figure
6.37(b). For small displacements behavior the column forms a collapse
mechanism when the P-M point reaches the yield surface. The figure shows
a P-M surface of steel type.
For a P-~ analysis the P-M path is not a straight line, because the mid-height
bending moment is amplified and increases more rapidly than the end
moment. Figure 6.37(b) shows three P-M paths, for short, mtermediate and
long columns. At one extreme, a short column has a large elastic buckling
load and orily a small amount of moment amplification. Hence, the P-M
path is nearly linear. At the other extreme, a long column has a smaller
elastic buckling load, more moment amplification, and a P-M path that is
strongly norilinear. In each case, the column buckles when the amplified
P-M point reaches the yield surface.
Based on this behavior, the following procedure could be used to check the
strength of a column with end moments.
(1)
(2)
(3)
Calculate the axial force and end moments.
Calculate the maximum amplified moment, using Equation 6.3.
If the amplified P-M point lies within the P-M interaction surface for
the column cross section, the strength DIC ratio is smaller than 1.
Beam-Column Strength
259
This procedure is correct, however, only if the behavior is elastic-perfectly-.
plastic. Most real columns yield progressively, which increases the amount
of amplification. Consider this next.
6.12.3 Behavior with Progressive Yield
Figure 6.38 is similar to Figure 6.37, except that the plastic hinge yields and
strain hardens before it reaches its ultimate strength!
p
HingePorM
PuorMu·~
..
·.
PyorMyL__.
Py
Small strain
hardening
Hinge extension
or rotation
My
(a) Loads
(b) Hinge properties
Mu
M
(c) Possible P-M paths
Figure 6.38 Effect of Strain Hardening of P-M Path
Figure 6.38(a) shows the column. As before, assilme that the axial force and
end moments increase in the same proportion, and that the column buckles
when a P-M plastic hinge forms at mid-height. To account for progressive
yielding, let the hinge have trilinear behavior as shown in Figure 6.38(b).
For the case of a moderately long column, Figure 6.38(c) shows four possible
P-M paths, for different hinge properties. These paths are as follows.
(1)
For small displacements analysis the P-M path is straight. The column
becomes less stiff when the P-M point reaches the "Y" surface, and a
mechanism forms when the P-M point reaches the ''U" surface.
(2)
For elastic-perfectly-plastic behavior (an infinite strain hardening
stiffness) the behavior is as shown in Figure 6.37 for the intermediate
column.
(3)
For a hinge with moderately large strain hardening, the bending
stiffness of the column decreases when the Y surface is reached.
260
Chapter 6 P-t. Effects, Stability and Buckling
Hence, the deflections increase more rapidly than for the e-p-p case,
and fhere is more moment amplification. The axial force at buckling,
and hence also the end moment, is substantially smaller than for e-p~p
behavior.
(4)
For a hinge with small strain hardening, the bending stiffness reduces
so much when the Y surface is reached that the column becomes
unstable immediately. The axial force and end moment at buckling
are much smaller than for e'-p-p behavior. In this case the postbuckling P-M path is likely to be as shown (the post-buckling
behavior is not shown for the other cases).
This example is still over-simplified, but it indicates the type of behavior
that might be expected for an actual beam-column. The example shows that
for strength-based design, where the strength is required on the capacity
side, it is not a simple task to determine the combination of axial force and
.end moment that corresponds to buckling. For deformation-based design
where columns are allowed to buckle, it is even more difficult to model the
buckling strength and post-buckling behavior. For a compression member
that has only axial force, it is possible to model the geometric nonlinearity as
equivalent material nonlinearity (see Section 6.10.5, Figure 6.32). This can
not be done so easily when there is bending moment as well as axial force.
6.12.4 Amplification Using Tangent Modulus Theory
Section 6.12.2 considered amplification based on Equation 6.2, using the
ratio PIP"'. As shown in Section 6.12.3, however, when progressive yielding
is considered the amount of amplification can be much larger than that
obtained from Equation 6.2. To estimate the amplification accurately, it is
necessary to account for inelastic behavior.
There are many ways that an inelastic beam-column might be modeled for
analysis, most of which are likely to be complex. There is, however, a simple
model that may be useful and is worth noting. This model uses tangent
modulus theory. ·
For a column with only axial load; tangent modulus theory assumes that as
the axial force, P, increases the effective bending stiffness, EI, decreases. The
effective stiffness is assumed to be E), where E, is the tangent modulus for
the column material. For any value of E, the bucking strength for a pinended column with length L (and effective length factor= 1) is given by
Beam,Column Strength
p - 1f2E,I
ct -
261
(6.4)
I!
For a column with both axial force and bending, a reasonable assumption
may be that the tangent modulus depends only on the ii.xial force and is not
affected by the bending moment. Consider a steel column with a "squash"
strength Pu. Two possible relationships between axial force and tangent
/
modulus are shown in Figure 6.39(a).
Stress
E/E
Quadratic
1.0
Linear
Linear
0.5au
0.5
1.0
P/Pu
(b) Implied stress-strain
relationships
(a) Tangent modulus
. relationships
Figure 6.39 Tangent Modulus Relationships
The linear relationship is defined by the following equations.
ForPIPu<0.5, E,=E
For PIP.> 0.5, E, =2E(l -P/P,)
(6.5a)
The quadratic relationship is defined by the following equations.
ForPIP~<0.5,
E,=E
F~r PIP.> 0.5, E, = E(4PIP,)(1-PIP,)
(6.5b)
The corresponding stress-strain relationships are essentially as shown in
Figure 6.37(b).
These relationships are used only for the purfioses of this discussion, and
are not nece5sarily realistic. However, it is common to assume that the maximum residual stress in a steei section is roughly 0.5 times the steel strength,
so yield can be expected to begiri when P is roughly 0.5Pu·
262
Chapter 6
P-~
Effects, Stability and Buckling
With these assumptions, the strength of a beam-column can be estimated.
Consider a column with a length L, an elastic bending stiffness EI, a "squash"
strength PU' and a plastic moment capacity Mu.
As before, assume that the column buckles when the amplified P-M path
reaches the P-M interaction surface. However, base the moment amplification on Pct not Pce1 where Pct is given by Equation 6.4 and E, is given by
Equation 6.5. That is, for a column with axial fore~ P and equal end
moments M"' calculate the amplified moment, M, using
M
1
M0
(l-P/Pi,1 )
(6.6)
Given P and M"' the steps for checking the strength are as follows.
(1)
(2)
(3)
(4) .
Calculate E, from Equation 6.5.
Calculate Pct from Equation 6.4.
Calculate the amplified moment, M, from Equation 6.6.
Check whether the P-M point lies inside the P-M interaction surface
for the column cross section, based on Pu and Mu.
For progressively increasing P and MO' a similar procedure as before can be
used to calculate the point where the P-M path intersects the P-M surface
(see Section 6.12.2, Figure 6.37). Since Pct is smaller than Pce1 the amount of
moment amplification is larger than for an elastic column.
As an example ofthe possible effect, consider a column as follows.
(1)
The elastic buckling strength, Pce1 is 1.4 times the "squash" strength,
Pu· This corresponds to a steel column with a slenderness ratio of
roughly60.
(2)
The axial force, P, is 0.7 times Pu. This is rather a large axial force.
However, PIPu must be larger than 0.5 for the tangent modulus to be
smaller than the elastic modulus.
(3)
For Pf Pu= 0.7, Equation 6.5(a), for the linear relationship, gives E,/E
= 0.6. Hence, Pct = 0.6P= = 0.84Pu. The corresponding values from
Equation 6.S(b), for the quadratic relationship, are E,/E = 0.84 and Pct
= 1.18Pu·
(4).
The ratio PIPce = 0.7Pu/1.4Pu = 0.5. Hence, for amplification based on
Pee the ratio M/M0 is 2.0.
Beam-Column Strength
263
(5)
For the linear relationship, the ratio PIPct = 0.7Pu/0.84Pu
Hence, for amplification based on Pct the ratio M/M0 is 5.,.
= 0.83.
(6)
For the quadratic relationship, the ratio P/Pct = 0.7Pu/1.18Pu
Hence, for amplification based on Pct the ratio M/ M0 is 2.4,
=0.59.
Hence, the amplification based on Pct can be much larger than that based on
Pee· With an appropriate choice of an equation .for E, (not necessarily
Equation 6.5), the calculated strength might agree with experiment.
Note that this method ·predicts only the column strength. It says nothing
about the post-buckling behavior. Hence, this is not a useful model for an
inelastic analysis where the columns are allowed to buckle.
6.12.5 Tangent, Reduced or Secant Modulus?
The preceding method is mentioned not merely because it is interesting, but
also because a similar method is an option in the "Direct Analysis Method"
for the strength-based design of steel frames using elastic ·analysis, as
considered later in this chapter. However, the Direct Analysis Method uses
a reduced, or effective, modulus rather than a tangent modulus. The
distinction is important and warrants some discussion.
In tangent modulus theory, the buckling strength of a column depends on
the tangent modulus, and is given by Equation 6.4. This is why the
preceding section uses the tangent modulus, E,. It might, howeYer, be
assumed that because of yield the effective bending stiffness of a column is.
smaller than the initial elastic stiffness, and is defined by a reduced, or
effective, elastic modulus, ET, that is smaller than the initial modulus. Since a
smaller bending stiffness means a smaller buckling strength, Equation 6.4
might be assumed to apply, with ET rather E,.
From a modeling viewpoint this has some theoretical implications. Figure
6.40 shows a stress-strain relationship and three different moduli (initial,
tangent and secant).
For elastic analysts, a reduced, or effective, modulus is a secant modulus. In
Section 6.12.4, however, the theory is presented in terms of the tangent
modulus. In principle this is a major distinction, which raises some interest. ing theoretical issues. In practice; however, it makes no difference. The key
point is that once the relationship in Equation 6.5 is chosen, the same amplified moment is obtained whether the relationship is for a tangent modulus
264
Chapter6 P-.l\ Effects, Stability and Buckling
or a reduced (secant) modulus. The only difference is in the implied stressstrain curve. This curve hits the general shape shown in Figure 6.39(b), and
the shape of the curve is different if Equation 6.5 defines a secant modulus
rather than a tangent modulus. However, the shape of this curve is incidental,
and it does not affect the calculations.
Stress
Tangent modulus -
Initial
modulus
Strain
Figure 6.40 Tangent, Secant and Reduced Moduli
6.13 Strength-Basetfl>esign of Beam-Columns
6.13.1 Overview
The preceding sections have considered how amplified moments might be
calculated, and how the strength of a beam-column might be determined,
but only in general terms.
This section consider8 some specific procedures that are used .in· design
codes to calculate strength D/C ratios. In particular, this section contrasts
the approaches that are used in the AISC steel code and the AO concrete
code..
6.13.2 Steel Beam-Columns
First consider the AISC procedure for steel beam-columns.
For a beam-column there are two extreme load cases, namely axial farce
only with no bending. moment, and bending moment only with· no axial
force. For the case with axial force only, the strength can be obtained fronf a
column strength curve. This is the actual buckling strength of the column,
P"' which is smaller than the elastic buckling strength, Pce· For the case with
bending moment only, the strength is· the ultimate moment capacity of the
column section, Afu.
Strength-Based Design of Beam-Columns
265
. Between these. two extremes there is interaction between P and M. This is
shown in Figure 6.41..
Axial Force
Short column has no amplification
and Pc = Pu· Interaction surface
is P-M surface for cross section.
Longer column has amplification and
Pc< Pu· Interaction surface with
amplified moments is roughly straight. .
~-......-- Interaction surface with
non-amplified moments.
Mmax
Cm
-=--Moment
Figure 6.41 P-M Interaction in a Beam-Column
for
The· figure shows P-M interaction surfaces
two column lengths. For a
short column there is no moment amplification, the buckling strength P. is
equal to the "squash" strength, Pw and the interaction surface is the P-M
yield surface for the column cross section. For a longer column. there is
moment amplification, P. is smaller than Pu, and the interaction surface is
smaller. There are also two different P-M surfaces, one for the amplified
moment and one for the small displacements moment with no amplification.
For the longer coiumn it has be~ found, from analysis and experiment, that
the interaction relationship is roughly linear, provided M is the amplified
moment; M,,,.., calculiited using the formula shown in Figure 6.41. If M is the
small displacements moment, the interaction relationship is as shown by the
dashed line.
Hence, for the longer column the interaction surface is defined fairly
accurately by the linear interaction equation:
P
P.,
Cm
M =l
(1-P/Pa) Mu
(6.7)
There is no physical reason why this equation should apply (and for a short
column it is not very accurate), so it is essentially empirical. The equation
also mixes geometric and material nonlinearity, whereas ideally they should
be kept separate. As noted earlier, the amplification formula is based on the
elastic buckling strength, peel and it underestimates the amount of
~
'
266
Chapter 6 P-A Effects, Stability and Buckling
amplification. It can be argued that a more rational equation would account
accurately for the moment amplification and use only tha P-M interaction
surface for the column cross section (based on Pv rather than PJ
However, Equation Q.7 is convenient to use for strength-based design, and it
has been used in the AISC Specification for many years (the Specification
actually adjusts the equation somewhat, so that it more closely matches the
P-M interaction surface for a short column).
For biaxial bending, the equation has essentially the following form :
P
Pc
-+
Cmx .M"
Cmy
My.
+
=1
(1-P/Pcex)MUr (1-P/Pcey)MUy
(6.8)
where x and y are the principal axes of the column cross section. This
equation can be conservative for the following two reasons.
(1) It assumes linear strength interaction between M,; and M,.
(2) . · P. is the smaller of Pex and Per·
This second point could be importantfor a column with double curvature,
where moment amplification is unlikely, and any plastic hinges will be at
the member ends (see Section 6.11.4, Figure 6.35). In this case the strength
irlteradion surface should be the surface· for a short col1lmn, based on Pv·
For in-plane buckling, the buckling strength, P., is likely to be close to Pv·
For out-of-plane buckling, however, the slenderness ratio is likely to be
larger, and P. may be smaller than Pu· Jn Equation 6.8 this smaller value of P.
must presumably be used. This is conservative for strength-based design.
For deformation-based design it would not be accurate to use a P-M-M yield
surface based on Equation 6.8.
·
6.13.3 Reinforced Concrete Beam-Columns
For the strength of reinforced concrete columns the AO code uses
essentially th~ folloWing procedure; This is substantially different from the
procedure for steel columns.
(1)
The "elastic" nominal buckling strength for an axially loaded column,
P.,., is calculated using the formula,P.,. = ~EI/(kL)2 , where k =1 for a
pin-ended column. However, whereas the elastic EI has a clear
meaning for steel, there is no clear meaning for the "elastic" EI of a
reinforced concrete section. The code specifies procedures for
.....
-~.
Deformation-Based Design of Beam-Columns
267
calculating EI. This is roughly 0.25 times EI for the gross concrete
cross section.
(2)
The bending moment at the end of a column can not be smaller than a
specified value (the axial force multiplied by a minimum eccentricity
equal to 0.6 inches, or 15 mm, plus 0.03 times the cross section
dimension).
(3)
The amplified bending moment is calculated for each bending
direction using the following equation.
~ax
cm
M2
(1-P/0.75Pce)
(6.9)
For a given EI this gives more amplification than the equation for a
steel column, because of the 0.75 factor.
(4)
The strength interaction surface is the P-M-M surface for the cross
section, which accounts for material nonlinearity only and does not
depend on the slenderness ratio.
Unlike the procedure for a steel column, this separates material and
geometric nonlinearity. Geometric nonlinearity is accounted for by Equation
6.9, and material nonlinearity by the P-M-M interaction surface. The use of a
small EI value, and the 0.75 factor in Equation 6.9, accounts for inelastic
effects. The minimum bending moment accounts for imperfections.
6.14 Deformation-Based Design of Beam-Columns
If buckling of a beam-column is allowed,. this is a complex problem for
modeling and analysis. Even for a column that has only axial load , the postbuckling behavior is difficult to model directly, and the best modeling
method for such a column is usually to treat all of the nonlinear behavior as
a form of material nonlinearity (see Section 6.10.5, Figure 6.32). For a beam- column this is generally not possible, because of interaction between the
axial force and bending moment.
There is no simple answer. If buckling of a beam-column is permitted and
must be modeled, an element model that accounts for inelastic behavior,
moment amplification and P-M interaction must be used.
268
Chapter 6 P-A Effects, Stability and Buckling
.
6.1 S Compression Members in Braced Frames
.
6.15.1 Overview
Section 6.10 considered the academic example of a braced frame with
. pinned connections. Real braced frames usually have connections that can
transmit ·moments, gusset plates that can act as stiff end zones, initial
imperfections, members and connections that are inelastic, and other
complications. Hence, the frame members usually have both bending and
axial force, and are not simple pin-ended columns.
This section considers braced frame structures. The analysis process is
similar to that in Section 6.10, but the details are more complicated.
6.15.2 Frame Braced by a Wair
Figure 6.42(a) shows a frame that is braced by a shear wall.
f_I =nal
p
Gravity load on all floors
+llt + l l l l l t
Moment
from beams
..J.:.
t~
from beams
Rotational
stiffness .
Moment \
~
from beams "-.../'
(a) Structure
from beams
(b) Typical column
Figure 6.42 Frame Braced by a Shear Wall
. ' -·
.
.
Thewall and frame are connected at the floor levels. The wall is muCh stiffer
laterally than the frame, and resists essentially all of the lateral load, The
frame resists mainly gravity load. The beam-to-column connections in the
frame can transmit bending moments.
A frame of this .type is more likely to be concrete than steel. The columns
will often support a flat plate floor, and the beams in the analysis model are
likely to represent slab strips.
......
-~.
Compression Members in Braced Frames
269
Figure 6.42(b) shows a single column. Some aspects of the column behavior
are as follows.
(1)
As with the beam-columns in earlier examples, the ends of the column
are supported laterally, so only the P-0 contribution has an effect on
the buckling behavior.
(2)
The beams provide rotational restraint, which· reduces the effective
length of the column. If the beam bending stiffness is small relative to
the column stiffness, the effective length factor, k, is close to 1. If the
beams are very stiff, k approaches 0.5. However, the amount of
restraint provided by the beams may be uncertain. For example, if a
beam yields, its stiffness decreases and it offers less restraint. A
conservative, and simple, assumption is that k = 1 for all columns.
(3)
Because of the gravity loads on the beams, the column has end
moments. These depend on the relative bending stiffnesses of the
beams and columns and on the gravity load distribution (a checkerboard live load pattern causes different end moments than a pattern
with load on all beams).
(4)
The end moments on the column are not sustained loads, as noted
earlier (see Section 6.11.6, Figure 6.36(b)). Strictly speaking, therefore,
the equations for moment amplification (e.g., Section 6.11.4, Equation
6.3) do not apply. For design purposes, however, it should be reasonable to assume that the end moments on a column are sustained loads.
6.15.3 Strength-Based Design Using Elastic Analysis
Section 6.10 listed the steps for performance assessment of a frame with pinended columns, using strength-based design and elastic analysis. The steps
for the current example are similar. Two differences are as follows.
(1)
The columns must be checked for combined axial force and bending,
with amplification of the bending moments. The procedure is described
in Section 6.13. For a 3D structure it is necessary to consider biaxial
bending.
(2)
The effective length factor, k, could be smaller than 1.0. As already
noted, it is reasonable to use k = 1.
270
Chapter 6 P-6 Effects, Stability and Buckling
6.15.4 Deformation-Based Design Using Inelastic Analysis
The procedure for deformation-based design, using inelastic analysis, is also
similar to· that for the pin-ended example. If the columns are capacityprotected, and designed not to buckle, only their strengths need to be
checked. In this case, inelastic behavior needs to be considered only in the
beams (and in the walls that provide the lateral strength). If the columns are
allowed to buckle, the inelastic modeling is more difficult.
6.15.5 Diagonally Braced Frame
Figure 6.43 shows a frame with diagonal bracing.
Gravity load on all floors
tltltlt
Figure 6.43 Diagonally Braced Frame
The overall procedure and problems are similar to the preceding example.
·
Some differences are as follows.
(1)
Buckling is most likely to occur in the braces, but might occur in the
columns and possibly the beams. The connection details, such as
gusset plates for the diagonal braces, can have a substantial effect on
the buckling· behavior. This can make it difficult to estimate the
effective length factor.
(2)
The diagonal braces may be allowed to buckle, especially for
earthquake loads. Since diagonal braces can be expected to have little
lateral. load, they can usually be modeled as elements with only
material nonlinearity, as noted earlier. Braces will usually buckle out
of the plane of the frame.
Columns in Unbraced Frames
(3)
271
ff the columns are designed not to buckle, a conservative assumption
is to ignore the rotationalrestraining effects of the beams and braces,
and to calculate the buckling strength assuming the effective length
factor, k, is 1.0. A justification for this is that the braces may buckle,
and hence offer no restraint to the columns, and that the beams may
yield, and hence offer an uncertain amount
restraint. As before, if
the columns are allowed to buckle, the behavior is difficult to model.
/
of
6.16 Columns in Unbraced Frames
6.16.1 Overview
In a braced frame there are two distinct modes of buckllng behavior, namely
(a) overall buckling of the structure and (b) buckling of an individual
column within its o"Wn length. For a building structure, it is unlikely that
overall buckling will occur. H buckling does occur, it is· likely to be
individual column buckling, as shown in Figure 6.44(a).
'----·--··\o====:.f-==,_\.
·.
(a) Braced frame
(b) Unbraced frame
Figure 6.44 Budding Modes for Braced and Unbrac~ Frames
In contrast, an unbraced frame can be regarded as .having only a single
buckling mode, · combining overall buckling and individual coiumn
buckling. For gravity load alone this is in a side-sway mode as shown in
Figure 6.44(b).
Also, in a braced frame the effective length factor, k, for column budding is
smaller thari 1, and column buckling is governed mainly by the P~
contribution. In contrast, in an unbraced frame k is· larger than 1, and
column buckling is governed mainly by the P-.:l contribution.
