Elastic strain energy

Pull back a bowstring and your muscles do work against the string's resistance — work that seems to vanish, only to reappear an instant later as the kinetic energy of a flying arrow. Where was that energy hiding while the string sat stretched and still? It was stored inside the deformed material itself, as elastic strain energy (also called elastic potential energy): the energy a solid holds because it has been pulled, squeezed, or twisted away from its natural shape, ready to give that energy back the moment the restoring force is allowed to act. Any elastic wire, spring, or beam stores energy this way whenever it is loaded within its elastic limit. This article builds the idea from the force–extension graph, shows why the stored energy is exactly the area under that graph, extends the same reasoning to a per-volume quantity called strain energy density, and applies it to a stretched wire with numbers you can check by two independent routes.

The core insight

Elastic strain energy is the work done in deforming a solid within its elastic limit, equal to the area under the force–extension (or stress–strain) graph, so it grows with the square of the extension for a Hookean material rather than linearly.

The core idea

Why work becomes stored energy. Stretching a wire means pulling its atoms slightly farther apart against the interatomic restoring force. As long as the deformation stays elastic (Hooke's law holds), no energy is dissipated as heat or permanent rearrangement — every joule of work done by the stretching force is stored in the strained bonds and released again when the wire is allowed to relax. This is exactly why a stretched bowstring or a compressed spring can do work on something else: it is giving back energy that was put into it.

Work is the area under the graph, not force times extension. Because the wire obeys Hooke's law, the force needed to stretch it grows linearly with extension, from $0$ up to its final value $F$ as the extension grows from $0$ to $\Delta L$ . The force is not constant during the stretch, so the work done is not simply $F \times \Delta L$ ; it is the area of the triangle traced out on a force–extension graph:

W = \frac{1}{2} F\, \Delta L

where: $F$ is the final applied force and $\Delta L$ is the final extension. Since no energy is lost for an ideal elastic body, this work equals the elastic potential energy $U$ stored in the wire, $U = W$ .

A per-volume version: strain energy density. Dividing the stored energy by the wire's original volume $V = A L$ (cross-section area $A$ , natural length $L$ ) gives the energy stored per unit volume, written in terms of stress $\sigma = F/A$ and strain $\varepsilon = \Delta L / L$ :

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

where: $u$ is strain energy density (joule per cubic metre), $\sigma$ is the tensile stress and $\varepsilon$ is the tensile strain, both dimensionless-ratio-free quantities that describe the deformation independent of the wire's particular size. This is the same triangular-area idea as $W = \tfrac12 F \Delta L$ , just rewritten per unit volume — the stress–strain graph is simply the force–extension graph rescaled by $A$ and $L$ .

Figure

A force extension graph with a straight line from the origin to a marked point, the triangular area beneath the line shaded to represent stored elastic energy

Common facts you should recognise

Qualitatively, the elastic energy stored per unit volume is greater when the material is stiffer (a larger force is needed for the same extension) or when the wire is stretched further before the elastic limit is reached. Beyond the elastic limit, the force–extension curve bends away from the straight line, so the simple triangular-area rule no longer applies, and some of the work done goes into permanent, non-recoverable deformation rather than stored elastic energy.

Derivation AWork done stretching a Hookean wire equals half the final force times the extension

The goal is to show, using only the fact that a Hookean force grows in direct proportion to extension and the geometric formula for the area of a triangle, that the work done stretching a wire from natural length to extension $\Delta L$ under a linearly growing force is exactly $\tfrac12 F \Delta L$ — the result every strain-energy calculation in this article depends on, obtained here without calculus, matching the syllabus's qualitative-to-graphical treatment.

Step 1. Picture the stretching process broken into many tiny steps of extension, each of size $\delta x$ , starting from zero extension and ending at $\Delta L$ . At any stage where the extension so far is $x$ , Hooke's law says the force needed is proportional to $x$ : $F(x) = kx$ for some constant $k$ (the wire's effective stiffness), rising from $F(0) = 0$ to $F(\Delta L) = k\Delta L = F$ .

Step 2. In each tiny step, the force is nearly constant (since $\delta x$ is small), so the tiny bit of work done is approximately $F(x)\,\delta x$ — force times the small extension produced. Plotted on a force-versus-extension graph, this tiny work is the area of a thin vertical strip of height $F(x)$ and width $\delta x$ sitting under the line $F = kx$ .

Step 3. Adding up all these thin strips from $x = 0$ to $x = \Delta L$ builds up exactly the region under the straight line $F = kx$ , which is a right-angled triangle with base $\Delta L$ (the total extension) and height $F$ (the final force, since the force grows linearly and finishes at $F$ when $x = \Delta L$ ). The total work done is therefore the geometric area of this triangle.

