# Analytic Solutions of the Forces and Displacements for Multicentre Circular Arc Tunnels

Authors: Xingbo Han; Yongxu Xia
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

A complex variable method for solving the forces and displacements for multicentre circular arc (MCA) tunnels by using analytical solutions is presented. The governing equations for the stresses and displacements are expressed in terms of series expression. Combined with the boundary conditions at lining inner region, lining-surrounding rock-mass interface, and infinity area, the undetermined coefficients of the analytic functions are obtained by solving a linear equation set. Forces solutions for linings are also presented according to the tangential stress at the two boundaries. Conformal mapping function for the MCA tunnel is given from the optimization method. Calculation results show that solutions with high accuracy can be obtained when the number of terms of power series is greater than 30. The effects of the degrees from the tunnel crown to the invert, coefficient of the lateral earth pressure, and distance from the rock-mass to the interface on regulations of the lining forces and rock-mass stresses are also thoroughly investigated.


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## Body

## 1. Introduction

The design of the cross-section shape for transportation tunnels has undergone the development from a singer-centre circular arc to a multicentre circular arc (MCA). Though the singer-centred arc tunnel is simple for engineers to design, the MCA tunnel is widely used for its advantage in the reduction of the excavation volume. However, lining forces and displacements, which are essential parameters during every stages (i.e., design, construction, operation, and maintaining [1–3]) of the tunnel life circle, are difficult to analyse for a MCA tunnel. Therefore, suitable methods should be established to give access to the forces and displacements analysis of the MCA tunnels.Very extensive numerical simulations have been performed to determine stresses and deformations of tunnels in recent year. Möller and Vermeer (2008) applied FEM to simulate lining forces and the ground deformation of the Steinhaldenfeld tunnel and Heinenoord tunnel [4]. Liu et al. (2017) simulated the whole construction stages of the Gongti North Road tunnel of Beijing Metro Line 10 built using the PBA method [5]. Also, Avgerinos et al. (2017) performed a series of three-dimensional finite-element analyses to discuss the forces, bending moment, and lining deformation of the future Crossrail tunnel [6]. However, these numerical results are mostly applied to certain examples and are difficult to give an understanding of how final solutions are influenced by different parameters. Behnen et al. (2015) pointed out that a design strategy based on comprehensible analytical models in combination with the valuable knowledge of experienced engineers should be preferred instead of relying on the complex simulation models [7].Many researchers have treated the tunnel lining structures as shallow arches to present theoretical analysis models for its forces and displacements. Pi et al. (2001) applied the energy method to investigate the stresses and buckling of the shallow arches subjected to a radial load uniformly distributed around the arch axis [8]. Gao et al. (2014) combined the structure analysis and material uncertainties of arches and presented the results of the axial compressive force, bending moment, and axial displacement [9]. But these shallow arches are still not well satisfied with the tunnel lining boundary conditions in field.The theoretical methods for circular tunnels have also fully developed. Lekhnitskii (1968) proposed a complex variable-function method for the accurate solution of stresses around a circular hole under in-plane loading [10]. Hefny and Lo (1999) obtained analytical solutions for stresses and displacements around unlined circular tunnels excavated in elastic transversely isotropic [11]. Vu et al. (2013) developed a semianalytical solution for an unlined circular tunnel excavated in a transversely isotropic formation with nonlinear behaviour [12]. In addition, for lined circular tunnels [13–15], ITA report (2000) recommended the elastic equation method to determine the member forces and cross-section deformation of the segmental ring [16]. Lu et al. (2011) applied the complex method and presented the stress solutions for a circular lined tunnel [17]. Yasuda et al. (2017) established a two-dimensional elastic solution for a deep circular tunnel under the far-field static loading [18]. It is still difficult to determine the accurate solutions for the stresses and displacements for the other irregular shapes.An attempt is made in this study to find the forces and displacements solutions for MCA tunnel linings by applying Muskhelishvili’s complex variable method.


## 2. Complex Variable Method for Lined Noncircular Tunnels

### 2.1. General Considerations

The problem can be considered as a reinforced hole in an infinite plane subjected to a uniform stress state at infinity, while the tunnel is located at a great depth compared with the tunnel dimension. The infinite plane on the complex plane is divided into two isotropic homogenous regions considering both the lining and the surrounding rock-mass. As shown in Figure1, the two regions are bounded by contours γ1 and γ2, which represents the inner and outer boundaries of the lining, respectively. It is assumed that the conformal mapping function can be expressed as follows:(1)z=ωζ=Rζ+∑k=0nckζ-k,where R is a real constant presenting the scale of the mapping.Figure 1
Conformal mapping of tunnel cross-section inz-plane into ζ-plane.The stress components provided by Muskhelishvili and Radok (1953) [19] based on the complex variable method is as follows:(2)σx+σy=4Re⁡φ′zσy-σx+2iτxy=2z¯φ′′z+ψ′z,where σx and σy are the horizontal and vertical stress components, respectively, τxy is the shear stress. φ(z) and ψ(z) are two analytic complex functions, and Re denotes the real part of .The displacement can be described as follows:(3)2Gu+iv=κφz-zφ′z¯-ψz¯,where i=-1, G is the shear modulus, and κ is a parameter related to Poisson’s ratio υ and is as follows:(4)κ=3-4υplanestrian3-4υ1+υplanestress.The surface tractions along the boundaries satisfy the following equation:(5)i∫f¯x+if¯yds=φz+zφ′z¯+ψz¯,where f¯x and f¯y are components of the surface traction vector in x and y directions, respectively.The tunnel from thez-plane can be mapped into the ζ-plane using (1). The analytic functions in ζ-plane can be calculated by the following equations:(6)φζ=φ1z=φ1ωζψζ=ψ1z=ψ1ωζφ1′z=φ′ζω′ζψ1′z=ψ′ζω′ζ.Substituting (6) and (1) into (2), the stress components in the orthogonal curvilinear coordinate system can be rewritten as follows:(7)σρ+σθ=σx+σy=4Reφ′ζω′ζσθ-σρ+2iτρθ=2ζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζ.When the tunnel is unlined, the displacement of the surrounding rock-mass can be expressed as follows:(8)2G1u1R+iv1R=κ1φ1ζ-ωζω′ζ¯φ1′ζ¯-ψ1ζ¯,where u1R and v1R are the displacement components in the x and y directions and G1 and κ1 refer to the parameters of the surrounding rock-mass.Considering that the lining is installed when the displacement isη times the total displacement, the displacement of the rock u2R+iv2R occurs before the support of the lining and can be expressed as follows:(9)u2R+iv2R=ηu1R+iv1R.After the lining is applied, a part of the displacement of the surrounding rock-mass is restricted and can be then presented as follows:(10)2G1u3R+iv3R=κ1φ2ζ-ωζω′ζ¯φ2′ζ¯-ψ2ζ¯.From (8) to (10), the final displacement of the surrounding rock-mass uR+ivR can be obtained as follows:(11)uR+ivR=u1R+iv1R-u2R+iv2R+u3R+iv3R=1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯.The displacement of the lining can be derived by the following equation:(12)2G2uL+ivL=κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3ζ¯,where uL and vL are the displacement components of any point of the lining in the x and y directions and G2 and κ2 refer to the parameters of the lining concrete.The stress components of any point in the lining can be then calculated by substituting (13) into (7).(13)φζ=φ3ζψζ=ψ3ζ.In particular, the lineρ=1 denotes the interface of the surrounding rock mass and the lining. The line ρ=R0 denotes the lining inner boundary.Similarly, the stress components of any point in the surrounding rock mass can be solved by substituting (14) into (7).(14)φζ=Γωζ+φ1ζ+φ2ζψζ=Γ′ωζ+ψ1ζ+ψ2ζ.In (14), Γ=p(1+λ)/4 and Γ′=p(λ-1)/2. The first term is the corresponding complex potential function before excavation, and the remaining terms represent the effects of the excavation and the lining support on the surrounding rock-mass.The stresses of the lining and surrounding rock mass can be determined using (7) once the six analytic functions are determined. And then the displacement can be determined using (11) and (12).


