Page 33 - Applied Petroleum Geomechanics

Stresses and strains 23

1.5.2 Transversely isotropic elastic rocks
A TI model is a simplification of orthotropy. It is azimuthally symmetric
about a single axis. Examples of TI materials are layered rocks or isotropic
rocks with a single set of oriented fractures. There are five independent
stiffness coefficients required to completely describe the elastic properties
(Havens, 2012). If rock properties are uniform horizontally within a layer,
but vary vertically from layer to layer, then the formations can be treated as
vertical transverse isotropy (VTI). The vertical axis (axis 3 as shown in
Fig. 1.14) is the axis of rotational symmetry, which is perpendicular to the
symmetric isotropic plane (the horizontal plane). Because the rock has
isotropic properties in the horizontal plane, the plane is the plane of
transverse isotropy. Transverse isotropy requires that c 22 ¼ c 11 , c 23 ¼ c 13 ,
c 55 ¼ c 44 , so that the stiffness matrix has the following form:
2 3
c 11 c 12 c 13 0 0 0
c
6 0 0 0 7
6 12 c 11 c 13 7
6 7
6 c 13 c 13 c 33 0 0 0 7
C ¼ 6 7 (1.39)
0 0 0 c 44 0 0
6 7
6 7
6
7
4 0 0 0 0 c 44 0 5
0 0 0 0 0 c 66
where c 66 ¼ (c 11 e c 12 )/2. For the VTI rock, Young’s modulus and
Poisson’s ratio must satisfy E 1 ¼ E 2 , n 12 ¼ n 21 , n 31 ¼ n 32 , n 13 ¼ n 23 .
The elastic stiffnesses (c ij )inthe TI rock in Eq. (1.39) can be obtained
from the acoustic velocities (dynamic stiffnesses) and from lab compression
tests (static stiffnesses). The stiffnesses can also be related to Young’s
moduli, Poisson’s ratios, and bulk moduli (King, 1964; Mavko et al.,
2009).
For the VTI rock the compliance matrix has the following form
(Bower, 2010):

2 3
1=E 1 n 12 =E 1 n 31 =E 3 0 0 0
6 0 0 0 7
n 12 =E 1 1=E 1 n 31 =E 3
6 7
6 7
6 n 13 =E 1 n 13 =E 1 1=E 3 0 0 0 7
S ¼ 6 7 (1.40)
0 0 0 0 0
6 7
6 1=G 3 7
6 7
0 0 0 0
4 1=G 3 0 5
0 0 0 0 0 1=G 1