# Calculation Method of Loose Pressure in Surrounding Rock Mass
## 5 Upper Bound Solution for Loose Pressure in the Surrounding Rock Mass
### Upper Boundary Solution for Loose Pressure for Extra-Large Section Tunnels
#### 5.1.3. Derivation of the Upper Bound Solution for the Loose Pressure  
(1) Limit Analysis Based on H-B Strength Criterion  
Based on the assumptions mentioned above, the loose zone is regarded as an ideal elastic-plastic body which obeys the H-B criterion. The relation between \(c_{t}\) and \(\varphi_{t}\) can be obtained as below using the tangent method (Figure 14) [36]:  
\[\frac{c_{t}}{\varphi_{t}}=\frac{\cos\varphi_{t}}{2}\bigg{[}\frac{m_{b}(1-\sin \varphi_{t})}{2\sin\varphi_{t}}\bigg{]}^{\frac{a}{1-a}}-\frac{\tan\varphi_{t}}{ m_{b}}\bigg{(}1+\frac{\sin\varphi_{t}}{a}\bigg{)}\bigg{[}\frac{m_{b}(1-\sin \varphi_{t})}{2\sin\varphi_{t}}\bigg{]}^{\frac{a}{1-a}}+\frac{s}{m_{b}} \tag{11}\]  
The non-linear shear strength index \(\varphi_{t}\) in the above correlations serves as an unknown parameter in the upper-bound limit analysis, reflecting the instantaneous variability of the tangent, which is obtained by the optimization algorithm. \(c_{t}\) is obtained by Correlation 10 after \(\varphi_{t}\) is determined. (2) Calculation of geometric quantities  
Owing to the symmetry of the failure mechanism, this paper takes the left part of \(MN\) as an example for geometric quantity calculation. The point \(N\) is taken as the coordinate origin and a rectangular coordinate system is established. The vertical is the X-axis, where the down is positive, and the horizontal direction is the Y-axis, where the left is positive. According to the loose zone boundary equation, the expression of the curve \(NC\) is \(y~{}=~{}ax^{2}~{}+~{}bx~{}+~{}c\). According to the relevant description of failure mechanism and the research results of the loose zone in the surrounding rock mass, it can be known that \(|AM|=B/2\), \(|AC|=h\), \(|A_{m}M|=L/2\), \(|MN|=H\), \(|OM|=H-H_{0}\), \(|ON|=H_{1}-H+H_{0}\). Figure 14: The tangential method. Figure 13: The velocity field schematic diagram. It can be concluded from the trigonometric function:  
\[ \begin{cases}\theta_{1}=\arctan[2(H-H_{0})/L]\\ \theta_{4}=\arctan[2(H-H_{0})/B]\end{cases}  \tag{12}\]  
The linear equations of \(OA_{0},OA_{1},\ldots,OA_{m}\) can be obtained from the coordinate \((H_{1}-H+H_{0},O)\) of point \(O\) and the division of \(\angle Az_{m}ON\) as follows:  
\[\left\{ \begin{array}{l}y=\tan\left[\pi-\left(\frac{\pi}{4}-\frac{\varphi}{2} \right)\right]\cdot\left[x-\left(H_{1}-H+H_{0}\right)\right]\\ y=\tan\left[\pi-\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)-\beta_{1}\right] \cdot\left[x-\left(H_{1}-H+H_{0}\right)\right]\\ \cdots\\ y=\tan\left[\pi-\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)-\sum\limits_{i=1 }^{m}\beta_{i}\right]\cdot\left[x-\left(H_{1}-H+H_{0}\right)\right]\end{array} \right. \tag{13}\]  
The linear equations of \(AA_{m+1},AA_{m+2},\ldots,AA_{m+n-1}\) can be obtained from the coordinate \((H,B/2)\) of point \(A\) and the division of \(\angle Az_{m}C\) as follows:  
\[\left\{ \begin{array}{l}y=\tan\left(\frac{\pi}{2}-\beta_{m+1}\right)\cdot \left(x-H\right)+\frac{B}{2}\\ y=\tan\left(\frac{\pi}{2}-\beta_{m+1}-\beta_{m+2}\right)\cdot\left(x-H\right)+ \frac{B}{2}\\ \cdots\\ y=\tan\left(\frac{\pi}{2}-\sum\limits_{i=m+1}^{m+n-1}\beta_{i}\right)\cdot \left(x-H\right)+\frac{B}{2}\end{array} \right. \tag{14}\]  
The coordinates \((x_{0},y_{0}),(x_{1},y_{1}),\ldots,(x_{m+n-2},y_{m+n-2}),(x_{m+n-1},y_{m+n-1})\) of \(A_{0},A_{1},\ldots,A_{m},\ldots,A_{m+n-1}\) can be obtained by combining the loose zone boundary equation with the linear equations, and the following correlations calculate the distance of each line segment:  
\[\left\{ \begin{array}{l}|NA_{0}|=\sqrt{x_{0}^{2}+y_{0}^{2}}\\ |A_{i}A_{i+1}|=\sqrt{\left(x_{i+1}-x_{i}\right)^{2}+\left(y_{i+1}-y_{i}\right) ^{2}},(i=0,1,\ldots,m+n-2)\\ |A_{m+n-1}C|=\sqrt{\left[x_{m+n-1}-\left(H_{1}+h\right)\right]^{2}+\left(y_{m +n-1}-B/2\right)^{2}}\\ |OA_{1}|=\sqrt{\left[x_{i}-\left(H_{1}-H+H_{0}\right)\right]^{2}+y_{j}^{2}},(i= 0,1,\ldots,m)\\ \left|AA_{j}\right|=\sqrt{\left(x_{j}-H_{1}\right)^{2}+\left(y_{j}-B/2\right) ^{2}},(i=m+1,m+2,\ldots,m+n-1)\\ |OA|=\sqrt{\left(H-H_{1}\right)^{2}+B^{2}/4}\end{array} \right.