Question on Alignment Chart In AISC 13th

[leeStruct]

I guess I had better to start a new thread not to mix with the previous thread to ask this question.

Please see attached sketch.

Suppose we know that the bracing girder is very weak (negligeable), by inspection, we can take the column as if there were no bracing girder, therefore, the effective length of the column is 2*Lc.

Now, by using Alignment Chart in AISC 13th page 16.1-241. For a pinned column end "A", Ga=10 (see 2nd paragraph of page 16.1-241), for bracing point "B":

GB=(Ic/Lc +Ic/Lc)/(Ig/Lg +Ig/Lg)

because Ig is very small to compare with Ic, therefore, GB=infinite, from Alignment Chart of Fig. C-C2.3 in AISC 13th page 16.1-241, we have approximately K=1, therefore the effective length of column AB is K*Lc=Lc, which means the effective length of the whole column is Lc, which contradict with previous result of 2*Lc (which is the correct answer).

Can anybody tell me what's wrong in this logic?

Thanks.

 

[BAretired]

@slickdeals,

If the column shown on your sketch is unrestrained, it is unstable and will collapse. k becomes infinite. But in your sketch, it appears that it will be restrained by the frame elements to the right.

BA

 

[slickdeals]

I agree with you BA. The moment frame will provide lateral stability to this column. In the commentary in AISC, a pinned base is assumed to have a G = infinity, but for real purposes it is recommended to use G = 10. I am using the same logic for the top. The lateral deflection is restrained, but rotation is not prevented. Using G = 10 for top and bottom, you get a K = 3.0. It does not seem realistic, but that's what the numbers appear to say.

 

[Lion06]

slick-
Once you say that the column is pinned at the top and the bottom it is no longer a frame column and the nomographs don't apply. It's simply a gravity column at that point.