Wood Shear Wall Stiffness

Last updated December 4, 2023
By Ian Story

Reverse Engineering the NDS Deflection Formula

The 3-term formula for shear wall deflection in the NDS standard is: (equation 4.3-1)

    \[\delta = \frac{8vh^3}{EAb}+\frac{vh}{1000G_a}+\frac{h\Delta_a}{b}\]

Where:
\delta =\text{deflection, in}
v =\text{unit shear, lb/ft}
h=\text{height, ft}
b=\text{length, ft}
E=\text{end post modulus of elasticity, psi}
A=\text{end post cross-section area, in}^2
G_a=\text{apparent shear wall panel stiffness, kip/in}
\Delta_a=\text{vertical elongation of wall hold-down system at applied load, in}

There are several unit conversions baked into the formula, and the input shear is given in units of pounds per foot. To convert this to stiffness in units of pounds per inch, we need to make several adjustments:

First, let’s convert unit shear to total shear:

    \[v = \frac{V}{b}\]

Where:
V =\text{applied shear, lb}

And let’s replace \Delta_a with a term that’s easier to look up. Let’s call \Delta the hold-down elongation at the maximum rated load, and define:

    \[k_a = \frac{T_{a,max}}{\Delta}\]

    \[\Delta_a = \frac{T}{k_a}\]

Where:
k_a =\text{hold-down stiffness, lb/in}
T_{a,max} =\text{maximum rated load for hold-down, lb}
T =\text{applied tension force on hold-down, lb}
\Delta =\text{hold-down elongation at maximum rated load, in}

The maximum applied hold-down force (neglecting the effect of in-set hold-downs on shortening the moment arm) is:

    \[T = \frac{Vh}{b}\]

So:

    \[\Delta_a = \frac{Vh}{k_ab}\]

This gives:

    \[\delta = \frac{8Vh^3}{EAb^2}+\frac{Vh}{1000G_ab}+\frac{Vh^2}{k_ab^2}\]

Next, let’s define the overall shear wall stiffness as:

    \[k = \frac{V}{\delta}\]

    \[\delta = \frac{V}{k}\]

Where:
k =\text{shear wall stiffness, lb/in}

Plugging into the formula, this allows us to cancel out V:

    \[\frac{1}{k} = \frac{8h^3}{EAb^2}+\frac{h}{1000G_ab}+\frac{h^2}{k_ab^2}\]

Solving for k:

    \[k = \frac{1}{\cfrac{8h^3}{EAb^2}+\cfrac{h}{1000G_ab}+\cfrac{h^2}{k_ab^2}}\]

And finally, reversing all built-in unit conversions:

    \[k = \frac{1}{\cfrac{8h^3(1\text{ ft}/12\text{ in})^3}{EAb^2(1\text{ ft}/12\text{ in})^2}+\cfrac{h(1\text{ ft}/12\text{ in})}{1000G_a(1\text{ kip}/1000\text{ lb})b(1\text{ ft}/12\text{ in})}+\cfrac{h^2(1\text{ ft}/12\text{ in})^2}{k_ab^2(1\text{ ft}/12\text{ in})^2}}\]

    \[k = \frac{1}{\cfrac{2h^3}{3EAb^2}+\cfrac{h}{G_ab}+\cfrac{h^2}{k_ab^2}}\]