Dynamic Analysis: Stresses that cause deformation
Experimentally Observed Relationships Between Stress and Strain
Specimens are jacketed with weak material - copper or plastic.
brittle
slate
ductile
sandstone
limestone
Specimens are drilled out cores that are ‘machined’ to have
perfectly parallel and smooth ends.
They are then carefully measured to determine their initial
length (lo) and diameter (to get initial cross-sectional area, Ao)
slate
sandstone
limestone
Biaxial Test Rig
Experiments are carried
out in steel pressure
vessels.
Confining pressure (s2 =
s3) is often supplied by
fluid that surrounds the
specimen.
Temperature can be
varied.
Pore-fluid pressure can
also be varied.
Types of Tests
• Axial compression either unconfined or
confined. Results in
length-parallel shortening.
• Triaxial - specimen is
subjected to 3 different
loads (difficult), one may
be tensile.
• Tensile strength specimen is pulled (rare).
s1
Fluid supplies
confining pressure
(s2 = s3)
s2 = s3
s2 = s3
s1
Brazilian
Test
Axial
splitting
simulates
tension
Results in length-parallel
shortening.
Vertical compressive
stress (parallel to core)
is taken to a higher level
than the horizontal
compressive stresses
(confining pressure).
Results in lengthparallel shortening.
Horizontal
compressive stress,
perpendicular to core
(confining pressure) is
taken to a higher level
than the vertical
compressive stresses
(parallel to the core).
Instead of squeezing a rock sample, they are pulled
apart. Called tensile strength test.
The aim is to determine the smallest amount of stress
that will cause the rock to fail in tension. Rocks are
much weaker in tension.
The Donath deformation apparatus
In a natural setting confining pressure is the pressure exerted by the weight
of interconnected pore fluids (hydrostatic pressure) and the overlying
rockmass (lithostatic pressure). Here it is supplied by pumping up
hydrologic fluid pressure.
If we increase confining pressure the specimen simply gets
smaller - decreases volume. Confining does not cause distortion
of the sample.
Once the appropriate level of confining pressure has been
achieved, then we can apply the axial load.
Calculating Axial Stress
What we know is the load or force
that is being applied (F) and the
cross sectional area of the (A).
s=F/A
We keep track of a pair of
numbers:
E = axial strain and
s = axial stress
plot them in x,y space
(respectively).
Measuring Shortening
e = Dl / l and S = lf / lo
Measuring Strain Rate
.
Strain Rate = e = e / t
Generally there are 2 kinds of tests:
Constant Strain Rate & Constant Stress
A Standard Axial
Compression Test
• Measure initial specimen length
and diameter.
• Transform load into stress and
displacement into % strain.
• Plot differential stress (diameter
of the Mohr circle for stress)
against % strain.
Differential stress: sd = s1 - s3
Elastic loading
A Standard Axial
Compression Test
1)Subject rock samples to controlled
loading under known conditions.
2) Fracture strength of rocks provides
insight for structural geology
Rock strengths are measured by
subjecting cylindrical samples (small
rock core) to an axial load, applied by a
cylinder or hydraulic ram.
The confining pressure is applied radially
around the side by a fluid that is
isolated from the rock by a deformable
rubber or copper jacket.
Elastic loading
Elastic loading and hysteresis
What is the state of stress on a
Mohr diagram?
The state of stress plots as a
single point on the Mohr diagram,
because the axial stress equals
the confining pressure.
Differential stress: sd = s1 - s3
The state of stress appears on the Mohr
diagram as successively large circle, of
diameter s3-s1, sharing on the confining
pressure s3, as a common point.
Eventually the sample starts to
deform plastically.
Its elastic behavior is
surpassed, and nonrecoverable deformation
begins to accumulate in the
rock.
Plastic deformation produces
deformation in a rock without
failure by rupture.
The onset of plastic deformation
begins when the stress-strain
curve departs from the
straight line elastic mode.
Below its yield
strength the rock
behaves as an elastic
solid.
The point of departure from elastic
behavior is called the elastic limit.
Its value is known as yield strength.
Faulting finally takes place at
about 120 MPa and the stress
drops to zero.
Some of the elastic energy is
expended making the fracture,
some in sound, some in the
frictional heating due to sliding.
When we remove the sample,
we notice that the fracture lies
about 24° to the axis of the
cylinder.
Stress/Strain Test:
1) Brittle rocks first shorten
a elastically during these
tests.
2) Then they fail abruptly
by discrete fractures.
3) Sometimes plastic
deformation occurs before
failure, called strain
softening.
Just prior to failure (faulting), what if we raised the confining pressure and
repeated the experiment on the same sample? How would the limestone
respond?
