# An Improved Stress and Strain Increment Approaches for Circular Tunnel in Strain-Softening Surrounding Rock Considering Seepage Force

Authors: Ling Wang; Jin-feng Zou; Yu-ming Sheng
Journal: Advances in Materials Science and Engineering
 (2019)

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## Abstract

Considering the effect of seepage force, a dimensionless approach was introduced to improve the stress and strain increment approach on the stresses and radial displacement around a circular tunnel excavated in a strain-softening generalized Hoek–Brown or Mohr–Coulomb rock mass. The circular tunnel can be simplified as axisymmetric problem, and the plastic zone was divided into a finite number of concentric rings which satisfy the equilibrium and compatibility equations. Increments of stresses and strains for each ring were obtained by solving the equilibrium and compatibility equations. Then, the stresses and displacements in softening zone can be calculated. The correctness and reliability of the proposed approach were performed by the existing solutions.


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## Body

## 1. Introduction

Analysis of stresses and displacements around circular tunnel excavated in isotropic rock mass is one of the fundamental problems in geotechnical engineering. A nonlinear method is needed to obtain a reliable solution since the deformation response depends on the stress path. Numerical method, elastoplastic methods, and limit analysis methods are popularly used for cavity expansion or contraction according to the existing literature surveys [1–15]. In the past, Mohr–Coulomb failure criterion was widely used by many scholars for analysis of rock mass. However, experimental observations indicated that the strength envelope for most of rock materials is nonlinear. For the nonlinear failure criterion, the criterion by Hoek and Brown is generally accepted in rock mechanics community, as it can provide a reliable tool for predicting the strength of jointed rock mass. Subsequently, the Hoek–Brown criterion has been updated to the generalized form and has been widely used. Elastoplastic analysis of circular tunnels excavated in Hoek–Brow and Mohr–Coulomb rock mass was attempted by many scholars [13–16]. For the theoretical analysis, the expressions of stresses and displacements around the tunnel are mainly obtained based on the elastic-perfectly plastic model and elastic-brittle-plastic model under the nonassociated flow rule. Park and Kim [13] provided a procedure of strain-softening model for elastoplastic analysis of a circular opening considering elastoplastic coupling. Park et al. [14] presented the variation laws of deformations based on different softening indexes and dilatancy characteristics. Sharan [15] provided an analytical solution for stresses and displacements around a circular opening in a generalized Hoek–Brown rock mass. Theoretical formulation and solutions for poorly consolidated rocks surrounding a circular opening are presented by Wang [16]. Han et al. [17] analyzed variation regulations of the stresses and deformation around the tunnel for elastic-brittle-plastic behavior considering the load-bearing characteristics of the plastic zone. However, the variation laws of the stresses and deformation around the tunnel excavated in a strain-softening rock mass were studied by numerical analysis. Zheng et al. [18] pointed out that the convergent problem was existed in the finite-element analysis of strain-softening rock mass with a higher softening rate. Therefore, the plastic zone of a circular tunnel may be divided into many zones to be analyzed generally. For example, Brown et al. [2] thought that the elastic deformation of rock mass is a constant value; then, the stresses and displacement field can be calculated easily. Wang et al. [19, 20] presented an analytical approximation solution for a circular opening in an elasto-brittle-plastic rock. Lee and Pietruszczak [21] provided a new numerical procedure for elastoplastic analysis of a circular opening excavated in a strain-softening rock mass by using the difference method.It can be seen from the researches worldwide that the theoretical analysis method and numerical simulation are two main techniques for analysis of strain-softening rock mass. The theoretical approaches can reflect the innate character of strain-softening rock mass, but the parameters for calculation which are usually not consistent with the actual values may cause computational errors as well. The numerical simulation shows the development of the softening rock mass. However, there exists some error in the solutions compared with the theoretical values. In the recent reports about the elastoplastic analysis of a circular opening excavated in a strain-softening rock mass, it is unusual to find a solution considering both the axial stress and seepage force in practical engineering. But the axial stress and seepage force have obvious influence on the surrounding rock’s stability, especially in strain-softening rock mass. And most diseases are related to the seepage force directly or indirectly. Only analyzing the influence of axial stress on the stresses and displacement of circular tunnel without considering seepage force does not accord with the practical engineering at abundant area. Therefore, the seepage fore is of great important for softening rock mass.Therefore, on the basis of theoretical analysis, the dimensionless method is introduced in this study to eliminate the influence on the dimensions of variables and parameters based on the generalized Hoek–Brown and M-C failure criterion. Simultaneously, both the axial stress and seepage force are considered for reconstruction of the step-stress approach to analyze the strain-softening rock mass. Then, a new dimensionless method for the elastoplastic analysis of a circular opening excavated in a strain-softening rock mass considering seepage force is established in this study, which in hope of providing theoretical supports for the digging process and design of tunnel in a strain-softening rock mass.The innovations of this study are listed as follows:(1)
The dimensionless method pointed out by Carranza-Torres and Fairhurst [22] is only applicable to the original Hoek–Brown rock mass. However, the dimensionless solutions in this study are based on the generalized Hoek–Brow and M-C rock mass. Thus, the dimensionless method of Carranza-Torres and Fairhurst [22] is a special case of that in this study, and the dimensionless method in this study is the expansion and extension of that of Carranza-Torres and Fairhurst [22].(2)
Not only the generalized Hoek–Brown and M-C failure criteria but also the seepage force is considered in this study, and a new dimensionless method for analysis of the circular opening excavated in a strain-softening rock mass is presented, which is rarely studied from the existing literatures.(3)
The dimensionless method pointed out by Carranza-Torres and Fairhurst [22] is only suitable for the elastoplastic rock mass. However, the dimensionless method in this study is used in strain-softening rock mass and can also be simplified for the elastic-brittle-plastic and elastoplastic rock mass, which means the dimensionless method in Carranza-Torres and Fairhurst [22] is a special case of that in this study, and the dimensionless method in this study is the expansion and extension of that in Carranza-Torres and Fairhurst [22].


## 2. Definition of Problem

Figure1 illustrates that a circular tunnel with radius b is imposed by a stress field σ0 throughout the domain before the tunnel is excavated. As the internal support pressure pi is less than the critical value pic, a plastic zone is formed around the circular tunnel. Plastic radius can be derived for the elastic-brittle-plastic or elastic-perfectly plastic behavior [23, 24]. Furthermore, when considering the strain-softening behavior, the plastic zone can be divided into softening and residual zones whose interface is expressed by RS in Figure 1. However, it is difficult to obtain a closed-form solution for strain-softening rock mass, and the distributions for stresses and displacement should be solved by numerical method.Figure 1
Plastic zone formed around circular opening.


### 2.1. Yield Function

Assuming that the yielding of the rock mass is governed by the function,(1)Fσθ,σr,γp=σθ−σr−Hσr,γp,where σθ and σr are the major principal stress and minor principal stress, respectively, and γp is the strain-softening parameter which reflects the evolution of the strength parameters in strain-softening rock mass and can be expressed as follows:(2)γp=γθp−γrp.Alonso et al. [1] pointed out that no universal method was available to define the strain-softening parameter, but the definition in Equation (2) is widely accepted now.For the M-C rock mass,(3)Hσr,γp=Nγp−1σr+Yγp,where N and Y are respectively strength parameters defined according to the friction angle ϕγp and cohesion cγp:(4)Nγp=1+sinϕγp1−sinϕγp,Yγp=2cγpcosϕγp1−sinϕγp.For the generalized H-B rock mass,(5)HHBσr,γp=σcγpmγpσrσcγp+sγpaγp,where σc is the uniaxial compressive strength of rock and m, s, and a are also the strength parameters for the H-B surrounding rock.


### 2.2. Plastic Potential Function

The M-C criterion is regarded as the plastic potential function, and it may be written as follows:(6)Gσθ,σr,γp=σθ−kγpσr,where kγp is the coefficient of dilation and may be written as follows:(7)kγp=1+sinφγp1−sinφγp,where φ is the angle of dilation, and when φ is equal to internal frictional angle ϕ, the plastic flow rule is related. When kγp=1.0, no plastic volume change occurs. Therefore, the relationship between the radial and circumferential plastic strain increments can be obtained based on the plastic flow rule:(8)dεrp=−kγpdεθp.


### 2.3. Evolution of Strength Parameters

The strength and deformation parameters presented in Equations (3), (5), and (6) are functions of γp. In plastic regime, these parameters can be described by bilinear functions of deviatoric plastic strain γp, which are shown in the following equation:(9)ωγp=ωp−ωr−ωpγpγp∗,0<γp<γp∗,ωr,γp≥γp∗,where ω denotes one of the parameters ϕ, c, σc, m, s, φ, and a, and γp∗ is the critical deviatoric plastic strain from which the residual behavior is firstly observed. The value of γp∗ should be identified by experiments. This linear deterioration process of strength parameters is illustrated in Figure 2, where the subscripts p and r represent the peak and residual values of strength and deformation parameters, respectively.Figure 2
Evolution of parameters in plastic regime.


### 2.4. Critical Supporting Pressure

When the internal support pressurepi is lower than pic, the plastic zone develops. For M-C rock mass, pic can be calculated by the following equation:(10)picMC=2σ0−YpNp+1,where Yp=2cpcosϕp/1−sinϕp and Np=1+sinϕp/1−sinϕp.For H-B rock mass,pic can be obtained by solving the following nonlinear equation:(11)2σ0−pic=σcpmppicσcp+spap.Whena=0.5, pic is expressed as follows:(12)picHB=12β−β2+4βσ0+sσcp2+σ0,where β=mpσcp/4.Ifa>0.5, picHB can be calculated numerically based on suitable root-finding algorithm, such as Newton–Raphson method.When the plastic zone is formed, radial stressσR is equal to pic on the elastic-plastic interface and σR is independent of radius r:(13)σR=σrRp=pic.


