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Numerical Investigation of the Failure Mechanism of Transversely Isotropic Rocks with a Particle Flow Modeling Method Xu-Xu Yang 1,2, *, Hong-Wen Jing 2 and Wei-Guo Qiao 1 1 2

*

Shandong Provincial Key Laboratory of Civil Engineering Disaster Prevention and Mitigation, Shandong University of Science and Technology, Qingdao 266590, China; [email protected] State Key Laboratory for Geomechanics and Deep Underground Engineering, China University of Mining and Technology, Xuzhou 221116, China; [email protected] Correspondence: [email protected]; Tel.: +86-0532-8605-7646

Received: 30 August 2018; Accepted: 14 September 2018; Published: 17 September 2018







Abstract: Transversely isotropic rocks are commonly encountered in rock engineering practices, and their strength and failure behavior is often governed by the property of anisotropy. The particle flow modeling method was utilized to investigate the failure mechanism of transversely isotropic rocks subject to uniaxial compressive loading. The details for establishing transversely isotropic rock models were first presented, and then a parametric study was carried out to look into the effect of interface properties on the failure mode and strength of transversely isotropic rock models by varying the interface dip angle. The smooth joint model was incorporated to create interfaces for the completeness of establishing transversely isotropic rock models with the particle flow modeling method. Accordingly, three failure modes observed in transversely isotropic rock models with varying dip angles were tensile failure across interfaces, shear failure along interfaces, and tensile failure along interfaces. Furthermore, the interface mechanical parameters were found to differently influence the failure behavior of transversely isotropic rock models. The bonded joint cohesion and bonded joint friction angle that contribute to the shear strength of interfaces have considerable influence on the uniaxial compressive strength (UCS) values, while the joint coefficient of friction and joint tensile strength have a slight influence on the UCS values. The findings in this paper indicated the importance of interfaces in estimating failure behavior of transversely isotropic rocks. Keywords: transversely isotropic rocks; failure mechanism; particle flow modeling; interface

1. Introduction The dominant anisotropy or transverse isotropy of geological materials, especially of foliated metamorphic rocks, such as slates, gneisses, schist, and sedimentary rocks with bedding planes, leads to complicated failure behaviors [1]. Rock anisotropy is one of the most significant characteristics that should be taken into consideration for underground engineering. The design and stability analysis of underground structures excavated in anisotropic rock masses, for instance, require a complete understanding of the failure behavior of the rock materials. Thus, the failure mechanism of anisotropic or transversely isotropic rocks has been a significant topic in rock mechanics. Engineering practice has suggested that the rock anisotropy is of importance in the stabilization of underground excavations in bedded rock masses [2,3]. Previous studies indicate that drilling through bedding planes was quite dangerous. The boreholes might become unstable if the deviation angle of drilled well applied to sub-horizontal bedding planes is very high due to rock strength anisotropy [4–8]. Particularly, Okland et al. [4] carried out hollow cylinder tests to study the critical bedding inclination angle of shale that can induce severe borehole damage during extended reach Processes 2018, 6, 171; doi:10.3390/pr6090171

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drilling. Meier et al. [6] also claimed that the lowest stress is required to induce borehole breakout when the borehole is sub-parallel to the bedding planes. Moreover, Zhou et al. [9] reported that in the Wudongde hydropower station on the Jinsha River in China, instability of high sidewalls, particularly separation of bedding planes and the accompanying remarkable deformation, is likely to happen in those sections where strata are almost parallel to the cavern axis with a thickness less than 10 cm. For better engineering design, the mechanical properties of aforementioned anisotropic rock materials are suggested to be determined through the field tests or through rock mass classification [10]. However, due to the inherent difficulty and inaccuracy in conducting field tests or rock mass classification, many investigators tend to investigate the mechanical properties of anisotropic rock materials through laboratory tests. To be specific, Gatelier et al. [11], according to laboratory tests, stated a decrease in uniaxial compressive strength with increasing inclination angle in Adamswiller sandstone. Kim et al. [12] through X-ray CT (Computed Tomography) study and uniaxial compression tests found that Berea sandstone in northern Ohio is composed of cross-bedded loose layers and relatively thin tightly packed layers, and its uniaxial compressive strength (UCS) value decreases with increasing porosity as well as with increasing inclination of the bedding plane. Heng et al. [13] investigated the directional shale samples obtained from the Longmaxi Formation in Shizhu County, China, and three types of shear failure modes were identified that were dependent on the shearing angle, which were sliding failure across the bedding plane, sliding failure along the bedding planes, and sliding failure across the bedding planes combined with tensile splitting along the bedding planes. Besides these real rock materials prepared from engineering sites, some rock-like materials have also been utilized to investigate the anisotropic or transversely isotropic mechanism. Kulatilake et al. [14] carried out uniaxial compressive tests on the jointed rock-like material blocks having existing joint sets. Their findings show that for rock samples with dip angles of 0◦ to 15◦ , the failure mechanism was the tensile failure through the intact model material, while for rock samples with joint dip angles of 40◦ to 60◦ , the main failure mechanism was the combined shear and tensile failure through the joints. A mixed mechanism of the above two modes accounted for the failure of jointed blocks having dip angles of 20◦ to 35◦ . Yang et al. [15] performed a series of physical model tests for jointed rock masses with persistent discontinuities and indicated that the failure modes of these models with different dip angles can be divided into split mode, sliding mode, and mixed mode. Recently, an experimental investigation was implemented by Yang et al. [16] on jointed rock models made of rock-like materials. In accordance with the experimental results, the failure of simulated rock models with varied orientations is classified into one of four modes: (a) tensile failure across the joint plane, (b) shear failure along the joint plane, (c) tensile failure along the joint plane, and (d) intact material failure. These above-mentioned researchers explained the anisotropic or transversely isotropic mechanism of rock materials with weak planes/layers presence. These studies broaden our understanding of weak planes/layers geometry on the mechanical property of rock materials. As one important research methodology in rock mechanics as well as in geotechnical engineering, the numerical modeling method has significantly enriched the data bank of anisotropic rock behavior besides those obtained from laboratory experiments. Particularly, the particle flow modeling method with the basis of the discrete element theory has obtained a pronounced development from being applied to rock and soil mechanics to greatly broader applications. This approach is capable of deriving the rock mass response based on the relatively simple particle contact laws at joints and in rock instead of more complex constitutive models [17,18]. Therefore, a particle flow modeling approach could be utilized directly to extend the experimental results obtained in the laboratory. For example, Chiu et al. [19] put forward a modified smooth-joint model utilizing the particle flow modeling approach to mimic the anisotropic behavior of a rock mass, and the failure modes of the rock mass obtained by Yang et al. [15] in the laboratory were perfectly reproduced. The load-deformation curves in different joint orientations were also well mimicked. Bahaaddini et al. [18], by using particle flow modeling, thoroughly investigated the effect of geometric parameters of joints on the rock mass failure mechanism, deformation modulus, and uniaxial compression strength, and critically

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compared these with the physical experiments [20]. Park and Min [21] also conducted particle flow simulations with embedded smooth joints for mimicking the strength and failure behavior of transversely isotropic rocks with systematic verification and extensions to laboratory and field problems. They succeeded in capturing the failure modes observed in anisotropic rock in which weak planes significantly matter. Furthermore, the particle flow modeling has been utilized as well to study the failure behavior of transversely isotropic rock mass through Brazilian tests [1,22]. Although the particle flow modeling promotes the understanding of the anisotropic or transversely isotropic rock mass mechanical behavior, the concentration is still mainly put on the geometrically structural effect. On the other hand, the influence of bedded planes/layers’ mechanical properties on the failure response of anisotropic or transversely isotropic rock masses is not well understood. This study aims to better understand the influence of mechanical properties of bedded layers (interfaces) on the failure behavior of transversely isotropic rocks through a particle flow modeling method. Thus, this paper is organized in the following matter. First, the physical experiment on simulated transversely isotropic rocks is briefly introduced in Section 2.1 as a benchmark for particle flow modeling. After that, the procedure to establish particle flow models, which consists of intact materials and interfaces, and the calibration of their mechanical parameters, are illustrated in Section 2.2. With these numerically simulated models, the failure mechanism of transversely isotropic rocks is analyzed at the mesoscale under uniaxial compressive loading in Section 3.1. In Section 3.2, a sensitivity study is carried out systematically to estimate the influence of mechanical parameters of interfaces on the failure strength of transversely isotropic rocks. 2. Materials and Methods 2.1. Physical Experiment on Simulated Transversely Isotropic Rocks To investigate the failure mechanism of the transversely isotropic rocks through particle flow modeling, associated experimental results are required. The present modeling study refers to artificial transversely isotropic rocks, as implemented by Tien et al. [23], and utilized their test data for the base of the particle flow simulations. The physical experiment program and results are described as follows. Two model materials, referred to as material A and material B, respectively, were chosen to combine into the transversely isotropic rock samples. Material A was composed of cement, kaolinite, and water in a weight ratio of 4:1:1.2, while material B was composed of cement, kaolinite, and water in a weight ratio of 1:1:0.6. According to the test results of the materials’ mechanical properties, the uniaxial compressive strength (UCS), Young’s modulus, and Poisson’s ratio of material A are 104.2 MPa, 21.7 GPa, and 0.23, respectively, representing rocks with high strength and stiffness. On the other hand, the uniaxial compressive strength, Young’s modulus, and Poisson’s ratio of material B are 43.3 MPa, 11.9 GPa, and 0.21, respectively, representing rocks with relatively low strength and stiffness. Tien et al. [23] incorporated eight steps to develop these transversely isotropic rock samples. After the mold assembly and model material preparation steps (steps 1 and 2), the preliminary compaction step (step 3) was conducted by using the MTS (Mechanical Testing and Simulation) servo-controlled loading frame to apply an axial load of 20 kN on the model materials to make them into a preliminary square rock mass soft enough for cutting into slices. In the cutting into slices step (step 4), the preliminarily compacted model materials (material A and material B) were put on a cutting platform and cut with a slice cutter into slices with a thickness of 5 mm in each layer. After that, a suction lifter was utilized to pile material A slices and material B slices up in sequence, which is referred to as the stacking slices step (step 5). Following the repairing step (step 6), the final compaction step (step 7) was implemented to apply a final load of 114 kN on those sliced materials, and then it was held for one hour until a steady settlement was achieved. Finally, upon the curing and drilling step (step 8), the transversely isotropic rock samples having different dip angles (α = 0◦ , 15◦ , 30◦ , 45◦ , 60◦ , 75◦ , and 90◦ ) were created for the subsequent compression tests. These artificial transversely isotropic

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rock samples were developed rock to represent anisotropic behavior of sandstone-shale sandstone-shale interlayered masses.the Figure 1 shows the physical dimension interlayered of a typical rock masses. Figure 1 shows the physical dimension of a typical transversely isotropic rock sample. transversely isotropic rock sample. Material A Material B 100 mm

5 mm 5 mm Interface

50 mm Figure 1. A typical typical transversely transversely isotropic isotropic rock 0◦ ). Figure 1. A rock sample sample (α (α == 0°).

