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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: "Szuchuan Chang" <szuchuan(--nospam--at)gmail.com>
- Date: Mon, 24 Jul 2006 09:00:57 -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(--nospam--at)gmail.com]
Sent: Sunday, July 23, 2006 7:05 PM
To: SEAOC
Subject: Find the Lr for the new LRFD provisions
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)
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- From: Garner, Robert
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