Diaphragm Deflection

Last updated July 27, 2023
By Ian Story

Reverse Engineering the SDPWS Diaphragm Deflection Equation

The equation for deflection of a plywood diaphragm given in SDPWS (eq 4.2-1):

\delta_d_i_a = \frac{5vL^3}{8EAW} + \frac{0.25vL}{1000G_a} + \frac{\Sigma x\Delta_c}{2W}

This equation gives the maximum deflection (at midspan), assuming a uniformly distributed load along the diaphragm length. The equation also bakes in several unit conversions. While useful as a final check, this equation is less helpful for design situations which require assessing the deflection or stiffness at an arbitrary point.

Behind the scenes, this equation is in fact just a specific solution of the timoshenko beam equation. Let’s break it down.

First, the derivation of this equation assumes a simply supported beam with a uniform distributed load. The deflection formula for the maximum deflection of a timoshenko beam under this loading is:

\Delta_m_a_x = \frac{5wL^4}{384EI}+\frac{wL^2}{8k_sA_s_h_e_a_rG}

Loads in the SPDWS formula are input as unit shears (v). To convert from uniform distributed load (w) to unit shears, we will apply the following:

Reaction at supports, R = \frac{wL}{2}

Max unit shear v = \frac{R}{W} = \frac{wL}{2W}

w = \frac{2vW}{L}

Substituting this into our timoshenko beam formula gives

\Delta_m_a_x = \frac{5vWL^3}{192EI}+\frac{vWL}{4k_sA_s_h_e_a_rG}

Next, the SDPWS formula incorporates a calculation for the moment of inertia (I) directly into the formula. Let’s make this explicit. A plywood diaphragm consists of chords at either end separated by a plywood web. For this derivation, the chords are assumed to take all of the bending moment, and the plywood web is assumed to take all of the shear. Therefore the plywood web’s contribution to the moment of inertia of the diaphragm “beam” is assumed to be zero. Using the parallel axis theorem, and neglecting the moment of inertia of the chords (these are trivially small compared to their distance from the neutral axis):

I = \Sigma Ad^2 = A(\frac{W}{2})^2 + A(\frac{W}{2})^2 = \frac{AW^2}{2}

Substituting into the beam formula gives (note that the W in the numerator cancels out the square of the W2 in the denominator):

\Delta_m_a_x = \frac{5vL^3}{96EAW}+\frac{vWL}{4k_sA_s_h_e_a_rG}

For the second term, SDPWS provides tabulated values for the shear modulus (labeled G_a) which include the effect of nail slip and the shear coefficient (k_s). Because the plywood is assumed to resist all shear, the cross-sectional shear area A_s_h_e_a_r = tW. However, the tabulated G_a values also include the effect of shear panel thickness, so A_s_h_e_a_r simplifies to W when used with G_a. This simplifies our beam formula as follows (note that the W in the numerator cancels out the W in the denominator):

\Delta_m_a_x = \frac{5vL^3}{96EAW}+\frac{vL}{4G_a}

Finally, the SDPWS formula bakes in several unit conversions. L and W are input in feet, v is input in plf, and the deflection is output in inches. Making those unit conversions explicit gives:

\Delta_m_a_x = \frac{5(v/12)(L*12)^3}{96EA(W*12)}+\frac{(v/12)(L*12)}{4G_a}

\Delta_m_a_x = \frac{5vL^3}{8EAW}+\frac{vL}{4G_a}

Which exactly matches the first 2 terms of the SDPWS equation.

The final term in the SDPWS equation adds an additional component of deflection based on nail slip in the chord splices. Applying this term precisely requires knowing exactly where the chord splices will occur, which is impractical in practice. However, a simple substitution will make this idea easier to use. The effect of splices is to increase the stretch in the chords for a given force. If we abandon the idea of identifying specific splice points along the chord and instead average this effect along the entire chord length, this has the effect of reducing the effective axial stiffness of the chords (axial stiffness defined as k_a = \frac{EA}{L}. If we can calculate the axial stiffness of the chords based solely on nail slip, we can use this to calculate an “effective modulus of elasticity,” E_e (keeping A and L constant) to use in the bending term of the beam equation and eliminate the third term.