### 2.2. Solutions for the Analytic Functions

The six analytic functions can be expressed by power series as [20](15)φ1ζ=∑k=1nakζ-k(16)ψ1ζ=-ω1/ζ¯ω′ζφ1′ζ+∑k=1n-2Skζk+S0-pR21+λζ-1+pR21-λ∑k=1nckζ-k(17)φ2ζ=b0+∑k=1∞bkζ-k(18)ψ2ζ=d0+∑k=1∞dkζ-k(19)φ3ζ=p0+∑k=1∞ekζ-k+∑k=1∞fkζk(20)ψ3ζ=q0+∑k=1∞gkζ-k+∑k=1∞hkζk.Assume that(21)bζ=∑k=1∞bkζ-k,dζ=∑k=1∞dkζ-k,eζ=∑k=1∞ekζ-kfζ=∑k=1∞fkζk,gζ=∑k=1∞gkζ-k,hζ=∑k=1∞hkζk.The coefficientsak of the φ1(ζ) can be represented by the following equations:(22)ai=-2ΓRci,i=n,n-1.Leta=a1⋯an-2T; a1,…,an-2 can be determined as follows:(23)An-2×n-2a1×n-2=B1×n-2when i=j, Aij=jLi+j+1-1, and if i+j+1>n, Aij=-1; when i≠j, Aij=jLi+j+1, and if i+j+1>n, Aij=0.(24)Bi=2ΓRc1+Γ′R2ΓRc23ΓRc3⋯n-2ΓRcn-2T.The coefficientsLk can be computed by the following equations:(25)Ln=cnLn-1=cn-1Ln-j+1=∑k=1j-2j-1-kcj-1-kLn-k+1+cn-j-1,j=3,…,n.Thus, the coefficientsak of analytic function φ1(ζ) can be determined by combining (22)–(25).For the analytic functionψ1(ζ), the Sk in (16) can be calculated by the following equations:(26)Sk=-∑j=1n-1jLj+1aj,k=0-∑j=1n-k-1jLj+k+1aj,k=1,…,n-2.The three boundary conditions used to solve the undetermined coefficients (b0,bk,d0,dk,p0,q0,ek,fk,gk,hk) from the analytic functions (φ2(ζ),ψ2(ζ),φ3(ζ),ψ3(ζ)) are listing:(i)
The surface tractions along the inner boundary of lining atL1 are zero.(ii)
The surface tractions continue at the interfaceL2 from the both surrounding rock mass side and lining side.(iii)
The displacement continues at the interfaceL2 from the both surrounding rock mass side and lining side.The surface traction condition at inner boundaryL1 can be expressed as follows:(27)φ3σ1+ωσ1ω′σ1¯φ3′σ1¯+ψ3σ1¯=0.The surface condition at the interfaceL2 can be presented as follows:(28)φ2σ+ωσω′σ¯φ2′σ¯+ψ2σ¯=φ3σ+ωσω′σ¯φ3′σ¯+ψ3σ¯.The displacement condition at the interfaceL2 can be expressed as follows:(29)uR+ivR=uL+ivL.Substituting (11) and (12) into (29), the following equation can be obtained:(30)1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯=12G2κ2φ3σ-ωσω′σ¯φ3′σ¯-ψ3σ¯.Two functions can be derived using (27) and (28) based on the Cauchy integration [20]:(31)φ′¯31ζ=-ω′¯1/ζωR02ζp0+q0+p1R02ζ+p2R02ζ+q11ζ+q21ζ(32)φ′¯31ζ=ω′¯1/ζωζb0+d0-p0-q0+bζ-p1ζ-p2ζ-q11ζ-q21ζ+d1ζ+d′1ζ.Substituting (28) into (30), the following equation can be obtained:(33)1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+κ1G2+G1G1G2φ2σ+G1-G2G1G2ωσω′σ¯φ2′σ¯+G1-G2G1G2ψ2σ¯=1+κ2G2φ3σ.Using the Cauchy integration, (33) can be expressed as follows:(34)-1-ηκ1G1φ1ζ+1-ηG1∑k=1n-2Skζ-k-κ1G2+G1G1G2bζ-G1-G2G1G2∑k=1n-2Tkζ-k=-1+κ2G2p1ζ.Substitutingζby 1/ζ, the Cauchy integration of the conjugate of (34) can be expressed as follows:(35)1-ηG1ω¯ζω′1/ζφ′11ζ-∑k=1n-2Skζ-k-S0+ψ11ζ-G1-G2G1G2ω¯ζω′1/ζb′1ζ-∑k=1n-2Tkζ-k-T′+d1ζ=-1+κ2G2p2ζ.Substituting (16) into (35), the following equation can be derived:(36)pR1-η2G1-1+λζ+1-λ∑k=1nckζk-G1-G2G1G2ω¯ζω′1/ζb′1ζ-∑k=1n-2Tkζ-k-T′+d1ζ=-1+κ2G2p2ζ.Then (36) can be expressed as follows [20]:(37)pR1-η2G1-1+λζ+1-λ∑k=1nckζk-G1-G2G1G2∑k=2∞Tk′ζk-1+d1ζ=-1+κ2G2p2ζ.The following equation can be obtained by combining (33)–(35):(38)κ1G2+G1G1G2b0+G1-G2G1G2d0=1-ηG1S0′-G1-G2G1G2T1′+1+κ2G2p0.The coefficientsTk and Tk′ in (34)–(38) can be calculated using the following equations:(39)Tk=-∑j=1n-k-1jLj+k+1bj,k=1,…,n-2(40)T1′=-∑j=1n-1jLj+1bjT2′=-∑j=1njLjbjTk′=-∑j=1k-2jLk-j-1bj-∑j=k-1n+k-2jLj-k+2bj,k=3,…,∞.The undetermined coefficients inb(ζ),d(ζ),e(ζ),f(ζ),g(ζ), and h(ζ) can be determined based on (31), (32), (34), and (38). Linear simultaneous equations, which contain infinite number of equations, can be obtained by comparing the coefficient of the same order of variables ζk. But only finite terms of b(ζ),d(ζ),e(ζ),f(ζ),g(ζ), and h(ζ) are discussed here in order to get the problem solved. The number of terms of bk, ek, and hk is denoted as Ne, and number of terms of dk, fk, and gk are denoted as Nf.Letp0=x1, ej=xj+1 (j=1,…,Ne), q0=xNe+2, hj=xNe+2+j (j=1,…,Ne), fk=x2Ne+2+k, and gk=x2Ne+Nf+2+k (j=1,…,Nf); the following linear simultaneous equations can be obtained by comparing the constant and coefficients of the positive power of ζk in (31):(41)Cijxj+CiNe+1+kxNe+1+k+Ci2Ne+2+Lx2Ne+2+L+Ci2Ne+Nf+2+Lx2Ne+Nf+2+L=0.N e + 2 equations are included in (41).Comparing negative powers ofζk in (31), the following linear simultaneous equations can be obtained:(42)Dijxj+DiNe+1+kxNe+1+k+Di2Ne+2+Lx2Ne+2+L+Di2Ne+Nf+2+Lx2Ne+Nf+2+L=0.N f - 1 equations are included in (42).Letbj=x2Ne+2Nf+2+j(j=1,…,Ne),d0=x3Ne+2Nf+3, and dj=x3Ne+2Nf+3+j(j=1,…,Nf); the following linear simultaneous equations can be obtained by comparing the constant and the coefficients of the positive power of ζk in (32):(43)Eijxj+EiNe+1+kxNe+1+k+Ei2Ne+2+Lx2Ne+2+L+Ei2Ne+Nf+2+Lx2Ne+Nf+2+L+Ei2Ne+2Nf+2+kx2Ne+2Nf+2+k+Ei3Ne+2Nf+2+Mx3Ne+2Nf+2+M=0.N e + 2 equations are included in (43).Comparing negative powers ofζk in (32), the following linear simultaneous equations can be obtained:(44)Fijxj+FiNe+1+kxNe+1+k+Fi2Ne+2+Lx2Ne+2+L+Fi2Ne+Nf+2+Lx2Ne+Nf+2+L+Fi2Ne+2Nf+2+kx2Ne+2Nf+2+k+Fi3Ne+2Nf+2+Mx3Ne+2Nf+2+M=0.N f - 1 equations are included in (44).Comparing the coefficients of the negative powers ofζk in (34), the linear simultaneous equations can be derived as follows:(45)Gijxj+Gi2Ne+2Nf+2+kx2Ne+2Nf+2+k=Hi.N f - 1 equations are included in (45).H i in (45) can be expressed as(46)Hi=-1-ηSi+1-ηκ1aii=1,…,n-21-ηκ1aii=n-1,n.Comparing the coefficients of the negative powers ofζk in (37), the linear simultaneous equations can be obtained as(47)Pi2Ne+2+jx2Ne+2+j+Pi2Ne+Nf+2+kx2Ne+Nf+2+k+Pi3Ne+2Nf+2+Lx3Ne+2Nf+2+L=Qi.N f - 1 equations are included in (47), where(48)Qi=pR1-η21+λ-1-λc1i=1-pR1-η21-λcii=2,…,n.Therefore, there are totally3Ne+3Nf+2 equations in (41)–(48) for the 3Ne+3Nf+3 undetermined variables. One more equation is still needed to solve all the variables.The support effect of the lining to the rock-mass decreases as the distance between them increases. Thus, (10) can be expressed at infinity as(49)limζ→∞2G1u3R+iv3R=limζ→∞κ1φ2ζ-ωζω′ζ¯φ2′ζ¯-ψ2ζ¯=0.It can be obtained from (49) that(50)κ1limζ→∞φ2ζ-limζ→∞ωζω′ζ¯φ2′ζ¯-limζ→∞ψ2ζ¯=0.And then(51)κ1b0-d0=0.Substituting (51) and (40) into (38), the following equations can be obtained:(52)p0-1+κ11+κ2d0+G1-G2G11+κ2∑j=1n-1jLj+1bj=-G21-ηG11+κ2S0′Now, there are totally3Ne+3Nf+3 equations for the 3Ne+3Nf+3 variables. Briefly, the xj can be calculated using the following equation set:(53)CNe+2×2Ne+2Nf+2DNf-1×2Ne+2Nf+2ENe+2×3Ne+3Nf+3FNf-1×2Ne+2Nf+2GNe×3Ne+2Nf+2PNf×3Ne+3Nf+31+κ2xj×1=02Ne+2Nf+2×1HNe×1QNf×1-G21-ηS0′G1.The coefficients ofCij, Dij, Eij, Fij, Gij, and Pij have been listed in [16].Thus, all the undetermined coefficients of analytic functions can be solved.


