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RE: 1997 UBC Masonry Wall Design

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> Where are you getting the distance of 0.367H for the lower
> portion force?

I found this with the help of Steve Rex, S.E. (who apparently cannot post to
the list for some reason). If you set Eqn. 32-2 equal to the minimum cutoff
of 32-3, you can solve for hx in terms of hr:

0.7=(1/3)*(1+3*hx/hr)

2.1=1+3*hx/hr

1.1=3*hx/hr

hx=1.1*hr/3

hx=0.367*hr

> It may be in the code and I haven't found it yet, but an
> example from the
> Seismic Design manual Vol. I by ICBO, (Example 35, page 107),
> indicates you
> just average the force at base and at roof level.

This may have been (again) one engineer's interpretation of 32-2. Who knows?
But, if there is an "hx" term, why would you average?

Regards,

Bill Allen, S.E.
ALLEN DESIGNS
Laguna Niguel, CA