# A New Unified Solution for Circular Opening considering Different Strength Criteria and the Postpeak Elastic Strain Form

Authors: Xuyang Shi; Wei Zhou; Liang Chen; Qingxiang Cai; Ming Li; Zhaolin Li; Boyu Luan
Journal: Advances in Civil Engineering
 (2020)

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## Abstract

The strength criterion is an extremely important basis for evaluating the stability of surrounding rock and optimizing the support pressure design. In this paper, nine different strength criteria are summarized and simplified based on the reasonable assumption. Then, a new unified criterion equation is established, which includes all strength theories proposed by this paper. Meanwhile, a new unified closed-form solution for circular opening based on the newly proposed unified criterion equation is deduced with the infinite and finite external boundary combining with the nonassociative flow rule under plane strain conditions. In the plastic zone, four different elastic strain assumptions are applied to solving the plastic zone deformation considering the effect of rock mass damage. The solution’s validity is also verified by comparison with the traditional solution. Finally, the influences of strength criteria, dilation coefficient, elastic strain form of plastic zone, and rock mass damage on the mechanical response of surrounding rock are discussed in detail. The research result shows that TR and VM criteria give the largest plastic zone radius, followed by IDP, MC, and MDP criteria, and seem to underestimate the self-strength of rock mass; The CDP criterion gives the smallest plastic zone radius and may overestimate the self-strength of rock mass; UST0.5, GSMP, GMC, and GLD criteria that reasonably consider the effect of internal principal stresses give an intermediate range and can be strongly recommended for evaluating the mechanics and deformation behavior of surrounding rock; as the dilation coefficient gradually increases, the dimensionless surface displacement presents the nonlinear increase characteristics; the deformation of plastic zone and the ground response curve, which are closely related to the strength criteria, are also greatly influenced by the elastic strain assumption in the plastic zone and rock mass damage degree. The assumption that the elastic strain satisfies Hook’s law (Case 3) may be more reasonable compared with the continuous elastic strain (Case 1) and thick-walled cylinders (Case 2) assumptions; in addition, the Young’s modulus power function damage model seems to give more reasonable solution for the deformation of plastic zone and is suggested to be a preferred method for solving plastic displacement. The research results can provide very important theoretical bases for evaluating the tunnel stability and support design reliability of different lithology rock masses in underground engineering.


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## Body

## 1. Introduction

Accurate prediction for the stress and displacement distribution of circular opening plays a crucial role in evaluating the mechanics and deformation behavior of rock mass in civil, mine, and oil engineering and natural gas development engineering. The circular opening may include tunnel, vertical shaft, boreholes, and pile foundation holes. In the early stage, many closed-form solutions of circular opening are obtained by regarding the outer boundary as the infinite boundary with association or nonassociative flow rules [1–9]. However, the semianalytical solutions considering the finite boundary are rarely deduced because of the complexity of the calculation process [6, 10–13]. In practice, it is possible to calculate the outer boundary as an infinite boundary only if it is much larger than the diameter of circular opening. Therefore, this simplified calculation method has certain limitations in many geological engineering, such as shallow-buried tunnel.As for most of the analyses reported in the past, the solutions were given considering different yield criteria, such as linear Mohr–Coulomb (MC), Tresca (TR), and nonlinear Hoek–Brown (HB) criteria [2, 4–9, 14–17]. Nevertheless, the above studies did not consider the effect of intermediate principal stress on the mechanical response of surrounding rock. Many research results have shown that the intermediate principal stress exerts significant influence on the failure behavior of rock mass [6]. In addition, for rock mass as a natural geological material, the yield failure criterion is more complicated under the influence of internal crack and joint. Therefore, it is extremely difficult to reconcile the calculation results with the field measured results if only one or two yield criteria are used to predict the stresses and displacement behavior of surrounding rock. In this paper, as shown in Table 1, different yield criteria, such as Mogi–Coulomb (GMC) [9, 10], Drucker–Prager (CDP, MDP, and IDP) [3, 4, 14, 28–30], Generalized Lade–Duncan (GLD) [31, 35], Generalized SMP (GSMP) [18, 27, 31], and von Mises (VM) criteria [15, 26], will be summarized and simplified and then used to study the mechanical response of rock mass.Table 1
A summary on the yield criterion under plane strain conditions.
Failure criterionInitial yield criterionSimplified yield criterionσ1=ξjσ3+YjLinearityIntermediate principal stressMohr–Coulomb (MC) [1, 14]σ1=1+sinφ/1−sinφσ3+2ccosφ/1−sinφξMC=1+sinφ/1−sinφ;YMC=2ccosφ/1−sinφLinearNoTresca (TR) [15]σ1−σ3/2=τmax=cξTR=1YTR=2cLinearNoGeneralized Lade–Duncan (GLD) [18–23]I1′3/I3′=KLadeKLade=3−sinφ3/1+sinφ1−sinφ2;ξGLD=ηGLDYGLD=ηGLD−1σiNonlinearYesMogi–Coulomb (GMC) [2, 24, 25]13σ1−σ22+σ2−σ32+σ1−σ32=q1σ1+σ2+σ3/3+q2q1=22/3sinφq2=22/3ccosφξMGC=6+3q1/6−3q1YMGC=6q2/6−3q1LinearYesvon Mises (VM) [15, 26]σ1−σ22+σ2−σ32+σ1−σ32/6=J2=2c/3ξVM=1YVM=4c/3NonlinearYesGeneralized SMP criterion (GSMP) [18, 27]I11I21/I31=KGSMPKGSMP=8tan2φ+9ξGSMP=ηGSMPYGSMP=ηGSMP−1σiNonlinearYesCircumscribe Drucker–Prager (CDP) [1, 3, 4, 28–30]J2=k−αI1k=6ccosφ/33−sinφα=2sinφ/33−sinφξCDP=1+3α/1−3αYCDP=2k/1−3αNonlinearYesMiddle Circumscribe Drucker–Prager (MDP) [28]The expression form is the same as CDP.k=6ccosφ/33+sinφα=2sinφ/33+sinφYesInscribe Drucker–Prager (IDP) [31]The expression form is the same as CDP.k=3ccosφ/3+sin2φα=sinφ/9+3sin2φYesUnified strength theory (UST) [5, 6, 32–34]Ifσ2≤σ1+σ3/2−σ1−σ3sinφ/2F=1−sinφ/1+sinφσ1−bσ2+σ3/1+b=2ccosφ/1+sinφ;If σ2≥σ1+σ3/2−σ1−σ3sinφ/2F′=1−sinφ/1+b1+sinφσ1+bσ2−σ3=2ccosφ/1+sinφξUST=2+b+2+3bsinφ/2+b1−sinφYUST=41+bccosφ/2+b1−sinφLinearYesGeneralized twin shear stress criterion (GTSS) [5, 6]Whenb = 1, UST is equal to GTSS.ξGTSS=3+5sinφ/31−sinφYGTSS=8ccosφ/31−sinφLinearYesThe compressive stress is positive and the tensile stress is negativeσ1≥σ2≥σ3; b is the intermediate principal stress coefficient; ηLD=1/4KLade3−1+KLade3−12−42; ηSMP=1/4KSMP−1+KSMP−3−2KSMP2; σi=ccotφ.For an elastic-brittle plastic rock mass, the postpeak deformation is closely related to the assumed form of elastic strain in the plastic zone. Brown et al. researched the ground response curve for the rock tunnel by assuming that the elastic strain in the plastic region was equal to that on the elastic-plastic interface [36]. Sharan presented a series of new closed-form solutions for the prediction of displacements around circular openings in a brittle rock mass with nonlinear Hoek–Brown criterion by regarding the elastic strain of plastic zone as the thick-walled cylinder [16, 37]. Yu and Zhang and Reed et al. held that the elastic strain in the plastic zone satisfied the generalized Hooke’s theorem and then derived the deformation of plastic zone with nonassociative flow laws [9, 38–40]. In addition, Park summarized the above three different definitions for elastic strains in the plastic zone and analyzed the deformation law of plastic zone with Mohr–Coulomb and Hoek–Brown criteria under three different case conditions [9]. However, the postpeak Young’s modulus attenuation along the radii direction is ignored in this study.In this paper, different yield criteria of rock mass are firstly summarized and then a unified yield criterion form is derived by simplifying the above criteria. Next, a new closed-form solution for stresses and deformation distribution around a circular opening subjected to the hydrostatic pressure at the finite or infinite outer boundary is also obtained with the new proposed unified yield criterion. In the plastic zone, five different definitions for elastic strains in the plastic zone and nonassociated flow rule are adopted to establish the radii displacement solution. The correctness of the solution is also verified by comparison with a series of traditional solution and numerical simulation results. Finally, the influences of strength theory and elastic strain definitions of plastic zone on the mechanical response of surrounding rock are discussed in detail.


## 2. Brief Description of Yield Criterion

### 2.1. Mohr–Coulomb Criterion

The linear Mohr–Coulomb criterion is widely used in geotechnical engineering because of its simple expression form. However, the influence of intermediate principal stress on rock mass failure is ignored. In addition, the triaxial strength of rock mass obtained by MC criterion deviates greatly from the measured data under the high confining pressure conditions. The following is the governing equation for MC criterion based on the cohesionc and internal friction angle φ [1, 14, 39]:(1)σ1=Nσ3+σc,where N is a constant which is a function of the internal friction angle and σc is the uniaxial compressive strength (UCS). They can be expressed as follows:(2)N=1+sinφ1−sinφ,σc=2ccosφ1−sinφ.


### 2.2. Tresca Criterion

The Tresca (TR) criterion is a simplified form of MC criterion and assumes that the failure will occur if the maximum shear stressτmax inside any plane of rock mass reaches a critical value [1, 15]. As shown in Figure 1, the yield curve of TR criterion is a regular hexagon, which is inscribed in the circular yield curve of the von Mises (VM) criterion in π-plane. The expression form is(3)σ1−σ32=τmax=c.Figure 1
Tresca and von Mises criterion in theπ-plane.The TR criterion can be considered as a special case where the internal friction angle equals zero in the MC criterion. Equation (2) can be rewritten as(4)fσ1,σ3=σ1−σ3−2c=0.


### 2.3. Generalized Lade–Duncan Criterion

In early 1973, the Lade–Duncan criterion has been firstly proposed by considering the intermediate principal stresses based on the triaxial compression test for noncohesive soil. The initial expression form can be written as follows:(5)I13I3=KLade,where I1 and I3 are, respectively, the first and third principal stress invariants of friction material. KLade is the material constant which is closely related to internal friction angle:(6)KLade=3−sinφ31+sinφ1−sinφ2,(7)I1=σ1+σ2+σ3,(8)I3=σ1σ2σ3.In 1999, the Lade–Duncan criterion was modified by Ewy so that it could reasonably describe the strength characteristics of cohesive soil by introducing bound stress (σ0) [35]. Its generalized expression is also given as(9)I1′3I3′=KLade,where I1′ and I3′ are, respectively, the first and third principal stress invariants of cohesive-friction material:(10)I1′=σ1+σ0+σ2+σ0+σ3++σ0,I3′=σ1+σ0σ2+σ0σ3++σ0,where σ0 is the bound stress: σ0=ccotφ. For the noncohesive soils, under plane strain conditions, the internal principal stress can be expressed as(11)σ2=σ1σ3.By substituting equations (7), (8), and (11) into equation (6), the Lade–Duncan criterion can be rewritten as(12)σ1=14KLade3−1+KLade3−12−42σ3=ηLadeσ3.For the cohesive-friction material, according to the coordinate translation method [18, 19], the generalized Lade–Duncan (GLD) criterion will be obtained by integrating equation (12).(13)σ1+σ0=14KLade3−1+KLade3−12−42σ3+σ0=ηLadeσ3+σ0.fσ1,σ3=σ1−ηLadeσ3−ηLade−1σ0=0.


