Question

A solid circular shaft of length 4 m is to transmit 3.5 MW at 200 rpm. If permissible shear stress is 50 MPa, the radius of the shaft is:

1. 1.286 mm
2. 12.86 mm
3. 0.1286 mm
4. 128.6 mm

Answer 1

Correct Answer - Option 4 : 128.6 mm

Concept:

Torque transmitted by the shaft

\(P = \frac{{2 \times \pi \times N \times T}}{{60}}\;\)

Also, shear stress is given by:

\(\tau = \frac{{16 \times T}}{{\pi \times {d^3}}}\;\)

P = power, N = speed, T = torque, d = diameter of shaft, τ = shear stress

Calculation:

Given:

P = 3.5 MW = 3.5 × 10⁶ W, N = 200 rpm, τ = 50 MPa = 50 × 10⁶ Pa

We have,

\(P = \frac{{2 \times \pi \times N \times T}}{{60}} \Rightarrow T = \frac{{3.5 \times {{10}^6} \times 60}}{{2 \times \pi \times 200}} = 167.112 \times {10^3}\;N/m\;\)

Also, shear stress is:

\(\tau = \frac{{16 \times T}}{{\pi \times {d^3}}}\;\)

\(50 \times {10^6} = \frac{{16 \times 167.112 \times {{10}^3}}}{{\pi \times {d^3}}}\; \Rightarrow {d^3} = \frac{{16 \times 167.112 \times {{10}^3}}}{{50 \times {{10}^6}}}\)

d = 0.2572 m

The radius of the shaft 

\(r = \frac{{0.2572}}{2} = 0.1286\;m = 128.6\;mm\)