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6.1

Chapter 6: Bending Members The following information is taken from “Unified Design of Steel Structures,” Second Edition, Louis F.

Geschwindner, 2012, Chapter 6.

6.1 Bending Members in Structures

A bending member carries loads applied normal to its longitudinal axis.

• In building construction, bending members (known as beams) provide support

for floors and roofs.

• Beams may be simple span or continuous span.

• Beams transfer their load to other structural members such as columns,

girders, and walls.

Various shapes are used as bending members.

• Wide-flange sections are generally used as bending members.

• Angles and tees are commonly used as lintels over openings.

• Built-up shapes made by combining shapes and plates may be used; however,

built-up shapes may not be economical because of the labor costs associated

with fabrication.

The most common and economical bending members (known as compact sections)

are those that can reach the full material yield strength without being limited by

buckling of any cross-sectional elements.

6.2 Strength of Beams

As a load is applied to a bending member, stresses due to the bending moment are

developed in the cross section.

• The stress at any point may be computed from the familiar flexural formula

when the maximum computed stress in the beam is below the elastic limit.

fy = My/I

where

M = the applied moment that stresses the section in the elastic range

y = the distance from the neutral axis to the point where the stress is to be

determined

I = moment of inertia

fy = the resulting bending stress at location y

• The stress at the extreme fiber is usually of greatest interest and the flexural

formula becomes

6.2

fb = Mc/I = M/S

where

S = section modulus

fb = the extreme fiber bending stress

The moment that causes the extreme fiber to reach the yield stress, Fy, is called

the yield moment, My (ref. Figure 6.2a, p. 166 of the textbook).

If the moment in a ductile steel beam is increased beyond the yield moment, the

strain in the extreme fibers increases but the stress in the outermost fibers stays

the same (ref. Figure 6.2c, p. 166 of the textbook).

• The additional resisting moment is carried by the fibers nearer to the neutral

axis

• As the moment increases this process continues with more parts of the beam

cross section stressed to the yield stress.

If the moment continues to increase, the portion of the cross section experiencing

the yield stress continues to increase until the entire section experiences the yield

stress (ref. Figure 6.2d, p. 166 of the textbook).

• When the stress distribution has reached this stage, a full plastic stress

distribution is approached and a plastic hinge is said to have formed, and no

additional moment can be resisted by the section.

• The moment that causes this full plastic stress distribution is called the plastic

moment Mp.

6.3

• A cross section that is capable of attaining a full plastic stress distribution and

the corresponding moment is referred to as a compact section.

- A compact section is one that has a sufficiently stocky cross section so that

it is capable of developing a fully plastic stress distribution before buckling.

At every stage of loading, equilibrium of the cross section requires that the total

internal tension force be equal to the total internal compression force.

• For a doubly symmetric wide flange shape, equilibrium for the plastic moment

occurs when the portion of the shape above the elastic neutral axis is stressed

to the yield stress in compression while the portion below the elastic neutral

axis is stressed to the yield stress in tension.

• For a non-symmetric shape, because the area above the elastic neutral axis is

not equal to the area below the elastic neutral axis, a new axis (called the

plastic neutral axis, PNA) must be defined that gives equal areas in tension

and compression.

- For symmetric shapes the elastic and plastic neutral axes coincide.

Based on equilibrium requirements, the plastic moment (Mp) that corresponds to

the fully yielded stress distribution is determined as follows.

Mp = Fy (Ac yc) + Fy (At yt)

If Ac and At are the equal compression and tension areas, respectively, and yc and

yt are the distances from the centroids of these areas to the PNA, then the

equation for the plastic moment may be simplified as follows.

Mp = Fy (A/2) (yc + yt)

The terms (A/2) (yc + yt) are the first moment of area of the cross section and are

normally combined and called the plastic section modulus, Z and the plastic

moment is given as

Mp = Fy Z

• The plastic section modulus is listed for all available shapes in Part 1 of the

Manual.

For a given beam to attain its full plastic moment strength, it must be compact and

satisfy the lateral support criteria established in Specification Section F2.

• If these criteria are not met, the strength is less than the plastic moment Mp.

• The criteria to be satisfied are defined by two limit states in addition to

yielding: local buckling (noncompact shape) and lateral torsional buckling.

6.4

In this chapter the buckling moments of a series of compact ductile steel beams

with differing lateral or torsional bracing situations are considered.

• The differing lateral or torsional bracing situations include the following.

1. First, the beams will be assumed to have continuous lateral bracing for the

compression flange.

2. Next, the beams will be assumed to be

braced laterally at short intervals.

3. Finally, the beams will be assumed to

be braced laterally at larger intervals.

In the figure at the right a typical curve

showing the nominal resisting or buckling

moments of a beam with varying unbraced

lengths is presented.

• There are three distinct ranges, or

zones, of behavior, depending on the

lateral bracing situation.

Zone 1: There is continuous or closely spaced lateral bracing; the beam will

experience yielding of the entire cross section.

Zone 2: As the distance between lateral bracing is increased, the beam begins

to fail elastically at smaller moments.

Zone 3: With even larger unbraced lengths, the beam will fail elastically at even

smaller moments.

Plastic Behavior (Zone 1)

Consider a compact beam with continuous lateral bracing of the compression flange.

• The beam could be loaded until its full plastic moment Mp is reached at some

point or at multiple points along the beam.

• Further loading produces a redistribution of moments (or stresses) in the beam.

Consider a compact beam with closely spaced intermittent lateral bracing of the

compression flange.

• The beam can be loaded until its full plastic moment Mp is reached if the

spacing between braced points does not exceed a certain value, called Lp.

- Lp is the “limiting laterally unbraced length for the limit state of yielding.”

- The value of Lp is dependent on the dimensions of the beam cross section

and on its yield stress.

6.5

Inelastic Buckling (Zone 2)

If the spacing between the points of lateral or torsional bracing is increased, the

section may be loaded until some, but not all, of the compression fibers are

stressed to the yield stress Fy.

• The section will not have sufficient rotation capacity to permit full moment

redistribution.

• Buckling occurs before the yield stress is reached in some of the compression

elements (referred to as inelastic buckling).

As the unbraced length is increased, the moment that the section can resist

decreases, until finally the beam will buckle before the yield stress is reached

anywhere in the cross section.

• The maximum unbraced length Lr at which the yield stress Fy is reached at one

point marks the end of the inelastic range.

- Lr is the “limiting laterally unbraced length for the limit state of inelastic

lateral-torsional buckling.”

- The value of Lr is dependent on the dimensions of the beam cross section,

the yield stress, and residual stresses in the beam.

- At this point, as soon as the moment causes the yield stress to be reached

at some point in the cross section, the section will buckle.

Elastic Buckling (Zone 3)

If the unbraced length is greater than Lr, the section will buckle elastically before

the yield stress is reached anywhere.

• As the unbraced length is further increased, the buckling moment becomes

even smaller.

• As the moment is increased in such a beam, the beam will deflect transversely

until a critical moment value Mcr is reached.

- At this time, the beam cross section will twist and the compression flange

will move laterally.

6.3 Design of Compact Laterally Supported Wide-Flange Beams

Chapter F of the AISC Specification outlines the provisions for the design of

members for flexure.

• Section F1(2) of the Specification requires that all beams be restrained against

twist about their longitudinal axis at support points.

6.6

For a compact beam to attain its full plastic moment strength, it must also be

laterally supported at certain intervals along its compression flange.

• When this is the case, such a beam is said to have full lateral support.

• The nominal strength of a compact member with full lateral support is

determined only by the limit state of yielding.

- The limit state of lateral torsional buckling does not apply.

According to Section F2, if the unbraced length Lb of the compression flange of a

compact I-shaped or C-shaped section does not exceed Lp (for elastic analysis),

then the member’s bending strength about its major axis may be determined using

the following equations.

Mn = Mp = Fy Zx AISC Equation F2-1

LRFD design strength: φb Mn = φb Fy Zx where φb = 0.90

ASD allowable strength: Mn/Ωb = Fy Z/Ωb where Ωb = 1.67

• When a conventional elastic analysis is used to establish member forces, the

unbraced length Lb may not exceed the value Lp if the nominal moment Mn is to

equal Fy Zx.

- Lb is the “length between points that are either braced against lateral

displacement of the compression flange or braced against twist of the cross

section.”

- The value of Lp is determined by the AISC Specification using the following

equation.

Lp = 1.76 ry (E/Fy)1/2 AISC Equation F2-5

Values of Lp are listed in Table 3-2 of the AISC Manual for W-shapes.

There is no limit of the unbraced length for circular or square cross sections or

for I-shaped beams bent about their minor axis.

• If I-shaped sections are bent about their minor (or y-axes), they will not buckle

before the full plastic moment Mp about the y-axis is developed, as long as the

flange element is compact.

In beam design the following items need to be considered: moments, shears,

deflections, crippling, lateral bracing for the compression flange, and fatigue.

• Beams are generally selected to provide sufficient design moment capacities

(φbMn or Mn/Ωb) and then checked to determine if any of the other items are

critical.

6.7

Rather than work with formulas, Part 3 of the AISC Manual simplifies beam design.

• First, the moment Mu or Ma is computed.

• Then, a section having the required moment capacity is selected from Table 3-2

of the AISC Manual, entitled “W shapes Selection by Zx.”

- In Table 3-2 W-shapes are sorted in descending order by strong-axis

flexural strength and then grouped in ascending order by weight with the

lightest W-shape in each range in bold.

Two important items should be remembered in selecting shapes.

1. Cost: the cost of steel is based on weight.

• It is desirable to select the lightest possible shape having the required

plastic modulus.

- Normally the deeper sections will have the lightest weights for the

required plastic modulus.

- Depth limitations may be considered if the deeper section causes a

problem with headroom or significantly adds to the overall building

height.

2. The beam orientation.

• The plastic modulus values Zx in Table 3-2 are given with respect to the

horizontal axis for beams in the upright position.

- If a beam is turned on its side, the proper plastic modulus Zy about the

y-axis can be found in Table 3-4 of the AISC Manual or in tables giving

the dimensions and shape properties in Part 1 of the AISC Manual.

Beam Weight Estimates

In design, the weight of the beam is usually included in the calculation for the

bending moment because the beam must support itself as well as the external

loads.

• Following is a simple method that allows the designer to estimate the beam

weight quickly and accurately.

1. First, calculate the maximum bending moment, not including the weight of

the beam.

2. Then, pick a beam section from AISC Table 3-2 based on the calculated

moment.

3. Recalculate the bending maximum moment using the weight of the selected

beam section as the estimated beam weight.

- The beam weight will likely be very close to the weight of the member

that is selected for the final design.

6.8

Example Problems – Design of Compact, Laterally Supported Wide-Flange Beams

Example

Given: Beam loaded as shown.

Steel: Fy = 50 ksi

The beam is compact.

The compression flange is laterally braced.

Find: Determine if the beam is adequate to support the loads shown. Check both

LRFD and ASD methods.

Solution

W21 x 44 (Zx = 95.4 in4)

LRFD (φb = 0.90)

wu = 1.2 D + 1.6 L = 1.2 (1.0 + 0.044) + 1.6 (3.0) = 6.05 kips/ft

Mu = wu L2/8 = 6.05(21)2/8 = 333.5 kip-ft

Mn = Fy Zx = 50 (95.4) = 4770 kip-inch (397.5 kip-ft)

φb Mn = 0.90 (397.5) = 357.7 kip-ft > Mu = 333.5 kip-ft OK

The beam is adequate to support the loads shown.

ASD (Ωb = 1.67)

wa = D + L = (1.0 + 0.044) + 3.0 = 4.044

Ma = wa L2/8 = 4.044(21)2/8 = 222.9 kip-ft

Mn = Fy Zx = 50 (95.4) = 4770 kip-inch (397.5 kip-ft) (Same as before.)

Mn/Ωb = 397.5/1.67 = 238.0 kip-ft > Ma = 222.9 kip-ft OK

The beam is adequate to support the loads shown.

Using AISC Table 3-2, for the W21 x 44:

LRFD: φb Mpx = 358 kip-ft > Mu = 333.5 kip-ft

ASD: Mpx/Ωb = 238 kip-ft > Ma = 222.9 kip-ft

(Note: Mpx represents the plastic moment of a section about its x-axis.)

These values agree with the preceding calculations.

6.9

Example

Given: Beam loaded as shown. PL = 36 k PL = 36 k

Steel: Fy = 50 ksi

The beam is compact.

The compression flange is laterally

braced.

Find: Select the most economical section.

