University of Technology

Building and Construction Engineering Department

Water and Dams Engineering Division

STRUCTURAL STEEL
DESIGN
Dr. Ammar A. Ali
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

STEEL DESIGN
1. INTRODUCTION

This lecture is based mainly on:

Applied Structural Steel Design, 4th edition, by Spiegel, L. and Limbrunner, G.F.

- Steels used in construction are generally carbon steels: alloys of iron and carbon.
- The carbon content is ordinarily less than 1% by weight.
- The chemical composition of steel is varied, according to the properties desired,
such as: strength and corrosion resistance. This achieved by addition of other
alloying elements, such as: silicon, manganese, copper, nickel, chromium, and
vanadium, in very small amounts.
- When steel contains a significant amount of any of such alloying elements, it is
referred to as an alloy steel.
- The advantages of steel are:
 Uniformity of material,
 Predictability of properties,
 Dimensional stability,
 Ease of fabrication, and
 Speed of erection.
- The disadvantages of steel are:
 Susceptibility to corrosion (in most but not all steels)
 Loss of strength at elevated temperatures.

Specifications:

AISC: The American Institute of Steel Construction

ASD Manual of Steel LRFD Manual of Steel

Construction (9th edition) Construction (13th edition)

ASD: Allowable Steel Design LRFD: Load and Resistance

Factor Design

Steel Properties:

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

- The more apparent mechanical properties of steel in which the designer is

interested may be determined by a tension test. The specimen shape is depicted in
Fig.1.1 and the testing machine is in Fig. 1.2. The formulas for stress and strain are

stress = 𝑃𝑓𝑡 =
𝐴

∆𝐿0
strain = 𝜖 =
𝐿0

where

f t = computed tensile stress (ksi)

P = applied tensile load (kips)

A = cross-sectional area of the tensile specimen (in.2)

ϵ = unit strain, elongation (in./in.)

ΔL0 = elongation or the change in length between two reference points on the tensile
specimen (in.)

L0 = original length between two reference points (in.)

GW

GL

RS G W L L0

FR

Fig. 1.1 Test specimen

G = Gauge length.

W = Width or diameter of specimen.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

FR = Fillet radius.

L = Overall length.

RS = Reduced section.

GW = Grip section width.

GL = Grip section length.

Fig. 1.2 A universal testing machine

Ref.: http://www.testresources.net/by-test-application/tensile-testing-equipment/tensile-
testing-machine-steel/

From typical stress-strain diagrams shown in Figs. 1.3 and 1.4, the behavior of structural
steel can be described as following:

- The steel remains elastic as long as it is not stressed past a value slightly higher than
the proportional limit, called the elastic limit.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

- The proportional limit is the point at which the stress-strain relationship becomes
nonlinear.
- The proportional limit and the elastic limit are so close that they are often
considered to be the same value.

Fig. 1.3 Typical stress-strain diagram for structural steel.

Fig. 1.4 Partial stress-strain diagram for structural steel.

-
The yield stress, Fy, will be reached when the strain in the specimen increases rapidly
at constant stress.
- The slightly higher stress that exists just after the proportional limit, called the
upper yield. It exists only instantaneously and is unstable.
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

- According to the allowable stress design (ASD) method, most structural steel is
designed so that actual stresses in the structural member do not exceed allowable
stresses, well below Fy.
- As the steel continues to strain, it reaches a point at which its load-carrying capacity
increases. This phenomenon of increasing strength is termed strain hardening.
- The maximum stress to which the test of increasing strength is called the tensile

- strength, Fu (ksi).

In the ASD method stresses are limited to some fraction of Fy, but another design method,
called plastic design, allows small but definite areas of members to be
stressed to Fy and strained into the plastic range.
- For all practical purposes, in structural steel design, it is only the elastic range and
the plastic range that need be of interest, since the strains in the strain-hardening
range are of such magnitude that the deformation of the structure would be
unacceptable. Thus, an idealized diagram for structural steel is sufficient for
purposes of illustrating the steel stress-strain relationships, as shown in Fig. 1.5.

Fig. 1.5 Idealized stress-strain diagram for structural steel.

- The strain at the upper limit of the plastic range, ϵp, is approximately 10-15 times the strain

at the yield point, ϵy.

- The modulus of elasticity, E (or Young’s modulus), is the slope of the stress-strain
curve in the elastic range:

stress 𝑓𝑡
𝐸= =
strain 𝜖

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

- Young’s modulus E is reasonably constant for structural steel. AISC recommends the
value of E to be 29,000 kips/in2.
- The ability to undergo large deformations before failure is called the ductility.
- The ductility of steel is important in the safeguarding of lives and property in
unknown and uncertain loading situations such as earthquakes.

Products Available

- Steels are usually specified according to ASTM (American Society for Testing and
Materials) number.
- The ASD manual, Part 1, Table 1, contains a list of ASTM structural steels. Note that
minimum Fy ranges from 32 to 100 ksi and tensile strength Fu ranges from 58 to 130 ksi.
- The most widely used structural carbon steel is A36. A36 steel has a yield stress of
36 ksi except for cross sections in excess of 8 in thick.
- More recently, the trend has been toward the higher-strength steels, with A572
grade 50 being widely used.
- A recently developed steel specification designated A992 covers steel that is similar
to A572 grade 50 but has some important differences. In addition to differences in
chemical composition, A992 has a minimum Fu of 65 ksi, and Fy range of 50-65 ksi, and a
maximum ratio of Fy/Fu of 0.85.
- Steels designated as A588 and A242 are weathering steels. These steels are popular
for use in bridges and exposed building frames.
- Structural shapes are:

1- W and M shapes (Fig. 1.6) are used primarily as beam and column members.

bf

Flange
tf slope 0-5%

Web
d

tw

Fig. 1.6 Wide-flange (W) and Miscellaneous (M)

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

W36 × 300

depth (𝑑) ≈ 36 in.

weight (𝑤) = 300 lb/ft

𝑑=
overa
ll
dept
h,

𝑏𝑓 =
flang
e
widt
h,
𝑡𝑓 =
flange
thickn
ess,

𝑡𝑤 =
web
tf
thickn slope 1623%
ess
d
Web
M shapes are miscellaneous shapes. They have cross sections that appear to be
exactly like W shapes, but withtwdifferent dimensions and not widely available.

M8 × 6.5

depth (𝑑) ≈ 8 in.

Fig. 1.7 American standard beam (S).
weight (𝑤) = 6.5 lb/ft
S12 × 35
2- S shapes (Fig. 1.7) are American standard beams. They have sloping inner faces on
the flanges, relatively thicker webs, and depths that are mostly full inches. They find
depth (𝑑) = 12 in.
some application where heavy point loads are applied to the flanges, such as in
monorails for the support of hung cranes.
weight (𝑤) = 35 lb/ft
bf
3- HP shapes (Fig. 1.8) are bearing pile Flange
shapes and are characterized by a rather
square cross section with flanges and webs of nearly the same or equal thickness (so
that the web will withstand pile-driving
8 hammer blows).
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

bf

tf
F
Web l
d a
tw n
g
e

Fig. 1.8 Bearing pile (HP).

HP12 × 74

depth (𝑑) ≈ 12 in.

weight (𝑤) = 74 lb/ft

4- C and MC shapes (Fig. 1.9) are American standard channels and miscellaneous
channels. The channel shapes are characterized by short flanges that have sloping
inner surfaces and depths to full inches. Their applications are usually as
components of built-up cross sections, bracing and tie members, and members that
frame openings. bf tf

Flange
slope 1623%
d
tw Web

Fig. 1.9 American standard (C) and Miscellaneous (MC) channels.

C10 × 30
depth (𝑑) = 10 in.
weight (𝑤) = 30 lb/ft

MC18 × 58
depth (𝑑) = 18 in.
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

weight (𝑤) = 58 lb/ft

MC shapes cannot be classified as C shapes by dimensions.

5- Angles (Fig. 1.10) are designated by the letter (L), leg length of each leg, and
thickness. They may be equal or unequal leg angles. Angles are commonly used
singly or in pairs as bracing members and tension members. They are also used as
brackets and connecting members between beams and their supports. Light trusses
and open web joists may also utilize angles for component parts.
For unequal leg angles the longer leg is stated first. For example, L9 × 4 × 12 indicates
an angle with one leg 9-in. long and one leg 4-in. long, both having thickness of 12 in.

Note: Designation does not provide the unit weight of the angle as has been the case with
all shapes before. The weights in pounds per foot is tabulated and can also be calculated by
using the unit weight of steel of 490 lb/ft3.

Fig. 1.10 Angles (L).

6- Structural tees (Fig. 1.11) are shapes that are produced by splitting the webs of W,
M, or S shapes. The tees are then designated WT, MT, or ST, respectively. For
example WT18 × 105 (nominally 18 in deep, 105 lb/ft weight) is obtained from
W36 × 210.

bf

tf
Flange
d
tw
Web

Fig. 1.11 Structural tees (WT, MT, ST).

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Tees are used primarily for special beam applications and as components in
connections and trusses.

7- Structural Steel Pipe and Tubing (Fig. 1.12)

(a) Pipe and (b) Square (c) Rectangular

round HSS HSS HSS

Fig. 1.12 Structural steel pipe and tubing.

Steel pipe is covered by ASTM A53 grade B (Fy = 35 ksi and Fu = 60 ksi) and is available in standard,
extra-strong, and double-extra-strong weights, with wall thickness being the
influencing factor. These are abbreviated (for a 4-in. nominal diameter) as P4, PX4, and
PXX4, respectively.

The HSS (Hollow Structural Section) are given in the ASD manual, Part 1, Table 3 and in the
LRFD manual, Part 2, Table 2-3.

A round HSS is designated by nominal diameter and wall thickness, each expressed to three
decimal places, such as HSS5.563 × 0.258.

A square or rectangular HSS is designated by nominal outside dimensions and wall

thickness, each in rational numbers, such as HSS5 × 3 × 38.

Tubes make excellent compression members, although the connections usually involve
some welding. The most common use of tubes is in compression and tension members, but
they also used as beams in some situations. Structural members made of tubes are easier to
clean and maintain than their wide-flange counterparts.

8- Plates

Plates are used in assorted applications such as plate girder elements, gusset plates,
stiffenners, beam bearing plates, column base plates, …etc. Plates are avalible in various
thicknesses, although a minimum thickness increment of 18 − in. is recommended.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

The letters PL followed by thickness, width, and length commonly designates plates. In this
designation, thickness and width are fractional inches, and length in feet and inches.

1 ′
PL × 14 × 1 − 4
2

The weight of a plate can be taken from tables, when available, or computed by using a unit
weight of stee of 49 lb/ft3.
Design Considerations

1. Safety
2. Economy, beauty, fuctionality, maintainability, permanance, …etc.

Safety

The expression of safety is normally made in terms of a factor of safety. The factor of
safety is the ratio of the load (or stress) that causes failure to the maximum load (or stress)
actually allowed in the structure.

In allowable stress design (ASD) the attainment of yield stress in a member is considered to
be analogous to failure.

Although the steel will not actually fail (rupture) at yield, significant and unacceptable
deformations are on the verge of occuring, which may render the structure unusable.

As an example, assume that a member composed of a steel having yield stress Fy has as
specified allowable stress of 0.66 Fy. The factor of safety (F.S.) against yielding would then
be

"𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠" 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝐹𝑦

𝐹. 𝑆. = = = = 1.5
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑟𝑒𝑠𝑠 0.66𝐹 𝑦
The LRFD method uses series of factors of safety called load factors when applied to loads
and resistance factors when applied to member strength or resistance. Each factor is the
result of a statistical study of the variability of the particular quantity and reflects the
probability that the specific load or resistance is incorrect. In that design method strength
reduction factors are used to predict a practical strength and load factors ae used to modify
service loads and estabilish design loads for use in design considerations.

F.S. reasons:

1. Danger to life and property.

2. Confidence in the prediction of loads.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

3. Variation in material properties.

4. Confidence in the analysis methods.
5. Possible deterioration during the design life of the structure.

Loads

Dead loads are static loads that produce vertical forces due to gravity and include the
weight of the steel framework and all materials permanently attached to it and supported
by it.

Live loads include all vertical loads that may be either present on or absent from the
structure.

Generally, lateral loads are considered live loads whether they are permanent or not.

Example of live loads are: snow, people, furniture, stored materials, vheicles (on brodges or
warehouses), cranes, wind, lateral prssure due to earth or stored lquids, and earhquakes.

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Dr. Ammar A. Ali

2. TENSION MEMBERS
Although easy to proprtion, however, tension members, and structures in which main load-
carrying members are iin tension, require great care in the design and detailing of their
connections.

A tensile load applied along the longitudinal axis of the member tends to hold the member
in alignment, therby making instability a minor concern.

Of the most concern in the selection of tension members is the choice of the gonfiguration
of the cross section so that the connections will be simple and efficient. Also, the connection
should transmit the load to the member with as little eccentricty as possible.

Examples of tension members may be found in many structures. They include hangers for
catwalks and strorrage bins, truss web and chord members, cables for direct support of
roofs, sag rods, tie rods, and various types of braces.

Most of the common hot-rolled structural steel shapes may be used as tension members.

Tension Member Analysis

The direct stress formula is the basis for tension member analysis. It may be written for
stress,

𝑓𝑡 𝑃=
𝐴

or for tensile capacity,

𝑃𝑡 = 𝐹𝑡𝐴

where

Ft = computed tensile stress

P = applied axial load

Pt = axial tensile load capacity (or maximum allowable axial tensile load)

Ft = allowable axial tensile stress

A = cross-sectional area of axially loaded tension member (either gross area Ag, net area An, or effective

net area Ae)

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Gross area is the origonal, unaltered cross-sectional area of the member
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝐴𝑔(for plate) = 𝑏 ∙ 𝑡

where

b = width of plate

t = thickness of plate

An, net area, is the cross-sectional area actually available to be stressed in tension (Fig. 2.1).

𝐴𝑛 = 𝐴𝑔 − (area of holes)

= 𝑏 ∙ 𝑡 − 2(𝑑ℎ ∙ 𝑡)

where (B2)
𝑑ℎ (fordnet
h is hole
area,diameter for analysis
A n, calculations ) =purposes.
𝑑 ℎ(𝑛𝑜𝑚𝑖𝑛𝑎𝑙) +161 "

𝑑ℎ(𝑛𝑜𝑚𝑖𝑛𝑎𝑙) = 𝑑 +161 "

1 1
𝑑ℎ = (𝑑 + 16 ") + 16 " = 𝑑 + 18"

For purposes of analysis and design, hole diameters are taken as the fastener diameter plus
1
8
".

Fig. 2.1 Net area.

The direct stress formula is used basing on the assumption that the tensile stress is
uniformly distributed over the net section of the tension member, despite the fact that high
stress concentrations are known to exist (at working loads) around the holes in a tension
member. The commonly used structural steels are suffciently ductile that they undergo

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

yielding and stress distribution. This will result in a uniform stress distribution at ultimate
load.

The allowable tensile stress Ft takes into consideration two types of failure:

1. The mamber may rupture on the least net area. This is the classical and historical
approach to tension member analysis.
2. The tension member may undergo uncontrolled yielding of its gross area without
rupture. Excessive elongation of tension member is undesirable in that it normally
results in deformation of the structure and can lead to failure in other parts of the
structural system.

The types of failure above may be predicted using the following formulas:

1. 𝐹𝑡 = 0.50𝐹𝑢 … … … (on the net area)

D1
2. 𝐹𝑡 = 0.60𝐹𝑦 … … … (on the gross area)

These allowable stresses do not apply to pin-connected members (such as eye bars or
plates connected with relatively large pins), threaded steel rods, or flexible tension
members such as cables and wire ropes.

Block Shear

A tearing failure that can occure at end connections along the perimeter of welds or
along the perimeter of a group of bolt holes (Fig 2.2).
Tension area

Tension area
Pt
Pt
Shear area Shear area
(a) Bolted angle (b) Bolted plate
Pt Shear area
Shear area
Tension
area
Pt

(c) Bolted W shape

Tension area
(d) Welded angles

Fig.2.2 Block shear in end connections

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Dr. Ammar A. Ali

Depending on the end connection, the failurecould occur in either the tension member
itself or in the member to which is attached (e.g., a gusset plate).

Block shear is characterized by a combination of shear failure along a plane through the
welds or bolt holes and a simulatianeous tension failure along a perpendicular plane.

Block shear strength is calculated from the summation of net shear are Av times the

allowable shear stress Fv and net tension area At times the allowable tensile stress Ft, where

𝐹𝑣 = 0.30 𝐹𝑢
and }………
𝐹𝑡 = 0.50 𝐹𝑢
𝐽4
Mathematically, this is stated as

𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡

For purposes of block shear strength calculations for bolted connections, hole diameters
for the net area determination aretaken as the fastener diameter plus 18 in. This is the same
approach as that used for tensile net area calculations.

Definitions

In Fig. 2.3, the tensile load P is assumed to be applied parallel to and ciincident with the
longitudinal axis of the member.
Gage lines
Longitudinal axis

d3 g
d1
E.D.2

E.D.1 s1 s2
d2

Fig 2.3 Definitions

E.D. = Edage distance

g = gage

s = pitch (bolt spacing)

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

d = distance between bolts

The bolt holes are located on gage lines that are parallel to the longitudinal axis. The
dimension g between the gage lines is called the gage. The dimension s parallel to the gage
line and taken between centers of bolt holes is called the pitch (or the bolt spacing). The
distance between bolts is a straight-line distance between any two bolts. The edge
distance is the prependicular distance from the center of a hole to the nearest edge.

The analysis of a tension member involves the determination of the indiviual allowable
loads based on the various failure modes. Once these have been determined, the axial load
capacity of the member is taken as the smallest value.

Example:

Find the axial tensile load capacity Pt of the lapped, bolted tension member shown in Fig.
3
2.4. Bolts are 4–in. diameter, and the plate is A36 steel (Fu = 58-80 ksi from the ASD
manual, Part 1, Table 1 and the LRFD manual, Part 2, Table 2-3). Assume that the fasteners
are adequate and do not control the tensile capacity. Pitch, gage, and distance are as
shown.

7 " Pt
Pt 16

7 "
16
(a)

3"
F B
Pt 3" Pt
C
3"
G D
3"

E
1
12" 3" 3" 121"

(b)

Fig. 2.4 Tension member analysis

Solution:

𝑃𝑡 = 𝐹𝑡𝐴𝑛 or 𝐹𝑡𝐴𝑔

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1. Based on gross area,

𝑃𝑡 = 0.60𝐹𝑦𝐴𝑔
7
= 0.60(36) ( 16 ) (12) = 113.4 kips

2. Based on net area, visulazing a transverse fructure along line ABCDE,

𝐴𝑛 = 𝐴𝑔 − 𝐴ℎ𝑜𝑙𝑒𝑠
7 3 1 7
= (12) − 3 ( + ) ( ) = 4.10 in.2
16 4 8 16

Using the lower limit of the Fu range (conservative),

𝑃𝑡 = 0.50𝐹𝑢𝐴𝑛
𝑃𝑡 = 0.5(58)(4.1) = 118.9 kips
3. Check the block shear strength. Two possible cases are investigated. The block shear
strength is written as

𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡
𝑃𝑡 = 𝐴𝑣(0.30𝐹𝑢) + 𝐴𝑡(0.50𝐹𝑢)
where Av and At are the net shear area and net tension area, respectively. Note that the hole
diameter is taken as
3 + = 0.875 in. 1
4 8
Case I: failure line FBCDG:

7
𝐴𝑣 = 2 ( ) [7.5 − 2.5(0.875)] = 4.65 in.2
16
7
𝐴𝑡 = [6 − 2(0.875)] = 1.859 in.2
16
𝑃𝑡 = 4.65(0.30)(58) + 1.859(0.50)(58) = 134.8 kips

Case II: failure line ABCDG:

7
𝐴𝑣 = ( ) [7.5 − 2.5(0.875)] = 2.32 in.2
16
7
𝐴𝑡 = ( ) [9 − 2.5(0.875)] = 2.98 in.2
16
𝑃𝑡 = 2.32(0.30)(58) + 2.98(0.50)(58) = 126.8 kips

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 STRUCTURAL STEEL DESIGN
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Therefore, the capacity Pt of this tension member is 113.4 kips as controlled by general
yielding of the gross area.

Staggered Holes

In some cases the fasteners will be arranged so that the controlling fructure line will be
something other than transverse, as shown in Fig. 2.5. This situation can occur when
fasteners are staggered to accommodate a desired size or shape of connection. There are
two possible failure lines across the width of the plate. These may be defined as lines ABCD
and ABE. For large values of s, line ABE will be the more critical failure line (smaller net
area). For small values of s, line ABCD will be more critical.

B
Pt Pt
g

E D
s

Fig. 2.5 Staggered holes

A combination of shear and tensile stresses acts on the sloping line BC of failure line ABCD.
The interaction of these stresses presents a rather complicated theoritical problem.
The ASD specifications, section B2, stipulates that where a fructure line contains within it a
diagonal line, the net width of the part should be obtained by deducting from the gross
width the diameters of all the holes along the fructure line and adding, for each diagonal
𝑠2
line, the quantity .
4𝑔

An expression for net width wn may be

written as 𝑠2
𝑤𝑛 = 𝑤𝑔 − � 𝑑ℎ + �
4𝑔

Where wg represents gross width and dh represents the hole diameter to be used for design.

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Dr. Ammar A. Ali

The forgoing formula for wn is convenient to use with members of uniform thickness. If the formula is
multiplied by thickness t it becomes
𝑠2𝑤𝑡 𝑡 = 𝑤 𝑡 − � 𝑑ℎ 𝑡 + �
𝑛 𝑔
4𝑔

Or, since 𝑤𝑛𝑡 = 𝐴𝑛 and 𝑤𝑔𝑡 = 𝐴𝑔,

𝑠2 𝑡𝐴 = 𝐴 − � 𝑑ℎ 𝑡 + �
𝑛 𝑔
4𝑔

The latter formula for An is the more useful, since it provides net area directly and is also
applicable with members that do not have uniform thickness (i.e., channels). In a
determination of critical net area where multiple failure lines exist, the critical net area is
the least net area.

Example:
Determine the critical net width wn for the plate shown in Fig. 2.6. Fasteners will be 1-in.- diameter
bolts.
A E
221"
B
3"
Pt F Pt

4"

C
221"
D G
2"

Fig. 2.6 Net width calculation

Solution:

Use the formula for the net width,

𝑠2
𝑤𝑛 = 𝑤𝑔 − � 𝑑ℎ + �
4

𝑑ℎ = 1 1
8
" = 1.13
in. 21
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Dr. Ammar A. Ali

Line Net width (wn)

ABCD 12 − 2 (1.13) + 0 = 9.74 in.
EFG 12 − 1 (1.13) + 0 = 10.87 in.
22
ABFG 12 − 2 (1.13) + = 10.07 in.
4(3)
22
EFCD 12 − 2 (1.13) + = 9.99 in.
4(4)
2 2 2 2
ABFCD 12 − 3 (1.13) + + = 9.19
in.
wn = 9.19 in. 4(3) 4(4)

A zigzag failure line sometimes occurs in a member that has more than one element making
up its cross section, such as an angle.

Example:

Calculate the tensile capacity Pt of the angle shown in Fig. 2.7. Assume that the member to which the
angle is connected
connection does not
does not govern govern
capacity. capacity.
Assume Likewise,
A36 steel assume thatbolts.
and 34-in.-diameter the

3 L4X3X14
14 "

1
4 " 22 "

5 @ 112"
(a) (b)

D A
F E
114 "

g
B
1 G
12 "
C
1
5 @ 1 2"

(c)

Fig. 2.7 Zigzag failure lines in an angle

Solution:

22
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

1.
Calculate Pt based on yielding of the gross area. The gross width of the angle is
determined from
𝑤𝑔 = 4 + 3 − 1 = 6.75 in.
4

The gross area is determined from

𝐴𝑔 = 6.75(0.25) = 1.69 in.2

Therefore,

𝑃𝑡 = 𝐴𝑔𝐹𝑡 = 𝐴𝑔(0.60𝐹𝑦) = 1.69 (0.60)(36) = 36.5 kips

2.
Next, caclulate Pt based on tensile fracture of the net area. Two net width values are
determined:
Line ABC (transverse section):

𝑤𝑛 = 6.75 − 0.875 = 5.88 in.

