anisotropy amadei

Int. J. Rock Mech. Min. Sri’. & Geomech. Absrr. Vol. 33, No. 3, pp. 293-325, 1996 Copyright 0 1996 Elsevier Science Ltd

Printed in Great Britain. All rights reserved 0148~9062/96 Sl5.00 + 0.00

Pergamon

Importance of Anisotropy When Estimating and Measuring In Situ Stresses in Rock -

B. AMADEIT Foliated metamorphic rocks and laminated, stratljied or bedded sedimentary rocks have properties @hysical, dynamic, thermal, mechanical, hydraulic) that vary with direction and are said to be anisotropic. Rock mass anisotropy can be found in volcanic formations and sedimentary formations consisting of alternating layers or beds of direrent rock types. Rock masses cut by one or several regularly spaced joint sets are anisotropic in addition to being discontinuous. This paper deals with the subjects of rock anisotropy and rock stress. Both topics are important in rock engineering and are interrelated. Rock fabric controls the build-up in in situ stresses in the Earth’s crust, their magnitude and orientation. On the other hand, stresses and in particular compressive stresses tend to close microcracks or discontinuities thus making rock behavior non-linear and rock anisotropy pressure dependent. This paper is divided into three parts. In the jirst part, existing models to describe the deformability properties of anisotropic rocks as well as the laboratory and-field methods to determine those properties are reviewed. Then, it is shown how to account for both rock anisotropy (intact and joint induced) and the curvature of the Earth when estimating in situ stresses in rock masses. Finally, the effect of anisotropy on the analysis of overcoring measurements with cells similar to the CSIR Triaxial Strain Cell is discussed.

INTRODUCTION

Many rocks exposed near the Earth’s surface show well defined fabric elements in the form of bedding, stratification, layering, foliation, fissuring or jointing. In general, these rocks have properties (physical, dynamic, thermal, mechanical, hydraulic) that vary with direction and are said to be inherently anfiotropic. Anisotropy can be found at different scales in a rock mass ranging from intact specimens to the entire rock mass.

Anisotropy is a characteristic of intact foliated metamorphic rocks (slates, gneisses, phyllites, schists). In these rocks, the fabric can be expressed in different ways. Closely spaced fractures called cleavages are found, for instance, in slates and phyllites. These rocks tend to split into planes due to parallel orientation of micro- scopic grains of mica, chlorite or other platy minerals. In schists, the fabric is created by the parallel to sub-parallel arrangement of large platy minerals such as mica, chlorite and talc. Foliation can also be expressed by alternating layers of different mineral composition such as in gneisses. Non-foliated metamorphic rocks such as marble also show some anisotropy due to preferred orientation of calcite grains. Anisotropy is also the characteristic of intact laminated, stratified or bedded sedimentary rocks such as shales, sandstones, siltstones, limestones, coal,

TDepartment of Civil, Environmental, and Architectural Engineering, University of Colorado, Boulder, CO 80309-0428, U.S.A.

etc. Here, the anisotropy results from complex physical and chemical processes associated with transportation, deposition, compaction, cementation, etc. It is noteworthy that rocks which have undergone several formation processes may present more than one directions of planar anisotropy such as foliation and bedding planes in slates. These directions are not necessarily parallel to each other. Also, linear features such as lineations can be superposed on the planar features.

Rock mass anisotropy can be found in volcanic formations (basalt, tuff) and sedimentary formations consisting of alternating layers or beds of different (isotropic or anisotropic) rock types. Rock masses cut by one or several regularly spaced joint sets are anisotropic in addition to being discontinuous. The rock between the joints can be isotropic or anisotropic. It is not unusual to have several types of planar anisotropy in a rock mass such as joints and foliation planes or joints and bedding planes. If the joints develop parallel to the foliation or bedding planes, they are called foliation joints or bedding joints, respectively.

The rocks mentioned above show clear evidence of anisotropy and were classified as Class B anisotropic rocks by Barla [I J. On the other hand, Class A anisotropic rocks are those rocks that exhibit anisotropic properties despite apparent isotropy. Some intact granitic rocks belong to that group. This paper deals exclusively with class B rocks having intact or joint induced anisotropy.

293

294 AMADEI: 1992 SCHLUMBERGER LECTURE AWARD PAPER

Anisotropy is important in civil, mining and petroleum engineering. A list of rock engineering activities for which anisotropy is relevant and needs to be accounted for is given in Table 1. The importance of rock anisotropy and the need to account for rock anisotropy (along with rock heterogeneities and discontinuities) in engineering depend largely on the relative size of the problem of interest with respect to the size of the rock features such as strata or bed thickness, joint spacing, etc. Further, the type of rock anisotropy which is critical to the engineering project may vary from one scale to the next. This is illustrated in Fig. 1 for two holes driven in the same rock mass cut by a single joint set with a spacing of 1 ft (0.30 m). Consider first the case when the rock in between the joints is isotropic. In case A, the opening is small with a diameter of 2 in. (0.05 m). For this case, the size of the opening is so small compared to the joint

Table 1. Rock engineering activities for which anisotropy is relevant

Civil and mining engineering ??Stability of underground excavations ??Drilling and blasting ??Stability of surface excavation ??Stability of foundations ??Fluid flow and contaminant transportation

Peiroleum engineering ??Borehole stability and deviation ??Borehole deformation and failure ??Fracturing and fracture propagation ??Fluid flow

ft

Fig. 1. Two openings driven in the same rock mass cut by a single joint set with a spacing of I ft (0.30 m). In case A, the opening is small with a diameter of 2 in. (50 mm). In case B, the opening is large with

a diameter of loft (3.OOm).

spacing that the rock mass can be modeled for all practical purposes as being essentially isotropic and continuous. On the other hand, in case B, the opening has a diameter of 10 ft (3.00 m). In that case, the rock mass has to be modeled as being discontinuous and anisotropic. If the intact rock in between the joints is now anisotropic, both cases A and B need to account for the fabric when modeling the rock response to excavation. In case A, the intact anisotropy is important whereas in case B, the joint induced anisotropy is more critical. This example shows that, for the same rock mass, different modeling strategies will have to be followed depending on the size of the problem of interest. This example also shows that for smaller openings such as boreholes (which are mostly of interest in this paper), one should expect intact rock anisotropy to be more critical than that created by joint sets (unless the joint spacing is extremely small compared to the borehole diameter).

This paper deals with the subjects of rock anisotropy and rock stress. Both topics are important in rock engineering and are interrelated. The interaction between rock stress and rock anisotropy can take different forms. For instance, the build-up in in situ stresses is strongly correlated to the rock fabric. Stress distributions around excavations depend on the rock elastic properties if the rock is anisotropic. Elastic stresses in anisotropic media are quite different to those in isotropic media. On the other hand, stresses affect the deformability of rocks as confinement results in making rock anisotropy (intact or joint induced) less pronounced. This paper is divided into three parts. In the first part, existing models to describe the deformability properties of anisotropic rocks as well as the methods to determine those properties are reviewed. Then, it is shown how to account for rock anisotropy and the curvature of the Earth when estimating in situ stresses in rock masses. Finally, the effect of anisotropy on the analysis of overcoring measurements with cells similar to the CSIR Triaxial Strain Cell is discussed.

DEFORMABILITY OF ANISOTROPIC ROCKS

Constitutive modeling

The directional character of the deformability properties of anisotropic rocks and rock masses is usually assessed by field and laboratory testing. Deformability test results on anisotropic rocks are commonly analyzed in terms of the theory of elasticity for anisotropic media by assuming Hooke’s law. The latter implies that the rock has at most 21 independent elastic components. However, for most practical cases, anisotropic rocks are often modeled as orthotropic or transversely isotropic media in a coordinate system attached to their apparent structure or directions of symmetry. Orthotropy (orthorhombic symmetry) implies that three orthogonal planes of elastic symmetry exist at each point in the rock and that these planes have the same orientation throughout the rock. Transverse isotropy implies that at each point

AMADEI: 1992 SCHLUMBERGER LECTURE AWARD PAPER 295

Y

J- X

Z

Fig. 2. Orthotropic rock with three planes of symmetry normal to the n, s, I directions.

in the rock there is an axis of symmetry of rotation (n-fold axis of symmetry) and that the rock has isotropic properties in a plane normal to that axis. The plane is called the plane of transverse isotropy.

For a rock mass that is orthotropic in a local n, s, t Cartesian coordinate system (Fig. 2) attached to clearly defined planes of anisotropy, Hooke’s law can be expressed as follows [2]

1 V E,

-2 VW -- Es 4 0 0 0

V _-!!!

1 V

4, E, -2

E, 0 0 0

V -- ii

V,f 1 -_

Es E, 0 0 0

0 0 0 $0 0 S,

0 0 0 0 & 0 ?I,

0 0 oool G,, or in a more compact matrix form

(EL, =(H)(a),,. (2) Nine independent elastic constants are needed to describe the deformability of the medium in the n, s, t coordinate system. En, Es and E, are the Young’s moduli in the n, s and t (or 1, 2 and 3) directions, respectively. G,,,, G,,, and G,, are the shear moduli in planes parallel to the ns, nt and st planes, respectively. Finally, vii (i,j = n, s, t) are the Poisson’s ratios that characterize the normal strains in the symmetry directions j when a stress is applied in the symmetry directions i. Because of symmetry of the compliance matrix (H), Poisson’s ratios vu and vji are such that v,/Ei = vjilEj. The orthotropic

(1)

formulation has been used in the literature to characterize the deformability of rocks such as coal, schists, slates, gneisses, granites and sandstones. For instance, the cleat and bedding planes of coal are often assumed to be planes of elastic symmetry.

Equations (1) and (2) still apply if the rock is transversely isotropic in one of the three ns, nt or st planes of Fig. 2. In that case, only five independent elastic constants are needed to describe the deformability of the rock in the n, s, t coordinate system. Throughout this paper, these constants are called E, E’, v, v’ and G’ with the following definitions:

(9

(ii)

(iii)

E and E’ are Young’s moduli in the plane of transverse isotropy and in direction normal to it, respectively, v and v’ are Poisson’s ratios characterizing the lateral strain response in the plane of transverse isotropy to a stress acting parallel or normal to it, respectively and G’ is the shear modulus in planes normal to the plane of transverse isotropy.

Relationships exist between E, E’, v, v’, G and G’ and the coefficients of matrix (H) in equations (1) and (2). For instance, for transverse isotropy in the st plane

1 1 I 1 1 1 1 1 -=-. _=--=--; __=__=_ E,, E” Es E, E G,, G,, G’

v V”, v‘ V,l ;I En E”

VfS V 1 -=--_=-. -=-=-. _=2Q$$ (3) Es Et E ’ Gs,

The transverse isotropy formulation has been used to characterize the deformability of rocks such as schists, gneisses, phyllites, siltstones, mudstones, sandstones, shales and basalts. For such rocks, the plane of transverse isotropy is assumed to be parallel to foliation, schistosity or bedding planes. Note that some of the five or nine elastic constants of anisotropic rocks are sometimes assumed to be related. For instance, for transversely isotropic rocks, the modulus G’ is often expressed in terms of E, E’, v and v’ through the following empirical equation

-&=;+++2;. (4)

For orthotropic rocks, the shear moduli G,, G,, and G,, are related to the three Young’s moduli and Poisson’s ratios. These relations were first introduced by Saint-Venant [3]. In a recent survey of elastic constants of anisotropic rocks, Worotnicki [43 concluded that most of the published experimental data support the validity of the Saint-Venant approximation, with however major exceptions. Martin0 and Ribacchi [5] also found that the empirical relations are not always acceptable for many rocks.

The five and nine elastic properties of transversely isotropic and orthotropic rocks, respectively, cannot be randomly selected. Indeed, some inequalities associated with the thermodynamic constraints that the rock strain energy remains positive definite, must be satisfied [6,7].


For instance., for transverse isotropy, the five elastic properties E, E’, v, v’ and G’ must satisfy the following thermodynamic constraints

E,E’,G’>O (5)

-l<v<l (6)

Equations (6) and (7) reduce to - 1 < v < 0.5 if the rock is isotropic for which E = E’, G = G’ and v = v’. For orthotropy, the expressions for the thermodynamic constraints on the nine elastic properties are more complex and can be found in Amadei et al. [8].

The constitutive models presented above apply to intact anisotropic rocks. When rock mass anisotropy is created by systems of regularly spaced rock joints, the rock mass becomes discontinuous in addition to being anisotropic. The increase in deformability associated with the joints can be modeled using numerical models where the joints are represented as discrete discontinuous features. This approach works well as long as the number of joints is limited. On the other hand, when the number of discontinuities is large, it is not feasible to account for each joint plane when assessing the overall deformability of a regularly jointed rock mass. In that case, it is convenient to replace the rock mass by an equivalent anisotropic continuum [9-141. The rock mass is modeled as transversely isotropic if cut by a single joint set and orthotropic if cut by two or three orthogonal joint sets.

Consider for instance a rock mass cut by three joint sets, each set being normal to one of the three axes of the n, s, t coordinate system of Fig. 2. The intact rock between the joints is assumed to be linearly elastic and isotropic with Young’s modulus, E, Poisson’s ratio, v, and shear modulus G = E/[2(1 + v)]. Each joint set i (i = 1,2,3) is characterized by its spacing, &, and its normal and shear stiffnesses kni and ksi. As shown by Duncan and Goodman [9], the regularly jointed rock mass can be replaced by an equivalent orthotropic continuum whose constitutive relation in the rz, s, t coordinate system is given by equations (1) and (2) with

1 1 1 -_=-+- Ei E k”iSi

and

where i, j = 1,2,3 or n, s, t, respectively. All non-zero off-diagonal terms in equation (1) are now equal to -v/E. Since k,,i and k,i have units of stress/length or force/length3, the quantities k,iSi and ksiSi in equations (8) and (9) can be seen as normal and shear joint moduli. Note that the constitutive relation for a regularly jointed rock mass converges to that for an intact isotropic

medium when the joint spacings or joint stiffnesses in equations (8) and (9) approach infinity.

