Determining elastic constants of transversely isotropic rocks using ...

INTERNATIONAL JOURNAL FOR NUMERICAL AND ANALYTICAL METHODS IN GEOMECHANICS 

Int. J. Numer. Anal. Meth. Geomech. 2008; 32:219–234 

Published online 23 April 2007 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/nag.619 

Determining elastic constants of transversely isotropic rocks using 

Brazilian test and iterative procedure 

Yen-Chin Chou 1,‡ and Chao-Shi Chen 2,∗,†,§ 

1 Department of Resources Engineering, National Cheng Kung University, No. 1, Da-syue Road, Tainan City 

70101, Taiwan, Republic of China 

2 Sustainable Environment Research Center, National Cheng Kung University, No. 1, Da-syue Road, Tainan City 

70101, Taiwan, Republic of China 

SUMMARY 

The elastic constants of rocks are the basic parameters for rock mechanics, and play a very important role 

in engineering design. There are many laboratory methods to determine the elastic constants of transversely 

isotropic rocks, and the Brazilian test is a popular method. This paper presented a method combination 

of the Brazilian test, back calculation, and iterative procedure to evaluate the five independent elastic 

constants of transversely isotropic rocks in laboratory. The strain data at the centre of discs were obtained 

using Brazilian test. The stresses at the centre of discs were computed using numerical programs. By 

using back calculation, the temporary elastic constants were computed after the stresses and stains were 

substituted into elastic mechanics equations. After iterative procedure, the convergent values of the elastic 

constants can be obtained. One numerical example and three experimental cases were proposed to show 

the applicability of this method. The convergent values of the five independent elastic constants can be 

obtained in no more than 10 iterative cycles. The results coming from numerical analysis method exhibited 

satisfactory outcome in accordance with those of generalized reduced gradient method. The merits of 

this method include convenient specimen preparation of the Brazilian test, simple iterative procedure, 

and readily available commercially numerical programs, so that this method can be easily popularized in 

research and engineering analysis. Copyright q 2007 John Wiley & Sons, Ltd. 

Received 9 November 2006; Revised 9 March 2007; Accepted 9 March 2007 

KEY WORDS: transversely isotropic rock; elastic constants; Brazilian test; stress concentration factor; 

iterative procedure 

∗ Correspondence to: Chao-Shi Chen, Sustainable Environment Research Center, National Cheng Kung University, 

No. 1, Da-syue Road, Tainan City 70101, Taiwan, Republic of China. 

† E-mail: chencs@mail.ncku.edu.tw 

‡ Ph.D. Candidate. 

§ Associate Professor. 

Copyright q 2007 John Wiley & Sons, Ltd.

220 Y.-C. CHOU AND C.-S. CHEN 

INTRODUCTION 

Many rocks exposed near 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, and hydraulic) that vary with direction and are said to 

be inherently anisotropic. 

Rock anisotropic property plays an important role in various engineering activities. Evaluating 

anisotropic mechanical properties helps predicting the behaviour of rock materials in analysis, 

design, and construction, and improves the quality and safety. Elastic constants of anisotropic 

rocks affect deformation analysis and design in engineering. It is important to estimate the elastic 

constants of anisotropic rocks, rapidly and accurately. 

Anisotropy is generally simplified to be orthotropic or transversely isotropic in engineering 

analysis and research. In dealing with engineering analysis and research of transversely isotropic 

rocks, the following five constants, Young’s modulus E and Poisson ratio � of isotropic plane, and 

Young’s modulus E ′ , Poisson ratio � ′ , and shear modulus G ′ of anisotropy, should be taken into 

account. 

There are several methods including in situ tests, laboratory tests, and numerical analysis methods 

to evaluate elastic constants of rocks. The tests in laboratory can be divided into dynamic and 

static methods. The dynamic methods include the resonant bar method and the ultrasonic pulse 

method. Dynamic elastic constants Ed, �d, andGd of isotropic rocks can be evaluated by using the 

resonant bar method [1] and the ultrasonic pulse method [2]. The five independent elastic constants 

of transversely isotropic rocks can be evaluated by using the ultrasonic pulse method [3], andthe 

restriction is that the transversely isotropic plane has to be parallel to the longitudinal axis of the 

specimen. 