Line DEBC:
1.5
𝑤𝑛 = 6.75 − 2(0.875) + = 5.14 in.
4(4)

The zigzag line controls, therefore,

𝐴𝑛 = 5.14(0.25) = 1.285 in.2

The tensile load capacity Pt based on net area is

𝑃𝑡 = 𝐴𝑛𝐹𝑡 = 𝐴𝑛(0.50𝐹𝑢) = 1.285 (0.50)(58) = 37.3 kips

3. Last, check block shear.

2
Case I: in
tension theand
area, block
theshear
𝑠 failure
term lineasdefined by FEBG, the diagonal line considered to be
is used
4𝑔
previously.
𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡 = 𝐴𝑣(0.30𝐹𝑢) + 𝐴𝑡(0.50𝐹𝑢)
𝐴𝑣 = 0.25[9(1.5) − 4(0.875) ] = 2.5 in.2
1.52

𝐴𝑡 = 0.25 [4 − 0.875 + 4(4) ] = 0.816 in.2

Therfore,

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑃𝑡 = 2.5(0.30)(58) + 0.816(0.50)(58) = 67.2 kips

Case II: use the block shear failure line defined by FEBC, where EB and BC are considered
tension areas and FE is a shear area:

𝐴𝑣 = 0.25[6 − 1.5(0.875)] = 1.172 in.2

1.52
] = 1.082 in.2
𝐴𝑡 = 0.25[1.5 − 0.5(0.875)] + 0.25 [4 − 0.875 + 4(4)
From which

𝑃𝑡 = 1.172(0.30)(58) + 1.082(0.50)(58) = 51.8 kips

Failure line DEBG could also be checked. It would be found to have tensile capacity slightly
greater than that of line FEBC. Therfore, the tensile capacity Pt of the angle is 36.5 kips,
controlled by yielding of the gross area.

Effective Net Area

For some tension members, such as rolled shapes, that do not have all the elements of the
cross section connected to the supporting members (see Fig. 2.8), the failure load is less
than would be predicted by the product AnFu. The phenomenon to this situation is generally attributed
is called shear lag.

Fig. 2.8 Shear lag.

Note that the angle is connected along only one leg. This leads to a concentration of stress
along that leg and leaves part of the unconnected leg unstressed or stressed very little.
Studies have shown that the shear lag effect diminishes as the length of the connection
increased.

24
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

The ASD specifications, Section B3 and LRFD specifications, Section D3, accounts for the
effect of shear lag through the use of an effective net area, which is a function of how the
tension member is connected at its ends.

When the load is transmitted directly to each of the cross-section elements by either bolts
or welds, the effective net area Ae is equal to the net area An.

When the load is transmitted by bolts through some, but not all, of the cross-sectional
elemets of the member, the effective net area Ae shall be computed from

𝐴𝑒 = 𝑈𝐴𝑛 … … … … 𝐸𝑞. 𝐵3 − 1

where

An = the net area of the member (in.2)

U = reduction coefficient (Table 2.1)

Case I W, M, S shapes or their tees. U = 0.9
Table 2.1 Values for Reduction
Connection Coefficient, U
is to the flanges.
Minimum of three bolts
per line in the direction of
stress.
2 d ( m i n . )
2 d (min.)
3

d
d

Case II All shapes and built-up cross-sections not meeting U = 0.85

the requirements of case I. Minimum of three bolts
per line in the direction of stress.
Case III All members whose connections have only two U = 0.75
bolts per line in the direction of stress.

When the load is transmitted by welds through some, but not all, of the cross-sectional
elements of the member, the effective net area Ae shall be computed from

𝐴𝑒 = 𝑈𝐴𝑔 … … … … 𝐸𝑞. 𝐵3 − 2

where
25
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Ag = the gross area of the member (in.2)

The U values for a welded connection is the same as that for a bolted connection except that
Case III is not applicable and the condition as to the number of bolts in Case I and Case II
does not apply.

With respect to weld end connections, the ASD specifications, Section B3, furnishes
effective net area criteria for two special considerations:

a. When a load is transmitted by transverse welds to some, but not all, of the cross-
sectional elements of W, M, or S shapes and structural tees cut from these shapes,
the effective net area Ae shall be taken as that area of the directly connected
elements.
b. When a load is transmitted by longitudinal welds used alone along both edges of a
flat bar (or plate) axially loaded tension member, the length of each weld shall not
be less than the width of the plate. The effective net area Ae shall be computed by Eq. (B3-2)
using the reduction coefficient U shown in Table 2.2, where

l = weld length (in.)

w = plate width (distance between welds, in.)

Table 2.2 Value of U

(Longitudinal Welds on a Flat
Condition U
Bar or Plate)
l˃2w 1.0
2 w ˃ l ˃ 1.5 w 0.87
1.5 w ˃ l ˃ w 0.75

Additionally, for relatively short connection fittings such as splice plates, gusset plates, and
beam-to-column fitting subjected to tensile force, the effective net area shall be taken as the
actual net area except that it shall not be taken as greater than 85% of the gross area.
Therefore, for these short plates and fittings subjected to tension, U does not apply, and

𝐴𝑒 = 𝐴𝑛 (not to exceed 0.85 𝐴𝑔)

Example:

A tension member in a truss is to be composed of a W8 × 24 and is connected with two

lines of 78-in.-diameter bolts in each flange as shown in Fig. 2.9. There are three bolts per

26
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

1
line, 33-in. pitch, 1 2in. edge distance, and A36 steel. Find the tensile load capacity Pt.
Assume that the gusset plates and the bolt capacities are satisfactory.

Pt
Gusset
plates

W8 X 24

Bottom
chord

Fig. 2.9 Truss connection.

Solution:

Properties of the W8 × 24:

Ag = 7.08 in.2

d = 7.93 in.

bf = 6.495 in.

tf = 0.40 in.

1. Based on gross area:

𝑃𝑡 = 𝐹𝑡𝐴𝑔
𝑃𝑡 = 0.60𝐹𝑦𝐴𝑔

𝑃𝑡 = 0.60(36)(7.08) = 152.9 kips

2. Based on effective net area:

𝑃𝑡 = 𝐹𝑡𝐴𝑒
𝑃𝑡 = 0.50𝐹𝑢𝐴𝑒
𝐴𝑒 = 𝑈𝐴𝑛
27
𝐴𝑛 = 𝐴𝑔 − 4(1.0)(0.40) = 7.08 − 1.60 = 5.48 in.2
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

For evaluation U, the member is covered by Case I if 𝑏 𝑓 ≥ 23𝑑

2
3 𝑑 = 0.67(7.93) = 5.31 in.
𝑏𝑓 = 6.495 in. > 5.31 𝑖𝑛.

Therefore, U = 0.90, and

𝑃𝑡 = 0.50𝐹𝑢𝑈𝐴𝑛
𝑃𝑡 = 0.50(58)(0.90)(5.48) = 143.0 kips

3. The block shear consideration involves four “blocks” as shown in Fig. 2.10, two in
each flange.
𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡 = 𝐴𝑣(0.30𝐹𝑢) + 𝐴𝑡(0.50𝐹𝑢)

𝐴𝑣 = 4(0.4
1.0)[7.5 − 2.5(1.0)] = 8.0 in.2
𝑡 𝐴 = 4(0.4) [1.5 − ]=
2
1.6 in. 2

Therefore,

𝑃𝑡 = 8.00(0.30)(58) + 1.60(0.50)(58) = 185.6 kips

For this member, Pt = 143.0 kips as controlled by a rupture failure based on the least net area.

1 "
62
1 "
12

1 "
12

3"
3"

11 "

Fig. 2.10 Flange block shear

Example:

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Find the tensile load capacity Pt for the double-angle tension member shown in Fig. 2.11. All

structural steel is A36 (Fu = 58 ksi). Assume that the welds are adequate and do not control the
tensile capacity.
3
8" Gusset plate

6"
2Ls 3 12 X 3 X 38 (LLBB)
Ag = 4.59 in.2

3"
1"
32
Pt

Fig. 2.11 Double-angle tension member

welded end connection

Solution:

1. Based on gross area:

𝑃𝑡 = 𝐹𝑡𝐴𝑔
𝑃𝑡 = 0.60𝐹𝑦𝐴𝑔

𝑃𝑡 = 0.60(36)(4.59) = 99.1
kips

2. Based on effective net area:

Since only one leg of each angle is connected to the
gusset plate,
𝑃𝑡 = 𝐹𝑡 𝐴𝑒
𝑃𝑡 = 0.50𝐹𝑢 𝐴𝑒

Where 𝐴𝑒 = 𝑈𝐴𝑔 and U = 0.85 from Case II. Therefore,

𝑃𝑡 = 0.5(58)(0.85)(4.59) =
113.1 kips

3. Check block shear in the gusset plate along the

perimeter of the welds:

𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡 = 𝐴𝑣(0.30𝐹𝑢) +29𝐴𝑡(0.50𝐹𝑢)

𝐴𝑣 = 0.375(6 + 3) = 3.38 in.2
𝐴𝑡 = 3.5(0.375) = 1.313 in.2
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑃𝑡 = 3.38(0.30)(58) + 1.313(0.50)(58) = 96.9 kips

For this member, Pt = 96.9 kips as controlled by block shear failure in the gusset plate.

Length Effects

Tension members do not suffer from the problems of instability and buckling that
compression members and beams do. Therefore, length plays a minor role. The ASD
specifications, section B7, suggests upper limits for the slenderness ratios of tension
members.

The recommended upper limit for slenderness ratio l / r is 300 for all tension members.

The recommended upper limit on l / r is preferred, but is not mandatory, and applies to
tension members other than steel rods and cables.

If the slenderness ratio is within the recommended limit, however, there will be some
resistance to undesirable vibrations as well as some resistance to bending and deformation
during shipping and erection handling.

Example:

For W8 × 24 tension member with 20 ft long, determine whether the member’s

slenderness ratio is within the ASD specifications recommendations.

Solution:

Maximum preferred l / r = 300. Calculate actual l / r. use the least radius of gyration.
𝑙
20(12) = 149 < 300 OK
=
𝑟 1.61

DESIGN OF TENSION MEMBERS

The design, or selection, of adequate tension members involves provision of the following:

1. Adequate gross area (Ag).

2. Adequate radius of gyration r to meet the preferred l / r limits.
3. Adequate net area (An or Ae).
4. Adequate block shear strength.
5. A cross-sectional shape such that the connections can be simple.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

The minimum required ptoperties as governed by the first two foregoing items involve
strength and slenderness and are easily calculated by using the principles already
discussed.

The third required property (net area) is also easily calculated. Sections, however, are
tabukated on the basis of gross area. The relationship between gross area and net area
depends in part on the thicknessof the material, which is unknown at this point.

The fourth item (block shear strength) depends on the material thickness, the size and type
of fasteners, and the geometry of the connection. If the connection details can be
estabilshed or approximated, a required thickness can be calculated.

Alternatively, the block shear strength cn be checked once the member has been selected
and the connection has been designed.

The fifth consideratrion involves the way the member will fit into and be affected by the
structure of which it is a part.

The selection of a tension member (particularly in trusses) must be based on assumed end
conditions.

After a member is selected and the end connections have been designed, it may be
necessary to revise the selection.

Since block shear plays such an imprtant role, a preliminary understanding of simple
connections for tension members is essenatial.

Two types of connections will be considered here, and they are, lap and butt connections,
as shown in Fig. 2.12.

P
P

(a) Lap connection

P P

(b) Butt connection

Fig. 2.12 Types of connection.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

In most structural connections the bolt is required to prevent the movement of the
connected material in a direction perpendicular to the length of the bolt as shown in Fig.
2.13.

Shear plane

P P

Fig. 2.13 Bolt in single shear.

In the connection shown, the bolt has a tendency to shear off along the single contact plane
of the two plates. Since the bolt is resisting the tendency of the plates to slide past one
another along the contact surface and is being sheared on a single plane, the bolt is said to
be in single shear.

In a butt connection, there are two contact planes. Therefore, the bolt is offering resistance
along two planes and is said to be in double shear.

Table I-D from the ASD manual, Part 4, provides bolt strengths.

Three types of bolts are given: A307, A325, and A409. These designations are ASTM
material designations. The notations under “loading” refer to single shear (S) and double
shear (D).

In addition to considering shear failure in the bolts, the members being connected where
they bear on the bolts must be considered. If a material is overly thin, the hole will
elangoate into an oval shape and the connection will be said to have failed in bearing as
shown in Fig. 3.14.

P P

Localized deformation of plate

Bolt shank
Elongated hole

Fig. 2.14 Bearing failure.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Table I-E, part 4 shows the bearing strengths of various diameters of bolts.

The ASD Specifications recommends that the minimum distance between bolts be taken as
2 32 times the bolt diameter (and that 3 times the bolt diameter is preferred).

Table J3-5, Part 5 shows minimum edge distances for various bolt sizes and edge
conditions.

Example:

Select the lightest double-angle tension member for member BC, a web member in a light
truss, as shown in Fig. 2.15. The tensile load will be 46 kips. Use A36 steel. The length is
13.4 ft. fasteners will be 34-in.-diameter A325 bolts and will connect the double-angle
member to a 38-in.-thick gusset plate. Assume that the strength of the gusset plate will not
control.

3"
8
B

A
A
Holes for 34"
C diam.
bolts
Section A-A

Fig. 2.15 Tension member design

Solution:

As was previously discussed, the design solution involves providing

1. Adequate gross area.

2. Adequate radius of gyration r.
3. Adequate net area An or Ae.
4. Adequate block shear capacity.

The required gross area, based on general yielding of the member, is calculated from
𝑃 𝑃
𝑔 46
𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴 = 𝐹𝑡 = 0.60𝐹𝑦 = 060(36) = 2.13 in.2

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STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

34
STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

35
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

The minimum required radius of gyration is calculated from

𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑙
𝑟
= 300
13.4(12)
𝑙 = 0.54 in.
𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑, 𝑟 = 300
=
300
Remaining to be considered are (a) minimum required net area based on tensile fructure
and (b0 block shear, both of which depend on the thickness of the member and on
connection details.

Assume one row (one gage line) of 34-in.-diameter A325 bolts.

A check of double-angle properties shows that angles of 14-in. thickness could provide the
required Ag of 2.13 in2.

1
Proceeding farther with the -in. thick angles, compare the allowable load per bolt for
4
shear and bearing from Tables
I-D and I-E.
- The shear allowable load = 15 kips.
- The bearing allowable load = 2 × 13.1 = 26.2 kips.

Therefore, the required minimum number of bolts can be calculated as

36
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

46
Use 4 bolts. 𝑁= =
bolt allowable
Assuming a minimum bolt spacing of 15
= 3.07

2
2 (𝑑𝑏) = 2.0 in.
3

The minimum edge distance = 1 41 in.

(From Table J3.5)
The connection detail would appear as shown in Fig. 2.16.

1"
14
3 @ 2"

1"
14 P = 46 kips

2 Ls

Fig. 2.16 Connection detail.

Changes in the angle thickness or changes in the connection, which affect strength, may yet
occur. For instance, if the bolt spacing or edge distance changes in the final connection, the
block shear strength will be affected.

Assuming an angle thickness of 14 in., calculate the minimum required gross area based on

the required effective net area Ae.

𝑃𝑡 = 𝐹𝑡𝐴𝑒 = 0.50𝐹𝑢𝐴𝑒
𝑃𝑡 = 0.50𝐹𝑢𝑈(𝐴𝑔 − 𝐴ℎ𝑜𝑙𝑒𝑠)

Therefore,

𝑃
𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴𝑔 = + 𝐴ℎ𝑜𝑙𝑒𝑠
𝑈(0.5)𝐹𝑢

From Table 1 a reduction coeffecient U of 0.85 is selected, from which

37
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

46 3 1 1
𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴𝑔 = + ) ( ) = 2.30 in.2
+2(
0.85(0.5)(58) 4 8
4 on the preceding requirements.
Select a double-angle member based

2 L3 × 2 × 14 (LLBB) meets the requirements.

𝐴𝑔 = 2.38 in.2

OK

𝑟𝑦 = 0.891 in. OK
3 1
+ diameter
Check the block shear cap[acity. The hole = 0.875isin.taken as
4 8

𝑃𝑡 = 𝐴𝑣𝐹𝑣 + 𝐴𝑡𝐹𝑡 = 𝐴𝑣(0.30𝐹𝑢) + 𝐴𝑡(0.50𝐹𝑢)

𝐴𝑣 = 2(0.25)(7.25 − 3.5(0.875)) = 2.09 in.2
𝐴𝑡 = 2(0.25)(1.25 − 0.5(0.875)) = 0.406 in.2

Block shear capacity is calculated as

𝑃𝑡 = 2.09(0.3)(58) + 0.406(0.5)(58) = 48.1 kips

48.1 kips > 46 𝑘𝑖𝑝𝑠
Use 2 L3 × 2 × 14.
𝑂𝐾
Note that 2 L2 21 × 2 12 × 14 would also be satisfactory.

38
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Flowchart 1: Tension member analysis

Start

Known: shape, end connection

details, l, Ag, Fy, Fu, least r.

Find Pt based on yielding in the gross section (Pt =

0.60 Fy Ag)

Determine the reduction coefficient U

Determine the net area (An = Ag – Aholes)

Find Pt based on fracture in the effective net

section (Ae = UAn)

Find Pt based on block shear (Pt

= Av Fv + At Ftv)

Check 𝑙/𝑟 ≤ 300 (preferred)

39

Lowest Pt controls
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Flowchart 2: Tension member design

Determine
Determine P, l, Fy, Fu, details of
Start end connection (assume), 𝑙
𝑚𝑖𝑛. 𝑟𝑒𝑞′ 𝑑, 𝑟 =
shape type desired 300

Determine req’d are based on

Select trial yielding in the gross section
𝑃
section 𝑟𝑒𝑞′ 𝑑, 𝐴𝑔 = (Use this for
0.60𝐹𝑦
initial trial section selection)

Assume reduction
coefficient U based on shape
and end connection details

Determine 𝐴𝑛 = 𝐴𝑔 − 𝐴ℎ𝑜𝑙𝑒𝑠

Determine Pt based on fracture

in the effective net section

𝑃𝑡 = 0.50𝐹𝑢 𝑈𝐴𝑛

Reselect trial section. Determine Pt based

Ensure that min. req’d 𝑃𝑡 ≥ 𝑃 ? on block shear
r and Ag are provided 𝑃𝑡 = 𝐴𝑣 𝐹𝑣 + 𝐴𝑡 𝐹𝑡

Specify section Adjust connection

𝑃𝑡 ≥ 𝑃 ?
to use and/or reselect trial
section. Ensure that
min. req’d r and Ag
End are provided

40
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

3. Compression Members
Structural members that carry compressive loads are sometimes given names that identify
them as to their function. Compression members that serve as bracing are commonly called
struts. Other compression members may be called posts or pillars. Trusses are composed
of members that are in compression and members that are in tension. These may be either
chord or web members. The main vertical compression members in building frames are
called columns, which are of primary interest here.

Columns are compression members that have their length dimension considerabily longer
than their least cross-sectional dimensions.

Members considered here are subjected to axial loads (concentric loads), that is, the loads
are coincedent with the longitudenal centroidal axis of the member. This is a special case
and one that exists rarely. Where small eccentricities exist, however, it may be
assumedthat an appropraite factor of safety will compensate for the eccentricity, and the
column may be designed as ttough it were axially loaded. Columns may support varying
amounts of axial load and bending moment. If the range of possible combinations of load
and moment supported on columns, then at one end of the range is the axially loaded
column. This column carries no moment. At the other end of the range is the member that
carries only moment with no (or very little) axial load. (As a moment-carrying member, it
could be considered a beam). When a column carries both axial load and moment, it is
called a beam-column.

Commonly used cross sections for steel compression members include most of the rolled
shapes. These and other typical cross sections are shown in Fig. 3.1.

For the W shapes, the cross sections usually used are those that are rather square in shape
and that have nominal depths of 14 in. or less. These shapes are more efficient than others
for supporting compressive loads (the deeper shapes are more efficient when used as
bending members). For larger loads it is common to use a built-up cross section. In
addition to providing increased cross-sectional area, the built-up sections allow a designer
to tailor to specific needs the radius of gyration (r) values about x-x and y-y axes.

In dealing with compression members, the problem of stability is of great importance.

Unlike tension members, where the load tends to hold the members in alignment,
compression members are very sensitive to factors that may tend to cause lateral
displacements or buckling.

The buckling problem is intensified and the load-carrying capacity is affected by such
factors as eccentric load, imperfection of material, and initial crookedness of the member.
Also, reidual stresses play a role. These are the variable stresses that are “locked up” in

41
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

the member as a result of the method of manfacture, which involves inequal cooling rates
within the cross section.

(a) (b) (c) (d) (e)

Shapes Tubes and pipes

(f) (g) (h)

Built-up
sections
Fig. 3.1 Compression member cross sections.

Ideal Columns

Euler’s formula gives the buckling load Pe for a pin-ended, homogeneous, initially straight, long
column of an elastic material that is concentrically loaded. This is considered to be the ideal
column. The Euler buckling load is expressed as
𝜋2𝐸𝐼
𝑃𝑒 =
𝑙2
where

Pe = concentric load that will cause initial buckling

π = mathematical constant (3.1416)

E = material modulus of elasticity

I = least moment of inertia of the cross section

l = length of the column from pin end to pin end

42
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Tests have verified that Euler’s formula accurately prdicts buckling load, where the
buckling stress is less than (approximatly) the proportional limit of the material and
adherence to the basic assumptions is maintened. Since the buckling stress must be
compared with the proportional limit, Euler’s formula is commonly written in terms of
stress. This may easily be derived from the preceding buckling load formula, recocnizing
that I = A r2:

𝜋2𝐸𝐼
𝑓𝑒 =
(𝑙/ 𝑟)2
where

fe = uniform compressive stress at which initial buckling occurs

r = least radius of gyration of the cross section �𝐼⁄𝐴, where A is the cross-sectional area.

l / r is termed the slenderness ratio.

It is convenient to classify columns into three broad categories according to their modes of
failure as shown in Fig. 3.2.

(a) Short (b) Intermediate (c) Long (Slender)

Fig. 3.2 Columns types and failure modes.

Columns that fail by elastic buckling, where buckling occurs at compressive stresses within
the elastic range, are called long columns. A very short and stocky column will obivously
not fail by elastic buckling. It will crush or squash owing to general yielding, and
compressive stresses will be in the inelastic range. If yielding is the failure criterion, the
failure load may be determined as the product of Fy and cross-sectional area. This column is

43
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Dr. Ammar A. Ali

called a short column. A column that falls between these two extremes will fail by inelastic
buckling when a localized yielding occur. This will be initiated at some point of weakness
or crookedness. This type of column is called an intermediate column. Its failure strength
cannot be determined by using either the elastic buckling criterion of the long column or
the yielding criterion of the short column. It is designed and analyzed by using emperical
formulas based on extensive test results.

Example 1:

Determine the Euler buckling load Pe for an axially loaded W14×22 shown in Fig. 3.3. The column
has pinne ends. Assume A36 steel with a proportional limit of 34 ksi. The column length
is 12 ft.

Pe=?

Pinned

Pinned

Pe=?

