68402: Structural Design of Buildings II
61420: Design of Steel Structures
62323: Architectural Structures II
Introduction to Structural Design
of Steel
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
Contents

Structural Design

Design Loads

Structural Steel - Properties

Design philosophies

Determining load and resistance factors

Load and resistance factors
Introduction to Design of Steel
Structures

General Introduction
•
Structural design is a systematic & iterative process that involves:
•
•
•
•
•
•
•
•
•
Identification of intended use & occupancy of a structure – by owner
Development of architectural plans & layout – by architect
Identification of structural framework – by engineer
Estimation of structural loads depending on use & occupancy
Analysis of the structure to determine member & connection design
forces
Design of structural members & connections
Verification of design
Fabrication & Erection – by steel fabricator & contractor
Inspection & Approval – by state building official
Primary Responsibilities

The primary responsibilities are:
• Owner
- primary responsibility is deciding the use &
occupancy, & approving the arch. plans of the
building.
• Architect
- primary responsibility is ensuring that the
architectural plan of the building interior is appropriate
for the intended use & the overall building is
aesthetically pleasing.
• Engineer
– primary responsibility is ensuring the
safety & serviceability of the structure, i.e., designing
the building to carry the loads safely.
Primary Responsibilities
• Fabricator – primary responsibility is ensuring that the
designed members & connections are fabricated
economically in the shop or field as required.
• Contractor/Erector
- primary responsibility is ensuring
that the members & connections are economically
assembled in the field to build the structure.
• State
Building Official – primary responsibility is
ensuring that the built structure satisfies the
appropriate building codes accepted by the Govt.
Structural Design

Conceptually, from an engineering standpoint,
parameters that can be varied (somewhat) are:
•
•



the
The material of construction
The structural framing plan.
The choices for material include:
•
•
•
Steel
Reinforced concrete
Steel-concrete composite construction.
The choices for structural framing plan include:
•
•
•
•
Moment resisting frames.
Braced frames.
Dual frames
Shear wall frames, and so on.
The engineer can also innovate a new structural framing
plan for a particular structure if required.
Structural Design

All viable material + framing plan alternatives
must be considered & designed to compare the
individual material + fabrication / erection costs to
identify the most efficient & economical design for
the structure.

For each material + framing plan alternative
considered, designing the structure consists of
designing the individual structural components,
i.e., the members & the connections, of the
framing plan.
Structural Design


Determination of dimensions and selection of cross sections.
The design process is a loop:
Assume dimensions, structural conditions and cross sections
Structural Analysis
Selection of cross sections to satisfy structural requirements
Does the design violate the initial assumptions?
YES
NO
Final Design
Structural Design

Optimal structural design shall achieve balance between
the following requirements:
Strength
Serviceability
Optimal design
Economy
Roles and responsibilities of the
structural steel designer

Arrange and proportion the members of the structures,
using engineer’s intuition and sound engineering
principles, so that they can be practically erected, have
sufficient strength (safe), and are economical.
•
Practicality:
Ensure structures can be fabricated and erected
without problems
•
Safety:
Ensure structures can safely support the loads.
Ensure deflections and vibrations are
controlled for occupants comfort.
•
Cost:
Minimize costs without sacrifice of strength
(consider labor costs in fabrication and
erection, not just material costs)
Basic Structural Shapes

Trusses

Frames ( Beam-Column)
• Beams
• Girders
• Columns

Space trusses/frames
Steel Structures
Purlin
s
Columns
BeamsFrames
Bracing
Steel Structures
Industrial/Parking
structures “Frames”
Steel Structures
Joists/Trusses
Steel Structures
High rise buildings
Steel Structures

Girder bridges
Steel Structures

Truss
bridges
Steel Structures

Cable stayed
& suspended
bridges
Structural Members

Structural members are categorized based up on the
internal forces in them. For example:
•
•
Tension member –subjected to tensile axial force only
•
Tension/Compression member –subjected to tensile/compressive
axial forces
•
Beam member –subjected to flexural loads, i.e., shear force &
bending moment only. The
•
•
•
Column or compression member –subjected to compressive axial
force only
axial force in a beam member is negligible.
Beam-column member – member subjected to combined axial
force & flexural loads (shear
force, & bending moments)
Structural Members
•
•
In trusses:
• All the members are connected using pin/hinge connections.
• All external forces are applied at the pins/hinges.
• All truss members are subjected to axial forces (tension
or
compression) only.
In frames:
• The horizontal members (beams) are subjected to flexural loads
•
only.
In braced frames:
•
•
The vertical members (columns) are subjected to compressive
axial forces only.
The
diagonal
members
(braces)
are
subjected
to
tension/compression axial forces only.
• In moment frames
•
The vertical members (beam-columns) are subjected to combined
axial & flexural loads.
Structural Connections

