## 2 Plasticity Formulation
### Return Mapping Algorithm
#### 2.2.1 Return-Mapping Scheme for the 1D Model  
To illustrate the proposed approach, we consider a 1D model, where we define the yield function \(f\) as obeying the mixed hardening law:  
\[f(\sigma,q)=|\sigma-q|-(\sigma_{\gamma}+K\alpha) \tag{8}\]  
where \(\sigma_{\gamma}\) represents the yield strength, \(K\) the plastic modulus and \(\alpha\) the internal hardening variable, which has the evolution equation:  
\[\dot{\alpha}=\dot{\gamma} \tag{9}\]  
and \(q\) is the back stress defined by [PERSON]'s rule as:  
\[\dot{q}=\dot{\gamma}\;H\;\frac{\partial f}{\partial\sigma}=\dot{\gamma}\;H\; \text{sign}(\sigma-q), \tag{10}\]  
with \(H\) representing the kinematic hardening modulus. For a 1D problem, the return mapping algorithm is explicitly presented in the work of [PERSON] and [PERSON] [4]. It is reproduced here (Algorithm 1) with a reformulation as an algebraic system (Equation (11)) in order to highlight our numerical method. ```
1. \(\varepsilon_{n}^{p}\), \(\alpha_{n}\) and \(q_{n}\) are known in step \(n\)
2. Given a strain \(\varepsilon_{n+1}=\varepsilon_{n}+\Delta\varepsilon_{n}\)
3. Compute elastic trial stress \(\sigma_{n+1}^{trial}\) and test for plastic loading
4. \(\sigma_{n+1}^{trial}=E\Big{(}\varepsilon_{n+1}-\varepsilon_{n}^{p}\Big{)}\)
5. \(\sigma_{n+1}^{trial}=\sigma_{n+1}^{trial}-q_{n}\)
6. \(f_{n+1}^{trial}=\left| \begin{subarray}{c}\sigma_{n+1}^{trial}\\ \sigma_{n+1}^{trial}\end{subarray} -\left[\sigma_{\gamma}+K\alpha_{n}\right]\)
7. if\(\mathbf{\mathbf{\mathbf{\mathbf{\mathbf{\mathbf{\mathbf{\bm



The analytical resolution of the system Equation (11) gives [4]:

\[ \begin{array}{c}\Delta\gamma=\frac{f_{n+1}^{trial}}{E+[K+H]}>0\\ \sigma_{n+1}=\sigma_{n+1}^{trial}-\Delta\gamma\;E\;\text{sign}(\xi_{n+1})\\ \varepsilon_{n+1}^{p}=\varepsilon_{n}^{p}+\Delta\gamma\;\text{sign}(\xi_{n+1} )\\ q_{n+1}=q_{n}+\Delta\gamma\;H\;\text{sign}(\xi_{n+1})\\ \alpha_{n+1}=\alpha_{n}+\Delta\gamma\end{array}  \tag{12}\]

While there are several numerical methods for solving non-linear equation systems, we chose a simple fixed-point method. This choice is motivated by the fact that it does not require the calculation of the derivatives. Integrating Equation (11) amounts to calculating the mechanical state \(Z_{n+1}=\left[\sigma_{n+1},\;\alpha_{n+1},\Delta\gamma,\;q_{n+1},\;\varepsilon_ {n+1}^{p}\right]\) at the instant \(t_{n+1}\) from a given strain increment \(\Delta\varepsilon\) and a state \(Z_{n}\) at instant \(t_{n}\). The initial values for \(Z_{0}\) are \(\left[\sigma_{n+1}^{trial},\;\alpha_{n},0,\;q_{n},\;\varepsilon_{n}^{p}\right]\). The iteration is stopped for a time step \(n+1\leq N_{max}\) when \(\parallel Z_{n+1}-Z_{0}\parallel/\parallel Z_{0}\parallel\) is less then a fixed threshold. Then, once \(Z_{n+1}\) is calculated it is tested on the plasticity criterion as presented in [1]. If \(\text{f}_{n+1}^{\text{trial}}\leq 0\) then the predicted state is the solution of the problem. Otherwise, the predicted state is not admissible, and a plastic correction step is performed. Algorithm 2 illustrates this iterative process for the 1D model:
```
1.Initialization : \(Z_{0}=\left[\sigma_{n+1}^{trial},\;\alpha_{n},0,\;q_{n},\;\varepsilon_{n}^{p}\right]\)
2.\(tol\): tolerance and \(N_{max}\) maximum number of iterations are set
3.While \(\parallel Z-Z_{0}\parallel/\parallel Z_{0}\parallel>\) tol and n \(\leq\text{N}_{\max}=20\)
4.\(\left\{ \begin{array}{c}\Delta\gamma=-\frac{(\sigma_{n+1}-\sigma_{n+1}^{ trial})}{\varepsilon_{n+1}^{p}}\text{sign}\xi_{n+1}^{p}\\ \varepsilon_{n+1}^{p}=\varepsilon_{n}^{p}+\Delta\gamma\;\text{sign}(\xi_{n+1} )\\ \alpha_{n+1}=\alpha_{n}+\Delta\gamma\\ q_{n+1}=q_{n}+\Delta\gamma\;H\;\text{sign}(\xi_{n+1})\end{array}