Subtopic: Overload safety devices used in farm machinery
Question Statement:
A rotavator input PTO driveline is protected against excessive torque by a single shear pin located at a radial distance of $r = 60\text{ mm}$ from the shaft centerline. The shear pin is made of mild steel with an ultimate shear strength of $\tau_u = 300\text{ MPa}$. If the driveline is designed to shear when the transmitted torque reaches $T_{\text{overload}} = 1200\text{ N}\cdot\text{m}$, the required diameter of the shear pin in $\text{mm}$ is ________ (round off to 2 decimal places).
Answer Key:
9.21
Step-by-Step Solution & Explanation
```
[Shaft Center] =====( r = 60 mm )=====> [Shear Pin (d)]
F_shear
```
**Step-by-step Calculation:**
1. Tangential shearing force at the pin location:
$F_s = \frac{T}{r} = \frac{1200\text{ N}\cdot\text{m}}{0.060\text{ m}} = 20000\text{ N}$
2. The pin is under single shear across its cross-sectional area $A = \frac{\pi}{4} d^2$:
$\tau_u = \frac{F_s}{A} = \frac{20000}{\frac{\pi}{4} d^2}$
3. Solving for diameter $d$:
$d^2 = \frac{4 \times 20000}{\pi \times \tau_u} = \frac{80000}{\pi \times (300 \times 10^6\text{ N/m}^2)} = \frac{80000}{9.42478 \times 10^8} = 8.488 \times 10^{-5}\text{ m}^2$
$d = \sqrt{8.488 \times 10^{-5}} = 9.213 \times 10^{-3}\text{ m} = 9.21\text{ mm}$
Range: 9.00 to 9.40 mm.