## 2.1. Yield Function

Assuming that the yielding of the rock mass is governed by the function,(1)Fσθ,σr,γp=σθ−σr−Hσr,γp,where σθ and σr are the major principal stress and minor principal stress, respectively, and γp is the strain-softening parameter which reflects the evolution of the strength parameters in strain-softening rock mass and can be expressed as follows:(2)γp=γθp−γrp.Alonso et al. [1] pointed out that no universal method was available to define the strain-softening parameter, but the definition in Equation (2) is widely accepted now.For the M-C rock mass,(3)Hσr,γp=Nγp−1σr+Yγp,where N and Y are respectively strength parameters defined according to the friction angle ϕγp and cohesion cγp:(4)Nγp=1+sinϕγp1−sinϕγp,Yγp=2cγpcosϕγp1−sinϕγp.For the generalized H-B rock mass,(5)HHBσr,γp=σcγpmγpσrσcγp+sγpaγp,where σc is the uniaxial compressive strength of rock and m, s, and a are also the strength parameters for the H-B surrounding rock.


## 2.2. Plastic Potential Function

The M-C criterion is regarded as the plastic potential function, and it may be written as follows:(6)Gσθ,σr,γp=σθ−kγpσr,where kγp is the coefficient of dilation and may be written as follows:(7)kγp=1+sinφγp1−sinφγp,where φ is the angle of dilation, and when φ is equal to internal frictional angle ϕ, the plastic flow rule is related. When kγp=1.0, no plastic volume change occurs. Therefore, the relationship between the radial and circumferential plastic strain increments can be obtained based on the plastic flow rule:(8)dεrp=−kγpdεθp.


## 2.3. Evolution of Strength Parameters

The strength and deformation parameters presented in Equations (3), (5), and (6) are functions of γp. In plastic regime, these parameters can be described by bilinear functions of deviatoric plastic strain γp, which are shown in the following equation:(9)ωγp=ωp−ωr−ωpγpγp∗,0<γp<γp∗,ωr,γp≥γp∗,where ω denotes one of the parameters ϕ, c, σc, m, s, φ, and a, and γp∗ is the critical deviatoric plastic strain from which the residual behavior is firstly observed. The value of γp∗ should be identified by experiments. This linear deterioration process of strength parameters is illustrated in Figure 2, where the subscripts p and r represent the peak and residual values of strength and deformation parameters, respectively.Figure 2
Evolution of parameters in plastic regime.


## 2.4. Critical Supporting Pressure

When the internal support pressurepi is lower than pic, the plastic zone develops. For M-C rock mass, pic can be calculated by the following equation:(10)picMC=2σ0−YpNp+1,where Yp=2cpcosϕp/1−sinϕp and Np=1+sinϕp/1−sinϕp.For H-B rock mass,pic can be obtained by solving the following nonlinear equation:(11)2σ0−pic=σcpmppicσcp+spap.Whena=0.5, pic is expressed as follows:(12)picHB=12β−β2+4βσ0+sσcp2+σ0,where β=mpσcp/4.Ifa>0.5, picHB can be calculated numerically based on suitable root-finding algorithm, such as Newton–Raphson method.When the plastic zone is formed, radial stressσR is equal to pic on the elastic-plastic interface and σR is independent of radius r:(13)σR=σrRp=pic.


## 3. Approximation of Strain-Softening Behavior

### 3.1. Preliminaries

The plastic zone can be divided inton concentric annuli, and the ith annulus is bounded by two circles of normalized radii ri−1=Ri−1/Rp and ri=Ri/Rp in Figure 3. The thickness of each annulus is not uniform in general. On the outer boundary of plastic zone, where r0=1, stress and strain components under plane strain condition are obtained from Equations (14) and (15).(14)σr0σθ0=σR2σ0−σR,(15)εr0εθ0=1+υEσR−σ0σ0−σR.Figure 3
Normalized plastic zone with finite number of annuli.


### 3.2. Increments of Stresses and Elastic Strains

A dimensionless method is proposed for calculating the increments of stresses and elastic strains. The circular tunnel can be simplified as axisymmetric problem, and the plastic zone is divided into a finite number of concentric rings which satisfy the equilibrium and compatibility equations in strain-softening rock mass. The increments of stresses and strains for each ring are obtained by solving the equilibrium and compatibility equations. Then, the stresses and displacements in the softening zone can be calculated.Define a dimensionless variableρ, which maps the physical plane r,ξ into a plane of coordinate ρ based on the following transformation:(16)ρ=rbξ.Then, the plastic zone can be translated into a unit plane. In the unit plane, the position of the elastic-plastic interface is fixed byρ=1, and the wall of the cavity is defined by ρ=1/ξ.In order to simplify the calculation process, the forms of the generalized Hoek–Brown and Mohr–Coulomb criteria need to be translated into the dimensionless forms which are in the same order of magnitude in this study. The yield function is expressed as follows:(17)fσθ,σr=σθ−σr−Hσr,γp=0.For the Hoek–Brown criterion, the transformed radial and tangential stresses are expressed as follows:(18)Sθ=σθmbσci+smb2,Sr=σrmbσci+smb2.The transformed internal pressure and far-field stress are(19)Pi=pimbσci+smb2,S0=σ0mbσci+smb2.In order to be consistent with the definition of elastic strain rates, shear modulus can be scaled according to the following expression:(20)γ=Gmbσci.Then, the yield function for the generalized Hoek–Brown criteria may be rewritten by(21)fSθ,Sr=Sθ−Sr−mb2a−1Sra=0.For the M-C criterion, the transformed radial and tangential stresses are given by(22)Sθ=σθN−1,Sr=σrN−1.And the transformed internal pressure and far-field stress are(23)Pi=piN−1,S0=σ0N−1.Then, the yield function for the Mohr–Coulomb criteria can be rewritten as follows:(24)fSθ,Sr=Sθ+NSr+YN−1=0.Based on the dimensionless method in Carranza-Torres and Fairhurst [22], the stress magnitude, σ0−σr∗, is used to normalize the stresses by(25)S˜θ=SθS0−Sr∗,S˜r=SrS0−Sr∗.Based on the method pointed out by Brown et al. [2], radial stresses in the plastic zone may be divided into n parts, and the increments of radial stress can be expressed as follows:(26)Δσr=pi−σ0n.According to Equation (23), and Equation (26) can be written as follows:(27)ΔS˜r=Pi−S0n.And the stress components for theith annulus can be given as follows:(28)S˜ri=S˜ri−1+ΔS˜r.In fact, ifn is sufficiently large, the circumferential stress can be written as follows:(29)S˜θi=S˜ri+HS˜ri,γi−1p.


### 3.3. Approximation of Displacements

When the number of annulin is sufficiently large and assuming that the strength parameters of rock mass is kept constant in each centric annulus, equilibrium equation can be transformed into(30)dσrdr+σr−σθr=0.According to the transformation in Equation (25), the partial derivatives of the field functions with respect to the variables r and ξ are evaluated with the following operators:(31)∂∂r=1bξddρ,(32)∂∂ξ=ρξddρ.In the unit plane, the equilibrium condition may be written as(33)dS˜rdρ+S˜r−S˜θρ=0,or(34)dS˜rdρ+HS˜ri,γi−1pρ=0.Equation (34) can be approximated for the ith annulus as(35)S˜ri−S˜ri−1ρi−ρi−1+HS˜¯ri,γi−1pρi−ρi−1=0,where S˜¯ri=S˜ri+S˜ri−1/2. Then, the inner radius can be obtained as follows:(36)ρi=2HS˜¯ri,γi−1p+ΔS˜r2HS˜¯ri,γi−1p−ΔS˜rρi−1.To solve the plastic strain increments, the displacement compatibility equation is considered:(37)dεθdr+εθ−εrr=0.In order to work with dimensionless field quantities, the strains may be normalized as follows:(38)ε˜θ=2γS0−Sr∗εθ,ε˜r=2γS0−Sr∗εr.So that the displacement compatibility equation can be written as follows:(39)dε˜θdρ+ε˜θ−ε˜rρ=0.In the plastic zone, the total strains can be decomposed into elastic and plastic parts as follows:(40)ε˜rε˜θ=ε˜reε˜θe+ε˜rpε˜θp.Equation (38) may be reformulated as(41)dε˜θpdρ+ε˜θp−ε˜rpρ=−dε˜θedρ−ε˜θe−ε˜reρ.Then,(42)dε˜θpdρ+ε˜θp−ε˜rpρ=−dε˜θedρ−ε˜θe−ε˜reρ.According to Hooke’s law, under the plane strain condition,(43)ΔεriΔεθi=1+vE1−v−v−v1−vΔσriΔσθi.So, Equation (43) may be reformulated as follows:(44)Δε˜riΔε˜θi=2γ1+vE1−v−v−v1−vΔS˜riΔS˜θi.Combining Equations (36), (42), and (44), the following equation can be obtained:(45)1Δρi+1+ki−11ρ¯iΔε˜θip=−Δε˜θieΔρi−2γ1+υEHS˜¯ri,γi−1pρ¯i−1ρ¯iε˜θi−1p−ε˜ri−1p,where ρ¯i=ρ¯i−1+ρ¯i/2 and ki−1=1+sinφi−1/1−sinφi−1. The deviatoric plastic shear strain is updated as follows:(46)γip=γi−1p+Δε˜θip−Δε˜rip.The total strain at theith annulus can be given by(47)ε˜riε˜θi=ε˜ri−1ε˜θi−1+Δε˜rieΔε˜θie+Δε˜ripΔε˜θip.Recalling the relationship,(48)u˜=2γbS0−Sr∗ur,ε˜θ=2γS0−Sr∗εθ,εθ=uR,then,(49)ε˜θ=u˜bR=u˜ρξRp.Then, the displacement normalized by plastic radiusRp can be obtained as follows:(50)U˜i=ε˜θiri=ε˜θiρiξ,where U˜i=u˜i/Rp.By using the dimensionless method, the stresses and displacements in the softening zone can be calculated. The actual value of displacement can be calculated by the following equation:(51)ur=bS0−Sr∗2γu˜.The plastic radiusRp can be calculated from the following formula:(52)Rp=brn=1ρnξ.


