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Theoretical solutions of a circular tunnel with the inﬂuence of axial in situ stress in elastic–brittle–plastic rock Shuilin Wang ⇑, Zhenjun Wu, Mingwei Guo, Xiurun Ge State Key Laboratory of Geomechanics and Geotechnical Engineering, Institute of Rock and Soil Mechanics, Chinese Academy of Sciences, Wuhan 430071, PR China

a r t i c l e

i n f o

Article history: Received 17 July 2011 Received in revised form 30 November 2011 Accepted 7 February 2012 Available online 14 March 2012 Keywords: Elastic–brittle–plastic model Out-of-plane stress Theoretical solution Circular tunnel Mohr–Coulomb criterion

a b s t r a c t Theoretical solutions are presented for the prediction of stresses and displacements around a circular tunnel in Mohr–Coulomb (M–C) rock mass, which is subject to a hydrostatic stress ﬁeld in the cross section of the tunnel and out-of-plane stress in the axis of it. Elastic–brittle–plastic constitutive model with a non-associated ﬂow rule is used in the analysis. After the axial in situ stress is taken into account, the outof-plane stress is not always the intermediate one in principle stress space. Therefore, solutions are different from those in the normal plane strain problem where the out-of-plane stress is supposed to be the intermediate principle stress. Moreover, different patterns of plastic zones are formed with the various combinations of hydrostatic stress in the cross section and out-of-plane stress in the axis of the tunnel. Numerical examples illustrate that the distribution of stresses and displacements in the surrounding rock mass of the tunnel is signiﬁcantly inﬂuenced by the axial stress in the case where there is a large drop in strength at yield or a big difference between the uniform in-plan stress and the out-of-plane stress. Ó 2012 Elsevier Ltd. All rights reserved.

1. Introduction It is well known that the geological activity and the gravity of the geo-material result in the pre-existing stresses (in situ rock stresses) in the ground. During underground engineering, the pre-existing stresses in the rock mass are redistributed. Therefore, the problem of stability arises in geotechnical engineering. It is well known that the surrounding rock mass will suffer from great damage if the major principle stress is high and acts in the cross section of the tunnel. In order to alleviate the failure of rock mass around the tunnel, the direction of the major principle stress usually intersects the axis of the tunnel at a small angle or they overlap each other during the process of the lay-out of the underground structure. Very often, when long excavations are carried out, the plane strain condition is assumed and the problem is simpliﬁed as twodimensional plain strain problem. Excavation of a circular tunnel in isotropic rock masses under hydrostatic stress ﬁelds is a typical one of the plane strain problems. For the axisymmetrical case of circular tunnel, many approximate or closed-form solutions have been developed. Brown et al. (1983) made an excellent review of the previous work and obtained the close-form solution for elastic–brittle–plastic Hoek–Brown medium and numerical solution for strain-softening Hoek–Brown medium. Constant elastic strains are assumed in the solutions ⇑ Corresponding author. Tel.: +86 27 87199227; fax: +86 27 87199560. E-mail address: [email protected] (S. Wang). 0886-7798/$ - see front matter Ó 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.tust.2012.02.016

(Brown et al., 1983). Later on, Ogawa and Lo (1987) derived an analytical solution and studied the effects of dilatancy and yield criteria on displacements around tunnels. Wang (1996) presented a numerical solution for the mentioned problem. Carranza-Torres and Fairhurst (1999) studied the elasto-plastic response of the excavation of the cylindrical and spherical cavities in symmetrically loaded condition. Closed-form solutions are given in a compact form where the extent of plastic zone and the stress and displacement ﬁelds are expressed in the same formulations for both circular tunnel and spherical cavity. Yu (2000) studied the expansion of the circular and spherical cavity and proposed closed-form solutions in brittle– plastic M–C and Hoek–Brown rock. Sharan (2003, 2005) improved the elastic–brittle–plastic solutions and a simple and exact solution is presented for elastic–brittle–plastic Hoek–Brown rock (Sharan, 2005). Park and Kim (2006) made a concise review on the above work and derived the analytical solutions for a circular tunnel in an elastic–brittle–plastic rock mass governed by M–C yield criterion and Hoek–Brown yield criterion and compared the displacements among different assumptions of elastic strain in the plastic zone. Solutions presented by Park and Kim (2006) are theoretically consistent with some others. Alonso et al. (2003), Park et al. (2008), Lee and Pietruszczak (2008) and Wang et al. (2010) presented different numerical solutions for the excavation of a circular tunnel in strain-softening M–C and Hoek–Brown rock masses and obtained ground response curves. Their results are in good agreement with one another. So far, most of the literatures assume that the out-of-plane stress is the intermediate one in principle stress space. The

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Nomenclature A = Yr/(ar 1) constant B = pin + Yr/(ar 1) constant c cohesion cr residual cohesion E elastic modulus G shear modulus q axial in situ stress r0 radius of the circular tunnel r radial coordinate in the cylindrical polar coordinate system R radius of the interface of the inner plastic zone and the outer plastic zone R plastic radius pci the critical internal pressure. i = 1 means that the outof-plane in situ stress is the major principle stress; i = 2 means that the out-of-plane in situ stress is the intermediate principle stress; i = 3 means that the outof-plane in situ stress is the minor principle stress pin the internal pressure uci the displacement at the interface of the plastic zone and the elastic zone. i = 1 means that the out-of-plane in situ stress is the major principle stress; i = 2 means that the out-of-plane in situ stress is the intermediate principle stress; i = 3 means that the out-of-plane in situ stress is the minor principle stress. uci = R(r0 pci)/(2G) the displacement at the interface of the inner plastic uR zone and the outer plastic zone u radial displacement Y = 2c cos u/(1 sin u) constant Yr = 2cr cos ur/(1 sin ur) constant

