# Coupled Vibration Analysis of Multi-Span Continuous Cable Structure Considering Frictional Slip
## 2 Nonlinear Coupled Vibration Model
### Main Cable Subsystem Model
#### 2.2.2 Three\(-\)Node Sliding Cable Element  
The main cables are often capable of sliding at the saddle at the top of the cable tower during the operation of the cable hoisting system under load. In the main cable design, the total stress-free length of the three-span continuous cable is usually given. The three-node sliding cable unit is shown in Figure 6, with \(I\), \(O\) and \(J\) as the three nodes of the element, where point \(O\) is the sliding node. A pulley is usually designed to realize the sliding. The element is composed of the left \(IO\) cable section and the right \(OJ\) cable section. The element considers sliding friction. Figure 5: Two-node plane cable element. The three-node catenary cable element shown in Figure 6 is analyzed separately as shown in Figure 7, which is shown by the curve equations of the cable segments [7]:  
\[\left\{ \begin{array}{l}l_{x1}=-\frac{F_{1}s_{1}}{EA}-\frac{F_{1}}{q}\ln\left( \frac{T_{A}+F_{2}}{T_{B}-F_{4}}\right)\\ l_{y1}=-\frac{F_{2}s_{1}}{EA}+\frac{g_{2}^{s_{2}}}{2 EA}+\frac{T_{B}-T_{A}}{q} \end{array} \right.,\quad\left\{ \begin{array}{l}l_{x2}=-\frac{F_{1}^{s_{2}}}{ EA}-\frac{F_{1}^{\prime}}{q}\ln\left(\frac{T_{A}+F_{2}^{\prime}}{T_{B}-F_{4}^{ \prime}}\right)\\ l_{y2}=-\frac{F_{1}^{s_{2}}}{EA}+\frac{g_{2}^{s_{2}}}{2 EA}+\frac{T_{B}-T_{A}}{q} \end{array} \right. \tag{9}\]  
where \(s_{1}\) and \(s_{2}\) are the unstressed cable lengths of the \(IO\) and \(OI\) sections, respectively. \(q\) is the load set of the cable. \(f_{3}=-F_{1}\), \(F_{4}=qs_{1}-F_{2}\); \(F_{3}^{\prime}=-F_{1}^{\prime}\), \(F_{4}^{\prime}=qs_{2}-F_{2}^{\prime}\). \[\left\{ \begin{array}{l}T_{A}=\sqrt{F_{1}^{2}+F_{2}^{2}}\\ T_{B}=\sqrt{F_{3}^{2}+F_{4}^{2}}\end{array} \right.,\quad\left\{ \begin{array}[] {l}T_{A}^{\prime}=\sqrt{F_{1}^{\prime 2}+F_{2}^{\prime 2}}\\ T_{B}^{\prime}=\sqrt{F_{3}^{2}+F_{4}^{\prime 2}}\end{array} \right. \tag{10}\]  
It follows from the constant length of unstressed cords on either side of node \(O\) that  
\[s_{1}+s_{2}=s_{0} \tag{11}\]  
where \(s_{0}\) is the unstressed cable length of the cable element, which is given as a known quantity during the division of the element. The tension equilibrium of the two cable segments at the slip node \(O\) can be obtained by considering the dynamic friction:  
\[T_{A}^{\prime}=T_{B}e^{-\psi\theta} \tag{12}\]  
where \(\theta\) is the angle between the cable end forces on both sides of the slip node. \[\theta=\arccos\left(\frac{F_{1}F_{1}^{\prime}+F_{2}F_{2}^{\prime}+(F_{3}-q_{1 }s_{1})F_{3}^{\prime}}{T_{B}T_{A}^{\prime}}\right) \tag{13}\]  
Figure 6: Three–node sliding cable element. Figure 7: Force decomposition of three–node sliding cable element. The direction function \(\psi\) is  
\[\rho=\left\{ \begin{array}{l}1,\ T_{B}>T_{A}^{\prime}\\ 0,\ T_{B}=T_{A}^{\prime}\\ -1,\ T_{B}<T_{A}^{\prime}\end{array} \right. \tag{14}\]  
From Equations (10)-(14), the basic set of equations for the sliding cable element can be formed:  
\[\left\{ \begin{array}{l}G_{1}=x_{1}-x_{O}+l_{x1}\\ G_{2}=y_{1}-y_{O}+l_{y1}\\ G_{3}=x_{O}-x_{I}+l_{x2}\\ G_{4}=y_{O}-y_{I}+l_{y2}\\ G_{5}=s_{1}+s_{2}-s_{0}\\ G_{6}=T_{B}-T_{A}^{\prime}\end{array} \right. \tag{15}\]  
Here, \(\mathbf{Z}=\left[F_{1},F_{2},F_{1}^{\prime},F_{2}^{\prime},s_{1},s_{2}\right]^ {T}\), \(\mathbf{G}=\left[G_{1},G_{2},G_{3},G_{4},G_{5},G_{6}\right]^{T}\). Equation (15) can be solved by Newton-Raphson iteration, where the initial value of the iteration in \(F_{1}=\frac{dI_{1}}{2\lambda}\), \(F_{2}=\frac{d}{2}[s_{1}-h_{1}\text{coth}\lambda]\), \(F_{1}^{\prime}=\frac{dI_{2}}{2\lambda}\), \(s_{2}=s_{0}-s_{1}\), \(s_{1}=\left\{ \begin{array}{l}L_{1}\xi s_{0}/\left(L_{1}+L_{2}\right),L_{1} \geq L_{2}\\ s_{0}-L_{2}\xi s_{0}/\left(L_{1}+L_{2}\right),L_{1}<L_{2}\end{array} \right..\)  
Here, \(\lambda_{0}=\left\{ \begin{array}{l}0.2\quad if\quad s_{i}\leq L_{i},\ \ i=1,2\\ \sqrt{3\left(\frac{s_{i}^{2}-h_{i}^{2}}{l_{i}}-1\right)}\quad if\quad s_{i}>L_{ i},\ \ \ i=1,2\end{array} \right..\)  
\(L_{1}\) and \(L_{2}\) are the chord lengths of the cable segments \(IO\) and \(O\)/, respectively.