## 3.1. Preliminaries

The plastic zone can be divided inton concentric annuli, and the ith annulus is bounded by two circles of normalized radii ri−1=Ri−1/Rp and ri=Ri/Rp in Figure 3. The thickness of each annulus is not uniform in general. On the outer boundary of plastic zone, where r0=1, stress and strain components under plane strain condition are obtained from Equations (14) and (15).(14)σr0σθ0=σR2σ0−σR,(15)εr0εθ0=1+υEσR−σ0σ0−σR.Figure 3
Normalized plastic zone with finite number of annuli.


## 3.2. Increments of Stresses and Elastic Strains

A dimensionless method is proposed for calculating the increments of stresses and elastic strains. The circular tunnel can be simplified as axisymmetric problem, and the plastic zone is divided into a finite number of concentric rings which satisfy the equilibrium and compatibility equations in strain-softening rock mass. The increments of stresses and strains for each ring are obtained by solving the equilibrium and compatibility equations. Then, the stresses and displacements in the softening zone can be calculated.Define a dimensionless variableρ, which maps the physical plane r,ξ into a plane of coordinate ρ based on the following transformation:(16)ρ=rbξ.Then, the plastic zone can be translated into a unit plane. In the unit plane, the position of the elastic-plastic interface is fixed byρ=1, and the wall of the cavity is defined by ρ=1/ξ.In order to simplify the calculation process, the forms of the generalized Hoek–Brown and Mohr–Coulomb criteria need to be translated into the dimensionless forms which are in the same order of magnitude in this study. The yield function is expressed as follows:(17)fσθ,σr=σθ−σr−Hσr,γp=0.For the Hoek–Brown criterion, the transformed radial and tangential stresses are expressed as follows:(18)Sθ=σθmbσci+smb2,Sr=σrmbσci+smb2.The transformed internal pressure and far-field stress are(19)Pi=pimbσci+smb2,S0=σ0mbσci+smb2.In order to be consistent with the definition of elastic strain rates, shear modulus can be scaled according to the following expression:(20)γ=Gmbσci.Then, the yield function for the generalized Hoek–Brown criteria may be rewritten by(21)fSθ,Sr=Sθ−Sr−mb2a−1Sra=0.For the M-C criterion, the transformed radial and tangential stresses are given by(22)Sθ=σθN−1,Sr=σrN−1.And the transformed internal pressure and far-field stress are(23)Pi=piN−1,S0=σ0N−1.Then, the yield function for the Mohr–Coulomb criteria can be rewritten as follows:(24)fSθ,Sr=Sθ+NSr+YN−1=0.Based on the dimensionless method in Carranza-Torres and Fairhurst [22], the stress magnitude, σ0−σr∗, is used to normalize the stresses by(25)S˜θ=SθS0−Sr∗,S˜r=SrS0−Sr∗.Based on the method pointed out by Brown et al. [2], radial stresses in the plastic zone may be divided into n parts, and the increments of radial stress can be expressed as follows:(26)Δσr=pi−σ0n.According to Equation (23), and Equation (26) can be written as follows:(27)ΔS˜r=Pi−S0n.And the stress components for theith annulus can be given as follows:(28)S˜ri=S˜ri−1+ΔS˜r.In fact, ifn is sufficiently large, the circumferential stress can be written as follows:(29)S˜θi=S˜ri+HS˜ri,γi−1p.


## 3.3. Approximation of Displacements

When the number of annulin is sufficiently large and assuming that the strength parameters of rock mass is kept constant in each centric annulus, equilibrium equation can be transformed into(30)dσrdr+σr−σθr=0.According to the transformation in Equation (25), the partial derivatives of the field functions with respect to the variables r and ξ are evaluated with the following operators:(31)∂∂r=1bξddρ,(32)∂∂ξ=ρξddρ.In the unit plane, the equilibrium condition may be written as(33)dS˜rdρ+S˜r−S˜θρ=0,or(34)dS˜rdρ+HS˜ri,γi−1pρ=0.Equation (34) can be approximated for the ith annulus as(35)S˜ri−S˜ri−1ρi−ρi−1+HS˜¯ri,γi−1pρi−ρi−1=0,where S˜¯ri=S˜ri+S˜ri−1/2. Then, the inner radius can be obtained as follows:(36)ρi=2HS˜¯ri,γi−1p+ΔS˜r2HS˜¯ri,γi−1p−ΔS˜rρi−1.To solve the plastic strain increments, the displacement compatibility equation is considered:(37)dεθdr+εθ−εrr=0.In order to work with dimensionless field quantities, the strains may be normalized as follows:(38)ε˜θ=2γS0−Sr∗εθ,ε˜r=2γS0−Sr∗εr.So that the displacement compatibility equation can be written as follows:(39)dε˜θdρ+ε˜θ−ε˜rρ=0.In the plastic zone, the total strains can be decomposed into elastic and plastic parts as follows:(40)ε˜rε˜θ=ε˜reε˜θe+ε˜rpε˜θp.Equation (38) may be reformulated as(41)dε˜θpdρ+ε˜θp−ε˜rpρ=−dε˜θedρ−ε˜θe−ε˜reρ.Then,(42)dε˜θpdρ+ε˜θp−ε˜rpρ=−dε˜θedρ−ε˜θe−ε˜reρ.According to Hooke’s law, under the plane strain condition,(43)ΔεriΔεθi=1+vE1−v−v−v1−vΔσriΔσθi.So, Equation (43) may be reformulated as follows:(44)Δε˜riΔε˜θi=2γ1+vE1−v−v−v1−vΔS˜riΔS˜θi.Combining Equations (36), (42), and (44), the following equation can be obtained:(45)1Δρi+1+ki−11ρ¯iΔε˜θip=−Δε˜θieΔρi−2γ1+υEHS˜¯ri,γi−1pρ¯i−1ρ¯iε˜θi−1p−ε˜ri−1p,where ρ¯i=ρ¯i−1+ρ¯i/2 and ki−1=1+sinφi−1/1−sinφi−1. The deviatoric plastic shear strain is updated as follows:(46)γip=γi−1p+Δε˜θip−Δε˜rip.The total strain at theith annulus can be given by(47)ε˜riε˜θi=ε˜ri−1ε˜θi−1+Δε˜rieΔε˜θie+Δε˜ripΔε˜θip.Recalling the relationship,(48)u˜=2γbS0−Sr∗ur,ε˜θ=2γS0−Sr∗εθ,εθ=uR,then,(49)ε˜θ=u˜bR=u˜ρξRp.Then, the displacement normalized by plastic radiusRp can be obtained as follows:(50)U˜i=ε˜θiri=ε˜θiρiξ,where U˜i=u˜i/Rp.By using the dimensionless method, the stresses and displacements in the softening zone can be calculated. The actual value of displacement can be calculated by the following equation:(51)ur=bS0−Sr∗2γu˜.The plastic radiusRp can be calculated from the following formula:(52)Rp=brn=1ρnξ.


