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Re: Plywood shear wall, nailing, deflection

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Plywood shear wall, nailing, deflection
• From: Szuchuan Chang <szchang(--nospam--at)pacbell.net>
• Date: Thu, 16 Jul 1998 08:29:10 -0700

```Thanks

Sam

Mlcse(--nospam--at)aol.com wrote:

> In a message dated 98-07-15 11:26:05 EDT, you write:
>
> << Assuming the shear to the wall is 2.79 kips.Uplift does not exceed 4.5
>  kips and so I am planning to use one Simpson Strong-tie's  PHD5 holddown
>  at each end.  The header is a  5-1/8"x12 glulam beam.  To minimize the
>  drift I am planning to using with 8d at 3" BN o.c.
>
>  Here are the values I used to calculated the deflectoin:
>   max shear = 443 plf
>   height = 8 ft
>   width wall = 6.4 ft
>   chord member area = 19.25 in^2
>   da = 0.047 in
>   en = 0.023 in
>   effective thickness of plywood = 0.535 in
>   E, chord member = 1700000  psi
>   G, plywood = 90000 psi
>
>  Here is the result:
>   deflection = 8*v*h^3/E*A*b + v*h/G*t + 0.75*h*en + h/b*da
>   deflection from bending = 0.00866 in.   1.46%
>   deflection from shear = 0.0736 in.   12.4%
>   deflection from nail slip = 0.453 in.   76.2%   <<-------
>   deflection from holddown slip = 0.0587 in.   9.9%
>   sum of deflections = 0.594 inches
>
>  This deflection is still exceeding the drift limit.  But I used Rw=6 not
>  8 to calculate the base shear.
>
>  I have two questions:
>
>  1) Why is the nail slip accounts about 3/4 of the total deflection?
>  2) Do I need to require pre-drill the nail holes (4xs) to prevent the
>  nail splitting the 4xs?
>
>  Thanks
>
>  Sam Chang, SE
>  Cupertino, CA
>   >>
>
> Sam,
>
> Double check your math;  I get 0.75*h*en = 0.75 (8)(.023) = 0.138  <<< 0.453
> Revised drift = 0.2789" << than .005h = .005*8*12 = 0.48 inches.
>
> You may want to use 10d commons since they have less slip than the 8d and I
> would not worry to much about splitting the 4x6 posts.
>
> In your drift calculation, remember that the the equation is for a statically
> applied load.  Some sources recommend that you use a 1.25 multiplier to
> account for the approximate cyclic deflection.
>
> Michael Cochran
>

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