Shaft desgn calculation

Question

QN 5. A shaft made of mild steel is required to transmit 100 kW at 300 r.p.m. Thesupported length
of the shaft is 3 metres. It carries two pulleys each weighing 1500 N supported at a distance of 1 metre from the ends respectively. Assuming the safe value of stress, determine the
diameter of the shaft.

Answer 1

Given : P = 100 kW = 100 × 10³ W ; N = 300 r.p.m. ; L = 3 m ; W = 1500 N

We know that the torque transmitted by the shaft,

\(T = \frac{P \times 60}{2\pi N} = \frac{100 \times 10^3 \times 60}{2 \pi \times 300} = 3183 \ N-m\)

The shaft carrying the two pulleys is like a simply supported beam as shown in figure. The reaction at each support will be 1500 N, i.e.

R_A = R_B = 1500 N

A little consideration will show that the maximum bending moment lies at each pulley i.e. at C and D.

∴ Maximum bending moment

M = 1500 × 1 = 1500 N-m

Let d = Diameter of the shaft in mm.

We know that equivalent twisting moment,

\(T_e = \sqrt{M^2 + T^2}\)

\(= \sqrt{(1500)^2 + (3183)^2} \)

\(= 3519\ N-m\)

\(= 3519 \times 10^3 \ N-mm\)

We also know that equivalent twisting moment (T_e),

\(3519 \times 10^3 = \frac \pi {16} \times \tau \times d^3\)

\(= \frac\pi {16} \times 60 \times d^3\)

\(= 11.8 \ d^3\)   ...(Assuming τ = 60 N/mm²)

\(\therefore d^3 = 3519 \times 10^3 /11.8 \)

\(= 298 \times 10^3\)

or \(d = 66.8 \) say \(70 \) mm.