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RE: wood- biaxial bending

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Andrew,

That's one approach.....the other equation you could be looking for is:

(fc / Fc')^2  +  (fbx / F'bx(1 - fc / Fcex)) + (fby / F'by(1 - fc / Fcey
- (fbx / Fbe)^2)) < = 1.0

If (fc / Fc) goes to zero (no axial loading) then....

(fbx / F'bx) + { fby / (F'by (1-(fbx / Fbe)^2))) }

Wher Fbe is the elastic buckling stress considering lateral torsional
buckling of beam....

This is the equation for a beam-column .....letting axial compression go
to zero.....it is more conservative than the tension approach because
the effect of lateral torsional buckling is taken into account with the
beam stability factor Cl, but the volume factor does not apply
Cv........

Robert Rogers, PE

-----Original Message-----
From: Andrew D. Kester [mailto:andrew(--nospam--at)baeonline.com] 
Sent: Wednesday, April 16, 2003 10:39 AM
To: SE Web List (E-mail)
Subject: wood- biaxial bending


For a timber beam in bending, I cannot find an NDS formula specifically
for biaxial bending. However, (NDS C3.9-2, from the Commentary), gives
one for biaxial bending and tension. If there is no tension and that
term goes to zero, then what is left is the standard biaxial bending
equation that is used from AISC, etc.

So can I simply use this for wood:

fbx/Fbx + fby/Fby < 1

Is this correct?

TIA

Andrew D. Kester, EI
Structural Engineer
Bentley Architects & Engineers
665 W. Warren Ave.
Longwood, FL 32750
1-407-331-6116
andrew(--nospam--at)baeonline.com
www.baeonline.com






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