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RE: Collector element

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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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