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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
presented in the AISC Manual (p. 3-9).
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
Given: Beam section W21 x 73
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.
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 [645 – 19.4(12.0 – 6.39)] = 536.2 kip-ft < φbMpx = 645 kip-ft OK
Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is
presented in the AISC Manual (p. 3-9).
Mnx/Ωb = Cb [Mpx/Ωb – (BF/Ωb) (Lb – Lp)] ≤ Mpx/Ωb
Mnx/Ωb = 1.0 [429 – 12.9(12.0 – 6.39)] = 356.6 kip-ft
< Mpx/Ωb = 429 kip-ft OK
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
Given: Beam section W18 x 97
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
For beams in Zone 2, use AISC Table 3-2 to check the moment strength of the
selected section: that is, φbMn for LRFD or Mn/Ωb for ASD.
• 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.
• 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 may provide the required moment
strength, so that
For LRFD, φbMn = φbMpx ≥ Mu
For ASD, Mn/Ωb = Mpx/Ωb ≥ Ma
• 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.
For beams in Zone 3, use AISC Table 3-2 to check the moment strength of the
selected section: that is, φbMn for LRFD or Mn/Ωb for ASD.
• 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.
• 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 may provide the required moment
strength, so that
For LRFD, φbMn = φbMpx ≥ Mu
For ASD, Mn/Ωb = Mpx/Ωb ≥ Ma
• 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.
For LRFD, φbMn = φbMpx ≥ Mu
For ASD, Mn/Ωb = Mpx/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.
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
Check the next larger section.
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
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 = 1.67 [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.
From Table 3-2: W24 x 84
Mpx/Ωb = 559 kip-ft < Ma = 607.1 kip-ft NG
Check the next larger section.
From Table 3-2: W27 x 84
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.
Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is
presented in the AISC Manual (p. 3-9).
Mnx/Ωb = Cb [Mpx/Ωb – (BF/Ωb) (Lb – Lp)] ≤ Mpx/Ωb
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.
Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is
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
< Mpx/Ωb = 499 kip-ft OK
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.
- Based on a plastic analysis, the
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.
- Based on a plastic analysis, the
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.
- Based on a plastic analysis, the
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)
Pr/2Pc + (Mrx/Mcx + Mry/Mcy) = 0 + 232.0/664 + 58.0/91.9
= 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)
Pr/2Pc + (Mrx/Mcx + Mry/Mcy) = 0 + 160.0/382 + 40.0/39.2
= 0 + 0.419 + 1.020 = 1.439 > 1.0 NG
Trial 2: Using AISC Table 3-2 (next larger W24):
Try W24 x 68 (Mpx/Ωb = 442 kip-ft, Zy = 24.5 in3)
Mpy/Ωb = Fy Zy/Ωb = (50)(24.5)/1.67 = 733.5 kip-inch (61.1 kip-ft)
Pr/2Pc + (Mrx/Mcx + Mry/Mcy) = 0 + 160.0/442 + 40.0/61.1
= 0 + 0.362 + 0.655 = 1.017 > 1.0 NG
Trial 3: Using AISC Table 3-2 (next larger W24):
Try W24 x 76 (Mpx/Ωb = 499 kip-ft, Zy = 28.6 in3)
Mpy/Ωb = Fy Zy/Ωb = (50)(28.6)/1.67 = 856.3 kip-inch (71.3 kip-ft)
Pr/2Pc + (Mrx/Mcx + Mry/Mcy) = 0 + 160.0/499 + 40.0/71.3
= 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
Select a beam section based on flexure requirements.
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
Select a beam section based on flexure requirements.
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
Select a beam section based on flexure requirements.
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
tw = the web thickness
tf = the flange thickness
LRFD (φ = 0.75): φ Rn = design strength
ASD (Ω = 2.00): Rn/Ω = allowable 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
tf = the flange thickness
LRFD (φ = 0.90): φ Rn = design strength
ASD (Ω = 1.67): Rn/Ω = allowable strength
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.
LRFD (φ = 0.85): φ Rn = design strength
ASD (Ω = 1.76): Rn/Ω = allowable strength
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.
LRFD (φ = 0.85): φ Rn = design strength
ASD (Ω = 1.76): Rn/Ω = allowable strength
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
tw = the web thickness
6.66
tf = the flange thickness
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
LRFD (φ = 0.90): φ Rn = design strength
ASD (Ω = 1.67): Rn/Ω = allowable strength
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