Shear Stress Question (Rocker Arm & pin diameter)

[jojosg]

Homework Statement

The illustrated rocker arm is subjected to the action of loads F1 and F2. The given loads are F1=50kN and F2,=35.4kN. The allowable shear stress τ=100 MPa ,and the allowable bearing stress σ=240 MPa. Determine the diameter d of the pin B.

Relevant Equations

τ= V/A, σ = F/t*d

Hi, just need somebody to check my homework solution. Not entirely sure of my answer. My teacher used Mechanics of Materials by RC Hibbler, but I could not find an example question to solve this homework question.

Equilibrium at pin B

Take moments about B:
$F_1 \cdot 40 = F_2 \cdot (80\sin45^\circ)$
$50 \cdot 40 = 35.4 \cdot 56.57 = 2000\ \text{kN·mm}$

Force equilibrium:
$\sum F_x = 0:\quad R_{Bx} + 35.4\cos45^\circ - 50 = 0 \quad\Rightarrow\quad R_{Bx} = 25\ \text{kN}$
$\sum F_y = 0:\quad R_{By} - 35.4\sin45^\circ = 0 \quad\Rightarrow\quad R_{By} = 25\ \text{kN}$
$R_B = \sqrt{25^2+25^2}=25\sqrt{2}=35.36\ \text{kN}$

Pin design (double shear) – two outer plates t=6 mm, center plate t=10 mm

Shear:
$\tau = \frac{R_B/2}{\pi d^2/4} = \frac{2R_B}{\pi d^2} \le 100\ \text{MPa} \quad\Rightarrow\quad d \ge \sqrt{\frac{2\times35360}{\pi\times100}} = 15.0\ \text{mm}$

Bearing – center plate:
$\sigma_b = \frac{R_B}{d\cdot10} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2400} = 14.73\ \text{mm}$

Bearing – outer plates:
$\sigma_b = \frac{R_B/2}{d\cdot6} = \frac{R_B}{12d} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2880} = 12.28\ \text{mm}$

Result shear governs
$\boxed{d = 15\ \text{mm}}$