1

The Complete

Force-Displacement Curve

plus

Inhomogeneity and Anisotropy

John A Hudson

Lecture 4

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?

For example, beneath a foundation pillar

or within a mine pillar

3

Standardised laboratory tests

4

From González de Vallejo & Ferrer

5

The

Large rock core at the URL

Canada

l to r: Cook, Hudson, Hoek,

Morgenstern, Fairhurst

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

9 From González de Vallejo & Ferrer

10

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…

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.

The stiffness of

the applied

loading in situ

depends on rock

type and depth.

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13

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

16

The

So Class II curves

could not be obtained

in either old or new

standard testing

machines…

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

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

Unloading

stiffness, -K, of

surrounding rock

Time dependency

22

Illustrating how the stiffness of the surrounding strata

can affect the long-term stability of a mine pillar

23

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.

24

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.

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26

Graphical output of simulated rock specimen

throughout the complete stress-strain curve

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Variation in numerically tested specimens having

the same characteristics

Wuhan – Feb 2008 – Unsolved Problems in Rock Mechanics and Rock Engineering

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Computer modelling of rock failure

Complete stress-strain curve Acoustic emission

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29

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

Simulating the progressive failure of a simulated crystalline grain structure

shear stresses and fracturing

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Simulated rock cutting

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The effect of increasing

inhomogeneity

(decreasing the

parameter m increases the

inhomogeneity)

Thus we can

model statistically

uniformly

distributed

inhomogeneity

34

The

The state of the art…

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.

35

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?

36

Anisotropy: Stress and strain are similar tensors

37

To consider the general relation between stress

and strain, we just express each strain component

as a linear function of all the stress components

38

The elastic compliance matrix

The elastic

compliance

matrix is

symmetrical

and so, in the

general case,

we have 21

elastic

constants

characterising

anisotropy

39

But these constants can be reduced by

considerations of symmetry

40

Isotropic

Transversely

isotropic

Orthotropic

Random

So, even with

the 21 elastic

constants, the

modelling

would not be

correct…

41

This has led to many laboratory tests on anisotropic rock

specimens throughout the years

42

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

43

44

The

Modelling with the RFPA code 45

Step 50-7

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Step 50-12

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Step 51-9

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Step 52-4

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Step 52-18

50

Step 52-32

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Step 52-43

52

Step 52-51

53

Step 52-54

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Step 53-2

55

Step 53-5

56

Step 53-6

57

Step 55-3

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Step 55-15

59

Step 55-40

60

Step 57-3

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Step 59-6

62

Step 59-25

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Step 59-38

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Step 59-46

65

Step 59-75

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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

67

68

Foundations and

strata orientation

69

?

69

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

End of Lecture 4

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