# Stress and Displacement of Deep-Buried Composite Lining Tunnel under Different Contact Conditions

Authors: Shuhong Liu; Jiajun Duan; Jinlu Fan; Yongquan Zhu
Journal: Mathematical Problems in Engineering
 (2023)

---

## Abstract

Stress and displacement of the composite lining are important factors to be considered during tunnel design. By the complex variable method, analytical solutions for stress and displacement of surrounding rock, primary support and secondary lining satisfying the interface continuity and boundary conditions under far-field stresses are derived. Taking the railway composite lining tunnel as an example, the analytical distributions of stress and displacement along boundaries are given, which was in good agreement with the numerical solution calculated by finite element software. The results show that the maximum normal stress ratio (load sharing ratio) of the outer boundary between the secondary lining and the primary support is 0.74. The radial displacement of the inner boundary of surrounding rock, primary support, and secondary lining change consistently. The maximum settlement and uplift occur at the vault and bottom, respectively. The tangential stress of secondary lining is compressive stress, while the tangential stress of primary support is tensile stress and compressive stress. The maximum tangential stress of primary support and secondary lining is smaller than the allowable stress of concrete.


---

## Body

## 1. Introduction

The complex variable function is the most effective method to solve deep-buried tunnel problems. Obtaining the conformal mapping function of the tunnel is the critical step when solving the mechanical response. The conformal mapping functions of unlined tunnels can be obtained by different methods. Lu [1] and Wang [2] obtained the conformal mapping function of noncircular tunnels with a single lining.Exadaktylos and Stavropoulou [3], Exadaktylos et al. [4], and Guan et al. [5] obtained analytical solutions for stress and displacement of rock mass around unlined tunnels with different cross sections. Based on the analytical stress and displacement distribution, Xu et al. [6], Ma et al. [7–9], and Liu et al. [10, 11] studied plastic zones around tunnels with circular, elliptical, and horseshoe-shaped cross sections, respectively; Wu et al. [12] clarified failure mechanism of horseshoe-shaped tunnel; Liu et al. [13] studied viscoelastic displacements of a noncircular tunnel by using the elastic-viscoelastic correspondence principle.For circular tunnels with single-layer lining, Wang and Li [14] obtained stress and displacement around a circular tunnel at great depth subjected to uniform internal pressure and unequal biaxial in situ stresses. Li et al. [15] deduced the elastic-plastic analytical solution for the stress and displacement of circular tunnel subjected to uniform internal and external pressure. Kargar [16] proposed analytical solutions around lined and unlined circular tunnels in viscous rock mass. Guo et al. [17] studied the stress, displacement, and stability of a deep lined circular pressure tunnel by combining the complex variable method with the Biot theory. For noncircular tunnels with single-layer lining, Kargar et al. [18, 19] and Lu et al. [20] investigated analytical stress solutions of inverted U-shaped tunnels according to different contact conditions between rock mass and lining, respectively; analytical displacement solutions were also given by Lu et al. [21] and Wang et al. [22] in the isotropic and orthotropic rock mass, respectively. Li and Chen [23, 24] and Liu et al. [25] obtained the analytical solutions for stress and displacement around horseshoe-shaped tunnels in the isotropic and orthotropic rock mass, respectively. Chen et al. [26] and Fang et al. [27] obtained analytic solutions of two circular and multiple noncircular tunnels at great depth considering the mutual interaction between linings and rock mass, respectively.Zhou and Yang [28] and Zhou et al. [29, 30] calculated the support loads of circular composite lining tunnels consisting of primary support and secondary lining in elastic and rheological rock mass, respectively. Li et al. [31] proposed the analytical stress solution of circular water conveyance tunnel with composite lining subjected to uniform internal pressure. Ramadan et al. [32], Maleska and Beben [33], and Embaby et al. [34] analysed numerically large-span culverts and soil-steel bridges. Maleska et al. [35], Maleska and Beben [36], Shen et al. [37], Jiang et al. [38], and Chen et al. [39] obtained numerical seismic response of soil-steel bridges, a shield tunnel, and subway stations.For noncircular tunnels, it can be seen from the above literature that there are many theoretical research results for unlined tunnels and single-layer lining tunnels, while there are few research results for composite lining tunnels widely used in practice. In this paper, the analytical solutions of stress and displacement of surrounding rock, primary support, and secondary lining of the noncircular composite lining tunnel are derived by using the complex variable function method and verified by finite element software ANSYS, which provides a theoretical basis for safe and economical tunnel design.