### 2.4. Generalized SMP Criterion

Considering the effect of internal principal stress, H. Matsuoka and T. Nakai proposed the SMP criterion, which could be expressed by three principal stress invariants [31]:(14)I1I2I3=KSMP,where I2 is the second principal stress invariant andKSMP is the material constant in the SMP criterion:(15)KSMP=8tan2φ+9,(16)I2=σ1σ3+σ1σ2+σ2σ3.For the frictional material, by introducing equations (7), (8), (11), and (16) into equation (14), another form of SMP criterion will be deduced.(17)σ1=14KSMP−1+KSMP−3−2KSMP2σ3=ηGSMPσ3.As the GLD criterion, according to the coordinate translation method, the generalized SMP criterion will be given as [18](18)σ1+σ0=14KSMP−1+KSMP−3−2KSMP2σ3+σ0=ηGSMPσ3+σ0.fσ1,σ3=σ1−ηGSMPσ3−ηGSMP−1σ0=0.As shown in Figure2, compared with the yield curve of MC criterion in the π-plane, the GLD and GSMP criteria are nonlinear and the cross-sectional area of GLD criterion is the largest, followed by the GSMP criterion.Figure 2
Comparison of MC, GLD, and GSMP criteria in theπ-plane under plane strain condition.


### 2.5. Mogi–Coulomb Criterion

Based on Mog’s theory [24], Al-Ajmi and Zimmerman found that the polyaxial test data could be fitted by linear relationships in τoct−σm,2 space.(19)τoct=aσm,2+b,(20)τoct=13σ1−σ22+σ1−σ32+σ2−σ32,(21)σm,2=σ1+σ2+σ33,where σm,2 is the mean normal stress and τoct is octahedral shear stress. The parameters a and b can be related exactly to the Coulomb strength parameter:(22)a=223sinφ,b=223ccosφ.As the von Mises criterion, we takeσ2=σ1+σ3/2. Then, the Mogi–Coulomb (GMC) criterion can be obtained by integrating equations (19), (20), and (21).(23)fσ1,σ3=σ1−6+3a6−3aσ3−6b6−3a=0.


### 2.6. Unified Strength Theory

Based on the twin shear yield criterion, the unified strength theory (UST) is established by considering the influence of all the stress components on the material yield failure [32–34]. In geotechnical engineering, the cohesion and internal friction angle are usually used to represent this yield theory. The yield function can be expressed as follows [5, 6]:Ifσ2≤σ1+σ3/2−σ1−σ3/2sinφ,(24a)fσ1,σ3=1−sinφ1+sinφσ1−bσ2+σ31+b=2ccosφ1+sinφ.Ifσ2≤σ1+σ3/2−σ1−σ3/2sinφ,(24b)fσ1,σ3=1−sinφ1+b1+sinφσ1+bσ2−σ3=2ccosφ1+sinφ.where b represents the yield parameter related to the intermediate principal stress, which can reflect the effect of the intermediate principal shear stress and the positive stress on the yield failure of the rock material, and 0≤b≤1. As shown in Figure 3, if b=0, UST will translate into the MC criterion; if b=1, UST is converted into general twin shear strength (GTSS) criterion; if 0<b<1, UST is a series of other ordered new strength criteria.Figure 3
Unified strength theory yield curve in theπ-plane [16].Just like Sections2.3 and 2.6, if σ2=σ1+σ3/2, σ2≥σ1+σ3/2−σ1−σ3/2sinφ can be judged by substituting σ2=σ1+σ3/2 into equation (24a) or equation (24b). Therefore, UST can be rewritten as(25)fσ1,σ3=σ1−2+b+2+3bsinφ2+b1−sinφσ3−41+bccosφ2+b1−sinφ=0.


### 2.7. Drucker–Prager Criterion

The Drucker–Prager (DP) criterion is an extension of the von Mises criterion, which takes into account the effect of intermediate principal stress and hydrostatic pressure on yield failure of the materials. It can be expressed as(26)fI1,J2=J2+αI1−k=0,where the parameters α and k are the material constants, which can be determined from the slope and the intercept of the failure envelope. The parameter k is related to the cohesion and internal friction angle of rock mass. The parameter α is only related to the friction angle. Therefore, the Mohr–Coulomb parameter can be used to depict the DP criterion. By comparison with the yield curve of MC criterion in the π-plane (see Figure 4), the DP criteria can be divided into Circumscribe Drucker–Prager (CDP) criterion, Middle Circumscribe Drucker–Prager (MDP) criterion, and Inscribe Drucker–Prager (IDP) criterion.(i)
When the yield curve of DP criterion is the circumcircle of MC criterion, the material parametersα and k for the CDP criterion can be obtained by [4, 28](27a)α=2sinφ33−sinφ,k=6ccosφ33−sinφ.(ii)
When the yield curve of DP criterion is the inscribed circle of MC criterion, the solution of parametersα and k for the IDP criterion presented by Vekeens et al. [31] is(27b)α=sinφ9+3sin2φ,k=3ccosφ3+sin2φ.(iii)
When the yield curve of DP criterion is between CDP and IDP, the material parametersα and k for the MDP criterion are expressed as follows:(27c)α=2sinφ33+sinφ,k=6ccosφ33+sinφ.Figure 4
Drucker–Prager yield criterion curve in theπ-plane.As shown in Sections2.6 and 2.7, we take σ2=σ1+σ3/2. Then, the Drucker–Prager criterion can be deduced by solving equation (26).(28)fσ1,σ3=σ1−1+3α1−3ασ3−2k1−3α=0.


### 2.8. Establishment of Unified Criterion Equation

As shown in Table1, the unified equation of different yield criterion can be summarized based on the above analysis under plane strain conditions.(29)fσ1,σ3=σ1−ξjσ3−Yj=0,where the subscript “j” represents different yield criteria and ξj and Yj are the material constants of different criteria, which can be divided into two different cases. When ξj=1, equation (29) corresponds to the TR and VM criteria; when ξj≠1, equation (29) represents other criteria. Therefore, it can be regarded as a unified criterion equation to research the mechanical response of rock mass.


## 2.1. Mohr–Coulomb Criterion

The linear Mohr–Coulomb criterion is widely used in geotechnical engineering because of its simple expression form. However, the influence of intermediate principal stress on rock mass failure is ignored. In addition, the triaxial strength of rock mass obtained by MC criterion deviates greatly from the measured data under the high confining pressure conditions. The following is the governing equation for MC criterion based on the cohesionc and internal friction angle φ [1, 14, 39]:(1)σ1=Nσ3+σc,where N is a constant which is a function of the internal friction angle and σc is the uniaxial compressive strength (UCS). They can be expressed as follows:(2)N=1+sinφ1−sinφ,σc=2ccosφ1−sinφ.


## 2.2. Tresca Criterion

The Tresca (TR) criterion is a simplified form of MC criterion and assumes that the failure will occur if the maximum shear stressτmax inside any plane of rock mass reaches a critical value [1, 15]. As shown in Figure 1, the yield curve of TR criterion is a regular hexagon, which is inscribed in the circular yield curve of the von Mises (VM) criterion in π-plane. The expression form is(3)σ1−σ32=τmax=c.Figure 1
Tresca and von Mises criterion in theπ-plane.The TR criterion can be considered as a special case where the internal friction angle equals zero in the MC criterion. Equation (2) can be rewritten as(4)fσ1,σ3=σ1−σ3−2c=0.


## 2.3. Generalized Lade–Duncan Criterion

In early 1973, the Lade–Duncan criterion has been firstly proposed by considering the intermediate principal stresses based on the triaxial compression test for noncohesive soil. The initial expression form can be written as follows:(5)I13I3=KLade,where I1 and I3 are, respectively, the first and third principal stress invariants of friction material. KLade is the material constant which is closely related to internal friction angle:(6)KLade=3−sinφ31+sinφ1−sinφ2,(7)I1=σ1+σ2+σ3,(8)I3=σ1σ2σ3.In 1999, the Lade–Duncan criterion was modified by Ewy so that it could reasonably describe the strength characteristics of cohesive soil by introducing bound stress (σ0) [35]. Its generalized expression is also given as(9)I1′3I3′=KLade,where I1′ and I3′ are, respectively, the first and third principal stress invariants of cohesive-friction material:(10)I1′=σ1+σ0+σ2+σ0+σ3++σ0,I3′=σ1+σ0σ2+σ0σ3++σ0,where σ0 is the bound stress: σ0=ccotφ. For the noncohesive soils, under plane strain conditions, the internal principal stress can be expressed as(11)σ2=σ1σ3.By substituting equations (7), (8), and (11) into equation (6), the Lade–Duncan criterion can be rewritten as(12)σ1=14KLade3−1+KLade3−12−42σ3=ηLadeσ3.For the cohesive-friction material, according to the coordinate translation method [18, 19], the generalized Lade–Duncan (GLD) criterion will be obtained by integrating equation (12).(13)σ1+σ0=14KLade3−1+KLade3−12−42σ3+σ0=ηLadeσ3+σ0.fσ1,σ3=σ1−ηLadeσ3−ηLade−1σ0=0.


## 2.4. Generalized SMP Criterion

Considering the effect of internal principal stress, H. Matsuoka and T. Nakai proposed the SMP criterion, which could be expressed by three principal stress invariants [31]:(14)I1I2I3=KSMP,where I2 is the second principal stress invariant andKSMP is the material constant in the SMP criterion:(15)KSMP=8tan2φ+9,(16)I2=σ1σ3+σ1σ2+σ2σ3.For the frictional material, by introducing equations (7), (8), (11), and (16) into equation (14), another form of SMP criterion will be deduced.(17)σ1=14KSMP−1+KSMP−3−2KSMP2σ3=ηGSMPσ3.As the GLD criterion, according to the coordinate translation method, the generalized SMP criterion will be given as [18](18)σ1+σ0=14KSMP−1+KSMP−3−2KSMP2σ3+σ0=ηGSMPσ3+σ0.fσ1,σ3=σ1−ηGSMPσ3−ηGSMP−1σ0=0.As shown in Figure2, compared with the yield curve of MC criterion in the π-plane, the GLD and GSMP criteria are nonlinear and the cross-sectional area of GLD criterion is the largest, followed by the GSMP criterion.Figure 2
Comparison of MC, GLD, and GSMP criteria in theπ-plane under plane strain condition.


## 2.5. Mogi–Coulomb Criterion

Based on Mog’s theory [24], Al-Ajmi and Zimmerman found that the polyaxial test data could be fitted by linear relationships in τoct−σm,2 space.(19)τoct=aσm,2+b,(20)τoct=13σ1−σ22+σ1−σ32+σ2−σ32,(21)σm,2=σ1+σ2+σ33,where σm,2 is the mean normal stress and τoct is octahedral shear stress. The parameters a and b can be related exactly to the Coulomb strength parameter:(22)a=223sinφ,b=223ccosφ.As the von Mises criterion, we takeσ2=σ1+σ3/2. Then, the Mogi–Coulomb (GMC) criterion can be obtained by integrating equations (19), (20), and (21).(23)fσ1,σ3=σ1−6+3a6−3aσ3−6b6−3a=0.


## 2.6. Unified Strength Theory

Based on the twin shear yield criterion, the unified strength theory (UST) is established by considering the influence of all the stress components on the material yield failure [32–34]. In geotechnical engineering, the cohesion and internal friction angle are usually used to represent this yield theory. The yield function can be expressed as follows [5, 6]:Ifσ2≤σ1+σ3/2−σ1−σ3/2sinφ,(24a)fσ1,σ3=1−sinφ1+sinφσ1−bσ2+σ31+b=2ccosφ1+sinφ.Ifσ2≤σ1+σ3/2−σ1−σ3/2sinφ,(24b)fσ1,σ3=1−sinφ1+b1+sinφσ1+bσ2−σ3=2ccosφ1+sinφ.where b represents the yield parameter related to the intermediate principal stress, which can reflect the effect of the intermediate principal shear stress and the positive stress on the yield failure of the rock material, and 0≤b≤1. As shown in Figure 3, if b=0, UST will translate into the MC criterion; if b=1, UST is converted into general twin shear strength (GTSS) criterion; if 0<b<1, UST is a series of other ordered new strength criteria.Figure 3
Unified strength theory yield curve in theπ-plane [16].Just like Sections2.3 and 2.6, if σ2=σ1+σ3/2, σ2≥σ1+σ3/2−σ1−σ3/2sinφ can be judged by substituting σ2=σ1+σ3/2 into equation (24a) or equation (24b). Therefore, UST can be rewritten as(25)fσ1,σ3=σ1−2+b+2+3bsinφ2+b1−sinφσ3−41+bccosφ2+b1−sinφ=0.