These artificial transversely isotropic rock samples were then subjected to compression tests. These artificial transversely isotropic rock samples were then subjected to compression tests. In accordance with the test results [23], with zero confining pressure, three failure modes were In accordance with the test results [23], with zero confining pressure, three failure modes were identified. When the dip angles were relatively low (α = 0◦ , 15◦ , and 30◦ ), the artificial transversely identified. When the dip angles were relatively low (α = 0°, 15°, and 30°), the artificial transversely isotropic rock samples failed in a mode of tensile fracture across discontinuities; whereas, when the dip isotropic rock samples failed in a mode of tensile fracture across discontinuities; whereas, when the angles were medium (α = 45◦ , 60◦ , and 75◦ ), the artificial transversely isotropic rock samples failed in a dip angles were medium (α = 45°, 60°, and 75°), the artificial transversely isotropic rock samples failed mode of sliding failure alongalong discontinuities. Moreover, for artificial transversely isotropic rock sample in a mode of sliding failure discontinuities. Moreover, for artificial transversely isotropic rock having a vertical dip angle, its failure mode was defined as the tensile-split along discontinuities. sample having a vertical dip angle, its failure mode was defined as the tensile-split along

discontinuities. 2.2. Particle Flow Modeling of Material A and Material B The particle flow code three dimensions (PFC 2.2. Particle Flow Modeling of in Material A and Material B 3D ) is a commercial software package based on 3D was utilized to develop the particle flow the discrete element method [24,25]. In this study, PFC The particle flow code in three dimensions 3D (PFC3D) is a commercial software package based on models of transversely isotropic rocks. In PFC , the intact materials (material A and material B) the discrete element method [24,25]. In this study, PFC3D was utilized to develop the particle flow were mimicked through a composite of particles that interact with each other at contacts (Figure 2a). models of transversely isotropic rocks. In PFC3D, the intact materials (material A and material B) were A linear contact model provides an elastic relationship between the relative displacements and forces mimicked through a composite of particles that interact with each other at contacts (Figure 2a). of particles at the point contact (Figure 2b). This model consists of the contact normal force component, A linear contact model provides an elastic relationship between the relative displacements and forces Fn , contact overlap, Un , shear increment, ∆Fs , and shear displacement increment, ∆Us , as follows: of particles at the point contact (Figure 2b). This model consists of the contact normal force component, Fn , contact overlap, U n , shearFnincrement, = k n Un ΔFs , and shear displacement increment, (1) ΔU s , as follows: ∆Fs = −k s ∆Us (2) Fn = knU n (1) with k n and k s being the contact normal and shear stiffness. The frictional resistance of the contact is asΔF follows: = − k s ΔU s (2) s

with

k n and k s being the contact normal and shear Fs ≤ µFnstiffness.

(3)

The frictional resistance of the contact is as follows: with µ being the friction coefficient between particles. ≤ μ Fn it is required to stick these granular particles To numerically mimic a relatively rock-likeFsmaterial, (3) through a bonded model [26]. The bonded model, herein, is a parallel bond model that resists not with μ being the friction coefficient between particles. only the contact forces but also the moments between the particles at a cemented contact (Figure 2c). To numerically mimic a relatively rock-like it isas:required to stick these granular The function mechanism of the parallel bond modelmaterial, is described particles through a bonded model [26]. The bonded model, herein, is a parallel bond model that resists not only the contact forces but also ∆F then moments the particles at a cemented contact = kn A∆Ubetween (4) n (Figure 2c). The function mechanism of the parallel bond model is described as: ∆F s = −ks A∆Us (5) (4) Δ F n = k n A ΔU n

Δ F s = − k s AΔ U s

(5)

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Δ M n = − k s J Δθ n

(6)

(7) Δ M s = − k n I Δθ s ∆Mn = −ks J∆θn (6) with F n , F s , M n , and M s being the force components and moments about the center of the (7) ∆Ms = −kn I∆θs cemented-contact zone, respectively; k n and k s being the normal and shear bond stiffness per unit with the force components and moments about the and center ofand the I n , F s , M n , and s being θn M θ s being and the rotation angle components, respectively; A, J, area,Frespectively; cemented-contact zone, respectively; kn and ks being the normal and shear bond stiffness per unit area, being the area, polar moment of inertia, and moment of inertia of the bond contact cross-section, respectively; θn and θs being the rotation angle components, respectively; and A, J, and I being the respectively. The strength value of the bonded contact is given by: area, polar moment of inertia, and moment of inertia of the bond contact cross-section, respectively. The strength value of the bonded contact is given by: M s R −F n (8) + < σ c σ max = MsI R − FA n σmax = (8) + < σc A I M R n −F s (9) τ max =− F + Mn R < τ c s A J τ max = + < τc (9) A J R being withR being the radii ofcemented the cemented contact between particles c and with the radii of the contact planeplane between particles (Figure(Figure 2c); and2c); σ and and τσ being and

the shear strength of the bond of contact, respectively. τ ctensile beingand the tensile and shear strength the bond contact, respectively.

(a)

(b)

c

c

(c)

Figure model and parallel bond model: (a) (a) intact material, (b) (b) particle contact, and and (c) Figure2.2.The Thecontact contact model and parallel bond model: intact material, particle contact, parallel bond. (c) parallel bond.

Tensile Tensilecracks cracksoccur occurwhen whenthe theapplied appliednormal normalstress stressexceeds exceedsthe thespecified specifiedtensile tensilestrength strengthofofthe the parallel bond, σc , while shear cracks happen as the applied shear stress surplus the shear strength, parallel bond, σ c , while shear cracks happen as the applied shear stress surplus the shear strength, τ c , either through rotation or through sliding of particles. The tensile strength at the contact soon τ c , toeither or through sliding particles. The tensile strength at friction the contact soon falls zero through once the rotation cracks occur, while the shearofstrength decreases as the residual strength falls to zero once the cracks occur, while the shear strength decreases as the residual friction strength (Equation (3)). (Equation (3)). For presenting the aforesaid mechanical properties of material A and material B, this numerical For method presenting the aforesaid mechanical properties of material A and material B, thisdeforming numerical modeling requires a calibration of the microparameters to describe contact and bond modeling method requires a calibration of the microparameters to describe contact and bond and strength behavior, and the selected package of microparameters are expected to derive the macro deforming and strength behavior, and the selected package of microparameters are expected mechanical properties of simulated material used in physical experiments [23]. Therefore, a calibrationto derive thewas macro mechanical properties of simulated materialfor used in rock physical experiments procedure carried out to determine these microparameters intact materials (material[23]. A Therefore, a calibration procedure was carried out to determine these microparameters for intact rock and material B). Several steps [25,27–29] are followed to reproduce certain properties of the artificial materialsFirst, (material A andand material B).bond Several steps [25,27–29] followed to reproduce certain materials. the particle parallel moduli and the ratios are of normal to shear stiffness are set of the artificial materials. First, bonds, the particle bond the ratios of toproperties be equal between the particles and parallel whichand aimsparallel to decrease themoduli numberand of independent normal to shear stiffness are set to be equal the through particlessetting and parallel bonds, strengths which aims parameters. Second, the Young’s modulus is between determined the material to ato decrease parameters. Second, Young’s modulus determined large valuethe andnumber varyingofEc independent (particle Young’s modulus) and Ecthe (parallel bond Young’sismodulus) to through setting the material strengths to a large value and varying E (particle Young’s modulus) and c match the Young’s modulus between the particle flow modeling and laboratory samples (with the (parallel bond Young’s modulus) to match the Young’s modulus the particle flow E c of size ϕ 50 mm × 100 mm). Then, through varying k n /k normalbetween stiffness/shear stiffness) s (particle kn / ks modeling and laboratory samples (with the size of φ 50 mmthe × 100 mm). Then, varying and kn /ks (parallel bond normal stiffness/shear stiffness), Poisson’s ratiothrough of the intact synthetic cylindrical sample was matchedstiffness) to the laboratory thestiffness/shear strength between the (particle normal stiffness/shear and k n / ksamples. (parallelAfterwards, bond normal stiffness), s numerical and laboratory samples were matched by decreasing the normal and shear bond strengths the Poisson’s ratio of the intact synthetic cylindrical sample was matched to the laboratory samples. ofAfterwards, the parallelthe bonds. During this procedure, it is of importance tosamples fix the ratio ofmatched normal to bond strength between the numerical and laboratory were byshear decreasing strength (σc /τ c ) for the reason that it influences the failure pattern of the sample. The determined

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microparameters of the intact rock material were listed in Table 1. A comparison between particle flow modeling and experiment results in Table 2 claims a good capability of this numerical modeling method to reproduce these simulated mechanical properties under uniaxial compression loading. The deviations for uniaxial compressive strength (UCS), Young’s modulus (E), and Poisson’s ratio (ν) of material A and material B were all less than 5.0%. Table 1. Microparameter values of intact rock materials (material A and material B). Property

Parameter

Value1 (Material A)

Value2 (Material B)

Particle

ρ (kg/m3 ) k n /k s Ec (GPa) µ Rrat = Rmax /Rmin Rmin (mm)

2150 1.65 18.8 0.554 1.66 0.65

1760 1.55 10.3 0.466 1.66 0.65

Parallel bond

λ kn /ks Ec (GPa) σc (mean ± std.dev., MPa) τ c (mean ± std.dev., MPa)

1.0 1.65 18.8 76.0 ± 19.0 152.0 ± 38.0

1.0 1.55 10.3 31.7 ± 7.9 63.4 ± 15.9

Note that ρ is the density of the synthetic rock material; λ is the radius multiplier used to set the parallel bond radii; Rrat is the radius of the particle; Rmax and Rmin are the maximum radius and minimum radius of the particle, respectively.

Table 2. Comparison of mechanical properties of intact material between the physical experiment and the particle flow modeling results. Material

Macro Properties

Experimental Results

Numerical Results

Abs. Deviation

Material A

UCS (MPa) E (GPa) ν

104.2 21.7 0.230

105.8 20.8 0.224

1.53% 4.15% 2.61%

Material B

UCS (MPa) E (GPa) ν

43.3 11.9 0.210

44.1 11.4 0.202

1.85% 4.20% 3.81%

2.3. Particle Flow Modeling of Interface between Material A and Material B 2.3.1. Smooth Joint Model As aforementioned, the simulated transversely isotropic rock samples were prepared through piling up the material A slices and material B slices in sequence [23]. The interfaces between material A and material B (as shown in Figure 1) were discontinuities that behave quite differently from the adjacent intact materials. Hence, to better describe the mechanical behavior of transversely isotropic rocks the interfaces were intersected between material A and material B slices and represented with the smooth joint model. The numerically developed transversely isotropic rock samples having different interface dip angles are shown in Figure 3.

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(a) (a)

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(b) (b)

(c) (c)

(d) (d)

material A material A material B material B interface interface

(e) (e)

(f) (f)

(g) (g)

Figure 3.3. Numerically developed transversely isotropic rock samples having different interface dip Figure3. Numericallydeveloped developedtransversely transverselyisotropic isotropicrock rocksamples sampleshaving havingdifferent differentinterface interfacedip dip Figure Numerically angles: (a) α = 0°, (b) α = 15°, (c) α = 30°, (d) α = 45°, (e) α = 60°, (f) α = 75°, and (g) α = 90°. ◦ ◦ ◦ ◦ ◦ ◦ ◦ angles: (a) α = 0 , (b) α = 15 , (c) α = 30 , (d) α = 45 , (e) α = 60 , (f) α = 75 , and (g) α = 90 . angles: (a) α = 0°, (b) α = 15°, (c) α = 30°, (d) α = 45°, (e) α = 60°, (f) α = 75°, and (g) α = 90°.

The smooth joint model shown in Figure 4. Once joint plane defined, smooth joint Thesmooth smoothjoint jointmodel modelisisisshown shownin inFigure Figure4. 4.Once Onceaaajoint jointplane planeisisisdefined, defined,aaasmooth smoothjoint jointisisis The defined at contacts between the particles whose centers are lying on the opposite sides of the joint definedat atcontacts contactsbetween betweenthe theparticles particleswhose whosecenters centersare arelying lyingon onthe theopposite oppositesides sidesof ofthe thejoint joint defined plane. At the contacts, the existing parallel bonds will be removed first, and the smooth joints are plane.AtAtthe thecontacts, contacts,the theexisting existingparallel parallelbonds bondswill willbeberemoved removedfirst, first,and andthe thesmooth smoothjoints jointsare are plane. assigned in the direction parallel to the joint plane. These contacts will act in accordance with the assigned in the direction parallel to the joint plane. These contacts will act in accordance with the assigned in the direction parallel to the joint plane. These contacts will act in accordance with the rules defined through the smooth joint model with particular parameter values assigned by the user. rulesdefined definedthrough throughthe thesmooth smoothjoint jointmodel modelwith withparticular particularparameter parametervalues valuesassigned assignedby bythe the user. rules user. The particles having such contacts may overlap or overpass through each other rather than to roll Theparticles particleshaving havingsuch suchcontacts contactsmay mayoverlap overlapororoverpass overpassthrough througheach eachother otherrather ratherthan thantotoroll roll The around one another (Figure 4). around one another (Figure 4). around one another (Figure 4).

