Shaft calculation

Question

Find the diameter of a solid steel shaft to transmit 20 kW at 200 r.p.m. Theultimate shear
stress for the steel may be taken as 360 MPa and a factor of safety as 8.If a hollow shaft is to be used in place of the solid shaft, find the inside and outside diameter when the ratio of inside to
outside diameters is 0.5.

Answer 1

Given: P = 20 kW = 20 × 10³ W ; N = 200 r.p.m.; τ_u = 360 MPa = 360 N/mm² ; F.S. = 8 ; k = d_i / d_o = 0.5

We know that the allowable shear stress,

\(\tau = \frac{\tau _u}{F.S.} = \frac{360}8 = 45 \ N/mm^2\)

Diameter of the solid shaft

Let d = Diameter of the solid 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 solid shaft (T),

\(955 \times 10^3 = \frac {\pi }{16} \times \tau \times d^3 = \frac \pi {16} \times 45 \times d^3 = 8.84 d^3\)

∴ d³ = 955 × 10³ / 8.84 = 108 032 or d = 47.6 say 50 mm

Diameter of hollow shaft

Let

d_i = Inside diameter, and

d_o = Outside diameter.

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

\(955 \times 10^3 = \frac \pi {16} \times \tau (d_0)^3 (1 - k^4)\)

\(= \frac \pi {16} \times 45 (d_0)^3[ 1- (0.5)^4] = 8.3 (d_0)^3\)

∴ (d_o)³ = 955 × 10³ / 8.3 = 115 060 or do = 48.6 say 50 mm

and d_i = 0.5 d_o = 0.5 × 50 = 25 mm