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Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks
Accepted Manuscript Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks Morteza Nejati, Ali Aminzadeh, Martin O. Saar, Thomas Driesner PII: DOI: Reference:
S0013-7944(18)31359-6 https://doi.org/10.1016/j.engfracmech.2019.03.008 EFM 6385
To appear in:
Engineering Fracture Mechanics
Received Date: Accepted Date:
4 December 2018 9 March 2019
Please cite this article as: Nejati, M., Aminzadeh, A., Saar, M.O., Driesner, T., Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks, Engineering Fracture Mechanics (2019), doi: https://doi.org/ 10.1016/j.engfracmech.2019.03.008
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Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks
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Morteza Nejati∗a , Ali Aminzadehb,c , Martin O. Saarb , Thomas Driesnerc
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4 5 6
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a Department of Earth Sciences, ETH Zurich, Switzerland Geothermal Energy and Geofluids, Department of Earth Sciences, ETH Zurich, Switzerland c Institute of Geochemistry and Petrology, Department of Earth Sciences, ETH Zurich, Switzerland b
Abstract
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The conventional semi-circular bend (SCB) test of anisotropic rocks, with symmetric loading,
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generates a Mixed-Mode I/II crack tip loading when the crack is not aligned with one of the prin-
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cipal material directions. This paper presents a modified SCB test for anisotropic rocks to ensure
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a pure Mode I crack tip loading. It is demonstrated that the stress intensity factor (SIF) solution
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of an anisotropic SCB specimen depends on two dimensionless parameters, the anisotropy ratios
13
of the Young’s modulus and apparent shear modulus, as well as the anisotropy orientation. These
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two dimensionless parameters are selected since they have physical meaning, and are generally
15
bounded for rock materials. Based on these anisotropy parameters, the semi-circular test is modi-
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fied to configure pure Mode I, or tensile stress at the crack tip, by using an asymmetric three-point
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bend configuration. Extensive finite element analyses are performed to obtain the span ratio and
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normalized SIF of the SCB specimen with different anisotropy ratios and orientations.
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Keywords: Fracture toughness; semi-circular bend; anisotropy; Mode I
Email address: *[email protected] (Morteza Nejati∗ )
Preprint submitted to Engineering Fracture Mechanics
February 26, 2019
ν, ν 0 ξ σ, σi τij , γij φ, Φ1 , Φ2 , F1 , F2
Nomenclature Crack length Specimen thickness Errors of the magnitude and the argument of the approximate complex parameters µ ˜i , (i = 1, 2) Young’s moduli within and normal to the isotropy plane Shear moduli within and normal to the plane of isotropy Transverse shear modulus approximated from the Saint-Venant relation Shear modulus in xy plane which is formed by the rotation of x00 y 00 along z 00 by the angle −β Mode I stress intensity factor and fracture toughness Load and peak load Polar coordinates of a point near the crack tip Sample radius Span length for SCB specimen with asymmetric three-point bend Compliance matrix and its ij component Traction and displacement boundary conditions Displacements along x and y directions Cartesian coordinates Normalised Mode I stress intensity factor Angle between the anisotropy orientation and the horizontal plane, or crack plane in cracked bodies Strain vector and strain component ij Anisotropy ratio of apparent shear modulus, G0 /G0sv Conjugate pair of roots to the characteristic equation in the material coordinate system x00 y 00 , known as complex parameters Complex parameters in xy coordinate system Approximated complex parameters when neglecting the term 2ξν 0 (1 − η)/η in the characteristic equation Poisson’s ratio within and normal to the plane of isotropy Anisotropy ratio of Young’s modulus, E/E 0 Stress vector and normal stress component in i direction Shear stress and strain ij Complex potentials
Abbreviations CB CCNBD FPZ GTS ISRM LEFM SCB SR TCBD
Chevron bend Cracked chevron notched Brazilian disc Fracture process zone Grimsel Test Site International Society for Rock Mechanics Linear elastic fracture mechanics Semi-circular bend Short rod Through-thickness cracked Brazilian disk
a B a em µi , eµi
E, E 0 G, G0 G0sv G0β KI , KIc P , Pm r, θ R S1 , S2 S, Sij t¯, u ¯ ux , uy x, y YI β , ij η µi , µ¯i , i = 1, 2 µ∗i , µ ¯∗i ¯˜i µ ˜i , µ
2
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1. Introduction
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The mechanical behavior of many types of rocks is anisotropic due to their complex micro-
22
structure. Rock anisotropy is often introduced due to one of the following processes: (i) Foliation
23
in metamorphic rocks that results in the flattening of grains and the alignment of platy minerals;
24
(ii) Bedding and lamination in sedimentary rocks that introduce a layered structure. Taking rock
25
anisotropy into account is crucial in the accurate predictions of rock mass deformation, stability
26
and failure (Amadei, 1996; Dutler et al., 2018; Krietsch et al., 2018). Performing Mode I fracture
27
toughness experiments in anisotropic rocks provides very useful information about the fracturing
28
processes that can occur in a wide range of geomechanical and geophysical problems. An important
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application is related to enhanced geothermal systems, normally placed in the crystalline basement,
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where anisotropic rocks, such as granite or gneiss, are likely to be present. Recently Amann et al.
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(2018), Gischig et al. (2018), Jalali et al. (2018), and Doetsch et al. (2018) have demonstrated the
32
importance of anisotropy in the in-situ stimulation and circulation project in the deep underground
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laboratory at the Grimsel Test Site (GTS) in Switzerland.
34
Three main rock mechanical properties are used for geomechanical and geophysical analyses:
35
(i) elasticity; (ii) strength (both tensile and compressive); and (iii) fracture toughness (both tensile
36
and shearing). These properties are anisotropic in a large number of sedimentary and metamor-
37
phic rocks that exhibit features such as foliation and bedding. These features yield a preferred
38
orientation of the constituent minerals, pores, and cracks, which can be idealised geometrically
39
with an axis of symmetry normal to the foliation and bedding planes (Jaeger et al., 2007; Dambly
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et al., 2018). Such an axis of symmetry makes transverse isotropy a suitable model for predict-
41
ing the deformational behaviour of rocks. The transversely isotropic elasticity model requires five
42
independent material constants that need to be measured experimentally (Dambly et al., 2018;
43
Nejati et al., 2018). The dominant orientation of micro-cracks along features like bedding and
44
foliation yields a preferential path for failure and fracture growth. Therefore, strength and fracture
45
toughness of rocks also exhibit anisotropy. These two properties are closely related through the
46
fracture process zone (FPZ) which also exhibits anisotropy in size (Dutler et al., 2018).
47
Our literature survey in Section 2 describes the available schemes for conducting fracture tough-
48
ness experiments on rocks. This review shows that the overall attempt over the last four decades
49
has been to reduce the variability of the fracture toughness among different tests to improve the
50
reliability of the measurements. Applying constraints to the sample size has significantly helped
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towards reaching this goal, reducing the variation of measured values among different tests consid-
52
erably (Whittaker et al., 1992; Iqbal and Mohanty, 2007; Wei et al., 2018a). Most of these schemes
53
are designed based on assuming isotropic elasticity, where the material constants play no role in the
54
elasticity solutions of the specimens. However, in anisotropic rocks both the elasticity constants 3
55
and the material orientation influence the stress solution. Although some researches have evaluated
56
the effect of material orientation on fracture toughness, the literature appear to lack research on
57
the potential influence of the elasticity constants on the measured values of fracture toughness.
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Consequently, there remains a literature gap regarding the influence of elasticity anisotropy when
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conducting fracture toughness tests in anisotropic rocks.
60
Kuruppu et al. (2014) presented the fourth ISRM-suggested method to measure Mode I frac-
61
ture toughness using the semi-circular bend (SCB) specimen. The SCB specimen seems to be a
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great choice for conducting fracture toughness experiments in anisotropic rocks. However, despite
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addressing the applicability of the SCB test for anisotropic rocks in Kuruppu et al. (2014), the in-
64
fluence of the elasticity anisotropy on the fracture toughness measurement has not been evaluated.
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This study investigates the anisotropic elasticity solution of the SCB specimen, and modifies the
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standard SCB testing scheme by employing asymmetric loading to make it applicable to conduct-
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ing pure Mode I fracture toughness measurements in anisotropic rocks. Enforcing a pure Mode I
68
loading condition enables us to i) determine the Mode I fracture toughness in a specific direction if
69
the crack grows in a self-similar manner, and ii) investigate how the crack path is influenced by the
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anisotropy if a crack kinking occurs. Note that understanding the fracturing process under Mode
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I is a prerequisite to the more complex Mixed-Mode I/II loading condition.
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2. A review of fracture toughness tests
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Fracture toughness represents the critical value of the stress intensity factor (SIF), defined in
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the context of linear elastic fracture mechanics (LEFM), in a particular crack deformation mode,
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at which the onset of crack propagation occurs. Determining Mode I or tensile fracture toughness,
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referred to as KIc , is of importance due to the dominance of this loading type in many applications.
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Since the mid 1970s, numerous specimen geometries have been suggested to determine KIc . For
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metallic materials, standard methods include single-edge-notched and fatigue-pre-cracked plates
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under tension or three-point bending loads (ASTM, 2018). While suitable for metals, these tests
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are unfavorable for rocks because: (i) They require large and impracticably shaped specimens, and
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thus significant machining; (ii) Difficulties can arise when applying tensile loads to rocks, as rocks
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are weak under tension, so that failure may occur in places, where the load is applied (Whittaker
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et al., 1992). Such shortcomings of the ASTM (2018) standard directed efforts to introduce tests
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with core-based samples, readily prepared from drilled rocks, and preferably under compressive
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loads, to determine the fracture toughness of rocks.
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To obtain precise, consistent and reproducible values in KIc of rocks, the International Society
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for Rock Mechanics (ISRM) is recommending four core-based test procedures, as listed in Table
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1. These tests include the chevron-notched round bar in bending (CB), the chevron-notched 4
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short rod in splitting (SR), the cracked chevron-notched Brazilian disk in dimetrical compression
90
(CCNBD), and the semi-circular bend (SCB) specimen. Associated guidelines provide the details
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of the requirements for the samples’ preparation and dimensions, as well as the test procedure
92
requirements in terms of loading type and rate. Explicit formulae are also provided to calculate
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the fracture toughness from the failure load and geometrical factors. Despite standardized testing,
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the results from CB, SR and CCNBD can still exhibit deviations which are often explained by
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size effects, anisotropy of the rock and inaccuracy of the dimensionless parameters used in the
96
calculations. Iqbal and Mohanty (2007) compared CB and CCNBD methods on three different
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rock types with two-hundred specimens and concluded that the results of these two methods
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were comparable, when the correct equation for fracture toughness calculations was used and the
99
specimen size was selected carefully. Wei et al. (2016b, 2018a) showed that the inconsistency of
100
the results from CB, SR and CCNBD may still be present when using accurate equations for the
101
SIFs. However, they suggested that this inconsistency may be explained by the size of the FPZ
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developed in these different specimens.
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Table 1: ISRM-suggested methods for determining the Mode I fracture toughness of rocks.
Method
Test configuration
Chevron-notched short rod in splitting (SR)
R
2R
A
A
P
!
P
L
a a0 2R
A-A
2R > 10× grain size L > 3.5(2R) 2S1 > 3.33(2R) ± 0.02(2R) Ψ = 90◦ ± 1◦ a0 = 0.15(2R) ± 0.10(2R)
A P
Chevron-notched round bar in bending (CB)
! a a0 2S1 L
2R A-A
A
a
a0
P/B A-A B P/B
R
A
Notched semicircular specimen in bending (SCB specimen)
2R = 75.0 mm B/R = 0.80 a0 /R = 0.2637 a1 /R = 0.65
R
a1
A
a 2S1 A
Reference
Barker (1977); Ouchterlony (1988)
Ouchterlony (1980, 1988)
B P/B
A
Cracked ChevronNotched Brazilian Disk in dimetrical compression (CCNBD)
Size requirements 2R > 10× grain size L = 1.45(2R) ± 0.02(2R) Ψ = 54.6◦ ± 1◦ a0 = 0.48(2R) ± 0.02(2R)
P/B A-A
2R > 10× grain size or 76 mm B > 0.4D or 30 mm 0.40 ≤ a/R ≤ 0.60 0.50 ≤ S1 /R ≤ 0.80
Sheity et al. (1985); Fowell and Xu (1994); Fowell (1995)
Chong and Kuruppu (1984); Kuruppu and Chong (2012); Kuruppu et al. (2014) Kuruppu and Chong (2012)
103
Two methods are generally used to create cracks in test specimens: 1) naturally pre-fabricated
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cracks, produced through fatigue loading (applicable mainly to metals) or stable crack growth in 6
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chevron-notched samples; 2) artificially made cracks by sawing a through-thickness slit. Out of
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the four ISRM-suggested methods for fracture toughness measurements, the first three, i.e. SR,
107
CB and CCNBD, use chevron-notched specimens, while the SCB test uses a through-thickness
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crack. Chevron-notches provide a stable and sub-critical crack growth in the first stage of the
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crack extension. This is because the length of the crack front gradually increases, causing the SIF
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to drop and the load required for the crack extension to rise. The peak load marks the beginning
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of the second stage of the crack extension, i.e. the unstable crack growth, when the SIF starts to
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increase with the crack extension. The numerical results on the SIF of the chevron-notch clearly
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show these two stages (Wei et al., 2016a,b). The stable crack propagation in the first stage can
114
readily be controlled, so that self-pre-cracking is essentially achieved.
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As a material property, Mode I plane-strain fracture toughness is not expected to vary with the
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testing procedure as well as the specimen type and size. However, the measured fracture toughness
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seems to be independent of specimen type and size only when certain minimum specimen size
118
requirements are met. This behavior can be attributed to the large size of the FPZ, compared
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to the sample dimensions in small samples. Since the concept of stress intensity factor is valid in
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the context of LEFM, the FPZ must be sufficiently smaller than the dimensions of the specimen
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in order for the LEFM to be applicable. Therefore, ensuring representative fracture toughness
122
measurements requires the characteristic dimensions, such as the crack length, the crack ligament,
123
and the specimen thickness to be sufficiently large. Since the FPZ size depends on the grain size,
124
the minimum size of the specimen also depends on the grain size of the rock. Fracture toughness,
125
measured during tests that satisfy minimum sample size requirements, has been proven to be a
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material property (Whittaker et al., 1992). Suggested methods in Table 1 have guidelines on the
127
minimum dimensions required for the samples.
128
The first three suggested methods have the following disadvantages: (i) The SR and CB tests
129
require large amounts of intact rock core; (ii) The SR test is based on the splitting load configuration
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which requires a complicated loading frame; (iii) The SR, CB and CCNBD tests require complex
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sample preparation schemes to produce chevron notches. The SCB test, however, can be conducted
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with smaller samples and enable a straightforward preparation scheme as well as a simple three-
133
point bending configuration. The CCNBD and SCB tests have the advantage that they can be
134
applied on Mode II and Mixed-Mode I/II fracture toughness tests. During CCNBD tests, this
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is achieved by adjusting the diametrical loading so that it does not align with the crack plane.
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During SCB tests, this is achieved by either introducing a slanted edge crack (Chong and Kuruppu,
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1988; Lim et al., 1993) or an asymmetrical three-point bending load configuration (Ayatollahi
138
et al., 2011). Note that the use of CCNBD tests for Mixed-Mode I/II or Mode II loading may be
139
problematic since the stable pre-cracking obtained through a chevron-notch is expected to align in 7
140
the notch plane, whereas kinking is often expected during Mixed-Mode I/II or Mode II loading.
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Recent CCNBD Mode II fracture toughness tests show a complex concave or convex fracture
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surface initiating from the two edges of the chevron notched ligaments (Wei et al., 2018b). This
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complexity of the fracture path invalidates the II fracture toughness results from the CCNBD
144
specimen. In order to avoid this issue and also the complications in preparing chevron notches,
145
several researchers have suggested the through-thickness cracked Brazilian disc (TCBD) test, which
146
seems to be more suitable for mixed-mode experiments than the CCNBD (Krishnan et al., 1998;
147
Chen et al., 1998b; Ke et al., 2008).