·· · · -
--~-----·--~---·------
--
--- ·----------- -----·----------------------------· ---------------------------
Chapter 6
272
P-~
Effects, Stability and Buckling
Because of these differences, unbraced frames used to be regarded as
fundamentally different from braced frames. However, this is an unnecessary distinction, and current design procedures recognize that braced and
unbraced frames are not fundamentally different.
This section considers the .behavior of unbraced frames, and shows that
there are close similarities in the design procedures for unbraced and braced
frames. The emphasis is mainly on steel frames. As noted earlier, there are
differences in the way that moment amplification is considered for steel and
concrete columns (see Section 6.13).
6.16.2 Frame and Column Buckling
Figure 6.45 shows an unbraced frame.
"ITD•
p
Gravity load on all floors
tlJJJJt
-
· Moment
----..,..-.....,...-""""'T---i
from beams
,.[
ft~~- ~::~al
from beams
L
-0--0-·+--+---+---1
-o--e>-+--+---+---1
-O--C>-·l---+---+---1
·
(a) Structure
Moment \
~
from beams '-""'
Top ca.n move
horizontally
Rotational
stiffness
from beams
(b) Typical column
Figure 6.45 Column in an Unbraced Frame
Figure 6.45(a) shows the frame, consisting of a moment-resisting frame and
some gravity-only framing. Figure 6.45(b) shows an individual column. H
this column buckles in the plane of the frame it can sway, with ends that are
restrained rotationally by the beams. H the beams are rigid, the effective
length factor for in-plane buckling is k = 1. For flexi'ble beams k is greater
than 1.
·
This suggests that the column can buckle in a side-sway mode, with a
buckling strength based on k > 1. In a frame, however, a single column can
not buckle in this way, because it is connected to the other columns by the
floor system. H any column is to buckle in a side-sway mode, all columns in
an entire story must buckle (or all except one column in a 3D frame, since
the frame collld rotate torsionally about that column).
·
Columns in Unbraced Frames
273
For many years, the design of columns in unbraced frames was pased on
equations for the buckling of C\Il individual column in a side-sway mode,
using effective length factors larger than 1. However, it has been recognized
for· a long time that this is not a rational approach. One reason is that a
single column can not buckle in a side-sway mode, as noted above. A
second reason is that the individual column approach makes it difficult to
account for columns in gravitr·only framing, which contnbute to the P-A
effect but not to the lateral strength and stiffness of the structure. In Figure
6.45, side-sway buckling depends mainly on the P-A contribution. If the
buckling strength is checked considering only the columns in the momentresisting frame, the P-A contnbution of the gravity framing is ignored,
which is a major error.
Current methods of design recogniie that there are two separate
considerations for the buckling behavior, which apply to both braced and
unbraced frames. These are as follows.
a
(1)
The frame buckles as whole. Jn both braced and unbraced frames,
this type of buckling is affected mainly by the P-A contribution for the
structure as a whole.
·
· ·
(2)
An individual column bt,1.ckles in its own length. In both braced and
unbraced. frames, this type of ·buckling
contnbution for the individual column.
iS
affected by the P-5 .
There is, however, a major difference between a braced frame and· arr-un·. braced frame. A braced frame is likely to be weaker for individual column
buckling than for overall frame buckling. The reverse is the likely to be the
case for an unbraced frame, at least for in-plane column buckling.
6.16.3 Strength-Based Design Using Elastic Analysis
The overall steps for performance assessment of an unbraced fraine ·are
similar to those in Section 6.10. However, there are differences from a
braced frame, as follows.
(l}
a
Jn braced frame the lateral resistance is provided primarily by shear.
walls or truss action. The P-A contnbution can be significant, but it has
relatively little effect on the moments at the· ends of the columns. The
~ effect of geometric nonlinearity is the amplification of the end
moments by the P~ contnbution. This is "non-sway" amplification.
The maximum moment can occur within the column length, and
274
Chapter 6 P-L\ Effects, Stability and Buckling
column design can be governed by buckling of a column within its
·
own length.
(2)
For an unbraced frame the lateral resistance is provided by bending in
the columns. In this case the.main effect of geometric nonlinearity is
· the amplification of the moments ·by the P-A contribution, which
increases the moments at the column ends and can have a substantial
effect on the strength in a side-sway mode. This is "sway" amplification. The P-o contnbution tends to have little effect, and column
design is likely to be governed by hinge formation at the column
ends, not by buckling of a column within its own length.
6.16.4 Deformation-Based Design Using Inelastic Analysis
The inelastic behavior of simple unbraced frames was considered in
Sections 6.5 and 6.6. The behavior under combined gravity and lateral load
is usually the most important. An unbraced frame typically collapses wh~
enough plastic hinges have formed to create a mechanism, or until hinge
formation reduces the lateral stiffness so much that the structure becomes
unstable in a side-sway mode.
It is usually relatively easy to set up an inelastic analysis model for an
unbraced frame. The inelastic behavior can usually be modeled adequately
using plastic hinges. Capacity design is often used to prevent hinges from
forming in the columns. If column hinges do form, they will almost alway&
be at the column ends, not within the. column length. Buckling of an
individual column within its own length is unlikely. The .P-A contribution
can easily be accounted for in the analysis model, using a P-A column. The
P-5 contribution is usually negligible and can be ignored.
If hinges are allowed in the. columns of a steel frame, a possible modeling
issue is as follows. As noted earlier, in Section 6.13.2, the P-M interaction
surface that is used for strength-based design may not be a cross section
·property (based On U' the "squash" Strength) but a COlumn property (based
on P.,, the buckling strength). This is conservative for strength-based design
using elastic analysis, and also for deformation-based design where capacity
design is used to avoid column hinges. However, if column hinges ate
allowed, the P-M surface must be the surface for the cross section, based on
PU' not for the column as a whole, based on P,. It is also easier to specify
hinge strengths if they depend only on the cross section, and not on the
column slenderness ratio.
·
r
AComplication - Initial Drifts
275
6.17 A Complication - Initial Drifts
6.17. l Overview
If a structure is perfectly symmetrical, and if the gravity load is also
symmetrical, the lateral deflections under gravity load are zero. However, a
real structure can never be perfectly symmetrical~. and gravity load will
cause some lateral sway.
'
An important cause of sway under gravity load can be initial imperfections
of the structure, in the form of initial drifts, where the columns are not
perfectly vertical. This can have a significant effect on the results of a gravity
load analysis, especially for unbraced frames where sway forces must be
resisted by bending in the columns. The effect is smaller for lateral load
analysis because the initial drifts are a smaller proportion of the total drifts,
The effect can be substantial in a hackling analysis, as noted earlier in
Sections 6.6.6 and 6.6.7.
This section describes three methods that can be used to account for initial
drifts in an analysis model.
6.17 .2 Modeling Methods
A structure may have imperfettions in the form of initial drifts, and may not
be initially perfectly vertical. This is illustrated (with greatly exaggerated
drifts) in Figure 6.46(a).
......
......
......
(a) Initial drills
(b) Move all nodes
(c) Move only the nodes
in the P-a column
...
_____
_______
~-----
,_
,_
(d) Add "notional" loads
Figure 6.46Methods for Initial Deformations
Chapter 6 P-L\ Effects, Stability and Buckling
276
Design codes may require that analyses account for initial drifts,
particularly for gravity load analysis. There are three differt'ht ways to do
. this, as follows.
(1)
Move the no_des of the analysis model, as in Figure 6.46(b). In this
JI1.odel, the P-i:\ contribution is accounted for using P-i:\ struts (not
shown in the figure). When the gravity load is applied, the P-i:\ struts
exert lateral loads on the small displacements main structure.
(2)
H a P-i:\ column is used, move only the nodes in that column, as
shown in Figure 6.46(c). When the gravity load is applied, the P-i:\ ·
colunin exerts lateral loads on the small displacements structure.
(3)
Do not 'move any nodes. Instead, calculate horizontal loads that
correspond to 'the forces. from an mclined P-i:\ column, and apply
. these loads as shown in Figure 6.46(d). These are not real loads on the
structure, and they are usually called "notional loads". Figure 6.47
shows how the notional loads are calculated:
Lciad =P3
.
P3~IP-~=====~~
' I I I J
p
t
2
'
Load= P2
I I I J
1'>--d-=::::::::==~
Initial drift
ratio= 90
in all stories
(P3 + P2)9o - P39o
(P3 + P2 + P1)9 0
+--
-
(P3
+P2)90
=P190
The notional loads are the
opposite of these forces.
(a) Gravity loads and initial drifts .
(b) Forces on P-L\ column, and
hence notional loads
·Figure 6.47 Notional Loads for a Specified Initial Drift
·----
The first two of the above methods can be expected to give similar results. It
may be less obvious, but the third method gives the same results as the
second method. As an exercise, you might like to show that this is true.
. To account for imperfections of this type, initial drifts must be estimated. A
common. assumption for steel frames is 0.2% initial drift, which corresponds
ASecond Complication - Stiffness Reduction
277
to the commonly accepted construction tolerance for the initial· drift in a
story. The extra (notional) loads from these drifts can have a significant
effect on the analysis results. .
6.18 A Second Complication - Stiffness Reduction
6.18.1 Overview
For elastic analysis of steel structures it was common practice for many
years to base the element bending (El) and axial (EA) stiffnesses on the
initial elastic modulus for steel. It is possible, however, that a steel structure
can yield significantly under the design loads. For example, if a steel beam
has residual stresses equal to. 40% of the steel yield strength, there can be
inelastic behaVior at a cross section when the bending stress reaches 60% of
yield. This has ·the effect of increasing the deflections of the structure or, in
effect, decreasing the structure stiffness.
Any yielding is likely to be localized, so the amount of stiffness reduction is
likely to be modest for the structure as a whole. However, it can have a
significant effect on the behavior, by increasing the deflections, increasing
the P-~ effect, and decreasing the buckling strength.
An accurate analysis model would account for the amount and distribution
of yield, cracking, etc. in the different members of a structure, and account
for them element-by-element, or even cross section-by-cross section. This is
impossible, not only because it is impractical computationally but also
because the amount and distribution of yield· or cracking. can never be
known for a real structure.
For concrete structures it is necessary to estimate reasonable values for the
axial and bending stiffnesses, accounting for cracking, yield and creep, For
steel structures it has become a common practice to specify an effective
modulus that is smaller .than the initial elastic modulus. The "correct"
stiffnesses are highly uncertain. To ensure consistency, if not accuracy, the
stiffness values for analysis should be specified by design codes, not left to
the judgment of individual engineers.
This section examines the effect of a modest amounts of stiffness reduction,
considering strength-based design of unbraced frames using elastic lateral
load analysis.
278
Chapter 6 P-d Effects, Stability and Buckling
6.18.2 Effed on Lateral Load Analysis
Figure 6.48(a) shows a simple unbraced frame. Gravity load is applied first,
then lateral load is added. Consider whether a modest amount of yield can
have an influence on (a) the lateral deflection and (b) the strength demands
on the members.
0.5P+
Actual
behavior
Lateral Load
l,
P-acolumn
Stiffness = -Pih
Main structure
El8stic stiffness = K0
Stiffness after some yield = Kr
(a) Loads and stiffiiesses
_H_
K0- Pih
..J L_H_.
I!.
Kr-Pih
(b) Horizontal deflections
Figure 6.48 Effect of Yield in Lateral Load Analysis
The initial elastic stiffness of the main frame is ~ and the stiffness of the
P-A. column is -P/h. Hence the effective lateral stiffness assuming elastic
behavior is Ka- P/h. For the lateral design load, H, the lateral deflection is
HI (K.i- P/h), as shown in Figure 6.48(b ).
Figure 6.48(b) also shows possible actual behavior, with some yield. H the
yield reduces the effective stiffness of the main structure to K,, the lateral
deflection increases as shown. Since only the main structure stiffness reduces,
while the stiffness of the P-A column remains the same, the increase in
lateral deflection could be significantly larger than might be expected from
the stiffness reduction alone.
Consider the following case.
(1)
(2)
(3)
The lateral stiffness of the elastic structure is Ko·
The reduced lateral stiffness, allowing for some yield, is K, =O.SK.i.
The stiffness of the P-A column is --0.lK.i. That is, the P-A effect reduces
the elastic lateral stiffness of the structure by 10%.
For the elastic case, with stiffness~ the P-A contribution reduces the effective
stiffness from K.i to 0.9Ko. Hence the. P.;A contribution increases the lateral
displacement from H/I<.i to H/0.9~ or by a factor of 1/0.9 = 1.111.
ASecond Complication - Stiffness Reduction
279
For the yielding case, with stiffuess K,, the P~A contnbution reduces the
effective stiffness from 0,81'.i to 0.7.Ku. Hence the P-A.contribution increases
the lateral displacement from H/0.81'.i to H/0.7K.i, or by a factor of 0.8/0.7 =
1.143.
Hence, the reduction of 20% in the stiffness of the small displacements
struchire has the following effects.
(1)
For the small displacements structure the lat~ cieflecti0n is HI K.i for
the elastic case and H/Kr = 1.25 H/K.i for the yielding case. This is a
25% increase.
This does not increase the element force demands, since the. element
stiffnesses are decreased in the same proportion. For a small displacements analysis the increased deflection might affect serviceability, but
the element forces do not change.
(2)
. For the structure with P-A effects, the deflection is qu H/K.i for the
elastic case compared with 1.143 Hf Kr 1.429 Hf Ka for the yielding
case. This is a 29% increase, whiclt is significantly larger than the 25%
deflection increase.
·
=
In this cas_e the inc-ease in deflection does increase the eleinent force
iemands; bu.
. by very much. For the elastic case the P-A
contribution amplifies the force demands by 11.1%. For the case with
some yield the· P-A contnbution amplifies the force demands by
14.3%. This is a 29% increase in the P-A contribution. However, there
is a much smaller increase in the total force demands. Compared with
the elastic case, the total demands increase by 1.143/1.111 = 1.029, or
2.9%.
This indicates that stiffness· reduction to account for yield can have a
substantial effect on the deflections for lateral load analysis, but is likely to
have only a small effect on the strength demands (if P-A effects are ignored,
there is no change in the strength demands). The magnitude of the effect
depends on the amount of stiffness reduction. The above example considers
a 20o/o stiffness reduction due to yield, and a 10% reduction from the P-A
contnbution. Table 6.1 shows the effects for different amounts of .stiffness
reduction.
280
Chapter 6 P-d Effects, Stability and Buckling
TABLE6.1
Effect of Stiffness Reduction on Deflection and Strength Demand
Lateral Load Analysis with P-L\ Effect
Stiffness
Reduction
Increase in
Deflection vs.
Elastic Analysis
(Small Displs +
P-L\ =Total)
Increase in
Strength
Demand vs.
Elastic Small
Displs Analysis
Increase in
Strength
Demand vs.
Elastic P-L\
Analysis
P-L\ Effect Reduces Elastic Stiffness by 10% (Pih= 0.1 Ko) ·
0%
o+ 11 =11%
11%
0%
10%
11 +13=25%
12.5%
1.2%
20%
25+18=43%
14%
2.9%
30%
43+24=67%
17%
5.0%
40%
67+33= 100%
20%
8.0%
P-L\ Effect Reduces Elastic Stiffness by 5% (Pih = 0.05Ko)
0%
0+5=5%
5.3%
0%
10%
11+7= 18%
5.9%
0.6%
20%
25+8=33%
6.6%
1.2%
30%
43 +11 =54%
7.6%
2.2%
40%
67+ 15=82%
9.1%
3.6%
Some key points in Table 6.1 are as follows.
(1)
The first part of the table assumes that the P-L\ contribution reduces
the elastic stiffness by 10% (P /h = 0.1 K0). This is at the upper end of
the practical range. The second part of the table is for a 5% reduction
(P /h = 0.05 K.)
(2)
Column 1 shows the percentage stiffness reduction to allow for yield.
A reduction of 20% might be reasonable for a practical structure. A
reduction of 40% implies a substantial amount of yielding, which is
less likely under design loads.
It can be argued, incidentally, that a typical analysis model tends to
underestimate, rather than overestimate, the stiffness, because it
A Second Complication - Stiffness Reduction
281
ignores "nonstructural" components. This is another reminder that
structural analysis is at best approximate.
(3)
Column 2 shows the increase in the calculated deflection compared
with an elastic small displacements analysis. This shows that a small
displacements analysis can substantially underestimate the lateral
deflection. For no stiffness reduction for yield, the increase in deflection
is 11%, which is the P-A contribution. For larger stiffness reductions
the increase in the deflection is partly due to the stiffness reduction
and partly due to the P-A contribution. For P/h = 0.1 Ko and a 20%
stiffness reduction, 25% of the deflection increase is from the stiffness
reduction, and 18% from the P-A contribution. For P/h = 0.1 Ka and a
40% stiffness reduction, 67% of the increase is from the stiffness
reduction and 33% from the P-A contribution.
(4)
Column 3 shows the increase in strength demand (and hence in
strength DIC ratio) compared with a small displacements analysis.
This increase is entirely due to the P-A contribution (in a small
displacements analysis, stiffness reduction does not change the
strength demands). The increase in strength demand can be
substantial, which ·indicates that lateral load analyses generally
should account for the P-A effect.
(5)
Column 4 shows the increase in strength demand compared with a
P-A analysis with no stiffness reduction for yield. This increase is
entirely due to the stiffness reduction. The increase is small. In a P-A
analysis, stiffness reduction increases the deflections but has only a
small effect on the strength demands.
Keeping in mind that this section considers strength-based design using
elastic analysis, the following are the two main conclusions.
(1)
Compared with a small displacements analysis, the P-A contribution
can significantly affect both the deflections and the strength demands.
(2)
In a P-A analysis, stiffness reduction to account for modest amounts of
yield can significantly affect the deflections but has a relatively small
effect on the strength demands.
6.18.3 Modeling of Stiffness Reduction
Two ways to account for stiffness reduction due to yield are as follows.
282
Chapter 6 P-A Effects, Stability and Buckling
(1)
Reduce the element stiffnesses. For the above frame example, reduce
the bending (EI) and axial (EA) stiffnesses for every element by 20%,
or by a factor of 0.8, by specifying a reduced E. This is the most direct
method.
(2)
Increase the load on the P-L\ column (or alternatively the forces in the
P-A struts if they are used). This is an indirect method but may be
easier to use. For the above example, reducing the element stiffnesses
by 20% increases the small displacements deflection by 25%. Hence,
increase the gravity load on the P-A column by a factor of 1.25. ·
To see the effect of this increase, consider the case where the P-A
contribution originally reduces the effective stiffness by 10% (from Ko
to 0.9Ko). With the increased ioad on the P-L\ column the effective
stiffness is reduced by 12.5% (from Ko to 0.875Ko). Hence the P-A
contribution increases the lateral deflection from H!Ko to H/0.875Ko,
or by a factor of 1/ 0.875 = 1.143, as required.
6.18.4 Effect on the P-o Contribution
This section considers only the P-A contribution, and only "sway" amplification. Stiffness reduction also affects the P-:-o contribution and "non-sway"
amplification. The effect· of stiffness reduction on non-sway amplification
was considered earlier, in Section 6.12.4.
This topic is revisited later in this chapter, for both sway and non-sway
amplification, when methods for lateral load analysis are considered (see
Section 6.20).
6.18.~ Effect on Buckling Analysis
For most building frames it is unlikely that the structure will collapse under
gravity load alone, and the buckling load· capacity will usually be much
larger than the load demand. Hence, if a buckling analysis is carried out
there is likely to be plenty of room for error. For some structures, however,
the calculated buckling load capacity may be close to the load demand. In
such cases it is important to examine the modeling assumptions. A buckling
load calculation is a strength capacity analysis, which· is always difficult to
perform accurately. The buckling strength can be affected by stiffness reduction and also by initial imperfections (see Section 6.6). Itmay be important
to take these effects into account.
' ·'""
.........
'
Some Theory- Geometric Stiffness
283
6.18.6 Is a Reduced Stiffness Analysis Necessary?
An ~lastic analysis model can be modified to account for modest amounts of
yielding, and also to account for initial drifts. An engineer might question
whether such modifications are necessary, since yield and imperfections
might be covered by the usual load factors and capacity reduction factors.
This is a question to be answered by code writers. As shown later in this
chapter, code writers have concluded that such anaj.yses are necessary.
From the point of view of this book, an important point is that structural
analysis should not be applied blindly. An engineer should understand the
behavior of real structures, recognize the effects that are important, decide
whether these effects need to be considered in the analysis, and choose a
suitable analysis model. Considering these effects can make the analysis
more complex, but ignoring them (or worse, being unaware of them) does
not make them go away.
6.19 Some Theory - Geometric Stiffness
6.19.1 Overview
The preceding sections in this chapter have considered the effects of
geometric nonlinearity, how these effects can be modeled conceptually, and
how the analysis results can be used to assess structural performance. Little
has been- said about P-d theory, how a P-d strut or a P-1> cable might be
modeled mathematically, or how a P-d analysis might be carried out.
The primary concerns of this book are modeling and behavior, not theoretical
details and computational procedures. For material nonlinearity this is
sufficient, and for most engineers the details are ··not important. For
geometric nonlinearity, however, it is necessary to consider some theory,
and also some details of the analysis process:
This section explains (or tries to explain) the concept of a "geometric"
stiffness, which is an essential part of P-d theory. The formal theory for a
geometric stiffness is quite dense mathematically, and it can be difficult to
determine what it means physically. For. some elements, however, it is
possible to develop the geometric stiffness using an "engineering" approach,
based on physical reasoning rather than mathematical equations. This
approach is not a general one, but it can provide insight on what a
geometric stiffness means and how it can be used.
Chapter 6 P-8 Effects, Stability and Buckling
284
6.19.2 Geometric Stiffn~ss Matrix for P-.i:\ Strut
In earlier sections it was noted that a P-A column in a single story frame has
a negative stiffness, numerically equal to Pih where P is the compression
force in the column and h is the story height. This is a "geometric" or "stress"
stiffness ("geometric" because it is related to geometric nonlinearity, "stress"
because it depends on the axial force in the column). The term "initial stress
stiffness" may also be used.
