Question

A key of width and height of 6 mm each is used to fix a gear on a shaft of 20 mm diameter. The shaft is used to transmit 10 kW power at 600 rpm to the gear. Permissible shear stress in the key is 80 N/mm\(^2\), while compressive stress in the key is neglected. The minimum length of the key, in mm, is .................

Hint:

The minimum length of the key can be calculated by balancing the torque transmitted with the shear stress and the key's geometry.

Answer 1

- The power transmitted by the shaft is given by: \[ P = \frac{T \cdot \omega}{1000} \] where \( T \) is the torque and \( \omega \) is the angular velocity.
- The torque can be found using the formula \( T = \frac{P \cdot 1000}{\omega} \), where \( P = 10 \, \text{kW} \) and \( \omega = 2\pi \times 600 / 60 \).
- The torque \( T \) is then used to calculate the force on the key. The force is related to the shear stress by: \[ \tau = \frac{F}{A} \] where \( \tau \) is the shear stress and \( A \) is the cross-sectional area of the key.
- The minimum length of the key is found from the relationship between shear stress, torque, and dimensions of the key. The final result for the length of the key is 32.00 mm to 34.00 mm.