Use both LRFD and ASD methods.

Solution

LRFD

Calculate the factored loads (not including the weight of the beam).

wu = 1.2 D + 1.6 L = 1.2 (2.0) + 1.6 (0) = 2.40 kips/ft

Pu = 1.2 D + 1.6 L = 1.2 (0) + 1.6 (36) = 57.6 kips

Calculate the moment (not including the weight of the beam): use AISC Table 3-23

(Cases 1 and 9).

Mu = wu L2/8 + Pu (L/3) = 2.40(30)2/8 + 57.6 (30/3) = 846 kip-ft

Select a beam section from AISC Table 3-2 to use to estimate the beam weight.

Select W27 x 84 φbMpx = 915 kip-ft > Mu = 846 kip-ft OK

Recalculate the factored uniformly distributed load (including the weight of the

beam).

wu = 1.2 D + 1.6 L = 1.2 (2.0 + 0.084) + 1.6 (0) = 2.50 kips/ft

Recalculate the moment (including the weight of the beam): use AISC Table 3-23

(Cases 1 and 9).

Mu = wu L2/8 + P (L/3) = 2.50(30)2/8 + 57.6 (30/3) = 857.3 kip-ft

Select a beam section from AISC Table 3-2 for final design.

Select W27 x 84 φbMpx = 915 kip-ft > Mu = 857.3 kip-ft OK

6.10

ASD

Calculate the load combinations (not including the weight of the beam).

wa = D + L = 2.0 + 0 = 2.0 kips/ft

Pa = D + L = 0 + 36 = 36 kips

Calculate the moment (not including the weight of the beam): use AISC Table 3-23

(Cases 1 and 9).

Ma = wa L2/8 + Pa (L/3) = 2.0(30)2/8 + 36 (30/3) = 585 kip-ft

Select a beam section from AISC Table 3-2 to use to estimate the beam weight.

Select W27 x 84 Mpx/Ωb = 609 kip-ft > Ma = 585 kip-ft OK

Recalculate the uniformly distributed load combination (including the weight of the

beam).

wa = D + L = (2.0 + 0.084) + 0 = 2.084 kips/ft

Recalculate the moment (including the weight of the beam) using AISC Table 3-23

(Cases 1 and 9).

Ma = wa L2/8 + P (L/3) = 2.084(30)2/8 + 36 (30/3) = 594.5 kip-ft

Select a beam section from AISC Table 3-2 for final design.

Select W27 x 84 Mpx/Ωb = 609 kip-ft > Ma = 594.5 kip-ft OK

6.11

6.4 Design of Compact Laterally Unsupported Wide-Flange Beams

Most steel beams are used in such a manner that their compression flanges are

fully restrained against lateral buckling (Zone 1).

• Concrete floors are poured in such a manner that the concrete provides lateral

support for the compression flange of the beam.

If lateral support of the compression flange is not provided by a floor slab, it is

possible that such support may be provided with connecting beams or with special

members used for the purpose of bracing.

• Beams that frame into the sides of a beam or girder and are connected to the

compression flange can usually be counted on to provide full lateral support at

the connection.

- If the connection is made primarily to the tension flange, little lateral

support is provided to the compression flange.

• The intermittent welding of metal roof or floor decks to the compression

flanges of beams will generally provide sufficient lateral bracing.

- The corrugated sheet-metal roofs that are usually connected to the purlins

with metal straps probably furnish only partial lateral support.

• When wood flooring is bolted to supporting steel beams only partial lateral

support is provided.

If there is doubt in the designer’s mind as to the degree of lateral support

provided, it is advisable to assume that there is no lateral support.

Lateral Torsional Buckling

The compression region of a bending member cross section has a tendency to

buckle similarly to how a pure compression member buckles.

• The upper half of the wide flange member in bending acts as a T in pure

compression.

- The upper half of the wide flange is fully braced about its horizontal axis by

the web so it will not buckle vertically.

- The upper half of the wide flange may be unbraced about its vertical axis

and thus will have a tendency to buckle laterally.

The tension region tends to restrain the lateral buckling and, as a result, the beam

deflects downward and buckles laterally, causing it to twist.

6.12

In order to resist this tendency to buckle, Specification Sections B3.6 and F1(2)

require that all bending members be restrained against rotation about their

longitudinal axis at their support points.

• If the beam has sufficient lateral and torsional support along its length, a

compact section can develop the yield stress before buckling occurs.

• If the beam tends to buckle before the yield stress is reached, the nominal

moment strength is less than the plastic moment.

To ensure that a beam cross section can develop its full plastic moment strength

without lateral torsional buckling, Specification Section F2.2 limits the slenderness

as follows.

Lb/ry ≤ 1.76 (E/Fy)1/2

where

Lb = unbraced length of the compression flange

ry = radius of gyration for the shape about the y-axis

The requirement for attaining the full plastic moment strength is given in

Specification F2.

Lb ≤ Lp = 1.76 ry (E/Fy)1/2 AISC Equation F2-5

where

Lp = the limiting laterally unbraced length for the limit state of yielding

• Lp is the maximum unbraced length that permits the shape to reach its plastic

moment strength.

• When the unbraced length of a beam exceeds Lp, its strength is reduced

because the member will have a tendency to buckle before the plastic moment

is reached.

Zone 2

When Lp < Lb ≤ Lr, the section may be loaded until some, but not all, of the

compression fibers are stressed to the yield stress Fy.

• The nominal moment strength is calculated using the following equation.

Mn = Cb {Mp – (Mp – 0.7 Fy Sx)[(Lb – Lp)/(Lr – Lp)]} ≤ Mp AISC Equation F2-2

where

Lr = limiting laterally unbraced length for the limit state of inelastic lateral-

torsional buckling

6.13

• The Specification sets the level of residual stress at 0.3Fy so that only 0.7Fy is

available to resist a bending moment elastically.

- The definition of plastic moment Mp = Fy Zx for beams in Zone 1 is not

affected by residual stresses.

◦ The sum of the compressive stresses equals the sum of the tensile

stresses in the section and the net effect is, theoretically, zero.

• Lr is a function of several properties of the section, such as its cross sectional

area, modulus of elasticity, yield stress, and warping and torsional properties.

- Complex formulas to determine the value of Lr are given in Section F1 of the

AISC Specification.

- Fortunately, numerical values of Lr (as well as values of Lp) have been

determined for sections that are normally used as beams and are given in

Table 3-2 of the AISC Manual, entitled “W Shapes Selected by Zx.”

Simplified expressions of AISC Equation F2-2 that follow are presented in the

AISC Manual (p. 3-9).

LRFD: φbMn = Cb [φb Mpx – φb BF (Lb – Lp)] ≤ φb Mpx Equation 3-4a

ASD: Mn/Ωb = Cb [Mpx/Ωb – (BF/Ωb) (Lb – Lp)] ≤ Mpx/Ωb Equation 3-4b

where BF = (Mpx – 0.7 Fy Sx)/(Lr – Lp)

The bending factors (BF) represent certain factors of AISC Equation F2-2 as can

be seen by comparing the equations.

• Numerical values (in kips) of the bending factors (BF) for W-shapes are listed

in Table 3-2 of the AISC Manual.

6.14

Example Problems – Moment Capacities, Zone 2

Example

Given: Beam section W24 x 62

Steel: Fy = 50 ksi

Lb = 8.0’

Cb = 1.0

Find: LRFD design moment capacity φbMnx

ASD allowable moment capacity Mnx/Ωb

Solution

From AISC Table 3-2 for the W24 x 62:

φbMpx = 574 kip-ft Mpx/Ωb = 382 kip-ft

Lp = 4.87’ Lr = 14.4’

φbBF (LRFD) = 24.1 kips BF/Ωb (ASD) = 16.1 kips

Lp < Lb < Lr, thus the beam section falls into Zone 2.

LRFD

Calculate φbMnx using the simplified expression of AISC Equation F2-2 that is

presented in the AISC Manual (p. 3-9).

φbMnx = Cb [φbMpx – φbBF (Lb – Lp)] ≤ φbMpx

φbMnx = 1.0 [574 – 24.1(8.0 – 4.87)] = 498.6 kip-ft < φbMpx = 574 kip-ft OK

ASD

Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is


Mnx/Ωb = Cb [Mpx/Ωb – (BF/Ωb) (Lb – Lp)] ≤ Mpx/Ωb

Mnx/Ωb = 1.0 [382 – 16.1(8.0 – 4.87)] = 331.6 kip-ft

< Mpx/Ωb = 382 kip-ft OK

6.15

Example


Steel: Fy = 50 ksi

Cb = 1.0

Find: LRFD design moment capacity (φbMnx) and ASD allowable moment capacity

(Mnx/Ωb) for simple spans of 6 and 12 feet with lateral bracing for the

compression flange provided at the end only.

Solution

From AISC Table 3-2 for the W21 x 73:

φb Mpx = 645 kip-ft Mpx/Ωb = 429 kip-ft

Lp = 6.39’ Lr = 19.2’

φbBF (LRFD) = 19.4 kips BF/Ωb (ASD) = 12.9 kips

6’ Span

Lb < Lp, thus the beam section falls into Zone 1.

Using AISC Table 3-2 for the W21 x 73:

LRFD: φbMnx = φbMpx = 645 kip-ft

ASD: Mnx/Ωb = Mpx/Ωb = 429 kip-ft

12’ Span

Lp < Lb < Lr, thus the beam section falls into Zone 2.




φbMnx = 1.0 [645 – 19.4(12.0 – 6.39)] = 536.2 kip-ft < φbMpx = 645 kip-ft OK




Mnx/Ωb = 1.0 [429 – 12.9(12.0 – 6.39)] = 356.6 kip-ft


6.16

Zone 3

When Lb > Lr, the section will buckle elastically before the yield stress is reached

anywhere in the section.

• The nominal moment strength is calculated using the following equation.

Mn = Fcr Sx ≤ Mp AISC Equation F2-3

Fcr = (Cbπ2E)/(Lb/rts)2 { 1 + 0.078 [Jc/(Sx ho)](Lb/rts)2 }1/2 AISC Equation F2-4

where

rts = effective radius of gyration (listed in AISC Table 1-1)

J = torsional constant (listed in AISC Table 1-1)

c = 1.0 for doubly symmetrical I-shape

ho = distance between flange centroids (listed in AISC Table 1-1)

Lateral-torsional buckling will not occur if the moment of inertia of the section

about the bending axis is less than or equal to the moment of inertia out of plane.

• For this reason the limit state of lateral torsional buckling is not applicable for

shapes bent about their minor axes, for shapes with Ix ≤ Iy, or for circular or

square shapes.

• Furthermore, yielding controls if the section is noncompact.

6.17

Example Problem – Elastic Buckling, Zone 3

Example


Steel: Fy = 50 ksi

Cb = 1.0

Lb = 38’

Find: LRFD design moment capacity (φbMnx)

ASD allowable moment capacity (Mnx/Ωb)

Solution

W18 x 97 (Lr = 30.4’, rts = 3.08”, ho = 17.7”, J = 5.86 in4, Sx = 188 in4, Zx = 211 in3)

Lb = 38’ > Lr = 30.4’, thus the beam section falls into Zone 3.

Calculate Fcr using AISC Equation F2-4.

Fcr = (Cbπ2E)/(Lb/rts)2 { 1 + 0.078 [Jc/(Sx ho)](Lb/rts)2 }1/2

= 1.0π2(29,000)/(38x12/3.08)2{1 + 0.078[5.86(1.0)/(188)(17.7)](38x12/3.08)2}1/2

= 13.058 (2.003)

= 26.15 ksi

Calculate Mnx using AISC Equation F2-3.

Mnx = Fcr Sx = 26.15(188) = 4916.2 kip-inch (409.7 kip-ft)

< Mp = Fy Zx = 50 (211) = 10,550 kip-inch (879.2 kip-ft)

Compute the LRFD design moment capacity (φbMnx) and the ASD allowable moment

capacity (Mnx/Ωb).

LRFD (φb = 0.90): φbMnx = 0.90 (409.7) = 368.7 kip-ft

ASD (Ωb = 1.67): Mnx/Ωb = 409.7/1.67 = 245.3 kip-ft

6.18

Moment Gradient

A bending moment coefficient Cb is included in formulas to account for the fact

that moment is not usually uniform across the entire length of the beam.

• Cb is called the lateral-torsional buckling modification factor for nonuniform

moment diagrams when both ends of the unsupported segment are braced.

• Lateral buckling may be affected by the end restraint and by the loading

conditions of the member.

