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energies

Article Application of the Mohr-Coulomb Criterion for Rocks with Multiple Joint Sets Using Fast Lagrangian Analysis of Continua 2D (FLAC2D) Software

Lifu Chang 1,2,* ID and Heinz Konietzky 1

1 Institut für Geotechnik, Technische Universität Bergakademie Freiberg, Freiberg 09599, Germany; [email protected] 2 School of & Civil Engineering, China University of Mining & Technology, Beijing 100083, China * Correspondence: [email protected]; Tel.: +86-189-1102-2674

Received: 6 February 2018; Accepted: 6 March 2018; Published: 9 March 2018

Abstract: The anisotropic behavior of a mass with persistent and planar joint sets is mainly governed by the geometrical and mechanical characteristics of the joints. The aim of the study is to develop a continuum-based approach for simulation of multi jointed geomaterials. Within the continuum methods, the discontinuities are regarded as smeared cracks in an implicit manner and all the joint parameters are incorporated into the equivalent constitutive equations. A new equivalent continuum model, called multi-joint model, is developed for jointed rock masses which may contain up to three arbitrary persistent joint sets. The Mohr-Coulomb yield criterion is used to check failure of the intact rock and the joints. The proposed model has solved the issue of multiple surfaces involved in this approach combined with multiple failure mechanisms. The multi-joint model was implemented into Fast Lagrangian Analysis of Continua software (FLAC) and is verified against the strength anisotropy behavior of jointed rock. A case study considering a circular excavation under uniform and non-uniform in-situ stresses is used to illustrate the practical application. The multi-joint model is compared with the ubiquitous joint model.

Keywords: jointed rock masses; multi-joint constitutive model; strength anisotropy; tunnel excavation; numerical simulation

1. Introduction The behavior of a jointed rock (within this paper this term includes also jointed rock masses) is anisotropic, non-linear and path dependent due to the presence of discontinuities. Compared to intact rock, jointed rock generally exhibit reduced stiffness, higher permeability and lower strength. Various analytical, numerical and empirical methods are suggested in order to take into account the influence of discontinuities on the mechanical behavior of jointed rock [1–6]. In the past several decades, a lot of lab tests with samples containing a single or multiple joint sets under confined and unconfined conditions have been implemented to investigate the joint orientation effect [7–9]. With the ongoing development of computer based simulation techniques, more sophisticated models have been developed in recent years and it is possible to incorporate more aspects into the simulations [10–13]. Two numerical techniques are common in rock mechanics: continuum-based methods and discontinuum based methods. Discontinuum methods consider the rock mass as an assemblage of rigid or deformable blocks connected along discontinuities [14], and investigate the microscopic mechanisms in granular material and crack development in rocks [3,15]. Boon et al. [16] performed 2D discrete element analyses to identify potential failure mechanisms around a circular unsupported tunnel in a rock mass intersected by three independent joint sets. In continuum methods, the discontinuities are considered as smeared planes of weakness (joints) incorporated in each zone. The equivalent

Energies 2018, 11, 614; doi:10.3390/en11030614 www.mdpi.com/journal/energies Energies 2018, 11, x FOR PEER REVIEW 2 of 23 tunnelEnergies in2018 a, 11rock, 614 mass intersected by three independent joint sets. In continuum methods,2 ofthe 23 discontinuities are considered as smeared planes of weakness (joints) incorporated in each zone. The equivalentcontinuum continuum method contains method three contains parts: three the equivalent parts: the continuumequivalent compliancecontinuum partcompliance (incremental part (incrementalelastic law), theelastic yield law), criterion the yield and criterion the an correctionsd the plastic part. corrections part. WillWill [ [1717]] implemented implemented a a multi multi-surface-surface plasticity plasticity model model into into a a finite finite element element code code with with internal internal iterationsiterations in in case case that that several several plasticity plasticity surfaces surfaces are are touched touched at at the the same same time. time. Zhang Zhang [ [1818]] developed developed furtherfurther the the Goodman’s Goodman’s joint joint element element model model and and derived derived the the equivalent equivalent method method for for rock rock masses masses with with multimulti-set-set joints joints in in a a global global coordinate coordinate system. system. Rihai Rihai [19 [19,20,20] ]proposed proposed a a method method which which takes takes into into accountaccount the the influence influence of of micro micro moments moments on on the the behavior behavior of of the the equivalent equivalent continuum continuum based based on on the the CosseratCosserat theory. HurleyHurley etet al.al. [ 21[21]] presented presented a numericala numerical method method which which implemented implemented the the method method into intoGEODYN-L GEODYN code.-L code. This This method method uses uses a series a series of stress of stress and strain-rateand strain- updaterate update algorithms algorithms to rule tojoints rule jointsclosure, closure, slip, and slip, stress and distributionstress distribution inside the inside computational the computational cells which cells contain which multiple contain embedded multiple embeddedjoints. A model joints. with A model multi-sets with ofmulti ubiquitous-sets of jointsubiquitous with anjoints extended with an Barton extended empirical Barton formula empirical was formulaproposed was by proposed Wang [22] by and Wang pre- and[22] post-peakand pre- and post-peak characteristics deformation characteristics of the rock mass of werethe rock also massconsidered. were also Many considered. models haveMany been models formulated have been to simulateformulated the to effect simulate of joint the angle effect on of the joint strength angle onand the deformability strength and of deformability rock masses, butof rock only masses, a few consider but only several a few jointconsider sets several having differentjoint sets strengthhaving differentparameters strength and their parameters effects on and failure their characteristics.effects on failure characteristics. ThisThis paper paper presents presents a a new new equivalent equivalent continuum continuum constitutive constitutive model model named named multi multi-joint-joint model model toto simulate simulate the the behavior behavior of ofjointed jointed rock. rock. This This model model is suited is suited for jointed for jointed rock containing rock containing up to three up to arbitrarythree arbitrary joint sets joint with sets corresponding with corresponding spatial spatialdistribution. distribution. The equivalent The equivalent compliance compliance matrix matrixof the rockof the mass rock is mass establis is establishedhed and the and Mohr the Mohr-Coulomb-Coulomb yield yield criterion criterion is used is used to tocheck check the the failure failure characteristicscharacteristics of thethe intactintact rock rock and and the the joints. joints. A circular A circular tunnel tunnel subjected subjected to uniform to uniform and non-uniform and non- uniformin-situ stresses in-situ isstresses used to is illustrateused to illustrate the application the application of this model. of this model.

2.2. Methodo Methodologylogy for for Equivalent Equivalent Continuum Continuum Multi Multi-Joint-Joint Model Model

2.1.2.1. Ubiquitous Ubiquitous Joint Joint and and Related Related Approaches Approaches TheThe term term ‘ubiquitous ‘ubiquitous joint’ joint’ used used by by Goodman Goodman [ [2323]] implies implies that that the the joint joint sets sets may may occur occur at at any any pointpoint in in the the rock rock mass mass without without a a fixed fixed location. location. The The joints joints are are embedded embedded i inn a a Mohr Mohr-Coulomb-Coulomb solid (Figure(Figure 11aa).).

FigureFigure 1. 1. SchematicSchematic for for isotropic isotropic and and anisotropic anisotropic models: models: (a) ubiquitous (a) ubiquitous joint model, joint model, (b) multi (b )joint multi model, joint (model,c) anisotropic (c) anisotropic model with model transversely with transversely isotropic isotropicrock matrix rock and matrix representative and representative joint orientation joint orientation..

Failure may occur in either the rock matrix or along the joints, or both, according to the stress Failure may occur in either the rock matrix or along the joints, or both, according to the stress state, the orientation of the joints and the parameters of matrix and joints [24]. More advanced models state, the orientation of the joints and the parameters of matrix and joints [24]. More advanced models based on ubiquitous joint or strain-harding/softening ubiquitous joint (subiquitous) were developed based on ubiquitous joint or strain-harding/softening ubiquitous joint (subiquitous) were developed in recent years [13,22,25–27] as shown in Figure 1b,c. These approaches consider aspects like stiffness

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Energies 2018, 11, x FOR PEER REVIEW 3 of 23 in recent years [13,22,25–27] as shown in Figure1b,c. These approaches consider aspects like stiffness anisotropy in addition to the strength anisotropy, joint spacing, strain softening or bi bi-linear-linear strength envelopes. The The proposed proposed multi multi-joint-joint model model is is characterized by by different strength parameters and considers overall stiffness reduction due to the joints likelike shownshown inin FigureFigure2 2. .