## 2.1. General Considerations

The problem can be considered as a reinforced hole in an infinite plane subjected to a uniform stress state at infinity, while the tunnel is located at a great depth compared with the tunnel dimension. The infinite plane on the complex plane is divided into two isotropic homogenous regions considering both the lining and the surrounding rock-mass. As shown in Figure1, the two regions are bounded by contours γ1 and γ2, which represents the inner and outer boundaries of the lining, respectively. It is assumed that the conformal mapping function can be expressed as follows:(1)z=ωζ=Rζ+∑k=0nckζ-k,where R is a real constant presenting the scale of the mapping.Figure 1
Conformal mapping of tunnel cross-section inz-plane into ζ-plane.The stress components provided by Muskhelishvili and Radok (1953) [19] based on the complex variable method is as follows:(2)σx+σy=4Re⁡φ′zσy-σx+2iτxy=2z¯φ′′z+ψ′z,where σx and σy are the horizontal and vertical stress components, respectively, τxy is the shear stress. φ(z) and ψ(z) are two analytic complex functions, and Re denotes the real part of .The displacement can be described as follows:(3)2Gu+iv=κφz-zφ′z¯-ψz¯,where i=-1, G is the shear modulus, and κ is a parameter related to Poisson’s ratio υ and is as follows:(4)κ=3-4υplanestrian3-4υ1+υplanestress.The surface tractions along the boundaries satisfy the following equation:(5)i∫f¯x+if¯yds=φz+zφ′z¯+ψz¯,where f¯x and f¯y are components of the surface traction vector in x and y directions, respectively.The tunnel from thez-plane can be mapped into the ζ-plane using (1). The analytic functions in ζ-plane can be calculated by the following equations:(6)φζ=φ1z=φ1ωζψζ=ψ1z=ψ1ωζφ1′z=φ′ζω′ζψ1′z=ψ′ζω′ζ.Substituting (6) and (1) into (2), the stress components in the orthogonal curvilinear coordinate system can be rewritten as follows:(7)σρ+σθ=σx+σy=4Reφ′ζω′ζσθ-σρ+2iτρθ=2ζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζ.When the tunnel is unlined, the displacement of the surrounding rock-mass can be expressed as follows:(8)2G1u1R+iv1R=κ1φ1ζ-ωζω′ζ¯φ1′ζ¯-ψ1ζ¯,where u1R and v1R are the displacement components in the x and y directions and G1 and κ1 refer to the parameters of the surrounding rock-mass.Considering that the lining is installed when the displacement isη times the total displacement, the displacement of the rock u2R+iv2R occurs before the support of the lining and can be expressed as follows:(9)u2R+iv2R=ηu1R+iv1R.After the lining is applied, a part of the displacement of the surrounding rock-mass is restricted and can be then presented as follows:(10)2G1u3R+iv3R=κ1φ2ζ-ωζω′ζ¯φ2′ζ¯-ψ2ζ¯.From (8) to (10), the final displacement of the surrounding rock-mass uR+ivR can be obtained as follows:(11)uR+ivR=u1R+iv1R-u2R+iv2R+u3R+iv3R=1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯.The displacement of the lining can be derived by the following equation:(12)2G2uL+ivL=κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3ζ¯,where uL and vL are the displacement components of any point of the lining in the x and y directions and G2 and κ2 refer to the parameters of the lining concrete.The stress components of any point in the lining can be then calculated by substituting (13) into (7).(13)φζ=φ3ζψζ=ψ3ζ.In particular, the lineρ=1 denotes the interface of the surrounding rock mass and the lining. The line ρ=R0 denotes the lining inner boundary.Similarly, the stress components of any point in the surrounding rock mass can be solved by substituting (14) into (7).(14)φζ=Γωζ+φ1ζ+φ2ζψζ=Γ′ωζ+ψ1ζ+ψ2ζ.In (14), Γ=p(1+λ)/4 and Γ′=p(λ-1)/2. The first term is the corresponding complex potential function before excavation, and the remaining terms represent the effects of the excavation and the lining support on the surrounding rock-mass.The stresses of the lining and surrounding rock mass can be determined using (7) once the six analytic functions are determined. And then the displacement can be determined using (11) and (12).