## 2.7. Drucker–Prager Criterion

The Drucker–Prager (DP) criterion is an extension of the von Mises criterion, which takes into account the effect of intermediate principal stress and hydrostatic pressure on yield failure of the materials. It can be expressed as(26)fI1,J2=J2+αI1−k=0,where the parameters α and k are the material constants, which can be determined from the slope and the intercept of the failure envelope. The parameter k is related to the cohesion and internal friction angle of rock mass. The parameter α is only related to the friction angle. Therefore, the Mohr–Coulomb parameter can be used to depict the DP criterion. By comparison with the yield curve of MC criterion in the π-plane (see Figure 4), the DP criteria can be divided into Circumscribe Drucker–Prager (CDP) criterion, Middle Circumscribe Drucker–Prager (MDP) criterion, and Inscribe Drucker–Prager (IDP) criterion.(i)
When the yield curve of DP criterion is the circumcircle of MC criterion, the material parametersα and k for the CDP criterion can be obtained by [4, 28](27a)α=2sinφ33−sinφ,k=6ccosφ33−sinφ.(ii)
When the yield curve of DP criterion is the inscribed circle of MC criterion, the solution of parametersα and k for the IDP criterion presented by Vekeens et al. [31] is(27b)α=sinφ9+3sin2φ,k=3ccosφ3+sin2φ.(iii)
When the yield curve of DP criterion is between CDP and IDP, the material parametersα and k for the MDP criterion are expressed as follows:(27c)α=2sinφ33+sinφ,k=6ccosφ33+sinφ.Figure 4
Drucker–Prager yield criterion curve in theπ-plane.As shown in Sections2.6 and 2.7, we take σ2=σ1+σ3/2. Then, the Drucker–Prager criterion can be deduced by solving equation (26).(28)fσ1,σ3=σ1−1+3α1−3ασ3−2k1−3α=0.


## 2.8. Establishment of Unified Criterion Equation

As shown in Table1, the unified equation of different yield criterion can be summarized based on the above analysis under plane strain conditions.(29)fσ1,σ3=σ1−ξjσ3−Yj=0,where the subscript “j” represents different yield criteria and ξj and Yj are the material constants of different criteria, which can be divided into two different cases. When ξj=1, equation (29) corresponds to the TR and VM criteria; when ξj≠1, equation (29) represents other criteria. Therefore, it can be regarded as a unified criterion equation to research the mechanical response of rock mass.


## 3. Problem Description

### 3.1. Establishment of Calculating Model

Figure5 shows that a circular opening is excavated in a finite, isotropic, homogeneous elastic-brittle plastic rock mass subjected to an inner pressure σ0 at the inner radii R0 and a hydrostatic pressure p0 at the external radius R2. As σ0 gradually decreases, the displacement will occur and the plastic zone with the radii R1 will firstly develop around the circular opening when the maximum principal stress and minimum principal stress satisfy the initial yield condition. The influence of the rock mass weight in the plastic zone on the radial displacement and inner pressure is ignored. In this paper, the brittle plastic rock mass is introduced to research the postpeak mechanical behavior of rock material. As shown in Figure 6, the strength of the rock mass suddenly drops after peak load and then the postpeak softening behavior of strength parameters will occur. In other words, the postpeak cohesion cres, internal friction angle φres, Young’s modulus Eres, and Poisson’s ratio νres are used to solve the stress and displacement distributions in the plastic zone.Figure 5
Calculation model of the circular opening with finite or infinite boundary.Figure 6
Postpeak failure behavior of brittle plastic rock mass.
(a)(b)Under the axisymmetric plane strain conditions, whenp0>σ0, the hoop stress σθ and radial stress σr are, respectively, maximum principal stress and minimum principal stress; the tangential strain εθ and radial strain εr are, respectively, maximum principal strain and minimum principal strain. Accordingly, the unified criterion equation can be rewritten as follows:(30a)σθi=ξjiniσri+Yjinifor peak stage,(30b)σθi=ξjresσri+Yjresfor postpeak stage,where the subscript “i” represents different zones (i = 1, 2), ξjini and Yjini are the initial strength parameters, respectively, and ξjres and Yjres are the residual strength parameters, respectively.


### 3.2. Based Equation

For the axisymmetric plane strain problem, the equilibrium differential equation can be expressed as (ignoring the body force of rock masses) [29, 30, 39](31)dσridr+σri−σθir=0.The geometric equation, based on the small deformation assumption, can be denoted as(32)εri=∂uri∂r,εθi=urir,where uri is the radial displacement. Both the radial displacement uR1 and radial contact stress σR1 should be continuous at the elastic-plastic interface, respectively. Therefore, the boundary conditions around the circular opening can be summarized as(33)r=R0,σr1=σ0,r=R1,σr1=σR1,ur1=uR1,r=R2,σr2=p0.


## 3.1. Establishment of Calculating Model

Figure5 shows that a circular opening is excavated in a finite, isotropic, homogeneous elastic-brittle plastic rock mass subjected to an inner pressure σ0 at the inner radii R0 and a hydrostatic pressure p0 at the external radius R2. As σ0 gradually decreases, the displacement will occur and the plastic zone with the radii R1 will firstly develop around the circular opening when the maximum principal stress and minimum principal stress satisfy the initial yield condition. The influence of the rock mass weight in the plastic zone on the radial displacement and inner pressure is ignored. In this paper, the brittle plastic rock mass is introduced to research the postpeak mechanical behavior of rock material. As shown in Figure 6, the strength of the rock mass suddenly drops after peak load and then the postpeak softening behavior of strength parameters will occur. In other words, the postpeak cohesion cres, internal friction angle φres, Young’s modulus Eres, and Poisson’s ratio νres are used to solve the stress and displacement distributions in the plastic zone.Figure 5
Calculation model of the circular opening with finite or infinite boundary.Figure 6
Postpeak failure behavior of brittle plastic rock mass.
(a)(b)Under the axisymmetric plane strain conditions, whenp0>σ0, the hoop stress σθ and radial stress σr are, respectively, maximum principal stress and minimum principal stress; the tangential strain εθ and radial strain εr are, respectively, maximum principal strain and minimum principal strain. Accordingly, the unified criterion equation can be rewritten as follows:(30a)σθi=ξjiniσri+Yjinifor peak stage,(30b)σθi=ξjresσri+Yjresfor postpeak stage,where the subscript “i” represents different zones (i = 1, 2), ξjini and Yjini are the initial strength parameters, respectively, and ξjres and Yjres are the residual strength parameters, respectively.


## 3.2. Based Equation

For the axisymmetric plane strain problem, the equilibrium differential equation can be expressed as (ignoring the body force of rock masses) [29, 30, 39](31)dσridr+σri−σθir=0.The geometric equation, based on the small deformation assumption, can be denoted as(32)εri=∂uri∂r,εθi=urir,where uri is the radial displacement. Both the radial displacement uR1 and radial contact stress σR1 should be continuous at the elastic-plastic interface, respectively. Therefore, the boundary conditions around the circular opening can be summarized as(33)r=R0,σr1=σ0,r=R1,σr1=σR1,ur1=uR1,r=R2,σr2=p0.