Figure 2. JointedJointed rock rock mass mass containing containing three random joint sets sets..

2.2. Elastic Matrix of the Equivalent Continuum Model The elastic stiffness properties of the joints are denoted by the joint normal and shear stiffness parameters kn and ks, respectively. The spatial characteristics of the joint sets are represented by joint parameters kn and ks, respectively. The spatial characteristics of the joint sets are represented by orientationjoint orientation and andspacing. spacing. The Themulti multi-joint-joint model model incorporates incorporates the the joint joint parameters parameters by by an an overall downgrading of the rock matrix parameters, so that in terms of stiffness still isotropic behavior is assumed. The corresponding strain strain-stress-stress relationship is given by the following expression:  1 ν ν     1 − − ν −− ν 00 00 0 0  ε1  E Eeq E Eeq EEeq  σ1  ε    eq− ν eq1 − νeq 0 0 0   σ   2  ν Eeq Eeq Eνeq   2   ε    − ν −1 ν 1   σ  ε13  − E E − E 00 00 0 0 σ3 1   =  eq eq eq    (1)  ε4 Eeq Eeq Eeq 1  σ4  ε    0 0 0 G 0 0   σ   2       2  ε5  ν 0ν 01 0 0 1 0  σ5  ε  − − 0 G 0 0  σ  ε36  1  σ6 3   = Eeq 0Eeq 0E 0eq 0 0 G   (1) ε   σ  4   4   1  In this matrix, Eeq isε the equivalent0 deformation0 0 modulus of0 the0 rockσ mass, G is 5  G  5 and ν is Poison’s ratio of the intact rock. If n is the number of joint sets, Si is the number of joints per 1  1  ε 6  σ 6  meter for joint set i, Ki is the stiffness 0 of joint0 set i and0 EM means0 Young’s0  modulus of the intact rock matrix, then the equivalent deformation modulus is obtained: G  1   0 0 n 0 0 0   1 Si 1 G  = ∑ + (2) Eeq i=1 Ki EM In this matrix, Eeq is the equivalent deformation modulus of the rock mass, G is shear modulus and νConsequently, is Poison’s ratio the of corresponding the intact rock. Young’s If n is modulusthe number of the of joint rock masssets, S isi is related the number to the jointof joints spacing per 1and meter the for joint joint stiffness. set i, K Springsi is the stiffness in series of representing joint set i and matrix EM means and joint Young’s stiffness modulus will lead of the to intact significant rock matrix,reduction then in the overall equivalent stiffness deformation as shown in modulus Figure3. is obtained: n 1 Si 1 = ∑ + (2) Eeq i=1 Ki EM

Consequently, the corresponding Young’s modulus of the rock mass is related to the joint spacing and the joint stiffness. Springs in series representing matrix and joint stiffness will lead to significant reduction in overall stiffness as shown in Figure 3.

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FigureFigure 3. 3.Downgraded Downgraded stiffnessstiffness for for the the equivalent equivalent continua continua..

2.3. Failure Criterion of Multi-Joint Model 2.3. Failure Criterion of Multi-Joint Model Figure 3. Downgraded stiffness for the equivalent continua. Plasticity leads to irreversible deformations after reaching the yield condition. For the multi-joint Plasticitymodel2.3. Failure the leads returnCriterion to mappingirreversible of MultFigurei -procedureJoint 3. Downgraded deformations Model is used. stiffness afterFigure for reaching the 4 elucidatesequivalent the yield continuapossible condition.. geometric For conditions the multi-joint modelarising the return from the mapping intersection procedure of two isyield used. sur Figurefaces (F41 elucidatesand F2) representing possible geometrictensile (F1) conditionsand shear (F arising2) 2.3. FailurePlasticity Criterion leads ofto Mult irreversiblei-Joint Model deformations after reaching the yield condition. For the multi-joint fromfailure. the intersection Several different of two situations yield surfaces have (toF1 beand consideredF2) representing [28]: (1) tensileIf only (oneF1) yield and shearcondition (F2) failure.is model the return mapping procedure is used. Figure 4 elucidates possible geometric conditions Severalviolated differentPlasticity (region situations leads s4 and to sirreversible5 have), plasticity to be deformations correction considered is performed [after28]: reaching (1) If by only thethe one correspondingyield yield condition. condition potential For the is violated multi functions.-joint (region arising from the intersection of two yield surfaces (F1 and F2) representing tensile (F1) and shear (F2) (2)model If more the thanreturn one mapping yield condition procedure is violated, is used. the Figure situation 4 elucidates can be categorized possible geometric as follows: conditions s4 andfailure.s5), plasticity Several correctiondifferent situations is performed have to by be the considered corresponding [28]: (1) potential If only functions.one yield condition (2) If more is than arising fromtrial the intersectionα of two yield surfaces (F1 and F2) representing tensile (F1) and shear (F2) one yield(violateda) If condition F (region (σ,εpl) iss>4 and0 violated, and s5 ),λ plasticityi + 1the > 0, situationfor correction both αcan = 1is and beperformed categorized 2 (region by s the12) asboth corresponding follows: surfaces are potential active [19 functions.]. failure. Severaltrial different situationsα have to be considered [28]: (1) If only one yield condition is (b(2)) IftrialIf more F (thanσ,εpl) one> 0 andyield λα conditioni + 1 < 0, for is α violated, = 1 or 2 (region the situation s1 or s 2can) only be onecategorized of the surfaces as follows: are active. (a) violatedIf F ( (regionσ,εpl) > s4 0 and and s5λ), plasticityi + 1 > 0, correction for both α is= performed 1 and 2 (region by the correspondings12) both surfaces potential are functions. active [19 ]. trial where(2)(a) Iftrial Ifmore FFtrial isthan(σ the,εpl )one failure > 0 yieldand criterion λ αconditionαi + 1 > 0, for for isthe bothviolated, yield α = surface 1 the and situation 2 and (region λ is can thes12 )be bothplastic categorized surfaces multiplier areas follows: active[29,30 ][. 19 ]. (b) If F (σ,εpl) > 0 and λ i + 1 < 0, for α = 1 or 2 (region s1 or s2) only one of the surfaces are active. (b) If Ftrial (σ,εpl) > 0 and λαi + 1 < 0, for α = 1 or 2 (region s1 or s2) only one of the surfaces are active. trial α (a) trialIf F (σ,εpl) > 0 and λ i + 1 > 0, for both α = 1 and 2 (region s12) both surfaces are active [19]. where F istrial the failure criterion for the and λ is the plastic multiplier [29,30]. (bwhere) If FFtrial (isσ ,theεpl) >failure 0 and criterion λαi + 1 < 0, for for the α = yield 1 or 2 surface (region and s1 or λ sis2) theonly plastic one of multiplier the surfaces [29 ,are30] .active.

where Ftrial is the failure criterion for the yield surface and λ is the plastic multiplier [29,30].

Figure 4. Determination of active surfaces and definition the multi-surface regions.

Jaeger [2] introduced the plane of weakness model which focus on the shear failure of anisotropicFigure rocks.Figure 4. WithinDetermination 4. Determination the multi of- jointof active active model surfaces surfaces Mohr’s and stress definition definition circle the thecan multi multi-surfacebe used-surface to interpretregions regions.. the jointed rock strength for different constellations of joint orientation and direction of loading. In case of uniaxial Jaeger [Figure2] introduced 4. Determination the plane of active of surfacesweakness and model definition which the multi focus-surface on the regions shear. failure of compression with only one single joint set, Mohr’s stress circle representation is shown in Figure 5. Jaegeranisotropic [2] introduced rocks. Within the the plane multi of-joint weakness model Mohr’s model stress which circle focus can on be theused shear to interpret failure the of jointed anisotropic Jaeger [2] introduced the plane of weakness model which focus on the shear failure of rocks.rock Within strength the for multi-joint different constellations model Mohr’s of joint stress orientation circle and can direction be used of toloading. interpret In case the of jointeduniaxial rock anisotropic rocks. Within the multi-joint model Mohr’s stress circle can be used to interpret the jointed strengthcompression for different with only constellations one single joint of jointset, Mohr’s orientation stress circle and representation direction of is loading. shown in In Figure case 5. of uniaxial rock strength for different constellations of joint orientation and direction of loading. In case of uniaxial compressioncompression with with only only one one single single jointjoint set, set, Mohr’s Mohr’s stress stress circle circle representation representation is shown is shownin Figure in 5.Figure 5.

