Question

The maximum shear stress induced in a solid circular shaft of diameter 15 cm, when the shaft transmits 150 kW power at 180 rpm, will be
1. 16 N/mm²
2. 14 N/mm²
3. 12 N/mm²
4. 10 N/mm²

Answer 1

Correct Answer - Option 3 : 12 N/mm²

Concept:

Torsional equation is given by,

\(\frac{{{τ _{max}}}}{R} = \frac{{Gθ }}{L} = \frac{T}{J}\)

Power developed in the shaft will be,

P = T × ω 

ω = \(\frac{{2\;\pi \;N}}{{60}}\;\)

Where, τ_max = Maximum shear stress, R = Radius of shaft, G = Modulus of rigidity, θ = Angle of twist, L = length of shaft, J = Polar moment of inertia, P = power developed, T = Applied Torsion or torque & N = Speed of rotation of shaft (rpm)

Calculation:

Given, P = 150 kw, μ = 180 rpm, d = 15 cm and \(R = \frac{d}{2}\)

Now, \(T = \frac{\rho }{ω } = \frac{\rho }{{\frac{{2\pi N}}{{60}}}} = \frac{{190 × {{10}^3} × 60}}{{2\pi × 180}}\)

\(T = \left( {\frac{{25 × {{10}^3}}}{\pi }} \right)\)

Using Torsional formula,

\(\frac{{{τ _{max}}}}{R} = \frac{T}{J}\)

J \( = \frac{\pi }{{32}}{d^4}\)

τ_max = \(\frac{{T\frac{d}{2}}}{{\frac{\pi }{{32}}{d^4}}} = \frac{{16T}}{{\pi {d^3}}}\)

τmax \( = \frac{{16 × 25 × {{10}^3}}}{{\left( \pi \right)\left[ {\pi × {{\left( {15} \right)}^3}} \right]}} = 12.02 × {10^6}\;\;N/{m^2}\)

∴ Maximum shear stress, τ_max = 12 N/mm²