Question

A bar of length \( L \), area of cross-section \( A \), is subjected to a tensile force \( P \). If \( E \) is the Young's modulus of the bar, the strain energy \( U \) stored in the body is:

• A. \( U = \frac{PL}{AE} \)
• B. \( U = \frac{PL}{2AE} \)
• C. \( U = \frac{P^2L}{2AE} \)
• D. \( U = \frac{P^2L}{AE} \)

Hint:

Remember that strain energy in a linearly elastic material subjected to axial loading can also be expressed in terms of stress and strain: $$U = \frac{1}{2} \times \text{Volume} \times \text{Stress} \times \text{Strain} = \frac{1}{2} (AL) \sigma \epsilon$$Using \( \sigma = P/A \) and \( \epsilon = \sigma/E = P/(AE) \), we get:$$U = \frac{1}{2} (AL) \left( \frac{P}{A} \right) \left( \frac{P}{AE} \right) = \frac{P^2L}{2AE}$$

Answer 1

The Correct Option is C

Step 1: Understand the concept of strain energy.
Strain energy is the energy stored in a deformable body due to the work done by external forces in deforming the body. When the material is elastic, this energy can be recovered upon the removal of the loads. Step 2: Recall the basic definitions and relationships.
Stress \( (\sigma) \) is defined as the force per unit area: \( \sigma = \frac{P}{A} \).
Strain \( (\epsilon) \) is defined as the change in length per unit original length: \( \epsilon = \frac{\Delta L}{L} \).
Young's modulus \( (E) \) is the ratio of stress to strain in the elastic region: \( E = \frac{\sigma}{\epsilon} \).
Step 3: Derive the expression for strain energy.
Consider a small element of the bar undergoing deformation \( d(\Delta L) \) due to an applied force \( F \). The work done \( dW \) on this element is \( F \cdot d(\Delta L) \). The total work done in extending the bar from 0 to \( \Delta L \) is given by the integral of the force over the displacement. Since the force increases linearly from 0 to \( P \) as the elongation goes from 0 to \( \Delta L \), the work done (which is stored as strain energy \( U \)) is the average force multiplied by the displacement: $$U = \frac{1}{2} \times \text{Force} \times \text{Elongation} = \frac{1}{2} P \Delta L$$ Now, we need to express \( \Delta L \) in terms of the given parameters \( P \), \( L \), \( A \), and \( E \). From the definition of Young's modulus, \( E = \frac{\sigma}{\epsilon} = \frac{P/A}{\Delta L/L} \). Rearranging this equation to solve for \( \Delta L \): $$\Delta L = \frac{PL}{AE}$$ Substitute this expression for \( \Delta L \) back into the strain energy equation: $$U = \frac{1}{2} P \left( \frac{PL}{AE} \right) = \frac{P^2L}{2AE}$$