# Geometric Nonlinear Analysis of the Catenary Cable Element Based on UPFs of ANSYS
## 2 Tangential Stiffness Matrix of Two-Node Catenary Cable Element Based on UL Formulation
### Iterative Solution of a Catenary Cable Element with Known Pretension \(T_{j}\) or \(T_{i}\)  
As can be seen from the tangential stiffness matrix expression of the catenary cable element in the previous section, the most critical thing is to solve for the tangential stiffness matrix \(\beta\), and the need to determine the value of \(\beta\) requires the solution of the horizontal force \(H\). In the initial tensioning of the cables of a cable-stayed bridge or the buckled anchor cables of a cable-stayed arch bridge with buckled construction, the cable end force \(T_{j}\) or \(T_{i}\) is often known, so the initial state horizontal force \(H^{0}\) and the unstressed cable length \(s_{0}\) can be obtained by seeking the cable end. For the catenary cable element shown in Figure 2, its differential cable equation under uniformly distributed vertical loads along the ties is:  
\[y=\frac{2H^{0}}{q}\sin\hbar\Big{(}\frac{qx}{H^{0}}+D_{1}\Big{)}\sin\hbar\Big{(} \frac{qx}{2H^{0}}\Big{)} \tag{8}\]where \(D_{1}=\sin\text{h}^{-1}\left(\frac{\beta^{0}c/l}{\sin\beta^{0}}\right)+\beta^{0}\), \(\beta^{0}=ql/2H^{0}\), \(H^{0}\) is the horizontal force in the initial state of the cable, \(c\) and \(l\) are the vertical and horizontal lengths in the initial tensioned state of the cable. The formula for calculating the cable length is:  
\[s=\sqrt{c^{2}+\left(\frac{\text{$\sin\text{h}\beta$}}{\beta}\right)^{2}} \tag{9}\]  
The conditions for the equilibrium of forces along the \(x\)-axis lead to the following:  
\[T^{0}\cos\theta^{0}-H^{0}=0 \tag{10}\] \[T^{0}=\cos\text{h}\Big{(}\frac{ax}{H^{0}}+D_{1}\Big{)}H^{0}\]  
From the above equation, the tensile forces on the ties at points \(i\) and \(j\), respectively, are:  
\[T^{0}_{i}=H^{0}\text{cos}\,\text{h}D_{1} \tag{11}\] \[T^{0}_{j}=H^{0}\text{cos}\,\text{h}\Big{(}\frac{ql}{H^{0}}+D_{1} \Big{)}\]  
The derivation is now made using the known \(T_{i}\) as an example and is obtained from Equation (11):  
\[D_{1}=\pm\text{ln}\left(\frac{T^{0}_{i}}{H^{0}}+\sqrt{\left(\frac{T^{0}_{i}}{ H^{0}}\right)^{2}-1}\right) \tag{12}\]  
From Equation (12), we have:  
\[c=\frac{H^{0}}{q}\text{cos}\,\text{h}\left(\frac{ql}{H^{0}}\right)\text{cos} \,\text{h}D_{1}+\frac{H^{0}}{q}\text{sin}\,\text{h}\left(\frac{ql}{H^{0}} \right)\text{sin}\,\text{h}D_{1}-\frac{H^{0}}{q}\text{cos}\,\text{h}D_{1} \tag{13}\]  
Substituting Equation (12) into Equation (13) gives:  
\[\frac{l\sqrt{\left(\frac{T^{0}_{i}}{ql}\right)^{2}\left(\frac{ql}{H^{0}} \right)^{2}-1}}{\frac{ql}{H^{0}}}\text{sin}\,\text{h}\left(\frac{ql}{H^{0}} \right)+\frac{T^{0}_{i}}{q}\text{cos}\,\text{h}\left(\frac{ql}{H^{0}}\right) =c+\frac{T^{0}_{i}}{q} \tag{14}\]  
Let: \(a=ql/H^{0}\), the left-hand side of Equation (14) be \(f(a)\), then:  
\[f(a)=\frac{l}{a}\sqrt{\left(\frac{T^{0}_{i}}{ql}\right)^{2}a^{2}-1}\text{sin }\,\text{h}a+\frac{T^{0}_{i}}{q}\text{cos}\,\text{h}a \tag{15}\]  
\[s=\sqrt{c^{2}+\left(\frac{\text{$\sin\text{h}\beta$}}{\beta}\right)^{2}} \tag{9}\]0 Local coordinate system of the cable. When \(a\geq ql/T_{i}^{0}\) (that is \(T_{i}^{0}\geq H\)),_f(a)_ is a single increasing function (\(f^{\prime}(a)\geq 0\)). Therefore, the equation _f(a) = c + T_{i}^{0}/q\)_ has a single root and it can be obtained by the dichotomy method. For this purpose, it is first necessary to determine two lower and upper limits, \(a_{1}\) and \(a_{2}\), that will encompass the value of a. In this paper, the initial lower and upper limits \(a_{1}\) and \(a_{2}\) are determined in the following way.