The static methods include the uniaxial compression test, conventional triaxial compression 

test, true triaxial compression test, hollow cylinder test, bending test, torsion test, and diametral 

compression test (Brazilian test). The uniaxial compression test [4–6], true triaxial compression 

test [7, 8], hollow cylinder test [9, 10], and Brazilian test [11] have been employed to evaluate the 

elastic constants of anisotropic rocks. The elastic constants of anisotropic rocks can be computed 

by substituting loading force and strain data recorded in testing into stress–strain equations. To 

evaluate the five independent elastic constants of transversely isotropic rocks, two cylindrical specimens 

with one loading direction are needed in uniaxial compression test. Two cubic specimens 

with three loading directions are needed in true triaxial compression test. Two hollow cylindrical 

specimens with two types of loading condition are needed in hollow cylinder test. And two discs 

with one loading direction are needed in Brazilian test. The numbers and geometry of specimens 

and types of loading condition needed in these methods are listed in Table I for comparison. Under 

Table I. Comparison of specimens needed in tests applied to evaluate elastic constants 

of transversely isotropic rocks. 

Number of Type of loading for 

Test Type of specimen specimen each specimen 

Uniaxial compression Cylinder 2 1 

True triaxial compression Cube 2 3 

Hollow cylinder Hollow cylinder 2 2 

Brazilian Disc 2 1 

Copyright q 2007 John Wiley & Sons, Ltd. Int. J. Numer. Anal. Meth. Geomech. 2008; 32:219–234 

DOI: 10.1002/nag

ELASTIC CONSTANTS OF TRANSVERSELY ISOTROPIC ROCKS 221 

Figure 1. Brazilian disc under diametral loading. 

such scenario, the diametral compression test (Brazilian test) seems to be the easiest method for 

specimen preparation and experimental procedure, and for which the analysis of the test results is 

relatively straightforward among those testing methods. 

As far as deformability is concerned, Hondros [12] used the Brazilian test method to determine 

Young’s modulus and Poisson’s ratio of concrete material, which was assumed to be isotropic. In 

1979, Pinto [13] extended the method proposed by Hondros [12] to evaluate elastic constants of 

anisotropic rocks, and conducted the Brazilian test on discs of schist. Closed-form solutions were 

derived to relate the elastic constants of an anisotropic rock in a disc under diametral compression 

(Figure 1) to the strains at the disc centre. It was assumed that the stress concentration at the centre 

of a disc of an anisotropic rock was the same as that of isotropic rock. This assumption is correct 

only if the disc surface is parallel to a transversely isotropic plane. In all other cases, however, 

anisotropy needs to be taken into account. In 1983, Amadei [4] revised Pinto’s procedure and 

successfully calculated the stress distribution of a disc of anisotropic rock by using the complex 

variable function method. Amadei et al. [14] indicated that the stress concentration factors (SCFs) 

in a disc of anisotropic rock are affected by E/E ′ (ratio of isotropic and anisotropic Young’s 

modulus), E/G ′ (ratio of isotropic Young’s modulus and anisotropic shear modulus), � (halfloading 

angle, Figure 1) and � (inclination angle of transversely isotropic plane to horizontal 

plane, Figure 2). In 1998, Chen et al. [11] combined the Brazilian test, complex variable function 

method, and generalized reduced gradient (GRG) method to calculate the five independent elastic 


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Figure 2. Geometry of anisotropic disc. 

constants of transversely isotropic rocks. Chen et al. [11] developed a computer program and 

pointed out that the stress distributions of a disc are affected seriously by the inclination angle of 

the transversely isotropic plane to the horizontal plane. 