(a) (b) (a) (b) Figure Figure4.4.Smooth Smoothjoint jointmodel: model:(a) (a)numerical numericalanalog, analog,and and(b) (b)physical physicalanalog. analog. Figure 4. Smooth joint model: (a) numerical analog, and (b) physical analog.

The newly newly defined contacts act uniformly distributed on The act mechanically mechanicallylike likeaaset setofofelastic elasticsprings springs uniformly distributed The newly defined disc contacts act mechanically like a set of elastic springs uniformly distributed a circular cross-section with thethe center at at thethe contact point. The area AA ofof the on a circular cross-section disc with center contact point. The area thesmooth smoothjoint jointdisc discis on a circular cross-section disc with the center at the contact point. The area A of the smooth joint disc isdefined definedas:as: 2 is defined as: A = πR (10)

A=πR

2

(10)

where R is the disc radius: Processes 2018, 6, 171

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R = λ min( R1 , R2 )

(11)

is a disc radius multiplier (usually set as 1.0), and R1 and R2 are particle radii, as shown in where Rλis the where radius: R = λmin( R1 , R2 ) (11) Figure 4. The smooth joint model can behave with respect to the following basic modes: (i) not bonded where λ is a radius multiplier (usually set as 1.0), and R1 and R2 are particle radii, as shown in Figure 4. and never fail mode, (ii) not bonded and fail in tension mode, (iii) not bonded and fail in shear mode, The smooth joint model can behave with respect to the following basic modes: (i) not bonded and (iv) bonded mode. Through these modes, the smooth joint model could be used to simulate and never fail mode, (ii) not bonded and fail in tension mode, (iii) not bonded and fail in shear mode, various discontinuity conditions. In the present paper, we use the smooth joint model with the and (iv) bonded mode. Through these modes, the smooth joint model could be used to simulate bonded mode (mode (iv)) to mimic the interfaces between material A slices and material B slices. various discontinuity conditions. In the present paper, we use the smooth joint model with the bonded Besides the stiffness provided by the aforementioned springs, the bonded smooth joint model also mode (mode (iv)) to mimic the interfaces between material A slices and material B slices. Besides the possesses both the normal and shear strengths. The shear strength is given by: stiffness provided by the aforementioned springs, the bonded smooth joint model also possesses both the normal and shear strengths. The shear is given τ cj strength = σ nj tan ϕ + cby: (12) j

In Equation (12),

σ nj

j

τ cj = σnj tan ϕ j + c j implies the normal stress acted on the joint surface;

ϕj

(12) implies the

c j normal bonded joint friction while the impliesstress the bonded joint givenϕin MPa. Ifthe thebonded normal In Equation (12),angle; σnj implies acted on thecohesion, joint surface; j implies joint friction angle; while implies the bonded cohesion, given in MPa. If thethe normal shear or shear stress exceeds thec j corresponding bondjoint strength, the joint bond between joint or surfaces stress exceeds the corresponding bond strength, the joint bond between the joint surfaces fails and the fails and the bond stiffness will be removed. As long as the joint bond fails either in shear or tensile bond stiffness will strength be removed. As long asresidual the joint value bond fails in shear or tensile pattern, the shear reduces as its and either the tensile strength will pattern, be set tothe be shear zero. strength reduces as its residual value and the tensile strength will be set to be zero. The residual shear The residual shear strength is a function of the normal stress, σ nj , as well as the joint friction strength is a function of the normal stress, σnj , as well as the joint friction coefficient, µ j . Figure 5 shows Figure 5 shows thesmooth constitutive for the bonded smooth joint model. coefficient, μ j .law the constitutive for the bonded joint law model.

Figure 5. Constitutive law for a bonded smooth joint model: (a) normal force, Fnj , versus normal displacement, Unj ; (b) shear force, Fsj , versus shear displacement, Usj ; and (c) strength envelope.

2.3.2. Calibration and Validation of Smooth Joint Mechanical Parameters The smooth joint parameters consist of the joint normal stiffness, knj ; joint shear stiffness, ksj ; joint tensile strength, σcj ; joint shear strength, τ cj ; joint coefficient of friction, µ j ; and joint dilation angle,

2.3.2. Calibration and Validation of Smooth Joint Mechanical Parameters The smooth joint parameters consist of the joint normal stiffness, k nj ; joint shear stiffness, k sj ; Processes 2018, 6, 171

joint tensile strength,

9 of 26 σ cj ; joint shear strength, τ cj ; joint coefficient of friction, μ j ; and joint dilation

angle, ψ j . These mechanical parameters have been determined through a calibration procedure ψutilizing have beenout determined through calibration procedure utilizing the j . These mechanical the data ofparameters experiments carried by Tien et al. [23]a on simulated transversely isotropic data of experiments carried out by Tien et al. [23] on simulated transversely isotropic rock samples. rock samples. The calibration process is illustrated in Figure 6. During the calibration procedure, sets The process is illustrated in Figure 6.on During the calibration procedure, sets of trial-and-error of calibration trial-and-error tests were implemented numerically developed transversely isotropic rock tests were implemented on numerically developed transversely isotropic rock samples 3) to samples (Figure 3) to reproduce the uniaxial compressive strength data with those(Figure obtained by reproduce the uniaxial compressive strength data with those obtained by Tien et al. [23]. Tien et al. [23]. Start

A. Create transversely isotropic rock models having different interface dip angles from 0° to 90° with an interval of 15° using PFC3D

B. Initiate the smooth joint parameter values and assign them to the interfaces in numerical models

C. Perform a series of uniaxial compressive tests on the created numerical models to estimate the USC values

E. Adjust smooth joint parameters one by one D. Are the estimated UCS

values

satisfactory

in

agreement

with those obtained in

No

physical experiments?

Yes Determination of smooth joint parameters is completed. Figure Figure6.6.Calibration Calibrationprocess processfor fordetermination determinationofofsmooth smoothjoint jointparameters. parameters.

Through the above calibration procedure, the mechanical parameters of the smooth joint model Through the above calibration procedure, the mechanical parameters of the smooth joint model for simulating interfaces are determined as listed in Table 3. Note that the value for each smooth for simulating interfaces are determined as listed in Table 3. Note that the value for each smooth joint joint parameter is not unique. That is to say, each joint parameter in Table 3 is changeable along with parameter is not unique. That is to say, each joint parameter in Table 3 is changeable along with rere-adjusting other joint parameters. Strong coupling effects exist between various joint parameters such that the macro mechanical properties of the simulated transversely isotropic rock models is a result of the complicated interaction among these joint parameters [30]. Based on the calibrated smooth joint mechanical parameters, the UCS of transversely isotropic rock samples are matched well with those obtained by Tien et al. [23]. Table 4 shows a comparison of unconfined strength values (UCS) between the experiments and the particle flow simulations. The comparison given claims the capability of the particle flow modeling method to derive the strength property of transversely isotropic rock

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models under uniaxial compressive loading. The uniaxial compressive strength values obtained through particle flow simulations agree well with those obtained through physical experiments. The particle flow models with the determined intact material and smooth joint mechanical parameters are subsequently utilized to investigate the failure mechanism of transversely isotropic rocks. Table 3. Mechanical parameter values used for the smooth joint model. Mechanical Parameter (N/m3 )

knj ksj (N/m3 ) µj ψ

Determined Value

Mechanical Parameter

Determined Value

σcj (MPa) c j (MPa) ϕj -

23.8 13.0 22.5◦ -

1012

1.81 × 0.79 × 1012 0.78 0

Table 4. Comparison of UCS values between physical experiment and numerical simulation results. Comparison Condition Dip angle, α (deg.) Physical experiment (MPa) Numerical simulation (MPa)

UCS Value for Different Interface Dip Angle 0 53.0 51.6

15 47.0 56.8

30 50.0 51.5

45 38.0 36.7

60 30.0 32.0

75 38.0 36.2

90 66.0 52.4

3. Results and Discussion 3.1. Failure Mechanism of Transversely Isotropic Rocks For better understanding of the failure mechanism of transversely isotropic rocks under uniaxial compressive loading, the created particle flow models, of which the mechanical parameters are determined through the above calibration procedure, were investigated under differing interface dip angles. The interface dip angles were varied from 0◦ to 90◦ with an interval of 15◦ . In this section, failure modes of transversely isotropic rock models are analyzed in terms of newly generated cracks which present the breakage of parallel bonds as well as the smooth joint bonds. Figure 7 plots the failure process of transversely isotropic rock models having different interface dip angles. Note that the breakage of parallel bonds is shown as red (tensile) or blue (shear) crack items, whereas the breakage of smooth joint bonds is marked as magenta (tensile) or black (shear) crack items. As shown in Figure 7, when the interface dip angles, α, are 0◦ and 15◦ , under uniaxial compressive loading the cracks initiate in the soft material (material B) rather than in the hard material (material A). With subsequent compression, the cracks (tensile and shear) increasingly occur both in material A and material B, and eventually coalesce into a macro failure plane penetrating the transversely isotropic rock models. This macro failure plane comes across the simulated interfaces, the rock models fail, and resistance capability was lost. At α = 30◦ , the cracks also initiated in material B under uniaxial compressive loading, as shown in Figure 7. With increasing compression stress, the cracks develop not only in intact materials (material A and material B), but also in interfaces. As a result, the final failure planes were generated along the soft intact material as well as the interfaces. At α = 45◦ , although considerable cracks developed in the soft intact material and interfaces, the failure planes that penetrated and failed the whole rock samples only occurred along the interfaces. When the interface dip angle increased to 60◦ and 75◦ , similar failure processes were observed, as shown in Figure 7, and the final failure planes were generated along the interfaces. When the beddings were vertical (α = 90◦ ), numerous cracks occurred both in material A and material B, and were mainly concentrated in the top and bottom ends of the simulated rock sample at the final stage.

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vertical (α = 90°), numerous cracks occurred both in material A and material B, and were mainly 11 of 26 top and bottom ends of the simulated rock sample at the final stage.

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Figure Figure 7. 7. Failure Failure process process of of transversely transversely isotropic isotropic jointed jointed rock rock models. models.

