Middle bar is 0.045 inch shorter between holes than outer bars.

$\Sigma F_H = 0$

$P_{mid} = 2P_{outer}$

$\delta_{outer} + \delta_{mid} = 0.045$

$\left( \dfrac{PL}{AE} \right)_{outer} + \left( \dfrac{PL}{AE} \right)_{mid} = 0.045$

$\dfrac{P_{outer} (30 \times 12)}{[ \,1.0(4.0) \, ] E} + \dfrac{P_{mid} (30 \times 12 - 0.045)}{[ \,1.0(4.0) \, ] E} = 0.045$

$360P_{outer} + 359.955P_{mid} = 0.18E$

$360P_{outer} + 359.955(2P_{outer}) = 0.18E$

(For steel: E = 29 × 10^{6} psi)

$1079.91P_{outer} = 0.18(29 \times 10^6)$

$P_{outer} = 4833.74 \, \text{lb}$

$P_{mid} = 2(4833.74)$

$P_{mid} = 9667.48 \, \text{lb}$

Use shear force $V = P_{mid}$

Shearing stress of drip pins (double shear):

$\tau = \dfrac{V}{A} = \dfrac{9667.48}{2 \, [ \, \frac{1}{4} \pi \left( \frac{7}{8} \right)^2 \, ]}$

$\tau = 8038.54 \, \text{ psi}$ *answer*