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Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

Compression and stretch viewed on rotated faces. To isolate shear, take equal small principal strains ${ϵ}_{xx}=-e$, ${ϵ}_{yy}=+e$. A square with sides at $45^{\circ }$ to these directions becomes a rhombus: adjacent side lengths are unchanged to first order, but their right angle changes by $2e$. In axes rotated through $45^{\circ }$, the same strain tensor has zero diagonal entries and off-diagonal entries of magnitude $e$.

The pure shear from opposite principal strains therefore has engineering shear strain $\gamma =2e$, not $e$. Equal opposite principal stresses $(-S,+S)$ give purely tangential traction of magnitude $S$ on those rotated faces. HenceIf the compression and stretch are not equal and opposite, an additional dilatational strain remains; their deviatoric part still supplies shear.

The shaft's torsional stiffness. At radius $r$, an end twist $\varphi$ over length $L$ gives tangential displacement $r\varphi$ and shear angle $r\varphi /L$. The local shear stress is $Gr\varphi /L$. An annular area $dA=2\pi rdr$ contributes torque $r\tau dA$, soThus torsion of a solid circular shaft givesHere $J_p$ is the polar second moment of area, in ${\mathrm{m}}^4$, not a mass moment of inertia. This $C$ is the transmitted torque magnitude; the shaft's restoring couple opposes the twist. The formula assumes small elastic shear and a circular uniform section.

The shear modulus $G$ is the ratio of shear stress to small engineering shear strain in an isotropic linear elastic material:For a slab of thickness $h$ whose upper face moves sideways by $\Delta x$ while the lower face is fixed, $\gamma \simeq \Delta x/h$. A tangential force $F$ distributed over face area $A$ gives $\tau =F/A$. Thus $G=Fh/(A\Delta x)$ in the uniform-shear idealization. Its units are pressure, and the engineering convention gives $\gamma =2{\epsilon }_{xy}$ in terms of the infinitesimal strain tensor.

An experiment on the torsion of a solid circular shaft measures the same parameter through torque and end rotation: for shaft length $L$ and polar second moment of area $J_p$, $\mathcal{T}=G{J}_p\theta /L$. This separates material resistance to shear strain from the specimen's geometric resistance to twisting. In isotropic linear elasticity, $G=E/[2(1+\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio.