Question

A bar of length L and having its area of cross-section A, is subjected to a gradually applied tensile load W. Modulus of elasticity of the bar material is E. The strain energy stored in the bar is:

• A. $\frac{WL}{2AE}$
• B. $\frac{WL}{AE}$
• C. $\frac{W^2 L}{AE}$
• D. $\frac{W^2 L}{2AE}$

Hint:

Always check the "Square" and the "Half."

• Square: Energy is always proportional to the square of the force ($W^2$).

• Half: Gradually applied loads always involve a factor of $1/2$ because the force starts at zero.

Answer 1

The Correct Option is D

Strain energy ($U$) is the internal work done by a load as it deforms a body. For a gradually applied load, the energy is stored as elastic potential energy within the material. 1. Work Done by Gradually Applied Load: When a load is applied gradually from zero to its final value $W$, the average force is $\frac{1}{2}W$. The work done (and thus the strain energy stored) is: $$U = \frac{1}{2} \times \text{Load} \times \text{Extension} = \frac{1}{2} W \delta L$$

2. Extension Formula: From Hooke's Law, the extension ($\delta L$) of a bar under axial load is given by: $$\delta L = \frac{WL}{AE}$$

3. Deriving the Strain Energy Equation: Substituting the extension formula into the work equation: $$U = \frac{1}{2} W \left( \frac{WL}{AE} \right)$$ $$U = \frac{W^2 L}{2AE}$$ This energy can also be expressed in terms of stress ($\sigma = W/A$): $$U = \frac{\sigma^2}{2E} \times (AL) = \frac{\sigma^2}{2E} \times \text{Volume}$$ This shows that strain energy is proportional to the square of the applied load and the length, while being inversely proportional to the stiffness of the bar.