Stress/Strain Test:
When the load is reapplied
at to Point C, the elastic
limit is greater than during
the first test.
The yield strength is also
greater, because the
original fabric of the rock
was changed slightly by the
plastic deformation.
This rock has undergone
strain hardening.
The yield strength
increases due to
modification of original rock.
Applying more load, the
limestone displays an
increase of plastic behavior
before fracturing, unlike the
previous experiment.
This accelerated plastic
deformation is called strain
softening, because less
stress is required for each
new increment of strain.
Eventually the rock fractures,
but the rupture strength is
greater in this experiment.
Rupture strength is
the stress level of
failure by fracturing.
Rocks become
stronger at higher
levels of confining
pressure.
Rocks also undergo
greater plastic
deformation at higher
levels of confining
pressure, with T and
strain rate fixed.
Confining Pressure (or Temperature)
Low
High
As confining pressure (depth) increases, deformation
becomes less discrete, and more distributed.
We need greater differential
stress levels (s1 - s3) to
cause rupture or failure
when the confining pressure
is increased (s3).
This forms the basis for the
laws that describe the
conditions under which
rocks fail by fracturing and
faulting.
Each sample as different
confining pressures before
failure.
It becomes obvious that we
need greater differential
stress levels (s1 - s3) to
cause rupture or failure
when the confining pressure
is increased (s3).
This forms the basis for the
laws that describe the
conditions under which
rocks fail by fracturing and
faulting.
The slope of this failure envelope is a function
of the rock’s angle of internal friction
Confining pressure
Increasing confining
pressure on a rock
affects its strength.
Lithology & Strength
Competent (strong & deform
in a brittle manner)
&
Incompetent (weak & deform
in a ductile manner)
Limestone become progressively more ductile at
higher confining pressures
Confining pressure
or s3
For a given rock,
yield strength, rupture
strength, and ductility
have greater values
with increasing
confining pressures.
Effect of Pore-Fluid Pressure on Strength
Fluid pressure also affects
the strength of a rock
1) Increase of fluid pressure
can offset or partially offset
the effects of confining
pressure.
Pf = fluid pressure
4) Trapped fluids that are have high pressures from burial
and compaction
5) New term
Effective stress = confining pressure – fluid pressure. The
net effect of confining pressure and fluid pressure.
6) Allows deeply buried rocks to deform as if they were
deforming in a low confining pressure environment.
2) High fluid pressure
decreases the strength of a
rock. Water trapped in
sediments during deposition
may be overpressurized by
burial and loading.
So the pore fluid stress is
greater than the hydrostatic
stress at any given level in
the crust.
Effect of Temperature on Strength
1) An increase in
temperature of a rock can
also depress the yield
strength of a rock.
2) Igneous rocks are less
affected by low
temperature changes than
sedimentary rocks.
3) If heated sufficiently,
rocks will deform in a
plastic or viscous fashion.
Effect of Strain Rate on Strength
Rock strength also if affected
by rate of which stress is
applied.
2) Difficult to observe in
experiments. Compare to long
distance runners. Fractures are
easier, at low stresses if stress
is applied over a long time
period (e.g. training and
competition)
Increase of strain rate
1) A rock can deform plastically
at very low levels of stress if
the rate of loading is very slow.
Time vs. rock strength.
Creep - strains that result from low differential stresses
over a very long time
Primary creep: is slightly
delayed elastic strain - tapers
off with time.
Secondary creep: a steady
state plastic defm. Where
strain and loading time are
linearly related.
Tertiary creep: fatigue finally
sets in and strain accelerates leads to failure by rupture.
Time & Rheids
looks like Mohr space?
Depending on friction and
normal forces, pre-existing frxs
can be re-activated.
As it turns out, friction properties
for most rocks are the same.
Linear relation between normal
and shear stresses required to
overcome friction and initiate
movement.
Elastic, plastic, and viscous models of rock deformation
Most tests are conducted as axial compression, where rocks fail in brittle, semibrittle, or ductile manner. Although brittle and ductile are useful terms, less
delve into the interplay of stress and strain. We will discuss three basic models.
Elastic behavior
Think of springs under a truck. If we load a truck, the springs will shorten
by a amount directly related to the magnitude of load.
The relationship of the load of blocks to the displacement of the bed is an
equation that describe a straight line.
It works perfectly, whether one loads or unloads the truck.
Elastic, Plastic and Viscous Models of Rock Behavior
Hooke’s Law
s = Ee: stress is linearly related to strain by the
constant E, known as Young’s modulus
Elastic, plastic, and viscous models of rock deformation
Hookes Law
1) This straight line relation between stress
and strain is called Hookes law (e µ s).