Fig. 3.3 Euler column

analysis.
Solution:

Properties of the W14 × 22 are

𝐴 = 6.49 in.2 , 𝑟𝑦 = 1.04 in., 𝐼𝑦 = 7.00 in.4

Solve for the buckling stress:

𝜋2𝐸

𝜋2(29,000)
𝑓𝑒 = =
( 𝑙 ⁄𝑟 ) 2 [(12 ×
= 14.93 ksi
12)/1.04]2

14.93 ksi < 34 𝑘𝑠𝑖

44

(Euler′sformula applies)

𝑃 =𝑓𝐴=
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Or

𝜋2𝐸𝐼
= 𝜋 (29,000)(7.00) = 96.6 kips
2
𝑃𝑒 =
𝑙2 (12 × 12)2
Note that in the preceding example, the use of higher-strength steel would not change the
value of the buckling load. The yield stress Fy plays no role in the Euler formula.

Effective Lengths

Euler’s formula gives the buckling load for a column that has pinned ends. A practical
column, in addition to being nonperfect in other aspects may have end conditions (end
supports) that provide restraint of some magnitude and will not allow the column ends to
rotate freely.

The use of Euler’s formula may be extended to columns having other than pinned ends
through the use of an effective length. This concept is illustarted in Fig. 3.4, where a column
having rigid (or fixed) ends is shown. The deflected shape of the buckled column is shown
by dashed line.

Pe
l/4 Fixed end

l l/2 Inflection points l/2

Fixed end
l /4
Pe
(a) (b)

Fig. 3.4 Column effective length.

Inflection points exist at the quarter points of the column length. These points of zero
moment and may theoretically be replaced with pins without affecting the equilibrium or
the deflected shape of the column. If the central portion of the fixed-ended column is
considered separately, it is seen that it behaves as a pin-ended column of length l /2. The
Euler critical load for the fixed-ended is then seen to be the same as for a pin-ended column
of length l / 2. The length l / 2 is said to be the effective length of the fixed ended column,
and the effective length factor K is 12 or 0.50. The effective length is written as Kl, where l is

45
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Dr. Ammar A. Ali

the actual length of column. The Euler’s formula may be rewritten with the inclusion of the
effective length as

𝜋2𝐸𝐼
𝑃𝑒 =
(𝐾𝑙)2

and

𝜋2𝐸
𝑓𝑒 =
(𝐾𝑙/𝑟)2

For the fixed-ended column,

𝜋2𝐸𝐼
4𝜋2𝐸𝐼
𝑙2
𝑃 = =
𝑒
(0.5𝑙)2by a factor of 4 when rigid end supports are
It is seen that the buckling load is increased
furnished for a column.

Other combinations of column end conditions are covered in the ASD Specifications
Commentary. Table C-C2.1 provides theoretical K values for six idealized conditions in
which joint rotation and translation are either fully realized or nonexistent. Since there is
no perfectly rigid column support and no perfect pin support, the referenced table also
provides recommended design values for K where ideal conditions are approximated.
These values are slightly higher than the ideal values and therefore are conservative (the
predicted Pe will be on the low side).

Euler’s formula for buckling load may also be adapted to result in an expression for an
allowable compressive load capacity (Pa). A factor of safety (F.S.) is introduced

𝑃𝑒
𝑃𝑎 =
F. S.

𝜋2𝐸𝐼
𝑃𝑎 =
(𝐾𝑙)2(F. S. )

Example 2:

A W10 × 49 column of a36 steel has end conditions that approximate the fixed-pinned
condition (fixed at the bottom, pinned at the top, no sidesway). Assume a proportional limit
of 34 ksi.

allowable compressive load capacity, Pa, using

(a) If the length of the column is 26 ft, find the 46
Euler’s formula and factor of safety of 2.0.
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

(b) What is the minimum length of this column at which the Euler formula would still be
valid?

Solution:

From the ASD Specification Commentary, Table C-C2.1, K (for design)= 0.80. For the
W10 × 49, 𝐴 = 14.4 in.2, 𝐼𝑦 = 93.4 in.4, and 𝑟𝑦 = 2.54 in.

(a) Find fe first and check the applicability of Euler’s formula:

𝜋2𝐸 𝜋2(29,000)
𝑓𝑒 = = = 29.6 ksi
(𝐾𝑙/𝑟)2 [0.8(26 × 12)/2.54]2

29.6 ksi𝑃<𝑒 34 𝑘𝑠𝑖 𝑓𝑒𝐴 29.6(14.4) (Euler′s formula

=
𝑃applies) = = = 213 kips
𝑎 F. S. F. S. 2.0

(b) Find the length at which fe equals the proportional limit:

𝜋2𝐸
𝑓𝑒 =

𝜋2𝐸 (𝐾𝑙/𝑟)2
𝜋2(29,000)
𝑙=� 2 =� = 291 in. = 24.3 ft
𝑓𝑒 (𝐾⁄𝑟) 34(0.8⁄2.54 )2

Example 3:

Use Euler’s formula to select an A-shape column to support an axial load of 50 kips. The
length is 12 ft, and the ends are pinned. Use A36 steel with proportional limit assumed to
be 34 ksi. Check the applicability of Euler’s formula. Assume a factor of safety = 3.0 (Note:
this is not the AISC method of column selection).

Solution:

𝜋2𝐸𝐼
𝑃𝑎 =
(𝐾𝑙)2(F. S. )

𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐼 = 𝑃𝑎 (𝐾𝑙) (F. S. ) = 50(1.0 × 12 × 12) (3.0) = 10.9 in. 4

2 2

𝜋2𝐸 𝜋2(29,000)
Try W6 × 20:

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝐼𝑦 = 13.3 in.4 , 𝐴 = 5.87 in.2 , 𝑟𝑦 = 1.50 in.

Check the applicability of Euler’s formula and the capacity of the W6 × 20:

𝑓𝑒 = 𝜋 𝐸 = 31.06 ksi < 34 𝑘𝑠𝑖𝜋 (29,000)

2 2
=(
(𝐾𝑙/𝑟)2 144/1.50)2

Calculating the buckling load,

𝑃𝑒 = 31.06(5.87) = 182 kips

From which the allowable compressive load capacity is

�
�𝑒

182𝑃 = = = 60.8 kips >

𝑎
F. S. 3.0
50 𝑘𝑖𝑝𝑠 OK

ASD Specifications:
Allowable Stresses in Compression Members

Previously, Euler’s formula was used to analyze and

design columns. In each case the applicability of the appraoch was checked. In each case Euler’s
formula did result in fe less than the proportional limit. In effect, all the columns were
slender columns. Practical columns, however, generally do not fall into this category. The
practical analysis / design method must concern itself with the entire possible range of
slenderness ratio Kl / r. Theoritical formulas are not applicable for intermediate and
short columns because of many material and geometric uncertanities. The strength
of intermedaite and short columns cannot be predicted accurately theoritically;
therefore, the results of extensive testing and experience must be utilized.

A plot of the failure stresses of columns versus their Kl / r

ratios as determined by testing, is shown in Fig. 3.5.

Since no two practical columns are identical, the failure

stresses are expressed to fall within a range of values for a particular Kl / r value.
Columns with Kl / r values to the right of line A-A have theor failure stresses closely
predicted by Euler’s formula. They are subject to elastic buckling where the buckling
occurs at a stress less than the proportional limit. Columns with Kl / r values to the left of
line A-A fail by inelastic buckling (yielding occurs), and a departure of the test data from
the curve that represents euler’s formula is noted.
48
The ASD Specifications allowable stress for compression
members, as found in Section E2, may be shown as in Fig. 3.6.

The maximum Kl / r is prefeably limited to 200 for

 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

formula curve. It is esentially the same curve with a factoe of safety applied. Allowable axial
compressive stress section is denoted Fa.

Fig. 3.5 Column test data.

Fig. 3.6 ASDS Fa versus Kl / r.

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Dr. Ammar A. Ali

The value of Kl / r that seperates elastic buckling from inelastic buckling has been
arbitrarily established as that value at which the Euler buckling stress (fe) is equal to Fy / 2. This Kl /r

value is denotes as Cc. Its value may be determined as follows:

𝜋2𝐸
𝑓𝑒 =
(𝐾𝑙/𝑟)2

Let fe = Fy / 2 and let Kl / r = Cc; then 𝐹𝑦 = 𝜋2𝐸

2 (𝐶 )2 𝑐

From which

2𝜋2𝐸
𝐶𝑐 = �
𝐹𝑦

Table 4 in the Numerical Values Section of the ASD Specifications list values of Cc for various

values of Fy.

For column Kl / r values less than Cc, Fa is determined by

(𝐾𝑙⁄𝑟)2 𝐹𝑦
[1 − 2 ]
2𝐶
𝐹𝑎 = 𝑐 (𝐸2 − 1)
5 + 3(𝐾𝑙⁄𝑟) − (𝐾𝑙⁄𝑟 )3
3 8𝐶𝑐 8𝐶 𝑐
3

For column Kl / r values greater than Cc,

12𝜋2𝐸
(𝐸2 − 2)
𝐹𝑎 =
23(𝐾𝑙/𝑟)2
This is the famialiar Euler formula for buckling stress with a factor of safety of 23 / 12 or
1.92 incorporated.

ASD Manual contains tables that are useful in the determination of Fa; see Tables C-36 and C-50 in
Part 3 entitled “Allowable Stresses for Comprssion Members”.

Analysis Of Columns Using ASD Specifications:

Example 4:

Find the allowable compressive load capaity Pa for a W12 × 120 column that has a length of
16 ft. Use A36 steel. The ends are pinned.
50
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Solution:

For the W12 × 120,

𝐴 = 35.3 in.2 , 𝑟𝑦 = 3.13 in.

𝐾𝑙 1.0(16)(12)
= 3.13 =
𝑟
The Kl /r value has been rounded to61
the nearest whole number from table use.

ASD Manual, Part 3, Table C-36: Fa = 17.33 ksi.

Therefore,

𝑃𝑎 = 𝐹𝑎𝐴 =
17.33(35.3) = 612
kips

Example 5:

A W10 × 68 column of A572 grade 50 steel is to carry an axial load of 400 kips. The length
is 20 ft. Determine whether the column is adequate if

(a) The ends are pinned.

(b) The ends are fixed.

Solution:

For the W10 × 68,

𝐴 = 20.0𝐾𝑙 in.
2
, 𝑟𝑦 = 2.59 in.
= 1.0(20)(12)
𝑟 = 93
(a) K = 1.0 from the ASD Specifications, Table 2.59
C-C2.1; therefore,
And from Table C-50, Fa = 16.29 ksi.
N𝐍
𝑃𝑎 = 𝐹𝑎𝐴 = 16.29(20.0) = 326 kips < 400 𝑘𝑖𝑝𝑠
𝐍
(b) K = 0.65 from the ASD Specifications, Table C-C2.1; therefore,

𝐾𝑙 0.65(20)(12) = 60
= 2.59
𝑟
And from Table C-50, Fa = 22.72 ksi.

𝑃𝑎 = 𝐹𝑎𝐴 = 22.72(20.0) = 454 kips > 400 𝑘𝑖𝑝𝑠

51
OK
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Example 6:

Find the compressive axial load capacity for a built-up column that has a cross section as
shown in Fig. 3.7. The steel is A36, the length is 18 ft, and the ends are assumed to be fixed-
pinned (totally fixed at bottom; rotation free, translation fixed at top).

6.38" W12 X 65
6.00"
y

4" PL
3

x 18'
12.12" x

Fig. 3.7 Built-up column.

Solution:

Properties of the W12 × 65 are

𝐴 = 20.0 in.2 , 𝑟𝑦 = 2.59 in. , 𝑏𝑓 = 12.00 in.

𝐼𝑥 = 533 in.4 , 𝐼𝑦 = 174 in.4

Determine the least moment of inertia for the built-up cross section:

�
1 � = � 𝐼𝑐 + � 𝐴𝑑2
𝐼 = 533 + 2 ( )(0.75)(12.12)3 = 756 in.4
𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 12

𝐼𝑦 = 174 + 2(0.75)(12.12)(6.38)2 = 914 in.4

Notice in the Iy calculation the Ic terms for the plates have been neglected, since they are very
small.

Calculating the radious of gyration:

𝑡𝑜𝑡𝑎𝑙 𝐴 = 19.1 + 2(12.12)(0.75) = 37.3 in.2

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Dr. Ammar A. Ali

𝐼𝑥 756
𝑟𝑥 = � =� =
𝐴 37.3
4.50 in.
𝐾𝑙 0.8(18)(12)
=
𝑟 = 38.4 (use 38)
4.50

From Table C-36 of the ASD Manual, Part 3, gives Fa = 19.35 ksi.

𝑃𝑎 = 𝐴𝐹𝑎 = 37.28(19.35) = 721 kips

Columns are sometimes braced differently about the major and minor axes, as shown by
column AB in Fig. 3.8. If all connections to the column are assumed to be simple (pinned)
connections, the deflected shapes for buckling about the two axes will be as shown.

Fig. 3.8 Column unbraced lengths.

Note that the column is braced so that the unbraced length for weak axis buckling is less
than the unbraced length for buckling about the strong axis. In this situation either axis

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Dr. Ammar A. Ali

may control, depending on which has the associate longer Kl / r ratio. Naturally, if there is
no reasonable certainty that the bracing will not be removed, the columns should be
designed with the bracing neglected.

Example 7:

Find the allowable compressive axial load capacity for a W10 × 88 that has an unbraced
length of 24 ft with respect to axis x-x and 12 ft with respect to axis y-y. Assume an A572
grade 50 steel member, pin-connected at the top and fixed at the bottom. (Assume that the
column is pin-connected at mid-height).

Solution:

W10 × 88: 𝐴 = 25.9 in.2 𝑟𝑦 = 2.63 in. 𝑟𝑥 = 4.54 in.

𝐾𝑙 = 1(12)(12) = 54.8 (top part of column)

𝑟𝑦 2.63
𝐾𝑙 = 0.8(12)(12) = 43.8 (bottom part of column)
𝑟𝑦 2.63
𝐾𝑙 = 0.8(24)(12) = 50.8 (top part of column)
𝑟𝑥 4.54

For 𝐾𝑙 = 55, 𝐹𝑎 = 23.55 ksi

𝑟

𝑃𝑎 = 𝐴𝐹𝑎 = 25.9 (23.55) = 610 kips

A common type of column used in one-story commercial construction is the unfilled

circular steel pipe column, also available as round HSS. With equal stiffness in all directions,
it is an efficient compression member. Connections to the pipe column may require special
considerations, however.

Three categories of pipe for structural purposes are manufactured: standard, extra-strong,
and double-extra-strong. Steel pipe is covered by ASTM A53 grade B. The round HSS is
covered by ASTM A500 in various grades (see ASD Manual, Part I, Table 3).

Square and rectangular HSS are also commonly used as building columns. These are
manufactured under ASTM A500 in various grades (see ASD Manual, Part I, Table 3). The
tubular members are also relatively efficient and have an advantage in that end-connection
details are simpler than with the pipe columns.

Example 8:

54
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Find the allowable compressive load capacity for a 10-in standard steel pipe column that
has an unbraced length of 15 ft. ends are pin-connected (K = 1), and the steel is A53 grade B
(use Fy = 36 ksi).

Solution:

For the 10-in. standard steel pipe (ASD Manual, Part I),

𝐴 = 11.9 in.2 𝑟 = 3.67 in.

𝐾𝑙 1(15)(12) = 49
=
𝑟 3.67
𝐹𝑎 = 18.44 ksi

𝑃𝑎 = 𝐴𝐹𝑎 = 11.9 (18.44) = 219 kips

Example 9:

Find the allowable compressive load capacity for an HSS8.625 × 0.322 column that has
unbraced length of 20 ft. The ends are pin-connected and the steel is ASTM A500 grade B.

Solution:

For this round HSS, Fy = 42 ksi (see ASD Manual, Part 1, Table 3) From

the ASD Manual, Part 1,

𝐴 = 8.40 in.2 𝑟 = 2.94 in.

Therefore,
𝐾𝑙 1(20)(12)
=
𝑟 = 82
2.94

From ASD Manual, Part 5, Table 4, Cc = 116.7. Since Fy = 42 ksi and the ASD Manual contains Fa tables for

only Fy = 36 ksi and Fy = 50 ksi, Table 3 in Part 5 will be used. Obtain Cc by calculation or
fromTable in Part 5:

𝐾𝑙
𝐶𝑐 = 116.7
Therefore, since < 𝐶𝑐 , Table 3 will be utilized for an Fa calculation.
𝑟

𝐾𝑙⁄𝑟 82
= = 0.70
116.7
𝐶𝑐

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

From which, in Table 3, Ca = 0.400. Then Fa may be found from

𝐹𝑎 = 𝐶𝑎𝐹𝑦 = 0.400(42) = 16.80 ksi

𝑃𝑎 = 𝐴𝐹𝑎 = 8.40(16.80) = 141.1 kips Note

also that ASD Specification Equation (E2-1) could be used for Fa. Example 10:

Find the allowable compressive load capacity for a W8 × 40

with an unbraced length equal to 26 ft. The member is usedin a wind-bracing system and is
pin-connected. Use A36 steel.

Solution:

𝐴 = W8
For the 11.7× in.2
40: 𝑟𝑦 = 2.04 in.

𝐾𝑙 = 1(26)(12) = 153
𝑟𝑦 2.04

𝐹𝑎 = 6.38 ksi

𝑃𝑎 = 𝐴𝐹𝑎 = 11.7(6.38) = 74.6 kip

Design of Axially Loaded Columns

The selection of cross sections for columns is greatly facilitated by the available of design
aids. The allowable axial stress Fa depends on the effective slenderness ratio Kl / r of the
column provided. Therefore, there is no direct solution for a required area or moment of
inertia. If ASD Specifications Equation (E2-2) were knowon to control, a required least r
could be calculated. This is no a practical solution, however.

Most structural steel columns are composed of W shapes, structural tubing, and / or pipes.
The ASD Manual, Part 3, contains allowable axial load tables (referred to as the “column
load tables”) for the popular column shapes. Allowable loads (Pa) are tabulated as a
function of KL (in feet) and cover the common length L for columns in building frames is
normally taken as the floor-to-floor distance since the floor provide lateral bracing. The
effective length factor K may be determined by using the aids discussed before. For
multistory frames, steel is commonly erected in two-story sections for reason of safety,
convenience, and the need to maintain alignment of the structure. Column splices are
placed just above a floor level.

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Dr. Ammar A. Ali

The column load tables, previously discussed, may be used for analysis as well as for
design. For instance, in previous example the allowable compressive load capacity of a
W12 × 120 of A36 steel was computed to be 612 kips. From the ASD Manual, Part 3, column
load table for the W12 × 120, with 𝐾𝐿 = 1.0 × 16 ft, the allowable load of 611 kips may be
obtained directly.

The tubular values of allowable loads are with respect to the members’ minor (or weak)
axis. Although the column load tables are indispendable for the selection of the types of
cross section noted, if built-up sections are required, or a section is desired for which a
column load table is not available, trial-and-error calculation approach will have to be used.

Example 11:

Select the lightest W shape for a column that will support an axial load P of 200 kips. The
length of the column will be 20 ft, and the ends may be assumed to be pinned. Use A992
steel.

Solution:

Using the ASD Manual, Part 3, column load tables, with 𝐾𝐿 = 1 × 20 = 20 ft and
𝑃 = 200 kips, the following W shapes are observed to be adequate (𝑃𝑎 ≥ 𝑃):

W14 × 61 (𝑃𝑎 = 272 kips)

W12 × 53 (𝑃𝑎 = 241 kips)

W10 × 49 (𝑃𝑎 = 230 kips)

W8 × 67 (𝑃𝑎 = 230 kips)

The W10 × 49 is selected, since it is the lightest shape with

adequate capacity.

Example 12:

Select the lightest W10 for column AB previously discussed, P = 160 kips. The overall
length (Lx) is 30 ft. The weak axis is braced at midheight (Ly = 15 ft). Assume pinned ends (K = 1.0)
for box axes and A36 steel.

Solution:

Assume that weak axis (y-y axis)will control. Select the column using ASD Manual, Part 3,
column load tables, and then check whether the assumption is correct:

𝐾𝐿𝑦 = 15 ft 57 𝑃 = 160 kips

 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Select a W10 × 39 (Pa = 162 kips based on weak axus buckling). For the

W10 × 39, 𝑟𝑥 = 4.27 in. , 𝑟𝑦 = 1.98 in.

𝐾𝑙𝑥 = 1(30)(12) = 84.3
𝑟𝑥 4.27

𝐾𝑙𝑦 = 1(15)(12) = 90.9

𝑟𝑦 1.98

The larger Kl / r controls, and the assumption of the weak axis controlling was correct. A
W10 × 39 will be adequate.

Under different conditions it is possible that the strong axis will control and be the buckling
axis for the column. The column load tables then cannot be used directly. Once an initial
section has been selected (based on the assumption that the y-y axis controls), however, a
very rapid analysis check can be made by using the tabulated properties at the bottom of
the column load tables. The procedure is as follows:

𝑟𝑥
1. Divide the strong-axis effective length (KLx) by the ratio. 𝑟𝑦
2.
Compare with KLy. The larger of the two values becomes the controlling KL. With the
3.
controlling KL value, find Pa in the appropriate column load table.
Example 13:

Rework the previous example except this time, the weak axis is braced at the third points
so that Ly = 10 ft. Lx remian at 30 ft.

Solution:

As previously, select on the basis of the weak axis controlling:

𝐾𝐿𝑦 = 10 ft 𝑃 = 160 kips

Try a W10 × 33, Pa = 167 kips (based on weak axis controlling) and rx / ry = 2.16:
𝐾𝐿𝑥 30 ft
= = 13.89 ft
𝑟 𝑥 ⁄𝑟 2.16
𝑦
This is an equivalent weak-axis length (i.e., column length based on weak-axis buckling
that results in the same capacity as does the 30-ft strong-axis buckling length).

Since 13.89 ft ˃ Ly, the strong axis controls.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Rounding the KL of 13.89 ft to 14 ft and entring the column load table for the W10 × 33
gives Pa = 142 kips, and

142 kips ˂ 160 kips

Try W10 × 39:

𝐾𝐿𝑥 30 ft
= = 13.89 ft ≈ 14 ft
𝑟 𝑥 ⁄𝑟 2.16
𝑦
13.89 ft > 𝐿𝑦

Therefor, the strong axis controls. From the column load table,

𝑃𝑎 = 170 kips > 160 𝑘𝑖𝑝𝑠

Therefor, use a W10 × 39.

Column Base Plates (Axial Load)

Columns are usually supported on concrete supports such as footing or piers. Since the
steel of the column is a higher-strength material than the concrete, the column load must
be spread out over the support. This is accomplished by use of a rolled-steel base plate.

Base plates may be square or rectangular. They must be large enough to keep the actual
bearing pressure under the plate below an allowable bearing pressure Fp, which may be obtained
from ASD Specification, Section J9, as follows.

For a plate covering the full area of concrete support,

𝐹𝑝 = 0.35𝑓′𝑐

For a plate covering less than the full area of concrete support,

𝐴2
𝐹𝑝 = 0.35𝑓𝑐′ � ≤ ′
𝐴1
0.70𝑓𝑐
where

𝑓𝑐′ = specified compressive strength (ksi)

𝐴1 = area of steel concentrically bearing on a concrete support (in.2 )

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Dr. Ammar A. Ali

𝐴2 = maximum area of the portion of the supporting surface that is geometrically

similar to and concentric with the loaded arrea (in.2 ).

The base plate must be thick enough that bending in the plate itself will not be critical.
Two-way bending is involved, since as the column pushes down on the base plate, the parts
of the plate not directly under the column itself will tend to curl (or deflect) upward. For all
but the smallest base plates, the required plate thickness may be determined by
considering 1-in.-wide sections of the base plate to act as cantilever beams spanning in
each of two directions, fixed at the edges of a rectangle whose sides are 0.80bf and 0.95d, as shown in
Fig. 3.9.

Notation:

𝑃 = total column load

𝐴1 = 𝐵 × 𝑁, area of plate (in.2)

𝑚, 𝑛 = length of cantilever from assumed critical plane of bending, for thickness

determination (in.)