Figure 5. Mohr-Coulomb failure criterion with tension cut-off for matrix and joint.

Figure 5. Mohr-Coulomb failure criterion with tension cut-off for matrix and joint.

Figure 5. Mohr-Coulomb failure criterion with tension cut-off for matrix and joint. Figure 5. Mohr-Coulomb failure criterion with tension cut-off for matrix and joint.

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Energies 2018, 11, 614 5 of 23 Point A indicates the critical failure stage for a joint. Joint angle αj identifies the corresponding failure plane: Point A indicates the critical failure stage for a joint.φ Joint angle αj identifies the corresponding failure plane:  j (3) α j = 45 + φj α = 45◦ + 2 (3) j 2 where ϕj is the joint angle, αj is the critical joint angle measured from horizontal direction. where φ is the joint friction angle, α is the critical joint angle measured from horizontal direction. fs,0 jand ft,0 are the shear and tensionj failure criteria for the rock matrix. fs,1 and ft,1 are the shear fs,0 and ft,0 are the shear and tension failure criteria for the rock matrix. fs,1 and ft,1 are the shear and tension failure criteria of the joint set. The criterion for this rock mass can be and tension failure criteria of the joint set. The shear strength criterion for this rock mass can be expressed as: expressed as: 1 1 1 s,0 = σ − σ φ + σ + σ + σ − σ φ φ − f ( 11 3 )cos ( 1 ( 1 3 ) 1 ( 1 3 )sin ) tan c (4) f s,02= (σ − σ ) cos φ + (2(σ + σ ) + (2σ − σ ) sin φ) tan φ − c (4) 2 1 3 2 1 3 2 1 3 1 1 1 s,1 i i f = (σ11 − σ 3 ) cos 2α j + ( 1 (σ1 + σ 3 ) +1 (σ1 − σ 3 ) cos 2α j ) tanφ j − c j (5) f s,12= (σ − σ ) cos 2α + (2 (σ + σ ) + 2(σ − σ ) cos 2α ) tan φi − ci (5) 2 1 3 j 2 1 3 2 1 3 j j j where c and ϕ are cohesion and internal friction angle for the matrix. cj and ϕj are joint cohesion and where c and φ are cohesion and internal friction angle for the matrix. cj and φj are joint cohesion and jointjoint frictionfriction angle.angle. WithWith increasingincreasing verticalvertical stress,stress, thethe intersectionintersection areaarea betweenbetween Mohr’sMohr’s stressstress circlecircle andand jointjoint shearshear failurefailure envelopenvelop growsgrows asas illustratedillustrated in in Figure Figure6 .6.

FigureFigure 6.6. Failure criterion criterion for for intact intact rock rock and and joint joint set set in incombination combination with with different different stress stress states: states: (a) (Positiona) Position B and B and C, C,(b ()b Position) Position D D and and E, E, (c (c) )Position Position F F and and G, ((d)) illustrationillustration ofof correspondingcorresponding orientationsorientations ofof weakweak planes.planes.

The general expressions for the maximum and minimum failure angles are: The general expressions for the maximum and minimum failure angles are: (ci cotφ i −σ )(1− sinφ) 2α = φ i + sin −1((1+ ij i j 1 )sinφ i ) (6) min j (cj cot φj − σ1)(1 − sin φ) j 2α = φi + sin−1((1 + −σ sinφ + ccosφ ) sin φi) (6) min j −σ sin φ + c cos φ j i 2αmax = π + 2φ j − 2αmin (7) i 2αmax = π + 2φj − 2αmin (7) Point A indicates the orientation of the plane with minimum strength. Jaeger’s complete curve, whichPoint shows A indicates the relationship the orientation between of the the joint plane angle with and minimum uniaxial strength. compressive Jaeger’s strength complete is shown curve, in whichFigure shows7. If a jointed the relationship rock has more between than the one joint sets angleof joints, and the uniaxial potential compressive failure modes strength which is shownhave to inbe Figureconsidered7. If aincrease. jointed The rock different has more joints than can one have sets either of joints, identical the strength potential parameters failure modes but different which have to be considered increase. The different joints can have either identical strength parameters but

Energies 20182018,, 1111,, x 614 FOR PEER REVIEW 6 of 23 joint angles, or they have also different strength parameters. Eight typical joint parameters different joint angles, or they have also different strength parameters. Eight typical joint parameters constellations are shown in Table 1 and in Figure 8. constellations are shown in Table1 and in Figure8.

(a)

(b)

Figure 7. Schematic illustration for rock mass strength anisotropy: (a) Jaeger’s curve, (b) dominant Figure 7. Schematic illustration for rock mass strength anisotropy: (a) Jaeger’s curve, (b) dominant failure area for weak plane and for rock matrix. failure area for weak plane and for rock matrix.

Table 1. Selected possible joint strength parameter combinations for two joints (MC model). Table 1. Selected possible joint strength parameter combinations for two joints (MC model).

Constellation Joint Cohesion (cj) Joint Friction Angle (ϕj) Joint Tension (σtj) t Constellation Joint Cohesion1 (2cj) Joint Friction1 2 Angle (φj)t,1 Jointt,2 Tension (σ j) 1 c j = c j φ j = φ j σ j = σ j 1 = 2 1 = 2 t,1 t,2 1 cj c1j 2 1 φj 2 φj t,1 t,2 σj = σj 2 > φ = φ σ > σ 1 >c j2 c j j 1 =j 2 j j t,1 t,2 2 cj cj φj φj σj > σj c1 > c12 2 1 φ1 >2 φ2 t,1 t,2 t,1 > t,2 3 3 j c jj > c j φ j >jφ j j σ j > σ j σj σj 1 > 2 1 > 2 t,1 t,2 4 cj c1j 2 1 φj 2 φj t,1 t,2 σj < σj 4 > φ > φ σ < σ 1 =c j2 c j j 1 >j 2 j j t,1 t,2 5 cj cj φj φj σj < σj c1 > 1c2 2 1 φ1 <2 φ2 t,1 t,2 t,1 > t,2 6 5 j c j j = c j φ j >jφ j j σ j < σ j σj σj 1 = 2 1 < 2 t,1 t,2 7 cj c1j 2 1 φj 2 φj t,1 t,2 σj > σj 6 > φ < φ σ > σ 1 j 2 j j t,1 t,2 8 cj cj φj φj σj > σj 1 2 1 2 t,1 t,2 7 c j = c j φ j < φ j σ j > σ j 1 2 1 2 t,1 t,2 8 c j < c j φ j > φ j σ j > σ j

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FigureFigure 8. 8. SketchSketch of of potential potential failure envelops forfor twotwo jointsjoints setssetscorresponding corresponding to to the the constellations constellations 1–8 1in–8 Table in Table1. 1.

There are two constellations for a rock mass which has two joints with different strength There are two constellations for a rock mass which has two joints with different strength parameters: (I) one joint is weaker in all strength parameters (series 2–4 in Table 1) and (II) strength parameters: (I) one joint is weaker in all strength parameters (series 2–4 in Table1) and (II) strength parameter have different relations to each other (series 5–8). parameter have different relations to each other (series 5–8). In order to determine the failure region of the two joints, several Mohr’s circles and the In order to determine the failure region of the two joints, several Mohr’s circles and the corresponding schemes are drawn. Exemplarily, Figure 9 illustrates that joint set 1 has lower strength corresponding schemes are drawn. Exemplarily, Figure9 illustrates that joint set 1 has lower strength values compared to joint set 2. In Figure 9a, the point A represents the least failure angle for joint set 2. At this stage for joint set 1, there is a region from B to C that fails. When the stress increases towards the critical state in the rock matrix, the angle of the failure range for each joint is different (Figure 9b). If the jointed rock only has joint set 1, the FOG area in

Energies 2018, 11, 614 8 of 23 values compared to joint set 2. In Figure9a, the point A represents the least failure angle for joint set 2. At this stage for joint set 1, there is a region from B to C that fails. EnergiesWhen 2018, 11 the, x FOR stress PEER increases REVIEW towards the critical state in the rock matrix, the angle of the failure8 of 23 range for each joint is different (Figure9b). If the jointed rock only has joint set 1, the FOG area in Figure9 c9c is is the the joint joint set set failure failure range. range. If the If rock the massrock hasmass joint has set joint 2 only, set the 2 only, gray areathe DOEgray representsarea DOE reprethe failuresents area.the failure As can area. be seen As can in Figure be seen9c, in there Figure is a 9 significantc, there is differencea significant between difference the areabetween of DOE the areaand FOG.of DOE If both and joint FOG. sets If areboth in joint zone BOC,sets are failure in zone may BOC, easily failure occurs may on joint easily 1. Compared occurs on withjoint the 1. Comparedblack and grey with regions, the black the and joint grey set regions, 2 might the also joint reach set 2 the migh criticalt also state reach when the critical joint set state 1 is when located joint in setregion 1 is FOBlocated or COG.in region FOB or COG.

