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Find the Lr for the new LRFD provisions

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I try to find out how the "Lr" distance in the Example F.2-2b of the 13the edition AISC-LRFD was reached for channel flexural members.  These are new equations with not much disscussions so far.
 
On page F-18 of the examples provide by the CD, the Lr is 14.5 feet fro m Table 3-8 for Channel 15x33.9.
 
I use h_o = d - tf = 15 - 0.65 = 14.35 inches.
 
Iy = 8.07 in^4 and Cw = 358 in^6
 
To reach this Lr using the code provision on page 48, I got c= ho/2*sqrt(Iy/Cw) = 1.38   <--Eq 2-8b for a channel.
 
I also calculated the Eq F2-7 (r_ts)^2 = sqrt(Iy*Cw)/Sx = 1.077.
 
Plugging the above values in Eq F2-6, I got Lr = 15.05 ft instead of 14.5 ft.
 
How did AISC get their Lr for steel channels?
 
Thanks
 
Sam Chang
 
ps: in the 3rd edition of LRFD, the Lr list in table 5-9 is 14.1 feet. (see page 5-120)