## 4. The New Unified Solution of Circular Opening

### 4.1. A Semianalytical Solution with the Finite External Boundary

According to the elastic mechanics theory, the stresses and displacement in the elastic zone can be deduced by regarding it as a hollow thick-wall cylinder subjected to an inner pressureσR1 at the internal boundary r=R1 and a hydrostatic pressure p0 at the external boundary r=R2 as follows:(34a)σr2=R2R12σR1−p0R22−R121r2+R22p0−R12σR1R22−R12,(34b)σθ2=−R2R12σR1−p0R22−R121r2+R22p0−R12σR1R22−R12,(34c)ur2=1−ν−2ν2ER22p0−R12σR1R22−R12r+1+νER2R12p0−σR1R22−R121r,where E and ν are the initial Young’s modulus and initial Poisson’s ratio, respectively. The initial yield failure condition should be satisfied at the elastic-plastic interface. Therefore, the radial contact stress σR1 can be derived by substituting equations (34a) and (34b) into equation (30a) under different yield criteria:(35)σR1=2p0−YjiniR22+R12Yjini1+ξjiniR22+1−ξjiniR12.Obviously, the stresses in the plastic zone should satisfy the equilibrium differential equation and are easily deduced by submitting equation (30b) into equation (31) as well as combining with the boundary condition σr1=σ0 at r=R0. However, for ξjres=1 and ξjres≠1, the stress solutions in the plastic zone are inconsistent.(i)
fξjres≠1, the unified stresses solution in the plastic zone will be obtained as(36a)σr1=σ0+Yjresξjres−1rR0ξjres−1−Yjresξjres−1,(36b)σθ1=ξjresσ0+Yjresξjres−1rR0ξjres−1−Yjresξjres−1.Here, the subscript “j” represents MC, GMC, GLD, GSMP, CDP, MDP, IDP, UST, and GTSS criteria.Then, the radii of plastic zone can be determined by considering the boundary conditionσr1=σR1 at the elastic-plastic interface. By integrating equation (35) and equation (36a), the nonlinear function between R1 and R2 can be established as follows:(37)1+ξjiniR22+1−ξjiniR12σ0+Yjresξjres−1R1R0ξjres−1−Yjresξjres−1=2p0−YjiniR22+R12Yjini.(ii)
Ifξjres=1, the unified stress solutions of plastic zone based on the TR and VM criteria are(38a)σr1=σ0+YjreslnrR0,(38b)σθ1=σ0+Yjres1+lnrR0.As the solution for equation (41), the nonlinear function between R1 and R2 based on the VM and TR criterion can be also deduced as follows:(39)1+ξjiniR22+1−ξjiniR12σ0+YjreslnR1R0=2p0−YjiniR22+R12Yjini.In practice, the external radiusR2 is a constant. The approximate solution for R1 in equations (37) and (39) can be obtained by using the numerical calculation methods such as the least square method and Newton iteration method. Then, the stresses and displacement in elastic zone can be determined by equations (34a), (34b), and (34c).In the plastic zone, the total hoop strainεθ1 and radial strain εr1 are, respectively, composed of elastic strain and plastic strain. Therefore, the total strain can be expressed as(40)εθ1=εθ1e+εθ1p,εr1=εr1e+εr1p,where εθ1e and εr1e are, respectively, the hoop elastic strain and radial elastic strain in the plastic zone; εθ1p and εr1p are the hoop plastic strain and radial plastic strain of plastic zone, respectively.For axisymmetric plane strain problems, the plastic strain relationships can be established by considering the small strain theory and nonassociated flow rule [5, 9].(41)εr1p+βεθ1p=0,where β is the dilation coefficient: β=1+sinψ/1−sinψ. ψ is the dilation angle.By substituting equations (32) and (41) into equation (44), the following differential equation for the radial displacement in the plastic zone can be derived as(42)∂ur1∂r+βur1r=εr1e+βεθ1e=fr.From equation (42), it can be seen that the radial displacement is closely related to the elastic strain form in the plastic zone.Then, the following function can be obtained by solving equation (42) as follows:(43)ur1=1rβ∫R1rfrrβdr+uR1R1rβ,where uR1 is the radial displacement at the elastic-plastic interface. It can be easily determined by taking r=R1 in equation (34c) as follows:(44)uR1=1−ν−2ν2ER22p0−R12σR1R22−R12R1+1+νER22R1p0−σR1R22−R12.In order to obtain the radial displacement of plastic zone, the expression for elastic strain should be firstly determined. Generally, four different definitions for elastic strain can be used to research the deformation behavior of rock mass in the plastic zone.(1)
Case 1: it is assumed that the elastic strain in the plastic region is equal to that on the elastic-plastic interface. Then, the elastic strains can be expressed as [36](45)εθ1e=ur2rr=R1=1−ν−2ν2ER22p0−R12σR1R22−R12+1+νER22p0−σR1R22−R12.εr1e=∂ur2∂rr=R1=1−ν−2ν2ER22p0−R12σR1R22−R12−1+νER22p0−σR1R22−R12.Then, the functionfr is(46)fr=1−ν−2ν21+βER22p0−R12σR1R22−R12−1+ν1−βER22p0−σR1R22−R12=δcase1.By substituting equation (46) into equation (43), the radial displacement in the plastic zone can be derived as follows:(47)ur1case1=δcase11+βrβ+1−R1β+1rβ+uR1R1rβ.(2)
Case 2: by regarding the plastic zone as the thick-wall cylinder subjected to the inner pressureσ0 at r=R0 and radial contact stress σR1 at r=R0, the elastic strain in the plastic zone can be written as [19, 39](48)εθ1e=1+vresEres1−2vresC1−C2r2,εr1e=1+vresEres1−2vresC1+C2r2,whereC1=σR1−p0R12−σ0−p0R02/R12−R02 and C2=R12R02σ0−σR1/R12−R02. Therefore, the function fr is(49)fr=1+vresEres1+β1−2vresC1+1−βC2r2,=δ1Case21+β+δ2Case21−βr−2,whereδ1Case2=1+vres/Eres1−2vresC1 and δ2Case2=1+vres/EresC2.By submitting equation (49) into equation (43), we can obtain the radial displacement of plastic zone as follows:(50)ur1case2=1rβδ1Case2rβ+1−R1β+1−δ2Case2rβ−1−R1β−1+uR1R1rβ.(3)
Case 3: adopting Hooke’s law, the elastic strains in the plastic zone can be written as follows [27]:(51)εθ1e=1+vresEres1−vresσθ1−vresσr1.εr1e=1+vresEres1−vresσr1−vresσθ1.Then, the functionfr can be expressed as follows:(52)fr=1+vresEres1−νres−βνresσr1+β−νres−βνresσθ1.(i)
Ifξjres≠1, the function fr will be rewritten by substituting equations (36a) and (36b) into equation (52) as follows:(53)fr=1+vresEresδ1case3rR0ξjres−1−δ2case3,whereδ1case3=σ0+Yjres/ξjres−11+βξjres−1+ξjresνres+βνres and δ2case3=1−2νres1+βYjres/ξjres−1.By substituting equation (53) into equation (43), the radial displacement in the plastic zone can be derived as follows:(54)ur1case3=1+νresEres1rβδ1case3ξjres+βR0ξjres−1rξjres+β−R1ξjres+β−δ2case31+βr1+β−R11+β+uR1R1rβ.(ii)
Ifξjres=1, the function fr can be obtained by substituting equations (38a) and (38b) into equation (52) as follows:(55)fr=1+vresEresδ3case3lnrR0+δ4case3,whereδ3case3=1−2vres1+βYjres and δ4case3=1−νres−βνresσ0+β−νres−βνresσ0+Yjres.By submitting equation (55) to equation (43), the radial displacements for TR and VM criterion in the plastic zone are(56)ur1case3=1+νresEres1rβδ3case31+βr1+βlnrR0−R11+βlnR1R0+δ4case31+β−δ3case31+β2r1+β−R11+β+uR1R1rβ.(4)
Case 4: the mechanical behavior of rock mass is closely related to its damage degree in the plastic zone. The higher damage usually leads to larger deformation behavior. Then, the attenuation of Young’s modulus should be considered in the plastic zone. In this case, the power function attenuation model of Young’s modulus along the radius direction is introduced to research the deformation behavior in the plastic zone [6]:(57)Er=EresrR0m.m=logE/EreslogR1/R0.(i)
Ifξjres≠1, the function fr will be rewritten by substituting equation (57) into equation (53) as follows:(58)fr=1+vresEresR0rmδ1case3rR0ξjres−1−δ2case3.By substituting equation (58) into equation (43), the radial displacement in the plastic zone can be derived. In fact, the expression form is the same as that in equation (54):(59)ur1case3=1+νresEres1rβδ1case3ξjres+β−mR0ξjres−m−1rξjres+β−m−R1ξjres+β−m−δ2case31+β−mR0−mr1+β−m−R11+β−m+uR1R1rβ.(ii)
Ifξjres=1, the function fr can be obtained by substituting equation (57) into equation (55) as follows:(60)fr=1+vresEresR0rmδ3case3lnrR0+δ4case3.By introducing equation (60) to equation (43), the expression form for radial displacement is similar to equation (56):(61)ur1case3=1+νresEres1rβδ3case3R0m1+β−mr1+β−mlnrR0−R11+β−mlnR1R0+δ4case31+β−m−δ3case31+β−m2R0mr1+β−m−R11+β−m+uR1R1rβ.


### 4.2. The Closed-Form Solution with the Infinite External Boundary

For the deep underground engineering, a circular opening subjected to an inner support pressureσ0 at r=R0 and hydrostatic pressure p0 at infinite external boundary is a special case of the axisymmetric thick-wall cylinder. The stress and displacement solutions can be obtained through the same solving process as that in Section 4.1. When R2⟶∞, the stresses in the elastic zone can be easily deduced by equations (34a) and (34b) as follows:(62a)σr2=p0−p0−σR1R1r2,(62b)σθ2=p0+p0−σR1R1r2.Accordingly, the stress distribution for deep circular opening in the plastic zone can be also calculated by equations (36a) and (36b) for ξjres≠1 or equations (38a) and (38b) for ξjres=1. In addition, the unified radial contact stress σR1 for different criteria can be rewritten by taking R2⟶∞ in equation (35) as follows:(63)σR1=2p0−Yjini1+ξjini.As shown in equation (37), if ξjres≠1, the radius of plastic zone for deep circular opening can be easily deduced as follows:(64a)R1=R02p0−Yjini/1+ξjini+Yjres/ξjres−1σ0+Yjres/ξjres−1ξjres−1.As shown in equation (39), if ξjres=1, the unified radius solution of plastic zone based on TR and VM criteria is(64b)R1=R0e2p0−Yjini/1+ξjini−σ0Yjres.For the deep underground engineering, the real displacement far away from the tunnel surface is almost zero after the tunnel excavation. Hence, the initial displacement values, that is,1−ν−2ν2p0r/E, caused by the hydrostatic pressure should be ignored at the infinite boundary (R2⟶∞). Therefore, the radial displacement caused by tunnel excavation in the elastic zone should be rewritten as follows:(65)ur2=1+νp0−σR1ER12r.The radial displacement at the elastic-plastic interface can be determined by equations (64a) and (64b) as follows:(66)uR1=1+νp0−σR1ER1.Just as the calculation method of radial displacement in Section4.2, four different assumptions about the elastic strain forms are also used to research the deformation behavior of rock mass in the plastic zone.(1)
Case 1: with the assumption of the constant elastic strain at the elastic-plastic interface, as the calculation methods in equations (45) and (46), the function fr can be easily deduced by integrating equations (65) and (32) as follows:(67)fr=−1+ν1−βEp0−σR1=δcase1.Then, the radial displacement of plastic zone for deep circular opening will be obtained by introducing equations (67) and (66) into equation (43). The expression form is consistent with equation (47) except for the parameters δcase1 (taking equation (67)) and uR1(taking equation (66)).(2)
Case 2: as shown in equations (49) and (50), the radial displacement for deep circular opening can be also obtained by regarding the plastic zone as the thick-wall cylinder. The expression is the same as that in equation (50) except for uR1 (equation (66)) and σR1 (equation (63)).(3)
Case 3: by adopting generalized Hooke’s law for removing the effect of initial hydrostatic pressurep0, the elastic strain in the plastic zone can be expressed as follows:(68)εθ1e=1+vresEres1−vresσθ1−p0−vresσr1−p0.εr1e=1+vresEres1−vresσr1−p0−vresσθ1−p0.Then, the functionfr can be expressed as follows:(69)fr=1+vresEres1−νres−βνresσr1+β−νres−βνresσθ1+2νres−11+βp0.(i)
Ifξjres≠1, the radial displacement for deep circular opening can be deduced by calculation methods similar to those discussed in Section 4.1. Its expression is the same as that in equation (54), but the constant δ2case3 in equation (54) needs to be replaced by(70)δ22case3=1−2νres1+βp0−Yjresξjres−1.(ii)
Ifξjres=1, the radial displacement for TR and VM criteria can be also obtained by substituting equations (38a), (38b), and (69) into equation (43). The expression is in accordance with equation (56); after that the constant δ4case3 in equation (56) is replaced by(71)δ44case3=1−νres−βνresσ0+β−νres−βνresσ0+Yjres+2νres−11+βp0.(4)
Case 4: in this case, the influence of rock mass damage degree on the mechanical and deformation behavior of surrounding rock is taken into account by assuming Young’s modulus attenuation along the radius direction. As discussed in Section4.1, for ξjres≠1, the radial displacement for deep circular opening in the plastic zone can be determined through the constant δ2case3 in equation (59) replaced by the value δ22case3 in equation (70). In addition, for ξjres=1, the radial displacement based on TR and VM criteria will be calculated by the constant δ4case3 in equation (61) replaced by the value δ44case3 in equation (71).


