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RE: Collector element

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Samir

First, I am still curious how the 355 psf DL is derived? Is this slab
supporting a wood framed building above? if yes, that is still an excessive
amount, since wood framing is normally about 15-20 psf per square foot and
even with 4 stories of wood framing and a 12 inch slab you probably get
about 250 psf.

Second, this may be one of those occasions that designer may use the code
provision in section 1633.2.6 that says "the quantity Em need not exceed the
maximum force that can be transferred to the collector by the diaphragm or
other elements...."

Ben Yousefi, SE
San Jose, CA

-----Original Message-----
From: Samir Ghosn [mailto:sghosn(--nospam--at)harris-assoc.com]
Sent: Tuesday, December 17, 2002 8:26 AM
To: seaint(--nospam--at)seaint.org
Subject: RE: Collector element


Gerard
the size of the poduim slab is 300' x350' and has an estimated seismic load
of, V=0.25 x 355 x 300 x350 = 9320 kips.  The uniform load is  w=9320/350
=26.6 kip per foot and the collector design is dreived form shear diagram
at point of collector.  Samir Ghosn, PE
Harris & Associates
(800) 827-4901 xt 360
At 10:23 PM 12/16/2002 -0800, you wrote:
>Here's my stab at it...
>
>Unless you have some kind of building irregularity or discontinuity in
>the lateral system, this seems to be a mistake. A collector force that
>huge (780 kips) could never come from just the floor diaphragm. 
>
>Can you describe where all this force is coming from??? Maybe he hit x
>instead of + on his calculator when determining the drag force...
>
>-gerard
>Santa Clara, CA
>
>
>
>
>-----Original Message-----
>From: Samir Ghosn [mailto:sghosn(--nospam--at)harris-assoc.com] 
>Sent: Monday, December 16, 2002 4:38 PM
>To: seaint(--nospam--at)seaint.org
>Subject: Collector element
>
>Question to the experts:
>I am reviewing a set of plans with a collector beam element in an
>elevated
>poduim slab.  the design load, drag element, is 780 kips.  the designer
>computed omega level forces per 1633.2.6 to a design level force of 2184
>kips= (2.8)*780.  
>In the design of the steel reinforcing, the designer get a 10 # 5 bars,
>(no
>calculation).  I computed the As required based on T=0.9*As*Fy and got a
>required area of steel of 40.44 inches square which is equivalent to 40
># 9
>bars.  Uhmm. What is it that I am missing?.  I am starting to second
>guess
>myself when I think, how in the world would one get 40 # 9 bars for a
>collector in an 11 inch thick slab and distribute it over the wall or on
>each side of the wall.  Section 1921.6.12 does not seem to limit the
>width
>of section considered effective in the transmittal of such load.
>Your input is appreciated.  
>Samir Ghosn, PE
>Harris & Associates
>(800) 827-4901 xt 360
>"sghosn @harris-assoc.com"
>
>
>
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