Although the five independent elastic constants of transversely isotropic rocks can be successfully 

calculated by using the method proposed by Chen et al. [11], the theorem and mathematical 

computation procedure are really complex. There are no commercial programs developed by using 

the GRG method till now. It is hard to popularize this method without the original computer 

codes developed by Chen et al. [11]. It will take a long time and is uneconomical to code a new 

program for engineering analysis application because of the complexity of theorem. This paper 

developed a method of combining the Brazilian test, a numerical program (e.g. finite difference 

program, finite element program or boundary element program), and an iterative procedure to 

determine the elastic constants of anisotropic rocks in the laboratory. Merits of this approach 

are convenient preparation of the Brazilian test, simple iteration procedure, and readily available 

commercial numerical programs. It will be easier to popularize this procedure, in engineering 

analysis. 

THEORETICAL BACKGROUND 

Elastic anisotropic theorem 

Consider a thin disc of a linearly elastic, homogeneous, continuous, and transversely isotropic 

medium with the geometry shown in Figure 2. The disc has a thickness t and a diameter D. Let 

x, y, andzbe the three axes of a global Cartesian co-ordinate system and the z-axis is defined as 

the longitudinal axis of the disc. A local co-ordinate system x ′ − y ′ − z ′ is attached to the plane of 

the transversely isotropic medium. The x ′ -axis is perpendicular to the plane, the y ′ -andz ′ -axes 

are contained within the plane and the z ′ -axis coincides with the z-axis. The angle � is defined as 


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the inclination angle between the plane of transverse isotropy and the x-axis. Then a generalized 

plane stress formulation was used. The constitutive law is expressed as 

⎧ ⎫ ⎡ 

⎤ ⎧ ⎫ 

�x a11 a12 a16 �x 

⎪⎨ ⎪⎬ 

⎪⎨ ⎪⎬ 

⎢ 

⎥ 

�y = ⎢a12 

a22 a26 ⎥ �y 

⎣ 

⎦ 

(1) 

⎪⎩ ⎪⎭ 

⎪⎩ ⎪⎭ 

� xy 

a16 a26 a66 

where �x, �y, and� xy are the strain components of any point in the disc; �x, �y, and�xy are 

the stress components of any point in the disc; a11, a12,...,a66 are the compliance components 

calculated in the x, y co-ordinate system. These components vary with the angle � and the elastic 

constants in the x ′ –y ′ –z ′ co-ordinate system. For a transversely isotropic medium, E, �, E ′ , � ′ , 

and G ′ are the five independent elastic constants. E and E ′ are Young’s moduli in the plane of 

transverse isotropy and in a direction perpendicular to it, respectively. � and � ′ are Poisson’s ratios 

characterizing the lateral strain response in the plane of transverse isotropy to a stress acting parallel 

and perpendicular to it, respectively. G ′ is the shear modulus in the plane that is perpendicular to 

the plane of transverse isotropy. In Equation (1), a11, a12,...,a66 are independent of � and are 

affected by E, E ′ , � ′ ,andG ′ . For a Brazilian disc of transverse isotropy, Amadei [4] proposed the 

expression of aij as 

a11 = sin4 � 

E ′ + cos4 � 

E + sin2 2� 

4 

�xy 

� 

1 2�′ 

− 

G ′ E ′ 

� 

a12 = sin2 � 

2� 1 1 1 

+ − 

4 E ′ E G ′ 

� 

− �′ 

E ′ 

� 

cos 4 � + sin 4 � 

� 

�� 

sin 

a16 = sin 2� 

2 � 

E ′ − cos2 � � 

� 1 �′ 

+ − 

E 2G ′ E ′ 

� � 

cos 2� 

a22 = cos4 � 

E ′ 

+ sin4 � 

E + sin2 � 

2� 1 2�′ 

− 

4 G ′ E ′ 

� 

�� 

cos 

a26 = sin 2� 

2 � 

E ′ 

− sin2 � � 

� 1 �′ 

− − 

E 2G ′ E ′ 

� � 

cos 2� 

a66 = sin 2 � 

1 1 2�′ 

2� + + 

E ′ E E ′ 

� 

+ cos2 2� 

G ′ 

where � is the inclination angle of transverse isotropy; E and E ′ are Young’s moduli of isotropic 

plane and transversely isotropic plane, respectively; � ′ is Poisson’s ratio of transversely isotropic 

plane; and G ′ is the shear modulus of transversely isotropic plane. 

