## 3 Kinetostatic Model
### CBCM for Circular Beams  
When applying the CBCM to circular beams, the BCM of the circular beams should be first analyzed, as shown in Figure 4. The radius of the curved beam is \(R\) and the end of the curved beam is subjected to horizontal force \(P_{ij}\), vertical force \(F_{ij}\), and counterclockwise moment \(M_{ij}\), resulting in tangential and radial deflections \(\Lambda_{ij}\) and \(\Delta_{ij}\), respectively, and end slope \(\alpha_{ij}\). The geometric parameters of the curved beam include length \(L\) along the \(X_{ij}\) axis, in-plane thickness \(T_{i}\), out-of-plane thickness \(W\), and Young modulus \(E\) of the material. In addition, \(I_{ij}=WT_{i}^{3}/12\) represents the area moment of inertia of the beam cross-section. The dimensionless curvature is defined as  
\[k=\pm\frac{1}{R/L} \tag{1}\]  
The sign of \(k\) indicates the rotation direction of the unit tangent vector according to the parameter along the curve. If the unit tangent rotates counterclockwise, \(k>0\); if it  
Figure 4: Circular beam subjected to combined force and moment loads at its free end.  
Figure 3: Fully compliant CBBM.  
rotates clockwise, \(k<0\). For the flexible circular beam \(L_{i}\) (\(i=1,2\)), the curvature is constant and the CBCM can be used for equal discretization, as shown in Figure 5.  
After discretization, we use the BCM to describe each element independently. In the local coordinate system \(O_{ij}X_{ij}Y_{ij}\) for element \(j\) of flexible beam \(i\), \(F_{ij},P_{ij}\), and \(M_{ij}\) are the radial, tangential, and moment forces at the ends of the element, respectively, and \(\Lambda_{ij},\Delta_{ij}\), and \(\alpha_{ij}\) are the corresponding tangential and radial deflections and end slope, respectively, as shown in Figure 6. All the parameters are normalized as follows:  
\[\begin{split} m_{ij}&=\frac{M_{ij}L_{ij}}{EI}\ \ \ \ p_{ij}=\frac{P_{ij}L_{ij}^{2}}{EI}\ \ \ \ f_{ij}=\frac{F_{ij}  
The end of the beam is balanced with the load of the first element:  
\[p_{i1}=p_{i0},f_{i1}=f_{i0} \tag{5}\]  
The static equilibrium between elements is provided by  
\[\left[\begin{array}{ccc}\cos\theta_{ij}&\sin\theta_{ij}&0\\ -\sin\theta_{ij}&\cos\theta_{ij}&0\\ (1+\lambda_{ij})&-(0.5k+\delta_{ij})&1\end{array}\right]\left[\begin{array}{ c}f_{ij}\\ p_{ij}\\ m_{ij}\end{array}\right]=\left[\begin{array}{c}f_{i1}\\ p_{i1}\\ m_{ij-1}\end{array}\right] \tag{6}\]  
and the torque balance can be expressed as  
\[m_{iN}=m_{i0} \tag{7}\]  
The geometric constraints are expressed as  
\[\left\{\begin{array}{c}\sum\limits_{j=1}^{N}\left[(1+\lambda_{ij})\cos \theta_{ij}-(0.5k+\delta_{ij})\sin\theta_{ij}\right]=x_{0}\\ \sum\limits_{j=1}^{N}\left[(1+\lambda_{ij})\sin\theta_{ij}+(0.5k+\delta_{ij}) \cos\theta_{ij}\right]=y_{0}\\ \beta_{iN}+\sum\limits_{j=1}^{N}\alpha_{ij}=\theta_{i0}\end{array}\right. \tag{8}\]  
and  
\[\theta_{i1}=0,\theta_{ij}=\beta_{ij}+\sum\limits_{k=1}^{j-1}\alpha_{k}(i=2,3,...,N) \tag{9}\]  
where \(\beta_{ij}\) is the slope of the \(ij\)th element of the circular beam and \(\theta_{i1}\) is the rotation angle of the coordinate system of the \(ij\)th element relative to the global coordinate system.  
The force-deformation relationship of a single flexible segment can be obtained from Equations (1)-(9).