Question

A solid round uniform diameter shaft is transmitting 25 kW power at 540 rpm. The maximum allowable shear stress of the shaft material is 35 MPa. If the maximum torque exceeds the mean torque by 20%, neglecting the bending effect, the minimum shaft diameter is _________ mm. (Rounded off to 2 decimal places)

Hint:

When calculating the shaft diameter for torque transmission, use the formula \( T_{{max}} = \frac{\pi \times d^3 \times \tau_{{max}}}{16} \) and account for the increase in maximum torque.

Answer 1

We are given the following values:
Power (\(P\)) = 25 kW = 25,000 W,
Speed (\(N\)) = 540 rpm,
Shear stress (\(\tau_{{max}}\)) = 35 MPa = 35 \(\times 10^6\) Pa,
The maximum torque exceeds the mean torque by 20%.
Step 1: Calculate the mean torque (\(T_{{mean}}\))
Using the formula for power transmission: \[ P = \frac{T_{{mean}} \times N \times 2\pi}{60} \] Rearranging to solve for \(T_{{mean}}\): \[ T_{{mean}} = \frac{P \times 60}{N \times 2\pi} = \frac{25,000 \times 60}{540 \times 2\pi} = \frac{1,500,000}{3392.3} \approx 442.3 \, {Nm} \] Step 2: Calculate the maximum torque (\(T_{{max}}\))
Since the maximum torque exceeds the mean torque by 20%, the maximum torque is: \[ T_{{max}} = 1.2 \times T_{{mean}} = 1.2 \times 442.3 = 530.76 \, {Nm} \] Step 3: Calculate the shaft diameter
The torque is related to the shaft diameter by the formula: \[ T_{{max}} = \frac{\pi \times d^3 \times \tau_{{max}}}{16} \] Rearranging to solve for \(d\): \[ d^3 = \frac{16 \times T_{{max}}}{\pi \times \tau_{{max}}} \] \[ d^3 = \frac{16 \times 530.76}{\pi \times 35 \times 10^6} = \frac{8481.6}{1.099 \times 10^8} \approx 7.71 \times 10^{-5} \] Now, taking the cube root to solve for \(d\): \[ d = \sqrt[3]{7.71 \times 10^{-5}} \approx 0.0415 \, {m} = 41.50 \, {mm} \] Thus, the minimum shaft diameter is 41.50 mm.