If a disc with diameter D and thickness t is loaded by force W , the stresses �x, �y, and�xy 

can be written as the form of SCFs qxx, qyy, andqxy shown as [14]: 

�x = W 

�Dt qxx 


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(2)


�y = W 

�Dt qyy 

�xy = W 

�Dt qxy 

By substituting Equation (3) into Equation (1) and moving the constant from the right-hand side 

to the left-hand side, Equation (1) can be rewritten as 

�Dt 

W 

⎧ 

⎪⎨ 

⎪⎩ 

�x 

�y 

� xy 

⎫ ⎡ 

⎪⎬ ⎢ 

= ⎢ 

⎣ 

⎪⎭ 

a11 a12 a16 

a12 a22 a26 

a16 a26 a66 

⎤ ⎧ 

⎪⎨ ⎥ 

⎦ 

⎪⎩ 

Numerical program 

It is important in this paper to obtain strain values and SCFs. Strain values can be recorded from 

the strain gauge on Brazilian discs in test, and SCFs can be calculated from stresses obtained by 

using numerical simulation. 

The three well-known and popular numerical analysis methods are finite difference method 

(FDM), finite element method (FEM), and boundary element method (BEM). These methods are 

developed from elastic law, geometry law, constitutive law, and equilibrium equations. Through 

discrete procedure, the boundary value problems can be simplified to finite meshes that are made 

up of elements and nodes. The stresses of interest in this paper can thus be obtained through the 

numerical simulation. 

There are many useful commercial numerical programs like FDM software FLAC, FEM software 

ANSYS, and BEM software GPBEST. In this paper, FLAC and ANSYS were adopted, and a BEM 

code developed by one author of this paper, named BEM-code [11], was used to obtain the stress 

state at the centre of the Brazilian disc in numerical simulation. 

Iterative procedure 

To evaluate the five independent elastic constants of transversely isotropic rocks, back calculation 

and iterative procedure were used. The operation sequences of iterative procedure were described 

as follows, and the flowchart is shown in Figure 3. 

(a) Prepare a Brazilian disc specimen (disc type N) with central axis perpendicular to the plane 

of transversely isotropy. The disc had a thickness t and a diameter D, and was loaded by the force 

W . The strains �0, �45, and�90 at the centre of disc can be recorded from the 0–45–90 strain gauge 

rosettes in the Brazilian test, and were transformed into �x, �y, and�xy by using Equation (5). The 

elastic constants E and � of isotropic plane can be computed by substituting strains �x and �y into 

Equation (6) [12]: 

⎧ 

⎪⎨ 

⎪⎩ 

�x 

�y 

� xy 

⎫ ⎡ 

⎤ ⎧ 

1 0 0 

⎪⎬ 

⎪⎨ 

⎢ 

⎥ 

= ⎢ 

⎣ 

0 0 1 ⎥ 

⎦ 

⎪⎭ 

⎪⎩ 

−1 2 −1 


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�0 

�45 

�90 

qxx 

qyy 

qxy 

⎫ 

⎪⎬ 

⎪⎭ 

⎫ 

⎪⎬ 

⎪⎭ 

(3) 

(4) 

(5)


Figure 3. Flow chart of iteration. 

16W 

E = 

�Dt(3�y + �x) 

� =− 3�x + �y 

3�y + �x 

where W is the loading force; �x and �y are the strains in the x and y directions, respectively. 

