Understanding Maximum Torque on a Shaft

This problem asks us to find the maximum torque that can be safely applied to a solid circular shaft. The safety is limited by two conditions: the maximum allowable angle of twist and the maximum allowable shear stress. We need to calculate the maximum torque permitted by each condition separately and then choose the smaller value, as that is the torque that satisfies both constraints simultaneously.

Given Data for the Shaft

• Shaft Diameter, $D = 100 \text{ mm}$
• Shaft Radius, $R = D/2 = 50 \text{ mm}$
• Length of the shaft, $L = 3 \text{ m} = 3000 \text{ mm}$
• Permissible angle of twist, $\theta = 1 \text{ degree}$
• Permissible shear stress, $\tau_{max} = 30 \text{ N/mm}^2$
• Modulus of rigidity, $G = 0.8 \times 10^5 \text{ N/mm}^2$

Step 1: Convert Angle of Twist to Radians

The angle of twist formula uses radians. We convert the given angle from degrees to radians:

$\theta \text{ (in radians)} = \theta \text{ (in degrees)} \times \frac{\pi}{180}$

$\theta = 1 \times \frac{\pi}{180} \text{ radians}$

Step 2: Calculate Polar Moment of Inertia (J)

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

$J = \frac{\pi D^4}{32}$ or $J = \frac{\pi R^4}{2}$

Using $R = 50 \text{ mm}$:

$J = \frac{\pi (50 \text{ mm})^4}{2} = \frac{\pi \times 6,250,000}{2} \text{ mm}^4 = 3,125,000 \pi \text{ mm}^4$

Step 3: Calculate Maximum Torque Based on Angle of Twist

The torsion formula relating torque ($T$), angle of twist ($\theta$), modulus of rigidity ($G$), polar moment of inertia ($J$), and length ($L$) is:

$\frac{T}{J} = \frac{G\theta}{L}$

Rearranging for $T$:

$T_{twist} = \frac{G \theta J}{L}$

Substitute the values:

$T_{twist} = \frac{(0.8 \times 10^5 \text{ N/mm}^2) \times (\frac{\pi}{180} \text{ rad}) \times (3,125,000 \pi \text{ mm}^4)}{3000 \text{ mm}}$

$T_{twist} = \frac{0.8 \times 10^5 \times \pi^2 \times 3,125,000}{180 \times 3000} \text{ N.mm}$

$T_{twist} = \frac{0.8 \times 10^5 \times 9.8696 \times 3,125,000}{540,000} \text{ N.mm}$

$T_{twist} \approx \frac{2,467,400,000,000}{540,000} \text{ N.mm}$

$T_{twist} \approx 4,569,259 \text{ N.mm}$

Converting N.mm to kNm (1 kNm = $10^6$ N.m = $10^9$ N.mm):

$T_{twist} \approx 4,569,259 \times 10^{-9} \text{ kNm} \approx 4.569 \text{ kNm}$

Rounding to two decimal places, $T_{twist} \approx 4.57 \text{ kNm}$.

Step 4: Calculate Maximum Torque Based on Shear Stress

The torsion formula relating torque ($T$), shear stress ($\tau$), polar moment of inertia ($J$), and radius ($R$) is:

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

Rearranging for $T$:

$T_{stress} = \frac{\tau_{max} J}{R}$

Substitute the values:

$T_{stress} = \frac{(30 \text{ N/mm}^2) \times (3,125,000 \pi \text{ mm}^4)}{50 \text{ mm}}$

$T_{stress} = \frac{30 \times 3,125,000 \pi}{50} \text{ N.mm}$

$T_{stress} = 0.6 \times 3,125,000 \pi \text{ N.mm}$

$T_{stress} = 1,875,000 \pi \text{ N.mm}$

$T_{stress} \approx 1,875,000 \times 3.14159 \text{ N.mm}$

$T_{stress} \approx 5,890,487.5 \text{ N.mm}$

Converting N.mm to kNm:

$T_{stress} \approx 5,890,487.5 \times 10^{-9} \text{ kNm} \approx 5.890 \text{ kNm}$

Rounding to two decimal places, $T_{stress} \approx 5.89 \text{ kNm}$.

Step 5: Determine the Maximum Safe Torque

The shaft must satisfy both the angle of twist limit and the shear stress limit. Therefore, the maximum safe torque is the smaller of the two values calculated:

$T_{max} = \min(T_{twist}, T_{stress})$

$T_{max} = \min(4.57 \text{ kNm}, 5.89 \text{ kNm})$

$T_{max} = 4.57 \text{ kNm}$

Thus, the maximum torque that can be safely applied to the shaft is 4.57 kNm.