Question

The elastic energy stored per unit volume in terms of longitudinal strain \( e \) and Young’s modulus \( Y \) is:

• A. \( \frac{Y e^2}{2} \)
• B. \( \frac{1}{2} Y e \)
• C. \( 2 Y e^2 \)
• D. \( 2 Y e \)

Hint:

Elastic potential energy per unit volume is given by \( U = \frac{1}{2} \sigma e \). Using Hooke’s Law \( \sigma = Y e \), we derive \( U = \frac{Y e^2}{2} \).

Answer 1

The Correct Option is A

Step 1: Elastic Potential Energy Per Unit Volume
The elastic potential energy stored per unit volume (also called strain energy density) is given by: \[ U = \frac{1}{2} \sigma e. \] where: - \( \sigma \) is the stress,
- \( e \) is the strain.
Step 2: Expressing Stress in Terms of Young's Modulus
From Hooke’s Law, \[ \sigma = Y e. \] Substituting this into the strain energy equation: \[ U = \frac{1}{2} Y e \cdot e. \] \[ U = \frac{1}{2} Y e^2. \] Step 3: Conclusion
Thus, the elastic energy stored per unit volume is: \[ \frac{Y e^2}{2}. \]

Answer 2

To determine the elastic energy stored per unit volume in terms of longitudinal strain \( e \) and Young’s modulus \( Y \), consider the following:

When a material is subjected to a stress that leads to strain \( e \), it stores elastic energy. The elastic potential energy per unit volume, also known as energy density, is given by the formula:

\(\text{Energy Density} = \frac{1}{2} \times \text{Stress} \times \text{Strain}\)

We know the relationship between stress, \( \sigma \), and strain, \( e \), for a material given by Hooke’s Law is:

\(\sigma = Y \times e\)

Where \( Y \) is Young’s modulus. Substituting stress from Hooke’s Law into the energy density formula gives:

\(\text{Energy Density} = \frac{1}{2} \times (Y \times e) \times e\)

Simplifying this, we find:

\(\text{Energy Density} = \frac{Y e^2}{2}\)

Therefore, the correct expression for the elastic energy stored per unit volume in terms of longitudinal strain \( e \) and Young’s modulus \( Y \) is:

\(\frac{Y e^2}{2}\)