## 4 Solution of \(K_{1}\) Value  
Formula (12) is difficult to solve by conventional mathematical methods. Therefore, the dichotomy method is used to find the solution of the value of \(K_{1}\). When using dichotomy, the function \(f\), as shown in Formula (16), must be established, then the solution of \(K_{1}\) needs to be solved, when \(f=0\).  
\[f(\alpha,K_{1})=sinK_{1}\alpha cos\alpha-K_{1}sin\alpha cosK_{1}\alpha=0 \tag{16}\]  
In order to solve the value of \(K_{1}\) by dichotomy, a monotone interval needs to be given, and then iteration is carried out. Since the function \(f\) is not a monotone function, in order to ensure its correctness, we should give as many monotone intervals as possible for multiple roots [29]. According to the formula \({K_{1}}^{2}=1+\frac{PR^{3}}{EI}\) and its physical meaning, the definition domain of \(K_{1}\) is (1, \(\infty\)). Therefore, in order to find the value of \(K_{1}\), 18 intervals, (1.1, 1.6), (1.6, 2.1)... (59.6, 60.1), are given in the solving process.  
Because \(\alpha\) is also an uncertain parameter in Formula (16), an \(\alpha\) value must be given when solving \(K_{1}\). According to Figure 1, the definition domain of \(\alpha\) is (0, \(\pi\)); therefore, \(\alpha\) is given and varies from \(\pi/36\) (\(5^{\circ}\)) to \(\pi\) (180\({}^{\circ}\)) by every interval of \(\pi/72\) (\(2.5^{\circ}\)).  
According to the above monotonic interval and values of \(\alpha\), Formula (16) is solved by MATLAB software programming, and the obtained numerical points are plotted as the distribution map of \(K_{1}\) value, as shown with a square marker in Figure 3. Figure 3 shows that each \(\alpha\) corresponds to multiple \(K_{1}\) values. Multiple sets of curves with similar shapes can be obtained by connecting each \(K_{1}\) value with a line, reflecting that the function \(f\) is a set of periodic functions.  
According to the calculation results of dichotomy, when \(\alpha\) is 180\({}^{\circ}\), there is no grouting between the liner and the original pipeline, and the minimum value of \(K_{1}\) is 2.0. Substituting it into Equation (14), Formula (17) can be obtained.  
\[E_{L}I=\frac{P_{cr}ND^{3}}{24C} \tag{17}\]  
Formulas (1) and (17) of the critical buckling bearing capacity of an infinite-length independent tube under hydrostatic external pressure are similar in form and have the same numerical parameters--24. Therefore, the minimum \(K_{1}\) value corresponding to each \(\alpha\), that is, the set of values marked by the red circle in Figure 3, should be taken as the solution of Formula (16). The dichotomy calculation results of \(K_{1}\) are shown in Table 1.  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline \(\alpha\) (\({}^{\circ}\)) & \(K_{1}\) & \(\alpha\) (\({}^{\circ}\)) & \(K_{1}\) & \(\alpha\) (\({}^{\circ}\)) & \(K_{1}\) \\ \hline  
5 & 4.98 & 112.5 & 2.48 & 172.5 & 2.00 \\ \hline \hline \end{tabular}
\end{table}
Table 1: Dichotomy calculation result of \(K_{1}\).  
Figure 3: Distribution of \(K_{1}\) relative to \(\alpha\) calculated by MATLAB.  
It can be observed from Figure 5 that the 1/2 non-grouting angle \(\alpha\) has a nonlinear relationship with \(\alpha\), and is a decreasing function. Formula (14) shows that the stiffness coefficient is also a decreasing function of \(K_{1}\) value, so the stiffness coefficient is an increasing function of the 1/2 non-grouting angle \(\alpha\). When the grouting slurry is fully filled with voids, the 1/2 non-grouting angle \(\alpha\) tends to be zero, and the \(K_{1}\) value tends to be infinity. Therefore, the stiffness coefficient of the lining calculated by Formula (14) tends to be zero, and, in this case, Formula (14) will be meaningless.  
Therefore, Formula (14) is only applicable to the structural design of rigid-flexible composite lining with incomplete grouting. During construction, the non-grouting angle \(2\alpha\) should be reasonably evaluated according to the grouting technology to design the lining structure.