𝑑 = depth of the column section (in.)

𝑏𝑓 = flange width of the column section (in.)

bf
m 1" B or N
A
P
N 0.95d 1" d tp
A A

m fp
Section A-A
Assumed A
critical planes n 0.80bf n
for bending
B

Fig. 3.9 Column base plate design.

𝐹𝑝 = allowable bearing pressure on concrete support (ksi)

𝐹𝑏 = allowable bending stress in plate (ksi)

𝑓𝑝 = actual bearing pressure on concrete support (ksi)

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑓𝑐′ = compressive strebgth of concrete (ksi)

𝑡𝑝 = thickness of plate (in.)

The column load P is assumed to be uniformly distributed over the top of the base plate in
sufficiently rigid and will distribute the applied load so that the pressure underneath the
base plate is also uniformly distributed.

The most economical column base plate will result if the length N and the width B are
selected so that m = n. This condition is approached when

𝑁 ≈ �𝑟𝑒𝑞′𝑑 𝐴1 + ∆

where ∆= 0.5(0.95𝑑 − 0.80𝑏𝑓).

Once the length N of the plate has been determined, the required B can be computed from
required 𝐴1
required 𝐵 =
𝑁
After m and n can be determined.

The allowable bending stress in the plate is taken as 0.75 Fy (ASD Specifications, Section F2). The
required thickness of the base plate can be computed from

𝑓𝑝
𝑡𝑝 = 𝑓𝑝𝑦 𝑜𝑟
𝐹 𝑡𝑝 = 2𝑛�

2𝑚� 𝐹𝑦
Note that the largest required plate thickness will rsult from the larger value of m or n.

When the base plate is just large enough in area to accommodate the outside envlope of the
column (dimensions d and bf), the values of m and n will be small. The resulting plate
thickness will then be small also, and the assumption of uniform bearing pressure under
the plate is no longer valid. For light loads with this type of base plate, the colmun load is
assumed to be distributed on an H-shaped area of the footing under the plate. This area is
shaded in Fig. 3.10.

For small base plates that more heavily loaded, the required plate thickness may be taken
as
𝑓𝑝
𝑡 𝑝 = 2𝑛′ �
𝐹𝑦 (ASD Specification, Part 3)

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

where

�𝑑𝑏𝑓
𝑛′ =
4

𝑑 = depth of the column section (in.)

𝑏𝑓 = flange width of the column section (in.)

Column
2L Wide flange
column CL
Column
web
A A
Assumed
loaded area plate

L
L L

Plan view Section A-A

Fig. 3.10 Common load distribution.

To provide for a smooth design transition between small plates that are heavily loaded and
those that are lightly loaded, the following coefficients have been developed:

𝜆 = 2(1 − √1 − 𝑞) ≤ 1.0
√𝑞
4𝑓𝑝 𝑑𝑏𝑓
𝑞= 2 < 1.0
(𝑑 + 𝑏𝑓 ) 𝐹𝑝

When 𝜆 is less than 1.0 or q is less than 0.64, the design for lightly loaded base plates
governs. If 𝑞 > 0.64, take 𝜆 as 1.0.

After computing 𝜆, calculate 𝜆𝑛′ and determine the required plate thickness from
𝑓𝑝
𝑡𝑝 = 2(𝜆𝑛′)�
𝐹𝑦

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Dr. Ammar A. Ali

Since the expression for required tp is in the same from as the expression for tp for large plates,

the largest dimensions (𝑚, 𝑛, or 𝜆𝑛′) will control, and expression may be rewritten as
𝑓𝑝
required 𝑡𝑝 = 2𝑐�
𝐹𝑦

where c represents the maximum value of (𝑚, 𝑛, or 𝜆𝑛′).

Example 14:

Design a rectangular base plate for a W14 × 74 column that it is to carry an axial load of
350 kips. Assume that the base plate will cover the full area of a concrete pier of 𝑓′ 𝑐 = 3 ksi.
Use A36 steel.

Solution:

From the ASD Specifications, Section J9,

𝐹𝑝 = 0.35𝑓′𝑐 = 0.35(3) = 1.05 ksi

1. The required area is
𝑃
1
𝐹350
𝑝 1.05
required 𝐴 = = = 333 in2
2. For the W14 × 74, 𝑑 = 14.17 in. and 𝑏𝑓 = 10.07 in.

∆ = 0.5(0.95𝑑 − 0.80𝑏𝑓)

= 0.5[0.95(14.17) − 0.8(10.07)] = 2.70 in.

required
𝑁 ≈ √required 𝐴1 + ∆𝐴=1√333333
= + 2.70 = 20.95 in
= 15.89 in.
𝐵= 20.95
𝑁

Round these required plate dimensions to the next whole inch:

𝑁 = 21 in. 𝐵 = 16 in.

area furnished = 21(16) = 336 in.2 > 333 in.2

3. The actual bearing pressure under the plate is

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑓𝑝 = 𝑃 = 350 = 1.042 ksi < 1.05 𝑘𝑠𝑖

𝐵𝑁 16(21) OK

4. Calculate m, n, and n´:

𝑁 − 0.95𝑑 21 − 0.95(14.17)
𝑚= = = 3.77 in.
2 2
𝐵 − 0.8𝑏𝑓 16 −
𝑛= = = 3.97 in.
0.8(10.07)
2 2

�𝑑𝑏𝑓 �14.17(10.07)
𝑛′ = = = 2.99 in.
4

4
4𝑓𝑝 𝑑𝑏𝑓
𝑞=
5. Check for the case of a(𝑑lightly
+ 𝑏 ) loaded
2
𝐹 small base plate using
𝑓 𝑝
4(1.042)(14.17)(10.07)
𝑞= = 0.964 > 0.64
(14.17 + 10.07)2(1.05)

Therefore, take 𝜆 = 1.0. Alternativly, the upper limit (𝜆 = 1.0) can always be used as a
conservative assumption.

6. Calculate the required plate thickness using c as a larger 𝑚, 𝑛, and 𝜆𝑛′:

𝑓𝑝 1.042
= 1.351 in.
required 𝑡𝑝 = 2𝑐� 𝐹𝑦 = 2 3.97
( )�
36
7. Refer to the ASD Manual, Part 1, Bars and Plates – Products Available for
information on plate thickness available. Select a thickness of 1 83 in.

Use a base plate 1𝟏𝟏 × 1 𝟖𝟑 × 1′ −

𝟗𝟗. 𝟖
Example 15:

Design an economical rectangular column base plate for a W12 × 50 column that is to carry
an axial load of 65 kips. All steel is A36. Assume that the base plate will cover the full area
of the concrete support. Here 𝑓𝑐′ = 3.0 ksi; therefore,

𝐹𝑝 = 0.35𝑓′𝑐 = 0.35(3) = 1.05 ksi

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Solution:

1. The required area is

𝑃
1 = 61.9 in2
𝐹65
𝑝 1.05
required 𝐴 = =
2. For the W12 × 50, 𝑑 = 12.19 in. and 𝑏𝑓 = 8.08 in. The area of the rectangular
profile of the column is

𝑏𝑓𝑑 = 8.08(12.19) = 98.5 in.2

98.5 in.2 > 61.9 in.2

Therefore, to accommodate the envelope of the W12 × 50, B and N must be selected to the
full inch dimensions that are larger than bf and d. Use N = 13 in. and B = 9 in. This will furnish
an area of 117 in.2

3. The actual bearing pressure under the plate is

𝑓𝑝 = 𝑃 = 65 ksi < 1.05 𝑘𝑠𝑖
= 0.556
𝐵𝑁 9(13) OK

4. Calculate m, n, and n´:

𝑁 − 0.95𝑑 13 − 0.95(12.19)
𝑚= = = 0.710 in.
2 2
𝐵 − 0.8𝑏𝑓 9 − 0.80(8.08)
𝑛= = = 1.268 in.
2 2

�𝑑𝑏𝑓 �12.19(8.08)
𝑛′ = = = 2.48 in.
4

4
4𝑓𝑝 𝑑𝑏𝑓
𝑞=
5. Check for a lightly loaded
(𝑑 +plate
2
𝑏) 𝐹 𝑓 𝑝

4(0.556)(12.19)(8.08)
= = 0.508 < 0.64
(12.19 + 8.08)2(1.05)

When q ˂ 0.64, the design for a lightly loaded plate governs.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

2(1 − √1 − 0.508)
𝜆 = 2(1 − √1 − 𝑞) = = 0.838
√𝑞 √0.508

𝜆𝑛′ = 0.838(2.48) = 2.08 in.

6. Calculate the required plate thickness using c as the largest of 𝑚, 𝑛, and 𝜆𝑛′:

𝑓𝑝 0.556
= 0.517 in.
𝑡 𝑝 = 2𝑐 �𝐹𝑦 = 2(2.08)� 36

9
Refering to ASD Manual, Part 1, selct a plate thickness of 16 in.

9
7. Use a base plate 9 × 16 × 1′ − 1.

Summary procedure for column base plate design:

1. Determine the required base plate area:

𝑃
required 𝐴1 =
𝐹𝑝

2. Select B and N so that m and n are

approximately equal (if possible). Use

𝑁 ≈ √required 𝐴1 +

where
required 𝐴1
required 𝐵 =
𝑁
∆= 0.5(0.95𝑑 −
Select B and N (usually to full inches) such
0.80𝑏 ) that
𝑓

𝐵 × 𝑁 ≥ required 𝐴1

3. Calculate the actual bearing pressure under the plate:

𝑓𝑝 = 𝑃
𝐵𝑁

4. Calculate m, n, and n´:

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑁 − 0.95𝑑
𝑚=
2
𝐵 − 0.80𝑏𝑓
𝑛=
2

�𝑑𝑏𝑓
𝑛′ =
4
5. Check for the case of a lightly loaded
small base plate using 4𝑓𝑝 𝑑𝑏𝑓
𝑞= 2 < 1.0
(𝑑 + 𝑏𝑓 ) 𝐹𝑝

and, if necessary (when q ˂ 0.64),

𝜆 = 2(1 − √1 − 𝑞) ≤ 1.0
√𝑞

If q ˂ 0.64, the design for a lightly loaded base plate governs (λ should be determined). If
q
≥ 0.64, then λ = 1.0. The upper limit (λ = 1.0) can always be used as a conservative
assumption and will simplify the computations a bit.

6. Calculate the required plate thickness tp using c as𝑓 the larger of 𝑚, 𝑛, and 𝜆𝑛′:
𝑝
𝑡𝑝 = 𝐹𝑦
2𝑐�
7. Specify the base plate: width, thickness, and length.

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Dr. Ammar A. Ali

Column and Base Plate Design (Axial Load, W-Shape Column)

Column Design: Assume weak axis

controls. Select section
Start Determine L, P, Fy,
using ASD Manual column
bracing conditions,
load tables
and end conditions

Strong axis controls.

Verify that 𝑃𝑎 ≥ 𝑃. If
𝐾𝐿𝑥 No
𝐾𝐿 𝑦 ≥ ? not, reselect section sol
(𝑟𝑥 ⁄𝑟𝑦) that
𝑃 ≥ 𝑃. Make use of ( 𝑥 )
𝑟
𝑎
𝑟𝑦
for analysis.
Yes

Weak axis controls; Specify column

assumption is OK section to use

Select B and N (whole

Base plate design:
inches) so that 𝑚 ≈ 𝑛
and 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴 = 𝑃 Determine Fp, Fy,
𝐹
𝑝 dimension limits for the
𝐵 × 𝑁 ≥ req′ d 𝐴
plate, d, and bf for the
column.
Calculate 𝑓𝑝 , 𝑚, 𝑛, 𝑛′ Calculate

4𝑓𝑝 𝑑𝑏𝑓
𝑞= 2 ≤ 1.0
(𝑑 + 𝑏𝑓 ) 𝐹𝑝
Conservatively
take 𝜆 = 1.0
2(1 − √1 − 𝑞)
Yes 𝜆= ≤ 1.0
𝑞 ≤ 0.64?
Calculate 𝜆𝑛′ √𝑞
No
𝜆 = 1.0
𝑓𝑝
Calculate 𝑟𝑒𝑞′ 𝑑 𝑡𝑝 = 2𝑐 �
𝐹𝑦
Select tp and specify
End
Where c = the largest of m, n, λn´ the plate size to use

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Dr. Ammar A. Ali

4. BEAMS

Beams are among the most common members that one will find in structures. They are
sructural members that carry loads that are applied at right angles to the longitudinal axis
of the member. These lloads cause the beam to bend.

In the process of beam design, the bending moment, which is produced in the beam by the
loads it supports, will be of initial concern. Other effects, such as shear or deflection, may
eventualy control the design of the beam and will have to be checked.

Beams are sometimes called by other names that are indicative of some specialized
fucntion(s):

Girder: a major, or deep, beam that often provides support for other beams,

Stringer: a main longitudinal beam, usually in bridge floors.

Floor beam: a transverse beam in bridge floors,

Joist: a light beam that supports a floor,

Lintel: a beam spanning an opening (a door or a window), usually in masonary

construction,

Spandrel: a beam on the outside perimeter of a building that supports, among other loads,
the exterior wall,

Purlin: a beam thatsupports a roof and frames between or over supports, such as roof
trussesor rigid frames, and

Grit: generally, a light beam that supports only the lightweight exterior sides of a building
(typical in preengineered metal buildings).

Mechanics of Bending

The maximum stress due to flexure (bending) in the beam may be determined by use of the
flexure formula:

𝑀𝑐 𝑀
𝑓𝑏 = 𝐼 =
𝑆
where

𝑓𝑏 = computed bending stress (maximum at top and/or bottom)

𝑀 = maximum applied moment

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑐 = distance from the nuetral axis to the extreme outside of the cross section

𝐼 = moment of inertia of the cross section about the bending nuetral axis

𝑆 = section modulus (𝑆 = 𝐼⁄𝑐) of the cross section about the bending neutral axis.

Fig. 4.1 shows the cross section and depicts the resulting bending stress diagram. The
shape of the diagram is typical for bending stress at any pint along the beam. Several points
should be noted.
Flange fb
y
Web

c Compression
x x

Neutral axis c Tension

y
fb

(a) Cross section (a) Stress diagram

Fig. 4.1 Simple beam bending.

1.
For wide-flange beams, the moment of inertia about the x-x axis, Ix, is greater than the
moment of inertia about the y-y axis. The beam is oriented so that bending
occurs about the x-x axis. This is true except in very rare situation.
2. In this case, owing to symmetry, the neutral axis is at the center of the cross section.
The c distance is equal whether on the tension or compression.
3. The maximum stress occurs at the top and the bottom of the cross section.
4. Generally, only the maximum bending stress is of interst. Therefore, unless
otherwise stated, fb is assumed to be the maximum stress. The flexure formula may also be
used to find the stress at any level in the cross section by substituting in place of
c the apprporaite distance to that level from the neutral axis.
The allowable bending stress, Fb is specified by the ASD Specifications based on the type of steel that
is beong used and other conditions that affect the strength of the beam in bending.

For equilibrium the internal resisting moment at any point in the beam must be equal to
the external applied moment at the same.

If the external applied moment becomes the maximum allowed, the bending stresses at the

top and bottom of the beam will be equal to the allowable bending stress Fb. Additional

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

moment should not be applied, since this would cause the actual bending stress to exceed
Fb. The internal resisting moment that when the bending stress is Fb is termed MR.

From the flexural formula the resisting moment MR can be calculated by substituting Fb for
fb, and MR for M:
𝑀𝑅 𝑐 𝑀𝑅
𝐹𝑏 = =
𝐼 𝑆
Then

𝑀𝑅 = 𝐹𝑏𝑆

Allowable Bending Stress

In dealing with beam problems, it is necessary to have an understanding of the specified

allowable bending stress F b, the maximum bending stress to which a beam should
subjected. The ASD Specifications treats this topic in Section F1.1.

Neglecting later complications, the basic allowable bending stress (in both tension and
compression) to be used for most rolled shapes is

𝐹𝑏 = 0.66 𝐹𝑦 (𝐹1 − 1)

where Fy is the material yield stress. For a member to qualify for an alowable bending stress Fb

of 0.66Fy, it must have an axis of symmetry in, and be loaded in, the plane of the web. An important

condition associated with the use of this value for Fb is the lateral support of the
compression flange. The compression flange behaves somewhat like a column, and it
will tend to buckle to the side, or laterally, as the stress increases if it is not restrained in
some way. Varying amounts and types of lateral support may be present as shown in Fig.
4.2.

The distance between the points of lateral support in inches is denoted l. for convenience
this distance will be denoted Lb when it is in feet.

76𝑏 20,000
and of a beam must have adequate lateral support
𝑓
To qualify for Fb = 0.66Fy, the compression
𝑙≤ flange
such that √ 𝐹𝑦 ( 𝑑/𝐴 ) 𝑦
𝑓
𝐹
where

𝑏𝑓 = flange width of the beam (in.)

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝐹𝑦 = yield stress of the steel (ksi)

𝑑 = depth of the beam (in.)

𝐴𝑓 = area of compression flange (in.2)

Fig. 4.2 Lateral support conditions.

The smaller of the two values of l is a tabulated property for each W shape (dependent on
Fy) and is designated Lc (ft). See the ASD Manual, Part 2, Allowable Stress Design Selection Table.

If the compression flange of a baem has inadequate lateral support (l is too large), the
lateral buckling tendncy will be counteracted by reducing Fb.

Another important condition that must be met if the beam cross section is to qualify for Fb

= 0.66Fy deals with the response of the beam in an overload situation. Allowable stress design

assumes failure to occur when Fy is first reached. The beam will not fail at this point because it has a
substantial resrve of strength. If the cross section continues to strain under
increased moment, the outer fibers will further strain, but the stress will remain at Fy. Fy
72
will be reached by the fibers at levels progressively closer to the neutral axis until virtually
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

the entire cross section is stressed to Fy. When this occurs, the beam has achieved its plastic
moment capacity (this is the basis for plastic design). The cross section, however, must be
proportioned so that no local buckling of the flange or web occur before the plastic moment
capacity is achieved. A cross section that meets this criterion is said to be compact.
The 1989 ASD Specifications, Section B5.1, clasifies steel sections as compact,
noncompact and slender element sections. Only compact sections qualify for Fb = 0.66Fy.

The test for compactness is found in the ASD Specifications, section B5 and table B5.1. the
governing criteria are the width-thickness ratio of the compression flange and compression
web elements of the cross section. These are called the flange criterion and web
criterion, respectivly.

Assuming that there are no axial loads on the beam, and using the defentions for width and
thickness from the ASD Specifications, section B5.1, the two equations required may be
simplifed as follows.

For a section to be considered compact:

The flange criterion is

𝑏𝑓 ≤
2𝑡
65𝑓 𝐹
� 𝑦
The web criterion is

𝑑 640
≤
𝑡𝑤 𝐹𝑦
�
where

𝑏𝑓 = flange width of the beam (in.)

𝑡𝑓 = flange thickness of the beam (in.)

𝐹𝑦 = material yield stress (ksi)

𝑑 = depth of the beam (in.)

𝑡𝑤 = web thickness of the beam (in.)

Both the flange and the web criteria must be satisfied for a member to be considered
compact.

Example 1:
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Determine whether a W18 × 76 of A36 steel (Fy = 36 ksi) is compact.

Solution:
W18 × 76 properties are

𝑏𝑓 = 11.035 in. 𝑡𝑓 = 0.680 𝑑 = 18.21 in. 𝑡𝑤 = 0.425 in.

in. Check the flange criterion:

𝑏𝑓 11.035
= = 8.11
2𝑡𝑓 2(0.68)
65 65
= =
10.8
�𝐹𝑦 √36
8.11 < 10.8 OK

Check the web criterion:

𝑑 18.21
𝑡𝑤 = 0.425 = 42.8

640 640
= = 106.7
� 𝐹𝑦 √36
42.8 < 106.7 OK

Therefore, the W18 × 76 is compact.

The preceding could be shortened by using tabulated quantities from the ASD Manual, part
1, Properties of A shapes, or the Numerical Values furnished in Table 5 of the ASD
Specifications. The tabulted quantities are rounded slighly in some cases.

A faster way to determine compactness of cross section the rolled shapes is to calculate
the value of a hypothetical yield stress Fy that would cause equality in each of the two criteria.
For the flange criterion,
𝑏𝑓 =
2𝑡
65𝑓 𝐹𝑦
�
65
𝐹𝑦 =( )2
𝑏𝑓⁄2𝑡𝑓
For the W18 × 76

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Dr. Ammar A. Ali

65
𝐹𝑦 = ( )2 = 64.2 ksi
8.11
This shows that the flange criterion is satisfied provided that Fy does not exceed 64.2 ksi.
This value is termed 𝐹′𝑦 and is tabulated in the ASD manual, Part 1, as a property of the
W18 × 76. If the W18 × 76 is of a steel with Fy in excess of 64.2 ksi, it is not compact by the flange
criterion. Therefore, for compactness by the flange criterion, the following
condition should exist:
𝐹𝑦 ≤ 𝐹′𝑦

Should the calculated value of 𝐹′ be in excess of the heighst available 𝐹 , this is reflected by
𝑦
𝑦
the tabulation of a dash (‒) for the 𝐹′𝑦 value. For the web criterion,

𝑑 640
≤
𝑡𝑤 𝐹𝑦
�
640
𝐹𝑦 = ( )2
𝑑⁄𝑡𝑤
For the W18 × 76,
640
𝐹𝑦 = ( )2 = 224 ksi
42.8

This shows that the W18 × 76 is compact by the web criterion provide that Fy does not exceed

224 ksi. Reference to the ASD Manual, Part 1, table 1, shows that shapes with Fy in excess of 65 ksi
are currently not available. Therefore, the W18 × 76 is compsct by the web criterion in all
steels. This is a general rule. All rolled, W, M, an S shapes tabulated in the ASD Manual are
compact by the web criterion ( when fa = 0). This does not hold true for built-up section and plate

girders. Web noncompactness will cause Fb to be reduced to 0.60Fy (assuming adequate lateral

support).

If a shape does not satisfy the flange criterion, it is considered a noncompact shape, and Fb must be

reduced. The variation of Fb for rolled W shapes that have adequate lateral support is
summarized graphically in the Fig. 4.3.

75 95/�𝐹𝑦, the ASD Specifications provides for

For the range of 𝑏𝑓/2𝑡𝑓 between 65/�𝐹𝑦 and

a linear reduction in Fb to 0.60Fy according to the following equation:

 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑏𝑓
) (𝐹1 − 3)
𝐹𝑏 = 𝐹 [0.79 − 0.002 ( 2𝑡𝑓 �𝐹𝑦 ]
𝑦

Fig. 4.3 Fb for rolled W shapes with adquate lateral support.

This equation provides for a transition in allowable stress between the values of 0.66Fy and 0.60Fy for

noncompact shapes. This adequate lateral support. If 𝑏𝑓/2𝑡𝑓 exceeds 95/�𝐹𝑦, the shape is considered
to be a slender element section, and ASD specifications, Appendix B5, applies.

In determining Fb, the yield stress Fy must be known. The best source for the value of Fy is the ASD

Manual, Part 1, Tables 1 and 2. For a known shape, determine the appropriate group
from Table 2. Then, knowing the steel type, use Table 1 to determine Fy.

Example 2:

Find Fb for the following shapes. Assume adequate lateral support for the compression
flange.

(a) W30 × 132 of A36 steel.

(b) W12 × 65 of A242 steel.