inﬂuence of the out-of-plane stress on displacement and stress distribution around the tunnel is ignored when M–C or Hoek–Brown criterion is employed. Reed (1988) did not think that the axial stress remained the intermediate one as the pressure at the tunnel wall is progressively reduced. In the analysis, the in situ stress ﬁeld of equal magnitude in all directions is taken into account. Closedform solution for M–C rock with brittle–plastic model is presented. The axial stress and the tangential stresses become equal at certain stage while the tunnel support pressure is gradually reduced. The effect of the out-of-plane stress on the cavity wall displacement is studied. Lu et al. (2010) analyze the circular tunnel with different magnitude of axial in situ stress. Stresses and strains are obtained when M–C criterion with elasto-plastic model is used. Numerical results show that the stresses, strains and the plastic zone in the surrounding rock are much inﬂuenced by the out-of-plane stress. The objective of this paper is to develop theoretically consistent solutions for the prediction of stresses and displacements around a circular tunnel in M–C rock mass, which is subject to a uniform stress ﬁeld in the cross section of the tunnel and out-of-plane stress in the axis of it. Elastic–brittle–plastic constitutive model with a non-associated ﬂow rule is used in the derivation of the formulations. After the axial in situ stress is taken into account, the solutions are different from those in the normal plane strain problem. Different patterns of plastic zones are formed with the various combinations of hydrostatic stress in the cross section and out-of-plane stress in the axis of the tunnel. In addition, the formulations evolve into elasto-plastic solutions by setting the residual strength parameters equal to the peak ones. Numerical examples illustrate that the distribution of stresses and displacements in the surrounding rock mass of the tunnel is greatly inﬂuenced by the axial stress in the case where there is a large drop in strength at yield or a big difference between the uniform in-plan stress and the out-of-plane stress.

a = (1 + sin u)/(1 sin u) constant ar = (1 + sin ur)/(1 sin ur) constant b = (1 + sin w)/(1 sin w) constant er the radial strain eer the elastic part of radial strain epr the plastic part of radial strain eh the tangential strain eeh the elastic part of tangential strain eph the plastic part of tangential strain ez the axial strain eez the elastic part of axial strain epz the plastic part of axial strain u friction angle ur residual friction angle g parameter to indicate plastic strains m Poisson’s ratio h

r0 r1 r2 r3 rr rh rz

rr ðRÞ w

tangential coordinate in the cylindrical polar coordinate system uniform in-plane in situ stress major principle stress intermediate principle stress minor principle stress radial stress tangential stress axial stress the radial stress at the interface of the inner plastic zone and the outer plastic zone the dilation angle of the rock mass

2. Description of the problem 2.1. Statement of the problem A circular tunnel with radius r0 is excavated in a homogeneous inﬁnite isotropic rock mass subjected to a hydrostatic stress (r0) in the cross section of the tunnel and out-of-plane stress (q) in the axis of it as shown in Fig. 1a. An internal pressure pin uniformly acts on the tunnel wall surface after the excavation. The stress components at a point in the cylindrical polar coordinate system are shown in Fig. 1b. The problem illustrated in Fig. 1a can be considered to be axisymmetric, and displacement and stresses are only the functions of radius r when gravity is ignored. The internal pressure (pin) is gradually decreased as the excavation is modeled. Initially, the medium is in elastic state, and the Lame’s solution (Reed, 1988) for the elastic region is:

rr ¼ r0 ðr0 pin Þðr0 =rÞ2 rh ¼ r0 þ ðr0 pin Þðr0 =rÞ2 rz ¼ q u¼

1 ðr0 pin Þr 20 =r 2G

ð1Þ

ð2Þ

where rz, rh and rr are shown in Fig. 1a. As the internal pressure (pin) is decreasing, yielding will occur in the rock mass after the elastic limit is reached. Then the stress state is governed by yield criterion. Here M–C yield criterion is used to govern the stresses in the rock when it is in plastic state. M–C yield criterion is written in principle stress space as

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

157

Fig. 1. (a) The axi-symmetrical problem of excavation of a cylindrical tunnel in geo-material subjected to a uniform stress (r0) in the cross section of the tunnel and out-ofplane stress (q) in the axis of it, (b) the stress components at a point in the surrounding rock mass in the cylindrical polar coordinate system.

Fðr1 ; r3 ; gÞ ¼ r1 a r3 Y ¼ 0

ð3Þ

Compression stress and the inward radial displacement are taken to be positive. In this problem, rz, rh and rr are the three principle stresses. Plastic zone will appear if Eq. (3) is met in principle stress space. According to Eq. (1), yielding occurs at the tunnel wall ﬁrstly when the internal pressure decreases to a critical magnitude. 2.2. Critical internal pressure of the tunnel The critical internal pressure pc is related with the order of the magnitude of rz, rh and rr. Eq. (1) shows that rh is always greater than rr if the internal pressure pin 2 (0, r0). At the tunnel wall (r = r0), rr = pin, rh = 2r0 pin, and rz = q in the elastic region. When the out-of-plane in situ stress is the major principle stress at the tunnel wall, there will be rz > rh > rr. By means of Eq. (3), the critical internal pressure is

pc1 ¼ ðq YÞ=a

ð4aÞ

This means that the surrounding rock will begin to enter plastic states if the out-of-plane stress is the major principle stress and internal pressure pin 6 pc1. If the out-of-plane stress is the intermediate principle stress, i.e. rh > rz > rr, we can get the critical internal pressure (pc2) in the same way.

pc2 ¼ ð2r0 YÞ=ð1 þ aÞ

ð4bÞ

At r = r0, when pin = pc2, rr = pin = (2r0 Y)/(1 + a) and rh = 2r0 pin = (2ar0 + Y)/(1 + a). Denote

q1 ¼ ð2r0 YÞ=ð1 þ aÞ

ð5Þ

and

q2 ¼ ð2ar0 þ YÞ=ð1 þ aÞ:

ð6Þ

From Eq. (1), we know that the out-of-plane stress is the intermediate principle stress when q2 > q > q1. In the case where the out-of-plane stress is the minor principle stress, the critical internal pressure is

pc3 ¼ 2r0 aq Y

ð4cÞ

zone, which depends on the combination of r0 and q in the surrounding rock mass of the tunnel. If the out-of-plane stress is the major principle stress and internal pressure pin < pc1. The plastic state occurs in the rock as shown in Fig. 2a. While the internal pressure pin continues to decrease, rz will decrease and rh increase. At certain stage, rh = rz at the tunnel wall. Thereafter, the plastic zone will be divided into two parts which are illustrated in Fig. 2b. In this case, the plastic zone shown in Fig. 2 is called type I. If the out-of-plane stress is the intermediate principle stress and internal pressure pin < pc2. The plastic state occurs in the rock as shown in Fig. 3a. As the internal pressure pin continues to decrease, rh will decrease and rz may increase. In the inner part of the plastic zone, rh = rz occurs. The plastic zone is divided into two parts which are illustrated in Fig. 3b. This kind of plastic zone shown in Fig. 3 is called type II. If the out-of-plane stress is the minor principle stress and internal pressure pin < pc3. The plastic state occurs in the rock as shown in Fig. 4a. If the internal pressure pin continues to decrease, rr will decrease. At certain moment, rr = rz. Thereafter, the plastic zone will be divided into two parts which are illustrated in Fig. 4b. In this case, the plastic zone shown in Fig. 4 is called type III. 4. Theoretical solutions Whether the medium is in elastic state or in plastic state, the equilibrium equation should be met for the tangential stress and radial stress in the cross section of the tunnel.

drr rr rh þ ¼0 dr r

ð7Þ

Eq. (7) can be rewritten as

rh ¼ r

drr þ rr dr

ð7aÞ

and this equation will be used to derive the tangential stress in terms of radial stress.

3. Types of plastic zone around the tunnel

4.1. Solution for the plastic zone with type I (rz – the major principle stress)

When the out-of-plane stress is considered in the plane strain problem as shown in Fig. 1, there will be several types of plastic

In this situation, the axial stress is the major principle stress, and the radial stress is the minor principle stress in the elastic

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Fig. 2. Evolution of plastic type I: (a) plastic zone with rz > rh > rr; (b) inner plastic zone with rz = rh > rr and outer plastic zone with rz > rh > rr.

Fig. 3. Evolution of plastic type II: (a) plastic zone with rh > rz > rr; (b) inner plastic zone with rh = rz > rr and outer plastic zone with rh > rz > rr.

Fig. 4. Evolution of plastic type III: (a) plastic zone with rh > rr > rz; (b) inner plastic zone with rh > rr = rz and outer plastic zone with rh > rr > rz.

er ¼ du=dr eh ¼ u=r

regime, and the radial stress is decreasing as the internal pressure is descending. When the internal pressure pin < pc1, plastic zone appears in the surrounding rock mass of the tunnel. The yield criterion govern the plastic region can be written as

Thereby, we can have

rz ¼ ar rr þ Y r

deh eh er þ ¼0 dr r

ð8Þ

With Eqs. (7) and (8), we have two equations but three unknown variables (i.e. rz, rh and rr). Another equation is needed. From the condition of compatibility under the inﬁnitesimal deformation assumption, strains and displacement have the following relationships:

ð9Þ

ð10Þ

If the plastic potential whose formulation is similar to the yield criterion is given by:

Gðrz ; rr ; gÞ ¼ rz b rr the plastic parts of radial and axial strain are related by

ð11Þ

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

epr þ bepz ¼ 0

ð12Þ

and the plastic part of tangential strain: p h

e ¼0

ð13Þ

The radial, tangential and axial strains can be decomposed into elastic and plastic components:

er ¼ eer þ epr eh ¼ eeh þ eph ez ¼ eez þ epz With

ð14Þ ð15Þ ð16Þ

p h

e ¼ 0 and ez = 0 considered, we can have:

e r e h

er ¼ e þ beez eh ¼ e

ð17Þ ð18Þ

The elastic strains should obey Hooke’s law, which is consistent with theory of solid mechanics. The elastic strains resulted from excavation can be expressed as:

1 E 1 ¼ ½rh mðrz þ rr Þ ðr0 mr0 mqÞ E 1 ¼ ½rz mðrr þ rh Þ ðq 2mr0 Þ E

eez

ð19Þ

By substituting Eq. (19) into Eqs. (17) and (18), and then into Eq. (10), we have:

drh drr drz r m m ¼ ð1 þ m þ mbÞrh þ ð1 þ m mbÞrr dr dr dr þ brz bðq 2mr0 Þ

ð20Þ

Eqs. (7), (8) and (20) will be used to get rz, rh and rr. Substitution of Eqs. (7a) and (8) into Eq. (20) produces a second-order differential equation for the radial stress which takes the following form. 2

d

rr

ð21Þ

where f1 = 3 + m(b ar), f2 = b(ar 2m) and f3 = b(Yr q + 2mr0). According to the theory of differential equation, the general solution for Eq. (21) is:

rr ¼ Frn1 þ Grn2 þ H 1f1 þ

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 2

ð1f1 Þ 4f 2 , 2

pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ 2

1f1

ð1f1 Þ 4f 2 2

ð22Þ ff32 .

with n1 ¼ n2 ¼ and H ¼ F and G are the unknown variables and can be determined by boundary conditions. In order to get F and G, we consider

u 1 ¼ eh ¼ eeh ¼ ½rh mðrz þ rr Þ ðr0 mr0 mqÞ r E

ð23Þ

From Eqs. (7a) and (8), Eq. (23) can be written as:

u 1 drr ¼ ðr þ f 4 rr þ f 5 Þ r E dr

ð24Þ

with f4 = 1 m(1 + ar), f5 = mq mYr (1 m)r0 By substituting Eq. (22) into Eq. (24), the radial displacement can be written as

r u ¼ ½F ðn1 þ f4 Þr n1 þ G ðn2 þ f4 Þr n2 þ f4 H þ f5 E

ð25Þ

At r = R, rr = pc1 and u = uc1, where uc1 ¼ RE ð1 þ mÞðr0 pc1 Þ, and at r = r0, rr = pin. Theses boundary conditions result in:

