# 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
### Tangential Stiffness Matrix of the Catenary Cable Element  
Research manuscripts reporting large datasets that are deposited in a publicly available database should specify where the data have been deposited and provide the relevant accession numbers. If the accession numbers have not yet been obtained at the time of submission, please state that they will be provided during review. They must be provided prior to publication. In order to highlight the advancement and accuracy of developing the cable element, this paper adopts the tangential stiffness matrix of the cable element derived based on the displacement pattern consistent with the cable properties and basic assumptions [26], as shown in Figure 1, in the right-angle coordinate system \(OXYZ\), with the suspension chain lead element as \(ij\) and the coordinates of the branch point \(i\) as \((x_{i},y_{i},z_{i})\) and the coordinates of pivot point \(j\) are \((x_{j},y_{j},z_{j})\). \(u_{i}\), \(v_{i}\), \(w_{i}\) are node \(i\) displacements, and \(u_{j}\), \(v_{j}\), \(w_{j}\) are node \(j\) displacements. Displacement pattern of the catenary cable element:  
\[\{u\}=[N][u_{ij}] \tag{1}\]  
where \([u]=\{u\ v\ w\}^{T},\{u_{ij}\}=\{u_{i}\ v_{i}\ w_{i}\ u_{j}\ v_{j}\ w_{j}\}^{T},[N]= \begin{bmatrix}\varphi_{1}&0&0&\varphi_{2}&0&0\\ 0&\varphi_{1}&0&0&\varphi_{2}&0\\ \varphi_{3}&0&\varphi_{1}&-\varphi_{3}&0&\varphi_{2}\end{bmatrix} \) is the form function. Figure 1: The general coordinate system of cable element. We have \(\varphi_{1}=1-x/L\); \(\varphi_{2}=x/L\); \(\varphi_{3}=\frac{D}{2\beta^{2}}\Big{[}\cos\hbar\beta-\cos\hbar\Big{(}\frac{2 \beta x}{L}-\beta\Big{)}\Big{]}-\beta\sin\hbar\beta+\big{(}\frac{2x}{L}-1\big{)} \beta\sin\hbar\Big{(}\frac{2\beta x}{L}-\beta\Big{)}\); \(L\) is the chord length of the cable element; \(\beta=ql/2H\); \(q\) is the mean load set along the cable curve; \(l\) is the length of the horizontal projection of the cable; \(H\) is the horizontal force of the cable. \[D=\frac{\beta(\beta-2\sin\hbar\beta+\sin\hbar\beta\cos\hbar\beta)}{2(\beta \cos\hbar\beta-\sin\hbar\beta)} \tag{2}\]  
The tangential stiffness matrix of the cable element is then obtained from the UL columnar imaginary work increment equation. \[[k]=[k_{0}][k_{\sigma}] \tag{3}\]  
where \([k_{0}]\) is the elastic stiffness matrix and \([k_{\sigma}]\) is the geometric stiffness matrix, expressed as follows [26]:  
\[[k_{0}]=\frac{EA}{L} \begin{bmatrix}k_{\tau}&-k_{\tau}\\ -k_{\tau}&k_{\tau}\end{bmatrix}  \tag{4}\]  
\[[k_{\sigma}]=\frac{N}{L} \begin{bmatrix}k_{\eta}&-k_{\eta}\\ -k_{\eta}&k_{\eta}\end{bmatrix}  \tag{5}\]  
where \([k_{\tau}]= \begin{bmatrix}\tau_{1}&0&0\\ 0&0&0\end{bmatrix} \), \([k_{\eta}]= \begin{bmatrix}\eta&0&0\\ 0&1&0\end{bmatrix} \), \(\eta=D^{2}\Big{(}\frac{1}{6}+\frac{\sin\hbar 2\beta-2\beta\cos\hbar 2\beta+2 \beta^{2}\sin\hbar 2\beta}{8\beta^{3}}\Big{)}\), \(\tau_{1}=1+D\Big{(}\frac{\sinh\hbar 2\beta}{4\beta^{2}}-\frac{\cos\hbar 2\beta}{2 \beta}\Big{)}+D^{2}\Big{(}-\frac{1}{4}+\frac{\cos\hbar 4\beta}{6\beta^{2}}+\frac{ \sin\hbar 4\beta}{25\beta^{3}}\Big{)}\), \(\tau_{2}=\sinh\!2\beta/4\beta-0.5\).