Strain Tensor and Deformation of Continua

A deformable continuum is described by a displacement field $\mathbf{u}(\mathbf{x})$ that maps each material point from its undeformed (reference) position $\mathbf{x}$ to its deformed position $\mathbf{x}+\mathbf{u}$ . The physically meaningful content of deformation is not the displacement itself — a uniform translation moves everything but deforms nothing — but how displacement varies from point to point. That information lives in the displacement-gradient tensor $\partial_i u_j$ . The gradient splits uniquely into a symmetric part and an antisymmetric part. The antisymmetric part is an infinitesimal local rotation that produces no change in lengths or angles; the symmetric part, the strain tensor $\varepsilon_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i)$ , is what stretches and shears material fibers. This decomposition is the continuum analogue of separating a velocity gradient into rate-of-strain and vorticity. Geometrically, strain answers the question: by how much does the squared distance between two nearby material points change under deformation? Expanding $|\mathrm{d}\mathbf{x}'|^2-|\mathrm{d}\mathbf{x}|^2$ yields exactly $2\,E_{ij}\,\mathrm{d}x_i\,\mathrm{d}x_j$ with the Green–Lagrange tensor; linearizing for small gradients gives $\varepsilon_{ij}$ . The diagonal entries measure elongation along coordinate directions, the off-diagonal entries measure the closing of right angles, and the trace measures volume change. Strain is purely kinematic — it knows nothing about forces or material properties. Its companion is the stress tensor, which describes internal forces. The relation between them is the constitutive law; for small deformations of an isotropic solid this is Hooke's law, $\sigma_{ij}=\lambda\,\varepsilon_{kk}\delta_{ij}+2\mu\,\varepsilon_{ij}$ . Combined with Newton's second law for a continuum, this closes the system and yields the equations of linear elasticity and the propagation of elastic waves.

Key takeaways

The strain tensor $\varepsilon_{ij}$ is the symmetric part of the displacement gradient; it captures local stretching and shearing while discarding rigid rotation.
Diagonal components $\varepsilon_{ii}$ are fractional length changes along the axes; off-diagonal $\varepsilon_{ij}$ ( $i\neq j$ ) are (half) the shear angles between fiber directions.
The trace $\varepsilon_{kk}=\nabla\cdot\mathbf{u}$ equals the fractional change of volume (dilatation) to first order.
The full nonlinear Green–Lagrange strain $E_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i+\partial_i u_k\,\partial_j u_k)$ reduces to the linear strain when gradients are small.
Strain is the kinematic input to constitutive laws: paired with the stress tensor through Hooke's law it governs linear elasticity and elastic waves.
Because $\varepsilon_{ij}$ is symmetric and real it can be diagonalized, giving three principal strains along mutually orthogonal principal axes.
Key idea: Separate the displacement gradient $\partial_i u_j$ into its symmetric part (strain, which deforms) and antisymmetric part (rotation, which does not). Strain alone changes lengths, angles, and volumes; everything mechanically meaningful about small deformation is encoded in the six independent components of $\varepsilon_{ij}$ .

Where this lives today

Strain-field measurement has gone from single gauges to full imaging. Digital image correlation (DIC) — cross-correlating speckle photographs of a deforming surface — now delivers complete $\varepsilon_{ij}(\mathbf{x})$ maps at sub-microstrain resolution and megahertz frame rates, and is the standard validation tool for finite-element models in aerospace and crash testing. Distributed fiber-optic sensing turns kilometers of embedded optical fiber into a continuous strain gauge: Rayleigh-backscatter interrogators resolve $\sim1\ \mu\varepsilon$ every few centimeters along the fiber, which is how modern bridges, pipelines, wind-turbine blades, and rocket tanks are monitored for damage in real time. The same tensor runs at planetary and atomic scales. Geodesists difference GPS and InSAR displacement fields to produce crustal strain-rate maps whose off-diagonal components locate locked fault segments; the 2026-generation seismic hazard models for California and Anatolia are built directly on such maps. At the other extreme, "straintronics" engineers band structure by design: a few percent of biaxial strain in a 2D semiconductor or a silicon transistor channel shifts band edges by hundreds of meV, and strain tensors extracted from high-resolution electron microscopy are now a routine chip-metrology product.

The physical picture

Picture a rubber block drawn on with a fine square grid. After you squeeze and twist it, each tiny square becomes a parallelogram. The strain tensor records exactly two things about each parallelogram: how much its sides lengthened or shortened (diagonal entries) and how much its corners departed from $90^\circ$ (off-diagonal entries). Any part of the motion that merely rotates a square without distorting it is rotation, not strain.

