Derivation

Elasticity: Hooke's Law, Young's Modulus and Elastic Energy

D-424 Home PU-307 Threads force · matter · waves Depends on Conservative Forces and Potential Energy

Statement

For a linearly elastic body of uniform cross-section \(A\), original length \(L\), and Young’s modulus \(Y\), the restoring force is proportional to the displacement from equilibrium: \(F = -kx\) for a spring, or \(F = -\frac{YA}{L}\Delta L\) for a rod, where \(k = YA/L\) is the effective spring constant. The elastic potential energy stored is \(U = \frac{1}{2}kx^2 = \frac{1}{2}F\Delta L\), and the energy density is \(u = \frac{\sigma^2}{2Y} = \frac{1}{2}Y\epsilon^2\), where \(\sigma = F/A\) is stress and \(\epsilon = \Delta L/L\) is strain.

Why it matters

Hooke’s law is the universal first-order approximation for any stable mechanical equilibrium. Whether you are analyzing a car suspension, a bridge cable, or the vibration of a crystal lattice, the linear force-displacement relationship \(F = -kx\) is the starting point. It allows us to model complex elastic behavior with simple harmonic oscillators and to calculate stored energy without solving differential equations.

Young’s modulus \(Y\) connects this macroscopic behavior to the material’s microscopic structure. By relating stress to strain, it provides a single number that characterizes the stiffness of a material, independent of its shape. This enables engineers to predict how structures will deform under load and to select appropriate materials for specific applications.

Assumptions

The material is linearly elastic.The stress is proportional to strain, and the material returns to its original shape when the load is removed. If the load exceeds the elastic limit, plastic deformation occurs, and Hooke’s law fails.

The strain is small.Typically \(\epsilon \ll 1\). For large strains, the relationship between stress and strain becomes nonlinear, and the simple formula \(F = (YA/L)\Delta L\) no longer holds.

The cross-section \(A\) is uniform.If the cross-section varies along the length, the stress and strain are not uniform, and the effective spring constant must be calculated by integrating over the length.

The material is isotropic.Young’s modulus is the same in all directions. For anisotropic materials (e.g., wood, composites), the modulus depends on the direction of the applied force, and a single value \(Y\) is insufficient.

The deformation is quasi-static.The load is applied slowly enough that inertial effects are negligible. For dynamic loading, the wave speed and frequency of the load matter, and the static stiffness \(k = YA/L\) is not the full picture.

Derivation

\[ U(x) = U(x_0) + \frac{1}{2}U''(x_0)(x-x_0)^2 + \cdots \]

Taylor expansion of the potential energy about the equilibrium position \(x_0\), where \(U'(x_0)=0\). A

\[ F = -\frac{dU}{dx} \approx -U''(x_0)(x-x_0) \]

The force is the negative gradient of the potential. Keeping only the linear term in \((x-x_0)\) gives Hooke’s law. B

\[ k \equiv U''(x_0) \]

The spring constant \(k\) is the curvature of the potential energy at the minimum. A

\[ \sigma = \frac{F}{A}, \quad \epsilon = \frac{\Delta L}{L} \]

Define stress \(\sigma\) (force per unit area) and strain \(\epsilon\) (fractional change in length). A

\[ Y = \frac{\sigma}{\epsilon} = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L} \]

Young’s modulus \(Y\) is the ratio of stress to strain in the linear regime. A

\[ F = \frac{Y A}{L} \Delta L \]

Rearranging the definition of \(Y\) gives the force in terms of the extension \(\Delta L\). A

\[ k = \frac{Y A}{L} \]

The effective spring constant of a rod is \(k = YA/L\). B

\[ U = \int_0^{\Delta L} F(\Delta L') \, d\Delta L' = \int_0^{\Delta L} \frac{Y A}{L} \Delta L' \, d\Delta L' \]

The elastic potential energy is the work done to stretch the rod from 0 to \(\Delta L\). B

\[ U = \frac{Y A}{2L} (\Delta L)^2 = \frac{1}{2} k (\Delta L)^2 \]

Evaluate the integral; the energy is proportional to the square of the extension. A

\[ u = \frac{U}{V} = \frac{U}{A L} = \frac{1}{2} \frac{Y A}{L} \frac{(\Delta L)^2}{A L} = \frac{1}{2} Y \epsilon^2 \]