(b) Prepare a Brazilian disc specimen (disc type P) with the central axis parallel to the transversely 

isotropic plane with inclination angle � that varied from the horizontal axis, thickness t, and 

diameter D. The disc was loaded under force W . The strains �x, �y, and� xy at the centre of the 

disc were obtained by using Equation (5), and the strains �0, �45, and�90 were recorded from the 

0–45–90 strain gauge rosettes in test. Then the strains �x, �y, and� xy can be transformed into 

�x�Dt/W , �y�Dt/W ,and� xy�Dt/W . 

(c) Assume an initial set of SCFs qxx, qyy, andqxy at the centre of disc. The initial values of 

SCFs can be any real number. As pointed out by Amadei [15] that in general intact rocks are not 

too strongly anisotropic compared to other engineering materials, the initial values of SCFs were 

respectively set to be −2, 6, and 0 which are identical to the SCFs of isotropic disc in this paper 

for efficient computation. 

(d) The temporary E ′ , � ′ ,andG ′ can be computed by using back calculation, after the normalized 

strains, �x�Dt/W , �y�Dt/W ,and� xy�Dt/W , and SCFs, qxx, qyy, andqxy were substituted into 

Equation (4). 

(e) The elastic constants E, �, E ′ , � ′ ,andG ′ obtained from previous steps were applied in 

numerical simulation of the Brazilian test by using commercial numerical programs. A new set of 

stresses �x, �y, and�xy at the centre of the disc was computed. 


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(6)


(f) By substituting the stresses �x, �y, and�xy at the centre of the disc obtained from last step 

into Equation (3), a new set of SCFs qxx, qyy, andqxy was calculated. 

(g) Repeat steps (d)–(f) until the E ′ , � ′ ,andG ′ were convergent. The different rate between two 

iterative cycles in less than 0.1% was adapted in this paper for accuracy. The final five independent 

constants E, �, E ′ , � ′ ,andG ′ of transversely isotropic rocks were thus obtained. 

NUMERICAL EXAMPLE 

Calculation of stress concentration factors 

It is important to know the stress distribution and stress states at the centre of a disc. A numerical 

example was employed to compare the SCFs of a Brazilian disc obtained through numerical 

methods and analytical solution. 

For an isotropic material, the elastic constants were E = 40 GPa and � = 0.25. The three programs 

FLAC, ANSYS, and BEM-code were employed to model the Brazilian disc. The stresses were 

computed from these programs and transformed into the SCFs by using Equation (3), and these 

factors are listed in Table II. 

In 1959, Hondros proposed the analytical solutions of stress state at any point of an isotropic 

Brazilian disc under diametral loading in 2D plane stress state, and the equations were as follows: 

�� =− W 

� 

[1 − (r/r0) 

�r0t� 

2 ] sin 2� 

1 − 2(r/r0) 2 � 

1 + (r/r0) 

− tan−1 

cos 2� + (r/r0) 4 2 �� 

− tan � 

1 − (r/r0) 2 

�r =+ W 

� 

[1 − (r/r0) 

�r0t� 

2 ] sin 2� 

1 − 2(r/r0) 2 � 

1 + (r/r0) 

+ tan−1 

cos 2� + (r/r0) 4 2 �� (7) 

− tan � 

1 − (r/r0) 2 

where r0 is the radius of the disc; r is the distance between any point inside the disc and centre, 

and 0 is applied when the stresses at centre of the disc are concerned; t is thickness of the disc; 

W is the loading force; � is the half-loading angle; �� is the tangential stress (considered as �x 

at the centre of disc); �r is the radius stress (considered as �y at the centre of disc); and for an 

isotropic medium, the shear SCF at centre of the Brazilian disc is 0 theoretically. 

The stresses obtained from analytical functions were transformed into SCFs by using Equation 

(3), and the factors are listed in Table II. The numerical results exhibited good agreement with 

each other and with the analytical solutions. 