In an analysis model, a P-A column consists of a stiff elastic bar plus a P-A
strut. The elastic bar provides vertical support for the gravity load, and the
P-A strut accounts for the P-A contribution. The bar has an elastic stiffness,
and the P-.!\ strut has a geometric stiffness. For a 2D element the elastic and
geometric stiffnesses are both 4-by-4 matrices. For a 3D element the stiffness
matrix is 6-by-6. For the special case of a 20 vertical bar, Figure 6.49 shows
how the elastic stiffness matrix can be formed.
EA
•
d04= Tdq4
-
dQ =-EAdq
4
L
2
-
f
L
f
d04=0
dq
3 d03= 0
d03= 0
EA
dF=Tdq2
compression
Modulus= E
Area=A
dq3= 0
_ __,_____
EA
dF=Tdq 4
tension
d01=0
~q1
t
d01=0
__.
f
~=O
Tdq2
f
EA
Q2,q2
d02= Tdq2
dQ2=0
(a) End forces and
displacements
{b) Impose
dq1 only
dQ2
rJ
dQ3
d04
=
(c) Impose
dq2 only
(d) Impose
dq3 only
~ [:
0
0
0
0
0
0
-1
0
L
EA
d02= -Tdq4
dQ2=0
0
(e) Impose
d14only
_: ] [=:]
0
dq3
1
dq4
t
t From
t From
t From
From
(b)
(c)
(d)
(e)
(f) Elastic stiffness matrix
Figure 6.49 Elastic Stiffness Matrix for a 2D Vertical Bar
....
·!~.
Some Theory- Geometric Stiffness
285
The bar has four end displacements, q1 through q,. (ignoring the end
rotations) and four corresponding end forces, Q1 through Q4• By definition,
if a vanishingly small displacement, d%, is imposed on the bar, with dq2 = dq3
= dq4 = 0, and if the corresponding end forces are calculated, these forces
define the stiffness coefficients ~.t' fs.1, 's.1 and k4.1' making up column 1 of the
stiffness matrix (see Section 3.1.2). Similarly, imposing dq2 with dq1 =dq3 =dq,
= 0 gives column 2 of the stiffness matrix, etc. Figure 6.49 illustrates the
procedure.
··
Using the same procedure, Figure 6.50 shows how the geometric stiffness
matrix can be built up for a vertical P-A strut.
llQ,=O
dQ 4 =0
·
·p
· . dQ 3=0
q dQ.= --dq. ~
-+
L
_..._dq 4
t
a.=-P, hence da,= o
a.= -P
f--+
a.=o
t~=tdq,
Force=P
dF=O
Force= P
compression
L
Force=P
dF=O
Q,=O
--+
dQ,=O
--+
t
=
Q2 =P
=
da2=0
Q 2 P, hence dQ 2 0
(a) Initial forces
(b) Impose
dq 1 only
(c) Impose
dq2 only
n·r
= ..f.
da2
dQ3
dQ,
.
L
0
0
0
0
1
0
-1
0
0
0
t
t
t
From From From
.(b)
(c)
(d)
(d) Impose
dq3 only
t
dQ 2 =0
(e)lmpose
dq4 only
~] [~]
t
From
(e)
(f) Geometric stiffness matrix
Rgure 650 Geometric Stiffness Matrix for a 2D P-L\ Strut
This strut has no elastic stiffness, but it has an axial force. To Sa.tisfy
equilibrium there must be end forces that balance the axial force. These end
286
Chapter 6 P-i'l Effects, Stability and Buckling
forces are shown in Figure 6.SO(a). To get the geometric stiffness, impose
vamshingly small end displacements and calculate the end forces required
for equilibrium in the displaced position. The changes in the element end
forces define the stiffness coefficients in the geometric stiffness matrix, as
shown in the figure. Note that since the axial stiffness is zero, the axial force
in the bar does not change, even when the bar is deformed axially.
This procedure can be extended to a bar element with any orientation. For a
3D element there are three displacements at each end of the element, and
the elastic and geometric stiffnesses are 6-by-6 matrices.
It should be noted that the physical procedure used in Figure 6.50 is not
applicable to elements of all types. There is a general mathematical
procedure that can be used to obtain geometric stiffness matrices for
elements of many types. That procedure is beyond the scope of this book.
6.19.3 Structure Stiffness Matrix
For an element that consists of a bar and a P-i'l strut, the sum of the elastic
and geometric stiffnesses is the effective element stiffness. These element
stiffnesses can be assembled into a structure stiffness matrix, which then
accounts for the P-A contribution. In an analysis model with P-i'l struts, each
element in the structure has a geometric stiffness. In a model with a P-A
column, only the P-A column elements have geometric stiffnesses.
For a 2D frame structure the stiffness for a small displacements frame
element is a 6-by-6 matrix, and for a 3D frame it is a 12-by-12 matrix.
Corresponding geometric stiffnesses for a P-i'l strut can easily be added, and
hence a structure stiffness that accounts for the P-i'l contribution can be
assembled. The small displacements stiffness can be elastic, or it can account
for inelastic behavior.
For an elastic structure the small displacements stiffness for each element is
constant, but the geometric stiffness depends on the axial force. For gravity
load analysis, as the load on the structure increases the column axial forces
increase, and the effective stiffness of the structure decreases. If the gravity
load is large enough the effective stiffness becomes zero (i.e., the structure
stiffness matrix becomes singular), and buckling occurs..
For lateral load analysis, it is usual to apply the gravity load first, then add
lateral load. As the lateral load is added, the axial forces in the elements
change. If the analysis model has a P-A strut for every element, the element
Methods for Elastic Lateral Load Analysis
287
geometric stiffnesses change, and hence the structure stiffness changes, even
if the structure is elastic. The analysis is thus nonlinear. If, however, the
analysis model has a P-A column, the axial forces in the P-A column
elements stay constant as the lateral load is added. Hence the structure
stiffness.matrix stays constant, and the analysis is "linearized". This is more
efficient computationally.
6.19.4 P-C> Contribution
The geometric stiffness ·for a P-A strut accounts only for the P-L\
contribution. One way to account for the P-o contribution is to add the
geometric stiffness for a P-o cable. This can be done, but without going into
details it may be noted that it may not work very well. The P-A contnbution
is simple, and it can be accounted for accurately using the geometric
stiffness for a P-A strut. The P-0 contribution is more complex, and in general
it can not be accounted for accurately just by adding the geometric stiffness
matrix for a P-0 cable.
An analysis model that accounts for the P-o contribution is urually much
more complex than one that accounts only for the P~L\ contn1mtion.
6.20 Methods for Elastic Lateral Load Analysis
6.20.1 Overview
There is often a relationship between an analysis model and the method
used to analyze it. For example, different analysis models may be used for
static and dynamic analysis. A particularly important, and interesting, case
is elastic analysis for strength-based design of building frames \tnder
combined gravity and lateral load.
·
In this case, the elements in the analysis model are elastic, but they mitst
account for P-A effects and hence are not necessarily linear. The results of
the analysis must include amplified bending moments, shear forces and
axial forces, and the analysis method may be nonlinear (which means that
iteration may be needed). There are two different types of amplification
(sway and non-sway amplification), at least three different ways to account
for them in an analysis model, and several possible analysis methods. There
is. a certain· amount of mystery about what is the most appropriate
combination of analysis model and analysis method.
288
Chapter 6 P-8 Effects, Stability and Buckling
This section considers this topic in some detail.
An analysis method that .accounts for P-A effects, • and . hence for
amplification, is usually termed a"second order" method. A sntall displacements analysis, which ignores P-A effects, is a "first order" method. This
section identifies three alternative methods for second order analysis, which
·differ on how they account for the amplification. P-A effects are inherently
nonlinear. However, with some models it is possible to use linearized
analysis.
This book is not concerned with the mathematical details of element models,
or with the computational details of analysis methods, and this section does
not cover many details.
6.20.2· SWi!Y and Non-Sway Amplification
In a btiilding·structure there are tWo types of amplification, as follows.
(1)
Sway Amplification. The story drifts are increased by the P-A effect.
Hence, the bending moments and other forces aSsociated with story
drifts are amplified. This type of amplification is ·caused almost
entirely by the P-A contribution. It has the greatest effect in unbraced
frames, and it affects the bending moments at th~ ends of a column.
(2)
Non-sway Amplification. The bending deformations within the length of an .
individual column may be amplified, and hence also the bending moments.
This type of amplification is caused by the P-0 contn1mtion. It has the
greatest effect in braced frames, and it affects the bending moments within
the column length.
6.20.3 Methods for Sway Amplification
Sway amplification can be accounted for either directly or indirectly.
In the direct approa91 the analysis model includes, in effect, a P-A column
or a· number of P-A struts. When the structure sways, the P-A column or
struts exert forces on the main structure, increasing the ·drifts, bending
moments and other member forces. One way to do this computationally is
to use ageometric stiffness.
The analysis model could include P-0 cables, in effect if not explicitly.
However, this complicates the analysis model and usually has little effect on
the drifts.
Methods for Elastic Lateral Load Analysis
289
The indirect approach uses a small displacements analysis, with no P-~
effects. The calculated story drifts, and hence also the member forces, are
amplified as follows.
Figure 6.51 shows a single story frame, of the type used for earlier examples
in this chapter.
Smalldispls
Lateral Load
With
...._.P-a
P-a column
Stiffness -Pih
=
Main structure
Stiffness K
=
(a) Loads and Stiffnesses
(b) Drifts
Figure-6.51 Sway Amplification
If the small displacements stiffness of the frame for lateral load is K, the
reduced stiffness is K - Pih where Pis the gravity load, his the story height,
and Pih is the geometric stiffness of the P-~ column. If the horizontal load is
H, the small displacements drift is Ao = H/K and the amplified drift is !!.. =
H/(K - P/h). Hence, the drift is amplified by a sway amplification factor, B...
given by:
1
(6.10)
The small displacements drifts, moments, shear forces and axial forces are
all amplified by this factor.
For a multi-story frame, B;, must be calculated for each story. The force His
the story shear, equal to the sum of the horizontal loads above the story, the
force P is the sum of the gravity loads above the story, and i\ is the small
displacements story drift.
A problem with this method is that for combiiled gravity and lateral loads
there can be both sway moments and non-sway moments. The Bs amplification factor applies to sway moments only. Non-sway moments, which are
290
Chapter 6 P-Ll Effects; Stability and Buckling
the moments when there is no drift, are amplified differently. Hence, it is
necessary to separate sway :and non-sway effects.
The "exact" way to do this is to support the frame horizontally at all floors
and apply the gravity loads. The resulting moments are non-sway moments:
The horizontal supports are then removed, and loads are applied that are
equal and opposite to the support reactions. The resulting moments are
gravity sway moments. These are added to the moments caused by lateral
loads, to give the total sway moments, which are amplified by B8 •
This is awkward analytieally. A reasonable approximatkm may be to assume
that gravity load causes only non-sway effects, and lateral load causes only .
sway effects.
·
An advantage of the direct approach, with a P-A column or a number of P-A
struts, is that it automatically separates sway and non-sway effects, since
P-A colrimns or struts have an effect only when there is sway.
6.20A Methods for Non-Sway Amplification
Non-sway amplification can be accounted for either directly or indirectly.
Consider the indirect approach first
Non~ay amplification applies to a single column. The column is braced
against sway, and bending can be caused by lateral load on the column, or
more likely by bending moments applied at the column ends. The case with
end moments is shown in Figure 6.52.
·~
M1 fl~
M1
man
isplacements
Amplified
~
.
· Figure 652. Non-Sway Amplifkation
Since the column is braced, there is no P-A contribution and moment
amplification is caused entirely by the P-~ contn1>Ution. It is'usual to assume
'·
-~.
Methods for Elastic .Lateral Load Analysis
291
that the effective length factor for column buckling is 1.0. The end moments,
and-also the axial force, are already amplified by sway amplification.
The maximum moment at any point in the column is Mmax = B.M2, where B.
is the non-sway amplification factor, given by:
B
n
=
Cm
(1-~)
>1
Wll)
where C., = 0.6 + 0.4MJM,, usually with a lower limit of 0.4, and P"' =
buckling strength for amplification purposes. P"' can have the following
.values.
(1)
For steel, P"' = P" = the elastic buckling strength. As noted earlier, this
underestimates the amplification.
(2)
For reinforced concrete P"' =0.75 times the buckling strength based on
a bending stiffness that accounts for cracking and creep. This presumably provides a realistic. estimate of the amplification.
(3)
A possible alternative for steel is P"' =Pd= the buckling strength based
on a tangent modulus (see Section 6.12.4). This might provide a more
realistic estimate of the amplification.
Equation 6.11 is not exact, but it has long been accepted as accurate enough
for design purposes,
For the direct approach, non-sway moment amplification must be accounted
for in the analysis model, and the maximum moment must be included in ·
the analysis results. To do this, a computer program must have a frame
element that accounts for the P-1> contribution. There are many ways to do
this, and different computer programs are likely to use different procedures.
Figure 6.53 shows a.uniform elastic column with three different bending
conditions. For each of these conditions, and for two different values of
PIP"' Table 6.2 shows the ratio M.,,JM0 {i.e., the amplification factor) for the
following analysis methods.
(1)
(2)
The exact solution.
The SAP2000 computer program, using one element to model the
column.
•
292
(3)
(4)
Chapter 6 P-L\ Effects, Stability and Buckling
The SAP2000 computer program, using two elements to model the
column.
Equation6.11,withP"'=P".
Case A
CaseB
CaseC
Figure 6.53 Examples for Moment Amplification
TABLE6.2
Moment Amplification by Different Methods
For Elastic Columns in Figure 6.53
Case
Exact
SAP2000
One element
SAP2000
Two elements
Equation 6.11
P/P,.=0.25
A
1.34
1.26
1.34
1.33
B
1.41
1.39
1.41
1.33
c
1.00
1.00
1.00
1.00
A
2.03
1.70
2.02
2.00
B
2.26
2.06
2.25
2.00
c
1.25
1.18
1.24
1.20
P!Pce=0.5
Methods for Elastic Lateral Load Analysis
293
Table 6.2 shows the following.
(1)
(2)
SAP2000 gives accurate amplification factors for all three cases, provided that the column is divided into two or more elements. It may be
.noted that the SAP2000 analysis also gives the correct shapes for the
ainplified bending moment diagrams. With a single elemeJ1t,· the
maXimum error is a 16% underestimate, for Case A with the larger
.,
value of Pl P.,.. .
&luation 6'.11 is accurate for Case A but less ~ccurate for Cases B and
C. The maximum error is a 12% underestimate, for Case B with the
larger value of PIP,..
These examples are for ~orm elastic columns wiUi negligiole sh~ar
deformations and no stiff end· zones. It does not necessarily follow that
either SAP2000 or Equation 6.11 give accurate amplification factors for other
cases.
6.20.5 ~Sway and Nori-Sway Similarities
Sway and non-sway amplification Itave different causes .and different
effects, but the methods that
used to account for them are very similar.
are
Consider :Equation 6.10, for the sway factor B,., ·and compare it with
Equation 6.11, for the non-sway factor Bn. One form of Equation 6.10 is as
follows:
(6.12)
As shown earlier (e.g., Section 6.6.2), the buckling strerigth for a single story
frame is Kh, where K is the lateral stiffness and h is the story height Hence,
Equation 6.12 cari be written as
·
1
B,=-(
p).
1--
(6.13)
P,,
where P. is the buckling strength. This has the same form as Equation 6.11, ,
for B..
.
./
294
Chapter 6 P-6 Effects, Stability and Buckling
One consequence of this concerns the effects of stiffness reduction caused by
a modest amount of yield. Stiffness reduction affects the buckling strength,
and hence has fundamentally similar effects for sway and non-sway
amplification. Section 6.12.4 considers the effect of stiffness reduction on
non-sway amplification in beam-columns. Section 6.18 considers the same
thing for sway amplification in frames. These two sections may appear to be
different, but they have many similarities. It follows, for reasons of
consistency, that if stiffness reduction is considered in an analysis method,
the same amount of reduction should be assumed for both sway and nonsway amplification. Depending on the details of the analysis method, this
may or may not be done.
A further point concerns analyses where amplification is accounted for
directly in the analysis, rather than through amplification factors. The
procedures for· sway and non-sway amplification are similar in concept but
different in detail. Sway amplifica~on
be accounted for by including P-A
struts in the analysis model, and non-sway amplification by including P-o
cables. Conceptually these are similar, but they are different computationally. In an analysis, sway amplification can be accounted for quite easily,
using P-A geometric stiffnesses. It is not so easy, however, for non-sway
amplification. Although P-o geometric stiffnesses.can be formed, the details
of calCulating and using them are complex; As n.?ted earlier, different
computer programs are likely to use different methods to account. directly
for non-sway amplification.
can
6.20.6 Analysis Methods
There are three different analysis methods that can be used for second order
analysis, each with a different way of accounting for sway and non-sway
amplification. These methods are as follows.
(1)
"Double-B" Analysis. Small displacements ·analysis, with sway and
non-sway amplification accounted for using both the Bs and Bn
amplification factors.
(2)
"Single-B" Analysis. P-A analysis, with only sway amplification
accounted for directly in the analysis model, and non-sway
amplification accounted for using the Bs amplification factor.
(3)
"Zero-B" Analysis. P-A analysis, with both sway and non-sway
amplification accounted for directly in the analysis mooel.
Methods for Elastic Lateral Load Analysis
295
An alternative "Single B" analysis is possible, with sway amplification
accounted for using Bsand non-sway amplification accounted for directly. It
is unlikely that this method would ever be used.
For these methods, the following sections consider the analysis steps. The
purpose of the analysis is to calculate member force demands (moments,
shears and axial forces) which account for sway and non-sway amplification. The emphasis is on unbraced frames, although the procedures for
braced frames are similar.
In all cases the analyses are elastic, However, they are not necessarily linear.
First consider the case with no initial imperfections and no stiffness
reduction to account for yield. These are considered later.
6.20.7 "Double-B" Analysis
This is the least sophisticated method. The usual procedure is as follows.
(1)
The analysis is a small displacements analysis. This has the advantage
that it is simple, and the structure stiffness matrix is constant.
(2)
Analyze the structure for gravity loads alone and calculate the
member forces. These are the "non-sway" moments, Mn, and shear
and axial forces, Fn· There may be several gravity load cases. If there is
significant sway under gravity load, it may be necessary to add lateral
supports and run separate non-sway and sway analyses, and to add
the sway forces to those in the following step.
(3)
Analyze the structure for lateralloads alone and calculate the member
forces. These are the "sway" moments, M., and forces, F•. There may
be several lateral load cases.
(4)
For each required combination of gravity and lateral load, calculate
the shear force and axial force demands as F = Fn + B•.F., where B, is
obtained from Equation 6.10.
(5)
For each required combination of gravity and lateral load, calculate
the bending moment demands as M = B.Mn + B.Ms1 where B. is
obtained from Equation 6.11.
This method has the advantages that it requires only a small displacements
analysis and that the analyses for gravity and lateral loads can be carried
296
Chapter 6 P-8 Effects, Stability and Buckling
out separately and superimposed. It has the disadvantage that it can be
difficult to separate sway and non-sway member forces.
6.20.8 "Single-B" Analysis
This is the simplest method computationally. The procedure is as follows.
(1)
The analysis model accounts directly for the P-A contribution, but not
the P-o contribution. The P-A contribution can be modeled using a P-A
column or a number of P-A struts, and can be accounted for using
geometric stiffnesses.
(2)
Analyze the structure for gravity loads alone and calculate the
member forces. Since the axial forces in the columns are initially not
known, if a P-A strut is used for each column, iteration is needed to
obtain the geometric stiffness matrix, since this matrix depends on the
column axial forces. If a P-A column is used, a two-cycle process can
be used. In the first cycle the small displacements stiffness is formed,
the gravity loads are applied to the P-A column only, and the
structure is analyzed to obtain the axial forces in the P-A column
elements. In the second cycle the structure. stiffness is modified by
adding the geometric stiffness for the P-A column, and the gravity
loads are added to the main structure. Alternatively, if the axial forces
in the P-A column can be specified directly, no iteration is needed.
There may be several gravity load cases.
As an alternative in this step, the element forces might be calculated
using a small displacements analysis, since the P-A contribution is
often small for gravity loads.
(3)
Analyze the structure for lateral loads alone and calculate the member
forces. If a P-A column is used, and if only horizontal loads are
applied, the geometric stiffness is -constant and i~eration is not needed
(there may, however, be a different geometric stiffness for each
gravity load case). This is a linearized analysis. If P-A struts are used
the geometric stiffness will change, and iteration is generally needed..
The axial force for any P-A strut must be the sum of the axial forces
from a previous gravity analysis and the current lateral load analysis.
It may be necessary to run analyses for combined gravity and lateral
loads.
Methods for Elastic Lateral Load Analysis
(4)
297
If separate analyses have been carried out for gravity and lateral
loads, for each . required load combination. calculate the bending
moment, shear force and axial force demands by adding the forces
from the gravity and lateral load analyses. This accounts for sway
amplification. Superposition generally can not be used when the
behavior is nonlinear. In this case, however, the results for separate
gravity and lateral load analyses can be superimposed. If analyses
have been carried out for combined gravity atJ.d lateral loads, use the
·
analysis results directly.
(5)
For each column, amplify the bending moments to account for nonsway amplification, using Bn from Equation 6.11.
This method compares as follows with the Double-B Method.
(1)
In Step 1 a model with a P-8 column can be expected to give
essentially the same results as the Double-B Method, since the sway
amplification in the Double-B Method is based on a model with a P-ii
column.
(2)
In the Double-B method, the sway and non-sway member forces are
calculated separately. The sway forces are then amplified by B,, and
the non-sway moments are amplified by B•. The moment demands are
usually calculated as M = B,.Mn + B,M,. The Single-B Method is
different because sway amplification is accounted for in the analysis,
and hence the non-sway amplification factor is applied to moments
that already account for sway amplification. The Double-B Method
can be made more closely equivalent to the Single-B Method by
calculating the amplified moments using M = Bn (Mn + B,M,).
