Bi-Axial Bending + Tension

[JoshPlumSE]

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.

 

[civilperson]

The net compression due to bending will be reduced from the uniform tension. A conservative method would ignore this lessening for the compression side. The tension would have to be permanent and unchanging to use the lesser compression.

 

[JAE]

Josh,

Wouldn't it be something like this:

(f[sub]b[/sub] - f[sub]t[/sub]) / F[sub]b[/sub]** < 1.0

is equivalent to:

f[sub]b[/sub] / F[sub]b[/sub]** - f[sub]t[/sub]/F[sub]b[/sub]** < 1.0

So with biaxial bending, you have an f[sub]b1[/sub] and an f[sub]b2[/sub], 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?:

f[sub]b1[/sub] / F[sub]b[/sub]** + f[sub]b2[/sub] / F[sub]b[/sub]** - f[sub]t[/sub]/F[sub]b[/sub]** < 1.0

 

[JoshPlumSE]

JAE -

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.