Pin in double shear and bearing

Given data

$P = 60\ \text{kN}$, $d = 25\ \text{mm}$, $t = 12\ \text{mm}$, double shear (two planes), $\tau_u = 300\ \text{MPa}$.

 outer plate  <-- 30 kN ===[ pin ]
 centre plate      60 kN --->[ pin ]
 outer plate  <-- 30 kN ===[ pin ]
   (two shear planes in the pin)

(a) Shear and bearing stress

Area of pin in one plane:

$$A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(25)^2 = 490.9\ \text{mm}^2$$

The pin has two shear planes, each carrying $P/2$:

$$\tau = \frac{P}{2A} = \frac{60\times10^3}{2\times 490.9} = 61.1\ \text{MPa}$$

Bearing on the central plate (it carries the whole 60 kN):

$$\sigma_b = \frac{P}{d\,t} = \frac{60\times10^3}{25\times 12} = 200\ \text{MPa}$$

(b) Factor of safety

$$\text{FOS} = \frac{\tau_u}{\tau} = \frac{300}{61.1} = 4.91$$

(c) Allowable load for FOS = 3

$$\tau_{all} = \frac{300}{3} = 100\ \text{MPa}, \qquad P_{all} = \tau_{all}\,(2A) = 100\times 2\times 490.9 = 98.2\times10^3\ \text{N}$$

Answer: $\tau = 61.1$ MPa, $\sigma_b = 200$ MPa; FOS $= 4.91$; $P_{all} = 98.2$ kN (bearing and plate tension should also be checked).