A potential problem with the Single-B method should be noted. This
method assumes that the P-o contribution is ignored in the analysis model.
In some cases a column member may be subdivided into two or more
elements. In such a case, if the model uses P-ii struts, these .struts exert
forces within the length of the column member. These forces approximate
the distributed forces from a P-o cable. Hence, the analysis accounts directly
for at least some non-sway amplification. if .the .moments are further
amplified using the amplification factor Bn' the non-sway amplification may
be overestimated. Hence, P-.i\ struts must be used carefully. This problem
does not arise when a P-8 column is used.
298
Chapter 6 P-A Effects, Stability and Buckling
6.20.9 "ZerQ-8" Analysis
This is the most complex method. In this method both sway and non-sway
amplification are accounted for directly in the analysis model, and
amplification factors are not used. The main complications are as follows.
(1)
The analysis model mtist account directly for both the P-A and
N> contributions. The P-A contribution can be mqdeled using a P-A
column or a number of P-A struts, with corresponding geometric
stiffnesses. As already noted, the P-o contribution is more diffiCult to
·model, and .different computer programs are likely to use different
models for column elements. As shown in Section 6.20.4 and Table
. 6.2, the SAP2000 element can account accurately for the P-o
contribution, provided each. column is divided ·into at least two ·
elements.
(2)
Superposition can not be used. Hence, for combined gravity and
lateral loads a separate analysis is required for ea<:h load combination.
Also, since the axial forces in the columns change as lateral load is
applied, and since the P-o contribution in any column depends on its
axial force, iteration is generally needed during the analysis.
With this method, the calculated force demands account for both sway and .
non-sway amplification, and the maximum bending moments are not
necessarily at the element ends. The member end moments may also be
different from those calculated using the Double-B and Single-B methods, as
shown by the following example.
6.20.10 End MQments in ZerQ-8 Analysis
In a braced frame, the end moments in the members depend on the loads on
the beams, and also the relative bending stiffnesses of the beams and
columns. Usually, the stiffer the beams the smaller the end moments on the
columns.
Because of non-sway amplification (the P-0 contribution), the effective
bending stiffness of an elastic column depends on its axial force. As the axial ·
force in the column increases, it becomes progressivelyless stiff in bending,
and hence the bending moment at the end of the column progressively
decreases. At the same time, the amount of moment amplification increases,
and the maximum moment is a progressively larger mtiltiple of the end
moment. This was noted earlier, in Section 6.11.6.
Methods for Elastic Lateral Load Analysis
299
For a very simple frame this is shown in Figure 6.54.
F~
El
L
(a) Frame and loads
(b) Bending Moment Diagrams
Figure 6.54 Effect of Analysis Method on Column Moments
Figure 6.54(a) shows the frame. The dimensions and loads are used below
for analyses of the frame. Figure 6.54(b) shows the shape of the bending
moment diagram. There are three different moment diagrams for the ·
column, for three different analysis methods, as follows.
(1)
(2)
(3)
·Small displacements analysis, with no P-o contnbution.
Second order analysis, with the P-o contribution considered directly
in the analysis. The P-o contribution reduces the bending stiffness of
the column, and hence reduces the moment at the top of the column.
The P-o contribution then amplifies this end moment, as shown. This
corresponds to analysis _!>y the Zero-B method.
Sma!l displacements analysis to obtain the moment at the top of the
·column, then use of Equation 6.11 to calculate the maximum moment.
This corresponds to analysis using the Single-B and Double-B methods.
To sfudy the behavior numerically, consider the frame in Figure 6.54(a),
with the following analysis steps.
(1)
· Apply the load P first, putting the column in compression. This causes
essentially no bending. Since the column has some rotational restraint
at the top, its effective length factor, k, is 0.92. Consider two values of
P, namely 25% and 50% of the elastic buckling load for this effective
length.
(2)
Add a small load, F, and calculate the bending moment diagram,
accounting directly for the P-o contribution. This gives the column
end moment and the maximum moment for the Zero-B method.
300
(3)
Chapter 6 P-il Effects, Stability and Buckling
For the same load, F, calculate the moment diagram ignoring the P-o
contribution. Using Equation 6.11, calculate the maximum moment.
Since the effective length factor for the column is 0.92, calculate Pca for
Equation 6.11usingk=0.92. Also, since it is usual practice to assume
that k = 1, calculate a second maximum moment where Pca for Equation
6.11 is based on k = 1.
Table 6.3 shows the results of the analyses. The moments in this table are
shown as· multiples of the column end moments that would be calculated
using small displacements analysis.
TABLE6.3
Moments in Column for Frame in Figure 6.54(b)
Multiples of Small Displacements Moment at Top of Column
P, multiple of elastic buckling strength for frame
0.25
0.50
Moment at top of column, Single or Double-B
method
1.00
1.00
Moment at top of column, Zero-B method
0.93
0.77
Maximum moment, Single or Double-B method,
k=0.92
1.00
1.20
Maximum moment, Single or Double-B method,
k=1D
.
1.00
1.46
Maximum moment, Zero-B method
0.94
1.04
The following are some points from this table.
(1) . For PIP"'= 0.25, the moment at the top of the column from the Zero-B
method is 7% smaller than that from the other methods, indicating a
modest, but significant, reduction in the effective bending stiffness of
the column. The maximum moment from the Zero-B method shows a
very small amount of moment amplification. Equation 6.11 predicts
no moment amplification for this case. An "exact" analysis of a colwrin
with an applied end moment also shows no amplification (see Section
6.20.4, Table 6.2).
(2)
For PIP"'= 0.5, the moment at the top of the column from the Zero-B
method is 23% smaller than that from the other methods, indicating a
substantial reduction in the effective bending stiffness of the column.
The maximum moment from the Zero-B method shows substantial
Methods for Elastic Lateral Load Analysis
301
moment., amplification, with an implied amplification factor of
1.04/0.77 =1.35.
(3)
For PIP"= 0.5, the amplification factor from Equation 6.11, assuming k
= 0.92, is 1.20. Even though this is smaller than the implied factor of
1.35 from the Zero-B method, the maximum moment using Equation
6.11 is still 15% larger than that from the Zero-B method (1.20 versus
1.04).
i
(4)
For PIP"= 0.5, the amplification factor from Equation 6.11, assuming k
= 1.0; is 1.46. The maximum moment in this case is 40% larger than
that from the Zero-B method (1.46 ver_sus 1.04).
(5)
For PIP"= 0.5, it is interesting that the implied amplification factor of
1.35 for the Zero-B method is larger than the factor of 1.20 from
Equation 6.11. Equation 6.11 is not exact, but earlier analyses show
that the "exact" amplification factor for this case is 1.25 (see Section
6.20.4, Table 6.2).
This indicates that for large axial forces in the columns of an unbraced frame
it is possible for the columns to become significantly less stiff in bending.
This, in tum, can significantly reduce the bending moments at the column
ends, and can also reduce the maximum moments in the columns.
It is, of course, not reasonable to draw conclusions from the analysis of one
simple structure. However, this analysis suggests that the Zero-B method
c:an be expected to give more accurate bending moment demands. It also
suggests that the Single-B and Double-B methods tend to give larger
maximum moments than the Zero-B method, and hence tend to be
conservative.
Some other points to consider are as follows.
(1)
It appears that the axial force in a column must be rather large before
there is a substantial effect. Unless a column is very slender, it seems
unlikely that the axial force will be a substantial proportion of the
elastic buckling strength.
(2)
Changes in the relative stiffness of the beams and columns have a
smaller effect in an unbraced frame than in a braced frame. For lateral
load analysis of an unbraced frame, the bending moments at the
column ends depend mainly on the story shears (for ari inflection
point at column mid-height, the bending moment caused by lateral
302
Chapter 6 P-A Effects, Stability and Buckling
load is the shear in the column multiplied by half the story height,
regardless of the relative beam and column stiffnesses).
(3)
Axial force decreases the bending stiffness of an elastic column.
Hence, the behavior in the above example can be expected to apply to
a steel column. For a reinforced concrete column, an axial
compression force suppresses cracking, which increases the bending
stiffness. Hence, the calculated behavior in the above example is
·
probably not accurate for concrete columns.
(4)
The behavior is complex, and any method of analysis is inevitably
approximate. The important thing is to use a method that is appropriate
for design, considering simplicity and practicality as well as theoretical
accuracy. An advantage of the Zero-B method is that it is theoretically
more accurate than the other methods. This does not mean, however,
that it is the only reasonable method, or necessarily the most practical.
(5)
An analysis.method should account for all important effects. One of
the difficulties in choosing an analysis method is deciding which
effects are important and which can be ignored. Two potentially
important effects that may need to be considered are (a) initial drifts
and (b) stiffness reduction caused by a modest amount of yield. These
have been considered earlier, in Sections 6.17 and 6.18. They are
considered again in the following sections.
6.20.11 Initial Imperfections
Initial drifts in the structure as a whole, and initial imperfections in
individual columns, can affect the behavior and may need to be accounted
for in the analysis.
Initial drifts can be accounted for by changing the structure geometry or by
applying notional loads (see Section 6.17). For the Double-B Method, initial
drifts must be accounted ·for using notional loads. For the Single-B and
Zero-B methods, initial drifts could be accounted for using notional loads or
by changing the structure geometry. H the geometry is changed, the
simplest method is to use a P-A column and specify initial drifts only for this
column.
Initial out"Of-straightnt
.,, li•:idual column can have an important
effect. on the column l uckling strength (see Section 6.9.3). For non-sway
moment amplification in· the Double-B and Single-B methods, out-ofstraightness is implicitly included in the formula for the non-sway amplifi-
Methods for Elastic Lateral Load Analysis
303
cation factor B•. Out-of-straightness does not need to be included in the
analysis model for these methods.
For the Zero-B Method the situation is less clear. Different computer
programs use different element models, and while it is likely that most
computer programs will assume .that the elements are initially straight,
some programs may account for initial lack of straightness.
/
6.20.12 Stiffness Reduction and Sway Amplification
A structure may yield significantly under the design loads. This reduces the
effective stiffness of the structure, and hence increases the deflections. This
can, in turn, affect the amount of amplification.
Consider sway amplification first. The effect of· stiffness reduction on the
sway behavior was considered in Section 6.18.
For analysis purposes it may be reasonable to assume the same amount of
stiffness reduction for all elements in the analysis model. In this case the
stiffness reduction does not directly affect the calculated force demands (for
a small displacements analysis the stiffness reduction increases the
deflections, but has no effect ori the member forces). However, when the
deflections increase, this has the indirect effect of increasing the sway
amplification.
To see the magnitude of this effect,_ consider the following form of the
equation for the sway amplification factor (from Equation 6.10):
(6.14)
· where K is the lateral stiffness of a single story frame with ·no stiffness
reduction, and P/his the P-£\ geometric stiffness.
As one example, if P/h = O.lK, B. = 1/0.9=1.111. If K is reduced by 20%, B.
increases to 1/0.875 = 1.143. This is a 2.9% increase in the tot;U force demands.
As a more extreme example, if P/h = 0.15K, B. = 1.176. If K is reduced by
40%, B. increases to 1.333. This is a 13.3% increase in the total force demands.
Hence, the increase in the calculated member forces is much smaller than
the amount of stiffness reduction. Nevertheless, stiffness reduction can have
304
Chapter 6
P-~
Effects, Stability and Buckling
a significant effect, and it can be argued that it should be accounted for in
the calculation.
6.20.13 Stiffness Reduction and Non-Sway Amplification
For the Double-B and Single-B analysis methods, the non-sway amplification factor for a beam-column is calculated using Equation 6.11, which can
be written as
(6.15)
where M0 is the maximum bending moment in the column accounting for
sway amplification, and Mmax is the maximum moment accounting also for
non-sway amplification.
·
The key parameter in this equation is p ca' which is a buckling strength for
the column under axial load alone. This is "a" buckling strength, not "the"
buckling strength. It can be calculated as
(6.16)
where th~ key property is the bending stiffness EI•.
As one example, if P = 0.25P"' where Pce is the elastic buckling strength with
no stiffness reduction, and if P"' = P"' then PIP"' = 0.25 and Equation 6.15
gives Bn = 1.33. For a 20% stiffness reduction, PaJ =0.8Pa· Hence, for the same
value of P the effective ratio P/Pca increases from 0.25 to 0.31, and Bn
increases to 1.45. This is a 9% increase in the calculated moment demand.
As a second example, if P = O.SPce and there is no stiffness reduction, PI Pc•=
0.5 and Bn =2.0. For a 20% stiffness reduction and the same P, the ratio PIPca
increases to 0.625, and Bn increases to 2.67. This is a 33% increase.
It is anybody's guess what El. and Pca might be in reality, and in practice the
value must be specified by a design code. As noted earlier, for steel design it
is common to use the elastic EI, so that Pca = Pce· A later section in this
chapter considers the Direct Analysis Method for steel design, which
requires a smaller EI and gives greater amplification. For reinforced concrete
Methods for Elastic: Lateral Load Analysis
305
design it is common to use an estimated EI that allows for cracking and load
duration.
·
It is also anybody's guess what k might be in reality. In practice it is usual to
usek=l.
In the Zero-B method the non-sway amplification is accounted for directly
in the analysis. The amount of. amplification depends on the assumed
bending stiffness. The real stiffness is uncertain, and in practice the value
must be specified by a design code.
6.20.14 Demand/Capacity Calculation
Up to this point this section has considered only the calculation of member
force demands, particularly the amplified bending moments. To assess
performance, DIC ratios must be calculated. For columns it is usually
necessary to use a P-M-M strength interaction surface.
The P and M demands are the amplified values. On the capacity side, for
steel design it is common to use a P-M interaction surface for the column as
a whole, not for the column cross section (see 5ection 6.13.2, Figure 6.41).
For a reinforced concrete column it is common to use the strength
interaction surface for the column cross section.
6.20.15 Conclusion for this Section
This section has considered general procedures only. In practice there are
many details that must be decided, including (a) which analysis method to
use, (b) the magnitudes and types of initial imperfection, and how to
account for them in the analysis, (c) the amount and type of stiffness
reduction, and how to account for it, and (d) how to calculate DIC ratios.
There are so many complications and uncertainties that "exact" analysis is
impossible. The goal of the analysis must be more modest, ·namely to
account for sway and non-sway amplification with sufficient accuracy for
design, while keeping the analysis as simple as possible.
Also, the details of the modeling and analysis are too complex to be left to
the judgment of individual engineers - they must be specified by a design
code. The next section considers such details, for a specific method that is
permitted for the design of steel frames.
.
'
306
Chapter 6 P-8 Effects, Stability and Buckling
6.21 Direct Analysis Method for Steel Frames
· 6.21.1 Overview
The preceding section considers second-order analysis in general terms,
with no. specific rules. This section considers a method for which specific
rules have been established, namely the Direct Analysis Method in the AISC
360-05 specification. The emphasis in this section is what the rules mean (or
appear to mean) in terms of modeling and analysis. This includes the choice
of an analysis method, and of the procedures for considering initial
imperfections and stiffness reduction.
The Direct Analysis Method is covered in Appendix 7 of AISC 360-05, with
interpretation and background information provided in a separate Commentary for Appendix 7. This section refers to these documents as "the AISC
DAM Specification" and "the AISC DAM Commentary", or more briefly as
"the Specification" and,"the Commentary".
6.21.2· Acceptable Second-Order Analysis
The AISC DAM Specification requires a second-order analysis, accounting
for both the P-A and P-0 contnbutions. The Specification says little about
how the P-A and P-o contributions should be modeled. However, the
Commentary uses two "benchmark" problems for which exact solutions are
known. An analysis method is a "rigorous second-order analysis" if it gives
results within 3% of the exact solution for these two benchmarks.
"'
Although the Commentary does not explicitly say so, one of the benchmark
problems deals with sway amplification and the second with non-sway
amplification. Consider these two benchmarks, for the Zero-B, Single-B and
Double-B analysis methods.
6.21.3 Benchmark for Sway Amplification
The benchmark problem for sway amplification is an elastic cantilever
column with a fixed base, constant EI and no shear deformation, as shown
· in Figure 6.55(a).
Direct Analysis Method for Steel Frames
(a) Sway amplification
307
. (b) Non-sway amplification
Figure 655 Benchmark Examples for Sway and Non-Sway Amplification
For the Zero-B method, consider the SAP2000 frame element. H the column
in Figure 6.SS{a) is modeled using a single element, the analysis is accurate
for a wide range of PIP"' values, and the benchmark is met.
·
For the Single-B method, if the column. is modeled using a P-Li strut, and no
P 6 cable, the benchmark. ts not met. !'or example, for PIP.. "" 0.5 the
calculated deflection ratio, M~ is 1.70, compared with the exact value of
1.82. This is a 7% error. However, a cantilever column overestimates the P-o
contribution (see Section 6.3). For a column in an unbraced frame the error
is smaller than 7% and almost certainly smaller than 3%. For the Single-B
analysis method it is accurate enough to use an analysis model with only a
P-Li column, even though this model does not meet the benchmark.
0
The same is true for the Double-B method since the results using the B,
amplification factor are essentially the same as those for the Single-B method.
6.21.4 Benchmark for Non-Sway Amplification
The benchmark problem for non-sway amplification is an elastic beamcolumn with constant EI and a uniform lateral load, as shown in Figure
6.55(b).
.
·For the Zero-B method the result can depend on the computer program that
is used. For SAP20001 if the column in Figure 6.SS(b) is modeled using two
or more elements, the analysis is accurate for a wide range of PIP"' values
(see Section 6.20.4). Hence, the benchmark is met.
308
Chapter 6
P-~ Effects, Stability and Buckling
For both the Single-B method and the Double-B method, non-sway amplification is accounted for using the non-sway amplification factor, Bn· For the
beam-column in Figure 6.55(b), Equation 6.11 gives amplified moments that
are within 3% of the exact value. For example, for PIP" = 0.5 the exact
amount of moment amplification is 2.03, and EqtiatiQn 6.11 gives B,, = 2.00.
Hence, the Double-B, Single-B anq z.ero-B methods are all acceptable second
order methods based on this benchmark.
6.21.5 Acceptable Methods
The Zero-B method, with the SAP2000 element, meets both the sway and
non-sway benchmarks, and hence qualifies as an acceptable analysis
method. The Double-B and Single-B methods meet the non-sway benchmark but not the sway ~chmark. Strictly speaking, therefore, these methods
are not acceptable.
This indicates that the two benchmark examples in the AISC DAM
· Commentary are not well chosen. H an analysis method meets these benchmarks it does not necessarily mean that the analysis is accurate for a real
structure, and if it fails to meet .them it does not necessarily mean that it is
inaccurate. Jn particular:
meet
(1)
An analysis that accounts for only the P-A contribution does·not
the "sway" benchmark, yet aceounts · for sway amplification with·
sllfficient accuracy for almost ariy building frame.
(2)
Equation 6.11, for Bn1 satisfies the "non-sway" benchmark. However,
this equation· assumes that the end moments on a column are fixed
loads, which may not be correct. As shown in Section 6.20.10, the end
moments can be affected by column aXial forces, and Equation 6.11
may not give accurate amplified moments~
Taking everything into account, a reasonable conclusion is that the Double"
B, Single-B and Zero-B methods are all acceptable for the Direct Analysis
Method.
6.21.6 · Initial Drifts
The AISC DAM Specification requires that initial drifts be accounted fot by
adding notional loads to all load combinations.·
Direct Analysis Method for Steel Frames
309
The default notional loads are based on.an initial story drift ratio of 0.002
(i.e., 0.2%), which corresponds to the usual construction tolerance. A different value can be used if this default value is not appropriate.
For any load combination the notional loads all act in the same direction,
and need to be applied along only one axis of the building (i.e., not along
both axes simultaneously). Along any direction, both positive and negative
notional loads must be considered for gravity load' analysis. For combined
gravity and lateral loads the notional loads act in the same direction as the
lateral loads.
The notional loads for an initial drift ratio of 0.2% are the loads that would
be applied by a P-L\ column with that initial drift. This can have a significant
effect for gravity load analysis. For lateral load analysis the effect is smaller.
For a lateral load analysis, the drift ratio under the design loads for a steel
unbraced frame might be roughly 0.4%. Hence, the notional loads increase
the P-~ contribution for a lateral load analysis by about 50%. For example, if
the P-~ contribution reduces the structure stiffness by 10%, and if there are
no initial drifts, the sway amplification factor is 1.111. If the initial drifts
increase the P-L\ contnbution by 50%, the sway amplification factor increases
to 1.167 (i.e., to 1.0 + 1.5 x 0,111). This is a 5% increase in the total element
forces.
This increase can, however, be ignored. If the sway amplification factor is
smaller than 1.5, the AISC DAM Specification allows the notional loads to be
omitted for combined gravity and lateral loads. Since a sway amplification
factor of 1.5 is a very large number for a practical structure, it is likely that
notional loads will be applied only for gravity load analysis.
6.21.7 Basic Stiffness Reduction
The AISC DAM Specification requires a basic stiffness reduction of 20%,
applied to the bending (EI) and axial (EA) stiffnesses of all elements. This is ·
easy to implement, as a 20% reduction in the value of E.
As noted earlier, this reduction cart cause a significant increase in the sway
amplification (see Sections 6.18.2 and 6.20.12). When E is reduced, its effect
on the sway amplification is automatically accounted for in all three of the
Double-B, Single-B andZero-B analysis methods.
This stiffness reduction also affects the non-sway amplification, and it tends
to have a larger effect than on non:..sway amplification (see Section 6.20.13).
310
Chapter 6 P-A Effects, Stability and Buckling
The stiffness reduction is automatically accounted for in all three of the
Double-B, Single-B and Zero-B analysis methods.
6.21.8 Advanced Stiffness Reduction
The AISC DAM Specification allows, as an option, a stiffness reduction in
addition to the basic 20%. This additional reduction applies to El only (EA is
reduced by the basic 20%). The amount of reduction varies from member to
member, depending on the ratio PIPui where P is the axial force in the
member and pu is the "squash" strength for the column cross section (section
area multiplied by steel yield strength).
