Solution to Problem 129 Bearing Stress

$P = 10.82 \, \text{kips}$
 

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