1. Introduction

2. Problem Description

3. Derivation of Stress/Displacement Equations in the Plastic Region

3.4. Derivation of Displacement in the Plastic Region

In the elastic region, the radial and tangential strains, ${\epsilon }_r^e$ and ${\epsilon }_{\theta }^e$, can be expressed by the equations of constitutive law as

$$\left\{\begin{array}{c}{\epsilon }_{\theta }^e\\ {\epsilon }_r^e\end{array}\right\}={\displaystyle \frac{1}{2G}}\left[\begin{array}{cc}1-\nu & \nu \\ \nu & 1-\nu \end{array}\right]\left\{\begin{array}{c}{\sigma }_{\theta }\\ {\sigma }_r\end{array}\right\}$$

(33)

where is the Poisson’s ratio and G is the shear modulus of the rock masses.

For the small strain problem, the relationship between strain and displacement at any point in the rock masses can be expressed by the equations of compatibility as

$${\epsilon }_r={\displaystyle \frac{\partial u_r}{\partial r}}$$

(34)

$${\epsilon }_{\theta }={\displaystyle \frac{\partial u_{\theta }}{r\partial \theta }}+{\displaystyle \frac{u_r}{r}}\cong {\displaystyle \frac{u_r}{r}}$$

(35)

Additionally, to determine the displacement field in the plastic region, a plastic flow rule is needed. By assuming that the elastic strains are relatively small in comparison to the plastic strains and that a non-associated flow rule is valid as shown in Figure 5, the plastic parts of the radial and tangential strains may be related to the plane strain condition.

$${\epsilon }_r^p+{K_{\psi }\epsilon }_{\theta }^p=0$$

(36)

where the coefficient of passive pressure K_p can be expressed as

$$K_{\psi }={tan}^2\left(45+{\displaystyle \frac{\psi }{2}}\right)$$

(37)

where $\psi$ is the dilation angle of the intact rock. The above equation with another form can be expressed as the following,

$$\left({\epsilon }_r-{\epsilon }_r^e\right)+K_{\psi }\left({\epsilon }_{\theta }-{\epsilon }_{\theta }^e\right)=0$$

(38)

where the subscripts e and p represent the elastic and plastic parts, respectively. Using Eq. (34), (35), and (38) leads to the following differential equation:

$$f\left(r\right)={\epsilon }_r^e+{K_{\psi }\epsilon }_{\theta }^e={\displaystyle \frac{\partial u_r}{\partial r}}+K_{\psi }{\displaystyle \frac{u_r}{r}}$$

(39)

By substituting the equation of constitutive law, equation (33), into the above equation, then

$$f\left(r\right)={\displaystyle \frac{\partial u_r}{\partial r}}+K_{\psi }{\displaystyle \frac{u_r}{r}}={\displaystyle \frac{1}{2G}}\left\{\left[\left(1-\upsilon \right)K_{\psi }-\nu \right]\left({\sigma }_{\theta }\right)+\left[\left(1-\upsilon \right)-\nu K_{\psi }\right]\left({\sigma }_r\right)\right\}$$

(40)

This differential equation may be solved by engineering mathematics with the homogeneous solution and the particular solution, and by using the following boundary condition for the radial displacement (u_r) in the plastic region.

(1) Homogeneous solution ($f\left(r\right)=0$):

$$f\left(r\right)={\displaystyle \frac{\partial u_r}{\partial r}}+K_{\psi }{\displaystyle \frac{u_r}{r}}=0$$

(41)

By integrating the above equation with the boundary condition (r = R_p and $u_r=u_{R_p}$), it can be represented as

$${\int }_{u_r}^{u_{R_p}}{\displaystyle \frac{\partial u_r}{u_r}}=-K_{\psi }{\int }_r^{R_p}{\displaystyle \frac{\partial r}{r}}$$

(42)

Finally, the radial displacement with the homogeneous solution can be expressed as

$${\displaystyle \frac{u_r}{r}}={\displaystyle \frac{1}{r^{K_{\psi }+1}}}{u}_{R_p}{\left(R_p\right)}^{K_{\psi }}$$

