Linear Elasticity and Hooke's Law for Solids

Continuum mechanics splits into kinematics, dynamics, and constitution. The strain tensor describes how a body deforms (kinematics); the stress tensor describes the internal forces transmitted across surfaces (dynamics, via Cauchy's relation). Neither alone determines how a real material responds to loading: that link is a constitutive relation, a statement about the material itself rather than about geometry or Newton's laws. Linear elasticity is the simplest physically grounded constitutive law, asserting that for small deformations the stress is a linear function of the strain. The historical anchor is Hooke's 1678 'ut tensio, sic vis' — as the extension, so the force. In its one-dimensional form this is the familiar spring law $F=-k\,x$ or, in continuum language, $\sigma=E\,\varepsilon$ for a rod under uniaxial tension. The generalization to three dimensions replaces the scalar spring constant by the rank-four stiffness tensor $C_{ijkl}$ relating the nine components of stress to the nine components of strain. The structure of this tensor — how many of its $81$ entries are truly independent — is dictated by tensor symmetries and by the existence of a strain-energy function. Three layers of symmetry collapse the tensor. The symmetry of the stress tensor ( $\sigma_{ij}=\sigma_{ji}$ , from angular-momentum balance) gives $C_{ijkl}=C_{jikl}$ ; the symmetry of the strain tensor gives $C_{ijkl}=C_{ijlk}$ ; and the existence of an elastic potential energy whose second derivatives commute gives the major symmetry $C_{ijkl}=C_{klij}$ . Together these reduce $81$ to $21$ independent moduli — the full anisotropic (triclinic) case. Material point-group symmetry reduces this further: cubic crystals have three constants, and full isotropy leaves only two. Linear elasticity is the leading term in an expansion of the energy in powers of strain, valid when strains are small (typically $<10^{-3}$ for metals) and well below yield. It is the bridge between the discrete loaded string and the continuum field theory: combining Hooke's law with Newton's second law per unit volume yields the elastic wave equation, whose Lagrangian density and Euler–Lagrange field equations place elasticity squarely within continuum classical field theory.

Key takeaways

Linear elasticity posits a linear, instantaneous, local relation between the stress tensor $\sigma_{ij}$ and the strain tensor $\varepsilon_{kl}$ : $\sigma_{ij}=C_{ijkl}\varepsilon_{kl}$ , where $C_{ijkl}$ is the rank-four elastic stiffness tensor.
Symmetries of stress, strain, and the existence of an elastic energy density reduce the $81$ components of $C_{ijkl}$ to at most $21$ independent constants for the most general (triclinic) anisotropic crystal.
For an isotropic solid only two independent moduli survive — the Lamé parameters $\lambda$ and $\mu$ — giving $\sigma_{ij}=\lambda\,\varepsilon_{kk}\delta_{ij}+2\mu\,\varepsilon_{ij}$ .
Engineering moduli (Young's modulus $E$ , Poisson's ratio $\nu$ , bulk modulus $K$ , shear modulus $G=\mu$ ) are algebraic combinations of $\lambda$ and $\mu$ ; thermodynamic stability requires $K>0$ and $\mu>0$ , equivalently $-1<\nu<1/2$ .
The stored elastic energy density is the quadratic form $u=\tfrac12 C_{ijkl}\varepsilon_{ij}\varepsilon_{kl}$ ; its positive-definiteness underpins both stability and the reality of elastic wave speeds.
Hooke's law is the constitutive closure that turns the kinematic strain tensor and the dynamical stress tensor into a complete field theory whose Euler–Lagrange equations are the elastic wave equations.
Key idea: Hooke's law $\sigma_{ij}=C_{ijkl}\varepsilon_{kl}$ is the constitutive closure of elasticity: kinematics gives strain, dynamics gives stress, and a single positive-definite quadratic energy $u=\tfrac12 C_{ijkl}\varepsilon_{ij}\varepsilon_{kl}$ ties them together, reducing the material response to at most $21$ — and for isotropy just $2$ — independent constants.

Where this lives today

Linear elasticity is the default physics inside every finite-element solver, and the frontier has moved from analysing the stiffness tensor to designing it. Mechanical metamaterials — 3D-printed lattices whose unit cells bend and rotate in engineered ways — now realize effective Poisson ratios below zero (auxetics that fatten when pulled, prized for impact padding and stents) and approach the theoretical corners of the $(K,\mu)$ plane: pentamode lattices with $\mu/K\sim10^{-2}$ behave like 'meta-fluids' that flow around shear while resisting compression. Soft robotics lives at the opposite corner, exploiting near-incompressible elastomers ( $\nu\approx0.4999$ ) whose motion is almost pure shape change. The anisotropic side of the theory earns its keep in silicon. Single-crystal silicon is cubic — three constants, $C_{11}=166$ , $C_{12}=64$ , $C_{44}=80$ GPa — so the Young's modulus of a MEMS resonator varies from $130$ GPa along $[100]$ to $188$ GPa along $[111]$ depending on how the die is cut; timing-grade MEMS oscillators budget this orientation dependence to parts per million. And whenever a seismometer inverts P- and S-wave speeds for $\lambda$ and $\mu$ deep in the Earth, it is running this article's algebra backwards — the constitutive law as a remote-sensing instrument.

