439.91 in 440 in
L For the total load,
( )
Therefore, Ireq is governed by the live load deflection, and the minimum moment of inertia is 440 in4.
Because the beam is laterally supported for its full length, the unbraced length of the beam is 0 ft and the full plastic moment, Mp,will be obtained. Therefore, Cb is 1.0, and the beam is in the plastic range, zone 1 bending.
Calculate the required plastic section modulus.
166.50 ft-kips 12 in ft
112.50 ft-kips 12 in 1.67 ft
Basing the selection on strength, for a W14 × 30 (see AISC Manual Table 3-2),
LRFD ASD
[ ]
3 3
4 4 no good
47.3 in 44.40 in
177 ft-kips 166.50 ft-kips 291 in 440 in
118 ft-kips 112.50 ft-kips 291 in 440 in
249 ft-kips 166.50 ft-kips 510 in 440 in
166 ft-kips 112.50 ft-kips 510 in 440 in
The design of the member is controlled by deflection rather than by yielding.
8. ZONE 2, INELASTIC BENDING: Lp< Lb ≤ Lr
In zone 2, the flexural member is subjected to inelastic lateral-torsional buckling, and this limit state is applicable in the design or analysis of the member. Therefore, the governing nominal flexural strength, as calculated with Eq. 5.9, must be less than or equal to the full plastic moment.
( 0.7 )
b p[
AISC Eq. F2-2]
n b p p y x p
r p
L L
M C M M F S M
L L
−
= − − − ≤
5.9
Calculating the nominal flexural strength using Eq. 5.9 can be simplified with Eq. 5.10 and Eq. 5.11.
( )
( BF ) [
LRFD]
bMn Cb bMp x Lb Lp bMp x
φ = φ − − ≤ φ
5.10( ) [
ASD]
BF
p x p x
n
b b p
b b b
M M
M C
L L
= − − ≤
Ω Ω Ω
5.11The bending factor (BF) for a specific beam depends on the beam’s properties and on whether the LRFD or ASD method is being used. Values for bending factors are given in the following AISC Manual tables.
• AISC Table 3-2 and Table 3-6: wide-flange (W) shapes
• AISC Table 3-7: I-shaped (S) shapes
• AISC Table 3-8: channel (C) shapes
• AISC Table 3-9: channel (MC) shapes
When the unbraced length of a beam exceeds the limiting length for plastic bending, Lp, but is no greater than the limiting length for inelastic bending, Lr, the beam is subject to inelastic lateral-torsional buckling. It is not possible to select such a member by calculating a required plastic section modulus. The easiest way to select such a member is to calculate the required flexural strength (Mu for LRFD or Ma for ASD) and use AISC Manual Table 3-10 with the required flexural strength and the unbraced length.
Example 5.2 ____________________________________________________
Zone 2 Bending
A W21 × 50c beam is 40 ft in length, and is laterally supported at its ends and quarter points.1 The beam supports a uniform dead load including the beam weight of 0.30 kip/ft and a uniform live load of 0.70 kip/ft.
1The superscript c on the beam designation in the AISC Manual indicates that the member is slender for compression with Fy = 50 ksi. Either the flange width-to-thickness or web height-to-thickness ratio exceeds the upper limit for a noncompact element, λr, for uniform compression as specified in AISC Specification Table B4.1b.
Section properties ASTM A992 steel Fy = 50 ksi Fu = 65 ksi
Determine whether the W21 × 50 steel beam is satisfactory and whether it meets the deflection criteria for floor beams set forth in the IBC.
Solution
Calculate the design loads.
LRFD ASD
Calculate the required strengths.
LRFD ASD
length, Lb, is 10 ft. Therefore, Lp < Lb ≤ Lr is true, and the beam’s failure mode is zone 2. This means that it is subject to inelastic bending and lateral-torsional buckling.
Calculate the available flexural strength. For LRFD, from Eq. 5.10,
( )
1.0 413 ft-kips 18.3 kips 10 ft 4.59 ft 314 ft-kips
1.0 274 ft-kips 12.1 kips 10 ft 4.59 ft 209 ft-kips
The preceding calculations are based on a beam bending coefficient, Cb, of 1.0. The beam is satisfactory. The beam bending coefficient for a uniformly loaded beam braced at the quarter points is 1.06 for the two quarter lengths adjacent to the midspan of the beam. (See Table 5.2.)