(a) All W shapes in a36 steel are compact except for the W6 × 15, since its 𝐹′𝑦 is less than
Solution:
36 ksi. From the ASD Manulal, Part 1, Table 1, all shapes in A36 steel have Fy = 36 ksi. Thus

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝐹𝑏 = 0.66𝐹𝑦 = 0.66(36) = 23.8 ksi

(b) From the ASD Manual, Part 1, Table 2, W12 × 65 is found in Group 2. From Table
1, Fy = 50 ksi. From the ASD Manual properties for the W12 × 65, 𝐹′ = 43.0 𝑦ksi. Since
𝐹𝑦 > 𝐹′𝑦, the member is not a compact shape in A242 steel. Another check for
compactness is as follows. The following quantities can also be found in the ASD
Manual, Part 1, Properties of W Shapes, and Table5 of the ASD Specifications
(Numerical Values):

𝑏𝑓
= 9.9
2𝑡𝑓
65
65
=

= 9.2
� 𝐹𝑦

√50

95
65 < 𝑏𝑓 <
95 � 𝐹
95 2𝑡 �𝐹𝑦
= 𝑦 𝑓

Calculate Fb from ASD Specifications Equation (F1-3):

= 13.4
� 𝐹𝑦 𝑏𝑓
)
𝐹 = 𝐹 [50
𝑏 0.79 − 0.002 ( 2𝑡𝑓 �𝐹𝑦 ]
𝑦 √

Therefore, this is a𝐹noncompact shape, since

𝑏 = 50[0.79 − 0.002(9.9)√50] = 32.5 ksi

Analysis Of Beams For Moment

The analysis problem is generally considered to be the investigation of a beam whose

cross section is known. One may be concerned with checking the adequacy of a given
beam, determining an allowable load, or finding the maximum computed bending stress in
the beam. All these problems are related. All make use of the flexure formula and require
an understanding of allowable bending stress Fb.

Example 3:

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A W21 × 44 beam is to span 24 ft on simple supports as shown in Fig. 4.4. Assume full
lateral support and A36 steel. The load shown is superimposed load, meaning that it does
not include the weight of the beam. Determine whether the beam is adequate by

(a) Comparing the computed bending stress with the allowable bending sress.
(b)Comparing the computed applied moment with the resisting moment MR.
1.75 kips/ft
(superimposed load)

24'-0"

Fig. 4.4 Load

diagram.
Solution:

(a) Determine the computed bending stress from the flexure formula

𝑀𝑐 𝑀
𝑓𝑏 = 𝐼 =
𝑆
From the properties tables, for the W21 × 44, 𝑆𝑥 = 81.6 in.3 Moment may be determined by
shear force and moment diagram or by formula (see the ASD Manual, Part 2, Beam
Diagrams and Formulas) for review).

The total load should include the weight of the beam:

1.75 (applied load)
kips/ft
+0.044 (beam weight)
kip/ft
1.794 (total uniform load w)
kips/ft
𝑤𝐿2 1.794(24)2
Applied moment 𝑀 = = = 129.2 ft. kips
8 8
𝑀 129.2(12)
= 19.00 ksi
𝑓𝑏 = 𝑥 = 81.6
𝑆

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Dr. Ammar A. Ali

From the ASD Manual, Part 1, 𝐹′ = (−). Therefore, since 𝐹′ > 𝐹 , the
member is
𝑦
𝑦
𝑦

compact in A36 steel, and the allowable bending stress Fb is 24.0 ksi. Therefore,

(b) The applied moment m has been determined to be 129.2 ft.kips. The resisting
𝑓𝑏 < 𝐹𝑏
moment MR may be calculated from the flexure formula:
𝑀𝑅 = 𝐹𝑏𝑆𝑥24(81.6)
=
OK = 163.2 ft.
12
kips
Therefore,

𝑀 < 𝑀𝑅 OK

Example 4:

A W18 × 40 beam spans 20 ft span, as shown in Fig. 4.5. Assume A992 steel. The
comprssion flange is supported laterally at the quarter points where equal concentrated
load P are applied. Therefore, Lb = 5 ft. Determine the allowable value for each load P (kips).
indicates lateral support
for compression flange

P P

5'-0" 5'-0" 5'-0" 5'-0"

Fig. 4.5 Load diagram.

Solution:

Determine Fb and MR. W18

× 40 properties are
𝑑
𝑏𝑓 = 6.015 in. = 5.67 𝑦𝐹 = 50 𝑆𝑥 = 68.4 in.3 𝐹′𝑦 = (−)
𝐴𝑓 ksi

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Dr. Ammar A. Ali

Test for adequate lateral support by comparing Lb with Lc. From the ASD Manual, Part 2, Allowable

Stress Design Selection Table, Lc = 5.4 ft. Lb = 5 ft. Since Lb ˂ Lc, this beam has
adequate lateral support. 𝐹′ = (−), implying that 𝐹′ is high enough that this shape is
always
𝑦 𝑦
compact. Therefore,

𝐹𝑏 = 33.0(68.4)
0.66𝐹𝑦 = 33.0 ksi
𝑀𝑅 = 𝐹𝑏𝑆𝑥 = = 188.1 ft. kips
12
The resisting moment is 188.1 ft.kips. The applied moment due to the beam’s own weight
and the moment due to the three equal loads P cannot exceed the resisting moment MR. the applied
moment due to beamweight is
𝑤𝐿2 0.04(20)2
𝑀= = = 2.0 ft.
8 8
kips
The moment due to the concentrated loads may be determined by using the shear and
moment diagrams shown in Fig. 4.6 , or aids such as those found in the ASD Manual, Part 2,
Beam Diagrams and Formulas.

P P P

3P 3P
2 5'-0" 5'-0" 5'-0" 5'-0" 2

3P
2 P
2
SFD
P
2 3P
2
10 P

BMD

Fig. 4.6 Load, shear and moment diagrams.

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Dr. Ammar A. Ali

Thus,

𝑀 = 10𝑃 ft. kips

The resisting moment remaining to support the concentrated loads is

188.1 − 2.0 = 186.1 ft. kips

Equating:

10𝑃 = 186.1
186.1
𝑃= = 18.61 kips
10
Therefore, the maximum allowable value of each concentrated load P on this beam is 18.61
kips.

It should be noted that the weight of the beam itself in both of the forgoing examples has
been a very minor part of the total load carried. This is generally true; nevertheless, it
should always be considered. As one becomes more experienced, the effect of beam weight
is easier to estimate. Various shortcuts and rules of thumbs are used by designers to
simplfy the inclusion of weight of structure in design problems. In analysis problems,
where the cross sectionis known, inclusion of the beam weight is a simple matter.

Summary of Procedure: Beam Analysis for Moment Only

The following procedure for beam analysis for moment only is general and typical for the
various shapes that may be used for beams, primarily wide-flange sections and, to a lesser
extent, other sections. The precise method of solution will depend on the nature of the
particular problem, the known conditions, and the information sought.

1. Determine Fy. Use the ASD Manual, Part 1, Tables 1 and 2.

2. Check the adequacy of lateral support. See the ASD Specifications, Section F1. If
lateral support is inadequate, F will be calculated as in next section.
3. Check the compactness of bthe cross section. Use 𝐹′𝑦 from the table of properties in
the ASD Manual, Part 1.
4. Using the preceding information, determine Fb.
5. If the applied loads are known, the applied moment can be found. Draw shear and
moment diagrams or use beam formulas from the ASD Manual, Part 2.
6. If the magnitude of the applied loads is unknown, write an expression for the
applied moment in terms of the unknown loads. This can then be equated to the
resisting moment of the beam.

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Dr. Ammar A. Ali

7. The flexure formula for use in analysis is

𝑀𝑐 𝑀
𝑓𝑏 = 𝐼 =
𝑆

𝑀𝑅 = 𝐹𝑏𝑆

Inadquate Lateral Support

As the distance between points of lateral support on the compression flange (l) becomes
larger, there is a tendancy for the compression flange to buckle laterally. There is no upper
limit for l. To guard against the buckling tendancy as l becomes larger, however, the ASD
Specifications provides that Fb be reduced. This, in effect, reserves some of the beam
strength to resist the lateral buckling.

Fig. 4.7 shows a beam that has deflected vertically with a compression flange that has
buckled laterally. The result is a twisting of the member. This called lateral-torsional
buckling. For simplicity this buckling mode of the beam will be refered to as lateral
buckling.

indicates lateral support

for compression flange Unloaded
position
Deflected
position
A Deflected and
buckled
position

(a) (b) Section A-A

Fig. 4.7 Beam deflection and lateral buckling.

Two general resistance are available to counteract lateral buckling torsional resistance of
the cross section and lateral bending resistance of the compression flange. The total
resistance to lateral buckling is the sum of the two. The ASD Specifications conservatively
considers only the larger of the two in the determination of a reduced Fb.

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Dr. Ammar A. Ali

The ASD Specifications, Section F1.3, establishes empirical expressions for Fb for the
inadequate lateral support situation. Tension and compression allowable bending stresses
are treated separately. The tension Fb is reduced. For typical rolled shapes, this of no

consequence because the shapes are symmetrical and the lower Fb of the two values will control.

The ASD Specifications provides three empirical equations for the reduced compression Fb. The
mathematical expressions that give an exact prediction of the buckling strength of
beams are too complex for general use. Therefore, the ASD specifications equations only
approximate this strength for purposes of determining a reasonable Fb. The Fb that is finally

used is the larger of the Fb values determined from the applicable equations. The first two ASD
specifications Equations (F1-6) and (F1-7), gives the Fb value when the lateral
bending resistance of the compression flange provides the lateral buckling
resistance. The third, ASD specifications Equation (F1-8), gives Fb when the torsional
resistance of the beam section provides the primary resistance to lateral buckling. In no
case should Fb be greater than 0.60Fy for beams that have inadequate lateral support. The equations

that will be applicable depend on the value of the ratio 𝑙⁄𝑟𝑇, where

𝑙 = distance between points of lateral support for the compression flange (in.)

𝑟𝑇 = radius of gyration of a section comprising the compression flange plus one-third of the
compression web area taken about an axis in the plane of the web (in.) as shown in Fig. 4.8.

y
d d Compression
2 tf 1 d
2 - tf 3(2-tf)

x x
Neutral axis

Fig. 4.8 rT determination.

Here 𝑟𝑇 is tabulated quantity for rolled shapes (see the ASD Manual, Part 1), and 𝑙⁄𝑟𝑇 may
be considered a slenderness ratio of the compression portion of the beam with respect to
the y-y axis.

The equations for Fb are as follows:

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Dr. Ammar A. Ali

2 𝐹𝑦(𝑙⁄𝑟𝑇)2 (F1-6)
𝐹𝑏 = [ 3 − 1530 × 103 𝐶𝑏] 𝐹𝑦

170 × 103𝐶𝑏
𝐹𝑏 = (F1-7)
(𝑙⁄𝑟 𝑇)2

12 × 103𝐶𝑏
𝐹𝑏 = 𝑙 (𝑑/ ) (F1-8)
𝐴
𝑓

where

𝐶𝑏 = a liberalizing modifying factor whose value is between 1.0 and 2.3 that accounts for a
moment gradient over the span and a decrease in the lateral buckling tendancy; Cb may be
conservatively taken as 1.0; see ASD Specifications, Section F1.3, for details.

𝑑 = depth of cross section (in.)

𝐴𝑓 = area of compression flange (in.2)

The decision-making process for the value of Fb is depicted here.

𝑙 102 × 103 𝐶𝑏 Yes 𝐹𝑏 = 0.66𝐹𝑦

≤ ??
𝑟𝑇 � 𝐹𝑦

No

𝑙 510 × 103 𝐶𝑏 Yes Fb is larger of (F1-6) or (F1-8), but

≤ ??
𝑟𝑇 𝐹𝑦 not greater than 0.60Fy
�

No
Fb is larger of (F1-7) or (F1-8), but
not greater than 0.60Fy

Note that one will use ASD Specifications Equations (F1-6) and (F1-8) or ASD Specifications
Equations (F1-7) and (F1-8). The larger resulting Fb is used. Note that Table 5 of the
Numerical Values Section of the ASD Specifications provides the following numerical
equivelants for A36 steel (Fy = 36 ksi):
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

102 × 103𝐶𝑏
�
𝐹𝑦 = 53√𝐶𝑏

510 × 103𝐶𝑏
�
𝐹𝑦 = 119√𝐶𝑏

Example 5:

A W21 × 50 shown in Fig. 4.9, spans 36 ft on simple span. The compression flange is
laterally supported at the third points. A36 steel is used. Determine Fb for this beam.

indicates lateral support

for compression flange

12'-0" 12'-0" 12'-0"

Fig. 4.9 Beam diagram.

Solution:

The shape is compact, and Fy = 36 ksi. Check for adequacy of lateral support (l = 144 in. or

Lb = 12 ft).

For the W21 × 50:

𝑑
𝑏𝑓 = 6.53 in. = 5.96 𝑟𝑇 = 1.60 in.
𝐴𝑓

To qualify for Fb = 0.66Fy, the unbraced length of the compression flange must be equal, or less than,

76𝑏𝑓 and 20,000

(𝑑
�𝐹𝑦 / 𝐴𝑓 ) 𝐹
𝑦
As required in the ASD Specifications, Section F1.1. For A36 steel, these expressions
become

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Dr. Ammar A. Ali

556
12.7𝑏𝑓 and 𝑑
/ 𝐴𝑓

As was mentioned previously, the lesser of these two expressions is designated Lc (in feet). If Lb ≤ Lc, the

beam will qualify for Fb = 0.66Fy.

𝐿𝑏 = 12 ft

𝐿𝑐 = 6.9 ft (ASD Manual, Part 2)

Since Lb ˃ Lc, Fb must be reduced. The compression flange is therefore said to be

inadequatily braced.

The value of 𝑙⁄𝑟𝑇, which determines the applicable ASD Specifications formulas for Fb, is compared
with

53�𝐶𝑏 and 119�𝐶𝑏

𝑙 144of the ASD Specifications). Next, compute the

(from Table 5 of the Numerical Values section
𝑟𝑇 = 1.60 = 90.0
slenderness ratio of the compression flange:

Conservatively assuming that Cb = 1.0, having

53�𝐶𝑏 = 53

119�𝐶𝑙𝑏 = 119
53 < < 119
𝑟𝑇

Therefore, the applicable Fb equations are ASD

Specifications Equation (F1-6):

2
2 𝐹𝑦(𝑙⁄𝑟𝑇)2 36(90)2
𝐹𝑏 = [ 3 − 1530 × 103 𝐶𝑏] 𝐹𝑦 = [ − ] 36 = 17.14 ksi
3 1530(103)(1.0)
And ASD Specifications Equation (F1-8):

12 × 103𝐶𝑏 12(103)(1.0)
𝐹𝑏 = 𝑙 (𝑑/ ) = = 14.0 ksi
𝐴 144(5.96)
𝑓

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Dr. Ammar A. Ali

The larger Fb is used:

𝐹𝑏 = 17.14 ksi

The ASD Specifications equations for Fb for beams that have inadequate lateral support are also
applicable to built-up members and plate girders, provide that they have an axis of
symmetry in the plane of the web.

Example 6:

Determine the allowable superimposed uniformly distributed load that may be placed on
the W21 × 50 of previous example.

Solution:

Fb was determnied to be 17.41 ksi. Thus 17.41(94.5)

𝑀𝑅 = 𝐹𝑏𝑆𝑥 = = 135.0 ft. kips
12
Since the beam is a simple span member with a 36-ft span,

𝑤𝐿2 8𝑀
𝑀=
8 and 𝑤 =
𝐿2

Since, as a limit, the applied moment M can equal MR,

8𝑀𝑅 8(135.0)
𝑤= = = 0.833 kip/ft
𝐿2 362
When the beam’s own weight is subtracted, the allowable superimposed load is

833 − 50 = 783 lb/ft = 0.783 kip/ft

Design of Beams for Moment

The basis for moment design is to provide a beam that has a moment capacity (MR) equal
to or greater than the anticipated maximum applied moment M. The flexure formula is used
to determine a required sectiom modulus S:

𝑀
required 𝑆 =
𝐹𝑏

The section modulus on which the selection will be based is assumed to be the strong-axis
section Sx. The allowable Stress Design Selection Table (Sx Table) in the ASD Manual, Part 2, can be use
to make this selection. It lists common beam shapes in order odf decreasing
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Dr. Ammar A. Ali

section modulus. This table also lists the resisting moment MR of each section. The value of MR is
calculated using an allowable stress Fb of 23.8 ksi (or 23.76 ksi) rather than the
rounded value of 24.0 ksi. This may cause some small inconsistencies in calculations and
results.

Example 7:

Select the lightest W shape for the beam shown in Fig. 4.10. Assume full lateral support (Lb
= 0) and A36 stee. Consider moment only.

Solution:

The beam reactions are determined from the load diagram; the shear and moment
diagrams are drawn.

6k
3 kips/ft

14'- 0" 14'- 0"

(a)

45 k

6k
V

(b)

336 ft-kips

M
(c)

Fig. 4.10 Load, shear and moment diagrams.

Note that no beam weight is included, since the beam is unknown. An estimate of the beam
weight could be made and its effect on the moment included. This is optional at this point.
The beam weight must be included before the final selection is made, however.

𝐹𝑏 = 0.66𝐹𝑦 = 0.66(36) = 24 ksi

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Dr. Ammar A. Ali

𝑀
𝑥 336(12)
required 𝑆 = 𝐹𝑏 = = 168.0 in.3
24
From the ASD Manual, Part 2 (Allowable Stress Design Selection Table), select a W24 × 76
with an Sx = 176 in3. It is the lightest W shape that will furnish the required section
modulus. Note that the section selected weighs 76 lb/ft. Add in the effect of the beam
weight:

additional 𝑀 = 𝑤𝐿 = 0.076(28) = 7.45 ft. kips

2 2
8 8
new total 𝑀 = 336 + 7.45 = 343 ft. kips
𝑀
𝑥 343(12)
new required 𝑆 = 𝐹𝑏 = = 171.5 in3
24
The W24 × 76 is satisfactory, since 176 in3 ˃ 171.5 in3 required. Also, now check the

assumed Fb; the W24 × 76 is compact, since 𝐹′ > 𝐹𝑦 , and has adequate lateral support;
𝑦
therefore the assumed Fb is satisfactory. Use W24 × 76.

Example 8:

The beam shown in Fig. 4.11 is to be of A36 steel. Note the lateral support conditions. Select
the lightest W shape. Consider moment only.

Solution:

For this design, an estimated beam weight of 40 lb/ft (0.04 kip/ft) has been added to the
given uniform load. (This estimate may be based on anything from an educated guess to a
rough design worked quickly on scrap paper).

Establish Fb.

Fy is 36 ksi, and a compact shape that has adequate lateral support will be assumed.

𝐹𝑏 = 0.66𝐹𝑦 = 0.66(36) = 24 ksi

𝑀 129.0(12)
From which required 𝑆𝑥 = = = 64.5 in.3
𝐹𝑏 24

Select a W16 × 40 with Sx = 64.7 in3. This is a compact shape, since 𝐹′ > 𝐹 . 𝑦Check the
𝑦
adequacy of lateral support (l = 60 in or Lb = 5 ft) by comparing Lb with Lc. From the ASD

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Dr. Ammar A. Ali

Manual, Part 2, Lc = 7.4 ft. Therefore, since Lb ˂ Lc, the beam has adequate lateral support, and the

assumed Fb is satisfactory. The assumed beam weight is satisfactory.

Use 𝐖𝐖1𝟏𝟏 × 𝟒𝟒𝟒𝟒.

indicates points of lateral

support for the compression
flange

10 k 10 5k
k
4.04
kips/ft

5' 5' 5' 5'

(a)
35.27
25.20
15.07
5.0
5.07
V
(kips)
15.13
25.13

(b) 45.33

129.0

M
(ft.kips)
(c) 75.4

Fig. 4.11 Load, shear and moment diagrams.

The rapid solutions of previous examples have depended on the correct assumption of Fb. Since Fb

depends in par on the section to be selected, it cannot always be predetermined. Various design

aids have been developed to speed the design process. In prvious example, Fb was
determined for a W21 × 50 beam of A36 steel that had inadequate lateral support. For l of 144 in.

(Lb = 12 ft), Fb = 17.1 ksi. For each value of Lb, a value for Fb could be determined. A plot for Fb
90
versus Lb is shown in Fig. 4.12.
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Fig. 4.12 Fb versus Lb for a W𝟐1

𝟐 × 𝟓𝟒 (A36).

The shape of plot is typical for a compact cross section. A plot of MR versus Lb, would have the same

form, since MR = FbSx and Sx is constant for the cross section.

A family of these curves is found in the ASD Manual, Part 2. These curves make up a very
valuable design aid for beams and should be used whenever possible. The lightest beam
section may be selected directly by using only the applied moment M (ft-kips) and the
unbraced length of compression flange Lb (ft). Note that the vertical axis for these curves is allowable

moment. This is the same as resisting moment MR. The curve values of maximum MR (for

the case where Lb ≤ Lc) are calculated using an allowable bending stress Fb of 24.0 ksi. Therefore,

these maximum M values do not correlate exactly with those tabulated in the Allowable
Stress Design Selection Table. The differences are very small. The two other terms are:

Lc = Maximum unbraced length (ft) of the compression flange at which the allowable bending

stress may be taken at 0.66Fy (for compact shapes), or as determined by ASD Specifications
76𝑏𝑓 and 20,000
Equation (F1-3) or (F2-3) (when applicable), 𝑑
L𝑦c is the smaller value obtained from
�𝐹 (
𝐴𝑓 ) 𝐹𝑦

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Lu = Maximum unbraced length (ft) of the compression flange at which the allowable bending

stress may be taken at 0.60Fy.

With Fb = 0.60Fy and Cb = 1.0, the value of Lu (in feet) for most shapes is given as

20,000
𝑑
( 𝐴𝑓) 𝐹 𝑦

from the ASD Specifications Equation (F1-8). For a few shapes, Lu (in feet) is given as

102,000 𝑟𝑇
� ( )
𝐹𝑦 12

from ASD Specifications Equation (F1-6). Lu is taken as the larger value obtained from these two

expressions. For lengths greater than Lc but not greater than Lu, Fb may be taken as 0.60Fy. In no

case is Lc taken as greater than Lu. Lc and Lu are unique for each shape and also vary with Fy. They are

tabulated properties and may be found in the ASD Manual, Part 2, Allowable Stress Design Selection

Table, and in the Beam Tables of Part 2, for Fy of 36 ksi and 50 ksi. Their use greatly facilitates

the determination of whether or not lateral support is adequate. The precise values of Lc and
Lu, however, are of no consequence when using the beam curves.

When using the curves, any shape represented by a curve that is found above or to the
right of a particular MR and Lb combination is a shape that is satisfcatory. The lightest
adequate shape is represented by a solid line. The dashed lines represent shapes that are
also adequate but are heavier. The ASD Manual contains beam curves for Fy = 36 ksi and Fy

= 50 ksi. These curves may be used for the range of Lb from 0 up to the maximum value indicated
in the curves.

Example 9:

Select the lightest W shape for the beam shown in Fig. 4.13. The depth of the beam is
limited to 36 in. maximum. Consider the following cases:

(a)A36 steel, Lb = 4 ft.

(b)A36 steel, Lb = 16 ft.
(c)A572 grade 50 steel, Lb = 20 ft.
92
Solution:

(a)
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Fy = 36 ksi. Determine the applied moment for a simply supported, uniformly loaded, single-
span beam. Neglect the beam weight temporarly. Thus
𝑤𝐿2 7.0(32)2
𝑀= = = 896 ft. kips
8 8

(excludes beam weight)

32'

Fig. 4.13 Load diagram.

From the beam curves, ASD Manual, Part 2, with M = 896 ft-kips and unbraced length (Lb) = 4 ft, select

a W36 × 150. Note that all lines are horizontal at the left vertical axis. For a moment of
896 ft-kips, the W36 × 150 would be satisfactory for any Lb from 0 to 14.6 ft. Add the moment
due to the beam weight: 𝑤𝐿2 0.150(32)2
additional 𝑀 = = = 19.2 ft. kips
8 8

total 𝑀 = 896 + 19.2 = 915 ft. kips

From the beam curves, MR for W36 × 150 with Lb = 4 ft is 1008 ft-kips:

1008 ft. kips > 915 𝑓𝑡. 𝑘𝑖𝑝𝑠 OK

Use W 𝟑𝟑𝟏𝟏 × 1𝟓𝟓𝟒𝟒

(c) Fy = 36 ksi. M = 896 ft-kips from part (a). Lb = 16 ft. From the beam curves, select a W36 × 160.

