(a) Maximum thickness of plate:
Based on puncher strength:
$P = \sigma A$
$P = 50 \, [ \, \frac{1}{4} \pi (2.5^2) \, ]$
$P = 78.125 \pi \, \text{kips}$ → Equivalent shear force of the plate
Based on shear strength of plate:
$V = \tau A$ → $V = P$
$78.125\pi = 40 \, [ \, \pi(2.5t) \, ]$
$t = 0.781 \, \text{inch}$ answer
(b) Diameter of smallest hole:
Based on compression of puncher:
$P = \sigma A$
$P = 50(\frac{1}{4} \pi d^2)$
$P = 12.5 \pi d^2$ → Equivalent shear force for plate
Based on shearing of plate:
$V = \tau A$ → $V = P$
$12.5 \pi d^2 = 40 [ \, \pi d(0.25) \, ]$
$d = 0.8 \, \text{in}$ answer