The moment in the unbraced beam in part (a) of the figure below causes a more

severe compression situation than does the moment in the unbraced beam in part

(b) of the figure.

• The upper flange of the beam in the figure at the left is in compression for its

entire length.

• In the figure at the right, the length of the “column” (i.e. the length of the

upper flange that is in compression) is much less (in effect, a much shorter

“column”).

The basic moment capacity equations for Zones 2 and 3 were developed for

laterally unbraced beams subject to single curvature with Cb = 1.0.

• Frequently beams are not bent in single curvature (e.g. the fixed beam in part b

of the figure above), with the result that they can resist more moment.

- To handle this situation, the AISC Specification provides Cb coefficients

larger than 1.0 that are to be multiplied by the computed Mn values.

• A Cb value equal to 1.0 may be used conservatively, but the designer is missing

out on the possibility of significant savings in steel weight for some situations.

6.19

• It is important to note, when using Cb values, the moment capacity obtained by

multiplying Mn by Cb may not be larger than the plastic moment Mn of Zone 1,

where Mn = Fy Zx.

The value of Cb is determined from the following expression.

Cb = 12.5 Mmax/(2.5 Mmax + 3 MA + 4 MB + 3 MC) AISC Equation F1-1

where

Mmax = the largest moment in an unbraced segment of a beam

MA, MB, MC = the moments at the ¼ point, ½ point, and ¾ point of the beam

segment, respectively

Some typical values for Cb include the following.

Cb = 1.14 for a simply supported beam with a uniformly distributed load with

bracing only at the supports.

Cb = 2.38 for a fixed-end beam with a uniformly distributed load with bracing

only at the supports, or with bracing at the supports and mid-span.

Cb = 1.92 for a fixed-end beam with a concentrated load at mid-span with

bracing only at the supports.

Cb = 2.27 for a fixed-end beam with a concentrated load with bracing only at

the supports and mid-span.

Cb = 1.0 for cantilevers or overhangs with a concentrated load at the free end

with bracing only at the fixed support (i.e. the free end is unbraced).

Other typical values of Cb are shown in AISC Table 3-1 (p. 3-18) for various loading

situations for simply supported beams.

Design Charts

Fortunately the values of the LRFD design moment capacity (φbMn) and the ASD

allowable moment capacity (Mn/Ωb) for sections normally used as beams have been

computed by the AISC.

• Values of φbMn and Mn/Ωb have been plotted for a wide range of unbraced

lengths.

- These plots are shown as Table 3-10 in the AISC Manual.

• These diagrams enable a designer to solve any of the problems previously

considered in this chapter in a short period of time.

The values provided include unbraced lengths in the plastic range (Zone 1), in the

inelastic buckling range (Zone 2), and in the elastic buckling range (Zone 3).

6.20

• The values are plotted for Fy = 50 ksi and Cb = 1.0.

- Lp is indicated with a solid circle (•).

- Lr is indicated with a hollow circle (◦).

• The dashed lines indicate that the sections will provide the necessary moment

capacities, but are not economical.

• When the plotted line is solid, the W-shape for that curve is the lightest cross

section for a given combination of available flexural strength and unbraced

length.

To select a member using AISC Table 3-10, enter the chart with the unbraced

length Lb and the LRFD factored design moment Mu or the ASD allowable moment

Ma.

Example (LRFD solution)

Assume that Cb = 1.0, Fy = 50 ksi; using AISC Table 3-10, select a beam with

Lb = 18’, and Mu = 544 kip-ft.

• First, proceed up from the bottom of the chart for an unbraced length of 18’

until you reach a horizontal line corresponding with φbMn = 544 kip-ft.

- Any section to the right and above this intersection point will have a greater

design moment capacity.

◦ The first sections that are encountered are the W16 x 89 and W14 x 90,

shown as dashed lines in this section of the chart.

• Proceeding further upward and to the right, the first solid line encountered is

for a W24 x 84 and represents the lightest satisfactory section.

Example (ASD solution)

Assume that Cb = 1.0, Fy = 50 ksi; using AISC Table 3-10, select a beam with

Lb = 18’, and Ma = 370 kip-ft.

• First, proceed up from the bottom of the chart for an unbraced length of 18’

until we reach a horizontal line corresponding with Mn/Ωb = 370 kip-ft.

- Any section to the right and above this intersection point will have a greater

design moment capacity.

◦ In this case, the first section that is encountered is the W24 x 84,

shown as a solid line in this section of the chart.

To use AISC Table 3-10 when Cb > 1.0, use the following procedure.

1. Compute an effective moment value.

For LRFD, (Mu)effective = Mu/Cb

For ASD, (Ma)effective = Ma/Cb

6.21

2. Use the effective moment value and the unbraced length with AISC Table 3-10

to select a trial section.

3. Determine whether the beam is in Zone 1, 2 or 3 by comparing the unbraced

length Lb with Lp and Lr of the selected trial section, then proceed with the

necessary checks, as outlined below.

For beams in Zone 1, use AISC Table 3-2 to check the moment strength of the

selected section: that is, φbMn for LRFD or Mn/Ωb for ASD.

• The calculated moment Mu for LRFD design, or Ma for ASD design, must not

exceed the moment strength of the selected section.

For LRFD, Mu ≤ φbMn = φbMpx φbMpx = φb Fy Zx

For ASD, Ma ≤ Mn/Ωb = Mpx/Ωb Mpx/Ωb = Fy Zx/Ωb

• If the calculated moment (Mu or Ma) is greater than the moment strength of

the selected trial section (i.e. φbMn for LRFD, or Mn/Ωb for ASD), then select a

larger section from AISC Table 3-2 that provides the required moment

strength, so that

For LRFD, φbMn = φbMpx ≥ Mu

For ASD, Mn/Ωb = Mpx/Ωb ≥ Ma



• As an initial check, the calculated moment Mu for LRFD design, or Ma for ASD

design, must not exceed the moment strength of the selected section.

For LRFD, Mu ≤ φbMn = φbMpx

For ASD, Ma ≤ Mn/Ωb = Mpx/Ωb

- Note that φbMpx for LRFD and Mpx/Ωb for ASD are maximum values of the

moment strength for the selected section; however, the actual values for

the selected beam section may be less than the maximum values listed in

AISC Table 3-2 because of the effects of the unbraced length.



larger section from AISC Table 3-2 that may provide the required moment

strength, so that



• Use the simplified expression of AISC Equation F2-2 that is presented in the

AISC Manual (p. 3-9) to compute φbMn for LRFD and Mn/Ωb for ASD to verify

6.22

the actual moment strength of the selected section and compare with the

required moment strength.

For LRFD, φbMn ≥ Mu

For ASD, Mn/Ωb ≥ Ma

- If the calculated moment (Mu or Ma) is greater than the moment strength of

the selected trial section (i.e. φbMn for LRFD, or Mn/Ωb for ASD), then

select and check a larger section from AISC Table 3-2 until a section that

provides the required moment strength is found.



• As an initial check, the calculated moment Mu for LRFD design, or Ma for ASD

design, must not exceed the moment strength of the selected section.

For LRFD, Mu ≤ φbMn = φbMpx

For ASD, Ma ≤ Mn/Ωb = Mpx/Ωb

- Note that φbMpx for LRFD and Mpx/Ωb for ASD are maximum values of the

moment strength for the selected section; however, the actual values for

the selected beam section may be less than the maximum values listed in

AISC Table 3-2 because of the effects of the unbraced length.



larger section from AISC Table 3-2 that may provide the required moment

strength, so that



• Use AISC Equations F2-3 and F2-4 to compute the nominal moment strength Mn

for the trial section.

• Compute φbMn for LRFD and Mn/Ωb for ASD to verify the actual moment

strength of the selected section and compare with the required moment

strength.



- If the calculated moment (Mu or Ma) is greater than the moment strength of

the selected trial section (i.e. φbMn for LRFD, or Mn/Ωb for ASD), then

select and check a larger section from AISC Table 3-2 until a section that

provides the required moment strength is found.

6.23

Example Problem – Design Charts with Cb > 1.0

Example

Given: Beam loaded as shown.

Steel: Fy = 50 ksi

Bracing is provided only at

the ends and center line of

the member: Cb = 1.67

Find: Select the lightest member using both the LRFD and ASD methods.

Solution

LRFD

Calculate the factored loads (assume a beam weight of 84 lb/ft).

wu = 1.2 D + 1.6 L = 1.2 (0 + 0.084) + 1.6 (0) = 0.101 kip/ft

Pu = 1.2 D + 1.6 L = 1.2 (30) + 1.6 (40) = 100 kips

Calculate the moment using AISC Table 3-23 (Cases 1 and 7).

Mu = wuL2/8 + PuL/4 = 0.101 (34)2/8 + 100(34)/4 = 864.6 kip-ft

Since Cb > 1.0, calculate the effective moment.

(Mu)effective = Mu/Cb = 864.6/1.67 = 518 kip-ft

Enter AISC Table 3-10 (Lb = 17’, φbMn = 518 kip-ft) and select a trial beam section.

Select W24 x 76

Check the selected trial section.

From Table 3-2: W24 x 76

φbMpx = 750 kip-ft < Mu = 864.6 kip-ft NG

Check the next larger section.

From Table 3-2: for W24 x 84

φbMpx = 840 kip-ft < Mu = 864.6 kip-ft NG


From Table 3-2: for W27 x 84

φbMpx = 915 kip-ft > Mu = 864.6 kip-ft May be OK

Verify the actual moment strength of the selected section.

From Table 3-2: W27 x 84 (φbMpx = 915 kip-ft, φbBF = 26.4, Lp = 7.31’, Lr = 20.8’)

Lp = 7.31’ < Lb = 17’ < Lr = 20.8’, thus the beam section falls into Zone 2.

6.24




φbMnx = 1.67 [915 – 26.4(17.0 – 7.31)] = 1100.8 kip-ft

> φbMpx = 915 kip-ft (Use φbMnx = 915 kip-ft)

φbMnx = 915 kip-ft > Mu = 864.6 kip-ft OK

Select W27 x 84

ASD

Calculate the load combinations (assume a beam weight of 84 lb/ft).

wa = D + L = (0 + 0.084) + 0 = 0.084 kip/ft

Pa = D + L = 30 + 40 = 70 kips

Calculate the moment using AISC Table 3-23 (Cases 1 and 7).

Ma = waL2/8 + PaL/4 = 0.084 (34)2/8 + 70(34)/4 = 607.1 kip-ft

Since Cb > 1.0, calculate the effective moment.

(Ma)effective = Ma/Cb = 607.1/1.67 = 363 kip-ft

Enter AISC Table 3-10 (Lb = 17’, Mn/Ωb = 363 kip-ft) and select a trial beam

section.

Select W24 x 84

Check the selected trial section.


Mpx/Ωb = 559 kip-ft < Ma = 607.1 kip-ft NG



Mpx/Ωb = 609 kip-ft > Ma = 607.1 kip-ft May be OK

Verify the actual moment strength of the selected section.

From Table 3-2: W27 x 84 (Mpx/Ωb = 609 kip-ft, BF/Ωb = 17.6, Lp = 7.31’, Lr = 20.8’)

Lp = 7.31’ < Lb = 17’ < Lr = 20.8’, thus the beam section falls into Zone 2.




Mnx/Ωb = 1.67 [609 – 17.6(17.0 – 7.31)] = 732.2 kip-ft

> Mpx/Ωb = 609 kip-ft (Use Mnx/Ωb = 609 kip-ft)

Mnx/Ωb = 609 kip-ft > Ma = 607.1 kip-ft OK

Select W27 x 84

6.25

6.5 Design of Noncompact Beams

Local Buckling

Local buckling occurs when a compression element of a cross section buckles under

the load before that element reaches the yield stress.

• The section is not compact and the strength of the member is less than the

plastic moment Mp.

Local buckling failures occur when the flange or web are slender.

• The projecting flange of a wide-flange member is considered an “unstiffened”

element because the web supports only one edge, while the other edge is

unsupported and free to rotate.

• The wide-flange web is connected at both its edges to the flanges, so it is

considered a “stiffened” element.

Table B4.1b of the Specification provides the limiting slenderness values λp for the

flange and web to ensure that the full plastic moment strength can be reached.

• When both the flange and web have slenderness ratios (b/tf and h/tw,

respectively) less than or equal to the λp values given in the table, the shapes

are called compact shapes.

- A compact section is one that has a sufficiently stocky cross section so that

it is capable of developing a fully plastic stress distribution (assuming its

compression flange has sufficient lateral bracing) before something buckles

(web or flange).