For transversely isotropic medium, the elastic constants of the numerical example were set to be 

E/E ′ = 2, E/G ′ = 10, � = 0.25, and � ′ = 0.2, and the inclination angle of the transversely isotropic 

plane to horizontal plane, �, was30 ◦ . The half-loading angle � was 7.5 ◦ . The results of SCFs qxx, 

qyy, andqxy at the centre of disc can be transferred from stresses �x, �y, and�xy obtained by 

Table II. Comparison of SCF isotropic medium at centre of disc. 

Methods qxx qyy qxy 

FLAC −1.943958 5.943312 −0.000071 

ANSYS −1.961952 5.957760 −0.000022 

BEM-code −1.953259 5.951952 −0.003986 

Analytical −1.954464 5.954464 0 


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Figure 4. Distribution of qxx using FLAC. 

using FLAC, ANSYS, and BEM-code. The distributions of SCFs qxx, qyy, andqxy obtained 

from FLAC are shown in Figures 4–6. The distributions of SCFs obtained from the other two 

programs were similar to the previous results. The SCFs at centre of the Brazilian disc are listed in 

Table III. There existed only few differences between each other. 

Calculation of elastic constants 

In this section, a numerical example was employed to illustrate the iterative procedure of calculating 

elastic constants of a transversely isotropic rock, and the results were compared with those presented 

by Chen et al. [11]. 

For the sandstone Young’s modulus was E = 40 GPa and Poisson’s ratio was � = 0.25 presented 

by Chen et al. [11]. The normalized strains �x�Dt/W , �y�Dt/W, and �xy�Dt/W at the centre 

of a type P’s Brazilian disc with an inclination angle � = 30◦ were −0.3454, 0.4501, and 

−0.2627 GPa−1 , respectively. 

First of all, the initial value of qxx, qyy,andqxy were set to be −2, 6, and 0, respectively. These 

SCFs and normalized strains were substituted into Equation (4). The initial solutions of temporary 

E ′ , � ′ ,andG ′ were 29.097 GPa, 0.238, and 4.223 GPa, respectively. These initial temporary 

transversely isotropic elastic constants were employed in numerical simulation by using FLAC, 

and the stresses at the centre of disc were extracted. After these stresses were substituted into 

Equation (3), the second set of the SCFs can be found to be −2.341, 4.643, and −0.569. With the 

same steps used previously, the second set of E ′ , � ′ ,andG ′ can be computed by substituting these 

second SCFs and normalized strains into Equation (4), and the new temporary E ′ , � ′ ,andG ′ were 

found to be 17.652 GPa, 0.222, and 4.034 GPa. After 10 cycles of iterative procedure, the results 

were convergent that the different rate of temporary E ′ , � ′ ,andG ′ between two steps was less than 


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Figure 5. Distribution of qyy using FLAC. 

Figure 6. Distribution of qxy using FLAC. 


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Table III. Comparison of SCF of anisotropic medium at centre of disc. 

Methods qxx qyy qxy 

FLAC −2.333666 4.800917 −0.360942 

ANSYS −2.350128 4.765200 −0.400824 

BEM-code −2.337643 4.800264 −0.370366 

Table IV. E ′ (GPa) in each iterative cycle. 

Cycles FLAC ANSYS BEM-code 

1 29.097 29.097 29.097 

2 17.652 17.191 17.536 

3 21.002 20.719 21.014 

4 19.516 19.172 19.436 

5 20.190 19.854 20.167 

6 19.886 19.559 19.841 

7 20.031 19.688 19.990 

8 19.963 19.632 19.923 

9 19.993 19.656 19.953 

10 19.990 19.646 19.939 

Table V. � ′ in each iterative cycle. 

