A shear pin of diameter $d = 6.0\text{ mm}$ is installed in a PTO drive shaft at a radius of $R = 40\text{ mm}$ from the shaft centerline. If the pin material fails at a shear stress of $\tau = 250\text{ MPa}$, calculate the critical torque in $\text{N}\cdot\text{m}$ that causes the pin to shear. (Take $\pi = 3.1416$)
Answer Key:
282.74
Step-by-Step Solution & Explanation
Cross-sectional area of shear pin:
$A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (6)^2 = 9\pi \approx 28.274\text{ mm}^2$
Shear force required:
$F_s = \tau \times A = 250\text{ N/mm}^2 \times 28.274\text{ mm}^2 = 7068.58\text{ N}$
Failure torque:
$T = F_s \times R = 7068.58\text{ N} \times 0.040\text{ m} \approx 282.74\text{ N}\cdot\text{m}$