AISC LRFD Deisgn of Slender HSS Column

[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

 

[JAE]

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

b_e = 10.22" - which is greater than b = 9.29" (39.9 x .233")

So you'd use b_e = 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

F_cr = 0.8268(0.658^{(.8268)([λ]c^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.

 

[UcfSE]

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.

 

[tslewis]

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!

 

[JAE]

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.

 

[tslewis]

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

 

[UcfSE]

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 members.

When you think about it, a larger stress should cause more of a section susceptible to local instability to buckle more. At very low stresses no instability should be observed. After all, if there is little or no stress on an element there is nothing to make it buckle. A larger stress should cause more of an element to buckle, starting farthest from its support. In the case of a rectangular HSS, that is in the middle of the element. Given that, using a larger stress to calculate effective properties should be more conservative. The draft version of the 4th edition LRFD gives those limits on Fy and b/t and states that within these parameters, using f=Fy yields conservative results that are about 9% off. Using f=Fy is also much easier than iterating using f equal to its real value. Once you have your effective section properties, you have to go back and re-evaluate your Qa with a new f. That can potentially take a long time and for 9% I doubt it worth it unless it really has to work.

 

[JAE]

I sent a question to the AISC technical help center - may take a day but we'll see what they say.

 

[tslewis]

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 member.

So I guess in reality I should have used the limiting non-compact value for compression AND flexure acting simultaneously, the provosions in the HSS specificaiotns of the LRFD 3rd edition gives a higher value. Something like 3 times more. So, I guess the "blowing up" of the equaiton due to a snall value of f will not occur as my wall slenderness limit is not exceeded for case of compression and flexurem and I dont need to reduce Q.

cheers

 

[chichuck]

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

 

[tslewis]

Hey Chichuck,

can you direct to where that note for the substitution of F?

cheers

 

[chichuck]

tslewis,

It is in the new combined spec, section E7.2(b). I found it on page 43 of the new spec, right after they have Eq. (E7-18). Its in a grey bar near the top of the page.


regards,


chichuck

 

[tslewis]

Thanks. The new AISC LRFD specificaitons have significantly more equations and details regaridning design of compression members for Chapter E compared to teh 3rd Edition of the manual.

F has also been revised and is now given by Pn/Ae isnatead of teh previous HSS specifcaitons of Pu/Ag. f is now an iterative calculaitons as be and f are dependant on one another.

Unfortuantelt when i set f to Fy I still got teh equaiton to blow up for the calculaiton of be

 

[JAE]

I got a response from AISC:

[blue]That expression assumes you are designing an HSS compression member for near its full design capacity. For very small loads over large gross areas, local buckling of stiffened slender elements may not occur and the expression begins to fall apart.

However, further guidance is available in Section E7.2 of the 2005 AISC Specification (a free download from

http://www.aisc.org/2005).

One can use f = Fy, and that will result in a slightly conservative estimate of column capacity.[/blue]

 

[tslewis]

Thanks for the respones JAE. Unfortunately I have already tried setting f to Fy as per the latetst AISC Specification - see my post above. Setting f to Fy still does not solve the problem and "be" is still miscalculated. You can try for yourslef - all my assuumpations and loads are given in the original post.

I guess the answer is to only use this expression for "be" when you have a member being designed purely in compression, where here the wall limits would not requries you to calculate "be" for such a small axial load.

 

[UcfSE]

When you say "blows up", tslewis, do you mean you get a negative answer or some absurdly large answer? I would check the math to be sure all the numbers were entered as they should be. With f=FY the equation should give a reasonable answer. It may still be greater than b, which then it gets limited to b, but it shouldn't be something like 10^6 or -5000.

 

[tslewis]

Yes I have checked the numbers sveral times. The assumptions and reults are fully described in my original post. Yes origianlly i was getting a large nehative number. Putting F = Fy does not solve the problem - it is still negative. Try putting f = Fy in my original calc, the magnitude of the value for be is reduced but in my particular exmaple it is still negative.

 

[UcfSE]

I put your numbers in the equation you reference in your original post and get be = 9.12". That is with lambda_h for b/t = 39.92. Using lambda_b=b/t=5.6, I get a negative number, but then lambda_b=5.6 is less than the limit for checking slenderness therefore using lambda_b=5.6 does not make sense.