solid steel shaft is made of steel has 100-mm diameter. The shaft is subjected to the distributed and centrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft. Gsteel = 75 GPa B 0.5 m 3 kN.m 0.7 m 4 kN.m A

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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### Problem Statement
4. A solid steel shaft with a 100-mm diameter is subjected to distributed and concentrated torsional loadings. Determine the angle of twist at the free end \( A \) of the shaft. The modulus of rigidity for steel is given as \( G_{\text{steel}} = 75 \text{ GPa} \).

### Diagram Description
The diagram shows a horizontal solid shaft fixed at one end, labeled \( B \), while the other end, labeled \( A \), is free. The shaft has a diameter of 100 mm. Two torsional loads are applied along the shaft:

- A 3 kN·m torque is applied at a distance of 0.5 m from point \( B \).
- A 4 kN·m torque is applied at the free end \( A \), at a distance of 0.7 m from \( B \).

These torsional loads are depicted using arrows and circular indicators to show the direction and position of the torques along the shaft.

### Calculation Requirements
To find the angle of twist at point \( A \), it involves using the torsion formula for circular shafts:

\[
\theta = \frac{TL}{GJ}
\]

Where:
- \(\theta\) = angle of twist (radians)
- \(T\) = torque applied (Nm)
- \(L\) = length of the shaft segment (m)
- \(G\) = modulus of rigidity (Pa)
- \(J\) = polar moment of inertia for a circular shaft section (m\(^4\)), calculated as \(J = \frac{\pi d^4}{32}\), where \(d\) is the diameter of the shaft.

### Conclusion
Use the formula and given values to solve for \(\theta\) at point \( A \) considering the torques applied and the shaft's material properties.
Transcribed Image Text:### Problem Statement 4. A solid steel shaft with a 100-mm diameter is subjected to distributed and concentrated torsional loadings. Determine the angle of twist at the free end \( A \) of the shaft. The modulus of rigidity for steel is given as \( G_{\text{steel}} = 75 \text{ GPa} \). ### Diagram Description The diagram shows a horizontal solid shaft fixed at one end, labeled \( B \), while the other end, labeled \( A \), is free. The shaft has a diameter of 100 mm. Two torsional loads are applied along the shaft: - A 3 kN·m torque is applied at a distance of 0.5 m from point \( B \). - A 4 kN·m torque is applied at the free end \( A \), at a distance of 0.7 m from \( B \). These torsional loads are depicted using arrows and circular indicators to show the direction and position of the torques along the shaft. ### Calculation Requirements To find the angle of twist at point \( A \), it involves using the torsion formula for circular shafts: \[ \theta = \frac{TL}{GJ} \] Where: - \(\theta\) = angle of twist (radians) - \(T\) = torque applied (Nm) - \(L\) = length of the shaft segment (m) - \(G\) = modulus of rigidity (Pa) - \(J\) = polar moment of inertia for a circular shaft section (m\(^4\)), calculated as \(J = \frac{\pi d^4}{32}\), where \(d\) is the diameter of the shaft. ### Conclusion Use the formula and given values to solve for \(\theta\) at point \( A \) considering the torques applied and the shaft's material properties.
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