Although convenient, the equivalent continuum formulation has certain limitations that limit its range of application. For instance, Duncan and Goodman [9] made three main assumptions in the equivalent continuum approach. First, the normal and shear joint stiffnesses are constant and independent of the stress level acting across the joints. Second, the joint response remains in the elastic domain (pre-slip condition). Third, the joints are assumed to have negligible thicknesses and not to create any Poisson’s effect upon loading of the rock mass. In other words, intact rock and joints are assumed to undergo equal strains in directions parallel to the contact planes. This assumption makes all non- zero off-diagonal terms in equation (1) equal to the same value, -v/E. It also results in reducing the number of elastic constants necessary to describe the deformability of the rock mass when cut by a single joint set to four, namely, E, v, k,S and k,S. For a rock mass cut by three joint sets the number of elastic constants is equal to eight: E, v, kniSi and ksiSi with i = 1,2,3.

The equivalent approach has also been used in the past for modeling the deformability of thinly layered, laminated or stratified rock masses that are clearly heterogeneous in addition to being anisotropic. Since it is not possible to account for each layer on an individual basis, one approach has been proposed by Pinto [15], Salamon [16] and Wardle and Gerrard [17] whereby the rock mass is replaced by an equivalent homogeneous transversely isotropic continuum. Again several conditions need to be satisfied when using such an equivalent model. First, the rock mass consists of isotropic and/or transversely isotropic layers whose thickness and elastic properties vary randomly with depth. Second, the rock is a continuum and remains a continuum when subjected to stresses. Third, the elastic properties of the equivalent continuum are derived by examining the behavior of two cubes both having the same edge dimension L. One cube is a representative sample of the rock mass whereas the other cube is cut from the equivalent continuum and is subject to homogeneous stress and strain distributions. The representative sample of the rock mass contains a large number of layers with known thicknesses and deformability properties and its volume is sufficiently small to make negligible in the equivalent continuum the variations of stresses and strains across it.

Consider, for instance, m horizontal layers forming a representative sample of the rock mass. If the thickness of the jth layer is hi, its relative thickness is @j = hi/L. The deformability of each layer is defined by five elastic constants Ej, El, vj, v( and Gj. As shown by Salamon [16], the five elastic properties of the equivalent homogeneous continuum E, E’, v, v’ and G’ are equal to

I x @jEi

-= i=q

E 7 VI x ~14 1 + vj 1 - vj


c OjEjVj xm. Ei ‘( V i=q V’ ‘E, 1 -vi - == ajEj z QjEi E E’= --

1 + Vj 1 - Vj

,s 1 - vj

~&A_ 2

+2 ‘E; l-vi

BDj Ej 1 - vi

1 1 -_=- G C@jGj (10)

in which the summation from j = l-m is implied. The shear modulus, G, is also equal to E/[2(1 + v)].

Degrees of rock anisotropy In general, intact rocks are not too strongly aniso-

tropic compared to other engineering materials such as wood or composites. Typical values of the nine or five elastic constants (static and dynamic) for different types of intact anisotropic rocks can be found in the literature [l&52]. The elastic constants were mostly determined from the results of static or dynamic laboratory tests (see below) and assuming linear elastic theory. Compre- hensive surveys of elastic properties can be found in Batugin and Nirenburg [27], Gerrard [33], Lama and Vutukuri [35], Amadei et al. [8] and Worotnicki [4]. For

4 70- (a)

8 60-‘_

% 50-

; 40- cn

1.0 1.5 2.0 2.5 3.0 3.5

E max lE min

instance, Amadei et al. [8] analyzed 98 measurements of elastic properties. They found that for most intact transversely isotropic rocks, the ratio E/E’ varies between 1 and 4. Several cases of rocks with E/E’ less than unity were found but did not fall below 0.7. The ratio G/G’ was found to vary between 1 and 3, the Poisson’s ratio, v, between 0.1 and 0.35 and v’E/E’ between 0.1 and 0.7. In a more recent paper, Worotnicki [4] classified anisotropic rocks into four groups similar to those of Gerrard [33]:

(1)

(2)

(3)

(4)

Quartzofeldspathic rocks (e.g. granites; quartz and arkose sandstones, granulites and gneisses). Basicjithic rocks (e.g. basic igneous rocks such as basalt; lithic and greywacke sandstones and amphibolites). Pelitic (clay) and pelitic (micas) rocks (e.g. mudstones, slates, phyllites and schists). Carbonate rocks (e.g. limestones, marbles and dolomites).

Based on 200 sets of test results, Worotnicki [4] concluded that quartzofeldspathic and basic/lithic rocks show low to moderate degrees of anisotropy with a maximum to minimum Young’s modulus ratio E,,,,, /Emin less than 1.3 for about 70% of the rocks analyzed and less than 1.5 in about 80%. This ratio was found not to exceed 3.5 [Fig. 3(a)]. Pelitic clay and pelitic micas rocks show the highest degree of anisotropy with E_/Emin less

(b)

1.5 2.0 2.5 3.0 3.5 4.0 5 6

E max lE min

$6 (d $5 04

B3

22

:1

40 hlL 1.0 1.2 1.4 1.6 1.6 2.0

Fig. 3. Histograms of

RMMS 33/3-F

E max lE min

ratios for: (a) quartzofeldspathic and basicjithic rocks; and (c) carbonate rocks (after Worotnicki [4]).

(b) pelitic clay and pelitic mica rocks;


than 1.5 for about 33% of the rocks analyzed and less than two in about 50%. The modulus ratio was found not to exceed six with most cases below four [Fig. 3(b)]. Finally, carbonate rocks were found to show an inter- mediate degree of rock anisotropy with Emax/Emin not exceeding 1.7 [Fig. 3(c)].

If rock mass anisotropy is induced by joints, the ratio of anisotropy can be much larger than for intact rock and depends on the stress level acting across the joint planes. As an illustrative example consider a rock mass cut by a single joint set. Using again the model of Duncan and Goodman [9], the ratio E/E’ is equal to

$=1+X. k,S

(11)

Laboratory tests on rock joints have shown that the joint normal stiffness, k,, depends on the normal stress, cm, acting across the joint planes. Using, for instance, the expression for the tangential normal stiffness, k,, proposed by Bandis et al. [53], equation (11) becomes

where k,i is the initial joint normal stiffness and I’,,, is the maximum joint closure. According to equation (12), at zero normal stress (0, = 0), the ratio E/E’ is equal to 1 + E/(k”iS) which can be large for joints with small values of the spacing and/or initial stiffness. As more compression is applied across the joint surfaces, the rock joints become stiffer and the ratio E/E’ approaches unity as less anisotropy is induced by the joints.

From a physical point of view, one should expect intact rock anisotropy to be also stress dependent with a decrease in anisotropy associated with an increase in confinement [33]. It is most likely that pores and in particular microcracks in rock would close under confinement, thus inducing an increase in the overall rock stiffness. This phenomenon should be more critical for rocks with microcracks such as slates and phyllites than

0 I I I

0 0.2 0.4 0.6 0.8

L (Kb)

for other intact anisotropic rocks. The effect of stress on rock anisotropy has been observed experimentally and has been reported by many authors including Podio et al. [19], Berry et al. [30], Simonson et al. [32], Lerau et al. [38], Gonano and Sharp [41], Lin [43,44], Guyader and Denis [45], Yu et al. [47], Lama and Vuturiki [35] and Homand et al. [49]. For instance, Fig. 4(a) and (b) shows the variation of E, E’, v, v’ and G’ with confining pressure for a slate tested under triaxial compression by Lerau et al. [38] at confining pressures ranging between 0 and 0.8 kb (80 kPa). It can be seen that E’ and v’ increase linearly with confinement and that the ratio E/E’ decreases from 2.tl4 to 1.69, thus indicating a decrease in anisotropy with an increase in stress. This phenomenon was also observed by Homand et al. [49] who conducted ultrasonic tests on slates during triaxial compression tests at confining pressures up to 40 MPa. They found that the rock’s Young and shear moduli could be expressed as power function of (1 + Us) where c3 is the applied confining pressure. In general, it has been found that the stress dependency of rock anisotropy obtained during static experiments is greater than that measured under dynamic conditions.

The stress dependency of rock anisotropy implies that linear elasticity may indeed be of limited value when describing the deformability of anisotropic rocks and that it should be replaced by non-linear elasticity or more complex constitutive behavior if permanent deformation occurs. This would, of course, lead to more complex models for rocks and rock masses. However, it is the author’s opinion that from an engineering point of view, acceptable predictions of rock behavior can still be achieved assuming linear anisotropic elasticity as long as the selected rock properties are determined in a stress range comparable to what is expected in situ. Being able to account for the directional character of anisotropic rocks instead of assuming them to be isotropic (as has often been done in the past) is certainly a step in the right direction.

0 I I I A 0 0.2 0.4 0.6 0.8

---% (Kb) Fig. 4. Variation of E, E’, Y, v’ and G’ with confining pressure for a slate tested under triaxial compression at confining

pressures ranging between 0 and 0.8 kb (80 kPa) (adapted from Lerau et al. 1381). 1 kb = 0.1 MPa.


Fig. 5. Transversely isotropic rock sample tested under uniaxial compression.

Laboratory testing The elastic properties discussed in the previous section

can be determined on laboratory specimens cut at different angles with respect to the apparent directions of rock symmetry. The testing methods can be divided into static and dynamic [ 18-521. Static methods include uniaxial compression, triaxial compression, multiaxial compression, diametral compression (Brazilian tests), torsion and bending. For all these tests, the specimens are instrumented with strain gages or displacement transducers. The dynamic methods include the resonant bar method and the ultrasonic pulse method. Note that several types of tests (static and/or dynamic) should be conducted on a given rock to determine its anisotropic elastic properties.

The type of loading test and the number of tests required to determine the elastic constants of a given rock depend largely on the degree and type of symmetry assumed for the rock. Consider for instance a tran.werseIy isotropic rock sample tested under uniaxial compression (Fig. 5). An x, y, z coordinate system is attached to the specimen with the z axis parallel to the plane of transverse isotropy dipping at an angle 8 with respect to the xz plane. The rock is transversely isotropic in the st plane of Fig. 5 and has five elastic properties E, E’, v, v’ and G’ as defined in equation (3). Strains are measured in the x, y and z directions using strain gages. Using the theory of elasticity for anisotropic media and assuming uniform stresses and strains in the test specimen, the strains L,, EZ,,, 6r and yXu can be related to the applied stress, 6, as follows

6, = a,,a; cY = a,,a; 5 = a2,a; yxv = a2d (13)

with

Fig. 6. Three specimens of transversely isotropic rock tested in uniaxial compression with: (a) 0 = 0”; (b) 0 = 90” and (c) 0 # 0 and 90”.

cos4tI sin40 sin220 1 a22=~+~+4 F-2$

( )

a23 = --$cos2tl -f sin28

a26 = sin 28 [ (++-$)-sin2*(k+g)] cos2 0

sin 28 cos 28 - 2G’ * (14)

By conducting uniaxial compression tests on three specimens cut at different angles with respect to the plane of transverse isotropy, the five elastic properties of the rock can theoretically be determined from the linear portion of the corresponding stress-strain curves. In Fig. 6, three specimens of the same rock are tested in uniaxial compression with 8 = O”, 8 = 90” and an inclined


Table 2. Comparison of elastic properties determined on Loveland sandstone I using uniaxial compression and diametral compression tests

Size of No. of Testing method spec. tests (GFa) (G:a) ” V’ (&a) E/E' G/G' Uniaxial camp. Diametral leading

NX 5 29.3 23.9 0.18 0.13 6.2 1.23 2.00 NX 9 28.9 24.9 0.13 0.13 12.5 1.16 1.02

angle 8 different from 0 or 90”. Using equations (13) and (14), the strains measured on the first specimen [Fig. 6(a)] allow the determination of E’ and v’ whereas those on the second specimen [Fig. 6(b)] are used to determine E and v. The strains measured on the third specimen [Fig. 6(c)] are used to determine the shear modulus G’. A procedure that has been found by the author to work well is to instrument each one of the three specimens in Fig. 6 with two to four 45” strain rosettes. All strain measurements are then analyzed simultaneously. Let N be the total number of strain measurements (with N > 5) for all three specimens in Fig. 6. According to equations (13) and (14), each strain measurement is linearly related to the five unknown compliances l/E, l/E’, v/E, v’/E’ and l/G’. In matrix form, the strain measurements can then be expressed in terms of the five compliances as follows

(15) where [c] is a (N x 1) matrix of strain measurements, [T] is a (N x 5) matrix and [Cl’ = (l/E l/E’ v/E v//E’ l/G’). Equation (15) is then solved for the least square (best fit) estimate of the five compliance terms by multilinear

90” *

0 2 4 6 6

0 0.1 0.2 0.3 0 0.1 0.2 0.3 Fig. 7. Variation of the apparent Young’s modulus, $ (x 10’ kg/cm*) in (a), and apparent Poisson’s ratios v,, and v,,._ in (b) and (c) with the

angle 0 for a schist (adapted from Pinto [24]).

regression analysis [54]. The advantage of this approach is that all strain measurements are taken into account when determining the compliance terms. Furthermore, the method can be extended to more than three specimens.

As an illustrative example, the method described above was applied to a local Colorado sandstone called Loveland sandstone I that showed well defined planes of rock anisotropy. A total of five NX-core specimens instrumented with 45” strain rosettes were tested in uniaxial compression: two with 8 = O”, two with 8 = 90 and one with 0 = 63”. The strains used in the analysis were all determined at 50% of the peak loads. Analysis of 45 strain measurements (N = 45) yielded the following best fit (secant) elastic properties: E = 29.3 GPa, E’=23.9GPa, v =0.18, v’=O.l3 and G’=6.2GPa which gave E/E’ = 1.23 and G/G’ = 2.00. These elastic properties are also reported in Table 2.

Equations (13) and (14) show that shear strains can develop under uniaxial compression as long as the dip angle, 8, of the planes of rock anisotropy differs from 0 or 90” or in other words as long as the applied stress does not coincide with the fabric. In that case, the principal strain directions do not coincide with the principal stress directions as for isotropic media.

Equations (13) and (14) can also be used to calculate the apparent Young’s modulus, E,, and apparent Poisson’s ratios vJx and v.yZ of the rock in the x, y, z coordinate system with

E,=i; aI2 a23 a22

vyx= --; a22

vyr= --. a22

(16)

These three quantities depend on the angle 0. Figure 7(a)-(c) shows an example of variation of E,,, vyx and vyr with the angle 8 for a schist tested by Pinto [24].

800 I-

0 15 30 45 60 75 90

0 (degrees) Fig. 8. Variation of ultimate strength, R,,, and apparent Young’s modulus, E,, = E,., for a diatomite tested under uniaxial compression

(adapted from Allirot and Boehler [36]).