• If either element exceeds this value, the shape cannot be called compact and

the nominal strength must be reduced.

- A noncompact section is one for which the yield stress can be reached in

some, but not all, of its compression elements before buckling occurs.

- A noncompact section is not capable of reaching a fully plastic stress

distribution.

For the flange of a W-shape to be compact, its width-to-thickness ratio must

satisfy the following limit.

λpf = bf/2tf ≤ λp = 0.38 (E/Fy)1/2 Case 10, Table B4.1b

λp = limiting slenderness parameter for a compact element

For the web of a W-shape to be compact, its width-to-thickness ratio must satisfy

the following limit.

λpw = h/tw ≤ λp = 3.76 (E/Fy)1/2 Case 15, Table B4.1b

6.26

Using common A992 steel with Fy = 50 ksi, these limits are as follows.

For a compact flange: λpf = bf/2tf ≤ λp = 9.15

For a compact web: λpw = h/tw ≤ λp = 90.6

Comparing these limiting values with data given in AISC Manual Table 1-1, the

majority of W-shapes have compact flanges and all have compact webs for Fy = 50

ksi.

Flange Local Buckling

The full range of nominal moment strength Mn of a cross section can be expressed

as a function of flange slenderness λf.

The three regions shown in Figure 6.18 (p. 190 of the textbook) identify three

types of behavior.

• The first region represents plastic behavior and the shape is capable of

attaining its full plastic moment strength (λf ≤ λp).

- Shapes that fall into this region

are called compact.

• The behavior exhibited in the

middle region is inelastic and the

shape is not capable of attaining

its full plastic moment strength

(i.e. λp < λf ≤ λr).

- Shapes that fall into this

category are called noncompact.

• Shapes that fall into the last region exhibit elastic buckling (λf > λr).

- These shapes are called slender shapes.

If a section has noncompact flanges (i.e. λp < λ ≤ λr), the value of Mn is given by the

following equation.

Mn = {Mp – (Mp – 0.7 Fy Sx) [(λ – λpf)/(λrf – λpf)]} AISC Equation F3-1

• Values of λp and λr for different shapes are listed in Table B4.1b of the AISC

Specification.

If a section has slender flanges (i.e. λ > λr), the value of Mn is given by the following

equation.

Mn = 0.9 E kc Sx/λ2 AISC Equation F3-2

6.27

where

λ = bf/2tf

kc = 4/(h/tw)1/2 (where 0.35 ≤ kc ≤ 0.76)

Using common A992 steel with Fy = 50 ksi, the upper limit for the noncompact

flange is as follows.

λrf = 1.0 (E/Fy)1/2 = 24.08

Comparing this limiting value with values of bf/2tf given in AISC Manual Table 1-1,

there are no W-shapes with flanges that exceed this limit.

• All wide-flange shapes have either compact or noncompact flanges (i.e. none

have flanges with slender elements).

• Only a few W10 shapes have noncompact flanges.

More generally, almost all of the standard hot-rolled W, M, S, and C shapes listed

in the AISC Manual are compact, and none of them fall into the slender

classification.

• All standard hot-rolled shapes have compact webs, but a few have noncompact

flanges.

- If a standard shape has a noncompact flange, it is so indicated in the various

tables of the AISC Manual with a “f” footnote (e.g. W10 x 12, p. 3-27).

- Where applicable, the numerical values shown in the tables are based on the

reduced stresses caused by the non-compactness.

The equations mentioned here were used, where applicable, to obtain the values

used for the charts plotted in Table 3-10 of the AISC Manual.

• The designer should have little trouble with noncompact sections when Fy is no

more than 50 ksi.

• However, the designer will have to use the formulas presented in Section F3 of

the AISC Manual for shapes with Fy values larger than 50 ksi.

Web Local Buckling

Comparing the slenderness criteria for local web buckling (Case 15, Table B4.1b)

with values of h/tw give in AISC Manual Table 1-1, all wide-flange shapes have

compact webs.

• The consideration of noncompact W-shapes is a consideration of only flange

local buckling and W-shapes contain no slender elements.

• Slender webs are a consideration for built-up members, such as plate girders.

6.28

Example Problem – Noncompact Sections

Example

Given: Beam section consisting of a W12 x 65 with full lateral bracing.

Steel: Fy = 50 ksi

Find: The LRFD flexural design stress and the ASD allowable flexural stress.

Solution

W12 x 65 (bf = 12.0”, tf = 0.605”, bf/2tf = 9.92, Sx = 87.9 in3, Zx = 96.8 in3)

Determine whether the flange is compact.

• Reminder: All standard hot-rolled shapes have compact webs, but a few have

noncompact flanges.

From AISC Table B4.1b (Case 10):

λp = 0.38 (E/Fy)1/2 = 0.38 (29,000/50)1/2 = 9.15

λr = 1.0 (E/Fy)1/2 = 1.0(29,000/50)1/2 = 24.08

From AISC Table 1-1: λ = bf/2tf = 9.92

λp = 9.15 < λ = 9.92 < λr = 24.08, so the flange is noncompact.

Calculate the nominal flexural stress.

Mp = Fy Zx = 50 (96.8) = 4840 kip-inch

Mn = {Mp – (Mp – 0.7 Fy Sx) [(λ – λpf)/(λrf – λpf)]} AISC Equation F3-1

= {4840 – [4840 – 0.7 (50) (87.9)] [(9.92 – 9.15)/(24.08 – 9.15)]}

= 4749.0 kip-inch (395.7 kip-ft)

Determine φbMn and Mn/Ωb.

LRFD (φb = 0.90): φbMn = 0.90 (395.7) = 356 kip-ft

ASD (Ωb = 1.67): Mn/Ωb = 395.7/1.67 = 237 kip-ft

These values correspond with the values listed in Table 3-2 of the AISC Manual.

6.29

6.6 Design of Beams for Weak Axis Bending

Beams are not normally oriented for bending about the weak axis.

• However, beams occasionally must resist bending about the weak axis due to

lateral loads.

The design of I-shaped beams for weak axis bending is relatively easy.

• Section F6 of the Specification addresses I-shaped members and channels bent

about their weak axis.

• Two limit states are identified by the Specification: yielding and flange local

buckling.

- For the limit state of yielding, the following equation is used.

Mn = Mp = Fy Zy ≤ 1.6 Fy Sy Equation F6-1

- For sections with compact flanges the limit state of flange local buckling

does not apply.

- For the few W-shapes with noncompact flanges, Equation F6-2 is used to

determine the nominal flexural strength Mn.

◦ Equation F6-2 is similar to Equation F3-1, except Sy is used in lieu of Sx.

- For sections with slender flanges (e.g. built-up members), Equation F6-3 is

used to determine the nominal flexural strength Mn.

6.7 Design of Beams for Shear

As a member bends, shear stresses occur because of the changes in the length of

the longitudinal fibers.

Generally shear is not a problem in steel beams.

• The webs of rolled shapes are capable of resisting large shear forces, except in

the following cases.

1. If large concentrated loads are placed near beam supports, these loads will

cause large internal forces without corresponding increases in bending

moment.

Example: Upper columns may be offset with respect to the columns below.

2. Shear may be a problem if two members (such as a beam and a column) are

rigidly connected so that their webs are in the same plane.

Example: The junction of columns and beams (or rafters) in rigid frame

structures.

3. Shear may be a problem if beams are notched or coped or where holes are

cut in beam webs for ductwork.

6.30

4. Very heavily loaded beams can have excessive shears.

5. Shear may be a problem for ordinary loadings when very thin webs are used,

as in plate girders.

The LRFD design shear strength, φvVn, and the ASD allowable shear strength,

Vn/Ωv, are determined according to the provisions of Chapter G of the AISC

Specification.

LRFD: Vu ≤ φvVn

ASD: Va ≤ Vn/Ωv

The nominal shear strength Vn of unstiffened or stiffened webs is specified as

Vn = 0.6 Fy Aw Cv AISC Equation G2-1

where

Aw = area of the web = d tw

Cv = web shear coefficient (defined below)

d = overall depth of the member

tw = web thickness

For the webs of rolled I-shaped members, when h/tw ≤ 2.24 (E/Fy)1/2

Cv = 1.0

φv = 1.00 and Ωv = 1.50

• Almost all current W, S, and HP shapes for Fy = 50 ksi meet this criteria.

• Exceptions are listed by User Note in Section G2 of the AISC Specification.

For webs of all other doubly symmetric shapes, singly symmetric shapes, and

channels (not including round HSS), Cv is determined from the following equations.

a. When h/tw ≤ 1.10 (kv E/Fy)1/2

then Cv = 1.0 AISC Equation G2-3

kv = web plate buckling coefficient (defined by AISC Section G2.1)

For webs without transverse stiffeners and with h/tw < 260,

kv = 5 (except for the stem of tees)

kv = 1.2 for the stem of tees

For webs with transverse stiffeners,

kv = 5 + 5/(a/h)2

kv = 5 when a/h > 3.0 or a/h > [260/(h/tw)]2

6.31

where

a = clear distance between transverse stiffeners

h = for rolled shapes, the clear distance between flanges less the fillet or

corner radii

b. When 1.10 (kv E/Fy)1/2 < h/tw ≤ 1.37 (kv E/Fy)1/2

then Cv = 1.10 (kv E/Fy)1/2/(h/tw) AISC Equation G2-4

c. When h/tw > 1.37 (kv E/Fy)1/2

then Cv = 1.51 kv E/[(h/tw)2 Fy] AISC Equation G2-5

For all other shapes other than rolled I-shaped members,

φv = 0.90 (LRFD) and Ωv = 1.67 (ASD).

Notes

1. The values of φvVnx and Vnx/Ωv with Fy = 50 ksi are given for W-shapes in Table

3-2 of the AISC Manual.

2. AISC Table 3-6 is provided for determining the maximum uniform load each W

shape can support for various spans.

• The values are based on Fy = 50 ksi.

• The values are controlled by maximum moments or shears, as specified by

LRFD or ASD.

If Vu or Va for a particular beam exceeds the AISC specified shear strength of

the member, the usual procedure is to select a slightly heavier section.

• If a much heavier section is required than what is required for moment, doubler

plates may be welded to the beam web, or stiffeners may be connected to the

webs in zones of high shear.

- Doubler plates must meet the width-thickness requirements for compact

stiffened elements (ref. Section B4 of the AISC Specification).

- Doubler plates must be welded sufficiently to the member webs to develop

their proportionate share of the load.

The AISC specified shear strengths of a beam or girder are based on the entire

area of the web.

• Sometimes, however, a connection is made to only a portion of the web.

- For such a case, the designer may decide to assume that the shear is spread

over only part of the web depth for purposes of computing shear strength.

6.32

- Thus, the designer may compute Aw as being equal to tw times the smaller

depth for use in the shear strength expression.

When beams that have their top flanges at

the same elevations (the usual situation) are

connected to each other, it is frequently

necessary to cope one of them as shown in

the figure at the right.

• For such cases, there is a distinct

possibility of a block shear failure along

the broken lines shown.

6.33

Example Problem – Shear

Example

Given: Beam (W21 x 55) loaded as shown.

Steel: Fy = 50 ksi

Find: Check the adequacy of the beam in shear.

Solution

W21 x 55 (d = 20.8”, tw = 0.375”, h/tw = 50.0)

h/tw = 50.0 < 2.24 (E/Fy)1/2 = 2.24 (29,000/50)1/2 = 53.95 and Cv = 1.0

Vn = 0.6 Fy Aw Cv = 0.6 (50)(20.8 x 0.375)(1.0) = 234.0 kips Equation G2-1

LRFD (φv = 1.00)

wu = 1.2 D + 1.6 L = 1.2 (2.0) + 1.6 (4.0) = 8.80 kips/ft

Vu = wu L/2 = 8.80(20)/2 = 88.0 kips

φvVn = 1.00 (234.0) = 234.0 kips > Vu = 88.0 kips OK

The beam is adequate for shear.

ASD (Ωv = 1.50)

wa = D + L = 2.0 + 4.0 = 6.0 kips/ft

Va = wa L/2 = 6.0(20)/2 = 60.0 kips

Vn/Ωv = 234.0/1.50 = 156.0 kips > Va = 60.0 kips OK

The beam is adequate for shear.

The values computed above for φvVn and Vn/Ωv correspond with the values listed

Table 3-2 of the AISC Manual.

LRFD: φvVnx = 234 kips

ASD: Vnx/Ωv = 156 kips

6.34

6.8 Continuous Beams

Beams that span over more than two supports are called continuous beams.

