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
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
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9496fb66-15f0-4d2e-bc7e-ca3bb9b70577%2F1f304840-3b77-4b5e-9e70-f1d4705e0ffa%2Fx888itc_processed.png&w=3840&q=75)
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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