A 30 in. tubular shaft is in torsion under a torque of 29 lb-in.. Its cross-section shown below as diameter d1 = 3.0 in., diameter d2 = 1.0 in. Calculate the polar moment of inertia of the tube cross-section about its axis perpendicular to the screen and same time passing through its center: J = ____ in4Calculate your answer to 3 significant figures. ### Problem Description:A 30-inch tubular shaft is in torsion under a torque of 29 lb-in. Its cross-section is shown below with the following dimensions:- Diameter \( d_1 = 3.0 \) inches- Diameter \( d_2 = 1.0 \) inch### Diagram Explanation:The diagram illustrates a cross-sectional view of a tubular shaft:- The outer circle represents the external diameter \( d_1 \).- The inner circle represents the internal diameter \( d_2 \).- The center is marked as **B**.- The axes are labeled as \( y \) and \( z \).### Task:Calculate the polar moment of inertia of the tube's cross-section about its axis perpendicular to the screen and passing through its center. Use the formula for the polar moment of inertia of a hollow circular section:\[J = \frac{\pi}{32} (d_1^4 - d_2^4)\]Provide your answer in \( \text{in}^4 \) to 3 significant figures.### Answer Box:\[ \boxed{} \]

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Answered: A 30 in. tubular shaft is in torsion…

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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A 30 in. tubular shaft is in torsion under a torque of 29 lb-in.. Its cross-section shown below as diameter d1 = 3.0 in., diameter d2 = 1.0 in.

Calculate the polar moment of inertia of the tube cross-section about its axis perpendicular to the screen and same time passing through its center: J = ____ in⁴

Calculate your answer to 3 significant figures.

### Problem Description:

A 30-inch tubular shaft is in torsion under a torque of 29 lb-in. Its cross-section is shown below with the following dimensions:
- Diameter \( d_1 = 3.0 \) inches
- Diameter \( d_2 = 1.0 \) inch

### Diagram Explanation:

The diagram illustrates a cross-sectional view of a tubular shaft:
- The outer circle represents the external diameter \( d_1 \).
- The inner circle represents the internal diameter \( d_2 \).
- The center is marked as **B**.
- The axes are labeled as \( y \) and \( z \).

### Task:

Calculate the polar moment of inertia of the tube's cross-section about its axis perpendicular to the screen and passing through its center. Use the formula for the polar moment of inertia of a hollow circular section:

\[
J = \frac{\pi}{32} (d_1^4 - d_2^4)
\]

Provide your answer in \( \text{in}^4 \) to 3 significant figures.

### Answer Box:

\[ \boxed{} \]

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