Non Uniform Bending
Transcript of Non Uniform Bending
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NON-UNIFORM BENDING BETWEEN LATERAL SUPPORTS
For uniform moment (elastic behavior ):
M n @ LTB = M cr
For non-uniform moment, i.e., when a moment gradient exists:
M n @ LTB ? M cr
X: denoteslateralsupport
B.M.D.
P1 P2
Lb1 Lb2 Lb3
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Consider the case of uniform moment again:
With moment gradient:
M n with moment gradient > M n with uniform moment
entire length of top flange incompression
max. compression occurs on yat one location
M
M
MM
M1
M1
σcomp.
decreases
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AISC PROVISIONS - NON-UNIFORM MOMENT
M n [non-uniform moment] = C b M n
[uniform moment]
where:
(F1-1)C M
M M M M R
b
A B C
m=
+ + +
≤12 5
2 5 3 4 33 0
.
..
max
max
M max = absolute value of maximum moment in the unbraced segment
M A = absolute value of moment at quarter point of the unbraced segment
M B = absolute value of moment at center line of the unbraced segment
M C = absolute value of moment at three-quarter point of the unbraced segment
Rm = 1.0 for doubly-symmetric members (e.g., W-shapes)
See AISC - F1 for singly symmetric members.
M C
M A M max
M B
L b / 4 L b / 4 L b / 4 L b / 4
NOTE:Cb is has no units and is a function of only the shape of the B.M.D.
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COMMON L b and C bVALUES
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EXCEPTION:
For cantilevers with an unbraced free end, C b = 1, irrespective of the shape of the
bending moment diagram.
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INELASTIC LTB ( L b
< Lr )
• AISC uses the same factor C b
• Good fit to experimental data and easy to use.
Design based on LTB including moment gradient (elastic and inelastic):
(for any Lb)( )[ ] M C M C n LTB b n LTB b= ⋅ =1
where
= lateral torsional buckling strength under uniform moment.( ) M C n LTB
b =1
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Mn
Mp
LbLr LpLpd
Mr
Plastic
designCb=1
Cb>1
CbMr
CbMp
Lm
AISC PROVISIONS
(COMPACT SECTIONS WITH MOMENT GRADIENT)
where
Lm = maximum unbraced length that will allow the beam
to reach M p (Lp no longer has physical significance)
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CHECKING BEAM FLEXURAL STRENGTH FOR LTB
Factored loads
X: lateral support
4 spans to check!
PROBLEM
Given Mu , W-shape, and Lb Find bMn
SOLUTION
For each span check that:
Mu ≤ bMn
Cb1
Lb1
Cb2
Lb2
Cb3
Lb3
Cb4
Lb4
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Note:
Generally, need to check only one span
Largest Mu
Controlling span Smallest C b
Largest Lb
HOW?
Use equations cumbersome; avoid if
possible
Use beam design charts:
o Enter with Lb and read [ bMn]C b =1 for constant moment
o Adjust for moment gradient, i.e., compute C b
o Compute bMn = C b [ bMn]C b =1
o Check that bMn ≤ bM p
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