A bar produces a lateral strain of magnitude 60x105 m/m when subjected to a tensile stress of magnitude 300 MPa along the axial direction. What is the elastic modulus of the material if the Poisson's ratio is 0.3?
A bar produces a lateral strain of magnitude 60x105 m/m when subjected to a tensile stress of magnitude 300 MPa along the axial direction. What is the elastic modulus of the material if the Poisson's ratio is 0.3?
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:**
A bar produces a lateral strain of magnitude \(60 \times 10^{-5} \, \text{m/m}\) when subjected to a tensile stress of magnitude \(300 \, \text{MPa}\) along the axial direction. What is the elastic modulus of the material if the Poisson’s ratio is 0.3?
**Explanation:**
In this problem, we are given the following:
1. Lateral Strain, \(\epsilon_l = 60 \times 10^{-5} \, \text{m/m}\).
2. Tensile Stress, \(\sigma = 300 \, \text{MPa}\).
3. Poisson’s Ratio, \(\nu = 0.3\).
We need to determine the elastic modulus, \(E\), of the material.
**Key Concepts:**
1. **Lateral Strain** - The deformation per unit length in the perpendicular direction to the applied stress.
2. **Tensile Stress** - The force applied per unit area in the axial direction.
3. **Poisson's Ratio** - A constant that describes the ratio of lateral strain to axial strain in a material.
4. **Elastic Modulus (Young’s Modulus, \(E\))** - A measure of the stiffness of a material, defined as the ratio of tensile stress to axial strain.
**Solution:**
The relationship between lateral strain (\(\epsilon_l \)), axial strain (\(\epsilon_a \)), and Poisson’s ratio (\(\nu\)) is given by:
\[ \epsilon_l = -\nu \epsilon_a \]
Rearranging for axial strain:
\[ \epsilon_a = -\frac{\epsilon_l}{\nu} \]
Substitute the known values:
\[ \epsilon_a = -\frac{60 \times 10^{-5} \, \text{m/m}}{0.3} = -200 \times 10^{-5} \, \text{m/m} \]
Now using Hooke's Law:
\[ \sigma = E \epsilon_a \]
\[ 300 \, \text{MPa} = E \times -200 \times 10^{-5} \, \text{m/m} \]
Solving for \(E\):
\[ E = \frac{300 \, \text{MPa}}{-200 \times 10^{-5}} \]
\[ E](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6edee462-3c73-4638-b361-1b9d1ce3b653%2Fc83ede67-b826-4d61-9d4c-bcec16b8c8d7%2Fwuo9hs_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A bar produces a lateral strain of magnitude \(60 \times 10^{-5} \, \text{m/m}\) when subjected to a tensile stress of magnitude \(300 \, \text{MPa}\) along the axial direction. What is the elastic modulus of the material if the Poisson’s ratio is 0.3?
**Explanation:**
In this problem, we are given the following:
1. Lateral Strain, \(\epsilon_l = 60 \times 10^{-5} \, \text{m/m}\).
2. Tensile Stress, \(\sigma = 300 \, \text{MPa}\).
3. Poisson’s Ratio, \(\nu = 0.3\).
We need to determine the elastic modulus, \(E\), of the material.
**Key Concepts:**
1. **Lateral Strain** - The deformation per unit length in the perpendicular direction to the applied stress.
2. **Tensile Stress** - The force applied per unit area in the axial direction.
3. **Poisson's Ratio** - A constant that describes the ratio of lateral strain to axial strain in a material.
4. **Elastic Modulus (Young’s Modulus, \(E\))** - A measure of the stiffness of a material, defined as the ratio of tensile stress to axial strain.
**Solution:**
The relationship between lateral strain (\(\epsilon_l \)), axial strain (\(\epsilon_a \)), and Poisson’s ratio (\(\nu\)) is given by:
\[ \epsilon_l = -\nu \epsilon_a \]
Rearranging for axial strain:
\[ \epsilon_a = -\frac{\epsilon_l}{\nu} \]
Substitute the known values:
\[ \epsilon_a = -\frac{60 \times 10^{-5} \, \text{m/m}}{0.3} = -200 \times 10^{-5} \, \text{m/m} \]
Now using Hooke's Law:
\[ \sigma = E \epsilon_a \]
\[ 300 \, \text{MPa} = E \times -200 \times 10^{-5} \, \text{m/m} \]
Solving for \(E\):
\[ E = \frac{300 \, \text{MPa}}{-200 \times 10^{-5}} \]
\[ E
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