Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A spring-loaded cultivator tine safety release mechanism uses a helical compression spring made of oil-tempered spring wire of diameter $d = 10\text{ mm}$. The mean coil diameter of the spring is $D = 60\text{ mm}$. The spring index is $C = D/d = 6$. When tripped by an underground obstacle, the spring experiences a peak compressive load of $F = 2000\text{ N}$. Using Wahl's stress concentration factor $K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}$, the maximum torsional shear stress in the spring wire in $\text{MPa}$ is ________ (round off to 1 decimal place).
Answer Key:
382.6
Step-by-Step Solution & Explanation
```
Helical Spring Torsion Model:
Mean Dia D = 60 mm, Wire Dia d = 10 mm
Spring Index C = D/d = 6
```
**Step-by-step Calculation:**
1. Wahl factor ($K_w$):
$K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C} = \frac{4(6) - 1}{4(6) - 4} + \frac{0.615}{6} = \frac{23}{20} + 0.1025 = 1.15 + 0.1025 = 1.2525$
2. Maximum shear stress formula with Wahl correction:
$\tau_{\text{max}} = K_w \times \frac{8 F D}{\pi d^3}$
$F = 2000\text{ N}, \quad D = 0.060\text{ m}, \quad d = 0.010\text{ m}$
$\tau_{\text{max}} = 1.2525 \times \frac{8 \times 2000 \times 0.060}{\pi \times (0.010)^3} = 1.2525 \times \frac{960}{\pi \times 10^{-6}} = 1.2525 \times 3.05577 \times 10^8\text{ Pa}$
$\tau_{\text{max}} = 3.82635 \times 10^8\text{ Pa} = 382.64\text{ MPa} \approx 382.6\text{ MPa}$