Question

Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the maximum allowable shear stress is 15 kpsi.

Answer 1

Answer:46.05 mm

Explanation:

Given

$Power=260\ hp\approx 260\times 746=193.96\ KW$

speed $N=550\ rev/min$

allowable shear stress $(\tau )_{max}=15\ kpsi\approx 103.421\ MPa$

Power is given by

$P=\frac{2\pi NT}{60}$

$193.96=\frac{2\pi 550\times T}{60}$

$T=3367.6\ N-m$

From Torsion Formula

$\frac{T}{J}=\frac{\tau }{r}-----1$

where J=Polar section modulus

T=Torque

$\tau$=shear stress

For square cross section

$r=\frac{a}{2}$

where a=side of square

$J=\frac{a^4}{6}$

Substituting the values in equation 1

$\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}$

$a=0.04605\ m$

$a=46.05\ mm$