Step 4. Apply the standard triangle-area formula, area equals half base times height: $W = \frac{1}{2}(\Delta L)(F) = \frac{1}{2} F \Delta L$ . Because the deformation is elastic and no energy is dissipated, this work is stored entirely as elastic potential energy, $U = \frac{1}{2} F \Delta L$ . Dividing both sides by the wire's volume $AL$ and writing $F/A = \sigma$ , $\Delta L/L = \varepsilon$ converts this directly into the energy-density form $u = \tfrac12 \sigma \varepsilon$ . Verified: for $F = 400\ \text{N}$ , $\Delta L = 2.0\times10^{-3}\ \text{m}$ , the triangle-area formula gives $W = \tfrac12(400)(2.0\times10^{-3}) = 0.40\ \text{J}$ , matching the direct area-under-the-graph calculation.

Worked example

A wire requires a force of $F = 400\ \text{N}$ to stretch it by $\Delta L = 2.0\ \text{mm}$ , obeying Hooke's law throughout. The wire's natural length is $L = 2.0\ \text{m}$ and its cross-sectional area is $A = 1.0\times10^{-6}\ \text{m}^2$ . Find the elastic potential energy stored, and check the result using the energy-density formula.

Step 1. Elastic potential energy from the area under the force–extension graph. With $\Delta L = 2.0\times10^{-3}\ \text{m}$ , $U = \tfrac12 F \Delta L = \tfrac12 (400)(2.0\times10^{-3}) = 0.40\ \text{J}$ .

Step 2. Find the wire's volume. $V = AL = (1.0\times10^{-6})(2.0) = 2.0\times10^{-6}\ \text{m}^3$ . The energy density is therefore $u = U/V = 0.40 / (2.0\times10^{-6}) = 2.0\times10^{5}\ \text{J/m}^3$ .

Step 3. Check by computing $u$ directly from stress and strain. Stress $\sigma = F/A = 400/(1.0\times10^{-6}) = 4.0\times10^{8}\ \text{Pa}$ , and strain $\varepsilon = \Delta L/L = (2.0\times10^{-3})/2.0 = 1.0\times10^{-3}$ . Then $u = \tfrac12 \sigma \varepsilon = \tfrac12 (4.0\times10^{8})(1.0\times10^{-3}) = 2.0\times10^{5}\ \text{J/m}^3$ — matching Step 2 exactly, confirming $U = 0.40\ \text{J}$ stored in the wire.

Competitive stretch

A spring stores $2.0\ \text{J}$ of elastic potential energy when stretched by $5.0\ \text{cm}$ from its natural length. If instead it is stretched to $10.0\ \text{cm}$ (still elastic), how much energy is stored? Since $U = \tfrac12 k x^2$ for a Hookean spring, energy scales with $x^2$ ; doubling $x$ multiplies $U$ by $2^2 = 4$ , so $U_2 = 4 \times 2.0 = \mathbf{8.0\ J}$ — quadrupling, not doubling, because stretching further both increases the force and the distance it acts over.

Common traps

Using $W = F \times \Delta L$ (force times the full extension) instead of $\tfrac12 F \Delta L$ — this overcounts because the force grows from zero and is not constant throughout the stretch; only the area under the graph gives the correct work.
Forgetting that stored elastic energy scales with the square of extension or strain, not linearly — doubling the stretch quadruples the stored energy, as long as the elastic limit is not exceeded.
Assuming all deformation work is recoverable — true only within the elastic limit; beyond it, part of the work becomes permanent, non-recoverable deformation and is not returned when the load is removed.
Mixing up $U = \tfrac12 F \Delta L$ (total energy, joules) with $u = \tfrac12 \sigma \varepsilon$ (energy per unit volume, joule per cubic metre) — the second needs dividing by volume, or equivalently using stress and strain instead of force and extension.

Ask the tutor: Want to work through more force–extension or stress–strain graph problems, or see how this connects to spring potential energy from Work, Energy and Power?

Why is the work done not simply force times extension?

Because the force itself is not constant while the wire stretches. At the very start of stretching there is almost no resistance, and the force needed grows steadily as the extension grows, reaching its final value $F$ only at the final extension $\Delta L$ . Multiplying the final force by the full extension would overcount the work, since that force was not acting for the whole stretch. The area under the force–extension graph correctly accounts for the force being smaller throughout most of the process, and for a straight-line (Hookean) graph that area works out to exactly half of $F \times \Delta L$ .

Does all the energy put into stretching a wire always come back out?

Only if the wire stays within its elastic limit. Inside that limit, the deformation is fully recoverable and essentially no energy is lost, so releasing the wire returns (very nearly) all the stored elastic potential energy, for example as kinetic energy of a released mass. Beyond the elastic limit, however, the material undergoes some permanent rearrangement of its structure; part of the work done goes into that permanent change and cannot be recovered, which is why an overstretched wire does not spring back to its original length.

What is the practical difference between U and u?

$U$ is the total elastic potential energy stored in a specific object, measured in joules, and depends on that object's actual size, since $U = \tfrac12 F \Delta L$ uses the actual force and extension. $u$ is the energy stored per unit volume, measured in joule per cubic metre, and depends only on the material's stress and strain, $u = \tfrac12 \sigma \varepsilon$ — it lets you compare how "energy-dense" different materials are regardless of how big or small the sample is. Multiplying $u$ by the actual volume of a given wire recovers its total $U$ , as shown in the worked example.