Question

In a hollow circular shaft of outer and inner diameters of 20 cm and 10 cm respectively, the shear stress is not to exceed 40 MPa. The maximum torque which the shaft can safely transmit is *
A) 29452 Nm
B) 40820 Nm
C) 58904 Nm
D) 32610 Nm

Answer 1

Answer:

Using the assumptions above, we have, at any point r inside the shaft, the shear stress is τr = r/c τmax.

Answer 2

Answer:

The maximum torque that can be transmitted by the shaft, T, calculated is $58875N-m$.

Therefore, option c) $58904N-m$ is correct. ( closest to $58875N-m$ ).

Explanation:

Given data,

The outer diameter of the hollow circular shaft, D = $20cm$ = $0.2m$

The inner diameter of the hollow circular shaft, d = $10cm$ = $0.1m$

The shear stress of the hollow circular shaft, $\tau$ = $40MPa$ = $40\times 10^{6}Pa$

The maximum torque that can be transmitted by the shaft, T =?

From the formula given below, we can find out the maximum torque:

• T = $\frac{\pi }{16} \times [\frac{D^{2} -d^{4} }{D} ] \tau$

After putting the given values of diameters and the shear stress in the formula, we get:

• T = $\frac{3.14 }{16} \times [\frac{(0.2)^{2} -(0.1)^{4} }{0.2} ] \times 40\times 10^{6}$
• T = $\frac{3.14 }{4} \times [\frac{0.0016 -0.0001 }{0.2} ] \times 10\times 10^{6}$
• T = $\frac{3.14 }{4} \times [\frac{0.0015 }{0.2} ] \times 10^{7}$
• T = $\frac{471 }{0.8} \times 10^{2}$
• T = $58875N-m$

Hence, the maximum torque that can be transmitted by the shaft, T = $58875N-m$.