## 4. Verification Examples

### 4.1. Verifications for H-B and M-C Rock Mass

In order to verify the correctness of the dimensionless solutions, the generalized Hoek–Brown criterion is firstly considered in this study. It is obvious that the strain-softening solutions converge to the brittle-plastic solutions when the critical deviatoric plastic strainγp∗ is equal to 0. However, the elastic-brittle-plastic case, γp∗=0, is the special case of strain-softening behavior. Comparing with the exact solutions for elastic-brittle-plastic rock mass, the dimensionless solutions can be verified.The stresses and displacement obtained by Sharan [15] are compared with the results in this study, and the equations given by Sharan [15] are obtained from the following equations:(53)σr=p0+14mbσcilnra2+lnrambσcip0+srσci2,(54)σθ=σr+mbσciσr+srσci2,(55)u=1r1+vEσR−σ0+∫rRrεr+εθdr.According to the dimensionless method, compared with the results in this study in the same order of magnitude, Equations (53)–(55) can be normalized as follows:(56)S˜r=1S0−Sr∗P0+14lnρξ2+lnρξP0,(57)S˜θ=S˜r+P0S0−Sr∗,(58)u˜=2γrbS0−Sr∗1+vEσR−σ0+∫rRrεr+εθdr.In addition, compared with the results based on the generalized H-B and M-C failure criteria, the technique of equivalent M-C and generalized H-B strength parameters was adopted for the comparison between the developed method and the exact solutions in Sharan [15]. The equations for the friction angle (ϕ) and cohesive (i) are given by Yang and Pan [25]:(59)ϕ=sin−16ambs+mbσ3na−121+a2+a+6ambs+mbσ3na−1,(60)c=σci1+2as+1−ambσ3ns+mbσ3na−11+a2+a1+6ambs+mbσ3na−1/1+a2+a,where σ3n=σ3max/σci, σ3max/σcm=0.47σcm/γH−0.94, σcm=σcimb+4s−amb−8smb/4+sa−1/21+a2+a, σcm is the rock mass strength, γ is the unit weight of the rock mass, and H is the depth of the tunnel below the surface.The above formulas can be used to transfer the parameters from the H-B yield criterion to M-C yield criterion. But that may create some errors in practical calculations.The data sets appearing in Sharan [15] were chosen as input parameters for calculation in this study: b=5m, σ0=30MPa, pi=5MPa, E=5GPa, v=0.25, σcp=σcr=30MPa, mp=1.7, sp=0.0039, mr=1.0, sr=0.0, and ap=ar=0.5. And two dilation angles, φr=0∘ and 30∘, were used to evaluate the influence of plastic volume change.Stresses and displacements in this study can be obtained by writing a program at the case ofγp=0. Then, based on Equations (56)–(58), the elastic-brittle-plastic solutions and exact solutions can be compared. The results are shown in Figures 4 and 5.Figure 4
Comparison between dimensionless stresses and exact stresses for Hoek–Brown rock mass.Figure 5
Comparison between dimensionless displacement and exact displacement for Hoek–Brown rock mass.Figure4 shows the comparison between dimensionless solutions and exact solutions for Hoek–Brown rock mass. The ordinate S stands for the dimensionless stresses.From Figure4, when the number of annuli n=500, the distributions of radial and circumferential stresses match the exact solutions well, which validate the accuracy of the dimensionless solutions. Figure 5 gives the comparison between dimensionless displacement and exact displacement for Hoek–Brown rock mass. As can be seen in Figure 5, the distribution of radial displacement obtained by the proposed approach shows a good agreement with the exact solution. So, the dimensionless solutions prove to be accurate.Figures6 and 7 show the distribution of errors between dimensionless solutions and exact solutions. The abscissa denotes ρξ. The ordinate stands for the error percentage between dimensionless solutions and exact solutions. Moreover, in order to examine the influence of mesh density on dimensionless solutions, two values of n are considered in this example, which are 10 and 500.Figure 6
Distribution of errors between dimensionless stresses and exact stresses.Figure 7
Distribution of errors between dimensionless displacement and exact displacement.From Figures6 and 7, radial stress has a little error to the exact solution with the maximum error 2.13% and the minimum error 0.06%. The circumferential stresses obtained in this study can finely match the exact solution with the maximum error 1.55% and minimum error 4.1×10−4%. Furthermore, the value of n has great influence on the dimensionless solutions, and the dimensionless solutions for n=500 are obviously more accurate than that for n=10.More different values ofn have been studied in this example, and we can find that the accuracy of dimensionless solutions will increase when the value of n grows. However, when value of n is larger than 500, the value of n has little influence on the dimensionless solutions. Therefore, to increase computational efficiency, the results of stresses and displacement for n=500 can be selected.To analyze the reliability of solutions under the M-C criterion, the stresses and displacement for the M-C surrounding rock obtained by the proposed dimensionless approach, and the approach by Lee and Pietruszczak [21] is compared. The parameters are presented as follows: b=5.35m, σ0=3.31MPa, E=1.38GPa, v=0.25, cp=1MPa, cr=0.7MPa, ϕp=30°, ϕr=22°, α=3.5γp∗=α−1εθ1e, φp=φr=3.75°, and pi=0MPa. The computed stresses and displacement using two numerical approaches are illustrated in Figures 8 and 9.Figure 8
Comparison of dimensionless stresses for M-C rock mass.Figure 9
Comparison of dimensionless displacement for M-C rock mass.The finite difference approach is also adopted by Lee and Pietruszczak [21]. The stress increment approach for the mechanical states of strain-softening rock mass is developed in their study. It can be seen from Figures 8 and 9, when n = 10 and 100, the determined circumferential stresses are respectively 10.2% and 4.5% larger than the circumferential stress in Lee and Pietruszczak [21]; meanwhile, the calculated displacements are respectively 11.1% and 3.2% smaller than the result in Lee and Pietruszczak [21]. When n = 500 and 1000, the obtained stresses and displacements are almost the same as the counterparts in Lee and Pietruszczak [21]. Therefore, n = 500 is selected in this study in order to reduce the calculation load, which can also meet the safety requirements in practical engineering. The comparisons in Figures 8 and 9 also verify the reliability of the proposed dimensionless approach for the M-C surrounding rock.


### 4.2. Influence of the Deviatoric Plastic Shear Strainγp∗ on the Dimensionless Solutions of the Generalized H-B and M-C Rock Mass

For the elastoplastic analysis in a strain-softening rock mass, the stresses and displacement are calculated in this study. The parameters which have great effects on the dimensionless solutions are chosen to be analyzed. Therefore, five values of the deviatoric plastic shear strainγp∗ considered in this section are 0, 0.004, 0.008, 0.012, and 100. Where γp∗=0 denotes the elastic-brittle-plastic behavior, and γp∗=100 can be approximately regarded as elastic-perfectly plastic behavior. Furthermore, the data sets appearing in Carranza-Torres [26] was chosen as input parameters for the calculation in this section: b=2m, σ0=15MPa, pi=2.5MPa, E=5.7GPa, v=0.3, σcp=σcr=30MPa, pi=2.5MPa, mp=1.7, sp=0.0039, mr=0.85, sr=0.0019, ap=0.55, and ar=0.6. The results are shown in Figures 10–13.Figure 10
Variation of the dimensionless radial and circumferential stresses with the change ofγp∗ based on the generalized Hoek–Brown rock mass.Figure 11
Variation of the dimensionless radial displacement with the change ofγp∗ based on the generalized Hoek–Brown rock mass.Figure 12
Variation of the dimensionless radial and circumferential stresses with the change ofγp∗ based on the M-C rock mass.Figure 13
Variation of the dimensionless radial displacement with the change ofγp∗ based on the M-C rock mass.Figures10 and 11 show the distributions of stresses and radial displacement with the change of γp∗ based on the generalized Hoek–Brown rock mass, respectively. It should be noted that the solutions finely match the brittle-plastic solutions for the case of γp∗=0. With the decrease of deviatoric plastic shear strain γp∗, the strain-softening solutions converge to the brittle-plastic solutions. The plastic radius Rp will grow with the decrease of deviatoric plastic shear strain γp∗, which means the plastic zone becomes larger. For γp∗=0.0012, the whole plastic zone is strain-softening zone.Figures12 and 13 show the distributions of stresses and radial displacement with the change of γp∗ based on the generalized M-C rock mass, respectively. The results are similar to the generalized Hoek–Brown rock mass. With the decrease of deviatoric plastic shear strain γp∗, the strain-softening solutions converge to the brittle-plastic solutions. The plastic radius Rp will reduce with the increase of deviatoric plastic shear strain γp∗, which means the plastic zone becomes smaller. For γp∗=0.0012, the whole plastic zone is strain-softening zone. Generally, the results reflect that the strain-softening behavior causes a smaller plastic radius and the strain-softening behavior will become obvious with the decrease of plastic radius.


### 4.3. Influence of the Strength Parameters on the Dimensionless Solutions for the Generalized H-B Rock Mass

Alonso et al. [1] presented self-similar solutions in a strain-softening rock mass based on the generalized Hoek–Brown failure criterion for a=0.5 However, the general strain-softening solutions are still not available now. Many scholars have made deep studies of the strength parameter a and find that it varies in accordance with the geological conditions. The strength parameter a can be formulated by(61)a=12+16e−GSI/15−e−20/3,where GSI is the geological strength index which indicates the degree of fracturing and the condition of fracture surfaces of rock mass. Since the value of GSI is in the range from 10 to 100, a can take a value in the range of 0.5–0.6. Therefore, it is necessary to find how the strength parameter a affects the dimensionless solutions of the generalized H-B rock mass. In this study, different values of a are analyzed, and the results are shown from Figures 14–19.Figure 14
Distribution of radial and circumferential stresses for different values ofa.Figure 15
Distribution of radial displacement for different values ofa.Figure 16
Variation of radial displacement on the opening surface with different values ofa.Figure 17
Variation of plastic radiusRp on the opening surface with the change of a.Figure 18
Variation of radial displacement on the opening surface with the change ofσ0 based on generalized Hoek–Brown failure criterion.Figure 19
Variation of radial displacement on the opening surface with the change ofP0 based on generalized Hoek–Brown failure criterion.Figure14 shows the distribution of the dimensionless stresses with different values of a. When the softening zone is much thinner than the residual zone, the plastic behavior will be shifted quickly to residual regime.It can be noted that the strength parametersap and ar have an obvious influence on the dimensionless solutions. The plastic radius Rp will grow when ap and ar become larger, which means the plastic zone becomes larger. When ap remains unchanged and ar becomes larger, the plastic radius will increase. In addition, the residual zone will increase with the increase of ap and ar.Figure15 shows the influence of different values of ap and ar on the radial displacement based on the generalized Hoek–brown failure criterion. Besides Figure 16 shows the variation of radial displacement on the opening surface with different values of a.As can be seen in Figure15, the strength parameters ap and ar have an obvious influence on the dimensionless displacement. The radial displacement grows rapidly with the increase of ap and ar. From Figure 16, the dimensionless displacement for ap=ar=0.6 is about 1.4 times larger than that for ap=ar=0.5. When translating the dimensionless displacement into actual displacement, we can find that the displacement for ap=ar=0.6 is about 3 times larger than that for ap=ar=0.5.Figure17 shows the variation of plastic radius Rp on the opening surface with the change of the parameter a, based on the generalized Hoek–brown failure criterion. The strength parameters ap and ar have an obvious influence on the plastic radius Rp. The radial displacement grows rapidly with the increase of ap and ar. The plastic Rp for ap=ar=0.6 is about 1.5 times larger than that for ap=ar=0.5.Figures18 and 19 show the variation of dimensionless radial displacement on the opening surface with the change of σ0 and P0 based on the generalized Hoek–brown failure criterion, respectively. From Figure 18, the radial displacement for brittle-plastic behavior rapidly grows, and the radial displacement for elastic-perfectly plastic behavior raises slowly with the increase of σ0. As can be seen from Figure 19, the radial displacement for brittle-plastic behavior gradually grows with the increase of P0 and the radial displacement for elastic-perfectly plastic behavior increases slowly, which reflects that σ0 has little influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes larger.Overall, we can draw some conclusions: different values ofa have an obvious influence on the dimensionless solutions. The plastic radius Rp and the radial displacement increase with the increase of a. The radial displacement for brittle-plastic behavior rapidly grows with σ0 increasing and the radial displacement for elastic-perfectly plastic behavior raises slowly, which reflects that σ0 has great influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes smaller. As can be seen from Figure 19, the radial displacement for brittle-plastic behavior gradually grows with the increase of P0 and the radial displacement for elastic-perfectly plastic behavior increases slowly, which reflects that P0 has little influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes larger.