(43)

where $u_{R_p}$is the radial displacement at the elastic-plastic interface (r = R_p) and can be given as

$$u_{R_p}={\displaystyle \frac{{\lambda }_e}{2}}\left[k_1+k_2\left(3-4v\right)\right]\left({\displaystyle \frac{{\sigma }_v{R}_p}{2G}}\right)$$

(44)

To consider in the normalized form, it can be obtained as

$${\displaystyle \frac{2G}{{\sigma }_v}}{\displaystyle \frac{u_r}{R}}={\displaystyle \frac{{\lambda }_e}{2}}\left[k_1+k_2\left(3-4v\right)\right]\left({\displaystyle \frac{r}{R}}\right){\left({\displaystyle \frac{R_p}{r}}\right)}^{K_{\psi }+1}$$

(45)

(2) Particular solution ($f\left(r\right)={\displaystyle \frac{\partial u_r}{\partial r}}+K_{\psi }{\displaystyle \frac{u_r}{r}}$):

According to the stresses obtained in the plastic region, equations (31) and (32), can be written in a new form as

$$f\left(r\right)={\displaystyle \frac{\partial u_r}{\partial r}}+K_{\psi }{\displaystyle \frac{u_r}{r}}={\displaystyle \frac{1}{2G}}\left[\left(1-v-\nu K_{\psi }\right)\left({\sigma }_r\right)+\left(K_{\psi }-v{K}_{\psi }-\nu \right)\left({\sigma }_{\theta }\right)\right]$$

(46)

By rearranging the above equation, therefore it can be represented as

$$f\left(r\right)={\displaystyle \frac{1}{2G}}\left[D_1+D_{2}ln\left({\displaystyle \frac{r}{R_p}}\right)+D_3{ln}^2\left({\displaystyle \frac{r}{R_p}}\right)\right]$$

(47)

where the coefficients D₁, D₂, and D₃ can be given as

$$D_1=\left(1-2v\right)\left(1+K_{\psi }\right){\displaystyle \frac{\left(1-{\lambda }_e\right)}{2}}\left(k_1+k_2\right)+\left(K_{\psi }-v{K}_{\psi }-\nu \right)\left[\left(k_1-k_2\right){\lambda }_e-k_2\right]$$

(48)

$$D_2=\left(1+K_{\psi }\right)\left(1-2v\right)\left[\left(k_1-k_2\right){\lambda }_e-k_2\right]+\left(K_{\psi }-v{K}_{\psi }-\nu \right)N{m}_b$$

(49)

$$D_3=\left(1+K_{\psi }\right)\left(1-2v\right)\left({\displaystyle \frac{N{m}_b}{2}}\right)$$

(50)

By combining the homogeneous solution equation (43) and the particular solution equation (47), the radial displacement in the plastic region can be expressed as

$${\displaystyle \frac{u}{r}}={\displaystyle \frac{1}{r^{K_{\psi }+1}}}{\int }_{R_p}^r{r}^{K_{\psi }}f\left(r\right)dr+{\displaystyle \frac{u_r}{r}}$$

(51)

In addition, the radial displacement in the plastic region can be obtained as

$${\displaystyle \frac{u}{r}}={\displaystyle \frac{1}{2G}}{\displaystyle \frac{1}{r^{K_{\psi }+1}}}\left[D_1{f}_1\left(r\right)+D_2{f}_2\left(r\right)+D_3{f}_3\left(r\right)+2G{u}_{R_p}{\left(R_p\right)}^{K_{\psi }}-D_1{f}_1\left(R_p\right)-D_2{f}_2\left(R_p\right)-D_3{f}_3\left(R_p\right)\right]$$

(52)

where $f_1\left(r\right)$, $f_2\left(r\right)$, and $f_3\left(r\right)$ can be given as

$$f_1\left(r\right)=\int r^{K_{\psi }}dr={\displaystyle \frac{r^{K_{\psi }+1}}{K_{\psi }+1}}$$

(53)

$$f_2\left(r\right)=\int r^{K_{\psi }}ln\left({\displaystyle \frac{r}{R_p}}\right)dr={\displaystyle \frac{r^{K_{\psi }+1}}{K_{\psi }+1}}\left[ln\left({\displaystyle \frac{r}{R_p}}\right)-{\displaystyle \frac{1}{K_{\psi }+1}}\right]$$

