Solution to Problem 125 Bearing Stress

Part (a):
From shearing of rivet:
$P = \tau A_{rivets}$

$P = 60 [ \, \frac{1}{4} \pi (20^2) \, ]$

$P = 6000\pi \, \text{N}$
 

From bearing of plate material:
$P = \sigma_b A_b$

$6000\pi = 120(20t)$

$t = 7.85 \, \text{mm}$           answer
 

Part (b): Largest average tensile stress in the plate:
$P = \sigma A$

$6000\pi = \sigma [ \, 7.85(110 - 20) \, ]$

$\sigma = 26.67 \, \text{MPa}$           answer