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Collector element

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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

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