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Fracture toughness measures resistance to crack advance under specified loading and environmental conditions. In linear elasticity, $K_c$ is the critical stress intensity factor. For opening-mode loading, a valid plane strain measurement is denoted $K_{Ic}$, with units $\mathrm{Pa}\sqrt{\mathrm{m}}$, commonly $\mathrm{MPa}\sqrt{\mathrm{m}}$. It differs from a tensile yield strength: a crack changes the near-tip stress concentration.

For a fixed geometry with $K_I=Y\sigma \sqrt{\pi a}$, the threshold $K_I=K_{Ic}$ givesFor example, $K_{Ic}=10\mathrm{MPa}\sqrt{\mathrm{m}}$, $\sigma =100\mathrm{MPa}$ and $Y=1$ give $a_c\simeq 3.18\mathrm{mm}$ in this idealized geometry. The crack-tip plastic zone must remain small for the elastic intensity description to apply; extensive plasticity in continuum mechanics requires a different characterization.^[ref]

A thin elastic plate model may instead prescribe a critical front curvature ${\kappa }_f$. Such a model parameter cannot be substituted numerically for $K_{Ic}$ without a relation specific to that fracture model.

In plane strain perfect plasticity with ${\sigma }_{11}=-p+k\sin 2\varphi$, ${\sigma }_{22}=-p-k\sin 2\varphi$ and ${\sigma }_{12}=-k\cos 2\varphi$, equilibrium givesThe slip line tangent directions $(\cos \varphi ,\sin \varphi )$ and $(-\sin \varphi ,\cos \varphi )$ are eigenvectors with eigenvalues $+1,-1$. Taking directional derivatives proves the two Hencky stress relations. These sign choices must be changed consistently if the shear-stress convention changes.

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.

Plane strain sets all strains involving $z$ to zero, including $e_{zz}=0$, while $e_{xx}=u_x$, $e_{yy}=v_y$ and $e_{xy}=(u_y+v_x)/2$. ThusIn particular, ${\tau }_{zz}$ generally does not vanish: the suppressed out-of-plane strain requires a reaction stress. Setting that entry to zero would instead be a plane-stress assumption.

Write $B=A+H$ for the reference outer radius and $b=a+h$ for the current outer radius. The inner material normal is $-e_r$. A fluid pressure $P$ applies traction $-Pn=P{e}_r$ there, so the radial Cauchy stress tensor component must be negative:The second condition uses zero external gauge pressure. Both faces also have zero shear traction; the axisymmetric diagonal stress field satisfies those conditions automatically. The axial constraint is plane strain, so supports may supply an axial reaction; it does not imply zero axial stress.

Use orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

In infinitesimal elasticity and plasticity in continuum mechanics, decompose the symmetric strain as $e=e^e+e^p$. For incompressible isotropic elasticity, ${\sigma }^{\prime }=2G{e}^e$, with $G$ the shear modulus and both strains traceless. The associated flow rule for the Von Mises yield criterion has ${\dot{e}}^p=\Lambda {\sigma }^{\prime }$ after absorbing a positive factor into $\Lambda \ge 0$. Plane strain has $e_{33}=0$, soIf ${\sigma }_{33}^{\prime }=0$ initially, it remains zero by this homogeneous scalar differential equation. Then $e_{33}^e=0$ and ${\dot{e}}_{33}^p=0$, so $e_{33}^p=0$ is consistent with zero initial plastic component. This proves consistency, not a claim that arbitrary prestressed plane-strain histories must have zero separate components. The out-of-plane shear components can consistently be zero by the same argument.

Write $p=-\mathrm{tr}\sigma /3$, so ${\sigma }_{33}=-p$. The proposed in-plane parameterization hasIts squared contraction is $2k^2({\sin }^{2}2\varphi +{\cos }^{2}2\varphi )=2k^2$, proving the yield condition identically. Its principal deviatoric stresses are $k,-k,0$. Define the two slip line families by tangent unit vectorsThus $\alpha$ lines have tangent angle $\varphi$ and $\beta$ lines have tangent angle $\varphi +\pi /2$. Their normal extensions from the associated plastic strain rate vanish, while their shear stresses have maximum magnitude $k$; they are not the principal stress directions.

In the absence of body force, differentiate the stress parameterization in the two equilibrium equations. One obtainsThe symmetric matrix multiplying $2k\nabla \varphi$ has eigenvectors $a,b$ with eigenvalues $+1,-1$. Taking its contraction with these vectors provesHence the Hencky stress relations areConstants can differ between lines, and these signs depend on the specified negative shear-stress convention.