(a)

(b)

(c)

Figure 9. Failure areas for weak planes and rock matrix. (a) Increasing vertical stress reached a critical Figure 9. Failure areas for weak planes and rock matrix. (a) Increasing vertical stress reached a critical state for joint set 2; (b) Critical state for rock matrix and failure areas for weak planes; (c) Failure areas state for joint set 2; (b) Critical state for rock matrix and failure areas for weak planes; (c) Failure areas for weak planes and for rock matrix. for weak planes and for rock matrix.

Two types of joint failure criteria for a rock mass are discussed here. For constellations 1–4 in Table 1, a new function h is introduced into the σ-τ-plane in order to solve the multi-surface plasticity problem. This function h is represented by the diagonal between the representation of fs,1 = 0 and ft,1 = 0. According to Figure 10a, the failure areas are divided into four sections. If shear failure is reached on the plane in Sections 1 and 2, the stress point will brought back to the curve fs,1 = 0. If the

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stress state belongs to the Section 3 or 4, local tensile failure takes place and the stress point will brought back to ft,1 = 0. EnergiesFor2018 the, 11 constellations, 614 5–8 in Table 1, two new functions h1 and h2 were introduced in the9 ofτ-σ 23- plane in order to solve the multi-surface plasticity problem. Function h1 represents the diagonal between fs,1 = 0 and ft = 0. Function h2 represents the diagonal between fs,1 = 0 and fs,2 = 0. The stress stateTwo which types violates of joint the failure joint failure criteria criterion for a rock will mass be arelocated discussed in sections here. For1 to constellations 8 corresponding 1–4 into Tablepositive1, a or new negative function domainh is introduced of ft = 0, fs,1 into = 0, thefs,2 =σ 0,-τ -planeh1 = 0 and in order h2 = 0. to According solve the multi-surfaceto Figure 10c, plasticityif for the s,1 problem.second joint This set function shear failureh is represented is detected byon the diagonalplane in section betweens 1 theand representation 2, the stress will of fbe brought= 0 and t,1 fback= 0to. the According curve fs,2 to = 0. Figure If for 10thea, first the joint failure set areas shear arefailure divided is detected into four in section sections.s 3, If4 and shear 5, the failure stress is s,1 reachedwill be brought on the plane back into Sectionsthe curve 1 andfs,1 = 2, 0. the If the stress stress point state will belongs brought to back section to the 6, curve7 or 8,f local= 0. tensile If the stressfailure state takes belongs place and to thethe Sectionstress will 3 or be 4, brought local tensile back to failure ft = 0. takesft = 0 placeis the andminimum the stress tension point failure will t,1 broughtcriterion backfor the to ftwo= joints 0. inside a rock mass.

(a)

(b)

(c)

Figure 10. Two kinds of joint failure criteria for a jointed rock mass. (a) Joint failure criterion for Figure 10. Two kinds of joint failure criteria for a jointed rock mass. (a) Joint failure criterion for constellations 1–4 (see Table 1); (b) Multi-stage failure criteria for joints having different parameters; constellations 1–4 (see Table1); ( b) Multi-stage failure criteria for joints having different parameters; (c) Eight sections for solving the multi-surface plasticity problem. (c) Eight sections for solving the multi-surface plasticity problem.

2.4. Numerical Implementation For the constellations 5–8 in Table1, two new functions h1 and h2 were introduced in the τ-σ-plane in orderThe toabove solve-mentioned the multi-surface Mohr-Coulomb plasticity yield problem. criteria Function for the multi h1 represents-joint model the diagonalis programmed between as s,1 t s,1 s,2 fa user= 0 definedand f = model 0. Function and implementedh2 represents into the the diagonal 2-dimensional between explicitf = 0 andfinitef difference= 0. The code stress FLAC state which[31]. The violates joints the are joint embedded failurecriterion in a Mohr will-Coulomb be located matrix. in sections The failure 1 to 8 corresponding criterion of the to intact positive rock or t s,1 s,2 negative domain of f = 0, f = 0, f = 0, h1 = 0 and h2 = 0. According to Figure 10c, if for the second joint set shear failure is detected on the plane in sections 1 and 2, the stress will be brought back to the curve fs,2 = 0. If for the first joint set shear failure is detected in sections 3, 4 and 5, the stress will Energies 2018, 11, 614 10 of 23 be brought back to the curve fs,1 = 0. If the stress state belongs to section 6, 7 or 8, local tensile failure takes place and the stress will be brought back to ft = 0. ft = 0 is the minimum tension failure criterion for the two joints inside a rock mass.

2.4. Numerical Implementation The above-mentioned Mohr-Coulomb yield criteria for the multi-joint model is programmed as a userEnergies defined 2018, 11 model, x FOR PEER and implementedREVIEW into the 2-dimensional explicit finite difference code FLAC10 [of31 23]. The joints are embedded in a Mohr-Coulomb matrix. The failure criterion of the intact rock (matrix) (matrix) is represented in the plane (σ, τ) and shown in Figure 11a. Equations (8) and (9) shows the is represented in the plane (σ, τ) and shown in Figure 11a. Equations (8) and (9) shows the failure failure envelops and plastic potentials for rock matrix and joints. envelops and plastic potentials for rock matrix and joints.

s,0  s,0 ff ==−−−ττ − σ22 tantanφφ+ +c  22  t,0 tt,0t  f = σ −=σ22 −  f σ σ22 (8) gs,0 = −τ − σ tan ψ (8)  s,0 22  t,0g =−−τ σ22 tan ψ  g = −σ22  t,0  g = −σ22

(a)

(b)

Figure 11. (a()a )Illustration Illustration of ofthe the intact intact rock rock failure failure criterion; criterion; (b) Three (b) Three joint failure joint failure criteria criteria in a multi in a- multi-jointjoint model model..

The yield criterion for each joint is a Mohr-Coulomb failure criterion with tension cut-off. Flow The yield criterion for each joint is a Mohr-Coulomb failure criterion with tension cut-off. Flow rule for shear failure is non-associated, and the tension flow rule is associated. For the joint set i (i = rule for shear failure is non-associated, and the tension flow rule is associated. For the joint set i (i = 1, 1, 3), the local failure envelops are defined as fs,i and ft,i according to the following equations: 3), the local failure envelops are defined as fs,i and ft,i according to the following equations:  f si, =−−τ σ′ tanφ i +c i  s,i j022 jji i  f = −τj − σ 22 tan φ + c  ti,,= ti − ′ j j  ft,i σt,i σ0 22  f = σ − σ 22 (9) ssi,i, =−−i 0′ ii (9)  g = −τjjj− σ 2222tantanψψj   gt,iti, = −σ0′  g = −σ22

i i t,i i where ϕj , c j and σ represent the friction, cohesion and tensile strength for joint set i. If ϕj ≠ 0, the

t,i i i tensile strength cannot be larger than σ max. τ j = |σ′j | stand for the magnitude of the tangential stress

component of a joint, and the corresponding strain variable is γji. For the multi-joint model, three joint

sets have six failure criterions in the ( σ22′ , τ) plane, as illustrated in Figure 11b. In Figure 11, strength envelops for three joint sets with independent strength parameters are presented. Obviously, various types of failure can occur in each computational step, including single or simultaneous shear yielding on 1, 2 or 3 weak planes, tensile failure on one or more joints as well as combined shear and tensile failure. In each calculation step the failure of the intact rock and the joint sets are checked, the code also checks for multiple active yield surfaces. After each stress increment, general failure through the intact rock is first checked and if there is any violation of the

