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Looking for formula for Unbraced Length, Lu

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MrFurleyEIT

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Apr 21, 2007
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Does anyone know the formula for Lu, please? Lu is the maximum unbraced length of compression flange at which the allowable bending stress may be taken as 0.60 Fy.

My Civil Engineering Reference Manual, 8th Ed. on Page 59-3 provides Lu as the larger of 20000*Cb/((d/Af)Fy) or rT*(102000*Cb/Fy)^1/2. However, I think there is a typo in the first equation as I can't seem to get the right Lu.

For instance, with a W14X120 shape, AISC Manual 9th Edition provides an Lu of 44.1 feet. I am not able to replicate it with the formulas from CERM. I am using a Cb value of 1.0.

Hope someone can help. Please provide references, too, in AISC or online, if possible, so I can look them up myself. Thanks a bunch.
 
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Lu is the unbraced length of your steel member (the compression flange). You may be thinking about Lc. The value of Lc (in ASD 9th) is given on p.5-45 of the green book. It doesn't have the same meaning from code to code.
 
Thanks for the prompt response, 3doorsdwn. No, I am looking for the formula for Lu, which is the maximum unbraced length of compression flange for which Fb=0.60Fy.

Lc is the maximum lengthn of compression flange for which Fb=0.66Fy.

Does anyone know of an accurate formula for Lu, please? I found a reference to it in my CERM 8th Ed, but can't seem to get the same Lu as published in AISC Manual, 9th Ed. Appreciate any help.
 
You may be`a victim of a typo.

Lc is the braced length at which a compact section can have an allowable stress of .67 Fy in ASD.
Lc = the smaller of (76bf)/(sqrt Fy) or 20,000/((d/Af)Fy)

Lu is the unbraced length where the allowable stress is less than .6 Fy.

This is governed by 3 equations,of which the
sqrt (102000*Cb/Fy) listed above is one.

Best bet is to get a copy of the green AISC book.
 
In ASD, when Lb exceeds Lc, then Fb <= 0.66Fy. The formulas for such case are a function "L" and "rt". You could set Fb = 0.6Fy and solve each condition of l/rt for what you refer to as "Lu".
 
Both equations look correct to me. Cb must be 1.0. These are from equations F1-6 & F1-8 from the 1989 ASD "Green Book".

Fb = {2/3 - (Fy(l/rT)^2 / 1,530,000Cb)} x Fy

Fb = 12 x 10^3Cb / (l x d/Af)

In each case l must be solved for and then equals Lu.
 
Thank you so much for your responses, people; RoadTech, Bagman2524, et. al. and esp. jike for giving the reference in the "green book" as Page 2-30. The formulas on that page for Lu are exactly the same ones that I put in my original question, i.e. Lu as the larger of 20000*Cb/((d/Af)Fy) or rT*(102000*Cb/Fy)^1/2. I got mine from the Civil Enginereing Reference Manul, 8th Ed. though.

Here's my thing, though. I think I can illustrate better with a Wide flange section, say W14x120. For this column shape, the Lu is 44.1 feet (page 3-23). rT=4.04, d=14.48, A=35.3. I am using A36 steel. I am using a Cb value of 1.0.

Plugging these variables into the two formulas for Lu, I get:

(i) 20000/[12(14.48/35.3)36] = 112.86, or
(ii) [SQRT(102000/36)]*(4.04/12) = 17.92

Since Lu is the larger or either (i) or (ii), I am getting an Lu value of 112.86 feet.

The Lu value given in AISC, 8th Ed for W14x120 is 44.1 feet. What gives? I must be doing something wrong. Am I doing something wrong?

How do I get an Lu of 44.1 feet from the formulas for Lu. Thanks in advance, and sorry, I may get some egg on my face, but I hope to find a resolution for this. Please provide your calculation, also.
 
You should be using d/Af, not d/A. Af is the Area of the flange. For W14 x 120, d/Af = 1.05 (from green book).
20000/(12 x 1.05 x 36) = 44.1 ft.
 
Thanks for helping out, people. The discussion and feedback was very useful and I hope, useful, to others, as well. My oversight, as I see it, is to use A, instead of Af, in my calcs. The other parts of the formulas were ok, but discussing and going to and fro about it reinforces the understand of this particular subject, which I think is very useful for the long term.

Now, if you'll excuse, I've got some egg I need to wipe off my face :)
 
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