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hold downs at perforated shear wall

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Joe,

Both ICC's International Building Code (IBC) and AF&PA's Special Design
Provisions for Wind and Seismic (SDPWS)
http://www.awc.org/Standards/SDPWS.html specify anchoring requirements for
perforated shear walls (PSW) based on shear in the wall. Accounting for
other effects such as gravity (dead load) and wind uplift are handled as
part of the load path. 

In SDPWS, there is a load path statement in the shear wall section. If there
isn't a load path statement in IBC Chapter 23, the general criteria for load
path can be found in Chapter 16. 

Note that there are 2 sources of uplift that must be resisted... First, the
global overturning of the entire wall, which should be calculated as:
T=(V*h)/(C_o*SumL) where, V=total shear load, h=height of the shear wall,
C_o=the opening factor that reduces the shear resistance based on the
percent openings, SumL = sum of the full-height segment lengths in the PSW
(see SDPWS 4.3.6.1.2). 

The second source of uplift that must be resisted is the uplift of the
bottom plate that provides resistance in those full-height segments that
don't have overturning restraint (i.e. those full-height segments away from
the loaded end of the PSW). The contribution of these segments is a function
of the dead load on each segment and the capacity of the bottom attachment
at that segment. In the 2005 SDPWS standard, we require the bottom plate to
be able to resist both uplift and shear (see SDPWS 4.3.6.4.1.1 &
4.3.6.4.2.1).  We realize the requirements in the SDPWS are excessive, but
until we finish a mechanics-based model to explain how to calculate the
proper bottom plate anchorage requirements, it is a requirement that the
design engineers wanted to include in the method. Dead load reductions
should not be taken twice. 

In the AF&PA Wood Frame Construction Manual (WFCM) for One- and Two-Family
Dwellings, 2001 Editions, it is only taken on the uplift loads that are
usually resisted by the anchor bolts or straps. The dead load reduction is
not taken on the hold downs that resist overturning. The WFCM doesn't
utilize the dead load to resist the shear wall overturning because it is
used to resist part of the uplift on the walls... There is no dead load left
at the bottom of the wall, just a reduced uplift load.

There is a Wood Design Focus (WDF) article that includes a design example
that suggests that dead load can be used to offset some of the
overturning/uplift requirements. The example states that dead load is
ignored to simplify example calculations.
http://www.awc.org/Publications/papers/WDFPLine.pdf

Hope this information helps.

Buddy

John "Buddy" Showalter, P.E. 
Director, Technical Media 
AF&PA/American Wood Council 
1111 19th Street, NW, Suite 800 
Washington, DC 20036 
P: 202-463-2769 
F: 202-463-2791 
http://www.awc.org 

The American Wood Council (AWC) is the wood products division of the
American Forest & Paper Association (AF&PA). AWC develops internationally
recognized standards for wood design and construction. Its efforts with
building codes and standards, engineering and research, and technology
transfer ensure proper application for engineered and traditional wood
products.

********************* 
The guidance provided herein is not a formal interpretation of any AF&PA
standard.  Interpretations of AF&PA standards are only available through a
formal process outlined in AF&PA's standards development procedures.

********************* 



From: Joe Grill [mailto:jgrill(--nospam--at)swiaz.com]=20
Sent: Thursday, May 18, 2006 4:40 PM
To: seaint
Subject: hold downs at perforated shear wall

When using the "perforated shear wall" method of design. The hold =
down force at the ends of the wall is calculated by dividing the =
overturning moment to the wall by an adjusted sum of the lengths of the =
individual lengths of full height segments. Is this hold down force =
required even if there is adequate dl to restrain overturning? And then =
how is the resisting moment calculated? Seems kind of like a dumb =
question, but the hold down force does not use the total length of the =
wall, but resisting moments would, as I see it anyway.


Can someone explain this to me?
Thanks,
Joe


Joseph R. Grill, P.E. (Structural)

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