## 3 Differential Equilibrium Equations
### Hybrid Section  
Figure 2 shows the force diagram of the infinitesimal element for the hybrid section. In the figure, \(N\), \(Q\), and \(M\) are the axial force, shear force, and moment of the main beam, respectively; \(w\) and \(v\) are the deflection of the main cable and main beam, respectively; \(q\) and \(p\) are the distributed load caused by the dead and the live load, respectively; \(c\) is the distributed force of the stay cable based on the homogeneous membrane assumption of stayed cables; \(c_{y}\) is the vertical component of \(c\); \(c_{yd}\) and \(c_{yp}\) are the vertical components of the stay cable generated by the dead and live loads, respectively, and \(c_{y}=c_{yd}+c_{yp}\). The horizontal force \(H\) of the main cable is composed of two parts: \(H_{q}\) and \(H_{p}\), which are generated by the dead and live loads, respectively, \(H=H_{q}+H_{p}\). Taking the bending moment at the left end point of the main beam, it gives the following:  
\[H(dy+dw)+dM-Ndv-(V+Q)dx=0 \tag{1}\]  
where V is the vertical component of the suspension cable. Based on the equilibrium in the vertical direction, it yields the following equation:  
\[d(V+Q)+\big{(}p+q-c_{y}\big{)}dx=0 \tag{2}\]  
Meanwhile, for the beam, the differential equation for the deformed shape is as follows:  
\[M=-EIv^{\prime\prime} \tag{3}\]  
Figure 2: Force diagram of the infinitesimal element of the hybrid section. where \(E\) is the elastic modulus of the main beam, and \(I\) is the bending moment of inertia of the main beam. Taking the first derivative of Equation (1) and substituting Equations (2) and (3) into Equation (1), it gives the following:  
\[EIv^{(4)}-\big{(}H_{q}+H_{p}\big{)}(y^{\prime\prime}+w^{\prime\prime})+Nv^{\prime \prime}-\big{(}p+q-c_{y}\big{)}=0 \tag{4}\]  
Based on the assumption that the hangers have negligible elongation, \(w=v\), and \(H_{q}y^{\prime\prime}=c_{yd}-q\)[31], then, the Equation (4) can be simplified as follows:  
\[EIv^{(4)}-\big{(}H_{q}-N\big{)}v^{\prime\prime}-H_{p}v^{\prime\prime}+c_{yp}=p +H_{p}y^{\prime\prime} \tag{5}\]  
Equation (5) is the differential equilibrium equation of the hybrid section of the bridge system under the assumption of deflection theory. Considering that the main beam of the bridge is supported by the cable in a large area and the area of the pure suspension section in the middle span is generally small, the ratio of the stiffness of the main beam part to the gravity stiffness of the main cable is larger than that of the conventional suspension bridge. In addition, for the bridge of more than 1000 m, \(H_{p}\leq 0.2H_{q}\). Therefore, Equation (5) is further simplified according to the linear deflection theory and the term \(H_{p}v^{\prime\prime}\) is ignored. The vertical component force generated by the stay cable due to the deflection of the main beam is a function of the deflection and position coordinates. To facilitate the solution, it is simplified according to Winkler-type elastic foundation. Assuming the vertical elastic support stiffness provided by the stay cable is \(k\), then, the vertical component force of the stay cable is \(c_{yp}=k\
u\). Substituting this into Equation (5) gives the following:  
\[EIv^{(4)}-\big{(}H_{q}-N\big{)}v^{\prime\prime}+k\
u=p+H_{p}y^{\prime\prime} \tag{6}\]  
The general solution of the corresponding linear homogeneous differential Equation (6) with constant coefficients is as follows:  
\[\
u=\left\{ \begin{array}{cc}Ash(\gamma x)\text{cos}(\delta x)+Bsh(\gamma x) \text{sin}(\delta x)+Cch(\gamma x)\text{cos}(\delta x)+Dch(\gamma x)\text{sin}( \delta x)&H_{q}<2\sqrt{EIk}\\ Axsh(\eta x)+Bxch(\eta x)+Csh(\eta x)+Dch(\eta x)&H_{q}=2\sqrt{EIk}\\ Ash(\eta x)+Bch(\eta x)+Csh(\varepsilon x)+Dch(\varepsilon x)&H_{q}>2\sqrt{ EIk}\end{array} \right. \tag{7}\]  
where  
\[\left\{