## 2.2. Solutions for the Analytic Functions

The six analytic functions can be expressed by power series as [20](15)φ1ζ=∑k=1nakζ-k(16)ψ1ζ=-ω1/ζ¯ω′ζφ1′ζ+∑k=1n-2Skζk+S0-pR21+λζ-1+pR21-λ∑k=1nckζ-k(17)φ2ζ=b0+∑k=1∞bkζ-k(18)ψ2ζ=d0+∑k=1∞dkζ-k(19)φ3ζ=p0+∑k=1∞ekζ-k+∑k=1∞fkζk(20)ψ3ζ=q0+∑k=1∞gkζ-k+∑k=1∞hkζk.Assume that(21)bζ=∑k=1∞bkζ-k,dζ=∑k=1∞dkζ-k,eζ=∑k=1∞ekζ-kfζ=∑k=1∞fkζk,gζ=∑k=1∞gkζ-k,hζ=∑k=1∞hkζk.The coefficientsak of the φ1(ζ) can be represented by the following equations:(22)ai=-2ΓRci,i=n,n-1.Leta=a1⋯an-2T; a1,…,an-2 can be determined as follows:(23)An-2×n-2a1×n-2=B1×n-2when i=j, Aij=jLi+j+1-1, and if i+j+1>n, Aij=-1; when i≠j, Aij=jLi+j+1, and if i+j+1>n, Aij=0.(24)Bi=2ΓRc1+Γ′R2ΓRc23ΓRc3⋯n-2ΓRcn-2T.The coefficientsLk can be computed by the following equations:(25)Ln=cnLn-1=cn-1Ln-j+1=∑k=1j-2j-1-kcj-1-kLn-k+1+cn-j-1,j=3,…,n.Thus, the coefficientsak of analytic function φ1(ζ) can be determined by combining (22)–(25).For the analytic functionψ1(ζ), the Sk in (16) can be calculated by the following equations:(26)Sk=-∑j=1n-1jLj+1aj,k=0-∑j=1n-k-1jLj+k+1aj,k=1,…,n-2.The three boundary conditions used to solve the undetermined coefficients (b0,bk,d0,dk,p0,q0,ek,fk,gk,hk) from the analytic functions (φ2(ζ),ψ2(ζ),φ3(ζ),ψ3(ζ)) are listing:(i)
The surface tractions along the inner boundary of lining atL1 are zero.(ii)
The surface tractions continue at the interfaceL2 from the both surrounding rock mass side and lining side.(iii)
The displacement continues at the interfaceL2 from the both surrounding rock mass side and lining side.The surface traction condition at inner boundaryL1 can be expressed as follows:(27)φ3σ1+ωσ1ω′σ1¯φ3′σ1¯+ψ3σ1¯=0.The surface condition at the interfaceL2 can be presented as follows:(28)φ2σ+ωσω′σ¯φ2′σ¯+ψ2σ¯=φ3σ+ωσω′σ¯φ3′σ¯+ψ3σ¯.The displacement condition at the interfaceL2 can be expressed as follows:(29)uR+ivR=uL+ivL.Substituting (11) and (12) into (29), the following equation can be obtained:(30)1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯=12G2κ2φ3σ-ωσω′σ¯φ3′σ¯-ψ3σ¯.Two functions can be derived using (27) and (28) based on the Cauchy integration [20]:(31)φ′¯31ζ=-ω′¯1/ζωR02ζp0+q0+p1R02ζ+p2R02ζ+q11ζ+q21ζ(32)φ′¯31ζ=ω′¯1/ζωζb0+d0-p0-q0+bζ-p1ζ-p2ζ-q11ζ-q21ζ+d1ζ+d′1ζ.Substituting (28) into (30), the following equation can be obtained:(33)1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+κ1G2+G1G1G2φ2σ+G1-G2G1G2ωσω′σ¯φ2′σ¯+G1-G2G1G2ψ2σ¯=1+κ2G2φ3σ.Using the Cauchy integration, (33) can be expressed as follows:(34)-1-ηκ1G1φ1ζ+1-ηG1∑k=1n-2Skζ-k-κ1G2+G1G1G2bζ-G1-G2G1G2∑k=1n-2Tkζ-k=-1+κ2G2p1ζ.Substitutingζby 1/ζ, the Cauchy integration of the conjugate of (34) can be expressed as follows:(35)1-ηG1ω¯ζω′1/ζφ′11ζ-∑k=1n-2Skζ-k-S0+ψ11ζ-G1-G2G1G2ω¯ζω′1/ζb′1ζ-∑k=1n-2Tkζ-k-T′+d1ζ=-1+κ2G2p2ζ.Substituting (16) into (35), the following equation can be derived:(36)pR1-η2G1-1+λζ+1-λ∑k=1nckζk-G1-G2G1G2ω¯ζω′1/ζb′1ζ-∑k=1n-2Tkζ-k-T′+d1ζ=-1+κ2G2p2ζ.Then (36) can be expressed as follows [20]:(37)pR1-η2G1-1+λζ+1-λ∑k=1nckζk-G1-G2G1G2∑k=2∞Tk′ζk-1+d1ζ=-1+κ2G2p2ζ.The following equation can be obtained by combining (33)–(35):(38)κ1G2+G1G1G2b0+G1-G2G1G2d0=1-ηG1S0′-G1-G2G1G2T1′+1+κ2G2p0.The coefficientsTk and Tk′ in (34)–(38) can be calculated using the following equations:(39)Tk=-∑j=1n-k-1jLj+k+1bj,k=1,…,n-2(40)T1′=-∑j=1n-1jLj+1bjT2′=-∑j=1njLjbjTk′=-∑j=1k-2jLk-j-1bj-∑j=k-1n+k-2jLj-k+2bj,k=3,…,∞.The