Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A rectangular sunk key of width $w = 12\text{ mm}$ and height $h = 8\text{ mm}$ is fitted to a shaft of diameter $d = 40\text{ mm}$. The key transmits a torque of $320\text{ N}\cdot\text{m}$. If the permissible shear stress for key material is $\tau = 60\text{ MPa}$, calculate the minimum required length of the key in $\text{mm}$.
Answer Key:
22.22
Step-by-Step Solution & Explanation
Tangential shear force acting at the shaft surface:
$F = \frac{T}{d/2} = \frac{320}{0.020} = 16000\text{ N}$
Shear area of the key is $A_s = w \times l = 12 \times l\text{ mm}^2$.
Shear stress relation:
$\tau = \frac{F}{w \times l} \le 60\text{ N/mm}^2$
$l = \frac{16000}{12 \times 60} = \frac{16000}{720} \approx 22.22\text{ mm}$