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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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