Members of a structural frame are connected together
using connections. Prominent connection types include:
•
•
•
Truss / bracing member connections are used to connect two or
more truss members together. Only the axial forces in the
members have to be transferred through the connection for
continuity.
Simple shear connections are the pin connections used to
connect beam to column members. Only the shear forces are
transferred through the connection for continuity. The bending
moments are not transferred through the connection.
Moment connections are fix connections used to connect beam to
column members. Both the shear forces & bending moments are
transferred through the connections with very small deformations
(full restraint).
Structural Connections
Truss connection
Simple Shear
connection
Moment resisting
connection
Structural Loads

The building structure must be designed to carry or resist
the loads that are applied to it over its design-life. The
building structure will be subjected to loads that have been
categorized as follows:
•
•
•
Dead Loads (D): are permanent loads acting on the structure.
These include the self-weight of structural & non-structural
components. They are usually gravity loads.
Live Loads (L): are non-permanent loads acting on the structure
due to its use & occupancy. The magnitude & location of live loads
changes frequently over the design life. Hence, they cannot be
estimated with the same accuracy as dead loads.
Wind Loads (W): are in the form of pressure or suction on the
exterior surfaces of the building. They cause horizontal lateral
loads (forces) on the structure, which can be critical for tall
buildings. Wind loads also cause uplift of light roof systems.
Structural Loads
• Snow Loads (S): are vertical gravity loads due to snow,
which are subjected to variability due to seasons &
drift.
• Roof Live Load (Lr): are live loads on the roof caused
during the design life by planters, people, or by
workers, equipment, & materials during maintenance.
• Values of structural loads can be computed based on
the design code.
Dead Loads (D)

Dead loads consist of the weight of all materials of
construction incorporated into the building including but
not limited to walls, floors, roofs, ceilings, stairways, builtin partitions, finishes, cladding & other similarly
incorporated architectural & structural items, & fixed
service equipment such as plumbing stacks & risers,
electrical feeders, & heating, ventilating, & air conditioning
systems.

In some cases, the structural dead load can be estimated
satisfactorily from simple formulas based in the weights &
sizes of similar structures. For example, the average
weight of steel framed buildings is 3 - 3.6 kPa, & the
average weight for reinforced concrete buildings is 5 - 6
kPa.
Dead Loads (D)

From an engineering standpoint, once the materials and
sizes of the various components of the structure are
determined, their weights can be found from tables that
list their densities. See Tables 1.2 & 1.3, which are taken
from Hibbeler, R.C. (1999), Structural Analysis, 4th
Edition.
Dead Loads (D)
Live Loads – Summary Table

Building floors are usually subjected to uniform live loads or
concentrated live loads. They have to be designed to safely support
these loads.
Type of occupancy
kPa
Offices
2.5 - 5
Corridors
5
Residential
2
Stairs and exit ways
5
Stadiums
5
Sidewalks
12
Wind Loads

Design wind loads for buildings can be based on: (a) simplified
procedure; (b) analytical procedure; & (c) wind tunnel or smallscale procedure.

Refer to ASCE 7-05 for the simplified procedure. This simplified
procedure is applicable only to buildings with mean roof height
less than 18 m or the least dimension of the building.

The wind tunnel procedure consists of developing a small-scale
model of the building & testing it in a wind tunnel to determine
the expected wind pressures etc. It is expensive & may be
utilized for difficult or special situations.

The analytical procedure is used in most design offices. It is
fairly systematic but somewhat complicated to account for the
various situations that can occur:
Wind Loads

Wind velocity will cause pressure on any surface in its
path. The wind velocity & hence the velocity pressure
depend on the height from the ground level. Equation 1.3
is recommended by ASCE 7-05 for calculating the velocity
pressure (qz) in SI
qz = 0.613 Kz KztKd V2 I (N/m2)
Wind Loads
qz – Static wind pressure
V - the wind velocity in m/s
Kd - a directionality factor (= 0.85 see Table 6.4 page 80)
Kzt - a topographic factor (= 1.0)
I - the importance factor (=1.0)
Kz - varies with height z above the ground level (see Table 6.3
page 79)
exposure B structure surrounded by buildings/forests/…
at least 6m height
exposure C open terrain
Wind Loads

