# Re: Diaphragm design forces

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Diaphragm design forces
• Date: Wed, 20 Sep 2000 08:15:02 -0700
```Peter,

1. The inter story shears used in the design of the vertical lateral
load resistting elements, UBC 97 Eq. 30-15, is different than the shears
used in the design of the diaphragms, UBC 97 Eq. 33-1.

2. The diaphragm force is a function of the interstory shear which is a
function of the story height.

3. I don't recall ever getting interstory shears greater than diaphragm
shears.  I think if you do the math, you can show that the diaphragm
force will match the intertory shear at the roof, and will exceed the
interstory shear at every other level.

West Covina, CA

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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