The physical picture

Think of the material at each point as a tiny set of springs connecting neighbouring atoms. Stretch the springs a little and they pull back with a force proportional to the stretch — that proportionality, summed over directions and orientations, is the stiffness tensor. The smallness of the stretch is what makes the response linear: the energy of a spring near equilibrium is quadratic in displacement, so the force is linear. This is the tensor-valued grandchild of the scalar spring law $F=-kx$ from Spring Force and Hooke's Law; nothing about the logic changes, only the bookkeeping. Isotropy means the springs look the same in every direction, so the response can depend only on two scalars: one resisting volume change (compression) and one resisting shape change (shear). The bulk modulus $K$ measures the first, the shear modulus $\mu$ the second. Any deformation splits into a pure volume change plus a volume-preserving shape change, and these two channels respond independently — this orthogonal decomposition is the deep reason isotropy needs exactly two constants.

Figure

Every small strain splits into two independent channels — a pure volume change resisted by the bulk modulus K, and a volume-preserving shape change resisted by the shear modulus mu. Isotropy prices each channel with one number, which is why two constants suffice

Poisson's ratio captures a non-obvious coupling: stretch a rubber band lengthwise and it thins sideways. A material with $\nu\to1/2$ (rubber) preserves volume — all strain goes into shape change. A material with $\nu\to0$ (cork) does not contract laterally at all. The bound $-1<\nu<1/2$ is not arbitrary; it is exactly the condition that the elastic energy be positive for every possible deformation.

Figure

The two engineering moduli read off one experiment — pull a bar along its axis and it stretches by sigma over E while thinning sideways by nu times the axial strain

The $21$ versus $2$ constant counting is a symmetry story. Each material symmetry operation must leave $C_{ijkl}$ invariant, and the more symmetric the crystal, the fewer independent numbers survive. This is the elastic analogue of how a more symmetric Hamiltonian has fewer independent coupling constants — symmetry is the great economizer. One material, several slopes: the same block of steel answers a pull, a twist, and a squeeze with different stiffnesses, because each experiment excites a different mixture of the two channels. The figure below shows the stress–strain lines a single steel specimen traces in four canonical loading paths — this is the picture to hold when the modulus zoo starts to blur.

Figure

One material, four slopes — the same steel answers each loading path with a different effective modulus, every one an algebraic combination of just lambda and mu

The core equations

The whole theory is one linear map plus its symmetries. We state the general law, count its independent constants, specialize to isotropy, and then extract the moduli an engineer actually measures. The most general linear, local, instantaneous relation between two symmetric rank-two tensors is a rank-four tensor contraction. That single sentence, applied to stress and strain, is the generalized Hooke's law:

\sigma_{ij}=C_{ijkl}\,\varepsilon_{kl} \qquad (1)

where: $\sigma_{ij}$ — Cauchy stress tensor (force per area), symmetric; $\varepsilon_{kl}$ — infinitesimal strain tensor, symmetric; $C_{ijkl}$ — elastic stiffness tensor, rank four, $81$ components. Read it as a material fingerprint: geometry supplies $\varepsilon$ , Newton supplies the balance law for $\sigma$ , and $C_{ijkl}$ is everything the material itself gets to say. For a spring it was one number $k$ ; for a crystal it is a tensor. Not all $81$ entries carry information. Because stress and strain are symmetric and because the response is derivable from an energy, the stiffness tensor obeys three symmetry families:

C_{ijkl}=C_{jikl}=C_{ijlk}=C_{klij} \qquad (2)

where: $C_{jikl}$ — minor symmetry inherited from $\sigma_{ij}=\sigma_{ji}$ ; $C_{ijlk}$ — minor symmetry inherited from $\varepsilon_{kl}=\varepsilon_{lk}$ ; $C_{klij}$ — major symmetry from the existence of a strain-energy function. The counting goes in three moves. Stress symmetry collapses the first index pair from nine values to six, leaving $6\times9=54$ ; strain symmetry does the same to the second pair, leaving $6\times6=36$ ; and the major symmetry $C_{ijkl}=\partial^2 u/\partial\varepsilon_{ij}\partial\varepsilon_{kl}=C_{klij}$ — mixed partial derivatives of a smooth energy commute — makes the $6\times6$ Voigt matrix symmetric, leaving $(36+6)/2=21$ independent constants. Twenty-one is the ceiling; every material symmetry lowers it. For an isotropic solid, rotational invariance does almost all the remaining work. The only isotropic rank-four tensors available are products of Kronecker deltas, $a\,\delta_{ij}\delta_{kl}+b\,\delta_{ik}\delta_{jl}+c\,\delta_{il}\delta_{jk}$ , and the minor symmetry in $(kl)$ forces $b=c$ . Two constants survive, and they are the Lamé parameters:

C_{ijkl}=\lambda\,\delta_{ij}\delta_{kl}+\mu\,(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}) \qquad (3)

where: $\lambda$ — first Lamé parameter, coupling the traces; $\mu$ — shear modulus (second Lamé parameter), coupling the shears. Contracting with $\varepsilon_{kl}$ gives the isotropic Hooke's law in the form used everywhere downstream:

\sigma_{ij}=\lambda\,\varepsilon_{kk}\,\delta_{ij}+2\mu\,\varepsilon_{ij} \qquad (4)

where: $\varepsilon_{kk}$ — trace of strain, the fractional volume change (dilatation); $\delta_{ij}$ — Kronecker delta. The first term is a pressure-like response to volume change; the second resists any change of shape. A pure shear ( $\varepsilon_{kk}=0$ ) feels only $\mu$ ; a uniform compression feels only the combination $\lambda+\tfrac23\mu$ — the two channels of the hero figure, now in symbols. Nobody's testing machine reads out $\lambda$ . Laboratories pull rods and measure how much they stretch and thin, which defines Young's modulus and Poisson's ratio; Derivation A below extracts both from the tensor law:

E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu},\qquad \nu=\frac{\lambda}{2(\lambda+\mu)} \qquad (5)

where: $E$ — Young's modulus, uniaxial stress over uniaxial strain; $\nu$ — Poisson's ratio, lateral contraction over axial extension, dimensionless. These two are what a data sheet quotes; the inverse relations $\lambda=E\nu/[(1+\nu)(1-2\nu)]$ and $\mu=E/[2(1+\nu)]$ convert back whenever the tensor form is needed. Two more combinations complete the working set. Squeezing a body uniformly probes the bulk modulus; twisting it probes the shear modulus directly:

K=\lambda+\tfrac{2}{3}\mu=\frac{E}{3(1-2\nu)},\qquad \mu=G=\frac{E}{2(1+\nu)} \qquad (6)

where: $K$ — bulk modulus, pressure over fractional volume decrease; $G$ — engineering shear modulus, identical to $\mu$ . Any two of $\{\lambda,\mu,E,\nu,K\}$ determine the rest; the practical question is always which loading path is my experiment on, answered by the table below.

Loading path	Effective modulus	Slope for steel ( $E=200$ GPa, $\nu=0.29$ )
Thin rod pulled axially (uniaxial stress)	$E$	$200$ GPa
Pure shear, $\sigma_{xy}$ vs $\gamma_{xy}$	$\mu=G$	$78$ GPa
Hydrostatic pressure vs volume strain	$K$	$159$ GPa
Confined compression (uniaxial strain)	$M=\lambda+2\mu$	$262$ GPa

Behind all of this sits a single scalar bookkeeper: the elastic energy density. Loading a body quasi-statically stores work $\mathrm{d}u=\sigma_{ij}\,\mathrm{d}\varepsilon_{ij}$ per unit volume; with the linear law this integrates — thanks to the major symmetry, which makes the integrand exact — to a quadratic form:

u=\tfrac12\,C_{ijkl}\,\varepsilon_{ij}\varepsilon_{kl}=\tfrac12\,\sigma_{ij}\varepsilon_{ij} \qquad (7)

where: $u$ — stored strain energy per unit volume. This is the potential-energy term of the elastic Lagrangian density, and it is the object whose positivity disciplines the whole theory. Splitting the strain into trace and deviatoric parts diagonalizes the energy into the two physical channels:

u=\tfrac12 K\,(\varepsilon_{kk})^2+\mu\,e_{ij}e_{ij},\qquad e_{ij}=\varepsilon_{ij}-\tfrac13\varepsilon_{kk}\delta_{ij} \qquad (8)

where: $\varepsilon_{kk}$ — volumetric strain; $e_{ij}$ — deviatoric (traceless) strain governing shape change. No cross term survives — compression energy and shear energy add independently, which is why $K$ and $\mu$ can be constrained separately. Demanding $u>0$ for every nonzero strain — otherwise the unloaded state is not a stable minimum and the material would spontaneously deform — bounds the constants:

