62 Strain and the theory of elasticity
material - why? Hence explain why five elastic constants are re-
quired for a transversely isotropic material rather than six.
A5.4 We explain this using the diagram below for the case of pure
shear, in which a square element of rock is subjected to compression
in the xdirection and tension in the y-direction of equal amounts.
However, the analysis is general in that it is applicable to all coupled
stress and strain states.
J/
Pure shear on .,o+
E
”+z
The two Mohr’s circle diagrams show the stress and strain states for
the square element. Because the circles are both centred at the origin, we
can see that t = u1 and y/2 = €1. Now, the shear modulus, G, is defined
as G = r/y which, on substituting for the two terms, gives G = ~((T~/E~).
For a 2-D stress state such as the one illustrated, we can write Hooke’s
Law as E, = (l/E)(ux - vu,). In this case, we have E, = EI, ox = ul, and
a, = u2 = -81. Substituting these values gives c1 = ( l/E)(al + uq), from
which we find that ~/EI = E/(l+ u).
Substituting these into the earlier expression for the shear modulus
gives
1 UI E
2El 2(1 +u)
G=--=-
as required.
This approach is only valid if the Young’s moduli and Poisson’s ratios
in the u1 and 02 directions are the same, i.e. for an isotropic material
having the same properties in all directions. If this were not the case, the
strain EI could not be expressed as simply as
81 = - (01 + vu]).
In the case of a transversely isotropic rock, say an unjointed rock
with a set of bedding planes perpendicular to the z-axis, as shown in
the adjacent sketch, the relation G = E/2(1 + u) will apply in the x-y
plane but not in the x-z and y-z planes, because the rock has different
properties in different directions in these planes. Thus, the six elastic
constants required to characterize a transversely isotropic rock, E, = E,,
E, , u,,, uxz = uyz , G,, , and G ,, = G,, , are reduced to five, E, , E,, u,, , v,, ,
and G,, because in the x-y plane of isotropy G,, = E,/2(1 + uxy) (see
diagram below).
1
E