FRn1 þ GRn2 þ H ¼ pc1

ð26aÞ

F ðn1 þ f4 ÞRn1 þ G ðn2 þ f4 ÞRn2 þ f4 H þ f5 ¼ ð1 þ mÞðr0 pc1 Þ

ð26bÞ

ð27aÞ

F ðn1 þ f4 ÞRn1 þ G ðn2 þ f4 ÞRn2 þ f4 H þ f5 ¼ ð1 þ mÞðr0 pc1 Þ

ð27bÞ

FRn1 þ GRn2 þ H ¼ rr ðRÞ

ð27cÞ

Now the unknown variables F; G; R and R are needed to obtain, but the above three equations are not enough to solve for them. At the interface ðr ¼ RÞ of the inner plastic zone and outer plastic zone, rh = rz. From Eqs. (7a) and (8), we have

rr þ r

drr ¼ ar rr þ Y r dr

ð28Þ

Substitution of Eq. (22) into Eq. (28) gives

Fð1 þ n1 ar ÞRn1 þ Gð1 þ n2 ar ÞRn2 ¼ ðar 1ÞH þ Y r

drr r þ f1 r þ f 2 rr þ f 3 ¼ 0 2 dr dr 2

ð26cÞ

The above three equations form a group of non-linear equations by which F, G and R can be easily solved through iterative methods such as Newton method. So far, the outer radius R of the plastic zone is determined. With the solved variables F and G, the radial stress rr and displacement u can be obtained by Eqs. (22) and (25), respectively. rh can be obtained by substituting rr into Eq. (7a) and rz can be obtained by substituting rr into Eq. (8). As it is mentioned in the above section, rz will decrease and rh increase when the pressure pin decreases further. Once rh = rz occurs, the plastic zone is divided into two parts. One is the inner plastic part with rz = rh > rr and the other is the outer plastic part with rz > rh > rr. In outer plastic zone with rz > rh > rr, the radial stress and displacement are in the same form as those in Eqs. (22) and (25). The boundary conditions are rr = pc1 and u = uc1 at r = R and rr ¼ rr ðRÞ at r ¼ R. Then, we have:

FRn1 þ GRn2 þ H ¼ pc1

eer ¼ ½rr mðrh þ rz Þ ðr0 mr0 mqÞ eeh

Frn01 þ Grn02 þ H ¼ pin

159

ð29Þ

Eqs. (27a), (27b), (27c) and (29) form a group of non-linear equations which will be used to solve for F; G; R and R. It should be noted that rr ðRÞ, the radial stress at the interface of the inner plastic zone and the outer plastic zone, is unknown at present. But it is the function of R, and it is obtained when the inner plastic zone is discussed. Iterative method is used for solving the equations and the variables can be easily obtained. So far, the outer radius R and the interface radius R of the two plastic zones have been determined. With the variables F and G known, rr can be easily obtained by Eq. (22) and u obtained by Eq. (25). Then rh can be gotten by Eq. (7a) and rz gotten by Eq. (8). In inner plastic zone, once rh = rz, the yield criteria governing the plastic region are:

rz ¼ ar rr þ Y r rh ¼ ar rr þ Y r

ð30aÞ ð30bÞ

Differentiate the above two equations, we have

drz ¼ ar drr drh ¼ ar drr

ð31aÞ ð31bÞ

From Eqs. (31a) and (31b), we know that the increments of tangential stress and axial stress are equal, i.e. drh = drz. This means that rh + drh = rz + drz once rh = rz. Eqs. (30a)–(31b) indicate rh = rz while the pressure pin continues decreasing. In the inner plastic zone ðr 0 < r < RÞ, we have rz = rh > rr. Eqs. (7) and (8) are used to get rh and rr. With the boundary condition rr = pin at r = r0, the closed form solution (Park and Kim, 2006) for the stresses in the inner plastic region is:

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rr ¼ A þ Bðr=r0 Þðar 1Þ rh ¼ A þ ar Bðr=r0 Þðar 1Þ

ð32aÞ ð32bÞ

and

rz ¼ rh

ð32cÞ

Because rz = rh in the inner plastic zone, the stresses lie on the intersection of the two yield surfaces (Eqs. (30a) and (30b)), which is one of the singularities of yield surface generated by M–C criterion in three-dimensional principal stress space. The correct ﬂow rule is obtained by summing the contributions from the two plastic potentials (Reed, 1988)

Eqs. (7) and (40) are used to get rh and rr. With the boundary condition rr = pin at r = r0, the closed form solutions for the stresses in the plastic region are given by:

rr ¼ A þ Bðr=r0 Þðar 1Þ rh ¼ A þ ar Bðr=r0 Þðar 1Þ

ð41aÞ ð41bÞ

and

rz ¼ mðrh þ rr Þ 2mr0 þ q

ð41cÞ

If the plastic potential

Gðrh ; rr ; gÞ ¼ rh b rr

ð42Þ

G1 ðrz ; rr ; gÞ ¼ rz b rr

ð33aÞ

is used, the plastic parts of radial and tangential strain are related by

G2 ðrh ; rr ; gÞ ¼ rh b rr

ð33bÞ

epr þ beph ¼ 0

The increments relationship:

of

plastic

strains

have

the

following

depr þ bdeph þ bdepz ¼ 0

ð34Þ

At the early stage of plastic deformation, the plastic strains satisfy Eqs. (12) and (13). Integration of the increments of plastic strains yields p h

p r

p z

e þ be þ be ¼ 0

ð35Þ

By using Eqs. (14), (15), (16) and (9), the differential equation for displacement in the inner plastic zone is given by

du u þ b ¼ f ðrÞ dr r

ð36Þ

where

f ðrÞ ¼ eer þ beeh þ beez

ð37Þ

By means of Eqs. (19), (32a), (32b) and (32c), f(r) can be written as

" ar 1 # 1 r T1A þ T2 þ T1B f ðrÞ ¼ E r0

ð38Þ

T 2 ¼ 2ðb bm mÞY r ð1 þ b m 3bmÞr0 ðb bm mÞq Integration of Eq. (36) gives the solution for u:

