Question

A shaft is used to transmit a power of 10 kW. The shear yield stress of the material is 150 MPa and factor of safety is 2. The shaft rotates at 1440 revolutions per minute. The diameter of the shaft (in mm) based on static strength is \(\underline{\hspace{2cm}}\). [round off to two decimal places]

Hint:

For power-transmitting shafts, always compute torque from $P=2\pi NT/60$ before applying the torsion equation.

Answer 1

Correct Answer: 16 - 17

Given: Power: \[ P = 10\ \text{kW} = 10{,}000\ \text{W} \] Speed: \[ N = 1440\ \text{rpm} \] Torque transmitted: \[ P = 2\pi N T/60 \] \[ T = \frac{\text{P} \cdot 60}{2\pi N} = \frac{10{,}000 \times 60}{2\pi \times 1440} \approx 66.15\ \text{N} \cdot \text{m} \] Shear yield stress: \[ \tau_y = 150\ \text{MPa} \] Factor of safety = 2 → allowable shear stress: \[ \tau_{allow} = \frac{150}{2} = 75\ \text{MPa} \] Torsion formula for solid shaft: \[ T = \frac{\pi d^3}{16}\,\tau_{allow} \] Solving for \(d\): \[ d^3 = \frac{16T}{\pi\tau_{allow}} \] \[ d^3 = \frac{16(66.15\times10^3)}{3.1416 \times 75\times10^6} \] \[ d^3 = 4.49 \times 10^{-6} \] \[ d = (4.49 \times 10^{-6})^{1/3} \approx 0.0165\ \text{m} \] \[ d \approx 16.5\ \text{mm} \] Thus the shaft diameter lies in: \[ \boxed{16.00\text{ to }17.00\ \text{mm}} \]