Geoscience Reference
In-Depth Information
t
Longitudinal
strain
g = tan c
G = t / g
s
c
m
E
Lateral
strain
G = 2
m
+ 1
E =
s
/
e
Fig. 3.95 Shear or rigidity modulus ( G ) and its relation to Young's
modulus ( E ) and Poisson's number ( m ).
Dilation
Shortening
e
Fig. 3.94 Longitudinal and lateral strain experienced by a rock
sample when an uniaxial compression is applied. The relation
between both strains may not be linear as in this case, and the
Poisson's ratio is not constant, it varies slightly for different stress
values.
(a)
Solid elastic body
(b)
Fluid viscous body
elastic stresses have to be accumulated somehow. Rock
samples will fragment at the elastic limit after experiencing
very little lateral strain when the Poisson's ratio is very
small (close to zero). The reciprocal to the Poisson's coef-
ficient is called the Poisson's number m
. This num-
ber is also constant for any material, and so the relation
between the longitudinal and lateral strains have a linear
relation. Nonetheless, as in the case of Young's modulus
there may be slight variations in the linear trend of
Poisson's coefficient (Fig. 3.94). It is important to remem-
ber that experiments to establish elasticity relationships
under unconfined uniaxial stress conditions allow the rock
samples to expand laterally. In the crust, any cube of rock
that we can define is not only subject to a vertical load due
to gravity but also due to adjacent cubes of rock in every
direction and is not free to expand laterally; in such cases
complex stress/strain relations can develop.
Other elastic parameters are the rigidity modulus ( G )
and the bulk modulus ( K ). The rigidity modulus or shear
modulus is the ratio between the shear stress (
1/
Fig. 3.96 (a) Solid elastic bodies are strained proportionally to the
applied forces. If the intensity of the force is maintained there is not a
further increase in strain. When the force is released the object
recovers the initial shape; (b) The viscous fluid will be deformed
when a shear force is exerted, but even when the intensity of the
force is maintained, an increment in deformation will occur, defining
a strain rate. That is why strain rate is used in rheological plots
instead of strain as in solids. The fluid body will remain deformed
permanently once the force is removed.
stresses, even infinitesimal, are applied. One of the chief
differences between an elastic solid and a viscous fluid is
that when a shear stress is applied to a piece of elastic mate-
rial it causes an increment of strain proportional to the
stress, if the same level of stress is maintained no further
deformation is achieved (Fig. 3.96a). In fluids when a
shear stress (
) and the
shear strain (
) in a cube of isotropic material subjected to
simple shear: G
) is applied the material suffers certain
amount of strain but the fluid keeps deforming with time
even when the stress is maintained with the same value
(Fig. 3.96b). In this case a level of stress gives way to a
strain rate (d
(Fig. 3.95). G is another measure of
the resistance to deformation by shear stress, in a way
equivalent to the viscosity in fluids. The bulk modulus ( K )
relates the change in hydrostatic pressure ( P ) in a block of
isotropic material and the change in volume ( V ) that it
experiences consequently: K
/
/d t ), not a simple increment of strain as in
the elastic solids. Higher stress values will give way to
higher strain rates, so the fluid will deform at more speed.
As in elastic materials there is no initial resistance to
deformation even when stresses acting are very small, but
the deformations are permanent in the viscous fluid case
(Fig. 3.97a,b).
As we have seen earlier (Sections 3.9 and 3.10) the parame-
ter relating stress to strain rate is the coefficient of dynamic vis-
cosity
d P /d V . The reverse to the
bulk modulus is the compressibility (1/ K ).
3.15.3
Viscous model
Viscous deformation occurs in fluids (Sections 3.9 and 3.10);
fluids have no shear strength and will flow when shear
or simply viscosity
(
):
/(d
/d t ), which is
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