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Diaphragm design forces

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Question:
Is the design seismic load for an individual diaphragm (roof or floor) 
different than the design seismic load for the building as a whole? Is that 
design load for a diaphragm independent of the height of that diaphragm (its 
vertical location in the building)?

Specifically, the Standard Building Code (1997) and BOCA include a paragraph 
(1607.3.6.2.7) which states:
"Floor and roof diaphragms shall be designed to resist the following seismic 
forces: A minimum force equal to 50% the effective peak velocity-related 
acceleration Av, times the weight of the diaphragm and other elements of the 
building attached thereto..".

(UBC has a similar paragraph with different equations and coefficients).
Also "Lateral Force Requirements" (SEAOC 1990) Section 1H2j provides an 
equation for diaphragm design forces.

Typically, I had been using the force Fx determined from 1607.4.2 "Vertical 
distribution of seismic forces" to get the force required to be distributed 
by the diaphragm.  The calculation of Fx is dependent on the individual floor 
weights and their vertical location.

In my case, I have a regular building, with shear walls lining up from floor 
to floor.  Using the calculation of F = 0.50 Av W, instead of the vertical 
distribution force, results in a significantly lower design force at the roof 
diaphragm.  However, I still need to design the shear walls for the (higher) 
vertical distributed force.

I cannot resolve that the "diaphragm design force" could be less than the 
force required to be resisted by the shear walls.  As a co-worker stated, "So 
you're designing the shear walls for a load that can't get there?"

Can anyone shed some light on this apparent discrepancy, or point out any 
flaws with my reasoning?

Peter Allen, PE
pallen(--nospam--at)clarknexsen.com
Norfolk VA / Charlotte NC



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