Figure

A deformed grid square decomposes into stretch plus shear plus rotation — only the first two change lengths or angles, so only the first two are strain

Strain is a local, relative quantity. The displacement of a single point tells you nothing — only the difference in displacement between neighboring points, divided by their separation, deforms the material. This is why strain is a gradient of $\mathbf{u}$ , and why a uniform translation (constant $\mathbf{u}$ ) produces zero strain. Because $\varepsilon_{ij}$ is a real symmetric tensor, at every point you can rotate to axes — the principal axes — along which the deformation is pure stretching with no shear. In those axes a material cube simply becomes a rectangular box with edges scaled by the three principal strains $\varepsilon_1,\varepsilon_2,\varepsilon_3$ . You have met this spectral logic before: the moment-of-inertia tensor plays the identical game with principal axes of rotation, because both are real symmetric rank-2 tensors.

Figure

Pure shear seen twice — in the original axes the square shears and right angles close, but in axes rotated by 45 degrees the same deformation is pure stretching along one diagonal and pure shrinking along the other, with no shear at all

The trace is special: $\varepsilon_{kk}=\nabla\cdot\mathbf{u}$ is rotation-invariant and equals the fractional volume change. Splitting strain into its trace (dilatation) and traceless part (pure shear) mirrors the way bulk modulus and shear modulus separately govern volume and shape changes in an elastic solid.

The core equations

Everything begins with the deformation map. Take a body, mark every material point by its undeformed coordinate $\mathbf{x}$ , deform the body, and record where each marked point lands. The displacement field is the bookkeeping device for this map:

x_i' = x_i + u_i(\mathbf{x}) \qquad (1)

where: $x_i$ — reference (undeformed) coordinate of a material point; $x_i'$ — deformed (current) coordinate of the same material point; $u_i(\mathbf{x})$ — displacement field, the fundamental kinematic variable. Note that $u_i$ is a field: each material point carries its own displacement, and it is precisely the spatial variation of $u_i$ — not its value — that will carry physical content. To first order, all local information about the deformation sits in the nine numbers $\partial_i u_j$ evaluated at the point of interest. This gradient contains both genuine deformation and mere rigid rotation, and the first structural move of continuum mechanics is to separate the two. Adding and subtracting the transpose splits any second-rank tensor uniquely into symmetric and antisymmetric parts:

\partial_i u_j = \varepsilon_{ij} + \omega_{ij} \qquad (2)

where: $\partial_i u_j$ — displacement-gradient tensor; $\varepsilon_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i)$ — symmetric strain tensor (deforms); $\omega_{ij}=\tfrac12(\partial_i u_j-\partial_j u_i)$ — antisymmetric rotation tensor (does not deform). The antisymmetric piece has only three independent components, equivalent to a rotation vector $\boldsymbol{\omega}=\tfrac12\nabla\times\mathbf{u}$ : to first order it turns every fiber through the same small angle, preserving all lengths. The symmetric piece is where stretching and shearing live. Why is the symmetric part the deformation? Because the honest, coordinate-free measure of deformation is the change in distance between neighboring material points, and (as Derivation A shows) that change is controlled by a manifestly symmetric object:

\mathrm{d}s'^2-\mathrm{d}s^2 = 2\,E_{ij}\,\mathrm{d}x_i\,\mathrm{d}x_j \qquad (3)

where: $\mathrm{d}s^2$ — squared length of a material line element before deformation; $\mathrm{d}s'^2$ — squared length of the same element after deformation; $E_{ij}$ — Green–Lagrange strain tensor, exact to all orders in the displacement gradient. If $E_{ij}=0$ everywhere, no fiber changed length and the motion was rigid — no matter how large the displacements were. For the small deformations that dominate engineering practice and linear elasticity, the quadratic term inside $E_{ij}$ is negligible and we work with the linear (Cauchy) strain:

\varepsilon_{ij}=\frac12\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) \qquad (4)

where: $\varepsilon_{ij}$ — symmetric $3\times3$ strain tensor, 6 independent components; $u_i$ — $i$ -th component of displacement; $x_j$ — $j$ -th reference coordinate. Strain is dimensionless: steel structures typically operate near $\varepsilon\sim10^{-3}$ (a "millistrain"), and experimental stress analysis quotes microstrain, $1\,\mu\varepsilon=10^{-6}$ . The trace of the strain tensor answers the most basic geometric question of all — did the material locally grow or shrink? Derivation C establishes the identification with the divergence of displacement:

\frac{\Delta V}{V}=\varepsilon_{kk}=\nabla\cdot\mathbf{u} \qquad (5)

where: $\varepsilon_{kk}$ — trace of the strain tensor (summed over $k$ ); $\Delta V/V$ — fractional change of volume to first order. The trace is invariant under rotations of the coordinate axes, as a volume change must be; the traceless remainder $\varepsilon_{ij}-\tfrac13\varepsilon_{kk}\delta_{ij}$ is the pure shape change (deviatoric strain). One notational trap costs more grief than any other in this subject. Engineers and finite-element codes describe shear not by the tensor component but by the total closing angle between two fibers:

