## 3 Results and Discussion
### An Analytical Model from [PERSON]'s Theory and Comparision Analysis  
For the purpose of comparison, an analytical model is derived from [PERSON]'s theory. Taking an infinitesimal segment ds on an arbitrary wire as a research object, as shown in Figure 15, the slippage criterion governing the tension distribution for each wire is expressed as follows:  
\[|dF_{w}|\leq\mu dN_{w}, \tag{34}\]  
where \(F_{w}\) is the tension of the wire and \(dN_{w}\) is the normal contact force on an infinitesimal segment. Based on the principles of force balance, there exists a relation between the tension and the normal contact force. \[dN_{w}=F_{w}d\varphi \tag{35}\]  
According to the geometric relation, the length of the infinitesimal segment is calculated:  
\[ds=\rho d\varphi=\frac{R_{h}d\theta_{w}}{\sin\alpha} \tag{36}\]  
where \(\rho\) represents the radius of the curvature of the outside wire and \(\theta_{w}\) represents the included angle between the neutral axis and the vector point from the centroid of the strand to the centroid of the wire \(w\) (\(w\) takes a value from 1 to 6 here) on a specified cross-section as shown in Figure 15.  
Assuming that the bending deformation is small, the radius of the curvature of the current deformed wire is identical to its initial value \(\kappa_{0}\). Thus,  
\[\rho=\frac{1}{\kappa_{0}}=\frac{R_{h}}{\sin^{2}\alpha} \tag{37}\]  
Substituting Equations (35)-(37) into Equation (34), the slippage criterion is rewritten as:  
\[|dF_{w}|\leq\mu F_{w}\sin\alpha d\theta_{w} \tag{38}\]  
The angular position of the centroid of the wire \(\theta_{w}\) locates in the range \([-\frac{\pi}{2},\frac{\pi}{2}]\). The axial tension is assumed to increase with the increasing angular position, and thus \(dF_{w}>0\). For the slip region, the solution of the differential Equation (38) is:  
\[F_{w,slip}=T_{0}e^{\mu\theta_{w}\sin\alpha} \tag{39}\]  
where \(T_{0}\) is the tension at the angular position \(\theta_{w}=0\). \[T_{0}=EA_{w}\varepsilon_{c}\cos^{2}\alpha \tag{40}\]  
The axial strain \(\varepsilon_{c}\) of the strand and depends on the total tensile force of the strand \(T\) as approximated by:  
\[\varepsilon_{c}=\frac{T}{EA_{c}+6 EA_{w}\cos^{3}\alpha} \tag{41}\]  
Figure 15: The schematic of mechanic model for each wire in the strand. Before the inside slippage happens, the tension of the individual wire satisfies:  
\[dN_{w}=F_{w}d\varphi \tag{35}\]0  
With the curvature of the strand increasing from 0 to a certain value, the contact state between adjacent wires changes from the original stick state to a slip state. Assuming the critical curvature is \(\kappa_{c}\), the maximum slip tension is equal to the static tension at the critical position:  
\[dN_{w}=F_{w}d\varphi \tag{35}\]1  
Therefore, the critical curvature can be solved as given below:  
\[dN_{w}=F_{w}d\varphi \tag{35}\]2  
The axial tension for each wire on a specified cross-section can thus be given as follows:  
\[F_{w}(\theta_{w})=\left\{ \begin{array}{l}min(\epsilon_{c}e^{\mu\theta_{w} \sin\alpha},R_{h}\kappa\sin\theta_{w}+\epsilon_{c})\cdot EA_{w}\cos^{2}\alpha,\theta_{w}\geq 0\\ max(\epsilon_{c}e^{\mu\theta_{w}\sin\alpha},R_{h}\kappa\sin\theta_{w}+ \epsilon_{c})\cdot EA_{w}\cos^{2}\alpha,\theta_{w}<0\end{array} \right. \tag{45}\]  
For ease of presentation, the outside wires of the strand are anticlockwise numbered as presented in Figure 16, the angular position for each wire depends on an initial angular position \(\theta_{0}\) as given by:  
\[\boldsymbol{\Phi}=[\theta_{1},\theta_{2},\cdots,\theta_{6}]=\left[\theta_{0 },\frac{\pi}{2}-\left|\theta_{0}-\frac{\pi}{6}\right|,\frac{\pi}{3}-\theta_{ 0},-\theta_{0},\left|\theta_{0}-\frac{\pi}{6}\right|-\frac{\pi}{2},\theta_{0} -\frac{\pi}{3}\right] \tag{46}\]  
and \(0\leq\theta_{0}<\frac{\pi}{3}\). Consequently, the whole bending moment of the strand with respect to the curvature could be obtained by superposing the bending moment generated by each wire referring to the neutral axis of the strand.