Shaft calculation

Question

A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The shaft may be assumedto be
made of mild steel with an allowable shear stress of 42 MPa. Determine the diameter of the shaft,
neglecting the bending moment on the shaft.

Answer 1

Given : N = 200 r.p.m.; P = 20 kW = 20 × 10³ W; τ = 42 MPa = 42 N/mm²

Let d = Diameter of the shaft.

We know that torque transmitted by the shaft,

\(T = \frac{P \times 60}{2\pi N} = \frac{20 \times 10^3 \times 60} {2 \pi \times 200} = 955 \ N-m = 955 \times 10^3\ N-mm\)

We also know that torque transmitted by the shaft ( T ),

\(955 \times 10^3 = \frac \pi {16} \times \tau \times d^3 \)

\(= \frac \pi {16} \times 42 \times d^3 \)

\(= 8.25 \ d^3\)

∴ d³ = 955 × 10³ / 8.25 = 115 733 or d = 48.7 say 50 mm