Strain ($\mathbf{\epsilon }$)

Strain measures local deformation relative to a reference configuration. In small-deformation theory it is the symmetric part of the displacement gradient.

Elastic strain is the reversible part of strain that determines stress through the elastic constitutive equation. In a small-strain model with plastic and thermal effects, subtract the plastic strain and thermal strain from total strain to obtain it. A stress-free heated body can therefore have nonzero total strain but zero elastic strain; after plastic loading, unloading can likewise remove elastic strain while leaving permanent deformation.

The time rate of strain; a scalar may describe an idealized deformation component, while general deformation uses a strain-rate tensor.

The first-order relative change of volume for a small displacement field. It is the trace of the infinitesimal strain tensor. For a radial displacement field $u(r){\mathbf{e}}_r$ in spherical coordinates, $e=u^{\prime }+2u/r$.

The fractional change in the length of a material line in a specified direction. For infinitesimal deformation with unit direction $\mathbf{n}$, it is $\mathbf{n}\cdot \mathbf{\epsilon }\mathbf{n}$, where $\mathbf{\epsilon }$ is the small strain tensor.

Engineering shear strain is the first-order change of the right angle between material lines initially directed along two perpendicular axes. For a displacement $\mathbf{u}$, ${\gamma }_{xy}={\partial }_y{u}_x+{\partial }_x{u}_y=2{\epsilon }_{xy}$, where ${\epsilon }_{xy}$ is the tensor strain component. Slender Euler-Bernoulli beam equation models neglect the transverse shear strain contribution to deflection.

The time derivative of engineering shear strain. A Newtonian simple-shear stress is $\eta \dot{\gamma }$. This scalar is twice the off-diagonal symmetric rate-of-strain-tensor component.

The infinitesimal strain tensor $\mathrm{diag}(-e,e)$ compresses one principal direction and stretches the perpendicular one by the same amount. Rotating the coordinate axes through $45^{\circ }$ turns it into a tensor with zero normal strains and off-diagonal entries of magnitude $e$. The engineering shear strain is twice the tensor shear component, so its shear angle has magnitude $\gamma =2e$. Likewise principal stresses $(-S,S)$ give zero normal stress and shear stress of magnitude $S$ on those rotated faces. The shear modulus law $S=G\gamma$ therefore gives $S=2Ge$.

Engineering strain is the change in length divided by that segment's initial length. For an extensometer, use its own initial gauge length and extension.

The infinitesimal strain tensor is the symmetric part of a displacement gradient tensor. It is the linear approximation to a finite strain measure for small displacement gradients. An infinitesimal rigid rotation contributes no strain, so the elastic energy depends on the symmetric part rather than the entire gradient.

The deviatoric strain is the trace-free part of the infinitesimal strain tensor. It removes the isotropic volume change. The decomposition is orthogonal for the Frobenius inner product, because a trace-free matrix has zero Frobenius inner product with $I$. ConsequentlyFor $\epsilon =\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$, its trace is $\mathrm{div}v$.