undetermined coefficients inb(ζ),d(ζ),e(ζ),f(ζ),g(ζ), and h(ζ) can be determined based on (31), (32), (34), and (38). Linear simultaneous equations, which contain infinite number of equations, can be obtained by comparing the coefficient of the same order of variables ζk. But only finite terms of b(ζ),d(ζ),e(ζ),f(ζ),g(ζ), and h(ζ) are discussed here in order to get the problem solved. The number of terms of bk, ek, and hk is denoted as Ne, and number of terms of dk, fk, and gk are denoted as Nf.Letp0=x1, ej=xj+1 (j=1,…,Ne), q0=xNe+2, hj=xNe+2+j (j=1,…,Ne), fk=x2Ne+2+k, and gk=x2Ne+Nf+2+k (j=1,…,Nf); the following linear simultaneous equations can be obtained by comparing the constant and coefficients of the positive power of ζk in (31):(41)Cijxj+CiNe+1+kxNe+1+k+Ci2Ne+2+Lx2Ne+2+L+Ci2Ne+Nf+2+Lx2Ne+Nf+2+L=0.N e + 2 equations are included in (41).Comparing negative powers ofζk in (31), the following linear simultaneous equations can be obtained:(42)Dijxj+DiNe+1+kxNe+1+k+Di2Ne+2+Lx2Ne+2+L+Di2Ne+Nf+2+Lx2Ne+Nf+2+L=0.N f - 1 equations are included in (42).Letbj=x2Ne+2Nf+2+j(j=1,…,Ne),d0=x3Ne+2Nf+3, and dj=x3Ne+2Nf+3+j(j=1,…,Nf); the following linear simultaneous equations can be obtained by comparing the constant and the coefficients of the positive power of ζk in (32):(43)Eijxj+EiNe+1+kxNe+1+k+Ei2Ne+2+Lx2Ne+2+L+Ei2Ne+Nf+2+Lx2Ne+Nf+2+L+Ei2Ne+2Nf+2+kx2Ne+2Nf+2+k+Ei3Ne+2Nf+2+Mx3Ne+2Nf+2+M=0.N e + 2 equations are included in (43).Comparing negative powers ofζk in (32), the following linear simultaneous equations can be obtained:(44)Fijxj+FiNe+1+kxNe+1+k+Fi2Ne+2+Lx2Ne+2+L+Fi2Ne+Nf+2+Lx2Ne+Nf+2+L+Fi2Ne+2Nf+2+kx2Ne+2Nf+2+k+Fi3Ne+2Nf+2+Mx3Ne+2Nf+2+M=0.N f - 1 equations are included in (44).Comparing the coefficients of the negative powers ofζk in (34), the linear simultaneous equations can be derived as follows:(45)Gijxj+Gi2Ne+2Nf+2+kx2Ne+2Nf+2+k=Hi.N f - 1 equations are included in (45).H i in (45) can be expressed as(46)Hi=-1-ηSi+1-ηκ1aii=1,…,n-21-ηκ1aii=n-1,n.Comparing the coefficients of the negative powers ofζk in (37), the linear simultaneous equations can be obtained as(47)Pi2Ne+2+jx2Ne+2+j+Pi2Ne+Nf+2+kx2Ne+Nf+2+k+Pi3Ne+2Nf+2+Lx3Ne+2Nf+2+L=Qi.N f - 1 equations are included in (47), where(48)Qi=pR1-η21+λ-1-λc1i=1-pR1-η21-λcii=2,…,n.Therefore, there are totally3Ne+3Nf+2 equations in (41)–(48) for the 3Ne+3Nf+3 undetermined variables. One more equation is still needed to solve all the variables.The support effect of the lining to the rock-mass decreases as the distance between them increases. Thus, (10) can be expressed at infinity as(49)limζ→∞2G1u3R+iv3R=limζ→∞κ1φ2ζ-ωζω′ζ¯φ2′ζ¯-ψ2ζ¯=0.It can be obtained from (49) that(50)κ1limζ→∞φ2ζ-limζ→∞ωζω′ζ¯φ2′ζ¯-limζ→∞ψ2ζ¯=0.And then(51)κ1b0-d0=0.Substituting (51) and (40) into (38), the following equations can be obtained:(52)p0-1+κ11+κ2d0+G1-G2G11+κ2∑j=1n-1jLj+1bj=-G21-ηG11+κ2S0′Now, there are totally3Ne+3Nf+3 equations for the 3Ne+3Nf+3 variables. Briefly, the xj can be calculated using the following equation set:(53)CNe+2×2Ne+2Nf+2DNf-1×2Ne+2Nf+2ENe+2×3Ne+3Nf+3FNf-1×2Ne+2Nf+2GNe×3Ne+2Nf+2PNf×3Ne+3Nf+31+κ2xj×1=02Ne+2Nf+2×1HNe×1QNf×1-G21-ηS0′G1.The coefficients ofCij, Dij, Eij, Fij, Gij, and Pij have been listed in [16].Thus, all the undetermined coefficients of analytic functions can be solved.