This approach is essentially the same as the "tangent modulus" procedure in
Section 6.12.4, except that the modulus is a "reduced modulus". As noted in
Section 6.12.4, this has no effect on the result.
The reduced modulus relationship is shown in Figure 6.56.
E,/E
0.8----..:
Modulus for El
0.5
1.0
P/Pu
Figure 6.56 Reduced Modulus Relationship
The basic 20% reduction in stiffness applies to both the bending and axial
stiffnesses. The additional reduction applies only to the bending stiffness.
The relationship for the reduced modulus is
For PIPu < 0.5, Et = 0.8E
For PIPu > 0.5, Et= 0.8E(4PIPu )(1- PIPu)
(6.17)
H this relationship is used (see later for an alternative option), some possible
effects are as follows.
(1)
When loads are applied, the aXial forces in the columns change. If the
axial force in any column exceeds 0.5Pu, the bending stiffness of the
column changes, and hence the structure stiffness matrix changes. As
a result, the analysis will generally require iteration (estimate El
Direct Analysis Method for Steel Frames
311
values, apply load and analyze, calculate P and hence E,., and if any E,
value has changed significantly, repeat the analysis). This can be a
substantial complication.
(2)
Stiffness reduction increases · the amount of sway amplification.
However, compared with an analysis that uses only the basic 20%
reduction, the effect is likely to be small. For lateral load analysis, the
amount of sway amplification in a story dep~ds on the story stiffness.
When lateral load is applied, the axial forces in some columns will
increase and in others will decrease. The axial forces tend to remain
roughly constant for the columns in the interior bays of a frame, and
to change substantially only for columns in the end bays. It is likely
that EI will change in only a few columns, and any change in story
stiffness is likely to be small. Hence, there is unlikely to be much
variation in the amount of sway amplification.
(3)
Stiffness reduction also increases the amount of non-sway amplification. This is calculated separately for .each column, and non-sway
amplification is more sensitive to stiffness change than sway amplification. Hence, compared with an analysis with only the basic 20%
reduction, there could be significant increases in the calculated force
demands for some members.
(4)
For calculating the strength D/C ratios for beam-columns, the AISC
Specification uses a P-M surface based on the axial buckling strength
of the column, not the "squash" strength of the column section (see
Section 6.13.2 and Figure 6.41). This assumes that non-sway amplification is based on the elastic buckling strength, Pce' not a smaller v~ue
based on a reduced bending stiffness. When a reduced stiffness is
used, the amplification is larger, and if this same P~M surface is used
(as the DAM seems to require), the strength capacity may be underestimated.
(5)
Although the Direct Analysis Method emphasizes static lateral load
analysis, earthquake loading may require dynamic analysis. Conceivably
this could be done using step-by-step analysis, changing the element
stiffnesses in each step as needed. However, this is a nonlinear
analysis that could be expensive computationally. Modal analysis is
likely to be preferred, in which case the structure stiffness matrix must
be constant. The obvious approach is to base the reduced modulus for
any column on the axial force under gravity load, and to assume that
this modulus is constant for the dynamic analysis.
312
Chapter 6
P-~
Effects, Stability and Buckling
6.21.9 Alternative to Advanced Stiffness Reduction
An alternative option is to use only·the basic 20% stiffness reduction, and to
apply notional loads corresponding to an additional 0.1 % drift ratio in each
story. This is simpler than adjusting the bending stiffness on an element-byelement basis, and it seems likely that this option will usually be chosen.
It may be noted that this alternative option does not appear to be logical,
since the purpose of notional loads is to account for initial drift, not for
stiffness reduction.
6.21.10 Conclusion for this Section
The key question for second order analysis is whether it gives member force
demands that are accurate enough for strength-based design, accounting for
the effects of geometric nonlinearity.
Three analysis methods may be used for the Direct Analysis Method,
namely the Double-B, Single-B and Zero-B methods. Given the inevitable
approximations in modeling and analysis, all three methods should be
accurate enough.
For sway amplification using the Single-B method, an analysis model that
accounts for the P-~ contribution, and. ignores the P-o contribution, is
accurate enough, even though it does not meet the AISC benchmark. It
follows·that the Double-B method, using a sway amplification factor, is also
accurate enough, although it may be oifficult to apply consistently. The
Zero-B method is sufficiently accurate.
Non-sway amplification is more complex, for the following reasons.
(1)
The end moments on a column can depend on the relative bending
stiffnesses of the beams and columns, and the stiffness for a column
can depend on the axial force in the column. Hence, the end moments
may not be sustained loads, as is usually assumed for calculating nonsway amplification. The Zero-B method is potentially more accurate
than the other two methods because it can account for this effect.
(2)
The effective length factor for calculating the amount of non-sway
amplification is uncertain. For the Double-B and Single-B methods it
is reasonable, an.d conservative, to assume k = 1. TheZero-B method is
potentially more accurate than the other two methods, because it
accounts directly for the rotational restraint· offered by adjacent
Inelastic Lateral Load Analysis of Frames
313
members. However, this method may overestimate the amoUI\t of
rotational restraint, since it assumes that the adjacent members are
elastic. One reason for using k = 1 is that the amount of rotational
restraint is uncertain, and may be less than expected from an elastic
analysis.
Jn these two points the words "potentially more accurate" are used
deliberately for the Zero-B method. The P-A effect,.is tompleX, and accounting for it in analysis is difficult. This is especially true for the P-5
contribution (i.e., non-sway amplification). Viewed strictly as methods of
analysis, the Doubl~B and Singl~B methods are less accurate. Viewed as
methods to calculate member force demands for strength-based design, they
look reasonable. Viewed strictly as a method of analysis, the Zero-B method
is more accurate. Whether it is better as a design tool is less obvious.
6.22 Inelastic Lateral Load Analysis of Frames
For strength-based design using elastic analysis, there are significant and .
complex effects from amplification, stiffness reduction and imperfections.
These can also affect inelastic analysis.
Imperfections, in the form of initial drifts, have the most effect for gravity
load analysis. For inelastic lateral load analysis, any initial drifts are likely to
be small compared with the total drifts, and they probably do not need to be
considered. For .gravity load analysis, it is unlikely that there will be
significant inelastic behavior. Hence~ the analysis is likely to be the same as
an elastic analysis for. strength-based design; and· initial drifts can be
considered using notional loads.
Stiffness- reduction may have significant effects for .inelastic lateral load
analysis. Jn an elastic analysis, stiffness reduction is used. to account for
yield under the design loads, caused, for example, by residual stress in steel
members. Ideally, such inelastic behavior would be accounted for directly in
. the inelastic analysis. Jn practice, however, inelastic elements are often
modeled assuming elastic-perfectly-plastic behavior, which does not
account for such things as residual stress. Hence, it might be reasonable to
reduce the elastic modulus of the material, for example by 20% as in the
Direct Analysis Method. Alternatively, inelastic components could be
mqdeled with initial yield followed by strairi hardening, rather than as
elastic-perfectly-plastic.
314
Chapter 6 P-i:\ Effects, Stability and Buckling
Amplification is potentially more complicated, mainly because of non-sway
amplification. In most cases, however, amplification can be considered
reasonably easily, as follows.
(1)
For an unbraced frame, sway amplification can be accounted for using
P-Li columns or struts. Non-sway amplification is likely to be negligible
and can be ignored. The maximum moments in a column will almost
always be at the column ends.
(2)
For a braced frame the columns are likely to be designed not to
buckle, using capacity design. Hence they need ·to be checked for
strength, using the same procedures as for strength-based design to
account for non-sway amplification.
(3)
Diagonal braces may be allowed to buckle. Since braces tend to be
axially loaded columns, rather than beam-columns, they· can be
modeled as bars with only material nonlinearity (see Section 6.10.5).
For other cases, inelastic modeling can be difficult. For example, if a· column
has both axial force and bending, and is allowed to buckle, there is no
simple way to account for the buckling strength and the post-buckling
behavior.
The more complex the analysis, or the theory on which it is based, the more
important it is to keep in mind the goal of the analysis. The main reason for
using Capacity Design is that it can lead to better behavior under extreme
loads. However, an important secondary reason is that when inelastic
behavior is limited to a few components, it is easier to set up an inelastic
analysis model, and hence easier to get reliable analysis results. In theory it
may be possible to use "advanced analysis", where all significant aspects of
behavior are accounted for directly in the analysis model, and where the
behavior of the structure can be predicted accurately by analysis. In practice
this is impossible. Even if it were possible, "design by analysis" - where
everything is modeled and the computer program tells everything about the
behavior - is a bad idea. There is much more to design than analysis,
Analysis is merely a useful tool for design, not an end in itself.
Buckling Analysis
315
6.23 Buckling Analysis
6.23.1 Overview
For a building structure it is unlikely that a structure will buckle under
gravity load alone (unless the structure is badly damaged or very badly
constructed). Hence, buckling analysis is usually not required. However, it
is useful to consider methods for buckling analysis. '
There are two main types of buckling analysis, namely with and without
equilibrium bifurcation. This section considers the analysis methods.
6.23.2 Analysis Method with Equilibrium Bifurcation
For an ideal elastic structure, such as a column with no imperfections or a
symmetrical frame with only vertical load, the deflections in the buckling
mode are zero until the structure beeomes unstable. At that point there is a ·
bifurcation of equilibrium. It is relatively easy, computationally, to calculate
the buckling load for such a structure.
The analysis requires that two separate stiffness matrices be set up, namely
an elastic stiffness matrix for the small displacements structure and a
geometric stiffness matrix for the P-A columns, P-A struts and P-o cables that
account for P-A effects. As already noted, the geometric stiffness for a .
P-A column or a P-A strut is easy to formulate, and the geometric stiffness for
a P-0 cable is not so easy. If buckling is expected in a side-sway mode, only
the P-A contribution usually needs to be modeled, and the P-o contribution
can be ignored. If individual columns can buckle within their own length as
part of the buckling mode, the P-o contribution must be modeled.
The steps in the analysis are as follows.
K.i· This
(1)
The stiffness matrix for the small displacements structure is
cart be assembled in the usual way.
(2)
The loads are applied to the small displacements structure, with a unit
load factor (e.g., 1.0 times the design gravity load). The element forces,
particularly the axial forces in the columns, are calculated using small
displacements analysis. The structure deflects, but there is (or is
assumed to be) no deflection in the buckling mode. Usually the load is
vertical, the buckling mode is horizontal, and the load causes little or
no horizontal deflection.
316
(3)
Chapter 6 P-A Effects, Stability and Buckling
Given the element forces, the element geometric stiffnesses are
calculated, and assembled into a geometric stiffness matrix, Ket· This
is the structure geometric stiffness matrix for a unit load factor. Since
the structure is elastic, the geometric stiffness matrix for a load factor
Ji. is ll\;r The stiffness Ka is constant.
(4) . For any load factor, Ji., the effective stiffness is Ka+ AI\;r lf this
stiffness is positive the structure is stable, if it is negative the structure is unstable, and if it is zero the structure is in a state of neutral
equilibrium. The value of Ji. for which the stiffness is zero defines the
·buckling load.
(5)
.
. .
The stiffness matrix will not, of course, have a zero value. ·Rather it
will have a zero determinant. When a stiffness matrix has a zero .
determinant, it means, physically, that there is some deflected
that. can .be imposed on the structure and the forces required to.
impose this shape are all zero. The deflected Shape is the buckling
mode shape. Mathematically it is necessary to solve the following
equation for Ji.:
Shape
Det(K0 +A.KG1)=0
(6~18)
For a multi-DOF structure this equation will have multiple solutions,
one for each buckling mode shape. The important solution is the
smallest one, for the lowest buckling mode.
There are a number of methods for solving Equation 6.18. One method is
"determinant search'; where, in effect, the ·value of Ji. is progressively
changed and the determinant of the effective stiffness matrix is monitored
until it .is very close to zero. Another method is formal eigenvalue extrac- .
tion, for which there are several numerical methods (one of which is based
on determinant search).
The details of the numerical computation are not important. The important
point is that this method is limited to ideal elastic structures. Since such
structures do not exist in practice, the analysis results are usually of only
academic interest.
There is an extension of this analysis method that allows it to be applied to
certain inelastic structures. The procedure is as follows.
(1)
Establish a relationship between the tangent modulus for the material
in the structure and the load factor. For example, for a column .
Buckling Analysis
317
establish a relationship between the tangent modulus and the axial
force, as in Section 6.12.4.
(2)
At each step of the determinant search, scale the small displacements
stiffness, Kw so that it is based on the tangent modulus not the original
elastic modulus.
This method can be useful, although it is not a general method for inelastic
buckling.
,
6.23.3 Buckling Analysis With Imperfections
If a column or a structure has imperfections, it will usually buckle progressively, with no equilibrium bifurcation. An exception is if the imperfection
contains no contribution from the lowest buckling mode, such as a symmetrical imperfection in a structure that buckles in an anti~symmetrical mode.
In this case there can still be an equilibrium bifurcatioi:t.
In general, if a structure has imperfections it will have some initial deflection
in the buckling mode. As the load increases, the deflection in the buckling
mode increases, in part because of forces exerted by the P-.!\ struts and/ or
P-o cables. If the imperfection is introduced only in a P-.!\ column (see
Section 6.6.6, Figure 6.20), the deflections in the buckling mode are caused
entirely by the P-.!\ column. There is a similar effect when notional loads are
used to account for imperfections.
As the load approaches the buckling load, the -effective stiffness of the
structure decreases and the deflections in the buckling mode get
progressively larger. If the structure remains elastic, its deflection becomes
theoretically very large as the load approaches the buckling load, but the
structure remains stable. In practice, however, the structure will yield and
lose stiffness as the deflection increases, and at some point it will become
unstable (see Section 6.6.7, Figure 6.21). This is the buckling load.
P-A theory is usually accurate enough for this type of behavior, since the
structure will usually become unstable before the deflections are large.
Hence, the effective structure stiffness is still the sum of a small
displacements stiffness and a geometric stiffness. If the structure remains
elastic (which is unlikely), the buckling load is the same as that for an elastic
structure with no imperfections, and the analysis method from the
preceding section can be used to calculate this load. However, if the
structure yields it is necessary to use some sort of incremental analysis,
accounting for progressive yield.
318
Chapter 6 P-A Effects, Stability and Buc;kling
This type of analysis is more difficult than that in the preceding section. The
details of the analysis method are outside the scope of this book. ff a
computer program provides this analysis option, its documentation should
explain the procedure.
6.24 Some Other Strudures
6.24.1 Overview
This chapter has considered· mainly building structures. There are, of course,
many other structures where P-A effects are important. This section
considers some examples.
6.24.2 Walls With Out-of-Plane Bending ·
A model for a shear wall will have at least one element per story (see, for
example, Figure 6.42). For in-plane loads the P-A contribution can be
signifi~t, especiallyfor tall walls. The in-plane P-0 contnbution ts negligible,
For out--Of-plane behavior, however, the P-0 contribution can be substantial,
·. and out-of·plane buckling can occur. Usually there will be beam-column
behavior; with a combination of axial (vertical) stress and transverse load
normal to the wall The transverse· load can be wind load or inertia load
associated with earthquake acceleration. The wall spans vertically between
floors, and possibly horizontally between cross-walls. The transverse load
causes sinall-displacements bending moments, which are amplified by the
axial stress. The behavior of a vertical strip of.wall is similar to the beamcolumn behavior described in Section 6.11.2, Figure 6.33.
As with a column in a frame structure, it is usually· not practical to account
for moment amplification by including a P-o cable in the analysis model.
The amplified moments will usually be calculated in the same way as for a
beam-columri.
6.24.3 Long Span Roofs
Long span roofs can be i:elatively light and flexible, and buckling under
gravity load may be an important consideration. As an example, consider
in-plane buckling of the shallow arch in Figure 6.57(a).
Some Other Structures
Load
Buckling based on
ideal undeformed shape
Buckling based on
symmetrical deflected shape
Symmetrical load
~lllllll
Undeformed shape
Strength reduces
after buckling
Deflected shape before
~
Buckling mode shape
(a) Arch showing gravity load
and deflected shapes
319
;
Displacement in buckling mode
(b) Relationship between gravity load
and buckling displacement
Figure 6.57 Buckling Behavior of a Shallow Arch
Some aspects of the buckling behavior are as follows.
(1)
Even if the arch is symmetrical and is loaded symmetrically, its
buckling mode is unsymmetrical.
(2)
If the arch and the loading are perfectly symmetrical, there is no
displacement in the buckling mode until the arch becomes unstable.
There is then a bifurcation of equilibrium, as shown in Figure 6.57(b).
(3)
The ability of an arch to support load depends on its curvature. As the
arch deflects under gravity load, its curvature decreases. This could
have a significant effect on the buckling strength. As shown in Figure
6.57(b}, if the arch remains symmetrical as it is loaded, the load at
which equilibrium bifurcation occurs gets smaller as the arch deflects
and flattens. Stiffness reduction caused by small amounts of yielding
or by spreading of the supports can also reduce the buckling strength.
(4)
The strength of an arch is likely to decrease after buckling occurs, as
shown in Figure 6.57(b). One consequence of this is that the buckling
strength is likely to be sensitive to small imperfections, in the form of
departures from symmetry in the arch geometry or the load.
(5)
If the structure is not perfectly symmetrical or (more likely) if the load
is not symmetrical, there will be displacements in the buckling mode
as the gravity load is applied. The displacements in the buckling
mode will tend to increase progressively, without equilibrium
bifurcation, as shown in Figure 6.57(b). The buckling load with this
behavior can be much smaller than the bifurcation load.
•· -
320
Chapter 6 P-d Effects, Stability and Buckling
For the design of a structure of this type, it is necessary to calculate the
buckling load capacity. It is not sufficient to calculate the bifurcation load
for an elastic, symmetrical arch with symmetrical loading. The analysis
model must account for 3D behavior, initial imperfections, unsymmetrical
loading, yield, and other effects. The analysis procedure must also account
for the progressive change in shape as the structure deflects.
It is usually sufficient to perform a P-L\ analysis. True large displacement
effects are important only in the post-buckling range, which is mainly of
academic interest. Unless the structure is extremely flexible, P-L\ analysis is
sufficiently accurate up to buckling.
6.24.4 Bridge Columns
For columns in a building frame, it is usually neither feasible nor desirable
to account for the P.:O contribution directly in the analysis model. This may
not be the case for bridge columns.
Bridge columns can be slender, with large gravity loads and significant
bending from lateral loads or end moments. It is unlikely that a column will
buckle. However, there can be significant amplification of the bending
moments, and amplification formulas based on beam-column behavior may
not be accurate. In addition, since there are far fewer columns in a bridge
than in a building, it is feasible to use detailed analysis models, allowing for
both material and geometric nonlinearity.
The most likely model for a tall colwm\ is to divide it into a number of
elements. This may be necessary to model lateral loads on the column,
particularly inertia forces for earthquake loads (with several masses over the
column height). With this type of model, each element is short, and only the
P-L\ contribution needs to be accounted for in the model.
6.24.5 Buckling of Pipe With Zero Axial Force
Figure 6.58 shows a fictitious column, consisting of a length of pipe with
frictionless pistons.
~
Some Other Structures
321
Frictionless
piston
Pipe
l
Fluid··
L:t~
Figure 658 Does This Pipe Bucklei
The axial load on the column is P. Because the pistons are frictionless, there
is no axial force in the pipe, and P is resisted by pressure in the contained
fluid.
Since there is no axial force in the pipe, does it buckle? ·
6.24.6 ·Buckling of Buried Pipe
The answer to the preceding question is that the pipe d\Jes buckle, and that
the buckling load is x2fi/L2, where EI is the bending stiffness of the pipe
and L is the length. Essentially, the fluid column acts like a P-~ cable and the ·
pipe acts like a small ·displacements column.
A more practical structure is a buried pipe, as shown in Figure 6.59..
.
.
i i1
Axial
anch.orage
~
....
~~ri?N4 I I
Downward soil
resistance
Figure 6.59 Model of a Buried Pipe
The figure shows a model for a straight buried pipe. The pipe is modeled
using short column-type elements, allowing for axial forces in the contained
fluid and axial and other forces in the. pipe. The soil is modeled using
discrete springs. Only a few of the soil springs are shown in the figure. A
typical spring spacing is of the order of the pipe diameter. There will
usually be axial sprit;tgs at the ends, to model the anchoring effects of the
i'
'·
322
Chapter 6
P-~
Effects, Stability and Buckling
adjacent pipe. There may also be axial springs to model longitudinal cohesion
between the pipe and the soil.
The soil spring stiffness in the upward direction depends on the burial
depth, and is usually smaller than the stiffness in the downward direction. H
the upward soil stiffness is below a threshold value, the pipe can buckle as
shown in the figure.
The axial force causing buckling is the force in the fluid plus the force in the
pipe. The fluid pressure causes tension
in the hoop direction of the
pipe, which by Poisson's effect causes longitudinal shortening of the pipe,
and hence causes a tension force. H the pipe content& are hot, the pipe wants
to expand longitudinally, and tends to be in compre'SSion because of
restrained thermal expansion. H the contents are cold, the pipe tends to be in
tension.
stress
These effects can be accounted for in an analysis model, and the buckling
load can be calculated. Since·the pipe is modeled using short elements, it is
sufficient to account for the P-.i\ contribution (using P-.i\ struts) and to ignore
the P-0 contribution. As with most structures, the ideal elastic buckling
strength is likely to be much larger than the actual strength. A realistic
estimate of the buckling strength should account for imperfections and yield.
-
.
6.24.7 Large Displacements of Buried Pipe ·
A more practical case for a buried pipe is shown in Figure 6.60.
Pipe tends to uplift progressively
1-N•
Ii
-r~.· f
~
·fl'
iI
Undeformea position, with overbend
Figure 6.60 Pipe With Overbend
_ Jn this case the pipe is not straight, but has an overbend. When the pipe is
pressurized, it tends to push upward. H the strength of the soil overburden
is exceeded, the pipe can push upward out of the ground. The behavior is
similar to buckling of a pipe that has a large initial imperfection.