" # T 1 A þ T 2 bþ1 T1B bþ1 bþar bþar R Þ þ ðr R Þ ðr bþ1 ðb þ ar Þr a0r 1 Er b !b R þ uR r 1

ð39Þ ðar 1Þ

By means of Eq. (32a), rr ¼ A þ BðR=r0 Þ at r ¼ R. In Eq. (27c), rr ðRÞ is replaced by rr ðRÞ ¼ A þ BðR=r0 Þðar 1Þ when F; G; R and R are being solved by Eqs. (27a), (27b), (27c) and (29) in the outer plastic zone. After u is obtained by Eq. (25) in the outer plastic zone, uR , the displacement at r ¼ R, becomes a known variable. The displacement in the inner plastic zone is determined by substituted uR into Eq. (39). 4.2. Solution for the plastic zone with type II (rz – the intermediate principle stress) In this case, the M–C criterion govern the plastic region can be written as

rh ¼ ar rr þ Y r

and the plastic part of axial strain:

epz ¼ 0

ð44Þ

By using Eqs. (14), (15) and (9), the differential equation for displacement in the outer plastic zone is expressed in Eq. (36) with

f ðrÞ ¼ eer þ beeh

ð45Þ

By means of Eqs. (19), (41a) and (41b), integration of Eq. (36) and complex manipulation produce

u¼

1 1 ½U 1 f1 ðrÞ þ U 2 f2 ðrÞ U 1 f1 ðRÞ U 2 f2 ðRÞ þ 2GuR Rb 2G r bþ1

ð46Þ

with U1 = (1 + b)(1 2m)(A r0),U2 = [(1 m b m) + (b bm m)ar] B,

f1 ðrÞ ¼

rbþ1 bþ1

and f 2 ðrÞ ¼

1 ar 1

r0

rbþar b þ ar

At the interface (r = R) of the plastic and elastic zones, rr = pc2. By means of Eq. (41a),

R ¼ r0

1 pc2 A ar 1 B

ð47Þ

Eqs. (41a), (41b), (41c), (46) and (47) are the solution which is the same as those in Park and Kim (2006). During the process of relaxation of the internal pressure, the plastic zone may be divided into two parts as shown in Fig. 3b. In the outer plastic zone ðR < r < RÞ, the solution for stresses and displacement is the same as those formulated in Eqs. (41a), (41b), (41c) and (46). At the interface ðr ¼ RÞ between the outer and inner plastic zones, rh = rz. By means of Eqs. (41b) and (41c), the interface radius ðRÞ of the outer and inner plastic zones are determined by

with T1 = 1 + 2(arb bm arm arbm),

u¼

ð43Þ

ð40Þ

R ¼ r0

a 11 U4 r U3

ð48Þ

where U4 = 2m(r0 A) + A q and U3 = (m + m ar ar)B. In the inner plastic zone ðr0 < r < RÞ, the solution for radial and tangential stresses is the same as in Eqs. (41a) and (41b), but rz = rh. The solution for displacement is the same as in Eq. (39) in Section 4.1. uR in Eq. (39) is calculated by Eq. (46). 4.3. Solution for the plastic zone with type III (rz – the minor principle stress) In this case, the M–C criterion govern the plastic region can be written as

rh ¼ ar rz þ Y r

ð49Þ

Eqs. (7) and (49) provide two equations for solving rz, rh and rr. So another equation is needed.

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

The plastic potential corresponding to the yield criterion (Eq. (49)) is

Gðrh ; rz ; gÞ ¼ rh b rz

ð50Þ

The plastic parts of tangential and axial strain are related by

epz þ beph ¼ 0

ð51Þ

and the plastic part of radial strain: p r

e ¼0

ð52Þ

Considering Eqs. (14)–(16), we can have:

er ¼ eer eh

ð53Þ

1 ¼ eeh þ eez b

ð54Þ

By substituting Eq. (19) into Eqs. (53) and (54), and then into Eq. (10), we have

r

m drh 1 drr 1 drz m 1 m 1þ þ b dr b b dr dr m m 1 1 ¼ 1 m rh þ 1 þ m þ rr rz þ ðq 2mr0 Þ b b b b

ð55Þ

Eqs. (7), (49) and (55) will be used to obtain rz, rh and rr. Substitution of Eqs. (7a) and (49) into Eq. (55) produces a second-order differential equation for the radial stress. 2

r2

d

rr

dr

2

þ g1 r

drr þ g 2 rr þ g 3 ¼ 0 dr

ð56Þ

3 m 2m 1 m g 1 ¼ ð3 þ 4 Þ T; g 2 ¼ 2 T; ar b b ar ar b b mr0 Y r q 1 m m T and T ¼ 1 þ g3 ¼ 2 b ar b b ar b b ar

n2 ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð1 g 1 Þ2 4g 2

n1 ¼ 2 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 1 g 1 ð1 g 1 Þ 4g 2

and

2

u 1 ¼ eeh þ eez r b

ð57Þ

By means of Eqs. (19), (7a) and (49), Eq. (57) can be written as:

u T drr ¼ r þ g 4 rr þ g 5 r E dr

ð58Þ

with g 4 ¼ 1 þ a1r b m 2 bm amr =T,

1 1 m 2m Yr 1 m m q r0 =T b ar b ar b

By substituting Eq. (22) into Eq. (58), the radial displacement can be written as

u¼

¼ ð1 þ mÞðr0 pc3 Þ Frn01 þ Grn02 þ H ¼ pin

T r ½F ðn1 þ g 4 Þr n1 þ G ðn2 þ g 4 Þr n2 þ g 4 H þ g 5 E

ð60bÞ ð60cÞ

The above three equations form a group of non-linear equations by which F, G and R can be readily solved by iterative methods. So far, the outer radius R of the plastic zone is determined. With the solved variables F and G, the radial stress rr and displacement u can be obtained by Eqs. (22) and (59), respectively. rh can be obtained by substituting rr into Eq. (7a) and rz can be obtained by substituting rh into Eq. (49). While the internal pressure continues to decrease, there will be two plastic zones. One is the inner plastic part with rh > rr = rz and the other is the outer plastic zone with rh > rr > rz. In outer plastic zone with rh > rr > rz, the radial stress and displacement are in the same form as those in Eqs. (22) and (59). The boundary conditions are rr = pc3 and u = uc3 at r = R and rr ¼ rr ðRÞ at r ¼ R. Then,

