Subtopic: Overload safety devices used in farm machinery
Question Statement:
A mild steel shear pin of diameter $6\text{ mm}$ is used in a PTO driveline at a radius of $40\text{ mm}$ from the shaft axis. If the ultimate shear strength of the pin material is $320\text{ N/mm}^2$, calculate the torque at which the pin shears in $\text{N}\cdot\text{m}$ (round off to 1 decimal place).
Answer Key:
361.9
Step-by-Step Solution & Explanation
Cross-sectional area of shear pin:
$A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (6)^2 = 28.274\text{ mm}^2$
Shearing force:
$F_s = \tau_u \times A = 320\text{ N/mm}^2 \times 28.274\text{ mm}^2 = 9047.8\text{ N}$
Radius $r = 40\text{ mm} = 0.04\text{ m}$.
Torque required to shear the pin:
$T = F_s \times r = 9047.8\text{ N} \times 0.04\text{ m} = 361.9\text{ N}\cdot\text{m}$