For LRFD, the beam is capable of supporting a maximum factored bending moment of
( ) (
1.06 314 ft-kips)( )
332.84 ft-kipsb b n bMp x
C
φ
M = = <φ For ASD, the beam is capable of supporting a maximum allowable bending moment due to service loads of
Calculate the moment of inertia required to comply with the IBC. Deflections are calculated on service loads and not factored loads. The allowable deflection for the live load is
The required moment of inertia for the live load is For the total load,
( )
The required moment of inertia for the specified serviceability criteria is controlled by the larger of the required values, 1045.37 in4. The moment of inertia of a W21 × 50 is 984 in4. This is not enough to meet the serviceability criteria. There are three possible resolutions to this problem.
• Select a section with a larger moment of inertia.
• Specify that the beam be fabricated with a camber (the deflection should be calculated to determine the amount of camber required).
• Accept the beam based on engineering judgment.
Calculate the deflections for a W21 × 50. For the live load,
( ) ( )
For the total load,
Specify a one inch camber, thereby reducing the live load deflection to 0.41 in and the total load deflection to 1.02 in.
9. ZONE 3, ELASTIC BENDING: Lb > Lr
When the unbraced length of a beam exceeds the limiting length for inelastic bending, Lr, the beam is subject to elastic lateral-torsional buckling, and this is the applicable limit state in the design and analysis of the beam.
It is not possible to select a beam for which Lb > Lr by calculating a required plastic section modulus. The easiest way to select such a beam is by calculating the required flexural strength (Mu for LRFD, Ma for ASD) and using AISC Manual Table 3-10 with the required flexural strength and unbraced length.
The governing nominal flexural strength will be less than or equal to the full plastic moment.
In Eq. 5.13, the square root factor can be conservatively taken as equal to 1.0. rts is the effective radius of gyration and is
[ ]
However, rts can be approximated accurately and conservatively as the radius of
For a doubly symmetrical I-shaped member, c = 1.0. Fora channel,
[
AISC Eq. F2-8b]
Example 5.3 _____________________________________________________
Zone 3 Bending
The W8 × 18 steel beam shown is 14.5 ft long and is laterally supported only at the beam ends.
Material properties ASTM A992 steel Fy = 50 ksi Fu = 65 ksi
Determine the available moment capacity of the beam.
Solution
The unbraced length, Lb, is 14.5 ft, which exceeds Lr. The beam is thus in zone 3 bending and is subject to elastic lateral-torsional buckling. Calculate the nominal moment capacity. From Eq. 5.18,
4
From Eq. 5.13, the critical stress is
( )
From Eq. 5.12, the nominal moment capacity is the lesser of
( )
Calculate the available flexural strength using LRFD and ASD.
These calculated moment capacities are comparable to the 36.0 ft-kips and 24.0 ft-kips obtained from AISC Manual Table 3-11. The lateral-torsional buckling modification factor, Cb, for a uniformly loaded beam braced at the ends only is 1.14. Taking the modification factor into consideration, the calculated available strength is as follows.
LRFD ASD
The calculated available strengths are less than φbMpx and Mpx/Ωb, respectively.
10. WEAK AXIS BENDING: I- AND C-SHAPED MEMBERS
When a beam is bent about its weak axis, lateral-torsional buckling will not occur.
Therefore, the beam will fail in yielding or flange local buckling. The nominal flexural strength, Mn, is the lower value obtained according to the limit states of yielding and flange local buckling. The yielding limit state will govern the design, provided the flanges are compact.
All current ASTM A6 W, S, M, C, and MC shapes except the following have compact flanges at Fy ≤ 50 ksi: W21 × 48, W14 × 99, W14 × 90, W12 × 65, W10 × 12, W8 × 31, W8 × 10, W6 × 15, W6 × 9, W6 × 8.5, and M4 × 6.
The nominal flexural strength for yielding is
[
AISC Eq. F6-1]
n p y y 1.6 y y
M =M =F Z ≤ F S 5.19
For sections with noncompact flanges, the nominal flexural strength is
(
0.7)
p f[
AISC Eq. F6-2]
is the ratio b/tf where bis the full nominal dimension of the flange.For sections with slender flanges, the nominal flexural strength is
[ ]
cr AISC Eq. F6-3
n y
M =F S 5.21
For Eq. 5.21, the critical flexural stress is