\gamma_{ij}=2\,\varepsilon_{ij}\quad(i\neq j) \qquad (6)

where: $\gamma_{ij}$ — engineering shear strain — the total angle decrease between two initially perpendicular fibers; $\varepsilon_{ij}$ — tensor shear component — half that angle. Both conventions are internally consistent, but mixing them silently corrupts stiffness matrices and energy expressions by factors of 2 or 4 (see the pitfall below). Because $\varepsilon_{ij}$ is real and symmetric, the spectral theorem applies at every point of the body:

\varepsilon_{ij}=\varepsilon_a\,n^{(a)}_i n^{(a)}_j,\qquad \varepsilon_{ij}n^{(a)}_j=\varepsilon_a n^{(a)}_i \qquad (7)

where: $\varepsilon_a$ — $a$ -th principal strain (eigenvalue), $a=1,2,3$ ; $n^{(a)}_i$ — $a$ -th principal direction (orthonormal eigenvector). Along principal axes the deformation is pure stretch: a cube aligned with $n^{(1)},n^{(2)},n^{(3)}$ becomes a rectangular box with edge ratios $1+\varepsilon_1$ , $1+\varepsilon_2$ , $1+\varepsilon_3$ and no shear whatsoever. Strain is kinematics; to make dynamics you must say how the material responds to strain. For an isotropic linear solid the most general response built from $\varepsilon_{ij}$ and $\delta_{ij}$ is Hooke's law — the rank-2 generalization of the scalar spring law $F=-kx$ you know from the spring-force article:

\sigma_{ij}=\lambda\,\varepsilon_{kk}\,\delta_{ij}+2\mu\,\varepsilon_{ij} \qquad (8)

where: $\sigma_{ij}$ — Cauchy stress tensor; $\lambda$ — first Lamé parameter; $\mu$ — shear modulus (second Lamé parameter); $\delta_{ij}$ — Kronecker delta. The two-constant structure is forced by isotropy: one constant prices volume change, the other prices shape change. The full story of this law — moduli, Poisson's ratio, energy positivity — belongs to the linear-elasticity article; here it marks where kinematics hands off to constitutive physics. Finally, straining a Hookean material stores energy. Incremental work per unit volume is $\mathrm{d}w=\sigma_{ij}\,\mathrm{d}\varepsilon_{ij}$ ; integrating from zero strain with the linear law (8) gives $w=\tfrac12\sigma_{ij}\varepsilon_{ij}$ , and inserting Hooke's law once more produces a quadratic form in strain alone:

U=\int\left(\tfrac12\lambda\,\varepsilon_{kk}^2+\mu\,\varepsilon_{ij}\varepsilon_{ij}\right)\mathrm{d}V \qquad (9)

where: $U$ — total stored elastic potential energy; $\varepsilon_{ij}\varepsilon_{ij}$ — full contraction (summed over both indices). This strain-energy function is George Green's great legacy: postulating that elastic forces derive from a potential quadratic in strain is what reduced the constant-count of general anisotropic elasticity to 21 and put the subject on an energy footing.

Derivation AStrain from the change in squared length

We derive the exact (Green–Lagrange) strain tensor by computing how the squared distance between two nearby material points changes under the deformation map, then linearize to recover the Cauchy strain. The strategy is to push a material line element $\mathrm{d}x_i$ through the map $x_i'=x_i+u_i$ , square its deformed length, and subtract the reference length; whatever multiplies $\mathrm{d}x_i\,\mathrm{d}x_j$ in the difference is, by construction, the complete and orientation-resolved measure of deformation. Watch two things: the quadratic term that survives exactly, and how symmetry of the result is automatic rather than assumed.

Step 1. Differentiate the deformation map to transport the line element:

\mathrm{d}x_i'=\left(\delta_{ij}+\partial_j u_i\right)\mathrm{d}x_j

Why: two material points separated by $\mathrm{d}x_j$ in the reference state land at positions differing by the map's Jacobian acting on $\mathrm{d}x_j$ ; the identity $\delta_{ij}$ carries the point, the gradient carries the deformation.

Step 2. Square the deformed element, summing over the repeated index $k$ :

\mathrm{d}s'^2=\mathrm{d}x_k'\,\mathrm{d}x_k'=\left(\delta_{ki}+\partial_i u_k\right)\left(\delta_{kj}+\partial_j u_k\right)\mathrm{d}x_i\,\mathrm{d}x_j

Step 3. Subtract the reference length $\mathrm{d}s^2=\delta_{ij}\,\mathrm{d}x_i\,\mathrm{d}x_j$ . The $\delta_{ki}\delta_{kj}$ term cancels it exactly, leaving

\mathrm{d}s'^2-\mathrm{d}s^2=\left(\partial_i u_j+\partial_j u_i+\partial_i u_k\,\partial_j u_k\right)\mathrm{d}x_i\,\mathrm{d}x_j

Why: the two cross terms give the linear pieces $\partial_i u_j$ and $\partial_j u_i$ ; the product of the two gradients gives the quadratic piece — no approximation has been made yet.

Step 4. Define the strain tensor by the convention $\mathrm{d}s'^2-\mathrm{d}s^2=2E_{ij}\,\mathrm{d}x_i\,\mathrm{d}x_j$ , which fixes

E_{ij}=\tfrac12\left(\partial_i u_j+\partial_j u_i+\partial_i u_k\,\partial_j u_k\right)

Why the factor of 2: it is chosen so that for a uniaxial stretch $\mathrm{d}s'=(1+\varepsilon)\mathrm{d}s$ the tensor component equals $\varepsilon$ itself at leading order, matching the elementary definition of fractional elongation.

Step 5. Linearize. For $|\partial_i u_j|\ll1$ the quadratic product is negligible and

E_{ij}\to\varepsilon_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i)

Why this matters: a rigid rotation $u_i=(R_{ij}-\delta_{ij})x_j$ with $R$ orthogonal gives $\mathrm{d}s'^2=\mathrm{d}s^2$ exactly, and indeed $E_{ij}=\tfrac12(R_{ki}R_{kj}-\delta_{ij})=0$ by $R^{\mathsf T}R=I$ — the exact strain vanishes for any rotation, however large, while the linear $\varepsilon_{ij}$ vanishes only for infinitesimal ones.

Go deeper · Derivation B · 1 minGeometric meaning of diagonal and off-diagonal components

Derivation BGeometric meaning of diagonal and off-diagonal components

We show that $\varepsilon_{11}$ is the fractional elongation of a fiber initially along $x_1$ and that $2\varepsilon_{12}$ is the decrease of the angle between fibers initially along $x_1$ and $x_2$ . The route is to feed specific line elements into the master formula $\mathrm{d}s'^2-\mathrm{d}s^2=2\varepsilon_{ij}\,\mathrm{d}x_i\,\mathrm{d}x_j$ from Derivation A — first a single axis-aligned fiber for the stretch, then a perpendicular pair for the angle. The factor of two between the tensor component and the engineering shear angle appears in the last step; it is a definition colliding with a geometry, not an error.

Step 1. Put $\mathrm{d}\mathbf{x}=(\mathrm{d}\ell,0,0)$ into the master formula:

\mathrm{d}s'^2=(1+2\varepsilon_{11})\,\mathrm{d}\ell^2

Step 2. Take the square root and expand to first order:

\frac{\mathrm{d}s'}{\mathrm{d}\ell}=\sqrt{1+2\varepsilon_{11}}\approx1+\varepsilon_{11}

Why: this identifies the diagonal component directly as relative stretch — a fiber laid along $x_1$ lengthens by the fraction $\varepsilon_{11}$ , with no reference to any other component.

Step 3. Now take two fibers, initially along $x_1$ and $x_2$ , and form the dot product of their deformed images using Step 1 of Derivation A:

\mathrm{d}\mathbf{x}'^{(1)}\!\cdot\mathrm{d}\mathbf{x}'^{(2)}=(\delta_{k1}+\partial_1 u_k)(\delta_{k2}+\partial_2 u_k)\,\mathrm{d}\ell_1\,\mathrm{d}\ell_2\approx(\partial_1 u_2+\partial_2 u_1)\,\mathrm{d}\ell_1\,\mathrm{d}\ell_2=2\varepsilon_{12}\,\mathrm{d}\ell_1\,\mathrm{d}\ell_2

Why: the reference fibers are perpendicular, so their reference dot product vanishes; the leading surviving term is exactly twice the tensor shear component.

Step 4. Convert the dot product into the angle $\theta'$ between the deformed fibers. With deformed lengths equal to the reference lengths at leading order,

\cos\theta'\approx 2\varepsilon_{12}

Step 5. Since $\cos\theta'=\sin(\pi/2-\theta')\approx\pi/2-\theta'$ for small deviations, the right angle closes by

\frac{\pi}{2}-\theta'\approx2\varepsilon_{12}=\gamma_{12}

Why: symmetric shear $u_1=\varepsilon_{12}x_2$ , $u_2=\varepsilon_{12}x_1$ tilts each fiber by $\varepsilon_{12}$ toward the other; the two tilts add to a total angle change of $2\varepsilon_{12}$ , confirming both the geometry and the engineering convention $\gamma=2\varepsilon$ .