## 4.1. Verifications for H-B and M-C Rock Mass

In order to verify the correctness of the dimensionless solutions, the generalized Hoek–Brown criterion is firstly considered in this study. It is obvious that the strain-softening solutions converge to the brittle-plastic solutions when the critical deviatoric plastic strainγp∗ is equal to 0. However, the elastic-brittle-plastic case, γp∗=0, is the special case of strain-softening behavior. Comparing with the exact solutions for elastic-brittle-plastic rock mass, the dimensionless solutions can be verified.The stresses and displacement obtained by Sharan [15] are compared with the results in this study, and the equations given by Sharan [15] are obtained from the following equations:(53)σr=p0+14mbσcilnra2+lnrambσcip0+srσci2,(54)σθ=σr+mbσciσr+srσci2,(55)u=1r1+vEσR−σ0+∫rRrεr+εθdr.According to the dimensionless method, compared with the results in this study in the same order of magnitude, Equations (53)–(55) can be normalized as follows:(56)S˜r=1S0−Sr∗P0+14lnρξ2+lnρξP0,(57)S˜θ=S˜r+P0S0−Sr∗,(58)u˜=2γrbS0−Sr∗1+vEσR−σ0+∫rRrεr+εθdr.In addition, compared with the results based on the generalized H-B and M-C failure criteria, the technique of equivalent M-C and generalized H-B strength parameters was adopted for the comparison between the developed method and the exact solutions in Sharan [15]. The equations for the friction angle (ϕ) and cohesive (i) are given by Yang and Pan [25]:(59)ϕ=sin−16ambs+mbσ3na−121+a2+a+6ambs+mbσ3na−1,(60)c=σci1+2as+1−ambσ3ns+mbσ3na−11+a2+a1+6ambs+mbσ3na−1/1+a2+a,where σ3n=σ3max/σci, σ3max/σcm=0.47σcm/γH−0.94, σcm=σcimb+4s−amb−8smb/4+sa−1/21+a2+a, σcm is the rock mass strength, γ is the unit weight of the rock mass, and H is the depth of the tunnel below the surface.The above formulas can be used to transfer the parameters from the H-B yield criterion to M-C yield criterion. But that may create some errors in practical calculations.The data sets appearing in Sharan [15] were chosen as input parameters for calculation in this study: b=5m, σ0=30MPa, pi=5MPa, E=5GPa, v=0.25, σcp=σcr=30MPa, mp=1.7, sp=0.0039, mr=1.0, sr=0.0, and ap=ar=0.5. And two dilation angles, φr=0∘ and 30∘, were used to evaluate the influence of plastic volume change.Stresses and displacements in this study can be obtained by writing a program at the case ofγp=0. Then, based on Equations (56)–(58), the elastic-brittle-plastic solutions and exact solutions can be compared. The results are shown in Figures 4 and 5.Figure 4
Comparison between dimensionless stresses and exact stresses for Hoek–Brown rock mass.Figure 5
Comparison between dimensionless displacement and exact displacement for Hoek–Brown rock mass.Figure4 shows the comparison between dimensionless solutions and exact solutions for Hoek–Brown rock mass. The ordinate S stands for the dimensionless stresses.From Figure4, when the number of annuli n=500, the distributions of radial and circumferential stresses match the exact solutions well, which validate the accuracy of the dimensionless solutions. Figure 5 gives the comparison between dimensionless displacement and exact displacement for Hoek–Brown rock mass. As can be seen in Figure 5, the distribution of radial displacement obtained by the proposed approach shows a good agreement with the exact solution. So, the dimensionless solutions prove to be accurate.Figures6 and 7 show the distribution of errors between dimensionless solutions and exact solutions. The abscissa denotes ρξ. The ordinate stands for the error percentage between dimensionless solutions and exact solutions. Moreover, in order to examine the influence of mesh density on dimensionless solutions, two values of n are considered in this example, which are 10 and 500.Figure 6
Distribution of errors between dimensionless stresses and exact stresses.Figure 7
Distribution of errors between dimensionless displacement and exact displacement.From Figures6 and 7, radial stress has a little error to the exact solution with the maximum error 2.13% and the minimum error 0.06%. The circumferential stresses obtained in this study can finely match the exact solution with the maximum error 1.55% and minimum error 4.1×10−4%. Furthermore, the value of n has great influence on the dimensionless solutions, and the dimensionless solutions for n=500 are obviously more accurate than that for n=10.More different values ofn have been studied in this example, and we can find that the accuracy of dimensionless solutions will increase when the value of n grows. However, when value of n is larger than 500, the value of n has little influence on the dimensionless solutions. Therefore, to increase computational efficiency, the results of stresses and displacement for n=500 can be selected.To analyze the reliability of solutions under the M-C criterion, the stresses and displacement for the M-C surrounding rock obtained by the proposed dimensionless approach, and the approach by Lee and Pietruszczak [21] is compared. The parameters are presented as follows: b=5.35m, σ0=3.31MPa, E=1.38GPa, v=0.25, cp=1MPa, cr=0.7MPa, ϕp=30°, ϕr=22°, α=3.5γp∗=α−1εθ1e, φp=φr=3.75°, and pi=0MPa. The computed stresses and displacement using two numerical approaches are illustrated in Figures 8 and 9.Figure 8
Comparison of dimensionless stresses for M-C rock mass.Figure 9
Comparison of dimensionless displacement for M-C rock mass.The finite difference approach is also adopted by Lee and Pietruszczak [21]. The stress increment approach for the mechanical states of strain-softening rock mass is developed in their study. It can be seen from Figures 8 and 9, when n = 10 and 100, the determined circumferential stresses are respectively 10.2% and 4.5% larger than the circumferential stress in Lee and Pietruszczak [21]; meanwhile, the calculated displacements are respectively 11.1% and 3.2% smaller than the result in Lee and Pietruszczak [21]. When n = 500 and 1000, the obtained stresses and displacements are almost the same as the counterparts in Lee and Pietruszczak [21]. Therefore, n = 500 is selected in this study in order to reduce the calculation load, which can also meet the safety requirements in practical engineering. The comparisons in Figures 8 and 9 also verify the reliability of the proposed dimensionless approach for the M-C surrounding rock.


## 4.2. Influence of the Deviatoric Plastic Shear Strainγp∗ on the Dimensionless Solutions of the Generalized H-B and M-C Rock Mass

For the elastoplastic analysis in a strain-softening rock mass, the stresses and displacement are calculated in this study. The parameters which have great effects on the dimensionless solutions are chosen to be analyzed. Therefore, five values of the deviatoric plastic shear strainγp∗ considered in this section are 0, 0.004, 0.008, 0.012, and 100. Where γp∗=0 denotes the elastic-brittle-plastic behavior, and γp∗=100 can be approximately regarded as elastic-perfectly plastic behavior. Furthermore, the data sets appearing in Carranza-Torres [26] was chosen as input parameters for the calculation in this section: b=2m, σ0=15MPa, pi=2.5MPa, E=5.7GPa, v=0.3, σcp=σcr=30MPa, pi=2.5MPa, mp=1.7, sp=0.0039, mr=0.85, sr=0.0019, ap=0.55, and ar=0.6. The results are shown in Figures 10–13.Figure 10
Variation of the dimensionless radial and circumferential stresses with the change ofγp∗ based on the generalized Hoek–Brown rock mass.Figure 11
Variation of the dimensionless radial displacement with the change ofγp∗ based on the generalized Hoek–Brown rock mass.Figure 12
Variation of the dimensionless radial and circumferential stresses with the change ofγp∗ based on the M-C rock mass.Figure 13
Variation of the dimensionless radial displacement with the change ofγp∗ based on the M-C rock mass.Figures10 and 11 show the distributions of stresses and radial displacement with the change of γp∗ based on the generalized Hoek–Brown rock mass, respectively. It should be noted that the solutions finely match the brittle-plastic solutions for the case of γp∗=0. With the decrease of deviatoric plastic shear strain γp∗, the strain-softening solutions converge to the brittle-plastic solutions. The plastic radius Rp will grow with the decrease of deviatoric plastic shear strain γp∗, which means the plastic zone becomes larger. For γp∗=0.0012, the whole plastic zone is strain-softening zone.Figures12 and 13 show the distributions of stresses and radial displacement with the change of γp∗ based on the generalized M-C rock mass, respectively. The results are similar to the generalized Hoek–Brown rock mass. With the decrease of deviatoric plastic shear strain γp∗, the strain-softening solutions converge to the brittle-plastic solutions. The plastic radius Rp will reduce with the increase of deviatoric plastic shear strain γp∗, which means the plastic zone becomes smaller. For γp∗=0.0012, the whole plastic zone is strain-softening zone. Generally, the results reflect that the strain-softening behavior causes a smaller plastic radius and the strain-softening behavior will become obvious with the decrease of plastic radius.


