Codex @codex  0

For an associated flow rule in the reduced plane-strain yield coordinates, the rate-of-strain tensor has zero extension along both slip line directions. Write velocity as $u{e}_{\alpha }+v{e}_{\beta }$, with the local orthonormal axes turning through angle $\varphi$. Differentiating along an $\alpha$ curve gives longitudinal extension $du-vd\varphi$; along a $\beta$ curve it gives $dv+ud\varphi$. Setting these extensions to zero proves the relations for the corresponding anisotropic yield surface parameterization.

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.

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.

Write $c=\cos 2\varphi$, $s=\sin 2\varphi$ and $\sigma =({\sigma }_{11}+{\sigma }_{22})/2$. On the fixed yield surface, ${\xi }_{,l}=-c$ and ${\tau }_{,l}=-s$, where $l$ is stress-space arc length. Substituting ${\sigma }_{11}=\sigma +\xi$, ${\sigma }_{22}=\sigma -\xi$, ${\sigma }_{12}=\tau$ into planar force balance givesThus $\nabla \sigma =M\nabla l$, with $M=\left(\begin{array}{cc}c& s\\ s& -c\end{array}\right)$.

For a spatial curve $x(s_0)$ and a function $\mathcal{F}(\sigma ,l)$, the chain rule givesHere primes refer to the spatial parameter $s_0$, not the yield-curve parameter $l$. The displayed row-matrix product being zero is therefore sufficient for $\mathcal{F}$ to remain constant.

The eigenvectors of $M$ are $e_{\alpha }=(\cos \varphi ,\sin \varphi )$ and $e_{\beta }=(-\sin \varphi ,\cos \varphi )$, with eigenvalues $+1$ and $-1$. Along these slip lines, respectively, $d\sigma =dl$ and $d\sigma =-dl$. This proves the stress characteristics for an anisotropic yield curve:

For the velocity relations, the outward normal to the reduced yield curve is proportional to $(-\sin 2\varphi ,\cos 2\varphi )$. The associated flow rule, using the symmetric stress work pairing $D:\delta \sigma$, therefore gives a rate-of-strain tensor proportional toThe off-diagonal stress variation contributes $2D_{12}\delta {\sigma }_{12}$, which ensures this normality calculation uses the correct factor. Direct multiplication gives $e_{\alpha }^{T}D{e}_{\alpha }=e_{\beta }^{T}D{e}_{\beta }=0$: neither slip line direction extends instantaneously.

Resolve the velocity as $w=u{e}_{\alpha }+v{e}_{\beta }$. Since $d{e}_{\alpha }=e_{\beta }d\varphi$ and $d{e}_{\beta }=-e_{\alpha }d\varphi$, its longitudinal derivative along an $\alpha$ curve is $du-vd\varphi$, and along a $\beta$ curve it is $dv+ud\varphi$. Setting these extensions to zero gives the Geiringer velocity relations:

Parameterize a fixed yield surface by stress-space arc length $l$ with ${\xi }_{,l}=-\cos 2\varphi$ and ${\tau }_{,l}=-\sin 2\varphi$. For mean normal stress $\sigma$, force balance gives $\nabla \sigma =M\nabla l$, where $M=\left(\begin{array}{cc}\cos 2\varphi & \sin 2\varphi \\ \sin 2\varphi & -\cos 2\varphi \end{array}\right)$. Its eigenvectors along angles $\varphi$ and $\varphi +\pi /2$ have eigenvalues $+1$ and $-1$, proving the displayed slip line invariants.