7.2 Strain: the deformation of a material element

Stress measures the internal forces; strain measures the deformation those forces accompany. This lesson builds the strain tensor from the displacement field and shows why only its symmetric part represents genuine deformation — the antisymmetric part is mere rotation, which stores no elastic energy.

The displacement field

When a body deforms, the material point originally at position $\mathbf{r}$ moves to $\mathbf{r} + \mathbf{u}(\mathbf{r})$ . The vector field $\mathbf{u}(\mathbf{r})$ is the displacement field, and it contains everything about the deformation. But $\mathbf{u}$ itself is not strain: a body can be picked up and carried across the room, giving a large uniform $\mathbf{u}$ with no deformation at all. What deforms the material is the way $\mathbf{u}$ varies from point to point — its gradient.

Separating stretch from rotation

▶ Why only the symmetric gradient is strain Derivation

The relative displacement between a point and a near neighbour a small vector $d\mathbf{r}$ away is, to first order, the displacement gradient $\partial u_i/\partial x_j$ contracted with $d\mathbf{r}$ . Split that gradient into symmetric and antisymmetric parts:

\frac{\partial u_i}{\partial x_j} \;=\; \underbrace{\tfrac12\!\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right)}_{\varepsilon_{ij}} \;+\; \underbrace{\tfrac12\!\left(\frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i}\right)}_{\omega_{ij}}.

The antisymmetric part $\omega_{ij}$ is an infinitesimal rotation: it turns the element rigidly without changing any length or angle, and stores no energy. Only the symmetric part changes distances between material points, so only it is the deformation.

The symmetric part is the strain tensor:

\varepsilon_{ij} \;=\; \tfrac12\!\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).

where

$\varepsilon_{ij}$: small-strain tensor (dimensionless) —
$u_i$: components of the displacement field m
$x_j$: reference (undeformed) coordinates m

This is the small-deformation, or linearised, strain — valid when displacement gradients are small, which covers nearly all of engineering and most of the elastic response of stiff materials.

ε₁₁ (stretch in x) = 0.150 ε₂₂ (stretch in y) = -0.080 ε₁₂ (shear) = 0.050

Each grid line maps under the displacement field u = ε·r. Diagonal terms ε₁₁, ε₂₂ stretch the grid in the corresponding direction. Off-diagonal ε₁₂ shears it — turning squares into parallelograms. The strain tensor records all of this with six independent components (in 3-D); positive trace is dilation, traceless part is pure shear.

Reading the components

The strain tensor decomposes deformation exactly as the stress tensor decomposes force:

Normal strains $\varepsilon_{ii}$ — the diagonal — are fractional stretches along each axis. $\varepsilon_{11} = \partial u_1/\partial x_1$ is the fractional elongation in the $x_1$ direction.
Shear strains $\varepsilon_{ij}$ with $i\ne j$ measure the change in angle between two lines that were initially perpendicular; each equals half the engineering shear $\gamma_{ij}$ .

The trace has a direct meaning. Summing the diagonal,

\varepsilon_{kk} \;=\; \nabla\cdot\mathbf{u},

is the volumetric strain — the fractional change in volume of the element. A deformation that preserves volume (pure shear) is traceless; a uniform expansion is pure trace. This split of strain into a volume-changing part and a shape-changing part is exactly the split that the elastic moduli of the next lesson respond to, through two independent constants.