The predictions (solid lines) agree quite well with the observed behavior. Figure 8 shows another example of variation of the apparent Young’s modulus for a diatomite tested by Allirot and Boehler [36]. The simple model based on equations (13) (14) and (16) indicates that the overall (apparent) Young’s modulus and Poisson’s ratio of an anisotropic rock can be expected to vary widely if different specimens of the same rock cut in different directions with respect to the anisotropy planes are tested under uniaxial compression. It is important to note that such variations should not be interpreted as experimental errors or as the end result of variations in rock composition.

As discussed by Amadei [%I, the uniaxial compression test can also be used to determine the nine elastic constants of an orthotropic rock by testing three samples inclined at different angles with respect to the planes of symmetry. However, the analysis of the test results is not as straightforward as for transversely isotropic rocks.

The five and nine elastic properties of transversely isotropic and orthotropic rocks can also be determined by diametral compression (line or strip load) of thin discs of rocks (Brazilian loading). Strains are measured using strain gage rosettes (usually 45” rosettes) glued at the center of the discs. This method was first used by Hondros [56] to determine the Young’s modulus and Poisson’s ratio of concrete assuming the material to be isotropic. Pinto [57] extended the method to anisotropic rocks and tested the method on discs of schist. Closed- form solutions were derived to relate the elastic constants of an anisotropic rock in a disc under diametral compression to strains at the disc center. In the closed-form solutions, Pinto [57] assumed that the stress concentration at the center of a disc in anisotropic rock is the same as for an isotropic rock. This assumption is only correct if the disc is parallel to a plane of transverse isotropy. In all other cases, the anisotropy needs to be taken into account. Based on correct closed-form solutions derived by Okubo [58] and Lekhnitskii [59], Amadei et al. [60] and Amadei and Jonsson [61] revised Pinto’s procedure and proposed a methodology to determine the nine and five elastic properties of orthotropic and transversely isotropic rocks, respectively. This methodology is

I! 0

. P +t t- c-

Fig. 9. Diametral loading of a disc of anisotropic rock over an angular width 2a (after Amadei et al. [60]).

currently being explored further by the author and co-workers and is summarized below.

Consider the geometry of Fig. 9. A disc of rock of diameter, D, and thickness, t, is subject to a diametral load, W, applied over an angular width 2a (assumed here to be small). The rock is assumed to be orthotropic with one of its three planes of elastic symmetry (defined as ns in Fig. 9) parallel to the disc xy plane. The n and s axes are inclined at an angle, $, with respect to the x and y axes. The applied pressure, p, is equal to W/(aDt). As shown by Amadei et al. [60], the stress components at the center of the disc are equal to

W W W flx=qxx--; nDt =y = qyy’ ~ ; Txy = qxy -.

nDt (17)

For the geometry of Fig. 9, it can be shown that the stress concentration factors q_, qyu and qxy have complex expressions which depend on the rock compliances l/E,, l/E,, v,/E, and l/G, and the dip angle $. If the rock were isotropic, q.x = -2, qyy = 6 and qxy = 0. The strains cxr cy and y,.. at the disc center are related to the compliances l/E,,, l/E,, v,/E,, and l/G, by combining equation (17) with the rock’s constitutive equations.

Consider the three diametral compression tests (j = 1,3) shown in Fig. 10. The rock is again assumed to be orthotropic. Each test consists of loading a disc whose middle plane is parallel to one of the three planes of elastic symmetry of the rock (ns, st or nt). In each test,

Test 1 Test 2 Test 3 Fig. 10. Measurement of the elastic properties of an orthotropic rock using three diametral compression tests. Each disc has

a middle plane parallel to a plane of elastic symmetry.


the strains at the center of the disc are measured in three directions caj, cbj and ccj (j = 1,3) using a 45” strain rosette, at an arbitrary load, W, and in the linear elastic range of rock behavior. To simplify the presentation, the three discs are assumed to have the same geometry, same strain gage orientation, the same loading angle, 2a, and the strains are measured at the same load level W. For the geometry of test 1, the stress concentration factors qXXl 9 qyyl and qxvl depend on the rock compliances l/E,, l/E,, v,/E,, and l/G,. Similarly, for test 2, the stress concentration factors qxx2, qvvz and qxuZ depend on the rock compliances 1 /Es, 1 /E, , v,,/E, and 1 /G,, . Finally, for test 3, the stress concentration factors qxx3, qyu3 and qxy3 depend on the rock compliances l/E,, , l/E,, v,,/E, and l/G,, . In addition, the nine stress concentration factors depend on the angle 2a and the orientation angles $j (j = 1,3). Combining equation (17) and the rock’s constitutive equations for each disc in Fig. 10, it can be shown that the nine compliances are related to the nine strain gage measurements in matrix form as follows

kl = g [TI[C] w-9

where [c] is a (9 x 1) matrix of strain measurements, [T] is a (9 x 9) matrix and [Cl’ = (l/E” l/E, l/E, v,/E” v,,/Es v,,/E~ l/G, l/G,, l/G,,). The components of matrix [T] depend on the nine stress concentration factors, which themselves depend on the nine compliances. The system of equations (18) is therefore highly non-linear and is furthermore constrained since the nine elastic constants must satisfy some thermodynamic constraints [8]. As shown recently by the author and co-workers, a solution to this constrained problem can be obtained using the Generalized Reduced Gradient Method which is essentially a constrained optimization technique.

The same methodology applies if the rock is trans- verseZy isotropic. The five elastic properties of the rock can be determined by conducting two instead of three diametral compression tests: one in the plane of transverse isotropy and another perpendicular to the plane of transverse isotropy. If the plane of transverse isotropy coincides, for instance, with plane st, then the five elastic constants can be determined by conducting tests 1 and 2 in Fig. 10. The strains measured in test 2 are then analyzed with qxx = -2, qYv = 6 and qxu = 0. This gives the Young’s modulus E = Es = E, and the Poisson’s ratio v = v,~. On the other hand, the strains measured in test 1 lead to a system of equations similar to equation (18) with only three equations and three unknowns E’ = E, , v ’ = v,, = v,, and G’ = G, = G,, . This system of equations is then solved using the generalized reduced gradient method for the three unknowns taking into account the constraints defined in inequalities (5)-(7).

As an illustrative example, Fig. 11 (a) and (b) shows the response curves of two discs of the same Loveland sandstone I, mentioned earlier in this paper. Two NX-size discs (D = 54.4 mm) with thickness, t = 26.2 mm and t = 26.8 mm were tested with the geometry of tests 1 and 2 in Fig. 10, respectively. The plane of transverse isotropy was assumed to be parallel to the apparent

-0.6 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.6 1.0 1.2

STRAIN (x 10-3)

I I I I 1 I I I I

10.6 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.6 1.0 1.2

STRAIN (x 10-3)

Fig. 11. Diametral compression of two discs of Loveland sandstone I: (a) load-strain curves for the three strain gages glued to a disc with middle plane parallel to the plane of transverse isotropy; (b) load-strain curves for the three strain gages glued to a disc with middle plane perpendicular to the plane of transverse isotropy.

1 psi = 0.007 MPa.

sandstone layers. Table 3 gives the strains measured at 50% of the peak loads which were used in the analysis. The rock was found to be anisotropic with E = 26.2 GPa, E’=22.7GPa, v =0.146, v/=0.161 and G’= 11.9GPa which gives E/E’ = 1.15 and G/G’ = 0.95. The average values of those elastic properties obtained on a total of nine discs (seven with the geometry of test 1 and two with the geometry of test 2 in Fig. 10) are given in Table 2.

The use of diametral compression to determine the anisotropic properties of rocks has the advantage of being simple. The tests can be carried out on core samples of different sizes, the specimen preparation is not as stringent as with other techniques and the analysis of the test results is relatively straightforward. The methodology assumes that the strains are measured at

Table 3. Results of diametral compression tests on Loveland sandstone I

Disc parallel to plane of transerse Disc perpendicular to plane of isotropy [Fig. 1 l(a)] transverse isotropy [Fig. 1 l(b)]

c,@r/W = -0.1098 GPa-’ a,YxDf/W = -0.1015 GPa-’ cy,nDrl W = 0.2403 GPa-’ ~nDt/ W = 0.0578 GPa-’

~~nDt/ W = 0.2384 GPa-’ c,,nDt/ W = 0.0841 GPa-’

E = 26.2 GPa E’ = 22.7 GPa v = 0.146 v/=0.161

G = 11.4GPa G’= 11.9GPa


the center of the discs which is not exactly true since the strain gages have a finite length. This is, however, not too critical since as shown by Amadei and Jonsson [61] the stresses and therefore the strains are constant over at least 60% of the disc diameter. Note that the strain gages must also be long enough to cover a representative sample of the rock. This last remark applies as well when strain gages are glued to rock samples tested under uniaxial compression. It is noteworthy that the elastic properties measured in diametral compression are obtained under a mixed compression-tension stress field. It is well known that rocks have different deformational properties when loaded in tension or compression [62]. Tests assuming the rock to be isotropic have shown that the Young’s modulus in tension is smaller than in compression [63]. This property often referred to as bimodularity in the literature, may affect the deformability properties of anisotropic rocks. A similar concern arises when analyzing the results of bending tests on anisotropic rock beams. Further research on the effect of bimodularity is needed.

Existing methods for determining the deformability properties of anisotropic rocks in the laboratory have two major problems. The first problem is that the symmetric character of the rock and the orientation of its planes of symmetry must be assumed before testing. Although a rock may show preferred directions of layering, the principal deformability directions of the rock may not necessarily coincide with those layers. This could happen if the rock contains two or three non-parallel fabric directions but only one is clearly visible and dominant. To be correct, the principal directions of rock symmetry (if any) should be determined from the test results at the same time as the rock deformability properties. This has been done for instance using dynamic methods by Homand et al. [49]. They were able to determine the 21 dynamic elastic constants of slate by passing ultrasonic waves in 18face polyhedric samples. The only assumption was that the rock’s compliance matrix was symmetric. It was found that for all practical purposes the slate could be described as orthotropic with nine elastic constants. Determination of the 21 elastic constants can also be done by using a multiaxial load cell in which three independent stresses are applied. For instance, using a cubical cell available at the University of Colorado, Ko and Sture [64] were able to determine all 36 elastic constants of a composite called Scotchply by testing two cubical specimens 4 in. (10 cm) in size. The symmetric character of the stress-strain matrix for Scotchply could thus be verified. One specimen (oblique

cube) had layers inclined with respect to the loading directions whereas the second specimen (normal cube) had layers parallel and normal to the applied stresses. Deformations were measured in all three principal stress directions using three proximitors mounted on each wall of the cubical cell. Thus, for each specimen, 18 strain measurements were made. One major advantage of the cubical cell is its ability to apply three independently principal stresses which makes it possible to perform three separate tests on a given specimen without retrieving it from the apparatus and reorienting it.

The second problem with existing laboratory methods is that, to date, no standards or suggested methods are available specifically for the sampling, preparation and testing of anisotropic rocks. Major difficulties can be encountered for instance during coring and preparation of samples of weak anisotropic rocks with planes of weakness. For such rocks obtaining samples might be extremely difficult and the current ISRM or ASTM methods no longer apply.

Existing testing methods on a same rock can lead to different values of its anisotropic elastic properties. Several reasons may be invoked to explain this variability such as rock composition, heterogeneities, testing method, etc. As an example, the test results in Table 2 show that the uniaxial compression and diametral compression tests on Loveland sandstone I, yield comparable values of the five elastic properties except for the shear modulus G’. As another example, the author and co-workers recently tested another sandstone (defined here as Loveland sandstone II) under different loading conditions: uniaxial compression, diametral compression (Brazilian loading), true triaxial compression and hollow cylinder tests. The test results are summarized, in Table 4. The procedures followed for the uniaxial compression and diametral compression tests were the same as those presented above. Six NX samples were tested in uniaxial compression with the orientation of Fig. 6(a)-(c) (with two samples per orientation). Each sample was instrumented with four 45” strain rosettes. The elastic properties were determined by least square analysis using equation (15) with N = 72. Six NX core samples were tested under diametral loading. Two true triaxial compression tests were carried out using the cubical cell at the University of Colorado following the procedure proposed by Ko and Sture [64]. Two hollow cylinder tests were conducted by Talesnick at the Technion, Israel using a new methodology discussed in one of his recent papers [52]. For all tests, the sandstone was assumed to be transversely isotropic in a plane parallel to its apparent layers.

Table 4. Comparison of elastic properties determined on Loveland sandstone II using different types of laboratory testing methods

Testing method Size of No. of spec. tests (G:a) (G?a) Y V’ (&a) E/E’ G/G’

Uniaxial NX 6 33.5 44.6 0.08 0.13 19.1 0.75 0.81 Diametral loading NX 6 43.2 54.8 0.08 0.14 24.9 0.79 0.80

- True triaxial Cube 2 40.5 42.9 0.29 0.20 21.6 0.94 0.72 100 mm in size

Hollow cylinder o.d. NX 2 52.3 44.7 0.17 0.15 - 1.17 - id. 37.5 mm


All samples originated from the same rock block approx. 3 x 3 x 1 ft (0.9 x 0.9 x 0.3 m) in size. The elastic properties were determined in the linear portions of the load (or stress) vs strain curves between 0 and 50% of the peak loads. Table 4 shows that, despite the proximity of the rock specimens and the fact that they essentially have the same volume, E/E’ varies between 0.75 and 1.17, G/G’ varies between 0.72 and 0.81, v varies between 0.08 and 0.29 and v’ varies between 0.13 and 0.20.

Field testing Because of scale effect, static field tests such as plate

loading, borehole expansion and gallery tests have been used in the past to characterize the deformability of anisotropic rock masses. Dynamic tests have also been used to assess the degree of rock mass anisotropy but in a more qualitative manner. As for the laboratory tests, field tests are conducted in different directions with respect to the apparent rock mass fabric. The analysis

of field tests is usually done assuming that the rock mass behaves linear elastically when subjected to the applied loads. Also, because of the complexity of the tests and the more sophisticated nature of the equations that are necessary in the analysis of the test results, very often, assumptions are made about the rock deformability in order to reduce the number of elastic constants that need to be determined. Sometimes, test results in anisotropic ground are analyzed using equations derived from the theory of linear elasticity for isotropic media. The Young’s modulus is determined for different loading directions.