FRn1 þ GRn2 þ H ¼ pc3

ð61aÞ

T ½F ðn1 þ g 4 ÞRn1 þ G ðn2 þ g 4 ÞRn2 þ g 4 H þ g 5 ¼ ð1 þ mÞðr0 pc3 Þ

ð61bÞ

FRn1 þ GRn2 þ H ¼ rr ðRÞ

ð61cÞ

drr r þ rr Y r ar dr 1

ð59Þ

At r = R, rr = pc3 and u = uc3, where uc3 ¼ RE ð1 þ mÞðr0 pc3 Þ, and at r = r0, rr = pin. Theses boundary conditions result in:

ð62Þ

Substitution of Eq. (22) into Eq. (62) produces

Fð1 þ n1 ar ÞRn1 þ Gð1 þ n2 ar ÞRn2 ¼ ðar 1ÞH þ Y r

In Eq. (22), F and G are the unknown variables and can be determined by introducing boundary conditions. From Eqs. (9) and (54), we have

g5 ¼

T ½F ðn1 þ g 4 ÞRn1 þ G ðn2 þ g 4 ÞRn2 þ g 4 H þ g 5

rr ¼

The general solution for Eq. (56) can be written in the same form as in Eq. (22) with

1 g1 þ

ð60aÞ

Another equation is needed for solving the four unknown variables F; G; R and R. At the interface ðr ¼ RÞ of the inner plastic zone and outer plastic zone, rr = rz. From Eqs. (7a) and (49), we have

where

g H ¼ 3; g2

FRn1 þ GRn2 þ H ¼ pc3

161

ð63Þ

Eqs. (61a), (61b), (61c) and (63) will be used to solve for F; G; R and R. In Eq. (61c), rr ðRÞ, the radial stress at the interface of the inner plastic zone and the outer plastic zone, is unknown at present. But it is the function of R, and it is obtained when the inner plastic zone is discussed. Thereby the outer radius R and the interface radius R of the plastic zone are determined, and rr can be easily obtained by Eq. (22) and u obtained by Eq. (59). Then rh can be deduced by Eq. (7a) and rz deduced by Eq. (49). In the inner plastic zone ðr0 < r < RÞ; rh > rr ¼ rz . The yield criteria governing the plastic region are:

rh ¼ ar rz þ Y r rh ¼ ar rr þ Y r

ð64aÞ ð64bÞ

The radial and tangential stresses are expressed in Eqs. (32a) and (32b), and rz = rr. In this case, the stresses lie on the singularities of yield surfaces (Eqs. (64a) and (64b)) generated by M–C criterion in three-dimensional principal stress space. The ﬂow rule is obtained by summing the contributions from the two plastic potentials

G1 ðrh ; rz ; gÞ ¼ rh b rz

ð65aÞ

G2 ðrh ; rr ; gÞ ¼ rh b rr

ð65bÞ

The increments of plastic strains have the following relationship

bdeph þ depr þ depz ¼ 0

ð66Þ

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At the early stage of plastic deformation, the plastic strains satisfy Eqs. (51) and (52), integration of the increments of plastic strains gives p h

p r

p z

be þ e þ e ¼ 0

ð67Þ

By using Eqs. (14), (15), (16) and (9), the differential equation for displacement in the inner plastic zone is written in the same form as in Eq. (36) with

f ðrÞ ¼ beeh þ eer þ beez

ð68Þ

With Eqs. (19), (32a), (32b) and rz = rr considered, f(r) can be written as

f ðrÞ ¼

" ar 1 # 1 r V 1A þ V 2 þ V 1B E r0

Rock parameter

Value

Radius of tunnel, r0 (m) Initial in-plane stress, r0 (MPa) Initial out-of-plane stress, q (MPa) Internal pressure, pin (MPa) Young’s modulus, E (GPa) Poisson’s ratio, m c (MPa) cr (MPa) u (deg) ur (deg) w (deg)

3 30 5–60 0 27.0 0.22 1.5 0.7 50.9 39 7.5

ð69Þ

with V1 = 2 2m + arb 2bm 2arm and

V 2 ¼ ðb 2mÞY r ð1 þ b 3m bmÞr0 ð1 bm mÞq: The solution for u is

u¼

Table 1 Material parameters used in the study.

" # V 1 A þ V 2 bþ1 V 1B bþ1 bþar bþar R Þ þ ðr R Þ ðr bþ1 ðb þ ar Þr0ar 1 Er b !b R þ uR r 1

ð70Þ ðar 1Þ

rr ðRÞ in Eq. (61c) is replaced by rr ðRÞ ¼ A þ BðR=r0 Þ when F; G; R and R are being solved by Eqs. (61a), (61b), (61c) and (63) in the outer plastic zone. uR , which is obtained in the outer plastic zone, is substituted into Eq. (70) to get the displacement in the inner plastic zone. 5. Numerical examples To illustrate some of the characteristics of the solutions, the derived formulations in the last section are programmed into a computer code, and a set of hard rock parameters is used to study the effect of axial in situ stress on displacements and plastic zone in the surrounding rock mass. Brittle–plastic and elasto-plastic models are considered in the following analysis. Meanwhile, different dilatant angles are taken into account. 5.1. Numerical results for elastic–brittle–plastic constitutive model The rock properties used in the analysis are summarized in Table 1. Several different values of the axial in situ stress (q) will be chosen to show the effect of q on displacements and plastic zone in the surrounding rock mass. When parameters presented in Table 1 are used, we can obtain q1 (=5.773 MPa) and q2 (=54.227 MPa) from Eqs. (5) and (6). So the out-of-plane stress is the intermediate principle stress when q2 > q > q1. The problem is the normal plane strain one when q = 2m r0 (i.e. q = 13.2 MPa). Normalized plastic radius and displacement at the tunnel wall versus the different axial in situ stress are presented in Table 2. In the case where q2 > q > q1, the plastic radius is not dependent on the out-of-plane stress, but the displacement increase as the out-of-plane stress increase. When q = 13.2 MPa, the corresponding displacement and plastic radius are used to be reference values. Compared with the reference displacement (corresponding to q = 13.2 MPa), the difference of the displacements increases as the out-of-plane stress is ascending. For example, the relative error of the displacements, which correspond to q = 54.22 MPa and q = 13.2 MPa, reaches 19.86%.