A significant portion of Palestine has V = 100 km/h. At these
location
qz = 402 Kz (N/m2)
The velocity pressure qz is used to calculate the design
wind pressure (p) for the building structure conservatively
as follows:
p = q GCp (N/m2)
ASCE 7-05 pg. 79
Kz - varies with height z above the ground level
A – large city centers
B – urban/ suburban area
C – open terrain with scattered obstructions
D – Flat unobstructed surface
Wind Loads
G - gust effect factor (= 0.85)
Cp - external pressure coefficient from Figure 6-6 page 48-49
in ASCE 7-05 or
Cp = 0.8 windward
Cp = -0.5 leeward
Cp = -0.7 sidewalls
Cp = -0.7 slope<0.75
(1.5)
• Note that:
• A positive sign indicates pressure acting towards a surface.
• Negative sign indicates pressure away from the surface
Example 1.1 – Wind Load

Consider the building structure with the structural floor plan & elevation
shown below. Estimate the wind loads acting on the structure when the
wind blows in the east-west direction. The structure is located in
Nablus.
15 m
15 m
15 m
15 m
Plan
6 @ 3m
6 @ 3m
Example 1.1 – Wind Load
Example 1.1 – Wind Load

Velocity pressure (qz)
•
•
•
•
Kd - directionality factor = 0.85
Kzt - topographic factor = 1.0
I - importance factor = 1.0
V = 100 kph in Nablus
qz = 402 Kz (N/m2)
•
•
Kz - varies with height z above the ground level
Kz values for Exposure B, Case 2
Example 1.1 – Wind Load

Wind pressure (p)
•
•
•
•
•
•
•
•
•
Gust factor = G = 0.85 for rigid structures
External pressure coefficient = Cp = +0.8 for windward walls
Cp = -0.5 for leeward walls
Cp = -0.7 for side walls
External pressure = q G Cp
External pressure on windward wall = qz GCp = 402 Kz x 0.85 x 0.8 =
273.4 Kz Pa toward surface
External pressure on leeward wall = qh GCp = 402 K18 x 0.85 x (-0.5)
= 145.2 Pa away from surface
External pressure on side wall = qh GCp = 402 K18 x 0.85 x (-0.7) =
203.3 Pa away from surface
The external pressures on the structure are shown in the following
two figures.
Example 1.1 – Wind Load
203.3
273.4 Kz
145.2
203.3
Example 1.1 – Wind Load
3m
232.4
3m
221.5
3m
207.8
3m
191.4
180.4
169.5
3m
155.8
3m
145.2
Background of Structural Steel

Economical production in large volume not available until mid 19th
century and the introduction of the Bessemer process. Steel became
the principal metallic structural material by 1890.

Steels consists almost entirely of iron (over 98%) and small quantities
of carbon, silicon, manganese, sulfur, phosphorus, and other
elements.

The quantities of carbon affect properties of steel the most.

Increase of carbon content increases hardness and strength

Alloy steel – has additional amounts of alloy elements such chronium,
vanadium, nickel, manganese, copper, or zirconium.

The American Society for Testing of Materials (ASTM) specifies exact
maximum percentages of carbon content and other additions for a
number of structural steels. Consult Manual, Part 2, Table 2-1 to 2-3
for availability of steel in structural shapes, plate products, and
structural fasteners.
ASTM classifications of structural
steels

Carbon steels – A36, A53, A500, A501, A529, A570. Have
well-defined yield point. Divided into four categories:
•
•
•
•

High-Strength Low-Alloy steels – A242, A572, A588,
A606, A607, A618, A709
•
•

Low-carbon steel (< 0.15%)
Mild steel (0.15 to 0.29%, structural carbon steels)
Medium-carbon steel (0.3 to 0.59%)
High-carbon steel (0.6 to 1.7%)
Well-defined yield point
Higher strengths and other properties
Alloy Steels – A514, A709, A852, A913.
•
•
Yield point defined as the stress at 0.2% offset strain
Low-alloy steels quenched and tempered → 550 to 760 MPa yield
strengths
Advantages and disadvantages of
steel as a structural material

Advantages
•
•
•
•
•
•
•
•
•
•
High strength per unit of weight → smaller weight of structures
Uniformity
Elasticity
Long lasting
Ductility
Toughness
Easy connection
Speed of erection
Ability to be rolled into various sizes and shapes
Possible reuse and recyclable
Advantages and disadvantages of
steel as a structural material