## 2. Conformal Mapping Function of Composite Lining Tunnel

Figure1(a) shows the load structure diagram of a composite lining tunnel with noncircular cross sections, where R, L1, and L2 represent the area of surrounding rock, primary support, and secondary lining, respectively. According to the theory of elastic mechanics, deep-buried tunnels can be simplified as a plane strain model with holes in infinite surrounding rock. Assuming that the buried depth is deep enough, the effect of gravity can be neglected. The surrounding rock is subjected to far-field stresses P and λP along Ox- and Oy-axes, respectively, and λ is the lateral pressure coefficient.Figure 1
Conformal mapping of the composite lining tunnel. (a) Schematic diagram of the tunnel in the physicalz plane. (b) Three concentric circles in the image ζ plane.
(a)(b)The conformal mapping function of the composite lining tunnel is(1)z=ωζ=∑k=1mCkζ−k−2,where z=x+iy, ζ=ρexpiθ, x and y are the rectangular coordinates in the physical z plane, and ρ and θ are the polar coordinates in the image ζ plane. The three concentric circles with radius ρ = 1, R1, and R2 in the image ζ plane are transformed into the outer boundary of primary support, the inner boundary of primary support (the outer boundary of secondary lining), and the inner boundary of secondary lining, respectively, as shown in Figure 1(b). m is the number of terms of the conformal mapping function. Ck, R1, and R2 are the coefficients related to the shape and size of the composite lining, which can be solved by Fan [40].


## 3. Basic Equations

The stress and displacement components of surrounding rock and composite lining are obtained by stress functions. The stress functions of surrounding rock, primary support, and secondary lining and the stress and displacement components in both Cartesian and curvilinear coordinates are given.


### 3.1. Stress Functions

The stress functionsφ1ζ and ψ1ζ of surrounding rock can be expressed as follows:(2)φ1ζ=Bωζ+φ0ζ,(3)ψ1ζ=B′+iC′ωζ+ψ0ζ,where constants B, B′, and C′ are related to far-field stress P and λP, B=P1+λ/4, B′=−P1−λ/2, C′=0, and the functions φ0 (ζ) and ψ0 (ζ) are analytic outside the unit circle in the ζ plane and can be written as follows:(4)ϕ0ζ=∑j=0najζ−j,ψ0ζ=∑j=0nbjζ−j.The stress functionsϕ2ζ and ψ2ζ of primary support are analytic in the annular region with radius R1≤ρ≤1 in the ζ image plane. The stress functions ϕ3ζ and ψ3ζ of secondary lining are analytic in the annular region with radius R2≤ρ≤R1 in the ζ image plane. They can be written as the following series expansions:(5)ϕ2ζ=∑j=1ndjζ−j+∑j=0nfjζj,(6)ψ2ζ=∑j=1ngjζ−j+∑j=0nhjζj,ϕ3ζ=∑j=1npjζ−j+∑j=0nqjζj,ψ3ζ=∑j=1nsjζ−j+∑j=0ntjζj,where 10n + 6 unknown real coefficients aj, bj, dj, fj, gj, hj, pj, qj, sj, and tj can be determined by the interface continuity and boundary conditions.


### 3.2. Stresses and Displacements

The stress components areσx, σy, and τxy; the displacement components ux and uy in the Cartesian coordinates xoy can be expressed as follows:(7)σx+σy=4ReΦζ,(8)σy−σx+2iτxy=2ω′ζωζ¯Φ′ζ+ω′ζΨζ,(9)2Gux+iuy=κϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯,where the notation Φζ=ϕ′ζ/ω′ζ and Ψζ=ψ′ζ/ω′ζ are introduced for simple expression. ϕζ and ψζ are the stress functions. Constant κ=3−4μ for plane strain problem, the shear modulus G=E/21+μ, and E and μ is Young’s modulus and Poisson’s ratio, respectively.The stress and displacement components in the curved coordinates are as follows:(10)σρ+σθ=4ReΦζ,(11)σρ−σθ+2iτρθ=2ζ2ρ2ω′ζ¯ωζ¯Φ′ζ+ω′ζΨζ,(12)uρ+iuθ=12Gζ¯ρω′ζ¯ω′ζκϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯,where σρ, σθ, and τρθ are the radial, tangential, and shear stress components, respectively; uρ and uθ are the radial and tangential displacements, respectively.The stress functionsφ (ζ) and ψ (ζ) in equations (7)–(12) are replaced by φ1 (ζ) and ψ1 (ζ) for surrounding rock, φ2 (ζ) and ψ2 (ζ) for primary support, and φ3 (ζ) and ψ3 (ζ) for secondary lining. The shear modulus G and constant κ in equations (9) and (12) are replaced by shear modulus G1 and κ1 of surrounding rock, G2 and κ2 of primary lining, and G3 and κ3 of secondary lining. Then, the stress and displacement components of surrounding rock, primary support, and secondary lining can be obtained.The relationship between surface force and stress functions is as follows:(13)ϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯=i∫fx¯+ify¯ds,where fx¯ and fy¯ are the surface force components in x and y directions, respectively.


## 3.1. Stress Functions

The stress functionsφ1ζ and ψ1ζ of surrounding rock can be expressed as follows:(2)φ1ζ=Bωζ+φ0ζ,(3)ψ1ζ=B′+iC′ωζ+ψ0ζ,where constants B, B′, and C′ are related to far-field stress P and λP, B=P1+λ/4, B′=−P1−λ/2, C′=0, and the functions φ0 (ζ) and ψ0 (ζ) are analytic outside the unit circle in the ζ plane and can be written as follows:(4)ϕ0ζ=∑j=0najζ−j,ψ0ζ=∑j=0nbjζ−j.The stress functionsϕ2ζ and ψ2ζ of primary support are analytic in the annular region with radius R1≤ρ≤1 in the ζ image plane. The stress functions ϕ3ζ and ψ3ζ of secondary lining are analytic in the annular region with radius R2≤ρ≤R1 in the ζ image plane. They can be written as the following series expansions:(5)ϕ2ζ=∑j=1ndjζ−j+∑j=0nfjζj,(6)ψ2ζ=∑j=1ngjζ−j+∑j=0nhjζj,ϕ3ζ=∑j=1npjζ−j+∑j=0nqjζj,ψ3ζ=∑j=1nsjζ−j+∑j=0ntjζj,where 10n + 6 unknown real coefficients aj, bj, dj, fj, gj, hj, pj, qj, sj, and tj can be determined by the interface continuity and boundary conditions.


## 3.2. Stresses and Displacements

The stress components areσx, σy, and τxy; the displacement components ux and uy in the Cartesian coordinates xoy can be expressed as follows:(7)σx+σy=4ReΦζ,(8)σy−σx+2iτxy=2ω′ζωζ¯Φ′ζ+ω′ζΨζ,(9)2Gux+iuy=κϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯,where the notation Φζ=ϕ′ζ/ω′ζ and Ψζ=ψ′ζ/ω′ζ are introduced for simple expression. ϕζ and ψζ are the stress functions. Constant κ=3−4μ for plane strain problem, the shear modulus G=E/21+μ, and E and μ is Young’s modulus and Poisson’s ratio, respectively.The stress and displacement components in the curved coordinates are as follows:(10)σρ+σθ=4ReΦζ,(11)σρ−σθ+2iτρθ=2ζ2ρ2ω′ζ¯ωζ¯Φ′ζ+ω′ζΨζ,(12)uρ+iuθ=12Gζ¯ρω′ζ¯ω′ζκϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯,where σρ, σθ, and τρθ are the radial, tangential, and shear stress components, respectively; uρ and uθ are the radial and tangential displacements, respectively.The stress functionsφ (ζ) and ψ (ζ) in equations (7)–(12) are replaced by φ1 (ζ) and ψ1 (ζ) for surrounding rock, φ2 (ζ) and ψ2 (ζ) for primary support, and φ3 (ζ) and ψ3 (ζ) for secondary lining. The shear modulus G and constant κ in equations (9) and (12) are replaced by shear modulus G1 and κ1 of surrounding rock, G2 and κ2 of primary lining, and G3 and κ3 of secondary lining. Then, the stress and displacement components of surrounding rock, primary support, and secondary lining can be obtained.The relationship between surface force and stress functions is as follows:(13)ϕζ−ωζω′ζ¯ϕ′ζ¯−ψζ¯=i∫fx¯+ify¯ds,where fx¯ and fy¯ are the surface force components in x and y directions, respectively.


