# Proposed New Analytical Method of Tower Load in Large-Span Arch Bridge Cable Lifting Construction
## 2 Proposed Calculation Method
### The Acting Force of Main Cable to Tower  
Figure 3[31] is a calculation schematic diagram of the forces acting on the tower from the main cable. A steel cable is arranged at the top of the tower as the main transport cable, which is supported on the tops of the towers at the left bank and right bank, the ends of which are fixed to the anchor, forming a three-span continuous structure. Generally, two sliding trolleys are rolling along the main cable, and their wheels act on the main cable with concentrated load P/2. When the cable lifting construction scheme is determined, see Figure 3a, the installation sag in mid-span of the main cable is known and the corresponding moment \(M_{L/2}\) in mid-span of the simply supported beam can be formulated as:  
\[M_{L/2}=\frac{1}{8}\cdot\frac{Q\cdot L^{2}}{cos\beta}=\frac{Q\cdot L\cdot\sqrt{ L^{2}+\Delta H^{2}}}{8} \tag{1}\]  
where \(Q\) represents the unit weight of the main rope (kN/m); \(L\) represents the lifting span (m); and \(\Delta H\) represents the height gap between the tower tops at both banks (m). Figure 3: Schematic diagram for calculating the forces acting on the towers from the main cable: (a) empty cable; (b) cable carrying load. Figure 2: Simplified model of cable lifting system. The horizontal tension \(H_{0}\) between the span of empty cable can be formulated as:  
\[H_{0}=\frac{M_{L/2}}{f_{0}}=\frac{Q\cdot L\cdot\sqrt{L\,^{2}+\Delta H^{2}}}{8f_{0 }} \tag{2}\]  
where \(f_{0}\) represents the installation sag in mid-span of the main cable (m). The vertical forces acting on the towers at the left and right banks by the empty cable during the span are \(R_{10}\) and \(R_{20}\), respectively, which can be formulated as:  
\[R_{10}=\frac{1}{2}\cdot\frac{Q\cdot L}{cos\beta}-H_{0}\cdot tan\beta=\frac{Q \cdot\sqrt{L\,^{2}+\Delta H^{2}}}{2}-\frac{H_{0}\cdot\Delta H}{L}=\frac{Q\cdot \sqrt{L\,^{2}+\Delta H^{2}}}{2}-\frac{Q\cdot L\cdot\sqrt{L\,^{2}+\Delta H^{2}} \cdot\Delta H}{8f_{0}\cdot L} \tag{3}\]  
\[R_{20}=\frac{1}{2}\cdot\frac{Q\cdot L}{cos\beta}-H_{0}\cdot tan\beta=\frac{Q \cdot\sqrt{L\,^{2}+\Delta H^{2}}}{2}-\frac{H_{0}\cdot\Delta H}{L}=\frac{Q\cdot \sqrt{L\,^{2}+\Delta H^{2}}}{2}-\frac{Q\cdot L\cdot\sqrt{L\,^{2}+\Delta H^{2}} \cdot\Delta H}{8f_{0}\cdot L} \tag{4}\]  
The tension of the main cable located behind the towers at the left and right banks is \(T_{10}\) and \(T_{30}\), respectively, which can be formulated as:  
\[T_{10}=\sqrt{R_{10}^{2}+H_{0}^{2}} \tag{5}\]  
\[T_{30}=\sqrt{R_{20}^{2}+H_{0}^{2}} \tag{6}\]  
Taking Equations (2)-(4) into Equations (5) and (6), \(T_{10}\) and \(T_{30}\) can be calculated. The horizontal and vertical forces on the tower tops at the left and right bank for the case of an empty cable are \(H_{10}\), \(H_{20}\), \(V_{10}\), and \(V_{20}\), respectively, which can be formulated as:  
\[H_{10}=H_{0}-T_{10}\cdot cos\alpha_{1}=\frac{Q\cdot L\cdot\sqrt{L\,^{2}+\Delta H ^{2}}}{8f_{0}}-\sqrt{R_{10}^{2}+H_{0}^{2}}\cdot cos\alpha_{1} \tag{7}\]  
\[H_{20}=H_{0}-T_{30}\cdot cos\alpha_{2}=\frac{Q\cdot L\cdot\sqrt{L\,^{2}+\Delta H ^{2}}}{8f_{0}}-\sqrt{R_{20}^{2}+H_{0}^{2}}\cdot cos\alpha_{2} \tag{8}\]  
Figure 3:0  
Figure 3:1  
Taking \(H_{0}\), \(R_{10}\), and \(R_{20}\) from Equations (2)-(4) into Equations (7)-(10), \(H_{10}\), \(H_{20}\), \(V_{10}\), and \(V_{20}\) can be calculated. When the rated load is transported by the main cable, see Figure 3b, set the sag of the cable carrying load at the front lifting point as \(f_{2}\), then the bending moment at the front and rear lifting points of the corresponding simply supported beam are \(M_{1}\) and \(M_{2}\), respectively, which can be formulated as:  
Figure 3:2  
Figure 3:3  
where \(f_{1}\) represents sag at the back lifting point of the main cable carrying load (m); \(f_{2}\) represents sag at the front lifting point of the main cable carrying load (m).