Energies 2018, 11, 614 11 of 23

i i t,i i where φj, cj and σ represent the friction, cohesion and tensile strength for joint set i. If φj 6= 0, the t,i i 0i tensile strength cannot be larger than σ max. τj = |σ j| stand for the magnitude of the tangential stress component of a joint, and the corresponding strain variable is γji. For the multi-joint model, three joint 0 sets have six failure criterions in the (σ 22, τ) plane, as illustrated in Figure 11b. In Figure 11, strength envelops for three joint sets with independent strength parameters are presented. Obviously, various types of failure can occur in each computational step, including single or simultaneous shear yielding on 1, 2 or 3 weak planes, tensile failure on one or more joints as well as combined shear and tensile failure. In each calculation step the failure of the intact rock and the joint Energiessets 2018 are, 11 checked,, x FOR PEER the REVIEW code also checks for multiple active yield surfaces. After each stress11 increment, of 23 general failure through the intact rock is first checked and if there is any violation of the failure criterion, failurethe crite correspondingrion, the corresponding plastic stress correctionplastic stress is applied. correction According is applied. to the According updated stressto the state, updated potential stressfailure state, ofpotential each joint failure set is of checked. each joint The set flowchart is checked. shown The inflowchart Figure 12 shown illustrates in Figure the operating 12 illustrates principle the operatingof this law. principle A detailed of this description law. A detailed is provided description by Chang is provided [32]. by Chang [32].

FigureFigure 12. Flowchart 12. Flowchart for equivalent for equivalent continuum continuum multi- multi-jointjoint model model..

3. Verification3. Verification of Multi of Multi-Joint-Joint Model Model To validate the proposed multi-joint model, a series of calculation cases including triaxial and To validate the proposed multi-joint model, a series of calculation cases including triaxial and uniaxial compressive tests with two perpendicular or three joints were selected and compared with uniaxial compressive tests with two perpendicular or three joints were selected and compared with analytical solutions in this section. analytical solutions in this section.

3.1. Analytical Solution for a Jointed Rock Sample The analytical solution for a jointed rock can be expressed using the shear and normal stresses acting on the plane as illustrated in Figure 13. Formulas are given below:  1 1 σ = σ + σ + σ − σ θ  θ ( 1 3 ) ( 1 3 )cos2  2 2 (8) 1  τ = (σ − σ )sin 2θ  θ 2 1 3 Slip will occur along a joint in a specimen when:

2 (c + σ tanφ ) j 3 j (9) σ 1 ≥ σ 3 + (1 − tanφ j tan β ) sin 2β

Energies 2018, 11, 614 12 of 23

3.1. Analytical Solution for a Jointed Rock Sample The analytical solution for a jointed rock can be expressed using the shear and normal stresses acting on the fracture plane as illustrated in Figure 13. Formulas are given below:

( 1 1 σθ = 2 (σ1 + σ3) + 2 (σ1 − σ3) cos 2θ 1 (10) τθ = 2 (σ1 − σ3) sin 2θ

Slip will occur along a joint in a specimen when: Energies 2018, 11, x FOR PEER REVIEW 12 of 23

2 (cj + σ3| tan φj) σ ≥ σ + (11) where cj is the joint cohesion, ϕj is1 the joint3 friction angle and β is the joint angle formed by vertical (1 − tan φj tan β) sin 2β stress and the joint. In an uniaxial compressive test, the maximum stress of a jointed rock can be whereexpressedcj is as: the joint cohesion, φj is the joint friction angle and β is the joint angle formed by vertical stress and the joint. In an uniaxial compressive test, the maximum stress of a jointed rock can be  expressed as: −2c j min 2c k , if (1−> tanφ tanβ ) 0 σ = √ (1− tanφ tanβ ) sin 2β (12) c  ( n −j 2cj o  min 2c k, (1−tan φ tan β) sin 2β i f (1 − tan φ tan β) > 0 σc = √ j −<(12)  2c2c k k i f (if1 −(1tan tanφ tanφ tanβ) <β )0 0 Bray [6] suggested that the overall strength of a rock mass containing multiple joints is given by Bray [6] suggested that the overall strength of a rock mass containing multiple joints is given the lowest strength of the individual strength relations. Final equations defining the uniaxial by the lowest strength of the individual strength relations. Final equations defining the uniaxial of a rock specimen containing multiple joints can be expressed as: compressive strength of a rock specimen containing multiple joints can be expressed as: 22cc √ 2c j122c j σc = min (,2c k j1 , j2 ) (13) σc = min (2c k, (1−− tanφ tanβ ) sin 2β, (1 tan φ tanβ ) sin 2)β (13) (1 − tan φjj111tan β1) sin 2β1 1(1 − tan φj2jtan 2β2) 2sin 2β2 2

The uniaxial compressive strength σc calculated by Equation (13) determines the minimum The uniaxial compressive strength σc calculated by Equation (13) determines the minimum strength of a rockrock massmass forfor variousvarious jointjoint combinationscombinations and used forfor analyticalanalytical solutions in the following section.section.

Figure 13. Mohr’s circle and stress state on failure plane.

3.2.3.2. Strength Anisotropy of a Rock Mass Containing TwoTwo PerpendicularPerpendicular JointsJoints

3.2.1. Uniaxial Compressive Tests The variation of strength for aa jointedjointed rockrock containingcontaining twotwo perpendicularperpendicular joints under uniaxial compression is shown in Figure 1414.. Mechanical properties of the rock matrix and the joints are listed in Table2 2..

Table 2. Mechanical properties of the jointed rock mass.

Material Parameters Intact Rock Joint 1 Joint 2 Density 1810 kg/m³ - - Young’s modulus (E) 20.03 MPa - - Poisson’s ratio (ν) 0.24 - - Cohesion (c) 2 kPa - - Friction angle (ϕ) 40° - - Dilation angle (ψ) 0° - - Joint cohesion (cj) - 1 kPa 1 kPa Joint friction angle (ϕj) - 30° 30° Joint tensile strength - 2 kPa 2 kPa Joint angle - α α + 90°

Figure 14a illustrates the analytical solution for the strength envelope considering both, the effective and the non-active joint. The colored curves in Figure 14a correspond to the colored joints shown in Figure 14b. In Point A and E the two joints are horizontal and vertical, respectively. Points

Energies 2018, 11, 614 13 of 23

Table 2. Mechanical properties of the jointed rock mass.

Material Parameters Intact Rock Joint 1 Joint 2 Density 1810 kg/m3 -- Young’s modulus (E) 20.03 MPa - - Poisson’s ratio (ν) 0.24 - - Cohesion (c) 2 kPa - - Friction angle (φ) 40◦ -- Dilation angle (ψ) 0◦ -- Joint cohesion (cj) - 1 kPa 1 kPa ◦ ◦ Joint friction angle (φj) - 30 30 Joint tensile strength - 2 kPa 2 kPa Joint angle - α α + 90◦

Figure 14a illustrates the analytical solution for the strength envelope considering both, the Energieseffective 2018 and, 11, thex FOR non-active PEER REVIEW joint. The colored curves in Figure 14a correspond to the colored13 joints of 23 shown in Figure 14b. In Point A and E the two joints are horizontal and vertical, respectively. Points B Band and D describeD describe the the critical critical joint joint position. position. Point Point C is theC is inflection the inflection point forpoint the twofor the curves. two Fromcurves. position From positionA to E, the A to two E, jointsthe two with joints fixed with inclination fixed inclination to each other to each are other rotated are by rotated 90 degrees. by 90 degrees.

(a)

(b)

Figure 14. (a) Failure envelop for sample with two perpendicular joints under uniaxial compression Figure 14. (a) Failure envelop for sample with two perpendicular joints under uniaxial compression (see Table 2 2);); ( (b)) SchematicSchematic ofof 55 samplesample withwith twotwo perpendicularperpendicular jointsjoints (A(A toto E).E).

In this section, two more situations for a sample with two perpendicular joints having different In this section, two more situations for a sample with two perpendicular joints having different strength parameters are discussed. The two joints can have only different joint friction angles or have variousstrength joint parameters cohesion are values. discussed. Specific The twoparameters joints can are have listed only in different Tables joint3 and friction 4. The angles jointed or haverock strength behavior is shown in Figures 15 and 16.

Table 3. Mechanical properties of jointed rock mass (different joint friction angle).

Material Parameters Primary Joint Secondary Joint Joint cohesion (cj) 1 kPa 1 kPa Joint friction angle (ϕj) 10° 40° Joint tensile strength 2 kPa 2 kPa Joint angle α + 90° α

Table 4. Mechanical properties of jointed rock mass (different joint cohesion).