## 4. Continuity and Boundary Conditions

The surface between surrounding rock and primary supportζ=expiθ=σ is assumed to be full contact. The corresponding stress and displacement components are equal, respectively [18, 19, 25]. From equations (9) and (13), the interface continuity conditions can be written as follows:(14)κ1ϕ1σ−ωσ/ω′σ¯ϕ1′σ¯−ψ1σ¯G1=κ2ϕ2σ−ωσ/ω′σ¯ϕ2′σ¯−ψ2σ¯G2,(15)ϕ1σ−ωσω′σ¯ϕ1′σ¯−ψ1σ¯=ϕ2σ−ωσω′σ¯ϕ2′σ¯−ψ2σ¯.Because the waterproof layer between primary support and secondary lining cannot bear the shear force, this interface (ζ = R1σ) is assumed to be slip contact. The corresponding normal displacement and stress are equal, respectively, and the shear stress is equal to 0. From equation (12), the displacement continuity condition can be expressed as follows:(16)Reσ¯ω′R1σ¯G2κ2ϕ2R1σ−ωR1σω′R1σ¯ϕ2′R1σ¯−ψ2R1σ¯=Reσ¯ω′R1σ¯G3κ3ϕ3R1σ−ωR1σω′R1σ¯ϕ3′R1σ¯−ψ3R1σ¯.From equations (11) and (12), the stress continuity condition can be expressed as follows:(17)Imσ2ω′R1σ¯ωR1σ¯Φ3′R1σ+ω′R1σΨ3R1σ=0,(18)ϕ2R1σ−ωR1σω′R1σ¯ϕ2′R1σ¯−ψ2R1σ¯=ϕ3R1σ−ωR1σω′R1σ¯ϕ3′R1σ¯−ψ3R1σ¯.The inner boundary of secondary lining (ζ = R2σ) is free; the radial stress and tangent stress are equal to 0. From equation (12), the stress boundary condition is as follows:(19)ϕ3R2σ−ωR2σω′R2σ¯ϕ3′R2σ¯−ψ3R2σ¯=0.


## 5. Solution Process of Stress Function

By applying series solution to solve unknown coefficientsaj, bj, dj, fj, gj, hj, pj, qj, sj, and tj of stress functions, the stress and displacement components can be obtained.In order to ensure the accuracy of the results,ωσ/ω′σ¯, ωR1σ/ω′R1σ¯, and ωR2σ/ω′R2σ¯ in equations (14)–(18) need to be expanded to a sufficient number of terms. From equations (2) and (5), the minimum exponent of ϕ1′σ¯ and ϕ2′σ¯ in equations (14) and (15) is 0 and −n−1, respectively. In order to multiply them by the maximum exponent of ωσ/ω′σ¯ to n, the positive exponent term of ωσ/ω′σ¯ should be 2n−1, and the expansion is as follows:(20)ωσω′σ¯=∑j=1m−2Wjσ−j+∑j=02n−1Wj′σj,where Wj and Wj′ are the known constants related to Ck [10, 11, 13].Similarly,ωR1σ/ω′R1σ¯ and ωR2σ/ω′R2σ¯ are expanded as follows:(21)ωR1σω′R1σ¯=∑j=1m−2Mjσ−j+∑j=02n−1Mj′σj,ωR2σω′R2σ¯=∑j=1m−2Ljσ−j+∑j=02n−1Lj′σj,where Mj, Mj′, Lj, and Lj′ are the known constants related to Ck, R1, and R2.Substituting stress functions into continuity and boundary conditions and equaling multipliers of the same order of variables, 2n + 1, 2n + 1, n + 1, n, 2n + 1, and 2n + 1 equations are obtained from equations (14)–(19), respectively. Thus, a total of 10n + 5 equations is obtained. But there is a total of 10n + 6 unknown coefficients, and an equation must be added.The displacement of the inner boundary of surrounding rock caused by tunnel excavation can be written as follows [25]:(22)κ1ϕ0σ−ωσ/ω′σ¯ϕ0′σ¯−ψ0σ¯2G1=−∑j=1m−2Wjσ−j+∑j=02n−1Wj′σj∑j=1njajσj+1+κ1∑j=0najσ−j−∑j=0nbjσj2G1.The displacement at infinity caused by tunnel excavation is equal to 0, that is, the constant term in the right side of equation (22) is 0. Since the minimum positive exponent term of ϕ0′σ¯ is σ2, the maximum negative exponent term of ωσ/ω′σ¯ is σ−2. The supplementary equation is expressed as follows:(23)κ1a0−∑k=2m−2Wkk−1ak−1+b0=0.


