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Elastic energy

Elastic energy is recoverable energy stored when a body deforms. For an ideal spring it depends only on extension; for a continuous solid it is distributed through the material. The energy must be derived from the same force or stress law used to describe the deformation.

Ideal linear spring

Let extension \(x\) be measured from the spring's unstretched equilibrium length. Hooke's law is

\[F_s=-kx\]

where stiffness \(k\) has SI units of newtons per metre. The minus sign means the spring force opposes displacement. Potential-energy difference is defined by

\[U(x)-U(x_0)=-\int_{x_0}^{x}F_s(s)\,ds\]

Choosing zero energy at \(x_0=0\) gives

\[U_s(x)=\int_0^xks\,ds=\frac12kx^2\]

Energy is nonnegative for positive stiffness and is the same for equal compression and extension. Doubling displacement quadruples energy. Force follows from the negative energy slope:

\[F_s=-\frac{dU_s}{dx}=-kx\]

This derivative check catches a sign mismatch between force and energy that a positive energy value alone would miss.

Work and system boundary

If an external agent stretches the spring quasistatically, the agent applies approximately \(F_{\mathrm{agent}}=kx\). Its work from zero to extension is

\[W_{\mathrm{agent}}=\int_0^xF_{\mathrm{agent}}\,dx=\frac12kx^2\]

The spring's work over the same outward motion is negative. These signs are compatible: energy enters the spring system from the agent. If the mass moves rapidly, kinetic energy also changes and external work need not equal spring-energy change at every instant.

The zero of potential energy is arbitrary. Adding a constant changes no force or energy difference.

import numpy as np

reference_length, current_length, stiffness = 1.0, 1.05, 200.0
extension = current_length - reference_length
force = -stiffness * extension
stored_energy = 0.5 * stiffness * extension**2
print(f"Restoring force: {force:.2f} N; stored energy: {stored_energy:.3f} J")

Keep length in metres and stiffness in newtons per metre so energy is in joules.

Force-extension curves

For any one-dimensional conservative elastic element, energy change is the signed area under the opposing force-extension curve. If restoring force is nonlinear, do not retain the quadratic formula. For example,

\[F_s(x)=-kx-\beta x^3\]

corresponds to

\[U(x)=\frac12kx^2+\frac14\beta x^4+C\]

when the coefficients and range make the potential appropriate. Differentiating must recover the complete force, including nonlinear terms.

For tabulated force data, numerical quadrature can estimate work. Sample order, loading direction, units, and integration method must remain attached to the result.

extensions = np.linspace(0, 0.05, 101)
opposing_forces = stiffness * extensions
energy_change = np.trapezoid(opposing_forces, extensions)
assert np.isclose(energy_change, stored_energy)
print(f"Integrated loading work: {energy_change:.3f} J")

Refine sampling and compare with an analytic linear-spring fixture before trusting an irregular material curve.

Axial strain energy in a uniform bar

For a uniform bar of original length \(L\), area \(A\), and Young's modulus \(E\), small linear axial extension is

\[\Delta L=\frac{FL}{EA}\]

The effective axial stiffness is \(k=EA/L\). Therefore its strain energy is

\[U=\frac12F\Delta L=\frac{F^2L}{2EA}=\frac{EA}{2L}(\Delta L)^2\]

These equivalent forms assume the force rises quasistatically from zero under a linear law. Using simply \(F\Delta L\) would double the energy because force is not at its final value throughout loading.

Strain-energy density

For uniaxial linear elasticity, energy per original volume is

\[u=\int_0^{\varepsilon}\sigma(\varepsilon')\,d\varepsilon'=\frac12\sigma\varepsilon=\frac12E\varepsilon^2\]

Stress has units of energy per volume because \(1\,\mathrm{Pa}=1\,\mathrm{J/m^3}\). Integrating density over volume gives total strain energy:

\[U=\int_Vu\,dV\]

If stress and strain vary along the body, one final average force and extension may not describe the local distribution, while the volume integral can.

Shear and bulk energy

In small linear shear, \(\tau=G\gamma\). The energy density is

\[u_{\mathrm{shear}}=\frac12\tau\gamma=\frac12G\gamma^2\]\n+ For small uniform volume change, bulk modulus \(B\) gives an analogous quadratic energy density in fractional volume strain. Sign conventions for pressure and volumetric strain must be handled consistently; stored energy remains nonnegative for a stable positive modulus.

