aisc lrfd deisgn of slender hss column

huangyhg

aisc lrfd deisgn of slender hss column
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
for the b/t = 39.9, it appears that the equation zeros out when f becomes 2.64 ksi.  
2.64 ksi is a very low stress and probably runs off the chart for the equation as most stresses in economical designs are closer to the 0.4 to 0.6 times fy.  for hss with fy = 46 ksi, a "normal" design would be in the range of 18.4 ksi to 27.6 ksi.  your stress of 0.166 ksi is very very low.
if you calculated f = 25 ksi (as an example), then the equation for the b/t = 39.9 results in
be = 10.22" - which is greater than b = 9.29" (39.9 x .233")
so you'd use be = 9.29" for the long sides.
for the short sides, the equation zeros out so the effective width of those is zero.
so your total effective widths would be 2 x 9.29 = 18.59 inches.  taking that times 0.233" thick - effective area = 4.33 sq. in.
q = 4.33 / 5.24 = 0.8268
fcr = 0.8268(0.658(.8268)(λc^2)46
this is my initial take on it.  it does look that the equation goes negative pretty easy so i guess you assume that you can't have a negative and so you get 0 in those cases.
i think your specific problem is the low stress - 0.166 ksi.
i don't think the short sides should be zero.  they have a low b/t ratio so they are not slender elements.  
the way this equation "blows up" for small values of "f", which doesn't really make sense, suggests to me that the be equation is empirical.  does any know if this is true?  i would just follow the note and for b/t<50 and fy<50ksi use f=fy.
hey guys
thanks for the response. it seems to me that aisc needs to add an extra clause to this equaiton to stop it blowing up. unless, what ucfse is referring to some additional clause/note that would allow me to bypass this slender wall requirement. ucfse; where is the note re: b/t<50 and fy <50ksi you are referring to in the aisc 3rd manual? i didnt see such a note in the manual for the section 4.2., eq 4.2-7
cheers!
note that this equation is also used in the non-hss lrfd specification (see appendix b5.3b in the lrfd 3rd edition).
ucfse - i agree - the short sides going to 0 and the long sides counting for something doesn't seem to be logical.
just to add to ucfse, the walls are slender. lambda_b is greater than the limit for a non-compact section. in accordance, the reduction factor q must be used. you can check for the provisions given in the table in the hss specificaiton section  included in lrfd 3 ed,
cheers
after looking through the 3rd edition lrfd, i can't find that note either.  i took it from the draft version of the 4th edition.  the closest i can find is on page 16.2-34, the second full paragraph where it is mentioned "the stress is taken as the yield stress".  that paragraph refers to flexural   
i sent a question to the aisc technical help center - may take a day but we'll see what they say.
some other clarification which might now clear things up. the limiting non compact value i calculated,  lambda_r=1.4 sqrt(e/fy) = 35.2, used came from the limits assuming the cloumn was acting only in compression and no flexure, which in fact is not true and in fact the major demand is flexure in may case; i htink it may be something like 30 times the compresison demand. which explains previous comments about the factor f (pc/ag) being way too small for my   
i don't know if this will help or confuse the issue.  in the new aisc 360, this same equation appears as eq. e7-18.  but in this code, it has a limit, in that be is required to be <= b.  so, i think the code writers have recognized that for some cases be calculates out to be > b.  i'm not sure but i suspect that this indicates that this equation for be is not applicable for any case in which it yields be>= b.  there is also a user note which says that it is acceptable to substitute fy for the calculated value of f in this equation, and this substitution will result in a slightly conservative value for column capacity.  (this is in section  e7 of the new aisc combined spec.)
regards,
chichuck
hey chichuck,
can you direct to where that note for the substitution of f?
cheers