Material Parameters Primary Joint Secondary Joint Joint cohesion (cj) 0.5 kPa 1 kPa Joint friction angle (ϕj) 30° 30° Joint tensile strength 2 kPa 2 kPa Joint angle α + 90° α

Energies 2018, 11, 614 14 of 23 various joint cohesion values. Specific parameters are listed in Tables3 and4. The jointed rock strength behavior is shown in Figures 15 and 16.

Table 3. Mechanical properties of jointed rock mass (different joint friction angle).

Material Parameters Primary Joint Secondary Joint

Joint cohesion (cj) 1 kPa 1 kPa ◦ ◦ Joint friction angle (ϕj) 10 40 Joint tensile strength 2 kPa 2 kPa Joint angle α + 90◦ α

Table 4. Mechanical properties of jointed rock mass (different joint cohesion).

Material Parameters Primary Joint Secondary Joint

Joint cohesion (cj) 0.5 kPa 1 kPa ◦ ◦ Joint friction angle (ϕj) 30 30 Joint tensile strength 2 kPa 2 kPa α ◦ α Energies 2018, 11, xJoint FOR PEER angle REVIEW + 90 14 of 23 Energies 2018, 11, x FOR PEER REVIEW 14 of 23

FigureFigureFigure 15. Failure 15. 15. Failure Failure envelope envelope envelope for for for sample sample sample with with different different friction friction friction values values values (see (see Table (seeTable Table3) 3). . 3).

FigureFigure 16. 16. Failure Failure envelope envelope for for sample sample with with different different joint joint cohesion cohesion values values (see (see Table Table 4) 4). . Figure 16. Failure envelope for sample with different joint cohesion values (see Table4). 3.2.2.3.2.2. Triaxial Triaxial Compressive Compressive Tests Tests SpecimensSpecimens withwith differentdifferent jointjoint orientationsorientations areare subjectedsubjected toto triaxialtriaxial compressioncompression atat variousvarious confiningconfining . pressures. Figure Figure 17 17 compares compares the the multi multi-joint-joint model model simulation simulation results results for for samples samples with with twotwo perpendicular perpendicular joints joints with with corresponding corresponding analytical analytical solutions solutions (see (see Section Section 3.1). 3.1). The The jointed jointed rock rock propertiesproperties are are listed listed in in Table Table 2 2 (see (see Figure Figure 17) 17). .The The match match is is excellent, excellent, with with a a relative relative error error smaller smaller thanthan 11%% forfor allall jointjoint angles.angles. TheThe blackblack curvecurve demonstratesdemonstrates thethe anisotropicanisotropic behaviorbehavior ofof thethe modelmodel withoutwithout confiningconfining pressurepressure (uniaxial(uniaxial compressioncompression testtest asas shownshown inin SectionSection 3.2.1).3.2.1). AA strengthstrength reductionreduction is is observed observed for for joint joint angles angles from from 5° 5° to to 30°. 30°. A A local local strength strength increase increase is is observed observed fo forr joint joint anglesangles at at 45° 45° ± ± 15°. 15°.

Energies 2018, 11, 614 15 of 23

3.2.2. Triaxial Compressive Tests Specimens with different joint orientations are subjected to triaxial compression at various confining pressures. Figure 17 compares the multi-joint model simulation results for samples with two perpendicular joints with corresponding analytical solutions (see Section 3.1). The jointed rock properties are listed in Table2 (see Figure 17). The match is excellent, with a relative error smaller than 1% for all joint angles. The black curve demonstrates the anisotropic behavior of the model without confining (uniaxial compression test as shown in Section 3.2.1). A strength reduction is observed for joint angles from 5◦ to 30◦. A local strength increase is observed for joint angles at Energies 2018, 11, x FOR PEER REVIEW 15 of 23 45◦ ± 15◦.

Figure 17. Peak strength versus orientation of joint system under various confiningconfining pressures.pressures.

Finally, the strength increases for joint angles from 60° to 85°. Increasing confining pressure Finally, the strength increases for joint angles from 60◦ to 85◦. Increasing confining pressure shifts the failure envelope upwards and the curve shape becomes more pronounced. Increasing shifts the failure envelope upwards and the curve shape becomes more pronounced. Increasing confining pressures leads to a broader curve shoulder (enlargement from 5° to 15° and from 80° to 90°). confining pressures leads to a broader curve shoulder (enlargement from 5◦ to 15◦ and from 80◦ to The strength anisotropy curve for two perpendicular joints has a W shape and is consistent with 90◦). The strength anisotropy curve for two perpendicular joints has a W shape and is consistent with experiments of Ghazvinian et al. [9] experiments of Ghazvinian et al. [9]

3.3. Strength Prediction for Three Joints Behavior of a sample with three joints under uniaxial compression i iss documented in in Figure 18.18. The minimum angle between each joint is 3030°.◦. Point A, B and D describes the critical joint positions. Point C represents the peak strength value.value. UCS varies from 3.5 kPa to 4.73 kPa, while the maximum UCS for one oror twotwo jointsjoints isis aroundaround 8.58.5 kPa.kPa. ComparedCompared withwith thethe analyticalanalytical solution,solution, thethe modelingmodeling results showshow aa good good match match for for all all joint joint angles. angles. The The strength strength anisotropy anisotropy behavior behavior for rockfor rock with with one jointone isjoint characterized is characterized by U orby VU shapeor V shape (Jaeger’s (Jaeger’s curve). curve). The curve The curve for sample for sample with twowith perpendicular two perpendicular joints hasjoints a Whas form. a W Inform. case In of case three of joints,three joints, three localthree peak local values peak values appear appear (see Figure (see Figure 18b). 18b).

(a)

(b)

Energies 2018, 11, x FOR PEER REVIEW 15 of 23

Figure 17. Peak strength versus orientation of joint system under various confining pressures.

Finally, the strength increases for joint angles from 60° to 85°. Increasing confining pressure shifts the failure envelope upwards and the curve shape becomes more pronounced. Increasing confining pressures leads to a broader curve shoulder (enlargement from 5° to 15° and from 80° to 90°). The strength anisotropy curve for two perpendicular joints has a W shape and is consistent with experiments of Ghazvinian et al. [9]

3.3. Strength Prediction for Three Joints Behavior of a sample with three joints under uniaxial compression is documented in Figure 18. The minimum angle between each joint is 30°. Point A, B and D describes the critical joint positions. Point C represents the peak strength value. UCS varies from 3.5 kPa to 4.73 kPa, while the maximum UCS for one or two joints is around 8.5 kPa. Compared with the analytical solution, the modeling results show a good match for all joint angles. The strength anisotropy behavior for rock with one jointEnergies is characterized2018, 11, 614 by U or V shape (Jaeger’s curve). The curve for sample with two perpendicular16 of 23 joints has a W form. In case of three joints, three local peak values appear (see Figure 18b).

(a)

Energies 2018, 11, x FOR PEER REVIEW (b) 16 of 23

Figure 18. (a) Schematic diagram for samples with three joints; ( b) UCS for sample with three joints:

multi-jointmulti-joint model (line) and analytical solution (triangular) versusversus jointjoint setset angle.angle.

4. A Case Study: Tunnel Excavation A circular tunnel tunnel subjected subjected to to an an initial initial stress stress state state is isdepicted depicted in Figure in Figure 19. 19The. Theobjective objective of this of thismodel model is to isinvestigate to investigate the mechanical the mechanical response response of a circular of a circular excavation excavation under under either either uniform uniform or non or- non-uniformuniform in-situ in-situ stresses. stresses. The Thegeometry geometry of the of the numerical numerical model model is ischaracterized characterized by by a a tunnel tunnel with diameter of D = 1.5 m and a quadratic modelmodel sizesize ofof WW == 1010 m.m. σyy isis set set to to 1 1 MPa MPa and and the the other input parameters are shown in Table5 5.. Three Three cases cases are are studied studied under under differentdifferent in-situ in-situ stresses.stresses. TheThe verticalvertical earth pressure coefficientcoefficient κκ isis setset toto 1.0, 0.5 and 2.0, respectively (Table(Table6 6).). A A value value of of κκ >> 1 means that in-situin-situ horizontal stress is greater than in-situin-situ verticalvertical stressstress andand vicevice versa.versa. Constellations with no joint, oneone jointjoint and and two two joints joints were were investigated. investigated. The The developed developed multi-joint multi-joint model model containing containing two jointtwo setsjoint is sets compared is compared with thewith ubiquitous the ubiquitous joint modeljoint model with onlywith oneonly joint one set.joint set.