It should be kept in mind that, in general, field tests involve larger rock volumes than laboratory tests. Compared to laboratory tests, the stress distributions in those volumes are more complex and the measured elastic properties are more likely to be average properties and to be affected by the level of applied load (in particular if the rock mass anisotropy is induced by joint

Fig. 12. Plate load tests at the Ridracoli dam (after Oberti er al. [67]).

sets). Also, discrepancies can arise between rock mass properties determined with one field testing method and another because different rock volumes are involved.

Rocha [65], Rocha and Silva [66], Oberti el al. [67], Akai et al. [68], Hata et al. [69] and Fishman and Iofis [70] have used plate loading tests to characterize the deformability of various anisotropic rock masses (stratified and/or jointed). Figure 12 shows for instance the set-up used by Oberti et al. [67] at the Ridracoli dam in Italy. Here, the rock mass consists of alternate layers of sandstone and marl and was modeled as orthotropic. Load tests with circular plates 0.6m in dia were conducted in directions normal and parallel to the layers in an exploratory tunnel with axis parallel to the stratas. Figure 12 shows the deformations measured in the rock mass with multiple point extensometers for different levels of applied load (between 0 and 6 MPa). The anisotropic response of the rock mass is quite clear from the deformation curves. The rock mass moduli in directions normal and parallel to the stratas were calculated using the combined results of the plate loading tests shown in Fig. 12 and additional sonic tests giving a modulus ratio for the rock of 1.33 (at an applied stress level of 3 MPa). Hydraulic pressure chamber tests (gallery tests) also conducted by Oberti et al. [67] on the same rock gave a modulus ratio equal to 1.125 (at an applied stress level of 3 MPa).

Rocha and Silva [66] used a combination of large flat jack tests and dilatometer tests to study the foundation of an arch dam located in a sedimentary formation consisting of layers of limestone, shale and marl, with limestone prevailing. The testing program consisted of determining the rock mass deformability in directions normal and parallel to the layers using equations based on the theory of elasticity for isotropic media. Aniso- tropy in the different rock layers was observed. Also dilatometer tests gave values of the rock modulus of deformation much less (and as low as 25%) than those determined with the flat jack tests.

Expansion tests in boreholes such as dilatometer and Goodman jack tests and gallery tests (hydraulic pressure chamber tests) in tunnels can also be used to determine the elastic properties of anisotropic rock masses in situ.


Kawamoto [71] suggested using a pressurized borehole to calculate the elastic constants of anisotropic rocks. However, his method of analysis was limited to certain directions of rock anisotropy with respect to the holes. Rocha et al. [72] also proposed using the dilatometer to assess rock anisotropy by plotting the variation of borehole diameter change during pressurization. Note that rock anisotropy can only be captured with radial displacement measuring types of dilatometers since volume change types can only give an average value for the rock mass modulus of deformation [73]. Note also that the method proposed by Rocha et al. [72] is essentially qualitative and can be used to identify the major and minor axes of rock mass anisotropy and to get an insight into the anisotropy of the rock mass being studied. Finally, it is noteworthy that the analysis of the dilatometer tests in anisotropic rock by Rocha et al. [72], Hughes and Ervin [73] and others was carried out using equations based on the theory of elasticity for isotropic media.

In two recent papers, Amadei and Savage [74,75] showed that the variation of hole diameter change measured along the circumference of a hole during in situ expansion tests in an anisotropic rock mass could be explained analytically by using closed-form solutions based on the theory of elasticity for anisotropic media. For dilatometer or gallery tests with the geometry of Fig. 13(a), an apparent modulus or modulus of deformation, E,, is introduced for each diametral measurement, U,, such that

1 u, (1 +v) __-I C---*

,. ^. q 2a 4

For the NX-borehole jack test introduced by Goodman et al. [76] and shown in Fig. 13(b), the modulus of deformation, E,, and the borehole deformation U,, in the direction of loading are related as follows

1 Ud K(v, Bc 1 -- =--- q 2a E, (20)

where 28, is an estimated value of the contact angle between the jack and the rock [77]. In equations (19) and (20), 2a is the diameter of the hole (borehole or tunnel)

Fig. 13. (a) Radial loading of a hole; (b) loading of a hole over two opposed arcs of angle 2/l, (after Amadei and Savage [TX]).


(b) ((3

APPARENT

Fig. 14. (a) Problem geometry; (b) orientation of the hole; (c) orientation of the anisotropy.

and q is the applied pressure. Equations (19) and (20) can also be used in terms of increments of diametral deformation, AU,, and increments of applied pressure, Aq, when pressure vs hole diametral deformation response curves are non-linear.

General expressions were derived for the modulus of deformation, E,, for the geometry of Fig. 14(a)-(c). The orientation of the n, s, t coordinate system in Fig. 14(a) attached to the rock anisotropy is defined with respect to a global coordinate system X, Y, 2 by a dip azimuth angle /I, and a dip angle J/, [Fig. 14(c)]. The borehole or tunnel orientation with respect to the X, Y, Z coordinate system is defined by two angles #I,, and b,, [Fig. 14(b)]. For an orthotropic rock mass, it can be shown that the ratio between the modulus of deformation, E,, and one of the Young’s moduli, say Es, of the rock depends on the following eight dimensionless quantities

2. 5. E E E,. Es En’ E,’ v,; v,; v,,; 6; G,,’ M G. (21)

If the medium is transversely isotropic with, for instance, transverse isotropy in the it plane, and using equation (3), the ratio EJE, = EJE depends only on four dimensionless terms

E G E’; v; V’; 1, G (22)

Finally, if the rock mass is cut by three orthogonal joint sets normal to the n, S, t coordinate axes of Fig. 14(a), the ratio E, /E depends on the following three dimensionless terms

E E k,iSi’ k,Si’ v (23)

Fig. 15. Variation of the ratio E,/E with the angle 0 for a jointed rock mass with one joint set with $, = 60” and 1, = 0, 30, 60 and 90 (E/k,S = 20, E/k,S = 40 and v = 0.25). The hole is oriented such that

@,, = 90” and 6, = 0” (after Amadei and Savage [75]).

with i = 1,2,3. If there is only one joint set, the first two ratios in equation (23) reduce to E/k,S and E/k,S. Note that EJE, and E,/E also depend on the orientation of the planes of anisotropy with respect to the hole in which the expansion tests are conducted. In addition, for the dilatometer and gallery tests, these ratios also depend on the angle 8 of Fig. 13(a).

Figure 15 shows an example of variation of the ratio E,/E for a regularly jointed rock mass with one joint set dipping at an angle J/a = 60” and oriented at different angles /I, = 0, 30, 60 and 90”. The hole is parallel to the Z-axis of Fig. 14(b) (/I,, = go”, B,, = 00). In this example, EIk,S = 20, E/ksS = 40 and v = 0.25. Each curve in Fig. 15 can be seen as an “anisotropic figure” using

1.2

I.0

0.8

0.6

0.4

0.2

0.0 0.2 0.4 0.6 0.8 1 .o

Fig. 16. Polar variation of E,/E with the angle 0 for transverse isotropy parallel to the yz plane (& = 90”, 6, = O”). E/E' = 1, 1.5, 2, 3 and 4, v = v’ = 0.25 and G/G’ = 1. Isotropy corresponds to E/E' = G/G' = 1

(after Amadei and Savage [74]).


the terminology of Rocha et al. [72] and shows a major and a minor axis. The major axis is always parallel to the apparent dip of the joints in the plane normal to the hole cross section.

As another numerical example, Fig. 16 shows the variation of EJE with the polar angle 0 of Fig. 13(a) (0 < 0 < 7r/2) for several degrees of intact rock anisotropy with E/E’ = 1, 1.5, 2, 3 and 4, v = v’ = 0.25, when the plane of transverse isotropy is parallel to the yz plane (/I, = 0” and +, = 90’) and for G/G’ = 1. Here again &, = 90” and 6, = 0”. The isotropic case corresponds to E/E’ = G/G’ = 1 or when the plane of transverse isotropy is parallel to the xy plane. Figure 16 indicates that as E/E’ increases, the rock mass becomes more deformable in a direction normal to the plane of transverse isotropy and the variation of E,/E with the angle 8 becomes more non-uniform as expected. Note that the value of the modulus of deformation in the direction parallel to the plane of transverse isotropy is not much affected by the value of E/E’ unlike the value of the modulus in the direction normal to the plane of transverse isotropy.

The variation in the modulus of deformation measured during expansion tests along the circumference of a borehole or tunnel in an anisotropic (intact or jointed) rock mass can be used to determine its elastic properties. A method was proposed by Amadei and Savage [74] for transversely isotropic rock masses for which the unknown elastic constants are E, E’, v, v’ and G’ for intact rock and E, v, k,S and k,S if the rock is cut by one joint set. In this method, values for the Poisson’s ratios v and v’ are assumed as in the conventional isotropic analysis of expansion tests and the other three elastic constants E, E’ and G’ or E, k,S and k,S are determined from the test data. Applied pressure, q, vs hole diametral deformation, U,, response curves measured in situ are assumed to be non-linear and are analyzed in terms of increments of pressure and deformation Aq and AU,. Thus, the elastic constants can be calculated over each increment of applied pressure and their stress dependency can be assessed.

The three elastic constants E, E’, G’ of an intact anisotropic rock can be determined by conducting dilatometer or gallery tests in two holes; one hole normal to the plane of transverse isotropy and another hole parallel to that plane. First, expansion tests are conducted in a hole having an axis normal to the plane of transverse isotropy. For this anisotropy orientation, the rock mass behaves as if it were isotropic and all hole diametral deformations are equal [Fig. 17(a)]. Over each increment of pressure, Aq, the slope of the q vs Ud/2a curve is then equal to E/( 1 + v) [Fig. 17(b)]. Assuming a value for the Poisson’s ratio, v, the in-plane Young’s modulus, E, and shear modulus, G = E/[2(1 + v)], can then be determined.

Next, expansion tests are conducted in a hole parallel to the plane of transverse isotropy. Let x, y, z be an arbitrary coordinate system attached to that plane and consider for instance the case when the plane of transverse isotropy is parallel to the yz plane [Fig. 18(a)]. Two

(4

__-___ m “L!L A _---

i I

+k! ud 2a

04 Fig. 17. (a) Dilatometer or gallery test in a hole with transverse isotropy normal to the hole axis; (b) corresponding q vs Ud/2a curve.

(a)

84

Fig. 18. (a) Dilatometer or gallery test in a hole with transverse isotropy parallel to the hole axis and yz plane. Hole diametral deformations are measured at 0, = 0 (point 1) and O2 = x/2 (point 2);

(b) corresponding q vs U,,/2a curves.

diametral diameter change measurements are conducted, one in the x direction, AU,, , and one in the y direction AU,,. For each direction of diametral measurement, a curve q vs U,/2a is obtained [Fig. 18(b)]. The respective slopes of those two curves over the same pressure increment, Aq, are equal to E,, /( 1 + v) and E,,/( 1 + v). Using the mathematical expressions for the apparent moduli E,, and E,>, charts were proposed by Amadei and Savage [74] to determine the modulus ratios E/E’ and G/G’. An example is given in Fig. 19 where (E/2a)

1.0 I I I 0.0 2 4 6

E '"d, -.- 2a Aq

Fig. 19. Variation of (E/Za)(AU,,/Aq) with (E/2a)(AC/,,,/Aq) for different values of E/E’ and G/G ’ ranging between 1 and 4, v = v’ = 0.25 and for plane strain condition. Point I represents the isotropic case with E/E’ = G/G ’ = 1 (after Amadei and Savage [74]).


(AU,,,/Aq) and (E/2a) (AU,,,/Aq) have been plotted for different values of E/E’ and G/G’ ranging between 1 and 4 and v = v’ = 0.25. The isotropic case corresponds toE/E’ = G/G’ = 1 and is represented by point I with coordinates (1 + v, 1 + v). Knowing the values of AU,,, AU,,, and Aq from the field tests and E from the first hole, Fig. 19 can be used to determine the two ratios E/E’ and G/G’ and therefore E’ and G’. Charts similar to Fig. 19 can be generated for other values of v and v’.

Note that for a regularly jointed rock, the three moduli E, k,S and k,S could be determined from expansion tests in only one hole parallel to the joint planes by first loading the jointed rock by an amount Aq followed by a cycle of unloading and reloading over the same pressure increment. If the slope of the curve q vs U,/2a during the unloading-reloading cycle can be used as an estimate to E/(1 + v) for intact rock (as suggested by Goodman [78] for static tests on fractured rock) then E,,IE and E,,IE can be obtained directly as the ratios between the slopes of the loading and unloading- reloading curves over the same pressure increment Aq.

The three elastic constants E, E’, G’ can also be determined by conducting borehole jack tests in two boreholes; one borehole normal to the plane of transverse isotropy and another parallel to that plane. Charts similar to that shown in Fig. 19 can be found in Amadei and Savage [74].

As for laboratory tests, field tests require that some assumptions be made regarding the type of rock mass anisotropy and the orientation of the planes of rock anisotropy with respect to the loading surfaces or boreholes. Anisotropy figures such as those shown in Fig. 15 are, in general, not sufficient to determine the orientation of the anisotropy since the theory shows that the apparent dip of the major axis of a given anisotropy figure may be associated with planes of rock anisotropy with various combinations of strike and dip angles.

The method summarized in Figs 17 and 18 assumes that the holes in which expansion tests are conducted, are either perpendicular or parallel to the plane of transverse isotropy. If the holes are inclined with respect to the anisotropy, the procedure for determining the rock elastic properties in situ becomes more complex. One approach, that is being currently investigated by the author and co-workers for the analysis of expansion test results, is to use equations (19) or (20) for each diametral measurement. Recall that in these equations, the apparent modulus, E,, is a non-linear function of the five or nine elastic properties of the rock. If N is the total number of diametral measurements recorded by conducting expansion tests in two or more holes (with N 2 5 for transverse isotropy or N > 9 for orthotropy), a system of non-linear equations similar to equation (15) can be written where matrix [c] is now a N x 1 matrix of diametral measurements. This system of equation can then be solved for the five or nine rock elastic properties with the same generalized reduced gradient method used for the diametral compression tests discussed in the section on laboratory testing.

ESTIMATING IN SITU STRESSES IN ANISOTROPIC ROCK

Unlike man made materials such as concrete, steel, etc. natural materials such as rocks and soils are subject to natural (virgin) stresses called in situ stresses. Knowledge of the in situ stress field is very important in many problems dealing with rocks in civil, mining and petroleum engineering, energy development as well as in geology and geophysics.

