shaft calculation

Question

A solid shaft is transmitting 1 MW at 240 r.p.m. Determine the diameter of theshaft if the
maximum torque transmitted exceeds the mean torque by 20%. Take the maximum allowable
shear stress as 60 MPa.

Answer 1

Given: P = 1 MW = 1 X 10⁶ W;

N = 240 r.p.m.;

T_max = 1.2 T_mean;

τ = 60 MPa = 60 N/mm²

Let d = Diameter of the shaft

Mean torque transmitted by the shaft,

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

\(= \frac{1 \times 10^6 \times 60}{2\pi \times 240}\)

\(= 39784 \ N-m\)

\(= 39784 \times 10^3 \ N-mm\)

Therefore, maximum torque transmitted,

T_max = 1.2 T_mean

= 1.2 x 39784 x 10³

= 47741 x 10³ N-mm

Also, maximum torque transmitted,

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

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

\(= 11.78\ d^3\)

Therefore, 47741 x 10³ N-mm = 11.78d³

d³ = \(\frac{47741 \times 10^3 N-mm}{11.78}\)

\(= 4053 \times 10^3\)

\(d = 159.4\) say 160 mm