148
The SCB test has become popular recently due to its advantages in terms of testing require-
149
ments, and also its potential to test anisotropic rocks and Mixed-Mode I/II tests. The SCB test
150
was suggested by Chong and Kuruppu (1984), and since then, it has been extended and improved
151
to accurately determine the fracture toughness not only under pure Mode I (Lim et al., 1994a;
152
Adamson et al., 1996) but also Mixed-Mode I/II conditions (Lim et al., 1994b; Khan and Al-
153
Shayea, 2000; Ayatollahi et al., 2006; Aliha et al., 2010, 2012). A lot of effort has been devoted
154
to accurately obtain the stress intensity factor formula for the entire range of pure Mode I to
155
pure Mode II conditions (Lim et al., 1993; Ayatollahi and Aliha, 2007). Researchers have also
156
studied various factors influencing fracture toughness, including the effect of specimen size (Lim
157
et al., 1994a; Khan and Al-Shayea, 2000), loading rate (Lim et al., 1994a; Dai and Xia, 2013), rock
158
anisotropy (Chong et al., 1987; Kataoka et al., 2015; Lee et al., 2015), chemical solutions (Karfakis
159
and Akram, 1993), water content (Lim et al., 1994a), temperature, confining pressure and water
160
vapor pressure (Funatsu et al., 2004; Kataoka et al., 2015), and possible effects of T-stress (Wei
161
et al., 2017). The SCB test has the following advantages:
162 163
164 165
166 167
168 169
• Small compact specimens can be produced directly from standard rock cores requiring minimal machining, thanks to its simple and practical cylindrical geometry. • The three-point bend loading configuration is easy to implement, using any compressive test frame. • Fracture toughness is determined simply from the maximum load and the geometric dimensions, employing well-established formulae. • Slight modifications of the specimen or loading specifications makes it possible to conduct the entire range of Mixed-Mode I/II fracture toughness tests.
170
• Fracture toughness testing of anisotropic rocks is conducted simply by cutting specimens from
171
a single core and by generating notches with desirable angles, with respect to the anisotropy
172
orientation. 8
173
The effect of the rock micro-structure anisotropy on the fracture toughness has been investigated
174
in a number of studies. Table 2 lists the fracture toughness experiments performed on different
175
anisotropic rocks using the aforementioned test setups. Apart from these tests, the cracked-ring
176
disk (Chen et al., 2008) and the notched deep beam (Luo et al., 2018) tests have also been employed
177
for fracture toughness experiments. Most of these cases consider the end-member configurations,
178
where the crack is oriented along one of the principal material directions (normally referred to
179
as arrester, divider or short-transverse (Chong et al., 1987)) of anisotropic rocks. An important
180
aspect during these tests is the proper identification of the role of the micro-structure in the fracture
181
toughness anisotropy. Fracture toughness is in fact closely linked to the presence of micro-cracks
182
and their orientation. Of the works listed in Table 2, only Chen et al. (1998b), Ke et al. (2008)
183
and Dai and Xia (2013) have considered the elasticity anisotropy of the specimens, while the
184
remaining studies have simply assumed an isotropic elasticity solution. Elasticity anisotropy of
185
rocks often accompanies fracture toughness anisotropy. Therefore, one needs to use an anisotropic
186
elasticity solution to obtain the stress intensity factor for fracture toughness calculations. This
187
article discusses how the elasticity anisotropy must be taken into account for fracture toughness
188
determination employing the SCB test. Table 2: A summary of tests used to measure fracture toughness of anisotropic rocks.
189
Test SR
Rock type Shale
Loading mode Mode I
CCNBD
Granite
Mode I
SCB
Granite, Shale
Mode I
TCBD
Sandstone, Marble , Shale
Mixed-Mode I/II
Reference Chandler et al. (2016, 2017) Nasseri et al. (2005, 2006); Nasseri and Mohanty (2008); Nasseri et al. (2009, 2010, 2011) Chong et al. (1987); Dai and Xia (2013); Kataoka et al. (2015); Lee et al. (2015); Dutler et al. (2018); Shi et al. (2019) Krishnan et al. (1998); Chen et al. (1998b); Ke et al. (2008)
3. Modified semi-circular bend test for anisotropic rocks
190
Although the semi-circular bend test has been a common test to determine fracture toughness
191
during the last two decades, it has been only recently listed in the ISRM-suggested methods
192
(Kuruppu et al., 2014). This test is based on a symmetric three-point bend configuration, i.e.
193
two symmetric supports hold the sample at the base and a compressive line load is applied at
194
the top (S1 = S2 in Figure 1). The Mode I fracture toughness is then determined employing
195
the normalized stress intensity factor, which is a function of the loading configuration and the
196
geometry. In anisotropic rocks, however, the stress intensity factor also depends on the anisotropy 9
197
paramaters, as will be shown in the next sections. In fact, a symmetrical loading (S1 = S2 ) of
198
materials, that do not show symmetry with respect to the load direction (β 6= 0, π/2), yields a
199
Mixed-Mode I/II crack tip loading.
C P/B β R
S1
R a
B
S2 C
Section CC
Figure 1: Schematics of the SCB test with asymmetric loading.
200
In order to resolve this issue, we suggest an asymmetric loading condition for SCB testing of
201
anisotropic rocks, as shown in Figure 1. This configuration was originally proposed by Ayatollahi
202
et al. (2011) to obtain the Mixed-Mode I/II loading of isotropic SCB specimens. For any geometrical
203
and material configuration, S2 can be obtained in such a way that the Mode II stress intensity
204
factor, KII , vanishes and only a pure Mode I condition is applied. We investigate the dependency
205
of the stress solution on the anisotropy parameters in the next three sections, before reporting the
206
details of our proposed modified SCB test configuration in Section 7.
207
4. Transverse isotropy material model
208
Anisotropic elasticity implies a directional dependency of the material deformability. Depend-
209
ing on the number of symmetry planes, different numbers of material constants are required to
210
describe the response of anisotropic materials: twenty-one constants for the triclinic model with
211
no symmetry plane; thirteen constants for the monoclinic model with one symmetry plane; nine
212
constants for the orthotropic model with three symmetry planes; five constants for the transversely
213
isotropic model with an axis of symmetry; and, ultimately, two elasticity constants for the isotropic
214
material with an infinite number of symmetry planes. This section gives some details about the
215
transversely isotropic material model. 10
216
4.1. Transverse isotropy
217
Due to the presence of foliation or bedding, many rock types exhibit a geometrical micro-
218
structural axis of symmetry, which makes the transverse isotropy a suitable model to describe
219
their response. This model defines an isotropy plane, often assumed to coincide with features
220
such as foliation and bedding, and an infinite number of symmetry planes, parallel to the axis of
221
symmetry. Consider the Cartesian coordinate system x0 y 0 z 0 , aligned with the material coordinate
222
system, where the isotropy plane coincides with the plane x0 y 0 , while z 0 denotes the axis of symmetry
223
(Figure 2a). The well-known generalized Hooke’s law defines the constitutive law as 1 E x0 y0 z 0 0 0 = γy z γ 0 0 xz γ x0 y 0 | {z }
ν E 1 E
−
ν0 E0 ν0 − 0 E 1 E0 −
0
0
0
0
0 1 G0
0 0 1 G0
0
|
{z S0
0
σ x0 0 σy0 σ 0 0 z τy0 z 0 0 τx0 z 0 0 τ 0 0 xy 1 | {z } G } σ0
(1)
224
Here, 0 and σ 0 are the strain and stress vectors, respectively, where the Vigot’s notation is
225
used, and S 0 is the well-known compliance matrix. Due to the requirement of positive strain
226
energy, S 0 must be a symmetric positive-definite matrix. Five independent engineering constants
227
characterize the elasticity of the transversely isotropic model in principal coordinates: E and E 0 ,
228
in-plane and transverse Young’s moduli, are applied within and normal to the isotropy plane x0 y 0 ;
229
ν and ν 0 , in-plane and transverse Poisson’s ratios, are applied within and normal to the isotropy
230
plane x0 y 0 ; and G0 , the transverse shear modulus, is applied transverse to the isotropy plane x0 y 0 .
231
The in-plane shear modulus, G, is applied within the isotropy plane x0 y 0 and depends on E and ν
232
through the relation G = E/[2(1 + ν)]. Defining directions 1, 2 and 3 along axes x0 , y 0 and z 0 , the
233
following equalities hold: E = E1 = E2 , E 0 = E3 , ν = ν12 = ν21 , ν 0 = ν31 = ν32 , G0 = G13 = G23
234
and G = G12 (see illustrative details in Nejati et al. (2018)).
235
We point out that the Poisson’s ratio is generally defined as νij = −j /i for a stress in
236
the i direction. Due to the symmetry of the elasticity matrix, νij and νji are related through
237
νij /Ei = νji /Ej . By this definition, the transverse Poisson’s ratio ν 0 = ν31 = ν32 characterizes
238
the relative strain in Directions 1 or 2 when a normal stress is applied along Direction 3. One
239
can alternatively define the transverse Poisson’s ratio as ν ∗ = ν13 = ν23 , which implies relative
240
strain in Direction 3 when normal stress is applied in Directions 1 or 2. The two definitions are 11
y
x
n
t
z
Гt y''
y'' x'' z''
y'
z'
y,u y x''
x' β
β
x,u x
Гu ū
a)
b)
Figure 2: a) Schematics of the material coordinate systems x0 y 0 z 0 and x00 y 00 z 00 with respect to the global coordinate system xyz. x00 y 00 forms an angle β with xy. b) The anisotropic deformation problem in the plane xy, with traction and displacement boundary conditions.
241
dependent on the relation ν 0 /E 0 = ν ∗ /E. In this paper, we choose to work with ν 0 . This is of
242
particular importance, as some commercial finite element packages may import ν13 (or ν23 ) as
243
inputs, requiring the correct calculation of these values employing ν ∗ = ν13 = ν23 = ν 0 E/E 0 .
244
4.2. Two-dimensional compliance matrix
245
Let us assume a coordinate system x00 y 00 z 00 in such a way that the x00 y 00 plane is orthogonal to
246
the material isotropy plane x0 y 0 , and z 00 is oriented along y 0 as shown in Figure 2a. We wish to
247
investigate the two-dimensional deformation within the plane x00 y 00 , where the maximum variation
248
of the deformability occurs. Note that x00 y 00 forms a symmetry plane of the material deformation.
249
The plane-stress deformation within x00 y 00 is based on the assumptions σz 00 = τx00 z 00 = τy00 z 00 = 0,
250
and depends only on four elastic constants: the Young’s moduli along x00 and y 00 (E and E 0 ), and
251
the Poisson’s ratio and shear modulus in the plane x00 y 00 (ν 0 and G0 ). Applying the transformation
252
formula in Appendix A to Eq. (1), the stress-strain relationship within x00 y 00 reads 1 E ν0 = − E0
x00
y00 γx00 y00
ν0 − 0 E 1 E0
0
0
0 00 σ x . 00 σ 0 y 00 00 τ 1 x y G0
(2)
253
The plane-strain deformation is based on the assumptions z 00 = γx00 z 00 = γy00 z 00 = 0, and also
254
depends on the Poisson’s ratio within the isotropy plane, ν. The 2D plane-strain Hooke’s law reads
12
1 − ν2 E x00 ν 0 (1 + ν) y00 = − E0 γx00 y00 0
ν 0 (1 + ν) E0 1 − (E/E 0 )ν 02 E0 −
0
0 σx00 0 σy00 . τx00 y00 1
(3)
G0
255
Let us assume that the xyz coordinates are formed by rotating x00 y 00 z 00 along the axis z 00 by an
256
angle −β, as shown in Figure 2, so that the isotropy plane has an angle β with respect to the x-
257
axis in the xyz coordinate system. Analyzing the deformation in xy coordinate system requires
258
the transformation of the compliance matrices in Eqs. (2) and (3). Using the transformation rules,
259
given in Appendix A, Hooke’s law in xy coordinate reads
S11 S12 S16 σx y = S12 S22 S26 σy , γxy S16 S26 S66 τxy | {z } | {z } | {z } x
260
S
(4)
σ
where Sij , i, j = 1, 2, 6 are the components of the compliance matrix, given by S11 S12 S16 S22 S26 S66
cos4 β sin4 β 1 2ν 0 sin2 2β = + + − 0 E E0 G0 E 4 2 1 1 1 sin 2β ν0 = + 0− 0 − 0 cos4 β + sin4 β E E G 4 E 2 cos β sin2 β 1 ν0 = − − − cos 2β sin 2β E E0 2G0 E 0 sin4 β cos4 β 1 2ν 0 sin2 2β = + + − E E0 G0 E0 4 2 sin β cos2 β 1 ν0 = − + − cos 2β sin 2β E E0 2G0 E 0 1 1 2ν 0 cos2 2β = + 0 + 0 sin2 2β + E E E G0
(5)
261
for the plane-stress state. For the plane-strain condition, Sij are readily obtaind by replacing 1/E,
262
1/E 0 and ν 0 /E 0 with (1 − ν 2 )/E, (1 − (E/E 0 )ν 02 )/E 0 and ν 0 (1 + ν)/E 0 , respectively.
263
4.3. Saint-Venant relation
264
The measurement of the elastic constants in rocks is not an easy task. This has frequently
265
led researchers to approximate some of the constants. For example, the well-known Saint-Venant
266
relation provides an approximation of the shear moduli in orthotropic materials, based on the 13
267
values of the Young’s moduli and Poisson’s ratios (Saint-Venant, 1863): 1 + 2νji 1 1 = + Gij Ei Ej
i, j = 1, 2, 3 .
(6)
268
Within the isotropic plane ij (Ei = Ej ), this relation simplifies to 1/G = 2(1 + ν)/E, which
269
indicates the exact dependency between the shear and the Young’s moduli and the Poisson’s ratio.
270
This dependency holds within the isotropy plane of a transversely isotropic material. However, the
271
transverse shear modulus, obtained from the Saint-Venant relation, 1/G0sv = 1/E + (1 + 2ν 0 )/E 0 ,
272
is only an approximation. In reality, G0 is an independent constant and can deviate from the
273
approximated G0sv value.
274
Despite being known in the literature as an empirical relation, the Saint-Venant approximation
275
has in fact a theoretical basis. Let us consider the shear deformation in the xy plane, which is
276
normal to the isotropy plane, as shown in Figure 3a. Using the compliance matrix component S66
277
in Eq. (5), the relation between the shear stress and the shear strain in this plane is given by
278
τxz = G0β γxz , where 1 1 + 2ν 0 2 1 sin2 2β cos2 2β 1 2 = + sin 2β + cos 2β = + G0β E E0 G0 G0sv G0
(7)
279
The variation of the ratio G0β /G0sv against the angle β is plotted in Figure 3b for three ratios of
280
G0 /G0sv . It can be seen that at β = π/4 the apparent shear modulus is equal to the Saint-Venant
281
approximation: 1 G0β β=π/4
=
1 1 1 + 2ν 0 = + G0sv E E0
(8)
282
This equation shows that the Saint-Venant approximation is in fact equal to the apparent shear
283
modulus G0β in a coordinate system rotated by π/4 with respect to the isotropy plane. The higher
284
the deviation of β from π/4, the higher the deviation of G0β from G0sv .
14
τxy
1.3 G'= 1.2 G' sv
G' / G' sv
y''
x''
1.2
τxy
y β
G'= G' sv G'= 0.8 G' sv
1.1 1 0.9 0.8
x
0.7 0
a)
/4
/2
b)
Figure 3: a) Schematics of a solid element in the xy coordinate system which forms an angle β with respect to the material coordinate system x00 y 00 , under shear stress τxy . b) The variation of the apparent shear modulus, G0β , against the angle β (τxz = G0β γxz ).
285
A question raised by many researchers is how accurate the Saint-Venant approximation is.
