Topic: Machine Design
A solid steel shaft transmits 40kW of power at a speed of 75/pi Hz. The internal torque needed in the shaft is
- A. 812/3 Nm
- B. 800/3 Nm
- C. 541/2 Nm
- D. 400/3 Nm
Answer: B. 800/3 Nm
Explanation: Power transmitted by a shaft is given by P = T * omega, where omega is the angular velocity in rad/s. Given frequency f = 75/pi Hz, omega = 2 * pi * f = 2 * pi * (75/pi) = 150 rad/s. Given power P = 40 kW = 40,000 W. The torque T is calculated as T = P / omega = 40,000 / 150 = 4000 / 15 = 800/3 Nm.