Polar Moment of Inertia of Fastener Group

Last updated May 31, 2022
By Ian Story

$I_i_j = A_i_jd_i_j^2$

$I = \sum_{i,j}A_i_jd_i_j^2$

Assuming all the fasteners are the same size, this simplifies to

$I = A\sum_{i,j}d_i_j^2$

If all the fasteners are the same size, we can factor out the A and replace I with a simplified expression

$I_a = \frac{I}{A} = \sum_{i,j}d_i_j^2$

The shear reaction at an individual fastener is

$V_i_j = \frac{M*r_i_j}{I}A_i_j$

Assuming all the fasteners are the same size, this simplifies to

$V = \frac{M*r}{I_aA}A = \frac{M*r}{I_a}$

For fasteners arranged in a rectangular grid, the solution is

$I_a = \frac{cr}{12}[s^2(r^2-1)+t^2(c^2-1)]$

c = number of columns
r = number of rows
s = spacing between columns
t = spacing between rows

(derivation to come)