Some Other Structures
323
This behavior can be modeled. Two considerations can be important, as
follows.
(I)
As the pipe deflects upward, a substantial catenary effect can develop,
which stiffens the pipe. P-a analysis does not account for this, and it is
usually necessary to consider true large displacement effects.
(2)
As.the pipe deflects it will tend to yield. For 1example, a plastic hinge
might form at the point of maximum deflection. This can dramatically
change the behavior compared with an elastic analysis.
This is an example of a structure where a detailed analysis that accounts
directly for all important aspects of material and geometrical nonlinearity is
feasible, and is useful for design.
6.24.8 Structures With "Follower'' Forces
Figure 6.61 shows a cantilever column. In Figure 6.6I(a) the load is vertical,
as is usually the case. In Figure 6.6I(b) the load is always directed at the
base of the column. (Imagine that a cable is attached to the top of the
column and passed through a pulley at the base, and that the column is
loaded by pulling on the cable.)
h
Effective
length= 2h
(a) Verticalforce
(b) Follower force
Figure 6.61 Cantilever Column with Follower Force
As shown in the figure, the effective lengths for the two columns are
different. The buckling strength for the column in Figure 6.6I(b) is four
times larger than that in Figure 6.6I(a).
-
,
324
Chapter 6 P-Ll Effects, Stability and Buckling
The load in Figure 6.6l(b) is usually termed a "follower" force, with a
direction that is not fixed but depends on the deformation of the structure.
Figure 6.62 shows a second example of a follower force.
.0.:- ,
Pressure load, normal
to buckled shape / /
I/ Fixed load, normal to
undeformed shape
\
.
l
~~-''
(a) Undeformed shape
.
(b) Buckled shape
Figure 6.62 Ring With External Pressure Load
Figure 6.62(a) shows a ring (i.e., a curved beam) loaded by external
pressure. This might represent, for example, a ring stiffener in a submarine
hull. At a critical pressure, which depends on the bending stiffness of the
beam, the ring will buckle. Figure 6.62(b) shows the buckling mode shape.
Figure 6.62(b) also shows two possible assumptions for the loads. In one
assumption the load direction is fixed, actirig through the center of the ring.
In the second assumption the load is normal to the deformed shape of the
ring, and hence changes direction as the ring buckles. The second
assumption is a follower force, and corresponds to an actual pressure load.
As with the preceding example, the buckling loads are different. In this case,
the buckling strength with a pressure load is smaller than with a fixed load.
The reason is as follows. ·Consider two cross sections through the ring in
Figure 6.62(b), one horizontal and one vertical. For the fixed load, the
norrflal forces on these cross sections have the same magnitudes. For the
pressure load, however, the normal force on the horizontal section is larger
than that on the vertical section, because the deformed horizontal section is
longer. This increases the deformation in the buckling mode, and
substantially reduces the buckling strength.
As noted earlier, the buckling strength of a structure can be obtained by
solving the equation Det(K0 +A.Kc,)= 0, where K0 and Kc, are the small
displacements and geometric stiffnesses (see Section 6.23.2). For the
structures in Figures 6.61 and 6.62, the matrices Ko and Ket are the same
Lateral-Torsional Buckling of Beams
325
whether the load has a fixed direction or is a follower force. Since the
buckling strengths are· different, there must be some additional effect. This
is a '1oad stiffness", Ki, which accounts for the effects of structure displacements on the external loads. For fixed loads, ~ is zero,· The buckling
strength with follower forces can,,be obtained by solving the equation
Det(K;. + ll<c1 + lI<i,1) =0.
.
6.25 Lateral-Torsional Buckling of Beams .·
i
6.25•.1 Overview
The preceding sections have considered column instability and buckling
caused by axial forces. This section collSiders instability caused by bending
moments and shear forces, and explains this instability in physical terms.
This is done using an extension of the P-.& strut and P-B cable con<::epts. The
purpose is to explain why lateral-torsional buckling occurs, and to' consider
how it can be accounted for in the design process.
6.25.2 Causes of Lateral-Torsional Buckling
A beam can buckle in a lateral-torsional deformation mode. The behavior
can be explained using a variation of the P-5 cable concept.
First" consider an elastic beam with uniform bending momeJ\t along its
length, as Shown in Figure 6.63(a).
M~-------~M
Cl
1)
(a) Beam with constant bending moment
P=~~
-----------~._!'.
f"1te
P ;!_ ~ _________________
.;.; _P l_d1
- (b) Equivalent forces and P-li cable5
Figure 6.63 Beam with Compression and Tension P-~ Column
Figure 6.63(b) shows an _analysis model, with two P-5 cables. The assumptions for this model, and its behavior, are as follows.
I
' -
326
Chapter 6 P-L\ Effects, Stability and Buckling
(1)
The end moments are replaced by an equivalent tension-compression
couple, with forces, P, acting at the centroids of the tension and
compression stress blocks.
(2)
There are two P-0 cables, one in compression and one in tension, as
shown by dashed lines in the figure.
(3)
The main structure (the small displacements beam) is the elastic
beam, with uniform bending moment and no P-~ effects. This beam
has bending and torsional stiffness.
(4)
The P-o cables are imbedded in the main structure, and they exert
forces on the main structure based on their curvatures.
(5)
For vertical bending of the beam, the P-o cables have vertical
curvature, and they exert vertical forces on the mam structure. Since
one P-o cable is in tension and the other is in compression, these
forces cancel each other out.
(6)
Ha small lateral deflected shape is assumed for the beam, so that the
beam bends laterally and/ or twists, the P-o cables are curved
laterally, and they exert lateral forces on the main structure. These
forces cause the main structure to deflect (bend laterally and/ or
twist). If the deflected shape is the same as the assumed deflected
shape, this shape is a buckling mode shape. If the deflection magni-"
tudes are the same, the bending moment is the buckling moment for
the beam.
The buckling mode shape for a beam usually involves both lateral bending
plus torsional twist. The reason for this is illustrated in Figure 6.64.
Figure 6.64(a) shows the axis of the beam and the initial positions of three
cross section, at the two ends and at mid-span. Consider some possible
buckling modes, as follows ..
(1)
Figure 6.64(b) shows a possible buckling mode with only lateral
bending. For this mode the two P-o cables have the same curvatures
but opposite axial forces.· Hence, the upper cable exerts a distributed
force to the right on the main structure, and the lower cable exerts an
equal and opposite distributed force to the left. These forces cause the
main. structure to twist, with no lateral deflection. This is different
from the assumed deflected shape, so this shape can not be a buckling
mode.
Lateral-Torsional Buckling of Beams
327
(2)
Figure 6.64(c) shows a second possible buckling mode, with only
twist. For this mode the P-o cables exert equal distributed forces on
the main structure, causing it to bend laterally with no twist. Again,
this is different from the assumed deflected shape, so this shape can
not be a buckling mode.
(3)
Figure 6.64(d) shows a third possible buckling mode, with both lateral
deflection and twist. For this mode the P-o cilbles cause both lateral
deflection and twist in the main structure. Hence, this can be a
buckling mode.
(a) Undeformed beam
Note lateral supports at ends
Forces exerted Deflection caused
by P-0 cables
by these forces
(b) Lateral bending only
Forces exerted Deflection caused
by P-0 cables
by these forces
(c) Torsional twistonly
Forces exerted Deflection caused
by P~li cables
by these forces
(d) Both bending and twist
Figure 6.64 Physical Explanation of Beam Buckling
328
Chapter 6
P-~ Effects, Stability and
Buckling
Since the buckling mode shape involves both lateral bending and torsional
twist, an equation for the buckling moment (critical moment) must include
both the lateral bending stiffness and the torsional stiffness of the beam.
This means that the lateral support conditions at the ends of the beam are
important. Figure 6.64(a) shows lateral supports that restrain both lateral
displacement and torsional twist at the ends of the beam. With this
assumption the lateral bending stiffness and the torsional stiffness are well
defined. For this case it is not difficult to develop an equation for the
buckling moment The details of this equation are not important.
However, the beam in Figure 6.64 is simpler than most beams. The
following are some possible complications.
(1)
In a real beam the end restraints may be less certain, and hence the
lateral and torsional stiffnesses may also be less certain. For example,
if torsional rotation is not restrained at the beam ends, the torsional
stiffness is reduced, and hence also the buckling load.
. (2)
Beams are often braced laterally along their length. The lateral bracing
may restrain only lateral deflection, .or only twist, or both lateral
deflection and twist. The type, location and stiffness of the lateral
bracing can have a large effect on the buckling load.
(3)
A beam
will rarely have uniform bending moment along its length.
·This means that the axial forces in the P:-0 cables will usually vary
along the beam length.
(4)
If loads are applied on the top of the beam, they tend to have a destabilizing effect as the beam twists, as shown in Figure 6.65~ This can
decrease the buckling load. Conversely, loads hung from the bottom
of a beam can increase the buckling load.
~::::::::=::::::::::::::::::::::::::::::~.
. .
!iii
.
(a) Bending moment varies along length.
Hence forces in P-0 cables also vary.
(b) Loads applied to top of
beam. apply extra torsion.
Figure 6.65 Some Complications
\,
-~
Lateral-Torsional Buckling of Beams
329
(5)
A beam may have biaxial bending, and may also have axial force.
(6)
For I-section beams the torsional stiffness can be uncertain. This can
be a complex issue, as considered later.
6.25.3 Analysis Models
For the lateral-torsional buckling analysis of a beafu such as that in Figure
6.65, a continuum model with P-15 cables is impractical. If the lateral
torsional buckling strength is to be calculated, it is more practical to divide
the beam into a number of short elements, and to replace each P-3 cable by a
number of P-~ struts. It is possible to calculate the buckling strength using
such a model. As with a column, however, the buckling strength can be
affected by initial imperfections and inelastic behavior, and it is easy to
overestimate the strength.
Also, as with column buckling it is generally not practical to account for
lateral torsional buckling directly in the analysis model for a complete
structure. In almost all cases, structures are designed so that lateral-torsional
buckling does not occur. This is done by providing lateral bracing, and
calculating the buckling strength using conservative equations. If lateraltorsional buckling is allowed to occur, this is a complex analysis problem,
especially if the post-buckling behavior must be considered.
6.25.4 Torsional Behavior of I-Section Beams
In steel structures, most beams have I-shaped sections. Such beams tend to
be flexible in torsion, and lateral-torsional buckling can be a concern.
The torsional stiffness of a beam section is usually expressed as GJ, where G
is. the material shear modulus and J is the torsional "inertia", or torsional
constant, for the section. For a cir~lar section, J. is the polar moment of
inertia (second moment of area) of the section. For other sections Jis smaller
than the polar moment of inertia, and for an I-shaped section it is much
smaller. The reason can be found in any textbook on structural mechanics.
This is "St. Venant" torsion.
This torsional stiffness is, however, based on the assumption that the cross
section is free to warp when an I-section beam is twisted. This is shown in
Figure 6.66.
330
Chapter 6 P-t. Effects, Stability and Buckling
Bottom flange
Top flange
Cross section
+-- warps (does not
.
remain plane).
(a) Twist of an
I-section beam
(b) Plan view
Figure 6.66 Torsion of I-Section Beam
In an actual beam, however, warping of the cross section is almost always
restrained. This is shown in Figure 6.67.
i
~======:::;(1 ~
(a) Elevation
Bottom flange
Top flange
(b) Plan. Unrestrained warping at beam ends.
(b) Plan. Warping restrained at beam ends.
Figure 6,67 Restrained Warping off-Section Beam
In Figure 6.67(b) the flanges are simply supported laterally, and the beam
cross section can warp. At mid-span, however, the cross section must
remain plane. Consequently, each flange bends laterally, behaving like a
simply supported beam. This can greatly increase the effective torsional
Lateral-Torsional Buckling of Beams.
331
stiffness, and hence can also increase the beam strength in lateral-torsional
buckling.
In Figure 6.67{c) the flangeS are also restrained against rotation at the ends
of the beam. Hence the flanges bend laterally like beams with fixed ends,
with an even greater increase in torsio~ stiffness.
·
This type of behavior is usually termed ''.warping,tOrsion". It can have a
substantial effect on the buckling strength.
6.25.5 Compression Flange as a Column
A relatively simple method for estimating the buckling strength of an
I-section beam is tO regard the compression flange as a· column that buckles
· laterally. This method ignores the tension flange and the St. Venant
torsional stiffness, so it tends to be conservative..
For the simple case of a beam with constant bending moment and lateral
support of the compression flange at the beain ends, the flange behaves like
a pin-ended column. The· general case is more complicated, because the
axial force in the flange can vary along the beam length and the lateral
supports can have different types and locations.
6.25.6 Effect of Shear Force on Beam Buckling
The shear forces in a beam can also contribute P-A effects. This section
attempts a reasonably simple physical explanation.
Figure 6.68{a) shows a beam with both bending and shear force, Consider
the forces in a length of beam equal to the distance between the twc> P~
cables. A different length cowd be used, and this length is chosen because it
simplifies the mathematics. Since the length is short, the P-8 cables can be
replaced by P-A struts,
Figure 6.68{b) shows the bending ·moment and shear force acting cin the
length of beam. The moment M is the moment at the middle of the length.
In Figure 6.68{c) the moments are replaced by equivalent couples; As shown
in Figure 6.68{d), the moment and shear are equivalent to forces in the P~A
struts plus a self-equilibrating set of shear force5. The forces in the P-A struts
account for the P-A contribution from the bending moment. There is also a,
P-A contribution from the set of shear forces.
332
Chapter 6
P-~ Effects,
Stability and Buckling
!
r··-;---1.
__ £_ -- -----------e
(a) Length of beam with P-a columns
(b) Moments and shear forces
(c) Equivalent forces for moments
.Y....
v
p
~1[:][ ·tlf~ ll·
p
l.e.I
p
(d) Forces in P-a Columns
•
7
v
'(2
72/
/.Y....
.Y_'\,.
><
12
12
(e) Shear forces, and equivalent
forces in diagonal P-.i Columns
Figure 6.68 P-~ Columns for Shear Force
As shown in Figure 6.68(e), the set of shear forces is equivalent to a pair of
diagonal tension and compression forces. These forces can be applied to a
pair of diagonal P-A struts, as shown. When the beain is bent or twisted,
there are P-A contributions from both the horizontal and diagonal P-A struts.
H the beam bends laterally, or if the beam twists, the horizontal P-A struts
exert forces on the main structure cable. These forces account for the P-A
contribution of the bending moment. The diagonal P-A struts also exert
forces, which account for the P-A contribution of the shear force. Figure 6.69
shows these forces when the beam twists.
Figure 6.69(a) shows the forces on the P-A struts. Figure 6.69(b) shows how
twisting of the beam causes these struts to exert forces on the main
structure. As an exercise, you might like to see what forces are associated
with lateral bending of the beam.
Bracing to Prevent Buckling
v
.121" ..Y.i;
333
/
2e
(a) Diagonal P-~ columns
(b) P-A forces when beam twists
Figure 6.69 P-A Contribution of Shear Force
6.25.7 Pradical Modeling
In an analysis model, P-A effects for beam buckling are similar to the P-o
contribution for column buckling. As With a column, the P~ contnbution is
not easy to model.
La~al-torsional effects can be modeled fairly easily, by dividing a beam .
into short elements and using the equivalent of P-L\ struts in each element.
This is feasible for long span structures, but not for building frames. For
most structures, lateral-torsional buckling is not accounted for directly in
the analysis model. Rather, -it is accounted for using. simplified, and
generally conservative, formulas for the buckling strength, and by
providing lateral bracing at appropriate locations.
6.26 Bracing to Prevent Buckling
Columns, beams and other components are often braced laterally to prevent
buckling. The locations, stiffnesses and strengths of braces can be an
important aspect of design.
Consider stiffness first Figure 6.70(a) shows an elastic, straight column with
a brace at mid-height H there is no brace, the effective length of the column
is L. H the brace is rigid, the buckling mode shape is as shown in Figure
6.70(b), and the effective length is 0.5L. This is also the case if the brace is not
rigid but has a. stiffness that is larger ·than some threshold value. H the brace·
stiffness is smaller than this threshold, the buckling mode shape has the
334
Chapter 6 P-d. Effects, Stability and Buckling
form shown in Figure 6.70(c), and the effective length is between Land O.SL.
The threshold stiffness for an ideal column is not difficult to calculate.
O.SL
O.SL
(a) Column with
(b) Buckling with
lateral brace '
stiff brace
(c) Buckling with
flexible brace
Figure 6.70 Buckling of a Column With a Lateral Brace
H the column is perfectly straight, and if it buckles anti-symmetrically as in
Figure 6.70(b), there is theoretically no force in the brace. In this case,
therefore, the brace stiffness is important but not its strength. This is not the
case in Figure 6.70(c). In this case the strength of the brace can affect the
post-buckling behavior, with behavior similar to that considered in Section
6.6.4. The strength of the brace can also ·affect the buckling strength when
there are imperfections in the column. In this case the behavior is similar to
that considered in Section 6.6.7. In general both the stiffness and the
strength of a bra.ce are important.
An indication of the required stiffness can be obtained from analysis, for
example from analysis of a column such as that in Figure 6.70. An indication
of the required strength can be obtained from analyses of columns with
initial imperfections. The required strength for a brace can be similar to a
notional load.
Brace location can also be important. In some cases a brace must be at a
specific location. In other cases a brace must perform its function but its
location may be flexible. An example is a brace to prevent lateral-torsional
buckling of a beam, which may not need to be at a specific location. Brace
locations may be determined by where they can be most effective. For
example, when a plastic hinge forms in a beam, there is a loss of stiffness in
the hinge region and an increased chance of lateral-torsional buckling in
78BP-· Effects in Seismic Isolators
335
that region. Hence, braces may be more effective close to the hinge region
than near mid-span.
Adequate bracing is important, and even critical, for design, but it is
unlikely that the brace behavior will be modeled directly in an analysis.
Rather, braces are usually proportioned and located based on rules in
design codes or on a designer's judgment.
6.27 P-A Effects in Seismic Isolators
6.27 .1 Overview
Seismic isolators can have large deformations, and the resulting change in
shape can cause geometric nonlinearity, with P-A effects that are both
interesting and substantial.
This section considers friction-pendulum isolators first. In this case the P-A
effect is straightforward. Rubber-type isolators are considered second. In
this case the P-A effect is not so straightforward.
Although it is not considered here, it is possible to develop geometric
stiffness matrices for isolator elements. It is also possible for an isolator to
become unstable and buckle.
6.27 .2 Friction-Pendulum Isolator with Flat Sliding Surface
Figure 6.71 shows a simplified frictio~-pendulum isolator.
Shear Force
Node
~
·-·-·-·f::Rigid link
.. Zero length
I--'----_,_~
1------4
(a) Isolator
sliding "hinge•
·-·-·-· · ·Node.
Increasing compression
,.------+-
I
Rigid link
(b) Element
Sliding Displacement
(c) Hinge F-0 relationship
Figure 6.71 Model for a Friction-Pendulum Isolator
336
Chapter 6 P-d Effects, Stability and Buckling
Figure 6.71(a) shows the isolator, located between a foundation structure
below and a. building frame above. The isolator in this figure is simplified
because the sliding surface is flat, whereas real isolators have curved sliding
surfaces. This is considered later.
Figure 6.71(b) shows a model of the isolator, considering it as a single
element made up of three components. This model has the following
features.
(1) · Nodes are located at the axes of the beams above and below the
isolator.
(2)
A zero length sliding hinge is located at the level of the sliding surface.
(3)
The hinge force-deformation relationship for sliding is shown in
Figure 6.7l(c). The hinge is very stiff before sliding begins. The
strength in sliding depends on the bearing (vertical} force -and the
friction coefficient.
(4)
The hinge has a small amount of vertical flexibility in bearing, and no
tension strength.
·
(5)
In between the nodes and the sliding hinge there are rigid links.
It is not difficult to form a small displacements stiffness matrix for this
. element. However, the concern here is the p..:L\ effect.
For simplicity consider the 2D case, where each node has three possible
displacements, namely rotation, vertical displacement and horizontal
displacement. Usually the rotations are small, and unless there is lift-off the
vertical deformations are also small. Figure 6.72(a) shows the most
important deformation, corresponding to sliding of the hinge.
ff P-L\ effects are ignored, the bending moment diagram is as· shown in
Figure 6.72(b). The bending moment at the element end is the horizontal
(slip) force multiplied by the appropriate lever arm. If P-L\ effects are
considered, there is an additional moment at the base of the element equal
to PL\. In this case the bending moment diagram is as shown in Figure
6.72(c).
The additional moment at the base can significantly increase the bending
moment demand on the supporting structure, and should be considered in
the design.
P-A Effects in Seismic Isolators
337
~Mb
(b) Moment diagram
(a) Element end
forces
with no P-A
(c) Moment diagram
with P-A
Figure 6.72 P-A Effect in Friction-Pendulum Isolator
6.27.3 Fridion-Pendulum Isolator with Curved Surface
Figure 6.73(a) shows an isolator with a curved (spherical) sliding surface.
Shear Force
Node
Hardening stiffness
depends on curvature
of sliding surface
·-·-1·-
Sliding hinge is
..............__ _..,._-+
flat, not curved
·-·-· . ·i--------+
Node
(a) Isolator and element
.
Sliding Displacement
(b) Hinge F-D relationship
Figure 6.73 Isolator With Curved Sliding Surface
The model is the same as before, except that the hinge force-deformation
relationship for sliding has a hardening stiffness, as shown in Figure 6.73(b).
Hardening occurs because the slope of the sliding surface gets larger as the
slip increases. This is a geometric effect. However, it is usually modeled as
material nonlini:!arity.
·
As the slip. increases the sliding part of the isolator also moves vertically
upward. This is also a geometric effect, and it is associated with true large
displacements. For small displacements analysiS, and also for P-a analysis,
this vertical displacement is ignored.
338
Chapter 6 P-ti Effects, Stability and Buckling
6.27.4 Rubber-Type Isolator
Figure 6.74(a) sh~ws a rubber-type isolator (plain rubber or rubber with a
lead core).