In general three-dimensional linear elasticity,

\[u=\frac12\boldsymbol\sigma:\boldsymbol\varepsilon\]

where the double contraction sums matched tensor components. This formula requires compatible stress and strain conventions; engineering shear strain introduces factors that must not be mixed with tensor shear components.

Series and parallel springs

For ideal massless springs in series, the force is common and extensions add:

\[\frac1{k_{\mathrm{series}}}=\sum_i\frac1{k_i}\]

For springs in parallel with common extension, forces add:

\[k_{\mathrm{parallel}}=\sum_i k_i\]

Total stored energy equals the sum over individual springs. Calculating each term is a useful check on equivalent-stiffness energy.

Hysteresis and dissipation

If loading and unloading forces differ at the same extension, force is not a single-valued function of extension alone. The closed-loop work

\[W_{\mathrm{loop}}=\oint F\,dx\]

measures energy dissipated per cycle under the chosen signs. A single conservative potential cannot represent hysteresis without additional internal state variables.

Plastic deformation, viscoelasticity, dry friction, damage, and temperature-dependent behavior can all make work path-dependent. Report recoverable unloading energy separately from total work supplied and dissipated energy.

Stability and positive energy

Near a stable unloaded configuration, small deformation energy should increase for every allowed nonzero deformation. In one dimension this requires positive stiffness. In a multi-coordinate linear model

\[U=\frac12\mathbf q^{\mathsf T}\mathbf K\mathbf q\]

the stiffness matrix must be positive definite on unconstrained physical modes for a strict stable minimum. Zero modes can represent rigid translation or an unconstrained mechanism; negative modes indicate instability or an inconsistent tangent model.

Energy is not a failure criterion

Stored elastic energy alone does not determine whether an object yields, buckles, cracks, or fractures. Failure can depend on local stress, flaw geometry, loading rate, cyclic history, and fracture toughness. A body with modest total energy can still have a severe local stress concentration.

Hooke's law ceases to be valid outside its elastic range. Extrapolating the quadratic energy beyond yielding can greatly misrepresent both recoverable energy and force.

Numerical validation

For a conservative discrete model, compare force with the negative gradient of energy by finite differences. Halve the displacement used for the derivative and confirm convergence before roundoff dominates. Test zero deformation, sign reversal, quadratic scaling, and superposition where linearity is assumed.

For a deforming mesh, refine spatial resolution and compare both total energy and local energy density. A stable total can hide poorly resolved concentrations.

Reliability checks

Elastic energy connects force-displacement behavior with stress-strain behavior. It is a recoverable state function only for a conservative elastic model; nonlinear paths, hysteresis, plasticity, and failure require a richer ledger and additional material state.

A worked Python notebook

Stretch a spring with stiffness 200 newtons per metre from its unstretched length. The question for this calculation is precise: how does stored spring energy change when we change extension, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The loading is elastic and follows a linear force-extension law. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 0.1 m. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for extension, expressed in m. The output is stored spring energy, expressed in J. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 100 q^{2}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 0.05 m, the expression gives 0.25 J. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("100*q**2", locals={"q": q})
reference_input = 0.05
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted stored spring energy:", reference_output, "J")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 J. The midpoint, at 0.05 m, gives 0.25 J. The upper endpoint gives 1 J. Read these together with the assumptions: they describe different choices of extension within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 0.05, 0.1])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m -> {output_value:.6g} J")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows extension; the vertical axis shows stored spring energy. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 0.1, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Extension (m)")
ax.set_ylabel("Stored spring energy (J)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does stored spring energy change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.05 m is 10 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 200 q\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 0.0505
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "J")
print("Full expression:", full_prediction, "J")
print("Approximation difference:", full_prediction - linear_prediction, "J")

Here the changed input is 0.0505 m. The tangent estimate is 0.255 J, while the full expression gives 0.255025 J. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 0.1
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of extension inside the stated interval, predict whether stored spring energy will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The loading is elastic and follows a linear force-extension law. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of J: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(0.1) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 0.1, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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