Figure 19. Sketch of model geometry and boundary conditions.conditions.

Table 5. Parameters for rock mass. Table 5. Parameters for rock mass. 3 t t E [GPa] ν ρ [kg/m ] c [MPa] Ψ [°] ϕ [°] Σ [MPa] ϕj [°] cj [MPa] σj [MPa] 3 ◦ ◦ t ◦ œt [MPa] E [GPa] 0.2ν 0.23ρ [kg/m ]1810 c [MPa] 1 Ψ [ ] 10 ϕ [ 21.2] Σ [MPa]0.1 32.8ϕj [ ] 0.1cj [MPa] 0.05 j 0.2 0.23 1810 1 10 21.2 0.1 32.8 0.1 0.05 Table 6. Stress and joint constellations for tunnel model.

Joint Orientation Stress Ratio No Joint −60 45/−45 κ = 1 σxx = κσyy = σzz κ = 2 σxx = κσyy = σzz κ = 0.5 σxx = κσyy = σzz

4.1. Simulation Results for Ubiquitous Joint Model This simulations are based on the standard ubiquitous joint model with joint angle of −60° (measured anticlockwise from vertical direction). Figures 20–22 show contour plots of displacement magnitude and stress components as well as the plasticity state for different stress fields.

Energies 2018, 11, x FOR PEER REVIEW 16 of 23

Figure 18. (a) Schematic diagram for samples with three joints; (b) UCS for sample with three joints: multi-joint model (line) and analytical solution (triangular) versus joint set angle.

4. A Case Study: Tunnel Excavation A circular tunnel subjected to an initial stress state is depicted in Figure 19. The objective of this model is to investigate the mechanical response of a circular excavation under either uniform or non- uniform in-situ stresses. The geometry of the numerical model is characterized by a tunnel with diameter of D = 1.5 m and a quadratic model size of W = 10 m. σyy is set to 1 MPa and the other input parameters are shown in Table 5. Three cases are studied under different in-situ stresses. The vertical earth pressure coefficient κ is set to 1.0, 0.5 and 2.0, respectively (Table 6). A value of κ > 1 means that in-situ horizontal stress is greater than in-situ vertical stress and vice versa. Constellations with no joint, one joint and two joints were investigated. The developed multi-joint model containing two joint sets is compared with the ubiquitous joint model with only one joint set.

Figure 19. Sketch of model geometry and boundary conditions. Energies 2018, 11, 614 17 of 23 Table 5. Parameters for rock mass.

3 t t E [GPa] ν ρ [kg/m ] c [MPa] Ψ [°] ϕ [°] Σ [MPa] ϕj [°] cj [MPa] σj [MPa] Table 6. Stress and joint constellations for tunnel model. 0.2 0.23 1810 1 10 21.2 0.1 32.8 0.1 0.05

Joint Orientation StressTable Ratio 6. Stress and joint constellations for tunnel model. No Joint −60 45/−45 Joint Orientation Stress Ratio κ = 1 No Jointσ xx−=60κσ yy45/=−45σzz κ = 2 κ = 1 σxx σ=xx κσ=yy κσ= σyyzz = σzz κ = 0.5 κ = 2 σxx σ=xx κσ=yy κσ= σyyzz = σzz κ = 0.5 σxx = κσyy = σzz 4.1. Simulation Results for Ubiquitous Joint Model 4.1. Simulation Results for Ubiquitous Joint Model − ◦ ThisThis simulations simulations are are based based on on the the standard standard ubiquitousubiquitous joint joint model model with with joint joint angle angle of − of60° 60 (measured(measured anticlockwise anticlockwise from from vertical vertical direction). direction). FiguresFigures 2020––2222 show show contour contour plots plots of displacement of displacement magnitudemagnitude and and stress stress components components as as well well as as the the plasticityplasticity state state fo forr different different stress stress fields. fields.

Energies 2018, 11, x FOR PEER REVIEW 17 of 23

FigureFigure 20. Simulation 20. Simulation results results for for tunnel tunnel model model with with jointjoint orientation of of −60°−60 and◦ and κ = κ1: =(a 1:) displacement (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa]. vertical stresses [Pa]. Figure 20 shows maximum displacements of 0.12 m at the tunnel boundary around 30° inclined Figureto the vertical. 20 shows Plasticity maximum pattern displacements follows the displacement of 0.12 m at field. the tunnel(Figure boundary20a,b). The aroundinfluence 30 of◦ inclinedthe to thejoint vertical. on the Plasticitysecondary patternstress fiel followsd is illust therated displacement by Figure 20c, field.d. (Figure 20a,b). The influence of the joint on the secondary stress field is illustrated by Figure 20c,d. As Figure 21 shows with κ = 2.00 the maximum displacements increase to 0.20 m and plasticity pattern changes. The maximum horizontal stress component becomes 1.60 MPa at the tunnel roof and invert. The maximum vertical stress component is 3.25 MPa and occurs at the tunnel sidewalls (Figure 22d). It also has the maximum plasticity area in this situation. Figure 22 illustrates the situation for κ = 0.5. Due to the lower vertical stress component, the maximum contour displacement is only 0.10 m. The plasticity area is similar to that of the isotropic stress case shown in Figure 22b. The maximum horizontal stress value of 1.80 MPa is observed at the tunnel roof and invert. The maximum vertical component of 0.70 MPa is found at the tunnel sidewalls (Figure 22d).

Figure 21. Simulation results for tunnel model with joint orientation of −60° and κ = 2: (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa].

As Figure 21 shows with κ = 2.00 the maximum displacements increase to 0.20 m and plasticity pattern changes. The maximum horizontal stress component becomes 1.60 MPa at the tunnel roof and invert. The maximum vertical stress component is 3.25 MPa and occurs at the tunnel sidewalls (Figure 22d). It also has the maximum plasticity area in this situation.

Energies 2018, 11, x FOR PEER REVIEW 17 of 23

Figure 20. Simulation results for tunnel model with joint orientation of −60° and κ = 1: (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa].

Figure 20 shows maximum displacements of 0.12 m at the tunnel boundary around 30° inclined

Energiesto2018 the, vertical.11, 614 Plasticity pattern follows the displacement field. (Figure 20a,b). The influence of the18 of 23 joint on the secondary stress field is illustrated by Figure 20c,d.

Figure 21. Simulation results for tunnel model with joint orientation of −60° and κ = 2: (a) displacement Figure 21. Simulation results for tunnel model with joint orientation of −60◦ and κ = 2: (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa]. Energiesvertical 2018, 11 stresses, x FOR PEER [Pa]. REVIEW 18 of 23 As Figure 21 shows with κ = 2.00 the maximum displacements increase to 0.20 m and plasticity pattern changes. The maximum horizontal stress component becomes 1.60 MPa at the tunnel roof and invert. The maximum vertical stress component is 3.25 MPa and occurs at the tunnel sidewalls (Figure 22d). It also has the maximum plasticity area in this situation.

Figure 22. Simulation results for tunnel model with joint orientation of −60° and κ = 0.5: (a) Figure 22. Simulation results for tunnel model with joint orientation of −60◦ and κ = 0.5: displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and contour plot of vertical stresses [Pa]. (d) contour plot of vertical stresses [Pa]. Figure 22 illustrates the situation for κ = 0.5. Due to the lower vertical stress component, the maximum contour displacement is only 0.10 m. The plasticity area is similar to that of the isotropic stress case shown in Figure 22b. The maximum horizontal stress value of 1.80 MPa is observed at the tunnel roof and invert. The maximum vertical component of 0.70 MPa is found at the tunnel sidewalls (Figure 22d).