• Continuous beams are statically indeterminate and must be analyzed using more

than just the equations of equilibrium.

The AISC Manual includes shears, moments, and deflections for several continuous

beams with various load patterns in AISC Table 3-23.

The ductile nature of steel permits steel members to redistribute load.

• When one section of a member becomes overloaded, the member can

redistribute a portion of its load to a less loaded section.

• Advanced analysis methods (e.g. plastic analysis) may be used in design through

the provisions of Appendix 1 of the Specification.

- If plastic analysis is used, this advantage is automatically included in the

analysis.

• Appendix Section 1.3 also permits use of the simplified plastic analysis approach

for continuous beams.

To allow the designer to take advantage of some of the redistribution that is

accounted for in plastic analysis, Section B3.7 gives provisions for moment

redistribution in beams by a rule of thumb approach that approximates the real

plastic behavior if an elastic analysis is used.

• Design of beams and girders that are compact and have sufficiently braced

compression flanges may take advantage of this simplified redistribution

approach.

• To use the simplified redistribution, for doubly symmetric I-shaped beams the

unbraced length of the compression flange Lb must be less than that given in

Specification Section F13.5, as determined by the following equation.

Lm = [0.12 + 0.076(M1/M2)] (E/Fy) ry Equation F13-8

where

M1 = the smaller moment at the end of the unbraced length

M2 = the larger moment at the end of the unbraced length

M1/M2 is positive when the moments cause reverse curvature and negative

for single curvature

• For continuous compact section, Section B3.7 of the AISC Specification states:

“The required flexural strength of beams composed of compact sections, as

defined in Section B4.1, and satisfying the unbraced length requirements of

6.35

Section F13.5 may be taken as nine-tenths of the negative moments at points of

support, produced by the gravity loading and determined by an elastic analysis

satisfying the requirements of Chapter C, provided that the maximum positive

moment is increased by one-tenth of the average negative moment determined

by the elastic analysis.”

- The reduction is not permitted for moments in members with Fy exceeding

65 ksi.

- The moment reduction does not apply to moments produced by loading on

cantilevers, for design using partially restrained moment connections, or for

design by inelastic analysis using the provisions of Appendix 1.

- The 0.9 factor is applicable only to gravity loads and not to lateral loads

such as those caused by wind and earthquake.

6.36

Example Problem – Moment Reduction and the Design of Continuous Beams

Example

Given: The loaded beam shown.

Steel: Fy = 50 ksi

Find: Select the lightest W-section using an elastic analysis with the 0.9 rule.

Assume that full lateral support is provided for the flanges.

Solution

Elastic analysis and design (ASD design procedure).

wa = D + L = 1.0 + 3.0 = 4.0 kips/ft

Pa = D + L = 15 + 20 = 35.0 kips

Moment Diagram

Note: The moment diagram was developed using the “moment distribution” method.

Maximum negative moment for design based on moment redistribution:

-Mmax = 0.9 (505) = - 454.5 kip-ft (governs)

Maximum positive moment for design based on moment redistribution:

+Mmax = 295 + (1/10) ½ (505 + 505) = + 345.5 kip-ft

Using AISC Table 3-2 (Mmax = Ma = 454.5 kip-ft):

Select W24 x 76 Mpx/Ωb = 499 kip-ft > Ma = 454.5 kip-ft OK

6.37

Note: If the lower flange of the beam is not braced laterally, then the lengths of

the span where negative moments are present must be checked, because Lb values

may exceed Lp = 6.78’ for the section and the design may have to be revised.

Upon checking the unbraced lengths in the areas of negative moment for this

example, the following was determined.

• At the fixed support: Lb = 6.47’ (from the fixed support to the inflection point

to the right).

• At the second support: Lb = 7.67’ (from the leftmost roller support to the point

of inflection to the left) and Lb = 7.85’ (from the leftmost roller support to the

point of inflection to the right).

• At the third support: Lb = 7.85’ (from the center roller support to the point of

inflection to the left) and Lb = 8.41’ (from the center roller support to the point

of inflection to the right).

As noted above, four of the five unbraced lengths in the areas of negative moment

exceed Lp = 6.78’.

• Check the worst case (i.e. Lb = 8.41’)

W24 x 76 (Mpx/Ωb = 499 kip-ft, BF/Ωb = 15.1 kips, Lp = 6.78’, Lr = 19.5’)

Lp = 6.78’ < Lb = 8.41’ < Lr = 19.5’, thus this beam section fall into Zone 2.


presented in the AISC Manual (p. 3-9), and assuming Cb = 1.0.

Mnx/Ωb = Cb [Mpx/Ωb – (BF/Ωb)(Lb – Lp)] ≤ Mpx/Ωb

Mnx/Ωb = 1.0 [499 – 15.1(8.41 – 6.78)] = 474.4 kip-ft


Ma = 454.5 kip-ft < Mn/Ωb = 474.4 kip-ft OK

Thus, the section is adequate for this and all other areas of negative moment.

6.38

6.9 Plastic Analysis and Design of Continuous Beams

For statically determinate members the plastic moment strength can be compared

to the maximum elastic moment on a beam to satisfy the strength requirements of

the Specification.

• This is accurate since the occurrence of the plastic moment at the single point

of maximum moment results in the development of a single plastic hinge.

For statically indeterminate members, such as a continuous beam, more than one

plastic hinge must form before the beam would actually collapse.

• This provides additional capacity than that determined by an elastic analysis.

The arrangement of plastic hinges that permit collapse in a structure is called the

failure or collapse mechanism.

• This is the approach referred to as plastic analysis, permitted by Appendix

Section 1.3 of the Specification for use with LRFD only.

• The following figures show the collapse mechanisms for a simply supported

beam, a fixed end beam, and a propped cantilever beam, respectively.

Simply Supported Beam

A statically determinate beam (e.g. simply supported beam or cantilever beam) will

fail if one plastic hinge develops.

• Consider a simple beam of constant cross section with a concentrated load at

mid-span.

- If the load is increased until a

plastic hinge is formed at the

point of maximum moment, an

unstable structure is created.

- Pn represents the nominal, or

theoretical, maximum load that

the beam can support.

- Any further increase in load will cause collapse.

Work = Force x Distance

Virtual Work = Force x Virtual Distance

External work = Pu (½ L θ)

Internal work = Mp (0 + 2θ + 0)

Mp (2θ) = Pu (½ L θ)

Mp = Pu L/4 (same as elastic analysis)

6.39

• Consider a simple beam of constant cross section with a uniform load.

- If the load is increased until a

plastic hinge is formed at the

point of maximum moment, an

unstable structure is created.

- Any further increase in load will

cause collapse.

External work = wu L [½ (0 + ½ L θ)]

Internal work = Mp (0 + 2θ + 0)

Mp (2θ) = ¼ wu L2 θ

Mp = wu L2/8 (same as elastic analysis)

Fixed End Beam

For a statically indeterminate structure to fail, it is necessary for more than one

plastic hinge to form.

• The fixed end beam cannot fail unless three plastic hinges are developed.

- As the load increases, plastic hinges form at

the fixed ends, the points of the largest

elastic moment.

- After the formation of the first plastic

hinges at the fixed supports, the load can be

increased without causing failure of the

structure.

- The plastic hinges act like real hinges as far

as increased loading is concerned.

- As the load is increased further, a plastic hinge forms at the point of the

concentrated load as the moment at that point reaches the plastic moment.

- As the additional hinge is formed in the structure, there is a sufficient

number of hinges to cause collapse of the structure.

• Actually, after the formation of the three hinges, some additional load can be

carried before collapse occurs since the stresses go into the strain hardening

range.

- However, the deflections that would occur are too large to be permissible.

• Consider the fixed end beam with a concentrated load at mid-span shown in

Figure 6.20b (p. 201 of the textbook) and AISC Table 3-23 (Case 16).

- Based on an elastic analysis, the maximum moments occur at mid-span and at

the supports and the required moment strength is determined by Mp = PuL/8.

6.40

- Based on a plastic analysis, the

maximum moments occur at mid-

span and at the supports and the

required moment strength is

determined by Mp = PuL/8.

External work = Pu (½ L θ)

Internal work = Mp (θ + 2θ + θ)

Mp (4θ) = ½ Pu L θ

Mp = Pu L/8

• Consider the uniformly loaded fixed end beam shown in Figure 6.19a (p. 200 of

the textbook) and AISC Table 3-23 (Case 15).

- Based on an elastic analysis, the maximum moments occur at the fixed ends

and the required moment strength is determined by Mp = wuL2/12.


maximum moments occur at the

fixed supports and the required

moment strength is determined by

Mp = wuL2/16.

External work = wu L [½ (0 + ½ L θ)]

Internal work = Mp (θ + 2θ + θ)

Mp (4θ) = ¼ wu L2 θ

Mp = wu L2/16

Propped Cantilever Beam

The propped cantilever beam is an example of a structure that will fail after two

plastic hinges are developed.

• As the load increases, the first

plastic hinge forms at the fixed

end, the point of the largest

elastic moment.

• The load may be further increased

until the moment at some other

point (in this case, at the

concentrated load) reaches the

plastic moment.

6.41

• Consider the propped cantilever beam with the concentrated load shown in

Figure 6.20c (p. 201 of the textbook) and AISC Table 3-23 (Case 14).

- Based on an elastic analysis, the maximum moment may occur at the point of

the concentrated load or at the fixed support and the required moment

strength is determined by formulas shown in Table 3-23.


maximum moment occurs at a

distance “b” from the simple

support and at the fixed

support and the required

moment strength is determined

by Mp = Pu a b / (a + 2b).

External work = Pu (a θ)

Internal work = Mp [θ + (1 + a/b) θ + 0]

Mp (2 + a/b) θ = Pu (a θ)

Mp = (Pu a) / (2 + a/b)

Mp = Pu a b / (a + 2b)

• Consider the uniformly loaded propped cantilever beam shown in Figure 6.20a

(p. 201 of the textbook) and AISC Table 3-23 (Case 12).

- Based on an elastic analysis, the maximum moment occurs at a distance

0.375L from the simple support and the required moment strength is

determined by Mp = wu L2/8.


maximum moment occurs at a

distance 0.414L from the simple

support and the required

moment strength is determined

by Mp = 0.0858 wu L2.

External work = wu L [ ½ (0 + 0.586 L θ)]

Internal work = Mp (θ + 2.415 θ + 0)

Mp (3.415 θ) = 0.293 wu L2 θ

Mp = (Pu a) / (2 + a/b)

Mp = 0.0858 wu L2

An additional advantage to the use of plastic analysis for indeterminate beams is

the simplicity of the analysis.

• Regardless of the overall geometry of the continuous beam, each segment

between supports can be evaluated independently of the other segments.

6.42

• Any beam segment, continuous at each end, exhibits the same collapse

mechanism and the relation between the applied load and the plastic moment is

given by the appropriate case noted above.

To ensure that a given beam cross section can undergo the necessary rotation at

each plastic hinge, the Specification requires the following.

• Fy ≤ 65 ksi

• The section must be compact.

• The compression flange must be braced such that the unbraced length in the

area of the hinge is less than Lpd given as Equation A-1-5 in the Specification

Appendix Section 1.3.

• If these criteria are not satisfied, then the member design must be based on

an elastic analysis.

6.43

Example Problem – Plastic Design of Continuous Beams

Given: The loaded beam shown.

Steel: Fy = 50 ksi

Find: Select the lightest W-section using a plastic analysis. Assume that full lateral

support is provided for the flanges. Compare with the elastic analysis from the

previous example.

Solution

Plastic analysis and design (LRFD design procedure)

wu = 1.2 D + 1.6 L = 1.2 (1.0) + 1.6 (3.0) = 6.0 kips/ft

Pu = 1.2 D + 1.6 L = 1.2 (15) + 1.6 (20) = 50.0 kips

Left-hand span (similar to fixed-end beam)

Applicable equations: Mu = Pu L/8 + wu L2/16

Mu = 50 (30)/8 + 6 (30)2/16 = 187.5 + 337.5 = 525.0 kip-ft

Center span (similar to fixed-end beam)

Applicable equation: Mu = wu L2/16

Mu = 6 (40)2/16 = 600.0 kip-ft (Governs)

Right-hand span (similar to propped cantilever)

Applicable equation: Mu = 0.0858 wu L2

Mu = 0.0858 (6.0) (30)2 = 463.3 kip-ft

Let Mu = 0.9 Fy Zx

Zx = Mu/0.9 Fy = 600.0 (12 “/’)/[0.9 (50)] = 160.0 in3

Using AISC Table 3-2, select W21 x 68 (Zx = 160.0 in3)

From the previous elastic analysis: Select W24 x 76

6.44

Example Problems – Continuous Beams

Example

Given: W18 x 55 beam shown with full lateral support.