K>0,\quad \mu>0\;\Longleftrightarrow\; -1<\nu<\tfrac12 \qquad (9)

where: $K>0$ — resistance to compression; $\mu>0$ — resistance to shear. Derivation C makes the equivalence explicit. Real materials cluster between $\nu\approx0.2$ and $0.35$ ; rubber sits near $0.4999$ , and engineered auxetic lattices genuinely reach $\nu<0$ . Finally, closing Newton's second law per unit volume, $\rho\,\ddot u_i=\partial_j\sigma_{ij}$ , with the isotropic law turns statics into dynamics — the Navier (Navier–Cauchy) equation:

\rho\,\ddot{u}_i=(\lambda+\mu)\,\partial_i(\partial_j u_j)+\mu\,\nabla^2 u_i \qquad (10)

where: $\rho$ — mass density; $u_i$ — displacement field. Derivation B assembles it. Its plane-wave solutions travel at two speeds, $\sqrt{(\lambda+2\mu)/\rho}$ and $\sqrt{\mu/\rho}$ — the P and S waves whose full story belongs to Elastic Waves in Solids.

Derivation AFrom isotropic Hooke's law to the engineering moduli

We derive Young's modulus and Poisson's ratio — the two numbers a tensile test reports — from the two-parameter tensor law, by analysing a thin rod pulled along its axis. The strategy is to impose the stress state the experiment actually creates ( $\sigma_{xx}=\sigma$ , every other component zero, because the lateral faces are free) and solve the isotropic Hooke's law backwards for the strains. The one subtlety worth watching: uniaxial stress does not mean uniaxial strain — the rod thins sideways, and that lateral strain is exactly what Poisson's ratio measures.

Step 1. Contract Hooke's law $\sigma_{ij}=\lambda\varepsilon_{kk}\delta_{ij}+2\mu\varepsilon_{ij}$ on $i=j$ , using $\delta_{ii}=3$ :

\sigma_{kk}=(3\lambda+2\mu)\,\varepsilon_{kk}

Why this first: the trace isolates the volumetric response in one scalar equation, and in our stress state the trace is simply $\sigma_{kk}=\sigma$ .

Step 2. Solve for the dilatation:

\varepsilon_{kk}=\frac{\sigma}{3\lambda+2\mu}

Step 3. Write the $xx$ component of Hooke's law and substitute $\varepsilon_{kk}$ :

\sigma=\lambda\,\varepsilon_{kk}+2\mu\,\varepsilon_{xx}\;\Rightarrow\; \varepsilon_{xx}=\frac{\lambda+\mu}{\mu(3\lambda+2\mu)}\,\sigma

Step 4. Young's modulus is stress over axial strain, by definition:

E\equiv\frac{\sigma}{\varepsilon_{xx}}=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu}

Step 5. The $yy$ component has zero stress but not zero strain:

0=\lambda\,\varepsilon_{kk}+2\mu\,\varepsilon_{yy}\;\Rightarrow\; \varepsilon_{yy}=-\frac{\lambda}{2\mu}\,\varepsilon_{kk}

Why the minus sign matters: the free faces cannot carry stress, so the material trades the forbidden lateral stress for a lateral contraction — the sideways thinning every rubber band demonstrates.

Step 6. Poisson's ratio is the ratio of contraction to extension:

\nu\equiv-\frac{\varepsilon_{yy}}{\varepsilon_{xx}}=\frac{\lambda}{2(\lambda+\mu)}

As a self-consistency check: inverting the pair gives $\lambda=E\nu/[(1+\nu)(1-2\nu)]$ and $\mu=E/[2(1+\nu)]$ ; substituting these back into the expressions for $E$ and $\nu$ reproduces the originals identically.

Go deeper · Derivation B · 1 minThe Navier equation from Hooke plus Newton

Derivation BThe Navier equation from Hooke plus Newton

We combine the continuum form of Newton's second law, $\rho\,\ddot u_i=\partial_j\sigma_{ij}$ (no body force), with the isotropic constitutive law to get a closed equation of motion for the displacement field alone. The route is pure substitution — express strain through displacement gradients, insert into Hooke's law, take the divergence — but the payoff is structural: the result contains two stiffness combinations, $(\lambda+\mu)$ multiplying a gradient-of-divergence and $\mu$ multiplying a Laplacian, and that split is precisely what will separate compression waves from shear waves. Watch the index gymnastics in Step 3; commuting partial derivatives does the real work.

