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# Re: 97 UBC Lateral force on elements of structures

• To: seaint(--nospam--at)seaint.org
• Subject: Re: 97 UBC Lateral force on elements of structures
• From: Robert Shaffer <rkdn(--nospam--at)cruzio.com>
• Date: Tue, 19 Oct 1999 10:07:03 -0700

```There is a good example of this calculation in the Seismic Design Manuel,
Vol 1 by SEAOC and ICBO.

They calculate the value of Fp for each diaphragm elevation and then average
them
w = [(Fp1x + Fp2x)/2] * W  .

It seems that there is contradictory language in the code as to the
application of Rho to Fp.  In 1630.1 the definition of Eh includes Fp,  but
then in 1632.2 it indicates to set Rho equal to 1.0.

Robert Shaffer, P.E.

Sam young wrote:

> Section 1632.2 of 97 UBC gives two formulas for calculating the total
> lateral force --- (32-1)  and alternatively, (32-2). Using the second
> formula we need to determine hx, which by definition is the element or
> component attachment elevation w/ respect to grade (hx>=0). A recent
> discussion in our office yielded three different values for hx when
> designing an exterior masonry wall:
>
> 1. hx = 0
>
> 2. hx = 0.5 * wall height
>
> 3. hx = wall height.
>
> notice that ap/Rp per table 16-o is 1/3 in this case, differences betw
> 32-1 and 32-2 are:
>
> 1. 4.0 vs 1/3
> 2. 4.0 vs 5/6
> 3. 4.0 vs 4/3
>
> since code also states Fp>= 0.7 Ca Ip Wp, it seems that choice 1 is out,
> and #3 is the most conservative.
>
> Any discussions is highly appreciated.
>
> Sam Young
> CBH Consulting Engineers
>
>
>
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