## 6. Results and Discussion

Taking the double-line railway composite lining tunnel in grade IV surrounding rock as an example, the analytical distributions of stress and displacement along boundaries are given and compared with the numerical distributions. The primary support is made of C25 concrete with a thickness of 0.25 m, and the secondary lining is made of C35 concrete with a thickness of 0.45 m. The material parameters of surrounding rock, primary support, and secondary lining are as follows: Young’s elastic modules:E1 = 3.7 Gpa, E2 = 23 Gpa, and E3 = 31.5 Gpa; and Poisson’s ratios: μ1 = 0.325, μ2 = 0.2, and μ3 = 0.2, respectively. The vertical infinity pressure P = 0.148762 Mpa, and the lateral pressure coefficient λ = 0.2 [41].


### 6.1. Conformal Mapping Function

The conformal mapping function of composite lining tunnel is obtained as follows:ωζ=0.7268/ζ+6.7015−0.5140ζ+0.2968ζ2−0.1282ζ3+0.0290ζ4+0.0105ζ5−0.0152ζ6+0.0083ζ7−0.0024ζ8−0.0002ζ9+0.0001ζ10−0.0004ζ11+0.0009ζ12 and R1 = 0.9632 and R2 = 0.8975. The composite lining after transformation is shown in Figure 2.Figure 2
Composite lining after transformation.


### 6.2. Comparison of Analytical Solution with Numerical Solution

In order to verify the correctness of the above analytical solution, the finite element software ANSYS is used for numerical simulation, and the analytical solution of stress and displacement is compared with the numerical solution. The composite lining adopts the transformed shape. Because the structure and load are symmetrical, the left half structure is adopted. The size of the plane strain model is180m×360m, far greater than the size of the tunnel. The upper, left, and lower boundaries are free, and normal constraints are imposed on the right boundary. The upper and lower boundaries apply pressure of 0.148762 Mpa, and the left boundary applies pressure of 0.0297524 Mpa. The above measures ensure that the model, boundary conditions, and loads in the numerical solution are consistent with those in the analytical solution, thus ensuring the comparability of the two solutions. Because the model is large and the mesh is dense, the finite element mesh of surrounding rock near the tunnel is shown in Figure 3(a), with a total of 24516 units and 49573 nodes. The primary support grid is shown in Figure 3(b), with 1140 elements and 2286 nodes. The secondary lining grid is shown in Figure 3(c), with 1484 elements and 3333 nodes. The contact surfaces of surrounding rock, primary support, and secondary lining are full contact and slip contact, respectively. Slip contact is realized by establishing contact pairs, which are composed of target surface and contact surface. The contact type is nonseparation contact, and the friction coefficient is zero.Figure 3
Finite element mesh. (a) Surrounding rock. (b) Primary support. (c) Secondary lining.
(a)(b)(c)Figures4(a) and 4(b) show the contours of stress σx and displacements ux around the tunnel, respectively. It can be seen that stress σx changes obviously in the composite lining and gradually approaches the applied load of 0.148762 Mpa at a distance from the tunnel in surrounding rock. The displacement ux changes consistently. The changes of stress and displacement of surrounding rock are mainly concentrated near the tunnel.Figure 4
Stress and displacement contours. (a) Stressσx. (b) Displacement ux.
(a)(b)It is convenient to read the calculation results under the rectangular coordinate system in the finite element software ANSYS. In order to clearly see the difference between the numerical solution and the analytical solution, as an example, Figures5(a) and 5(b) show the numerical solution and the analytical solution of the stress and displacement of the inner boundary of secondary lining (ρ = 0.8975). A positive angle α turns counterclockwise from the positive x-axis to the positive y-axis (Figure 1(a)). The analytical solution is calculated by taking n = 100 in this study. From Figure 5(a), it can be seen that the numerical solution and the analytical solution of stresses σxL2 and σyL2 are almost identical. When α=0° and 180°, the stress σxL2 is zero, satisfying the boundary condition. From Figure 5(b), it can be seen that the numerical solution and the analytical solution of the displacement uyL2 are almost identical. While, there is a small rigid body translation for the displacement uxL2 because the finite element model has no constraints in the vertical direction. When α=0° and 180°, the displacement uyL2 is zero, satisfying the symmetry condition.Figure 5
Analytical and numerical solutions (ρ = 0.8975). (a) Stresses. (b) Displacements.
(a)(b)


