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The displayed scalar stress potential automatically satisfies the two in-plane static momentum equations without body force, because the mixed third derivatives cancel. For homogeneous isotropic linear elasticity, strain compatibility further makes $A$ biharmonic. Since stresses depend only on second derivatives, $A$ is unique only up to $ax+by+c$ on a connected domain. Conversely a function with all second derivatives zero is affine, proving that this is exactly the gauge freedom. It does not license adding an arbitrary biharmonic function, which generally changes stress.

The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

A small arc of a rotating thin hoop needs inward force per length $\rho A{\omega }^{2}r$. Its circumferential tension supplies this through curvature, giving tension $\rho A{\omega }^2{r}^2$ and tensile stress $\rho {\omega }^2{r}^2$. Hooke's law converts that stress to circumferential strain, which equals fractional radius change. Small strain, a slender cross-section and uniaxial linear elasticity are required; the cross-sectional area cancels at leading order.

A damped torsional oscillation is rotation about a fiber's fixed reference orientation with a restoring torque and dissipative torque. In the ideal viscous model, damping contributes $-b\dot{\theta }$ with constant $b\ge 0$. This is a rotational damped harmonic oscillator. For known constant moment of inertia $I$ and torsion constant $\kappa$, define $\gamma =b/(2I)$. In the oscillatory regime ${\gamma }^2<\kappa /I$,$A$ and $\phi$ are set by the initial state. Successive positive peaks measured from the same equilibrium have magnitudes $p_n$ and $p_{n+1}$; their ratio gives $\delta =\ln (p_n/p_{n+1})=\gamma T_d$. Consequently a measured damped oscillation period and decay together recoverUsing only $4{\pi }^{2}I/T_d^2$ underestimates the torsion constant in this model. The toy curve in torsion constant sets $I=2.0\times 10^{-6}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, and $\gamma =0.020{\mathrm{s}}^{-1}$. It gives $T_0\simeq 88.86\mathrm{s}$, $T_d\simeq 92.64\mathrm{s}$, and $\delta \simeq 1.853$. Ignoring the decay recovers $9.2\times 10^{-9}\mathrm{N}m/rad$, an $8\%$ underestimate; including it recovers the stipulated torsion constant.

These equations assume a single rotational mode, a stationary zero, linear elasticity, and viscous damping. Internal fiber losses need not behave as a constant viscous coefficient: Fleischer and colleagues model their precision apparatus using a loss angle. Their response is evidence for specifying a damping model, not for applying the viscous correction universally. ^[ref]

For a time-independent symmetric negative semidefinite boundary matrix $k$, the difference of two linear elasticity solutions with the same data has traction $kw$. Integration by parts conserves the displayed nonnegative energy. Zero initial data force zero velocity and then zero displacement. The matrix $-k$ is the stiffness of boundary springs to a prescribed support; null directions are traction-free.

The elastic polarization is the excess stress relative to a homogeneous comparison elastic stiffness tensor. In heterogeneous linear elasticity, $\tau =(C-C^0)\epsilon$. The comparison static elastic equilibrium equation gives the integral equation $\epsilon ={\epsilon }^0-\Gamma \tau$, where ${\epsilon }^0$ is the reference strain under the same external loading.

For homogeneous stable linear elasticity, the three positive eigenvalues of the acoustic tensor give three phase velocities for each wavefront normal. Plotting the corresponding three vectors $\mathbf{p}=\mathbf{n}/v$ gives the three sheets of the elastic slowness surface. Sheets can meet at degeneracies. The elastic-wave energy flux is normal to each regular sheet: differentiating its eigenvalue equation along a sheet tangent $d\mathbf{p}$ gives $c_{ijkl}{a}_i{a}_k{p}_{l}d{p}_j=0$.

Reduction of an applied load with purely elastic strain increments while accumulated plastic strain is fixed. The superposition principle applies to the increment, so the current stress equals the previous stress plus the linear elasticity response to the load change. The resulting stress must still satisfy the yield surface constraint; reverse yielding ends the purely elastic stage.

An elastic wave propagates a disturbance through an elastic solid. Small displacements in linear elasticity satisfy a vector wave equation governed by mass density and the two Lamé parameters.

The causal Green function for isotropic linear elasticity is tensor-valued: column $k$ is the displacement caused by a point impulsive body force in direction $k$. Its radiative terms are longitudinal P waves at $t=R/\alpha$ and transverse S waves at $t=R/\beta$, proportional to $R^{-1}$. A term supported between the arrivals carries the intermediate/near-field response. Convolution in time and space constructs the displacement for a general source.

