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# RE: Collector element

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Collector element
• From: Samir Ghosn <sghosn(--nospam--at)harris-assoc.com>
• Date: Tue, 17 Dec 2002 09:56:40 -0800

Thanks Gerard for your input.  You are correct, the large foot print uses
only two lines of resistance, that is why such magnitude is shown.  I Agree
with you, perhaps the way out is to add additional lines or resistance to
bring this to a rasonable level.
Samir Ghosn, PE
Harris & Associates
At 09:33 AM 12/17/2002 -0800, you wrote:
>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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