AISC LRFD Rectangular HSS Column Effective Area Mistake?

[tslewis]

Has anyone come across a problem with calculating the reduced effective width be of a rectnagular section neede for deriving the Q factor used in assessing the compression capacty of a rectangular HSS section. I cannot find any literature which addresses this thoroughly enough.

I rever to:

HSS 10x2x1/4
lambda_b (b/t)wall slenerness of 5.6
lambda_h (h/t) wall slenderness of 39.92
Pu = 0.87 kip
Ag = 5.24 in*in

Non Compact wall slenderness lambda_r=1.4 SQRT(E/Fy) = 35.2

hence, Lambda_h > lambda_r

hence, in acccordancwe with HSS provisions for LRFD design - see p.16.2-6 LRFD 3rd Edtion, Section 4.2., EQ 4.2-7

be= 1.91*t*SQRT(E/f)*((1-(0.381/lambda_b)*SQRT(E/f))

f = Pu/Ag = 0.166ksi
E= 29000 ksi
I always get a NEGATIVE value of be that is 100 times or so larger than b. I can only get teh equation to work if the lambda_b value i ssomething like 200 which is crazy.

For the current values

lambda_b gives -5118.001 in !!!!!
lambda_h gives -5555.917 in !!!

I ahve checked noth the 2000 and 1996 HSS specificaitons produced by AISC and they are the same for this equation!

Any help would be apreciated!

thanks

 

[rday]

Well, my first thought was why such a large column for an 870 pound load?

Note that while lambda_h > lambda_r
lambda_b < lambda_r, so using lambda_b Q=1

I would guess that has something to do with why the calc is blowing up, but I would have to look deeper to be sure. Since f is the applied stress the equation would seem to work out better when the section is near capacity and buckling would become an issue in non-compact sections. The b/t of 200 could be a thin wall tube.

Without more info I would say reduce the column to something more reasonable to the loading and redo the calcs.

Rik