286
Based on the experimental measurements from two-hundred static and dynamic tests, Worotnicki
287
(1993) confirmed the validity of the Saint-Venant approximation for many rock types, such as
288
granite, gneiss, sandstone and basalt. Comparisons made by Talesnick and Ringel (1999) and Cho
289
et al. (2012) confirm this finding as they show variations of only up to 20% between the measured
290
values of the transverse shear modulus and the Saint-Venant approximation for anisotropy ratios of
291
up to E/E 0 = 2. It is worth noting that there are error sources related to the measurement schemes
292
employed, nonlinear response of the rock, as well as the rock heterogeneity where multiple samples
293
are used to determine the constants. For example, the tests conducted by Cho et al. (2012)
294
determine the transverse shear modulus indirectly based on a third sample, while the Young’s
295
moduli and Poisson’s ratios, used to calculate G0sv , are obtained from two other samples. Generally,
296
there is consensus that the Saint-Venant relation yields a good approximation for rocks with low to
297
moderate anisotropy ratios. Note that the majority of rock types show low to moderate anisotropy
298
ratios (Worotnicki, 1993; Amadei, 1996).
299
5. Elasticity solution of plane deformation
300
This section investigates the dependency of the stress solution on the material properties in an
301
anisotropic plane. Let us consider the plane deformation of an anisotropic material, schematically
302
illustrated in Figure 2b. We wish to solve the spatial deformation within the plane, u = (ux , uy ),
303
with the strain-displacement dependency governed by the infinitesimal strain theory, Eq. (9a),
304
where the strains satisfy the compatibility relation in Eq. (9b). Stress is related to deformation 15
305
in Eq. (9c) through the generalized Hooke’s law for a linear elastic anisotropic material, and
306
conserves momentum in Eq. (9d). Stresses and displacements must satisfy boundary conditions
307
given by Eq. (9e).
∂uy ∂ux ∂ux ∂uy , y = , γxy = + ∂x ∂y ∂y ∂x 2 2 2 ∂ xy ∂ x ∂ y + −2 =0 2 2 ∂y ∂x ∂x∂y = Sσ ∂σxy ∂σy ∂σx ∂σyx + = 0, + =0 ∂x ∂y ∂x ∂y σn = t¯ on Γt , u = u ¯ on Γu
x =
(9a) (9b) (9c) (9d) (9e)
308
The Airy stress function , φ(x, y), is introduced, where the stresses, defined through this po-
309
tential, automatically satisfy the equilibrium equation, Eq. (9d), for the case where body forces
310
are zero: σx =
311 312
∂2φ , ∂y 2
σy =
∂2φ , ∂x2
τxy = −
∂2φ . ∂x∂y
(10)
Inserting these stresses into the compatibility equation, Eq. (9b), employing the stress-strain relation given by Eq. (4), yields: S22
∂4φ ∂4φ ∂4φ ∂4φ ∂4φ − 2S + (2S + S ) − 2S + S =0, 12 66 26 16 11 ∂x4 ∂x3 ∂y ∂x2 ∂y 2 ∂x∂y 3 ∂y 4
(11)
313
where Sij are defined by Eq. (5). The general solution to this equation can be found using the
314
method of characteristics, looking for solutions of the form φ = φ(x+µ∗ y), where µ∗ is a parameter.
315
Substituting this stress function into Eq. (11) yields the characteristic equation: S11 µ∗ 4 − 2S16 µ∗ 3 + 2S12 + S66 µ∗ 2 − 2S26 µ∗ + S22 = 0 .
316 317 318
(12)
This equation governs the dependency of the elasticity solution on the material constants. It can be shown that the roots of the characteristic equation are complex, and always occur in conjugate pairs: µ∗1 = a1 + ib1 , µ∗2 = a2 + ib2 , µ∗3 = µ¯∗1 , µ∗4 = µ¯∗2 (Lekhnitskii, 1968). The roots
319
of this characteristic equation, called complex parameters, characterize the influence of elasticity
320
anisotropy on the plane deformation solution. Based on the values of the complex parameters,
321
we can evaluate how much the solution of an anisotropic body differs from an isotropic one for
322
which µ∗1 = µ∗2 = i. For the isotropic case, Eq. (11) becomes the well-known biharmonic equation, 16
323
∇4 φ = 0.
324
The components Sij in Eq. (5) are lengthy relations that make it difficult to analyze the
325
dependency of the parameter µ∗ on the elastic constants in the off-axis coordinate xy. The simplest
326
relations are obtained for Sij in the material coordinate system x00 y 00 . We will, therefore, evaluate
327
the complex parameters in this coordinate system. It is worth noting that, as the components of
328
the compliance matrix Sij change, based on the transformation rules of a fourth-order tensor, the
329
complex parameters must also change, based on a transformation rule. It can be shown that once
330
the complex parameters µ1 and µ2 are obtained in the x00 y 00 coordinate system, the transformed
331
complex parameters in xy are given by (Lekhnitskii, 1968) µ∗1 =
µ1 cos β + sin β , cos β − µ1 sin β
µ∗2 =
µ2 cos β + sin β . cos β − µ2 sin β
(13)
332
Here, µi and µ∗i (i = 1, 2) are the complex parameters in x00 y 00 and xy, respectively. Thus, the
333
solution of plane deformation depends on the complex parameters µ1 and µ2 , which are the material
334
properties on the material coordinate system, and on β which is the material orientation. By
335
substituting the plane-stress compliance matrix components of Eq. (2) into Eq. (12) and multiplying
336
by E, we obtain µ4 +
E G0
− 2ν 0
E 2 E µ + 0 =0. 0 E E
(14)
337
This shows that three physical parameters, E/E 0 and E/G0 as well as ν 0 , define the dependency
338
of the complex parameters on the elastic constants. Chen et al. (1998b) used these parameters to
339
study the influence of anisotropy on the stress intensity factors of the TCBD specimen.
340
Another approach to determine the dependency of the complex parameters on the elastic con-
341
stants is to define two non-dimensional parameters, ξ = E/E 0 and η = E/G0 −2ν 0 E/E 0 . In this defi-
342
nition, ξ indicates the anisotropy ratio of Young’s modulus, however η appears to be a rather mathe-
345
matical expression as opposed to being a parameter that has a clear physical meaning. Claesson and √ Bohloli (2002) suggested another pair of parameters, ξ = E/E 0 and η = (1/G0 − 2ν 0 /E 0 ) EE 0 /2, √ and rewrote Eq. (14) as µ4 + 2η ξµ2 + ξ = 0. However, this definition also results in η being a
346
parameter that is more of a mathematical nature for which defining bounds is not straightforward.
347
We thus propose here an alternative for η, which has a clear physical meaning, and for which
348
proper bounds and also an average can be defined. To do so, we define the two non-dimensional
349
parameters as
343 344
ξ=
E E0
and η =
17
G0 . G0sv
(15)
350 351
Here, ξ and η indicate the anisotropy ratios of the Young’s modulus and the apparent shear modulus, as discussed in Section 4.3. Rewriting Eq. (14), based on these parameters, yields µ4 +
352 353
h1 + ξ η
+ 2ξν 0
1 − ηi 2 µ +ξ =0 . η
(16)
As ν 0 1 and η ≈ 1, the contribution of the term 2ξν 0 (1 − η)/η may be negligible. This assumption follows the simplification of Eq. 16 to µ4 +
1+ξ 2 µ +ξ =0 . η
(17)
354
In order to evaluate this simplification, we compiled the reported values of the elastic constants
355
for different rock types in the literature, and plotted the transverse Poisson’s ratio and the pa-
356
rameter η against the Young’s modulus anisotropy ratio, ξ, in Figure 4. These plots show the
357
experimental values from about forty experiments conducted on different rock types. Figure 4
358
indeed shows that the transverse Poisson’s ratio is small and that η is close to unity for most rocks.
359
Note that an average value for η is one, according to Figure 4. This means that the Saint-Venant
360
relation can yield a good approximation for the transverse shear modulus. Based on these reported
361
experimental results, we define the following bounds: 0 < ν 0 < 0.3,
0.5 < η < 1.5 .
(18)
2
0.5 Cho et al., 2012 Chen et al., 1998a Amadei, 1996 Talesnick and Ringel, 1999 Gholami and Rasouli, 2014 Wang and Laio, 1998 Liao et al., 1997b other available data
0.4 0.3 0.2
1.5
1
G'=G' SV
0.5 0.1 0
0 1
1.5
2
2.5
3
3.5
4
1
1.5
2
2.5
3
3.5
4
Figure 4: Experimental data for ν 0 and η = G0 /G0sv versus the anisotropy ratio ξ = E/E 0 . The values are taken from Cho et al. (2012); Chen et al. (1998a); Amadei (1996); Talesnick and Ringel (1999); Gholami and Rasouli (2014); Wang and Liao (1998); Liao et al. (1997b). Other available data include the values reported in Liao et al. (1997a); Chen et al. (2008); Chou and Chen (2008); Espada and Lamas (2016); Togashi et al. (2018a,b).
362
Let us now evaluate the error induced in the complex parameters due to neglecting the term
363
2ξν 0 (1
364
(16), given by
− η)/η in the characteristic equation. The exact complex parameters are the roots of Eq.
18
v s u u 1 t 1+ξ 1−η 1+ξ 1−η 2 0 0 √ µ1 = − − 2ξν − + 2ξν − 4ξ, η η η η 2 v s u 1 u 1 + ξ 1 − η 1+ξ 1−η 2 t 0 0 µ2 = √ − − 2ξν + + 2ξν − 4ξ, η η η η 2 365 366
368
(19) µ4 = µ ¯2 .
In contrast, the approximated complex parameters (˜ µi , i = 1, 2) are obtained from the roots of Eq. (17): v s u 1 u 1 + ξ 1+ξ 2 t µ ˜1 = √ − − − 4ξ, η η 2 v s u u 1 t 1+ξ 1+ξ 2 √ µ ˜2 = − + − 4ξ, η η 2
367
µ3 = µ ¯1 ,
¯˜1 , µ ˜3 = µ (20) ¯˜2 . µ ˜4 = µ
We now define two error measures to evaluate the error induced by using Eq. (20) instead of Eq. (19): |µ1 | − |˜ µ1 | , |µ1 | |µ2 | − |˜ µ2 | =− , |µ2 |
Arg(µ1 ) − Arg(˜ µ1 ) , Arg(µ1 ) Arg(µ2 ) − Arg(˜ µ2 ) = . Arg(µ2 )
em µ1 = +
eaµ1 =
em µ2
eaµ2
(21)
369
Here |µ| and Arg(µ) respectively denote the magnitude and the argument of the complex
370
parameter µ in polar form. Figure 5 illustrates the errors bounds defined by Eq. (21), considering
371
a transverse Poisson’s ratio of ν 0 = 0.3 and two cases of η = 0.5 and η = 1.5. Note that the error
372
is zero for the case η = 1. Since any Poisson’s ratio smaller than 0.3 yields lower errors, Figure 5
373
demonstrates the errors bounds in the complex parameters with the constraints given by Eq. (18).
374
Figure 5 shows that considering realistic rock properties, a maximum error of about 10% occurs in
375
the complex parameters’ magnitudes and arguments. Note that 10% represents the error bound,
376
whereas for most rocks, the error is much smaller than 10% as the Saint-Venant relation results in
377
a good approximation given that most data are close to the line η = 1. The transverse Poisson’s
378
ratio is also much smaller than 0.3 in Figure 5. We therefore conclude that the complex parameters
379
can be practically given by Eq. (20) using the two dimensionless parameters ξ and η.
19
15
15 10
e
1
=1.5
0 =0.5
Error in
Error in
1
5
m
ea
-5
1
=1.5
-10 -15
=0.5
em
5
2
(%)
=0.5
(%)
10
2
=1.5
0
=0.5
ea
-5
2
=1.5
-10 1
1.5
2
2.5
3
3.5
-15
4
1
1.5
2
2.5
3
3.5
4
Figure 5: The range of errors in the complex parameters, defined by Eq. (21), due to neglecting the term 2ξν 0 (1−η)/η in the characteristic equation.
380
The above discussion leads to the following conclusion: when assuming an anisotropic plane
381
deformation with only traction boundary conditions, the elasticity solution depends, with a good
382
approximation for rocks, on two non-dimensional material parameters, ξ = E/E 0 and η = G0 /G0sv ,
383
and the material orientation, β.
384
The proof of this conclusion is as follows: for unequal complex conjugate roots, that occur in
385
most anisotropic elasticity problems, the general solution in the xy coordinate system becomes
386
(Lekhnitskii, 1968): φ(x, y) = 2Re[F1 (z1 ) + F2 (z2 )],
387 388
z1 = x + µ∗1 y,
z2 = x + µ∗2 y .
(22)
By introducing the new complex potentials Φ1 (z1 ) = dF1 /dz1 , Φ2 (zz ) = dF2 /dz2 , the in-plane stress and displacements are given by σx = 2Re µ∗1 2 Φ01 (z1 ) + µ∗2 2 Φ02 (z2 ) σy = 2Re Φ01 (z1 ) + Φ02 (z2 ) τxy = −2Re µ∗1 Φ01 (z1 ) + µ∗2 Φ02 (z2 )
(23)
u = 2Re p1 Φ1 (z1 ) + p2 Φ2 (z2 ) v = 2Re q1 Φ1 (z1 ) + q2 Φ2 (z2 ) . 389
where pi = S11 µ∗i 2 − S16 µ∗i + S12 (24) qi =
S12 µ∗i
− S26 + 20
S22 /µ∗i
.
390
The complex potentials Φ1 (x + µ∗1 y) and Φ2 (x + µ∗2 y) satisfy the equilibrium and compatibility
391
equations. Such potentials define the elasticity solution, if they also satisfy the traction and
392
displacement boundary conditions. When only traction-based boundary conditions are given, the
393
dependency of the elasticity solution on the material is due to µ∗1 and µ∗2 only. From Eqs. (13)
394
and (13), the complex parameters in any coordinate system are functions of the anisotropy ratios
395
ξ and η, and the anisotropy orientation β. It was shown that the error due to neglecting the
396
term 2ξν 0 (1 − η)/η in the characteristic equation, Eq. (16), is trivial for rocks. Note that when
397
displacement constraints are necessary to define the boundary conditions, the actual value of the
398
elasticity constants also play a role in defining the solution, according to Eq. (24).
399
We suggest that the two parameters ξ and η are good choices to work with for determining the
400
elasticity solution of anisotropic rocks since:
401 402
• ξ and η are non-dimensional parameters with clear physical meanings. ξ and η denote the anisotropy ratios of the Young’s modulus and the apparent shear modulus, respectively.
403
• Clear upper and lower bounds can be defined for ξ and η in rocks. The anisotropy of the
404
Young’s modulus varies by up to about four in most rocks (1 < ξ < 4) (Amadei, 1996),
405
and the anisotropy of the apparent shear modulus can be bounded between 0.5 < η < 1.5
406
according to the available data in the literature. These bounds are beneficial and helpful
407
when analysing the importance of the anisotropy in the elasticity solutions of rock samples.
408
• The Saint-Venant relation yields a good approximation of the transverse shear modulus,
409
leading to η = 1. Therefore, in cases, where not enough data are available for the material
410
constants, the choice of η = 1 can be reasonably employed to characterize the elasticity
411
solution. In such cases, the solutions only depend on the anisotropy ratio ξ, fixing η to the
412
representative value of η = 1.
413
6. Crack tip fields in anisotropic media
414
Let us consider the near-tip region of a crack in an anisotropic plane as shown in Figure 6.
415
Assuming that this plane is a symmetry plane of the material, and using the crack-tip coordinate
416
system xy, the stress field adjacent to the crack tip under pure Mode I loading is given by (Sih
417
et al., 1965)
21
" KI µ ∗ µ∗ σx = √ Re ∗ 1 2 ∗ µ1 − µ2 2πr
µ∗2 µ∗1 p p − cos θ + µ∗2 sin θ cos θ + µ∗1 sin θ
!#
" KI 1 σy = √ Re ∗ µ1 − µ∗2 2πr
µ∗1 µ∗2 p p − cos θ + µ∗2 sin θ cos θ + µ∗1 sin θ
!#
τxy
" KI µ∗ µ∗ =√ Re ∗ 1 2 ∗ µ1 − µ2 2πr
,
1 1 p −p ∗ cos θ + µ1 sin θ cos θ + µ∗2 sin θ
σy
(25)
!# .
