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.
RA = RB = 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 (Te),
\(3519 \times 10^3 = \frac \pi {16} \times \tau \times d^3\)
\(= \frac\pi {16} \times 60 \times d^3\)
\(= 11.8 \ d^3\) ...(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.