### 6.3. Analytical Distributions of Stress and Displacement along Boundaries

The analytical distributions of stress and displacement along boundaries are given, in order to verify the continuity conditions and boundary conditions and find the maximum stress and displacement.On the interface between surrounding rock and primary support (ρ = 1), the radial stresses σρR, σρL1, shear stresses τρθR, τρθL1, the radial displacements uρR, uρL1, and tangential displacements uθR, uθL1 of the inner boundary of surrounding rock and the outer boundary of primary support are shown in Figures 6(a) and 6(b), respectively. It can be seen that the corresponding stress and displacement are equal to satisfy the full contact condition. The radial stress is all compressive, and the maximum value of 0.146 MPa occurs at point A (Figure 2).Figure 6
Distributions on the interface between surrounding rock and primary support. (a) Stresses. (b) Displacements.
(a)(b)On the interface between primary support and secondary lining (ρ = 0.9632), the radial stresses σρL1, σρL2 and shear stresses τρθL1, τρθL2 and the radial displacements uρL1, uρL2 and tangential displacements uθL1, uθL2 of the inner boundary of primary support and the outer boundary of secondary lining are shown in Figures 7(a) and 7(b), respectively. The corresponding radial stress is equal, while the shear stress is zero. The corresponding radial displacement is equal, while the corresponding tangential displacement is not equal. From these results, it can be seen that the slip contact condition is satisfied. The radial stress is compressive stress, and the maximum value of 0.108 MPa occurs at point B (Figure 2). The load sharing ratio concerned in the design, that is, the maximum normal stress ratio of the outer boundary between the secondary lining and the primary support is 0.74.Figure 7
Distributions on the interface between primary support and secondary lining. (a) Stresses. (b) Displacements.
(a)(b)The radial stressσρL2 and shear stress τρθL2 and the radial displacement uρL2 and tangential displacement uθL2 of the inner boundary of secondary lining (ρ = 0.8975) are shown in Figures 8(a) and 8(b), respectively. Since the radial stress and shear stress are equal to zero, the stress boundary conditions are satisfied.Figure 8
Distributions on the inner boundary of secondary lining. (a) Stresses. (b) Displacements.
(a)(b)From Figures6–8, it can be seen that the stress change trend is different on each boundary, while the radial displacement change trend is consistent. The maximum settlement, uplift, and peripheral displacement occur at the vault, bottom, and waist, respectively. On the vertical axis, the shear stress and tangential displacement are equal to zero, satisfying the symmetry condition.Figure9 shows the tangential stress of the inner boundary of surrounding rock (ρ = 1), the inner and outer boundaries of primary support (ρ = 1 and ρ = 0.9632), and the inner and outer boundaries of secondary lining (ρ = 0.9632 and ρ = 0.8975), respectively. The tangential stress σθR value of the inner boundary of surrounding rock is the smallest. For composite lining, the maximum compressive stress and tensile stress occur at the inner boundary of primary support. The maximum compressive stress is 1.425 MPa at point C (Figure 2), and the maximum tensile stress is 0.472 MPa at the bottom. The maximum values of compressive stress and tensile stress are much smaller than the allowable compressive design strength of 13 MPa and tensile design strength of 1.3 MPa of C25 shotcrete [41]. The tangential normal stress σθL2 of secondary lining is compressive stress, and the maximum compressive stress is 1.338 MPa at point D (Figure 2) of the inner boundary of secondary lining, which is much smaller than the allowable compressive stress of 13 MPa of C35 concrete [41]. Therefore, the thickness of primary support and secondary lining can be appropriately reduced.Figure 9
Tangential stress distributions along boundaries.


## 6.1. Conformal Mapping Function

The conformal mapping function of composite lining tunnel is obtained as follows:ωζ=0.7268/ζ+6.7015−0.5140ζ+0.2968ζ2−0.1282ζ3+0.0290ζ4+0.0105ζ5−0.0152ζ6+0.0083ζ7−0.0024ζ8−0.0002ζ9+0.0001ζ10−0.0004ζ11+0.0009ζ12 and R1 = 0.9632 and R2 = 0.8975. The composite lining after transformation is shown in Figure 2.Figure 2
Composite lining after transformation.


