## 2 Plasticity Formulation
### Return Mapping Algorithm
#### 2.2.1 Return-Mapping Scheme for the 1D Model
\begin{array}{c}d\sigma_{1}^{tr}\\ d\sigma_{2}^{tr}\\ d\sigma_{3}^{2 tr}\end{array} \right\}-2\Delta\gamma\left\{ \begin{array}{c} \left(K+\frac{G}{3}\right)\sin\psi+G\\ \left(K-\frac{2G}{3}\right)\sin\psi\\ \left(K+\frac{G}{3}\right)\sin\psi-G\end{array} \right\}\] (A1) Knowing that \[d\sigma_{j}^{tr}=Dd\epsilon_{j}\] (A2) and because the trial stress is assumed to be elastic, Equation (A2) becomes: \[d\sigma_{j}=Dd\epsilon_{j}-\frac{2\,df_{1}^{trial}}{a}\mathcal{V}_{pm}\] (A3) with \[\mathcal{V}_{pm}=\left\{ \begin{array}{c}\left(K+\frac{G}{3}\right)\sin\psi+G \\ \left(K-\frac{2G}{3}\right)\sin\psi\\ \left(K+\frac{G}{3}\right)\sin\psi-G\end{array} \right\}\] (A4) representing the normal vector applied to the flow potential of the main plane, and the differentiation of the consistency condition given in Equation (28) is defined by \[df_{1}^{tr}\big{(}\sigma^{tr},c\big{)}=2\mathcal{V}_{cm}d\epsilon_{j},\;\;j=1,2,3\] (A5) with \[\mathcal{V}_{cm}=\left\{ \begin{array}{c}\left(K+\frac{G}{3}\right)\sin\psi+G \\ \left(K-\frac{2G}{3}\right)\sin\psi\\ \left(K+\frac{G}{3}\right)\sin\psi-G\end{array} \right\}\] (A6) representing the normal vector applied to the yield function of the main plane, so that \[d\sigma_{j}=\left(\mathbf{D}-\frac{4}{a}\mathcal{V}_{pm}\cdot V_{cm}^{T} \right)d\epsilon_{j},\;\;j=1,2,3\] (A7)Finally, the consistent tangent operator in the principal basis and for the main plane \(\big{(}D_{p,m}^{ep}\big{)}\) obtained analytically can be written as:

\[\mathbf{D}_{p,m}^{ep}=\mathbf{D}-\frac{4}{a}\mathcal{V}_{pm}\cdot\mathcal{V}_{cm}^{T} \tag{10}\]

_Right edge_

We replace Equation (38) in Equation (30), and then differentiate to find:

\[\left\{ \begin{array}{c}d\sigma_{1}\\ d\sigma_{2}\\ d\sigma_{3}\end{array} \right\}\ =\ \left\{ \begin{array}{c}d\sigma_{1}^{tr}\\ d\sigma_{2}^{tr}\\ d\sigma_{3}^{tr}\end{array} \right\}-2\ \Delta\gamma^{a1}\underbrace{\left\{ \begin{array}{c}\left(K+\frac{C}{3}\right)\sin\psi+G\\ \left(K-\frac{C}{3}G\right)\sin\psi\\ \left(K+\frac{C}{3}\right)\sin\psi-G\end{array} \right\}}_{\mathcal{V}_{pm}}-2 \ \Delta\gamma^{b1}\underbrace{\left\{ \begin{array}{c}\left(K+\frac{C}{3} \right)\sin\psi+G\\ \left(K+\frac{C}{3}\right)\sin\psi-G\\ \left(K-\frac{C}{3}G\right)\sin\psi\end{array} \right\}}_{\mathcal{V}_{pr}} \tag{11}\]

where \(\mathcal{V}_{pr}\) represents the normal vector applied to the flow potential of the right edge, and

\[d\sigma_{j}=\mathbf{D}d\epsilon_{j}-2\ d\Delta\gamma^{a1}\ \mathcal{V}_{pm}-2\ d \Delta\gamma^{b1}\ \mathcal{V}_{pr} \tag{12}\]

In this last relationship, we still have to express the incremental plastic multiplier as a function of the incremental plastic strain. Therefore, after differentiating the consistency conditions given in Equations (28) and (35) defined by:

\[\left\{ \begin{array}{c}d\mathbf{f}_{1}^{tr}\big{(}\sigma^{tr},c\big{)}=2\mathcal{ V}_{cm}d\epsilon_{j}\\ d\mathbf{f}_{6}^{trial}\big{(}\sigma^{trial},c\big{)}=2\mathcal{V}_{cr}d\epsilon_{j} \end{array} \right. \tag{13}\]

with

\[\mathcal{V}_{cr}=\left\{ \begin{array}{c}\left(K+\frac{C}{3}\right)\sin\psi+ G\\ \left(K+\frac{C}{3}\right)\sin\psi-G\\ \left(K-\frac{2C}{3}\right)\sin\psi\end{array} \right\} \tag{14}\]

representing the normal vector applied to the yield function of the right edge and the plastic multipliers defined by

\[\left\{ \begin{array}{c}d\Delta\gamma^{a1}=\frac{a\ d\mathbf{f}_{1}^{trial}-b_{ 1}\ d\mathbf{f}_{6}^{trial}}{Det_{1}}\\ d\Delta\gamma^{b1}=\frac{a\ d\mathbf{f}_{6}^{trial}-b_{1}\ d\mathbf{f}_{1}^{trial}}{Det _{1}}\end{array}