Analytic stress solution for a circular tunnel in half plane with a concentrated force

Abstract

An analytic stress solution is presented for a circular tunnel problem in a half plane with a concentrated force acting on any position in the field under gravity. The solution uses the complex variable method and the power series method. The influence of the unbalanced force system on the tunnel boundary is considered. The relationship between two analytic functions is established by using surface stress boundary condition. The analytic functions can be determined from the tunnel stress boundary condition. Based on the principle of superposition, the stresses of the surrounding rock can be calculated by superimposing three partial solutions which are obtained separately. The examples give contour plots of the principal stresses in the surrounding rock, focus on the stress distribution on the ground surface and the tunnel boundary and analyze the effect on the stress distribution of some main parameters.

Keywords

Analytic solution half plane concentrated force complex variable method stress tunnel

1. Introduction

In tunnel engineering, as an economical and practical support method, rockbolts can effectively improve the bearing capacity and control the deformation of surrounding rock, which is widely used in practical engineering [1 –5]. When addressing the issue of a tunnel with point anchored rockbolts, the rockbolts are usually simplified as two identical concentrated forces of opposite direction applied at the head and at the anchor points [1]. Moreover, when a tunnel is deeply buried, the problem can usually be simplified to an infinite plane strain problem. On this premise, Bobet [1] obtained the analytic solution for a circular tunnel in an infinite plane with a concentrated force, and then obtained the approximate solution of point anchored rockbolts in deep tunnels. When the depth of the tunnel is shallow, the problems become more complex to solve.

When the tunnel is buried at a shallow depth, the body force of rock mass should be considered, and the ground surface must be considered as a boundary. This kind of problem can only be simplified to a half plane problem. For the problem of shallow circular tunnels, Mindlin [6] first used a bipolar coordinate system to obtain the stress solutions of this problem under certain lateral stress coefficients. Verruijt et al. [7 –9] obtained an approximate solution by mapping the region in the z plane onto an annulus region in the ζ plane with the help of conformal mapping method. In order to reflect the unbalanced force system caused by excavation, certain terms were added into the complex potential functions. But the problem was solved based on the displacements specified along the tunnel boundary, which cannot reflect the true situation of tunnel excavation. On the basis of Verruijt’s researches, Lu et al. [10] obtained the analytic solution of the problem by using the stress boundary condition along the tunnel boundary, which is more in line with the actual situation.

However, the above studies [6 –10] only considered the body force of the rock mass without other loads. The situation becomes more complicated when loads are applied in the ground surface or the half plane domain. Flamant [11] firstly gave the stress expression for the case of a concentrated force acting on the half plane surface, namely Flamant’s solution. The classical stress solution for an elastic half plane loaded by a concentrated force was given by Melan as early as in 1932 [12]. For the problem of half plane weakened by a circular hole and with a concentrated force applied to the surface boundary, Proskura et al. [13] gave the stress and displacement fields for this problem using the bipolar coordinate system. Furthermore, for an elastic half plane with a tunnel loaded uniformly on the surface boundary, Yu et al. [14,15] and Fang et al. [16] presented an analytic solution for a circular tunnel to calculate the stress field and the settlement of the ground surface by using complex variable method. Kuliyev [17] presented an analytic solution for an elliptic hole with two rectilinear cracks based on the complex variable method and the principle of superposition. However, when a concentrated force acts on any position in the half plane, the solution process becomes more complicated and the exact analytical solutions for the problem are still unavailable in the current literatures.

Therefore, in this paper, an analytic solution is presented for a circular tunnel with a concentrated force acting at any position in the half plane. The solution considered the interaction of gravity and concentrated force in rock mass and the influence of the unbalanced force system caused by excavation based on the complex variable method and the principle of superposition. The mapping function and complex potential function proposed by Verruijt et al. [7 –9] were adopted. The proposed analytical solution has great value for the conceptual understanding of the mechanical behavior of a shallow circular tunnel with a concentrated force. Furthermore, it can also be applied in solving the problem of point anchored rockbolts in circular shallow tunnels.

2. Statement of problem and basic equations

2.1. Statement of problem

There is a circular tunnel with radius $R_{0}$ in the half plane, and a concentrated force $F = F_{x} + i F_{y}$ acts at any point $z_{0} = x_{0} + i y_{0}$ outside the tunnel (Figure 1), where $L_{s}$ is the ground surface, $L_{h}$ is the tunnel boundary, $γ$ is uniform volumetric weight of rock mass, $H_{1}$ is the depth of the tunnel, $H_{2}$ is the depth of the action point of concentrated force, and L is the abscissa difference between the action point of concentrated force and the center of tunnel. In this paper, the stress distribution in the half plane under this condition will be solved.

Figure 1.

Circular tunnel in a half plane with a concentrated force.

2.2. Relationship between $φ (z)$ and $ψ (z)$

For the two-dimensional elasticity problem, the stress boundary conditions can be described by two analytic functions, $φ (z)$ and $ψ (z)$ :

φ (z) + z \bar{φ' (z)} + \bar{ψ (z)} = i \int_{z_{1}}^{z} (X_{n} + i Y_{n}) ds + C

(1)

where $z_{1}$ and z are two points on the boundary, $X_{n}$ and $Y_{n}$ are the surface force components along the x and y directions, respectively, and C is an undetermined constant.

