Without the loads P:
$\tau = \dfrac{T\rho}{J}$
Where:
$T = 14(10) = 140 \, \text{kip}\cdot\text{in}$
$\rho = \sqrt{13} \, \text{in}$
$J = \Sigma A\rho^2 = \frac{1}{4}\pi (\frac{7}{8})^2 \, [ \, 4(\sqrt{13})^2 + 2(2)^2 \, ] = 36.08 \, \text{in}^4$
$\tau_{maximum} = \dfrac{140\sqrt{13}}{36.08} = 14.0 \, \text{ ksi}$ answer
$\tau_{minimum} = \dfrac{140(2)}{36.08} = 7.76 \, \text{ ksi}$ answer
Note:
The τmaximum is carried by the corner rivets (4 rivets in all) while the τminimum is carried by the middle two rivets.
With the loads P, two cases will arise:
1st case (P 14 kips)
$T = 10(14) - 6P = (140 - 6P) \, \text{kip}\cdot\text{in}$
$\tau = \dfrac{T \rho}{J}$
$8000 = \dfrac{(140 - 6P)(1000)(\sqrt{13})}{36.08}$
$80.05 = 140 - 6P$
$P = 10.0 \, \text{ kips}$ answer
Note:
Without the load P = 10 kips, the shear stress at corner rivets is 14 ksi (see τmaximum), which is way above 8 ksi. Thus, this minimum value of P is necessary to prevent stressing the corner rivets beyond 8 ksi.
2nd case (P > 14 kips)
$T = 6P - 10(14) = (6P - 140) \, \text{kip}\cdot\text{in}$
$\tau = \dfrac{T \rho}{J}$
$8000 = \dfrac{(6P - 140)(1000)(\sqrt{13})}{36.08}$
$80.05 = 6P - 140$
$P = 36.68 \, \text{kips}$ answer
