Benchmark Mohr Coulomb (MC) viscosplastic material is sheared by a constant velocity

Plastic Simple Shear Test

In this benchmark, a square block (1 m x 1 m) of Mohr-Coulomb (MC) visocoplastic material is simply sheared with a constant velocity.

• $v_{x}(y) = v_{x}y$ in ${(x,y)| (x,y) \in [0,1] \times [0,1]}.$
• $v_{y} = 0$ and $0 \le y \le 1$
• $u(y,t) = (v_{x}yt, 0)$ $$\varepsilon_{xy} = \frac{1}{2}\frac{\partial (v_{x}yt)}{\partial y} = \frac{1}{2} v_{x} t.$$

$$\sigma_{xy} = 2\mu\varepsilon_{xy} = \mu v_{x} t,$$ where $\mu$ is a shear modulus.

$$\Delta\boldsymbol{\varepsilon} = \begin{pmatrix} 0 & \frac{1}{2} v_{x} \Delta t \\ \frac{1}{2} v_{x} \Delta t & 0 \end{pmatrix} \Rightarrow \Delta \mathbf{e} = \begin{pmatrix} -\frac{1}{2} v_{x} \Delta t & 0 \\ 0 & \frac{1}{2} v_{x} \Delta t \end{pmatrix}$$

$$\Delta\boldsymbol{\sigma} = \begin{pmatrix} 0 & \mu v_{x} \Delta t \\ \mu v_{x} \Delta t & 0 \end{pmatrix} \Rightarrow \Delta\mathbf{S} = \begin{pmatrix} -\mu v_{x} \Delta t & 0 \\ 0 & \mu v_{x} \Delta t \end{pmatrix}$$

Here $S_{1} < S_{2}$ and $e_{1} < e_{2}$ and the tension positive convention is followed.

The principal directions are the columns of the following matrix:

$$\mathbf{V} = \begin{pmatrix} \cos(\pi/4) & \sin(\pi/4) \\ -\sin(\pi/4) & \cos(\pi/4) \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$$

$$\begin{split} f_{s}(S_{1}^{n+1}, S_{2}^{n+1}) &= (S_{1}^{n} + 2\mu\Delta\varepsilon_{1}^{e}) - N_{\phi}(S_{2}^{n} + 2\mu\Delta\varepsilon_{2}^{e}) + 2c\sqrt{N_{\phi}}, \\ g_{s}(S_{1}^{n+1}, S_{2}^{n+1}) &= S_{1}^{n+1} - N_{\psi}S_{2}^{n+1},\\ \end{split}$$

$$ \begin{split} \Delta\varepsilon_{1}^{e} &= \Delta\varepsilon_{1} - \Delta\beta \frac{\partial g}{\partial S_{1}} = \Delta\varepsilon_{1} - \Delta\beta,\\ \Delta\varepsilon_{2}^{e} &= \Delta\varepsilon_{2} - \Delta\beta \frac{\partial g}{\partial S_{2}} = \Delta\varepsilon_{2} + N_{\psi}\Delta\beta. \end{split}$$

From $f_{s}(S_{1}^{n+1}, S_{2}^{n+1}) = 0$,

$$(S_{1}^{n} + 2\mu(\Delta\varepsilon_{1} - \Delta\beta)) - N_{\phi}(S_{2}^{n} + 2\mu (\Delta\varepsilon_{3} + N_{\psi}\Delta\beta)) + 2c\sqrt{N_{\phi}} = 0.$$

$$ (S_{1}^{n+1,el} - 2\mu \Delta\beta) - N_{\phi}(S_{2}^{n+1,el} + 2 \mu N_{\psi} \Delta\beta) + 2c\sqrt{N_{\phi}} = 0. $$

$$ \Delta\beta = \frac{f_{s}(S_{1}^{n+1,el},S_{2}^{n+1,el})}{2\mu(1+N_{\phi}N_{\psi})}. $$

When $\psi=0$ and $N_{\psi}=1$, the plastic corrections for $S_{1}$ and $S_{2}$ are identical in magnitude and opposite in sign. Together with the shape of the elastic stress tensor, this property allows the plastic Cauchy stress tensor have the same shape: zero-diagonals and symmetric. Also, $\sigma_{xy}^{n+1}$ is always equal to either $S_{2}^{n+1,el}$ or $S_{2}^{n+1,pl}$.

Summarized below is the algorithm for the analytic solution for the simple shear case:

1. $\sigma_{xy}^{n+1,el} = \sigma_{xy}^{n,el} + \mu v_{x} \Delta t$.
2. $S_{1}^{n+1,el} = -\sigma_{xy}^{n+1,el}$ and $S_{2}^{n+1,el} = \sigma_{xy}^{n+1,el}$.
3. If $f_{s}(\mathbf{S}^{n+1,el}) > 0$, $\sigma_{xy}^{n+1} = \sigma_{xy}^{n+1,el}$; continue
4. If $f_{s}(\mathbf{S}^{n+1,el}) <= 0$,
   1. $\Delta\beta = \frac{f_{s}(S_{1}^{n+1,el},S_{2}^{n+1,el})}{2\mu(1+N_{\phi}N_{\psi})}$.
   2. $\bar{S}_{1}^{n+1} = S_1^{n+1,el} - 2\mu \Delta\beta$.
   3. $\bar{S}_{2}^{n+1} = S_2^{n+1,el} + 2\mu \Delta\beta$.
   4. $\bar{\sigma}_{xy}^{n+1} = \bar{S}_2^{n+1}$.
   5. $\sigma_{xy}^{n+1} = (\sigma_{xy}^{n+1,el} + \frac{\Delta t}{\eta} \bar{\sigma}_{xy}^{n+1}) / (1 + \frac{\Delta t}{\eta})$.

Comparison with LHOGST solution

See the benchmark notebook