# Mechanical Properties of the Functionally Graded Lining for a Deep Buried Subway Tunnel
## 2 Analytical Solutions for the Mechanical Properties of the Functionally Graded Lining
### Solutions of the Internal Force Coefficient of the Functionally Graded Lining  
As shown in Figure 3d, the angular displacement is equal to zero at the position of \(\alpha\)= \(0^{\circ}\), and the force method equation is given by  
\[\delta_{11}X_{1}+\Delta_{1\text{p}}=0 \tag{5}\]  
where \(\delta_{11}\) is the displacement at \(X_{1}\) of \(\overline{X}_{1}\)= \(1\); \(\Delta_{1\text{p}}\) is the displacement at \(X_{1}\) of external forces;  
\[\delta_{11}=\int\frac{\overline{M}_{1}{}^{2}}{E(\alpha)I}\text{ds}=\int_{0}^{ \pi/2}\frac{\overline{M}_{1}{}^{2}}{E(\alpha)I}R_{\text{H}}\text{d}\alpha=\int _{0}^{\pi/2}\frac{R_{\text{H}}}{E(\alpha)I}\text{d}\alpha=\frac{R_{\text{H}}}{ E_{0}I}m_{1} \tag{6}\]  
where  
\[m_{1}=\int_{0}^{\pi/2}\frac{1}{E(\alpha)/E_{0}}\text{d}\alpha\]  
Moreover,  
\[\Delta_{1\text{p}}=\int\frac{\overline{M}_{1}M_{\text{p}}}{E(\alpha)I}\text{ ds}=\int_{0}^{\pi/2}\frac{M_{\text{p}}}{E(\alpha)I}R_{\text{H}}\text{d}\alpha\]  
where  
\[M_{p}=\frac{1}{2}(\lambda-1)p_{0}R_{\text{H}}^{2}\sin^{2}\alpha\]  
Therefore,  
\[\Delta_{1\text{p}}=\int\frac{\overline{M}_{1}M_{\text{p}}}{E(\alpha)I}\text{ ds}=\int_{0}^{\pi/2}\frac{M_{\text{p}}}{E(\alpha)I}R_{\text{H}}\text{d}\alpha= \frac{1}{2}(\lambda-1)p_{0}R_{\text{H}}^{3}\int_{0}^{\pi/2}\frac{\sin^{2} \alpha}{E(\alpha)I}\text{d}\alpha=\frac{1}{2}\frac{(\lambda-1)p_{0}R_{\text{H }}^{3}}{E_{0}I}m_{2} \tag{7}\]  
where \(m_{2}=\int_{0}^{\pi/2}\frac{\sin^{2}\alpha}{E(\alpha)/E_{0}}\text{d}\alpha\); \(\lambda\) is the coefficient of lateral pressure.  
So,  
\[X_{1}=\frac{-\Delta_{1\text{p}}}{\delta_{11}}=\frac{1}{2}(1-\lambda)p_{0}R_{ \text{H}}^{2}\frac{m_{2}}{m_{1}},\]  
\[M_{\alpha}=\overline{M}_{1}X_{1}+M_{\text{p}}=\frac{1}{2}p_{0}R_{\text{H}}^{2 }(1-\lambda)(\frac{m_{2}}{m_{1}}-\sin^{2}\alpha).\]  
Further, the moment nondimensionalized is given by  
\[\frac{M_{\alpha}}{p_{0}R_{\text{H}}^{2}}=\frac{1}{2}(1-\lambda)(\frac{m_{2}}{ m_{1}}-\sin^{2}\alpha). \tag{8}\]  
The axial force can be obtained by calculating moment at the center, and the axial force nondimensionalized is given by  
\[\frac{N}{p_{0}R_{\text{H}}}=\sin^{2}\alpha+\lambda\cos^{2}\alpha. \tag{9}\]  
Boundary conditions at \(\alpha=0\) and \(\alpha=\pi/2\) can be obtained by the principle of virtual work.  
\[\omega_{0} =-(\int_{0}^{\pi/2}\frac{\overline{M}_{2}M_{\text{p}}}{E(\alpha)I }R_{\text{H}}\text{d}\alpha+\int_{0}^{\pi/2}\frac{\overline{M}_{2}M_{1}}{E( \alpha)I}R_{\text{H}}\text{d}\alpha)\] \[=\frac{(\lambda-1)p_{0}R_{\text{H}}^{4}}{2E_{0}I}(\int_{0}^{\pi /2}\frac{\sin^{3}\alpha}{E(\alpha)/E_{0}}\text{d}\alpha-\frac{m_{2}}{m_{1}} \int_{0}^{\pi/2}\frac{\sin\alpha}{E(\alpha)/E_{0}}\text{d}\alpha)\ . \tag{10}\]where \(\overline{M}_{2}=1\) is exerted at \(\alpha=0^{\circ}\) vertically;  
\[m_{3}=\int_{0}^{\pi/2}\frac{\sin^{3}\alpha}{E(\alpha)/E_{0}}\mathrm{d}\alpha- \frac{m_{2}}{m_{1}}\int_{0}^{\pi/2}\frac{\sin\alpha}{E(\alpha)/E_{0}}\mathrm{d}\alpha.\]  
\[\omega_{\pi/2} =-(\int_{0}^{\pi/2}\frac{\overline{M}_{3}M_{\mathrm{P}}}{E(\alpha)I }R_{\mathrm{H}}\mathrm{d}\alpha+\int_{0}^{\pi/2}\frac{\overline{M}_{2}M_{1}}{ E(\alpha)I}R_{\mathrm{H}}\mathrm{d}\alpha) \tag{11}\] \[=\frac{(\lambda-1)p_{0}R_{\mathrm{H}}^{4}}{2E_{0}I}\frac{(m_{2} \int_{0}^{\pi/2}\frac{1-\cos\alpha}{E(\alpha)/E_{0}}\mathrm{d}\alpha-\int_{0}^ {\pi/2}\frac{\sin^{2}\alpha(1-\cos\alpha)}{E(\alpha)/E_{0}}\mathrm{d}\alpha)} \leavevmode\
obreak\ \leavevmode\
obreak\.\]  
where \(\overline{M}_{3}=1\) is exerted at \(\alpha=90^{\circ}\) horizontally to the right;  
\[m_{4}=\frac{m_{2}}{m_{1}}\int_{0}^{\pi/2}\frac{1-\cos\alpha}{E(\alpha)/E_{0}} \mathrm{d}\alpha-\int_{0}^{\pi/2}\frac{\sin^{2}\alpha(1-\cos\alpha)}{E(\alpha )/E_{0}}\mathrm{d}\alpha.\]  
From derivation, when external forces are determined, the internal force can be obtained, and the radial displacement can be obtained according to the governing equation and boundary conditions. The boundary conditions are concluded as follows.  
\[\alpha=0,\omega_{0}=\frac{(\lambda-1)p_{0}R_{\mathrm{H}}^{4}}{2E_{0}I}m_{3} \leavevmode\
obreak\ \leavevmode\
obreak. \tag{12}\]