Question

A cylindrical transmission shaft of length 1.5 m and diameter 100 mm is made of a linear elastic material with a shear modulus of 80 GPa. While operating at 500 rpm, the angle of twist across its length is found to be 0.5 degrees.
The power transmitted by the shaft at this speed is ............... kW. (Rounded off to two decimal places)
Take \(\pi\) = 3.14.

Hint:

Pay close attention to units. The most common error in torsion problems is forgetting to convert the angle of twist from degrees to radians. The torsion formula \(T/J = G\theta/L\) is only valid when \(\theta\) is in radians. Also, ensure all length units are consistent (e.g., all in meters).

Answer 1

Step 1: Understanding the Concept:
The power transmitted by a rotating shaft is the product of the torque it carries and its angular velocity. The torque can be determined from the material properties (shear modulus), geometry (length, diameter), and the angle of twist using the torsion equation.
Step 2: Key Formula or Approach:
1. Torsion Equation: Relates torque (\(T\)) to the angle of twist (\(\theta\)):
\[ \frac{T}{J} = \frac{G \theta}{L} \] where \(J\) is the polar moment of inertia, \(G\) is the shear modulus, and \(L\) is the length.
2. Polar Moment of Inertia (\(J\)): For a solid circular shaft:
\[ J = \frac{\pi}{32} D^4 \] 3. Angular Velocity (\(\omega\)): Converts rotational speed from rpm (\(N\)) to rad/s:
\[ \omega = \frac{2 \pi N}{60} \] 4. Power Transmission (\(P\)):
\[ P = T \cdot \omega \] Step 3: Detailed Explanation or Calculation:
Given data:
Length, \(L = 1.5\) m
Diameter, \(D = 100\) mm = 0.1 m
Shear modulus, \(G = 80\) GPa = \(80 \times 10^9\) Pa
Rotational speed, \(N = 500\) rpm
Angle of twist, \(\theta = 0.5\) degrees
\(\pi = 3.14\)
First, convert the angle of twist from degrees to radians:
\[ \theta = 0.5^{\circ} \times \frac{\pi}{180^{\circ}} = 0.5 \times \frac{3.14}{180} \approx 0.008722 \text{ rad} \] Next, calculate the polar moment of inertia \(J\):
\[ J = \frac{\pi}{32} D^4 = \frac{3.14}{32} (0.1)^4 = 0.098125 \times 10^{-4} = 9.8125 \times 10^{-6} \text{ m}^4 \] Now, calculate the torque \(T\) using the torsion equation:
\[ T = \frac{G \theta J}{L} = \frac{(80 \times 10^9 \text{ Pa}) \times (0.008722 \text{ rad}) \times (9.8125 \times 10^{-6} \text{ m}^4)}{1.5 \text{ m}} \] \[ T \approx \frac{6848.8}{1.5} \approx 4565.87 \text{ N-m} \] Next, calculate the angular velocity \(\omega\):
\[ \omega = \frac{2 \pi N}{60} = \frac{2 \times 3.14 \times 500}{60} = \frac{3140}{60} \approx 52.333 \text{ rad/s} \] Finally, calculate the power \(P\):
\[ P = T \times \omega = 4565.87 \text{ N-m} \times 52.333 \text{ rad/s} \approx 238910.8 \text{ W} \] Convert the power to kilowatts (kW):
\[ P = \frac{238910.8}{1000} \approx 238.91 \text{ kW} \] Step 4: Final Answer:
Rounding off to two decimal places, the power transmitted by the shaft is 238.91 kW.
Step 5: Why This is Correct:
The solution systematically applies the fundamental principles of power transmission in shafts. Each step, from unit conversion (degrees to radians, rpm to rad/s) to applying the torsion and power formulas, is correctly executed. The final calculated value of 238.91 kW lies within the accepted answer range of 237 to 240 kW.