τxy σx
r
y,v
,
β
θ
Crack
x,u y' '
x''
Figure 6: The near-tip stress field for a sharp crack in an anisotropic plane.
418
Here, r and θ are the polar coordinates of the material point, Re denotes the real parts of
420
complex numbers, and µ∗1 and µ∗2 are the complex parameters in the xy coordinate system. We assume that this plane is subjected to the remote traction boundary condition t¯ on Γt . The stress
421
variation along the crack ligament (θ = 0) is given by
419
KI σy = √ , 2πr
h i KI Re −µ∗1 µ∗2 , σx = √ 2πr
τxy = 0 ,
(26)
422
which shows that the shear component of the stresses vanishes along the crack ligament, irrespective
423
of the material orientation, β. Note that when β 6= 0, π/2, the Mode I loading also includes shear
424
deformation in addition to the opening mode. As explained in the previous section, the stress
425 426 427 428
solution of the anisotropic plane is a function of the boundary condition as well as the complex parameters KI (µ∗1 , µ∗2 , t¯). From Eqs. (20) and (13), the complex parameters in any coordinate system are functions of the anisotropy ratios ξ and η as well as the anisotropy orientation β. Therefore, the Mode I stress intensity factor is a function of KI (ξ, η, β, t¯). This means that the 22
429
stress intensity factor is not only dependent on geometry and loading, but also on the material
430
anisotropy ratios and the material orientation.
431
7. Results of the finite element analysis
432
Based on the results in the previous sections, the stress intensity factor for the modified SCB
433
test, introduced in Section 3, is a function of several parameters: Geometrical ones are R, S1 and
434
a; the anisotropy-related ones are ξ , η and β as well as the load, P . Note that S2 is dependent
435
on S1 to experience a pure Mode I loading condition. In fact, for any geometrical and material
436
configuration, S2 must be obtained by the constraint that the crack tip is subject to pure Mode
437
I loading. The stress intensity factor is therefore a geometrical-, material- and load-dependent
438
KI (R, S1 , a, ξ, η, β, P ). We define the normalized stress intensity factor YI , which depends on the
439
geometry ratios, and the anisotropy ratios and the orientation: YI (S1 /R, a/R, ξ, η, β) =
2RBKI √ . P πa
(27)
440
It is worth noting that although we use displacement boundary constraints in our models, all
441
boundary conditions can also be set as force boundary conditions, since the forces can be readily
442
obtained from the balance of force and momentum in the specimen. Therefore, our conclusion in
443
the previous section regarding the dependency of the stress intensity factor on two non-dimensional
444
parameters applies here. Once YI is known for a specific Mode I configuration, the fracture tough-
445
ness is calculated using the geometrical dimensions and the maximum load, Pm , employing √ Pm π a KIc = YI 2RB
(28)
446
Hence, YI depends on both the geometrical factors and the material anisotropy and is gener-
447
ally obtained from numerical methods. We used the finite element (FE) method to obtain this
448
value for different geometrical factors and anisotropy ratios. The SCB specimen was modelled
449
and analysed with the commercial finite element code ABAQUS. A two-dimensional model with
450
quadratic plane stress quadrilateral elements was employed. The mid-side nodes in the first row
451
of the elements adjacent to the crack tip were moved to the one-quarter position to reproduce
452
the crack square root stress singularity (Barsoum, 1976; Nejati et al., 2015b). The finite ele-
453
ment mesh and boundary conditions are shown in Figure 7. An anisotropic elasticity model was
454
used to define the anisotropy ratios ξ = 1, 2, 3, 4 and η = 0.7, 1, 1.3 as well as the anisotropy
455
orientations of β = 0◦ , 15◦ , 30◦ , 45◦ , 60◦ , 75◦ , 90◦ . The contour integral module of ABAQUS uses
456
cylindrical domains to calculate the interaction integrals and subsequently the stress intensity fac-
457
tors (ABAQUS/CAE, 2014). The domain integral method, used to calculate the stress intensity 23
458
factors, has been successfully employed for isotropic materials (Shih and Asaro, 1988; Nejati et al.,
459
2015a, 2016) as well as for anisotropic ones (Wang et al., 1980; Banks-sills et al., 2005, 2007). In all
460
the FE analyses, the transverse Poisson’s ratio of ν 0 = 0.1 is employed to calculate the transverse
461
shear modulus based on the Saint-Venant relation.
Figure 7: The finite element mesh and the boundary conditions used during the finite element analyses of the SCB specimen.
462
For each configuration, finite element analyses were first performed to obtain the ratio S2 /S1
463
at which the crack is subjected to pure Mode I condition, and its corresponding stress intensity
464
factor is taken to compute YI from Eq. (27). The ratio S2 /S1 is determined in such a way
465
that |YII /YI | < 0.01. Table 3 lists the normalised SIFs of the SCB specimen for the geometrical
466
configuration a/R = 0.4, S1 /R = 0.6 and different anisotropy parameters. Figure 8 also shows the
467
variation of S2 /S1 and YI versus the anisotropy orientation, β. More plots for other geometrical
468
factors a/R = 0.5, 0.6 and S1 /R = 0.8 are given in Appendix B. The isotropic solution of YI ,
469
to which symmetric loading (S2 = S1 ) applies, is also plotted for comparison. Our FE isotropic
470
solution for YI is in good agreement with the value obtained by the formula given in Kuruppu et al.
471
(2014), showing only 2% deviation. It can also be seen in the plots that the influence of anisotropy
472
on the stress intensity solution, YI , is negligible for β = 0◦ , 90◦ compared to the isotropic solutions.
24
Table 3: The normalized SIFs of the SCB specimen for the case a/R = 0.4, S1 /R = 0.6. ξ
β (degree) 0 15 30 45 60 75 90 0 15 30 45 60 75 90 0 15 30 45 60 75 90
2
3
4
η=1 YII 0.00 -0.02 -0.00 -0.01 0.00 -0.03 0.00 0.00 0.02 0.00 -0.02 0.01 -0.02 0.00 0.00 0.00 -0.01 0.01 -0.02 0.01 0.00
YI 3.84 3.01 2.64 2.64 2.80 3.22 3.66 3.88 2.49 2.25 2.37 2.57 3.03 3.63 3.92 2.27 2.08 2.19 2.52 2.90 3.63
ξ=2
1
η = 0.7 YII 0.00 -0.02 -0.00 -0.02 0.01 0.02 0.00 0.00 0.01 0.01 -0.02 0.02 0.01 0.00 0.00 0.01 0.00 0.02 -0.01 0.03 0.00
YI 3.84 2.73 2.49 2.74 3.02 3.35 3.72 3.89 2.29 2.22 2.46 2.74 3.19 3.72 3.92 2.05 2.03 2.26 2.71 3.08 3.74
ξ=3
ξ=4
η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
η = 1.3 YII 0.00 -0.02 -0.02 0.02 0.01 -0.03 0.00 0.00 0.00 0.00 0.01 0.00 -0.02 0.00 0.00 -0.01 -0.01 -0.01 -0.01 0.02 0.00
YI 3.84 3.17 2.78 2.54 2.63 3.05 3.64 3.89 2.69 2.27 2.27 2.45 2.89 3.59 3.92 2.42 2.11 2.17 2.40 2.74 3.56
0.7 0.6 0.5 0.4 0.3 4 Isotropic
3.5
YI
3 2.5 2 1.5 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure 8: The span ratio S2 /S1 and normalised Mode I stress intensity factor YI for the SCB test for a/R = 0.4, S1 /R = 0.6.
473 474
In order to perform fracture toughness experiments, using the modified SCB test, the following steps are taken: 25
475
• The elastic constants of the rock must be measured employing proper methods (see Nejati
476
et al. (2018) for a list of available methods). The non-dimensional parameters ξ and η are then
477
calculated and used for the test design. In case there is no data available for the transverse
478
shear modulus, the representative value of η = 1 should be employed.
479
• The SCB specimens are prepared in such a way that the isotropy plane forms a normal angle
480
with the specimen and a desirable angle of 0◦ < β < 90◦ with the initial crack orientation.
481
The span length, S1 , is also chosen for the test. Based on the anisotropy parameters ξ, η
482
and β and the geometrical factors S1 /R and a/R, the values of the span S2 and YI are
483
obtained from Figure (8), or Figures (B.1-B.5) given in Appendix B. The fracture toughness
484
experiment is then performed, using this span ratio.
485
• The fracture toughness is finally calculated, using Eq. (28). Note that the crack is likely
486
to kink towards the weak features of the rock, such as foliation or bedding planes. In such
487
cases, the measured value may not represent the fracture toughness in the direction of the
488
initial crack.
489
As an example, let us consider the Grimsel Test Site granodiorite with the elastic parameters
490
ξ = 2, η = 1 (Nejati et al., 2018). Assume that a fracture toughness experiment is planned for the
491
configuration a/R = 0.4, S1 /R = 0.6 and β = π/4. The reference plot in Figure 8 gives the values
492
S2 /S1 = 0.62 (equivalent to S2 /R = 0.37) and YI = 2.64 for such a configuration. Performing the
493
experiment with such a span ratio and measuring the maximum load, Pm , provides the necessary
494
data to calculate the Mode I fracture toughness from Eq. (28).
495
8. Conclusions
496
The conventional SCB test, with symmetric three-point bending, generates a Mixed-Mode I/II
497
loading, when the crack is not aligned with one of the principal material directions. We suggest
498
a SCB test with asymmetric three-point bending to guarantee pure Mode I loading at the crack
499
tip in anisotropic rocks. We first show that the elasticity solution of an anisotropic plane with
500
traction boundary conditions depends on two non-dimensional parameters, the anisotropy ratios of
501
the Young’s modulus and the apparent shear modulus, as well as the material orientation. For the
502
rocks that satisfy the Saint-Venant assumption, this dependency is reduced to only two parameters:
503
the anisotropy ratio of the Young’s modulus and the material orientation. We provide reference
504
plots from the finite element solutions for the span ratio and the normalized stress intensity factor
505
for the SCB test with different anisotropy ratios and material orientations. These plots can be
506
employed to conduct pure Mode I fracture growth experiments in anisotropic rocks. 26
507
Acknowledgement
508
This research project was financially supported by the Swiss Innovation Agency Innosuisse and
509
is part of the Swiss Competence Center for Energy Research - Supply of Electricity (SCCER-SoE).
510
The authors also thank the Werner Siemens-Stiftung for its financial support.
27
511
Appendices
512
Appendix A. Transformation of the stress, strain, and compliance matrix
513
Analyzing anisotropic deformation requires the rotation of elasticity matrix from the coordi-
514
nates of the material orientation to the global coordinate system used. Consider two Cartesian
515
coordinate systems, X 0 and X, with bases {x0 , y 0 , z 0 } and {x, y, z}, respectively. Assuming that the
516 517
material is oriented according to the X 0 coordinate system, while the global coordinate system is defined by X. Let us consider the rotation matrix, Ω, in a way that Ωij = ei · e0j = cos xi , x0j .
518
The transformation of the stress, strain and compliance matrix from X 0 to X is then obtained from
519
(Ting, 1996): σ = Kσ 0
520
= (K −1 )T 0
S = (K −1 )T S 0 K −1 ,
(A.1)
where
K=
" # K1 2K2 K3
K4
" (K −1 )T
=
K1
K2
#
2K3 K4
Ω211 Ω212 Ω213 2 2 2 K1 = Ω Ω Ω 22 23 21 Ω231 Ω232 Ω233 Ω12 Ω13 Ω13 Ω11 Ω11 Ω12 K2 = Ω22 Ω23 Ω23 Ω21 Ω21 Ω22 Ω32 Ω33 Ω33 Ω31 Ω31 Ω32 Ω21 Ω31 Ω22 Ω32 Ω23 Ω33 K3 = Ω31 Ω11 Ω32 Ω12 Ω33 Ω13 Ω11 Ω21 Ω12 Ω22 Ω13 Ω23 Ω22 Ω33 + Ω23 Ω32 Ω23 Ω31 + Ω21 Ω33 Ω21 Ω32 + Ω22 Ω31 . K4 = Ω Ω + Ω Ω Ω Ω + Ω Ω Ω Ω + Ω Ω 32 13 33 12 33 11 31 13 31 12 32 11 Ω12 Ω23 + Ω13 Ω22 Ω13 Ω21 + Ω11 Ω23 Ω11 Ω22 + Ω12 Ω21
28
(A.2)
521
Appendix B. Finite element results for YI
522
The normalized stress intensity factor in anisotropic rocks does not only depend on the geometry
523
of the cracked specimen and the loading but also on the anisotropy ratios of the Young’s modulus
524
and the apparent shear modulus as well as the anisotropy orientation, as discussed in this paper.
525
Applying an asymmetric loading to the SCB specimen, shown in Figure 1, makes it possible to
526
impose pure Mode I loading, even in the presence of material anisotropy in the SCB specimen.
527
Figures B.1-B.5 show the variation of the span ratio S2 /S1 that yields pure Mode I condition
528
versus the anisotropy orientation β for different values of the span S1 , the crack length a and the
529
anisotropy ratios ξ and η. The normalized Mode I stress intensity factor, YI , is also given for
530
such pure Mode I loading conditions. Such reference plots provide data to configure pure Mode I
531
fracture toughness experiments for anisotropic rocks. ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5 0.4 0.3 5.5 Isotropic
5
Isotropic
YI
4.5 4 3.5 3 2.5 2 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.1: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.4, S1 /R = 0.8).
29
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5 0.4 0.3 5.5 Isotropic
5
YI
4.5 4 3.5 3 2.5 0
15
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90
β (degrees)
Figure B.2: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.5, S1 /R = 0.6).
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5
YI
0.4 0.3 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3
Isotropic
0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.3: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.5, S1 /R = 0.8).
30
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5
YI
0.4 0.3 7.5 7 6.5 6 5.5 5 4.5 4 3.5
Isotropic
0
15
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90
β (degrees)
Figure B.4: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.6, S1 /R = 0.6).
ξ=2
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 10.5
ξ=3
ξ=4
S2/S1
η=1 η=0.7 η=1.3
Isotropic
9.5
YI
8.5 7.5 6.5 5.5 4.5 3.5 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.5: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB specimen (a/R = 0.6, S1 /R = 0.8).
31
532
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Journal of Applied Mechanics 55, 299–316. Sih, G. C., Paris, P. C., Irwin, G. R., 1965. On cracks in rectilinearly anisotropic bodies. International Journal of Fracture Mechanics 1 (3), 189–203. Talesnick, M. L., Ringel, M., 1999. Completing the hollow cylinder methodology for testing of transversely isotropic rocks: Torsion testing. International Journal of Rock Mechanics and Mining Sciences 36 (5), 627–639.
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Ting, T. C.-T., 1996. Anisotropic elasticity: theory and applications. Oxford University Press, New York.
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Togashi, Y., Kikumoto, M., Tani, K., 2018a. Determining anisotropic elastic parameters of transversely isotropic
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rocks through single torsional shear test and theoretical analysis. Journal of Petroleum Science and Engineering
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Togashi, Y., Kikumoto, M., Tani, K., Hosoda, K., Ogawa, K., 2018b. Detection of deformation anisotropy of tuff
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by a single triaxial test on a single specimen. International Journal of Rock Mechanics and Mining Sciences 108,
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23–36. Wang, C. D., Liao, J. J., 1998. Stress influence Charts for transversely isotropic rocks. International Journal of Rock Mechanics and Mining Sciences 35 (6), 771–785. Wang, S. S., Yau, J. F., Corten, H. T., 1980. A mixed-mode crack analysis of rectilinear anisotropic solids using
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Wei, M. D., Dai, F., Liu, Y., Xu, N. W., Zhao, T., 2018a. An experimental and theoretical comparison of CCNBD
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and CCNSCB specimens for determining mode I fracture toughness of rocks. Fatigue and Fracture of Engineering
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Materials and Structures 41 (5), 1002–1018.
