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Re: UBC 1633.2.5 clarification
- To: <seaint(--nospam--at)seaint.org>
- Subject: Re: UBC 1633.2.5 clarification
- From: "Pat Clark" <bcinc(--nospam--at)nanosecond.com>
- Date: Wed, 30 Oct 2002 09:55:15 -0800
I'm pretty familiar with 1633.2.6, and understand
those provisions. I'm questioning 1633.2.5, where they want an increase to
0.5 Ca*I*W, which is well above my system seismic load. Can this load be
reduced by 1.4 for ASD? Does this increase affect both the connectors and the
----- Original Message -----
Sent: Wednesday, October 30, 2002 9:01
Subject: RE: UBC 1633.2.5
1633.2.6 Talks about designing collectors = drag struts. It says use the
regular diaphragm formula 33-1 for wood and that you can multiply the
collector capacity by 1.7 to determine it's strength without a 1.33 ASD
increase. Dividing by 1.4 is in UBC 1612 ASD load combinations which are
not mentioned in 1633.2.6 so you can't divide by 1.4 if you are checking
collectors but you multiply the capacity by 1.7 to determine strength so it
comes out in the wash. 1.33*1.4=1.862>1.7, so there is a slight hit
for collectors in 1633.2.6.
1997 UBC notes the following:
1633.2.5 Ties and continuity. All parts of a
structure shall be interconnected and the connections shall be capable of
transmitting the seismic force induced by the parts being connected. As a
minimum, any smaller portion of the building shall be tied to the remainder
of the building with elements having at least a strength to resist 0.5
CaI times the weight of the smaller
A positive connection for resisting a
horizontal force acting parallel to the member shall be provided for each
beam, girder or truss.This force shall not be less than 0.5 CaI times the
dead plus live load.
For a wood framed structure (R=5.5), with a drag strut or
drag truss, would this minimum force include the drag truss or strut
capacity, the connectors/fasteners, or both. Can it be divided by 1.4
to get back to ASD level forces?
Thanks in advance,