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# RE: Find the Lr for the new LRFD provisions

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Find the Lr for the new LRFD provisions
• From: "Garner, Robert" <rgarner(--nospam--at)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(--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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