Step 1. Express strain kinematically:

\varepsilon_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i),\qquad \varepsilon_{kk}=\partial_k u_k=\nabla\!\cdot\!\mathbf u

Step 2. Insert into the isotropic Hooke's law:

\sigma_{ij}=\lambda\,(\partial_k u_k)\,\delta_{ij}+\mu\,(\partial_i u_j+\partial_j u_i)

Step 3. Take the divergence $\partial_j$ term by term, using $\partial_j\delta_{ij}=\partial_i$ and $\partial_j\partial_i u_j=\partial_i(\partial_j u_j)$ :

\partial_j\sigma_{ij}=\lambda\,\partial_i(\partial_k u_k)+\mu\,\partial_i(\partial_j u_j)+\mu\,\nabla^2 u_i

Why derivatives may be freely commuted: the displacement field is assumed twice continuously differentiable — the same smoothness that let us define strain in the first place.

Step 4. Collect the two gradient-of-divergence terms:

\partial_j\sigma_{ij}=(\lambda+\mu)\,\partial_i(\nabla\!\cdot\!\mathbf u)+\mu\,\nabla^2 u_i

Step 5. Equate to $\rho\,\ddot u_i$ and write the vector form:

\rho\,\ddot{\mathbf u}=(\lambda+\mu)\,\nabla(\nabla\!\cdot\!\mathbf u)+\mu\,\nabla^2\mathbf u

Step 6. Read off the two propagation speeds. A curl-free field ( $\nabla\times\mathbf u=0$ ) has $\nabla^2\mathbf u=\nabla(\nabla\cdot\mathbf u)$ , so it obeys a wave equation with speed $c_L=\sqrt{(\lambda+2\mu)/\rho}$ ; a divergence-free field kills the first term, leaving speed $c_T=\sqrt{\mu/\rho}$ . Setting $\nabla\cdot\mathbf u=0$ or $\nabla\times\mathbf u=0$ reproduces the transverse and longitudinal wave speeds respectively; the full Helmholtz treatment lives in Elastic Waves in Solids.

Go deeper · Derivation C · 1 minPositive-definite energy and the bounds on Poisson's ratio

Derivation CPositive-definite energy and the bounds on Poisson's ratio

We show that demanding the strain energy be positive for every possible deformation forces $K>0$ and $\mu>0$ , and translate those into the famous bounds $-1<\nu<1/2$ . The strategy is to decompose an arbitrary strain into its trace and deviatoric parts and watch the energy split into two independent quadratic forms — independence is the crux, because it lets us drive each channel separately and conclude each coefficient must be positive on its own. The step to watch is the vanishing of the cross term.

Step 1. Define the deviatoric strain and note it is traceless:

e_{ij}=\varepsilon_{ij}-\tfrac13\varepsilon_{kk}\delta_{ij},\qquad e_{kk}=0

Step 2. Start from $u=\tfrac12\sigma_{ij}\varepsilon_{ij}$ with the isotropic law:

u=\tfrac12\lambda\,(\varepsilon_{kk})^2+\mu\,\varepsilon_{ij}\varepsilon_{ij}

Step 3. Substitute $\varepsilon_{ij}=e_{ij}+\tfrac13\varepsilon_{kk}\delta_{ij}$ into the quadratic:

\varepsilon_{ij}\varepsilon_{ij}=e_{ij}e_{ij}+\tfrac13(\varepsilon_{kk})^2

Why the cross term dies: it is proportional to $e_{ij}\delta_{ij}=e_{kk}=0$ — the deviatoric part is orthogonal to the identity under the tensor inner product.

Step 4. Collect and recognize the bulk modulus:

u=\tfrac12\Big(\lambda+\tfrac{2}{3}\mu\Big)(\varepsilon_{kk})^2+\mu\,e_{ij}e_{ij}=\tfrac12 K\,(\varepsilon_{kk})^2+\mu\,e_{ij}e_{ij}

Step 5. Demand positivity channel by channel. A pure dilation has $e_{ij}=0$ but $\varepsilon_{kk}\ne0$ , forcing $K>0$ ; a pure shear has $\varepsilon_{kk}=0$ but $e_{ij}\ne0$ , forcing $\mu>0$ . Neither channel can subsidize the other.

Step 6. Translate through $K=E/[3(1-2\nu)]$ and $\mu=E/[2(1+\nu)]$ with $E>0$ : the first demands $\nu<\tfrac12$ , the second $\nu>-1$ :

-1<\nu<\tfrac12

As expected from thermodynamic stability, rubber approaching $\nu=1/2$ and auxetic foams with $\nu<0$ both sit safely inside the open interval $-1<\nu<1/2$ .