Place the notch tip at the origin, with the opening towards negative $x_1$ and tensile loading in the transverse direction. For a usual reentrant notch, $0\le \gamma \le \pi /2$. The upper face has tangent angle ${\vartheta }_f=\pi -\gamma$. Choose its tangent $t=(\cos {\vartheta }_f,\sin {\vartheta }_f)$ directed away from the tip and outward material normal $n=(-\sin {\vartheta }_f,\cos {\vartheta }_f)$ pointing into the notch. Direct rotation of the stress gives the normal and tangential traction componentsTake the notch faces traction-free, as intended by the notch problem. Then the cosine vanishes. Select the tensile branch $\sin (2\varphi -2{\vartheta }_f)=1$, for which the face-tangential normal stress is $2k$. Thus the upper face has $\boxed{p=-k,{\varphi }_f=\pi /4-\gamma (\mathrm{mod}\pi )}$. Choosing the opposite branch would describe compression rather than the tensile field.

Construct the centred slip-line fan at a traction-free V-notch to connect this surface state to the forward symmetry axis. Near the upper face there is a constant-stress region. It joins a fan bounded by raysInside the fan choose $\beta$ lines radial and $\alpha$ lines circular about the tip, so their angle parameter is $\varphi =\vartheta -\pi /2$. Along a circular $\alpha$ line the first invariant givesThe second invariant is automatically constant along each radial $\beta$ line because $p,\varphi$ are independent of radius. Thus this is an actual equilibrium fan, not just an assumed transport of a surface value. At ${\vartheta }_2$ it matches the free-face constant state. At ${\vartheta }_1$ it matches a forward constant-stress region withReflect the upper field across the symmetry axis to obtain the lower field. The forward region has zero shear, respects this symmetry, and contains the positive $x_1$ axis. There $\sin 2\varphi =-1$, and henceThis value holds in the fully plastic notch-tip region described by the field. It is not a uniform stress assertion for the entire axis of an arbitrary finite specimen, including any elastic or far-field region.

The inextensible fibre constraint gives $|{\mathbf{x}}_{,1}|=1$. Write the fibre direction and its transverse unit direction asThen $x_{1,1}=\cos \theta$ and $x_{2,1}=\sin \theta$. In plane strain, incompressibility gives $\det F=1$. Decomposing the second column of the deformation gradient along $e,n$ therefore gives ${\mathbf{x}}_{,2}=\gamma e+n$, where $\gamma ={\mathbf{x}}_{,2}\cdot e$. HenceThis is a local rotation followed by simple shear.

The compatibility of the deformation map requires ${\mathbf{x}}_{,12}={\mathbf{x}}_{,21}$. Since $e_{,j}={\theta }_{,j}n$ and $n_{,j}=-{\theta }_{,j}e$, this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus $\theta$ is constant when $d{X}_2/d{X}_1=-1/\gamma$. The equivalent condition $d{X}_1+\gamma d{X}_2=0$ also covers $\gamma =0$. Its current tangent is proportional to $F(-\gamma ,1{)}^T=n$, so these transverse curves are straight wherever the deformation is regular.

To construct the stress with zero body force, specify the local shear stress $h$ as the material's function or functional of $\gamma$ under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has $\nabla \theta =\kappa e$ with $\kappa ={\theta }_{,1}$, so $\nabla \cdot e=0$, $\nabla \cdot n=-\kappa$, $(e\cdot \nabla )e=\kappa n$ and $(n\cdot \nabla )e=0$. Resolving $\nabla \cdot \sigma =0$ along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines $s$ by integration along fibres and then $t$ along transverse curves. Thus any regular compatible deformation has a local zero-body-force stress construction once the shear law and suitable boundary integration data are supplied. There is no additional constitutive normal-stress law to specify: the two normal components are fixed by force balance and boundary tractions. This does not assert that arbitrary separately prescribed boundary tractions can accompany an arbitrary deformation.

In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.

Slip-line field theory solves certain two-dimensional problems of plastic deformation by constructing two orthogonal families of characteristic curves, called slip lines. Its classical assumptions include plane strain, negligible inertia and body force, and a rigid, perfectly plastic material. The curves are aligned with the directions of maximum in-plane shear stress, rather than the principal stress directions.^[ref]

The curve network encodes both the stress field and constraints on admissible velocity. Indentation of a thick workpiece by a rigid flat punch provides a motivating application: the construction predicts a plastic flow region and the load needed to sustain that flow. It is useful because geometry can turn coupled field equations into relations integrated along curves. A slip line is a continuum characteristic, however, not necessarily a microscopic crystal slip system or the actual path of a moving material particle.

The reduction in a mechanical system's potential energy per incremental crack area under specified loading conditions. In small-scale-yielding linear elasticity, mode-I plane strain gives $G=K_I^2/E^{\prime }$ with $E^{\prime }=E/(1-{\nu }^2)$. The critical value includes the dissipative processes associated with crack advance.