When no force acts on the surface boundary and let C = 0, the stress boundary condition on the ground surface is

φ (z) + z \bar{φ' (z)} + \bar{ψ (z)} = 0, z \in L_{s}

(2)

It can be obtained from equation (2) that

ψ (z) = - \bar{φ (z)} - \bar{z} φ' (z)

(3)

On the ground surface (y = 0), the imaginary part of any point z is zero, so $\bar{z} = z$ ; substituting this into equation (3), we have

ψ (z) = - \bar{φ} (z) - z φ' (z)

(4)

Equation (4) satisfies the stress boundary condition of the ground surface and the expression of $ψ (z)$ is an analytic function. Therefore, we can assume that equation (4) establishes in the entire half plane ( $y \leq 0$ ) [18]. On the basis of this assumption, it only needs to solve the expression of $φ (z)$ , which greatly simplifies the solving process. The expression of the stress boundary condition on the tunnel boundary can be obtained by substituting equation (4) into equation (1):

φ (z) - φ (\bar{z}) + (z - \bar{z}) \bar{φ' (z)} = i \int_{z_{1}}^{z} (X_{n} + i Y_{n}) ds + C, z \in L_{h}

(5)

The analytic function $φ (z)$ can be written in the following form [9]:

φ (z) = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [κ \ln (z - {\bar{z}}_{c}) + \ln (z - z_{c})] + φ_{0} (z)

(6)

where $F_{x}$ and $F_{y}$ are the resultant forces along the x and y directions on the tunnel boundary, $z_{c}$ is any point within the tunnel, and $φ_{0} (z)$ is single valued analytic function. For the plane strain problem discussed in this paper, $κ = 3 - 4 μ$ .

2.3. Conformal mapping to an annulus

Using the conformal mapping in the complex variable method, the region in the z plane can be mapped onto an annulus in the ζ plane, bounded by the circles $| ζ | = 1$ and $| ζ | = α$ , where $α < 1$ (Figure 2). The circle $| ζ | = 1$ corresponds to the ground surface and the circle $| ζ | = α$ corresponds to the tunnel boundary. The conformal mapping function is [7,8]

z = ω (ζ) = - i a \frac{1 + ζ}{1 - ζ}

(7)

Figure 2.

Plane of conformal transformation.

where $a = H (1 - α^{2}) / (1 + α^{2})$ ; $α$ can be obtained by

\frac{R_{0}}{H} = \frac{2 α}{1 + α^{2}}

(8)

Through mapping function $z = ω (ζ)$ , the analytic function $φ_{0} (z)$ can be written as a function of $ζ$ and expressed by Laurent series:

φ_{0} (z) = φ_{0} (ζ) = \sum_{k = 0}^{\infty} a_{k} ζ^{k} + \sum_{k = 1}^{\infty} b_{k} ζ^{- k}

(9)

3. The solving process

The solution of this problem can be obtained by the superposition of three partial solutions: The first part is the solution of a shallow circular tunnel under the action of gravity (Figure 3(a)). The second part is the solution of a concentrated force F acting at any point in the half plane without considering gravity and tunnel (Figure 3(b)). The process of excavation can be regarded as eliminating the forces along the boundary, or applying the equal but opposite surface traction on the boundary. Let $X_{n}$ and $Y_{n}$ denote the surface force components on the boundary of the proposed tunnel, which is caused by the concentrated force F of the second part (Figure 4(a)). Then the third part is the solution of a circular tunnel in the half plane without considering the action of gravity, and the boundary of the circular tunnel is applied with the surface force components which is equal but opposite to $X_{n}$ and $Y_{n}$ (Figure 4(b)). The latter two parts can be superposed to obtain the solution of a circular tunnel in half plane with a concentrated force acting at any point of the field, which is the focus of the solution in this paper. Then the solution of the problem in this paper can be calculated by superimposing the solutions of the three parts.

Figure 3.

(a) Half plane with a circular tunnel. (b) Half plane with a concentrated force.

Figure 4.

(a) Surface force components caused by the concentrated force acting on the proposed excavated boundary. (b) Applied equal and opposite surface force components on the boundary.

3.1. First partial solution

Before excavation, the stress distribution in the half plane is

{\begin{matrix} σ_{x}^{0} = k_{0} γ y \\ σ_{y}^{0} = γ y \\ τ_{xy}^{0} = 0 \end{matrix}

(10)

where $k_{0}$ is the lateral stress coefficient.

Excavation of a tunnel is equivalent to applying the equal but opposite surface traction on the boundary. Under the action of the surface traction without considering the gravity, the analytic function $φ_{1} (z)$ can be determined by equation (5). It can be known from equation (6) that the expression of $φ_{1} (z)$ is

φ_{1} (z) = - \frac{i R_{0}^{2} γ}{2 (1 + κ)} [κ \ln (z - i a) + \ln (z + i a)] + \sum_{k = 0}^{\infty} a_{k} ζ^{k} + \sum_{k = 1}^{\infty} b_{k} ζ^{- k}

(11)

where $a = H_{1} (1 - α^{2}) / (1 + α^{2})$ , $α$ can be obtained from equation (8). For the solution of $φ_{1} (z)$ , Lu et al. [10] gave a reasonable solution.

3.2. Second partial solution

For the second part, Melan [12] has given the stress expressions for the problem. However, Melan didn’t use the complex variable method so that the analytic function $φ_{2} (z)$ for the second part, which is needed for the third part, was not provided in Melan’s paper. To obtain $φ_{2} (z)$ , the problem of the second part is solved by the complex variable method in this section.

It’s difficult to solve the problem directly by the complex variable method. Therefore, assume that there is a virtual circular hole with the point $z_{0}$ (the action point of the concentrated force) as the center in the half plane (Figure 5). The radius of the virtual circular hole is $R'$ . Then applying uniform surface forces $X_{n}' + i Y_{n}'$ parallel to the concentrated force F on the boundary of the virtual circular hole, where $X_{n}' = F_{x} / (2 π R')$ , $Y_{n}' = F_{y} / (2 π R')$ . When the radius $R'$ is equal to zero and $F_{x}$ and $F_{y}$ remain unchanged, the solution required in the second part can be obtained.

Figure 5.

Parallel surface forces acting on the virtual circular hole.

A new coordinate system $x' o' y'$ is established with the ground surface as the x-axis and the vertical line passing through the point $z_{0}$ as the y-axis. In the new coordinate system $x' o' y'$ , mapping the region in the z plane onto an annulus. From equation (7), the mapping function can be obtained as follows:

z' = ω (ζ') = - i a' \frac{1 + ζ'}{1 - ζ'}

(12)

where $z'$ indicates the point in the new coordinate system, $a' = H_{2} (1 - α'^{2}) / (1 + α'^{2})$ , and $α'$ can be obtained by equation (8).