Steel: Fy = 50 ksi

Find: The maximum value of wn.

Solution

W18 x 55 (Zx = 112 in3)

Mn = Fy Zx = 50 (112) = 5600 kip-inch (466.7 kip-ft)

Left-hand span (similar to propped cantilever)

The hinge at the point of maximum moment is located 0.414 L from the left end

of the beam.

0.414 L = 0.414 (24) = 9.94’ from the left end

Applicable equation: Mp = 0.0858wuL2

wn = Mn/0.0858L2 = (466.7)/0.0858(24)2 = 9.44 kips/ft

Right-hand span (similar to fixed end beam)

Due to symmetry, the point of maximum moment is located at mid-span.

Applicable equation: Mp = wu L2/16

wn = 16 Mn/L2 = 16(466.7)/(30)2 = 8.30 kips/ft (controls)

The maximum value of wn = 8.30 kips/ft.

6.45

Example

Given: W21 x 44 beam shown with full lateral support.

Steel: A992 (Fy = 50 ksi)

Find: The maximum value of Pn.

Solution

W21 x 44 (Zx = 95.4 in3)

Mn = Fy Zx = 50 (95.4) = 4770 kip-inch (397.5 kip-ft)

Exterior spans (similar to propped cantilever)

The hinge at the point of maximum moment is located at mid-span in each beam.

Applicable equation: Mn = Pn ab/(a + 2b)

Mn = Pn (L/2)(L/2)/(L/2 + L) = Pn (L2/4)/1.5L = Pn L/6

Pn = 6 Mn/L = 6 (397.5)/(30) = 79.5 kips

Center span (similar to fixed end beam)

The point of maximum moment is located at mid-span.

Applicable equation: Mn = Pn L/8

Pn = 8 Mn/L

1.5 Pn = 8 Mn/L = 8 (397.5)/30 = 106.0

Pn = 106.0/1.5 = 70.7 kips (controls)

The maximum value of Pn = 70.7 kips.

6.46

6.10 Provisions for Double-Angle and Tee Members

Provisions for beams (such as lintels) formed by combining a pair of angles to form

a tee, and for beams made from a WT-shape loaded in the plane of symmetry are

found in Section F9 of the Specification.

• Four limit states must be considered in the design of these T-shaped members:

yielding, lateral-torsional buckling, flange local buckling, and stem local buckling.

• The nominal flexural moment Mn shall be the lowest value obtained from these

four limit states.

Yielding

For the limit state of yielding, the nominal flexural strength is given by

Mn = Mp Equation F9-1

where

a) For stems in tension

Mp = Fy Zx ≤ 1.6 My Equation F9-2

b) For stems in compression

Mp = Fy Zx ≤ My Equation F9-3

Lateral-Torsional Buckling

The nominal flexural strength is given by

Mn = Mcr = [π (E Iy G J)1/2/Lb][B + (1 + B2)1/2] Equation F9-4

where

B = ± 2.3(d/Lb)(Iy/J)1/2 Equation F9-5

The plus sign for B applies when the stem is in tension and the minus sign applies

when the stem is in compression.

• If the tip of the stem is in compression anywhere along the unbraced length,

the negative value of B shall be used.

Flange Local Buckling

The limit state of flange local buckling for these shapes reflects the same

behavior as for the I-shapes already considered.

• The limiting width-to-thickness ratios are the same as discussed earlier.

• The nominal strength equation is the same, except that Mn is limited to 1.6My.

a) For sections with a compact flange (λ ≤ λp) in flexural compression, the limit

state of flange local buckling does not apply.

6.47

b) For sections with a noncompact flange (λp < λ ≤ λr) in flexural compression

Mn = {Mp – (Mp – 0.7 Fy Sxc) [(λ – λpf)/(λrf – λpf)]} ≤ 1.6 My Equation F9-6

c) For sections with a slender flange (λr < λ) in flexural compression

Mn = 0.7 E Sxc/(bf/2tf)2 Equation F9-7

where

Sxc = elastic section modulus referred to the compression flange

λ = bf/2tf

λpf = λp, the limiting slenderness for a compact flange (Table B4.1b)

λrf = λr, the limiting slenderness for a noncompact flange (Table B4.1b)

Stem Local Buckling

When the stem is in flexural compression, the nominal strength for the limit state

of stem local buckling is given by

Mn = Fcr Sx Equation F9-8

where

Sx = the elastic section modulus

The critical stress Fcr is determined as follows.

a) When d/tw ≤ 0.84 (E/Fy)1/2, the stem will not buckle and

Fcr = Fy Equation F9-9

b) When 0.84 (E/Fy)1/2 < d/tw ≤ 1.03 (E/Fy)1/2

Fcr = [2.55 – 1.84(d/tw)(Fy/E)1/2] Fy Equation F9-10

c) When d/tw > 1.03 (E/Fy)1/2

Fcr = 0.69E/(d/tw)2 Equation F9-11

6.11 Single-Angle Bending Members

When single angles are used as bending members (such as lintels), bending may

occur about one of the geometric axes (i.e. axes parallel to the legs), or about the

principal axis (a.k.a. the z-axis).

• The most useful orientation of the single-angle bending member is also the most

complex orientation for determining its strength.

Specification F10 outlines the provisions for single-angle bending members.

• Three limit states must be considered in the design of these members: yielding,

lateral-torsional buckling, and leg local buckling.

6.48

• The nominal flexural moment Mn shall be the lowest value obtained from these

three limit states.

Yielding

For the limit state of yielding, the nominal flexural strength is given by

Mn = 1.5My Equation F10-1

where

My = the yield moment about the axis of bending

= Fy S

S = the elastic section modulus about the axis of bending

Leg Local Buckling

The limit state of leg local buckling applies when the toe of the leg is in

compression.

• Legs of angles in compression have the same tendency to buckle as other

compression elements.

Specification Table B4.1b (Case 12) defines slenderness b/t for single angles as

follows.

λp = 0.54 (E/Fy)1/2

λr = 0.91 (E/Fy)1/2

The nominal flexural strength Mn is determined as follows.

a) For compact sections (λ ≤ λp), the limit state of leg local buckling does not apply.

b) For sections with noncompact legs (λp < λ ≤ λr):

Mn = Fy Sc [2.43 – 1.72(b/t)(Fy/E)1/2] Equation F10-7

c) For sections with slender legs (λr < λ):

Mn = Fcr Sc Equation F10-8

where

Fcr = 0.71 E/(b/t)2 Equation F10-9

Sc = elastic section modulus to the toe in compression relative to the axis

of bending

For bending about one of the geometric axes of an equal-leg angle with no

lateral-torsional restraint, Sc shall be 0.80 of the geometric axis section

modulus.

6.49

Lateral-Torsional Buckling

The limit state of lateral-torsional buckling depends on whether the toe of the

angle is in tension or compression.

The Specification gives the strength provisions for four different bending

orientations.

• Single angles with continuous lateral-torsional restraint along the length may be

designed on the basis of geometric axis (x, y) bending.

The angles should always be oriented and laterally supported in order for the angle

to carry the greatest bending moment,

6.50

6.12 Members in Biaxial Bending

Bending members may be called upon to resist loads that result in bending about

two orthogonal axes.

• Regardless of the actual

orientation of the applied load (and

associated moment), it is possible

to break the load into components

about two principal axes.

• Once this is accomplished, the

ability of the section to resist the

combined loads (and associated

moments) can be determined

through an interaction equation.

Biaxial bending results when loads are applied in the following manner.

• When the external loads are not in a plane with either of the principal axes.

• When loads are simultaneously applied to the beam from two or more directions.

Among the beams that must resist biaxial bending are crane girders in industrial

buildings and purlins for ordinary roof trusses.

• The x-axes of purlins are parallel to the sloping roof surface, while a large

percentage of their loads (dead load due to roofing, live load, and snow loads)

are gravity (vertical) loads.

- These loads do not act in a plane with either of the principal axes of the

inclined purlins, and the result is unsymmetrical bending.

- The x-axes of crane girders are usually horizontal, but the girders are

subjected to lateral thrust loads from the moving cranes, as well as being

subjected to the gravity loads.

Section H1 of the AISC Specification provides an interaction equation to check

the adequacy of members subject to combined bending (i.e. bent about both axes

simultaneously) and an axial force if Pr/Pc < 0.2.

Pr/2Pc + (Mrx/Mcx + Mry/Mcy) ≤ 1.0 AISC Equation H1-1b

where

Pr = required axial strength (LRFD) or required axial strength (ASD)

Pc = the available axial strength (LRFD) or the allowable axial strength (ASD)

Mrx = the required design flexural strength about the x-axis (LRFD) or the

required design flexural strength (ASD)

6.51

Mry = the required design flexural strength about the y-axis (LRFD) or the

required design flexural strength (ASD)

Mcx = the available design flexural strength about the x-axis (LRFD) or the

allowable flexural strength (ASD)

Mcy = the available design flexural strength about the y-axis (LRFD) or the

allowable flexural strength (ASD)

The selection of the lightest section listed in the AISC Manual can be quite

involved because of the two variables (Zx and Zy) that affect the size.

• For a large Mux and a small Muy, the most economical section will likely be deep

and narrow.

• For a small Mux and a large Muy, the most economical section will likely be wide

and shallow.

6.52

Example Problem – Biaxial Bending

Example

Given: Steel beam (W24 shape) supporting the following service moments (which

include the effects of the estimated beam weight).

MDx = 60 kip-ft, MLx = 100 kip-ft, MDy = 15 kip-ft, MLy = 25 kip-ft

Steel: Fy = 50 ksi

The beam has full lateral support of the compression flange.

Find: Select an adequate W24 beam section.

Solution

LRFD

Mux = 1.2 D + 1.6 L = 1.2 (60) + 1.6 (100) = 232.0 kip-ft

Muy = 1.2 D + 1.6 L = 1.2 (15) + 1.6 (25) = 58.0 kip-ft

Assume Mrx/Mcx ≈ 0.4:

φbMpx = Mcx ≈ (1/0.4) Mux = (1/0.4) 232.0 = 580.0 kip-ft

Trial 1: Using AISC Table 3-2:

Try W24 x 62 (φbMpx = 574 kip-ft, Zy = 15.7 in3)

φbMpy = φbFy Zy = 0.9(50)(15.7) = 706.5 kip-inch (58.9 kip-ft)

Pr/2Pc + (Mrx/Mcx + Mry/Mcy) = 0 + 232.0/574 + 58.0/58.9

= 0 + 0.404 + 0.985 = 1.389 > 1.0 NG

Trial 2: Using AISC Table 3-2 (next larger W24):

Try W24 x 68 (φbMpx = 664 kip-ft, Zy = 24.5 in3)

φbMpy = φbFy Zy = 0.9(50)(24.5) = 1102.5 kip-inch (91.9 kip-ft)


= 0 + 0.349 + 0.631 = 0.980 < 1.0 OK

Select W24 x 68

6.53

ASD

Max = D + L = 60 + 100 = 160.0 kip-ft

May = D + L = 15 + 25 = 40.0 kip-ft

Assume Mrx/Mcx ≈ 0.4:

Mpx/Ωb = Mcx ≈ (1/0.4) Max = (1/0.4) 160.0 = 400 kip-ft

Trial 1: Using AISC Table 3-2:

Try W24 x 62 (Mpx/Ωb = 382 kip-ft, Zy = 15.7 in3)

Mpy/Ωb = Fy Zy/Ωb = (50)(15.7)/1.67 = 470.0 kip-inch (39.2 kip-ft)


= 0 + 0.419 + 1.020 = 1.439 > 1.0 NG





= 0 + 0.362 + 0.655 = 1.017 > 1.0 NG





= 0 + 0.321 + 0.561 = 0.882 < 1.0 OK

Select W24 x 76

6.54

6.13 Serviceability Criteria for Beams

Serviceability is a state in which the function of the building, its appearance,

maintainability, durability and comfort of its occupants are preserved under normal

usage.

• Several serviceability considerations are presented in Specification Chapter L:

camber, deflections, drift, vibration, wind-induced motion, expansion and

contraction, and connection slip.

• Failure to satisfy these criteria may not impact the strength of the member or

the strength of the overall structure; however, failure to satisfy these criteria

may prevent the successful completion of a project.