For Lb of 16 ft, MR = 960 ft-kips for this shape. Add the moment due to the beam weight:

𝑤𝐿2 0.16(32)2
additional 𝑀 = = = 20.5 ft. kips
8 8

total 𝑀 = 896 + 20.5 = 917 ft. kips < 960 𝑓𝑡. 𝑘𝑖𝑝𝑠

OK

Use W 𝟑𝟑𝟏𝟏 × 1𝟏𝟏𝟒𝟒

(c) Assume that Fy = 50 ksi. M = 896 ft-kips from part (a). Lb = 20 ft. From the
93

beam curves, select a W36 × 150 with MR = 960 ft-kips.

 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Add the beam weight moment:

𝑤𝐿2 0.15(32)2
additional 𝑀 = = = 19.2 ft. kips
8 8

total 𝑀 = 896 + 19.2 = 915 ft. kips < 960 𝑓𝑡. 𝑘𝑖𝑝𝑠 OK

Check Fy. From the ASD Manual, Part 1, Table 1 and 2, the W36 × 150 is a Group 2 shape and Fy = 50

ksi. OK

Use W 𝟑𝟑𝟏𝟏 × 1𝟓𝟓𝟒𝟒

When the curves are not applicable (e.g., when Fy equals some value other than 36 ksi or 50 ksi), Fb must

be assumed and subesquently verified.

Example 10:

Rework the previous example. Select the lightest W shape for the beam shown. The framing
system used indicates the use of an Lb = 9.0 ft. The steel is to be A572 Grade 60 (Fy = 60 ksi).

Solution:
𝑀
Assume that Fb = 0.66Fy = 0.66(60) = 40
𝑥
ksi. Using the moment calculated in previous example
915(12)
required
(a), which included a 150-lb/ft = 𝐹𝑏 =
beam𝑆weight, = 275 in.3
40
From the ASD Manual, Part 2 (Allowable Stress Design Selection Table), select a W30 × 108
(Sx = 299 in.3).

Now verify Fb. The section is comapct since 𝐹′ = (−)𝑦 (ASD Manual, Part 1), which indicates
that it is in excess of 65 ksi. Therefore, 𝐹′ > 𝐹 . Since Lc is not tabulated for a stell with Fy =
𝑦 𝑦

60 ksi, it must be computed and compared with Lb = 9 ft. With reference to the ASD

Specifications, Section F1.1, Lc is the smaller of

76𝑏𝑓 76(10.475)
= =
� 𝐹𝑦 √102.8
60 in

and
20,000 = 20,000= 88.9 in
3.75(60)
𝑑
(
𝐴𝑓 ) 𝐹𝑦

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Dr. Ammar A. Ali

Therefore,

88.9
𝐿𝑐 = = 7.41 ft
12

Since Lb ˃ Lc, Fb must be reduced to at least 0.60Fy and possibly lower. This will affect the required Sx

with a subsequent change in the section selected. Assume the new Fb to be 0.60(60) = 36 ksi.

915(12)
𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑆𝑥 = = 305 in.3
36

From the ASD manual, Part 2, select a W30 × 116 (Sx = 329 in.3). For this shape, a

calculation for Lc (similar to the preceding) will show that Lc = 8.3 ft.

Therefore, Lb is still greater than Lc. Now determine Lu as the larger of

20,000 20,000
𝑑 = = 8.3 ft
( 12(3.75)(60)
𝐴𝑓 ) 𝐹𝑦

and

102,000 𝑟 𝑇 102,000 2.64

� 𝐹𝑦 ( 12) = � 60 ( ) = 9.1 ft
12
Therefore, Lu = 9.1 ft and Fb = 0.60Fy (as assumed), since

𝐿𝑐 < 𝐿𝑏 < 𝐿𝑢

The beam weight included in the design moment is on the conservative side, since the
assumed 150 lb/ft is greater than the actual beam weight of 116 lb/ft. No modifications of
the calculations are necessary, however.

Use W 𝟑𝟑𝟒𝟒 × 11𝟏𝟏

Note 1: Other beam design aids are available, such as the table of Allowable Loads on
Beams also found in the ASD Manual, Part 2.

Note 2: The curves found in the ASD Manual, Part 2, may also be used for analysis problems
as well as for design. If the beam size and unbraced length (Lb) are known, the allowable moment
may be obtained from the curves.

Example 11:
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Compute the allowable superimposed uniformly distributed load that may be on a

W36 × 150 spanning 24 ft-0 in. The beam is laterally supported at its supports only (Lb = 24 ft) and is
A36 steel.

Solution:

Entering the curves for a W36 × 150 with an Lb = 24 ft, the total allowable moment = 680
ft-kips. The moment due to beam’s own weight is
0.15(24)2
𝑀= = 10.8 ft. kips
8
The allowable moment for superimposed load is

680 − 10.8 = 669 ft. kips

Since the applied moment can equal the allowable moment, as a limit,

𝑤𝐿2 𝑤 = 8𝑀
2 =
8(669)
= 9.29
𝑀= 𝑜𝑟 𝐿 242
8 kips/ft

Summary of Procedure: Beam design for Moment

1. Establish the conditions of load, span, and lateral support. This is the best done with
a sketch. Establish the steel type.
2. Determine the design moment. If necessary, complete shear and moment diagrams
should be drawn. An estamated beam weight may be included in the applied load.
3. The beam curves should be used to select an appropraite section when possible. As
an alternative, Fb must be estimated and the required section modulus determined:

𝑀
required 𝑆𝑥 =
𝐹𝑏

the section is then selected using the Sx table.

4. After the section has been selected, recompute the design moment, including the
effect of the weight of the section. Check to ensure that the section selected is still
adequate.
5. Check any assumptions that may have been made concerning Fy or Fb.
6. Be sure that the solution to the design problem is plainly stated.

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Dr. Ammar A. Ali

Shear in Beams

Except under very special loading conditions, all beams are subjected to shear as well as
moment. In the normal process of design, beams are selected on the basis of the moment to
be resisted and then checked for shear. Shear rarely controls a design unless loads are very
heavy (and, possibly, close to the supports) and/or spans are very short. From strength of
materials, the shear stress that exists within a beam may be determined from the general
shear formula

𝑓𝑉𝑉𝑉
𝑣 = 𝐼𝑏

where

𝑓𝑣 = shear stress on a horizontal plane located with reference to the nuetral axis (ksi).

𝑉 = vertical shear force at that particular section (kips).

𝑉𝑉 = statical moment of area between the plane under consideration and the outside of
the section, about the nuetral axis (in.3).

𝐼 = moment of inertial if the section about the nuetral axis (in.4)

𝑏 = thickness of the section at the plane being considered (in.)

This formula furnishes us with the horizontal shear stress at apoint, which is equal in
intensity to the vertical shear stress at the same point in a beam.

Example 12:

A W16 × 100 is subjected to a vertical shear of 80 kips. Determine the maximum shear
stress and plot the distribution of shear stress for the entire cross section.

Solution:

Design dimensions for the W16 × 100 are shown in Fig. 4.14. Other properties are

𝐴 = 29.4 in.2 𝐼𝑥 = 1490 in.4

The maximum shear stress will be at the neutral axis (where Q is maximum). Taking Q for
the shaded area about the neutral axis, having
0.985 7.5
𝑉𝑉 = 10.425(0.985) (7.5 + ( ) ) = 98.5 in.3
2 ) + 7.5 0.585 ( 2

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Dr. Ammar A. Ali

𝑉𝑉𝑉
80(98.5)
maximum 𝑓𝑣 = =
𝐼𝑏 ksi
= 9.04
1490(0.585)

10.425 "
0.42 ksi 7.53 ksi

0.985 "
7.500 "
9.04 ksi
16.970 "
0.585"

(a) Cross section (b) Shear

distribution

Fig. 4.14 Shear stress.

If shear stress values are calculated at other levels in the cross section and the results
plotted, the shear distribution will appear as in the figure.

Note that the flanges resist very low shear stresses. Even though the areas of the flanges
are large, it is the web that predominantly resists the shear in wide-flange beams. For this
reason the ASD Specifications allows the use of an average web shear approach for the
shear stress determination:
𝑉
𝑓𝑣 = 𝑤
𝑑𝑡

where

𝑓𝑣 = computed maximum shear stress (ksi).

𝑉 = vertical shear at the section considered (kips).

𝑑 = deoth of the beam (in.).

𝑡𝑤 = web thickness of the baem (in.)

This method is approximate compared with the theoritically correct general shear formula
and assumes that the shear is resisted by the rectangular area of the web extending the full
depth of the beam. For the W16 × 100 of previous example,
𝑉 80
𝑓𝑣 = = = 8.06 ksi
𝑑𝑡 𝑤 16.97(0.585)

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Dr. Ammar A. Ali

This is less than the shear stress of 9.04 ksi calculated by the general shear stress formula
and could be considered unsafe.

Allowable shear stresses are set intentially low to account for the fact the computed
average shear stress will be lower than the actual shear stress.

In rolled beams, at locations other than end connections, the ASD Specifications, Section F4,
establishes the allowable shear stress Fv as follows. For
ℎ 380
≤
𝑡 𝐹
𝑤 � 𝑦
where

ℎ = clear distance between flanges at the section under investigation (in.)

𝑡𝑤 = thickness of the web (in.)

The allowable shear stress is based on the overall depth of the beam d times the web
thickness and is taken as

𝐹𝑣 = 0.40𝐹𝑦 (F4 − 1)

ℎ 380
For rolled shapes tabulated in the ASD Manual, ratios
𝑡𝑤 do not exceed limit when
𝐹
Fy =
� 𝑦
36 ksi. However, for the higher-yield-strength steels, this is frequently not the case.

ℎ 380
For > , the allowable shear stress is based on the clear distance between flanges times
𝑡𝑤 � 𝐹𝑦
the web thickness and is taken as

𝐹𝑦
𝐹𝑣 =
2.89 (𝐶𝑣) ≤ 0.40𝐹𝑦 (F4 −
2)
where Cv is a function of the distance between transverse stiffners and is computed from
45,000𝑘𝑣
𝐶𝑣 = ℎ when 𝐶𝑣 < 0.8
𝐹𝑦 ( 𝑡𝑤 )2

or
190
𝑣 when 𝐶𝑣 > 0.8
ℎ𝑘⁄𝑣𝑡𝑤 𝐹𝑦
𝐶 = �

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Dr. Ammar A. Ali

𝑘𝑣 = 4.00 + ( 5.34
⁄ )2
𝑤ℎ𝑒𝑛 𝑎 < 1.0
𝑎 ℎ ℎ

𝑘𝑣 = 5.34 + ( 4.00
⁄ )2
𝑤ℎ𝑒𝑛 𝑎 > 1.0
𝑎 ℎ ℎ

where

𝑎 = clear distance between transverse stiffners (in.)

For a rolled section without transverse stiffners, kv = 5.34.

Example 13:

The W18 × 50 beam shown in Fig. 4.15, has been designed for moment. The uniform load
includes the beam weight. Check the beam for shear. Assume A36 steel and full lateral
support.
20 k 50 k 60 k
3.07 kips/ft

4' 3' 4' 3'

78.8
69.6
19.6
0 7.3
V 0
(kips) 20
32.3
52.7

61.9
Solution:
Fig. 4.15 Load and shear diagrams.
The shear diagram is drawn as shown in Fig. 4.15. The maximum shear to which the beam
is subjected is 78.8 kips. The W18 × 50 properties are

𝑑 = 17.99 in. 𝑡𝑤 = 0.355 in. 𝑡𝑓 = 0.570 in.

Calculating the shear stress,

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Dr. Ammar A. Ali

𝑉 78.8
𝑓𝑣 = = = 12.34 ksi
𝑑𝑡 𝑤 17.99(0.355)
Next, calculating the allowable shear stress (ASD Specifications, Section F4):

ℎ 17.99 − 2(0.570)
= = 47.5
𝑡𝑤 0.355

380 380
= = 63.3
� 𝐹𝑦 √36
47.5 < 63.3

Therefore,

𝐹𝑣 = 0.40𝐹𝑦 = 0.40(36) = 14.4 ksi

Which is rounded to 14.5 ksi in the ASD Specifications, Numerical Values. Thus

𝑓𝑣 < 𝐹𝑣

W1𝟖𝟖 × 𝟓𝟓𝟒𝟒 OK for shear.

The concept of shear capacity is also useful. The shear capacity may be determined by
multiplying its web area (𝑑 × 𝑡𝑤) by the allowable shear stress Fv. The ASD Manual calls this the
maximum web shear and designates it V.

For the W18 × 50 of previous example,

shear capacity = 𝐹𝑣𝑑𝑡𝑤 = 14.5(17.99)(0.355) = 92.6 kips

This may be readily computed with the maximum applied shear to verify that the beam is
satisfactory. The maximum permissible web shear (V) is a tabulated quantity. Refer to the
Allowable Uniform Load Tables in the ASD Manual, Part 2, where V for each section is
tabulated for Fy = 36 ksi and 50 ksi.
Deflection

When a beam is subjected to a load that creates bending, the beam must sag or deflect, as
shown in Fig. 4.16. Although a beam is safe for moment and shear, it may be
unsatisfactorily because it is too flexible. Therefore, the consideration of the deflection of
beams is another part of the beam design process.

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Deflection

Fig. 4.16 Beam deflection.

Excessive deflections are to be avoided for many reasons. Among these are the effects on
attached nonstructural elements such as windows and partitions, undesirable vibrations,
and the proper functioning of roof drainage systems.

To counteract the sag in a beam, and upward bend or camber may be given to the beam.
This is commonly done for longer beams to cancel out the dead load deflection and,
sometimes, part of the live load deflection. One production method involves cold bending of
the beam by applying a point load with a hydraulic press or ram. For shorter beams, which
are not intentionally cambered, the fabricator will process the beam so that any neutral
sweep within accepted tolerance will be placed so as to counteract expected deflection.

Deflection criteria are based on some maximum limit to which the deflection of the beam
must be held. This is generally in terms of some fraction of the span length. For the
designer this invloves a calculation of the expected deflection for the beam in question, a
determination of the appropraite limit of deflection, and a comparison of the two.

Various methods are available to calculate the deflection. For common beams and loadings,
the ASD Manual, Part 2, Beam Diagrams and Formulas, contains deflection formulas. The
use of some of these will be illustrated in subesquent examples.

The deflection limitations of specifications and codes are usually in the form of suggested
guidelines because the strength adequacy of the beam is not at stake. Traditionaly, beams
that have supported plastered cielings have been limited to maiximum live load
deflections of span/360. This is a requirement of the ASD Specifications, section L3.1. The
span/360 deflection limit is often used for live load deflections in other situations. It is
common practice, and in accordance with some codes, to limit maximum total deflection
(due to live load and dead load) to span/240 for roofs and floors that support other than
plastered cielings.

The ASD Specifications Commentary, Section L3.1, contains guidelines of another nature. It
suggests:

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Dr. Ammar A. Ali

1. The depth of fully stressed beams and girders in floors should, if practicable, be not
less than Fy/800 times the span.
2. The depth of fully stressed roof purlins should, if practicable, be not less than
Fy/1000 times the span, except in the case of flat roofs.

Further, it recommends that where human comfort is the criterion for limiting motion, as in
the case oforvibrations,
partitions the depth
other sources of a steel
of damping beam
should supporting
be not less thanlarge,
1 open
of the floor areas free of
span.
20

Since the moment of inertia increases with the square of the depth, the guidelines for
minimum beam depth limit deflections in a general way. The ASD Specifications
Commentary, Section K2, also contains a method for checking the flexibilty of roof systems
when ponding, the retention of water on flat roofs, is a consideration.

Example 14:

Select the lightest W shape for the beam shown in Fig. 4.17. Assume full lateral support and
A572 grade B steel. Consider moment, shear, and deflection. Maximum allowable deflection
for total load is to be span/360.

10 k 10
k
2.0 kips/ft

8'-0" 16'-0" 8'-0"

Fig. 4.17 Load diagram.

Solution:

The usual procedure is to design for moment and check for other effects. Select the section
for moment:

𝑤𝐿2 2.00(32)2
𝑀=
8 + 𝑃𝑎 = 8 + 10(8) = 336 ft. kips

From the beam curves, assuming that Fy = 50 ksi, select a W21 × 62 that has an MR of 349 ft- kips. From

the ASD Manual, Part 1, Tables 1 and 2, Fy = 50 ksi. Check the additional moment due to the beam’s
own weight:

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Dr. Ammar A. Ali

0.062(32)2
𝑀= = 7.94 ft. kips
8
Total 𝑀 = 336 + 7.94 = 3.44 ft. kips

344 < 349 OK

Check the shear: For this beam, the maximum shear occurs at, and is equal to, the reaction.
Therefore,

maximum shear = 2(10) + 32(2.062) = 43.0 kips

2
From the Allowable Uniform Load Tables, Part 2, maximum permissible web shear for the
W21 × 62 is 168 kips. Therefore, the shear strength is ok.

Check deflection (Δ): From the ASD Manual properties tables, I for the W21 × 62 is 1330
in4:

maximumspan
allowable ∆= = 32(12) = 1.07 in.
360 360

From formula in the ASD Manual, Part 2, Beam Diagrams and Formulas, the actual expected
deflection may be calculated. Note that the units are kips and inches.

5𝑤𝐿4 𝑃𝑎(3𝐿2 − 4𝑎2)

∆= +
384𝐸𝐼
24𝐸𝐼
∆= 5(2.062)(32 12 +
)4( 10(8)(12)3
)3
[3(32)2 − 4(8)2]
384(29,000)(1330) 24(29,000)(1330)
∆= 1.26 + 0.42 = 1.68 in. > 1.07 in.

N𝐍𝐍
1.68
The moment of inertia (I) must be increased.
(1330)Select
= 2088a larger
in.4 beam with an I value of
1.07
From the ASD Manual, Part 2, Moment of Inertia Selection Tables, select W24 × 76 (I = 2100
in.4). This shape has a higher Sx and greater shear capacity than the W21 × 62. Therefore, moment
and shear are satisfactory.

Use W 𝟐𝟐𝟒𝟒 × 𝟕𝟕𝟏𝟏.

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Holes in Beams

Beams are normally found as elements of a total structural system rather than as indivdual,
isolated entities. They are peretrated by mechanical and electrical systems, are envloped by
nonstructural elements, and must be connected to other structural members. They must
sometimes be cut to provide clear areas. The problem of holes in beams is a common one.

Among the more evident of holes in beams (or, generally, any decrease in cross-sectional
area) is capacity reduction. Two such reductions may be readily identified. Holes in beam
webs reduce the shear capacity. Holes in beam flanges reduce the moment capacity.

The ASD Specifications is not specific concerning a recommended design procedure for
beams with web holes. The common procedure of reducing shear capacity in direct
proportion to web area reduction is an oversimplification for other than small holes. Where
larger openings occur, it is common practice to reinforce the beam web by welding
stiffening members around the perimeter of the hole.

Some general rules may be atated with regard to web holes. They should be located away
from areas of higher shear. For uniformly loaded beams, web holes near the center of the
span will not be critical. Holes should be centered on the nuetral axis to avoid high bending
stresses. The holes should be round or have rounded corners (to rectangular holes) to
avoid stress concentations. The cutting of web holes in the field should not be allowed
without the approval of the designer.

With respect to holes in the flanges of beams, it is the moment capacity that is affected. The
cross-sectional property that giverns moment capacity is moment of inertia I. The web of a
wide-flange beam contributes very little to the moment of inertia, and the effect of web
holes on moment capacity may be neglected. The effect of flange holes, however, is to
reduce the moment of inertia. The calculation of the reduction is accomplished by
subtracting from the gross moment of inertia the quantity Ad 2 for each hole, where

A = cross-sectional area of the hole (diameter × flange thickness) (in.2)

d = distance from the nuetral axis to the centorid of the hole (in.)

The neutral axis shifts very little where holes exist is only one flange and may be assumed
to remain at the centroid of the gross cross section.

No distinction will be made as to whether the flange is tension or compression, even

though, beams are usually controlled by the strength of the compression flange.

It is the conservative practice of some designers to consider both flanges to have holes (in
symmetrical pattern) even though only one does.

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It is generally agreed that flange holes for bolts do not reduce the moment capacity of
beams to the extent indicated by the erduced moment of inertia as described before. ASD
Specifications, in Section B10, states that no reduction in moment of inertia shall be made
for bolt holes in either flange providedthat

0.5𝐹𝑢𝐴𝑓𝑛 ≥ 0.6𝐹𝑦𝐴𝑓𝑔 (B10 − 1)

𝐴𝑓𝑛 = net flange area

𝐴𝑓𝑔 = gross flange area

If
0.5𝐹𝑢𝐴𝑓𝑛 < 0.6𝐹𝑦𝐴𝑓𝑔

(B10 − 2)

Then the flexural properties of the member

5 𝐹𝑢 shall be based on an effective tension flange
(B10 − 3)
area Afe, where 𝐴𝑓𝑒 = 𝑦 𝐴𝑓𝑛
6𝐹

Example 15:

Using the ASD Specifications, determine the resisting moment MR for a W18 × 71 shown in Fig. 4.18,

that has two holes punched in each flange for 1-in.-diameter bolts. Fb = 24 ksi. Assume A36 steel. (Fy

= 36 ksi, Fu = 58 ksi).
1 81 " Ø hole
(for design)

8.83 "

17.47 "

9.24 "
0.81 "

7.635 "
Ix = 1170 in4
Sx = 127
in3

Fig.
Solution: 4.18
W18×71.
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First check whether the moment of inertia (I) must be reduced on the ASD Specifications,
Section B10. For one flange,

𝐴𝑓𝑔 = 7.635(0.81) = 6.18 in.2

𝐴𝑓𝑛 = 𝐴𝑓𝑔 − 𝐴ℎ𝑜𝑙𝑒𝑠

𝐴𝑓𝑛 = 6.18 − 2(1.125)(0.81) = 4.36 in.2

0.5𝐹𝑢𝐴𝑓𝑛 = 0.5(58)(4.36) = 126.4 kips

0.6𝐹𝑢𝐴𝑓𝑔 = 0.6(58)(6.18) = 133.5 kips

Since

0.5𝐹𝑢𝐴𝑓𝑛 < 0.6𝐹𝑦𝐴𝑓𝑔

5 𝐹𝑢 5 2
(
The reduction for holes must
𝐴𝑓𝑒 be considered,
= 58 𝐴 = 6 and
36 ) the
( effective
4.36) = 5.85 flange
in. area is
6 𝐹 𝑦 𝑓𝑛

This is a decrease of 6.18 – 5.85 = 0.33 in2 per flange.

net 𝐼𝑥 = 𝐼𝑥 − 𝐴𝑑2 = 1170 − 2(0.33)(8.83)2 = 1119 in.4

net 𝐼𝑥 1119
net 𝑆𝑥 = = = 121.1 in.3
𝑐 9.24
from which

24(121.1)
reduced 𝑀𝑅 = 𝐹𝑏𝑆𝑥 = = 242 ft. kips
12

calculating the percent reduction in the resisting moment, noting that MR for the gross section
(from the ASD Manual beam curves) is 254 ft-kips, yields
254 − 242
254 (100) = 4.7%

The resisiting moment has been reduced by 4.7%.

Web Yielding and Web Crippling

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A beam that is subjected to concentrated leads applied normal to the flange and symmetric
to the web must be checked to ensure that a localized failure of the web does not occur.

The ASD Specifications, Section K1, establishes requirements for beam webs under
compression due to concentrated loads. When the stapulated requirements are exceeded,
the webs of the beams should be reinforced or the length of bearing increased.

Two cinditions are considered: web yielding and web crippling. The type of deformation
failure expected is illustrated in Fig. 4.19. Practical and commonly used bearing lengths N
are usually large eniugh to prevent this type of failure from occuring.

