## 4 Deformation Mechanism
### Calculation of Potential Energy  
1. Strain energy  
Because the strain of the soil is derived from the deformation mechanism, and the deformation mechanism in this paper is expressed in a curvilinear coordinate system, we have to convert the strain in the curvilinear coordinate system to the strain in the Cartesian coordinate system. Equation (30) gives the relationship between the curvilinear coordinate system and the Cartesian coordinate system.  
\[\left\{\begin{array}{ll}x=-r\sin\theta\\ y=(R_{\beta}-r\cos\theta)\cos\beta\\ z=(R_{\beta}-r\cos\theta)\sin\beta\end{array}\right. \tag{30}\]  
\[R_{\beta}=\left(\cos^{2}\beta/R_{f}^{2}+\cos^{2}\beta/(0.75R_{f})^{2}\right)^{ -0.5} \tag{31}\]  
The conversion can be easily given by tensor calculation, as shown in Equations (32) and (33).  
\[\left[\begin{array}{ccc}\varepsilon_{x}&\varepsilon_{xy}&\varepsilon_{xz} \\ \varepsilon_{yx}&\varepsilon_{y}&\varepsilon_{yz}\\ \varepsilon_{zx}&\varepsilon_{zy}&\varepsilon_{z}\end{array}\right]=P\left[ \begin{array}{ccc}\varepsilon_{r}&\varepsilon_{ry}&\varepsilon_{rz}\\ \varepsilon_{\theta r}&\varepsilon_{\theta}&\varepsilon_{\theta\beta}\\ \varepsilon_{\beta r}&\varepsilon_{\beta\theta}&\varepsilon_{\beta}\end{array} \right]P^{T} \tag{32}\]  
\[P=\left[\begin{array}{ccc}-\sin\theta&\cos\beta\cos\theta&\cos\theta\sin \beta\\ \cos\theta&\cos\beta\sin\theta&-\sin\beta\sin\theta\\ 0&\sin\beta&\cos\beta\end{array}\right] \tag{33}\][PERSON] and [PERSON] [28] derived the geometrical equations in the curvilinear coordinate system. Based on the work performed by [PERSON] and [PERSON] [28], we give the geometrical equations modified for the coordinate in our method as follows.  
\[\left\{\begin{array}{l}\varepsilon_{rr}=\frac{\partial v_{r}}{ \partial r}\\ \varepsilon_{rr}=\frac{1}{2}\Big{(}\frac{\partial v_{r}}{\partial\theta}+r \frac{\partial v_{\theta}}{\partial r}-v_{\theta}\Big{)}\\ \varepsilon_{r\beta}=\frac{1}{2(R_{\beta}-r\cos\theta)}\Big{(}v_{\theta}\cos \theta+\frac{\partial v_{r}}{\partial\beta}+(R_{\beta}-r\cos\theta)\frac{ \partial v_{\theta}}{\partial r}\Big{]}\\ \varepsilon_{\theta\theta}=\frac{1}{\Gamma}\Big{(}v_{r}+\frac{\partial v_{ \theta}}{\partial\theta}\Big{)}\\ \varepsilon_{\theta\beta}=\frac{1}{2r(R_{\beta}-r\cos\theta)}\Big{[}r\frac{ \partial v_{\theta}}{\partial\theta}-v_{\theta}r\sin\theta+\frac{\partial v_{ r}}{\partial\theta}+(R_{\beta}-r\cos\theta)\frac{\partial v_{\beta}}{ \partial\theta}\Big{]}\\ \varepsilon_{\beta\beta}=\frac{1}{R_{\beta}-r\cos\theta}\Big{(}v_{\theta} \sin\theta-v_{r}\cos\theta+\frac{\partial v_{r}}{\partial\beta}\Big{)}\\ \end{array}\right. \tag{34}\]  
Equation (34) gives the train energy of the soil. Equation (4), the strain energy of the soil, can be calculated by combining Equation (4) with Equation (34).  
(b) Potential energy of external force  
According to the deformation mechanism, the vertical displacement can be given as follows:  
\[w=-v_{r}\cos\theta\cos\beta+v_{\theta}\cos\beta\sin\theta-v_{\beta}\sin\beta \tag{35}\]  
Substituting Equation (35) into Equation (6), the potential energy of gravity can be calculated.  
When \(\beta=0\), the orthoradial displacement \(v_{\beta}(r,\,\theta,\,\beta)\) is equal to the horizontal displacement on the tunnel face \(u_{f}\). Equation (36) gives the horizontal displacement on the tunnel face.  
\[u_{f}=v_{\beta}(r,\,\theta,\,0) \tag{36}\]  
Combining Equation (36) with Equation (5), the potential energy of support pressure can be calculated.