Various techniques for measuring in situ stresses have been developed and improved over the past 30 yr. The best known and most used techniques are the hydraulic methods (hydraulic fracturing, sleeve fracturing, hydraulic tests on pre-existing fractures) and overcoring methods. Before measuring virgin stresses with some of these methods, an attempt should be made to obtain an estimate of the in situ stress field. This can be done, for instance, from stress vs depth relationships or observations obtained from stress measurements made in the past in the region of interest or by extrapolation from regions with similar geological and tectonic settings. Information can also be derived from the topography, the geology, the rock fabric, the rock loading history, the first motion analysis of earthquakes, the occurrence of stress release phenomena (squeezing, pop-ups, buckling, etc.), breakouts in boreholes, tunnels and shafts, and the presence of stratification, heterogeneities or geologic structures (faults, folds, shear zones, unconformities, volcanic vents and dikes). Estimating in situ stresses can be useful in the early stage of engineering design, for planning process and when selecting stress measuring methods and the location of those measurements.

The exact prediction of in situ stresses in rock and their spatial variation is very difficult and for all practical purpose impossible, since the current stress state is the end product of a series of past geologic events and is the superposition of stress components of several diverse types: gravitational, tectonic (active and remanent), residual, terrestrial [79,80]. In situ stresses not only vary in space but also with time due to tectonic events, erosion, glaciation, etc. The problem is further complicated in that the present rock fabric may or may not be correlated at all with the current in situ stress field [81]. Further, assumptions can only be made about the load history and the rock’s constitutive model. To date, no rigorous methods are available to predict exactly in situ stresses. Estimating and measuring in situ stresses represent two alternatives. It is noteworthy that the process of estimating in situ stresses should not be considered as a substitute for their measurement.

In general, estimating in situ stresses requires a detailed characterization of the rock and considerable judgment. Models (physical or numerical) can be developed to explore the effect of such parameters as the constitutive model of the rock, its loading history, the topography, critical geologic structures and the boundary conditions on in situ stresses. In this section, we explore the effect of rock anisotropy, stratification and Earth curvature on the distribution of in situ stresses under

AMADEI: 1992 SCHLUMBERGER LECTURE AWARD PAPER

gravity only. The effect of topography on in situ stresses is not discussed here as it has been the subject of recent publications by the author and co-workers [82-841.

Classical approach to estimating in situ stresses and its limitations

It is common practice to make two basic assumptions when estimating the state of stress at any depth, z, in a rock mass. The first assumption is that the state of stress can be described by two components: a vertical component, c,,, equal due to the weight of overlying rock at that depth and equal to yz (where y = pg is the average unit weight of the rock) and a uniform horizontal component, (T,, , equal to K times 0”. The second assumption is that both cv and cr,, are principal stresses.

Different expressions have been proposed in the literature for the coefficient K. Talobre [85] suggested that K could be taken (as a working hypothesis) equal to unity, a proposal that has come to be known in the literature as Heim’s rule (based on the work published by the Swiss geologist Heim in 1878). A state of stress where all three principal stress components are equal to yz is often referred to as lithostatic in the rock mechanics literature.

Another expression that is often used in the literature for the coefficient K is K,, = v /( 1 - v) where v is the rock’s Poisson’s ratio. This expression was derived assuming: (I) that the rock mass is an ideal, homogeneous, linearly isotropic continuous half space with horizontal surface; (2) that the rock mass is under gravity alone with vanishing horizontal displacements; and (3) that the loading history has no influence on how in situ stresses build-up. It also implies that horizontal and vertical stresses vanish at the Earth’s surface. The coefficient, K,, is often called in the geotechnical literature the coeficient of earth pressure at rest.

A fundamental question that arises is “how realistic are the aforementioned assumptions compared to actual field measurements”. For comparison sake, Fig. 20(a) and 20(b) shows, respectively, typical variations of the vertical stress and the ratio between the average horizontal stress and the vertical stress with depth for different regions of the world, proposed by Brown and Hoek [87]. Other sources of in situ stress data are available in the literature and can be found, for instance, in Engelder [86] and Amadei & Stephansson [80].

The unit weight y = pg of rocks varies, in general, between 0.025 and 0.03 MN/m3. Thus, the gravitational vertical stress yz should increase linearly with depth with a gradient ranging between 0.025 and 0.03 MPa/m. An average value for the rock unit weight of 0.027 MN/m3 is often assumed giving an average vertical stress gradient of 0.027 MPa/m. Taking a value for Poisson’s ratio v = 0.25 gives K, = v/(1 -v) = l/3. In other words, if the K, condition were true, the horizontal stress should increase with a gradient of 0.009 MPa/m. Note that k?,, can only vary between 0 and 1 as the Poisson’s ratio v varies between 0 and 0.5.

Comparison of the vertical stress gradient of 0.027 MPa/m with what is shown in Fig. 20(a) and what

. AUSTRALIA

0 10 20 30 40 50 60 70

VERTICAL STRESS o,-MPa

&i?!J

- : z + 0.3

3000 ! ’ . :

I I I I 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

K= Ave. HORIZONTAL STRESS oh, a”

VERTICAL STRESS oz

Fig. 20. (a) Plot of vertical stresses against depth below surface; (b) variation of the ratio K between the average horizontal stress and the vertical stress with depth below surface (after Brown and Hoek [87]).

has been reported in the literature [80,86,87] indicates that, in most cases, the magnitude of the vertical stress can be explained by the overburden weight only. Local- ized departures from this assumption are not uncommon, however, and have been observed due to local geological features or active tectonic zones [88]. Bulin [89] reported values of the vertical stress measured at depths of 600 and 900 m (in the Donets-Makeyekva area in the former Soviet Union) three to four times higher than those predicted by gravity due to complex geologic structures. Localized departure can also be due to horizontal tectonic shear stresses as suggested by Voight [go].

Comparison of the assumed horizontal stress gradient of 0.009 MPa/m, the assumption of vanishing horizontal stress at the ground surface and the assumption of uniform horizontal stress with what is shown in Fig. 20(b) and what has been reported in the literature [80,86,87] shows much more discrepancy. In particular, measured horizontal stress levels at the surface of the Earth have been found to have an average maximum value of about


10 MPa [91]. The stress ratio, K, measured in the field is rarely equal to 1 or & = l/3 especially at shallow depths, and is often larger than unity. For instance, Herget [92] noted that 75% of his world stress data (at the time) showed higher horizontal than vertical stresses. Also, in situ stress measurements seem to indicate that K is rarely uniform in the horizontal plane.

The assumption that the vertical and horizontal stresses are principal stresses has been investigated in the literature. For instance, McGarr and Gay [93] plotted on a lower hemisphere stereographic projection net, the orientation of principal stresses measured in several mines in southern Africa. They found a loose cluster of points around the center of the net with most of the points falling within a circle of radius of 30” about the vertical axis, thus indicating some deviation from the vertical and horizontal assumption, McGarr and Gay suggested that the observed scatter could be attributed to complex geology in the areas of stress measurements. They also suggested that for sedimentary basins, the scatter could be less. The conclusions reached by McGarr and Gay [93] are consistent with the analysis of 165 stress measurements in the Canadian shield reported by Herget [94] and with the conclusions reached by Bulin [89] who analyzed a large amount of in situ stress measurements conducted around the world at depths ranging between 25 and 2700 m (with many measurements in the former Soviet Union). Bulin found that for over 60% of the cases, the principal stresses were inclined at less than 30 to the horizontal and vertical directions. Stress measurements in Fennoscandia [95,96] and in the U.K. [97] have also shown that in situ principal stresses are near vertical and horizontal. It must be kept in mind, that the simplifying assumption that the principal stresses are vertical and horizontal with depth breaks down when the ground surface is not horizontal. Topography (even gentle) can have a drastic effect on the magnitude and orientation of in situ stresses in particular close to the ground surface [82-841.

In summary, the classical approach for estimating in situ stresses has limitations in particular when predicting the magnitude of horizontal in situ stresses. In the literature, the difference between predicted and measured stresses has been attributed to the effects of such phenomena as tectonic, residual and thermal stresses, erosion, lateral straining, anisotropy, glaciation and deglaciation, topography, curvature of the Earth and other active geologic processes [62,93,98,99]. The reader should be aware that no agreement has yet been reached with certainty on this matter and that there is still room for discussion.

Estimating in situ stresses in homogeneous anisotropic rocks

The expression for the horizontal to vertical stress ratio & = v/(1 - v) applies to a semi-infinite continuum with linear elastic, homogeneous and isotropic properties. In this section, we discuss the effect of anisotropy on the value of K assuming the ground to be horizontal.

Fig. 21. Stress determination in anisotropic rock mass with inclined layers. Layers are parallel to plane P (after Amadei and Pan [IOO]).

The effect of anisotropy on gravitational stresses in homogeneous rock masses with a horizontal ground surface has been addressed by Amadei et al. [8] and Amadei and Pan [ 1001 who proposed expressions for the coefficient K in transversely isotropic, orthotropic and generally anisotropic rock masses. Consider, for instance, the geometry of Fig. 21 where a rock mass is assumed to be transversely isotropic in a plane P. Let n, s, t be a coordinate system attached to P and inclined with respect to a global x, y, z coordinate system such that the x and y axes are horizontal and the z-axis is vertical downward. Plane P dips at an angle $ and strikes parallel to the y-axis. The rock mass is subject to gravity only and the displacements components in the x and y directions are assumed to be independent of x and y and to depend on z only. This assumption leads to a condition of no-lateral strain where the normal strains 6 XV cY and the shear strain, yXu, vanish.

As shown by Amadei and Pan [loo], for the geometry of Fig. 21 and the condition of no-lateral strain, the stresses in the x, y and z directions are principal stresses with ar = pgz, ax = Kxpgz and aY = K,pgz. In general, the two stress ratios K, and KY are not equal and depend on the dip angle, II/, and the ratios E/E’, G/G’, v and v’. If Ic/ = 0” (horizontal planes of transverse isotropy), K, and KY take simpler forms and reduce to

1 K,=K,=-$-$J’~.__. E’ l-v (24)

If the planes of transverse isotropy are vertical ($ = go”), equation (24) is replaced by

.,. E

K,- ‘CX _ “(lfv); Ky=~=v+v“Z. (25) Pgz 1 -d_ Pgz 1 _ ,,'Z

1 I

E’ A ,

E’ For an isotropic rock mass, equations (24) and (25) reduce to K,= KY= & = v/(1 -v). When $ =0 or 90”, it can be shown that the shear strains yXZ and yur vanish in addition to cX, cg and yXY and that the condition of no-lateral strain reduces to a condition of no-lateral displacement.

As a numerical example, Fig. 22(a) and (b) shows, respectively, the variations of K, = a,/pgz and KY = a,/ pgz for a transversely isotropic rock mass with E/E’ and


0.60

GIG’

(4

0.20 g 0.50 1.00 1.50 i.00 2.50 3.00 3.50

E/E’

0.60 :

0.60

!

13’b.oo

0.20

GIG'

(W

00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

E/E’ Fig. 22. Variation of (a) o,/pgz and (b) uJpgz with E/E’ and G/G ’

for v = v’ = 0.25 and $ = 30” (after Amadei and Pan [loo]).

G/G’ ranging between 1 and 3, v = v’ = 0.25 and for a dip angle I/ equal to 30”. Compared to the isotropic solution, e.g. a,/pgz = u,,/pgz = 0.333 which is represented by point I in Fig. 22(a) and (b), both a, and a,, increase with E/E’ and G/G’. For a fixed value of G/G’, the stresses increase as E/E’ increases, that is as the rock mass becomes more deformable in a direction normal to the plane of transverse isotropy. Note that for a fixed value of E/E’, the stress a, parallel to the dip direction of the plane of transverse isotropy depends strongly on the value of G/G’. On the other hand, the stress au parallel to the strike of the plane of transverse isotropy is not much affected by the value of G/G’. An increase of G/G’ indicates that the rock mass becomes more deformable in shear in planes normal to the plane of transverse isotropy. For a fixed value of G/G’, the stress ratio a,/~,, decreases as E/E’ increases.

The models of Amadei et al. [S] and Amadei and Pan [loo] show that for anisotropic rock masses, the gravity induced stress field is multiaxial and is strongly correlated

to the rock mass structure. The vertical stress is always a principal stress and is equal to the weight of the overlying rock. Its magnitude is independent of anisotropy. The two horizontal principal stress components are, in general, not equal and their magnitude and orientation in the horizontal plane depend on the anisotropic character of the rock mass. Note that the solutions of Amadei et al. [8] and Amadei and Pan [loo] should not be used to estimate gravitational stresses in rock masses with rigid lateral boundaries (no horizontal lateral displacement) when the dip angle, tj, is not equal to 0 or 90”. For those cases, it has been shown by Dolezalova [loll, using the finite element method, that the principal stresses are inclined with respect to the vertical and horizontal directions.

The analyses of Amadei and Pan [ 1001 and Dolezalova [loll show the importance of lateral boundary conditions when estimating in situ stresses. Stresses in a rock mass not only depend on the rock constitutive behavior but also on the conditions along the boundaries of the domain under consideration. For example, adding lateral tectonic straining to the model presented above leads to different stress fields. As shown by Savage et al. [ 1021, if the rock mass of Fig. 21 is under gravity and is also strained (due to tectonism) in the x and y directions by an amount cX and cYy, respectively and for the case of horizontal transverse isotropy ($ = 0’) the horizontal stress components become equal to

u,= (1 _g> x E(t + vc,) +; &Pg’

These expressions for the horizontal stress components predict non-zero horizontal stresses at the Earth’s surface (even if the rock were isotropic). Various three- dimensional stress regimes can be predicted depending on the vanishing or non-vanishing character of the lateral strains and their respective values.

For isotropic rocks, it is known that the Poisson’s ratio must be less than 0.5 due to thermodynamic constraints. Thus, the ratio & = v/(1 - v) is always less than 1 and horizontal stresses larger than the vertical stress are not possible. As shown by Amadei et al. (81, the thermodynamic constraints for anisotropic materials are not as restrictive as for isotropic materials. By combining these constraints with the analytical expressions for gravity induced stresses in orthotropic and transversely isotropic rocks, it can be shown that the predicted in situ stresses can vary over a larger domain than in the isotropic case and that horizontal stresses larger than the vertical stress are admissible. This can easily be done by combining, for instance, the positive part of the domains of variations for the Poisson’s ratios v and v’ in inequalities (5)-(7) with equations (24) and (25).