## 3. Conformal Mapping for the MCA Tunnel

The tunnel lining and rock-mass inz-plane can be mapped into an annulus in theζ-plane by (1). The interface of the lining and surrounding rock-mass is reflected by the external radius (r1=1). The inner boundary of the lining is reflected by the inside radius (r0<1). The undetermined parameters in (1) are R and ck.Assume that any pointAj at the inner boundary of lining inz-plane is transferred to the point A′j at the inside radius inζ-plane (Figure 1). Any point Bj at the interface is similar to point B′j at the external radius. The conformal mapping function at the interface and the lining inner boundary can be expressed as follows:(54)rAj=Rr0eiβAj-αAj+∑k=0nckr0-ke-ikβAj+αAjrBj=Rr1eiβBj-αBj+∑k=0nckr1-ke-ikβBj+αBj.Departing the real part and imaginary part, (54) can be then expressed as(55)sin⁡βAj-αAj-∑k=0nckr0-ksin⁡kβAj+αAj=0rAj=Rcos⁡βAj-αAj+∑k=0nckr0-kcos⁡kβAj+αAjsin⁡βBj-αBj-∑k=0nckr1-ksin⁡kβBj+αBj=0rBj=Rcos⁡βBj-αBj+∑k=0nckr1-kcos⁡kβBj+αBj.Suppose thatA1 can be accurately mapped into the point A′1,R can be determined as follows:(56)R=r01+∑k=0nck/r1k.c k and r0 in (56) can be obtained by solving the following optimization problem:Objective function:(57)minf=∑j=0mr∗Aj-Rcos⁡βAj-αAj+∑k=0nckcos⁡kβAj+αAjr02+∑j=0mr∗Bj-Rcos⁡βBj-αBj+∑k=0nckcos⁡kβBj+αBjr1-k2.Boundary condition:(58)∑k=0nkck<1The cross-section of one common MCA highway tunnel is shown in Figure2(a). Using the above method, the conformal mapping function can be expressed as follows:(59)z=5.7577ζ+0.22694-0.13876ζ-1+0.0578ζ-2-0.00958ζ-3-0.00504ζ-4R0=0.935.Figure 2
Tunnel cross-section before and after mapping (unit m).
(a) (b)As shown in Figure2(b), the lining boundaries before and after mapping have a good agreement.


## 4. Solutions for the Forces and Displacement

### 4.1. Solutions for Stresses of Lining and Rock Mass

Combining (7) with (13), the stress components for any points of the lining can be obtained by the following equations:(60)σρ+σθ=4Re⁡φ3′ζω′ζσθ-σρ+2iτρθ=2ζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ′3ζω′′ζω′ζ2+ψ′3ζ.Moreover, (60) can be expanded as follows:(61)σθ=2Re⁡φ3′ζω′ζ+Reζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζσρ=2Re⁡φ3′ζω′ζ-Reζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζτρθ=Imζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζ.In particular, the stresses of the interface can be calculated whenρ=1 and the stresses of the inner boundary of the lining can be calculated when ρ=R0.Combining (7) with (14), the stress components for points in rock-mass can be obtained by the following equations:(62)σθ=2Re⁡φ′ζω′ζ+Reζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζσρ=2Re⁡φ′ζω′ζ-Reζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζτρθ=Imζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζ.The related terms in (61)-(62) can be calculated by the following equations:(63)ωζ¯=ω-ζ-=ω-ρ2ζ=Rρ2ζ-1+∑k=0nckρ-2kζkω-1ζ=Rζ-1+∑k=0nckζkω-1ζ′=R-ζ-2+∑k=1nkckζk-1ω′ζ=R1-∑k=1nkckζ-k-1ω′ζ¯=R1-∑k=1nkckρ-2k-2ζk+1ω′′ζ=R∑k=1nkk+1ckζ-k-2.


### 4.2. Solutions for Internal Forces

Assume that the tangential stress at different depth of the lining varies linearly along the lining thickness, the bending moment, and axial force of the lining can be obtained based onσθ of the inner and outer boundaries of the lining. The strains under the two boundaries of the lining caused by the bending moment and axial force are shown in Figure 3.Figure 3
Sketch map for lining forces.Denote that the strain is caused by the bending moment asεM and axial force as εN. Strains at the two boundaries can be expressed as(64)ε1=εN-εMε2=εN+εM.Furthermore,εM and εN can be calculated by the following equations:(65)εM=ε2-ε12εN=ε2+ε12.According to the material mechanics, the stress can be obtained as follows:(66)σ=MWσ=NA,where, W is the lining bending rigidity and A is the cross-section area.Combining (64)–(66), the bending moment and axial force can be expressed as(67)M=E2ε2-ε12W=σθ2-σθ12WN=E2ε2+ε12A=σθ2+σθ12A,where σθ1 is the Tangential stress on the tension side, while σθ2 is the Tangential stress on the compression side.


### 4.3. Solutions for Displacements

From (12), the displacement components of the lining can be expressed as follows:(68)uL=Re12G2κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3′ζ¯vL=Im12G2κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3′ζ¯.From (11), the displacement components of the surrounding rock-mass can be expressed as(69)uR=Re1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯vR=Im1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯.