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Wei, M. D., Dai, F., Xu, N. W., Liu, J. F., Xu, Y., 2016a. Experimental and Numerical Study on the Cracked
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Chevron Notched Semi-Circular Bend Method for Characterizing the Mode I Fracture Toughness of Rocks. Rock
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chevron notched specimens for mode I fracture toughness testing of rocks. Engineering Fracture Mechanics 168,
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notched Brazilian disc specimen for fracture toughness testing of rock. Fatigue & Fracture of Engineering Materials
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Highlights:
- The conventional semi-circular bend (SCB) test of anisotropic rocks generates a Mixed-Mode I/II crack tip loading. - We present a modified SCB test for anisotropic rocks to ensure a pure Mode I crack tip loading. - The stress intensity factor solution of an anisotropic SCB sample depends on two dimensionless parameters as well as the anisotropy orientation. - Extensive finite element analyses are performed to obtain the configuration of the SCB specimen for anisotropic rocks.
S0013-7944(18)31359-6 https://doi.org/10.1016/j.engfracmech.2019.03.008 EFM 6385
To appear in:
Engineering Fracture Mechanics
Received Date: Accepted Date:
4 December 2018 9 March 2019
Please cite this article as: Nejati, M., Aminzadeh, A., Saar, M.O., Driesner, T., Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks, Engineering Fracture Mechanics (2019), doi: https://doi.org/ 10.1016/j.engfracmech.2019.03.008
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2
Modified semi-circular bend test to determine the fracture toughness of anisotropic rocks
3
Morteza Nejati∗a , Ali Aminzadehb,c , Martin O. Saarb , Thomas Driesnerc
1
4 5 6
7
a Department of Earth Sciences, ETH Zurich, Switzerland Geothermal Energy and Geofluids, Department of Earth Sciences, ETH Zurich, Switzerland c Institute of Geochemistry and Petrology, Department of Earth Sciences, ETH Zurich, Switzerland b
Abstract
8
The conventional semi-circular bend (SCB) test of anisotropic rocks, with symmetric loading,
9
generates a Mixed-Mode I/II crack tip loading when the crack is not aligned with one of the prin-
10
cipal material directions. This paper presents a modified SCB test for anisotropic rocks to ensure
11
a pure Mode I crack tip loading. It is demonstrated that the stress intensity factor (SIF) solution
12
of an anisotropic SCB specimen depends on two dimensionless parameters, the anisotropy ratios
13
of the Young’s modulus and apparent shear modulus, as well as the anisotropy orientation. These
14
two dimensionless parameters are selected since they have physical meaning, and are generally
15
bounded for rock materials. Based on these anisotropy parameters, the semi-circular test is modi-
16
fied to configure pure Mode I, or tensile stress at the crack tip, by using an asymmetric three-point
17
bend configuration. Extensive finite element analyses are performed to obtain the span ratio and
18
normalized SIF of the SCB specimen with different anisotropy ratios and orientations.
19
Keywords: Fracture toughness; semi-circular bend; anisotropy; Mode I
Email address: *[email protected] (Morteza Nejati∗ )
Preprint submitted to Engineering Fracture Mechanics
February 26, 2019
ν, ν 0 ξ σ, σi τij , γij φ, Φ1 , Φ2 , F1 , F2
Nomenclature Crack length Specimen thickness Errors of the magnitude and the argument of the approximate complex parameters µ ˜i , (i = 1, 2) Young’s moduli within and normal to the isotropy plane Shear moduli within and normal to the plane of isotropy Transverse shear modulus approximated from the Saint-Venant relation Shear modulus in xy plane which is formed by the rotation of x00 y 00 along z 00 by the angle −β Mode I stress intensity factor and fracture toughness Load and peak load Polar coordinates of a point near the crack tip Sample radius Span length for SCB specimen with asymmetric three-point bend Compliance matrix and its ij component Traction and displacement boundary conditions Displacements along x and y directions Cartesian coordinates Normalised Mode I stress intensity factor Angle between the anisotropy orientation and the horizontal plane, or crack plane in cracked bodies Strain vector and strain component ij Anisotropy ratio of apparent shear modulus, G0 /G0sv Conjugate pair of roots to the characteristic equation in the material coordinate system x00 y 00 , known as complex parameters Complex parameters in xy coordinate system Approximated complex parameters when neglecting the term 2ξν 0 (1 − η)/η in the characteristic equation Poisson’s ratio within and normal to the plane of isotropy Anisotropy ratio of Young’s modulus, E/E 0 Stress vector and normal stress component in i direction Shear stress and strain ij Complex potentials
Abbreviations CB CCNBD FPZ GTS ISRM LEFM SCB SR TCBD
Chevron bend Cracked chevron notched Brazilian disc Fracture process zone Grimsel Test Site International Society for Rock Mechanics Linear elastic fracture mechanics Semi-circular bend Short rod Through-thickness cracked Brazilian disk
a B a em µi , eµi
E, E 0 G, G0 G0sv G0β KI , KIc P , Pm r, θ R S1 , S2 S, Sij t¯, u ¯ ux , uy x, y YI β , ij η µi , µ¯i , i = 1, 2 µ∗i , µ ¯∗i ¯˜i µ ˜i , µ
2
20
1. Introduction
21
The mechanical behavior of many types of rocks is anisotropic due to their complex micro-
22
structure. Rock anisotropy is often introduced due to one of the following processes: (i) Foliation
23
in metamorphic rocks that results in the flattening of grains and the alignment of platy minerals;
24
(ii) Bedding and lamination in sedimentary rocks that introduce a layered structure. Taking rock
25
anisotropy into account is crucial in the accurate predictions of rock mass deformation, stability
26
and failure (Amadei, 1996; Dutler et al., 2018; Krietsch et al., 2018). Performing Mode I fracture
27
toughness experiments in anisotropic rocks provides very useful information about the fracturing
28
processes that can occur in a wide range of geomechanical and geophysical problems. An important
29
application is related to enhanced geothermal systems, normally placed in the crystalline basement,
30
where anisotropic rocks, such as granite or gneiss, are likely to be present. Recently Amann et al.
31
(2018), Gischig et al. (2018), Jalali et al. (2018), and Doetsch et al. (2018) have demonstrated the
32
importance of anisotropy in the in-situ stimulation and circulation project in the deep underground
33
laboratory at the Grimsel Test Site (GTS) in Switzerland.
34
Three main rock mechanical properties are used for geomechanical and geophysical analyses:
35
(i) elasticity; (ii) strength (both tensile and compressive); and (iii) fracture toughness (both tensile
36
and shearing). These properties are anisotropic in a large number of sedimentary and metamor-
37
phic rocks that exhibit features such as foliation and bedding. These features yield a preferred
38
orientation of the constituent minerals, pores, and cracks, which can be idealised geometrically
39
with an axis of symmetry normal to the foliation and bedding planes (Jaeger et al., 2007; Dambly
40
et al., 2018). Such an axis of symmetry makes transverse isotropy a suitable model for predict-
41
ing the deformational behaviour of rocks. The transversely isotropic elasticity model requires five
42
independent material constants that need to be measured experimentally (Dambly et al., 2018;
43
Nejati et al., 2018). The dominant orientation of micro-cracks along features like bedding and
44
foliation yields a preferential path for failure and fracture growth. Therefore, strength and fracture
45
toughness of rocks also exhibit anisotropy. These two properties are closely related through the
46
fracture process zone (FPZ) which also exhibits anisotropy in size (Dutler et al., 2018).
47
Our literature survey in Section 2 describes the available schemes for conducting fracture tough-
48
ness experiments on rocks. This review shows that the overall attempt over the last four decades
49
has been to reduce the variability of the fracture toughness among different tests to improve the
50
reliability of the measurements. Applying constraints to the sample size has significantly helped
51
towards reaching this goal, reducing the variation of measured values among different tests consid-
52
erably (Whittaker et al., 1992; Iqbal and Mohanty, 2007; Wei et al., 2018a). Most of these schemes
53
are designed based on assuming isotropic elasticity, where the material constants play no role in the
54
elasticity solutions of the specimens. However, in anisotropic rocks both the elasticity constants 3
55
and the material orientation influence the stress solution. Although some researches have evaluated
56
the effect of material orientation on fracture toughness, the literature appear to lack research on
57
the potential influence of the elasticity constants on the measured values of fracture toughness.
58
Consequently, there remains a literature gap regarding the influence of elasticity anisotropy when
59
conducting fracture toughness tests in anisotropic rocks.
60
Kuruppu et al. (2014) presented the fourth ISRM-suggested method to measure Mode I frac-
61
ture toughness using the semi-circular bend (SCB) specimen. The SCB specimen seems to be a
62
great choice for conducting fracture toughness experiments in anisotropic rocks. However, despite
63
addressing the applicability of the SCB test for anisotropic rocks in Kuruppu et al. (2014), the in-
64
fluence of the elasticity anisotropy on the fracture toughness measurement has not been evaluated.
65
This study investigates the anisotropic elasticity solution of the SCB specimen, and modifies the
66
standard SCB testing scheme by employing asymmetric loading to make it applicable to conduct-
67
ing pure Mode I fracture toughness measurements in anisotropic rocks. Enforcing a pure Mode I
68
loading condition enables us to i) determine the Mode I fracture toughness in a specific direction if
69
the crack grows in a self-similar manner, and ii) investigate how the crack path is influenced by the
70
anisotropy if a crack kinking occurs. Note that understanding the fracturing process under Mode
71
I is a prerequisite to the more complex Mixed-Mode I/II loading condition.
72
2. A review of fracture toughness tests
73
Fracture toughness represents the critical value of the stress intensity factor (SIF), defined in
74
the context of linear elastic fracture mechanics (LEFM), in a particular crack deformation mode,
75
at which the onset of crack propagation occurs. Determining Mode I or tensile fracture toughness,
76
referred to as KIc , is of importance due to the dominance of this loading type in many applications.
77
Since the mid 1970s, numerous specimen geometries have been suggested to determine KIc . For
78
metallic materials, standard methods include single-edge-notched and fatigue-pre-cracked plates
79
under tension or three-point bending loads (ASTM, 2018). While suitable for metals, these tests
80
are unfavorable for rocks because: (i) They require large and impracticably shaped specimens, and
81
thus significant machining; (ii) Difficulties can arise when applying tensile loads to rocks, as rocks
82
are weak under tension, so that failure may occur in places, where the load is applied (Whittaker
83
et al., 1992). Such shortcomings of the ASTM (2018) standard directed efforts to introduce tests
84
with core-based samples, readily prepared from drilled rocks, and preferably under compressive
85
loads, to determine the fracture toughness of rocks.
86
To obtain precise, consistent and reproducible values in KIc of rocks, the International Society
87
for Rock Mechanics (ISRM) is recommending four core-based test procedures, as listed in Table
88
1. These tests include the chevron-notched round bar in bending (CB), the chevron-notched 4
89
short rod in splitting (SR), the cracked chevron-notched Brazilian disk in dimetrical compression
90
(CCNBD), and the semi-circular bend (SCB) specimen. Associated guidelines provide the details
91
of the requirements for the samples’ preparation and dimensions, as well as the test procedure
92
requirements in terms of loading type and rate. Explicit formulae are also provided to calculate
93
the fracture toughness from the failure load and geometrical factors. Despite standardized testing,
94
the results from CB, SR and CCNBD can still exhibit deviations which are often explained by
95
size effects, anisotropy of the rock and inaccuracy of the dimensionless parameters used in the
96
calculations. Iqbal and Mohanty (2007) compared CB and CCNBD methods on three different
97
rock types with two-hundred specimens and concluded that the results of these two methods
98
were comparable, when the correct equation for fracture toughness calculations was used and the
99
specimen size was selected carefully. Wei et al. (2016b, 2018a) showed that the inconsistency of
100
the results from CB, SR and CCNBD may still be present when using accurate equations for the
101
SIFs. However, they suggested that this inconsistency may be explained by the size of the FPZ
102
developed in these different specimens.
5
Table 1: ISRM-suggested methods for determining the Mode I fracture toughness of rocks.
Method
Test configuration
Chevron-notched short rod in splitting (SR)
R
2R
A
A
P
!
P
L
a a0 2R
A-A
2R > 10× grain size L > 3.5(2R) 2S1 > 3.33(2R) ± 0.02(2R) Ψ = 90◦ ± 1◦ a0 = 0.15(2R) ± 0.10(2R)
A P
Chevron-notched round bar in bending (CB)
! a a0 2S1 L
2R A-A
A
a
a0
P/B A-A B P/B
R
A
Notched semicircular specimen in bending (SCB specimen)
2R = 75.0 mm B/R = 0.80 a0 /R = 0.2637 a1 /R = 0.65
R
a1
A
a 2S1 A
Reference
Barker (1977); Ouchterlony (1988)
Ouchterlony (1980, 1988)
B P/B
A
Cracked ChevronNotched Brazilian Disk in dimetrical compression (CCNBD)
Size requirements 2R > 10× grain size L = 1.45(2R) ± 0.02(2R) Ψ = 54.6◦ ± 1◦ a0 = 0.48(2R) ± 0.02(2R)
P/B A-A
2R > 10× grain size or 76 mm B > 0.4D or 30 mm 0.40 ≤ a/R ≤ 0.60 0.50 ≤ S1 /R ≤ 0.80
Sheity et al. (1985); Fowell and Xu (1994); Fowell (1995)
Chong and Kuruppu (1984); Kuruppu and Chong (2012); Kuruppu et al. (2014) Kuruppu and Chong (2012)
103
Two methods are generally used to create cracks in test specimens: 1) naturally pre-fabricated
104
cracks, produced through fatigue loading (applicable mainly to metals) or stable crack growth in 6
105
chevron-notched samples; 2) artificially made cracks by sawing a through-thickness slit. Out of
106
the four ISRM-suggested methods for fracture toughness measurements, the first three, i.e. SR,
107
CB and CCNBD, use chevron-notched specimens, while the SCB test uses a through-thickness
108
crack. Chevron-notches provide a stable and sub-critical crack growth in the first stage of the
109
crack extension. This is because the length of the crack front gradually increases, causing the SIF
110
to drop and the load required for the crack extension to rise. The peak load marks the beginning
111
of the second stage of the crack extension, i.e. the unstable crack growth, when the SIF starts to
112
increase with the crack extension. The numerical results on the SIF of the chevron-notch clearly
113
show these two stages (Wei et al., 2016a,b). The stable crack propagation in the first stage can
114
readily be controlled, so that self-pre-cracking is essentially achieved.
115
As a material property, Mode I plane-strain fracture toughness is not expected to vary with the
116
testing procedure as well as the specimen type and size. However, the measured fracture toughness
117
seems to be independent of specimen type and size only when certain minimum specimen size
118
requirements are met. This behavior can be attributed to the large size of the FPZ, compared
119
to the sample dimensions in small samples. Since the concept of stress intensity factor is valid in
120
the context of LEFM, the FPZ must be sufficiently smaller than the dimensions of the specimen
121
in order for the LEFM to be applicable. Therefore, ensuring representative fracture toughness
122
measurements requires the characteristic dimensions, such as the crack length, the crack ligament,
123
and the specimen thickness to be sufficiently large. Since the FPZ size depends on the grain size,
124
the minimum size of the specimen also depends on the grain size of the rock. Fracture toughness,
125
measured during tests that satisfy minimum sample size requirements, has been proven to be a
126
material property (Whittaker et al., 1992). Suggested methods in Table 1 have guidelines on the
127
minimum dimensions required for the samples.
128
The first three suggested methods have the following disadvantages: (i) The SR and CB tests
129
require large amounts of intact rock core; (ii) The SR test is based on the splitting load configuration
130
which requires a complicated loading frame; (iii) The SR, CB and CCNBD tests require complex
131
sample preparation schemes to produce chevron notches. The SCB test, however, can be conducted
132
with smaller samples and enable a straightforward preparation scheme as well as a simple three-
133
point bending configuration. The CCNBD and SCB tests have the advantage that they can be
134
applied on Mode II and Mixed-Mode I/II fracture toughness tests. During CCNBD tests, this
135
is achieved by adjusting the diametrical loading so that it does not align with the crack plane.