(54)

$$f_3\left(r\right)=\int r^{K_{\psi }}{ln}^2\left({\displaystyle \frac{r}{R_p}}\right)dr={\displaystyle \frac{r^{K_{\psi }+1}}{K_{\psi }+1}}\left[{ln}^2\left({\displaystyle \frac{r}{R_p}}\right)-{\displaystyle \frac{2}{K_{\psi }+1}}ln\left({\displaystyle \frac{r}{R_p}}\right)+{\displaystyle \frac{2}{{\left(K_{\psi }+1\right)}^2}}\right]$$

(55)

Finally, the closed-form analytical solution for the radial displacement in the plastic region can be obtained. For validation of this equation, one can examine the radial displacement at the elastic-plastic interface (r = R_p), then the same result is obtained as shown by the equation (44).

4. Utilization of an Incremental Procedure for the Analytical Solution

5. Comparison of Results between the Findings of this Study and Published Data

5.2. Stress/Displacement in the Plastic Region

Considering a circular tunnel excavation under anisotropic stress conditions, the key factor is the distribution of stress and displacement in the surrounding rock, particularly at the crown (θ = $0^{°}$), inclination (θ = ${45}^{°}$), and side-wall (θ = ${90}^{°}$) of the tunnel. For stress and displacement at the tunnel intrados, the characteristic behavior in Case IV (about the plastic region) is illustrated in Figure 10a, Figure 11a and Figure 12a. These figures indicate that the rock masses exhibit non-linear elastic-perfectly-plastic behavior in the ground reaction curve. Additionally, the dimensionless radial displacement is observed in the order of the crown, inclination, and side-wall of the tunnel.

As shown in Figure 1, the interface between the elastic region (line AB) and the plastic region (curve BC) is represented by point B. For the stress path, the Hoek-Brown failure criterion clearly demonstrates characteristic non-linearity along curve BG. In terms of the order in which the stress in the surrounding rock reaches the plastic state, the positions are the side-wall, inclination, and crown of the tunnel, as shown in Figure 10b, Figure 11b and Figure 12b, respectively. The results obtained using the EAM under anisotropic stress conditions differ significantly from those obtained under isotropic stress fields.

For the distribution of stress/displacement on the cross-section of the tunnel in anisotropic stress fields, one is interested in the stress/displacement distribution over the tunnel cross-section. According to the stress distribution results of the tunnel, the following five stress states as shown in Figure 13a, Figure 14a and Figure 15a can be used to describe and verify the stress changes caused by the advancing excavation of the tunnel face:

(1) When λ = 0 ($r\to \infty$), it indicates that the stresses are in the initial stress state as shown in the horizontal line on the right side in Figures; the radial stress (σ_h/σ_v) and tangential stress (σ_h/σ_v) used in the calculation are 0.67 and 1.0, respectively.

(2) When 0$\le \lambda \le {\lambda }_e$ ($R_p\le r\le \infty$), it describes that the stresses are in the elastic region, and the radial and tangential stresses increase with the increase of the confinement loss, and both stresses are separated along the horizontal axis (r/R).

(3) When $\lambda ={\lambda }_e$ ($r=R_p$), the plastic radius appears and stresses are at the elastic-plastic interface. The radial stress begins to change the curvature, and the tangential stress attains the maximum value.

(4) When ${\lambda }_e\le \lambda \le 1$ ($R\le r\le R_p$), it indicates that the stresses are in the plastic region, and this leads to both the radial stress and tangential stress being decreased steeply.

(5) Until $\lambda =1$ ($r=R$), the radial stress becomes zero and the tangential stress is equal to the coefficient $\sqrt{4s{N}^2}$ proposed by the Hoek-Brown failure criterion.

According to the analysis results comparing the EAM with the listed articles (shown in Table 2), the percentage error for radial displacement ranges from 0.93% to 1.18% in the elastic region and from 3.28% to 5.51% in the plastic region. Additionally, the percentage error for the radius of the plastic zone ranges from 4.99% to 5.6%. When comparing the results calculated by the EAM in this study with published data, the approximate results align well with the observed trend. The analysis reveals that tunnels in anisotropic stress fields experience different failure mechanisms compared to those in isotropic fields. This has implications for the design and support systems of tunnels, suggesting that standard practices may need to be adjusted to account for anisotropic conditions.

6. Conclusions

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