## 4.3. Influence of the Strength Parameters on the Dimensionless Solutions for the Generalized H-B Rock Mass

Alonso et al. [1] presented self-similar solutions in a strain-softening rock mass based on the generalized Hoek–Brown failure criterion for a=0.5 However, the general strain-softening solutions are still not available now. Many scholars have made deep studies of the strength parameter a and find that it varies in accordance with the geological conditions. The strength parameter a can be formulated by(61)a=12+16e−GSI/15−e−20/3,where GSI is the geological strength index which indicates the degree of fracturing and the condition of fracture surfaces of rock mass. Since the value of GSI is in the range from 10 to 100, a can take a value in the range of 0.5–0.6. Therefore, it is necessary to find how the strength parameter a affects the dimensionless solutions of the generalized H-B rock mass. In this study, different values of a are analyzed, and the results are shown from Figures 14–19.Figure 14
Distribution of radial and circumferential stresses for different values ofa.Figure 15
Distribution of radial displacement for different values ofa.Figure 16
Variation of radial displacement on the opening surface with different values ofa.Figure 17
Variation of plastic radiusRp on the opening surface with the change of a.Figure 18
Variation of radial displacement on the opening surface with the change ofσ0 based on generalized Hoek–Brown failure criterion.Figure 19
Variation of radial displacement on the opening surface with the change ofP0 based on generalized Hoek–Brown failure criterion.Figure14 shows the distribution of the dimensionless stresses with different values of a. When the softening zone is much thinner than the residual zone, the plastic behavior will be shifted quickly to residual regime.It can be noted that the strength parametersap and ar have an obvious influence on the dimensionless solutions. The plastic radius Rp will grow when ap and ar become larger, which means the plastic zone becomes larger. When ap remains unchanged and ar becomes larger, the plastic radius will increase. In addition, the residual zone will increase with the increase of ap and ar.Figure15 shows the influence of different values of ap and ar on the radial displacement based on the generalized Hoek–brown failure criterion. Besides Figure 16 shows the variation of radial displacement on the opening surface with different values of a.As can be seen in Figure15, the strength parameters ap and ar have an obvious influence on the dimensionless displacement. The radial displacement grows rapidly with the increase of ap and ar. From Figure 16, the dimensionless displacement for ap=ar=0.6 is about 1.4 times larger than that for ap=ar=0.5. When translating the dimensionless displacement into actual displacement, we can find that the displacement for ap=ar=0.6 is about 3 times larger than that for ap=ar=0.5.Figure17 shows the variation of plastic radius Rp on the opening surface with the change of the parameter a, based on the generalized Hoek–brown failure criterion. The strength parameters ap and ar have an obvious influence on the plastic radius Rp. The radial displacement grows rapidly with the increase of ap and ar. The plastic Rp for ap=ar=0.6 is about 1.5 times larger than that for ap=ar=0.5.Figures18 and 19 show the variation of dimensionless radial displacement on the opening surface with the change of σ0 and P0 based on the generalized Hoek–brown failure criterion, respectively. From Figure 18, the radial displacement for brittle-plastic behavior rapidly grows, and the radial displacement for elastic-perfectly plastic behavior raises slowly with the increase of σ0. As can be seen from Figure 19, the radial displacement for brittle-plastic behavior gradually grows with the increase of P0 and the radial displacement for elastic-perfectly plastic behavior increases slowly, which reflects that σ0 has little influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes larger.Overall, we can draw some conclusions: different values ofa have an obvious influence on the dimensionless solutions. The plastic radius Rp and the radial displacement increase with the increase of a. The radial displacement for brittle-plastic behavior rapidly grows with σ0 increasing and the radial displacement for elastic-perfectly plastic behavior raises slowly, which reflects that σ0 has great influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes smaller. As can be seen from Figure 19, the radial displacement for brittle-plastic behavior gradually grows with the increase of P0 and the radial displacement for elastic-perfectly plastic behavior increases slowly, which reflects that P0 has little influence on dimensionless radial displacement when the deviatoric plastic shear strain γp∗ becomes larger.


## 5. Analysis of a Circular Opening Excavated in a Strain-Softening Rock Mass considering Seepage Force

In practical engineering, the seepage force has significant influence on the rock mass surrounding tunnel, especially when considering the strain-softening behavior. Most diseases are related to the seepage force directly or indirectly. However, the elastoplastic analysis of a circular opening excavated in a strain-softening rock mass considering seepage force may be obtained by the dimensionless method, and the exact solutions are difficult to accomplish. Therefore, based on the generalized Hoek–Brown failure criterion, the dimensionless method is used for reconstruction of the step-stress approach to analyze the strain-softening rock mass considering seepage force. Then, a new dimensionless method for the elastoplastic analysis of a circular opening excavated in a strain-softening rock mass considering seepage force is established in this study.


### 5.1. Seepage Force

For the circular tunnel with inner radius (r0), water head (ha) acts on the tunnel wall and water head of hydrostatic pressure (h0) is far from the tunnel wall. It is assumed that hydraulic conductivity of the surrounding rock is the same in all directions. The main seepage flow is along radial direction, and the continuity seepage differential equation can be written as follows:(62)∂2H∂r2+1r∂H∂r=0.In Equation (62), Hrr=r0=ha and Hrr=∞=h0 are the boundary conditions. Equation (61) can be solved by replacing the second condition (Hrr=∞=h0) with Hrr=αr0=h0. Therefore, the solution of Equation (63) is given by(63)H=1lnαhalnαr0r+h0lnrr0,where a is a constant of seepage force, and a conveniently large value can be taken if it meets the engineering accuracy need. In terms of Li et al. [24], α=30 is selected in this study.For the axisymmetric plane strain problem, seepage force is the volume force and is given by(64)Fr=−γwi=−γwdςHdr=γwςha−h0rlnα,where i means hydraulic gradient, ς is the rock effective coefficient of pore water pressure, H denotes the water level fluctuation, α is the constant of seepage force, γw is the unit weight of water, r is the radial distance from center of the opening, and h0 and ha are the initial and final water levels, respectively.


### 5.2. Solutions for Stresses and Displacement considering Seepage Force

Considering the influence of seepage force, stress equilibrium differential equation can be expressed as follows:(65)dσrdr+σr−σθr+Fr=0.As is mentioned before, it is assumed that the yielding of the rock mass is governed by the yielding function:(66)fσθ,σr=σθ−σr−Hσr,γp=0.In order to work with dimensionless field quantities, the stress magnitude,σ0−σr∗, is used to normalize the stresses:(67)σ˜θ=σθσo−σr∗,σ˜r=σrσo−σr∗.Strains are normalized accordingly, considering the extra term 2G:(68)ε˜θ=2Gσo−σr∗εθ,ε˜r=2Gσo−σr∗εr.Displacement is normalized in terms of the radiusb:(69)u˜r=2Gbσo−σr∗ur.Similarly, the step-stress approach is used to divide the plastic zone into a finite number of concentric rings which satisfy the equilibrium and compatibility equations in strain-softening rock mass. The increments of stresses and strains for each annulus are obtained by solving the equilibrium and compatibility equations. Then, the stresses and displacements in the softening zone can be calculated.When considering the effects of seepage force, a step-stress approach is used for solving the stresses and displacement in the plastic zone numerically. The whole plastic zone is divided into n annuli with a constant radial stress increment between the adjacent two annuli, which is defined as follows:(70)Δσ˜r=pi−σ0n.So that the stress components for thei th annulus may be presented by the following equation:(71)σ˜ri=σ˜ri−1+Δσ˜r.Then, the circumferential stress can be written as follows:(72)σ˜θi=σ˜ri+Hσ˜ri,γi−1p.According to Hooke’s law, the elastic strain increments are related to the stress increment. That is,(73)Δε˜riΔε˜θi=2G1+vE1−v−v−v1−vΔσ˜riΔσ˜θi.When the number of annulin is sufficiently large, substituting Equation (64) into Equation (65), the equilibrium equation can be transformed as follows:(74)dσ˜rdρ+σ˜r−σ˜θρ+γwςha−h0ρlnα=0,or(75)dσ˜rdρ+Hσ˜r,γpρ+γwςha−h0ρlnα=0.Equation (75) can be approximated for the ith annulus as(76)σ˜ri−σ˜ri−1ρi−ρi−1+Hσ˜¯ri,γi−1pρi+ρi−1+γwςha−h0ρi+ρi−1lnα=0,where σ˜¯ri=σ˜ri+σ˜ri−1/2.Then, the inner radius can be obtained by(77)ρi=2Hσ˜¯ri,γi−1p+γwςha−h0/lnα+Δσ˜r2Hσ˜¯ri,γi−1p+γwςha−h0/lnα−Δσ˜rρi−1.The strain components are given in Equation (36), and rearranging for ε˜θp gives, in view of Equations (8), (73), and (77), the following equation:(78)1Δρi+1+ki−11ρ¯iΔε˜θip=−Δε˜θieΔρi−2G1+υEHS˜¯ri,γi−1p+γwςha−h0/lnαρ¯i−1ρ¯iε˜θi−1p−ε˜ri−1p,where ρ¯i=ρ¯i−1+ρ¯i/2 and ki−1=1+sinφi−1/1−sinφi−1. The deviatoric plastic shear strain is updated as(79)γip=γi−1p+Δε˜θip−Δε˜rip.The total strain at theith annulus can be given as follows:(80)ε˜riε˜θi=ε˜ri−1ε˜θi−1+Δε˜rieΔε˜θie+Δε˜ripΔε˜θip.According to Equations (68) and (69), the relationship, εθ=u/R, can be recalled as(81)ε˜θ=u˜ρξRp.Then, the displacement normalized by plastic radiusRp can be obtained as follows:(82)U˜i=ε˜θiri=ε˜θiρiξ,where U˜i=u˜i/Rp.Plastic radiusRp can be calculated from the following relationship:(83)Rp=brn=1ρnξ.


