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RE: Find the Lr for the new LRFD provisions
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 Subject: RE: Find the Lr for the new LRFD provisions
 From: "Garner, Robert" <rgarner(nospamat)moffattnichol.com>
 Date: Mon, 24 Jul 2006 07:50:21 0700
Have you sent this in to "Ask
AISC"? I found a minor error (errata?) in one of their drawings for the
example problems and received a quick reply. Bob Garner From: Szuchuan Chang
[mailto:szuchuan(nospamat)gmail.com] I try to find out how the "Lr" distance in the
Example F.22b of the 13the edition AISCLRFD was reached for channel flexural
members. These are new equations with not much disscussions so far. On page F18 of the examples provide by the CD, the Lr is 14.5 feet fro
m Table 38 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 28b for a channel. I also calculated the Eq F27 (r_ts)^2 = sqrt(Iy*Cw)/Sx = 1.077. Plugging the above values in Eq F26, 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 59 is 14.1 feet.
(see page 5120)

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