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1
The Complete
Force-Displacement Curve
plus
Inhomogeneity and Anisotropy
John A Hudson
Lecture 4
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2
F1
F2
F3
Fn
Fractures
Intact rock
Boundary
conditions
Excavation
Water flow What happens
when a volume of
rock is subject to
increasing stress?
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For example, beneath a foundation pillar
or within a mine pillar
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Standardised laboratory tests
4
From González de Vallejo & Ferrer
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The
Large rock core at the URL
Canada
l to r: Cook, Hudson, Hoek,
Morgenstern, Fairhurst
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Displacement
Force
Pre-peak
region Post-peak region
The complete force-displacement curve
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7
The complete stress-
strain curve in uniaxial
compression
But, working with stresses, we have a variety of options
From González de Vallejo & Ferrer
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Complete stress-strain curve for marble
in uniaxial compression
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9 From González de Vallejo & Ferrer
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The But if we are
conducting a
laboratory test,
what do we choose
to control in order
to obtain the
complete curve?
the stress
or the strain?
It can’t be
a steadily
increasing stress…
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11
The
Soft
loading
system
Stiff loading system
The control must be
via the strain, but
there is still a problem,
because the testing
machine must be stiff
enough compared to
the rock sample.
If the testing machine
is too soft, it will
unload into the
specimen just after the
peak stress.
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The stiffness of
the applied
loading in situ
depends on rock
type and depth.
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The dependent variable
The independent (i.e. controlled) variable
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Complete stress-strain curves were obtained by Wolfgang
Wawersik at the University of Minnesota in the late 1960s
using a thermally controlled testing machine
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Complete stress-strain curves obtained by Wolfgang Wawersik
at the University of Minnesota in the late 1960s
Introducing another problem: that, for some rocks, the curves go backwards…
From
González de
Vallejo &
Ferrer
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The
So Class II curves
could not be obtained
in either old or new
standard testing
machines…
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17
Operation
of a servo-
controlled
testing
machine
Recording of
required
experimental data
EXPERIMENT
Program indicating
desired value of
test variable
Electronic
comparison of
feedback signal (f)
and
program signal (p)
Closed LoopAdjust until f = p
Feedback signal, f
Program signal, p
Correction signal
…but, luckily, servo-controlled testing machines arrived
and these are ideal for rock testing
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18
So, this feedback signal should be the most
sensitive indicator of failure in order to ensure the
greatest control in the experiment.
The use of such testing machines is only really
limited by the imagination
EXPERIMENT
Electronic
comparison of
feedback signal (f)
and
program signal (p)
Closed LoopAdjust until f = p
Feedback signal, f
Correction signal
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Optimal control by suitable choice of feedback, i.e.
the most sensitive indicator of rock failure
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Uniaxial compression test on a rock sample
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The
By suitable choice of feedback, in the laboratory we can
control rock failure in any loading configuration
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Unloading
stiffness, -K, of
surrounding rock
Time dependency
22
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Illustrating how the stiffness of the surrounding strata
can affect the long-term stability of a mine pillar
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So we can now have an excellent capability for testing
rocks in the laboratory under a variety of conditions.
We can obtain the complete force-displacement curve
for any laboratory experiment and study rock failure in
detail.
Let us now look at the matching capability in numerical
modelling.
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The next slides illustrate outputs
from the Chinese RFPA finite
element program
(Rock Failure Process Analysis)
developed by
Prof Chun’an Tang
and his team.
But it is also possible to
numerically simulate rock failure
using a variety of other
computer programs.
25
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26
Graphical output of simulated rock specimen
throughout the complete stress-strain curve
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27
Variation in numerically tested specimens having
the same characteristics
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Wuhan – Feb 2008 – Unsolved Problems in Rock Mechanics and Rock Engineering
28
Computer modelling of rock failure
Complete stress-strain curve Acoustic emission
28
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Numerical experiments can be conducted to
establish, for example, the influence of the
height:width ratio of the tested specimen
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The
0
5
10
15
20
25
30
35
40
0 0.5 1 1.5 2
Strain (0.001)
Str
es
s (
MP
a)
H/W=3
H/W=1.5
H/W=1
H/W=0.67
H/W=0.5
H/W=3 H/W=1.5 H/W=1 H/W=2 H/W=0.5
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Simulating the progressive failure of a simulated crystalline grain structure
shear stresses and fracturing
31
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Simulated rock cutting
32
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The effect of increasing
inhomogeneity
(decreasing the
parameter m increases the
inhomogeneity)
Thus we can
model statistically
uniformly
distributed
inhomogeneity
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The
The state of the art…
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So, not only do we now have an excellent capability for
testing rocks in the laboratory under a variety of
conditions,
but we also have an excellent numerical modelling
capability.
