When a material deforms elastically under load, mechanical work is stored as elastic strain energy. If the load is removed without dissipation or permanent deformation, that stored energy can be returned.

Consider a small material element under uniaxial stress $\sigma$ and strain $\varepsilon$. During a small additional strain $d\varepsilon$, the work stored per unit volume is

$$du=\sigma,d\varepsilon.$$

Accumulating from the undeformed state to strain $\varepsilon$ gives the strain-energy density

$$\boxed{u=\int_0^{\varepsilon}\sigma(\varepsilon'),d\varepsilon'}.$$

Geometrically, $u$ is the area under the stress-strain curve up to the current strain.

Because stress has units of pascals,

$$1,\mathrm{Pa}=1,\mathrm{J/m^3},$$

so strain-energy density has units of energy per volume.

Linear elastic material

If the material obeys Hooke's law in uniaxial loading,

$$\sigma=E\varepsilon,$$

then

$$u=\int_0^{\varepsilon}E\varepsilon',d\varepsilon' =\frac12E\varepsilon^2.$$

Using $\sigma=E\varepsilon$, equivalent forms are

$$\boxed{u=\frac12\sigma\varepsilon}$$

and

$$\boxed{u=\frac{\sigma^2}{2E}=\frac12E\varepsilon^2}.$$

The factor $1/2$ appears because the stress rises from zero to its final value as the material is loaded linearly.

Uniform axial bar

Consider a uniform bar of length $L$ and area $A$ carrying an axial force $F$ in the linear elastic range.

The stress is

$$\sigma=\frac{F}{A},$$

and the strain-energy density is

$$u=\frac{\sigma^2}{2E}.$$

The bar volume is $AL$, so its total strain energy is

$$U=uAL =\frac{1}{2E}\left(\frac{F}{A}\right)^2AL.$$

Therefore

$$\boxed{U=\frac{F^2L}{2EA}}.$$

The axial extension is

$$\delta=\frac{FL}{EA},$$

so the same result can be written

$$\boxed{U=\frac12F\delta}.$$

This is the area under the linear force-displacement curve, just as $u$ is the area under the linear stress-strain curve.

Worked example

A steel bar has

$$L=2.0,\mathrm m,$$

$$A=4.0\times10^{-4},\mathrm{m^2},$$

$$E=200,\mathrm{GPa},$$

and carries an axial tensile force

$$F=40,\mathrm{kN}.$$

Its extension is

$$\delta=\frac{FL}{EA} =\frac{(4.0\times10^4)(2.0)}{(2.0\times10^{11})(4.0\times10^{-4})} =1.0\times10^{-3},\mathrm m.$$

Thus

$$\delta=1.0,\mathrm{mm}.$$

The stored elastic energy is

$$U=\frac12F\delta =\frac12(4.0\times10^4)(1.0\times10^{-3}) =\boxed{20,\mathrm J}.$$

The same answer follows from $F^2L/(2EA)$.

Nonlinear elastic loading

The general relation

$$u=\int\sigma,d\varepsilon$$

still applies when the elastic stress-strain curve is nonlinear, but the triangular formula $u=\tfrac12\sigma\varepsilon$ does not unless the stress-strain relation is linear from the origin.

If plastic deformation, hysteresis, fracture, or other irreversible processes occur, not all work supplied during loading is recoverable elastic strain energy.

Strain energy provides a bridge between local material deformation and global structural energy methods: integrating appropriate strain-energy density over a body gives its total stored elastic energy.