Limiting cases

The general law $\sigma_{ij}=C_{ijkl}\varepsilon_{kl}$ specializes dramatically as material symmetry increases:

Case	Result	Meaning
Triclinic (no symmetry)	$\text{21 independent } C_{ijkl}$	The most general anisotropic linear elastic solid. The full Voigt $6\times6$ matrix is symmetric with $21$ free entries.
Cubic crystal	$C_{11},\; C_{12},\; C_{44}$	Three independent constants. Examples: copper, iron, silicon. Isotropy requires the Zener ratio $A=2C_{44}/(C_{11}-C_{12})=1$ .
Isotropic solid	$\sigma_{ij}=\lambda\,\varepsilon_{kk}\delta_{ij}+2\mu\,\varepsilon_{ij}$	Two constants $(\lambda,\mu)$ . All directions equivalent; the most common engineering idealization.
Uniaxial 1D limit	$\sigma=E\,\varepsilon$	A thin rod free to contract laterally. Recovers Hooke's original scalar law with Young's modulus $E$ .
Incompressible limit	$\nu\to\tfrac12,\quad K\to\infty$	Volume preserved; only shear deformations cost energy. Rubber and many soft tissues approach this; Lamé $\lambda\to\infty$ .
Fluid limit	$\mu\to0,\quad \sigma_{ij}=-p\,\delta_{ij}$	No resistance to shape change; stress reduces to a pressure with $p=-K\varepsilon_{kk}$ . Shear waves cease to exist.

Common trapEngineering shear strain is twice the tensor shear strain

The single most expensive factor of two in mechanics: the strain-tensor component $\varepsilon_{xy}$ equals half the engineering shear $\gamma_{xy}=\partial_x u_y+\partial_y u_x$ . Hooke's law in tensor form reads $\sigma_{xy}=2\mu\,\varepsilon_{xy}=\mu\,\gamma_{xy}$ — feed it $\gamma$ where it expects $\varepsilon$ and every shear stress doubles. The same bookkeeping haunts Voigt notation: the $6\times6$ stiffness and compliance matrices are true matrix inverses of each other only when the shear-strain slots carry their factors of $2$ (and the compliance shear diagonals their factors of $4$ ). Check the convention of any table before using it.

Worked examples

Example 1 — from the data sheet to the full modulus set (confidence builder). A structural-steel data sheet quotes $E=200$ GPa and $\nu=0.29$ . Find $\mu$ , $\lambda$ , $K$ , and the confined modulus $M=\lambda+2\mu$ .

Step 1. Shear modulus from $\mu=E/[2(1+\nu)]$ :

\mu=\frac{200}{2(1.29)}=77.5\ \text{GPa}

Step 2. First Lamé parameter from $\lambda=E\nu/[(1+\nu)(1-2\nu)]$ :

\lambda=\frac{200\times0.29}{1.29\times0.42}=\frac{58}{0.5418}=107.1\ \text{GPa}

Step 3. Bulk modulus, two independent ways as a cross-check. From $K=E/[3(1-2\nu)]=200/(3\times0.42)=158.7$ GPa, and from $K=\lambda+\tfrac23\mu=107.1+51.7=158.8$ GPa — agreement to rounding.

Step 4. Confined (P-wave) modulus:

M=\lambda+2\mu=107.1+155.0=262.1\ \text{GPa}

Answer: $\mu\approx77.5$ GPa, $\lambda\approx107$ GPa, $K\approx159$ GPa, $M\approx262$ GPa. Note the ordering $\mu<K<E<M$ — a useful sanity ladder for any metal with $\nu$ near $0.3$ .

Example 2 — a loaded steel column, in millimetres (exam level). A cylindrical steel column, diameter $100$ mm and length $3.00$ m, carries an axial compressive load of $400$ kN. Using $E=200$ GPa and $\nu=0.29$ , find the axial shortening, the change in diameter, the fractional volume change, and the stored elastic energy.

Step 1. Cross-section and stress. $A=\pi r^2=\pi(0.050)^2=7.854\times10^{-3}\ \text{m}^2$ , so

\sigma=\frac{4.00\times10^{5}}{7.854\times10^{-3}}=5.09\times10^{7}\ \text{Pa}=50.9\ \text{MPa}

compressive — comfortably inside the linear regime (yield for structural steel is $\sim250$ MPa).

Step 2. Axial strain and shortening. $\varepsilon_{\parallel}=\sigma/E=5.09\times10^{7}/2.00\times10^{11}=2.55\times10^{-4}$ , so the column shortens by

\Delta L=\varepsilon_{\parallel}L=2.55\times10^{-4}\times3000\ \text{mm}=0.764\ \text{mm}

Step 3. Lateral response. $\varepsilon_{\perp}=-\nu\,\varepsilon_{\parallel}$ ; under compression the axial strain is negative, so the lateral strain is positive — the column fattens:

\Delta d=0.29\times2.55\times10^{-4}\times100\ \text{mm}=7.4\times10^{-3}\ \text{mm}\approx7.4\ \mu\text{m}

Step 4. Volume change. $\varepsilon_{kk}=\varepsilon_{\parallel}(1-2\nu)=-2.55\times10^{-4}\times0.42=-1.07\times10^{-4}$ — the volume shrinks by about $0.011\%$ ; Poisson fattening compensates most, but not all, of the axial loss.