Figure showsa comparison a comparison of failure the failure modes of artificial and simulated transversely Figure 88shows of the modes of artificial and simulated transversely isotropic ◦ ◦ ◦ isotropic rock samples with dip angles of 0°, 15°, and 30°. This figure shows that the numerically rock samples with dip angles of 0 , 15 , and 30 . This figure shows that the numerically simulated simulated failure modes well matched with those obtained et the al. inphysical the physical experiments failure modes matched withwell those obtained by Tienby et Tien al. in experiments [23]. [23]. Therefore, the numerically simulated be referred to as tensile fractures Therefore, the numerically simulated failurefailure modesmodes couldcould be referred to as tensile fractures across across discontinuities as defined in physical experiments. more specific,ininthis thisstudy study the the failure discontinuities as defined in physical experiments. To To be be more specific, failure mechanism wasfurther further analyzed at mesoscale. the mesoscale. 9 illustrates the microcrack number mechanism was analyzed at the FigureFigure 9 illustrates the microcrack number evolution evolution of transversely isotropic rock models having of 30 0°,◦ . 15°, and 30°. of transversely isotropic rock models having interface dipinterface angles ofdip 0◦ ,angles 15◦ , and As shown in As shown in this figure, under uniaxial compressive loading, the crack number increased slowly at this figure, under uniaxial compressive loading, the crack number increased slowly at the beginning, the and increased sharply when reached the axialthe stress reached the peak. andbeginning, increased sharply when the axial stress peak. Additionally, theAdditionally, crack numberthe in crack intact number in intact materials was consistently larger than that in interfaces. That is to say, the intact materials was consistently larger than that in interfaces. That is to say, the intact materials rather than materials rather than the the interfaces dominated theofresistance of the whole rock transversely the interfaces dominated resistance capability the wholecapability transversely isotropic models. isotropic models. as intact shownmaterials, in Figurethe 10,tensile in intact materials, tensile crack number Moreover,rock as shown in Moreover, Figure 10, in crack numberthe was dramatically larger was dramatically larger than that of shear cracks. Therefore, tensile failure of an intact material is thought to be the main cause of the transversely isotropic rock models failing when the interface dip

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than that of6,shear Therefore, Processes 2018, x FORcracks. PEER REVIEW

tensile failure of an intact material is thought to be the main 13 cause of 26 of the transversely isotropic rock models failing when the interface dip angles are relatively low, e.g., Processes FOR PEER REVIEW 13 of 26 α= 0◦ and 156,◦ .x Taking the effect into consideration, this type of failure mode, herein, angles are2018, relatively low, e.g., α = of 0° interfaces and 15°. Taking the effect of interfaces into consideration, this ◦ was redefined as the tensile across interfaces. As the interface dipinterfaces. angle increased 0 todip 30◦ type of failure mode, herein,failure was redefined as the tensile failure across As thefrom interface angles are relatively low, e.g., α = 0° and 15°. Taking the effect of interfaces into consideration, this through 15◦ , thefrom difference of crack number in intact materials andnumber in interfaces decreased, as and shown angle increased 0° to 30° through 15°, the difference of crack in intact materials in type of failure mode, herein, was redefined as the tensile failure across interfaces. As the interface dip ◦ in Figure 9. Furthermore, at α =in 30Figure , the crack number in interfaces equaled that in in interfaces interfaces. interfaces decreased, as shown 9. Furthermore, at α = 30°,almost the crack number angle increased from 0° to 30° through 15°, the◦ difference of crack number in intact materials and in Therefore, whenthat the interface dip angle was 30when the transversely rock sample partially fails in almost equaled in interfaces. Therefore, the interfaceisotropic dip angle was 30° the transversely interfaces decreased, as shown in Figure 9. Furthermore, at α = 30°, the crack number in interfaces the moderock of tensile failure across interfaces. isotropic sample partially fails in the mode of tensile failure across interfaces. almost equaled that in interfaces. Therefore, when the interface dip angle was 30° the transversely isotropic rock sample partially fails in the mode of tensile failure across interfaces.

(a) (b) (c) (a) (b) (c) Figure Figure 8. 8. Failure Failuremodes modesof oftransversely transverselyisotropic isotropicjointed jointedrock rockmodels modelshaving havingdip dipangles anglesof of(a) (a)αα ==0°, 0◦ , ◦ ◦ Figure 8. Failure modes of transversely isotropic jointed rock models having dip angles of (a) α = 0°, (b) (c)(c) α =α30°. (Note that the failure imagesimages of physical samples samples were adopted Ref. [23]). (b)αα==15°, 15 and , and = 30 . (Note that the failure of physical werefrom adopted from (b)[23]). α = 15°, and (c) α = 30°. (Note that the failure images of physical samples were adopted from Ref. [23]). Ref.

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(c) (c) Figure Cracknumber numberevolution evolution of transversely transversely isotropic dip angles of (a) α =α0°, Figure 9.9.Crack isotropicrock rockmodels modelswith with dip angles of (a) = 0◦ , Figure 9. Crack number evolution of transversely isotropic rock models with dip angles of (a) α = 0°, ◦ , and ◦. α =1515°, and(c) (c)αα==30 30°. (b)(b) α= (b) α = 15°, and (c) α = 30°.

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(b) (c) (a) Figure 10. Intact material crack number evolution of transversely isotropic jointed rock models having Figure 10. crack evolution Figure 10.Intact Intact crack number evolution transverselyisotropic isotropicjointed jointedrock rockmodels modelshaving having dip angles of (a)material αmaterial = 0°, (b) α = number 15° and (c) α = 30°.ofoftransversely dip dipangles anglesofof(a) (a)αα==0°, 0◦(b) , (b)αα==15° 15◦and and(c)(c)α α= =30°. 30◦ .

Figure 11 displays the comparison of failure modes of artificial and simulated transversely Figure 11models displays thecomparison comparison failure modes of artificial and simulated transversely Figure 11 displays the ofof failure modes and transversely isotropic rock having dip angles of 45°, 60°, and 75°.of It artificial shows that thesimulated numerically simulated ◦ , 60◦ , and 75◦ . It shows that the numerically simulated isotropic rock models having dip angles of 45 isotropic rock models having dip agreement angles of 45°, 60°,those and obtained 75°. It shows that the numerically simulated failure modes were also in good with through physical experiments [23]. failure modes were also in good agreement with those obtained through physical experiments [23]. failure modes were also in good agreement with those obtained through physical experiments [23]. The numerically simulated failure modes could be defined as a sliding failure along discontinuities The numerically simulated failure modes could be defined as a sliding failure along discontinuities in The numericallywith simulated failureexperiments. could Furthermore, be defined as Figure a sliding along in accordance the physical 12failure displays thediscontinuities crack number accordance the the physical experiments. Furthermore, 12 displays theof crack number in accordance with physical experiments. Furthermore, Figure displays the crack number evolution ofwith transversely isotropic rock models having Figure interface dip12 angles 45°, 60°, andevolution 75°. As ◦ , 60◦ , and 75◦ . As shown in of transversely isotropic rock models having interface dip angles of 45 evolution of transversely isotropic rock models having interface dip angles of 45°, 60°, and 75°. As shown in this figure, under compressive loading, the crack number increases slowly for a relatively this figure, under loading,when the crack increases slowly aslowly relatively period, shown in this figure, under compressive loading, the crack number increases forContrary along relatively long period, andcompressive increases sharply the number axial stress arrives at for peak stress. to and increases sharply when the axial stress arrives at peak stress. Contrary to transversely isotropic long period, and increases sharply when the axial stress arrives at peak stress. Contrary to transversely isotropic rock samples having low dip angles (α = 0°, 15°, and 30°), the transversely ◦ ), the transversely isotropic rock samples rock samples having low dip angles (α dip = 0◦low , 15◦dip , and transversely isotropic samples having (α = and 0°, 15°, 30°), the transversely isotropic rock samplesrock having medium angles (αangles =30 45°, 60°, 75°)and generated more cracks in ◦ ◦ ◦ having medium angles (α medium = 45 ,That 60 dip , and ) generated more cracks in interfaces thancracks in intact isotropic rock having (αthe = 45°, 60°, and 75°) generated more in interfaces thansamples indip intact materials. is angles to75say, interfaces rather than the intact materials materials. That is to say, the interfaces rather than the intact materials dominated the resistance interfaces than in intact materials. That is to say, the interfaces rather than the intact materials dominated the resistance capability of the whole rock models. Moreover, as shown in Figure 13, in capability the of whole rock models. as shown inof Figure 13, incrack. interfaces, shear crack dominated thethe resistance capability ofMoreover, the whole rock models. Moreover, as shown inthe Figure in interfaces, shear crack number was much larger than that the tensile Therefore, the13, shear number was much larger than that of the tensile crack. Therefore, the shear failure of interfaces was the interfaces, the shear crack number was much larger than that of the tensile crack. Therefore, the shear failure of interfaces was the main mechanism of the transversely isotropic rock models failing when main mechanism ofwas thewere transversely isotropic rock models when therock interface dip angleswhen were failure of interfaces the main mechanism the60°, transversely isotropic models failing the interface dip angles medium, e.g., α =of 45°, andfailing 75°. To emphasize the failure mechanism ◦ , 60◦ , and 75◦ . To emphasize the failure mechanism of the transversely isotropic medium, e.g., α = 45 the interface dip angles were medium, e.g.,this α = 45°, and 75°. To emphasize failure mechanism of the transversely isotropic rock models, type60°, of failure mode, herein, is the redefined as the shear rock models, this type of failure mode, herein, is redefined as the shear failure along interfaces. of the transversely isotropic rock models, this type of failure mode, herein, is redefined as the shear failure along interfaces. failure along interfaces.

(a) (b) (c) (a) (b) (c) Figure11. 11.Failure Failuremodes modesof oftransversely transverselyisotropic isotropic jointed jointed rock rock models models having having dip dip angles angles of (a) (a) αα ==45°, 45◦ , Figure ◦ ◦ Figure 11. Failure modes of transversely isotropic jointed rock models having dip angles of (a) α = 45°, (b) αα == 60°, 60 , and and (c) (c) αα == 75°. 75 . (Note (Notethat thatthe the failure failure images images of of physical physical samples samples were were adopted adopted from from (b) (b) α[23]). = 60°, and (c) α = 75°. (Note that the failure images of physical samples were adopted from Ref. [23]). Ref. Ref. [23]).

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dip dip angles angles of of (a) (a) αα ==45°, 45 ,(b) (b)αα==60° 60 and and(c) (c)αα==75°. 75 .

Figure 14 plots the comparison of failure modes of transversely isotropic rock models with Figure 14 the of failure modes of transversely isotropic rock models with vertical Figure 14 plots plots thecomparison comparison of failure transversely isotropic rock models with vertical interfaces in numerical simulations and inmodes physicalofexperiments [23]. In the numerically interfaces in numerical simulations and in physical experiments [23]. In the numerically simulated vertical interfaces in numerical simulations andobserved in physical experiments [23]. In numerically simulated rock model, there were many more cracks than those in the physical rockthe model. rock there were many more cracks observed those in thethose physical rock model. Thismodel. might simulated model, there were many more cracksthan observed than in the physical Thismodel, mightrock be because the particle flow modeling method could recognize more microcracks thanrock the be because thethe particle flow modeling method could recognize more the eyes eyes [18]. On other the hand, Figure 15 displays that before the peakrecognize stress, microcracks the more number ofthan interface This might be because particle flow modeling method could microcracks than [18]. the cracks was larger than that of intact material crack, which accounts for the failure planes mainly On the other hand, Figure 15 displays that before the peak stress, the number of interface cracks eyes [18]. On the other hand, Figure 15 displays that before the peak stress, the number of interface observed vertical interfaces of artificial rock models having a dip angle 90°.the This failure mode,observed was larger that of intact material crack, which accounts for the of failure planes mainly cracks wasinthan larger than that of intact material crack, which accounts for failure planes mainly ◦ herein, is referred to as tensile failure along interfaces. in vertical interfaces of artificial rock models having a dip angle of 90 . This failure mode, herein,

observed in vertical interfaces of artificial rock models having a dip angle of 90°. This failure mode, is referred to as tensile herein, is referred to as failure tensile along failureinterfaces. along interfaces.

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(a) (b) (a) (b) Figure 14. Failure modes of transversely isotropic jointed rock model having dip angle of 90°: ◦ Figure 14. Failure modes modes of transversely isotropic isotropic jointed jointed rock rock model model having having dip angle angle of of 90 Figure 14. Failure 90°:: (a) experiment sample, andof(b)transversely numerical sample. (Note that the failure images ofdip physical samples (a) experiment sample, and (b) (Note that (a) experiment sample, (b) numerical numerical sample. sample. (Note that the the failure failure images images of of physical physical samples samples were adopted from Ref.and [23]). were adopted from Ref. [23]). were adopted from Ref. [23]).