## 4.1. A Semianalytical Solution with the Finite External Boundary

According to the elastic mechanics theory, the stresses and displacement in the elastic zone can be deduced by regarding it as a hollow thick-wall cylinder subjected to an inner pressureσR1 at the internal boundary r=R1 and a hydrostatic pressure p0 at the external boundary r=R2 as follows:(34a)σr2=R2R12σR1−p0R22−R121r2+R22p0−R12σR1R22−R12,(34b)σθ2=−R2R12σR1−p0R22−R121r2+R22p0−R12σR1R22−R12,(34c)ur2=1−ν−2ν2ER22p0−R12σR1R22−R12r+1+νER2R12p0−σR1R22−R121r,where E and ν are the initial Young’s modulus and initial Poisson’s ratio, respectively. The initial yield failure condition should be satisfied at the elastic-plastic interface. Therefore, the radial contact stress σR1 can be derived by substituting equations (34a) and (34b) into equation (30a) under different yield criteria:(35)σR1=2p0−YjiniR22+R12Yjini1+ξjiniR22+1−ξjiniR12.Obviously, the stresses in the plastic zone should satisfy the equilibrium differential equation and are easily deduced by submitting equation (30b) into equation (31) as well as combining with the boundary condition σr1=σ0 at r=R0. However, for ξjres=1 and ξjres≠1, the stress solutions in the plastic zone are inconsistent.(i)
fξjres≠1, the unified stresses solution in the plastic zone will be obtained as(36a)σr1=σ0+Yjresξjres−1rR0ξjres−1−Yjresξjres−1,(36b)σθ1=ξjresσ0+Yjresξjres−1rR0ξjres−1−Yjresξjres−1.Here, the subscript “j” represents MC, GMC, GLD, GSMP, CDP, MDP, IDP, UST, and GTSS criteria.Then, the radii of plastic zone can be determined by considering the boundary conditionσr1=σR1 at the elastic-plastic interface. By integrating equation (35) and equation (36a), the nonlinear function between R1 and R2 can be established as follows:(37)1+ξjiniR22+1−ξjiniR12σ0+Yjresξjres−1R1R0ξjres−1−Yjresξjres−1=2p0−YjiniR22+R12Yjini.(ii)
Ifξjres=1, the unified stress solutions of plastic zone based on the TR and VM criteria are(38a)σr1=σ0+YjreslnrR0,(38b)σθ1=σ0+Yjres1+lnrR0.As the solution for equation (41), the nonlinear function between R1 and R2 based on the VM and TR criterion can be also deduced as follows:(39)1+ξjiniR22+1−ξjiniR12σ0+YjreslnR1R0=2p0−YjiniR22+R12Yjini.In practice, the external radiusR2 is a constant. The approximate solution for R1 in equations (37) and (39) can be obtained by using the numerical calculation methods such as the least square method and Newton iteration method. Then, the stresses and displacement in elastic zone can be determined by equations (34a), (34b), and (34c).In the plastic zone, the total hoop strainεθ1 and radial strain εr1 are, respectively, composed of elastic strain and plastic strain. Therefore, the total strain can be expressed as(40)εθ1=εθ1e+εθ1p,εr1=εr1e+εr1p,where εθ1e and εr1e are, respectively, the hoop elastic strain and radial elastic strain in the plastic zone; εθ1p and εr1p are the hoop plastic strain and radial plastic strain of plastic zone, respectively.For axisymmetric plane strain problems, the plastic strain relationships can be established by considering the small strain theory and nonassociated flow rule [5, 9].(41)εr1p+βεθ1p=0,where β is the dilation coefficient: β=1+sinψ/1−sinψ. ψ is the dilation angle.By substituting equations (32) and (41) into equation (44), the following differential equation for the radial displacement in the plastic zone can be derived as(42)∂ur1∂r+βur1r=εr1e+βεθ1e=fr.From equation (42), it can be seen that the radial displacement is closely related to the elastic strain form in the plastic zone.Then, the following function can be obtained by solving equation (42) as follows:(43)ur1=1rβ∫R1rfrrβdr+uR1R1rβ,where uR1 is the radial displacement at the elastic-plastic interface. It can be easily determined by taking r=R1 in equation (34c) as follows:(44)uR1=1−ν−2ν2ER22p0−R12σR1R22−R12R1+1+νER22R1p0−σR1R22−R12.In order to obtain the radial displacement of plastic zone, the expression for elastic strain should be firstly determined. Generally, four different definitions for elastic strain can be used to research the deformation behavior of rock mass in the plastic zone.(1)
Case 1: it is assumed that the elastic strain in the plastic region is equal to that on the elastic-plastic interface. Then, the elastic strains can be expressed as [36](45)εθ1e=ur2rr=R1=1−ν−2ν2ER22p0−R12σR1R22−R12+1+νER22p0−σR1R22−R12.εr1e=∂ur2∂rr=R1=1−ν−2ν2ER22p0−R12σR1R22−R12−1+νER22p0−σR1R22−R12.Then, the functionfr is(46)fr=1−ν−2ν21+βER22p0−R12σR1R22−R12−1+ν1−βER22p0−σR1R22−R12=δcase1.By substituting equation (46) into equation (43), the radial displacement in the plastic zone can be derived as follows:(47)ur1case1=δcase11+βrβ+1−R1β+1rβ+uR1R1rβ.(2)
Case 2: by regarding the plastic zone as the thick-wall cylinder subjected to the inner pressureσ0 at r=R0 and radial contact stress σR1 at r=R0, the elastic strain in the plastic zone can be written as [19, 39](48)εθ1e=1+vresEres1−2vresC1−C2r2,εr1e=1+vresEres1−2vresC1+C2r2,whereC1=σR1−p0R12−σ0−p0R02/R12−R02 and C2=R12R02σ0−σR1/R12−R02. Therefore, the function fr is(49)fr=1+vresEres1+β1−2vresC1+1−βC2r2,=δ1Case21+β+δ2Case21−βr−2,whereδ1Case2=1+vres/Eres1−2vresC1 and δ2Case2=1+vres/EresC2.By submitting equation (49) into equation (43), we can obtain the radial displacement of plastic zone as follows:(50)ur1case2=1rβδ1Case2rβ+1−R1β+1−δ2Case2rβ−1−R1β−1+uR1R1rβ.(3)
Case 3: adopting Hooke’s law, the elastic strains in the plastic zone can be written as follows [27]:(51)εθ1e=1+vresEres1−vresσθ1−vresσr1.εr1e=1+vresEres1−vresσr1−vresσθ1.Then, the functionfr can be expressed as follows:(52)fr=1+vresEres1−νres−βνresσr1+β−νres−βνresσθ1.(i)
Ifξjres≠1, the function fr will be rewritten by substituting equations (36a) and (36b) into equation (52) as follows:(53)fr=1+vresEresδ1case3rR0ξjres−1−δ2case3,whereδ1case3=σ0+Yjres/ξjres−11+βξjres−1+ξjresνres+βνres and δ2case3=1−2νres1+βYjres/ξjres−1.By substituting equation (53) into equation (43), the radial displacement in the plastic zone can be derived as follows:(54)ur1case3=1+νresEres1rβδ1case3ξjres+βR0ξjres−1rξjres+β−R1ξjres+β−δ2case31+βr1+β−R11+β+uR1R1rβ.(ii)
Ifξjres=1, the function fr can be obtained by substituting equations (38a) and (38b) into equation (52) as follows:(55)fr=1+vresEresδ3case3lnrR0+δ4case3,whereδ3case3=1−2vres1+βYjres and δ4case3=1−νres−βνresσ0+β−νres−βνresσ0+Yjres.By submitting equation (55) to equation (43), the radial displacements for TR and VM criterion in the plastic zone are(56)ur1case3=1+νresEres1rβδ3case31+βr1+βlnrR0−R11+βlnR1R0+δ4case31+β−δ3case31+β2r1+β−R11+β+uR1R1rβ.(4)
Case 4: the mechanical behavior of rock mass is closely related to its damage degree in the plastic zone. The higher damage usually leads to larger deformation behavior. Then, the attenuation of Young’s modulus should be considered in the plastic zone. In this case, the power function attenuation model of Young’s modulus along the radius direction is introduced to research the deformation behavior in the plastic zone [6]:(57)Er=EresrR0m.m=logE/EreslogR1/R0.(i)
Ifξjres≠1, the function fr will be rewritten by substituting equation (57) into equation (53) as follows:(58)fr=1+vresEresR0rmδ1case3rR0ξjres−1−δ2case3.By substituting equation (58) into equation (43), the radial displacement in the plastic zone can be derived. In fact, the expression form is the same as that in equation (54):(59)ur1case3=1+νresEres1rβδ1case3ξjres+β−mR0ξjres−m−1rξjres+β−m−R1ξjres+β−m−δ2case31+β−mR0−mr1+β−m−R11+β−m+uR1R1rβ.(ii)
Ifξjres=1, the function fr can be obtained by substituting equation (57) into equation (55) as follows:(60)fr=1+vresEresR0rmδ3case3lnrR0+δ4case3.By introducing equation (60) to equation (43), the expression form for radial displacement is similar to equation (56):(61)ur1case3=1+νresEres1rβδ3case3R0m1+β−mr1+β−mlnrR0−R11+β−mlnR1R0+δ4case31+β−m−δ3case31+β−m2R0mr1+β−m−R11+β−m+uR1R1rβ.


## 4.2. The Closed-Form Solution with the Infinite External Boundary

For the deep underground engineering, a circular opening subjected to an inner support pressureσ0 at r=R0 and hydrostatic pressure p0 at infinite external boundary is a special case of the axisymmetric thick-wall cylinder. The stress and displacement solutions can be obtained through the same solving process as that in Section 4.1. When R2⟶∞, the stresses in the elastic zone can be easily deduced by equations (34a) and (34b) as follows:(62a)σr2=p0−p0−σR1R1r2,(62b)σθ2=p0+p0−σR1R1r2.Accordingly, the stress distribution for deep circular opening in the plastic zone can be also calculated by equations (36a) and (36b) for ξjres≠1 or equations (38a) and (38b) for ξjres=1. In addition, the unified radial contact stress σR1 for different criteria can be rewritten by taking R2⟶∞ in equation (35) as follows:(63)σR1=2p0−Yjini1+ξjini.As shown in equation (37), if ξjres≠1, the radius of plastic zone for deep circular opening can be easily deduced as follows:(64a)R1=R02p0−Yjini/1+ξjini+Yjres/ξjres−1σ0+Yjres/ξjres−1ξjres−1.As shown in equation (39), if ξjres=1, the unified radius solution of plastic zone based on TR and VM criteria is(64b)R1=R0e2p0−Yjini/1+ξjini−σ0Yjres.For the deep underground engineering, the real displacement far away from the tunnel surface is almost zero after the tunnel excavation. Hence, the initial displacement values, that is,1−ν−2ν2p0r/E, caused by the hydrostatic pressure should be ignored at the infinite boundary (R2⟶∞). Therefore, the radial displacement caused by tunnel excavation in the elastic zone should be rewritten as follows:(65)ur2=1+νp0−σR1ER12r.The radial displacement at the elastic-plastic interface can be determined by equations (64a) and (64b) as follows:(66)uR1=1+νp0−σR1ER1.Just as the calculation method of radial displacement in Section4.2, four different assumptions about the elastic strain forms are also used to research the deformation behavior of rock mass in the plastic zone.(1)
Case 1: with the assumption of the constant elastic strain at the elastic-plastic interface, as the calculation methods in equations (45) and (46), the function fr can be easily deduced by integrating equations (65) and (32) as follows:(67)fr=−1+ν1−βEp0−σR1=δcase1.Then, the radial displacement of plastic zone for deep circular opening will be obtained by introducing equations (67) and (66) into equation (43). The expression form is consistent with equation (47) except for the parameters δcase1 (taking equation (67)) and uR1(taking equation (66)).(2)
Case 2: as shown in equations (49) and (50), the radial displacement for deep circular opening can be also obtained by regarding the plastic zone as the thick-wall cylinder. The expression is the same as that in equation (50) except for uR1 (equation (66)) and σR1 (equation (63)).(3)
Case 3: by adopting generalized Hooke’s law for removing the effect of initial hydrostatic pressurep0, the elastic strain in the plastic zone can be expressed as follows:(68)εθ1e=1+vresEres1−vresσθ1−p0−vresσr1−p0.εr1e=1+vresEres1−vresσr1−p0−vresσθ1−p0.Then, the functionfr can be expressed as follows:(69)fr=1+vresEres1−νres−βνresσr1+β−νres−βνresσθ1+2νres−11+βp0.(i)
Ifξjres≠1, the radial displacement for deep circular opening can be deduced by calculation methods similar to those discussed in Section 4.1. Its expression is the same as that in equation (54), but the constant δ2case3 in equation (54) needs to be replaced by(70)δ22case3=1−2νres1+βp0−Yjresξjres−1.(ii)
Ifξjres=1, the radial displacement for TR and VM criteria can be also obtained by substituting equations (38a), (38b), and (69) into equation (43). The expression is in accordance with equation (56); after that the constant δ4case3 in equation (56) is replaced by(71)δ44case3=1−νres−βνresσ0+β−νres−βνresσ0+Yjres+2νres−11+βp0.(4)
Case 4: in this case, the influence of rock mass damage degree on the mechanical and deformation behavior of surrounding rock is taken into account by assuming Young’s modulus attenuation along the radius direction. As discussed in Section4.1, for ξjres≠1, the radial displacement for deep circular opening in the plastic zone can be determined through the constant δ2case3 in equation (59) replaced by the value δ22case3 in equation (70). In addition, for ξjres=1, the radial displacement based on TR and VM criteria will be calculated by the constant δ4case3 in equation (61) replaced by the value δ44case3 in equation (71).


## 5. The Theoretical and Practical Value

The new unified closed-form solution obtained in Section4.1 with a finite boundary can be regarded as a semianalytical solution due to solving equations (37) and (39) by a numerical calculation. In addition, the new closed-form solution for a deep circular opening in Section 4.2 with an infinite boundary is a true analytical solution. The new closed-form solution with the infinite and finite external boundary can be used to predict the stress and displacement of circular opening excavated in an elastic-brittle plastic rock mass under different yield criteria. Actually, the circular opening maybe includes a borehole, vertical shaft, and deep- or shallow-buried tunnel. In addition, the new unified solution proposed in this paper is a series of traditional solutions rather than one specific solution. The solution based on the MC, GMC, TR, VM, GLD, GSMP, DP (CDP, MDP, and IDP), UST, and GTSS yield criteria can be obtained from the new unified solution by making ξj and Yj take different parameters (as shown in Table 1). Therefore, the new unified solution has a broader engineering application scope and can be more appropriate for different lithology rock masses.Zhang et al. [6], Park and Kim [9], and Chen et al. [17] carried out relevant theoretical research on the new closed-form solution of circular opening as discussed in Sections 4.1 and 4.2. However, the solution by Zhang et al. was based on the unified strength theory and ignored the influence of the other criteria on the mechanical response of rock mass. Hence, the application of this solution has some limitations for different lithology rock masses. Chen’s solution, based on the elastic, perfectly plastic model and Mohr-Coulomb criterion, did not reflect the effect of internal principal stress and strength parameter attenuation. In addition, Park and Kim [9] analyzed the deformation law of plastic zone by using three different definitions for elastic strains in the plastic zone with Mohr-Coulomb (MC) and Hoke-Brown (HB) criteria. However, the solution based on the finite external boundary was not given. For the shallow-buried tunnel, it may not be reasonable to regard the outer boundary of circular tunnel as the infinite boundary. Moreover, the internal principal stress and Young’s modulus attenuation are also ignored in this paper. When j=UST, the new unified solution proposed by this paper will be converted to the results by Zhang et al. [6] except for Case 1 and Case 2. When j=MCvres=v, cres=c, φres=φ, and Eres=E, the solution based on this paper will be transformed into the solution by Chen et al. [17]. Meanwhile, Park and Kim’s solution based on MC criterion is a special case in Section 4.2 of this paper except for Case 4. Therefore, the new unified closed-form solution obtained in this paper is a series of results, and it is suitable for a wide range of rock masses and engineering backgrounds. Meanwhile, it can provide a broader yield criterion for the engineering application as well.