On a connected isotropic linear elastic body, assume $\mu >0$ and $3\lambda +2\mu >0$. Then $W(e)=\mu |e^{\mathrm{dev}}{|}^2+(3\lambda +2\mu )(\mathrm{tr}e{)}^2/6$ is positive for every nonzero symmetric strain tensor. If two equilibria have the same body force and boundary traction, their difference $v$ satisfies zero loading. Multiplication of equilibrium by $v$ and the divergence theorem yield ${\int }_{B}2W(e(v))dV=0$. Hence $e(v)=0$, and the kernel of the symmetric gradient consists of infinitesimal translations and rotations: $v=a+\omega \times x$.

The conclusion is uniqueness modulo these modes, not existence for arbitrary tractions. Existence additionally requires zero total resultant force and torque. On a disconnected body each connected component has its own rigid modes; fixing enough displacements removes those modes and can produce full uniqueness.

For a uniform traction-free sphere, first-order self-gravity is $b=-gx/a$. Static averaging gives $\overline{\sigma }=-\rho gaI/5$. Homogeneous linear elasticity then gives $\overline{e}=-\rho gaI/(15\kappa )$. The boundary strain identity identifies $\overline{e}=u(a)I/a$, proving the displayed positive radius decrease. The local strain is not uniform, and the cancellation of the shear modulus concerns only the mean contraction.

An eigenstrain is the deformation a material would prefer locally without applied stress. In a one-dimensional linear elasticity model, let $E$ be Young's modulus, take tension as positive stress, and write $\sigma =E(\epsilon -{\epsilon }^{\mathrm{mag}})$. A free rod has $\sigma =0$ and follows $\epsilon ={\epsilon }^{\mathrm{mag}}$. A perfectly blocked rod has $\epsilon =0$ and $\sigma =-E{\epsilon }^{\mathrm{mag}}$. Thus positive preferred elongation produces compressive stress when blocked.

For stipulated $E=100\mathrm{GPa}$, cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$, and ${\epsilon }^{\mathrm{mag}}=20\times 10^{-6}$, blocking gives $\sigma =-2\mathrm{MPa}$ and axial force $F=\sigma A=-2\mathrm{N}$. This idealization ignores changes of magnetization caused by the load itself. A real magnetostrictive actuator therefore needs both magnetic and mechanical constitutive information: a free-length measurement alone does not determine its loaded motion.

Write a quadratic small-strain energy as $\tilde{W}(F)=Q(\mathrm{sym}(F-I))$ with $Q$ homogeneous of degree two. If it were fully frame indifferent, the rotation $R=\mathrm{diag}(1,-1,-1)$ would have zero energy. Quadratic homogeneity would then make every $F_s=I+s(R-I)$ have zero energy. At $s=1/4$, $F_s=\mathrm{diag}(1,1/2,1/2)$ has determinant $1/4$, yet zero energy. Thus a nondegenerate linear elasticity energy cannot also be used as a fully objective finite-deformation energy. Its invariance concerns infinitesimal rotations around the reference state.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

The antiplane field $w=b\arg (x+iy)/(2\pi )$ has a smooth local gradient on the punctured plane and satisfies the differential strain compatibility condition there. Its circulation around the defect axis is nevertheless $b\ne 0$, excluding a globally single-valued displacement on that domain. A radial cut permits a branch with an axial jump $b$ between the two cut faces; joining faces with that mismatch is the Volterra construction of a screw dislocation with Burgers vector parallel to the axis. The stress has a singular core where linear elasticity requires microscopic or plastic regularization. This example separates a local differential integrability test from global closure around a hole.

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.

In a homogeneous isotropic elastic body without body forces, static equilibrium of the displacement field is

This is the static, homogeneous-isotropic displacement form of linear elasticity.

For the one-dimensional reduction of isotropic linear elasticity, nondegeneracy means that the effective longitudinal modulus $\lambda +2\mu$ is nonzero. The static Navier-Cauchy equation then forces the displacement to be affine.

In three-dimensional isotropic linear elasticity, decompose the symmetric strain as $e=pI+d$ with $p=\mathrm{tr}(e)/3$ and $\mathrm{tr}d=0$. The stored energy is $E=\mu {\sum }_{ij}{d}_{ij}^2+\frac{3}{2}(3\lambda +2\mu )p^2$. Its nonnegativity for every strain is equivalent to the displayed inequalities: the traceless and purely volumetric test strains prove necessity, and the sum of squares proves sufficiency.