Go deeper · Derivation C · 1 minVolume change equals the trace of the strain tensor

Derivation CVolume change equals the trace of the strain tensor

We prove the dilatation formula $\Delta V/V=\varepsilon_{kk}$ by tracking a material volume element through the deformation map. The tool is the Jacobian determinant: volumes transform by $\det(\partial x_i'/\partial x_j)$ , so the entire computation reduces to expanding a determinant near the identity. The punchline explains why only the symmetric part of the gradient can change volume — the trace of an antisymmetric tensor vanishes identically.

Step 1. The deformed volume element is the Jacobian of the map times the reference element:

\frac{\mathrm{d}V'}{\mathrm{d}V}=\det\!\left(\delta_{ij}+\partial_j u_i\right)

Why: this is the standard change-of-variables rule; the matrix $\delta_{ij}+\partial_j u_i$ is precisely the deformation gradient that transports line elements in Derivation A, and its determinant transports volumes.

Step 2. Expand the determinant to first order in the small gradients using $\det(I+A)\approx1+\mathrm{tr}\,A$ :

\det(\mathbf{I}+\nabla\mathbf{u})\approx 1+\mathrm{tr}(\nabla\mathbf{u})=1+\partial_k u_k

Why the expansion is legitimate: every neglected term is a product of two or more gradient components, second order in the same small parameter that already defines linear strain.

Step 3. Read off the fractional volume change and note which part of the gradient contributes:

\frac{\Delta V}{V}=\partial_k u_k=\varepsilon_{kk},\qquad \omega_{kk}=0

Why: for pure dilatation $u_i=\alpha x_i$ the formula gives $\Delta V/V=3\alpha$ , matching $(1+\alpha)^3\approx1+3\alpha$ directly; for any rotation or pure shear the trace vanishes and volume is preserved, as geometry demands.

Limiting cases

The strain tensor specializes cleanly in several physically important regimes:

Case	Result	Meaning
Small deformation (linear strain)	$\varepsilon_{ij}=\tfrac12\left(\partial_i u_j+\partial_j u_i\right)$	Valid when all displacement gradients $\vert \partial_i u_j\vert \ll1$ ; the quadratic Green–Lagrange term is dropped. This is the regime of linear elasticity.
Pure dilatation (uniform expansion)	$\varepsilon_{ij}=\tfrac13\,\theta\,\delta_{ij},\qquad \theta=\nabla\cdot\mathbf{u}$	All off-diagonal components vanish and the material scales isotropically; $\theta$ is the fractional volume change. Pure shape change is the complementary traceless part.
Plane strain (2D)	$\varepsilon_{33}=\varepsilon_{13}=\varepsilon_{23}=0$	Deformation confined to the $x_1$ – $x_2$ plane; appropriate for long prismatic bodies loaded uniformly along their axis. Strain reduces to a $2\times2$ symmetric tensor.
Finite deformation	$E_{ij}=\tfrac12\left(\partial_i u_j+\partial_j u_i+\partial_i u_k\,\partial_j u_k\right)$	The full Green–Lagrange strain. Necessary for large stretches or finite rotations, where the linear strain wrongly reports a rigid rotation as deformation.

Common trapRotating a body does not strain it — and the shear angle is twice the tensor component

Two classic factor traps. First, the engineering shear strain $\gamma_{ij}=2\varepsilon_{ij}$ is the total closing angle between two fibers, while the tensor component is half of it; forgetting the factor of two corrupts strain-energy densities and every stiffness matrix written in Voigt notation. Second, linear strain is only trustworthy when $|\partial_i u_j|\ll1$ : for finite rotations the linear $\varepsilon_{ij}$ spuriously reports strain (a rotation by angle $\theta$ shows fake $\varepsilon_{11}=\cos\theta-1$ ), and you must keep the quadratic term of the Green–Lagrange tensor, which cancels rotational contributions exactly.

Applications

Strain gauges and experimental stress analysis. A resistive strain gauge bonded to a structure measures the fractional elongation $\varepsilon_{nn}$ along its own axis $\mathbf{n}$ . Three gauges in a rosette at known angles overdetermine the in-plane strain components $\varepsilon_{11},\varepsilon_{22},\varepsilon_{12}$ , from which principal strains and (via Hooke's law) principal stresses are reconstructed. This is the everyday engineering use of the symmetric strain tensor's transformation law $\varepsilon_{nn}=\varepsilon_{ij}n_i n_j$ .

Elastic wave speeds in solids. Inserting Hooke's law into the continuum equation of motion $\rho\,\ddot{u}_i=\partial_j\sigma_{ij}$ yields $\rho\,\ddot{\mathbf{u}}=(\lambda+\mu)\nabla(\nabla\cdot\mathbf{u})+\mu\nabla^2\mathbf{u}$ . The dilatational (trace) part of strain propagates as longitudinal P-waves at $c_L=\sqrt{(\lambda+2\mu)/\rho}$ , the traceless shear part as transverse S-waves at $c_T=\sqrt{\mu/\rho}$ . The strain decomposition into dilatation and shear directly explains why two distinct wave speeds exist — the full derivation lives in the elastic-waves article.

