# Aeroelastic Optimization Design of the Global Stiffness for a Joined Wing Aircraft
## Appendix A  
As shown in Figure 2a. \[ \begin{split} I_{z1}&=2(\frac{1}{12}X_{3}\left(X_{6}- \frac{X_{1}+X_{2}}{2}\right)3+\text{A}(X_{1})(\frac{X_{6}}{2})^{2}+\text{A}(X _{2})(\frac{X_{6}}{2})^{2}+\text{A}(X_{4})(\frac{X_{6}}{2})^{2})\\ &=\frac{1}{2}A_{t}X_{6}^{2}+\frac{1}{6}\text{A}(X_{3})(L_{w}^{2} -3X_{6}^{2})\;=\;\frac{1}{2}A_{t}X_{6}^{2}+\frac{X_{3}(-X^{3}-2X_{6}^{3}+3X_{6 }X^{2})}{6})\\ \end{split}  \tag{1}\]  
\[ \begin{split} I_{y1}&=2(\frac{1}{12}X_{4}\left(X_ {5}-\frac{X_{1}+X_{2}}{2}\right)3+\text{A}(X_{1})(\frac{X_{5}}{2})^{2}+\text{A }(X_{2})(\frac{X_{5}}{2})^{2}+\text{A}(X_{3})(\frac{X_{5}}{2})^{2})\\ &=\frac{1}{2}A_{t}X_{5}^{2}+\frac{1}{6}\text{A}(X_{4})(L_{s}^{2} -3X_{5}^{2})=\frac{1}{2}A_{t}X_{5}^{2}+\frac{X_{4}(-X^{3}-2X_{6}^{3}+3X_{5}X^{2 })}{6})\end{split}  \tag{2}\]\[ \begin{array}{ll}I_{t1}&=\frac{4\Omega^{2}}{f}=\frac{4(X_{5}X_{6})^{2}}{2(\frac{1} {X_{1}}X_{1}+\frac{1}{X_{1}}I_{w}+\frac{1}{X_{2}}X_{2}+\frac{1}{X_{4}}I_{w})}= \frac{4(X_{5}X_{6})^{2}}{4+2(\frac{X_{6}}{X_{3}}+\frac{X_{5}}{X_{4}})-(X_{1}+X _{2})(\frac{1}{X_{3}}+\frac{1}{X_{4}})}\\ &\approx\frac{2(X_{5}X_{6})^{2}}{\frac{X_{6}}{X_{3}}+\frac{X_{5}}{X_{4}}}=\frac {2X_{3}X_{4}(X_{5}X_{6})^{2}}{X_{4}X_{6}+X_{3}X_{5}}\end{array}  \tag{11}\]  
The first item is directly taken for calculation in the program calculation about \(I_{t1}\)  
\[I_{yz1}=2(A(X_{1})\cdot(-\frac{X_{5}X_{6}}{4})+A(X_{2})\cdot(\frac{X_{5}X_{6}} {4}))=\frac{X_{5}X_{6}}{2}(A(X_{2})-A(X_{1})) \tag{12}\]  
where:  
\[ \begin{array}{ll}A(X_{1})=X_{1}^{2},\quad A(X_{2})=X_{2}^{2},\quad A(X_{3})= X_{3}L_{w},\quad A(X_{4})=X_{4}L_{s}\\ A_{t}=\sum\limits_{i=1}^{4}A(X_{1}),\quad L_{s}=X_{5}-\overline{X},\quad L_{w}= X_{6}-\overline{X},\quad\overline{X}=\frac{X_{1}+X_{2}}{2}\end{array}  \tag{13}\]  
In general, the flange size is larger than the size of the web thickness and skin thickness, and as mentioned earlier, to simulate the material distribution, the main diagonal flange size is larger than the secondary diagonal flange size in Figure 2a, so the relations of the cross section parameters are as in Equation (10):  
\[\max(X_{3},X_{4})<X_{2}<X_{1} \tag{14}\]  
According to the rotation axis formula of the moment of inertia:  
\[I_{u}=\frac{I_{z}+I_{y}}{2}+\frac{I_{z}-I_{y}}{2}\cos 2\alpha-I_{yz}\sin 2\alpha \tag{15}\]  
After Taylor expansion of the sine and cosine expression and discarding of the high-order small quantities, the moment of inertia of the beam section relative to the bending axis is:  
\[ \begin{array}{ll}I_{u1}&\approx\frac{I_{z1}+I_{y1}}{2}+(1-2\alpha^{2})\frac {I_{z1}-I_{y1}}{2}-2\alpha I_{yz1}\\ &=\big{(}1-\alpha^{2}\big{)}I_{z1}+\alpha^{2}I_{y1}-2\alpha I_{yz1}\end{array}  \tag{16}\]  
For the case of Figure 2b:  
\[I_{z2}=I_{z1}\quad I_{y2}=I_{y2}\quad I_{yz2}=0 \tag{17}\]  
then  
\[I_{u2}-I_{u1}=2\alpha I_{yz1}<0 \tag{18}\]  
For the case of Figure 2c, the four flange areas are equal (\(X=X_{1}=X_{2}\)), keeping parameters \(X_{3}\)-\(X_{6}\) and \(A_{t}\) unchanged, according to Equations (12)-(16):  
\[I_{y3}=\frac{1}{2}A_{t}X_{5}^{2}+\frac{X_{4}(-X^{3}-2X_{5}^{3}+3X_{5}X^{2})}{6} \tag{19}\]  
\[I_{z3}=\frac{1}{2}A_{t}X_{6}^{2}+\frac{X_{3}(-X^{3}-2X_{6}^{3}+3X_{6}X^{2})}{6} \tag{20}\]  
\[I_{yz3}=0 \tag{21}\]  
\[I_{u3}=\left(1-\alpha^{2}\right)I_{z3}+\alpha^{2}I_{y3}-2\alpha I_{yz3} \tag{22}\]  
\[\Delta I_{u}=\left.I_{u1}-I_{u3}=(1-\alpha^{2})(I_{z1}-I_{z3})+\alpha^{2}(I_{ y1}-I_{y3})-2\alpha I_{yz1}\right. \tag{23}\]  
According to Equations (12) and (19):  
\[ \begin{array}{ll}I_{z1}-I_{z3}&=\frac{X_{3}(-\overline{X}^{3}+3X_{6}\overline {X}^{2})}{6}-\frac{X_{3}(-X^{3}+3X_{6}X^{2})}{6}\\ &=\frac{X_{3}X_{6}(\overline{X}^{2}-X^{2})}{6}+\frac{X_{3}(-\overline{X}^{3}+ 3X_{6})}{6}\end{array}  \tag{24}\]  
To compare the magnitude of the moment of inertia between them, according to Equation (A20), it can be seen that  
\[