Question

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.

Answer 1

Answer:

25 - $\sqrt[4]{26.66*10^{-8} }$  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2$\pi$f = 2 * $\pi$ * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  $\frac{J}{c_{2} } = \frac{T}{t_{max} }$  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

$\frac{J}{c_{2} }$ = $\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )$ =  $\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ]$

therefore; 0.025^4 - $c^{4} _{1}$ = 0.050 / $\pi$ (7.8 *10^-6)

$c^{4} _{1}$ = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

$c_{1} = \sqrt[4]{26.66 * 10^{-8} }$  =

THE TUBE THICKNESS

$c_{2} - c_{1}$ = 25 - $\sqrt[4]{26.66*10^{-8} }$  mm