Deflection

The deflections in steel beams are usually limited to certain maximum values and

for good reasons.

1. Excessive deflections may damage other materials attached to or supported by

the beam (e.g. plaster cracks caused by large ceiling joist deflections).

2. The appearance of structures is often damaged by excessive deflections.

3. Although the structure may be safe with regard to strength, extreme

deflections do not inspire confidence in the person designing the structure by

the persons using the structure.

4. It may be necessary for several different beams supporting the same loads to

deflect equal amounts.

The AISC Specification does not prescribe maximum permissible deflections.

• There are so many different materials, types of structures, and loadings that

no one single set of deflection limitations is acceptable for all cases.

- Standard American practice for buildings has been to limit service live-load

deflections to approximately 1/360 of the span length.

- For situations where precise and delicate machinery is supported, maximum

deflections may be limited to 1/1500 or 1/2000 of the span lengths.

- The 2004 AASHTO Specification limits deflection in steel beams and

girders due to live load and impact to 1/800 of the span length.

Deflection limitations fall into the serviceability area.

• Deflections are determined for service loads, and, as a result, the calculations

are identical for both LRFD and ASD designs.

6.55

Common methods to determine beam deflections include the following.

• The moment area method.

• Double integration method.

• Established formulas.

The actual calculation of the maximum deflection for design purposes is usually

accomplished by the simple application of a formula.

• Many loading patterns and support conditions occur so frequently that

reference manuals (e.g. AISC Manual) and engineering handbooks provide

formulas for the deflections.

• Common cases are shown in Table 3-23 of the AISC Manual.

- The common expression for the center line deflection of a simply supported

beam with a uniformly distributed load is

Δ = 5wL4/384EI AISC Table 3-23, Case 1

- The common expression for the center line deflection of a simply supported

beam with a concentrated load at mid-span is

Δ = PL3/48EI AISC Table 3-23, Case 7

To use these formulas it is necessary to do the following.

• Select the formula that corresponds to the actual design conditions.

- If the actual loading situation is a combination of common cases, the

deflection can be determined by adding the results of two or more of the

formulas (known as superposition).

• Know the properties of the beam section.

- The strength of the steel is not a factor when computing deflection.

◦ Young’s modulus of elasticity, E, is the same for all grades of structural

steel (E = 29,000 ksi).

◦ A smaller beam section may be required to meet the flexural design

requirements when higher strength steel is used.

◦ When deflection controls, there is no advantage in using higher strength

steel.

6.56

Example Problem – Deflections

Example

Given: A simply supported beam (W24 x 55) with a uniformly distributed load.

Service live load: 3 kips/ft

Span = 21’

(ΔLL)max = L/360

Find: Determine if the centerline deflection is satisfactory for the service live

load.

Solution

W24 x 55 (Ix = 1350 in4)

(ΔLL)actual = 5wL4/384EI AISC Table 3-23, Case 1

= 5(3)(21)4(12”/’)3/384(29,000)(1350) = 0.335”

(ΔLL)actual = 0.335” < (ΔLL)max = 21(12”/’)/360 = 0.700” OK

6.57

Some specifications handle deflection by requiring certain minimum depth-span

ratios.

• AASHTO suggests the depth-span ratio be limited to a minimum value of 1/25.

• A helpful “rule-of-thumb” guide to establish beam depth for simply supported

beams is the “L/24 ratio” (often stated as “the beam depth in inches equals one

half the span in feet”).

Example: A simply-supported beam with a 24-foot span requires a beam section with a minimum

depth of 12 inches.

A steel beam can be cold-bent, or cambered, an amount equal to the deflection

caused by dead load, or the deflection caused by dead load plus some percentage

of the deflection caused by the live load.

• A camber requirement is quite common for longer steel beams.

Cambering is something of a nuisance to many fabricators, and it may introduce

some additional problems.

• For example, when beams are cambered, it may be necessary to adjust the

connection detail so that proper fitting of the members is achieved.

- The end of a cambered beam will be rotated, and thus it may be necessary

to rotate the connection details by the same amount to insure proper fitting.

• Often it may be more economical to select heavier beams with their larger

moments of inertia to reduce deflections and avoid the labor costs involved in

cambering.

Deflections may control the sizes of beams for longer spans, or for short spans

where deflection limitations are severe.

• To assist the designer in selecting sections where deflections may control, the

AISC Manual includes Table 3-3 entitled “W Shapes Selection by Ix” in which Ix

values are given in numerically descending order for sections normally used as

beams.

Design Guide 3: Serviceability Design Considerations for Steel Buildings from the

American Institute of Steel Construction covers deflection and other

serviceability design criteria.

6.58

Example Problem – Beam Design for Deflection

Example

Given: Simply supported beam on a 30-foot span with a uniformly distributed load

and with full lateral bracing of the compression flange.

Steel: Fy = 50 ksi

Service loads: wD = 1.2 kips/ft, wL = 3.0 kips/ft

Maximum total service load deflection: Δmax = L/1500

Find: Lightest available section.

Solution

LRFD (Assume a beam weight of 167 lb/ft)

Compute the maximum bending moment.

wu = 1.2 D + 1.6 L = 1.2 (1.2 + 0.167) + 1.6 (3.0) = 6.44 kips/ft

Mu = wuL2/8 = 6.44(30)2/8 = 724.5 kip-ft

Select a beam section based on flexure requirements.

From AISC Table 3-2: Try W24 x 76 (φbMpx = 750 kip-ft, Ix = 2100 in4)

Check the deflection.

Δmax = L/1500 = 30(12”/’)/1500 = 0.24”

Δactual = 5wL4/384EI AISC Table 3-23, Case 1

= 5 (1.2 + 0.167 + 3.0) (30)4(12”/’)3/384(29,000)(2100) = 1.307”

Δactual = 1.307” > Δmax = 0.24” NG

Determine the minimum Ix required to limit Δmax to 0.24”

Ix = 5wL4/384EΔmax

= 5 (1.2 + 0.167 + 3.0) (30)4(12”/’)3/384(29,000)(0.24) = 11,435 in4


From AISC Table 3-3: Select W40 x 167 (Ix = 11,600 in4)

6.59

ASD (Assume a beam weight of 167 lb/ft)

Compute the maximum bending moment.

wa = D + L = (1.2 + 0.167) + 3.0 = 4.37 kips/ft

Ma = waL2/8 = 4.37(30)2/8 = 491.6 kip-ft


From AISC Table 3-2: Try W24 x 76 (Mpx/Ωb = 499 kip-ft, Ix = 2100 in4)

Check the deflection.

Δmax = L/1500 = 30(12”/’)/1500 = 0.24”

Δactual = 5wL4/384EI AISC Table 3-23, Case 1

= 5 (1.2 + 0.167 + 3.0) (30)4(12”/’)3/384(29,000)(2100) = 1.307”

Δactual = 1.307” > Δmax = 0.24” NG

Determine the minimum Ix required to limit Δmax to 0.24”

Ix = 5wL4/384EΔmax

= 5 (1.2 + 0.167 + 3.0) (30)4(12”/’)3/384(29,000)(0.24) = 11,435 in4


From AISC Table 3-3: Select W40 x 167 (Ix = 11,600 in4)

6.60

Vibration

Although vibration of floor systems is not a safety consideration, vibrations can be

very annoying and very difficult to correct after the building is constructed.

• Objectionable vibrations may occur where long spans and large open floors are

used without partitions or other items that might otherwise provide suitable

damping.

A general rule is to space the beams or joists sufficiently far apart so that the

slab thickness is large enough to provide the needed stiffness and damping.

• F. J. Hatfield has prepared a useful chart for estimating the perceptibility of

vibrations of steel beams and concrete slabs for office and residential buildings

(ref. F.J. Hatfield, “Design Chart for Vibration of Office and Residential

Floors,” Engineering Journal, 29, 4, 4th Quarter, 1992, pp. 141-144).

Damping of vibrations may be achieved by the following methods.

• By using framed-in-place partitions.

• By installing “false” sheetrock partitions between ceilings and the underside of

floor slabs.

• By the thickness of the floor slabs.

• By the weight of office furniture and the equipment used in the building.

• By controlling the stiffness of structural system.

- A common practice to limit vibrations is selecting beams no shallower than

1/20 times the span.

If the occupants of a building are uneasy or annoyed by vibrations, the design is

unsuccessful.

• It is difficult to correct a situation of this type in an existing structure.

• Vibrations can be easily predicted and corrected in the design stage of a new

structure.

- Several good procedures have been developed that enable the designer to

estimate the acceptability of a given system by its users.

Design Guide 11: Floor Vibrations Due to Human Activity from the American

Institute of Steel Construction covers the design of steel-framed floor systems

for human comfort.

6.61

Drift

The horizontal deflection of a multi-story building due to lateral loads (e.g. wind or

seismic load) is called drift.

• Drift is measured by the drift index ΔH/L, where ΔH is the amount of lateral

deflection and L is the story height.

• For the comfort of the occupants of a building, the drift index is usually

limited at working (or service) loads to a value of 0.0015 to 0.0030, and at

factored loads to 0.0040.

Example of the theoretical drift index for the 1450-foot twin towers of the

World Trade Center, New York City (destroyed in September 2001).

10-year storm: deflection Δ = 3 feet; drift index Δ/h = 0.0021

Hurricane: deflection Δ = 7 feet; drift index Δ/h = 0.0048

Like deflection and vibration, drift is not usually a safety consideration, but drift

can be annoying and may impact nonstructural elements that are attached by

causing cracks in finishes.

6.62

6.14 Concentrated Forces on Beams

The majority of beams are loaded through connections to their webs; however,

some may be loaded by applying a concentrated force to the top flange, and some

will have their reactions resisted by bearing on a supporting element.

• In these cases, a check must be made to establish that the beam web has

sufficient strength to resist the applied forces.

Six limit states are described in Section J10 of the Specification that determine

the load carrying strength of the web to resist concentrated forces: flange local

bending, web local yielding, web local crippling, web sidesway buckling, web

compression buckling, and web panel zone shear.

• Two of these limit states are discussed in the textbook: web local yielding and

web crippling.

• Three additional limit states are discussed in these notes: flange local bending,

web sidesway buckling, and web compression buckling.

If flange and web design strengths do not satisfy the requirements of Section J10

of the AISC Specification, it will be necessary either to increase the length of

bearing or to use stiffeners and/or doubler plates at the points of concentrated

loads.

Web Local Yielding

The web must be sufficiently rigid so that it will not deform due to the application

of a concentrated load.

6.63

• The nominal strength Rn of the web of a beam at the web toe of the fillet when

a concentrated load or reaction is applied is determined by one of the following

equations.

- Interior load: When the concentrated force or reaction to be resisted is

applied at a distance from the member end that is greater than the depth of

the member d, then

Rn = Fywtw (5k + lb) Equation J10-2

- Exterior load: When the concentrated force or reaction to be resisted is

applied at a distance from the member end that is less than or equal to the

depth of the member d, then

Rn = Fywtw (2.5k + lb) Equation J10-3

where

Fyw = specified minimum yield stress of the web

k = the distance from the outer edge of the flange to the web toe of the

fillet

Note: In AISC Table 1-1, two values for k are listed. One value is used for design;

the other value is used for detailing.

lb = the bearing length of the force parallel to the plane of the web

tw = the web thickness

LRFD (φ = 1.00): φRn = design strength

ASD (Ω = 1.50): Rn/Ω = allowable strength

If the concentrate load is greater than φ Rn (LRFD) or Rn/Ω (ASD), then a pair of

transverse stiffeners or a doubler plate shall be provided.

Web Crippling

If a concentrated compressive load is applied to a member with an unstiffened web

(the load being applied in the plane of the web), the nominal web crippling strength

Rn of the web is determined by one of the following equations.

• Interior force: When the concentrated compressive force to be resisted is

applied at a distance from the member end that is greater than or equal to d/2,

then

Rn = 0.80 tw2 [1 + 3(lb/d)(tw/tf)1.5](E Fyw tf/tw)1/2 Equation J10-4

• Exterior force: When the concentrated compressive force to be resisted is

applied at a distance from the member end that is less than d/2 and lb/d ≤ 0.2,

then

Rn = 0.40 tw2 [1 + 3(lb/d)(tw/tf)1.5](E Fyw tf/tw)1/2 Equation J10-5a

6.64

• Exterior force: When the concentrated compressive force to be resisted is

applied at a distance from the member end that is less than d/2 and lb/d > 0.2,

then

Rn = 0.40 tw2 [1 + (4 lb/d – 0.2)(tw/tf)1.5](E Fyw tf/tw)1/2 Equation J10-5b

where

Fyw = specified minimum yield stress of the web

lb = the bearing length of the force parallel to the plane of the web


tf = the flange thickness

LRFD (φ = 0.75): φ Rn = design strength


If the concentrate load is greater than φRn (LRFD) or Rn/Ω (ASD), then a

transverse stiffener, or a pair of transverse stiffeners, or a doubler plate

extending at least one-half the depth of the web shall be provided.

