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Re: Redundancy Factor

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the problem is the 10/Lw factor.  If you have shear walls less than 10 feet in
length you start to pay a penalty in terms of a higher redundancy factor.  Per
Gerry Neville at ICBO, the wood industry proposed 4/Lw but lost.  See pages
129-130 of the the ICBO109 Seminar notes for a discussionIn the example I cited
all stories would have 1.5 factor in the short direction except the ground floor
which might have a steel frame or masonry lateral force system.

Seaintonln(--nospam--at) wrote:

> Bob,
> At what point do you believe that the redundancy factor begins to appear? In
> other words, does it appear in one story, two story or more than two story's
> on the example that you defined. I would tend to think that even on a narrow
> lot, if the structure is less than two stories and the shearwalls are more
> closely spaced (as would be more common in single family residence).
> I guess the conclusion I am hoping to reach is that the redundancy factor is
> more applicable to certain types of structures and less critical for others.
> To define what these structures are by having enough examples may help to
> work out  a simplification of the requirments. For every stage that can be
> simplified, the time to design can become more manageable.
> Thanks,
> Dennis
> In a message dated 7/29/99 10:32:21 PM Pacific Daylight Time,
> rjbossi(--nospam--at) writes:
> << In San Francisco, because of narrow deep lots (25x120) the redundancy
> factor in
>  the short direction will almost always be 1.5.  This type of lot is usually
>  occupied by 1-3 dwelling unit buildings of 3-4 stories.  It would be
> applicable
>  for both new construction and retrofit under the SF Building Code.  In the
> long
>  direction the redundancy factor could not exceed 1 because the side walls,
> on the
>  property line and are typically 60'+ long with no openings. >>