### 5.3. Verification Examples

The dimensionless stresses and displacements in this study can be obtained by programming the dimensionless method into MATLAB codes. In order to examine the difference between the solutions with and without considering seepage force, the parameters are adopted as follows:b=5m, σ0=30MPa, pi=5MPa, E=5GPa, v=0.25, σcp=σcr=30MPa, mp=1.7, sp=0.0039, mr=1.0, γp=0, sr=0.0, ap=ar=0.5, γw=9.8kN/m3, ς=1, ha=50m, h0=0, and α=30. The results are shown in Figures 20 and 21.Figure 20
Comparison between dimensionless stresses with and without considering seepage force for Hoek–Brown rock mass.Figure 21
Comparison between dimensionless displacement with and without considering seepage force for Hoek–Brown rock mass.Figures20 and 21 show the comparison between dimensionless solutions considering seepage force and solutions without considering seepage force. As can be seen from Figure 20, the seepage force has a great influence on the dimensionless stresses. The variation law of stresses considering seepage force is coincident with that without considering seepage force. But the plastic radius considering seepage force is larger than that without considering seepage force, which results from the increase of effective stress of the surrounding rock when the effects of seepage force are considered. The results reflect that seepage force can increase the effective stresses of the surrounding rock. From Figure 21, the dimensionless displacement is also be affected by the seepage force. The variation law of displacement considering seepage force is coincident with that without considering seepage force, and the plastic displacement considering seepage force is larger than that without considering seepage force.Figures22 and 23 show the distributions of stresses and radial displacement considering seepage force with different γp∗ for the generalized H-B rock mass, respectively. It should be noted that the variation law of stresses considering seepage force agrees with that without considering seepage force. The plastic radius Rp will reduce when the deviatoric plastic shear strain γp∗ increases. When the deviatoric plastic shear strain is close to 0, the solutions finely match the brittle-plastic solutions. The largest plastic zone appears in brittle-plastic case, and the thinnest plastic zone develops for the case of elastic-perfectly plastic rock mass.Figure 22
Variation of the dimensionless radial and circumferential stresses considering seepage force with the change ofγp∗ for the generalized Hoek–Brown rock mass.Figure 23
Variation of the dimensionless radial displacement considering seepage force with the change ofγp∗ for the generalized Hoek–Brown rock mass.


### 5.4. Analysis of Influencing Factors of the Dimensionless Solutions considering Seepage Force

Based on the generalized Hoek–Brown failure criterion, the parametera is selected as an influencing factor of the dimensionless solutions considering seepage force. The influence of different values of a on the dimensionless stresses and displacement is shown from Figures 24–27.Figure 24
Ground reaction curves for differenta for the generalized H-B rock mass.Figure 25
Distribution of radial and circumferential stresses for different values ofa for the generalized H-B rock mass.Figure 26
Evolution of plastic radii for different values ofa considering seepage force.Figure 27
Variation of radial displacements on the opening surface with the change of parametera considering seepage force.Figure24 shows four ground reactions curves based on different values of a. Three curves among them postulate that the peak and residual values of a are equal to 0.5, 0.55, and 0.6, respectively. The differences in the shape of those curves seem substantial, especially when pi is very low. The curves for ap=0.5 and ap=0.6 is close to that for ap=ar=0.6, this is because the thickness of the strain-softening zone in this problem is very narrow as is inferred in Figure 24, in which the radial and circumferential stresses for pi=0 are displayed.Figure25 shows the distribution of the dimensionless stresses for different values of a. When the strain-softening zone is much thinner than the residual zone, the plastic behavior will be shifted quickly to residual regime.As can be seen from Figure25, when seepage force is considered, the variation law is coincident with that without considering seepage force. It can also be noted that the strength parameters ap and ar have an obvious influence on the dimensionless solutions when considering seepage force. When ap=ar, the plastic radius Rp will grow when the ap and ar become larger. In addition, the stresses for ap=ar=0.5 are very close to that for ap=0.5 and ap=0.6, which denotes that the difference of ap and ar has little effects on the stresses when considering seepage force.Figures26 and 27 show the influence of different values of ap and ar on the radial radius based on the generalized Hoek–brown failure criterion and the variation of radial displacements on the opening surface with different values of parameter a.As are illustrated in Figure26, the strength parameters ap and ar have an obvious influence on the plastic radius Rp. The plastic radius grows with the increase of ap and ar. However, the plastic radius decreases rapidly with the increase of the support pressure pi. Furthermore, the influence will become more obvious when the support pressure pi is lower; that is, the influence of ap and ar on the solutions reduces with the growth of support pressure pi, when considering seepage force.From Figure27, the strength parameters ap and ar have great influence on the displacement when considering seepage force. The radial displacement grows rapidly with the increase of ap and ar. But the influence of ap and ar on the displacement when considering seepage force is weaker than that without considering seepage force.


## 5.1. Seepage Force

For the circular tunnel with inner radius (r0), water head (ha) acts on the tunnel wall and water head of hydrostatic pressure (h0) is far from the tunnel wall. It is assumed that hydraulic conductivity of the surrounding rock is the same in all directions. The main seepage flow is along radial direction, and the continuity seepage differential equation can be written as follows:(62)∂2H∂r2+1r∂H∂r=0.In Equation (62), Hrr=r0=ha and Hrr=∞=h0 are the boundary conditions. Equation (61) can be solved by replacing the second condition (Hrr=∞=h0) with Hrr=αr0=h0. Therefore, the solution of Equation (63) is given by(63)H=1lnαhalnαr0r+h0lnrr0,where a is a constant of seepage force, and a conveniently large value can be taken if it meets the engineering accuracy need. In terms of Li et al. [24], α=30 is selected in this study.For the axisymmetric plane strain problem, seepage force is the volume force and is given by(64)Fr=−γwi=−γwdςHdr=γwςha−h0rlnα,where i means hydraulic gradient, ς is the rock effective coefficient of pore water pressure, H denotes the water level fluctuation, α is the constant of seepage force, γw is the unit weight of water, r is the radial distance from center of the opening, and h0 and ha are the initial and final water levels, respectively.


## 5.2. Solutions for Stresses and Displacement considering Seepage Force

Considering the influence of seepage force, stress equilibrium differential equation can be expressed as follows:(65)dσrdr+σr−σθr+Fr=0.As is mentioned before, it is assumed that the yielding of the rock mass is governed by the yielding function:(66)fσθ,σr=σθ−σr−Hσr,γp=0.In order to work with dimensionless field quantities, the stress magnitude,σ0−σr∗, is used to normalize the stresses:(67)σ˜θ=σθσo−σr∗,σ˜r=σrσo−σr∗.Strains are normalized accordingly, considering the extra term 2G:(68)ε˜θ=2Gσo−σr∗εθ,ε˜r=2Gσo−σr∗εr.Displacement is normalized in terms of the radiusb:(69)u˜r=2Gbσo−σr∗ur.Similarly, the step-stress approach is used to divide the plastic zone into a finite number of concentric rings which satisfy the equilibrium and compatibility equations in strain-softening rock mass. The increments of stresses and strains for each annulus are obtained by solving the equilibrium and compatibility equations. Then, the stresses and displacements in the softening zone can be calculated.When considering the effects of seepage force, a step-stress approach is used for solving the stresses and displacement in the plastic zone numerically. The whole plastic zone is divided into n annuli with a constant radial stress increment between the adjacent two annuli, which is defined as follows:(70)Δσ˜r=pi−σ0n.So that the stress components for thei th annulus may be presented by the following equation:(71)σ˜ri=σ˜ri−1+Δσ˜r.Then, the circumferential stress can be written as follows:(72)σ˜θi=σ˜ri+Hσ˜ri,γi−1p.According to Hooke’s law, the elastic strain increments are related to the stress increment. That is,(73)Δε˜riΔε˜θi=2G1+vE1−v−v−v1−vΔσ˜riΔσ˜θi.When the number of annulin is sufficiently large, substituting Equation (64) into Equation (65), the equilibrium equation can be transformed as follows:(74)dσ˜rdρ+σ˜r−σ˜θρ+γwςha−h0ρlnα=0,or(75)dσ˜rdρ+Hσ˜r,γpρ+γwςha−h0ρlnα=0.Equation (75) can be approximated for the ith annulus as(76)σ˜ri−σ˜ri−1ρi−ρi−1+Hσ˜¯ri,γi−1pρi+ρi−1+γwςha−h0ρi+ρi−1lnα=0,where σ˜¯ri=σ˜ri+σ˜ri−1/2.Then, the inner radius can be obtained by(77)ρi=2Hσ˜¯ri,γi−1p+γwςha−h0/lnα+Δσ˜r2Hσ˜¯ri,γi−1p+γwςha−h0/lnα−Δσ˜rρi−1.The strain components are given in Equation (36), and rearranging for ε˜θp gives, in view of Equations (8), (73), and (77), the following equation:(78)1Δρi+1+ki−11ρ¯iΔε˜θip=−Δε˜θieΔρi−2G1+υEHS˜¯ri,γi−1p+γwςha−h0/lnαρ¯i−1ρ¯iε˜θi−1p−ε˜ri−1p,where ρ¯i=ρ¯i−1+ρ¯i/2 and ki−1=1+sinφi−1/1−sinφi−1. The deviatoric plastic shear strain is updated as(79)γip=γi−1p+Δε˜θip−Δε˜rip.The total strain at theith annulus can be given as follows:(80)ε˜riε˜θi=ε˜ri−1ε˜θi−1+Δε˜rieΔε˜θie+Δε˜ripΔε˜θip.According to Equations (68) and (69), the relationship, εθ=u/R, can be recalled as(81)ε˜θ=u˜ρξRp.Then, the displacement normalized by plastic radiusRp can be obtained as follows:(82)U˜i=ε˜θiri=ε˜θiρiξ,where U˜i=u˜i/Rp.Plastic radiusRp can be calculated from the following relationship:(83)Rp=brn=1ρnξ.


