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Re: Find the Lr for the new LRFD provisions

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No, I have not.
 
"Ask AISC" allows only 500 characters in a question.  It is so hard to fit a question in.
 
Mr. Charlies Carter browses our list off and on.  He might forward it to AISC if he think it is necessary.
 
Are there an official erratas for the 13th edition yet?
 
Thanks 
 
Sam Chang

 
On 7/24/06, Garner, Robert <rgarner(--nospam--at)moffattnichol.com> wrote:

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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