Volumetric and bulk response. Because $\varepsilon_{kk}=\Delta V/V$ , the bulk modulus is defined by $p=-K\,\varepsilon_{kk}$ with $K=\lambda+\tfrac23\mu$ . Splitting strain into its trace and deviatoric (traceless) parts cleanly separates a material's resistance to volume change ( $K$ ) from its resistance to shape change ( $\mu$ ) — and explains why fluids, which have $\mu=0$ , support no static shear strain and hence no static shear stress.

Compatibility and continuity of deformation. Not every symmetric tensor field is a valid strain: the six $\varepsilon_{ij}$ derive from only three displacement components, so they must satisfy the Saint-Venant compatibility conditions $\partial_k\partial_l\varepsilon_{ij}+\partial_i\partial_j\varepsilon_{kl}-\partial_i\partial_l\varepsilon_{jk}-\partial_k\partial_j\varepsilon_{il}=0$ . These guarantee a single-valued, continuous displacement field — essential when solving elasticity problems directly in terms of stress or strain rather than displacement.

Worked examples

Example 1 — a steel tie rod under load (confidence builder). A cylindrical steel tie rod of diameter $d=20\ \mathrm{mm}$ and length $L=2.0\ \mathrm{m}$ carries an axial tension $F=60\ \mathrm{kN}$ . Take $E=200\ \mathrm{GPa}$ and Poisson's ratio $\nu=0.30$ . Find the axial strain, the elongation, the lateral strain, and the fractional volume change.

Step 1. Cross-sectional area: $A=\pi d^2/4=\pi(0.020)^2/4=3.14\times10^{-4}\ \mathrm{m^2}$ .

Step 2. Axial stress: $\sigma=F/A=60\times10^{3}/3.14\times10^{-4}=1.91\times10^{8}\ \mathrm{Pa}=191\ \mathrm{MPa}$ — safely below the $\sim250\ \mathrm{MPa}$ yield of mild steel, so linear elasticity applies.

Step 3. Axial strain: $\varepsilon_{zz}=\sigma/E=1.91\times10^{8}/2.00\times10^{11}=9.55\times10^{-4}$ , i.e. about $955\ \mu\varepsilon$ . The elongation is $\Delta L=\varepsilon_{zz}L=9.55\times10^{-4}\times2.0=1.91\times10^{-3}\ \mathrm{m}\approx1.9\ \mathrm{mm}$ .

Step 4. The lateral strains follow from Poisson contraction: $\varepsilon_{xx}=\varepsilon_{yy}=-\nu\,\varepsilon_{zz}=-0.30\times9.55\times10^{-4}=-2.87\times10^{-4}$ . The strain tensor is diagonal in these axes — the load axis and any two transverse axes are principal.

Step 5. Fractional volume change is the trace: $\varepsilon_{kk}=\varepsilon_{zz}(1-2\nu)=9.55\times10^{-4}\times0.40=3.82\times10^{-4}$ . The rod stretches 1.9 mm, its diameter shrinks by about $5.7\ \mu\mathrm{m}$ , and its volume grows by 0.038% — tension at $\nu<\tfrac12$ always dilates.

Example 2 — simple shear of an elastomer bearing pad (exam level). A square elastomeric bridge bearing, $300\ \mathrm{mm}\times300\ \mathrm{mm}$ in plan and $h=200\ \mathrm{mm}$ thick, has its top plate displaced horizontally by $6.0\ \mathrm{mm}$ while the bottom is fixed: $u_x=\kappa y$ with $\kappa=6.0/200=0.030$ , all other displacement components zero. The shear modulus is $G=0.90\ \mathrm{MPa}$ . Find the strain and rotation tensors, the principal strains and their directions, the volume change, and the horizontal force.

Step 1. The only nonzero gradient is $\partial u_x/\partial y=\kappa=0.030$ . Symmetrize and antisymmetrize: $\varepsilon_{xy}=\varepsilon_{yx}=\kappa/2=0.015$ and $\omega_{xy}=-\omega_{yx}=\kappa/2=0.015$ ; all diagonal strain components vanish.

Step 2. The rotation tensor corresponds to $\boldsymbol{\omega}=\tfrac12\nabla\times\mathbf{u}=(0,0,-\kappa/2)$ : the element rotates clockwise by $0.015\ \mathrm{rad}\approx0.86^\circ$ . Simple shear is therefore pure shear plus rotation — half the gradient deforms, half merely turns.