4.2. Simulation Results for Multi-Joint Model In this section, the behavior of a rock mass with two joint sets is investigated by using the new developed multi-joint model. Symmetric interconnected joint sets are considered. Figures 23–25 show stress and displacement fields for given values of initial stress, strength parameters and joint orientation. These figures clearly illustrate the anisotropic behavior due to the presence of joints especially under non-uniform stress states. The numerical results for κ = 1.0 and joint angle of 45°/−45° are illustrated in Figure 23. Maximum displacement is 0.05 m, plasticity is restricted to the immediate tunnel contour (Figure 23b). As can be seen from Figure 23c,d, stress components show symmetric pattern similar to displacements and plasticity. Results for κ = 2 and joint angles of 45°/−45° are shown in Figure 24. The maximum displacement is 0.14 m as shown in Figure 22a. Locally plasticity extends deeper into the rock mass following the two joint orientations (Figure 24b). The maximum horizontal stress component is 1.70 MPa at the tunnel roof and invert, the maximum vertical stress component is 3.50 MPa and observed at the tunnel sidewalls (Figure 24d).

Energies 2018, 11, 614 19 of 23

4.2. Simulation Results for Multi-Joint Model In this section, the behavior of a rock mass with two joint sets is investigated by using the new developed multi-joint model. Symmetric interconnected joint sets are considered. Figures 23–25 show stress and displacement fields for given values of initial stress, strength parameters and joint orientation. These figures clearly illustrate the anisotropic behavior due to the presence of joints especially under non-uniform stress states. The numerical results for κ = 1.0 and joint angle of 45◦/−45◦ are illustrated in Figure 23. Maximum displacement is 0.05 m, plasticity is restricted to the immediate tunnel contour (Figure 23b). As can be seen from Figure 23c,d, stress components show symmetric pattern similar to displacements and plasticity. Results for κ = 2 and joint angles of 45◦/−45◦ are shown in Figure 24. The maximum displacement is 0.14 m as shown in Figure 22a. Locally plasticity extends deeper into the rock mass following the two joint orientations (Figure 24b). The maximum horizontal stress component is 1.70 MPa at the tunnel roof and invert, the maximum vertical stress component is 3.50 MPa and observed at the tunnel sidewallsEnergies 2018 (Figure, 11, x FOR 24 PEERd). REVIEW 19 of 23

Figure 23. SimulationSimulation results results for for tunnel tunnel model model with with two two joint joint sets sets (45° (45 an◦ andd −45°)−45 and◦) and κ = κ1:= (a 1:) (displacementa) displacement contours contours [m], [m], (b) (plasticityb) plasticity state, state, (c) contour (c) contour plot plot of horizontal of horizontal stresses stresses [Pa] [Pa] and and (d) (contourd) contour plot plot of vertical of vertical stresses stresses [Pa] [Pa]..

Figure 24. Simulation results for tunnel model with two joint sets (45° and −45°) and κ = 2: (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa].

Energies 2018, 11, x FOR PEER REVIEW 19 of 23

Figure 23. Simulation results for tunnel model with two joint sets (45° and −45°) and κ = 1: (a) Energies 2018, 11displacement, 614 contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) 20 of 23 contour plot of vertical stresses [Pa].

Figure 24. Simulation results for tunnel model with two joint sets (45° and −45°) and κ = 2: (a) Figure 24. Simulation results for tunnel model with two joint sets (45◦ and −45◦) and κ = 2: displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) (a) displacementcontour plot contours of vertical [m],stresses (b )[Pa] plasticity. state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa].

(a) (b)

(c) (d)

Figure 25. Simulation results for tunnel model with two joint sets (45◦ and −45◦) and κ = 0.5: (a) displacement contours [m], (b) plasticity state, (c) contour plot of horizontal stresses [Pa] and (d) contour plot of vertical stresses [Pa].

The simulation results for κ = 0.50 and joint angles of 45◦/−45◦ are shown in Figure 25. Maximum displacement is 0.06 m in the sidewalls. The secondary horizontal stress is larger than the vertical stress. The failure area is symmetric as illustrate in Figure 25b. The maximum horizontal stress components

Energies 2018, 11, x; doi: FOR PEER REVIEW www.mdpi.com/journal/energies Energies 2018, 11, 614 21 of 23 is 1.80 MPa and situated at the tunnel roof and invert, the vertical stress component is 0.75 MPa and observed at the tunnel sidewalls (Figure 25d).

5. Discussion and Conclusions In this paper, an equivalent continuum based anisotropic model has been implemented into FLAC to simulate the behavior of jointed rock masses containing up to three persistent joint sets. The equivalent compliance matrix of the rock mass has been deduced and the Mohr-Coulomb yield criterion has been used to check the failure characteristics of the intact rock and the joints. Thus, through several verifications of the model, the following conclusions can be drawn: The potential functions and flow rules for yield in the matrix and the joints are considered. Since violation of multiple plasticity surfaces can occur within one calculation step, a consistent elasto-plastic algorithm which automatically identifies the activated surfaces is applied. Joint stiffness and spacing are considered and lead to an overall softer behavior in the elastic stage. The strength anisotropy behavior of the jointed rock mass is closely related to the direction of loading relative to the orientation of the joints. The failure envelop for uniaxial compressive loaded sample with two perpendicular joints is W shaped. By extensive testing of interconnected joints, it is shown that the multi-joint formulation can be used in predicting the strength anisotropy of rock mass with two or three joints. A circular tunnel in a fractured rock mass is investigated under uniform and non-uniform in-situ stresses. The mechanical response of a rock mass shows different characteristics in dependence on stress state and joint orientation. In the stability analysis of a circular tunnel excavation, matrix failure, single joint and double joint plane shear or tensile failure are three typical failure modes. For κ = 1 condition, the displacement and plasticity area of the tunnel were mainly influenced by the joint orientation. Increased vertical earth pressure coefficient (κ change from 0.5 to 2), lead to larger tunnel deformation. For a tunnel in a rock mass with two joint sets, plastic failure can occurred at both joint directions as simulated correctly by the multi-joint model. The current analysis is based on the developed multi-joint model where more than twenty parameters have been introduced to consider the influence of the second or third joints. Further development is necessary to incorporate anisotropic behavior even in the elastic phase. Nevertheless, the proposed multi-joint model can be regarded as an effective method to solve anisotropic stability problems in rock engineering.

Acknowledgments: The work was supported by the China Scholarship Council (CSC) and Deutscher Akademischer Austausch Dienst (DAAD) for financially supporting Lifu Chang’s study in Germany. Author Contributions: Lifu Chang and Heinz Konietzky proposed the idea for the extended multi-joint model; Lifu Chang performed the theoretical analysis and numerical simulation; he wrote the manuscript corrected and improved by Heinz Konietzky. Conflicts of Interest: The authors declare no conflict of interest.

Notation c, cji (i = 1, 2, 3) cohesion and joint cohesion, respectively D0 diameter of tunnel E elastic modulus Em, Eeq elastic modulus of intact rock and for the equivalent rock mass, respectively fs,i (i = 0, 1, 2, 3) shear failure envelops for rock matrix and joint set, respectively ft,i (i = 0, 1, 2, 3) tensile failure envelops for rock matrix and joint set, respectively Ftrial strength reduction factor gs,i (i = 0, 1, 2, 3) shear potential functions for rock matrix and joint set, respectively gt,i (i = 0, 1, 2, 3) tensile potential functions for rock matrix and joint set, respectively Energies 2018, 11, 614 22 of 23

G, G0 shear modulus of rock and for any plane normal to the plane of isotropy h, h1, h2 functions to separate the failure envelope jαi (i = 1, 2, 3) critical failure angle for joint set i kn, ks joint normal stiffness and shear stiffness Ki (i = 1, 2, 3) stiffness of joint set i κ vertical earth pressure coefficient Si (i = 1, 2, 3) number of joints per 1 meter for joint set i v, ν0 Poisson’s ratio and Poisson’s ratio in different directions, respectively αmin, αmax maximum and minimum failure angle, respectively αj critical joint angle measure from horizontal direction β joint inclination angle measured from vertical βi (i = 1, 2, 3) inclination angle of weak plane ϕ, ϕ0 friction angle and effective friction angle of the intact rock, respectively ϕji (i = 1, 2, 3) friction angle for the joint set i ψ, ψj dilation angle for rock matrix and joints, respectively λ plastic multiplier ρ density of rock σ, σn, σt tensile stress, normal stress and intact rock tensile strength, respectively σ1, σ3 maximum and minimum principal stresses, respectively t,i σj (i = 1, 2, 3) joint tensile strength for joint set i σci unconfined compressive strength of the intact rock σc, σu uniaxial compressive strength and uniaxial compression stress trial σi trial elastic stress 0 0 τ, τ , τji shear stress components in the global and local coordinates, respectively

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