# Stress Analysis and Spalling Failure Simulation on Surrounding Rock of Deep Arch Tunnel
## 2 Stress Analysis of Surrounding Rock
### Stress Solution of the Surrounding Rock
\begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \(k\) & \(h\) Value & \(C_{0}\) & \(C_{1}\) & \(C_{2}\) & \(C_{3}\) & \(C_{4}\) & \(C_{5}\) & \(C_{6}\) \\ \hline \multirow{3}{*}{6} & 0.6193 & \(-0.1089\) & \(-0.0067\) & \(0.0845\) & \(-0.0796\) & \(0.0333\) & \(0.0075\) & \(-0.0083\) \\  & 0.6419 & \(-0.0915\) & \(-0.0947\) & \(0.0749\) & \(-0.0963\) & \(0.0285\) & \(0.0135\) & \(-0.0109\) \\  & 0.4767 & \(-0.0717\) & \(-0.1468\) & \(0.0543\) & \(-0.1043\) & \(0.0265\) & \(0.0192\) & \(-0.0088\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The results of \(C_{k}\) obtained by the optimization method. Figure 3: Boundary contour drawing of arch hole mapping function. In the above equation, \(\omega(\sigma)/\overline{\omega^{\prime}(\sigma)}\) can be expressed in the form of a Laurent series. The real constant coefficients \(D_{6}\), \(D_{5}\), \(D_{4}\), \(D_{3}\), \(D_{2}\), \(D_{1}\), \(D_{0}\), and \(d_{k}\) (\(k\) = 1, 2, 3, \(\cdots\), \(\infty\)) in Equation (22) can be obtained by multiplying the denominator on the left side of the equation by the term on the right side of the equation, given that the coefficients of the same power on both sides of the equation are equal. The results are as following: \(D_{6}=-0.00848\), \(D_{5}=0.00737\), \(D_{4}=0.03324\), \(D_{3}=-0.08113\), \(D_{2}=0.08749\), \(D_{1}=-0.00337\), \(D_{0}=-0.13033\), \(d_{1}=1.0393\), \(d_{2}=-0.03053\), \(\cdots\), \(d_{100}=-3.74\times 10^{-8}\). In this paper, the maximum value of \(k\) is set to 100, which has little effect on the stress solutions. This is because when \(k\) is sufficiently large, the value of \(d_{k}\) is approximately zero. Substituting Equations (4) and (5) into Equation (10), comparing the same power coefficients on both sides of the equation to be equal gives the values of \(a_{n}\) and \(b_{n}\) in conjunction with Equation (21). The \(a_{n}\) and \(b_{n}\) in Equation (8) are calculated as follows: \(a_{6}=0.115p(\lambda+1)\), \(a_{5}=-0.10p(\lambda+1)\), \(a_{4}=(-0.344\lambda+0.557)p\), \(a_{3}=(1.105\lambda+1.186)p\), \(a_{2}=-(1.611\lambda+0.767)p\), \(a_{1}=(-12.527\lambda+12.620)p\), \(a_{7}=a_{8}=\cdots=a_{n}=0\), \(b_{1}=-(14.012\lambda+14.199)p\), \(b_{2}=(0.865\lambda-0.014)p\), \(b_{3}=(-11.586\lambda+12.093)p\), \(\cdots\), \(b_{99}=-(6.176\times 10^{-7}\lambda+9.919\times 10^{-7})p\), \(b_{100}=(9.743\times 10^{-7}\lambda+5.615\times 10^{-8})p\). Note that the number of terms in \(a_{n}\) is 6, while the number of terms in \(b_{n}\) is infinite. However, we find that the \(b_{n}\) is approximately zero when \(n\) is sufficiently large. Therefore, \(n\) in \(b_{n}\) is also set to 100 in this study. Substituting \(a_{n}\) and \(b_{n}\) into Equation (5) yields the following two analytic functions:  
\[\left\{ \begin{array}{l}\varphi_{1}(\zeta)=p[6.78p(\lambda+1)\zeta-0.738p( \lambda+1)+\frac{p(-12.572\lambda+12.575)}{\zeta}-\frac{p(1.039\lambda+0.194)} {\zeta^{2}}+\\ \frac{p(0.475\lambda+0.646)}{\zeta^{3}}-\frac{p(0.119\lambda+0.332)}{\zeta^{4 }}-\frac{0.05p(\lambda+1)}{\zeta^{5}}+\frac{0.0575p(\lambda+1)}{\zeta^{6}} ]\\ \psi_{1}(\zeta)=p[13.56p(\lambda-1)\zeta-1.475p(\lambda-1)-\frac{p(14.102 \lambda+14.109)}{\zeta}+\frac{p(2.010\lambda-1.159)}{\zeta^{2}}+\\ \frac{p(-12.666\lambda+13.173)}{\zeta^{2}}+\ldots-\frac{p(6.176\times 10^{-7} \lambda+9.919\times 10^{-7})}{\zeta^{99}}+\frac{p(9.743\times 10^{-7} \lambda+5.615\times 10^{-8})}{\zeta^{100}}]\end{array} \right. \tag{23}\]  
Similarly, for a three-centered arch with a rise-span ratio of \(f/B=1/3\) and an aspect ratio of \(B/H=1.20\), \(D_{6}\), \(D_{5}\), \(D_{4}\), \(D_{3}\), \(D_{2}\), \(D_{1}\), \(D_{0}\) and \(d_{k}\) (\(k\) = 1, 2, 3, \(\cdots\), \(\infty\)) can be calculated as follows: \(D_{6}=-0.01107\), \(D_{5}=0.01344\), \(D_{4}=0.02951\), \(D_{3}=-0.09937\), \(D_{2}=0.07711\), \(D_{1}=-0.08617\), \(D_{0}=-0.12121\), \(d_{1}=1.05344\), \(d_{2}=-0.03391\), \(\cdots\), \(d_{100}=-3.119\times 10^{-7}\).