Topic: Strength of Materials
A tubular shaft, having an inner diameter of 30 mm and an outer diameter of 40 mm, is to be used to transmit 80 kW of power. The speed of rotation of the shaft so that the shear stress will not exceed 50 MPa is
- A. 29.6 rpm
- B. 3557.4 rpm
- C. 1778.7 rpm
- D. 59.2 rpm
Answer: C. 1778.7 rpm
Explanation: Power P = (2 * pi * N * T) / 60, so Torque T = (P * 60) / (2 * pi * N). For a hollow shaft, maximum shear stress tau = 16 * T / (pi * do³ * (1 - k^4)), where k is the ratio di/do = 30/40 = 0.75. Rearranging to solve for N: N = (16 * 80000 * 60) / (50e6 * pi² * 0.04^3 * (1 - 0.75^4)). After computation, N calculates to roughly 1778.7 rpm.