Localized deformation
at toe of fillet Web deformation

Reaction or
concentrated load

(a) Web yielding (b) Web crippling

Fig. 4.19 Web yielding/web criplling.

With respect to web yielding, the ASD Specifications, Section K1.3, requires that the
compressive stress at the toe of the fillet, shown in Fig. 4.20, not exceed 0.66Fy.

The assumption is made that the load “spreads out” so that the critical area for stress,
which occurs at the toe of the fillet, has a length of (N + 2.5k) or (N + 5k) for end reactions
and interior loads, respectively, and a width of tw. The dimension k, which locates the toe of the fillet, is
tabulated for various shapes in the ASD Manual,Part 1.

The controlling equations for web yielding are as follows:

1. For interior loads (defined as applied at a distance from the end of the member that
is greater than the depth d of the member),
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Dr. Ammar A. Ali

𝑅
≤ 0.66𝐹 𝑦 (K1 − 2)
𝑡𝑤(𝑁 + 5𝑘)
2. For end reactions,

𝑅
≤ 0.66𝐹 𝑦 (K1 − 3)
𝑡𝑤(𝑁 + 2.5𝑘)
Where are R is the applied concentrated load and N is the length of bearing.

N+5k
Toe of fillet
tw N+2.5k

Fig. 4.20 Web yielding.

Should the web yielding stress be excessive, the problem may be corrected by increasing
the bearing length, by designing bearing stiffeners, or by selecting a beam with a thicker
web.

With respect to web crippling, the ASD Specifictaions, Section K1, places limits on the
compressive concentated load.

1. For interior loads (defined as applied at a distance from the end of the member that
is greater than d/2), the limiting load R may be taken as
1.5
𝑁 𝑡 𝑡𝑓
(K1 − 4)
𝑅 = 67.5 𝑡 �1 + 3 (𝑑 ) ( 𝑤 )
𝑤 2
� 𝑦𝑤
𝑡𝑤
�𝐹 𝑡𝑓
2. For end reactions,
1.5
𝑁 𝑡 𝑡𝑓
(K1 − 5)
𝑅 = 34 𝑡 �1 + 3 (𝑑 ) ( 𝑤 )
𝑤 2
� 𝑦𝑤
𝑡𝑤
�𝐹 𝑡𝑓
where
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R = maximum concenttated load or reaction (kips).

Fyw = specified minimum yield stress of beam web (ksi).

For unreinforced webs, both web yielding and web crippling shouls be checked under all
concentrated loads and at points where the beam is supported by walls or pedestals or at
columns when the connection is a seated type. If web stiffeners ae providid and extend at
least one-half the web depth, Equations (K1-4) and (K1-5) need not be checked.

Example 16:

A W24 × 55 beam of A36 steel has an end reaction of 70 kips and is supported on a plate
such that N = 6 in. Check the beam for web yielding and web crippling.

Solution:

For this shape

𝑡𝑤 = 0.395 in. 5
𝑘 = 1 16 in.

𝑡𝑓 = 0.505 in. 𝑑 = 23.57 in.

𝐹𝑦𝑤 = 36 ksi

Check web yielding [using Equation (K1-3) for end reactions]. The compressive stress at
the toe of the fillet is

𝑅 70
= = 19.1 ksi
𝑡𝑤(𝑁 + 2.5𝑘) 0.395[6 + 2.5(1.31)]
The web yielding allowable stress is

0.66𝐹𝑦 = 0.66(36) = 23.76 ksi

19.1 ksi < 23.76 𝑘𝑠𝑖

OK

Check web crippling [using Equation (K1-5)1.5 for end reactions]. The maximum compressive
force is 𝑁 𝑡 𝑡𝑓
𝑤
𝑅 = 34 𝑡 𝑤 2
� 1 + 3 (𝑑 ) ( ) � 𝑦𝑤
𝑡𝑤
�𝐹 𝑡𝑓
6 0.395 1.5 0.505
) = 55 kips
𝑅 = 34 (0.395) �1 + 3 (23.57 ) ( 0.505 ) � � 360.395
2

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Since 55 kips ˂ 70 kips, the beam is inadequate with respect to web crippling. Bearing
stiffeners must be provided, or the length of bearing must be increased.

The web yielding equations [ASD Specifications, Equations (K1-2) and (K1-3)] may be
expressed in different forms. To determine allowable load (based on allowable web
yielding stress):

For end reactions,

𝑅 = 0.66𝐹𝑦𝑤 (𝑡𝑤 )(𝑁 + 2.5𝑘)

For interior loads,

𝑅 = 0.66𝐹𝑦𝑤 (𝑡𝑤 )(𝑁 + 5𝑘)

To determine minimum length of bearing required (based on allowable web yielding

stress):

For end reactions,

𝑅
minimum 𝑁 = − 2.5𝑘
0.66𝐹 𝑦𝑤 (𝑡𝑤 )

For interior loads,

𝑅
minimum 𝑁 = − 5𝑘
0.66𝐹 𝑦𝑤 (𝑡𝑤 )

Rather than use the web yielding and web criplling equations as previously described, this
checking process can be significantly simplified by using the ASD Manual, Part 2, Allowable
Uniform Load Tables. The data furnished in the tables are summarized as follows:

1. The tabulated value R (kips) is the maximum end reaction for 3 12 in. of bearing
length (N = 3 12 in). For other values of N, the maximum end reaction with respect to
web yielding is computed
from
𝑅 = 𝑅1 + 𝑁𝑅2

where R1 and R2 are costants tabulated for each shape in the Allowable Uniform Load
Tables.

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2. In a similar manner the maximum end reaction with respwct to web crippling is
computed from
𝑅 = 𝑅3 + 𝑁𝑅4

where R3 and R4 are also tabulated constants.

Using the tabular values greatly simp;ifies the calculations involving web yielding and web
crippling.

Example 17:

Rework previous example using the tabukated values to determine the maximum end
reaction and compare with the end reaction of 70 kips.

Solution:

From the Alowable Uniform LoadTables, for the W24 × 55 of A36 steel, for a bearing length N
of 6 in., the maximum end reaction is computed with respect to web yielding and web
crippling.

Web yielding:
𝑅 = 𝑅1 + 𝑁𝑅2
OK
𝑅 = 30.8 + 6(9.39) = 87.1 kips > 70 𝑘𝑖𝑝𝑠
Web crippling:

𝑅 = 𝑅3 + 𝑁𝑅4
N𝐍
𝑅 = 36 + 6(3.17) = 55 kips < 70 𝑘𝑖𝑝𝑠
𝐍
The beam is not satisfactory with respect to web crippling.

Beam Bearing Plates

Beams may be supported by connection to other structural members, or they may rest on
concrete or masonary supports such as walls or pilasters. When the support is of some
material that is weaker than steel (such as concrete), it is usually necessary to spread the
load over a larger area so as not to exceed the allowable bearing stress Fp. This is
achieved through the use of a bearing plate. The plate must be large enough so that the
actual bearing pressure fp under plate is less than Fp. Also, the plate must be thick enough so that
the bending stress in the plate at the assumed critical section is less than the

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Dr. Ammar A. Ali

allowable bending stress Fb. An assumption is made that the pressure developed under the plate is
uniformly distributed.

Fig. 4.21 Beam bearing plate.

Fb, from the ASD Specifications, Section F2, is 0.75Fy. The allowable bearing pressure, Fp, for
masonary or concrete may be obtained from the ASD Specifications, Section J9, as
follows:

For a plate covering the full area of concrete support,

𝐹𝑝 = 0.35𝑓′𝑐

For a plate covering less than the full area of concrete support,
′ 𝐴2 ′
𝐹𝑝 = 0.35𝑓𝑐� ≤ 0.7𝑓𝑐
𝐴1

where

𝑓𝑐′ = specified compressive strength of concrete (ksi).

𝐴1 = area of concentrically bearing on a concrete support (in.2).

𝐴2 = maximum area of the portion of the supporting surface that is geometrically similar to
and concentric with the loaded area (in.2).

The moment at the critical section for a 1-in.-wide strip of depth tp (in.), which acts like a
cantilever beam is

𝑀 = (actual bearing pressure)×(area)×(moment arm)

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Dr. Ammar A. Ali

𝑛 𝑓𝑝
𝑀 = 𝑓𝑝 × (𝑛 + 1) × 𝑛2= 2
2

The bending stress is determined from the formula:

𝑓 𝑝𝑛2 𝑡
𝑀𝑐 ( ) 3𝑓𝑝 𝑛2
𝑓𝑏 = 2 2 =
𝐼 = ( ) 3
𝑝

1(𝑡𝑝 ) 𝑡𝑝2
12
As a limit, 𝑓𝑏 = 𝐹𝑏. Solving for the required thickness:

3𝑓𝑝 𝑛2
required𝑡𝑝 = � 𝐹𝑏

since 𝐹𝑏 = 0.75𝐹𝑦, this may be rewritten as

3𝑓𝑝 𝑛2 𝑓𝑝
required𝑡𝑝 = �
0.75𝐹𝑦 =
𝐹𝑦 2𝑛�
A procedure for the design of beam bearing plates is given in the ASD Manual, Part 2.

Example 18:

A W16 × 50 is to be supported on a concrete wall, as shown in Fig. 4.22. 𝑓′ 𝑐 = 3000 psi. The
beam reaction is 55 kips. Design a bearing plate for the beam. Assume a 2-
in. edge distance
from the edge of the plate to the edge of the wall (maximum N = 6 in.). All
steel is A36.

R = 55 kips

N+2.5k
k

(N)
k 6" max.

n 10" wall
B
Fig. 4.22 Beam bearing plate design.

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Dr. Ammar A. Ali

Solution:

5
𝑘 = 1 16 in. = 1.31 in.

From the Allowable Uniform Load Tables, for the W16 × 50,

𝑅1 = 29.6 kips

𝑅2 = 9.03 kips/in.

𝑅3 = 37.9 kips

𝑅4 = 3.28
𝑅 −kips/in.
𝑅1 55 −
𝑁= = = 2.81 in.
29.6
𝑅2 9.03
1. Calculate the minimum bearing length N based on
(a) Web yielding:
(b) crippling:
𝑅 − 𝑅3 55 − 37.9
𝑁= = = 5.21 in.
𝑅4 3.28

Therefore, use N = 6 in.

2.
Since the area of the support and the bearing area (A1 and A2) are unknown,
conservatively assume that
𝐹𝑝 = 0.35𝑓′𝑐 = 0.35(3) = 1.05 ksi
3. The required support area is

𝑅
1 = 52.4 in.2
55𝑝
𝐹 1.05
required 𝐴 = =
4. The required B dimension is calculated from

𝐴1
= 8.73 in.
required 𝐵 = 52.4= 6
𝑁
Use B = 9.0 in.
5. The actula bearing pressure is

𝑓𝑝 = 𝑅 = 55 = 1.02 ksi < 1.05 𝑘𝑠𝑖

𝐵𝑁 6(9)

OK

6. Calculate the cantilever length

𝐵 n: 9.0
𝑛 = − 𝑘 = 2 − 1.31 = 3.19 in.
2
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Dr. Ammar A. Ali

7. Calculate the required plate thickness:

𝑓𝑝 1.02
required𝑡𝑝 = 2𝑛� 𝐹𝑦 = 2(3.19)� = 1.07 in.

See the ASD Manual, Part 1, Bars and36

Plates - Product Availibilty, for information on
plate availablity.
8. Use a bearing plate 1 𝟖1 × 𝟏𝟏 × 𝟒𝟒′
− 𝟗𝟗. 𝟖
It may be possible, if reactions are small, to support a beam in a bearing situation without
the use of a bearing plate. Bearing pressure, web yielding, web crippling, and flange
bending are the considerations. The critical section for flange bending is again assumed to
be at a distance k from the center of the section.

Example 19:

A W24 × 76 is to be supported on a 12-in.-wide concrete wall such that there is bearing 8

in. wide, as shown in Fig. 4.23. 𝑓′𝑐 = 3000 psi. The beam reaction is 25 kips. Determine
whether a bearing plate is required. Assume A36 steel.

R = 25 kips

8"
k n

12"

Fig. 4.23 Beam without bearing plate.

Solution:

Beam properties and dimensions are

7
𝐹𝑦 = 36 ksi 𝑏𝑓 = 8.99 in. 𝑡𝑓 = 0.68 in. 𝑘=1
16
= 1.44 in.
1. Check the bearing pressure:
25
𝑓𝑝 =
=
8.99(8)
0.35 ksi
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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝐹𝑝 = 0.35𝑓′𝑐 = 0.35(3) = 1.05 ksi > 0.35 𝑘𝑠𝑖 OK

2. Check the maximum end reaction using data from the Allowable Uniform Load
Tables. Based on web yielding,
𝑅 = 𝑅1 + 𝑁𝑅2
OK
= 37.6 + 8(10.5) = 121.6 kips > 25 𝑘𝑖𝑝𝑠
Based on web crippling:
𝑅 = 𝑅3 + 𝑁𝑅4
OK
= 49.1
3. The bending stress in + 8(3.21
the flange ) =be
may 74.8 kips > 25by
determined 𝑘𝑖𝑝𝑠
using the formula developed
previously for fb in the bearing plate (the flange acts exactly as the plate does):
𝑏𝑓
– 1.44 = 3.06 in.
𝑛 = 8.99
−𝑘= 2
2
3𝑓𝑝 𝑛2
= 3(0.35)(3.06)
2
𝑓𝑏 = = 21.26 ksi
𝑡𝑝2 (0.68)2

𝐹𝑏 = 0.75𝐹𝑦 = 0.75(36) = 27 ksi > 21.26 𝑘𝑠𝑖

OK

Therefore, this beam may be used on a bearing length of 8 in. without a bearing plate.

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Flowchart 4: Beam analysis for Fb and MR (W shapes, strong axis bending)

Known:
Start Shape, type of span,
Fy, Fu, Lb, Lc, Lu, Sx, Fy’

Yes
(Compact)
Yes
𝐹𝑦 ≤ 𝐹′𝑦 ? 𝐿𝑏 ≤ 𝐿𝑐 ? 𝐹𝑏 = 0.66𝐹𝑦

No
(Not Compact)

See ASDS No 65 𝑏𝑓 95
≤ ≤
Appendix B 2𝑡𝑓
√ 𝐹𝑦 √ 𝐹𝑦

Yes
(Non-Compact)
No Yes
𝐿𝑏 ≤ 𝐿𝑐 ? 𝐿𝑏 ≤ 𝐿𝑢 ? 𝐹𝑏 = 0.66𝐹𝑦

Yes No

Fb from ASD Fb from the larger of ASD

Specifications Eq. Specifications Eqs. (F1-6)
(F1-3) or (F1-7) and (F1-8)

𝑀𝑅 = 𝐹𝑏 𝑆𝑥

Note: May also use beam curves, as applicable

End

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Flowchart 5: Beam design (W and M shapes; moment, shear and deflection)

Determine Fy, L, Lb, type of

Start span, superimposed loads

Estimate the Calculate V & M; may

1 need V & M diagrams
beam weight

Select trial Assume Fb. Calculate

Yes Use ASDM req'd 𝑆𝑥
section so that No Use F b =
beam 0.66Fy for 𝑀
M≤ =
allowable M curves? initial trial 𝐹𝑏

Select trial
Determine section
No based on
Lc, Lu, Fy’
latest req’d
Sx
Estimated
Reselect beam 𝐹𝑏 Yes 𝐿𝑏 Yes
trial 2
weight = 0.66𝐹𝑦 ≤ 𝐿𝑐 ? 𝐹𝑦 ≤ 𝐹𝑦′ ?
section Compact
based on ok? No
req’d No Noncompact
shear
Yes 𝐹𝑏 Yes 𝐿𝑏 Slender
capacity
65 element
and req’d = 0.60𝐹𝑦 ≤ 𝐿𝑢 ? 𝑏𝑓
≤ 2𝑡𝑓 No section.
moment No V ≤ Shear √ 𝐹𝑦
capacity 95 See ASDS
capacity? ≤
√𝐹𝑦 Appendix
B
No
1
Yes
Reselect Yes
trial Fb from the larger of
Fb from ASDS Eq.
section No Deflection ASDS Eqs. (F1-6) or
based on (F1-7) and (F1-8) (F1-3)
ok?
req’d I,
req’d
shear Yes
capacity Specify 𝑀 𝑆𝑥 No
and req’d Calculate req 'd 𝑆𝑥 =
section to 𝐹𝑏 > req′d 𝑆𝑥
moment use
capacity
Yes

1 End

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5. BEAM COLUMNS
A structural member that is subjected to varying amounts of both axial compression and
bending moment is commonly termed a beam-column.

It is generally accepted that axially (concentrically) loaded compression members are

nonexistent in actual structures and that all compression members ae subjected to some
amount of bending moment. The bending moment may be induced by an eccentric load, as
shown in Fig. 5.1a. The interior column of Fig. 5.1b, shown with a concentric load, will not
be concentrically loaded if the live loads are not symmetrical.

Concentric
load
Symmetrical
Eccentric dead load
load e

(a) (b)

Fig. 5.1 Column loadings.

The combined stresses for a short beam-column subjected to an axial load and bending
moment with respect to one axis only may be expressed as
𝑃𝑓 ±𝑀𝑐
𝑚𝑎𝑥 =
𝐴 𝐼

where

𝑓𝑚𝑎𝑥 = computed maximum stress

𝑃 = axial load

𝐴 = gross cross-sectional area

𝑀 = applied moment

𝑐 = distance from the neutral axis to the extreme outside of the cross section.

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Dr. Ammar A. Ali

𝐼 = moment of inertial of the cross section about the bendong nuetral axis.

If bending occurs with respect to both axes, the expression becomes

𝑃 𝑀𝑥 𝑐 𝑀𝑦 𝑐
𝑓𝑚𝑎𝑥 =𝐴 ± ( 𝐼𝑥 ) ± ( 𝑦 )
𝐼

The previous expression may be rewritten as

𝑓𝑚𝑎𝑥 = 𝑓𝑎 + 𝑓𝑏𝑥 + 𝑓𝑏𝑦

With the negative signs neglected. Dividing both sieds by fmax, yields
𝑓𝑎 𝑓𝑏𝑥
1= + +
𝑓𝑚𝑎𝑥 𝑓 𝑓
𝑓𝑏𝑦
𝑚𝑎𝑥

𝑚𝑎𝑥
𝑓𝑎
𝑓𝑏𝑥
1 =𝑎 + +
𝑏𝑥the 𝑏𝑦 terms:
By substititing the applicable allowavle stresses𝐹in place
𝐹 of 𝐹 fmax
𝑓𝑏𝑦
An equation of theis type is commonly called an interaction equation.

With this arrangement, if any two of the computed stresses become zero, the correct
allowable stress is approached either as an axially loaded column or as a beam subjected to
bending about either axis.

Analysis of Beam-Columns

The previous expression developed is the basis for ASD Specifications Equation (H1-3):

𝑓𝑎 𝑓𝑏𝑥
+ + ≤ 1.0 (H1-
𝐹𝑎 𝐹𝑓 𝑏𝑥 𝐹 𝑏𝑦
𝑏𝑦
3)
where

𝑓𝑎 = computed axial compressive stress

𝑓𝑏 = computed maximum compressive bending stress

𝐹𝑎 = allowable axial compressive stress for axial force alone

𝐹𝑏 = allowable compressive bending stress for bending moment alone

It applies to members subjected to both axial compression and bending stresses when
𝑓𝑎 /𝐹𝑎 is less than or equal to 0.15.
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Dr. Ammar A. Ali

When 𝑓𝑎 /𝐹𝑎 > 0.15, the secondary moment due to the member deflection may be of
a significant magnitude. This secondary moment results from a lateral deflection
initially caused by the bending moment, as shown in Fig. 5.2. The product of this
deflection and axial load (𝑃 × ∆, sometimes called P-delta moment) causes further
bending and creates secondary stresses that normally are not considered in individual
beam or column analysis and design.

The effect of the secondary moment may be approximated by multiplying fbx and fby by an
amplification factor
1
(C-H1-1)
1 − (𝑓𝑎 ⁄𝐹 ) ′𝑒

23
where 𝐹𝑒′ is the Euler stress divided by a factor of safety of and is expressed as follows:
12

12𝜋2𝐸
𝐹𝑒′ =
𝐾𝑙 2
23 ( 𝑟 𝑏 )
𝑏

In this expression, K is the effective length factor in the plane of bending, lb is the unbraced length in the

plane of bending, and rb is the corresponding raduis of gyration.

?
Δ

Fig. 5.2 Column lateral deflection.

𝐾𝑙
Values of 𝐹𝑒′ (as functions of 𝑟 𝑏 ) may be obtained from table 8 in the Numerical Values
𝑏
section of the ASD Specifications or through the use of the properties tabulated at the
bottom of the column load tables in the ASD Manual, Part3.

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Dr. Ammar A. Ali

Under some combinations of loading, it was found that this amplification factor
overetimated the effect of secondary moment. To compensate for this condition, the
ampification factor was modofied by a reduction factor Cm.

With the introduction of the two factors, ASD Specifications Equation (H1-3) was modified
for the case when 𝑓𝑎 /𝐹𝑎 > 0.15 and expressed as follows:
𝑓𝑎 𝐶𝑚 𝑓𝑏𝑥 + ≤ 1.0 (H1-1)
+
𝐹𝑎 (1 − 𝑓′𝑎 𝑏𝑥
𝑓𝑎
(1 − 𝐹 ′ 𝑏𝑦
𝐹
𝑓𝑏𝑦
) 𝐹 𝑒𝑥 ) 𝐹 𝑒𝑥

where Cm is a coefficient defined as follows:

𝑀1
𝐶𝑚 = 0.6 − 0.4 ( )
𝑀2
𝑀1 𝑀1
In which is the ratio of the smaller end moment to the larger end moment. is taken as
𝑀2 𝑀2
positive if the moments tend to cause reverse curavature and negative if they tend to cause
single curvature. Examples of Cm values are shown in Fig. 5.3.

P P P

M2 M2 M2

?Δ

M1 M1

P P P
M1 = M2 M1 = 0 M1 = 1 M2
2
Cm = 1.0 Cm = 0.6 Cm = 0.4

(a) (b) (c)

Fig. 5.3 Values of Cm.

If the column bending moment is a result of a lateral load placed between column support
points, Cm may be conservativly taken as unity. If frame is not prevented by adequate bracing

or other means, Cm should not be taken as less than 0.85, since in this case the
123
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Dr. Ammar A. Ali

column ends move out of align,ent, causing an additional secondary moment from the axial
load.

The question as to whether adequate bracing exists to prevent sidesway is difficult to

answer and is usually a judgment factor. Sidesway itself may be described as a kind of
deformation whereby one end of a memebr moves laterally with respect to the other. A
simple example is a column fixed at one end and entirely free at the other (cantilever
column or flagploe). Such a column will buckle as shown in Fig. 5.4. The upper end will
move laterally with respect to the lower end.

Fig. 5.4 Sidesway for a flagpole-type column.

To guard against overstressing at one end of a member where no buckling action is

present, stresses are limited by a modified iteraction expression,

𝑓𝑎 𝑓𝑏𝑥
+ + ≤ 1.0 (H1-
0.6𝐹𝑦 𝐹𝑓 𝑏𝑥 𝐹 𝑏𝑦
𝑏𝑦
2)
If only one acis of bendign is involved in a problem, one of the terms will equal zero with
the remaining formula still applicable.

In determining Fb for the interaction equations, the compactness of the beam-column must be

established. As be discussed before, the web compactness of a beam is based on fa = 0. This applies

to a beam subjected to bending only, with no axial load. With a baem-column (fa is not
zero), the web compactness must be checked by using equations from ASD Specifications,
table B5.1, for “Webs in combined flexural 124
and axial compression”. This is
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

simplified in the ASD Manual through the use of 𝐹𝑦′′′ , which is defined as the theoritical
maximum yield stress (ksi) based on the depth-thickness ratio of the web below which a
particular shape may be considered compact for any condition of combined bending and
axial stresses.
𝐹′′′ is tabulated in the properties table of part 1. It is determined in the same way that 𝐹′ is
𝑦 𝑦
determined, as discussed before. If 𝐹′′′ > 𝐹 , the member is not comact based on the web
𝑦 𝑦

criteron, and Fb cannot exceed 0.60Fy.