Step 5. Stored energy. Energy density $u=\sigma^2/2E=(5.09\times10^{7})^2/(4.00\times10^{11})=6.49\times10^{3}\ \text{J/m}^3$ ; volume $V=AL=0.02356\ \text{m}^3$ , so

U=uV=6.49\times10^{3}\times0.02356=153\ \text{J}

Answer: $\Delta L\approx0.76$ mm shortening, diameter grows $\approx7.4\ \mu$ m, volume shrinks by $1.1\times10^{-4}$ , and the column stores $\approx153$ J — about the kinetic energy of a cricket ball bowled at $44$ m/s, held silently in the steel.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Gresham College and the Royal Society, London

Hooke was the Royal Society's first Curator of Experiments — a professional demonstrator required to produce new experiments weekly, which made him fluent in springs, watches, and instruments of every kind. Wary of rivals, he published his law of the spring in 1676 as the anagram ceiiinosssttuv, unscrambling it two years later as ut tensio, sic vis — as the extension, so the force. The one-line proportionality he stated for a coiled spring took another century and a half, and the labors of Young, Navier, and Cauchy, to grow into the twenty-one-constant tensor law of this article. His quarrels with Newton over priority left him nearly erased from official portraiture; his law, however, is quoted more often than either man's name.

History

1678

— Hooke's De Potentia Restitutiva reveals the law hidden in his 1676 anagram ceiiinosssttuv: ut tensio, sic vis — as the extension, so the force

1807

— Thomas Young identifies the stiffness of a material, as opposed to a structure, introducing the modulus that now carries his name

1821

— Navier writes the first general equations of elastic motion from a molecular spring model — one constant short of the full isotropic truth

1822

— Cauchy replaces molecular hypotheses with the stress and strain tensors, giving linear elasticity its modern two-constant form

1852

— Lamé's Leçons on elasticity names the pair of constants, lambda and mu, that parameterize every isotropic solid

Going further

References. Landau & Lifshitz, Theory of Elasticity (Vol. 7), Ch. 1 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 13 · Morin, Introduction to Classical Mechanics, Ch. on oscillations and waves · Sokolnikoff, Mathematical Theory of Elasticity, Ch. 3 · Timoshenko & Goodier, Theory of Elasticity, Ch. 1 The 81→36→21 constant counting follows from the three tensor symmetries by direct index counting. The Navier equation yields wave speeds $c_L=\sqrt{(\lambda+2\mu)/\rho}$ and $c_T=\sqrt{\mu/\rho}$ . The stability bounds $-1<\nu<1/2$ follow from positive-definiteness of the energy. The algebraic inversions $\lambda=E\nu/[(1+\nu)(1-2\nu)]$ , $\mu=E/[2(1+\nu)]$ , and $K=E/[3(1-2\nu)]$ are consistent with all moduli relations above. Quoted material values: steel $E\approx200$ GPa, aluminium $E\approx70$ GPa.

Builds on: Strain Tensor and Deformation of Continua · Stress Tensor and Cauchy's Relation · Lagrangian Density and Continuum Lagrangian Formalism Leads to: Elastic Waves in Solids · Euler–Lagrange Equations for Fields · The One-Dimensional Wave Equation

Why must a strain-energy function exist at all? Could a material have 36 constants instead of 21?

The major symmetry $C_{ijkl}=C_{klij}$ follows only if the stress derives from a potential, $\sigma_{ij}=\partial u/\partial\varepsilon_{ij}$ — so-called hyperelasticity. A 'Cauchy-elastic' material defined merely by a linear stress–strain map could in principle have $36$ constants, but the extra $15$ would let you extract net work from closed strain cycles: load along one path, unload along another, and harvest energy from nothing. Thermodynamics forbids perpetual motion, so any material in equilibrium with well-defined internal energy must be hyperelastic, and $21$ is the true ceiling. The nineteenth-century 'rari-constant versus multi-constant' controversy — whether isotropic solids need one constant or two — was settled the same way: by experiment (Poisson's ratio is not universally $1/4$ ) and by energy accounting.

Where does the factor of 2 between engineering shear and tensor shear actually come from?

