Elastic Potential Energy

When a solid is elastically deformed, work done by the external force is stored as elastic potential energy in the body. This energy is fully recovered when the load is removed (within the elastic limit).

Concept

Elastic PE per unit volume (energy density) is:

$u = \frac{1}{2}\,\sigma\,\varepsilon = \frac{1}{2}\,Y\,\varepsilon^2 = \frac{1}{2}\,\frac{\sigma^2}{Y}$

Total elastic PE in a wire of volume $V$ is:

$U = u \cdot V = \frac{1}{2}\,\sigma\,\varepsilon\,V$

For a spring obeying $F = kx$:

$U = \frac{1}{2}\,k\,x^2$

Physical picture. The factor $1/2$ appears because the stress grows linearly from zero to its final value as the strain is built up — the average force over the displacement is half the final force.

Derivation

Consider a wire of length $L$, area $A$, stretched from natural length to extension $\Delta L$ by a slowly applied longitudinal force.

Step 1 — At an intermediate extension $x$ (where $0 \le x \le \Delta L$), the instantaneous tension is:

$F(x) = \frac{Y A}{L}\,x$

(This follows from Hooke's law applied at extension $x$.)

Step 2 — Work done by this force in extending the wire by an additional $dx$:

$dW = F(x)\,dx = \frac{Y A}{L}\,x\,dx$

Step 3 — Total work done in stretching from 0 to $\Delta L$:

$W = \int_0^{\Delta L} \frac{Y A}{L}\,x\,dx = \frac{Y A}{L}\,\frac{(\Delta L)^2}{2}$

Step 4 — This work is stored as elastic potential energy:

$U = \frac{1}{2}\,\frac{Y A (\Delta L)^2}{L}$

Step 5 — Rewrite in terms of stress $\sigma = F_\text{final}/A$ and strain $\varepsilon = \Delta L/L$:

$U = \frac{1}{2}\,Y \varepsilon^2 \cdot (AL) = \frac{1}{2}\,Y\,\varepsilon^2\,V$

Step 6 — Equivalently, using $\sigma = Y\varepsilon$:

$\boxed{u = \frac{U}{V} = \frac{1}{2}\,\sigma\,\varepsilon = \frac{1}{2}\,\frac{\sigma^2}{Y} = \frac{1}{2}\,Y\,\varepsilon^2}$

Worked Example

A steel wire of length $2\,\text{m}$ and area $1\,\text{mm}^2 = 10^{-6}\,\text{m}^2$ is stretched by $0.1\,\text{mm}$. Find the elastic PE stored. ($Y = 2\times 10^{11}\,\text{Pa}$.)

Strain: $\varepsilon = 10^{-4}/2 = 5\times 10^{-5}$.

Energy density: $u = \tfrac{1}{2}\,Y\,\varepsilon^2 = \tfrac{1}{2}\,(2\times 10^{11})(2.5\times 10^{-9}) = 250\,\text{J/m}^3$.

Volume: $V = AL = 10^{-6} \times 2 = 2\times 10^{-6}\,\text{m}^3$.

Total energy: $U = uV = 250 \times 2\times 10^{-6} = 5\times 10^{-4}\,\text{J}$.

Sanity check via the spring form. Effective spring constant $k = YA/L = (2\times 10^{11})(10^{-6})/2 = 10^5\,\text{N/m}$. Then $U = \tfrac{1}{2}\,k\,x^2 = \tfrac{1}{2}(10^5)(10^{-4})^2 = 5\times 10^{-4}\,\text{J}$. Matches.

Common Confusions

• The factor $1/2$ is not because we go up to half the force; it is because the force grows linearly with extension, so the average force is half the maximum.
• Energy density $u$ has units of J/m^3 = Pa, the same as stress.
• For a spring, $U = \tfrac{1}{2}\,k\,x^2$, not $kx^2$.
• Beyond the elastic limit, some of the work is dissipated as heat or used in permanent deformation; not all is recoverable.

Key Takeaways

• Elastic PE density: $u = \tfrac{1}{2}\,\sigma\,\varepsilon$ (J/m^3).
• Total PE in wire: $U = \tfrac{1}{2}\,\sigma\,\varepsilon\,V = \tfrac{1}{2}\,k\,(\Delta L)^2$ with $k = YA/L$.
• Recoverable only within the elastic limit.
• The $1/2$ factor comes from the linear build-up of stress with strain.