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# FW: Compression Capacities for WFs with Webs Not Compact

• To: "Seaint (E-mail)" <seaint(--nospam--at)seaint.org>
• Subject: FW: Compression Capacities for WFs with Webs Not Compact
• From: "LaCount, Curt" <Curt.LaCount(--nospam--at)jacobs.com>
• Date: Tue, 11 Jan 2005 11:57:34 -0500

```Sam,

It looks to me like you are using f=36 ksi in equation A-B5-12.  The value
should be the computed elastic stress, which in this case is limited by
column buckling to 7.13 ksi.  At that stress, there is no reduction for be.

I hope that helps.

Curt La Count
Jacobs Engineering
Portland, OR

-----Original Message-----
From: Szuchuan Chang [mailto:szuchuan(--nospam--at)gmail.com]
Sent: Monday, January 10, 2005 8:28 PM
To: seaint(--nospam--at)seaint.org
Subject: Compression Capacities for WFs with Webs Not Compact

Hi,

I am reviewing an existing building with LRFD steel design.  Along the
drag lines of a roof diaphragm are the existing W14x22 beams that
suppose to pass on the collected diaphragm shear to the brace frame at
the end of the building.  I am treating these beams as compression
members just to see if it can take the loads to the brace frame.
(Ignore bending for the moment.)

Fy=36 ksi;
bf/2tf of the W14x22 is 7.46 which meets the flange compactness of 15.9;
h/tw of the W14x22 is 53.3 which is bigger than the compactness of 42.3;
assuming the beam is braced at 15 feet along the weak axis.
Kl/r = 173.1;
Table 3-36 on the manual page 16.1-143 give phi*Fcr = 7.13 ksi.
area = 6.49 sq in
Pall (without reducing for no-compact web) = 7.13*6.49 = 46.3 kips

To account for the reduction on the effectiveness on the web,
I apply the Eq A-B5-12 on pg 16.1-91
(using b=13.7 & t=0.23)
I get "b sub e" = 10.2 inches.

I than reduce the cross section area form 6.49 - 0.23*(13.7-10.2) =
5.69 sq inches

The new capacity I got is reduced to 40.5 kips.

Question? Am I doing this correctly?

Thanks

Sam

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