The energy density \(u\) is the energy per unit volume. B

\[ u = \frac{\sigma^2}{2Y} \]

Using \(\sigma = Y \epsilon\), we can also express the energy density in terms of stress. C

\[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \]

For springs in series, the total extension is the sum of the individual extensions, and the force is the same. B

\[ k = k_1 + k_2 \]

For springs in parallel, the total force is the sum of the individual forces, and the extension is the same. A

\[ v = \sqrt{\frac{Y}{\rho}} \]

Longitudinal waves in a thin rod: for a slice of length \(dx\) with displacement \(\xi(x,t)\), the tension is \(F = YA\,\partial\xi/\partial x\) (Step 6 applied locally), and Newton’s second law \(\rho A\,dx\,\partial_t^2\xi = (\partial F/\partial x)\,dx\) gives \(\partial_t^2\xi = (Y/\rho)\,\partial_x^2\xi\), a wave equation with speed \(\sqrt{Y/\rho}\). For steel, \(\sqrt{2.0\times10^{11}/7850} \approx 5.0\ \mathrm{km/s}\). B

Result

\[ F = -kx, \quad k = \frac{YA}{L}, \quad U = \frac{1}{2}kx^2 = \frac{1}{2}Y\epsilon^2 V \]

Reading. The force is linear in displacement, the spring constant depends on the material and geometry, and the stored energy is quadratic in the extension.

Units check. \([Y] = \mathrm{Pa = N/m^2}\), \([A] = \mathrm{m^2}\), \([L] = \mathrm{m}\). Thus \([k] = \mathrm{(N/m^2)(m^2)/m = N/m}\). \([U] = \mathrm{(N/m)(m)^2 = N\cdot m = J}\). \([u] = \mathrm{J/m^3 = N/m^2 = Pa}\). Consistent.

Limiting cases

Small strain. As \(\epsilon \to 0\), the linear approximation \(F = (YA/L)\Delta L\) becomes exact, and the energy is purely quadratic.
Very stiff material. As \(Y \to \infty\), the spring constant \(k \to \infty\), and the extension \(\Delta L \to 0\) for a fixed force.
Very long rod. As \(L \to \infty\), the spring constant \(k \to 0\), and the rod becomes very compliant.
Series vs. parallel. Springs in series are softer (\(1/k_{\text{series}} = \sum 1/k_i\)), while springs in parallel are stiffer (\(k_{\text{parallel}} = \sum k_i\)).

Breaks when

Beyond the elastic limit. If the stress exceeds the yield strength, the material undergoes plastic deformation, and the force-displacement relationship is no longer linear or reversible.
Large strains. Metals leave the linear range at strains of order \(10^{-3}\) (steel yields near \(\epsilon \approx 0.1\%\)\(-\)\(0.5\%\)); rubbers stay elastic to strains of several hundred per cent but are strongly nonlinear. In both cases \(F \propto \Delta L\) no longer holds.
Anisotropic materials. For materials like wood or composites, the modulus depends on the direction of the load, and a single value \(Y\) is insufficient.
Rubber and polymers. These materials exhibit entropic elasticity, where the restoring force arises from the entropy of the polymer chains, not from the curvature of the interatomic potential.

Failure modes

Confusing stress and force. Stress is force per unit area (\(\sigma = F/A\)), not force. Forgetting the area factor leads to incorrect calculations of strain and energy.
Using the wrong spring constant. For a rod, \(k = YA/L\), not \(Y\). Forgetting the geometric factors \(A\) and \(L\) leads to errors in the force and energy.
Assuming linearity for large strains. Hooke’s law is only valid for small strains. For large deformations, the force-displacement relationship is nonlinear, and the energy is not quadratic.
Ignoring the direction of the load. For anisotropic materials, the modulus depends on the direction of the load. Using a single value \(Y\) for all directions is incorrect.

Discussion

Hooke’s law is a consequence of the stability of equilibrium. Any smooth potential energy function has a minimum, and near that minimum, the potential is approximately quadratic. This leads to a linear restoring force, which is the essence of Hooke’s law. The spring constant \(k\) is the curvature of the potential, and it determines the frequency of small oscillations around the equilibrium.

Young’s modulus \(Y\) is a material property that connects the microscopic interatomic forces to the macroscopic stiffness of a solid. It is determined by the curvature of the interatomic potential and the density of atoms in the material. For a typical solid, \(Y \sim k_{\text{bond}}/a\), where \(k_{\text{bond}}\) is the spring constant of a single bond and \(a\) is the interatomic spacing.

