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
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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
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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