Bi-Axial Bending + Tension
Bi-Axial Bending + Tension
(OP)
A friend recently pointed out what seems to be a flaw (albeit a small one) in the NDS.
Setion 3.9.1 deals with bending + Axial Tension.
Equation 3.9-1 deals with the side of the member that is in tension. The commmentary has some good commentary on how to extend this equation to include the effects of bi-axial bending.
ft/Ft' + fb /Fb* < 1.0 can easily become:
ft/Ft' + fb1/Fb1* + fb2/Fb2* < 1.0
Unfortunately, equation 3.9-2 really doesn't address the issue of bi-axial bending at all. This equation covers the potential failure on the compression side of the member. However, the equation assumes uni-axial bending.
(fb-ft) / Fb** < 1.0
This is the root of my question. How should we address the compression side failure of a member that is in tension and bi-axial bending?
I've got some definite opinions on how to address this, but I don't want to bias the discussion towards my point of view.
Setion 3.9.1 deals with bending + Axial Tension.
Equation 3.9-1 deals with the side of the member that is in tension. The commmentary has some good commentary on how to extend this equation to include the effects of bi-axial bending.
ft/Ft' + fb /Fb* < 1.0 can easily become:
ft/Ft' + fb1/Fb1* + fb2/Fb2* < 1.0
Unfortunately, equation 3.9-2 really doesn't address the issue of bi-axial bending at all. This equation covers the potential failure on the compression side of the member. However, the equation assumes uni-axial bending.
(fb-ft) / Fb** < 1.0
This is the root of my question. How should we address the compression side failure of a member that is in tension and bi-axial bending?
I've got some definite opinions on how to address this, but I don't want to bias the discussion towards my point of view.





RE: Bi-Axial Bending + Tension
RE: Bi-Axial Bending + Tension
Wouldn't it be something like this:
(fb - ft) / Fb** < 1.0
is equivalent to:
fb / Fb** - ft/Fb** < 1.0
So with biaxial bending, you have an fb1 and an fb2, each creating compression. The combination of the two of them creates a maximum compression on one corner of the section.
So would you just use them in an additive fashion like this?:
fb1 / Fb** + fb2 / Fb** - ft/Fb** < 1.0
RE: Bi-Axial Bending + Tension
I have two thoughts on this one. The first is the equation that you wrote down:
fb1 / Fb1** + fb2 / Fb2** - ft/Fb** < 1.0
Except, that we need to decide whether ft is divided by Fb1** or Fb2**.
The second would be a variation of the compression equation. Something like:
fb1/Fb1` + fb2 /(Fb`*(1-fb1/Fb1E))
Or, maybe....
fb1/Fb1` + fb2 /(Fb`*(1-(fb1/Fb1E)^2))
My thoughts are that if I've got 0.0001 lbs of compression or 0.0001 lbs of tension, then I need to get essentially the same code check for a bi-axial loaded wood member.
It's sort of an academic issue. But, since we're involved in structural software, my company ends up spending a lot of time on the "weird" cases that one would normally ignore in hand calculations.