## 3 Differential Equilibrium Equations
### Cable-Stayed Section  
Figure 4 shows the force diagram of the infinitesimal element for the cable-stayed section. When compared with the hybrid section in Figure 2, there is no contribution from the main cable in the cable-stayed section. The derivation of the equilibrium differential equation of the cable-stayed section can be derived from the Equation (4) by removing the main cable-related terms and meanwhile substituting \(q-c_{yd}=0\). It gives the following:  
\[EIv^{(4)}+Nv^{\prime\prime}-\left(p-c_{yp}\right)=0 \tag{13}\]  
Similarly, assuming that the cable is equivalent to a beam supported on an elastic foundation and the vertical elastic support stiffness \(k\) provided by the cable, then, the Equation (13) can be simplified as follows:  
Figure 3: Force diagram of the infinitesimal element of the pure suspension section. The general solution of the linear homogeneous differential Equation (14) with constant coefficients is as follows:  
\[v=\left\{ \begin{array}{cl}Ash(\beta x)\cos(\lambda x)+Bsh(\beta x)\sin(\lambda x )+Ch(\beta x)\cos(\lambda x)+Dch(\beta x)\sin(\lambda x)&N<2\sqrt{EIk}\\ Axsin(\phi x)+Bxcos(\phi x)+Csin(\phi x)+Dcos(\phi x)&N=2\sqrt{EIk}\\ Asin(\phi x)+Bcos(\phi x)+Csin(\phi x)+Dcos(\phi x)&N>2\sqrt{EIk}\\ \end{array} \right. \tag{15}\]  
where  
\[\beta=\sqrt{\frac{2\sqrt{EIk}-N}{4 EI}},\;\lambda=\sqrt{\frac{2\sqrt{EIk}+N}{4 EI}}\;N<2\sqrt{EIk}\]  
\[\eta=\sqrt{\frac{N+\sqrt{N^{2}-4 EIk}}{2 EI}},\;\varepsilon=\sqrt{\frac{N-\sqrt{ N^{2}-4 EIk}}{2 EI}}\;N\geq 2\sqrt{EIk}\]  
Then, the relationship of the characteristic mechanical vector \(T\) and the coefficient vector \(S\) is as follows:  
\[T=K_{xl}S \tag{16}\]  
where \(K_{xl}\) is the eigen mechanical matrix for the cable-stayed section. (1) When \(N<2\sqrt{EIk}\)  
\[K_{xl}= \begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\\ \end{bmatrix} \]  
and each element is computed as follows:  
\[ \begin{array}{l}a_{11}=sh(\beta x)\cos(\lambda x),\;a_{12}=sh(\beta x)\sin( \lambda x),\;a_{13}=ch(\beta x)\cos(\lambda x),\;a_{14}=ch(\beta x)\sin( \lambda x)\\ a_{21}=\beta ch(\beta x)\cos(\lambda x)-\lambda sh(\beta x)\sin(\lambda x),\;a_{ 22}=\beta ch(\beta x)\sin(\lambda x)+\lambda sh(\beta x)\cos(\lambda x)\\ a_{23}=\beta sh(\beta x)\cos(\lambda x)-\lambda ch(\beta x)\sin(\lambda x),\;a_{ 24}=\beta sh(\beta x)\sin(\lambda x)+\lambda ch(\beta x)\cos(\lambda x)\\ a_{31}=EI\big{\{}(\lambda^{2}-\beta^{2})sh(\beta x)cos(\lambda x)+2\beta ch( \beta x)\sin(\lambda x)\big{\}}\\ a_{32}=EI\big{\{}(\lambda^{2}-\beta^{2})sh(\beta x)sin(\lambda x)-2\beta ch( \beta x)\cos(\lambda x)\big{\}}\\ a_{33}=EI\big{\{}(\lambda^{2}-\beta^{2})ch(\beta x)cos(\lambda x)+2\beta sh( \beta x)\sin(\lambda x)\big{\}}\\ a_{34}=EI\big{\{}(\lambda^{2}-\beta^{2})ch(\beta x)sin(\lambda x)-2\beta sh( \beta x)\cos(\lambda x)\big{\}}\\ a_{41}=EI\big{\{}(3\lambda^{2}-\beta^{2})\beta ch(\beta x)\cos(\lambda x)+ \big{(}3\beta^{2}-\lambda^{2})\lambda sh(\beta x)\sin(\lambda x)\big{\}}\\ a_{42}=EI\big{\{}(3\lambda^{2}-\beta^{2})\beta ch(\beta x)sin(\lambda x)- \big{(}3\beta^{2}-\lambda^{2})\lambda sh(\beta x)\cos(\lambda x)\big{\}}\\ a_{43}=EI\big{\{}(3\lambda^{2}-\beta^{2})\beta sh(\beta x)cos(\lambda x)+ \big{(}3\beta^{2}-\lambda^{2})\lambda ch(\beta x)\sin(\lambda x)\big{\}}\\ a_{44}=EI\big{\{}(3\lambda^{2}-\beta^{2})\beta sh(\beta x)sin(\lambda x)- \big{(}3\beta^{2}-\lambda^{2})\lambda ch(\beta x)\cos(\lambda x)\big{\}}\\ \end{array} \]  
Figure 4: Force diagram of the infinitesimal element of the cable-stayed section. Its inverse matrix is \(K_{xl}^{-1}= \begin{bmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\ b_{21}&b_{22}&b_{23}&b_{24}\\ b_{31}&b_{32}&b_{33}&b_{34}\\ b_{41}&b_{42}&b_{43}&b_{44}\end{bmatrix}