Estimating in situ stresses in strat$ed rock masses The expressions for K, = q,/pgz and KY = o,,/pgz

nronosed bv Amadei et al. 181 and Amadei and Pan 11001


only apply to rock masses that are homogeneous. in horizontal stress can take place across a stratum with Stratification which is common in sedimentary as well as properties that differ to those in the surrounding rock. volcanic rock masses creates heterogeneities. Depending The influence of lithology on the distribution of on the lithology and the relative stiffness between the horizontal stresses at depth has been demonstrated by different layers, in situ stresses may vary substantially numerous measurements in sedimentary rocks [91, from one layer to another. In general, abrupt changes 103-I 1 I], and volcanic rocks [I 12,113]. As an example,

(a)

i I I

~EsnMATEDovm- I I sumENsTNEM \ (1.05 #I/n) I I -2200

0 1 I

b I - 2250 1 I m I I I I -2400

o \ I I

01 I - 2420

01 I !

STRESS ( hwo) QHyhx DIRECTION 5 IO IS 1 I 1

QEoLoQle SECTION

Fig. 23. (a) In situ stress measurements in lower Mesaverde formation (after Warpinski et al. [104]); (b) in situ stress measurements in layered lava flows in Iceland (after Haimson and Rummel [I 121).


Fig. 23(a) shows the variation of the minimum in situ horizontal stress measured by hydraulic fracturing through perforations by Warpinski et al. [104] at the DOE’s Multiwell Experiment site in the Mesaverde sedimentary formation of Western Colorado. This figure shows higher stresses in the shale layers compared with the surrounding sandstones and siltstones. Another example, shown in Fig. 23(b), corresponds to hydraulic stress measurements conducted by Haimson and Rummel [112] for lava flows in Iceland. Here, both horizontal stress components vary from one lava flow to the next. Stress measurements by hydraulic fracturing conducted by Warpinski and Teufel [113] in welded tuff at the Nevada Test Site have also revealed large stress contrasts due to changes in material properties, beddings and faults. Such contrasts were found to occur sometimes on a scale less than one meter if the contrast in rock properties was sufficient. Finally, an analysis of stress measurements in sedimentary basins by Swolfs [89] revealed that at depths greater than 600m, the effect of lithology on stresses is apparent and that at more shallow depths, other near-surface phenomena may predominate in affecting the stress distribution.

Figure 23(a) and (b) clearly indicates that when lithology affects the distribution of in situ stresses, stress differences (sometimes large) should be expected across layers of different rock types. Thus, for such geologic environments, using linear regression analyses to describe the variation of individual stress components with depth (as has often been done in the literature) becomes meaningless.

Amadei et al. [114] proposed analytical solutions based on the theory of linear elasticity for the stresses in horizontal strata under a condition of no lateral displacement. Each stratum can be isotropic or horizontally layered with moduli Ei, El, G,, G; and Poisson’s ratios vi and vi. For the geometry of Fig. 24, and assuming continuity along the stratum contacts, the state of stress at depth z in the ith stratum is given by

X.Y Qround surface

azi = pigz + f_ (Pj - Pikhj.

j= 1 (27)

In each stratum, as for a homogeneous horizontally layered rock mass, the nature and magnitude of the stress field depend on the anisotropic character of the stratum deformability. Jumps in the magnitude of the horizontal stress occur across stratum contacts due to contrasts in deformability from one stratum to the next. The ratio ahi/ari can be larger than, less than or equal to unity. If a stratum is isotropic, that ratio can only vary between 0 and 1. If a stratum is layered and a representative sample consisting of m layers can be identified, then the ratio ahi/aZi in that stratum is equal to

(28)

where 3 = hi/L with hi is equal to the thickness of layer j in the representative sample of edge dimension L. In writing equation (28) it is assumed that the multi- layered stratum can be replaced by an equivalent anisotropic continuum using the model proposed by Salamon Ml-

Equations (27) and (28) were derived for transversely isotropic rock masses with anisotropic elastic properties that are constant. As discussed earlier in this paper, laboratory and field tests have shown that rock anisotropy is affected by confinement. Thus, the degree of rock mass anisotropy must decrease with depth as more confinement takes place in the rock mass. For intact anisotropic rocks, the increase in confinement may close preferred oriented microcracks and make the material more isotropic with depth. For regularly jointed rock masses, the stiffness of the joints that create the rock mass anisotropy increases with the normal stress acting across their surfaces. As shown in equation (12), the ratio E/E’ approaches a value of unity as the normal stress increases. Thus, as the elastic properties change with the state of stress, so does the state of stress change because of the reduction in rock mass anisotropy. This closely interrelated phenomenon was investigated by Amadei and Savage [115] for the stress distribution in horizontally and vertically regularly jointed rocks. Using the equivalent concept proposed by Duncan and Goodman [9] and the expression for the variation of joint normal stiffness with stress proposed by Bandis et al. [53], it was shown that stress distributions similar to those observed by Brown and Hoek [87] and others could be generated. For instance, for a jointed rock mass cut by horizontal joints with spacing S, combining equations (12) and (24) the horizontal to vertical stress ratio can be expressed as follows

-~ u Pl t

1 hl

1%’ 1 Z*

i-th unit - ohi 4 I hi

: pn T h, n-th unit

Fig. 24. Horizontally layered rock mass with different stratas (after Amadei ef al. [I 141).

As z approaches infinity, the stress ratio defined in equation (29) approaches its value for the isotropic case (assuming that v’ converges toward v). As a numerical

RMMS 33,3-a


0 2 4 6 8

B ' 600

800

1000 Fig. 25. Variation of u,/pgz with depth z for V,,,kmi = 1.71 MPa, v = v’ = 0.25 and E/B,,.9 varying between 0 and 20. The isotropic case corresponds to E/k,iS = 0. Jointed rock mass cut by horizontal joints

with spacing S.

example, Fig. 25 shows the variation of K with depth z for V,,,k,, = 1.71 MPa, v = v’ = 0.25 and E/k,iS varying between 0 and 20. The isotropic case corresponds to E/k,iS = 0. AS Elk,iS increases, or in other words as E increases or k”i and/or S decreases, the joints affect the stress field near the ground surface only. Stress distributions similar to those shown in Fig. 25 could also be obtained by making the moduli E and E’ appearing in equation (24) functions of depth. Linear variations of modulus with depth were proposed by Gibson [116] for soils and could be applied to rock masses as well.

Efect of earth curvature when estimating in situ stresses in stratljied rock

The models discussed above do not account for the effect of the curvature of the Earth on in situ stresses. Some recent models proposed by McCutchen [117] and Sheorey [118] have addressed this problem by modeling the Earth as a self-gravitating spherical shell consisting of one or several concentric slices or layers. As shown below, the curvature of the Earth seems to have a strong effect on in situ stress.

McCutchen [ 1171 considered an isotropic spherical shell representing the Earth’s crust, consisting of material with constant unit weight and subject to gravity. The shell was assumed to be situated on an unyielding massive interior body. No displacement was assumed at the crust-mantle interface. Using the equations of equilibrium, the stress-strain relations and the constitutive equations, expressions for the radial stress, cr,, (also equal to the vertical stress), the tangential stress, og, (assumed the same in all tangential directions and equal to the horizontal stress) and the tangential strain were derived. McCutchen showed that by using the upper and lower bounds of the horizontal to vertical stress ratio, K = ag/q, determined by Brown and Hoek [87] and shown in Fig. 20(b), the depth corresponding to the base of the crust would vary between 33.73