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Figure 15. Evolution of crack number in transversely isotropic jointed rock models having dip angle Figure 15. Evolution of crack number in transversely isotropic jointed rock models having dip angle of 90◦ . Figure of 90°. 15. Evolution of crack number in transversely isotropic jointed rock models having dip angle of 90°.

3.2. Effect of Interfaces on Strength of Transversely Isotropic Rocks 3.2. Effect of Interfaces on Strength of Transversely Isotropic Rocks 3.2. Effect of Interfaces Strength of Transversely Rockson the failure behavior of transversely As aforesaid, theonrole of interfaces played Isotropic an influence As aforesaid, thefailure role ofmode interfaces played an influence on thewere failure behavior of transversely isotropic rocks. The of transversely isotropic rocks classified into three types, As aforesaid, the role of interfaces played an influence on the failure behavior of transversely isotropic rocks. The failure mode of transversely isotropic rocks were classified into three along types, which are tensile failure across interfaces, shear failure along interfaces, and tensile failure isotropic rocks. The failure mode of transversely isotropic rocks were classified into failure three types, which are tensile failure across interfaces, shear failure along interfaces, and tensile along interfaces. The failure mode are significantly dependent on the interface dip angle. However, which are tensile failure across interfaces, shear failure along interfaces, and tensile failure along interfaces. The failure mode are significantly dependent on the interface dip angle. However, in in laboratory experiments, the interfaces between material A and material B were somehow ignored interfaces. The failure mode are significantly dependent on the interface dip angle. However, in laboratory experiments, the interfaces between material A This and material B were somehow ignored without taking their mechanical properties into account [23]. section thus focuses on the effect of laboratory experiments, the interfaces between material A and material Bthus werefocuses somehow ignored without taking their mechanical properties into account [23]. This section on the effect interfaces on the strength behavior of transversely isotropic rock models. without taking mechanical properties into account [23]. This section thus focuses on the effect of interfaces ontheir the strength of transversely models. For comparison, Figurebehavior 16 displays the failure isotropic modes ofrock transversely isotropic rock models of interfaces on the strength behavior of transversely isotropic rock models. Forinterfaces. comparison, displays therock failure modes of transversely isotropicofrock models without ThisFigure figure16 shows that the models having only inter-beddings material A For interfaces. comparison, Figure 16shows displays the failure modes of transversely isotropic of rock models without This figure that the rock models having only inter-beddings material A and material B still behaved anisotropically. The microcracks mainly concentrated in the soft material, without interfaces. This figure shows that the rock models having only inter-beddings of material A andmaterial materialB,B and still behaved anisotropically. microcracks mainly the softetmaterial, i.e., most of them were tensileThe cracks (red), which was concentrated also observedinby Kim al. [12] and material B still behaved anisotropically. The microcracks mainly concentrated in by theKim soft et material, i.e., material B, and most of them were tensile cracks (red), which was also observed al. [12] through experimental tests on Berea sandstone. However, as shown in Figure 17, which illustrates the i.e., material B, and most of them weresandstone. tensile cracks (red), which was also observed by Kimillustrates et al. [12] through experimental tests on Berea However, as shown in Figure 17, which comparison of UCS of transversely isotropic rocks with and without weak interfaces, the UCS values through experimental tests on Berea sandstone. However, asand shown in Figure 17, which illustrates therock comparison of UCS of transversely isotropic rocks with interfaces, the UCS of models without weak interfaces were highly beyond those without obtainedweak in physical experiments. the comparison of UCS of transversely isotropic rocks with and without weak interfaces, the UCS values of rock models without weak interfaces were highly beyond those obtained in physical On the contrary, the UCS values of rock models with weak interfaces agreed very well with the values of rockOnmodels without weak interfaces weremodels highlywith beyond those obtained in very physical experiments. the contrary, the UCS values of rock weak interfaces agreed well physical experiment results. In addition, the shape of the compressive strength interface dip angle experiments. On the contrary,results. the UCS of rock models weak interfaces agreed very well with the physical experiment Invalues addition, the shape of with the compressive strength interface dip with the physical experiment results. In addition, the shape of the compressive strength interface dip angle curve obtained in physical experiments could be classified as a shoulder type [31], which means angle curve obtained in physical experiments could be classified as a shoulder type [31], which means

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curve obtained in physical experiments could be classified as a shoulder type [31], which means that the compressive strength was relatively high with low high and high diplow angles, while the compressive strength thatthe thecompressive compressive strength was relatively high with low and high dip angles,while while the that strength was relatively with and high dip angles, the was relatively low with medium dip angles. Some typical examples of such curves are verified by compressive strength was relatively low with medium dip angles. Some typical examples of such compressive strength was relatively low with medium dip angles. Some typical examples of such Yasar [32], Autio etby al. [33], Nasseri et al.etet [34], Cho et al. [35], and Fjaer and Nes [36]. InFjaer these studies, curves areverified verified byYasar Yasar [32],Autio Autio [33],Nasseri Nasseri al.[34], [34], Cho al.[35], [35], and Fjaerand andNes Nes curves are [32], al.al.[33], etetal. Cho etetal. and the research objects included various rocks such as sandstone, siltstone, shale, gneiss, and schist. [36].InInthese thesestudies, studies,the theresearch researchobjects objectsincluded includedvarious variousrocks rockssuch suchasassandstone, sandstone,siltstone, siltstone,shale, shale, [36]. In most of these works, it was noted that bedding planes acted as planes of weakness. The particle gneiss, and schist. In most of these works, it was noted that bedding acted as planes of weakness. gneiss, and schist. In most of these works, it was noted that bedding planes acted as planes of weakness. flow modeling results of transversely isotropic rocks with interfaces fit interfaces this type of the compressive Theparticle particleflow flow modeling resultsofoftransversely transversely isotropic rockswith with interfaces this typeofofthe the The modeling results isotropic rocks fitfit this type strength interface dip angle curve. That is to say, the interfaces were very necessary elements to build compressive strength interface dip angle curve. That say, the interfaces were verynecessary necessary compressive strength interface dip angle curve. That isistotosay, the interfaces were very transversely isotropic rock models with particle flow modeling method. To further understand the elements to build transversely isotropic rock models with particle flow modeling method. To further elements to build transversely isotropic rock models with particle flow modeling method. To further effect of interfaces on the failure behavior of transversely isotropic rocks, the mechanical properties of understandthe theeffect effectofofinterfaces interfaceson onthe thefailure failurebehavior behaviorofoftransversely transverselyisotropic isotropicrocks, rocks,the the understand interfaces were systematically investigated as follows. mechanical properties of interfaces were systematically investigated as follows. mechanical properties of interfaces were systematically investigated as follows.

(a) (a)

(b) (b)

(c)(c)

(d) (d)

(e) (e)

(f)(f)

(g) (g)

Figure16. 16.Failure Failure modesofoftransversely transverselyisotropic isotropicrocks rocks without weak interfaces: (a)α α =0°, 0°, (b) α= =15°, 15°, Figure Failure modes modes without weak interfaces: (a) α = 0◦ , (b) α α 15◦ , (c) α = 30°, (d) α = 45°, (e) α = 60°, (f) α = 75°, and (g) α = 90°. ◦ ◦ ◦ ◦ ◦ (c) α α = 30°, 30 , (d) (d) αα == 45°, 45 ,(e) (e)αα ==60°, 60 ,(f) (f)αα==75°, 75 and , and(g) (g)αα==90°. 90 . 7575

Labexperiment experiment Lab With weakinterfaces interfaces With weak

6565 UCS (MPa) UCS (MPa)

Withoutweak weakinterfaces interfaces Without

5555 4545 3535 2525 00

1515

3030

4545 6060 Dip angle (deg.) Dip angle (deg.)

7575

9090

Figure 17. Comparison of UCS of transversely isotropic rocks with and without weak interfaces. Figure17. 17.Comparison ComparisonofofUCS UCSofoftransversely transverselyisotropic isotropicrocks rockswith withand andwithout withoutweak weakinterfaces. interfaces. Figure

3.2.1. Joint Normal Stiffness 3.2.1.Joint JointNormal NormalStiffness Stiffness 3.2.1. To look into the effect of the interface property on the failure behavior of transversely isotropic Tolook lookinto intojoint theeffect effectofofthe theinterface interface propertyon on,the thefailure failure behaviorofoftransversely transversely isotropic rock To models, the normal stiffness of interfaces, knj was first investigated. A set of jointisotropic normal the property behavior 13 3 13 3 13 3 13 3 were stiffness values ofjoint 0.15 ×normal 10 N/m , 0.50 10 N/mk,knj1.50 × 10 N/m , and 5.00 10ofofjoint N/m , was first investigated. set jointnormal normal rockmodels, models, the jointnormal stiffness interfaces, nj, was first investigated. AA× set rock the stiffness ofof× interfaces, numerically tested, respectively, for the transversely isotropic rock samples having various interface 13 3 13 3 13 3 13 3 were stiffnessvalues valuesofof0.15 0.15× ×1010 N/m 0.50× ×1010 N/m 1.50× ×1010 N/m and5.00 5.00× ×1010 N/m 13 N/m 3, ,0.50 13 N/m 3, ,1.50 13 N/m 3, ,and 13 N/m 3 were stiffness ◦ ◦ ◦ dip angles from 0 torespectively, 90 with an for interval of 15 (Figure 3). Meanwhile, the joint shear stiffness, ksj , numerically tested, thetransversely transversely isotropic rocksamples samples having various interface numerically tested, respectively, for the isotropic rock having various interface was set to be 1/3knj for all tests. Furthermore, the other smooth joint parameters were kept constant, dip anglesfrom from0°0°toto90° 90°with withanan intervalofof15° 15°(Figure (Figure3).3).Meanwhile, Meanwhile, thejoint jointshear shearstiffness, stiffness, ksj sj, , dip i.e., angles σcj = 23.8 MPa, ϕj = 22.5◦ ,interval c j = 13.0 MPa, and µ j = 0.40. Duethe to the planar surface ofkthe interfaces inbethe experiments, the joint dilation angle, ψjoint , was always set to be 0. Figure 18 1 /13/physical k3knj nj for wasset settotobe forallall tests.Furthermore, Furthermore, the othersmooth smooth parameters were kept constant, jjoint was tests. the other parameters were kept constant, shows the uniaxial compressive strength variation of transversely isotropic rock samples with different c = 13 . 0 MPa 22.5° °, , c j =j 13.0 MPa , and 23.8MPa MPa, ,ϕϕ j==22.5 0.40. Due i.e., σσcj cj==23.8 , and μμ j==0.40 . Duetotothe theplanar planarsurface surfaceofof i.e., j j

, wasalways alwaysset settotobebe0.0.Figure Figure1818 theinterfaces interfacesininthe thephysical physicalexperiments, experiments,the thejoint jointdilation dilationangle, angle,ψψ,j was the j showsthe theuniaxial uniaxialcompressive compressivestrength strengthvariation variationofoftransversely transverselyisotropic isotropicrock rocksamples sampleswith with shows

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different joint normal stiffness. It is shown that the joint normal stiffness had a dramatic effect on the joint normal stiffness. It is shown that the joint normal stiffness had a dramatic effect on the uniaxial uniaxial compressive strength (UCS) of transversely isotropic rocks. Particularly, when the interface compressive strength (UCS) of transversely isotropic rocks. Particularly, when the interface dip angle dip angle was less than 30° (i.e., α = 0° and 15°), the UCS basically increased with the increment of was less than 30◦ (i.e., α = 0◦ and 15◦ ), the UCS basically increased with the increment of joint normal joint normal stiffness. Whereas, when the interface dip angle was equal or more than 30° (α ≥ 30°), stiffness. Whereas, when the interface dip angle was equal or more than 30◦ (α ≥ 30◦ ), the UCS the UCS decreased with the increment of joint normal stiffness. Moreover, as the joint normal stiffness, decreased with the increment of joint normal stiffness. Moreover, as the joint normal stiffness, knj , 3, the numerically obtained set of UCS values of transversely 13 N/m k njset , was to × be100.50 × 10313, the N/m was to beset 0.50 numerically obtained set of UCS values of transversely isotropic isotropic rock samples close to thoseinobtained physical experiments [23]. rock samples were close were to those obtained physicalin experiments [23]. 75

Lab experiment 13 knstiffness×10 N/m 3 nj = 5.0 × 10

65

UCS (MPa)

13 knstiffness×3 N/m 3 nj = 1.5 × 10 13 kStandard N/m 3 nj = 0.5 × 10

55

13 k nstiffness×0.3 N/m 3 nj = 0.15 × 10

45 35 25 15 0

15

30

45 Dip angle (deg.)

60

75

90

Figure 18. UCS variation of transversely isotropic rock samples with different joint normal stiffness. Figure 18. UCS variation of transversely isotropic rock samples with different joint normal stiffness.