Add proportionality constant to get
Hookes law: s = Ee
strain (e) is linearly proportional to stress (s)
where
E = Young’s modulus
E = s/e = stress/strain
The value of E, or Young’s modulus
describes the slope of a straight line,
stress-strain curve.
Stress and strain are directly and linearly
related = the slope of the line.
Young’s Modulus - a measure of elastic stiffness
“a spring constant”
E = s/e = stress/strain an
elastic modulus.
Young’s modulus, How
much stress is required to
achieve a given amount of
length-parallel elastic
shortening of a rock.
Hookes Law and Youngs Modulus
For Hookes law relations, we plot
stress (s) versus strain (e).
The slope of the straight line is a
measure of stiffness of a rock.
The higher the value, the stiffer
the rock, given in units of stress
(MPa),
Think of Young’s modulus as an
elastic modulus, that describes
how much stress is required to
achieve a given amount of
length-parallel elastic shortening.
Significance of Young’s Modulus
The relations between stress and strain (E = s/e ). E, or Young’s modulus
equals force per unit area divided by changes in line length (strain).
1) Use steel railroad track example. Changes of expansion with respect
to changes in temperature (thermal expansion). Steel track expands with
increasing temperatures.
2) Its coefficient of linear thermal expansion (a = 11 x 10-6). Thus a
track 30 m long at 0°C will become 0.013m (i.e., 1.3 cm) longer when the
steel warms to a temperature of 40°C.
Young’s Modulus
3) The changes of track
length may not be much
(centimeters), but if we look
up Young’s modulus for steel
(E = 8.67x107 N/m2)
and
4) Calculate the stress and
then the force (s = F/A), we
get very high forces (29
tons).
A track can not respond
elastically to such forces, so
it deforms. This was from
thermal expansion, not
tectonic deformation.
Poisson’s Ratio (n)
Describes the relationship between lateral strain and longitudinal
strain.
n = elat / elong
n, another elastic modulus.
Vertical loading will produce horizontal stresses because of the
Poisson effect.
The degree to which a specimen will widen upon shortening is
a function of it’s Poisson’s ratio.
s2 = s3 = (n / (1 - n)) s1
For common rocks, Poisson’s ratio tends to be around n = 0.25
Poisson’s ratio, Greek letter
nu (n).
This describes the amount
that a rock bulges as it
shortens.
The ratio describes the ratio
of lateral strain to longitudinal
strain:
n = elat/elong
Poisson’s ratio is unitless,
since it is a ratio of extension.
What does this ratio mean?
Typical values for n are:
Fine-grained limestone: 0.25
Apilite: 0.2
Oolitic limestone: 0.18
Granite: 0.11
Walcareous shale: 0.02
Biotite schist: 0.01
Young’s Modulus
If we shorten a granite and
measure how much it bulges,
we see that we can shorten a
granite, but it may not be
compensated by an increase
in rock diameter.
So stress did not produce the
expected lateral bulging.
Somehow volume decreases
and stress was stored until the
rock exploded!
Thus low values of Poisson’s
ratio are significant.
Typical values for n are:
Fine-grained limestone: 0.25
Apilite: 0.2
Oolitic limestone: 0.18
Granite: 0.11
Walcareous shale: 0.02
Biotite schist: 0.01
Bulk and Shear Moduli
Bulk modulus (K): K = Dhydrostatic stress / Ddilation
Shear modulus (G): G = ss / g
The two other parameters that describe the elastic relationship between stress
and strain are:
1) bulk modulus (K): resistance that elastic solids to changes in volume.
Divide the change of hydrostatic pressure by the amount of dilation produced
by pressure changes.
K = bulk modulus = hyrdostatic stress /dilation
2) Shear modulus (G): resistance that elastic solids to shearing:
Divide shear stress (ss) by shear strain (g)
G = shear modulus = ss/g
Elastic Stress - Strain Equations
Applies to very small deformations in rocks
Given si, E and n - we can solve for the strains
Plastic Strain
No strains, below its yield
stress, then when this
level is achieved, the
body will strain plastically
as long as the yield stress
magnitude is maintained.
If we begin to deform
(push) an ideal plastic
body, nothing happens
until the yield stress is
overcome.
Strain accumulates with no
increase in stress.
Viscous Strain
Think of shock absorbers.
Viscous bodies are fluids,
similar to a shock
absorber. Even a tiny
amount of stress will
produce flow.
These flows are
permanent and not
recoverable. Even the
smallest amount of stress
will displace a piston
within a shock cylinder.
No yield stress must be
overcome in viscous
behavior.
We commonly think that the magnitude of the largest stresses than can be
generated is due to the forces that create the stresses.
However, the really important limitation is the strength of the rocks, the weaker
the rock, easier to deform.