This is a superb combination for rock mechanics
studies.
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The context of inhomogeneity and anisotropy
Inhomogeneity: different properties at different locations
Anisotropy: different properties in different directions
The ideal material is:
CHILE: Continuous, Homogeneous, Isotropic and
Linearly Elastic
The actual rock material and the rock mass is:
DIANE: Discontinuous, Inhomogeneous, Anisotropic and
Not Elastic
We have seen some examples of inhomogeneity. What
about anisotropy?
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Anisotropy: Stress and strain are similar tensors
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To consider the general relation between stress
and strain, we just express each strain component
as a linear function of all the stress components
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The elastic compliance matrix
The elastic
compliance
matrix is
symmetrical
and so, in the
general case,
we have 21
elastic
constants
characterising
anisotropy
39
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But these constants can be reduced by
considerations of symmetry
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Isotropic
Transversely
isotropic
Orthotropic
Random
So, even with
the 21 elastic
constants, the
modelling
would not be
correct…
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This has led to many laboratory tests on anisotropic rock
specimens throughout the years
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from “Geological Engineering” by de Vallejo and Ferrer, 2011
In particular, sudying the
strength of a specimen
with planes of weakness
at different angles to the
loading direction
Idealised
Actual
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The
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Modelling with the RFPA code 45
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β=0° β=15° β=30° β= 45° β=60° β=75° β=90°
Seven transversely isotropic rock samples composed
of two rock materials with different dip angles
0
20
40
60
80
100
0 20 40 60 80 100
uniaxial compressive peak
strength
angle
Numerical test
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Foundations and
strata orientation
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?
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19622012
Do we know the strength of rock?
Müller replied: “For rock (specimens)
tested in the laboratory, yes.
For a rock mass, no.
This is what we need to determine.
This is why we need an
International Society for Rock Mechanics.”
For a billion years the patient earth amassed documents and inscribed them with signs and pictures which lay unnoticed and unused. Today, at last, they are waking up, because man has come to rouse them. Stones have begun to speak, because an ear is there to hear them. …..
Hans Cloos. Conversations with the Earth , 4 (1885-1951)
“Geomechanics“
Prof. F. Stini.Engineering Geology
(1883- 1958)Geologie und Bauwesen 1929
Prof. Karl von Terzaghi.Soil Mechanics
(1883- 1963)
Prof. Leopold Müller. Rock Mechanics.
(1908-1988)
Geomechanics - Austrian Heritage
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Stress
Strain, %
0 0.1 0.2
200
MPa
The complete stress-strain curve
has been obtained for a cylindrical
specimen of intact granite tested in
uniaxial compression.
The specimen is 100 mm long and
50 mm in diameter. Assume that,
for the purposes of estimation, the
curve can be approximated to the
form opposite.
The uniaxial compressive strength
is reached at 0.1 % strain and 200
MPa stress.
When the curve reaches 0.2 %
strain, the rock microstructure has
been destroyed and all that remains
are small flakes of crushed mineral
grains.
For how long would you
have to switch on a
domestic 100 W light bulb
to use up the same amount
of energy as required to
destroy the rock specimen?
Interesting Question…the energy required to break rock
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a) The area under the complete stress-strain curve is equivalent
to the product of the stress at failure and the strain at failure —
because the two triangles have the same area as the dotted
square. Similarly, the area under the complete load-
displacement curve is the product of the load at failure and the
displacement at failure.
The load at failure = stress at failure x cross-sectional area = 200,000,000 x x
(50/2 x 1000)2 = 392,699 N. The displacement at failure = strain at failure x
specimen length = 0.001 x 0.1 = 0.0001 m. Thus, the energy used = 3.93x105 = 39
N-m. A joule is defined as the energy expended by 1 N moving through 1 m, or 1 J
= 1 N-m, so the energy required = 39 joules.
b) To establish for how long a domestic 100 W light bulb has to be illuminated in
order to expend the same amount of energy, we need to express the energy in watt-
seconds. 1 joule is also the energy expended by 1 W for one second, or 1 J = 1 W-
s. Hence, 39 joules is equivalent to 39 W-s, and so the 100 W light bulb has to be
illuminated for 39/100 s = 0.4 s.
Stress
Strain, % 0 0.1 0.2
200
MPa
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End of Lecture 4
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