Let the complex potential function corresponding to the case shown in Figure 5 is $φ_{2} (z)$ in the original coordinate system xoy and $φ_{(2)} (z')$ in the new coordinate system $x' o' y'$ . Then the stress boundary condition on the whole boundary can be obtained by equation (5):

φ_{(2)} (z') - φ_{(2)} (\bar{z'}) + (z' - \bar{z'}) \bar{{φ'}_{(2)} (z')} = i \int_{z_{1}'}^{z'} (X_{n}' + i Y_{n}') ds + C, z' \in L_{h}'

(13)

where the expression of $φ_{(2)} (z')$ is

φ_{(2)} (z') = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [κ \ln (z' - i a') + \ln (z' + i a')] + \sum_{k = 0}^{\infty} c_{k} ζ'^{k} + \sum_{k = 1}^{\infty} d_{k} ζ'^{- k}

(14)

Substituting equations (12) and (14) into equation (13) gives

\begin{matrix} \sum_{k = 0}^{\infty} c_{k} ({α'}^{k} - {α'}^{- k} {σ'}^{k} + \sum_{k = 1}^{\infty} d_{k} ({α'}^{- k} - {α'}^{k}) {σ'}^{- k} \\ - \frac{(1 - {α'}^{2}) (σ' - α')}{1 - α' σ'} \sum_{k = 1}^{\infty} (k \bar{c_{k}} {α'}^{k - 1} {σ'}^{- k} - k \bar{d_{k}} {α'}^{- k - 1} {σ'}^{k}) \\ = i \int_{{z'}_{1}}^{z'} (X_{n}' + i Y_{n}') ds - \frac{W (κ α' + σ')}{α_{1} (1 - α' σ')} + \frac{F_{x} + i F_{y}}{2 π} \ln \frac{α' - σ'}{1 - α' σ'} + C \end{matrix}

(15)

where $W = \frac{(F_{x} - i F_{y}) (1 - {α'}^{2})}{2 π (1 + κ)}$ .

From Figure 5, we have

{\begin{matrix} θ_{2}' = θ_{1}' - θ' \\ R' e^{i θ'} = z' + i H_{2} \\ R' e^{i θ_{1}'} = z_{1}' + i H_{2} \end{matrix}

(16)

From equation (16), we have

θ_{2}' = i \ln (z' + i H_{2}) - i \ln (z_{1}' + i H_{2})

(17)

If $z'$ is at the boundary of the circular hole, then

z' = - i H_{2} \frac{1 - {α'}^{2}}{1 + {α'}^{2}} \frac{1 + α' σ'}{1 - α' σ'}

(18)

Substituting equation (18) into equation (17) gives

θ_{2}' = i [\ln \frac{α' - σ'}{1 - α' σ'} + \ln \frac{2 i α' H_{2}}{1 + {α'}^{2}} - \ln (z_{1}' + i H_{2})] = i \ln \frac{α' - σ'}{1 - α' σ'} + C

(19)

where C is a complex constant. From equation (19), we have

i \int_{z_{1}'}^{z'} (X_{n}' + i Y_{n}') ds = i (X_{n}' + i Y_{n}') R' θ_{2}' = - \frac{F_{x} + i F_{y}}{2 π} (\ln \frac{α' - σ'}{1 - α' σ'} + iC)

(20)

Substituting equation (20) into equation (15) gives

\begin{matrix} \sum_{k = 0}^{\infty} c_{k} ({α'}^{k} - {α'}^{- k}) {σ'}^{k} + \sum_{k = 1}^{\infty} d_{k} ({α'}^{- k} - {α'}^{k} {σ'}^{- k} \\ - \frac{(1 - {α'}^{2}) (σ' - α')}{1 - α' σ'} \sum_{k = 1}^{\infty} (k \bar{c_{k}} {α'}^{k - 1} {σ'}^{- k} - k \bar{d_{k}} {α'}^{- k - 1} {σ'}^{k}) \\ = - \frac{W (κ α' + σ')}{α' (1 - α' σ')} + C' \end{matrix}

(21)

In order to solve the problem more easily, multiplying both sides of equation (21) by $(1 - α' σ')$ gives

\begin{matrix} (1 - α' σ') [\sum_{k = 0}^{\infty} c_{k} ({α'}^{k} - {α'}^{- k}) {σ'}^{k} + \sum_{k = 1}^{\infty} d_{k} ({α'}^{- k} - {α'}^{k}) {σ'}^{- k}] \\ - (1 - {α'}^{2}) (σ' - α') [\sum_{k = 1}^{\infty} (k \bar{c_{k}} {α'}^{k - 1} {σ'}^{- k} - k \bar{d_{k}} {α'}^{- k - 1} {σ'}^{k})] \\ = - \frac{W (κ α' + σ')}{α'} + C' (1 - α' σ') \end{matrix}

(22)

Comparing the coefficients of $σ'$ in both sides of equation (22), the following equations can be obtained:

σ'^{k} \to (α'^{2} - α'^{2 k}) c_{k - 1} + (k - 1) (1 - α'^{2}) \bar{d_{k - 1}} + (α'^{2 k} - 1) c_{k} - k (1 - α'^{2}) \bar{d_{k}} = 0, k \geq 3

(23)

σ'^{- k} \to k (1 - α'^{2}) α'^{2 k} c_{k} + (1 - α'^{2 k}) \bar{d_{k}} - (k + 1) (1 - α'^{2}) α'^{2 k} c_{k + 1} + (α'^{2 k + 2} - 1) \bar{d_{k + 1}} = 0, k \geq 2

(24)

