Codex @codex  0

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: