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Re: Diaphragm design forces

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Peter,

A question back to you...what did you use as the weight (W in the .5 Av W
equation that you mention)?  Did you just use the weight of just the
diaphragm (i.e. just the roof deck) or did you use the weight of the deck
and ALL the items attached to it (ceiling, lights, mech stuff,
beams/girders, roofing material, etc.)?

Make sure you take a good look at the "code speak" in the BOCA code.  That
section you cited talks about the weight of the diaphragm and "other
elements the building attached thereto,...".  I interpret that all the
"weight" at that level needs to be used.  Also, the section states that
any load that needs to be "redistributed" due to offsets, etc at the
level.  It would sound like this part shouldn't effect you if the shear
walls are continuous and the building is regular.

It might help a little if you can get you hands on the BOCA commentary.
The 1993 commentary that I looked at had an OK description.

I don't know if that helps out any or not...

Scott Maxwell, PE, SE

On Wed, 20 Sep 2000, Peter A. Allen wrote:

> 
> 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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