Cycles FLAC ANSYS BEM-code 

1 0.238 0.238 0.238 

2 0.222 0.221 0.223 

3 0.186 0.190 0.186 

4 0.203 0.204 0.204 

5 0.196 0.198 0.197 

6 0.199 0.201 0.200 

7 0.198 0.200 0.199 

8 0.199 0.200 0.200 

9 0.198 0.200 0.199 

10 0.199 0.200 0.199 

0.1%. The final solutions were 19.990 GPa, 0.199, and 3.987 GPa, respectively. The approximate 

solutions of elastic constants can be found by using the other two programs ANSYS and BEMcode, 

and iterative procedures. The elastic constants E ′ , � ′ ,andG ′ in each cycle computed by 

using these three programs are listed in Tables IV–VI. Figures 7–9 show the variations of E ′ , � ′ , 

and G ′ with the number of iterations. It is obvious that the elastic constants converged quickly in 

only 10 cycles. The final results are listed in Table VII. 

The solutions of elastic constants computed by using GRG method were E ′ = 20 GPa, � ′ = 0.2, 

and G ′ = 4 GPa proposed by Chen et al. [11], and are listed in Table VII. It is obvious that the 

elastic constants computed by using the method proposed in this paper exhibited good agreement 

with the results computed using GRG method. 


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Table VI. G ′ (GPa) in each iterative cycle. 

Cycle FLAC ANSYS BEM-code 

1 4.223 4.223 4.223 

2 4.034 4.046 4.042 

3 3.961 3.971 3.966 

4 3.996 4.006 4.002 

5 3.982 3.993 3.989 

6 3.988 3.999 3.995 

7 3.985 3.997 3.992 

8 3.987 3.998 3.994 

9 3.987 3.997 3.993 

10 3.987 3.998 3.993 

Figure 7. Variation of E ′ with number of iterations. 

EXPERIMENTAL RESULT 

Several diametral compression tests were constructed on discs of marble mined in Taiwan, and 

the iterative procedure was used to measure their anisotropic elastic constants. The rocks of 

marble were composed of significant foliations, which may suggest a transversely isotropic type 

of symmetry. The plane of transverse isotropy was parallel to the foliation. 

To evaluate the elastic constants E and �, a disc of type N was cut with its z-axis which was 

taken perpendicular to the foliation. Three discs of type P with � = 15, 30, and 45 ◦ were cut 

with their z-axis parallel to the foliation to evaluate the elastic constants E ′ , � ′ ,andG ′ . The discs 

were prepared following the procedure suggested by the International Society for Rock Mechanics 

(ISRM). 

The discs were loaded up to failure by using a MTS loading system under a constant loading 

rate of 200 N/s. In order to apply the load over an arc of constant angle 2� = 15 ◦ , two steel jaws 


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Figure 8. Variation of � ′ with number of iterations. 

Figure 9. Variation of G ′ with number of iterations. 

were inserted in between the discs and the platens of loading frame, like what has been suggested 

by the ISRM. 

For the type N disc, the diameter D and thickness t of the disc, loading force W and strains 

at the centre of the disc are listed in Table VIII. E and � of the marble can then be computed by 

substituting the strains into Equation (6), and the results were found to be E = 78.302 GPa and 

� = 0.267. 

For the type P discs with inclination angles � = 15, 30, and 45 ◦ , the strains of these three discs 

were recorded by the strain gauge rosettes at the centre of discs. �x, �y, and� xy were computed 

from these strains. The diameter D and thickness t of discs, loading force W and strains are listed 


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Table VII. Comparison of back-calculated moduli 

using different approaches. 

Methods E ′ (GPa) � ′ G ′ (GPa) 

FLAC 19.990 0.199 3.987 

ANSYS 19.646 0.200 3.998 

BEM-code 19.939 0.199 3.993 

GRG 20.000 0.200 4.000 

Table VIII. Experimental data of N type disc. 

D (mm) t (mm) W (kN) �x (10 −3 ) �y (10 −3 ) 

74.0 12.0 12.553 −0.207006 0.375507 

Table IX. Experimental data of P type discs. 

� ( ◦ ) D (mm) t (mm) W (kN) �x (10 −3 ) �y (10 −3 ) � xy (10 −3 ) 

15 74.0 11.0 8.055 −0.147168 0.291564 −0.087318 

30 74.0 10.9 7.425 −0.139497 0.274014 −0.002608 

45 74.0 10.9 7.957 −0.163374 0.283919 0.012089 

Table X. Normalized strains of marble discs. 