3.2.2. Joint Shear Stiffness 3.2.2. Joint Shear Stiffness Keeping the joint normal stiffness constant (i.e., knj = 0.5 × 1013 N/m3 ), the joint shear stiffness, ksj , is subsequently adjusted fromstiffness 1/10knj constant to 1.0knj through Furthermore, other smooth nj . × = 0.5 1013 N / m3the ), the joint shear Keeping the joint normal (i.e., k nj1/3k joint parameters were set to be constant, that is to say, σcj = 23.8 MPa, ϕ j = 22.5◦ , c j = 13.0 MPa, / 10k nj to 1.0k nj through 1 / 3k nj . Furthermore, the k sj Figure , is subsequently adjusted from 1of stiffness, and µ j = 0.40. 19 shows the UCS variation transversely isotropic rock samples with different joint shear stiffness. shown in this figure, the be joint shear stiffness influence other smooth jointAsparameters were set to constant, that ispossessed to say, aσsimilar cj = 23.8 MPa , to the joint normal stiffness with respect to the UCS values of transversely isotropic rock samples. 13.0angle MPawas ϕ j =the 22.5 ° , c j =dip , and Figure shows the UCS variation transversely j = 0.40 When interface lessμthan 30◦ , .the UCS19 basically increased with theofincrement of ◦ ), joint shear rock stiffness. Meanwhile, when the interface angle was equal to more than the 30◦ joint (α ≥ 30 isotropic samples with different joint shear dip stiffness. As shown inorthis figure, shear stiffness possessedwith a similar influenceoftojoint the shear joint normal stiffness with the UCS values the UCS decreased the increment stiffness. Specially, asrespect the jointtoshear stiffness, ksjof , transversely isotropic samples.obtained When the dip angle was less thanisotropic 30°, the rock UCS samples basically was set to be 1/3k numerically setinterface of UCS values of transversely nj , therock increased increment Meanwhile, when the agreed wellwith withthe those obtainedofbyjoint Tienshear et al. stiffness. in the physical experiments [23].interface dip angle was equal tojoint or more than 30° (α and ≥ 30°), theshear UCSstiffness decreased the increment shearvariation stiffness. The normal stiffness joint hadwith a similar influenceof onjoint the UCS ◦ of transversely isotropic rock models. When the interface dip angle was low (e.g., α = 0 and 15◦ ), Specially, as the joint shear stiffness, k sj , was set to be 1 / 3k nj , the numerically obtained set of UCS the uniaxial loading compressed the inter-layered materials (material A, material B, and interfaces) to values of transversely isotropic rock samples agreed well with those obtained by Tien et al. in the deform normally, which was dominated by the normal stiffness, and further to deform tangentially, physical experiments [23]. which was dominated by the shear stiffness because of the Poisson’s ratio effect, which means that The joint normal stiffness and joint shear stiffness had a similar influence on the UCS variation the normal deformation would induce the tangential deformation. If the joint stiffness was low, of transversely isotropic rock models. When the interface dip angle was low (e.g., α = 0° and 15°), the the interfaces would have a large stiffness difference compared with the adjacent intact materials uniaxial loading compressed the inter-layered materials (material A, material B, and interfaces) to (material A and material B), which resulted in a significant deforming difference between them. deform normally, which was dominated by the normal stiffness, and further to deform tangentially, Figure 20 shows the force chains in the transversely isotropic rock models having horizontal which was dominated by the shear stiffness because of the Poisson’s ratio effect, which means that interfaces (α = 0◦ ) with different joint normal stiffness values at the peak stress. It shows that the normal deformation would induce the tangential deformation. If the joint stiffness was low, the the deforming difference caused tensile force chains in the interfaces, and the lower joint stiffness interfaces would have a large stiffness difference compared with the adjacent intact materials (e.g., knj = 0.15 × 1013 N/m3 ) introduced more tensile force chains. These tensile force chains further (material A and material B), which resulted in a significant deforming difference between them. cause tensile cracks and eventually lower the UCS values of transversely isotropic rock samples. Figure 20 shows the force chains in the transversely isotropic rock models having horizontal Therefore, the UCS decreased with the reduction of joint stiffness at low dip angles (i.e., α = 0◦ and 15◦ ), interfaces (α = 0°) with different joint normal stiffness values at the peak stress. It shows that the deforming difference caused tensile force chains in the interfaces, and the lower joint stiffness (e.g.,

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k nj = 0.15 × 1013 N/m 3 ) introduced more tensile force chains. These tensile force chains further cause

tensile cracks lower thetensile UCS values transversely isotropic rockchains samples. Therefore, ) introduced more force of chains. These tensile force further cause k nj = 0.15 × 1013 and N/m 3eventually 19 of 26 the UCS decreased with the reduction of joint stiffness at low dip angles (i.e., α = 0° and 15°), as shown tensile cracks and eventually lower the UCS values of transversely isotropic rock samples. Therefore, in Figures 18 and 19. Onthe thereduction other hand, when the interface dip angles angle was (α ≥15°), 30°),asboth the the UCS decreased with of joint stiffness at low dip (i.e., higher α = 0° and shown ◦ high joint stiffness and the low joint stiffness induced substantial tensile force chains in the interfaces. Figures and 19. hand, interface dip angle higher (α ≥ 30°), both asinshown in 18 Figures 18 On andthe 19. other On the otherwhen hand,the when the interface dipwas angle was higher (α ≥ 30 the ), 13 However, the lower stiffness (e.g., increased the deforming capability k njstiffness = 0.15 × 10 N/m 3 ) substantial high joint stiffness andjoint the and low jointlow stiffness induced substantial tensile force chains inchains the interfaces. both the high joint stiffness the joint induced tensile force in theof 3 13 N/m these interfaces andthe postponed theirstiffness failure. This process motivated more3the material to contribute However, the lower joint stiffness (e.g., ) 10 increased deforming of k nj(e.g., = 0.15 N/m interfaces. However, lower joint k×nj10=13 0.15 × )intact increased the capability deforming to theinterfaces bearing capability ofand the their whole rocktheir samples. Therefore, the UCS increased with the reduction capability of these interfaces postponed This process motivated intact material these and postponed failure. Thisfailure. process motivated more intact more material to contribute of joint stiffness at higher dip angles (i.e., α ≥ 30°). toto contribute to the bearing capability of the whole rock samples. Therefore, the UCS increased with the bearing capability of the whole rock samples. Therefore, the UCS increased with the reduction ◦ the of joint stiffness higher(i.e., dip αangles of reduction joint stiffness at higher dipatangles ≥ 30°).(i.e., α ≥ 30 ).

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70

Lab experiment

UCSUCS (MPa) (MPa)

70 60

Lab sstiffness×3 k sj =experiment k nj sstiffness×3 kStandard k /nj3k nj sj = 1 sj kStandard k njk nj sjsj = ksstiffness×0.3 =11//310 ksstiffness×0.3 sj = 1 / 10 k nj

60 50 50 40 40 30 30 20

45 60 75 90 Dip angle (deg.) 0 15 30 45 60 75 90 Dip angle (deg.) Figure 19. UCS variation of transversely isotropic rock samples with different joint shear stiffness. Figure 19. UCS variation of transversely isotropic rock samples with different joint shear stiffness. Figure 19. UCS variation of transversely isotropic rock samples with different joint shear stiffness. 20

0

15

(a)

30

(b)

(a) (b)having horizontal interfaces with Figure 20. Force chains in the transversely isotropic rock models 3 13 Figurejoint 20. Force chains in the transversely horizontal interfaces with different normal stiffness at peak stress: (a)isotropic knj = 0.15rock × 10models N/m3having , and (b) knj = 5.0 × 1013 N/m . 3 Note that 20. the Force black and chain imply force force, different joint chain normal stiffness at compression peak stress: (a)andktension , and and the (b) = 0.15 ×1013respectively, N/minterfaces Figure chains in red the transversely isotropic rock models horizontal with njhaving 13 3 force quantity is to the chain thickness. 13proportional 3 different joint normal at peak stress: (a) imply and (b) k nj = 0compression .15 ×10 N/m . Notestiffness that the black chain and red chain force, and tension

k nj = 5.0 ×10 N/m

3 force, the force quantity proportional to theimply chain compression thickness. force and tension . Note that the black is chain and red chain k nj = 5respectively, .0Joint ×1013Cohesion N/mand 3.2.3. Bonded force, respectively, and the force quantity is proportional to the chain thickness.

Further investigation was implemented to look into the effect of interface strength parameters, 3.2.3. Bonded Joint Cohesion including joint shear strength, τ cj , joint coefficient of friction, µ j , and joint tensile strength, σcj , on the 3.2.3.Further Bondedinvestigation Joint Cohesion was implemented tomodels. look intoInthe effect of interface strength strength behavior of transversely isotropic rock accordance with Equation (9),parameters, the joint Further investigation was implemented to look into the effect of interface strength parameters, shear strength consisted of bonded c j ,ofand bonded friction angle, ϕ j . Herein, τ cj joint σ cj , , jointcohesion, coefficient friction, μ joint including joint shear strength, j , and joint tensile strength, the bonded joint cohesion was first adjusted to be 4.0 MPa, 8.0 MPa, 13.0 MPa, and 16.0 MPa, while the τ cj , joint isotropic coefficient of friction, μ accordance tensile strength, including joint shear strength, cj , on the strength behavior of transversely rock models. In Equation (9),σthe j , and joint with other joint parameters were kept constant, i.e., knj = 0.5 × 1013 N/m3 , ksj = 1/3knj , σcj = 23.8 MPa, on the strength behavior of transversely isotropic rock models. In accordance with Equation (9), the ϕ j = 22.5◦ , and µ j = 0.40.

joint shear strength consisted of bonded joint cohesion, c j , and bonded joint friction angle,

ϕj .

Herein, the bonded joint cohesion was first adjusted to be 4.0 MPa, 8.0 MPa, 13.0 MPa, and 16.0 MPa, while the other joint parameters were kept constant, i.e.,

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k nj = 0.5 × 1013 N / m3 , k sj = 1 / 3k nj , 20 of 26

σ cj = 23.8 MPa , ϕ j = 22.5° , and μ j = 0.40 .

