## Appendix A  
In this section, the internal forces of the seven bars are derived, and then their corresponding performance functions are given.  
As shown in Figure A1, one force equilibrium equation and two moment equilibrium equations can be given as follows:  
\[\begin{cases}\sum F_{x}=0&F_{B_{x}}-F_{2}=0\\ \sum M_{B}=0&LF_{1}+\frac{\sqrt{3}L}{2}F_{2}+\frac{3L}{2}F_{3}-2 LF_{C_{y}}=0\\ \sum M_{C}=0&LF_{1}+\frac{1}{2}F_{3}-\frac{\sqrt{3}L}{2}F_{2}-2 LF_{B_{y}}=0 \end{cases}\] (A1)  
from which we have  
\[\begin{cases}\begin{array}{l}F_{B_{x}}=F_{2}\\ F_{C_{y}}=\frac{F_{1}}{2}+\frac{\sqrt{3}F_{2}}{4}+\frac{3F_{3}}{4}\\ F_{B_{y}}=\frac{F_{1}}{2}-\frac{\sqrt{3}F_{2}}{4}+\frac{F_{3}}{4}\end{array}\end{cases}\] (A2)  
where \(F_{B_{x}}\) denotes the orthogonal component of the hinge support in the x-direction; \(F_{C_{x}}\) and \(F_{C_{y}}\) denote the orthogonal components of the rolling support in the x and y directions, respectively.  
Let \(F_{N_{i}}\) be the internal force of the \(i\)th bar. As shown in Figure A2a, \(F_{N_{1}}\) and \(F_{N_{3}}\) can be obtained by the following force equilibrium equation:  
\[\begin{cases}\begin{array}{l}F_{N_{1}}+\frac{1}{2}F_{N_{3}}-F_{B_{x}}=0\\ F_{B_{y}}+\frac{\sqrt{3}}{2}F_{N_{3}}=0\end{array}\end{cases}\] (A3)  
Substituting Equation (A2) into Equation (A3) yields  
\[\begin{cases}\begin{array}{l}F_{N_{1}}=\frac{F_{1}}{2}\gamma_{3}+\frac{3F_{2 }}{4}+\frac{F_{3}}{4}\\ F_{N_{3}}=\frac{F_{1}}{\sqrt{3}}-\frac{F_{2}}{2}+\frac{F_{3}}{2\sqrt{3}}\end{array} \end{cases}\] (A4)  
As shown in Figure A2b, the following force equilibrium equation can be given:  
\[\begin{cases}\begin{array}{l}F_{N_{2}}+\frac{1}{2}F_{N_{4}}+F_{N_{1}}-F_{B_{ x}}=0\\ \frac{\sqrt{3}}{2}F_{N_{4}}-F_{B_{y}}=0\end{array}\end{cases}\] (A5)  
Substituting Equations (A2) and (A4) into Equation (A5) yields  
\[\begin{cases}\begin{array}{l}F_{N_{4}}=\frac{F_{1}}{\sqrt{3}}-\frac{F_{2}}{ 2}+\frac{F_{3}}{2\sqrt{3}}\\ F_{N_{7}}=\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2}+\frac{F_{3}}{2\sqrt{3}}\end{array} \end{cases}\] (A6)  
From Figure A2c, the force equilibrium condition is constructed as follows:  
\[\begin{cases}\begin{array}{l}-F_{N_{2}}-\frac{1}{2}F_{N_{6}}=0\\ F_{C_{y}}+\frac{\sqrt{3}}{2}F_{N_{6}}=0\end{array}\end{cases}\] (A7)  
Upon substituting Equation (A2) into Equation (A7), we have  
\[\begin{cases}\begin{array}{l}F_{N_{2}}=\frac{F_{1}}{2\sqrt{3}}+\frac{F_{2}} {4}+\frac{\sqrt{3}F_{3}}{4}\\ F_{N_{6}}=\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2}+\frac{\sqrt{3}F_{3}}{2}\end{array} \end{cases}\] (A8)  
The force equilibrium equation from Figure A2d is given as follows:  
\[F_{C_{y}}-\frac{\sqrt{3}}{2}F_{N_{5}}=0\] (A9)Substituting Equation (A2) into Equation (A9) yields  
\[F_{\text{N}_{5}}=\frac{F_{1}}{\sqrt{3}}+\frac{F_{2}}{2}-\frac{F_{3}}{2\sqrt{3}}\] (A10)  
Then, the corresponding performance functions are as follows:  
\[\begin{array}{l}g_{1}=S-\frac{F_{\text{N}_{1}}}{A_{1}}=S-\left(\frac{F_{1}}{2 \sqrt{3}}+\frac{3F_{2}}{4}+\frac{F_{3}}{4\sqrt{3}}\right)/A_{1}\\ g_{2}=S-\frac{F_{\text{N}_{2}}}{A_{2}}=S-\left(\frac{F_{1}}{2\sqrt{3}}+\frac{F_{ 2}}{4}+\frac{F_{3}}{4}\right)/A_{2}\\ g_{3}=S-\frac{F_{\text{N}_{3}}}{A_{3}}=S-\left(\frac{F_{1}}{\sqrt{3}}-\frac{F_{ 2}}{2}+\frac{F_{3}}{2\sqrt{3}}\right)/A_{3}\\ g_{4}=S-\frac{F_{\text{N}_{4}}}{A_{4}}=S-\left(\frac{F_{1}}{\sqrt{3}}-\frac{F_{ 2}}{2}+\frac{F_{3}}{2\sqrt{3}}\right)/A_{4}\\ g_{5}=S-\frac{F_{\text{N}_{5}}}{A_{5}}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{ 2}}{2}-\frac{F_{3}}{2\sqrt{3}}\right)/A_{5}\\ g_{6}=S-\frac{F_{\text{N}_{6}}}{A_{6}}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{ 2}}{2}+\frac{\sqrt{3}F_{3}}{2}\right)/A_{6}\\ g_{7}=S-\frac{F_{\text{N}_{7}}}{A_{7}}=S-\left(\frac{F_{1}}{\sqrt{3}}+\frac{F_{ 2}}{2}+\frac{F_{3}}{2\sqrt{3}}\right)/A_{7}\end{array}\] (A11)