-T- __----_ -R

~~~_=_=E~~ =_-----_ ‘h2

.- 4 __-__-_ - f%

8 ------ -R5

? w -b

N L 2 5 4 5 ------ ----- ---- - ‘R ‘R ‘h 2 3

LIQUID RI CORE

(b) (c)

0 1 2 3

K= HORIZONTAL STRESS se STRESS (MPa)

VERTICAL STRESS 0,

Fig. 26. Spherical shell model of Sheorey: (a) geometry of the Earth consisting of 12 annular slices; (b) predicted variation of K = uo/a, with depth, H, and comparison with upper and lower bounds of Brown and Hoek [87]; and (c) predicted variation of vertical stress, Q,, and horizontal stress, u,,, with depth (modified after Sheorey [118]).

and 138.37 km which is considerably greater than the accepted value of about 15 km over young oceanic areas and 40-50 km over shield areas.

An interesting aspect of the model of McCutchen [ 1171 is that, despite its relative simplicity, the model leads to a predicted variation of K with depth which is consistent with the expressions reported in the literature. In particular, at shallow depths (less than 3 km), K is found to be proportional to l/z. The model also shows that the horizontal stress depends on the thickness of the crust, producing larger stresses in a thicker crust. The main drawback of the model of McCutchen is that the elastic constant and density of the rock in the crust do not vary with depth and the model does not account for the effect of the geothermal gradient.

Sheorey [118] extended McCutchen’s model to account for the effect of the geothermal gradient, the variations of the coefficient of thermal expansion, the unit weight and the elastic properties with depth, and possible displacement within the mantle, on in situ stresses. Figure 26(a) shows the geometry of the Earth modeled by Sheorey. It consists of a series of 12 annular slices, six in the mantle and six in the crust. The crust has an average thickness of 35 km and the radius of the Earth is equal to 6370 km. No displacement is allowed to occur at the mantle-core interface taken at a depth of 2900 km. The state of stress in the mantle is assumed to be hydrostatic. Table 5 gives the values of the coefficient of thermal expansion, cli, the Young’s modulus, Ei, the radius, Ri, and the unit weight, yi, for each slice i = 1, 12. The temperature in the Earth is assumed to vary between 0°C at the ground surface and 3961°C at the base of the


Table 5. Values of the coefficient of thermal expansion, xi, the Young’s modulus, E,, the radius, Ri, and the unit weight, yi, for each slice i = 1,12 in the spherical shell model of Sheorey [I 181. Slices i = 16 correspond to the mantle and slices i = 7-12 correspond to the crust

Slice no.

Radius Rj a, x 10-s x lo6 (m) (/Cl

Unit weight yi E, @PaI (MPa/m)

1 3.470 2.4 760 0.052

8 9

10 11 12

3.870 1.9 700 0.048 4.370 1.6 610 0.045 4.870 1.35 520 0.043 5.370 1.25 360 0.040 5.958 1.2 200 0.037 6.335 0.77 20 0.027 6.340 0 30 0.027 6.436 2.2 40 0.027 6.352 1.5 45 0.027 6.358 0.9 50 0.027 6.364 0.6 50 0.027

mantle, with three temperature gradients of O.O008”C/m (for slices l-5), O.O03”C/m (for slice 6) and O.O24”C/m for slices (7-12). The mantle is assumed to have a uniform Poisson’s ratio, v,, equal to 0.27 and the crust has a uniform Poisson’s ratio, v, equal to 0.2.

Figure 26(b) shows the variation of the stress ratio K = ~~/a, with depth and Fig. 26(c) shows the variation of the horizontal stress, eO, and the vertical stress, (TV, with depth predicted by Sheorey. Figure 26(b) and (c) indicates that Sheorey’s model predicts large values of K at shallow depths and a value of 11 MPa for the horizontal stress at the ground surface, which is in agreement with maximum in situ stress values of about 10 MPa measured at the surface of the Earth [91].

A parametric study conducted by Sheorey reveals several important trends. First, the magnitude of the horizontal stress depends on the elastic modulus with softer slices producing less horizontal stresses than harder slices which is consistent with what has been reported in the literature [91,103-l 141. Variations in the modulus of the top slice of Fig. 26(a) show that the stress ratio K in that slice is essentially proportional to the value of the modulus. Another trend found by Sheorey is that inclusion of the thermal gradient keeps the magnitude of horizontal stresses within reasonable limits. For instance, if the coefficients of thermal expansion are assumed to vanish, the model of Sheorey gives an unrealistic horizontal stress of 132.4 MPa at the ground surface. Inclusion of the coefficient of thermal expansion reduces that stress component to a more reasonable value of 11 MPa. Finally, the model of Sheorey seems to indicate that larger horizontal stresses could be expected in areas where the crust is thicker, e.g. in the continental crust.

Both models of McCutchen and Sheorey reveal that the curvature of the Earth could be responsible for large values of K and of the horizontal stress near the Earth’s surface. The model of Sheorey predicts that horizontal stresses vary between 0 if the curvature of the Earth is ignored to 11 MPa if the curvature is taken into account.

In his paper, Sheorey explored the importance of anisotropy in the top slice [slice 12 in Fig. 26(a)] on the stress distribution in the crust, This slice is 7 km thick

and was assumed to be transversely isotropic with a plane of transverse isotropy (plane of layering) parallel to the ground surface. This part of Sheorey’s model was explored further to investigate the combined effect of Earth’s curvature and intact or joint induced anisotropy (which depends on depth) on in situ stress.

Let r be the distance from the center of the Earth and Z be the depth below the ground surface with Z = R - r. The top slice (i = 12) is assumed to be transversely isotropic in a ~(3 plane parallel to the ground surface with E, = E, = E, E, = E’, vor = vzo = v, v, = vro = v’, and G,, = GIO = G’. In this slice, the rock has two coefficients of thermal expansion 8, and /$ in directions parallel and normal to the plane of transverse isotropy, respectively. As shown by Sheorey [118], the vertical stress (T, and the horizontal stress crO = a, in the top slice are equal to (compressive stresses being positive)

a,=y(R -r)

+1,G3(R -r) (30)

where y is the unit weight and G, = O.O24”C/m is the thermal gradient of the rock in slices i = 7-12. Also,

a+b= E Ev’

* c=

1 -v -2v”;’ 1 --v -2v$’

d= E’(1 -v)

l-v-2v’$

&=&(a+b)+&c; &=2&c+/?& (31)

The constant C,, depends on the coefficients of thermal expansion, ai, the Young’s moduli, Ei, the radii, Ri, and the unit weights, yi, of the other slices i = 1, 11. It also depends on the Poisson’s ratio of the mantle, v,, and the two temperature gradients G, = O.O008”C/m (for slices l-5) and G2 = O.O03”C/m (for slice 6) as follows

+C,,R,%

“‘i,2v’)( R - T R12)]. (32a)

For i = 8-l 1, Ci is equal to


For i = 7, C, is equal to

- g7G&$))(R -TR7)] (32~) 7

with

P~=YW-&)+Y&; b=GW-&I.

For i = 6, C, is equal to

C,=C,+R,(l -2+-g-$($-;)]

a,t,-aol,t,-aor,G, RI--$ ( )

+a,G,$ 1

(324

with

PS = ~6 + (~5 - MRe; t, = G3(R - R7) i- G,(R, - Rd.

For i = 2-5, Ci is equal to

Ci=Ci_I

Pi-1 Pi Yi- 14 YiRi -z+E,+~E,_,-- 2Ei 1 - Ri(ai-arl)[tl +Gl(Rh-T)] We)

with

Finally, C, is equal to

(320

Analysis of equations (32a)-(32f) shows that the dimensionless stress ratio a&R depends on the following dimensionless quantities

r E EE R’E” vv VI, vrn9 yR’ Ei - -(i = 1, ll),? (i = 1, 12)

z,:,a,RG,(i = 1, ll),/?,RG,,$,~ (i = 136). (33) I I I

The stresses in the top slice of Fig. 26(a) were determined for different degrees of anisotropy of that slice. The properties of slices 1-11 were the same as those used by Sheorey [118] and reported in Table 5. For the top slice, E = 50 GPa, v = 0.2, 8, = Q2 = 0.6 x 10-5/“C

V’

Fig. 27. Variation of the horizontal stress c,, at the ground surface with E/E’ and v’. The rock in the top slice is transversely isotropic with E = 50GPa, v =0.2, B, =&=0.6 x 1O-5/“C and E/E’= l-4. The thermodynamic constraint corresponding to inequality (7) is

represented by a dashed line.

E/E’, the Poisson’s ratio v’ was varied between 0 and its maximum thermodynamically admissible value defined by the right hand side of inequality (7). Figure 27 shows the variation of crO at the ground surface (r = R or Z = 0) for different values of E/E’ and v’. This figure shows that the horizontal stress at the ground surface is not equal to zero and that, compared to its value of 11.02 MPa when the rock is isotropic (represented by point I in Fig. 27), cB is more sensitive to the value of v’ as E/E’ increases. Nevertheless, Fig. 27 indicates that the magnitude of the horizontal stress at the ground surface is not much affected by the anisotropy of the top slice. This conclusion can be generalized for the entire top slice as shown in Fig. 28, where the variation of the stress ratio K = be/b, with depth has been plotted for several cases of rock anisotropy including isotropy.

Figures 27 and 28 were obtained assuming that the anisotropy in the top slice is uniform with depth. The effect of joint induced anisotropy on the stress distribution in the Earth’s crust was investigated by combining equation (12) which is based on the models of Duncan

Isotropy & E/E’=2

1600

Fig. 28. Variation of K = uO/ur with depth for isotropic solution (E/E’ = 1, v = v’ = 0.2) and three cases of anisotropy with E/E' = 2, 3 and 4 and v’ = 0.1. In all cases, E = 50 GPa, v = 0.2, 8, = /I2 = 0.6 __ z__ and E/E’ was varied between 1 and 4. For each value of x lO_‘/“C.


3 10

8 9

’ 8

E/k,,,S Fig. 29. Variation of o0 at the ground surface for E = 50 GPa, v = v’ = 0.2, /?, = fiz = 0.6 x 10-s~C and Y,,,k,, = 1.71 MPa (V&&R = lo-‘). The ratio E&S varies between 0 and 30 with the isotropic

case corresponding to E/kniS = 0.

and Goodman [9] and Bandis et al. [53] with the model of Sheorey [118]. Assuming the joints to be parallel to the ground surface with a spacing S, the normal stress in equation (12) becomes equal to or = 72. Combining equation (12) with equations (30)-(32), it can be shown

0, (MPG

0 4 8 12 16 20 24 28

200

g 400

s 8 600

800

K

1 2 3 o” u , I , I , I , 4

200-

g 400-

5 E 600-

SOO-

lOOO- Fig. 30. (a) Variation of u0 with depth; and (b) variation of K = uo/ur with depth, for E = 50 GPa, v = v’ = 0.2, 8, = b2 = 0.6 x lO-s/“C and V,,,k,, = 1.71 MPa (V,k,,/yR = lo-‘). The ratio E/knit3 = 0, 10, 20

and 30. The isotropic case corresponds to E/kniS = 0.

that, for joint induced anisotropy, a&R depends on E/k,iS and V,,,k,,/yR, in addition to the other dimensionless quantities in (33).

Figure 29 shows the variation of a0 at the ground surface for E = 50 GPa, v = v’ = 0.2, p, = & = 0.6 x 10-s/C and V,kni = 1.71 MPa (V,k,i/yR = lo-‘). The ratio Elk,iS varies between 0 and 30 with the isotropic case corresponding to E/k,iS = 0. Figure 29 indicates that as Elk,i S increases, or in other words as E increases or k,i and/or S decreases, the magnitude of the horizontal stress at the ground surface decreases in particular for values of E/k”iS larger than five. It can be shown that for larger values of E/k,iS, the horizontal stress at the ground surface becomes tensile (negative). Figure 30(a) and (b) shows, respectively, the variation of the horizontal stress, ati, and the stress ratio K = aB/a, with depth as Elk,iS varies between 0 and 30. As expected, the joints affect the in situ stress field the most near the ground surface. With depth, the confinement decreases joint induced anisotropy and the in situ stress field approaches its value for an isotropic medium.

Figures 28 and 30(b) indicate that the combined effect of rock anisotropy (intact or jointed) and Earth’s curvature can yield variations of K with depth that are quite comparable to those reported in the literature. It is noteworthy that no unique solution is available as various stress variations can be generated by simply changing how the various rock properties vary with depth.

EFFECT OF ANISOTROPY ON THE ANALYSIS OF OVERCORING MEASUREMENTS

Overcoring techniques can be classified as borehole relief methods, i.e. methods and procedures that isolate a rock specimen from the stress field in the surrounding rock by coring. Strain or displacement measurements on the specimen thus isolated are recorded in the vicinity of the point at which the state of stress has to be determined. This requires the stress field to be homogeneous throughout the zone of interest before the measurements are performed which is a reasonable assumption in the absence of heterogeneities or major geological features in the rock mass.

Figure 3 1 illustrates the three steps commonly involved when measuring in situ stresses by overcoring with instrumented devices such as the CSIR Triaxial Strain cell [119], the CSIRO Hollow Inclusion cell [ 1201, the USBM gage [ 1211 and other CSIR type strain cells [122-1251. First, a large diameter hole is drilled to the required depth in the volume of rock in which stresses have to be determined. Then, a small pilot hole is drilled at the end of the previous hole. An instrumented device is inserted into the pilot hole. Finally, the large diameter hole is resumed and resulting changes in strain or displacement within the instrumented device are recorded.

The successful interpretation of overcoring tests depends to a great extent on the ability (1) to establish a stress-strain (or displacement) relationship for the


LARGE DIAMETER HOLE

INSTRUMENTED DEVICE

PILOT HOLE

Fig. 31. Steps commonly involved in overcoring techniques with instrumented devices such as the CSIR cell, the CSIRO cell, the USBM

gage and other CSIR type strain cells.

rock and (2) to be able to determine rock mass properties from tests on core samples. It is common practice to relate strains or displacements to the in situ stress field components through equations derived from the theory of linear elasticity for isotropic media. The purpose of this section is to show the role of rock anisotropy in the analysis of results of overcoring tests with CSIR type strain cells.

Problem geometry and assumptions Consider again the geometry of Fig. 14(a)-(c). The

global X, Y, 2 coordinate system is now chosen such that the +X and +Z axes point in the North and East directions, respectively. The Y axis is positive upward. The hole in Fig. 14(a) now represents a pilot hole of diameter D = 2a. The orientation of the pilot hole, and that of the x, y, z axes attached to that hole, are defined with respect to the X, Y, Z coordinate system by two angles /l,, (borehole azimuth) and d,, (borehole rise) as shown in Fig. 14(b) such that the x-axis lies in the XZ plane.

The rock is assumed to be orthotropic or transversely isotropic in the n, s, t coordinate system of Fig. 14(a). Transverse isotropy is assumed to be in the st plane. The orientation of the n, s, t coordinate system with respect to the X, Y and Z axes is defined by the dip direction angle, fl,, and dip angle, $, , of a clearly defined plane of rock anisotropy. The constitutive equation of the rock in the n, s, t coordinate system is given by equations (1) and (2). Using second order tensor coordinate transformation rules [55], the rock constitutive equation in the x, y, z coordinate system can be written as follows

where (6 )i,, = (G y cu 9 6,) yyz, ~~~~~ yxy) and 6~ X,, = (ox, oy, 02, zp, Txr, zXy) are the strain and stress matrices, respectively. As shown by Amadei [55], (A) is a (6 x 6) symmetric compliance matrix whose components ay (i,j = 1,6) depend on the elastic properties of the rock in the n, s, t coordinate system and the four orientation angles b,,, 6,, & and $, defined in Fig. 14(b) and (c).

The rock is assumed to be subjected to a three- dimensional in situ stress field with components uxo, afi,

~203 ~)a09 ~no and z,,,, acting at infinity and defined in the x, y, z coordinate system. Let (a,,) be the stress matrix such that

(0,)’ = (OHI e$~za ~yzo rx_?a rxyo ). (35)

In the global X, Y, Z coordinate system, the in situ stress field components are cxo, tag, cm, zyZO, zxzo and rXyO and the stress matrix (G,,)~,,~ is related to (a,) as follows

(all) = (T,)(%)X, (36)

where (T,) is a (6 x 6) stress transformation matrix whose components depend on the direction cosines of the x, y and z axes with respect to the global X, Y, Z coordinate system, and therefore on the two angles )6,, and 8,, of Fig. 14(b).

Stresses and strains around a borehole drilled in a stressed medium

As shown by Amadei [55], the stress field at a point P(a, 0) along the contour of the hole of Fig. 14(a) can be expressed in matrix form as follows

(a), = (F)(%) (37)

where (F) is a (6 x 6) matrix with components that depend on the elastic properties of the medium in the n, s, t coordinate system, the four orientation angles /l,,, Bh, & and +a and the angle 8. The latter assumes all values from 0 to 2x for a complete circuit along the contour of the hole. The corresponding strain field at P(a, 0) is obtained by combining equations (37) and (34) which gives

(c).ryz = (A)(F)(%). (38)

Let r, 8, z be a cylindrical coordinate system attached to the borehole. The strain components in that system are related to those in the x, y, z coordinate system as follows

(C)r& = (T&)(c)Xy; (39)

where (TreZ) is a (6 x 6) coordinate transformation matrix for strains [55]. Combining equations (38) and (39), the strain field at P(a, 0) can be written as

(c)r& = (T,,Z)(AW)(%). (40)

Finally, the longitudinal strain, E,, at P(a, 0) in a direction inclined at an angle, J/, with respect to the z-axis in the 8z plane is equal to

c, = (0 sin’ $ cos’ JI cos $sin JI OO)(T,,:)(A)(F)(a,).

(41)


CABLE

\

Fig. 32. (a) Geometry of a CSIR type of strain cell; (b) orientation of strain gages.

Analysis of in situ stress measurements with CUR type strain cells

Recalling the general procedure described in Fig. 31, the process of overcoring can be seen as canceling the components of stress acting across a cylindrical surface in the anisotropic medium. Hence, this is equivalent to add tractions equal in magnitude but opposite in sign to those that existed across the cylindrical overcore surface before it was drilled.

In general, CSIR type strain cells glued directly on the wall of a pilot hole do not interfere much with the deformation of the medium during overcoring. Except for a small region near the contact between the rock and the cells, the overcore is free of stress and strain at the end of the process of overcoring. Thus, the size of the overcore should not affect the analysis of the field measurements. Overcoring releases displacements, strains and stresses not only due to the in situ state of stress but also those induced by introduction of the hole associated with overcoring.