## 6. Correctness Verification and Parameter Analysis

### 6.1. A Comparison with the Traditional Solution

Park et al. summarized three different definitions for elastic strains in the plastic zone (see Cases 1∼3 in Section4.2) and analyzed the deformation law of plastic zone with Mohr-Coulomb criterion [9]. However, the Young’s modulus and Poisson’s ratio attenuation were ignored. In fact, it can be obtained by taking Eres=E, νres=ν, ξj=ξMC, and Yj=YMC in this paper. To further verify the correctness of the calculation results for deep circular opening in Section 4.2, the solution by Park et al. will be presented as a comparison with the solution proposed by this paper. The geometrical and physical parameters for circular opening are shown in Table 2.Table 2
Geometrical and physical parameters of circular opening.
ParametersRock type by Park and Kim (2006) [10]Hard rockSoft rockRadius of opening,R0 (m)11Initial stress,σ0 (MPa)11Internal pressure,pin (MPa)00Young’s modulus,E (MPa)50,0005000Poisson’s ratio,ν0.20.2Shear modulus,G (MPa)208332083c (MPa)0.1730.276φ (deg)5535cres (MPa)0.0610.055φres (deg)5230The radial displacements obtained by Park are shown in Figure7 in this paper. It can be seen that the calculation results in this paper are in accordance with Park’s closed-form solution. Therefore, the closed-form solution proposed by this paper is correct and can be regarded as an extension of Park’s solution. In other words, this paper’s solution has a more wide application in practice engineering.Figure 7
Comparison with the traditional solution. (a) Radial displacement (data: hard rock,β = 1.0). (b) Radial displacement (data: hard rock, β = 3.0). (c) Radial displacement (data: soft rock, β = 1.0). (d) Radial displacement (data: soft rock, β = 3.0).
(a)(b)(c)(d)


### 6.2. Parameters Analysis

#### 6.2.1. The Effect of the Strength Yield Criteria

As previously mentioned, the strength criteria exert an extremely important effect for evaluating the mechanical response and deformation behavior of surrounding rock. For studying the influence of strength theories on the stresses and displacement of surrounding rock, the mechanical and geometrical parameters of circular tunnel are shown in Table3.Table 3
The mechanical and geometrical parameters of circular opening.
p0 (MPa)σ0 (MPa)R0 (m)Eres=E (GPa)νres=νcres=c (MPa)φres=φ (°)β250330.257.218.31.0The stresses distribution laws under different strength criteria are shown in Figure8. Meanwhile, the dimensionless radius of plastic zone and critical inner pressure at the elastic-plastic interface are presented in Table 4. It can be seen that the dimensionless values (R1/R0 and σR1/p0) show the characteristics of TR > VM > IDP > MC > MDP > UST0.5 > GSMP > GMC > GLD > GTSS > CDP. Compared with the traditional solution obtained by MC criterion, the dimensionless value (R1/R0) calculated by TR and VM criteria obviously increases by 125.26% and 78.53%, respectively. This is mainly because the TR and VM criteria only regard the rock mass as friction-less bonding material and ignore the impact of friction effect on the mechanical properties of the rock mass so that the bearing capacity of rock mass is underestimated. Therefore, the calculation result is obviously larger than the solution obtained by other criteria. The MC criterion does not take the effect of the intermediate principal stress into account and it is easy to underestimate the bearing capacity of rock mass. Therefore, the deformation of surrounding rock calculated by MC criterion may be slightly larger. The IDP and MDP criteria underestimate the influence of internal principal stresses, so the calculation results may also be larger than the solution obtained by UST0.5, GSMP, GMC, GLD, CDP, and GTSS criteria. Meanwhile, the result obtained by CDP criterion is minimal compared with other criteria’s solutions. In fact, the criterion overestimates the effect of intermediate principal stresses on rock mass strength and may not be reasonable in practical engineering. In addition, the calculation results (R1/R0) obtained by UST0.5, GSMP, GMC, and GLD criteria are close to each other within the range of 1.294–1.347. The above four criteria seem to be more reasonable considering the effect of intermediate principal stress on yield strength of rock masses.Figure 8
The stresses distributions law under different criteria.
(a)(b)(c)(d)Table 4
The radius of plastic zone and critical inner pressures under different criteria.
Strength criterionR1/R0σR1/p0Strength criterionR1/R0σR1/p0MC1.528 (0.413)VM2.728 (0.667)GMC1.318 (0.322)IDP1.556 (0.422)GLD1.294 (0.308)MDP1.456 (0.386)GSMP1.332 (0.329)CDP1.198 (0.242)TR3.442 (0.712)UST0.51.347 (0.337)GTSS1.266 (0.291)——From the above analysis, it can be seen that the UST0.5, GSMP, GMC, and GLD criteria can be strongly recommended for evaluating the mechanics and deformation behavior of surrounding rock, followed by IDP, MDP, GTSS, and MC criteria. The TR, VM, and CDP criteria are not recommended to be used for underground engineering.


#### 6.2.2. The Effect of Dilation Coefficient

As shown in Figure9, the dilation coefficient has an extremely important influence on the surface displacement of surrounding rock. As the parameter β gradually increases, the dimensionless surface displacement u0E/p0R0 presents the nonlinear increase characteristics. However, the increasing rate of surface displacement under different yield criteria is significantly different. For instance, as shown in Table 5, ß increases from 1.0 to 3.0, and the dimensionless value u0E/p0R0, respectively, increases by 38.12% for MC criterion, 16.94% for GSMP criterion, 18.30% for UST0.5 criterion, and 29.58% for MDP criterion under Case 1 condition. Therefore, the effect of dilation coefficient should be considered in the design of tunnel support parameters and strength.Figure 9
The effect of dilation coefficient on the surface displacement.
(a)(b)Table 5
The surface displacement valueu0E/p0R0 of the circular tunnel.
Parameter (β)MCGSMPUST0.5MDPCase 11.01.7131.4881.5031.6262.01.9901.6011.6261.8343.02.3661.7401.7782.107Case 21.01.9451.6071.6301.8152.02.58671.8791.9242.3023.03.5672.2412.3203.011


#### 6.2.3. The Effect of Elastic Strain Form and Rock Mass Damage

In this study, the geometrical and mechanical parameters of circular tunnel are shown in Table3 (soft rock). Figure 10 presents the influence of elastic strain form in the plastic zone on the surface displacement of surrounding rock. From the above analysis, the relevant conclusions can be summarized as follows:(i)
The radial displacement of plastic zone is closely related to the selection of the elastic strain form. Case 2 has the greatest effect on the radial displacement of plastic zone, followed by Case 3. Then, the results obtained by Case 2 are minimal. For example, whenr=R0 (tunnel surface), compared with Case 1, the dimensionless surface displacement u0E/p0R0, respectively, increases by 0.272 for Case 2 and 0.206 for Case 3 with an increasing rate of 16.47% for Case 2 and 12.46% for Case 3 under GLD criterion.(ii)
Meanwhile, the influence of elastic strain form on the radial displacement of plastic zone is also closely related to the strength criterion. For example, the dimensionless surface displacementsu0E/p0R0 are, respectively, 1.926 for Case 2 and 1.860 for Case 3 under GLD criterion. However, the value significantly increases by 18.32% for Case 2 and 31.71% for Case 3 under GTSS criterion.Figure 10
The effect of elastic strain form on the radial displacement.
(a)(b)When the rock mass enters the plastic zone, its mechanical and deformation behavior are closely related to the rock damage degree. Generally, Young’s modulus attenuation could be used to indicate the damage degree of rock mass. The influence of rock mass damage degree on the surface displacement of surrounding rock is shown in Figure11.(iii)
It can be seen that the radial displacement of plastic zone is closely related to the selection of Young’s modulus attenuation model. Compared with Case 3Eres=E, Case 4 has the greatest effect on the radial displacement of plastic zone. Case 3 (Eres=0.65E) is the second. For example, as shown in Table 6, compared with Case 3 (Eres=E), the dimensionless surface displacements u0E/p0R0 of Case 3 Eres=0.65E and Case 4 increase by 0.163 and 0.074 with an increasing rate of 7.49% and 3.40%, respectively, under the GTSS criterion.(iv)
In addition, the influence of Young’s modulus attenuation on the radial displacement of plastic zone is closely related to the strength criterion. For Case 3Eres=0.65E and Case 4, the dimensionless surface displacements u0E/p0R0 of surrounding rock are 1.971 and 1.911, respectively, under the GLD criterion. However, when taking the GSMP criterion, the value significantly increases by 25.82% and 24.54% compared with the GLD criterion, respectively.Figure 11
The effect of rock mass damage on the radial displacement.
(a)(b)Table 6
The dimensionless surface displacementu0E/p0R0 of the circular tunnel.
GLDGTSSGMCGSMPCase 3m=0,Eres=E1.8602.1771.9372.297m≠0,Eres=0.65E1.9712.3402.0612.480Case 4m≠0,Eres=0.65E1.9112.2511.9932.380Figure12 illustrates the effect of elastic strain form on the ground response curve. From Figure 12, it can be seen that the surface displacement of surrounding rock presents the change characteristic of Case 2 solution > Case 3 solution > Case 1 solution when the support pressures are equal under the same strength criterion. For example, when σ0/p0=0, compared with Case 2, the surface displacements of surrounding rock decrease, respectively, by 26.37% for Case 1 and 7.13% for Case 3 under the MC strength criterion or by 18.84% for Case 1 and 4.76% for Case 3 under the GSMP strength criterion. From the above analysis, we can see that the support strength design should satisfy the relationship of Case 2 > Case 3 > Case 1 under the same deformation conditions. In fact, the support strength predicted by using Case 2 is relatively large, while that predicted by using Case 1 is relatively small.Figure 12
The effect of elastic strain form on the ground response curve (note: Young’s modulus attenuation is ignored in Case 3).
(a)(b)(c)(d)In addition, the change characteristic of the ground response curve is closely related to the section of strength criterion. For instance, whenσ0/p0=0, the surface displacement (u0/R0) increases, respectively, by 74.46% for the MC criterion, 14.68% for GSMP criterion, and 28.17% for UST0.5 criterion by comparison with the GMC criterion under Case 1 condition. The main reason for the above results is that the different strength criteria are different in evaluating the bearing capacity of rock mass. For example, the MC criterion does not consider the effect of intermediate principal stress and then it is easy to underestimate the bearing capacity of rock mass. Therefore, the surface displacement obtained by MC criterion is larger than that obtained by other criteria under the same support pressure.The ground and support response curves under GMC and GSMP criteria are shown in Figure13 considering different Young’s modulus attenuation models. It is shown that the damage degree of rock mass has an extremely important influence on the ground response curve. For instance, when σ0/p0=0, compared with Case 3 (Eres=E), the surface displacements of surrounding rock increase, respectively, by 7.96% for Case 3 (Eres=0.65E) and 3.59% for Case 4 under GSMP criterion. In other words, when the Young’s modulus attenuation is ignored (Case 3 (Eres=E)), the deformation of the surrounding rock and support force will be underestimated; however, when a residual value of Young’s modulus is assumed in the plastic zone (Case 3 (Eres=0.65E)), the deformation of the surrounding rock and support force will be overestimated. So, the Young’s modulus power function attenuation model (Case 4) seems to present more reasonable results and is recommended for optimizing support parameters design and evaluating stability of surrounding rock in deep underground engineering.Figure 13
The effect of rock mass damage on the ground response curve.
(a)(b)