The energy stored in an elastic body is proportional to the square of the strain. This quadratic dependence is a direct consequence of the linear force-displacement relationship. The energy density \(u = \frac{1}{2}Y\epsilon^2\) is a measure of the work done per unit volume to deform the material. It is a key quantity in the analysis of stress waves and vibrations in solids.

At a deeper level, the same quadratic expansion of the interatomic potential is the harmonic approximation of lattice dynamics, and quantising it gives phonons. The vibrational modes of a crystal lattice are quantized as phonons, and the dispersion relation for acoustic phonons is linear at low frequencies, corresponding to the elastic regime. The speed of longitudinal sound in a thin rod, \(v = \sqrt{Y/\rho}\), is the long-wavelength slope \(d\omega/dk\) of that acoustic branch. This connection between elasticity and quantum mechanics is a beautiful example of how macroscopic properties emerge from microscopic physics.

Common misconceptions. Hooke’s law is not a fundamental law of nature; it is an approximation that holds for small strains. The force is not always proportional to the displacement; for large strains, the relationship is nonlinear. Also, the spring constant \(k\) is not a material property; it depends on the geometry of the object. Young’s modulus \(Y\), on the other hand, is a material property and is independent of the geometry.

Worked examples

Example 1. A steel rod of length \(L = 2.0\ \mathrm{m}\) and cross-sectional area \(A = 1.0 \times 10^{-4}\ \mathrm{m^2}\) is stretched by a force \(F = 1000\ \mathrm{N}\). Given \(Y = 200\ \mathrm{GPa}\), find the extension \(\Delta L\) and the elastic energy stored.

\[ \Delta L = \frac{F L}{Y A} \]

Rearrange the definition of Young’s modulus to solve for the extension. A

\[ \Delta L = \frac{(1000\ \mathrm{N})(2.0\ \mathrm{m})}{(200 \times 10^9\ \mathrm{Pa})(1.0 \times 10^{-4}\ \mathrm{m^2})} \]

Substitute the given values. A

\[ \Delta L = \frac{2000}{2.0 \times 10^7}\ \mathrm{m} = 1.0 \times 10^{-4}\ \mathrm{m} = 0.10\ \mathrm{mm} \]

Calculate the extension. A

\[ U = \frac{1}{2} F \Delta L = \frac{1}{2} (1000\ \mathrm{N})(1.0 \times 10^{-4}\ \mathrm{m}) \]

Calculate the elastic energy stored. A

\[ U = 0.050\ \mathrm{J} \]

Final answer for the energy. A

\[ \Delta L = 0.10\ \mathrm{mm}, \quad U = 0.050\ \mathrm{J} \]

Reading. The rod extends by 0.10 mm and stores 0.050 J of elastic energy.

Units check. \([F L / Y A] = \mathrm{(N\cdot m)/(Pa\cdot m^2) = (N\cdot m)/(N/m^2 \cdot m^2) = m}\). \([U] = \mathrm{N\cdot m = J}\). Consistent.

Example 2. Two springs with spring constants \(k_1 = 100\ \mathrm{N/m}\) and \(k_2 = 200\ \mathrm{N/m}\) are connected in series. Find the effective spring constant \(k_{\text{eff}}\) and the extension if a force \(F = 30\ \mathrm{N}\) is applied.

\[ \frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} \]

For springs in series, the reciprocal of the effective spring constant is the sum of the reciprocals. A

\[ \frac{1}{k_{\text{eff}}} = \frac{1}{100} + \frac{1}{200} = \frac{2}{200} + \frac{1}{200} = \frac{3}{200} \]

Calculate the sum of the reciprocals. A

\[ k_{\text{eff}} = \frac{200}{3}\ \mathrm{N/m} \approx 66.7\ \mathrm{N/m} \]

Invert to find the effective spring constant. A

\[ x = \frac{F}{k_{\text{eff}}} = \frac{30\ \mathrm{N}}{66.7\ \mathrm{N/m}} \]

Calculate the total extension. A

\[ x \approx 0.45\ \mathrm{m} \]

Final answer for the extension. A

\[ k_{\text{eff}} \approx 66.7\ \mathrm{N/m}, \quad x \approx 0.45\ \mathrm{m} \]

Reading. The effective spring constant is 66.7 N/m, and the total extension is 0.45 m.

