Question

A shaft is required to transmit a power of 25 kW at 360 rpm. The force analysis due to attached parts results in Bending Moment of 830 Nm at a section between bearings. If permissible stresses in the shaft are 60 N/mm\(^2\) in bending and 40 N/mm\(^2\) in shear, calculate the diameter of the shaft.

• A. 51 mm
• B. 41 mm
• C. 36 mm
• D. 60 mm

Hint:

In combined loading shaft design, always calculate using the equivalent torque formula.
Ensure that you convert torque and bending moment to $\text{N}\cdot\text{mm}$ and stress to $\text{N/mm}^2$ so that the diameter is obtained directly in millimeters.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine the diameter of a transmission shaft subjected to combined bending and torsional (twisting) loads. We are given the transmitted power, rotational speed, the bending moment, and the allowable bending and shear stress limits.

Step 2: Key Formula or Approach:
First, calculate the transmitted torque (\(T\)) using the power and speed relation:
\[ P = \frac{2 \pi N T}{60} \implies T = \frac{60 P}{2 \pi N} \] For combined loading, calculate the equivalent twisting moment (\(T_{\text{e}}\)) using Guest's (Maximum Shear Stress) Theory:
\[ T_{\text{e}} = \sqrt{M^{2} + T^{2}} \] The shaft diameter (\(d\)) based on shear stress limit (\(\tau\)) is:
\[ d = \left( \frac{16 T_{\text{e}}}{\pi \tau} \right)^{1/3} \]

Step 3: Detailed Explanation:

•
Step 3.1: Calculate Torsional Moment (Torque):
Given: Power, \(P = 25\text{ kW} = 25 \times 10^{3}\text{ W}\).
Speed, \(N = 360\text{ rpm}\).
\[ T = \frac{60 \times 25 \times 10^{3}}{2 \pi \times 360} = \frac{1500000}{2261.95} \approx 663.15\text{ N}\cdot\text{m} = 663150\text{ N}\cdot\text{mm} \]
•
Step 3.2: Calculate Equivalent Twisting Moment:
Given Bending Moment, \(M = 830\text{ N}\cdot\text{m} = 830000\text{ N}\cdot\text{mm}\).
\[ T_{\text{e}} = \sqrt{M^{2} + T^{2}} = \sqrt{830^{2} + 663.15^{2}} \] \[ T_{\text{e}} = \sqrt{688900 + 439768} = \sqrt{1128668} \approx 1062.39\text{ N}\cdot\text{m} = 1.0624 \times 10^{6}\text{ N}\cdot\text{mm} \]
•
Step 3.3: Calculate Shaft Diameter based on Shear Stress limit (\(\tau = 40\text{ N/mm}^{2}\)):
\[ d^{3} = \frac{16 T_{\text{e}}}{\pi \tau} \] \[ d^{3} = \frac{16 \times 1.0624 \times 10^{6}}{\pi \times 40} \approx 135269.8\text{ mm}^{3} \] Taking the cube root:
\[ d = (135269.8)^{1/3} \approx 51.33\text{ mm} \] Rounding to the nearest standard size or available options gives \(51\text{ mm}\).

Step 4: Final Answer:
The calculated diameter of the shaft is \(51\text{ mm}\).