Shear Force
le
Shear Displacement
(a) Isolator
(b) Element
(c) Hinge F-D relationship
Figure 6.74 Model for a Rubber-Type Isolator
Figure_ 6.74(b) shows a possible model. This is similar to the model for.a
friction pendulum isolator, with the following differences.
(1)
The sliding hinge is replaced with a shear hinge. This hinge has a
force-deformation relationship that depends on the isolator dimensions
and material properties.
(2)
The shear hinge is located at mid-height of the isolator component.
Figure, 6.75(a) shows_ the most important deformation of the element,
corresponding to shear deformation of the isolator.
H
~Mb
(a) Element,end
forces
Ma= Ha
Ma= Ha+ 0.5P.1
\h
(b) Moment diagram
(c) Moment diagram
with no P-ti
with P-ti
Figure 6.75 P-ti Effect in Rubber-Type Isolator
P-A Effects in Seismic Isolators
339
Figures 6.75(b) and {c) show the bending moment diagrams for small displacements and P-A analyses; respectively. Figure 6.75{c) is similar to Figure
6.72{c), except that one half of the P~A mo~ent acts at the top of the element
and one half at the bottom.
6.27.5 Alternative Model for Rubber-Type Isolator
A rubber-type isolator can also be·.modeled as ati equivalent beam, rather.
than a zero-length hinge. In this case the beam is'a."shear beam", with a very
large bending i;tiffness and a shear stiffness to match the stiffness of the
isolator. The beam is inelastic in shear, to match the force-deformation
relationship of the isolator.
Figure 6.76{a) shows the element model, with a deformation co~g
to that in Figure 6.75{a). Note that the shear beam has shear deformation
only, with negligible bending deformation.
·
M8
=Ha+0.5PA
7
+-.
V=H+ PA/h
~Mb
(a) Element end
forces
(b) Model with
P-A strut
Mb= Hb+0.5PA
(c) Shear and moment
diagrams
Figure 6.76 P-A Effect in Rubber-Type Isolator
Figure 6.76{b) shows one way that the P-A effect might be accotinted for,
using a single P-A strut extending over the full length of the element. Figure
6.76{c) shows the shear force and bending moment diagrams for the small
displacements element, accounting for the P-A effect. The following are the
key points.
(1)
The P-.1 strµt exerts horizontal forces on the small displacements
element, increasing the shear force as shown, and hence also increasing
the bending moments.
340
Chapter 6 P-i\ Effects, Stability and Buckling
(2)
The end moments are the same as in Figure 6.75(c). Hence, the P-fl
moment effect is the same as in the model with a zero-length shear
hinge.
(3)
The shear force in the isolator is larger than the shear force in the
model with a zero-length shear hinge.
Hence, the model with a shear beam and a P-fl strut is different from that
with a shear hinge. To provide the same relationship between Hand fl for
the element, the isolator in the shear beam model must be stronger than in
the shear hinge model.
·
Figure 6.77(a) shows the same basic model as in Figure 6.76(a), but with a
different model for the P-fl effect, using three P-fl struts rather than one.
H .
M8
=Ha + 0.5Pil
~•P~~
H
~Mb
(a) Element end
forces
(b) Model with
P-Li struts
Mb= Hb + 0.5Pil
(c) Shear and moment
diagrams
Figure 6.77 P-i\ Effect in Rubber-Type Isolator
This is a more accurate model, since it accounts more accurately for the P-fl
~ffect. Figure 6.77(c) shows the shear force and bending moment diagrams
for this model. The bending moments at the element ends are the same as in
the preceding two models. However, the shear force in the isolator is
increased even further.
6.27.6 Which Model is Correct?
Assuming that P-fl effects in rubber type isolators are significant, the models
in Figures 6.74 and 6.77 both seem to be reasonable. However, they give
different shear force demands on the isolator. As an exercise you might like
to determine which model is more correct.
P-~ Effects in
Seismic Isolators
341
The following are some points to consider, .
(1)
(2)
Are P.;i\ effects significant, or are they small enough to be ignored?
The relationship between shear force and shear deformation is usually
. determined experimentally, with a test specimen that consists of two
identical isolators, as shown in Figure 6.78(a). The vertical load, P, is .
applied first and held constant, and the h~rizontal load, H in each
isolator, is applied. The upper platen of tlle test machine does not
rotate, so there must also be a moment, M. Assuming zero rotation at
the top and bottom, the shear force and bending moment diagrams
are as shown in Figure 6.78(b). There are P-L\ effects in the test. In
particular, for an external load as shown the shear force demand per
isolator is larger than H.
(a) Test procedure
(b) Shears and moments
Figure 6.78 Test Method for Isolator F-0 Relationship
(3)
There are bending moments, and hence bending stresses, in the
isolators. However, an isolator is not a slender beam, and the bending
stresses almost certainly do not vary linearly· across the width of the
isolator. The distnbutiori of shear stresses is also likely to be complex.
(4)
The P-i\ effect increases the shear force on an isolator, or alternatively
decreases its.shear stiffness and strength. Other things being equal, if,
two tests are carried out with different P values, the test with the
smaller P should give larger values for stiffness and strength.
However, other things are not necessarily equal. In particular there .
may be P-M-V interaction in an isolator, and a larger P may increase
the shear stiffness and strength. If so, this could counteract ariy
stiffness and strength reduction from the P-i\ effect.
342
Chapter 6
P-~
Effects, Stability and Buckling
6.28 Some Other Types of Buckling
Some other types of buckling are as follows.
(1)
Torsional buckling under axial load. Some members, notably compression members with angle sections, have a small torsional stiffness and
may buckle in a torsional, rather than a flexural, mode.
(2)
Local buckling of a slender flange or web in a steel member.
(3)
Buckling of an unstiffened edge in a concrete shear wall.
(4)
Buckling of steel bars in a reinforced concrete member. This can occur
under large cyclic deformations, where a bar yields in tension, then
buckles in compression when the deformation is reversed.
These types of behavior can rarely be accounted for directly in an analysis
model.
6.29 True Large Displacements
P-8 theory is accurate enough for the vast majority of second order analyses~
However, a true large displacements analysis may be needed in some cases,
mainly when it is necessary to account for the "catenary effect" (~ Section
1.7). Some such cases are as follows.
(1)
Cable structures, such as cable roofs.and cable-stayed masts.
(2)
"Incremental collapse" or "disproportionate collapse" analysis of buildings that have been damaged by explosions, where there can be
substantial catenary effects in the floor systems.
(3)
Buried pipelines, where the catenary effect can be. substantial (see
Section 6.24.7). In this case it is possible, and important, to model the
post-buckling behavior.
(4)
Post-buckling analysis of columns, beams or complete structures. P-8
. theory is usually adequate until buckling occurs. Any post-buckling
analysis is likely to be very approximate, and may be of only
academic interest
. (4)
Possibly the collapse of building structures under earthquake loads.
P-8 theory is almost certainly adequate up to initiation of collapse.
Beyond that point the displacements can be large, but any analysis is
Conclusion for this Chapter
343
likely to be extremely approximate, and the analysis results are
· unlikely to be of much practical value. •
A large displacements analysis must account for the nonlinear relationship
between node displacements and element deformations (P-A analysis
assumes a linear relationship). Because of the change in shape of the
structure, the structure stiffuess continuously changes, and linearized
analysis is not possible. Also, in a structure where iteration would be
required for a P-A analysis, a large displacements analysis is likely to
converge more slowly, with more iterations and a longer execution time.
Large displacements analyses also tend to be more sensitive numerically,
and if there is material nonlinearity as well as geometric nonlinearity it can
be more difficult to obtain a solution.
Since it is computationally more expensive, large displacement analysis
·should be used only when necessary. Some computer programs may automatically use large displacements theory for second order analysis, even
.though it is not needed.
6.30 Conclusion for this Chapter
From a design viewpoint, the most important aspect of geometric nonlinearity
is that it can increase deflections, increase member strength and deformation demands, and decrease structure strength. In some cases it does the
opposite, but these cases are relatively rare.
Geometric nonlinearity can affect structural b~vior at multiple levels,
from the overall structure to single member to a cross section. Some
aspects of behavior can be considered directly in an analysis model, but
most can not, and they must be accounted for indirectly.
a
Direct modeling of P-A effects can be complex, and the analysis results can
be sensitive to the modeling assumptions. It is important for design_codes to
provide modeling guidelines, so that the analysis models used by different
engineers are consistent.
.P-A analysis is interesting academically. However, the goal is design, not
analysis. For strength-based design using elastic analysis, the purpose of a
P-A analysis is not to account exactly for P-A effects but to calculate them
with sufficient accuracy for d~sign purposes, using as simple a procedure as
possible. Complex models and analysis methods can give a false impression
of accuracy. Over-reliance on analysis, and especially "design by analysis",
'is not a good idea.
CHAPTER
7
Some Other Aspects of Behavior
The preceding chapters have considered analysis models, .the' Direct
Stiffness Method, and material and geometric nonlinearity. There are many
other aspects of structural behavior, and many other tools that can be used
for analysis and design. This chapter considers a few of them.
7.1 Plastic Mechanisms
7.1.1
~ollapse Mechanism vs. Plastic Mechanism
Figure 7.l(a) shows a building frame with gravity and lateral loads.
... + ***•
(a) Frame and hinge locations
(b) Collapse mechanism
(c) Plastic mechanism
Figure 7.1 Collapse arid Plastic Mechanisms
H the gravity load is applied, and then the lateral load is progressively
increased, the frame will yield and ·ultimately collapse. Assume that yielding can be modeled using plastic hinges; and ignore P-~ effects.
345
346
Chapter 7 Some Other Aspects of Behavior
As the lateral load is increased, plastic hinges will form progressively, in the
beams and/ or in the columns. The older of hinge formation is not important for this discussion. Suppose that hinges form at the beam ends and at
the column bases, as shown in Figure 7.l(a).
When all of the hinges have formed, the frame collapses. In effect, it forms a
mechanism, as shown in Figure 7.l(b). This is commonly called a "collapse
mechanism". Note that in this mechanism the beams and columns are
straight. This does not mean that the beams and columns are actually
straight when collapse occurs - they are likely to have substantial
curvatures, as shown in Figure 7.l(c). Rather, it means that after a
mechanism forms, all additional deformations are in the hinges, with no
increase in the beam and column curvatures (this assumes that there is zero
strain hardening in the hinges).
A collapse mechanism is a useful concept, but it can give a false impression
of the behavior. The reasons are as follows.
(1)
A mechanism with straight beams and columns is a rigid-plastic
mechanism, where all hinge rotation occurs after the mechanism
forms. This ·can give a false impression of the relative amounts of
rotation at different hinges. In an actual mechanism the hinges form
progressively. Some hinges can have relatively more rotation than
indicated by a collapse mechanism, and some less.
(2)
It is possible that some hinges will reach their ductile capacities before
a mechanism forms, and begin losing strength. A collapse mechanism
does not take this-into account.
(3)
Because of P-A effects, a structure can become unstable before a complete mechanism forms. A collapse mechanism does not take this into
account.
It follows that if a collapse mechanism is used to calculate the strength of a
structure, the calculation will tend to overestimate the strength, possibly by
a substantial amount. A collapse mechanism can be useful conceptually, but
it is not a reliable method for estimating the strength of a structure.
An alternative is to think in terms of a "plastic mechanism" that includes
both hinge rotations ~d curvatures in the beams and columns, as shown m
Figure 7.l(c). This type of mechanism develops when a nonlinear push-over
analysis is tised to calculate the strength of a structure. The analysis accounts
Plastic Mechanisms
347
for progressive hinge formation, can consider limited hinge ductility, and can
account for P-~ effects.
7.1.2
Push-Over Analysis for Earthquake Loads
For a structure subjected to a large earthquake, substantial inelastic behavior
is usually allowed, and the structure may form a plastic mechanism.
Earthquake loads are dynamic, but static push-o\rer analysis can still be a
useful tool for performance evaluation. The 'steps in the analysis are
essentially as follows.
(1)
(2)
(3)
(4)
(5)
Apply the gravity load.
Choose a distribution for the lateral load, but not a magnitude. Apply
the lateral load, progressively increase its magnitude until a plastic
mechanism forms, and hence calculate the nonlinear relationship
between lateral load and lateral displacement. This is the "push-over
curve". It provides a measure of the lateral strength capacity of the
structure.
Assume that for dynamic earthquake loads the structure will form the
same plastic mechanism. Given the push-over curve and a response
spectrum for the earthquake ground motion, estimate the lateral
displacement of the structure. This is the displacement demand on the
structure, or the "target displacement". There are procedures for
calculating this displacement. The details are not important for this
discussion.
At the target displacement, get the hinge rotations from the push-over
analysis. These are hinge rotation demands.
At each hinge, compare the rotation demand with the rotation capacity.
If the demand is smaller than the capacity for all hinges, the·
performance is satisfactory.
·
For a low rise building or a short span bridge, it can be reasonable to
assume that the plastic mechanism for dynamic earthquake loads is the
same as that calculated in a push-over analysis. In such cases, push-over
analysis can be an effective tool for performance evaluation.
In a tall building, however, or in a. long span bridge, the dynamic response
is usually more complex, involving substantial "higher mode" displacement
patterns, not just the displacements associated with a single plastic mechanism.
Hence, if a plastic mechanism for a tall frame building is calculated using
pushover analysis, this mechanism may never form, when the building is
subjected to dynamic earthquake load. Plastic hinges will form, but the
348
Chapter 7 Some Other Aspects of Behavior
hinge distnbution and hinge rotations may be very different from those in
the static plastic mechanism. In such cases, static push-over analysis is not
an accurate tool for formal performance evaluation. However, it can still
give useful information on the inelastic behavior of a structure.
7.1.3
Desirable and Undesirable Mechanisms
In :the plastic mechanism for a frame structure, the plastic hinge locations
depend on the beam and column strengths, and on the graVity and lateral
loads. Figure 71 shows two possible mechanisms for a simple frame.
c!I
.(a) Beam mechanism
E;
Ell
(b) Story mechanism
Figure 7.2 Desirable and Undesirable Mechanisms
Figure 7.2(a) shows a "beam" mechanism. In this case there are plastic
hinges in all beams, in .the columns at the base, and in the center column at
the roof (where there are two beams and only one column). Figure 7.2(b)
shows a "story" mechanism, where plastic hinges form at the top and
bottom of all columns in a single story.
The beam mechanism is more desirable, for the following reasons.
(1).
Plastic hinges in beams tend to be more ductile than hinges in columns,
mainly because beams have small axial forces.
(2)
There are many hinges, which provides more energy dissipation as
the hinges rotate. In effect this provides damping, and tends to reduce
the deflection of the structure.
(3)
A beam mechanism is less sensitive to P-A effects· than a story
mechanism.
Hplastic hinges are to form in the beam8, the columns must be stronger than
the beams (i.e., this is a "strong column, weak beam" mechanism). Also, the
Plastic Mechanisms
349
beam-to-column connections must be stronger than the beams. To calculate
the required column strength, it must be possible to calculate the bending
strength fairly accurately at each hinge location. This may not be easy if.a
beam has a composite slab. In addition, the column cross sections at the base
must be designed to provide reliable plastic hinge action under cyclic load.
The story mechanism is less desirable, and should usually be avoided in
building frames. However, the equivalent of a &tory mechanism is often
allowed in bridge structures.
7.1.4 ·Lateral Strength Calculation Given a Mechanism
An interesting property of a mechanism is that the collapse load ·can be
calculated using only equilibrium. This is illustrated in Figure 7.3.
Load=P2
Total load = P
:ri.i * *.1
(a)
(b)
Figure 7.3 Equilibrium in Collapse Mechanisms
Figure 7.3(a) shows the loads, mechanism and hinge strengths for a simple
frame. From equilibrium of each column as a free body, the shear force in
each column is 2M/h, where M is the hinge moment and h is the story
height. Hence, the lateral load,H, is 4M/h. This is the opposite of a usual
analysis, where the external loads are known and the element forces are .
calculated. In this case the element forces are known and the external loads
are calculated.
The calculation can be extended to account for the P-L\ contribution,
provided the drift can be estimated. For a gravity load P and a drift L\, the
lateral load is H =. 4M/h-P!J./h.
The concept is the same for the frame in Figure 7.3(b), but the calculation is
more complicated. The hinge moments M 1 through M,4 are known. The
350
Chapter 7 Some Other Aspects of Behavior
gravity loads are known, and the lateral load distribution is knoWll, with
loads of relative magnitude H1, H2 and H31 but the load factor, '). , , is not
known. The value of ')..,,can be found using free bodies and equilibrium
equations, but this can be tedious. It is much better to use the Virtual
Displacements Principle. The steps are as follows.
(1)
(2)
(3)
Impose a small displacement on the mechanism. For this displacement the beams and columns remain straight and the plastic hinges
rotate. This can be done because the displacement is virtual (or
imaginary), not real. Using geometric compatibility, calculate the
hinge rotations.
Since the mechanism is in equilibrium and the displacements are
small, the extemai work done by the loads is equal to the internal
work done by the hinge moments moving through the hinge rotations.
This is the Virtual Displacements Principle. The bending moments
along the beam and column lengths do no work because there are no
curvatures. Knowing the moment at each hinge, and knowing the
lateral load distribution, use this principle to calculate the lateral load
magnitude.
To account for P-d effects, estimate the story drifts. Knowing the
gravity loads, calculate the horizontal P-d loads at each floor. These
are exactly the "notional loads" used·in P-d analysis. Include the work
done by these loads in the external work.
A complication for a column hinge is that the plastic moment is affected by
P-M interaction, and as lateral load is applied the axial force, P, can change.
For practical purposes the hinge moment can be based on the axial force for
gravity load alone.
After the lateral loads are known, the bending moment and shear force
diagrams can be constructed. These define the strength demands within the
beam and column lengths. The strength capacities at all locations other than
the hinges must exceed these strength demands. If this condition is not
satisfied, hinges will form at locations other than those assumed, and the
assumed mechanism is not correct.
This property of a mechanism is used in the "Direct Displacement-Based
Design" method for earthquake loads, where "direct" means that a final
design is obtained with little or no iteration. .This method is dramatically
different from conventional strength-based design and is well worth
studying for engineers who are interested in earthquake resistant design.
The method is described in the book Displacement-Based Seismic Design Of
Structures by Priestley, Calvi and Kowalsky, IUSS Press, Pavia, Italy, 2007.
Mechanism Control Using Capacity Design
7.1.5
351
Other Causes of a Mechanism
The plastic mechanism in a real structure can involve much more than
moment hinges. There could be shear hinges, connection failure, column
buckling, beam lateral-torsional buckling, etc. If such modes of behavior are
allowed, they can often be difficult to model. Undesirable modes of behavior,
or modes that are difficult to model, can be avoided. by using capacity
design.
7.2 Mechanism Control Using Capacity Design
7.2.1
Concept
As considered in earlier chapters, one way to obtain better structural
performance (and also to simplify the modeling and analysis) is to prevent
undesirable behavior by using capacity design. This can be regarded as
mechanism control, where the designer chooses the desired plastic mechanism, then designs the structure to provide that mechanism.
For example, if the desired mechanism for a frame is a beam mechanism, as
in Figure 7.2(a), the columns are deliberately designed to be stronger than
the beams (a strong column, weak beam design). This simplifies the
modeling because the column elements in the analysis model can be elastic,
with no column hinge components. The major design issues are to provide
enough ductility in the beam hinges (and in the column hinges at thefoundation level), and to provide enough column strength to make sure that
hinges form only in the beams. There must also be no connection failure, no
lateral-torsional buckling, etc.
7.2.2
Examples of Mechanism Control
Some examples of mechanism control are as follows.
(1)
Jn an unbraced frame building, a strong column weak beam mechanism
can be desirable. The columns should not yield significantly, except at
the base and top of the frame. It may be reasonable to allow a few
columns to hinge, but a story mechanism must be prevented. A key
problem is to determine how strong the columns need to be, without
being too conservative. The sum of the column strengths at a beam-tocolumn connection must be some multiple of the sum of the beam
strengths. A multiple of 1.0 will usually allow a column hinge to form,
352
Chapter 7 Some Other Aspects of Behavior
since the beams to left and right of a column cari bOth yield· and the
column moments above an!i below the beam·are unlikely to be equal.
A multiple of 2.0 allows one of the column moments to be zero, and
intuitively would seem to be conservative. However, some analyses
suggest that this may not always be· the case. Design codes may
specify the multiple, based in part on inelastic dynamic analyses of
typical frames. A more direct approach is to carry out an inelastic
dynamic analysis.
(2)
In an eccentrically braced frame, the plastic mechanism usually allows
only the link beams to yield, usually in shear and not in bending.
There should be_ little or no yield in the other beams and the columns.
The diagonals should remain essentially elastic and not buckle.
(3)
In a frame with buckling restrained braces, the plastic mechanism
usually allows only the J?races to yield. The columns must be strong
enough to force yield into the braces. H the brace strengths are matched to the story shear demands, there is less demand on the columns.
However, story shear demands may be difficult to estimate
accurately.
(4)
In a base isolated building, the plastic mechariism usually allows in- .
elastic behavior only in the isolators. The btniding above the isolators
remains _essentially elastic.
(5)
In ~bridge Witji ~lation at the tops of the columns, the plastic mechaniS,m usually allows inelastic behavior only in the isolators. The goal
of the isolation is usually to reduce the forces on the columns and
foundations. The structure above and bel<>W the isolators should remain
essentially elastic.
It may not be easy to identify a desirable and practical mechanism for a .
complex structure. It can be particularly difficult for retrofitting an existing
building. However, controlling the mechanism should be a goal. Relying on
analysis to identify the mechanism is not a good idea.
7.2.3
Higher Mode Effects in Tall Buildings
~ noted earlier, for a low tise building it can be reasonable to assume that .
the plastic mechanism for dynamic earthqllake loads is the same as that
calculated in a push-over analysis. This means that the designer can select
the plastic mechanism, and when the building is loaded dynamic8.lly it is
likely to deform in the shape defined by that mechanism. In a tall building,
Static Indeterminacy and Redundancy
353
however, the dynamic response is usually more complex, with substantial
higher mode effects. If a plastic mechanism is chos~ for a tall frame, the
frame may deform primarily in that mecharusm, but there can be other
deformations that contribute substantial element forces. These forces are not
considered when static push-over analysis is used.