�x�Dt/W �y�Dt/W � xy�Dt/W 

� ( ◦ ) (GPa −1 ) (GPa −1 ) (GPa −1 ) 

15 −0.04672 0.09256 −0.02772 

30 −0.04761 0.09352 −0.00089 

45 −0.05203 0.09042 0.00385 

in Table IX. Then the strains �x, �y, and� xy, were transferred into normalized form, �x�Dt/W , 

�y�Dt/W ,and� xy�Dt/W . The normalized strains of these three discs are listed in Table X. 

By using the iterative procedure proposed in this paper, the elastic constants E ′ , � ′ ,andG ′ of 

the marble discs with inclination angle � = 15, 30, and 45 ◦ can then be easily computed. The final 

results can be obtained in no more than 10 cycles by using three numerical programs and are listed 

in Table XI. The GRGs solutions computed by using the program developed with GRG method 

proposed by Chen et al. [11] are also listed in Table XI. It can be seen that numerical results 

were in good agreement with the GRGs solutions. Theoretically, the elastic constants evaluated 

from these three marble discs with the same material should be identical or close; however, the 

results may be affected by the non-homogeneity in nature and experimental error. Even though 

the results listed in Table XI were affected by the non-homogeneity in nature and experimental 

error, these three experimental cases showed the applicability of evaluating the elastic constants 

of transversely isotropic rocks by using the Brazilian test and iterative procedure. 


DOI: 10.1002/nag


Table XI. Anisotropic elastic constants of marble. 

� ( ◦ ) Program E ′ (GPa) � ′ G ′ (GPa) 

15 

30 

45 

FLAC 76.420 0.202 22.921 

ANSYS 76.540 0.200 23.075 

BEM-code 76.406 0.201 23.016 

GRG 75.619 0.186 23.383 

FLAC 66.498 0.187 27.343 

ANSYS 66.259 0.185 27.434 

BEM-code 66.516 0.187 27.409 

GRG 67.773 0.195 28.047 

FLAC 71.238 0.230 27.322 

ANSYS 71.195 0.219 27.328 

BEM-code 71.076 0.233 27.396 

GRG 72.189 0.163 27.425 

CONCLUSION 

This paper presented a method of combining Brazilian test and numerical procedure to determine 

the elastic constants of transversely isotropic rocks. Brazilian test was used to obtain the strains 

at the centre of discs. Numerical programs were applied to simulate and compute the SCFs at the 

centre of disc. By using back calculation, the temporary elastic constants were computed, and the 

final convergent values can be found after iterative procedure. 

A numerical example with sandstone disc was proposed. It can easily be seen that the convergent 

elastic constants E ′ , � ′ ,andG ′ were obtained in no more than 10 iterative cycles. The final results 

were satisfactory in comparison with the GRGs solutions. Three experimental cases with marble 

discs mined in Taiwan were presented. Through Brazilian test, back calculation and iterative 

procedure, the elastic constants E ′ , � ′ ,andG ′ of the marble disc can also be easily obtained. 

Although there were discrepancies existing in the results of these three discs, the numerical 

results were close to the GRGs solutions. It has to be mentioned again that the results may be 

affected by the non-homogeneity in nature and experimental error. The numerical example and 

three experimental cases showed the applicability of this method combining the Brazilian test and 

numerical procedure to determine the five independent elastic constants of transversely isotropic 

rocks. 

The procedure of preparing Brazilian specimen is easier than that in other laboratory tests. The 

iterative procedure is simple without complex mathematical functions. The commercially numerical 

programs are easy to obtain. There is no need to develop programs with complex theorems. It is 

evident that the method can be easily popularized and employed to evaluate the five independent 

elastic constants of transversely isotropic rocks in laboratory. 

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DOI: 10.1002/nag

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