**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*

## Comments

## Good evening, good sir; i was

Good evening, good sir; i was going to ask that where did u get the 20t? Does the diameter of the rivet have that initial thickness of the two plates? Thats y u use it 20t?.