σ'^{2} \to (α'^{2} - α'^{4}) c_{1} + (1 - α'^{2}) \bar{d_{1}} + (α'^{2} - 1) c_{2} - 2 (1 - α'^{2}) \bar{d_{2}} = 0

(25)

σ'^{1} \to (1 - α'^{2}) (c_{1} + \bar{d_{1}}) = W + α'^{2} C'

(26)

σ'^{0} \to (1 - α'^{2}) (c_{1} + \bar{d_{1}}) = κ \bar{W} - \bar{C'}

(27)

σ'^{1} \to (α'^{2} - α'^{4}) c_{1} + (1 - α'^{2}) \bar{d_{1}} - 2 (α'^{2} - α'^{4}) c_{2} + (α'^{4} - 1) \bar{d_{2}} = 0

(28)

From equations (25) and (28), the equation without the unknowns $c_{2}$ and $d_{2}$ can be obtained;

c_{2} + \bar{d_{2}} = 0

(29)

Assuming that the highest positive and negative powers of $ζ'$ in $φ_{(2)} (z')$ is N, then combining equations (23), (24), and (29), there are $2 N - 2$ equations for $2 N - 2$ undetermined coefficients ( $c_{k}, d_{k}, k = 2, 3, \dots, N$ ). The resulting system of equations is homogeneous. Homogeneous systems of equations must have zero solutions and the analytic function is unique, so we have

c_{k} = d_{k} = 0, 2 \leq k \leq N

(30)

Therefore, only the following equations need to be solved:

{\begin{matrix} ({α'}^{2} - {α'}^{4}) c_{1} + (1 - {α'}^{2}) \bar{d_{1}} = 0 \\ (1 - {α'}^{2}) (c_{1} + \bar{d_{1}}) = W + {α'}^{2} C' \\ (1 - {α'}^{2}) (c_{1} + \bar{d_{1}}) = κ \bar{W} - \bar{C'} \end{matrix}

(31)

From equation (31), we have

{\begin{matrix} Re (C') = - \frac{(1 - κ) Re (W)}{1 + {α'}^{2}} \\ Im (C') = \frac{(1 - κ) Im (W)}{1 - {α'}^{2}} \\ c_{1} = \frac{W + {α'}^{2} C'}{{(1 - {α'}^{2})}^{2}} \\ \bar{d_{1}} = - \frac{{α'}^{2} (W + {α'}^{2} C')}{{(1 - {α'}^{2})}^{2}} \end{matrix}

(32)

When the radius $R'$ is equal to zero, then $α' = 0$ . Letting $α' = 0$ and substituting equations (12), (30), and (32) into equation (14):

φ_{(2)} (z') = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [κ \ln (z' - i a') + \ln (z' + i a')] + \frac{F_{x} - i F_{y}}{2 π (1 + κ)} \frac{z' + i H_{2}}{z' - i H_{2}}

(33)

Therefore, the derivative of $φ_{(2)} (z')$ is

φ_{(2)}' (z') = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [\frac{κ}{z' - i a'} + \frac{1}{z' + i a'}] - \frac{F_{x} - i F_{y}}{2 π (1 + κ)} \frac{2 i H_{2}}{{(z' - i H_{2})}^{2}}

(34)

Due to the stress components at each point are independent of the translation of the coordinate system, the expression of $φ_{2}' (z)$ can be derived as follows:

φ_{2}' (z) = φ_{(2)}' (z') = φ_{(2)}' (z - L) = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [\frac{κ}{z - L - i a'} + \frac{1}{z - L + i a'}] - \frac{F_{x} - i F_{y}}{2 π (1 + κ)} \frac{2 i H_{2}}{{(z - L - i H_{2})}^{2}}

(35)

Integrating equation (35) and neglecting the integral constant which does not affect the stress, we have

φ_{2} (z) = - \frac{F_{x} + i F_{y}}{2 π (1 + κ)} [κ \ln (z - L - i a') + \ln (z - L + i a')] + \frac{F_{x} - i F_{y}}{2 π (1 + κ)} \frac{2 i H_{2}}{z - L - i H_{2}}

(36)

Equation (36) is the complex potential function in the second part.

3.3. Third partial solution

For the second part, the following equation is established on the boundary $L_{h}$ of the proposed excavated tunnel:

φ_{2} (z) - φ_{2} (\bar{z}) + (z - \bar{z}) \bar{φ_{2}' (z)} = i \int_{z_{1}}^{z} (X_{n} + i Y_{n}) ds + C_{1}, z \in L_{h}

(37)

$φ_{3} (z)$ is used to represent the complex potential function corresponding to the third part. The third part does not consider the effect of gravity, and the force components $X_{n}$ and $Y_{n}$ caused by the concentrated force F on the tunnel boundary are a set of equilibrium force systems, so $φ_{3} (z)$ does not contain the logarithmic term and the expression of $φ_{3} (z)$ is

φ_{3} (z) = φ_{3} (ζ) = \sum_{k = 0}^{\infty} e_{k} ζ^{k} + \sum_{k = 1}^{\infty} f_{k} ζ^{- k}

(38)

Moreover, $φ_{3} (z)$ should satisfy the following equation:

φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)} = i \int_{z_{1}}^{z} (- X_{n} - i Y_{n}) ds + C_{2}, z \in L_{h}

(39)

Substituting equation (38) into equation (39), we have

φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)} = - [φ_{2} (z) - φ_{2} (\bar{z}) + (z - \bar{z}) \bar{φ_{2}' (z)} + C_{3}], z \in L_{h}

(40)

Multiplying both sides of equation (21) by $(1 - α σ)$ gives

(1 - α σ) [φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)}] = - (1 - α σ) [φ_{2} (z) - φ_{2} (\bar{z}) + (z - \bar{z}) \bar{φ_{2}' (z)} + C_{3}], z \in L_{h}

(41)

Substituting equations (7) and (36) into equation (41), we have

(1 - α σ) [φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)}] = \sum_{k = 0}^{\infty} A_{k} σ^{k} + \sum_{k = 1}^{\infty} B_{k} σ^{- k} + (1 - α σ) C_{3}

(42)

The expressions for $A_{k}$ and $B_{k}$ in the equation (42) are given in the Appendix.

