Question 9.1: Design of a Shaft for Steady Torsion Loading A solid circula......
Design of a Shaft for Steady Torsion Loading
A solid circular shaft is to transmit 500 kW at n =1200 rpm without exceeding the yield strength in shear of S_{ys} or a twisting through more than 4° in a length of 2 m. Calculate the required diameter of the shaft.
Design Decisions: The shaft is made of steel having S_{ys} =300 MPa and G =80 GPa. A safety factor of 1.5 is used.
The torque, applying Equation (1.15),
kW =\frac{F V}{1000}=\frac{T n}{9549} (1.15)
T=\frac{9549 kW }{n}=\frac{9549(500)}{1200}=3979 N \cdot m
Strength specification. Through the use of Equation (9.1), we have
\frac{J}{c}=\frac{T}{\tau_{\text {all }}} (9.1)
\frac{\pi}{2} c^3=\frac{3979(1.5)}{300\left(10^6\right)}
The foregoing gives c=23.3 mm.
Distortion specification. The size of the shaft is now obtained from Equation (4.9):
\phi=\frac{T L}{G J} (4.9)
\frac{\phi_{\text {all }}}{L}=\frac{T}{G J} (9.2)
Substituting the given numerical values,
\frac{4^{\circ}(\pi / 180)}{2}=\frac{3979}{\left(80 \times 10^9\right) \pi c^4 / 2}
This yields c =30.9 mm.
Comment: The minimum allowable diameter of the shaft must be 61.8 mm. A 62 mm shaft should be used.