## 3 Hybrid Analysis
### Symmetric Model of Rigid-Body Displacements  
The segmental tunnel linings analyzed in the present paper have six degrees of freedom (DOFs), as shown in Figure 1. Three overall DOFs, i.e., one rotational DOF and two translational DOFs, do not result in convergences because they refer to displacements of the tunnel structure as a rigid body. Three internal DOFs refer to rigid-body displacements of the segments relative to each other. Herein, three overall DOFs are set equal to zero, see in Equation (14). Thus, the remaining types of rigid-body displacements represent quantities relative to the segment at the top, which serves as the reference. As for the symmetric mode of rigid-body displacements, see Figure 5a for the half-ring model, the values of \(\Delta\theta_{j=1,2,\dots,6}\), referring to rigid-body displacements, must fulfill the following three conditions, expressing that the analyzed rings remain closed [35]:  
\[\sum_{j=1}^{6}u^{L}(2\pi,\Delta\theta_{j})=0 \Rightarrow \sum_{j=1}^{6}\Delta\theta_{j}\sin\varphi_{j}=0\,, \tag{30}\] \[\sum_{j=1}^{6}v^{L}(2\pi,\Delta\theta_{j})=0 \Rightarrow \sum_{j=1}^{6}\Delta\theta_{j}(1-\cos\varphi_{j})=0\,,\] (31) \[\sum_{j=1}^{6}\theta^{L}(2\pi,\Delta\theta_{j})=0 \Rightarrow \sum_{j=1}^{6}\Delta\theta_{j}=0\,. \tag{32}\]  
Figure 5: Symmetric mode of rigid-body displacements: (**a**) half-ring models for the analysis and (**b**) graphical representation. Symmetric relative rotations, \(\Delta\theta_{j=1,2,\ldots,\delta}^{sym}\), fulfill the conditions \(\Delta\theta_{1}^{sym}=\Delta\theta_{6}^{sym}\), \(\Delta\theta_{2}^{sym}=\Delta\theta_{5}^{sym}\), and \(\Delta\theta_{3}^{sym}=\Delta\theta_{4}^{sym}\). Inserting them into Equations (30)-(32), it is observed that Equation (30) is automatically satisfied. The reason is that the sine function is asymmetric with respect to the vertical axis, whereas the relative rotations are symmetric. The other two conditions, i.e., (31)-(32), read as  
\[\Delta\theta_{1}^{sym}\left(1-\cos\varphi_{1}\right)+\Delta \theta_{2}^{sym}\left(1-\cos\varphi_{2}\right)+\Delta\theta_{3}^{sym}\left(1- \cos\varphi_{3}\right) = 0\,, \tag{33}\] \[\Delta\theta_{1}^{sym}+\Delta\theta_{2}^{sym}+\Delta\theta_{3}^{ sym} = 0\,. \tag{34}\]  
For the purpose of solving these two equations with three unknowns, one of the unknowns must be prescribed. The other two follow in the sense of an inevitable kinematic chain. This provides the motivation to express the angles \(\Delta\theta_{j=2,3,\ldots,6}^{sym}\) as functions of \(\Delta\theta_{1}^{sym}\), yielding the general solution  
\[\Delta\theta_{2}^{sym} = \Delta\theta_{5}^{sym}\ =\ \frac{\cos\varphi_{3}-\cos\varphi_{1}}{\cos \varphi_{2}-\cos\varphi_{3}}\,\Delta\theta_{1}^{sym}\,, \tag{35}\] \[\Delta\theta_{3}^{sym} = \Delta\theta_{4}^{sym}\ =\ \frac{\cos\varphi_{1}-\cos\varphi_{2}}{\cos \varphi_{2}-\cos\varphi_{3}}\,\Delta\theta_{1}^{sym}\,,\] (36) \[\Delta\theta_{6}^{sym} = \Delta\theta_{1}^{sym}\,. \tag{37}\]  
The graphical representation of symmetric rigid-body displacements, see Figure 5b, is obtained with the help of the transfer relations (2), where the load integrals for \(\Delta\theta_{j}\) are calculated by inserting Equations (35)-(37) into Equations (10)-(13). The six integration constants at the initial cross-section are set equal to zero because rigid-body displacements do not generate internal forces.