Comparing the coefficients of σ in both sides of equation (42), the following equations can be obtained:

σ^{k} \to (α^{2} - α^{2 k}) e_{k - 1} + (k - 1) (1 - α^{2}) \bar{f_{k - 1}} + (α^{2 k} - 1) e_{k} - k (1 - α^{2}) \bar{f_{k}} = α^{k} A_{k}, k \geq 2

(43)

σ^{- k} \to k (1 - α^{2}) α^{2 k} e_{k} + (1 - α^{2 k}) \bar{f_{k}} - (k + 1) (1 - α^{2}) α^{2 k} e_{k + 1} + (α^{2 k + 2} - 1) \bar{f_{k + 1}} = α^{k} \bar{B_{k}}, k \geq 1

(44)

σ^{1} \to (1 - α^{2}) (e_{1} + \bar{f_{1}}) = - α A_{1} + α^{2} C_{3}

(45)

σ^{0} \to (1 - α^{2}) (e_{1} + \bar{f_{1}}) = - \bar{A_{0}} - \bar{C_{3}}

(46)

From equations (45) and (46), we have

Re (C_{3}) = Re (\frac{α A_{1} - \bar{A_{0}}}{α^{2} + 1}), Im (C_{3}) = Im (\frac{α A_{1} - \bar{A_{0}}}{α^{2} - 1})

(47)

Assuming that the highest positive and negative powers of ζ in $φ_{3} (z)$ is N, then combining equations (43)–(45) and (47), there are $2 N$ equations for $2 N$ undetermined coefficients ( $e_{k}, f_{k}, k = 1, 2, \dots, N$ ). These undetermined coefficients can be solved by the following complex linear equation set:

AX = B

(48)

where

X = {(e_{1}, \bar{f_{1}}, e_{2}, \bar{f_{2}}, . . ., e_{N}, \bar{f_{N}})}^{T}

(49)

4. Solving formulas for stresses

Superimposing three parts of the complex potential functions:

φ (z) = φ_{1} (z) + φ_{2} (z) + φ_{3} (z)

(50)

The stress components $σ_{x}^{*}$ , $σ_{y}^{*}$ , and $τ_{xy}^{*}$ at any point in the half plane can be determined by

σ_{x}^{*} + σ_{y}^{*} = 4 Re [φ' (z)]

(51)

σ_{y}^{*} - σ_{x}^{*} + 2 i τ_{xy}^{*} = 2 [\bar{z} φ ″ z) + ψ' z]

(52)

Substituting equation (4) into equation (52) gives

σ_{y}^{*} - σ_{x}^{*} + 2 i τ_{xy}^{*} = 2 [(\bar{z} - z) φ ″ z) - φ' z) - \bar{φ'} z)]

(53)

Superimposing $σ_{x}^{*}$ , $σ_{y}^{*}$ , and $τ_{xy}^{*}$ with $σ_{x}^{0}$ , $σ_{y}^{0}$ , and $τ_{xy}^{0}$ given by equation (10), the stress solution in Cartesian coordinates for a circular tunnel in a half plane with a concentrated force acting at any position in the field under gravity is obtained:

{\begin{matrix} σ_{x} = σ_{x}^{0} + σ_{x}^{*} \\ σ_{y} = σ_{y}^{0} + σ_{y}^{*} \\ τ_{xy} = τ_{xy}^{0} + τ_{xy}^{*} \end{matrix}

(54)

The stress components $σ_{ρ}$ , $σ_{θ}$ , and $τ_{ρ θ}$ in orthogonal curvilinear coordinates can be obtained by [19]

{\begin{matrix} σ_{θ} + σ_{ρ} = σ_{x} + σ_{y} \\ σ_{θ} - σ_{ρ} + 2 i τ_{ρ θ} = (σ_{y} - σ_{x} + 2 i τ_{xy}) e^{2 i α} \end{matrix}

(55)

where $e^{2 i α} = \frac{ζ^{2} ω' (ζ)}{ρ^{2} \bar{ω' (ζ)}}$ .

5. Discussion on accuracy

The analytic function $φ (z)$ is obtained on the basis of equation (4), so it must satisfy the stress boundary condition on the ground surface L_s. However, for the series in the analytic function $φ (z)$ , infinite terms are needed theoretically to accurately satisfy the stress boundary condition on the tunnel boundary L_h, but only finite terms can be employed in practical calculation. The accuracy of the first part has been discussed by Lu et al. [10] and the second part is an exact solution. Therefore, this section focuses on the value of N in the analytic function $φ_{3} (z)$ required to satisfy the stress boundary condition on the tunnel boundary L_h. If the obtained stresses $σ_{ρ}$ and $τ_{ρ θ}$ on the tunnel boundary L_h are equal to zero, it suggests that the stress boundary condition on L_h is satisfied accurately. The following equation is used as the criterion for determining the value of N:

max_{θ \in [0, 2 π]} {| σ_{ρ} |, | τ_{ρ θ} |} < 0.001 MPa

(56)

Let μ = 0.3, $γ = 25 kN / m^{3}$ , R₀ = 5 m, k₀ = 0.5, F_x = F_y = 500 kN, and L = 0 m (Figure 6). Table 1 shows the values of N required for different H₁ and $H_{2} - H_{1}$ . From Table 1 we can see that the value of N decreases gradually with the increase of H₁ and $H_{2} - H_{1}$ .

Figure 6.

Concentrated force acting in the half plane.

Table 1.

Values of N required for different H₁ and $H_{2} - H_{1}$ .