From work conjugacy. The increment of stored energy is $\mathrm{d}u=\sigma_{ij}\,\mathrm{d}\varepsilon_{ij}$ , and the off-diagonal pair $(\varepsilon_{xy},\varepsilon_{yx})$ contributes twice — once for each index order — giving $2\sigma_{xy}\,\mathrm{d}\varepsilon_{xy}=\sigma_{xy}\,\mathrm{d}\gamma_{xy}$ . Engineers who prefer to write energy as (stress) × (one strain measure) per shear component are therefore forced to define $\gamma_{xy}=2\varepsilon_{xy}$ . Both conventions are self-consistent; mixing them is what produces silent factor-of-two errors, especially in Voigt-notation stiffness tables.

Does the Lamé parameter lambda mean anything physical on its own?

Directly, almost nothing — no simple experiment reads out $\lambda$ alone, which is why data sheets quote $E$ and $\nu$ instead. Its cleanest reading: $\lambda$ is the stress that appears perpendicular to a stretch when volume changes, the cross-talk between orthogonal directions ( $\sigma_{yy}=\lambda\,\varepsilon_{xx}$ if only $\varepsilon_{xx}\neq0$ ). It earns its keep algebraically: formulas like $C_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})$ and the P-wave modulus $M=\lambda+2\mu$ are shortest in the $(\lambda,\mu)$ language. For a fluid, though, $\lambda$ becomes concrete: with $\mu=0$ it is the bulk modulus.

Is a negative Poisson ratio physically real, or just a mathematical allowance of the bound?

Real and purchasable. The stability bound only requires $\nu>-1$ , and engineered auxetic materials sit firmly in the negative range: re-entrant polymer foams reach $\nu\approx-0.7$ , and certain crystalline phases (alpha-cristobalite) are naturally auxetic along some directions. The trick is architecture, not chemistry — unit cells shaped like bow-ties unfold when pulled, expanding sideways. Auxetics resist indentation (material flows toward the impact), which is why they appear in padding, and they curve synclastically, making dome-shaped panels without crumpling.

Why is the linear regime so small for metals — strains below one part in a thousand?

Two separate ceilings. First, anharmonicity: the interatomic potential is only approximately quadratic, so at strains of a few percent the modulus itself would drift. But metals never get there, because of the second ceiling: dislocations start to move at shear stresses of order $10^{-3}\mu$ , producing plastic (permanent) deformation long before the lattice springs leave their quadratic regime. Yield, not anharmonicity, ends Hooke's law in structural practice. Materials without mobile dislocations push further: fused silica fibres and graphene stay elastic to strains of several percent.

A cubic crystal has three constants while isotropy needs two — what physical difference does the third constant encode?

Direction dependence of shear stiffness. In a cubic crystal, shearing on a cube face (resisted by $C_{44}$ ) and shearing at 45 degrees to it (resisted by $(C_{11}-C_{12})/2$ ) are physically different experiments with independent answers. The Zener ratio $A=2C_{44}/(C_{11}-C_{12})$ measures the mismatch: $A=1$ is isotropy, tungsten sits almost exactly there ( $A\approx1.0$ ), copper is strongly anisotropic ( $A\approx3.2$ ). Polycrystals average over grain orientations, which is why bulk copper behaves isotropically even though each grain is not.

How does this tensor law connect back to the scalar spring law F equals minus k x?

The scalar law is the one-dimensional shadow of Eq. (1). For a uniform rod of cross-section $A$ and length $L$ under uniaxial stress, $\sigma=E\varepsilon$ becomes $F/A=E\,(x/L)$ , i.e. $F=(EA/L)\,x$ — a spring with stiffness $k=EA/L$ . The tensor apparatus exists because a 3D body can also be sheared, twisted, and squeezed, and because the sideways response ( $\nu$ ) has no 1D analogue. Conversely, any complicated elastic structure, once linearized about equilibrium, reduces to effective spring constants computed from $C_{ijkl}$ and geometry — which is exactly what finite-element software does.

Do the elastic constants depend on temperature, and does Hooke's law survive heating?

The linear form survives; the constants drift. Moduli soften as temperature rises — for steel, $E$ drops roughly $10\%$ from room temperature to $500\,°$ C, and much faster approaching melt — because thermal vibration samples the anharmonic parts of the interatomic potential. Thermal expansion adds a strain offset: the stress-free state shifts, so the law becomes $\sigma_{ij}=C_{ijkl}(\varepsilon_{kl}-\alpha\,\Delta T\,\delta_{kl})$ with $\alpha$ the expansion coefficient. Constrained heating therefore generates stress ( $\sim2.4$ MPa per degree for confined steel), the origin of buckled rails and the expansion joints in every bridge.

Linear Elasticity and Hooke's Law for Solids

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

Going further

Before this

Where this leads