Calculating Shaft Diameter Under Torque

This question asks us to find the maximum permissible diameter of a circular shaft that is subjected to a specific torque, given the maximum allowable shear stress in the shaft material. This is a common problem in mechanical engineering design, focusing on the strength of materials under torsional load.

Understanding Torsional Stress in a Circular Shaft

When a circular shaft is subjected to a torque, it experiences shear stress. The shear stress is not uniform across the shaft's cross-section; it is zero at the center and maximum at the outer surface. The relationship between torque (T), shear stress ($\tau$), polar moment of inertia (J), and the radial distance (r) from the center is given by the torsion formula:

$\frac{\tau}{r} = \frac{T}{J}$

We are interested in the maximum shear stress ($\tau_{max}$), which occurs at the outer radius of the shaft. If the diameter of the shaft is $d$, the outer radius is $r = d/2$.

For a solid circular shaft, the polar moment of inertia (J) is given by:

$J = \frac{\pi d^4}{32}$

Substituting $r = d/2$ and $J = \frac{\pi d^4}{32}$ into the torsion formula and solving for the maximum shear stress $\tau_{max}$:

$\frac{\tau_{max}}{d/2} = \frac{T}{\pi d^4/32}$

$\tau_{max} = \frac{T \cdot (d/2)}{\pi d^4/32} = \frac{T \cdot d/2}{\pi d^4/32} = \frac{16T}{\pi d^3}$

Finding the Maximum Permissible Diameter

We are given the permissible shear stress ($\tau_{perm}$) which is the maximum allowable shear stress ($\tau_{max}$). We can rearrange the formula to solve for the diameter $d$:

$d^3 = \frac{16T}{\pi \tau_{perm}}$

$d = \left( \frac{16T}{\pi \tau_{perm}} \right)^{1/3}$

Step-by-Step Calculation

Given values:

• Torque, $T = 50 \text{ kN-m}$
• Permissible shear stress, $\tau_{perm} = 40 \text{ MPa}$

First, let's convert the given values into consistent units, such as N and mm.

• $T = 50 \text{ kN-m} = 50 \times 10^3 \text{ N} \times 10^3 \text{ mm} = 50 \times 10^6 \text{ N-mm}$
• $\tau_{perm} = 40 \text{ MPa} = 40 \text{ N/mm}^2$

Now substitute these values into the formula for $d^3$:

$d^3 = \frac{16 \times (50 \times 10^6 \text{ N-mm})}{\pi \times (40 \text{ N/mm}^2)}$

$d^3 = \frac{800 \times 10^6}{\pi \times 40} \text{ mm}^3$

$d^3 = \frac{20 \times 10^6}{\pi} \text{ mm}^3$

$d^3 \approx \frac{20 \times 10^6}{3.14159} \text{ mm}^3$

$d^3 \approx 6.3662 \times 10^6 \text{ mm}^3$

Now, calculate $d$ by taking the cube root:

$d = (6.3662 \times 10^6)^{1/3} \text{ mm}$

$d \approx (6.3662)^{1/3} \times (10^6)^{1/3} \text{ mm}$

$d \approx 1.8533 \times 10^2 \text{ mm}$

$d \approx 185.33 \text{ mm}$

This calculated value represents the minimum diameter required to withstand the given torque without exceeding the permissible shear stress. Therefore, the maximum permissible diameter would be this value or any value larger than this, but typically in such questions, the calculated value is considered the maximum permissible diameter based on the shear stress limit.

Comparing this result with the given options, the value 185.33 mm matches one of the options.