## 3 Numerical Simulations Using the ANN Flow Law
### Numerical Simulations and Comparisons  
To compare the influence of choosing different activation functions on the numerical results using Abaqus explicit, we have made the choice to model the compression test presented earlier in Section 1.2. We consider therefore a medium-carbon steel, type P20 cylinder in compression with the initial dimensions \(\phi_{0}=10\) mm and \(h_{0}=15\) mm as reported in Figure 8, where only the superior half part of cylinder is represented, as the solution is symmetrical on either side of a cutting plane located halfway up the cylinder. At the end of the process, the height of the cylinder is \(h_{f}=9\) mm, i.e., the top edge displacement is \(d=6\) mm and the reduction is 40% of the total height. The displacement is applied with a constant speed and the simulation time is fixed to \(t=1\) s, i.e., the strain rate is in the range \(\dot{\epsilon}=[0.5,1.0]\) s\({}^{-1}\) at the center of the specimen. The mesh comprises 600 axis-symmetric thermomechanical quadrilateral finite elements (CAX4 RT) featuring four nodes and reduced integration. It includes 20 elements along the radial direction and 30 elements along the axis. The element size is \(0.25\times 0.25\) mm\({}^{2}\). Only reduced integration is available in Abaqus explicit for an axis-symmetric structure. The anvils are modeled as two rigid surfaces and a Coulomb friction law with \(\mu=0.15\) is used. To reduce the computing time, a global mass scaling with a value of \(M_{s}=1000\) is used. The initial temperature of the material is set to \(T_{0}=1150\)\({}^{\circ}\)C, and we use an explicit adiabatic solver for the simulation of the compression process. All simulations are performed using the 2022 version of the Abaqus explicit solver on the same computer as the one used for the learning of the ANNs in Section 2.2. Simulations are performed with the double precision version of the solver without any parallelization directive to better compare the CPU times. Figure 9 shows, at the end of the simulation (when the displacement of the top edge is \(d=6\) mm), a comparison of the equivalent plastic strain \(\overline{\varepsilon}^{p}\) contourplot for the six activation functions. From the latter, we can clearly see that all activation functions gives almost the exact same results and the choice of any of the available ANN has no influence on the values and isovalues contourplot reported in Figure 9.  
Figure 10 shows a comparison of the von Mises equivalent stress \(\overline{\sigma}\) contourplot. In this figure, we can see that the solutions differ slightly both in terms of maximum stress value and stress isovalues distribution. In order to compare the different models quantitatively, Table 3 reports the values of the plastic strain \(\overline{\varepsilon}^{p}\), the von Mises equivalent stress \(\overline{\sigma}\) and the temperature \(T\) for the element located at the center of the specimen at the end of the simulation. Figure 10: Von Mises equivalent stress \(\overline{\sigma}\) contourplot for the six activation functions. Figure 9: Equivalent plastic strain \(\overline{\epsilon}\) contourplot for the six activation functions. From the values reported in this table, we can see that the models differ a little concerning the equivalent stress (below 1%), the equivalent plastic strain (below 1%) and the temperature (below 0.02%). It is important to note that one origin of the differences between the models comes from the fact that in the end of the simulation, the plastic strain \(\overline{e}^{p}\) is greater than 0.7, so the model has to extrapolate the yield stress with respect to the data used for the training phase. This increases the discrepancy between the models, since each extrapolates the results from the training domain differently with respect to its internal formulation. In Table 3, we also have reported the number of increments \(N_{inc}\) needed to complete the simulation along with the total CPU time.