## 4.1. Solutions for Stresses of Lining and Rock Mass

Combining (7) with (13), the stress components for any points of the lining can be obtained by the following equations:(60)σρ+σθ=4Re⁡φ3′ζω′ζσθ-σρ+2iτρθ=2ζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ′3ζω′′ζω′ζ2+ψ′3ζ.Moreover, (60) can be expanded as follows:(61)σθ=2Re⁡φ3′ζω′ζ+Reζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζσρ=2Re⁡φ3′ζω′ζ-Reζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζτρθ=Imζ2ρ21ω′ζ¯ωζ¯φ′′3ζω′ζ-φ3′ζω′′ζω′ζ2+ψ′3ζ.In particular, the stresses of the interface can be calculated whenρ=1 and the stresses of the inner boundary of the lining can be calculated when ρ=R0.Combining (7) with (14), the stress components for points in rock-mass can be obtained by the following equations:(62)σθ=2Re⁡φ′ζω′ζ+Reζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζσρ=2Re⁡φ′ζω′ζ-Reζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζτρθ=Imζ2ρ21ω′ζ¯ωζ¯φ′′ζω′ζ-φ′ζω′′ζω′ζ2+ψ′ζ.The related terms in (61)-(62) can be calculated by the following equations:(63)ωζ¯=ω-ζ-=ω-ρ2ζ=Rρ2ζ-1+∑k=0nckρ-2kζkω-1ζ=Rζ-1+∑k=0nckζkω-1ζ′=R-ζ-2+∑k=1nkckζk-1ω′ζ=R1-∑k=1nkckζ-k-1ω′ζ¯=R1-∑k=1nkckρ-2k-2ζk+1ω′′ζ=R∑k=1nkk+1ckζ-k-2.


## 4.2. Solutions for Internal Forces

Assume that the tangential stress at different depth of the lining varies linearly along the lining thickness, the bending moment, and axial force of the lining can be obtained based onσθ of the inner and outer boundaries of the lining. The strains under the two boundaries of the lining caused by the bending moment and axial force are shown in Figure 3.Figure 3
Sketch map for lining forces.Denote that the strain is caused by the bending moment asεM and axial force as εN. Strains at the two boundaries can be expressed as(64)ε1=εN-εMε2=εN+εM.Furthermore,εM and εN can be calculated by the following equations:(65)εM=ε2-ε12εN=ε2+ε12.According to the material mechanics, the stress can be obtained as follows:(66)σ=MWσ=NA,where, W is the lining bending rigidity and A is the cross-section area.Combining (64)–(66), the bending moment and axial force can be expressed as(67)M=E2ε2-ε12W=σθ2-σθ12WN=E2ε2+ε12A=σθ2+σθ12A,where σθ1 is the Tangential stress on the tension side, while σθ2 is the Tangential stress on the compression side.


## 4.3. Solutions for Displacements

From (12), the displacement components of the lining can be expressed as follows:(68)uL=Re12G2κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3′ζ¯vL=Im12G2κ2φ3ζ-ωζω′ζ¯φ3′ζ¯-ψ3′ζ¯.From (11), the displacement components of the surrounding rock-mass can be expressed as(69)uR=Re1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯vR=Im1-η2G1κ1φ1σ-ωσω′σ¯φ1′σ¯-ψ1σ¯+12G1κ1φ2σ-ωσω′σ¯φ2′σ¯-ψ2σ¯.


## 5. Example and Discussion

Take a highway tunnel as an example. The tunnel cross-section is shown in Figure2. The parameters for the calculation are E1 = 15 GPa; E2 = 20 GPa; μ1 = 0.28; μ2 = 0.2; η = 0.6; p = 6 MPa; λ=1.


### 5.1. Accuracy of the Complex Function Method

The accuracy of the complex function method was highly related to the number ofNe andNf. The reality stress components σρ and τρθ of the lining inner boundary equal zero. Thus, the accuracy of the proposed method can be examined by comparing the calculated σρ and τρθ of the lining inner boundary with zero. The results of σρ and τρθ at L1 were shown in Figures 4–9. As shown in Figures 4–9, the calculation precision was improved with the increase of the number ofNe. WhenNe equals to 10 or 20, the boundary condition at L1 was not well satisfied with the reality. WhileNe equals 30, the absolute results of calculated normal stress σρ and shear stress τρθ are far less than 0.001. This means the complex variable method could be accurate enough to satisfy the boundary condition at L1 when theNe is greater than 30.Figure 4
Normal stress of lining inner boundary (Ne=10).Figure 5
Shear stress of lining inner boundary (Ne=10).Figure 6
Normal stress of lining inner boundary (Ne=20).Figure 7
Shear stress of lining inner boundary (Ne=20).Figure 8
Normal stress of lining inner boundary (Ne=30).Figure 9
Shear stress of lining inner boundary (Ne=30).The displacement components of lining and surrounding rock-mass were also presented in Figures10 and 11. It can be concluded from the two figures that the displacement continuity condition was also well satisfied whenNe equals 30. Thus, 30 were chosen as the number ofNe in the following calculation.Figure 10
Displacementu along L2 (Ne=30).Figure 11
Displacementv along L2 (Ne=30).


### 5.2. Lining Forces and Discussion

The tangential stresses of the two boundaries of the lining were shown in Figure12. Both tangential stresses showed the same trend. The left or right half of the tunnel lining possessed two peaks and two troughs in the tangential stress curve. The maximum value occurred at 115° from the tunnel crown to the side wall. Another peak occurred at the tunnel crown. The troughs occurred at about 70° from the tunnel crown to the side wall as well as the centre part of the tunnel invert. From Figure 13, both positive and negative bending moment existed in the lining. The three turning points of the positive and negative of the bending moment are 40°, 95°, and 135°. Figure 14 is the 3D surface plot of the tangential stress in different depth of the lining. From the projection of the 3D surface on the x-o-z plane (θ-o-σθ plane), the tangential stress of different layers of lining intersected each other at the sameθ where the bending moment was zero. From the projection of the 3D surface on the y-o-z plane (t-o-σθ plane), a linear relationship between lining depth and tangential stress could be found and provided evidence for the assumption in Section 4.2.Figure 12
Tangential stress at the lining boundaries.Figure 13
Axial force and bending moment.Figure 14
Tangential stress of the lining in different depth.Figures15 and 16 showed the relationship between the tangential stresses and the parametersθ andλ at the two lining boundaries.It can be obtained that λ had great influence on tangential stresses of the lining. From the projection on thex-o-yplane, negative tangential stress occurred whenλ was smaller than 0.6. Thus, for tunnels to be built in rock with small lateral earth pressure coefficient (e.g., loess with a coefficient of lateral earth pressure equals 0.5 [21, 22]), reinforced concrete, and fibre concrete linings rather than plain concrete linings are recommended for the higher durability performance of the tunnel. From the projection on thex-o-zplane, negative tangential stress occurred whenθ was within 40°~90°. Therefore, tension-resistant design should be applied mainly to this part of lining.Figure 15
Tangential stress of inner boundary of the lining withθ and λ.Figure 16
Tangential stress of outer boundary of the lining withθ and λ.The 3D surface projection on they-o-zplane showed a various linear relationship between the coefficient of lateral earth pressure and the tangential stress with differentθ.