One example· is a· tall shear wall building. It is usual to allow flexural
hinging. at the base, and to require that the wall otherwise remains
· essentially elastic. For a tall rectangular wall the chosen plastic mechanism
is usually simple, with a single P-M hinge at the base. (This may not be a
simple hinge for a shear core with a complex cross section, but the design
concept is the same.) If the earthquake loads are assumed to be static, for
any given gravity and lateral loads the bending moment and shear force
diagrams for the wall can be calculated, using static push-over analysis.
These are the calculated bending moment and shear force demands,
including the shear force demand in the plastic hinge region. It is possible,
however, for these demands to be substantially smaller than those
calculated using inelastic dynamic analysis, The reason is that they do not
account for higher mode effects.
·
One way to account for higher mode effects might be to run an elastic
dynamic analysis, and scale the results so that the bending moment at the
base .of the wall is equal to the hinge strength. However, this procedure can
also underestimate the moment and shear demands. The reason is that the
higher mode shapes change when a hinge forms at the base of the wall, arid
elastic
these changes can cause larger forces than the higher modes in
model. To calculate the strength demands for a tall building it may be
necessary to run an inelastic dynamic analysis
an
7.3 Static Indeterminacy and Redundancy
7.3.1
Statically Determinate Structure
If a structure is statically determinate, for any set of external loads there is
only one set of element forces that satisfies equilibrium. Some other aspects
of behavior are as follows.
(1) · The element forces can be calculated using only equilibrium (ignoring
P-8 effects). Hence, the element forces do not depend on the element
stiffnesses, and the element forces are the same whether the elements
·
have linear or nonlinear behavior.
354
Chapter 7 Some Other Aspects of Behavior
(2)
The calculated displacements do depend on the element stiffnesses,
and they are also affected by nonlinear element behavior. •
(3)
Movement of the supports causes the structure to displace, but does
not cause any element forces.
(4)
Thermal expansion loads cause the structure to displace, but do not
cause any element forces.
(5)
If any element factures, the structure becomes a mechanism and can
collapse.
7.3.2 Statically Indeterminate Structure
If a structure is statically indeterminate, for any set of external loads there is
an infinite number of element force distributions that satisfy equilibrium.
Some other aspects of behavior are as follows:
(1)
To calculate the element forces it is necessary to consider compatibility (continuity) and the element F-D relationships as well as equilibrium.
Hence, the element forces depend on the element stiffnesses. If the
elements have inelastic behavior, the element forces will change as the
elements progressively yield (i.e., there will usually be a redistribution of forces within the structure).
(2)
The element forces depend on the relative values of the element
stiffnesses, not their absolute values. If the element .stiffnesses all
change in the same proportion, the elemenflorces do not change. If
the element stiffnesses change in different proportions, the element
forces will change. Some structures may be relatively insensitive to
changes in relative stiffness. For example, iii. an unbraced frame the
shear in a story is resisted by column, bending, and the column moments
depend on the lateral loads rather than the column stiffnesses. In
general, however, uncertaiii.ty in estimating the stiffness should be
considered. As with a statically determinate structure, the calculated
displacements depend on the absolute stiffnesses.
(3)
If an element fractures, the element forces may be able to redistribute,
developing an alternative load path, and collapse does not necessarily
occur, Statically indeterminate structures· tend to be more resilient
than statically determinate structures
Static Indeterminacy and Redundancy
355
(4)
Movement of the supports generally causes element forces. However,
these forces are different from those caused by external loads. If a set
of external loads is large enough, it can cause the structure to form a
mechanism, by yieldiilg or buckling. Support movements can also
cause yield or buckling, and might form a mechanism, but collapse
does not necessarily occur. The difference is that the amount of yield
caused by a sustained external load can continue to increase after a
mechanism forms, leading to collapse. For support displacements,
however, once the amount of yield has accommodated the support
displacement, yielding stops, and collapse need not occur even if
there is a mechanism. Support displacement is a "self limiting" load,
not a "sustained" load.
(5)
Similarly, thermal expansion generally causes element forces, but
thermal expansion loads are self limitmg, not sustained.
(6)
In a statically determinate structure, for a given set of external loads
there is only one set of .element forces that satisfies equilibrium
(assuming small displacements). In a statically indeterminate structure
there is an infinite number of element force sets that satisfy equilibrium.
The result of any structural analysis must satisfy equilibrium.
However, there can be substantial uncertainty about the element
stiffnesses, support movements can affect the element forces, thermal
expansion and material shrinkage or creep can. have significant
effects, and it can be argued that it is not essential to satisfy
compatibility exactly. If two different sets of element forces both
satisfy equilibrium, it can be impossible to say whether one set is
more accurate than the other.
(7)
If the external loads on a statically determinate structure are zero, the
element forces are zero. This is not the case for a statically
indeterminate structure. A statically indeterminate structure can have
an initial force state where the external loads are zero but the element
forces are noJ - they form a self-equilibrating set of internal forces.
There is an infinite number of possible states. If a structure has initial
forces, these forces usually will have little or no effect on the elastic
behavior of the structure. If external loads are applied and the
structure is linear, the element forces caused by the loads will be the
same as if there. were no initial forces. If, however, the structure
yields, the initial forces can have a substantial effect. When an
external load is applied the total element forces are the forces caused
by this load plus any initial forces that may be present. If the initial
forces are large, the external load required to cause yield in an
356
Chapter 7 Some Other Aspects of Behavior
element can be substantially different from the load when there are no
initial forces. Some elements will yield earlier, and others later.
The effect of initial forces in a statically indeterminate structure is illustrated
by the simple example in: Figure 7.4.
FA
8 -i--r----,i.. _.
4
H, A.~.
BarA
2 4+---4--'--4--"--
~
~....,_
Ds
_ _ Initial state
(a) Element F-D relationships
123456
(b) Structure H-A relationship
Figure 7.4 Simple Example to Illustrate Effect of Initial Forces
The structure consists of two nonlinear bars in parallel, as shown in Figure
7.4(a). The force-deformation relationships for the bars are as shown. The
H-A relationship is obtained by adding the forces in the two bars (i.e., H = FA
+ FJ. Both bars shorten the same amount. Hen~e A= DA =D8 •
Consider two cases, one With no initial forces and the second with initial
forces as shown in Figure 7.4(a). For the case with initial forces, Bar A has an
initial compression force of 4 units, and Bar B has an initial tension force of
the same amount. These are self-equilibrating forces.
When a bar has an initial force, the effect is to change the origin of the F-D
relationship. For Bar A the effective origin is at F = 4, D =1 in compression,
and for Bar B the effective origin is at F = 4, D = 2 in tension.
Figure 7.4(b) shows the H-A relationships for the two cases. The dashed line
is for the case with no initial forces, and the solid line is for the case with
initial forces. As the figure shows, the initial forces have a dramatic effect on
the behavior of the structure, reducing both its strength and its ductile limit.
It is an interesting exercise to consider inelastic F-D relationships with
hysteresis loops, and to see how initial forces can affect the cyclic behavior.
Nonstructural Components
357
In this example the initial forces have a larger effect than might be expected
in a real structure. :Nevertheless, initial forces can have significant effects on
the inelastic behavior (an example is the effect of residual stresses on the
buckling behavior of a steel column). For strength-based design using elastic
analysis it is usual to account for support movements, thermal expansion
and creep. For inelastic analysis, however, it is usual practice to ignore
them, arguing that after the structure yields (and especially after it yields
cyclically), the initial forces have little effect. Given the many other
assumptions this is probably reasonable, but it emphasizes· again that
structural analysis is at best approximate.
7.3.3
Redundancy
A statically indeterminate structure iS "redundant" ht the sense that it can
have multiple load paths, and can redistribute forces internally so that
failure of one member does not necessarily cause the structure to collapse.
In the Force Method of structural analysis the primary unknowns are forces
.rather than displacements, and the "degree of static indeterminacy" or the
"degree of redundancy" is the number of unknown forces.
For design, "redundancy" means the ability of a structure to develop alternative load paths when it is damaged, which makes it inherently more
resilient. Redundancy can be particularly important for lateral load behavior
under earthquake loads. Most structures are highly redundant in the sense
that they have a high degree of static indeterminacy. However, this does not
necessarily mean that they are resilient. Some design codes, notably ASCE 7,
have specific requirements for redundancy.
.·.
7.4 Nonstrudural CompOhents
A nonstructural component contributes (or is assumed to contribute) no
. significant stiffness or strength to the structure. Components that are often
assumed to be nonstructural include exterior cladding, interior partitions,
. piping systems, and mechanical equipment.
Nonstructural components can not simply be ignored, since they contri"bute
gravity load, and for earthquake loading they may contribute mass· and
damping. It may also be important to assess the performance of nonstructural components. For example, it may be necessary to consider story drifts
to assess the performance of cladding, and to consider floor accelerations for
sensitive mechanical '.equipment. When a structure is damaged by earth-
358
Chapter 7 Some Other Aspects of Behavior
quake or wind loads, the main cost of the damage is often in the nonstructural components, not in the structure itseH.
It is usually not difficult to account for nonstructural components in an
analysis model, including the calculation of DIC ratios for assessing performance. For a component where the demand-capacity measure is story
drift or floor acceleration, only the weight of the component needs to be
accounted for in the analysis model, and the demand values can be
calculated from the analysis results for existing nodes. For piping, the
piping system is often modeled as a separate structure, with gravity and
·inertia loads based on the weight of the pipe and its contents, and possibly
support movements based on the displacements of the main structure. For
major industrial piping, the piping system might be regarded as structural,
and included in the main model.
7.5 Work·and Energy
7.5.1
Energy Balance in a Real Structure
When loads are applied to a structure, they cause displacements and do
· work4 moving though those displacements. This is external work on the
strui:ture. Since energy must be conserved, there is an equal amount of
internal work done by forces acting on the components of the structure.
If the external load is static (i.e., applied very slowly) and if the structure is
elastic, all of the internal work is stored as recoverable strain energy. If the
structure is inelastic, some of the internal work is recoverable and some is
dissipated by inelastic mechanisms such as yield, friction and fracture. If the
. loading is dynamic, or if the structure responds dynamically to a static load
(as it may if sudden fracture occurs), the internal work is more complicated.
This work can consist of strain energy and kinetic energy, both of which
·recoverable, and energy that is dissipated by a variety of mechanisms, not
just inelastic behavior. These mechanisms can include such things as wave
propagation into the foundation.
are
7.5.2 Energy Balance in an Elastic Analysis Model
The preceding applies to a real structure. There must. also be· an energy
balance when a model of a structure is analyzed.
Work and Energy
359
For static analysis of an elastic structure, all of the work that is done by the
external loads must be stored as recoverable strain energy in the structural
components. In this case there is a simple energy balance that can be written
as
EW=SE
(7.1)
where EW = external work and SE = strain energy. /
For the analysis of an elastic structure that is loaded dynamically, the
internal energy includes kinetic energy in the masses as well as strain
energy in the structural components. In this case the energy balance
equation is
EW=KE+SE
(7.2)
where KE =kinetic energy.
Like strain energy, kinetic energy is recoverable. If the external dynamic
load ceases after a time, the structure continues to displace dynamically, in a
state of free vibration. As the structure vibrates, EW is constant and there is
transfer of energy back and forth between KE and SE. this assumes,
however, an ideal elastic structure. In a real structure there will always be
some energy dissipation, and the energy balance equation is
EW=KE+SE +DE
(7.3)
where DE = dissipated energy.
In a state of free vibration, DE progressively increases (by definition it can
never decrease). Hence, since EW is constant, KE + SE progressively decreases,
and the vibration is "damped''.
If a real structure is essentially elastic, the dissipation (damping) mechanism
is usually not known in any detail, and it can depend substantially on the
· structure type. Possible mechanisms include inelastic energy dissipation of
various types, in both the structural and nonstructural components, and
wave propagation into the foundation ("radiation damping"). In elastic
analysis the dissipation is almost always modeled as linear viscous
damping. This is not realistic, but it is accurate enough for practical
purposes and has the overwhelming advantage that it is simple
computationally. The energy balance is given by
EW=SE +KE +VE
(7.4)
360
Chapter 7 Some Other Aspects of Behavior
where VE =dissipated viscous energy.
For elastic analysis it is usual to assume "modal" viscous damping. An
undamped elastic structure has a number of vibration modes. These modes
are uncoupled, lli the sense that if a structure vibrates in only one mode, it
continues to vibrate in only that mode and there is n0 effect on the other
modes. A VI"bration mode can have a complex shape, but it has only a single
degree of freedom, which is its modal·amplitude. Usually a dynamic force
or an earthquake ground motion will excite many modes, and the dynamic
response is the sum of the modal responses. For modal damping, each mode
is assumed to be damped by some viscous mechanism, in such a way that
the mode shape is not affected and the modes remain uncoupled. The
amount of damping is usually expressed as a proportion of critical damping,
where a mode (i.e., a single OOF system) has 100% critical damping if, when· ·
it is displaced and released, it return5 to rest with no Vibration. H a typical
structure were displaced and released it would vibrate, and return essentially
to rest after several cycles, corresponding to much less than 100% of critical·
damping.
For elastic dynamic analysis of building structures undet earthquake loads,
it is common to assume 5% damping, usually the same in all mooes. It may
be noted that this corresponds to a substantial amount of energy dissipation.
H a structure is displaced in a mode shape and released, and if there is 5%
damping, after two cycles of vibration the amplitude is only one half of its
original value. This means that 75% of the original strain energy has been
dissipated.
7~5.3
Energy Balance in an Inelastic Analysis Model
In an inelastic analysis model, some or all of the structural components are
inelastic, and they dissipate energy. In the absence of ·any other type of
energy dissipation, the energy balance equation is
EW=SE +KE +IE
(7.5)
where IE is the dissipated inelastic energy.
Experience shows, however, that the inelastic energy in an analysis model
usually (but not always) underestimates the dissipated energy in the
corresponding real structure. That is, the inelastic components in an analysis
model usually do not capture all of the dissipated energy. It is common
practice to assume additional viscous damping to account for the difference,
and the energy balance equation is
Work and Energy
EW =SE +KE +IE +VE
361
(7.6)
In this case, howeV"er, viscotts damping is much more complex than
assuming, say, 5% modal damping. The main reason is that for elastic
analysis the structure stiffness matrix remains constant, and hence the mode
shapes are constant, whereas for inelastic analysis the stiffness continually
changes, and in effect the mode shapes chcµige. Theoretically it is possible to
recalculate the mode shapes l?ach time the stiffn~ss changes, but this is
impractical. Aiso, the structure damping matrix depends on the mode
shapes, and if these shapes change suddenly the damping mahix changes.
This means that the damping forces also change suddenly, which introduces
a sudden equilibrium unbalance, or a numerical "shock". Such shocks can
cause computational problems, and as a general rule they should be
avoided.
·
There are a number of viscous damping mechanisms that can be used for
inelastic analysis, other than modal damping. Different computer programs
may use different mechanisms, which should be explained in the program
do,cumentation. Whichever method is used, it is usually possible to express
the amount of damping as a roughly equivalent amount of modal damping.
A problem that is even more important than the damping mechanism is the
amount of damping. As noted above, for elastic analysis it is common to
assume 5% damping. It can be argued that elastic analysis assumes that the
structure is loaded significantly beyond yield, even for design loads, and
hence that 5% damping accounts for some inelastic behavior.Ha structure is
subjected to only small amplitude vibrations,1here is less inelastic behavior,
and the damping ratio is likely to be smaller. The amount of damping
depends on the type of structure, but the damping ratio for a building with ·
· small amplitude vibration might be roughly 2% rather than 5%. For inelastic
dynamic analysis it is not a simple problem to choose the "correct" amount
of viscous damping. It is probably not correct to use 5% damping, since this
accounts indirectly for some inelastic behavior, and in an inelastic analysis
this behavior is modeled directly. A more reasonable assumption may be to
assume a viscous damping ratio that corresponds to small amplitude vibration (i.e., of the order of 2% rather than 5%).
7.5.4
Energy Balance Check During A".'alysis
A computer program may check the energy balance at each step in the
analysis, by calculating the work done by the external loads and comparing
it with the sum of the internal energies.
362
Chapter 7 Some Other Aspects of Behavior
For static load on an actual structure, as the load is progressively increased
there is an energy balance at all points on the load path. In a static analysis
there should be an energy balance at the end of each load increment. For
dynamic load on an actual structure there is an energy balance at every
instant of time. In a dynamic analysis there should be an energy balance at
the end of each time step.
For an elastic analysis, if the external work and internal energy ·.are
calculated and compared, there should be a very close energy balance, with
differences caused only by numerical round-off. This is because there
should be only very small equilibrium unbalances. If equilibrium between
the external loads and internal forces is satisfied in an· analysis, and if
geometrical compatibility is satisfied, the work done by the external and
internal forces will be the same, and there will be an energy balance. If there
are equilibrium unbalances, for example because the analysis model is
poorly conditioned, there will also be energy unbalances.
For an inelastic analysis there will usually be temporary equilibrium
unbalances (see Section 3.7). Hence, there will usually not be an exact
energy balance. The amount of energy unbalance can be useful for checking
the accuracy of the numerical computation. If there is a substantial energy
unbalance the analysis may be substantially inaccurate, and it may be
necessary to change the analysis details, for example by choosing a smaller
time step for a dynamic analysis.
7.5.5
Amount and Distribution of Dissipated Energy
A computer program may calculate the amount of inelastic energy that is
dissipated in different parts of a structure. For example, in a frame building
a computer program may calculate, and compare, the amounts of energy
dissipated in the beams and the columns. ·If most of the inelastic energy is
dissipated in the beams, this indicates that the frame tends towards a strong
column design. If most of the inelastic energy is dissipated in the columns,
this indicates weak column behavior.
A computer program may also compare the inelastic dissipated energy with
the energy dissipated through viscous damping.
Comparisons of this type can be useful for understanding the overall
behavior of a structure.
·
Living With Uncertainty
7.5.6
363
Dissipated Energy as a Demand-Capacity Measure
The dissipated inelastic energy can also be calculated for each inelastic
component in an analysis ·model. This energy provides a measure of the
amount of cyclic deformation, and can be used as a demand-capacity
measure for assessing performance.
In ASCE 41, maximum deformation is used as. the demand-capacity
measure for inelastic components. This is relatively simple, but maximum
deformation is not necessarily a reliable demand-capacity measure. The
reason is that for most components a single excursion to a given deformation tends to be less damaging than cyclic deformation with a number of
excursions to the same deformation. This is not taken into account (at least
directly) when only the maximum deformation is considered.
A more sophisticated damage measure for an inelastic component might
consider both the maximum deformation and the accumulated deformation
from several inelastic cycles. One way to account for the accumulated
deformation is to add up the deformations from all inelastic excursions. An
alternative is to calculate the dissipated inelastic energy. The analysis gives
the demand value. Some method must be available to calculate the
corresponding capacity.
As noted earlier, the amount and distribution of the dissipated energy can
be useful for understanding the overall behavior of a structure. For
earthquake loading the total amount of dissipated energy can also provide
an indication of the damaging effect of the earthquake and the resilience of
the structure. It is doubtful, however, whether the total dissipated energy is
useful as a formal demand-capacity measure.
7 .6 Living With Uncertainty
There are many uncertainties in structural analysis and design. Some can be
accounted for more-or-less scientifically, for example using load factors to
account for uncertainty in load magnitudes, and resistance (capacity
reduction) factors to account for variations in material strength. Such factors
can be based largely on observed data, although tradition and guesswork
may be just as important.
It is not so straightforward to account for uncertainty in modeling. If a large
number of analyses of real structures were carried out, using a variety of
models, and if the results were correlated with full scale tests on a large
number of real structures, it might be possible to quantify the uncertainty.
364
Chapter 7 Some Other Aspects of Behavior
This would be prohibitively expensive, however, and we have to make do
with much less.
"
Structural analysis can give the illusion that it is precise. Elastic analysis
allows us to divide a structure into thousands of small elements, to calculate
stresses with great apparent accuracy, and to display them as colorful
contour plots that purport to show which parts of a structure are most likely
to be damaged. For most building and bridge structures, however, elastic
analyses are very approximate, and may even give misleading information.
Inelastic analysis can provide information that is more useful, but it is more
difficult to carry out and is also very approximate.
What can be done? The following are some suggestions.
(1)
Always keep in mind that design drives analysis, not vice versa.
Analysis is merely a tool. Be clear on the nature of the design problem,
and what results you need from the analysis.
{2)
Use appropriate analyses. A few simple analyses that provide bounds
on the expected behavior may be more useful than a single elaborate
analysis that gives only a single data point. It is easy to get seduced by
the promises of elaborate analysis models, but there are more
important things to think about.
(3)
Consider how to organize the analysis results to support decision
making for design. Graphical presentations can be extremely useful,
but justbecause a presentation method exists does not mean that it
has to be used. A contour plot of the stresses in a wall, obtained from
an elastic analysis, may be colorful, but it may not have much value
when the wall is made of reinforced concrete. An animation of the
dynamic response of an analysis model to an earthquake may be
entertaining, and it has value for selling services to a client, but its
technical value may be limited.
(4)
Use capacity design. It ensures better behavior, it helps to keep the
focus on what is really important, and it forces us to acknowledge that
we do not know some important things.
(5)
A void design by analysis. It is lazy engineering, and it can lead to
poor performance.
{6)
Engage in informal risk assessment. Ask what can go wrong, and
what would be the consequences. Identify the major areas of
uncertainty, and possibly provide extra resiliency as protection. Make
Living With Uncertainty
365
a design inherently more forgiving, so that if the analysis is inaecurate
the consequences are less severe.
·
(7)
Try presenting analysis results to two significant figures, not seven,
and see how it affects your thinking.
/