# Elastoplastic Analysis of Plates with Radial Point Interpolation Meshless Methods
## 3 Elastoplastic Analysis
### Nonlinear Solution Algorithm  
The elastoplastic analysis of structures involves a set of nonlinear equations that need to be solved in order to obtain the problem's corresponding solution. Generally, such nonlinear equations are solved by using iterative-incremental methods. In this work, the Newton-Raphson method is used as a nonlinear solution technique. Therefore, for a given increment \(i\) and iteration \(j\), the the displacement, \(\delta u_{i}^{j}\) for each load increment \(\delta f_{i}^{j}\) is given by  
\[\delta u_{i}^{j}=\left(\frac{df}{du}\right)^{-1}\cdot\delta f_{i}^{j}=K_{T}^{- 1}\cdot\delta f_{i}^{j} \tag{55}\]  
and the total displacements are given by  
\[u_{i}^{j}=u_{i}^{j-1}+\partial u_{i}^{j-1}. \tag{56}\]  
During the incremental-iterative process, the increment of load acts as a predictor. It provides a starting solution, \(u_{i}^{0}\), in order for the iterative process to have a starting point to begin the convergence process. In addition, the complete version of the Newton-Raphson method, in which the stiffness matrix is calculated in each iteration (of each increment), there are several other versions documented in the literature [23]. For instance, the KT0 approach, in which only the initial stiffness matrix, calculated in the first iteration of the first increment (the elastic stiffness matrix), is considered for the complete analysis. Another version is the KT1, in which the stiffness matrix is calculated only at the first iteration of each increment, being then used in all the iterations of such increments. Both KT0 and KT1 versions allow us to reduce the number of computations in slightly nonlinear problems; however, for highly nonlinear problems the best approach is to use the complete version of the Newton-Raphson method, which allows us to significantly reduce the number of iterations in each increment. The complete version of the Newton-Raphson used in this work can be summarised as follows. First, the problem domain is discretized with a nodal mesh, the background integration mesh is constructed and the nodal connectivity is imposed. Then, the RPI shape functions are constructed for each integration point. In this work, an incremental-iterative procedure was implemented for the elastoplastic deformation problem. Thus, the discretized system of equation represented in Equation (30) can be written as  
\[\mathbf{\mathrm{K}}_{T}\cdot\Delta\mathbf{\mathrm{u}}-\Delta\mathbf{\mathrm{f}}=\mathbf{ \mathrm{f}}_{res}, \tag{57}\]  
where \(\mathbf{\mathrm{K}}_{T}\) is the tangent stiffness matrix, \(\Delta\mathbf{\mathrm{u}}\) the incremental displacement field, \(\Delta\mathbf{\mathrm{f}}\) is the incremental load vector, and \(\mathbf{\mathrm{f}}_{res}\) is the residual force vector. Thus, the process can start for the first increment (\(i=1\)). 1. Set the incremental load, \(\Delta\mathbf{\mathrm{f}}=\mathbf{\mathrm{f}}^{i}\). 2. Loop over all integration points, \(\mathbf{x}_{I}\). 1. Construct \(\mathbf{c}(\mathbf{x}_{I})_{p}\), Equation (45). 2. Construct \(\mathbf{K}(\mathbf{x}_{I})_{T}\), Equation (31). 3. Assemble the global tangent stiffness matrix: \(\mathbf{K}_{T}=A_{I=1}^{n_{Q}}\mathbf{K}(\mathbf{x}_{I})_{T}\). 4. Solve \(\Delta\mathbf{u}^{i}=\mathbf{K}_{T}^{-1}\cdot\Delta\mathbf{f}\). 5. Update the displacement field: \(\mathbf{u}^{i}=\mathbf{u}^{i-1}+\Delta\mathbf{u}^{i}\). 6. Loop over all integration points, \(\mathbf{x}_{I}\). 1. Evaluate the strain increment: \(\Delta\mathbf{\varepsilon}(\mathbf{x}_{I})^{i}=\mathbf{B}(\mathbf{x}_{I})\cdot \Delta\mathbf{u}^{i}\). 2. Considering the elastic constitutive matrix, Equation (46), obtain the incremental stress state: \(\Delta\mathbf{\sigma}(\mathbf{x}_{I})^{i}=\mathbf{c}(\mathbf{x}_{I})\cdot\Delta \mathbf{c}(\mathbf{x}_{I})^{i}\). 3. Calculate the trial stress state: \(\mathbf{\sigma}(\mathbf{x}_{I})^{i}_{trial}=\mathbf{\sigma}(\mathbf{x}_{I})^{i-1}+ \Delta\mathbf{\sigma}(\mathbf{x}_{I})^{i}\). 4. Verify: \(f\left(\mathbf{\sigma}(\mathbf{x}_{I})^{i}_{trial}\right)\leq\sigma_{Y}(\mathbf{ \varsigma})\).