Question

A hollow shaft of outer radius 140 mm and inner radius 125 mm is subjected to a compressive force of 200kN and a torque. If the allowable shearing stress is 100 MPa, what is the maximum torque that can be applied?

Answer 1

Answer on Question #80270 – Physics | Mechanics Relativity

Question:

A hollow shaft of outer radius 140 mm and inner radius 125 mm is subjected to a compressive force of 200kN and a torque. If the allowable shearing stress is 100 MPa, what is the maximum torque that can be applied?

Solution:

We have hollow shaft with outer radius $c_{2} = 140$ mm and inner radius $c_{1} = 125$ mm. Also we know the allowable shearing stress is $\tau_{al} = 100$ MPa. To find the maximum torque we can use equation for shear stress:

$$\tau = \frac{T \times c}{J},$$

where $J$ is the polar moment of inertia for the hollow shaft and $T$ is the torque. The maximum permissive stress will be observed for the outer radius $c_{2}$. So, the final formula for Torque will look like

$$T = \frac{\tau_{al} \times J}{c_{2}}$$

Now let's calculate J:

$$J = \frac{\pi}{2} (c_{2}^{4} - c_{1}^{4}) = 2.2 \times 10^{-4} \, \text{m}^{4}.$$

Then

$$T = \frac{100 \times 10^{6} \times 2.2 \times 10^{-4}}{0.14} = 1.57 \times 10^{5} \, \text{Nm} = 157 \, \text{kNm}$$

Answer: 157 kNm

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