Step 3. Diagonalize the $2\times2$ strain block $\begin{pmatrix}0&0.015\\0.015&0\end{pmatrix}$ : eigenvalues $\varepsilon_{1,2}=\pm0.015$ with eigenvectors at $\pm45^\circ$ to the $x$ -axis. The pad stretches by 1.5% along one diagonal and shortens by 1.5% along the other — exactly the geometry of the second figure.

Step 4. Volume change: $\varepsilon_{kk}=0+0+0=0$ . Simple shear preserves volume exactly to first order, consistent with the trace theorem of Derivation C.

Step 5. The engineering shear strain is $\gamma=2\varepsilon_{xy}=0.030$ , so the shear stress is $\tau=G\gamma=0.90\times10^{6}\times0.030=2.7\times10^{4}\ \mathrm{Pa}=27\ \mathrm{kPa}$ . Acting on the plan area $A=0.30\times0.30=0.090\ \mathrm{m^2}$ , the horizontal force is $F=\tau A=27\times10^{3}\times0.090\approx2.4\times10^{3}\ \mathrm{N}$ . A 2.4 kN push shears the pad by 6 mm — soft in shear yet nearly incompressible, which is precisely why elastomeric bearings isolate bridge decks.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · Sneinton mill, Nottingham; later Caius College, Cambridge

George Green was a miller's son who taught himself mathematics from library books while running the family windmill, with roughly one year of formal schooling. His self-published 1828 essay introduced the potential function and what are now Green's theorem and Green's functions; admitted to Cambridge at 40, he then transformed elasticity by postulating that elastic forces derive from a stored energy function quadratic in the strain components — replacing ad hoc molecular force models with an energy principle. The finite-deformation strain measure $E_{ij}$ and the Green elastic (hyperelastic) material class both carry his name, and his energy method is the direct ancestor of the Lagrangian-density formulation of continuum mechanics.

History

1678

— Hooke publishes the anagram-resolved law "ut tensio, sic vis": restoring force proportional to extension

1822

— Cauchy's memoir to the Paris Academy defines the infinitesimal strain and stress tensors and their linear relation

1839

— Green derives elasticity from a strain-energy function, reducing general anisotropic elasticity to 21 constants

1852

— Lamé publishes the first systematic treatise on elasticity, fixing the two-parameter isotropic law and the constants that bear his name

1952

— Truesdell's rational mechanics program rebuilds finite-strain continuum mechanics on the exact Green–Lagrange kinematics

Going further

References. Landau & Lifshitz, Theory of Elasticity (Vol. 7), §1–§4 · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 13 (Continuous Systems) · Morin, Introduction to Classical Mechanics, Ch. on waves and continuous systems · Taylor, Classical Mechanics, Ch. 16 · Fung, A First Course in Continuum Mechanics Key results collected here: the symmetric/antisymmetric split of $\partial_i u_j$ is unique and $\boldsymbol{\omega}=\tfrac12\nabla\times\mathbf{u}$ ; the Green–Lagrange derivation gives $\mathrm{d}s'^2-\mathrm{d}s^2=2E_{ij}\mathrm{d}x_i\mathrm{d}x_j$ with linearization to $\varepsilon_{ij}$ ; the dilatation identity is $\det(\mathbf{I}+\nabla\mathbf{u})\approx1+\partial_k u_k$ ; the elastic wave speeds are $c_L=\sqrt{(\lambda+2\mu)/\rho}$ , $c_T=\sqrt{\mu/\rho}$ , and the bulk modulus is $K=\lambda+\tfrac23\mu$ .

Builds on: Discrete Loaded String and the Continuum Limit · Lagrangian Density and Continuum Lagrangian Formalism · Stress Tensor and Cauchy's Relation Leads to: Linear Elasticity and Hooke's Law for Solids · Elastic Waves in Solids · Euler–Lagrange Equations for Fields

Why is there a factor of one half in the definition of the strain tensor?

Two reasons converge. Geometrically, the change of squared length is $\mathrm{d}s'^2-\mathrm{d}s^2=2E_{ij}\mathrm{d}x_i\mathrm{d}x_j$ , and dividing by 2 makes the diagonal component equal the plain fractional elongation: a 1% uniaxial stretch gives $\varepsilon_{11}=0.01$ exactly, not 0.02. Algebraically, $\tfrac12(\partial_i u_j+\partial_j u_i)$ is the unique symmetric projection of the gradient — without the half you would double-count the gradient when adding back the antisymmetric part, since $\partial_i u_j=\varepsilon_{ij}+\omega_{ij}$ must be an identity.

My FEA package uses a strain vector with six entries and gamma components. How does that map onto the tensor here?

