Calculate the minimum diameter of the shaft, such that the shearing stress does NOT exceed 50 N/mm² during torque transmission of 15000 N-m.

Concept:

We use the torsion formula for solid circular shafts to determine the minimum diameter required to ensure the shear stress does not exceed a specified limit during torque transmission.

Given:

• Maximum allowable shear stress, \( \tau = 50 \, \text{N/mm}^2 \)
• Torque transmitted, \( T = 15000 \, \text{N-m} = 15 \times 10^6 \, \text{N-mm} \)

Step 1: Recall the torsion formula

The shear stress in a solid shaft is given by:

\( \tau = \frac{16T}{\pi d^3} \)

where \( d \) is the diameter of the shaft.

Step 2: Rearrange the formula to solve for diameter

\( d^3 = \frac{16T}{\pi \tau} \)

Step 3: Substitute the given values

\( d^3 = \frac{16 \times 15 \times 10^6}{\pi \times 50} = \frac{240 \times 10^6}{157.08} \approx 1.528 \times 10^6 \, \text{mm}^3 \)

Step 4: Calculate the diameter

\( d = \sqrt[3]{1.528 \times 10^6} \approx 115.2 \, \text{mm} \)