## 6.1. A Comparison with the Traditional Solution

Park et al. summarized three different definitions for elastic strains in the plastic zone (see Cases 1∼3 in Section4.2) and analyzed the deformation law of plastic zone with Mohr-Coulomb criterion [9]. However, the Young’s modulus and Poisson’s ratio attenuation were ignored. In fact, it can be obtained by taking Eres=E, νres=ν, ξj=ξMC, and Yj=YMC in this paper. To further verify the correctness of the calculation results for deep circular opening in Section 4.2, the solution by Park et al. will be presented as a comparison with the solution proposed by this paper. The geometrical and physical parameters for circular opening are shown in Table 2.Table 2
Geometrical and physical parameters of circular opening.
ParametersRock type by Park and Kim (2006) [10]Hard rockSoft rockRadius of opening,R0 (m)11Initial stress,σ0 (MPa)11Internal pressure,pin (MPa)00Young’s modulus,E (MPa)50,0005000Poisson’s ratio,ν0.20.2Shear modulus,G (MPa)208332083c (MPa)0.1730.276φ (deg)5535cres (MPa)0.0610.055φres (deg)5230The radial displacements obtained by Park are shown in Figure7 in this paper. It can be seen that the calculation results in this paper are in accordance with Park’s closed-form solution. Therefore, the closed-form solution proposed by this paper is correct and can be regarded as an extension of Park’s solution. In other words, this paper’s solution has a more wide application in practice engineering.Figure 7
Comparison with the traditional solution. (a) Radial displacement (data: hard rock,β = 1.0). (b) Radial displacement (data: hard rock, β = 3.0). (c) Radial displacement (data: soft rock, β = 1.0). (d) Radial displacement (data: soft rock, β = 3.0).
(a)(b)(c)(d)


## 6.2. Parameters Analysis

### 6.2.1. The Effect of the Strength Yield Criteria

As previously mentioned, the strength criteria exert an extremely important effect for evaluating the mechanical response and deformation behavior of surrounding rock. For studying the influence of strength theories on the stresses and displacement of surrounding rock, the mechanical and geometrical parameters of circular tunnel are shown in Table3.Table 3
The mechanical and geometrical parameters of circular opening.
p0 (MPa)σ0 (MPa)R0 (m)Eres=E (GPa)νres=νcres=c (MPa)φres=φ (°)β250330.257.218.31.0The stresses distribution laws under different strength criteria are shown in Figure8. Meanwhile, the dimensionless radius of plastic zone and critical inner pressure at the elastic-plastic interface are presented in Table 4. It can be seen that the dimensionless values (R1/R0 and σR1/p0) show the characteristics of TR > VM > IDP > MC > MDP > UST0.5 > GSMP > GMC > GLD > GTSS > CDP. Compared with the traditional solution obtained by MC criterion, the dimensionless value (R1/R0) calculated by TR and VM criteria obviously increases by 125.26% and 78.53%, respectively. This is mainly because the TR and VM criteria only regard the rock mass as friction-less bonding material and ignore the impact of friction effect on the mechanical properties of the rock mass so that the bearing capacity of rock mass is underestimated. Therefore, the calculation result is obviously larger than the solution obtained by other criteria. The MC criterion does not take the effect of the intermediate principal stress into account and it is easy to underestimate the bearing capacity of rock mass. Therefore, the deformation of surrounding rock calculated by MC criterion may be slightly larger. The IDP and MDP criteria underestimate the influence of internal principal stresses, so the calculation results may also be larger than the solution obtained by UST0.5, GSMP, GMC, GLD, CDP, and GTSS criteria. Meanwhile, the result obtained by CDP criterion is minimal compared with other criteria’s solutions. In fact, the criterion overestimates the effect of intermediate principal stresses on rock mass strength and may not be reasonable in practical engineering. In addition, the calculation results (R1/R0) obtained by UST0.5, GSMP, GMC, and GLD criteria are close to each other within the range of 1.294–1.347. The above four criteria seem to be more reasonable considering the effect of intermediate principal stress on yield strength of rock masses.Figure 8
The stresses distributions law under different criteria.
(a)(b)(c)(d)Table 4
The radius of plastic zone and critical inner pressures under different criteria.
Strength criterionR1/R0σR1/p0Strength criterionR1/R0σR1/p0MC1.528 (0.413)VM2.728 (0.667)GMC1.318 (0.322)IDP1.556 (0.422)GLD1.294 (0.308)MDP1.456 (0.386)GSMP1.332 (0.329)CDP1.198 (0.242)TR3.442 (0.712)UST0.51.347 (0.337)GTSS1.266 (0.291)——From the above analysis, it can be seen that the UST0.5, GSMP, GMC, and GLD criteria can be strongly recommended for evaluating the mechanics and deformation behavior of surrounding rock, followed by IDP, MDP, GTSS, and MC criteria. The TR, VM, and CDP criteria are not recommended to be used for underground engineering.


### 6.2.2. The Effect of Dilation Coefficient

As shown in Figure9, the dilation coefficient has an extremely important influence on the surface displacement of surrounding rock. As the parameter β gradually increases, the dimensionless surface displacement u0E/p0R0 presents the nonlinear increase characteristics. However, the increasing rate of surface displacement under different yield criteria is significantly different. For instance, as shown in Table 5, ß increases from 1.0 to 3.0, and the dimensionless value u0E/p0R0, respectively, increases by 38.12% for MC criterion, 16.94% for GSMP criterion, 18.30% for UST0.5 criterion, and 29.58% for MDP criterion under Case 1 condition. Therefore, the effect of dilation coefficient should be considered in the design of tunnel support parameters and strength.Figure 9
The effect of dilation coefficient on the surface displacement.
(a)(b)Table 5
The surface displacement valueu0E/p0R0 of the circular tunnel.
Parameter (β)MCGSMPUST0.5MDPCase 11.01.7131.4881.5031.6262.01.9901.6011.6261.8343.02.3661.7401.7782.107Case 21.01.9451.6071.6301.8152.02.58671.8791.9242.3023.03.5672.2412.3203.011


### 6.2.3. The Effect of Elastic Strain Form and Rock Mass Damage

In this study, the geometrical and mechanical parameters of circular tunnel are shown in Table3 (soft rock). Figure 10 presents the influence of elastic strain form in the plastic zone on the surface displacement of surrounding rock. From the above analysis, the relevant conclusions can be summarized as follows:(i)
The radial displacement of plastic zone is closely related to the selection of the elastic strain form. Case 2 has the greatest effect on the radial displacement of plastic zone, followed by Case 3. Then, the results obtained by Case 2 are minimal. For example, whenr=R0 (tunnel surface), compared with Case 1, the dimensionless surface displacement u0E/p0R0, respectively, increases by 0.272 for Case 2 and 0.206 for Case 3 with an increasing rate of 16.47% for Case 2 and 12.46% for Case 3 under GLD criterion.(ii)
Meanwhile, the influence of elastic strain form on the radial displacement of plastic zone is also closely related to the strength criterion. For example, the dimensionless surface displacementsu0E/p0R0 are, respectively, 1.926 for Case 2 and 1.860 for Case 3 under GLD criterion. However, the value significantly increases by 18.32% for Case 2 and 31.71% for Case 3 under GTSS criterion.Figure 10
The effect of elastic strain form on the radial displacement.
(a)(b)When the rock mass enters the plastic zone, its mechanical and deformation behavior are closely related to the rock damage degree. Generally, Young’s modulus attenuation could be used to indicate the damage degree of rock mass. The influence of rock mass damage degree on the surface displacement of surrounding rock is shown in Figure11.(iii)
It can be seen that the radial displacement of plastic zone is closely related to the selection of Young’s modulus attenuation model. Compared with Case 3Eres=E, Case 4 has the greatest effect on the radial displacement of plastic zone. Case 3 (Eres=0.65E) is the second. For example, as shown in Table 6, compared with Case 3 (Eres=E), the dimensionless surface displacements u0E/p0R0 of Case 3 Eres=0.65E and Case 4 increase by 0.163 and 0.074 with an increasing rate of 7.49% and 3.40%, respectively, under the GTSS criterion.(iv)
In addition, the influence of Young’s modulus attenuation on the radial displacement of plastic zone is closely related to the strength criterion. For Case 3Eres=0.65E and Case 4, the dimensionless surface displacements u0E/p0R0 of surrounding rock are 1.971 and 1.911, respectively, under the GLD criterion. However, when taking the GSMP criterion, the value significantly increases by 25.82% and 24.54% compared with the GLD criterion, respectively.Figure 11
The effect of rock mass damage on the radial displacement.
(a)(b)Table 6
The dimensionless surface displacementu0E/p0R0 of the circular tunnel.
GLDGTSSGMCGSMPCase 3m=0,Eres=E1.8602.1771.9372.297m≠0,Eres=0.65E1.9712.3402.0612.480Case 4m≠0,Eres=0.65E1.9112.2511.9932.380Figure12 illustrates the effect of elastic strain form on the ground response curve. From Figure 12, it can be seen that the surface displacement of surrounding rock presents the change characteristic of Case 2 solution > Case 3 solution > Case 1 solution when the support pressures are equal under the same strength criterion. For example, when σ0/p0=0, compared with Case 2, the surface displacements of surrounding rock decrease, respectively, by 26.37% for Case 1 and 7.13% for Case 3 under the MC strength criterion or by 18.84% for Case 1 and 4.76% for Case 3 under the GSMP strength criterion. From the above analysis, we can see that the support strength design should satisfy the relationship of Case 2 > Case 3 > Case 1 under the same deformation conditions. In fact, the support strength predicted by using Case 2 is relatively large, while that predicted by using Case 1 is relatively small.Figure 12
The effect of elastic strain form on the ground response curve (note: Young’s modulus attenuation is ignored in Case 3).
(a)(b)(c)(d)In addition, the change characteristic of the ground response curve is closely related to the section of strength criterion. For instance, whenσ0/p0=0, the surface displacement (u0/R0) increases, respectively, by 74.46% for the MC criterion, 14.68% for GSMP criterion, and 28.17% for UST0.5 criterion by comparison with the GMC criterion under Case 1 condition. The main reason for the above results is that the different strength criteria are different in evaluating the bearing capacity of rock mass. For example, the MC criterion does not consider the effect of intermediate principal stress and then it is easy to underestimate the bearing capacity of rock mass. Therefore, the surface displacement obtained by MC criterion is larger than that obtained by other criteria under the same support pressure.The ground and support response curves under GMC and GSMP criteria are shown in Figure13 considering different Young’s modulus attenuation models. It is shown that the damage degree of rock mass has an extremely important influence on the ground response curve. For instance, when σ0/p0=0, compared with Case 3 (Eres=E), the surface displacements of surrounding rock increase, respectively, by 7.96% for Case 3 (Eres=0.65E) and 3.59% for Case 4 under GSMP criterion. In other words, when the Young’s modulus attenuation is ignored (Case 3 (Eres=E)), the deformation of the surrounding rock and support force will be underestimated; however, when a residual value of Young’s modulus is assumed in the plastic zone (Case 3 (Eres=0.65E)), the deformation of the surrounding rock and support force will be overestimated. So, the Young’s modulus power function attenuation model (Case 4) seems to present more reasonable results and is recommended for optimizing support parameters design and evaluating stability of surrounding rock in deep underground engineering.Figure 13
The effect of rock mass damage on the ground response curve.
(a)(b)


