## 3 Theoretical Derivations
### Bolts
#### 3.1.2. Bearing Capacity Derivation  
The static balance equation of the bolt:  
\[\sigma_{b}\cdot\frac{\pi}{4}{d_{b}}^{2}+\tau\cdot\pi\cdot d_{b}\cdot dr=(\sigma _{b}+d\sigma_{b})\cdot\frac{\pi}{4}{d_{b}}^{2} \tag{1}\]  
where \(d_{b}\) is the diameter of the bolt, \(\sigma_{b}\) is the axial stress of the bolt, and \(\tau\) is a function of \(r\), which can be written as \(\tau(r)\). By simplifying the above formula, we can obtain:  
\[\frac{d\sigma_{b}}{dr}=\frac{4\tau(r)}{d_{b}} \tag{2}\]  
It is assumed that the axial displacement \(u_{b}\) of the bolt is positive towards the inside of the tunnel, which is opposite to the direction of the polar coordinate system with the tunnel center as the origin. At this time, the axial strain of the bolt microsegment is  
\[\varepsilon_{b}=-\frac{du_{b}}{dr} \tag{3}\]  
The axial stress of the bolt is  
\[\sigma_{b}=-E_{b}\frac{du_{b}}{dr} \tag{4}\]  
where \(E_{b}\) is the elastic modulus of the bolt. The simultaneous Equations (2) and (4) can be obtained as  
\[\frac{d^{2}u_{b}}{dr^{2}}=-\frac{4}{E_{b}d_{b}}\tau(r) \tag{5}\]  
Use \(G_{b}\) to represent the shear modulus of the bolt agent, \(t_{b}\) to represent the thickness of the bolt agent, and \(u_{rb}\) to represent the radial displacement of the surrounding rock after the bolt is applied. At this time, the shear strain of the adhesive layer is \((u_{rb}-u_{b})/t_{b}\). It can be seen that the shear stress of the bolt agent is  
\[\tau(r)=\frac{G_{b}}{t_{b}}(u_{rb}-u_{b}) \tag{6}\]  
The simultaneous Equations (5) and (6) can be obtained as  
\[\frac{d^{2}u_{b}}{dr^{2}}-m^{2}u_{b}+m^{2}u_{rb}=0 \tag{7}\]  
which are obtained after solving  
\[ \begin{array}{l}u_{b}=C_{1}e^{Br}+C_{2}e^{-Br}-\frac{Be^{Br}}{2}\int e^{-Br} u_{rb}dr+\frac{Be^{-Br}}{2}\int e^{Br}u_{rb}dr\\ B=\sqrt{4G_{b}/E_{b}d_{b}t_{b}}\end{array}  \tag{8}\]  
Figure 4: The microsegmented force diagram of the bolt. To solve the above formula, it is necessary to solve the radial displacement of the surrounding rock. Assuming that the surrounding rock around the bolt is in a plastic state, according to the literature [28], the displacement of the axisymmetric plastic zone of the circular tunnel is  
\[u=\frac{A}{r} \tag{9}\]  
\(A\) is the integral constant. Here, \(u\) is replaced with \(u_{rb}\):  
\[u_{rb}=\frac{A_{rb}}{r} \tag{10}\]  
\(u_{rb}\) is the displacement of the tunnel wall after the bolt is applied, and \(A_{rb}\) is the integral constant after the bolt is applied. Substitute Equation (10) into Equation (8). \[u_{b}=C_{1}e^{Br}+C_{2}e^{-Br}-\frac{A_{rb}Be^{Br}}{2}\int\frac{e^{-Br}}{r}dr- \frac{A_{rb}B^{2}e^{-Br}}{2}\int\frac{e^{Br}}{r}dr \tag{11}\]  
According to the literature in [26], there are  
\[ \begin{array}{l}\int\frac{e^{-Br}}{r}dr\approx\frac{e^{-Br}}{Br}\sum\limits \limits_{n=0}^{\infty}\frac{(-1)^{n}nt}{(Br)^{n}}\\ \int\frac{e^{Br}}{r}dr\approx\frac{e^{br}}{Br}\sum\limits_{n=0}^{\infty}\frac {nt}{(Br)^{n}}\end{array}  \tag{12}\]  
Since the general fully bonded bolt does not need to apply a backing plate, and the axial force at both ends of the bolt is 0, there are the following boundary conditions:  
\[\left.\frac{du_{b}}{dr}\right|_{r=r_{0}}=0,\left.\frac{du_{b}}{dr}\right|_{r=r _{0}+L}=0 \tag{13}\]  
where \(r_{0}\) is the radius of the tunnel and \(L\) is the length of the bolt. The simultaneous Equations (11)-(13) can be obtained as  
\[C_{1}=A_{rb}D_{1},\ C_{2}=A_{rb}D_{2} \tag{14}\]  
where  
\[ \begin{array}{l}D_{1}=\frac{e^{-Br_{0}}}{2r_{0}}\frac{\gamma(r_{0})-\gamma(r _{0}+L)e^{Br}}{r_{0}^{2}}\\ D_{2}=\frac{e^{2(r_{0}+L)}}{2r_{0}}\frac{\gamma(r_{0})e^{Br}-\gamma(r_{0}+L)}{ e^{2 Br}-1}\\ \gamma(r)=\frac{1}{r}\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}nt-nt}{(Br)^{n}} \end{array}