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# RE: Collector element

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Collector element
• Date: Tue, 17 Dec 2002 09:33:42 -0800

```Why is the dead load so high? 355 PSF? Are there are only two lines of
shear resistance for such a large footprint? What is the RHO value? Does
the diaphragm work in shear and is it capable of transferring the shear
to collectors? The chord force must be checked as well, perhaps that's
all he provided steel for???

If you are comfortable with the numbers, then he has to increase the
number of bars  or add vertical resisting elements to reduce the
collector force. He could take a 10 foot wide strip of slab, but how
does it all get into the shearwall, brace, or frame where the collector
force is max? I think you are right to question him on this, but
something doesn't add up to me on the forces.

-gerard
santa clara

-----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
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.
>Samir Ghosn, PE
>Harris & Associates
>(800) 827-4901 xt 360
>"sghosn @harris-assoc.com"
>
>
>
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