### 5.3. Surrounding Rock-Mass Stress and Discussion

From Figure17, the tangential stress declined sharply whenρ was small than 2. In contrast, tangential stress was almost the same in regions whereρ was great than 3. Therefore, the tunnel excavation and lining support had a limited impact on the surrounding rock-mass.Figure 17
Relationship between tangential stress of rock-mass and parametersθ andρ.The surrounding rock-mass will be in an unstable state when the tangential stress equals zero [23]. The tangential stress of the interface from the rock side was calculated and shown in Figure 18. When the coefficient of the lateral pressure was between 0 and 0.6, the zero value of the tangential stress occurred whereθ is from 45° to 90°, 160° to 200°, and 270° to 315°. Therefore, supporting of the surrounding rock-mass in these regions should be enhanced to avoid the collapse during the tunnel excavation.Figure 18
Relationship betweenθ andλof the interface from the rock side, while σθ=0.


## 5.1. Accuracy of the Complex Function Method

The accuracy of the complex function method was highly related to the number ofNe andNf. The reality stress components σρ and τρθ of the lining inner boundary equal zero. Thus, the accuracy of the proposed method can be examined by comparing the calculated σρ and τρθ of the lining inner boundary with zero. The results of σρ and τρθ at L1 were shown in Figures 4–9. As shown in Figures 4–9, the calculation precision was improved with the increase of the number ofNe. WhenNe equals to 10 or 20, the boundary condition at L1 was not well satisfied with the reality. WhileNe equals 30, the absolute results of calculated normal stress σρ and shear stress τρθ are far less than 0.001. This means the complex variable method could be accurate enough to satisfy the boundary condition at L1 when theNe is greater than 30.Figure 4
Normal stress of lining inner boundary (Ne=10).Figure 5
Shear stress of lining inner boundary (Ne=10).Figure 6
Normal stress of lining inner boundary (Ne=20).Figure 7
Shear stress of lining inner boundary (Ne=20).Figure 8
Normal stress of lining inner boundary (Ne=30).Figure 9
Shear stress of lining inner boundary (Ne=30).The displacement components of lining and surrounding rock-mass were also presented in Figures10 and 11. It can be concluded from the two figures that the displacement continuity condition was also well satisfied whenNe equals 30. Thus, 30 were chosen as the number ofNe in the following calculation.Figure 10
Displacementu along L2 (Ne=30).Figure 11
Displacementv along L2 (Ne=30).


## 5.2. Lining Forces and Discussion

The tangential stresses of the two boundaries of the lining were shown in Figure12. Both tangential stresses showed the same trend. The left or right half of the tunnel lining possessed two peaks and two troughs in the tangential stress curve. The maximum value occurred at 115° from the tunnel crown to the side wall. Another peak occurred at the tunnel crown. The troughs occurred at about 70° from the tunnel crown to the side wall as well as the centre part of the tunnel invert. From Figure 13, both positive and negative bending moment existed in the lining. The three turning points of the positive and negative of the bending moment are 40°, 95°, and 135°. Figure 14 is the 3D surface plot of the tangential stress in different depth of the lining. From the projection of the 3D surface on the x-o-z plane (θ-o-σθ plane), the tangential stress of different layers of lining intersected each other at the sameθ where the bending moment was zero. From the projection of the 3D surface on the y-o-z plane (t-o-σθ plane), a linear relationship between lining depth and tangential stress could be found and provided evidence for the assumption in Section 4.2.Figure 12
Tangential stress at the lining boundaries.Figure 13
Axial force and bending moment.Figure 14
Tangential stress of the lining in different depth.Figures15 and 16 showed the relationship between the tangential stresses and the parametersθ andλ at the two lining boundaries.It can be obtained that λ had great influence on tangential stresses of the lining. From the projection on thex-o-yplane, negative tangential stress occurred whenλ was smaller than 0.6. Thus, for tunnels to be built in rock with small lateral earth pressure coefficient (e.g., loess with a coefficient of lateral earth pressure equals 0.5 [21, 22]), reinforced concrete, and fibre concrete linings rather than plain concrete linings are recommended for the higher durability performance of the tunnel. From the projection on thex-o-zplane, negative tangential stress occurred whenθ was within 40°~90°. Therefore, tension-resistant design should be applied mainly to this part of lining.Figure 15
Tangential stress of inner boundary of the lining withθ and λ.Figure 16
Tangential stress of outer boundary of the lining withθ and λ.The 3D surface projection on they-o-zplane showed a various linear relationship between the coefficient of lateral earth pressure and the tangential stress with differentθ.


## 5.3. Surrounding Rock-Mass Stress and Discussion

From Figure17, the tangential stress declined sharply whenρ was small than 2. In contrast, tangential stress was almost the same in regions whereρ was great than 3. Therefore, the tunnel excavation and lining support had a limited impact on the surrounding rock-mass.Figure 17
Relationship between tangential stress of rock-mass and parametersθ andρ.The surrounding rock-mass will be in an unstable state when the tangential stress equals zero [23]. The tangential stress of the interface from the rock side was calculated and shown in Figure 18. When the coefficient of the lateral pressure was between 0 and 0.6, the zero value of the tangential stress occurred whereθ is from 45° to 90°, 160° to 200°, and 270° to 315°. Therefore, supporting of the surrounding rock-mass in these regions should be enhanced to avoid the collapse during the tunnel excavation.Figure 18
Relationship betweenθ andλof the interface from the rock side, while σθ=0.


## 6. Conclusion

An elastic plane strain solution for MCA tunnels was presented based on the complex variable method. Stress and displacement components were predicted by employing complex potential functions which were consequently determined by Cauchy integration and series solution method. Force solutions for linings were also obtained according to the tangential stress at the two boundaries.The boundary conditions can be well satisfied with the reality when the number of terms of the analytic function is greater than 30. As to the linings, the maximum value of the tangential stress occurs at the corner of the side wall. The negative tangential stress occurs in tunnels with small lateral earth pressure coefficients, especially at regions from the tunnel shoulder to the side wall. A linear relationship was shown between the tangential stress and the lining depth. Also, the change of tangential stress is proportional to the coefficient of lateral earth pressure, but the slope varies with angles from the tunnel crown to the tunnel side wall.As to the surrounding rock-mass, the tunnel excavation and lining support have limited influence on the region whenρ is less than 2. When the coefficient of the lateral pressure is between 0 and 0.6, the zero value of the tangential stress occurs in the section from the tunnel shoulder to the side wall as well as the middle part of the invert. Thus, during the tunnel excavation, the addition of surrounding rock-mass in these regions should be considered to avoid the collapse accidents.The difference existing before and after the conformal mapping of the lining outline also has an influence on correctness of results. More attempts should be made in upcoming works to improve the accuracy of conformal mapping functions and preserve shape after mapping.



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*Source: 8409129-2018-05-16.xml*