## 3 Elastic Foundation Beam Model Used in the Small Radius TBM Interval
### Model Solution  
The uniform cross-section circular curved beam on the Winkler elastic foundation is depicted in Figure 4. The microelement body of the circular curved beam on the Winkler elastic foundation  
Figure 3: Schematic diagram of the circular curved beam on the Winkler elastic foundation [33]. is shown in Figure 5. The equivalent equation of the microelement body, ignoring the high order, is shown below [32]. \[\frac{dQ}{dx}=p-\frac{N}{R}-q_{r}, \tag{4}\]  
\[\frac{dN}{dx}=\frac{Q}{R}-q_{t}, \tag{5}\]  
\[\frac{dM}{dx}=Q, \tag{6}\]  
where \(Q\) is the shear force; \(N\) is the axial force. For simplification, the effect of axial force on the deformation is neglected. Then, the deflection differential equation of the circular curved beam is obtained [34],  
\[\left(\frac{d^{2}y}{dx^{2}}+\frac{y}{R^{2}}\right)+\frac{M}{EI}=0, \tag{7}\]  
Combining Equations from (4) to (7), we can get deflection differential equation of the circular curved beam, which ignores the tangential deformation [34],  
\[\frac{d^{5}y}{d\Phi^{5}}+2\frac{d^{3}y}{d\Phi^{3}}+\mu^{2}\frac{dy}{d\Phi}= \frac{R^{4}}{EI}\left(\frac{dq_{r}}{d\Phi}+q_{t}\right), \tag{8}\]  
where \(EI\) is the bending stiffness of the tunnel, \(\mu^{2}\) is the coefficient, \(\mu^{2}=I+R^{4}DK/EI\), \(K\) is the coefficient of foundation bedding, \(D\) is the width of the curved beam cross-section, \(R\) is the curvature radius of the curved beam. When there is no load on the beam span, the Equation (8) can be changed into,  
\[\frac{d^{5}y}{d\Phi^{5}}+2\frac{d^{3}y}{d\Phi^{3}}+\mu^{2}\frac{dy}{d\Phi}=0. \tag{9}\]  
This is the homogenous fifth-order differential equation with constant coefficients, whose general solution is as below [34],  
\[y(\Phi)=C_{1}+(C_{2}\mathrm{ch}\alpha\Phi+C_{3}\mathrm{sh}\alpha\Phi)\cos\beta \Phi+(C_{4}\mathrm{ch}\alpha\Phi+C_{5}\mathrm{sh}\alpha\Phi)sin\beta\Phi, \tag{10}\]  
where,  
Figure 3:0  
\(C_{1}\)-\(C_{5}\) are undetermined coefficients, obtained by loading and boundary conditions. With the successive derivation of Equation (10), we can obtain analytical expressions of the rotation angle \(\theta\), the bending moment \(M\), the shear force \(Q\), the axial force \(N\). With the initial condition of the beam, we can get the solutions of undetermined coefficients \(C_{1}\)-\(C_{5}\), the general solution of the defection differential equation can be converted into the below equation [34],  
Figure 3:1where  
\[\left\{ \begin{array}{l}F_{y1}(\[PERSON])=\frac{2a\beta}{1+2a\beta}+\frac{1}{1+2a\beta} W_{1}(\[PERSON])+\frac{R^{4}kD}{EI(1+2a\beta)}W_{2}(\[PERSON])\\ F_{y2}(\[PERSON])=\frac{\alpha}{\alpha^{2}+\beta^{2}}W_{2}(\[PERSON])+\frac{\beta}{ \alpha^{2}+\beta^{2}}W_{3}(\[PERSON])\\ F_{y3}(\[PERSON])=\frac{R^{2}}{(1+2a\beta)EI}W_{1}(\Phi)-\frac{R^{6}kD}{2a\beta(1+2 a\beta)(EI)^{2}}W_{4}(\Phi)\\ F_{y4}(\Phi)=\frac{R^{3}}{2 EIa(\alpha^{2}+\beta^{2})}W_{2}(\Phi)-\frac{R^{3}}{ 2 EI\beta(\alpha^{2}+\beta^{2})}W_{3}(\Phi)\\ F_{y5}(\Phi)=-\frac{R^{3}}{2\alpha\beta EI}W_{4}(\Phi)\\ \left\{\begin{array}{l}W_{1}(\Phi)=ch(a\Phi)\cos(\beta\Phi)\\ W_{2}(\Phi){=}sh(a\Phi)\cos(\beta\Phi)\\ W_{3}(\Phi)=ch(\alpha\Phi)\sin(\beta\Phi)\\ W_{4}(\Phi)=sh(\alpha\Phi)\sin(\beta\Phi)\end{array}\right.\end{array} \right.\]  
When the arbitrary point on the curved span has the concentrated moment \(M_{i}\), the concentrated force \(P_{i}\), the radial distribution load \(q_{ri}\), the tangential distribution load \(q_{ti}\), the external load on the beam span can be regarded as the partial initial parameters.