If q > q2, the out-of-plane stress is the major principle stress. In this situation, both the plastic radius and the displacement increase while out-of-plane stress (q) is increasing. The relative error of the displacements and plastic radius is increasing, as well. In the case where q < q1, both the plastic radius and the displacement also increase while out-of-plane stress (q) is decreasing. This condition is rare in geotechnical engineering. Figs. 5–10 show the effect of q on the distribution of the stresses and displacement in the plastic zones. The support pressure is completely released. Fig. 5 shows that only one plastic zone with rh > rz > rr is formed in the normal plane strain problem (q = 13.2 MPa). In Fig. 6, only one plastic zone with rz = rh > rr is formed because of the stress drop when elastic–brittle–plastic constitutive model is taken into account. In Fig. 7, only one plastic zone with rh = rz > rr is formed because rh is almost equal to rz when yield appears in the surrounding rock mass. In Fig. 9, only one plastic zone with rh > rr = rz is formed because rr is almost equal to rz when yield appears in the surrounding rock mass. Both the inner plastic zone with rh = rz > rr and the outer plastic zone with rh > rz > rr are shaped as shown in Fig. 8 when q = 30.0 MPa. At q = 5.0 MPa, rz is the minor principle stress when yield appears in the surrounding rock mass. After the support pressure is completely released, the plastic zone is divided into the inner one with rh > rr = rz and the outer one with rh > rr > rz. Fig. 10 shows the plastic zone and the distribution of stresses and displacement in the rock mass. In the above analysis, the radial stress is continuous. However, it is observed that the axial stress and tangential stress jump at the interface of plastic and elastic zones in Figs. 5–10 where the elastic–brittle–plastic constitutive model is employed. 5.2. Numerical results for elasto-plastic constitutive model By setting the residual cohesion and friction angle equal to the corresponding peak ones presented in Table 1, elastic–plastic

Table 2 Normalized plastic radius and displacement at the tunnel wall by elastic–brittle– plastic constitutive model. q (MPa)

R/r0 (relative error)a

2uG/(r0r0) (relative error)

60.00 54.22 30.00 13.20 5.77 5.00

1.88 1.82 1.82 1.82 1.82 2.22

5.60 5.19 4.52 4.33 4.22 4.80

(3.30%) (0.00%) (0.00%) (0.00%) (21.98%)

(29.33%) (19.86%) (4.39%) (2.54%) (10.85%)

a The normalized plastic radius and displacement corresponding to q = 13.2 MPa are used to be reference values. Relative errors are obtained by comparing the reference value with those corresponding to other out-of-plane stresses. For example, the relative error of the normalized displacements, which correspond to q = 54.22 MPa and q = 13.2 MPa, is calculated as follows: j5:194:33j 100% ¼ 19:86%. 4:33

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

163

Fig. 5. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 13.2 MPa and pin = 0.0 MPa.

Fig. 6. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 60.0 MPa and pin = 0.0 MPa.

Fig. 7. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 54.22 MPa and pin = 0.0 MPa.

solution can be easily obtained. The normalized plastic radius and displacement versus the different axial in situ stress are presented in Table 3. Similar results are observed.

When q2 > q > q1, the plastic radius is independent on the outof-plane stress. In the case where q > q2 or q < q1, the plastic radius increases. Compared with the displacement in normal plane strain

164

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

Fig. 8. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 30.0 MPa and pin = 0.0 MPa.

Fig. 9. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 5.77 MPa and pin = 0.0 MPa.

Fig. 10. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–brittle–plastic model when q = 5.0 MPa and pin = 0.0 MPa.

condition (q = 13.2 MPa), the difference of the displacements increases as the out-of-plane stress is ascending.

Figs. 11–16 illustrate the effect of q on the distribution of the stresses and displacement in the plastic zones when elasto-plastic

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168 Table 3 Normalized plastic radius and displacement at the tunnel wall by elasto-plastic constitutive model. q (MPa)

R/r0 (relative error)

2uG/(r0r0) (relative error)

60.00 54.22 30.00 13.20 5.77 5.00

1.31 1.29 1.29 1.29 1.29 1.49

1.82 1.78 1.64 1.62 1.61 1.61

(1.55%) (0.00%) (0.00%) (0.00%) (15.50%)

(12.35%) (9.88%) (1.23%) (0.62%) (0.62%)

constitutive model is considered. Only one plastic zone with rh > rz > rr is formed in the normal plane strain condition as shown in Fig. 11. When q = 60.0 MPa, Fig. 12 shows that the inner plastic zone with rz = rh > rr and the outer plastic zone with rz > rh > rr are formed. The pattern of the plastic zone is different from that shown in Fig. 6 where the axial in situ stress is also 60.0 MPa. In Fig. 13, only one plastic zone with rh = rz > rr is formed because rh is almost equal to rz when yield appears in the surrounding rock mass. In Fig. 15, only one plastic zone with rh > rr = rz is formed because rr is almost equal to rz when yield appears in the surrounding rock mass. Two plastic zones with rh = rz > rr in the inner one and rh > rz > rr in the outer one are

165

shaped as shown in Fig. 14. When q = 5.0 MPa, rz is the minor principle stress when yield appears in the surrounding rock mass. Fig. 16 shows that the plastic zone is divided into the inner one with rh > rr = rz and the outer one with rh > rr > rz. It is observed that the radial, axial and tangential stresses are all continuous at the interface of plastic and elastic zones in Figs. 11– 16 where elasto-plastic constitutive model is used.