## 6.2. Comparison of Analytical Solution with Numerical Solution

In order to verify the correctness of the above analytical solution, the finite element software ANSYS is used for numerical simulation, and the analytical solution of stress and displacement is compared with the numerical solution. The composite lining adopts the transformed shape. Because the structure and load are symmetrical, the left half structure is adopted. The size of the plane strain model is180m×360m, far greater than the size of the tunnel. The upper, left, and lower boundaries are free, and normal constraints are imposed on the right boundary. The upper and lower boundaries apply pressure of 0.148762 Mpa, and the left boundary applies pressure of 0.0297524 Mpa. The above measures ensure that the model, boundary conditions, and loads in the numerical solution are consistent with those in the analytical solution, thus ensuring the comparability of the two solutions. Because the model is large and the mesh is dense, the finite element mesh of surrounding rock near the tunnel is shown in Figure 3(a), with a total of 24516 units and 49573 nodes. The primary support grid is shown in Figure 3(b), with 1140 elements and 2286 nodes. The secondary lining grid is shown in Figure 3(c), with 1484 elements and 3333 nodes. The contact surfaces of surrounding rock, primary support, and secondary lining are full contact and slip contact, respectively. Slip contact is realized by establishing contact pairs, which are composed of target surface and contact surface. The contact type is nonseparation contact, and the friction coefficient is zero.Figure 3
Finite element mesh. (a) Surrounding rock. (b) Primary support. (c) Secondary lining.
(a)(b)(c)Figures4(a) and 4(b) show the contours of stress σx and displacements ux around the tunnel, respectively. It can be seen that stress σx changes obviously in the composite lining and gradually approaches the applied load of 0.148762 Mpa at a distance from the tunnel in surrounding rock. The displacement ux changes consistently. The changes of stress and displacement of surrounding rock are mainly concentrated near the tunnel.Figure 4
Stress and displacement contours. (a) Stressσx. (b) Displacement ux.
(a)(b)It is convenient to read the calculation results under the rectangular coordinate system in the finite element software ANSYS. In order to clearly see the difference between the numerical solution and the analytical solution, as an example, Figures5(a) and 5(b) show the numerical solution and the analytical solution of the stress and displacement of the inner boundary of secondary lining (ρ = 0.8975). A positive angle α turns counterclockwise from the positive x-axis to the positive y-axis (Figure 1(a)). The analytical solution is calculated by taking n = 100 in this study. From Figure 5(a), it can be seen that the numerical solution and the analytical solution of stresses σxL2 and σyL2 are almost identical. When α=0° and 180°, the stress σxL2 is zero, satisfying the boundary condition. From Figure 5(b), it can be seen that the numerical solution and the analytical solution of the displacement uyL2 are almost identical. While, there is a small rigid body translation for the displacement uxL2 because the finite element model has no constraints in the vertical direction. When α=0° and 180°, the displacement uyL2 is zero, satisfying the symmetry condition.Figure 5
Analytical and numerical solutions (ρ = 0.8975). (a) Stresses. (b) Displacements.
(a)(b)