Because of the above considerations, the strains recorded with CSIR type strain cells can be assumed to be equal in magnitude but opposite in sign to those that would be developed if the pre-existing in situ stresses were applied at a large distance from a central hole already drilled and instrumented in the medium. From an analytical point of view this means that, if G.~, afl, G,, rYfi, L, and 7xvo are the components of the in situ stress field to be determined with a CSIR type strain cell, equation (41) can be used to relate the strain measured in each strain gage of the cell with the components of

the in situ stress field by simply substituting -6, for E,. It is noteworthy that this is only applicable if the pilot hole is long and strain measurements take place in cross-sections remote from the ends of the pilot hole and the end of the large diameter hole. Also, perfect bonding between the strain gages and the rock is assumed.

Consider a CSIR type strain cell with the geometry of Fig. 32(a). For each strain gage, i, whose center is located at an angle, ei, from the x-axis (or ai from the y-axis) and whose longitudinal axis is inclined at an angle, $, , with respect to the z axis [Fig. 32(b)], the strain bli measured during overcoring is equal to

-cl, = (0 sin* tii cos* tji cos tii sin ei 0 0) (T,@,)

x (A)(F)@,). (42)

Equation (42) can also be rewritten in terms of the in situ stress field (aO)XYZ by using equation (36) e.g.

- cli = (0 sin’ ll/j cos* ei cos tii sin ((li 0 0) ( TrBZ)

x (A)(F)(T,)(e,),,. (43)

Equation (43) represents the basic equation for the analysis of overcoring measurements with CSIR type cells in anisotropic or isotropic rocks. It also implies that the strain gage length is neglected in the analysis.

Equation (43) shows that, in general, each strain is a linear function of all six components of the in situ stress field. Therefore, determination of those components requires that we set up a system of six independent equations from the results of six independent strain measurements. However, any additional measurements can be used to obtain a least square estimate of the stress components. This can be done by multilinear regression analysis [54]. Let N (N 2 6) be the number of strain gages in the cell. Equation (43) can be written for each one of the N strain gages. This leads to the following system of N equations and six unknowns

(6) = (T)(%)XW (44)

where (c) and (T) are (N x 1) and (N x 6) matrices, respectively. As shown in Draper and Smith [54], the least square solution of equation (44) is the solution of the following system of six equations and six unknowns

(WE) = (~)‘(W%),YZ~ (45)

Computer program CSIRA.FOR

The model presented above was implemented into a computer program called CSIRA.FOR. This program can be used to determine in situ stresses in isotropic, transversely isotropic or orthotropic rocks by overcoring of a CSIR type strain cell in a single borehole. The latter can be inclined with respect to the planes of rock anisotropy. The program has been written for CSIR type strain cells with a maximum number of four strain rosettes, with up tofour strain gages per rosette [N < 16 in equations (44) and (45)]. The program is generic enough in that the user specifies (1) the number of strain rosettes and strain gages in the cell, (2) the location of each one of the strain gages around the pilot hole [angle


Table 6. Orientation of strain gages and measured strains during overcoring using a CSIR type strain cell. The orientation angles ai and $,,

are defined in Fig. 32(b)

Rosette Strain a, 0 $, (“) 1 -0.4565 x lo-“ 300 0

0.6149 x 1O-4 300 90 0.7654 x 1O-4 300 45

-0.6070 x 1O-4 300 135 2 -0.1661 x 10-r 180 45

-0.9939 x 1o-4 180 135 -0.2198 x lo-’ 180 90 -0.4569 x 1O-4 180 0

3 -0.4565 x 1O-4 60 0 -0.1637 x 10-j 60 90 -0.1400 x 10-j 60 45 -0.6935 x 1O-4 60 135

ei, or cli in Fig. 32(b)], and (3) the orientation of each strain gage in the 8z plane of Fig. 32(b). The program can be used for the analysis of overcoring results with the CSIR cell of Leeman [119], the cells developed by Hiltscher et al. [122] and Hallbjiirn [123] for the Swedish State Power Board, the LuT cell developed by Leijon [124] and the Auckland New Zealand soft inclusion (ANZSI) cell of Mills and Pender [125].

The program calculates the least square estimates of the principal components of the three-dimensional in situ stress field and their orientation with respect to a fixed X, Y, 2 coordinate system. The +X and +Z axes point in the North and East directions, respectively. The orientation of the borehole and the planes of rock anisotropy in that coordinate system is input using the angles defined in Fig. 14(b) and (c). The program also calculates the domain of variation of each one of the in situ stress components for different levels of confidence limits as specified by the user. Finally, the program runs on IBM PC compatible computers (486 and above).

Numerical examples In all the numerical examples presented below, the

rock is assumed to be transversely isotropic and the borehole is assumed to be parallel to the + Z (East) direction of Fig. 14(a)-(c) with an azimuth angle /3,, = 90” and a rise angle 6, = 0”. The CSIR type strain cell is assumed to contain three rosettes at 120” from each other. Each rosette contains four strain gages. The orientation of the 12 strain gages and the magnitude of the strains recorded during overcoring are given, in Table 6. The in situ stress field was determined by solving equation (45) with N = 12, for different degrees of rock anisotropy and different anisotropy orientations with respect to the borehole.

In the first numerical example, the plane of rock anisotropy (or plane of transverse isotropy) is assumed

(a) G/G’ = 1

N

A: E/E’ E 1.5

??: E/E’ I 2.0

o: E/E’ = 3.0

(b) G/G’= 2

A: E/E’ = 1.5

0: E/E’ = 2.0

0: E/E’ = 3.0

Fig. 33. Orientation of principal stresses u,, u2 and c1 for E/E’ = I, 1.5, 2 and 3 and G/G’ = 1 and 2 in (a) and (b), respectively. The isotropic case corresponds to E/E’ = G/G ’ = I. Lower hemisphere

stereographic projection. First numerical example.

to be horizontal [I,+, = 0” in Fig. 14(c)], and to be parallel to the borehole. The rock elastic properties are such that E = 35 GPa, v = 0.25 and G = 14 GPa. The ratio E/E’ was taken equal to 1, 1.5, 2 or 3. The ratio G/G’, was taken equal to 1 or 2. The Poisson’s ratio v’ was taken equal to 0.25 (for the isotropic case) or 0.27 (for the anisotropic cases). Table 7 and Fig. 33 give, respectively, the magnitude and orientation of the three principal stresses for different values of E/E’ and G/G’.

Table 7 indicates that for a given value of G/G’, the magnitude of the stresses increases with E/E’ or in other

Table 7. Magnitude of the three in situ principal stresses (in MPa) for different values of E/E’ and G/G’. First numerical example. The case E/E’ = G/G’ = 1 corresponds to isotropy

G/G’ = 1 G/G’=2

E/E’ = 1 E/E’ = 1.5 E/E’ = 2 E/E’ = 3 E/E’ = I E/E’ = 1.5 E/E’ = 2 E/E’ = 3

=I 3.83 3.87 3.93 4.14 3.04 3.08 3.15 3.34 =2 3.07 3.26 3.42 3.78 2.57 2.65 2.74 2.93 Q3 0.24 0.32 0.38 0.51 0.33 0.37 0.41 0.48


Table 8. Magnitude of the three in situ principal stresses (in MPa) for different values of /?. ranging between 0 and 90”. Second numerical example with E/E’ = G/G’ = 2. The isotropic case is shown for comparison. The plane of rock anisotropy dips at an angle I/I. = 30

Isotropic /?,=OO 8, = 15” p, = 30 8, = 45” /?.=60” 8, = 75” /3, = 90

01 3.83 3.08 3.10 3.14 3.20 3.28 3.36 3.44 =2 3.07 2.38 2.47 2.58 2.71 2.84 2.96 3.06 03 0.24 0.29 0.33 0.38 0.43 0.47 0.49 0.49

Fig. 34. Orientation of principal stresses 6, , u2 and uj when the plane of rock anisotropy dips at an angle $. = 30” and its dip direction angle /3, varies between 0 and 90”. The isotropic solution is also shown for comparison. Lower hemisphere stereographic projection. Second

numerical example.

words as the rock becomes more deformable in the direction perpendicular to the plane of rock anisotropy (vertical direction in the present case). The error involved in neglecting anisotropy by assuming isotropy can be large. For instance, Table 7 shows that when E/E’ = 3 and G/G’ = 1, the errors in 6,) tag and cr3 are 8, 23 and 112%, respectively.

Figure 33(a) and (b) gives the orientation of the principal stresses using the lower hemisphere stereographic projection when G/G’ = 1 and 2, respectively. For a given value of G/G’, both cr, and o2 rotate as E/E’ increases whereas the orientation of cr3 is essentially not affected by the rock anisotropy. Figure 33(a) shows that when G/G’ = 1, neglecting rock anisotropy by assuming isotropy results in a maximum error of 15” for the bearing of 0,) 18” for u1 and 3” for u3.

In the second numerical example, the rock is transversely isotropic with E = 35 GPa, E’ = 17.5 GPa, v = 0.25, v’ = 0.27, G = 14 GPa and G’ = 7 GPa. Thus, both E/E’ and G/G’ are equal to 2. The plane of rock anisotropy (plane of transverse isotropy) dips at an angle $, = 30” and its dip direction angle /I, varies between 0” (anisotropy striking parallel to the borehole) and 90” (anisotropy striking perpendicular to the borehole). Table 8 and Fig. 34 give, respectively, the magnitude and orientation of the three principal stresses for different values of /I,,. The isotropic case is also shown for comparison.

Table 8 indicates that the magnitude of b,, a, and 0) depends greatly on the orientation of the plane of rock anisotropy with respect to the borehole. Neglecting rock anisotropy by assuming isotropy results in a maximum error of 19% for cr, (when /Ia = O’), 22% for err (when /I, = O’), and 104% for g3 (when /I, = 90’). Figure 34

indicates that both IT, and rrz rotate with the plane of rock anisotropy. The rotation takes place essentially in a plane perpendicular to (r3 which is represented by a great circle in Fig. 34. Note that the orientation of (TV is essentially unaffected by the value of pa. Neglecting rock anisotropy by assuming isotropy results in a maximum error of 120” for the bearing of 6, (when pa = 0’) 125” for c2 (when /?, = 60”), and 12” for cr3 (when /I, = 90”).

Field evidence at the URL site

The importance of anisotropy has been clearly shown in the analysis of overcoring tests conducted at the Underground Research Laboratory (URL) site located 100 km northeast of Winnipeg, Manitoba (Canada) in the Lac du Bonnet granite batholith. An overview of the geomechanics aspects of the URL site and a comprehensive literature review on this project can be found in Martin and Simmons [ 1261.

At the 240 Level of the URL site, the granite was found to be anisotropic due to pervasive microcracks which were found to be aligned with a major joint set. The rock was modeled as transversely isotropic with the plane of transverse isotropy parallel to the plane of microcracks. The rock’s Young’s modulus in the direction perpendicular to the plane of rock anisotropy was 30 GPa which was about 50% of that in the direction parallel to the plane. Poisson’s ratios parallel and perpendicular to the planes were equal to 0.25 and 0.15, respectively. Overcoring results obtained in two boreholes 0C2 and PH3, 20 m apart, were analyzed assuming isotropic and anisotropic conditions. The isotropic and anisotropic analyses are shown in Fig. 35(a) and (b), respectively. These two figures indicate that inclusion of anisotropy creates a rotation of the stress field and yields consistent results in both boreholes, and less scatter. As discussed by Martin and Simmons [126] and as shown in Fig. 35(c), the magnitude for each of the three principal stresses was found to be reduced if the anisotropy was taken into account.

CONCLUSION

This paper emphasizes the interaction that exists between rock anisotropy and rock stress. Rock fabric controls the build-up in in situ stresses in the Earth’s crust, their magnitude and orientation. On the other hand, stresses and in particular compressive stresses tend to close microcracks or planes of discontinuities in rock masses thus making rock behavior non-linear and rock anisotropy pressure dependent. Rock anisotropy decreases with an increase in confinement.

In rock engineering, rock anisotropy can be accounted for in various analytical and numerical tools such as the finite element, boundary element and discrete element


270” 90”

ISOTROPIC

180” ANISOTROPIC

40

1 ??ISOTROPIC

(c) * ??ANISOTROPIC 3

02 03 Ql 02 =3

PH3 oc2 Fig. 35. Comparison of principal stresses obtained from overcore tests in boreholes PH3 and 0C2 at the 240 level of the URL site. (a) Stress orientation assuming isotropic rock properties; (b) stress orientation assuming anisotropic rock properties; (c) comparison between stress magnitudes obtained with isotropic and anisotropic analyses (after Martin and Simmons [126]).

methods. Despite all those existing models, the rock engineer is still faced today with a challenge which is the exploration and characterization of anisotropic rocks in the laboratory and in situ.

The constitutive models used for the analysis of laboratory and field tests on anisotropic rocks rely heavily on the simplified orthotropic and transversely isotropic models of linear elasticity for anisotropic bodies. In most practical cases, the transversely isotropic model is retained in order to reduce the number of elastic properties to a more manageable number of five. For field tests, additional assumptions are often made in order to reduce the number of elastic properties further. Prior to testing, it is common practice to select the type of rock symmetry with the assumption that the planes of rock symmetry coincide with the apparent fabric of the rock. It is noteworthy that this assumption should be used with caution as the principal deformability directions of a rock may not necessarily coincide with its apparent fabric. This could happen if the rock contains two or three non-parallel fabric directions but only one is clearly visible and dominant. Slate is a good example of such a material where the original bedding planes of the rock prior to metamorphism can be overshadowed by the more pronounced cleavage planes created by metamorphism.

Also, linear features such as lineations can be superposed on planar features. To be correct, the principal directions of rock symmetry (if any) should be determined from the test results at the same time as the rock deformability properties. This requires more advanced testing, techniques and apparatuses (such as the multiaxial cell) which may not always be available to the practicing engineer on a regular basis.

Another practical problem which arises with anisotropic rocks is that, to date, no standards or suggested methods are available specifically for the sampling, preparation and testing of anisotropic rocks and rock masses. Major difficulties can be encountered for instance during coring and preparation of samples of weak anisotropic rocks as the rocks tend to break along the fabric. For such rocks obtaining samples might be extremely difficult and the current ISRM or ASTM suggested testing methods no longer apply. It is the opinion of the author that new standards and procedures are critically needed for the exploration and testing of anisotropic rocks in the laboratory and in the field.

Despite some of the problems associated with determining rock anisotropy quantitatively, rock anisotropy can also be assessed qualitatioely, by testing rock samples and rock masses in different directions. Analysis of the


test results, even with solutions based on linear elasticity for isotropic bodies, may reflect substantial changes in rock deformability with direction, thus indicating that the rock being investigated is anisotropic. This can be captured for instance by carrying out plate load tests or flat jack tests in different directions. The variation of the rock mass modulus around the wall of a borehole determined from LVDT measurements during dilatometer tests or jacking tests may also give an indication of rock anisotropy. Finally, the results of biaxial tests on overcores samples containing a CSIR type of strain cell may also indicate that the rock is anisotropic. This would be the case if for each strain cell rosette, the 45 or 135” strain gage measurement differs substantially from the average of the tangential and longitudinal strain measurements [ 1271.

This paper and other papers written by the author and co-workers have shown that the classical expression for the horizontal to vertical in situ stress ratio, e.g. K = v/(1 - v), no longer applies for anisotropic rock masses under gravity and no lateral strain condition. In particular, it has been found that the gravity induced in situ stress field is multiaxial with non-uniform horizontal stresses and that it is strongly correlated to the rock mass fabric. Horizontal in situ stresses larger than the vertical stress become admissible for certain values of the rock mass elastic properties. Anisotropy combined with heterogeneity (associated with stratification) may result in complex stress regimes, and scatter and perturbation in the stress field at all scales. When lithology affects the distribution of in situ stresses, stress differences (sometimes large) should be expected across layers of different rock types. Thus, for such geologic environments, using linear regression analyses to describe the variation of individual stress components with depth (as is often done in the literature) becomes meaningless.

This paper indicates that phenomena such as rock anisotropy (intact and joint induced) and the Earth curvature could result in high horizontal in situ stresses in particular near the ground surface. The predicted variations of K with depth have been found to be quite similar to the measurements reported in the literature. Such high stresses have often been thought to be associated with tectonism in the past. This is not to say that tectonic stresses do not exist but simply that their contribution to the measured stress fields may not be as large as previously thought.

Finally, this paper shows that the analysis of overcoring tests in anisotropic rocks depends greatly on the anisotropic character of the rock, the degree of rock anisotropy as well as the orientation of the planes of rock anisotropy with respect to the borehole in which overcoring takes place. Large errors in both in situ stress magnitude and orientation may result if rock anisotropy is not taken into account.

Acknowledgemenrs-This research is funded in part by National Science Foundation, Grant No. MS-9215397. The author is grateful to Professor Mark L. Talesnick from the Israel Institute of Technology (Technion) for conducting hollow cylinder tests on sandstone and to

Mr C.-S. Chen, graduate student at the University of Colorado at Boulder, for carrying out other laboratory tests on sandstone.

Accepted for publication 24 May 1995

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