## 5.3. Verification Examples

The dimensionless stresses and displacements in this study can be obtained by programming the dimensionless method into MATLAB codes. In order to examine the difference between the solutions with and without considering seepage force, the parameters are adopted as follows:b=5m, σ0=30MPa, pi=5MPa, E=5GPa, v=0.25, σcp=σcr=30MPa, mp=1.7, sp=0.0039, mr=1.0, γp=0, sr=0.0, ap=ar=0.5, γw=9.8kN/m3, ς=1, ha=50m, h0=0, and α=30. The results are shown in Figures 20 and 21.Figure 20
Comparison between dimensionless stresses with and without considering seepage force for Hoek–Brown rock mass.Figure 21
Comparison between dimensionless displacement with and without considering seepage force for Hoek–Brown rock mass.Figures20 and 21 show the comparison between dimensionless solutions considering seepage force and solutions without considering seepage force. As can be seen from Figure 20, the seepage force has a great influence on the dimensionless stresses. The variation law of stresses considering seepage force is coincident with that without considering seepage force. But the plastic radius considering seepage force is larger than that without considering seepage force, which results from the increase of effective stress of the surrounding rock when the effects of seepage force are considered. The results reflect that seepage force can increase the effective stresses of the surrounding rock. From Figure 21, the dimensionless displacement is also be affected by the seepage force. The variation law of displacement considering seepage force is coincident with that without considering seepage force, and the plastic displacement considering seepage force is larger than that without considering seepage force.Figures22 and 23 show the distributions of stresses and radial displacement considering seepage force with different γp∗ for the generalized H-B rock mass, respectively. It should be noted that the variation law of stresses considering seepage force agrees with that without considering seepage force. The plastic radius Rp will reduce when the deviatoric plastic shear strain γp∗ increases. When the deviatoric plastic shear strain is close to 0, the solutions finely match the brittle-plastic solutions. The largest plastic zone appears in brittle-plastic case, and the thinnest plastic zone develops for the case of elastic-perfectly plastic rock mass.Figure 22
Variation of the dimensionless radial and circumferential stresses considering seepage force with the change ofγp∗ for the generalized Hoek–Brown rock mass.Figure 23
Variation of the dimensionless radial displacement considering seepage force with the change ofγp∗ for the generalized Hoek–Brown rock mass.


## 5.4. Analysis of Influencing Factors of the Dimensionless Solutions considering Seepage Force

Based on the generalized Hoek–Brown failure criterion, the parametera is selected as an influencing factor of the dimensionless solutions considering seepage force. The influence of different values of a on the dimensionless stresses and displacement is shown from Figures 24–27.Figure 24
Ground reaction curves for differenta for the generalized H-B rock mass.Figure 25
Distribution of radial and circumferential stresses for different values ofa for the generalized H-B rock mass.Figure 26
Evolution of plastic radii for different values ofa considering seepage force.Figure 27
Variation of radial displacements on the opening surface with the change of parametera considering seepage force.Figure24 shows four ground reactions curves based on different values of a. Three curves among them postulate that the peak and residual values of a are equal to 0.5, 0.55, and 0.6, respectively. The differences in the shape of those curves seem substantial, especially when pi is very low. The curves for ap=0.5 and ap=0.6 is close to that for ap=ar=0.6, this is because the thickness of the strain-softening zone in this problem is very narrow as is inferred in Figure 24, in which the radial and circumferential stresses for pi=0 are displayed.Figure25 shows the distribution of the dimensionless stresses for different values of a. When the strain-softening zone is much thinner than the residual zone, the plastic behavior will be shifted quickly to residual regime.As can be seen from Figure25, when seepage force is considered, the variation law is coincident with that without considering seepage force. It can also be noted that the strength parameters ap and ar have an obvious influence on the dimensionless solutions when considering seepage force. When ap=ar, the plastic radius Rp will grow when the ap and ar become larger. In addition, the stresses for ap=ar=0.5 are very close to that for ap=0.5 and ap=0.6, which denotes that the difference of ap and ar has little effects on the stresses when considering seepage force.Figures26 and 27 show the influence of different values of ap and ar on the radial radius based on the generalized Hoek–brown failure criterion and the variation of radial displacements on the opening surface with different values of parameter a.As are illustrated in Figure26, the strength parameters ap and ar have an obvious influence on the plastic radius Rp. The plastic radius grows with the increase of ap and ar. However, the plastic radius decreases rapidly with the increase of the support pressure pi. Furthermore, the influence will become more obvious when the support pressure pi is lower; that is, the influence of ap and ar on the solutions reduces with the growth of support pressure pi, when considering seepage force.From Figure27, the strength parameters ap and ar have great influence on the displacement when considering seepage force. The radial displacement grows rapidly with the increase of ap and ar. But the influence of ap and ar on the displacement when considering seepage force is weaker than that without considering seepage force.


## 6. Conclusions

A dimensionless approach is developed for elastoplastic analysis of circular opening excavated in a strain-softening rock mass based on the generalized Hoek–Brown and M-C failure criterion. The plastic zone is divided into a finite number of concentric rings in this study. In order to solve the equilibrium and compatibility equations for each ring, the dimensionless method is used for calculating the stresses and displacement in the strain-softening zone. Through analysis of examples, some conclusions can be drawn:(1)
As the number of annulin is sufficiently large (n=500), the distribution of stresses and displacement of strain-soften rock mass obtained in this study show a good agreement with the existing solutions presented by Sharan [15] and Lee and Pietruszczak [21], which indicates that the dimensionless method developed in this study is reasonable and efficient.(2)
Five values of the deviatoric plastic shear strainγp∗ analyzed in this study are 0, 0.004, 0.008, 0.012, and 100. The plastic radius Rp will reduce with the increase of deviatoric plastic shear strain γp∗. When the deviatoric plastic shear strain is close to 0, the obtained solutions have a good agreement with the brittle-plastic solutions. The largest plastic zone appears in the brittle-plastic case, and the thinnest plastic zone takes place for the case of elastic-perfectly plastic surrounding rock.(3)
The strength parametersap and ar have an obvious influence on the dimensionless solutions. For ap=ar, the plastic radius Rp grows with the increase of ap and ar, which means the plastic zone becomes larger. When ap remains unchanged and ar becomes larger, the plastic radius will increase. In addition, the residual zone will increase with the increase of ap and ar. The dimensionless displacement for ap=ar=0.6 is about 1.4 times larger than that for ap=ar=0.5. The plastic radius Rp for ap=ar=0.6 is about 1.5 times larger than that for ap=ar=0.5. Furthermore, different values of σ0 and P0 have great impact on the dimensionless solutions under the generalized H-B criterion. The radial displacement for brittle-plastic rock mass rapidly grows with σ0 increasing and the radial displacement for elastic-perfectly plastic behavior raises slowly, which reflects that σ0 has great influence on dimensionless radial displacements when the deviatoric plastic shear strain γp∗ becomes smaller. The radial displacement for brittle-plastic behavior gradually grows with the increase of P0 and the radial displacement for elastic-perfectly plastic behavior increases very slowly, which reflects that P0 has little influence on dimensionless radial displacements when the deviatoric plastic shear strain γp∗ becomes larger.(4)
For the analysis of a circular opening excavated in a strain-softening rock mass considering seepage force, the stresses are smaller but the plastic radius and radial displacement are larger than the results without considering seepage force, which reflects that the seepage force can increase the effective stresses around a circular tunnel. The deviatoric plastic shear strainγp∗ has an obvious influence on the solutions when considering seepage force. The plastic radius Rp will reduce with the increase of deviatoric plastic shear strain γp∗. As the deviatoric plastic shear strain is close to 0, the solutions finely match the brittle-plastic solutions. The largest plastic zone appears in the brittle-plastic case, and the thinnest plastic zone takes place for the case of elastic-perfectly plastic surrounding rock. Moreover, the strength parameters ap and ar have an obvious influence on the plastic radius Rp. The plastic radius grows with the increase of ap and ar. But the plastic radius will decrease rapidly with the increase of support pressure pi.



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