What is the normal stress in the bar if P=13 kN and A = 559 mm2? Calculate your answer to 1 decimal place, and don't include the negative sign in your answer (already there). Answer: - (negative)_________ M Pa
What is the normal stress in the bar if P=13 kN and A = 559 mm2? Calculate your answer to 1 decimal place, and don't include the negative sign in your answer (already there). Answer: - (negative)_________ M Pa
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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Question
What is the normal stress in the bar if P=13 kN and A = 559 mm2? Calculate your answer to 1 decimal place, and don't include the negative sign in your answer (already there).
Answer: - (negative)_________ M Pa.
![**Problem Statement:**
What is the normal stress in the bar if \( P = 13 \, \text{kN} \) and \( A = 559 \, \text{mm}^2 \)? Calculate your answer to 1 decimal place, and don't include the negative sign in your answer (already there).
**Answer:**
- (negative) _______ MPa.
**Diagram Explanation:**
The diagram illustrates a cylindrical bar under axial compression. The bar is subjected to a compressive force \( P \) at both ends. The area of cross-section where the force applies is labeled \( A \), and the compressive force \( P \) applied results in normal stress denoted by \( \sigma \).
The normal stress (\( \sigma \)) is depicted as distributed across the cross-sectional area, calculated using the formula:
\[
\sigma = \frac{P}{A}
\]
Where:
- \( P \) is the compressive load (in this case, 13 kN),
- \( A \) is the cross-sectional area (559 mm\(^2\)).
The bar shows both top and bottom arrows pointing inward, indicating the direction of the compressive force.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0daff6b4-7b3b-4798-aa81-781ef422795b%2F174020fa-6b7f-4906-92d0-2c4ac292068e%2Fmj99s1r_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
What is the normal stress in the bar if \( P = 13 \, \text{kN} \) and \( A = 559 \, \text{mm}^2 \)? Calculate your answer to 1 decimal place, and don't include the negative sign in your answer (already there).
**Answer:**
- (negative) _______ MPa.
**Diagram Explanation:**
The diagram illustrates a cylindrical bar under axial compression. The bar is subjected to a compressive force \( P \) at both ends. The area of cross-section where the force applies is labeled \( A \), and the compressive force \( P \) applied results in normal stress denoted by \( \sigma \).
The normal stress (\( \sigma \)) is depicted as distributed across the cross-sectional area, calculated using the formula:
\[
\sigma = \frac{P}{A}
\]
Where:
- \( P \) is the compressive load (in this case, 13 kN),
- \( A \) is the cross-sectional area (559 mm\(^2\)).
The bar shows both top and bottom arrows pointing inward, indicating the direction of the compressive force.
Expert Solution
Step 1
Given data:-
P=13 kN and A = 559 mm2
To find:- Normal stress in the bar
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