Question

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
1. 29.6 rpm
2. 3557.4 rpm
3. 1778.7 rpm
4. 59.2 rpm

Answer 1

Correct Answer - Option 3 : 1778.7 rpm

Concept:

The shear stress for the hollow shaft is given as,

\(τ =\frac{{16\;T}}{{\pi \;d_o^3\left( {1 - {k^4}} \right)}}\) and 

Power transmitted by shaft, P = Tω 

where, k = \(\frac{{{d_i}}}{{{d_0}}} \), T = torque to be transmitted, d_o = outer diameter of tube, ω = angular velocity of shaft 

Calculation:

Given:

k = \(\frac{{{d_i}}}{{{d_0}}} = \frac{{30}}{{40}}\), d_o = 40 mm, τ = 50 MPa, P = 80 kW

Therefore, the Torque transmitted is

50 = \(\frac{{16\;T}}{{\pi \;{{40}^3}\left( {1 - {{0.75}^4}} \right)}}\)

T = 429514.62 N-mm = 429.51 N-m

And Power is

P = \(\frac{{2\pi NT}}{{60}}\)

80 × \({10^3}\) = \(\frac{{2\pi N\;\left( {429.51} \right)}}{{60}}\)

N = 1778.7 rpm