That is Voigt notation: the symmetric tensor is flattened to $(\varepsilon_{11},\varepsilon_{22},\varepsilon_{33},\gamma_{23},\gamma_{13},\gamma_{12})$ with engineering shears $\gamma_{ij}=2\varepsilon_{ij}$ in the last three slots. The factor of two is deliberate: it makes the work density a plain dot product, $w=\tfrac12\boldsymbol{\sigma}^{\mathsf T}\boldsymbol{\varepsilon}$ , without extra 2s. The price is that the $6\times6$ stiffness matrix entries mixing normal and shear slots are not raw tensor components of $C_{ijkl}$ . Convert once, at the boundary between your hand calculation and the code, and never mix conventions mid-derivation.

Why exactly does the antisymmetric part change no lengths? A rotated fiber clearly moves.

Moving is not straining. For $u_i=\omega_{ij}x_j$ with antisymmetric $\omega$ , a fiber $\mathrm{d}x_i$ acquires $\mathrm{d}u_i=\omega_{ij}\mathrm{d}x_j$ , which is perpendicular to $\mathrm{d}x_i$ because $\omega_{ij}\mathrm{d}x_i\mathrm{d}x_j=0$ by antisymmetry. Adding a perpendicular increment changes a vector's direction but its length only at second order — $|\mathrm{d}\mathbf{x}+\mathrm{d}\mathbf{u}|^2=\mathrm{d}s^2+O(\omega^2)$ . That second-order remainder is exactly what the quadratic term of the Green–Lagrange tensor removes, which is why finite rotations need $E_{ij}$ while infinitesimal ones are already handled by $\varepsilon_{ij}=0$ .

Strain has six components but displacement only three. Can I prescribe any symmetric tensor field as a strain?

No — that mismatch of counts is the whole story of compatibility. Six functions derived from three potentials must satisfy three independent constraints: the Saint-Venant compatibility conditions, schematically $\nabla\times(\nabla\times\boldsymbol{\varepsilon})^{\mathsf T}=0$ . Violate them and no continuous single-valued displacement field exists; the "body" you have described would have to tear or overlap itself. Compatibility is what makes stress-based and strain-based solution methods honest, and its failure is put to work deliberately: dislocations and thermal misfit strains are modeled precisely as sources of incompatibility.

Is the strain tensor really a tensor? What happens to the components when I rotate my axes?

Yes, rank 2 and genuinely tensorial: under a rotation $R$ the components transform as $\varepsilon'_{ij}=R_{ik}R_{jl}\varepsilon_{kl}$ . That is not pedantry — it is the working content of the transformation law $\varepsilon_{nn}=\varepsilon_{ij}n_in_j$ used every day to read a strain-gauge rosette: each gauge samples the same tensor along a different direction, and three directions suffice to reconstruct the three in-plane components. The eigenvalues (principal strains), the trace (dilatation), and the deviatoric invariants are the rotation-independent content.

What is the difference between the Green-Lagrange strain and the Almansi strain I sometimes see in finite-element manuals?

They answer the same question in different coordinates. Green–Lagrange $E_{ij}$ refers the change of squared length to the reference configuration (Lagrangian description); the Almansi tensor refers it to the deformed configuration (Eulerian description). Both vanish exactly for rigid motions and both linearize to the same Cauchy strain $\varepsilon_{ij}$ when gradients are small — the distinction only bites at finite deformation, where you must also pair each strain with its energy-conjugate stress (second Piola–Kirchhoff with Green–Lagrange, Cauchy with Almansi rates) to keep work expressions consistent.

Does a negative diagonal strain component just mean compression?

Yes: $\varepsilon_{11}<0$ means fibers along $x_1$ shortened by that fraction, and a negative trace means net volume loss. The sign convention (tension positive) is inherited from the deformation map with no freedom to choose. Two cautions: a negative off-diagonal component does not mean "compression", it means the right angle between the two fiber directions opened rather than closed; and principal strains of mixed sign are the generic case — the bearing-pad example has $\pm1.5\%$ simultaneously, stretch along one diagonal, shortening along the other.

At what strain magnitude do I actually have to abandon the linear tensor in practice?

There is no universal cliff, but useful landmarks exist. Metals yield near $\varepsilon\sim10^{-3}$ , far inside the linear-kinematics regime, so classical structural analysis almost never needs $E_{ij}$ for strain magnitude alone. The earlier breakdown is rotational: a slender beam or plate can rotate by several degrees while its material strains stay tiny, and at rotation $\theta$ the linear tensor fabricates spurious strains of order $\theta^2/2$ — already $1.5\times10^{-3}$ (a full yield-strain of error) at $\theta=3^\circ$ . Rule of thumb: trust $\varepsilon_{ij}$ when both strains and rotations are below about $10^{-2}$ ; switch to finite-deformation measures for rubber ( $\varepsilon\sim1$ ), biological tissue, metal forming, and any buckling or large-deflection problem.