determine the elastic potential energy of a material deformed within its limit of proportionality from the area under the force–extension graph

Elastic and Plastic Behaviour

Learning Objective

By the end of this lesson you should be able to determine the elastic potential energy of a material deformed within its limit of proportionality by calculating the area under its force–extension graph.

Key Concepts

• Elastic deformation: Deformation that is fully reversible when the applied force is removed.
• Plastic deformation: Permanent deformation that remains after the force is removed.
• Limit of proportionality (LoP): The maximum extension up to which the force–extension relationship is linear and obeys Hooke’s law.
• Hooke’s law: Within the LoP, the restoring force \$F\$ is proportional to the extension \$x\$,

  \$F = kx,\$

  where \$k\$ is the spring (or material) constant.

Force–Extension Graph

The typical force–extension graph for a uniform material is shown below. The straight line segment from the origin to the point \$A\$ represents elastic behaviour (within the LoP). Beyond \$A\$ the curve deviates, indicating the onset of plastic deformation.

Elastic Potential Energy

Elastic potential energy \$U{\text{e}}\$ is the energy stored in a material when it is deformed elastically. It is equal to the work done in stretching the material from \$x=0\$ to \$x=x{\text{max}}\$ (where \$x_{\text{max}}\$ lies within the LoP).

Mathematically, the work done is the area under the force–extension curve:

\$U{\text{e}} = \int{0}^{x_{\text{max}}} F\,dx.\$

Because the relationship is linear within the LoP (\$F = kx\$), the graph is a right‑angled triangle, and the area can be calculated using the triangle formula:

\$U{\text{e}} = \frac{1}{2} F{\text{max}} x{\text{max}} = \frac{1}{2} k x{\text{max}}^{2}.\$

Symbols and Units

Worked Example

1. For a steel wire the force–extension graph is linear up to \$F{\text{max}} = 120\ \text{N}\$ at an extension \$x{\text{max}} = 0.015\ \text{m}\$. Determine the elastic potential energy stored in the wire.

Solution:

\$U{\text{e}} = \frac{1}{2} F{\text{max}} x_{\text{max}} = \frac{1}{2} (120\ \text{N})(0.015\ \text{m}) = 0.90\ \text{J}.\$

Alternatively, first find the spring constant:

\$k = \frac{F{\text{max}}}{x{\text{max}}} = \frac{120\ \text{N}}{0.015\ \text{m}} = 8.0 \times 10^{3}\ \text{N m}^{-1}.\$

Then use \$U{\text{e}} = \tfrac12 k x{\text{max}}^{2}\$:

\$U_{\text{e}} = \frac{1}{2} (8.0 \times 10^{3}\ \text{N m}^{-1})(0.015\ \text{m})^{2} = 0.90\ \text{J}.\$

Common Misconceptions

• “All deformation stores energy.” Only deformation that is elastic (i.e., within the LoP) stores recoverable potential energy. Plastic deformation dissipates energy as heat.
• “The area under any part of the graph gives elastic energy.” The area must be taken only up to the point where the material is still obeying Hooke’s law.
• “The spring constant is the same for all extensions.” \$k\$ is constant only within the linear region; beyond the LoP the effective stiffness changes.

Summary

To determine the elastic potential energy of a material deformed within its limit of proportionality:

1. Confirm that the deformation lies entirely within the linear (Hooke’s law) region of the force–extension graph.
2. Identify \$F{\text{max}}\$ and \$x{\text{max}}\$ at the end of the linear region.
3. Calculate the area of the right‑angled triangle formed by the origin, \$F{\text{max}}\$, and \$x{\text{max}}\$ using \$U{\text{e}} = \tfrac12 F{\text{max}} x_{\text{max}}\$.
4. Alternatively, determine the spring constant \$k = F{\text{max}}/x{\text{max}}\$ and use \$U{\text{e}} = \tfrac12 k x{\text{max}}^{2}\$.

Practice Questions

1. A rubber band obeys Hooke’s law up to a force of \$30\ \text{N}\$, at which point its extension is \$0.05\ \text{m}\$. Calculate the elastic potential energy stored.
2. A metal rod has a spring constant of \$2.5 \times 10^{4}\ \text{N m}^{-1}\$. If it is stretched \$2.0\ \text{mm}\$ within the LoP, what is the energy stored?
3. Explain why the area under the force–extension curve beyond the limit of proportionality does not represent elastic potential energy.