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# Boggled Out By Load Combinations

• To: <seaint(--nospam--at)seaint.org>
• Subject: Boggled Out By Load Combinations
• Date: Mon, 12 Apr 2004 12:56:41 -0500

```Bill,
I agree with Bill Cain.  The correct multiplier is zero.  I know what
you said.  I just can't believe AE could be that small in a steel member
(sounds like you're talking about steel).  Besides, won't it buckle when
the buildings move the other way?

What is this mystery member, anyhow?  You've got me curious now.

TKDA
St. Paul, Minnesota

Bill Allen wrote, in response to Bill Cain:

Bill -

=20

Believe it or not, I think I can accommodate the force (think: rubber =
band - AE is very small). Allowing for the movement (sliding connection)
in my = case would create more problems than it solves.

=20

Sooo.If I understand your response to the original point, you're saying
= to take Delta_sub_s and divide by 1.4 to arrive at the ASD level =
displacement for design of this rubber band?

=20

Thanks,

=20

T. William (Bill) Allen, S.E. (CA #2607)

V/F (949) 248-8588

San Juan Capistrano, CA

http://www.AllenDesigns.com

-----Original Message-----
From: BCainse(--nospam--at)aol.com [mailto:BCainse(--nospam--at)aol.com]=20
Sent: Saturday, April 10, 2004 3:20 PM
To: seaint(--nospam--at)seaint.org
Subject: Re: Boggled Out By Load Combinations

=20

Bill-

The force E for ASD is 1/1.4 times the value calculated by code =
formulas. (See CBC Section 1612.3.1 Eqns 12-9, 12-10 and 12-11)

=20

But, unless the L is very big and/or the AE is very small, the force for
most calculated deflections will be huge.  Wouldn't it be better to =
provide a sliding connection on one end?  Then the force would be quite
small.  = But the bearing for the element will need to allow the full
range of = movement.

=20

Regards,

Bill Cain, SE

Berkeley  CA

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