Figure2121shows showsthe theUCS UCSvariation variationofoftransversely transverselyisotropic isotropicrock rocksamples sampleswith withdifferent differentbonded bonded Figure jointcohesion. cohesion. As As shown joint cohesion, influence onon the joint shown in inthis thisfigure, figure,the thebonded bonded joint cohesion,c jc,j ,had hada avarying varying influence the UCS values depending on the interface angle. Specifically, when the interface dip angles, α, UCS values depending on the interface dip dip angle. Specifically, when the interface dip angles, α, were were 0◦ and 15◦ ,UCS the UCS values decreased slightly the decrement of bonded cohesion. 0° and 15°, the values decreased slightly with with the decrement of bonded joint joint cohesion. Even Even though the transversely isotropic rock samples withdip dipangles anglesofof0° 0◦ and and 15° 15◦ failed in though the transversely isotropic rock samples with in the themode modeofof tensile tensilefailure failureacross acrossinterfaces interfaces(Figure (Figure8), 8),the theshear shearcracks cracksoccurred occurredinininterfaces, interfaces,as asshown shownin inFigure Figure22. 22. This Thisfigure figuredisplays displaysthe theinterface interfaceshear shearcrack crackproportion proportionatatpeak peakstress stressvarying varyingwith withbonded bondedjoint joint cohesion cohesionofof13.0 13.0MPa MPaand and4.0 4.0MPa. MPa.For Foraabetter betterunderstanding, understanding,the theshear shearcrack cracknumber numberinininterfaces interfaces was byby thethe total crack number in the rockrock model. WithWith the decrement of bonded joint wasnormalized normalized total crack number in whole the whole model. the decrement of bonded cohesion, the shear in the interfaces increased significantly. At α = 0◦At , when thewhen bonded joint cohesion, thecrack shearnumber crack number in the interfaces increased significantly. α = 0°, the joint cohesion 13.0 MPa to 13.0 4.0 MPa, crack proportion increased from bonded jointdecreased cohesion from decreased from MPathe to interface 4.0 MPa,shear the interface shear crack proportion increased from 5.1% to 31.8%.ofThe occurrence of more interface shear cracks accelerated the failure 5.1% to 31.8%. The occurrence more interface shear cracks accelerated the failure of the whole rock ◦ , when the of the whole rock at model. atbonded α = 15°, joint whencohesion the bonded joint cohesion decreased from 13.0 model. Moreover, α = 15Moreover, decreased from 13.0 MPa to 4.0 MPa, MPa to 4.0 shear MPa, crack the interface shear crack proportion from 5.9% the to 63.0%. Therefore, the the interface proportion increased from 5.9%increased to 63.0%. Therefore, shear failure, in this shear failure,dominated in this case, became dominated the was bonded joint cohesion wasmode small, and the case, became when the bonded jointwhen cohesion small, and the failure changed failure mode changed from tensile failure interfaces to shear failure along interfaces. from tensile failure across interfaces to shearacross failure along interfaces. 85

Lab experiment cCj=1.59e7 j = 16.0 MPa cCj=1.3e7 j = 13.0 MPa cCj=0.8e7 j = 8.0 MPa cCj=0.4e7 j = 4.0 MPa

75

UCS (MPa)

65 55 45 35 25 15 5 0

15

30

45 60 Dip angle (deg.)

75

90

Figure 21. Strength variation of transversely isotropic rock samples with different bonded joint cohesion. Figure 21. Strength variation of transversely isotropic rock samples with different bonded joint cohesion. When the interface dip angles were medium (α = 30◦ , 45◦ , 60◦ , and 75◦ ), the UCS values decreased

heavily with the decrement of the bonded joint cohesion (Figure 21). The decrement of bonded joint When the interface dip medium (α = 30°,of 45°, 60°, and 75°), the UCS values decreased cohesion directly reduced theangles shearwere resistance capability interfaces, and made them much easier heavily with the decrement of the bonded joint cohesion (Figure 21). The decrement of bonded joint to fail. As shown in Figure 22, with the bonded joint cohesion decreasing from 13.0 MPa to 4.0 MPa, cohesion directly reduced the shear resistance capability of interfaces, and made them much easier the shear cracks in the interfaces having medium dip angles were the main cracks occurring in the to fail.rock As models. shown in Figure with the(material bonded A joint 4.0 MPa, whole The intact22, materials andcohesion material decreasing B) could notfrom play13.0 theirMPa role to effectively shear cracks in resistance the interfaces havingofmedium dipmodels, angles were theaccounted main cracks in the tothe contribute to the capability the whole which for occurring the decrement whole rock models. The intact materials (material A and material B) could not play their role ◦ of UCS values in Figure 21. When the interfaces were vertical (α = 90 ), even though the interface effectively to contribute to the from resistance of the models, which accounted the shear crack proportion increased 57.8%capability to 93.8% with thewhole bonded joint cohesion decreasingfor from decrement of UCS values in Figure 21. When the interfaces were vertical (α = 90°), even though the 13.0 MPa to 4.0 MPa, the UCS value decreased less significantly, as shown in Figure 21. This was interface shear crack proportion increased from 57.8% to 93.8% with the bonded joint cohesion because the intact materials (material A and material B) could still resist the top and bottom loading decreasing fromtowards 13.0 MPa to 4.0 MPa, theeven UCStheir valueconnections decreased less significantly, shown in Figure 21. platens moving each other, and (interfaces) wereas very weak. This was because the intact materials (material A and material B) could still resist the top and bottom loading platens moving towards each other, and even their connections (interfaces) were very weak.

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Interface crack proportion Interface shear crackshear proportion

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1.0 1.2 0.8 1.0 0.6 0.8 0.4 0.6 0.2 0.4 0.0 0.2 0

ccj=4.0 = 4.0MPa MPa j ccj=13.0 = 13.0 MPa MPa j 15

30

ccj=4.0 = 4.0MPa MPa j 45 60 75 ccj=13.0 = 13.0 MPa MPa j Dip angle (deg.)

90

0.0 0 shear crack 15 proportion 30 45 60 different 75 bonded90joint cohesion. Figure 22. Interface variation with Dip angle (deg.)

3.2.4. Bonded Joint Friction Angle

Figure 22. Interface shear crack proportion variation with different bonded joint cohesion. Figure 22. Interface shear crack proportion variation with different bonded joint cohesion.

The bonded joint friction angle, ϕ j , was then adjusted to 15°, 22.5°, and 30°, respectively, while 3.2.4. Bonded Joint Friction Angle 3.2.4. Bonded Joint Friction Angle 13 3 ◦ × 10 ◦N / m , k sj = 1 / 3k nj , the The other jointjoint parameters wereϕkept constant, i.e., k nj◦ = 0.5 bonded friction angle, j , was then adjusted to 15 , 22.5 , and 30 , respectively, while the 3 k and ϕ j , was The bonded friction angle, adjusted to N/m 15°, 22.5°, 30°, respectively, while other parameters constant, i.e.,μthen knj= = 0.5 × 1013 , σcj =variation 23.8 MPa, sj = 1/3k σ cj =joint 23.8 MPajoint = 13kept .0 MPa 0.40 , cwere , and . Figure 23 ,shows the njUCS of j j c j = 13.0 MPa, and µ j = 0.40. Figure 23 shows the UCS variation of transversely rock samples 13 isotropic 3 = 0.5 × 10 angles. N / mAs, shown k sj = 1in / 3kthis the other joint parameters were kept i.e., k njjoint nj , transversely isotropic rock samples with constant, different bonded friction with different bonded joint friction angles. As shown in this figure, the modification of the bonded joint figure, the modification of the bonded joint fiction angle had certain influences on the UCS values of fiction certain on the UCS μ values of transversely when theof σ cj =angle 23.8had MPa = 13.0 MPa = 0.40 , cinfluences , and . Figure 23isotropic shows rock the samples UCS variation j transversely isotropic rock samples when the j interface joint angles were medium. However, this interface joint angles were medium. However, this influence was negligible when the interface transversely rock samples with different bonded(αjoint angles. As shown inwas this influence wasisotropic negligible when the interface was horizontal = 0°)friction or vertical (α = 90°). Like bonded ◦ ◦ horizontal (α = 0 ) or vertical (α = 90 ). Like bonded joint cohesion, the bonded joint friction angle figure, the modification ofjoint the bonded angle had certain the UCSHowever, values of joint cohesion, the bonded friction joint anglefiction contributed to the shear influences strength ofon interfaces. contributed to the shear strength of interfaces. However, the contribution of the bondedHowever, joint friction transversely isotropic rock samples when the interface joint angles were medium. the contribution of the bonded joint friction angle depended on the normal stress componentthis in angle depended on the normal stress component accordance with Equation bonded influence was negligible the interface wasin horizontal = 0°) or vertical(9). (α =Thus, 90°). the Like bonded accordance with Equationwhen (9). Thus, the bonded joint friction(αhad a negligible influence on the shear joint had a negligible influence on the shear resistance of the interfaces when the dipHowever, angles jointfriction cohesion, bondedwhen joint friction the strength of interfaces. resistance of thethe interfaces the dipangle anglescontributed were high, to i.e., α =shear 75° and 90°, due to the small normal ◦ and 90◦ , due were high, i.e., α = 75 to the small normal stress component. With respect to the case ofin the contribution ofWith the bonded joint friction angle depended on the normal stress component stress component. respect to the case of α = 0°, although the normal stress component was ◦ α accordance = 0 , although the normal stress component was considerable, which enhanced the interface Equation (9). Thus, the bonded friction had a negligible on theshear shear considerable,with which enhanced the interface shearjoint strength, the shear behaviorinfluence of the interfaces was strength, the shear behavior of the interfaces was not the main mechanism for the failure of the whole resistance of the interfaces when the dip angles were high, i.e., α = 75° and 90°, due to the small normal not the main mechanism for the failure of the whole rock model. This accounted for the negligible rock model. This accounted for the negligible variation of the UCS valuesnormal with thestress bonded joint friction stress component. respect to bonded the case of αfriction = 0°, although component was variation of the UCSWith values with the joint angle at αthe = 0° in Figure 23. Figure 24 further ◦ angle at α = 0 in Figure 23. Figure 24 further indicates that the change of bonded joint friction angle considerable, enhanced the interface shear strength, the shear behavior the interfaces indicates that which the change of bonded joint friction angle mainly influenced theoffailure behaviorwas of mainly influenced the failure behavior of of transversely isotropic rock samples whenfor thethe dipnegligible angles not the main mechanism for the failure the whole rock model. This accounted transversely isotropic rock samples when the dip angles were medium. were medium. variation of the UCS values with the bonded joint friction angle at α = 0° in Figure 23. Figure 24 further

UCS (MPa) UCS (MPa)

indicates that the change of bonded joint friction angle mainly influenced the failure behavior of transversely isotropic were medium. experiment 65rock samples when the dip anglesLab ϕφj=15 = 15.0° j ϕ = 22.5° φj=22.5 j Lab experiment ϕ φj=30 j = 30 .0° ϕφj=15 = 15.0° j ϕ = 22.5° φj=22.5 j

55 65 45 55

ϕ φj=30 j = 30 .0°

35 45 25 35 0

15

30

45 60 Dip angle (deg.)

75

90

25 Figure 23. Strength variation of transversely samples bonded joint friction angle. 0 30 isotropic 45rock 60 with 75 Figure 23. Strength variation of15 transversely isotropic rock samples with bonded 90 joint friction angle. Dip angle (deg.) Figure 23. Strength variation of transversely isotropic rock samples with bonded joint friction angle.

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0.8

Interface crack proportion Interface shear crackshear proportion

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0.6 0.8 0.4 0.6

ϕψj=15° j = 15 °

0.2 0.4 0.0 0.2

ϕcj=22.5° = 22.5° j 0

15

30

45 60 Dip angle (deg.)

ϕψj=15° j = 15 ° 75

90

ϕcj=22.5° = 22.5° j

0.0 Figure 24. Interface shear crack different 75 bonded joint 0 15proportion 30 variation 45 with60 90 friction angle. Dip angle (deg.)

3.2.5. Joint Coefficient of Friction

Figure 24. Interface shear crack proportion variation with different bonded joint friction angle.