## 6.2.1. The Effect of the Strength Yield Criteria

As previously mentioned, the strength criteria exert an extremely important effect for evaluating the mechanical response and deformation behavior of surrounding rock. For studying the influence of strength theories on the stresses and displacement of surrounding rock, the mechanical and geometrical parameters of circular tunnel are shown in Table3.Table 3
The mechanical and geometrical parameters of circular opening.
p0 (MPa)σ0 (MPa)R0 (m)Eres=E (GPa)νres=νcres=c (MPa)φres=φ (°)β250330.257.218.31.0The stresses distribution laws under different strength criteria are shown in Figure8. Meanwhile, the dimensionless radius of plastic zone and critical inner pressure at the elastic-plastic interface are presented in Table 4. It can be seen that the dimensionless values (R1/R0 and σR1/p0) show the characteristics of TR > VM > IDP > MC > MDP > UST0.5 > GSMP > GMC > GLD > GTSS > CDP. Compared with the traditional solution obtained by MC criterion, the dimensionless value (R1/R0) calculated by TR and VM criteria obviously increases by 125.26% and 78.53%, respectively. This is mainly because the TR and VM criteria only regard the rock mass as friction-less bonding material and ignore the impact of friction effect on the mechanical properties of the rock mass so that the bearing capacity of rock mass is underestimated. Therefore, the calculation result is obviously larger than the solution obtained by other criteria. The MC criterion does not take the effect of the intermediate principal stress into account and it is easy to underestimate the bearing capacity of rock mass. Therefore, the deformation of surrounding rock calculated by MC criterion may be slightly larger. The IDP and MDP criteria underestimate the influence of internal principal stresses, so the calculation results may also be larger than the solution obtained by UST0.5, GSMP, GMC, GLD, CDP, and GTSS criteria. Meanwhile, the result obtained by CDP criterion is minimal compared with other criteria’s solutions. In fact, the criterion overestimates the effect of intermediate principal stresses on rock mass strength and may not be reasonable in practical engineering. In addition, the calculation results (R1/R0) obtained by UST0.5, GSMP, GMC, and GLD criteria are close to each other within the range of 1.294–1.347. The above four criteria seem to be more reasonable considering the effect of intermediate principal stress on yield strength of rock masses.Figure 8
The stresses distributions law under different criteria.
(a)(b)(c)(d)Table 4
The radius of plastic zone and critical inner pressures under different criteria.
Strength criterionR1/R0σR1/p0Strength criterionR1/R0σR1/p0MC1.528 (0.413)VM2.728 (0.667)GMC1.318 (0.322)IDP1.556 (0.422)GLD1.294 (0.308)MDP1.456 (0.386)GSMP1.332 (0.329)CDP1.198 (0.242)TR3.442 (0.712)UST0.51.347 (0.337)GTSS1.266 (0.291)——From the above analysis, it can be seen that the UST0.5, GSMP, GMC, and GLD criteria can be strongly recommended for evaluating the mechanics and deformation behavior of surrounding rock, followed by IDP, MDP, GTSS, and MC criteria. The TR, VM, and CDP criteria are not recommended to be used for underground engineering.


## 6.2.2. The Effect of Dilation Coefficient

As shown in Figure9, the dilation coefficient has an extremely important influence on the surface displacement of surrounding rock. As the parameter β gradually increases, the dimensionless surface displacement u0E/p0R0 presents the nonlinear increase characteristics. However, the increasing rate of surface displacement under different yield criteria is significantly different. For instance, as shown in Table 5, ß increases from 1.0 to 3.0, and the dimensionless value u0E/p0R0, respectively, increases by 38.12% for MC criterion, 16.94% for GSMP criterion, 18.30% for UST0.5 criterion, and 29.58% for MDP criterion under Case 1 condition. Therefore, the effect of dilation coefficient should be considered in the design of tunnel support parameters and strength.Figure 9
The effect of dilation coefficient on the surface displacement.
(a)(b)Table 5
The surface displacement valueu0E/p0R0 of the circular tunnel.
Parameter (β)MCGSMPUST0.5MDPCase 11.01.7131.4881.5031.6262.01.9901.6011.6261.8343.02.3661.7401.7782.107Case 21.01.9451.6071.6301.8152.02.58671.8791.9242.3023.03.5672.2412.3203.011


## 6.2.3. The Effect of Elastic Strain Form and Rock Mass Damage

In this study, the geometrical and mechanical parameters of circular tunnel are shown in Table3 (soft rock). Figure 10 presents the influence of elastic strain form in the plastic zone on the surface displacement of surrounding rock. From the above analysis, the relevant conclusions can be summarized as follows:(i)
The radial displacement of plastic zone is closely related to the selection of the elastic strain form. Case 2 has the greatest effect on the radial displacement of plastic zone, followed by Case 3. Then, the results obtained by Case 2 are minimal. For example, whenr=R0 (tunnel surface), compared with Case 1, the dimensionless surface displacement u0E/p0R0, respectively, increases by 0.272 for Case 2 and 0.206 for Case 3 with an increasing rate of 16.47% for Case 2 and 12.46% for Case 3 under GLD criterion.(ii)
Meanwhile, the influence of elastic strain form on the radial displacement of plastic zone is also closely related to the strength criterion. For example, the dimensionless surface displacementsu0E/p0R0 are, respectively, 1.926 for Case 2 and 1.860 for Case 3 under GLD criterion. However, the value significantly increases by 18.32% for Case 2 and 31.71% for Case 3 under GTSS criterion.Figure 10
The effect of elastic strain form on the radial displacement.
(a)(b)When the rock mass enters the plastic zone, its mechanical and deformation behavior are closely related to the rock damage degree. Generally, Young’s modulus attenuation could be used to indicate the damage degree of rock mass. The influence of rock mass damage degree on the surface displacement of surrounding rock is shown in Figure11.(iii)
It can be seen that the radial displacement of plastic zone is closely related to the selection of Young’s modulus attenuation model. Compared with Case 3Eres=E, Case 4 has the greatest effect on the radial displacement of plastic zone. Case 3 (Eres=0.65E) is the second. For example, as shown in Table 6, compared with Case 3 (Eres=E), the dimensionless surface displacements u0E/p0R0 of Case 3 Eres=0.65E and Case 4 increase by 0.163 and 0.074 with an increasing rate of 7.49% and 3.40%, respectively, under the GTSS criterion.(iv)
In addition, the influence of Young’s modulus attenuation on the radial displacement of plastic zone is closely related to the strength criterion. For Case 3Eres=0.65E and Case 4, the dimensionless surface displacements u0E/p0R0 of surrounding rock are 1.971 and 1.911, respectively, under the GLD criterion. However, when taking the GSMP criterion, the value significantly increases by 25.82% and 24.54% compared with the GLD criterion, respectively.Figure 11
The effect of rock mass damage on the radial displacement.
(a)(b)Table 6
The dimensionless surface displacementu0E/p0R0 of the circular tunnel.
GLDGTSSGMCGSMPCase 3m=0,Eres=E1.8602.1771.9372.297m≠0,Eres=0.65E1.9712.3402.0612.480Case 4m≠0,Eres=0.65E1.9112.2511.9932.380Figure12 illustrates the effect of elastic strain form on the ground response curve. From Figure 12, it can be seen that the surface displacement of surrounding rock presents the change characteristic of Case 2 solution > Case 3 solution > Case 1 solution when the support pressures are equal under the same strength criterion. For example, when σ0/p0=0, compared with Case 2, the surface displacements of surrounding rock decrease, respectively, by 26.37% for Case 1 and 7.13% for Case 3 under the MC strength criterion or by 18.84% for Case 1 and 4.76% for Case 3 under the GSMP strength criterion. From the above analysis, we can see that the support strength design should satisfy the relationship of Case 2 > Case 3 > Case 1 under the same deformation conditions. In fact, the support strength predicted by using Case 2 is relatively large, while that predicted by using Case 1 is relatively small.Figure 12
The effect of elastic strain form on the ground response curve (note: Young’s modulus attenuation is ignored in Case 3).
(a)(b)(c)(d)In addition, the change characteristic of the ground response curve is closely related to the section of strength criterion. For instance, whenσ0/p0=0, the surface displacement (u0/R0) increases, respectively, by 74.46% for the MC criterion, 14.68% for GSMP criterion, and 28.17% for UST0.5 criterion by comparison with the GMC criterion under Case 1 condition. The main reason for the above results is that the different strength criteria are different in evaluating the bearing capacity of rock mass. For example, the MC criterion does not consider the effect of intermediate principal stress and then it is easy to underestimate the bearing capacity of rock mass. Therefore, the surface displacement obtained by MC criterion is larger than that obtained by other criteria under the same support pressure.The ground and support response curves under GMC and GSMP criteria are shown in Figure13 considering different Young’s modulus attenuation models. It is shown that the damage degree of rock mass has an extremely important influence on the ground response curve. For instance, when σ0/p0=0, compared with Case 3 (Eres=E), the surface displacements of surrounding rock increase, respectively, by 7.96% for Case 3 (Eres=0.65E) and 3.59% for Case 4 under GSMP criterion. In other words, when the Young’s modulus attenuation is ignored (Case 3 (Eres=E)), the deformation of the surrounding rock and support force will be underestimated; however, when a residual value of Young’s modulus is assumed in the plastic zone (Case 3 (Eres=0.65E)), the deformation of the surrounding rock and support force will be overestimated. So, the Young’s modulus power function attenuation model (Case 4) seems to present more reasonable results and is recommended for optimizing support parameters design and evaluating stability of surrounding rock in deep underground engineering.Figure 13
The effect of rock mass damage on the ground response curve.
(a)(b)


## 7. Conclusions

By summarizing and simplifying different strength theories, a new unified criterion equation is firstly proposed based on the certain assumption. Then, a new unified closed-form solution for circular opening based on the newly proposed unified criterion equation is deduced with the infinite and finite boundary combining with the nonassociative flow rule. In the plastic zone, four different elastic strain assumptions are applied to solving the plastic zone deformation considering the effect of rock mass damage. Then, the validity of the solution is also verified by comparison with the traditional solution. Finally, the influences of strength criterion effect, dilation coefficient, elastic strain form, and rock mass damage on the mechanical response of surrounding rock are discussed in detail. The primary conclusions can be summarized as follows:(1)
The selection of strength criterion has an extremely important effect on the stresses, plastic zone radius, and critical inner pressure distributions of surrounding rock. For the radius of plastic zone, the calculation results obtained by TR and VM criteria which ignore the impact of friction effect are obviously the largest, followed by IDP, MC, and MDP criteria. The result obtained by CDP criterion which overestimates the effect of intermediate principal stresses on rock mass strength is minimal compared with other criteria’s solutions. In addition, the calculation results obtained by UST0.5, GSMP, GMC, and GLD criteria which reasonably consider the effect of internal principal stresses give an intermediate range. Therefore, UST0.5, GSMP, GMC, and GLD criteria can be strongly recommended for evaluating the mechanics and deformation behavior of surrounding rock, followed by IDP, MDP, GTSS, and MC criteria. TR, VM, and CDP criteria are not recommended to be used.(2)
The influence of dilation coefficient on the surface displacement of surrounding rock is closely related not only to the assumption form of elastic strain in the plastic zone but also to the strength criteria. As the dilation coefficient gradually increases, the dimensionless surface displacement presents the nonlinear increase characteristics. Therefore, the support parameters design should take the influence of the dilation coefficient into account.(3)
The elastic strain assumption forms in the plastic zone have a significant important effect on the deformation of plastic zone and ground response curve by ignoring the effect of rock mass damage. Case 1 gives a smallest deformation and seems to overestimate the plastic bearing capacity of rock mass; however, Case 2 gives a largest deformation and maybe underestimates the plastic bearing capacity of rock mass. Therefore, Case 3 may be more reasonable for evaluating stability of surrounding rock and optimizing support strength design.(4)
The effects of Young’s modulus attenuation on the deformation of the plastic zone and ground response curve are related to the selection of the Young’s modulus attenuation model and also closely related to the strength criterion. When ignoring the continuity of Young’s modulus attenuation, the deformation of surrounding rock is easy to be overestimated or underestimated. So, the Young’s modulus power function attenuation seems to give more reasonable results.



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*Source: 8872201-2020-09-30.xml*