Tensile force on the bolt:
$P = \sigma A = 18 [ \, \frac{1}{4} \pi (\frac{7}{8})^2 \, ]$
$P = 10.82 \, \text{kips}$
Shearing stress in the head of the bolt:
$\tau = \dfrac{P}{A} = \dfrac{10.82}{\pi (\frac{7}{8})(\frac{1}{2})}$
$\tau = 7.872 \, \text{ ksi}$ answer
Shearing stress in the threads:
$\tau = \dfrac{P}{A} = \dfrac{10.82}{\pi (0.731)(\frac{5}{8})}$
$\tau = 7.538 \, \text{ ksi}$ answer
Outside diameter of washer:
$P = \sigma_b \, A_b$
$10.82(1000) = 800 \bigl\{ \, \frac{1}{4} \pi \, [ \, d^2 - (\frac{9}{8})^2 \, ] \, \bigr\}$
$d = 4.3 \, \text{inch}$ answer