H ₁	6 m	8 m	10 m	15 m	30 m
H ₂−H₁ = 6 m	269	159	130	105	89
H ₂−H₁ = 8 m	90	53	43	34	30
H ₂−H₁ = 10 m	58	34	27	22	19
H ₂−H₁ = 15 m	35	20	16	13	11
H ₂−H₁ = 30 m	22	12	10	8	7

6. Examples

6.1. The principal stresses of surrounding rock

The following example discusses the case of a vertical downward concentrated force acting at coordinate origin o. The specific parameters are shown in Table 2. The tensile stress is positive and the compressive stress is negative in this paper.

Table 2.

Specific parameters for calculation of principal stresses.

μ	γ/kN m⁻³	R₀/m	H₁/m	H₂/m	L/m	F_x/kN	F_y/kN
0.3	25	5	20	0	0	0	−1000

The principal stresses $σ_{1}$ and $σ_{2}$ can be obtained from the stress components $σ_{x}$ , $σ_{y}$ , and $τ_{xy}$ as follows:

\begin{matrix} σ_{1} \\ σ_{2} \end{matrix} = \frac{σ_{x} + σ_{y}}{2} \pm \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + τ_{xy}^{2}}

(57)

Figures 7 and 8 show contour plots of principal stresses $σ_{1}$ and $σ_{2}$ in surrounding rock when the lateral stress coefficient $k_{0}$ is 0.3 and 0.5, respectively. The range of the surrounding rock is 100 m and 70 m in the horizontal and vertical directions, respectively. It can be seen from the figures that the contours near the action point (original point o) of the concentrated force are dense; the stress concentration is obvious near the point. Besides the action point of the concentrated force, the stress concentration around the tunnel is also obvious. When $k_{0} = 0.3$ , the maximum of major principal stress $σ_{1}$ appears at the top (0.15 MPa) and bottom (0.12 MPa) of the tunnel, the minimum of minor principal stress $σ_{2}$ appears at the middle of side walls (−1.48 MPa). When $k_{0} = 0.5$ , the values of major principal stress $σ_{1}$ are all equal to zero on the tunnel boundary. The maximum of $σ_{2}$ appears at the top (−0.12 MPa) and bottom (−0.24 MPa) of the tunnel, the minimum of $σ_{2}$ appears at the middle of side walls (−1.38 MPa). At a distance from the tunnel, the distribution of principal stress is basically consistent with the original in-situ stress, which is not affected by tunnel excavation.

Figure 7.

For k₀ = 0.3, contour plots for the principal stresses: (a) major principal stress σ₁; (b) minor principal stress σ₂.

Figure 8.

For k₀ = 0.5, contour plots for the principal stresses: (a) major principal stress σ₁; (b) minor principal stress σ₂.

6.2. Effect of depth H₁ on stress

Figure 9 presents the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary for different values of H₁; the other parameters are the same as in Table 2. The tangential stress $σ_{θ}$ on the ground surface is the horizontal normal stress $σ_{x}$ as well. The tangential stress $σ_{θ}$ in the region of [−20 m, 20 m] is discussed. (The stress at the action point of concentrated force tends to be infinite, so the tangential stress at the action point is neglected in Figure 9. The distribution of the tangential stress on the ground surface is discussed except for the action point of concentrated force.).= Due to the load and the tunnel are symmetric about the y-axis, only the right half cavity $θ \in [0^{°}, 180^{°}]$ is discussed. $θ = 0^{°}$ corresponds to the middle point at the top of the tunnel and $θ = 180^{°}$ corresponds to the middle point at the bottom of the tunnel.

Figure 9.

Distribution of σ_θ for various H₁ values: (a) ground surface; (b) tunnel boundary.

Figure 9(a) presents the tangential stress $σ_{θ}$ on the ground surface for different value of H₁. We can see that stress concentration occurs near the original point o due to the concentrated force. The smaller the burial depth H₁, the more obvious the stress concentration is. The tangential stress $σ_{θ}$ on the ground surface is compressive stress near the original point o and turns to tensile stress in the near field. The absolute value of $σ_{θ}$ decreases with the increase of H₁ within the vicinity of the original point o. And as the distance from the origin point o increases, the tangential stress $σ_{θ}$ gradually approaches to zero.

Figure 9(b) gives the distribution of $σ_{θ}$ in the right half cavity for different value of H₁. It can be seen that, the tangential stress $σ_{θ}$ at the top of tunnel is larger than that at the bottom of tunnel. When H₁ is small, $σ_{θ}$ at the top of tunnel is tensile stress and turns to compressive stress in the near field, the value of $σ_{θ}$ at the top of tunnel decreases as H₁ increases. In particular, when H₁ is very small, the overlying rock mass of the tunnel is thin, the concentrated force and the ground surface boundary have a greater impact on the stress distribution of tunnel boundary. Therefore, the trend of the curve H₁=8m is different from others apparently.

6.3. Effect of concentrated force F on stress

Figure 10 presents the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary for different values of F_y; other parameters are the same as in Table 2.

Figure 10.

Distribution of σ_θ for various F_y values: (a) ground surface; (b) tunnel boundary.

Figure 10(a) shows that there is great compressive stress near the original point o due to the concentrated force. The maximum value of $σ_{θ}$ in tension stress zone increases with the increase of the concentrated force F_y (when F_y = 3000 kN, the maximum value of $σ_{θ}$ is 0.88 MPa). When F_y = 0 kN, i.e., there is no load acting on the ground surface, the absolute value of $σ_{θ}$ near the original point o is much smaller than other curves.

Figure 10(b) shows that all curves converge at one point at about 26° (0.35 MPa) and 147° (0.62 MPa). The tangential stress $σ_{θ}$ at the top of the tunnel increases as F_y increases, but $σ_{θ}$ at the middle of side walls increases as F_y decreases. The minimum of $σ_{θ}$ appears at the middle of side walls (when F_y = 3000 kN, the minimum value of $σ_{θ}$ is −1.55 MPa). The value of the concentrated force F_y has less effect on $σ_{θ}$ near the bottom of the tunnel.

