66 Strain and the theory of elasticity(c) A strong sandstone with almost no fractures present.
(d) A granodiorite with three fracture sets.
Fracture set 1 : dip direction 314O; dip 35O; frequency 1.2/m.
Fracture set 2: dip direction 048O; dip 43O; frequency 1.3/m.
Fracture set 3: dip direction 089O; dip 79O; frequency 0.9/m.
A5.7 (a) For the limestone, the presence of three orthogonal fracture
sets with significantly different fracture spacings (and probably different
normal and shear fracture stiffnesses) indicates that an orthotropic model
assumption with nine constants would be most appropriate. We would
try to estimate the nine constants from the intact rock and fracture
characteristics. This is the assumption from the elasticity point of view
but it will have to be tempered in a practical case by the availability
of analytical solutions and numerical codes that will support such an
assumption.
(b) For the volcanic tuff, with five fracture sets having widely different
orientations and frequencies, the first assumption will be an isotropic
model, with the two effective elastic constants approximating the contri-
butions from all fracture sets. There is no method for dealing with this
more complex type of anisotropy in the theory of elasticity, apart from
aggregating all the deformation contributions.
(c) For the intact sandstone, the first assumption would be the trans-
versely isotropic model. We know that sandstone is a sedimentary rock
and is likely to have different properties perpendicular and parallel to
the bedding. However, if these properties turned out to be similar, we
could then revert to the isotropic model.
(d) The granodiorite has three fracture sets which are not even close
to being mutually orthogonal. There is no directly suitable simplified
elasticity assumption for this type of rock mass because there is no
orthogonal symmetry. One could use an isotropic model, although there
is no coherent way to estimate the two effective elastic constants, and so
the results would be unreliable.
45.8 (a) At the time of writing this book, most elastic analyses
that have been conducted for rock engineering design purposes
have assumed that the rock is perfectly isotropic with two elastic
constants. why do you suppose that is, given that most rock masses
are clearly not isotropic?
(b) Conversely, no elastic analysis for rock mechanics has been
conducted assuming that the rock mass is fully anisotropic with 21
elastic constants? Why is that?
(c) In this context, what do you think will happen in future ana-
lyses?A5.8 (a) In the laboratory, the intact rock is studied, not the rock mass
with all the fractures of different types it contains. Also, it is difficult
to estimate, let alone measure, the in situ rock anisotropy. Moreover,
measuring the elastic properties is time-consuming and expensive. Fi-
nally, few elastic solutions exist for non-isotropic elastic rocks. Thus, it is