Question

A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 35 Hz. Determine the maximum power that the shaft can transmit, knowing that G = 77.2 GPa, that the allowable shearing stress is 50 MPa, and that the angle of twist must not exceed 7.5°.

Answer 1

`P=58.3kW`

Step-by-step explanation:

Given data

Length L=2.5 m

Radius R=d/2=30/2 = 15 mm

Torque based on allowable stress

Allowable shear stress τ=50 Mpa

Allowable torque T=(π/2)τc³

`T=(\pi )/(2)(50*10^(6) )(0.015)^(3) \\ T=264.94N.m`

Torque based on allowable angle of twist

Allowable Angle of twist

Ф=7.5°

Ф=7.5×(π/180)=130.90×10⁻³ rad

Allowable torque

T=(GJФ)/L

T=(G(π/2)c⁴)Ф)/L

T=(πGc⁴Ф)/2

`T=(\pi (77.2*10^(9) )(0.015)^(4)(130.90*10^(-3) ) )/(2)\\ T=265.07N.m`

Maximum Power Transmitted

Maximum power transmitted is given by

`P=2\pi fT\\P=2\pi (35)(265.07)\\P=58262.386watt\\P=58.3kW`