136
During SCB tests, this is achieved by either introducing a slanted edge crack (Chong and Kuruppu,
137
1988; Lim et al., 1993) or an asymmetrical three-point bending load configuration (Ayatollahi
138
et al., 2011). Note that the use of CCNBD tests for Mixed-Mode I/II or Mode II loading may be
139
problematic since the stable pre-cracking obtained through a chevron-notch is expected to align in 7
140
the notch plane, whereas kinking is often expected during Mixed-Mode I/II or Mode II loading.
141
Recent CCNBD Mode II fracture toughness tests show a complex concave or convex fracture
142
surface initiating from the two edges of the chevron notched ligaments (Wei et al., 2018b). This
143
complexity of the fracture path invalidates the II fracture toughness results from the CCNBD
144
specimen. In order to avoid this issue and also the complications in preparing chevron notches,
145
several researchers have suggested the through-thickness cracked Brazilian disc (TCBD) test, which
146
seems to be more suitable for mixed-mode experiments than the CCNBD (Krishnan et al., 1998;
147
Chen et al., 1998b; Ke et al., 2008).
148
The SCB test has become popular recently due to its advantages in terms of testing require-
149
ments, and also its potential to test anisotropic rocks and Mixed-Mode I/II tests. The SCB test
150
was suggested by Chong and Kuruppu (1984), and since then, it has been extended and improved
151
to accurately determine the fracture toughness not only under pure Mode I (Lim et al., 1994a;
152
Adamson et al., 1996) but also Mixed-Mode I/II conditions (Lim et al., 1994b; Khan and Al-
153
Shayea, 2000; Ayatollahi et al., 2006; Aliha et al., 2010, 2012). A lot of effort has been devoted
154
to accurately obtain the stress intensity factor formula for the entire range of pure Mode I to
155
pure Mode II conditions (Lim et al., 1993; Ayatollahi and Aliha, 2007). Researchers have also
156
studied various factors influencing fracture toughness, including the effect of specimen size (Lim
157
et al., 1994a; Khan and Al-Shayea, 2000), loading rate (Lim et al., 1994a; Dai and Xia, 2013), rock
158
anisotropy (Chong et al., 1987; Kataoka et al., 2015; Lee et al., 2015), chemical solutions (Karfakis
159
and Akram, 1993), water content (Lim et al., 1994a), temperature, confining pressure and water
160
vapor pressure (Funatsu et al., 2004; Kataoka et al., 2015), and possible effects of T-stress (Wei
161
et al., 2017). The SCB test has the following advantages:
162 163
164 165
166 167
168 169
• Small compact specimens can be produced directly from standard rock cores requiring minimal machining, thanks to its simple and practical cylindrical geometry. • The three-point bend loading configuration is easy to implement, using any compressive test frame. • Fracture toughness is determined simply from the maximum load and the geometric dimensions, employing well-established formulae. • Slight modifications of the specimen or loading specifications makes it possible to conduct the entire range of Mixed-Mode I/II fracture toughness tests.
170
• Fracture toughness testing of anisotropic rocks is conducted simply by cutting specimens from
171
a single core and by generating notches with desirable angles, with respect to the anisotropy
172
orientation. 8
173
The effect of the rock micro-structure anisotropy on the fracture toughness has been investigated
174
in a number of studies. Table 2 lists the fracture toughness experiments performed on different
175
anisotropic rocks using the aforementioned test setups. Apart from these tests, the cracked-ring
176
disk (Chen et al., 2008) and the notched deep beam (Luo et al., 2018) tests have also been employed
177
for fracture toughness experiments. Most of these cases consider the end-member configurations,
178
where the crack is oriented along one of the principal material directions (normally referred to
179
as arrester, divider or short-transverse (Chong et al., 1987)) of anisotropic rocks. An important
180
aspect during these tests is the proper identification of the role of the micro-structure in the fracture
181
toughness anisotropy. Fracture toughness is in fact closely linked to the presence of micro-cracks
182
and their orientation. Of the works listed in Table 2, only Chen et al. (1998b), Ke et al. (2008)
183
and Dai and Xia (2013) have considered the elasticity anisotropy of the specimens, while the
184
remaining studies have simply assumed an isotropic elasticity solution. Elasticity anisotropy of
185
rocks often accompanies fracture toughness anisotropy. Therefore, one needs to use an anisotropic
186
elasticity solution to obtain the stress intensity factor for fracture toughness calculations. This
187
article discusses how the elasticity anisotropy must be taken into account for fracture toughness
188
determination employing the SCB test. Table 2: A summary of tests used to measure fracture toughness of anisotropic rocks.
189
Test SR
Rock type Shale
Loading mode Mode I
CCNBD
Granite
Mode I
SCB
Granite, Shale
Mode I
TCBD
Sandstone, Marble , Shale
Mixed-Mode I/II
Reference Chandler et al. (2016, 2017) Nasseri et al. (2005, 2006); Nasseri and Mohanty (2008); Nasseri et al. (2009, 2010, 2011) Chong et al. (1987); Dai and Xia (2013); Kataoka et al. (2015); Lee et al. (2015); Dutler et al. (2018); Shi et al. (2019) Krishnan et al. (1998); Chen et al. (1998b); Ke et al. (2008)
3. Modified semi-circular bend test for anisotropic rocks
190
Although the semi-circular bend test has been a common test to determine fracture toughness
191
during the last two decades, it has been only recently listed in the ISRM-suggested methods
192
(Kuruppu et al., 2014). This test is based on a symmetric three-point bend configuration, i.e.
193
two symmetric supports hold the sample at the base and a compressive line load is applied at
194
the top (S1 = S2 in Figure 1). The Mode I fracture toughness is then determined employing
195
the normalized stress intensity factor, which is a function of the loading configuration and the
196
geometry. In anisotropic rocks, however, the stress intensity factor also depends on the anisotropy 9
197
paramaters, as will be shown in the next sections. In fact, a symmetrical loading (S1 = S2 ) of
198
materials, that do not show symmetry with respect to the load direction (β 6= 0, π/2), yields a
199
Mixed-Mode I/II crack tip loading.
C P/B β R
S1
R a
B
S2 C
Section CC
Figure 1: Schematics of the SCB test with asymmetric loading.
200
In order to resolve this issue, we suggest an asymmetric loading condition for SCB testing of
201
anisotropic rocks, as shown in Figure 1. This configuration was originally proposed by Ayatollahi
202
et al. (2011) to obtain the Mixed-Mode I/II loading of isotropic SCB specimens. For any geometrical
203
and material configuration, S2 can be obtained in such a way that the Mode II stress intensity
204
factor, KII , vanishes and only a pure Mode I condition is applied. We investigate the dependency
205
of the stress solution on the anisotropy parameters in the next three sections, before reporting the
206
details of our proposed modified SCB test configuration in Section 7.
207
4. Transverse isotropy material model
208
Anisotropic elasticity implies a directional dependency of the material deformability. Depend-
209
ing on the number of symmetry planes, different numbers of material constants are required to
210
describe the response of anisotropic materials: twenty-one constants for the triclinic model with
211
no symmetry plane; thirteen constants for the monoclinic model with one symmetry plane; nine
212
constants for the orthotropic model with three symmetry planes; five constants for the transversely
213
isotropic model with an axis of symmetry; and, ultimately, two elasticity constants for the isotropic
214
material with an infinite number of symmetry planes. This section gives some details about the
215
transversely isotropic material model. 10
216
4.1. Transverse isotropy
217
Due to the presence of foliation or bedding, many rock types exhibit a geometrical micro-
218
structural axis of symmetry, which makes the transverse isotropy a suitable model to describe
219
their response. This model defines an isotropy plane, often assumed to coincide with features
220
such as foliation and bedding, and an infinite number of symmetry planes, parallel to the axis of
221
symmetry. Consider the Cartesian coordinate system x0 y 0 z 0 , aligned with the material coordinate
222
system, where the isotropy plane coincides with the plane x0 y 0 , while z 0 denotes the axis of symmetry
223
(Figure 2a). The well-known generalized Hooke’s law defines the constitutive law as 1 E x0 y0 z 0 0 0 = γy z γ 0 0 xz γ x0 y 0 | {z }
ν E 1 E
−
ν0 E0 ν0 − 0 E 1 E0 −
0
0
0
0
0 1 G0
0 0 1 G0
0
|
{z S0
0
σ x0 0 σy0 σ 0 0 z τy0 z 0 0 τx0 z 0 0 τ 0 0 xy 1 | {z } G } σ0
(1)
224
Here, 0 and σ 0 are the strain and stress vectors, respectively, where the Vigot’s notation is
225
used, and S 0 is the well-known compliance matrix. Due to the requirement of positive strain
226
energy, S 0 must be a symmetric positive-definite matrix. Five independent engineering constants
227
characterize the elasticity of the transversely isotropic model in principal coordinates: E and E 0 ,
228
in-plane and transverse Young’s moduli, are applied within and normal to the isotropy plane x0 y 0 ;
229
ν and ν 0 , in-plane and transverse Poisson’s ratios, are applied within and normal to the isotropy
230
plane x0 y 0 ; and G0 , the transverse shear modulus, is applied transverse to the isotropy plane x0 y 0 .
231
The in-plane shear modulus, G, is applied within the isotropy plane x0 y 0 and depends on E and ν
232
through the relation G = E/[2(1 + ν)]. Defining directions 1, 2 and 3 along axes x0 , y 0 and z 0 , the
233
following equalities hold: E = E1 = E2 , E 0 = E3 , ν = ν12 = ν21 , ν 0 = ν31 = ν32 , G0 = G13 = G23
234
and G = G12 (see illustrative details in Nejati et al. (2018)).
235
We point out that the Poisson’s ratio is generally defined as νij = −j /i for a stress in
236
the i direction. Due to the symmetry of the elasticity matrix, νij and νji are related through
237
νij /Ei = νji /Ej . By this definition, the transverse Poisson’s ratio ν 0 = ν31 = ν32 characterizes
238
the relative strain in Directions 1 or 2 when a normal stress is applied along Direction 3. One
239
can alternatively define the transverse Poisson’s ratio as ν ∗ = ν13 = ν23 , which implies relative
240
strain in Direction 3 when normal stress is applied in Directions 1 or 2. The two definitions are 11
y
x
n
t
z
Гt y''
y'' x'' z''
y'
z'
y,u y x''
x' β
β
x,u x
Гu ū
a)
b)
Figure 2: a) Schematics of the material coordinate systems x0 y 0 z 0 and x00 y 00 z 00 with respect to the global coordinate system xyz. x00 y 00 forms an angle β with xy. b) The anisotropic deformation problem in the plane xy, with traction and displacement boundary conditions.
241
dependent on the relation ν 0 /E 0 = ν ∗ /E. In this paper, we choose to work with ν 0 . This is of
242
particular importance, as some commercial finite element packages may import ν13 (or ν23 ) as
243
inputs, requiring the correct calculation of these values employing ν ∗ = ν13 = ν23 = ν 0 E/E 0 .
244
4.2. Two-dimensional compliance matrix
245
Let us assume a coordinate system x00 y 00 z 00 in such a way that the x00 y 00 plane is orthogonal to
246
the material isotropy plane x0 y 0 , and z 00 is oriented along y 0 as shown in Figure 2a. We wish to
247
investigate the two-dimensional deformation within the plane x00 y 00 , where the maximum variation
248
of the deformability occurs. Note that x00 y 00 forms a symmetry plane of the material deformation.
249
The plane-stress deformation within x00 y 00 is based on the assumptions σz 00 = τx00 z 00 = τy00 z 00 = 0,
250
and depends only on four elastic constants: the Young’s moduli along x00 and y 00 (E and E 0 ), and
251
the Poisson’s ratio and shear modulus in the plane x00 y 00 (ν 0 and G0 ). Applying the transformation
252
formula in Appendix A to Eq. (1), the stress-strain relationship within x00 y 00 reads 1 E ν0 = − E0
x00
y00 γx00 y00
ν0 − 0 E 1 E0
0
0
0 00 σ x . 00 σ 0 y 00 00 τ 1 x y G0
(2)
253
The plane-strain deformation is based on the assumptions z 00 = γx00 z 00 = γy00 z 00 = 0, and also
254
depends on the Poisson’s ratio within the isotropy plane, ν. The 2D plane-strain Hooke’s law reads
12
1 − ν2 E x00 ν 0 (1 + ν) y00 = − E0 γx00 y00 0
ν 0 (1 + ν) E0 1 − (E/E 0 )ν 02 E0 −
0
0 σx00 0 σy00 . τx00 y00 1
(3)
G0
255
Let us assume that the xyz coordinates are formed by rotating x00 y 00 z 00 along the axis z 00 by an
256
angle −β, as shown in Figure 2, so that the isotropy plane has an angle β with respect to the x-
257
axis in the xyz coordinate system. Analyzing the deformation in xy coordinate system requires
258
the transformation of the compliance matrices in Eqs. (2) and (3). Using the transformation rules,
259
given in Appendix A, Hooke’s law in xy coordinate reads
S11 S12 S16 σx y = S12 S22 S26 σy , γxy S16 S26 S66 τxy | {z } | {z } | {z } x
260
S
(4)
σ
where Sij , i, j = 1, 2, 6 are the components of the compliance matrix, given by S11 S12 S16 S22 S26 S66
cos4 β sin4 β 1 2ν 0 sin2 2β = + + − 0 E E0 G0 E 4 2 1 1 1 sin 2β ν0 = + 0− 0 − 0 cos4 β + sin4 β E E G 4 E 2 cos β sin2 β 1 ν0 = − − − cos 2β sin 2β E E0 2G0 E 0 sin4 β cos4 β 1 2ν 0 sin2 2β = + + − E E0 G0 E0 4 2 sin β cos2 β 1 ν0 = − + − cos 2β sin 2β E E0 2G0 E 0 1 1 2ν 0 cos2 2β = + 0 + 0 sin2 2β + E E E G0
(5)
261
for the plane-stress state. For the plane-strain condition, Sij are readily obtaind by replacing 1/E,
262
1/E 0 and ν 0 /E 0 with (1 − ν 2 )/E, (1 − (E/E 0 )ν 02 )/E 0 and ν 0 (1 + ν)/E 0 , respectively.
263
4.3. Saint-Venant relation
264
The measurement of the elastic constants in rocks is not an easy task. This has frequently
265
led researchers to approximate some of the constants. For example, the well-known Saint-Venant
266
relation provides an approximation of the shear moduli in orthotropic materials, based on the 13
267
values of the Young’s moduli and Poisson’s ratios (Saint-Venant, 1863): 1 + 2νji 1 1 = + Gij Ei Ej
i, j = 1, 2, 3 .
(6)
268
Within the isotropic plane ij (Ei = Ej ), this relation simplifies to 1/G = 2(1 + ν)/E, which
269
indicates the exact dependency between the shear and the Young’s moduli and the Poisson’s ratio.
270
This dependency holds within the isotropy plane of a transversely isotropic material. However, the
271
transverse shear modulus, obtained from the Saint-Venant relation, 1/G0sv = 1/E + (1 + 2ν 0 )/E 0 ,
272
is only an approximation. In reality, G0 is an independent constant and can deviate from the
273
approximated G0sv value.
274
Despite being known in the literature as an empirical relation, the Saint-Venant approximation
275
has in fact a theoretical basis. Let us consider the shear deformation in the xy plane, which is
276
normal to the isotropy plane, as shown in Figure 3a. Using the compliance matrix component S66
277
in Eq. (5), the relation between the shear stress and the shear strain in this plane is given by
278
τxz = G0β γxz , where 1 1 + 2ν 0 2 1 sin2 2β cos2 2β 1 2 = + sin 2β + cos 2β = + G0β E E0 G0 G0sv G0
(7)
279
The variation of the ratio G0β /G0sv against the angle β is plotted in Figure 3b for three ratios of
280
G0 /G0sv . It can be seen that at β = π/4 the apparent shear modulus is equal to the Saint-Venant
281
approximation: 1 G0β β=π/4
=
1 1 1 + 2ν 0 = + G0sv E E0
(8)
282
This equation shows that the Saint-Venant approximation is in fact equal to the apparent shear
283
modulus G0β in a coordinate system rotated by π/4 with respect to the isotropy plane. The higher
284
the deviation of β from π/4, the higher the deviation of G0β from G0sv .
