## 5 Determination of Intermediate Cable Force Formula
### Derivation of Intermediate Cable Force Formula  
The tension value of the cable-stayed cable is not only related to the initial tension of the cable-stayed cable and the tension of the bridge cables, but also closely related to the construction process and the main beam construction plan. Different main beam e  
Figure 11: Steel beam hinge simulation. (**a**) Vertical displacement. (**b**) cable force. Figure 10: Steel beam rigid joint simulation. (**a**) Vertical displacement. (**b**) cable force. segment steel beam. The simplified simulation of the steel beam connection after lifting the main beam segment is modeled as a hinged support connection. At this time, the hinged supports at points A, B, and C each receive a horizontal force \(H_{i}\) and a vertical force \(R_{i}\). \(G_{i}\) is the weight of the lifted main beam segment, \(T_{i}\) is the tension force of the cable-stayed cable, and \(\alpha_{i}\) is the angle between the cable-stayed cable and the horizontal direction. \(D_{1}\) and \(D_{2}\) are the forces acting on the front support point and rear anchor point of the bridge deck crane, respectively; \(L_{Gi}\) is the horizontal distance between the weight action point of the beam segment and the hinged support; \(L_{T1}\) is the horizontal distance between the cable-stayed cable anchor point at the beam end and the hinged support; \(L_{R1}\) is the length of the lifted beam segment. It should be noted that the above formula symbols i = 1, 2, 3. To facilitate the structural analysis of the three-span continuous construction, it can be considered as a main structure and a sub-structure. As the sub-structure affects the main structure, while the main structure does not affect the sub-structure, it is necessary to divide them into three parts in the calculation of the structural force diagram. Specifically, in Figure 12, the completed beam section of the cable-stayed bridge can be regarded as the main structure, while the subsequent suspended beam sections can be viewed as sub-structures applied to the main structure. Therefore, the forces of the \(n+2\#\) sub-structure can be calculated first, followed by the forces of the \(n+1\#\) sub-structure, and finally the forces of the \(n\#\) sub-structure. The balance equation for the first part can be written as:  
\[H_{3}=T_{3}cos\alpha_{3} \tag{1}\]  
\[R_{3}+T_{3}sina_{3}=G_{3} \tag{2}\]  
The moment of the hinge support at point C is obtained as  
\[G_{3}\cdot L_{\text{G3}}=T_{3}sina_{3}\cdot L_{T3} \tag{3}\]  
The above equation can be solved to obtain  
\[\left\{ \begin{array}{c}H_{3}=\frac{G_{3}\cdot L_{G3}}{tan\alpha_{3}\cdot T_{ 3}}\\ R_{3}=\frac{G_{3}\cdot(L_{T3}-L_{G3})}{L_{T3}}\\ T_{3}=\frac{G_{3}\cdot L_{G3}}{sin\alpha_{3}\cdot L_{T3}}\end{array} \right. \tag{4}\]  
Once the horizontal force \(H_{3}\) and vertical force \(R_{3}\) in the first part of the \(n+2\#\) auxiliary structure are determined, the mechanical balance equations can be set up for the second part of the \(n+1\#\) auxiliary structure. \[H_{2}+H_{3}=T_{2}cos\alpha_{2} \tag{5}\]  
\[R_{2}+R_{3}+T_{2}sina_{2}=G_{2}+D_{1} \tag{6}\]  
Figure 12: Force diagram of three-segment cyclic structure. The moment of the hinge support at point B is obtained as  
\[G_{2}\cdot L_{G2}+D_{1}\cdot L_{T2}=T_{2}sin\alpha_{2}\cdot L_{T2}+R_{3}\cdot L_{ R2} \tag{7}\]  
The above equation can be solved to obtain  
\[\left\{ \begin{array}{l}H_{2}=\frac{G_{2}\cdot L_{G2}+D_{1}\cdot L_{T2}-R_{3} \cdot L_{R2}}{sin\alpha_{2}\cdot L_{T2}}-H_{3}\\ R_{2}=\frac{G_{2}\cdot(L_{T2}-L_{G2})\cdot R_{3}\cdot(L_{R2}-L_{T2})}{T_{2}= \frac{G_{2}\cdot L_{G2}+D_{1}\cdot L_{T2}-R_{3}\cdot L_{R2}}{sin\alpha_{2}\cdot L _{T2}}}\\ \end{array} \right. \tag{8}\]  
After obtaining the horizontal force \(H_{2}\) and vertical force \(R_{2}\) in the second part of the \(n+1\#\) auxiliary structure, the mechanical equilibrium equation can be established for the first part of the \(n\#\) basic structure.