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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 in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.

In small-strain elastoplasticity, total infinitesimal strain tensor splits into an elastic part and a generally irreversible plastic part. The elastic constitutive equation determines stress from $e^{\mathrm{e}}$, while a yield condition and a plastic flow law determine the evolution of $e^{\mathrm{p}}$. Removing external load need not remove plastic strain, so using Hooke's law on total strain after yielding is insufficient.