What is Displacement field and strain tensor? — Physics

Definition 3.1 University Physics — Year 3 · Chapter 3 — Continuum Mechanics and Elasticity

When a solid deforms, the particle initially at $\vect r$ moves to $\vect r + \vect u(\vect r)$ : $\vect u$ is the displacement field. For small deformations, the local distortion is measured by the strain tensor, the symmetric array

\varepsilon_{ij} = \frac12\Big( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \Big) , \qquad i, j \in \{x, y, z\} .

A diagonal component $\varepsilon_{xx}$ is the relative elongation of the material along $x$ (dimensionless, e.g. $10^{-3}$ for a steel cable in service); an off-diagonal component $\varepsilon_{xy}$ is half the closing of the angle between the $x$ and $y$ material directions — a shear. The trace is the relative change of volume, the dilatation:

\frac{\delta V}{V} = \varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz} = \operatorname{div}\vect u .

The strain tensor sees only true deformation: stretching (diagonal components), shearing (off-diagonal components) — and a rigid rotation not at all.

Examples

Example 3.2 (Three elementary deformations)

Uniform dilation $\vect u = \alpha\vect r$ : $\varepsilon_{ij} = \alpha\,\delta_{ij}$ , volume change $3\alpha$ , no shear. Simple shear $\vect u = (\gamma y, 0, 0)$ : $\varepsilon_{xy} = \gamma/2$ , all diagonal components zero — shape changes, volume does not. Rigid rotation $\vect u = (-\omega y, \omega x, 0)$ : $\varepsilon_{ij} = 0$ identically — the antisymmetric part of $\partial u_i/\partial x_j$ , discarded by the symmetrisation, is exactly the part that rotates without deforming. Strain measures deformation only.

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