The first step in calculating the axial stiffness of the chords is to calculate the stiffness of the fasteners. This is called the load slip modulus, and is defined in NDS for nailed wood to wood connections as follows (term definitions for eq. 11.3-1):

\gamma = 180,000(D)^1^.^5 lb/in/nail, where D is nail diameter in inches

We can then multiply this by the number of nails per splice to calculate the stiffness of a single splice in lb/in. If there are multiple splices in a chord, these act as springs in series: doubling the number of splices doubles the extension for a given load (or, equivalently, halves the stiffness). Therefore, we can calculate the overall stiffness of the chord by dividing the single-splice stiffness by the number of splices. If we calculate the number of splices as the total length (L) divided by the length between splices (s), less 1 (because there will always be one less splice than lengths of lumber), this gives a total chord stiffness of:

k_c_h_o_r_d = \frac{{\gamma}n}{\frac{L}{s} - 1}

However, before we apply this, one more adjustment is required. Because the chord stresses are greater near the center of the diaphragm beam, where the moment is greatest, splices located near the center will have a greater effect on the beam deflection than splices located near the edges. Because chords are generally constructed from standardized lumber sizes, splices are less likely near the ends of a wall. Assuming the deflection is caused by a point load, the difference in deflection between a splice at the center and a splice spread out over the whole length of the chord is a factor of 3. In reality, the actual difference will be somewhat less than this, but this provides a conservative estimate of chord slip and agrees reasonably well with the SDPWS nail slip term for typical diaphragm lengths.

Putting all the pieces together gives the following axial chord stiffness based solely on nail slip:

k_a_,_s_l_i_p = \frac{{\gamma}n}{3(\frac{L}{s} - 1)} (lb/in)

Treating A and L as constants, we can convert this stiffness to an equivalent modulus of elasticity:

E_s_l_i_p = \frac{k_a_,_s_l_i_pL}{A} = \frac{{\gamma}nL}{3A(\frac{L}{s} - 1)}

Combining moduli of elasticity works like springs in series, so the combined effective modulus of elasticity is

E_e = \frac{1}{\frac{1}{E} + \frac{1}{E_s_l_i_p}} = \frac{1}{\frac{1}{E} + \frac{3A(\frac{L}{s} - 1)}{{\gamma}nL}}

Plugging this into the SDPWS formula gives the following simplification:

\Delta_m_a_x = \frac{5vL^3}{8E_eAW}+\frac{vL}{4G_a}

\Delta_m_a_x = \frac{5vL^3}{8W}(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL})+\frac{vL}{4G_a}

Deriving General Formulas

As the above reverse engineering exercise shows, plywood diaphragms can be modeled as timoshenko beams for purposes of analyzing deflection. Using this knowledge, and the same simplification steps used by SDPWS, we can construct more generally-applicable formulas to handle alternative loadings (such as point loads) or to find deflections at points other than the midpoint.

Following are several such useful formulas:

Note: these formulas do not include unit conversions. All lengths should be input in inches, and G_a should be input in lb/in (multiply the table values by 1,000 before inputting). Also note that loads are input as distributed loads (w, lb/in) or point loads (P, lb), rather than unit shears.

Uniformly Distributed Load

\Delta_m_a_x (at center) = \frac{5wL^4}{192W^2}(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL})+\frac{wL^2}{8WG_a}

\Delta_x = \frac{wx}{12W^2}(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL})(L^3-2Lx^2+x^3)+\frac{wx}{2WG_a}(L-x)

Point Load at x

\Delta_m_a_x (does not necessarily occur at point of load application) = TBD

\Delta_a (at point of load) = \frac{2Pa^2b^2}{3LW^2}\left(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL}\right)+\frac{Pab}{LWG_a}

\Delta_x (when x < a) = \frac{Pbx}{3LW^2}\left(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL}\right)(L^2-b^2-x^2)+\frac{Pbx}{LWG_a}

\Delta_x (when x > a) = \frac{Pa(L-x)}{3LW^2}\left(\frac{1}{EA} + \frac{3(\frac{L}{s} - 1)}{{\gamma}nL}\right)(2Lx-a^2-x^2)+\frac{Pa(L-x)}{LWG_a}