14
τxy
1.3 G'= 1.2 G' sv
G' / G' sv
y''
x''
1.2
τxy
y β
G'= G' sv G'= 0.8 G' sv
1.1 1 0.9 0.8
x
0.7 0
a)
/4
/2
b)
Figure 3: a) Schematics of a solid element in the xy coordinate system which forms an angle β with respect to the material coordinate system x00 y 00 , under shear stress τxy . b) The variation of the apparent shear modulus, G0β , against the angle β (τxz = G0β γxz ).
285
A question raised by many researchers is how accurate the Saint-Venant approximation is.
286
Based on the experimental measurements from two-hundred static and dynamic tests, Worotnicki
287
(1993) confirmed the validity of the Saint-Venant approximation for many rock types, such as
288
granite, gneiss, sandstone and basalt. Comparisons made by Talesnick and Ringel (1999) and Cho
289
et al. (2012) confirm this finding as they show variations of only up to 20% between the measured
290
values of the transverse shear modulus and the Saint-Venant approximation for anisotropy ratios of
291
up to E/E 0 = 2. It is worth noting that there are error sources related to the measurement schemes
292
employed, nonlinear response of the rock, as well as the rock heterogeneity where multiple samples
293
are used to determine the constants. For example, the tests conducted by Cho et al. (2012)
294
determine the transverse shear modulus indirectly based on a third sample, while the Young’s
295
moduli and Poisson’s ratios, used to calculate G0sv , are obtained from two other samples. Generally,
296
there is consensus that the Saint-Venant relation yields a good approximation for rocks with low to
297
moderate anisotropy ratios. Note that the majority of rock types show low to moderate anisotropy
298
ratios (Worotnicki, 1993; Amadei, 1996).
299
5. Elasticity solution of plane deformation
300
This section investigates the dependency of the stress solution on the material properties in an
301
anisotropic plane. Let us consider the plane deformation of an anisotropic material, schematically
302
illustrated in Figure 2b. We wish to solve the spatial deformation within the plane, u = (ux , uy ),
303
with the strain-displacement dependency governed by the infinitesimal strain theory, Eq. (9a),
304
where the strains satisfy the compatibility relation in Eq. (9b). Stress is related to deformation 15
305
in Eq. (9c) through the generalized Hooke’s law for a linear elastic anisotropic material, and
306
conserves momentum in Eq. (9d). Stresses and displacements must satisfy boundary conditions
307
given by Eq. (9e).
∂uy ∂ux ∂ux ∂uy , y = , γxy = + ∂x ∂y ∂y ∂x 2 2 2 ∂ xy ∂ x ∂ y + −2 =0 2 2 ∂y ∂x ∂x∂y = Sσ ∂σxy ∂σy ∂σx ∂σyx + = 0, + =0 ∂x ∂y ∂x ∂y σn = t¯ on Γt , u = u ¯ on Γu
x =
(9a) (9b) (9c) (9d) (9e)
308
The Airy stress function , φ(x, y), is introduced, where the stresses, defined through this po-
309
tential, automatically satisfy the equilibrium equation, Eq. (9d), for the case where body forces
310
are zero: σx =
311 312
∂2φ , ∂y 2
σy =
∂2φ , ∂x2
τxy = −
∂2φ . ∂x∂y
(10)
Inserting these stresses into the compatibility equation, Eq. (9b), employing the stress-strain relation given by Eq. (4), yields: S22
∂4φ ∂4φ ∂4φ ∂4φ ∂4φ − 2S + (2S + S ) − 2S + S =0, 12 66 26 16 11 ∂x4 ∂x3 ∂y ∂x2 ∂y 2 ∂x∂y 3 ∂y 4
(11)
313
where Sij are defined by Eq. (5). The general solution to this equation can be found using the
314
method of characteristics, looking for solutions of the form φ = φ(x+µ∗ y), where µ∗ is a parameter.
315
Substituting this stress function into Eq. (11) yields the characteristic equation: S11 µ∗ 4 − 2S16 µ∗ 3 + 2S12 + S66 µ∗ 2 − 2S26 µ∗ + S22 = 0 .
316 317 318
(12)
This equation governs the dependency of the elasticity solution on the material constants. It can be shown that the roots of the characteristic equation are complex, and always occur in conjugate pairs: µ∗1 = a1 + ib1 , µ∗2 = a2 + ib2 , µ∗3 = µ¯∗1 , µ∗4 = µ¯∗2 (Lekhnitskii, 1968). The roots
319
of this characteristic equation, called complex parameters, characterize the influence of elasticity
320
anisotropy on the plane deformation solution. Based on the values of the complex parameters,
321
we can evaluate how much the solution of an anisotropic body differs from an isotropic one for
322
which µ∗1 = µ∗2 = i. For the isotropic case, Eq. (11) becomes the well-known biharmonic equation, 16
323
∇4 φ = 0.
324
The components Sij in Eq. (5) are lengthy relations that make it difficult to analyze the
325
dependency of the parameter µ∗ on the elastic constants in the off-axis coordinate xy. The simplest
326
relations are obtained for Sij in the material coordinate system x00 y 00 . We will, therefore, evaluate
327
the complex parameters in this coordinate system. It is worth noting that, as the components of
328
the compliance matrix Sij change, based on the transformation rules of a fourth-order tensor, the
329
complex parameters must also change, based on a transformation rule. It can be shown that once
330
the complex parameters µ1 and µ2 are obtained in the x00 y 00 coordinate system, the transformed
331
complex parameters in xy are given by (Lekhnitskii, 1968) µ∗1 =
µ1 cos β + sin β , cos β − µ1 sin β
µ∗2 =
µ2 cos β + sin β . cos β − µ2 sin β
(13)
332
Here, µi and µ∗i (i = 1, 2) are the complex parameters in x00 y 00 and xy, respectively. Thus, the
333
solution of plane deformation depends on the complex parameters µ1 and µ2 , which are the material
334
properties on the material coordinate system, and on β which is the material orientation. By
335
substituting the plane-stress compliance matrix components of Eq. (2) into Eq. (12) and multiplying
336
by E, we obtain µ4 +
E G0
− 2ν 0
E 2 E µ + 0 =0. 0 E E
(14)
337
This shows that three physical parameters, E/E 0 and E/G0 as well as ν 0 , define the dependency
338
of the complex parameters on the elastic constants. Chen et al. (1998b) used these parameters to
339
study the influence of anisotropy on the stress intensity factors of the TCBD specimen.
340
Another approach to determine the dependency of the complex parameters on the elastic con-
341
stants is to define two non-dimensional parameters, ξ = E/E 0 and η = E/G0 −2ν 0 E/E 0 . In this defi-
342
nition, ξ indicates the anisotropy ratio of Young’s modulus, however η appears to be a rather mathe-
345
matical expression as opposed to being a parameter that has a clear physical meaning. Claesson and √ Bohloli (2002) suggested another pair of parameters, ξ = E/E 0 and η = (1/G0 − 2ν 0 /E 0 ) EE 0 /2, √ and rewrote Eq. (14) as µ4 + 2η ξµ2 + ξ = 0. However, this definition also results in η being a
346
parameter that is more of a mathematical nature for which defining bounds is not straightforward.
347
We thus propose here an alternative for η, which has a clear physical meaning, and for which
348
proper bounds and also an average can be defined. To do so, we define the two non-dimensional
349
parameters as
343 344
ξ=
E E0
and η =
17
G0 . G0sv
(15)
350 351
Here, ξ and η indicate the anisotropy ratios of the Young’s modulus and the apparent shear modulus, as discussed in Section 4.3. Rewriting Eq. (14), based on these parameters, yields µ4 +
352 353
h1 + ξ η
+ 2ξν 0
1 − ηi 2 µ +ξ =0 . η
(16)
As ν 0 1 and η ≈ 1, the contribution of the term 2ξν 0 (1 − η)/η may be negligible. This assumption follows the simplification of Eq. 16 to µ4 +
1+ξ 2 µ +ξ =0 . η
(17)
354
In order to evaluate this simplification, we compiled the reported values of the elastic constants
355
for different rock types in the literature, and plotted the transverse Poisson’s ratio and the pa-
356
rameter η against the Young’s modulus anisotropy ratio, ξ, in Figure 4. These plots show the
357
experimental values from about forty experiments conducted on different rock types. Figure 4
358
indeed shows that the transverse Poisson’s ratio is small and that η is close to unity for most rocks.
359
Note that an average value for η is one, according to Figure 4. This means that the Saint-Venant
360
relation can yield a good approximation for the transverse shear modulus. Based on these reported
361
experimental results, we define the following bounds: 0 < ν 0 < 0.3,
0.5 < η < 1.5 .
(18)
2
0.5 Cho et al., 2012 Chen et al., 1998a Amadei, 1996 Talesnick and Ringel, 1999 Gholami and Rasouli, 2014 Wang and Laio, 1998 Liao et al., 1997b other available data
0.4 0.3 0.2
1.5
1
G'=G' SV
0.5 0.1 0
0 1
1.5
2
2.5
3
3.5
4
1
1.5
2
2.5
3
3.5
4
Figure 4: Experimental data for ν 0 and η = G0 /G0sv versus the anisotropy ratio ξ = E/E 0 . The values are taken from Cho et al. (2012); Chen et al. (1998a); Amadei (1996); Talesnick and Ringel (1999); Gholami and Rasouli (2014); Wang and Liao (1998); Liao et al. (1997b). Other available data include the values reported in Liao et al. (1997a); Chen et al. (2008); Chou and Chen (2008); Espada and Lamas (2016); Togashi et al. (2018a,b).
362
Let us now evaluate the error induced in the complex parameters due to neglecting the term
363
2ξν 0 (1
364
(16), given by
− η)/η in the characteristic equation. The exact complex parameters are the roots of Eq.
18
v s u u 1 t 1+ξ 1−η 1+ξ 1−η 2 0 0 √ µ1 = − − 2ξν − + 2ξν − 4ξ, η η η η 2 v s u 1 u 1 + ξ 1 − η 1+ξ 1−η 2 t 0 0 µ2 = √ − − 2ξν + + 2ξν − 4ξ, η η η η 2 365 366
368
(19) µ4 = µ ¯2 .
In contrast, the approximated complex parameters (˜ µi , i = 1, 2) are obtained from the roots of Eq. (17): v s u 1 u 1 + ξ 1+ξ 2 t µ ˜1 = √ − − − 4ξ, η η 2 v s u u 1 t 1+ξ 1+ξ 2 √ µ ˜2 = − + − 4ξ, η η 2
367
µ3 = µ ¯1 ,
¯˜1 , µ ˜3 = µ (20) ¯˜2 . µ ˜4 = µ
We now define two error measures to evaluate the error induced by using Eq. (20) instead of Eq. (19): |µ1 | − |˜ µ1 | , |µ1 | |µ2 | − |˜ µ2 | =− , |µ2 |
Arg(µ1 ) − Arg(˜ µ1 ) , Arg(µ1 ) Arg(µ2 ) − Arg(˜ µ2 ) = . Arg(µ2 )
em µ1 = +
eaµ1 =
em µ2
eaµ2
(21)
369
Here |µ| and Arg(µ) respectively denote the magnitude and the argument of the complex
370
parameter µ in polar form. Figure 5 illustrates the errors bounds defined by Eq. (21), considering
371
a transverse Poisson’s ratio of ν 0 = 0.3 and two cases of η = 0.5 and η = 1.5. Note that the error
372
is zero for the case η = 1. Since any Poisson’s ratio smaller than 0.3 yields lower errors, Figure 5
373
demonstrates the errors bounds in the complex parameters with the constraints given by Eq. (18).
374
Figure 5 shows that considering realistic rock properties, a maximum error of about 10% occurs in
375
the complex parameters’ magnitudes and arguments. Note that 10% represents the error bound,
376
whereas for most rocks, the error is much smaller than 10% as the Saint-Venant relation results in
377
a good approximation given that most data are close to the line η = 1. The transverse Poisson’s
378
ratio is also much smaller than 0.3 in Figure 5. We therefore conclude that the complex parameters
379
can be practically given by Eq. (20) using the two dimensionless parameters ξ and η.
19
15
15 10
e
1
=1.5
0 =0.5
Error in
Error in
1
5
m
ea
-5
1
=1.5
-10 -15
=0.5
em
5
2
(%)
=0.5
(%)
10
2
=1.5
0
=0.5
ea
-5
2
=1.5
-10 1
1.5
2
2.5
3
3.5
-15
4
1
1.5
2
2.5
3
3.5
4
Figure 5: The range of errors in the complex parameters, defined by Eq. (21), due to neglecting the term 2ξν 0 (1−η)/η in the characteristic equation.
380
The above discussion leads to the following conclusion: when assuming an anisotropic plane
381
deformation with only traction boundary conditions, the elasticity solution depends, with a good
382
approximation for rocks, on two non-dimensional material parameters, ξ = E/E 0 and η = G0 /G0sv ,
383
and the material orientation, β.
384
The proof of this conclusion is as follows: for unequal complex conjugate roots, that occur in
385
most anisotropic elasticity problems, the general solution in the xy coordinate system becomes
386
(Lekhnitskii, 1968): φ(x, y) = 2Re[F1 (z1 ) + F2 (z2 )],
387 388
z1 = x + µ∗1 y,
z2 = x + µ∗2 y .
(22)
By introducing the new complex potentials Φ1 (z1 ) = dF1 /dz1 , Φ2 (zz ) = dF2 /dz2 , the in-plane stress and displacements are given by σx = 2Re µ∗1 2 Φ01 (z1 ) + µ∗2 2 Φ02 (z2 ) σy = 2Re Φ01 (z1 ) + Φ02 (z2 ) τxy = −2Re µ∗1 Φ01 (z1 ) + µ∗2 Φ02 (z2 )
(23)
u = 2Re p1 Φ1 (z1 ) + p2 Φ2 (z2 ) v = 2Re q1 Φ1 (z1 ) + q2 Φ2 (z2 ) . 389
where pi = S11 µ∗i 2 − S16 µ∗i + S12 (24) qi =
S12 µ∗i
− S26 + 20
S22 /µ∗i
.
390
The complex potentials Φ1 (x + µ∗1 y) and Φ2 (x + µ∗2 y) satisfy the equilibrium and compatibility
391
equations. Such potentials define the elasticity solution, if they also satisfy the traction and
392
displacement boundary conditions. When only traction-based boundary conditions are given, the
393
dependency of the elasticity solution on the material is due to µ∗1 and µ∗2 only. From Eqs. (13)
394
and (13), the complex parameters in any coordinate system are functions of the anisotropy ratios
395
ξ and η, and the anisotropy orientation β. It was shown that the error due to neglecting the
396
term 2ξν 0 (1 − η)/η in the characteristic equation, Eq. (16), is trivial for rocks. Note that when
397
displacement constraints are necessary to define the boundary conditions, the actual value of the
398
elasticity constants also play a role in defining the solution, according to Eq. (24).
399
We suggest that the two parameters ξ and η are good choices to work with for determining the
400
elasticity solution of anisotropic rocks since:
401 402
• ξ and η are non-dimensional parameters with clear physical meanings. ξ and η denote the anisotropy ratios of the Young’s modulus and the apparent shear modulus, respectively.
403
• Clear upper and lower bounds can be defined for ξ and η in rocks. The anisotropy of the
404
Young’s modulus varies by up to about four in most rocks (1 < ξ < 4) (Amadei, 1996),
405
and the anisotropy of the apparent shear modulus can be bounded between 0.5 < η < 1.5
406
according to the available data in the literature. These bounds are beneficial and helpful
407
when analysing the importance of the anisotropy in the elasticity solutions of rock samples.
408
• The Saint-Venant relation yields a good approximation of the transverse shear modulus,
409
leading to η = 1. Therefore, in cases, where not enough data are available for the material
410
constants, the choice of η = 1 can be reasonably employed to characterize the elasticity
411
solution. In such cases, the solutions only depend on the anisotropy ratio ξ, fixing η to the
412
representative value of η = 1.
