## 3 Theoretical Derivations
### Bearing Capacity of the Primary Support Derivation  
The circular tunnel can be regarded as an axisymmetric plane strain problem without shear stress. At this time, the axial stress and tangential stress of the surrounding rock are \(\sigma_{1}\) and \(\sigma_{3}\), respectively. The surrounding rock under combined support follows the Mohr-Coulomb criterion, including  
\[\sigma_{1}=\sigma_{3}\frac{1+\sin\varphi_{b}}{1-\sin\varphi_{b}}+2C_{b}\frac{2 \cos\varphi_{b}}{1-\sin\varphi_{b}} \tag{48}\]  
Let \(K_{\rm r}=(1+\sin\varphi_{b})/(1-\sin\varphi_{b})\), there is \(P_{c}=K_{\rm r}P\), and the support resistance \(P\) can be decomposed into the support force \(P_{s}\) provided by the bolt and the support resistance \(P_{c}\) provided by the steel arch frame and shotcrete layer. \(P_{s}\) can be written as  
\[P_{\rm s}=\frac{Q_{s}}{D_{l}D_{a}} \tag{49}\]  
where \(Q_{s}\) is the anchoring force of the bolt and \(D_{l}\) and \(D_{a}\) are the spacing and row spacing of the system, respectively. It is generally assumed that these two values are equal. The support structure composed of the shotcrete and the steel arch frame provides support force by changing the circumferential axial force, so \(P_{l}\) can be written as  
\[P_{l}=\frac{T}{r_{0}L_{s}} \tag{50}\]  
where \(L_{s}\) is the length of the support formed by the shotcrete and the steel arch frame. Since the length of the semicircle arch can be expressed as  
\[L_{s}=\pi r_{0} \tag{51}\]  
The above equation can be written as  
\[P_{l}=\frac{T}{\pi r_{0}^{2}} \tag{52}\]  
Then, the support generated by the primary support can be written as Equation (53):  
\[P=\frac{Q_{s}}{D_{l}D_{a}}+\frac{T}{\pi r_{0}^{2}} \tag{53}\]  
where \(\sigma_{T}\) is the circumferential stress of the inner support, \(b_{T}\) is the longitudinal unit length of the tunnel, taken as \(1\)m, and \(h\) is the thickness of the inner support. The bearing capacity of the bolt is further analyzed below. The thickness \(b\) of the surrounding rock composite arch is taken as follows:  
\[b=\frac{L\tan\alpha-D_{b}}{\tan\alpha} \tag{54}\]  
where \(L\) is the effective length of the bolt; \(\alpha\) is the control angle of the bolt in the broken rock mass, usually 43-45\({}^{\circ}\); and \(D_{b}\) is the row spacing of the bolts. Along the longitudinal unit length of the tunnel, the combined bearing capacity \(N\) of the surrounding rock composite arch is  
\[N=Pb\frac{1+\sin\varphi_{b}}{1-\sin\varphi_{b}}+\frac{1}{2}kb^{2} \tag{55}\]  
where \(k\) is the increasing slope of radial stress. When the surrounding rock is in an unstable weak and broken state, \(k=0\); \(L\) is the effective length of the anchor rod, and the above formula is simplified as  
\[N=P\frac{1+\sin\varphi_{b}}{1-\sin\varphi_{b}}\frac{L\tan\alpha-D_{b}}{\tan \alpha} \tag{56}\]If the circumferential axial force of the surrounding rock composite arch is \(N_{0}\) and the differential segment of the composite arch along the circumferential direction is \(ds\), then \(ds=(r_{0}+b/2)\ d\theta\) and perform the static balance solution:  
\[ \begin{array}{l}\sum Y=0;\int_{0}^{\pi}q\sin\theta ds=2N_{0}\\ N_{0}=(r_{0}+\frac{b}{2})q\end{array}  \tag{57}\]  
If you want to ensure the stability of the composite arch, you need to meet \(N\geq N_{0}\) to ensure the safety of the surrounding rock. When \(N=N_{0}\), the composite arch reaches the limit equilibrium state. Bring this condition into Equation (57), and you can obtain  
\[q=(\frac{Q_{s}}{D_{i}D_{a}}+\frac{T}{\pi r_{0}{}^{2}})\frac{2(1+\sin\varphi_{b })(L\tan\alpha-D_{b})}{(1-\sin\varphi_{b})[(2r_{0}+L)\tan\alpha-D_{b}]} \tag{58}\]  
In the above equation, \(q\) is the support force, and \(L\) is the length of the bolt in the broken surrounding rock.