Disadvantages
•
•
•
•
•
Maintenance costs
Fire protection/Fireproofing costs
Susceptibility to buckling failure
Fatigue
Brittle fracture
Types of Steel


Three basic types of steel used for structural steel
• Plain Carbon Steel
• Low-alloy steel
• High-alloy “specialty steel”
The most commonly used is mild steel - ASTM A36
Fy  248 MPa (36 ksi)


Typical high strength steel:
Fu  400 MPa (58 ksi)
Fy
ASTM A242
 290  344 MPa (42  50 ksi)
Fy
ASTM A992
 344 MPa (50 ksi)
Fu
 444  482 MPa (63  70 ksi)
Fu
 448 MPa (65 ksi)
The higher the steel strength, the higher the carbon content and
the less ductile it is.
Stress-strain curve

Standard Plain Carbon Steel
Stress “f”
f 
P ( Load )
A ( Area )
Necking & Fracture
Strain Hardening
Fu
Fy
E
Elastic
Yield plateau
 
L ( Deformation) Strain “”
Lo (Original Length)
What is a Limit State

When a structure or structural element becomes
unfit for its intended purpose it has reached or
exceeded a limit state

Two categories of limit states:
• Strength limit states
• Serviceability limit states
Limit States
Strength Limit States
a) Loss of Equilibrium
b) Loss of load bearing capacity
c) Spread of local failure
d) Very large deformations
Serviceability Limit States
a) Excessive deflection
b) Excessive local damage
c) Unwanted vibration
Design Philosophies

Allowable Stress Design (ASD)

Plastic Design (PD)

Load and Resistance Factor Design (LRFD)
Allowable Stress Design





Service loads are calculated as expected during service
life.
Linear elastic analysis is performed.
A factor of safety (FOS) of the material strength is assumed
(usually 3-4)
Material Strength
Allowable Stress 
FOS
Design is satisfactory if (maximum stress < allowable
stress)
Limitations
•
•
Case specific, no guarantee that our design covers all cases
Arbitrary choice of FOS?!
Plastic Design





Service loads are factored by a “load factor”.
The structure is assumed to fail under these loads, thus,
plastic hinges will form under these loads “Plastic Analysis”.
The cross section is designed to resist bending moments
and shear forces from the plastic analysis.
Members are safe as they are designed to fail under these
factored loads while they will only experience service loads.
Limitations
•
•
No FOS of the material is considered, neglecting the uncertainty in
material strength!
Arbitrary choice of overall FOS?!
Load and Resistance Factor Design
(LRFD)




LRFD is similar to plastic design in that it performs design
with the assumption of failure! - Reliability Based Design
Service loads are multiplied by load factors (g) and linear
elastic analysis is performed.
Material strength is reduced by multiplying the nominal
material strength by a resistance factor (f)
The design rule is:
Load Effect < Resistance
 g i Q i  fi R n
•
This rule shall be attained
for all limit states!!
Where Rn is the nominal strength and Q is the load effect for the ith
limit state
Load and Resistance Factor Design
(LRFD)


Resistance: Shear, Bending, Axial Forces
Advantages of LRFD
•
•
Non-case specific, statistical calculations guarantee population
behavior.
Uniform factor of safety as both load and material factors are tied
by reliability analysis
Probabilistic Basis for LRFD

If we have the probability distribution of the load effect (Q)
and the material resistance (R) then:
•
•
The probability of failure can be represented by observing the
probability of the function (R-Q)
The probability of failure PF can be represented as the probability that
Q ≥ R:
Probability
of failure
AISC Load combinations
AISC considers the following load combinations
g
in design

i
Qi  fi Rn
1 1.4 D
2  1.2 D  1.6 L  0.5( Lr or S or R)
g
3  1.2 D  1.6 ( Lr or S or R)  0.5L or (0.8W )
i
Qi
4  1.2 D  1.6 W  0.5 L  0.5 ( Lr or S or R)
5  1.2 D  1.0 E  0.5 L  0.2 S


6 0.9 D  (1.6 Wor 1.0 E )
f  0.75  1.00
fi Rn

e.g. f for yield is 0.9 and for bolt shear is 0.75

Dead loads (D)
Live loads (LL)
• Occupancy load
(L)
• Roof load (Lr)
• Snow load (S)
• Rain loads (R)
• Trucks and
pedestrians
Wind Loads (W)
Earthquakes (E)