413
6. Crack tip fields in anisotropic media
414
Let us consider the near-tip region of a crack in an anisotropic plane as shown in Figure 6.
415
Assuming that this plane is a symmetry plane of the material, and using the crack-tip coordinate
416
system xy, the stress field adjacent to the crack tip under pure Mode I loading is given by (Sih
417
et al., 1965)
21
" KI µ ∗ µ∗ σx = √ Re ∗ 1 2 ∗ µ1 − µ2 2πr
µ∗2 µ∗1 p p − cos θ + µ∗2 sin θ cos θ + µ∗1 sin θ
!#
" KI 1 σy = √ Re ∗ µ1 − µ∗2 2πr
µ∗1 µ∗2 p p − cos θ + µ∗2 sin θ cos θ + µ∗1 sin θ
!#
τxy
" KI µ∗ µ∗ =√ Re ∗ 1 2 ∗ µ1 − µ2 2πr
,
1 1 p −p ∗ cos θ + µ1 sin θ cos θ + µ∗2 sin θ
σy
(25)
!# .
τxy σx
r
y,v
,
β
θ
Crack
x,u y' '
x''
Figure 6: The near-tip stress field for a sharp crack in an anisotropic plane.
418
Here, r and θ are the polar coordinates of the material point, Re denotes the real parts of
420
complex numbers, and µ∗1 and µ∗2 are the complex parameters in the xy coordinate system. We assume that this plane is subjected to the remote traction boundary condition t¯ on Γt . The stress
421
variation along the crack ligament (θ = 0) is given by
419
KI σy = √ , 2πr
h i KI Re −µ∗1 µ∗2 , σx = √ 2πr
τxy = 0 ,
(26)
422
which shows that the shear component of the stresses vanishes along the crack ligament, irrespective
423
of the material orientation, β. Note that when β 6= 0, π/2, the Mode I loading also includes shear
424
deformation in addition to the opening mode. As explained in the previous section, the stress
425 426 427 428
solution of the anisotropic plane is a function of the boundary condition as well as the complex parameters KI (µ∗1 , µ∗2 , t¯). From Eqs. (20) and (13), the complex parameters in any coordinate system are functions of the anisotropy ratios ξ and η as well as the anisotropy orientation β. Therefore, the Mode I stress intensity factor is a function of KI (ξ, η, β, t¯). This means that the 22
429
stress intensity factor is not only dependent on geometry and loading, but also on the material
430
anisotropy ratios and the material orientation.
431
7. Results of the finite element analysis
432
Based on the results in the previous sections, the stress intensity factor for the modified SCB
433
test, introduced in Section 3, is a function of several parameters: Geometrical ones are R, S1 and
434
a; the anisotropy-related ones are ξ , η and β as well as the load, P . Note that S2 is dependent
435
on S1 to experience a pure Mode I loading condition. In fact, for any geometrical and material
436
configuration, S2 must be obtained by the constraint that the crack tip is subject to pure Mode
437
I loading. The stress intensity factor is therefore a geometrical-, material- and load-dependent
438
KI (R, S1 , a, ξ, η, β, P ). We define the normalized stress intensity factor YI , which depends on the
439
geometry ratios, and the anisotropy ratios and the orientation: YI (S1 /R, a/R, ξ, η, β) =
2RBKI √ . P πa
(27)
440
It is worth noting that although we use displacement boundary constraints in our models, all
441
boundary conditions can also be set as force boundary conditions, since the forces can be readily
442
obtained from the balance of force and momentum in the specimen. Therefore, our conclusion in
443
the previous section regarding the dependency of the stress intensity factor on two non-dimensional
444
parameters applies here. Once YI is known for a specific Mode I configuration, the fracture tough-
445
ness is calculated using the geometrical dimensions and the maximum load, Pm , employing √ Pm π a KIc = YI 2RB
(28)
446
Hence, YI depends on both the geometrical factors and the material anisotropy and is gener-
447
ally obtained from numerical methods. We used the finite element (FE) method to obtain this
448
value for different geometrical factors and anisotropy ratios. The SCB specimen was modelled
449
and analysed with the commercial finite element code ABAQUS. A two-dimensional model with
450
quadratic plane stress quadrilateral elements was employed. The mid-side nodes in the first row
451
of the elements adjacent to the crack tip were moved to the one-quarter position to reproduce
452
the crack square root stress singularity (Barsoum, 1976; Nejati et al., 2015b). The finite ele-
453
ment mesh and boundary conditions are shown in Figure 7. An anisotropic elasticity model was
454
used to define the anisotropy ratios ξ = 1, 2, 3, 4 and η = 0.7, 1, 1.3 as well as the anisotropy
455
orientations of β = 0◦ , 15◦ , 30◦ , 45◦ , 60◦ , 75◦ , 90◦ . The contour integral module of ABAQUS uses
456
cylindrical domains to calculate the interaction integrals and subsequently the stress intensity fac-
457
tors (ABAQUS/CAE, 2014). The domain integral method, used to calculate the stress intensity 23
458
factors, has been successfully employed for isotropic materials (Shih and Asaro, 1988; Nejati et al.,
459
2015a, 2016) as well as for anisotropic ones (Wang et al., 1980; Banks-sills et al., 2005, 2007). In all
460
the FE analyses, the transverse Poisson’s ratio of ν 0 = 0.1 is employed to calculate the transverse
461
shear modulus based on the Saint-Venant relation.
Figure 7: The finite element mesh and the boundary conditions used during the finite element analyses of the SCB specimen.
462
For each configuration, finite element analyses were first performed to obtain the ratio S2 /S1
463
at which the crack is subjected to pure Mode I condition, and its corresponding stress intensity
464
factor is taken to compute YI from Eq. (27). The ratio S2 /S1 is determined in such a way
465
that |YII /YI | < 0.01. Table 3 lists the normalised SIFs of the SCB specimen for the geometrical
466
configuration a/R = 0.4, S1 /R = 0.6 and different anisotropy parameters. Figure 8 also shows the
467
variation of S2 /S1 and YI versus the anisotropy orientation, β. More plots for other geometrical
468
factors a/R = 0.5, 0.6 and S1 /R = 0.8 are given in Appendix B. The isotropic solution of YI ,
469
to which symmetric loading (S2 = S1 ) applies, is also plotted for comparison. Our FE isotropic
470
solution for YI is in good agreement with the value obtained by the formula given in Kuruppu et al.
471
(2014), showing only 2% deviation. It can also be seen in the plots that the influence of anisotropy
472
on the stress intensity solution, YI , is negligible for β = 0◦ , 90◦ compared to the isotropic solutions.
24
Table 3: The normalized SIFs of the SCB specimen for the case a/R = 0.4, S1 /R = 0.6. ξ
β (degree) 0 15 30 45 60 75 90 0 15 30 45 60 75 90 0 15 30 45 60 75 90
2
3
4
η=1 YII 0.00 -0.02 -0.00 -0.01 0.00 -0.03 0.00 0.00 0.02 0.00 -0.02 0.01 -0.02 0.00 0.00 0.00 -0.01 0.01 -0.02 0.01 0.00
YI 3.84 3.01 2.64 2.64 2.80 3.22 3.66 3.88 2.49 2.25 2.37 2.57 3.03 3.63 3.92 2.27 2.08 2.19 2.52 2.90 3.63
ξ=2
1
η = 0.7 YII 0.00 -0.02 -0.00 -0.02 0.01 0.02 0.00 0.00 0.01 0.01 -0.02 0.02 0.01 0.00 0.00 0.01 0.00 0.02 -0.01 0.03 0.00
YI 3.84 2.73 2.49 2.74 3.02 3.35 3.72 3.89 2.29 2.22 2.46 2.74 3.19 3.72 3.92 2.05 2.03 2.26 2.71 3.08 3.74
ξ=3
ξ=4
η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
η = 1.3 YII 0.00 -0.02 -0.02 0.02 0.01 -0.03 0.00 0.00 0.00 0.00 0.01 0.00 -0.02 0.00 0.00 -0.01 -0.01 -0.01 -0.01 0.02 0.00
YI 3.84 3.17 2.78 2.54 2.63 3.05 3.64 3.89 2.69 2.27 2.27 2.45 2.89 3.59 3.92 2.42 2.11 2.17 2.40 2.74 3.56
0.7 0.6 0.5 0.4 0.3 4 Isotropic
3.5
YI
3 2.5 2 1.5 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure 8: The span ratio S2 /S1 and normalised Mode I stress intensity factor YI for the SCB test for a/R = 0.4, S1 /R = 0.6.
473 474
In order to perform fracture toughness experiments, using the modified SCB test, the following steps are taken: 25
475
• The elastic constants of the rock must be measured employing proper methods (see Nejati
476
et al. (2018) for a list of available methods). The non-dimensional parameters ξ and η are then
477
calculated and used for the test design. In case there is no data available for the transverse
478
shear modulus, the representative value of η = 1 should be employed.
479
• The SCB specimens are prepared in such a way that the isotropy plane forms a normal angle
480
with the specimen and a desirable angle of 0◦ < β < 90◦ with the initial crack orientation.
481
The span length, S1 , is also chosen for the test. Based on the anisotropy parameters ξ, η
482
and β and the geometrical factors S1 /R and a/R, the values of the span S2 and YI are
483
obtained from Figure (8), or Figures (B.1-B.5) given in Appendix B. The fracture toughness
484
experiment is then performed, using this span ratio.
485
• The fracture toughness is finally calculated, using Eq. (28). Note that the crack is likely
486
to kink towards the weak features of the rock, such as foliation or bedding planes. In such
487
cases, the measured value may not represent the fracture toughness in the direction of the
488
initial crack.
489
As an example, let us consider the Grimsel Test Site granodiorite with the elastic parameters
490
ξ = 2, η = 1 (Nejati et al., 2018). Assume that a fracture toughness experiment is planned for the
491
configuration a/R = 0.4, S1 /R = 0.6 and β = π/4. The reference plot in Figure 8 gives the values
492
S2 /S1 = 0.62 (equivalent to S2 /R = 0.37) and YI = 2.64 for such a configuration. Performing the
493
experiment with such a span ratio and measuring the maximum load, Pm , provides the necessary
494
data to calculate the Mode I fracture toughness from Eq. (28).
495
8. Conclusions
496
The conventional SCB test, with symmetric three-point bending, generates a Mixed-Mode I/II
497
loading, when the crack is not aligned with one of the principal material directions. We suggest
498
a SCB test with asymmetric three-point bending to guarantee pure Mode I loading at the crack
499
tip in anisotropic rocks. We first show that the elasticity solution of an anisotropic plane with
500
traction boundary conditions depends on two non-dimensional parameters, the anisotropy ratios of
501
the Young’s modulus and the apparent shear modulus, as well as the material orientation. For the
502
rocks that satisfy the Saint-Venant assumption, this dependency is reduced to only two parameters:
503
the anisotropy ratio of the Young’s modulus and the material orientation. We provide reference
504
plots from the finite element solutions for the span ratio and the normalized stress intensity factor
505
for the SCB test with different anisotropy ratios and material orientations. These plots can be
506
employed to conduct pure Mode I fracture growth experiments in anisotropic rocks. 26
507
Acknowledgement
508
This research project was financially supported by the Swiss Innovation Agency Innosuisse and
509
is part of the Swiss Competence Center for Energy Research - Supply of Electricity (SCCER-SoE).
510
The authors also thank the Werner Siemens-Stiftung for its financial support.
27
511
Appendices
512
Appendix A. Transformation of the stress, strain, and compliance matrix
513
Analyzing anisotropic deformation requires the rotation of elasticity matrix from the coordi-
514
nates of the material orientation to the global coordinate system used. Consider two Cartesian
515
coordinate systems, X 0 and X, with bases {x0 , y 0 , z 0 } and {x, y, z}, respectively. Assuming that the
516 517
material is oriented according to the X 0 coordinate system, while the global coordinate system is defined by X. Let us consider the rotation matrix, Ω, in a way that Ωij = ei · e0j = cos xi , x0j .
518
The transformation of the stress, strain and compliance matrix from X 0 to X is then obtained from
519
(Ting, 1996): σ = Kσ 0
520
= (K −1 )T 0
S = (K −1 )T S 0 K −1 ,
(A.1)
where
K=
" # K1 2K2 K3
K4
" (K −1 )T
=
K1
K2
#
2K3 K4
Ω211 Ω212 Ω213 2 2 2 K1 = Ω Ω Ω 22 23 21 Ω231 Ω232 Ω233 Ω12 Ω13 Ω13 Ω11 Ω11 Ω12 K2 = Ω22 Ω23 Ω23 Ω21 Ω21 Ω22 Ω32 Ω33 Ω33 Ω31 Ω31 Ω32 Ω21 Ω31 Ω22 Ω32 Ω23 Ω33 K3 = Ω31 Ω11 Ω32 Ω12 Ω33 Ω13 Ω11 Ω21 Ω12 Ω22 Ω13 Ω23 Ω22 Ω33 + Ω23 Ω32 Ω23 Ω31 + Ω21 Ω33 Ω21 Ω32 + Ω22 Ω31 . K4 = Ω Ω + Ω Ω Ω Ω + Ω Ω Ω Ω + Ω Ω 32 13 33 12 33 11 31 13 31 12 32 11 Ω12 Ω23 + Ω13 Ω22 Ω13 Ω21 + Ω11 Ω23 Ω11 Ω22 + Ω12 Ω21
28
(A.2)
521
Appendix B. Finite element results for YI
522
The normalized stress intensity factor in anisotropic rocks does not only depend on the geometry
523
of the cracked specimen and the loading but also on the anisotropy ratios of the Young’s modulus
524
and the apparent shear modulus as well as the anisotropy orientation, as discussed in this paper.
525
Applying an asymmetric loading to the SCB specimen, shown in Figure 1, makes it possible to
526
impose pure Mode I loading, even in the presence of material anisotropy in the SCB specimen.
527
Figures B.1-B.5 show the variation of the span ratio S2 /S1 that yields pure Mode I condition
528
versus the anisotropy orientation β for different values of the span S1 , the crack length a and the
529
anisotropy ratios ξ and η. The normalized Mode I stress intensity factor, YI , is also given for
530
such pure Mode I loading conditions. Such reference plots provide data to configure pure Mode I
531
fracture toughness experiments for anisotropic rocks. ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5 0.4 0.3 5.5 Isotropic
5
Isotropic
YI
4.5 4 3.5 3 2.5 2 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.1: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.4, S1 /R = 0.8).
29
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5 0.4 0.3 5.5 Isotropic
5
YI
4.5 4 3.5 3 2.5 0
15
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90
β (degrees)
Figure B.2: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.5, S1 /R = 0.6).
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5
YI
0.4 0.3 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3
Isotropic
0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.3: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.5, S1 /R = 0.8).
30
ξ=2
1
ξ=3 η=1 η=0.7 η=1.3
0.9 0.8
S2/S1
ξ=4
0.7 0.6 0.5
YI
0.4 0.3 7.5 7 6.5 6 5.5 5 4.5 4 3.5
Isotropic
0
15
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90 0
15
β (degrees)
30
45
60
75
90
β (degrees)
Figure B.4: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB test (a/R = 0.6, S1 /R = 0.6).
ξ=2
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 10.5
ξ=3
ξ=4
S2/S1
η=1 η=0.7 η=1.3
Isotropic
9.5
YI
8.5 7.5 6.5 5.5 4.5 3.5 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
β (degrees)
75
90 0
15
30
45
60
75
90
β (degrees)
Figure B.5: The span ratio S2 /S1 and the normalised Mode I stress intensity factor YI for the SCB specimen (a/R = 0.6, S1 /R = 0.8).
31
532
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Highlights:
- The conventional semi-circular bend (SCB) test of anisotropic rocks generates a Mixed-Mode I/II crack tip loading. - We present a modified SCB test for anisotropic rocks to ensure a pure Mode I crack tip loading. - The stress intensity factor solution of an anisotropic SCB sample depends on two dimensionless parameters as well as the anisotropy orientation. - Extensive finite element analyses are performed to obtain the configuration of the SCB specimen for anisotropic rocks.