6.4. Effect of H₂ on stress

Figure 11 presents the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary for different value of H₂, other parameters are the same as in Table 2. H₂ reflects the vertical position of the concentrated force. When $0 m \leq H_{2} \leq 15 m$ , the concentrated force acts above the tunnel. When $H_{2} \geq 25 m$ , the concentrated force acts below the tunnel.

Figure 11.

Distribution of σ_θ for various H₂ values: (a) ground surface; (b) tunnel boundary.

Figure 11(a) shows that when the concentrated force acts above the tunnel ( $0 m \leq H_{2} < 15 m$ ), the compressive stress $σ_{θ}$ on the original point o increases as H₂ increases. The closer the position of the concentrated force is to the top of the tunnel, the more significant the stress concentration becomes. When the concentrated force acts below the tunnel ( $H_{2} > 25 m$ ), the absolute value of $σ_{θ}$ is much smaller than other curves, the concentrated force has little effect on the stress distribution on the ground surface.

Figure 11(b) shows that the maximum of the tangential stress $σ_{θ}$ appears at the top of the tunnel and the minimum of $σ_{θ}$ appears at the middle of side walls. When $0 m \leq H_{2} \leq 15 m$ , the curves are different in the small range among the top of the tunnel and the curves almost coincide in the remaining range. The tangential stress $σ_{θ}$ at the top of the tunnel increases as the value of H₂ increases, even when H₂ = 13 m, the tangential stress $σ_{θ}$ at the top of the tunnel is tensile stress (0.12 MPa). Instead, when $H_{2} \geq 25 m$ , the curves are only different in the small range among the bottom of the tunnel. The tangential stress $σ_{θ}$ at the bottom of the tunnel increases with the value of H₂ increases. The closer the concentration force is applied to the tunnel, the greater the effect on the tangential stress $σ_{θ}$ on the tunnel boundary.

6.5. Effect of depth L on stress

Figure 12 presents the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary for different value of L, other parameters are the same as in Table 2. L reflects the horizontal position of the concentrated force. When $L \neq 0 m$ , $σ_{θ}$ on the ground surface and the tunnel boundary are no longer symmetrical to the y-axis.

Figure 12.

Distribution of σ_θ for various L values: (a) ground surface; (b) tunnel boundary.

Figure 12 shows that when L = 0 m, the minimum value of $σ_{θ}$ appears at the action point of the concentrated force. However, when $L \neq 0 m$ , the minimum value of $σ_{θ}$ does not appear at the action point of the concentrated force, but between the action point of the concentrated force and the original point o. And the minimum value of $σ_{θ}$ increases with the increase of L. When $L > 20 m$ the curves get more closer, which suggests that the concentrated force has little effect on distribution of $σ_{θ}$ on the ground surface within the region [−10 m, 10 m] as L is more than four times of the radius R₀.

Figure 12 shows that when $L > 15 m$ , the concentrated force has little effect on the distribution of $σ_{θ}$ on the tunnel boundary. When k₀ = 1.0, the influence of L on the tunnel boundary is most obvious.

7. Conclusion

An analytical stress solution for an elastic half plane with a circular tunnel, which is subjected to a concentrated force acting at any position, is derived by using a complex variable method in this paper. The relationship between the two analytic functions $φ (z)$ and $ψ (z)$ is established by surface boundary condition, which simplifies the calculation process. The solution consists of three partial solutions. The analytic function expressions for each part are calculated and the final solution can be obtained by superimposing the three partial solutions. The examples presented the contour plots for the principal stresses $σ_{1}$ and $σ_{2}$ . Moreover, the examples show the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary for different parameters and analyzed the effect on the stress distribution of some main parameters.

The examples show that for the tangential stress $σ_{θ}$ on the ground surface, with the increase of H₁ and F_y, there is great compressive stress above the tunnel and great tensile stress near the compressive stress zone. The closer the position of the concentrated force is to the top of the tunnel, the larger the absolute value of $σ_{θ}$ . For the tangential stress $σ_{θ}$ on the tunnel boundary, $σ_{θ}$ at the top of tunnel is larger than that at the bottom of tunnel. The value of $σ_{θ}$ at the top and the bottom of the tunnel increases as the value of H₁ decreases or F_y increases. And the value of $σ_{θ}$ at the middle of side walls increases as the value of F_y decreases. The closer the concentrated force is to the tunnel, the greater the tangential stress $σ_{θ}$ on the ground surface and the tunnel boundary is affected.

Footnotes

Appendix

Let

$T_{1} = α \frac{L + i (H_{1} - a)}{L + i (H_{1} + a)}$

$T_{2} = α \frac{L - i (H_{1} + a)}{L - i (H_{1} - a)}$

$T_{3} = α \frac{L + i (H_{1} + a)}{L + i (H_{1} - a)}$

$T_{4} = α \frac{L - i (H_{1} - a)}{L - i (H_{1} + a)}$

$U_{1} = \frac{F_{x}' + i F_{y}'}{2 π (1 + κ)}$

$U_{2} = \frac{F_{x}' - i F_{y}'}{2 π (1 + κ)}$

$W_{1} = - \frac{2 i H_{1}}{L + i (H_{1} + a)}$

$W_{2} = - \frac{2 i H_{1}}{L + i (H_{1} - a)}$

$W_{3} = - \frac{1}{L - i (H_{1} + a)}$

$W_{4} = - \frac{1}{L + i (H_{1} - a)}$

$W_{5} = \frac{2 i H_{1}}{{[L - i (H_{1} + a)]}^{2}}$

Substituting equations (7) and (36) into equation (41) gives

(A-1)

\begin{matrix} (1 - α σ) [φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)}] \\ = - U_{1} (1 - α σ) {\begin{matrix} κ [\ln (1 - T_{1} σ) - \ln (1 - α σ)] + [\ln (1 - T_{2} σ) - \ln (1 - α σ)] \\ - κ [\ln (1 - \frac{T_{3}}{σ}) - \ln (1 - \frac{α}{σ})] - [\ln (1 - \frac{T_{4}}{σ}) - \ln (1 - \frac{α}{σ})] \end{matrix}} \\ + U_{2} (1 - α σ) (W_{1} \frac{1 - α σ}{1 - T_{1} σ} - W_{2} \frac{1 - \frac{α}{σ}}{1 - \frac{T_{3}}{σ}}) \\ + 2 i a (1 - α^{2}) [U_{2} (W_{3} \frac{κ}{1 - \frac{T_{4}}{σ}} + W_{4} \frac{1}{1 - \frac{T_{3}}{σ}}) - U_{1} W_{5} \frac{1 - \frac{α}{σ}}{{(1 - \frac{T_{4}}{σ})}^{2}}] \end{matrix}