## 6.3. Analytical Distributions of Stress and Displacement along Boundaries

The analytical distributions of stress and displacement along boundaries are given, in order to verify the continuity conditions and boundary conditions and find the maximum stress and displacement.On the interface between surrounding rock and primary support (ρ = 1), the radial stresses σρR, σρL1, shear stresses τρθR, τρθL1, the radial displacements uρR, uρL1, and tangential displacements uθR, uθL1 of the inner boundary of surrounding rock and the outer boundary of primary support are shown in Figures 6(a) and 6(b), respectively. It can be seen that the corresponding stress and displacement are equal to satisfy the full contact condition. The radial stress is all compressive, and the maximum value of 0.146 MPa occurs at point A (Figure 2).Figure 6
Distributions on the interface between surrounding rock and primary support. (a) Stresses. (b) Displacements.
(a)(b)On the interface between primary support and secondary lining (ρ = 0.9632), the radial stresses σρL1, σρL2 and shear stresses τρθL1, τρθL2 and the radial displacements uρL1, uρL2 and tangential displacements uθL1, uθL2 of the inner boundary of primary support and the outer boundary of secondary lining are shown in Figures 7(a) and 7(b), respectively. The corresponding radial stress is equal, while the shear stress is zero. The corresponding radial displacement is equal, while the corresponding tangential displacement is not equal. From these results, it can be seen that the slip contact condition is satisfied. The radial stress is compressive stress, and the maximum value of 0.108 MPa occurs at point B (Figure 2). The load sharing ratio concerned in the design, that is, the maximum normal stress ratio of the outer boundary between the secondary lining and the primary support is 0.74.Figure 7
Distributions on the interface between primary support and secondary lining. (a) Stresses. (b) Displacements.
(a)(b)The radial stressσρL2 and shear stress τρθL2 and the radial displacement uρL2 and tangential displacement uθL2 of the inner boundary of secondary lining (ρ = 0.8975) are shown in Figures 8(a) and 8(b), respectively. Since the radial stress and shear stress are equal to zero, the stress boundary conditions are satisfied.Figure 8
Distributions on the inner boundary of secondary lining. (a) Stresses. (b) Displacements.
(a)(b)From Figures6–8, it can be seen that the stress change trend is different on each boundary, while the radial displacement change trend is consistent. The maximum settlement, uplift, and peripheral displacement occur at the vault, bottom, and waist, respectively. On the vertical axis, the shear stress and tangential displacement are equal to zero, satisfying the symmetry condition.Figure9 shows the tangential stress of the inner boundary of surrounding rock (ρ = 1), the inner and outer boundaries of primary support (ρ = 1 and ρ = 0.9632), and the inner and outer boundaries of secondary lining (ρ = 0.9632 and ρ = 0.8975), respectively. The tangential stress σθR value of the inner boundary of surrounding rock is the smallest. For composite lining, the maximum compressive stress and tensile stress occur at the inner boundary of primary support. The maximum compressive stress is 1.425 MPa at point C (Figure 2), and the maximum tensile stress is 0.472 MPa at the bottom. The maximum values of compressive stress and tensile stress are much smaller than the allowable compressive design strength of 13 MPa and tensile design strength of 1.3 MPa of C25 shotcrete [41]. The tangential normal stress σθL2 of secondary lining is compressive stress, and the maximum compressive stress is 1.338 MPa at point D (Figure 2) of the inner boundary of secondary lining, which is much smaller than the allowable compressive stress of 13 MPa of C35 concrete [41]. Therefore, the thickness of primary support and secondary lining can be appropriately reduced.Figure 9
Tangential stress distributions along boundaries.


## 7. Conclusion

According to the complex variable function method, the analytical solutions for stress and displacement of composite lining tunnel are obtained, and compared with the numerical solutions obtained by finite element software, the results are in good agreement. The main conclusions are as follows:(1)
On the contact surface between surrounding rock and primary support, the corresponding radial stress, shear stress, radial displacement, and tangential displacement of the inner boundary of surrounding rock and the outer boundary of primary support are equal, respectively, which satisfy the full contact condition. On the contact surface between primary support and secondary lining, the corresponding radial stress and displacement of the inner boundary of primary support and the outer boundary of secondary lining are equal, respectively, and the shear stress is zero, which satisfies the slip contact condition. The radial stress and shear stress of the inner boundary of secondary lining are both zero, which satisfies the stress boundary condition. The shear stress and horizontal displacement on the vertical axis are zero, satisfying the symmetry condition.(2)
The maximum normal stress ratio (load sharing ratio) of the outer boundary between the secondary lining and the primary support is 0.74. The maximum compressive and tensile tangential stress occur at the inner boundary of primary support and the maximum compressive tangential stress of secondary lining occurs at the arch foot of inner boundary. Since they are far less than the allowable stress of concrete, the thickness of primary support and secondary lining can be appropriately reduced from an economic point of view.(3)
The radial displacement change trend of surrounding rock, primary support, and secondary lining is consistent. The maximum settlement, uplift, and peripheral displacement occur at the vault, bottom, and waist, respectively. Therefore, during the design and construction of the tunnel, attention should be paid to these locations.Subsequently, we will combine the on-site monitoring of specific projects to verify the applicability of our solutions, so as to better guide the safe and economical tunnel design.



---

*Source: 7467290-2023-01-23.xml*