## 2 Methods
### The Total Potential Energy of Plane Stress Members
\(\sigma_{x}\)\(\sigma_{y}\), and \(\tau_{xy}\) are the normal stress in the x-direction, normal stress in the y-direction, and shear stress, respectively. \[\sigma_{x}=\frac{E}{(1+\
u)(1-2\
u)}((1-\
u)\varepsilon_{x}+\
u\varepsilon_{y}) \tag{15}\]  
\[\sigma_{y}=\frac{E}{(1+\
u)(1-2\
u)}(\
u\varepsilon_{x}+(1-\
u)\varepsilon_{y}) \tag{16}\]  
\[\tau_{xy}=\frac{E}{(1+\
u)(1-2\
u)}(\frac{1-2\
u}{2}\tau_{xy}) \tag{17}\]  
The strains given as Equations (3)-(5) can be calculated via the constant values formulated as Equations (10)-(13), and then, both strains and stresses are shown in Equations (15)-(17) that used to obtain strain energies using Equation (14) for the linear case. As the main investigation of TPO/MA, three nonlinear stress-strain relationships are formulated as Equations (18)-(20) for Case 1, Equation (21) for Case 2, and Equations (22)-(24) for Case 3. It must be noted that these equations do not necessarily correspond to specific material, and the intention of using these formulations is to demonstrate and test the method that can handle easily. Case 1. Nonlinear stress-strain relation 1  
\[\sigma_{x}=\frac{E}{(1+\
u)(1-2\
u)}((1-\
u)\varepsilon_{x}^{3}+\
u \varepsilon_{y}) \tag{18}\]  
\[\sigma_{y}=\frac{E}{(1+\
u)(1-2\
u)}(\
u\varepsilon_{x}+(1-\
u)\varepsilon_{y }^{3}) \tag{19}\]  
\[\mathrm{C_{2}}=\frac{\mathrm{a_{k}}(\mathrm{u_{j}}-\mathrm{u_{i}})}{\mathrm{ a_{k}}\mathrm{b_{j}}-\mathrm{a_{j}}\mathrm{b_{k}}}+\frac{\mathrm{a_{j}}( \mathrm{u_{k}}-\mathrm{u_{i}})}{\mathrm{a_{j}}\mathrm{b_{k}}-\mathrm{a_{k}} \mathrm{b_{j}}} \tag{11}\]0  
Case 2. Nonlinear stress-strain relation 2  
\[ \begin{array}{l}\sigma_{x}=\frac{E}{(1+\
u)(1-2\
u)}((1-\
u)\varepsilon_{x} +\
u\varepsilon_{y})\\ \sigma_{y}=\frac{E}{(1+\
u)(1-2\
u)}(\
u\varepsilon_{x}+(1-\
u)\varepsilon_{y })\\ \tau_{xy}=\frac{E}{(1+\
u)(1-2\
u)}(\frac{1-2\
u}{2}\tau_{xy})\end{array} \ \ \ \ \begin{array}{l}\mbox{for }\mbox{abs}(\varepsilon_{x}+ \varepsilon_{y})<\varepsilon_{0},\ E=E\\ \mbox{abs}(\varepsilon_{x}+\varepsilon_{y})>\varepsilon_{0},\ E=0.5E\end{array} , \tag{21}\]  
where \(\varepsilon_{0}\) represents a value to control the linearity limit. In numerical examples, it was taken as 0.0001. Case 3. Nonlinear stress-strain relation 3  
\[\sigma_{x}=\frac{E}{(1+\
u)(1-2\
u)}((1-\
u)\varepsilon_{x}+\
u\varepsilon_{ y})^{3} \tag{22}\]  
\[\sigma_{y}=\frac{E}{(1+\
u)(1-2\
u)}(\
u\varepsilon_{x}+(1-\
u)\varepsilon_{y })^{3} \tag{23}\]  
\[\tau_{xy}=\frac{E}{(1+\
u)(1-2\
u)}(\frac{1-2\
u}{2}\tau_{xy}^{3}) \tag{24}\]  
In the formulation of the strain energy of the mth triangular element (\(U_{m}\)) given as Equation (25), strain energy density and volume of the mth triangular element are shown as \(\mbox{e}_{m}\) and \(\mbox{V}_{m}\), respectively. \[U_{m}=e_{m}V_{m}. \tag{25}\]  
Equation (26) defines the volume of mth triangular element including the constant thickness (t) of the plate under consideration. \[\mbox{V}_{m}=\frac{(\mbox{a}_{\mbox{b}}\mbox{b}_{\mbox{k}}-\mbox{a}_{\mbox{ k}}\mbox{b}_{\mbox{j}})\mbox{t}}{2} \tag{26}\]Then, the total strain energy of a system is calculated by summing the strain energies of all elements generating the system. By calculating the energy for all members defined with the three nodes, as done in examples given in this study, compatibility is automatically provided since the neighbor members are assigned with the corresponding node numbers of the main system, as in FEM applications. As a formulation, the strain energy can be written as in Equation (27) for a system consisting of n elements. \[\mathrm{U}=\sum_{\mathrm{m}=1}^{\mathrm{n}}\mathrm{U}_{\mathrm{m}} \tag{27}\]  
Finally, the total potential energy (\(\Pi_{\mathrm{p}}\)) that is the difference between work realized with external forces and the total strain energy is calculated via Equation (28) if \(\mathrm{P}_{\mathrm{xi}}\) and \(\mathrm{P}_{\mathrm{yi}}\) are the positioned point loads in x and y directions, respectively.