7.1 Stress: force per unit area on a plane

An elastic solid is a continuum that stores energy when deformed and springs back when released. To describe it we need two fields: the internal forces it carries, and the deformation it undergoes. This lesson builds the first — stress — and shows why the internal force at a point is not a vector but a tensor.

Force across an internal surface

Cut an imaginary plane through the interior of a loaded body. The material on one side pushes and pulls on the material on the other, across that cut, with a force per unit area. That force per unit area is the traction $\mathbf{t}$ . Crucially, it depends on how the plane is oriented: the same point carries a different traction on a horizontal cut than on a vertical one. A single vector cannot encode a quantity that varies with the orientation of the surface it acts on — so stress must be richer than a vector.

The resolution is a linear map. At any interior point there is a single object, the Cauchy stress tensor $\boldsymbol{\sigma}$ , that turns the plane’s unit normal $\hat{\mathbf{n}}$ into the traction across that plane:

\mathbf{t}(\hat{\mathbf{n}}) \;=\; \boldsymbol{\sigma}\,\hat{\mathbf{n}}.

where

$\mathbf{t}$: traction — force per unit area on the cut surface Pa
$\hat{\mathbf{n}}$: unit normal of the cut plane —
$\boldsymbol{\sigma}$: Cauchy stress tensor, components $\sigma_{ij}$ Pa

That one tensor carries all the information: feed it any normal and it returns the traction on that plane. This is Cauchy’s fundamental result, and it is what makes a field theory of solids possible.

σ₁₁ = 2.00 σ₂₂ = 1.00 σ₁₂ (shear) = 0.50 Plane normal angle θ = 30°

The stress tensor σ acts on a plane with normal n̂ to give the traction vector t = σ·n̂ — the force per unit area that the material on the +n̂ side exerts on the material on the −n̂ side. The traction has a normal component (pointing along n̂, the "pressure-like" part) and a shear component (perpendicular to n̂, the "drag-like" part). Rotating the plane changes how σ apportions itself between the two.

Rotate the plane’s normal: the same stress tensor produces a different traction vector on each orientation. The traction generally is not parallel to $\hat{\mathbf{n}}$ — it has a component along the normal (pulling or pushing on the face) and a component in the plane (dragging along it).

Reading the components

Write the tensor as a matrix $\sigma_{ij}$ . The index $j$ names the face (the plane whose normal is the $j$ -axis); the index $i$ names the direction of the force on that face. The two kinds of component have distinct physical meaning:

Normal stresses $\sigma_{ii}$ — the diagonal entries — act perpendicular to their face. A positive value is tension (the material is being pulled apart), a negative value is compression.
Shear stresses $\sigma_{ij}$ with $i\ne j$ — the off-diagonal entries — act tangent to their face, sliding one layer over the next.

The tensor is symmetric, $\sigma_{ij} = \sigma_{ji}$ , a consequence of the balance of angular momentum on a vanishing element: an antisymmetric part would spin the element with infinite angular acceleration. Symmetry leaves six independent components rather than nine.

A fluid at rest is the degenerate case. It cannot support shear, so every off-diagonal component vanishes and the diagonal ones are all equal to the negative pressure:

\sigma_{ij} \;=\; -p\,\delta_{ij}.

The stress tensor of a solid is the general object of which the fluid pressure of the previous chapters is the isotropic special case. With the internal forces now in hand, the next lesson describes the deformation they accompany.