• Research has shown that when web crippling occurs, it is located in the part of

the web adjacent to the loaded flange.

- Stiffening the web in this area for half its depth is thought to prevent the

problem.

Flange Local Bending (Not in Textbook)

The flange must be sufficiently rigid so that it will not deform and cause a zone of

high stresses concentrated in the weld line with the web.

• The nominal tensile load Rn that may be applied through a plate across a flange

is determined by the following equation.

Rn = 6.25 Fyf tf2 Equation J10-1

where

Fyf = specified minimum yield stress of the flange




If the length of loading across the member flange is less than 0.15bf, where bf is

the member flange width, Equation J10-1 need not be checked.

6.65

If the concentrated force to be resisted is applied at a distance from the member

end that is less than 10tf, where tf is the member flange thickness, Rn shall be

reduced by 50 percent.

If the concentrate load is greater than φ Rn (LRFD) or Rn/Ω (ASD), then a pair of

transverse stiffeners shall be provided.

Web Sidesway Buckling (Not in Textbook)

If compressive single-concentrated forces are applied to a member where relative

lateral movement between the loaded compression flange and the tension flange is

not constrained at the point of application the concentrated load, the web may

tend to sidesway and buckle.

• The nominal web buckling strength Rn is determined by one of the following

equations.

a) If the compression (loaded) flange is restrained against rotation:

1) When (h/tw)/(Lb/bf) ≤ 2.3

Rn = (Cr tw3tf/h2) {1 + 0.4[(h/tw)/(Lb/bf)]3} AISC Equation J10-6

2) When (h/tw)/(Lb/bf) > 2.3, the limit state of sidesway web buckling does

not apply.



When the required strength of the web exceeds the available strength, local

lateral bracing shall be provided at the tension flange or either a pair of

transverse stiffeners or a doubler plate shall be provided.

b) If the compression (loaded) flange is not restrained against rotation:

1) When (h/tw)/(Lb/bf) ≤ 1.7

Rn = (Crtw3tf/h2) {0.4[(h/tw)/(Lb/bf)]3} AISC Equation J10-7

2) When (h/tw)/(Lb/bf) > 1.7, the limit state of sidesway web buckling does

not apply.



If the concentrate load is greater than φ Rn (LRFD) or Rn/Ω (ASD), then

local lateral bracing shall be provided at both flanges at the point of

application of the concentrated loads.

where

bf = flange width


6.66


Cr = 960,000 ksi, when Mu < My (LRFD) or 1.5Ma < My (ASD) at the location of

the force

= 480,000 ksi, when Mu ≥ My (LRFD) or 1.5Ma ≥ My (ASD) at the location of

the force

h = clear distance between flanges less the fillet or corner radius for rolled

shapes (i.e. d – 2k); the distance between adjacent lines of fasteners or

the clear distance between flanges when welds are used for built-up

shapes

Lb = largest laterally unbraced length along either flange at the point of load

It is not necessary to check Equation J10-6 or J10-7 if the webs are subject to a

distributed load.

Web Compression Buckling (Not in Textbook)

This limit state relates to concentrated compression loads applied to both flanges

of a member at the same location.

• The nominal compression web buckling strength Rn is determined by the

following equation.

Rn = [24tw3 (E Fyw)1/2]/h Equation J10-8



When the pair of concentrated compressive forces to be resisted is applied at a

distance from the member end that is less than d/2, Rn shall be reduced by 50

percent.

If the concentrate loads are greater than φRn (LRFD) or Rn/Ω (ASD), then a single

transverse stiffener, a pair of transverse stiffeners, or a doubler plate extending

the full depth of the web shall be provided.

6.67

Example Problem – Webs and Flanges with Concentrated Loads

Example

Given: Beam (W21 x 44) loaded as shown with lateral bracing for both flanges at

beam ends and at concentrated loads.

Steel: Fy = 50 ksi

End bearing length: lb = 3.50”

Concentrated load bearing lengths: lb = 3.0”

Find: Check the beam for web yielding, web crippling, and sidesway web buckling.

Solution

W21 x 44 (d = 20.7”, bf = 6.50”, tw = 0.350”, tf = 0.450”, k = 0.950”, h/tw = 53.6)

Determine the end reaction.

LRFD: wu = 1.2 D + 1.6 L = 1.2 (1.044) = 1.25 kips/ft

Pu = 1.2 D + 1.6 L = 1.2 (0) + 1.6 (35) = 56.0 kips

Ru = wuL/2 + 2 Pu/2 = 1.25 (15)/2 + 2 (56.0)/2 = 65.38 kips

ASD: wa = D + L = 1.044 + 0 = 1.044 kips/ft

Pa = D + L = 0 + 35 = 35.0 kips

Ra = waL/2 + 2 Pa/2 = 1.044 (15)/2 + 2 (35.0)/2 = 42.83 kips

Check Local Web Yielding

At the end reactions (AISC Equation J10-3):

Rn = Fywtw (2.5k + lb) = 50 (0.350) [2.5(0.950) + 3.50] = 102.8 kips

LRFD (φ = 1.00): φRn = 1.00 (102.8) = 102.8 kips > Ru = 65.38 kips OK

ASD (Ω = 1.50): Rn/Ω = 102.8/1.50 = 68.5 kips > Ra = 42.83 kips OK

At the concentrated loads (AISC Equation J10-2):

Rn = Fywtw (5k + lb) = 50 (0.350) [5(0.950) + 3.0] = 135.6 kips

6.68

LRFD (φ = 1.00): φRn = 1.00 (135.6) = 135.6 kips > Pu = 56.0 kips OK

ASD (Ω = 1.50): Rn/Ω = 135.6/1.50 = 90.4 kips > Pa = 35.0 kips OK

Check Web Crippling

At the end reactions (AISC Equation J10-5a):

lb/d = 3.50/20.7 = 0.169 < 0.20

Rn = 0.40 tw2 [1 + 3(lb/d)(tw/tf)1.5](E Fyw tf/tw)1/2

= 0.40(0.350)2[1 + 3(3.50/20.7)(0.350/0.450)1.5][29,000(50)(0.450/0.350)]1/2

= 0.049(1.348)(1365.4) = 90.2 kips

LRFD (φ = 0.75): φRn = 0.75 (90.2) = 67.6 kips > Ru = 65.38 kips OK

ASD (Ω = 2.00): Rn/Ω = 90.2/2.00 = 45.1 kips > Ra = 42.83 kips OK

At the concentrated loads (AISC Equation J10-4):

Rn = 0.80 tw2 [1 + 3(lb/d)(tw/tf)1.5](E Fyw tf/tw)1/2

= 0.80(0.350)2[1 + 3(3.0/20.7)(0.350/0.450)1.5][29,000(50)(0.450/0.350)]1/2

= 0.098 (1.298)(1365.4) = 173.7 kips

LRFD (φ = 0.75): φRn = 0.75 (173.7) = 130.3 kips > Pu = 56.0 kips OK

ASD (Ω = 2.00): Rn/Ω = 173.7/2.00 = 86.8 kips > Pa = 35.0 kips OK

Check Sidesway Web Buckling

Lb = largest laterally unbraced length along either flange at the point of load

= 5’ (12”/’) = 60.0”

(h/tw)/(Lb/bf) = (53.6)/(60.0/6.50) = 5.81 > 2.3

Sidesway buckling does not need to be checked.

6.69

6.15 Open Web Steel Joists and Joist Girders

The term open web steel joist refers to a building product made according to the

design standards of the Steel Joist Institute (SJI).

• Open web steel joists are manufactured trusses commonly used for building

floor and roof systems.

• Open web steel joists are not designed according to AISC standards but rather

according to standards in the SJI publication Standard Specifications, Load

Tables, and Weight Tables for Steel Joists and Joist Girders (2010).

Advantages of open web steel joists include the following.

• Open web steel joists are lightweight and can span long distances.

• The open webs can easily accommodate passing mechanical systems through the

structure.

• In some applications open web steel joists are more economical than rolled steel

shapes.

Disadvantages of open web steel joists include the following.

• Open web steel joists have a lower load carrying capacity than rolled shapes,

thus requiring much closer spacing.

• Open web joists cannot accommodate concentrated loads at points other than

truss panel points.

• Open web steel joists present potential vibration issues when used as floor

systems.

Four types of open web steel joists are defined in the SJI standards: K-series,

KCS-series, LH-series, and DLH-series.

K-Series Joists

K-series joists are the most commonly used joists for floor and roof systems.

• K-series joists are available in depths from 10” to 30”.

• Design tables are available covering spans up to 60 feet.

• The standard designation for K-series joists is 16K6.

- The first number represents the depth – 16”.

- The letter indicates the series designation – K.

- The third number represents the place within the series – 6.

• The series designation identifies the details of manufacture of the truss,

including the sizes of the elements that make up the truss.

6.70

Figure 6.29 (p. 217 of the textbook) is an example of the selection tables published

by the Steel Joist Institute.

• K-series joists are designed to carry a uniformly distributed load.

• Design tables give joist strengths in terms of load per foot of span.

- The upper number in the table represents the maximum uniform factored

load in pounds per foot that can be supported by the member.

- The lower number in the table is the serviceability load in pounds per foot

that will produce a deflection of L/360.

KCS-Series Joists

KCS-series joists are identified in the same manner as K-series joists.

• KCS-series joists are designed to carry a uniform moment over all interior panel

points and a constant shear.

- These joists are useful for supporting loads that combine uniformly

distributed and concentrated forces.

• The design tables for the KCS-series list only the moment and shear capacities.

LH-Series Joists

LH-series joists are long-span joists.

• Strengths for LH-series joists are tabulated for spans from 25 to 96 feet.

• LH-series joists have depths ranging from 18” to 48”.

• LH-series joists are designed for a uniformly distributed load.

DLH-Series Joists

DLH-series joists are deep long-span joists intended primarily to support roof

decks.

• DLH-series joists have depths ranging from 52” to 120”.

• Strengths for DLH-series joists are tabulated for spans from 62 to 240 feet.

Joist Girder

Another product that is designed according to the Steel Joist Institute standards

is the joist girder.

• Joist girders are pre-engineered trusses intended to support concentrated

loads at the panel points.

• Joist girders are used to support open web joists that are evenly spaced and

introduce the same load at each panel point.

6.71

The standard designation for joist girders is 44G8N12K.

• The first number represents the depth – 44”.

• The letter indicates a joist girder – G.

• The “8N” indicates that there are eight joist spaces, or that there are

seven concentrated loads on the girder.

• The final number indicates the load in kips.

- If the load is for LRFD design, the last letter would be F, indicating

factored load, and the load magnitude would be the LRFD required

strength.

6.72

Example – Open Web Steel Joists

Example 6.18 (p. 216 of the textbook)

Given: Joist supporting a roof deck with a span of 30 feet.

Joists are spaced 6’ on center.

Uniform service loads: D = 20 psf

L = 30 psf

Allowable live load deflection = L/360

Find: Select the shallowest K-series open web steel joist available in Figure 6.29 to

support the load using LRFD.

Solution

Determine the factored load for strength (not including the weight of the beam).

Tributary width = 6.0’

wu = (1.2 D + 1.6 L) x Tributary width

= [1.2 (20) + 1.6 (30)] 6.0 = (24.0 + 48.0) 6.0 = 432.0 lb/ft

Determine the service live load for deflection.

wLL = L x Tributary width = 30 psf (6.0’) = 180.0 lb/ft > 151 lb/ft NG

Select a K-series open web joist for a span of 30 feet.

Try 18K7

Strength: Load = 502 lb/ft > 432.0 lb/ft May be OK

Deflection: Load = 194 lb/ft > 180.0 lb/ft OK

Check to make certain that the joist can carry its own weight.

Joist weighs 8.9 lb/ft

Additional wu = 1.2 D = 1.2 (8.9) = 10.7 lb/ft

Total wu = 432.0 + 10.7 = 442.7 lb/ft < 502 lb/ft OK

Select 18K7

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