## 4 Results  
The way to check the proposed model according to the pressure under the footing is shown in Table 2. The coordinates are substituted into Equation (8). Therefore, this model has been verified mathematically. For all cases, when the load, \(P\), decreases, the eccentricities (\(e_{x}=M_{y}/P\) and \(e_{y}=M_{xT}/P\)) increase since the moments remain constant. Therefore, the load, \(P\), tends to move away from the center of the footing. Table 11 (see Appendix A) for Problem 1 shows the following: For Problems 1.A and 1.B, the minimum area appears in Case V, and for Problem 1.C, the minimum area appears in Case II. For this problem, in all cases, the load, \(P\), is located outside the central nucleus. Table 12 (see Appendix A) for Problem 2 shows the following: For Problems 2.A and 2.B, the minimum area appears in Case I, and for Problem 2.C, the minimum area appears in Case III. For this problem, in the first two cases, the load, \(P\), is located inside the central nucleus (Case I), and in the last case, the load, \(P\), is located outside the central nucleus (Case I), and in the last case, the load, \(P\), is located outside the central nucleus (Case III). Table 13 (see Appendix A) for Problem 3 shows the following: For Problems 3.A and 3.B, the minimum area appears in Case I, and for Problem 3.C, the minimum area appears in Case IV. For this problem, in the first two cases, the load, \(P\), is located inside the central nucleus (Case I), and in the last case, the load, \(P\), is located outside the central nucleus (Case IV). Table 14 (see Appendix A) for Problem 4 shows the following: For Problem 4.A, the minimum area appears in Case I, and for Problems 4.B and 4.C, the minimum area appears in Case V. For this problem, in the first case, the load, \(P\), is located inside the central nucleus (Case I), and in the last two cases, the load, \(P\), is located outside the central nucleus (Case V). \begin{table}
\begin{tabular}{c c c c} \hline \hline \multirow{2}{*}{**Case**} & \multicolumn{3}{c}{**Coordinates**} & \multirow{2}{*}{\(\sigma_{Z}\)} \\ \cline{2-3} \cline{5-5}  & \(x\) & \(y\) & \\ \hline \multirow{3}{*}{II} & \(L_{x}/2\) & \(L_{y}/2\) & \(\sigma_{max}\) \\  & \(L_{x}/2\) & \(L_{y}/2-L_{y1}\) & 0 \\  & \(L_{x}/2-L_{x1}\) & \(L_{y}/2\) & 0 \\ \hline \multirow{3}{*}{III} & \(L_{x}/2\) & \(L_{y}/2\) & \(\sigma_{max}\) \\  & \(-L_{x}/2\) & \(L_{y}/2-L_{y2}\) (\(L_{y2}=L_{y1}(L_{x1}-L_{x})/L_{x1}\)) & 0 \\  & \(L_{x}/2\) & \(L_{y}/2-L_{y1}\) & 0 \\ \hline \multirow{3}{*}{IV} & \(L_{x}/2\) & \(L_{y}/2\) & \(\sigma_{max}\) \\  & \(L_{x}/2-L_{x1}\) & \(L_{y}/2\) & 0 \\  & \(L_{x}/2-L_{x2}\) (\(L_{x2}=L_{x1}(L_{y1}-L_{y})/L_{y1}\)) & \(-L_{y}/2\) & 0 \\ \hline \multirow{3}{*}{V} & \(L_{x}/2\) & \(L_{y}/2\) & \(\sigma_{max}\) \\  & \(-L_{x}/2\) & \(L_{y}/2-L_{y2}\) (\(L_{y2}=L_{y1}(L_{x1}-L_{x})/L_{x1}\)) & 0 \\ \cline{1-1}  & \(L_{x}/2-L_{x2}\) (\(L_{x2}=L_{x1}(L_{y1}-L_{y})/L_{y1}\)) & \(-L_{y}/2\) & 0 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Verification of the proposed model. One way to justify this study is to make some comparisons with the results of other authors. A comparison is made between the model proposed by [PERSON] (2017), which considers only Case I, and the new model, which takes into account five cases (I, II, III, IV, and V) for biaxial bending with an eccentric column. Table 3 shows example 1 from [PERSON] (2017) for \(P=1000\) kN (Problem 1.1), \(P=850\) kN (Problem 1.2), \(P=750\) kN (Problem 1.3), \(P=600\) kN (Problem 1.4), \(M_{x}=225\) kN-m, \(M_{y}=150\) kN-m, \(x_{c}=0\), \(y_{c}=0\), \(c_{1}=c_{2}=0.40\) m, and \(\sigma_{max}=180\) kN/m\({}^{2}\). Table 4 shows example 4 from Luevanos-Rojas (2017) for \(P=750\) kN (Problem 4.1), \(P=605\) kN (Problem 4.2), \(P=450\) kN (Problem 4.3), \(P=300\) kN (Problem 4.4), \(M_{x}=-750\) kN-m, \(M_{y}=-600\) kN-m, \(x_{c}=L_{x}/2-c_{2}/2\), \(y_{c}=L_{y}/2-c_{1}/2\), \(c_{1}=c_{2}=0.40\) m, and \(\sigma_{max}=180\) kN/m\({}^{2}\).