Example 1:

An A36 steel W6×25 column is subjected to an eccentric load of 32 kips, as

shown in Fig
5.5. The column has an unbraced length of 15 ft and may be assumed to have pinned ends.
Bracing prevents sidesway. Determine whether the column is adequate.

CL column

P = 32 k
W6 X 25 W 16 beam

Seat angle
e = 5 12"

Fig. 5.5

Solution:

For the W6×25

𝐴 = 7.34 in.2

𝑆𝑥 = 16.7 in.3

𝑟𝑦 = 1.52 in.

𝑟𝑥 = 2.70 in.

𝐾 = 1 (pinned ends)
𝑦
𝐹′′′ = −

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Dr. Ammar A. Ali

1. Replacing the eccentric load with a concentric load and a couple (moment),

𝑃 = 32 kips

𝑀𝑥 = 𝑃𝑒 = 32(5.5) = 176 in.-kips

2. Calculating actual axial compressive stress,

𝑓𝑎 = 𝑃 = 32 = 4.36 ksi
𝐴 7.34

3.
Fa is a function of Kl/ry. Thus

𝐾𝑙 = 1(15)(12) = 118.4
𝑟𝑦 1.52

Rounding to 118, from the ASD Manual, Part 3, Table C-36, obtain

𝐹𝑎 = 10.57 ksi

4. Calculating actual maximum compressive bending stress, have

𝑀𝑥
𝑓𝑏𝑥 = 176
𝑆𝑥 = 16.7 = 10.5 ksi

5.
Fbx is a function of the actal unbraced length. Determine whether the member is adequately or
inadequately braced. The actual unbraced length Lb = 15 ft. Lc and Lu may be
obtained from the column load tables in the ASD Manual, Part 3:
𝐿𝑐 = 6.4 ft

𝐿𝑢 = 20.0 ft
Therefore, 𝐿𝑐 < 𝐿𝑏 < 𝐿𝑢 . The W6×25 is compact, since 𝐹𝑦′′′ > 𝐹𝑦 , and the allowable
bending stress is

𝐹𝑏 = 0.60𝐹𝑦 = 21.6 ksi

6. 𝑓𝑎 ⁄𝐹𝑎 = 4.36⁄10.57 = 0.41 > 0.15. Therefore, use ASD Specifications Equations
7. (H1-1) and (H1-2).
Calculate 𝐶𝑚 𝑎𝑛𝑑 𝐹′ .𝑒𝑥Since M1 = 0 and M2 = 176 in.-kips, and sidesway is prevented
𝑀1
𝐶𝑚 = 0.6 − 0.4 ( )
𝑀2
= 0.6

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Dr. Ammar A. Ali

′
𝐹𝑒𝑥 is a function of Klb/rb, which in this case Kl/rx:

𝐾𝑙 = 1(15)(12) = 66.7
𝑟𝑥 2.70

Rounding to 67, from Table 8 in the Numerical Values section of the ASD
Specifications obtain

′
𝐹𝑒𝑥
= 33.27 ksi

8. Checking ASD Specifications Equation (H1-1)

𝑓𝑎 𝐶𝑚𝑥𝑓𝑏𝑥 ≤ 1.0
𝐹𝑎 +
(1 − 𝑓𝑎 ⁄𝐹 ′𝑒𝑥
)𝐹 𝑏𝑥
4.36

0.6(10.5)
+ ≤ 1.0
10.57 (1 − 4.36/33.27)(21.6)

0.41 + 0.34 = 0.75 < 1.0

𝑓𝑏𝑥
𝑓𝑎
OK + ≤ 1.0
0.6𝐹𝑦 𝐹
𝑏𝑥
9. Checking ASD Specifications Equation
4.36 (H1-2), 10.5
+ ≤ 1.0
21.3 21.6

0.20 + 0.49 = 0.69 < 1.0 OK

The beam-column is adequate.

Example 2:

An A572 (Fy = 50 ksi) W12×136 column supports beams framing into it, as shown in Fig.
5.6. The connections are moment connections. The column supports an axial load of 600
kips, which includes the beam reactions at its top. Due to unbalanced floor loading,
moments of 80 ft-kips rach are applied in opposite directions at the top and bottom of
columns as shown. Sidesway is prevented by a bracing system. Ky = 1.0 and Kx is estimated to be 0.9.
Determine whether the member is adequate.

Solution:

For the W12×136

𝐴 = 39.9 in.2
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Dr. Ammar A. Ali

𝑆𝑥 = 186 in.3

𝑟𝑦 = 3.16 in.

𝑟𝑥 = 5.58 in.

𝐾𝑥 = 0.9

𝐾𝑦 = 1.0
𝑦
𝐹′′′ = −

P = 600 kips

M = 80 ft-kips

14'-0"

M = 80 ft-kips

P = 600 kips

Fig. 5.6

1. P = 600 kips, M = 80 ft-kips.

2. 𝑓𝑎 = 𝑃⁄𝐴 = 600⁄39.9 = 15.0 ksi.
3. Fa is a function of the largest slenderness ratio:
𝐾𝑦𝑙 = 1(14)(12) = 53.2
𝑟𝑦 3.16
𝐾𝑥𝑙 = 0.9(14)(12) = 27.1
𝑟𝑥 5.58
Rounding to 53, from the ASD Manual, Part 3, Table C-50,
𝐹𝑎 = 23.88 ksi
4. 𝑓𝑏𝑥 = 𝑀𝑥 ⁄𝑆𝑥 = 80(12)⁄186 = 5.16 ksi.
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Dr. Ammar A. Ali

5. 𝐹𝑏𝑥 is a function of 𝐿𝑏 , 𝐿𝑐 and 𝐿𝑢 .

𝐿𝑏 = 14 ft 𝐿𝑐 =
11.1 ft 𝐿𝑢 =
Since 𝐿𝑐 < 𝐿𝑏 < 𝐿𝑢 , and since W12×136 is compact (𝐹𝑦′′′ > 𝐹𝑦 ), the allowable
38.3 ft
bending stress is

𝐹𝑏 = 0.60𝐹𝑦 = 30 ksi

6. 𝑓𝑎 ⁄𝐹𝑎 = 15.0⁄23.88 = 0.63 > 0.15. Therefore, use ASD Specifications Equations
(H1-1) and (H1-2).
7. ′ 𝑒𝑥
Calculate 𝐶𝑚 𝑎𝑛𝑑 𝐹 . Since M1 = M2 = 80 ft-kips, and causes single curvature, the
ratio M1/M2 is negative (see the ASD Specifications, Section H1). Thus
𝑀1
𝐶𝑚 = 0.6 − 0.4 ( )
𝑀2
= 0.6 − 0.4(−1)
= 1.0
′
𝐹𝑒𝑥 is a function of Klb/rb, which in this case Kxl/rx. Note that this is with respect to
the x-x axis. Using the properties tabulated in the ASD Manual, Part 3, the column
load table for the W12×136, gives

𝐹′𝑒𝑥 (𝐾 𝑥𝐿 𝑥)2 = 323

102
From which

323(10)2
′
𝐹𝑒𝑥
= = 203 ksi
[0.9(14) ]2

Note that 𝐹′ 𝑒𝑥may also be obtained using Kxl/rx and Table 8 of the Numerical Values
section of the ASD Specifications.

8. Checking ASD Specifications Equation (H1-1) gives us

𝑓𝑎 ≤ 1.0
𝐹𝑎 +
(1 − 𝑓𝑎 ⁄𝐹′𝑒𝑥 )𝐹 𝑏𝑥
15.0
𝐶𝑚𝑥𝑓𝑏𝑥
1.0(5.16)
+ ≤ 1.0
23.88 (1 − 15.0/203)(30.0)
0.63 + 0.19 = 0.82 < 1.0

OK 129
9. Checking ASD Specifications Equation (H1-2),
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

𝑓𝑏𝑥
𝑓𝑎
+ ≤ 1.0
0.6𝐹𝑦 𝐹
𝑏𝑥

15.0 5.16
+ ≤ 1.0
30.0 30.0
0.50 + 0.17 = 0.67 < 1.0 OK

The beam-column is adequate.

Design of Beam-Columns

The use of the interaction formulas furnishes a convenient means of beam-column analysis.
These may also be used for beam-column design. A trial section must first be selected,
however. After the selection is made, the problem becomes one of analysis. In essence, the
design process is one of trial and error, since no simple design procedure exists whereby a
most economical member can be selected in one quick step.

The ASD Manual furnishes a method of design whereby a trial section may be attained
using an equivelant axial load in conjunction with the ASD Manual axial load table of Part 3.

Using the ASD Manual approach to determine a trial section, it is referred to the ASD
Manual, Part 3, Table B. The equivalent axial load, for design purposes, is designated Peff:

𝑃𝑒𝑓𝑓 = 𝑃0 + 𝑀𝑥 𝑚 + 𝑀𝑦 𝑚𝑈

where

𝑃0 = actual axial load (kips)

𝑀𝑥 = bending moment about the strong axis (ft-kips)

𝑀𝑦 = bending moment about the weak axis (ft-kips)

𝑚 = factor taken from the ASD Manual, Part 3, Table B

𝑈 = factor taken from the ASD Manual, Part 3, column load tables

The procedure for selection of a trial sections is as follows:

1. With the known value of KL (in feet), select a value of m from the first
approximation section of Table B and assume that U = 3.
2. Solve for Peff.
3. From the column load table in Part 3 of the ASD Manual, select a trial section to
support Peff. 130
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

4. On the basis of the section selected, obtain a subsequent approximation value of

m from table B and a U value from the column load table. Solve for Peff again.
5. Select another section (if necessary) and continue the process untill the values of m
and U stabilize.

By using this trial section, the beam-column may then be analyzed in the manner discussed
previously using the ASD manula interaction equations.

Example 3:

Using A36 steel and the ASD Specifications, select a wide-flange column for the
conditionsshown in Fig. 5.7. The column has pinned ends and sidesway is prevented.
Bending occurs with respect to the strong (x-x) axis.

P = 85 kips

M = 20 ft-kips

13'-0"

M = 20 ft-kips

P = 85 kips

Fig. 5.7

Solution:

From the ASD Manual, Part 3, Table B, with KL = 13 ft, select a value of m = 2.25 from the
first approximation portion. Since My = 0, the expression for the effective axial load
becomes

𝑃eff = 𝑃0 + 𝑀𝑥 𝑚
= 85 + 20(2.25)
= 130 kips

131
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Dr. Ammar A. Ali

From the column load table of the ASD Manual, Part 3, select a W8×31 (allowable load Pa is 143 kips).
From Table B again, select a value of m = 2.85 from the Subsequent Approximations portion
of the table.

𝑃eff = 85 + 20(2.85)
= 142 kips

Since 142 kips ˂ 143 kips, the W8×31 remains as the trial section should be checked using
interaction equations. The checking procedure is identical to that in Examples 1 and 2. ASD
Specifications Equation (H1-1) results in a value of 0.77, and Equation (H1-2) results in a
value of 0.82. The W8×31 is therefore satisfactory.

Example 4:

Using A36 steel and the ASD Specifications, select a wide-flange column for the conditions
shown in Fig. 5.8. Architectural requirements indicate the use of a W8, if possible. The
column is pinned at both ends. Bending occurs with respect to both axes. Sidesway is
prevented in both directions.

Solution:

Replace the eccentric loads with concentric loads and couples (moments). A 4-in.
eccentricity is assumed for strong-axis bending:

𝑃 = 58 kips

𝑀𝑥 = 30(4) = 120 in. −kips

𝑀𝑦 = 28(3) = 84 in. −kips

From the ASD Manual, Part 3, Table B, with KL = 16 ft, select a value of m = 2.2 from the
first approximation portion. Let U = 3; tehrefore,
120 84
𝑃eff = 𝑃0 + 𝑀𝑥 𝑚 + 𝑀𝑦 𝑚𝑈
= 58 + 12 (2.2) + 12 (3)(2.2)
= 58 + 22 + 46.2 = 130 kips

From the ASD Manual, Part 3, column load table, select a W8×35 (allowab;e load Pa is 141
kips). From Table B again, select a value of m = 2.6 from the subsequent approximation
portion and a value of U = 2.59 from the column load table.

𝑃eff = 58 +120 (2.6) + 84 (2.6)(2.59)

12 12

132
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

= 58 + 26 + 47 = 131 kips

Since 131 kips ˂ 141 kips, the W8×35 remains as the trial section should be checked using
interaction equations.

Fig. 5.8 Corner beam-column.

For the W8×35,

𝐴 = 10.3 in.2
𝑆𝑥 = 31.2 in.3
𝑟𝑦 = 2.03 in.
𝑟𝑥 = 3.51 in.
𝐾 = 1.0
𝑆𝑦 = 10.6 in.3
𝐹𝑦′′′ = −

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Dr. Ammar A. Ali

1. P = 58 kips, Mx = 120 in.-kips, My = 84 in.-kips 2. 𝑓𝑎

= 𝑃⁄𝐴 = 58⁄10.3 = 5.63 ksi.

3. Fa is a function of Kl/ry: 𝐾𝑙 = 1(16)(12) = 94.6
𝑟𝑦 2.03
Rounding to 95, from the ASD Manual, Part 3, Table C-36,
𝐹𝑎 = 13.6 ksi
4. 𝑓𝑏𝑥 = 𝑀𝑥 ⁄𝑆𝑥 = 120⁄31.2 = 3.85 ksi, 𝑓𝑏𝑦 = 𝑀𝑦 ⁄𝑆𝑦 = 84⁄10.6 = 7.92 ksi.
5. 𝐹𝑏𝑥 is a function of 𝐿𝑏 , 𝐿𝑐 and 𝐿𝑢 .

𝐿𝑏 = 16 ft 𝐿𝑐 = 8.5 ft 𝐿𝑢
Since 𝐿𝑐 < 𝐿𝑏 < 𝐿𝑢=, and
22.6since
ft W8×35 is compact (𝐹𝑦 > 𝐹𝑦 ), the allowable bending
′′′

stress is

𝐹𝑏𝑥 = 0.60𝐹𝑦 = 21.6 ksi

From the ASD Specifications, Section F2:

𝐹𝑏𝑦 = 0.75𝐹𝑦 = 27.0 ksi

6. 𝑓𝑎 ⁄𝐹𝑎 = 5.63⁄13.60 = 0.41 > 0.15. Therefore, use ASD Specifications Equations
(H1-1) and (H1-2).
7. Calculate 𝐶𝑚 𝑎𝑛𝑑 𝐹𝑒 ′ . Sidesway is prevented in both directions; therefore,
𝑀1
𝐶𝑚 = 0.6 − 0.4 ( ) ≥ 0.4
𝑀2
Since M1 = 0,
𝐶𝑚𝑥 = 𝐶𝑚𝑦 = 0.6
Since bending occurs respect to both axes, 𝐹𝑒′ values must be obtained with respect
to each axis. Using properties for the W8×35
in the column table,
𝐹′𝑒𝑥 (𝐾 𝑥𝐿 𝑥)2 = 128
102

𝐹′𝑒𝑦 (𝐾 𝐿 )2 = 42.7
𝑦 𝑦
102
From which

128(10)2
′
𝐹𝑒𝑥
= = 50.0 ksi
162

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Dr. Ammar A. Ali

42.7(10)2
′
𝐹𝑒𝑦
= = 16.7 ksi
162
8. Checking ASD Specifications Equation (H1-1) gives
𝐶𝑚𝑦𝑓𝑏𝑦
𝑓𝑎 𝐶𝑚𝑥𝑓𝑏𝑥 + ≤ 1.0
𝐹𝑎 +
(1 − 𝑓𝑎 ⁄𝐹′𝑒𝑥 )𝐹 𝑏𝑥 (1 − 𝑓 ⁄𝐹 ′ 𝑒𝑦
)𝐹 𝑏𝑦
5.63 𝑎 0.60(3.85)
0.60(7.92)
+ + ≤ 1.0
13.60 (1 − 5.63/50.0)(21.6) (1 − 5.63/16.7)(27.0)
0.41 + 0.12 + 0.27 = 0.80 < 1.0
𝑓𝑏𝑥 𝑓
𝑓𝑎 𝑏𝑦
OK + + ≤ 1.0
9. Checking ASD Specifications0.6𝐹
Equation
𝑦 𝐹𝑏𝑥(H1-2)
𝐹𝑏𝑦 yields
5.63 3.85 7.92
+ + ≤ 1.0
21.6 21.6 27.0

0.26 + 0.18 + 0.29 = 0.73 < 1.0 OK

The W8×35 is satisfactory.

Effective Length Factor K

The effective length of the member depends on the restraints against relative rotation and
lateral movement (sidesway) imposed at the ends of the member. The portal frame shown
in Fig. 5.9 is having rigidly connected beam to the supporting columns, and no lateral
movement allowed. The effective length factor K for the columns can have values range
from 0.5 for ends fixed against rotation to 1.0 for pinned ends.

When the frame depends on its own stiffness for resistance to sidesway, as shown in Fig.
5.10, K will have a value larger than 1.0. Fig. 6.10a shows the deformed shape due to
vertical load. The frame will deflect to the side (sidesway) so as to equalize the moments at
the tops of the columns. Sidesway may also be caused by a laterally applied force, as shown
in Fig. 5.10b. As a rule, columns free to translate in a sidesway mode are appreciably
weaker than columns of equal length braced against sidesway. Also of importance is that
the magnitude of the sidesway of a column is directly affected by the stiffness of the other
members in the frame; or, the magnitude of joint rotation is directly affected by the
stiffness of the members framing into the joint.

To simplify the determination of the effective length factor K, alignment charts are
furnished in the ASD Manual, Part 3 (Figure 1), for the two cases of sidesway prevented
and sidesway not prevented.

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 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Fig. 5.9 Loaded braced frame.

(a) (b)

Fig. 5.10 Loaded unbraced frame.

The use of the charts requires an evaluation of the relative stiffness of the members of the
frame at each of the column. The stiffness ratio or relative stiffness of the members rigidly
connected at each joint may be expressed as

𝐺𝐴 = ∑(𝐼𝑐 ⁄𝑙𝑐 )𝐴
∑(𝐼𝑔⁄𝑙 𝑔) 𝐴

and

𝐺𝐵 = ∑(𝐼𝑐 ⁄𝑙𝑐 )𝐵
∑(𝐼𝑔⁄𝑙 𝑔) 𝐵

136
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Dr. Ammar A. Ali

where A and B subscripts refer to the joint at which the erlative stiffness is being
determined; the c and g subscripts refer to the column and girder or beam, respectivly; I is
the moment of inertia, and l is the unsupported length of member. The I / l terms are taken
with respect to an axis normal to the plane of buckling under consideration. Having
determined GA and GB, the appropraite chart may be used to determine K.

For column ends supported by, but not rigidly connected to, a footing or foundation, GB is

theoritically infinity, but unless the joint is designed as a true friction-free pin, GB may be taken as

10 for practical designs. If the column end is rigidly attached to a properly designed
footing, GB may be taken as 1.0. Fig. 5.11 shows how the G values would be determined for a
given column AB.
Ic1
𝐺𝐴 = (𝐼𝑐1⁄𝑙𝑐1) + (𝐼𝑐2⁄𝑙𝑐2)
(𝐼𝑔1 ⁄𝑙𝑔1) + (𝐼𝑔2 ⁄𝑙𝑔2 )
𝐺𝐵 = 1.0
A
Ig1 Ig2

Ic2

Rigidly
attached

Fig.
5.11
Colum
nG
values.

The use of the alignment charts requires prior knowledge of the column and beam sizes. In
other words, before the charts can be used, a trial design has to be made of each of the
members. To start the design, it is therefore necessary to assume a reasonable value of K,
choose a column section to support the axial load and moments, and then determine the
actual value of K. In addition, trial beam sizes must be reasonably estimated.

Example 5:

Compute the effective length factor K for each of the columns in the frame shown in Fig.
137
5.12 using ASD Manual alignment charts. Preliminary sizes of each member are furnished.
Sidesway is not prevented. Webs of the wide-flange shapes are in the plane of the frame.
 STRUCTURAL STEEL DESIGN
Dr. Ammar A. Ali

Solution:

Member Shape I (in.4) l (in.) I/l

AB W14×53 541 168 3.22
BC W14×53 541 144 3.76
CD W18×50 800 360 2.22
BE W21×62 1330 360 3.69
DE W14×53 541 144 3.76
EF W14×53 541 168 3.22

C W18X50 D
W14X53

W14X53
12'-0"

B W21X62 E
W14X53

W14X53
14'-0"

A F

30'-0"

Fig. 5.12 Two-story

rigid frame.

G factors for each joint are determined as follows:

∑(𝐼𝑐⁄𝑙𝑐)
Joint G
∑(𝐼𝑔⁄𝑙𝑔)

A Pinned end = 10.0

3.76 + 3.22
B = 1.89
3.69
3.76
C = 1.69
2.22
3.76
D = 1.69
2.22

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Dr. Ammar A. Ali

E 3.76 + 3.22 = 1.89

3.69
F Fixed end = 1.0
Column K factors from the chart (sidesway uninhibited) are as follows:

Column G values at column ends K

AB 10.0 1.89 2.08
BC 1.89 1.69 1.54
DE 1.69 1.89 1.54
EF 1.89 1.0 1.44

Because of the smaller effective length factors used for frames where sidesway is
prevented, it is advisable to provide lateral support wherever possible. This may be
accomplished with diagonal bracing, shear walls, or attachment to an adjacent structure
having adequate lateral stability or by floor slabs or roof decks secured horizontally by
walls or bracing systems.

The determination of the K factors utilizing the alignment charts based on several
assumptions. Two of the principal assumptions are that all columns in a story buckle
simultaneously and that all columns behavior is purely elastic. Either or both of these
conditions may not exist in an actual structure, and as a result, the use of the alignment
charts will produce overly conservative designs.

The ASD Manual, Part 3, contains a design procedure to reduce the K factor value by
multiplying the elastic G value by a stiffness reduction factor. Then, using the alignment
charts as discussed previously, an inelastic K factor is obtained. The stiffness reduction
factor is obtained from Table A in Part 3 of the ASD Manual.

Example 6:

Compute the inelastic K factors for columns BC and DC of previous example. Consider
behavior in the plane of the frame only. Use Figure 1 and Table A of the ASD Manual, Part ,
for the inelastic K factor procedure. Preliminary sizes are shown in Fig. 5.13. Sidesway in
the plane of the frame is not prevented. Webs of the W-shape members are in the plane of
the frame. Use A36 steel. Assume that the columns support loads of 245 kips as shown in
Fig. 5.13.

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Dr. Ammar A. Ali

245 kips 245 kips

C W18X50 D

W14X53

W14X53
12'-0"

B W21X62 E

30'-0"

Fig. 5.13 Inelastic K factor

determination.

Solution:

1. The trial column size is W14×53. A = 15.6

in.2
2. Compute fa:
𝑓𝑎 = 245
3. From Table A, the stiffness = 15.7 ksi
reduction factor 𝑓𝑎 /𝐹𝑒′ is 0.621.
15.6
4. From previous example, the elastic stiffness ratios are 1.89 at joint B and E (bottom)
and 1.69 at joints C and D (top).
5.
Calculate Ginelastic:

𝐺inelastic (top ) = 0.621(1.69) = 1.05

𝐺inelastic (bottom ) = 0.621(1.89) = 1.17

6. Determine K from Figure 1 of the ASD Manual, Part 3:

𝐾 = 1.35

This compares with a K of 1.54 as determined in previous example, indicating a

greater column capacity if inelastic behavior is considered.

140