Interface shear crack proportion variation with different bonded joint friction angle. TheFigure joint 24. coefficient of friction began to work only after the joint bond broke to provide the 3.2.5. Jointshear Coefficient of Friction residual strength. To learn about the effect of the interface residual shear strength on the failure 3.2.5. Joint Coefficient of Friction μ j , was to be 0.10, behavior of transversely rock models, the joint of friction, The joint coefficient isotropic of friction began to work onlycoefficient after the joint bond broke to set provide the The joint coefficient of friction began to work only after the joint bond broke to provide the residual shear strength. To learn about the effect of the interface residual shear strength on the 13 failure k nj = 0.5 ×10 N failure / m3 , 0.25, and shear 0.40, whereas, the other joint parameters were kept constant, i.e.,shear residual strength. To learn about the effect of the interface residual strength on the behavior of transversely isotropic rock models, the joint coefficient of friction, µ j , was set to be 0.10, 3, μ=j 0.5 ,displays was beUCS 0.10, transversely isotropic rock models, the were c j =parameters 13.0 MPa 0.25, and the other ,joint kept i.e., knj 25 × set 1013to N/m σ cj = 23.8 MPa kbehavior 1 / 30.40, kof ϕ j =constant, 22.5of° friction, ,joint andcoefficient . Figure the sj = nj , whereas, kvariation , transversely σcj = 23.8 MPa, c j = 13.0 and with ϕ j = different 22.5◦ . Figure displays the UCS variation of3 sj = 1/3k nj isotropic rockMPa, samples joint25 coefficients friction. = 0.5 ×1013AsNshown /m , 0.25, and of 0.40, whereas, the other joint parameters were kept constant, i.e., k nj of transversely isotropic rock samples with different joint coefficients of friction. As shown in this figure, in this figure, the joint coefficient of friction had a negligible influence on the resistance capability of cFor = rock 13.0 samples MPa = friction 23.8 MPa k sjjoint = 1 /coefficient 3k nj isotropic =resistance 22.5 , σ cj of , and . Figure 25 αdisplays the had ,a negligible influence onϕthe capability j j low transversely rock samples. having dip°angles, e.g., =of0°transversely andthe 15°,UCS the ◦ ◦ isotropic rock samples. Forisotropic rock having low dip angles, e.g., to αcoefficients =affect 0 and 15ofvalues ,friction. the increment of variation of rock samples with different joint As shown increment oftransversely the interface shearsamples strength was not effective enough UCS because the the interface shear strength was not effective enough to affect UCS values because the rock samples in this figure,tended the jointtocoefficient friction had a negligible influence on theFurthermore, resistance capability of rock samples fail in theof mode of tensile failure across interfaces. when the tended to fail in the mode of tensile failure across interfaces. Furthermore, when the dip angle increased transversely isotropic rock samples.such For as rock low dipeven angles, e.g.,the α = shear 0° andstrength 15°, the dip angle increased to be medium, α =samples 30°, 45°,having 60°, and 75°, though ◦ , 45◦ , 60◦ , and 75◦ , even though the shear strength of interfaces mattered, toof beinterfaces medium, such as α =the 30shear increment ofmattered, the interface strength was not effective to affecttoUCS because the joint coefficient of friction couldenough not contribute the values peak stress of the the joint coefficient ofThis friction not contribute to the peak across stress the whole rock This was rock samples tended towas failcould in the mode tensile failure interfaces. Furthermore, when the whole rock model. due to the factofthat the interface shearof cracks began to model. occur quite close due toangle the fact that stage, the interface shear cracks began to 60°, occur quite close to the peak stressstrength stage, dip increased to be medium, such as α =12. 30°,Therefore, 45°, and 75°, even though thefriction shear to the peak stress as shown in Figure the joint coefficient of mainly ascontributed shown in Figure 12. Therefore, the joint coefficient of friction mainly contributed to the residual of interfacestomattered, the strength joint coefficient of friction couldofnot the peak stressagreed of the the residual rather than to the UCS thecontribute whole rocktomodels, which strength rather than to the UCS of the whole rock models, which agreed very well with Ref. [25]. whole rock model. This was due to the fact that the interface shear cracks began to occur quite close very well with Ref. [25]. For α = 90°, transversely isotropic rock model failed according to tensile ◦ For α = 90 , the transversely isotropic rock model failed according to tensile failure along interfaces, to the along peak interfaces, stress stage, as shown in Figure Therefore, the joint coefficient of friction mainly failure which accounted for the12. negligible influence of the joint coefficient of friction which accounted forresidual the negligible influence thetojoint friction onmodels, UCS values. contributed to the strength rather of than the coefficient UCS of theofwhole rock which agreed on UCS values.

very well with Ref. [25]. For α = 90°, the transversely isotropic rock model failed according to tensile failure along interfaces, which accounted for the negligible influence of the joint coefficient of friction Lab experiment 65 on UCS values. μ = 0.40 μj=0.40 UCS (MPa)UCS (MPa)

j

μ j = 0.25 μj=0.25 μLab =experiment 0.10 j μj=0.10

55 65

μ j = 0.40 μj=0.40 μ j = 0.25 μj=0.25

45 55

μ j = 0.10 μj=0.10

35 45 25 35 0

15

30

45 60 Dip angle (deg.)

75

90

Figure 25. Strength25variation of transversely isotropic rock samples with different joint coefficient Figure 25. Strength variation of15 transversely with75different90 joint coefficient of 0 30 isotropic 45 rock samples 60 of friction. friction. Dip angle (deg.)

Figure 25. Strength variation of transversely isotropic rock samples with different joint coefficient of friction.

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3.2.6. Joint Tensile Strength 3.2.6. Joint Tensile Strength A parametric study on the influence of the joint tensile strength,

σ cj , on the strength behavior

of transversely models was carried out asstrength, well through this behavior joint microA parametricisotropic study onrock the influence of the joint tensile σcj , on varying the strength of mechanicalisotropic parameter 15.90 MPa to 31.80 23.80 MPa setting the other transversely rockfrom models was carried out asMPa well through through varying this while joint micro-mechanical parameters constant. accordance 26, 23.80 the joint tensile strength slight parameters influence on parameter from 15.90 In MPa to 31.80 with MPaFigure through MPa while settinghad theaother the UCS values of transversely isotropic rock models. For rock samples having low constant. In accordance with Figure 26, the joint tensile strength had a slight influence ondip the angles UCS ◦, (α = 0°, and 30°),isotropic although themodels. increment thesamples joint tensile strength somehow values of 15°, transversely rock Forofrock having low dip angles (αaffected = 0◦ , 15the strength behaviorthe of transversely models, thesomehow influenceaffected was notthe significant. When the and 30◦ ), although increment ofisotropic the joint rock tensile strength strength behavior dip angle was medium, because the rock models failed in the mode of shear failure along interfaces, of transversely isotropic rock models, the influence was not significant. When the dip angle was the increment of the tensile strength a negligible influence oninterfaces, the strength the whole medium, because the interface rock models failed in the had mode of shear failure along theofincrement rock models. Furthermore, although the rock model failed in the mode of tensile failure along of the interface tensile strength had a negligible influence on the strength of the whole rock models. interfaces, the UCS ofthe therock transversely isotropic sample wasfailure mainly contributed bythe theUCS vertical Furthermore, although model failed in therock mode of tensile along interfaces, of intact materials. Thus, the change of joint tensile strength also had a negligible influence on the UCS the transversely isotropic rock sample was mainly contributed by the vertical intact materials. Thus, value. the change of joint tensile strength also had a negligible influence on the UCS value.

Lab experiment

65

UCS (MPa)

σσcj=3.18e7 cj = 31 .80 MPa

55

σσcj=2.38e7 cj = 23 .80 MPa σσcj=1.59e7 cj = 15 .90 MPa

45 35 25 0

15

30

45 60 Dip angle (deg.)

75

90

Figure 26. Strength variation of transversely isotropic rock samples with different joint tensile strength. Figure 26. Strength variation of transversely isotropic rock samples with different joint tensile strength. 4. Conclusions

Particle flow modeling was undertaken to investigate the failure mechanism of transversely 4. Conclusions isotropic rocks under uniaxial compressive loading. After setting up a particle flow model with the modeling was to undertaken to investigate the failure mechanism PFC3DParticle softwareflow package according the conceptual model proposed by Tien et al. [23],ofa transversely parametric isotropic rocks under uniaxial compressive loading. After setting up a particle flow model with the study was then carried out to investigate the effect of the interface dip angle and interface mechanical 3D PFC software package the conceptual model proposed by transversely Tien et al. [23], a parametric parameters on the failure according mode and to uniaxial compressive strength of the isotropic rock study was then carried out to investigate the effect of the interface dip angle and interface models. The following conclusions can be drawn from the obtained particle flow modelingmechanical results: parameters on the failure mode and uniaxial compressive strength of the transversely isotropic rock (1) The The interfaces interspaced in can intact materials were pivotal elements to successfully build models. following conclusions be drawn from the obtained particle flow modeling results: transversely isotropic rock models with a particle flow modeling method. With careful calibrations (1) The interfaces interspaced in intact materials were pivotal elements to successfully build of the interface mechanical parameters, these simulated transversely isotropic rock models transversely isotropic rock models with a particle flow modeling method. With careful derived quite similar failure modes and UCS values to those obtained in physical experiments. calibrations of the interface mechanical parameters, these simulated transversely isotropic rock (2) To highlight the effect of interfaces, the failure mode of transversely isotropic rock models was models derived quite similar failure modes and UCS values to those obtained in physical redefined according to the observed crack revolution at the meso level. Three basic failure experiments. modes were identified in the transversely isotropic rock models under uniaxial compressive (2) To highlight the effect of interfaces, the failure mode of transversely isotropic rock models was loading: (a) tensile failure across interfaces, (b) shear failure along interfaces, and (c) tensile redefined according to the observed crack revolution at the meso level. Three basic failure modes failure along interfaces. were identified in the transversely isotropic rock models under uniaxial compressive loading: (3) The joint normal stiffness and joint shear stiffness had a dramatic influence on the failure (a) tensile failure across interfaces, (b) shear failure along interfaces, and (c) tensile failure along strength of transversely isotropic rock models. The difference of mechanical response to uniaxial interfaces. compressive loading for each layered material accounted for the UCS variation with varying (3) The joint normal stiffness and joint shear stiffness had a dramatic influence on the failure stiffness values. strength of transversely isotropic rock models. The difference of mechanical response to uniaxial

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The mechanical parameters for the bonded joint shear strength property had quite a different influence on the failure behavior of transversely isotropic rock models. The bonded joint cohesion and bonded joint friction angle, which contributed to the shear strength of interfaces, had a considerable influence on the UCS values, while the joint coefficient of friction, which contributed to the residual strength of interfaces, had a negligible influence on the UCS values. The shear failure of interfaces was the dominant mechanism for anisotropic behavior of layered rock models, the change of joint tensile strength had a negligible influence on the UCS values of transversely isotropic rock models.

The present study put the emphasis on the failure mechanism of transversely isotropic rocks under the unconfined compression condition, whilst a further study on the anisotropic behavior of transversely isotropic rocks under the triaxial stress condition is undergoing. Moreover, the hydromechanical behavior of the transversely isotropic rocks is also a challenge topic that is of our interest. The following studies would provide useful guidelines for the stabilization control of engineering structures constructed in or on the transversely isotropic rocks. Author Contributions: Funding acquisition, X.-X.Y. and W.-G.Q.; Investigation, X.-X.Y.; Methodology, X.-X.Y.; Supervision, H.-W.J. and W.-G.Q. Funding: This research was funded by the National Natural Science Foundation of China, grant numbers 51704183, 51774192, and 51474135. This research was also partially funded by the Natural Science Foundation of Shandong Province, grant number ZR2017BEE020. Acknowledgments: The authors would like to express sincere appreciation to the editor and the anonymous reviewers for their valuable comments and suggestions for improving the presentation of the manuscript. Conflicts of Interest: The authors declare no conflict of interest.

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