Each item in equation (A-1) can be expanded by using the following Taylor expansion formula:

(A-2)

{\begin{matrix} \ln (1 - z) = - \sum_{k = 1}^{\infty} \frac{z^{k}}{k} \\ \frac{1}{1 - z} = \sum_{k = 0}^{\infty} z^{k} \\ \frac{1}{{(1 - z)}^{2}} = \sum_{k = 1}^{\infty} k z^{k - 1} \end{matrix}, | z | < 1

Expanding equation (A-1), we have

(A-3)

\begin{matrix} (1 - α σ) [φ_{3} (z) - φ_{3} (\bar{z}) + (z - \bar{z}) \bar{φ_{3}' (z)}] \\ = - U_{1} (1 - α σ) [\begin{matrix} κ (- \sum_{k = 1}^{\infty} \frac{T_{1}^{k} σ^{k}}{k} + \sum_{k = 1}^{\infty} \frac{α^{k} σ^{k}}{k}) + (- \sum_{k = 1}^{\infty} \frac{T_{2}^{k} σ^{k}}{k} + \sum_{k = 1}^{\infty} \frac{α^{k} σ^{k}}{k}) \\ - κ (- \sum_{k = 1}^{\infty} \frac{T_{3}^{k} σ^{- k}}{k} + \sum_{k = 1}^{\infty} \frac{α^{k} σ^{- k}}{k}) - (- \sum_{k = 1}^{\infty} \frac{T_{4}^{k} σ^{- k}}{k} + \sum_{k = 1}^{\infty} \frac{α^{k} σ^{- k}}{k}) \end{matrix}] \\ + U_{2} (1 - α σ) [W_{1} (1 - α σ) \sum_{k = 0}^{\infty} T_{1}^{k} σ^{k} - W_{2} (1 - \frac{α}{σ}) \sum_{k = 0}^{\infty} T_{3}^{k} σ^{- k}] \\ + 2 i a (1 - α^{2}) [U_{2} (κ W_{3} \sum_{k = 0}^{\infty} T_{4}^{k} σ^{- k} + W_{4} \sum_{k = 0}^{\infty} T_{3}^{k} σ^{- k}) - U_{1} W_{5} (1 - \frac{α}{σ}) \sum_{k = 1}^{\infty} T_{4}^{k - 1} σ^{- k + 1}] \end{matrix}

From equations (A-3) and (42), we have

(A-4)

\begin{matrix} A_{k} = - U_{1} [κ (\frac{T_{1}^{k} - α^{k}}{k} - \frac{α T_{1}^{k - 1} - α^{k}}{k - 1}) + (\frac{T_{2}^{k} - α^{k}}{k} - \frac{α T_{2}^{k - 1} - α^{k}}{k - 1})] \\ - U_{2} W_{1} (T_{1}^{k} - 2 α T_{1}^{k - 1} + α^{2} T_{1}^{k - 2}), k \geq 2 \end{matrix}

(A-5)

\begin{matrix} B_{k} = U_{1} [κ (\frac{T_{3}^{k} - α^{k}}{k} - \frac{α T_{3}^{k + 1} - α^{k + 2}}{k + 1}) + (\frac{T_{4}^{k} - α^{k}}{k} - \frac{α T_{4}^{k + 1} - α^{k + 2}}{k + 1})] \\ - U_{2} W_{2} (- α T_{3}^{k + 1} + (1 + α^{2}) T_{3}^{k} - α T_{3}^{k - 1}) \\ + 2 i a (1 - α^{2}) {U_{1} (κ W_{3} T_{4}^{k} + W_{4} T_{3}^{k}) - U_{2} W_{4} [(k + 1) T_{4}^{k} - k α T_{4}^{k - 1}]}, k \geq 1 \end{matrix}

(A-6)

A_{1} = - U_{1} [κ (T_{1} - α) + (T_{2} - α)] - U_{2} [W_{1} (T_{1} - 2 α) + W_{2} α]

(A-7)

A_{0} = - U_{1} [κ (α T_{3} - α^{2}) + (α T_{4} - α^{2})] - U_{2} [W_{1} - W_{2} (1 + α^{2} - α T_{3})]

Funding

The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This work was supported by the National Natural Science Foundation of China (grant numbers 11572126 and 51704117).

ORCID iD

Hui Cai

Ai-zhong Lu

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Analytic stress solution for a circular tunnel in half plane with a concentrated force

Abstract

Keywords

1. Introduction

2. Statement of problem and basic equations

2.1. Statement of problem

2.2. Relationship between φ ( z ) and ψ ( z )

2.3. Conformal mapping to an annulus

3. The solving process

3.1. First partial solution

3.2. Second partial solution

3.3. Third partial solution

4. Solving formulas for stresses

5. Discussion on accuracy

6. Examples

6.1. The principal stresses of surrounding rock

6.2. Effect of depth H1 on stress

6.3. Effect of concentrated force F on stress

6.4. Effect of H2 on stress

6.5. Effect of depth L on stress

7. Conclusion

Footnotes

Appendix

Funding

ORCID iD

References

2.2. Relationship between $φ (z)$ and $ψ (z)$

6.2. Effect of depth H₁ on stress

6.4. Effect of H₂ on stress