5.3. Discussions of the numerical results In Sections 5.1 and 5.2, the plastic radius and the distribution of the stresses and displacement in the plastic zones are presented when elastic–brittle–plastic and elasto-plastic constitutive model are taken into account in the analysis. Numerical results indicate that the axial in situ stress affect the displacement and the plastic radius in the surrounding rock mass of the tunnel for both models. If the relative errors are compared in Tables 2 and 3, the results are much more inﬂuenced by the axial in situ stress when the elastic– brittle–plastic constitutive model is used. In Table 1, a small dilatant angle (w = 7.5°) is considered. In order to understand the inﬂuence of the axial in situ stress on displacement when the dilatant angle is big, the same parameters are employed except for dilatant angle (w = 19.5°). Numerical results are presented in Table 4 where elastic–brittle–plastic and

Fig. 11. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 13.2 MPa and pin = 0.0 MPa.

Fig. 12. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 60.0 MPa and pin = 0.0 MPa.

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Fig. 13. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 54.22 MPa and pin = 0.0 MPa.

Fig. 14. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 30.0 MPa and pin = 0.0 MPa.

Fig. 15. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 5.77 MPa and pin = 0.0 MPa.

elasto-plastic constitutive models are used. Plastic radius does not change because the plastic zone is not dependent on dilatant angle. If corresponding displacement and the relative error are compared

in Tables 2–4, we can say that the bigger the dilatant angle is, the greater the difference of displacement is for both elastic– brittle–plastic model and elasto-plastic model.

S. Wang et al. / Tunnelling and Underground Space Technology 30 (2012) 155–168

167

Fig. 16. (a) Normalized radial displacement, (b) normalized radial, tangential and axial stress in the surrounding rock mass in elastic–plastic model when q = 5.0 MPa and pin = 0.0 MPa.

Table 4 Normalized displacement at the tunnel wall by elastic–brittle–plastic and elastoplastic constitutive models with w = 19.5°.

a b

q (MPa)

2uG/(r0r0)a (relative error)

2uG/(r0r0)b (relative error)

60.00 54.22 30.00 13.20 5.77 5.00

9.27 8.38 6.90 6.53 6.45 7.66

2.24 2.15 1.91 1.88 1.87 1.86

(41.93%) (28.19%) (5.65%) (1.33%) (17.25%)

(2)

(19.19%) (14.66%) (1.65%) (0.59%) (0.64%)

Obtained by elastic–brittle–plastic constitutive models. Obtained by elastio-plastic constitutive models.

When the axial in situ stress q 2 [q1, q2], the displacement will increase as q increases, but the plastic radius is not dependent on q. When q > q2, both the displacement and the plastic radius are much affected by the axial in situ stress. When q < q1, both the displacement and the plastic radius are inﬂuenced by the axial in situ stress, as well. However, this case is rare in geotechnical engineering. Generally speaking, the inﬂuence of the out-of-plane stress should be evaluated when the plane strain condition is assumed in geotechnical engineering. In the case where the problem is simpliﬁed to be two-dimensional one, the inﬂuence of the out-ofplane stress cannot be ignored if the difference of the out-of-plane stress and the in-plane stress is big and brittle–plastic characteristics of the rock mass is observed. For example, in a part of the drainage tunnel in Jinping hydropower station located in Sichuan province of China, the cross section of the tunnel is circular and the surrounding rock is marble with good quality. The three principle stresses are 64 MPa, 36 MPa and 36 Mpa (Liu et al., 2011). This case is similar to the problem discussed in the paper.

6. Concluding remarks (1) In this paper, the excavation of a circular tunnel is studied. The surrounding rock mass, obeying M–C criterion, is treated as an inﬁnite isotropic medium subject to a uniform stress ﬁeld in the cross section of the tunnel and out-of-plane stress in the axis of it. Theoretical solutions are presented for displacement, stresses and plastic radius with the different axial in situ stress taken into account. Elastic– brittle–plastic constitutive model with a non-associated ﬂow rule is used in the analysis. The elasto-plastic solutions

(3)

(4)

(5)

(6)

are easily derived by setting the residual strength parameters equal to the peak ones. When the axial stress is the intermediate stress, i.e. q2 > q > q1, the closed-form solutions are obtained, and the plastic zone is not dependent on the axial stress. When the axial stress is the major principle stress (q > q2) or the minor principle stress (q < q1), iterative methods have to be used for obtaining the solution. Different types of plastic zone will be formed in the surrounding rock mass of the tunnel according to the magnitude of the axial in situ stress. When the elasto-plastic constitutive model is considered, the results show that the displacement will increase as the axial stress is ascending in the case where axial stress is the intermediate stress (q2 > q > q1). Both the plastic zone and the deformation are affected by the axial stress in the case where the axial stress is the major principle stress or the minor principle stress. The bigger the difference of q and r0 is, the greater the inﬂuences are. Compared with the results obtained by elasto-plastic model, the displacement, stresses and the plastic radius are much affected by the axial stress when elastic–brittle–plastic constitutive model is used. It can be inferred that the inﬂuence will be increased when the difference of peak and residual strength parameters increases. When geo-material exhibits signiﬁcant dilatancy in post failure stage, the inﬂuence of the axial in situ stress on the tunnel deformation will increase. In geotechnical engineering, the direction of the major principle stress usually intersects the axis of the tunnel at small angle to alleviate the failure of rock mass around the tunnel. The inﬂuence of the axial stress should be included in the plane strain analysis when the difference of the out-of-plane stress and the in-plane stress is big. Meanwhile, the out-ofplane stress should be taken into account if the strength drops signiﬁcantly at yielding stage.

Acknowledgements The authors thank State Key Laboratory of Geomechanics and Geotechnical Engineering (SKLQ005, SKLZ0802) and National Natural Science Foundation of China (Grant Nos. 51179185 and 41130742) for supporting the present work. Thanks are also given to the editor and the two reviewers for their valuable comments.

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