Units check. \([1/k] = \mathrm{m/N}\), \([k] = \mathrm{N/m}\). \([x] = \mathrm{N/(N/m) = m}\). Consistent.

Problems

A spring with spring constant \(k = 50\ \mathrm{N/m}\) is stretched by \(x = 0.20\ \mathrm{m}\). Find the force and the elastic energy stored.
Solution
Force: \(F = kx = (50\ \mathrm{N/m})(0.20\ \mathrm{m}) = 10\ \mathrm{N}\). Energy: \(U = \frac{1}{2}kx^2 = \frac{1}{2}(50\ \mathrm{N/m})(0.20\ \mathrm{m})^2 = \frac{1}{2}(50)(0.04) = 1.0\ \mathrm{J}\).
A copper wire of length \(L = 1.0\ \mathrm{m}\) and diameter \(d = 1.0\ \mathrm{mm}\) is stretched by a force \(F = 10\ \mathrm{N}\). Given \(Y = 110\ \mathrm{GPa}\), find the extension \(\Delta L\).
Solution
Area: \(A = \pi (d/2)^2 = \pi (0.5 \times 10^{-3}\ \mathrm{m})^2 = \pi (0.25 \times 10^{-6}) \approx 7.85 \times 10^{-7}\ \mathrm{m^2}\). Extension: \(\Delta L = \frac{F L}{Y A} = \frac{(10\ \mathrm{N})(1.0\ \mathrm{m})}{(110 \times 10^9\ \mathrm{Pa})(7.85 \times 10^{-7}\ \mathrm{m^2})} = \frac{10}{8.635 \times 10^4} \approx 1.16 \times 10^{-4}\ \mathrm{m} = 0.116\ \mathrm{mm}\).
Two springs with spring constants \(k_1 = 100\ \mathrm{N/m}\) and \(k_2 = 200\ \mathrm{N/m}\) are connected in parallel. Find the effective spring constant and the extension if a force \(F = 30\ \mathrm{N}\) is applied.
Solution
Effective spring constant: \(k_{\text{eff}} = k_1 + k_2 = 100 + 200 = 300\ \mathrm{N/m}\). Extension: \(x = \frac{F}{k_{\text{eff}}} = \frac{30\ \mathrm{N}}{300\ \mathrm{N/m}} = 0.10\ \mathrm{m}\).
A steel rod of length \(L = 1.0\ \mathrm{m}\) and cross-sectional area \(A = 1.0 \times 10^{-4}\ \mathrm{m^2}\) is subjected to a stress \(\sigma = 100\ \mathrm{MPa}\). Given \(Y = 200\ \mathrm{GPa}\), find the strain \(\epsilon\) and the energy density \(u\).
Solution
Strain: \(\epsilon = \frac{\sigma}{Y} = \frac{100 \times 10^6\ \mathrm{Pa}}{200 \times 10^9\ \mathrm{Pa}} = 5.0 \times 10^{-4}\). Energy density: \(u = \frac{1}{2} Y \epsilon^2 = \frac{1}{2} (200 \times 10^9\ \mathrm{Pa}) (5.0 \times 10^{-4})^2 = \frac{1}{2} (200 \times 10^9) (25 \times 10^{-8}) = 2.5 \times 10^4\ \mathrm{J/m^3}\). Check: \(\sigma^2/2Y = (10^8)^2/(4\times10^{11}) = 2.5\times10^4\ \mathrm{J/m^3}\); the whole rod (volume \(10^{-4}\ \mathrm{m^3}\)) stores \(2.5\ \mathrm{J}\).
A longitudinal wave travels through a thin aluminum rod with speed \(v = 5100\ \mathrm{m/s}\). Given the density \(\rho = 2700\ \mathrm{kg/m^3}\), find Young’s modulus \(Y\) for aluminum.
Solution
Using \(v = \sqrt{Y/\rho}\), we have \(Y = \rho v^2\). Substitute the values: \(Y = (2700\ \mathrm{kg/m^3}) (5100\ \mathrm{m/s})^2 = 2700 \times 2.601 \times 10^7 = 7.02 \times 10^{10}\ \mathrm{Pa} = 70.2\ \mathrm{GPa}\).