## 2 Methodology
### Theoretical Calculation of the Opening of Circumferential Joints
#### 2.3.2 Steel Channel Reinforcement Condition  
As for the segment ring load model established in Figure 4 under the condition of steel channel reinforcement, due to the presence of a steel channel, the longitudinal displacement of any segment ring will change the overall force. The longitudinal displacement of a segment ring must overcome both \(k_{c}\), the circumferential joint spring stiffness, and \(K\), the equivalent spring stiffness of the steel channel under the action of tension \(F_{t}\), so the structural stress analysis is complicated. Figure 5: Relationship curve of the circumferential joint variation with different numbers of segment rings. Assuming that the reinforcement range is m, the segment lining is divided into the reinforced area and unreinforced area. The longitudinal mechanical balance of each block in the reinforcement area is analyzed. Firstly, determine n and n\({}_{\rm c}\) after thrust unloading under the condition of steel channel reinforcement:  
\[\rm n=int(\frac{F_{k}(\Delta x_{1})}{f}) \tag{7}\]  
\[\rm n_{c}=int(\frac{F_{k}(\Delta x_{1})-F_{c}}{f}) \tag{8}\]  
When the reinforcement range m \(\geq\) n\({}_{0}\), establish the longitudinal mechanical balance equation matrix:  
\[\rm K\bullet X=F \tag{9}\]  
\[\left(\left[ \begin{array}{cccc}-F_{k}(\Delta x)&&&0\\ F_{k}(\Delta x)&-F_{k}(\Delta x)&&&\\ &\ddots&\ddots&\\ 0&&&F_{k}(\Delta x)&-F_{k}(\Delta x)\end{array} \right]+\left[ \begin{array}{ c}K_{1\times n}\\ 0_{(n-2)\times n}\\ \vdots\\ K_{1\times n}\end{array} \right]\right)\bullet\left[ \begin{array}{c}\Delta x _{1}\\ \Delta x_{2}\\ \vdots\\ \Delta x_{n}\end{array} \right]=\left[ \begin{array}{c}f-F_{t}\\ \mathrm{f}\\ \vdots\\ \mathrm{f}\end{array} \right] \tag{10}\]  
The variation of circumferential joints \(\Delta x\) is:  
\[\Delta x_{i-1}-\Delta x_{i}=\frac{f}{k_{c1}}\qquad\qquad(2\leq i\leq n_{c}) \tag{11}\]  
\[\Delta x_{n_{c}+1}=\Delta x_{1}-\frac{F_{c}-f}{k_{c2}}-\frac{f(n_{c}+1)-F_{c}} {k_{c1}} \tag{12}\]  
\[\Delta x_{i}-\Delta x_{i+1}=\frac{f}{k_{c1}}\qquad\quad(n_{c}+1\leq i\leq n-1) \tag{13}\]  
\[\Delta x_{i}=0\qquad\quad(i>n) \tag{14}\]  
\[\sum_{i=1}^{n-1}\Delta x_{i}=\Delta x_{1}n_{c}+\Delta x_{n_{c}+1}(n-n_{c})- \frac{n_{c}(n_{c}-1)f}{2k_{c2}}-\frac{(n-n_{c})(n-n_{c}-1)f}{2k_{c1}} \tag{15}\]  
When the reinforcement range 0 \(<\) m \(<\) n\({}_{0}\), establish the longitudinal mechanical balance equation matrix:  
\[\rm K\bullet X=F \tag{16}\]  
\[\left(\left[ \begin{array}{cccc}-F_{k}(\Delta x)&&&0\\ F_{k}(\Delta x)&-F_{k}(\Delta x)&&&\\ &\ddots&\ddots&\\ 0&&&F_{k}(\Delta x)&-F_{k}(\Delta x)\end{array} \right]+\left[ \begin{array}{ c}K_{1\times n}\\ 0_{(m-2)\times n}\\ K_{1\times n}\\ 0_{(n-m)\times n}\end{array} \right]\right)\bullet\left[ \begin{array}{c} \Delta x_{1}\\ \Delta x_{2}\\ \vdots\\ \mathrm{f}\end{array} \right]=\left[ \begin{array}{c}f-F_{t}\\ \mathrm{f}\\ \vdots\\ \mathrm{f}\end{array} \right] \tag{17}\]  
When m \(<\) n\({}_{\rm c}\), the variation of circumferential joints \(\Delta x\) is:  
\[\Delta x_{i-1}-\Delta x_{i}=\frac{f}{k_{c1}}\qquad\qquad(2\leq i\leq n_{c}) \tag{18}\]  
\[\Delta x_{n_{c}+1}=\Delta x_{1}-\frac{F_{c}-f}{k_{c2}}-\frac{f(n_{c}+1)-F_{c} }{k_{c1}} \tag{20}\]\[\Delta\mathrm{x_{i}}-\Delta\mathrm{x_{i+1}}=\frac{\mathrm{f}}{\mathrm{k_{c1}}} \qquad\qquad(\mathrm{n_{c}}+1\leq\mathrm{i}\leq\mathrm{m}-1) \tag{21}\]  
It can be seen from the relationship curve in Figure 6 that steel channel reinforcement can effectively reduce the variation of circumferential joints within the reinforcement range, and the reduction is related to the equivalent stiffness of the steel channel. There is a distinct change at the m-ring, and the variation of circumferential joints after m is equal to that no reinforcement. It is worth noting that the curve slope of the opening is the same before and after reinforcement, but the variation of circumferential joints in the unreinforced area does not change.