The strength of fractured rock
Erling Fjær
SINTEF Petroleum Research
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1
Challenge: Estimation of rock strength from log data
Porosity,
Density, Sonic, .
...
Strength
Traditional
approach:
correlations
UCS  a tPb
Wanted
Available
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2
Challenge: Estimation of rock strength from log data
Porosity,
Density, Sonic, .
...
Available
Strength
Brandås et al. (2012)
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Wanted
3
Alternative approach:
1. Establish a constitutive model for static and dynamic
moduli of rocks
2. Use the measured dynamic moduli (i.e. velocities) to
calibrate the model
3. Use the calibrated model to simulate a test where
strength can be measured
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Dry, weak sandstone
static moduli vs dynamic moduli
160
140
Axial
Stress (MPa)
Rock mechanical test
including acoustic
measurements
on a dry sandstone
120
Peak
stress
100
80
60
40
Radial
20
0
static moduli  dynamic moduli
The differences changes with stress
and strain
Young's modulus (GPa)
20
0.005
Dynamic
0.010
0.015
0.020
0.025
0.015
0.020
0.025
15
10
5
Static
0
-5
0.005
0.010
Axial strain
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Dry, weak sandstone
static moduli vs dynamic moduli
160
140
120
Rock mechanical test
including acoustic
measurements
on a dry sandstone
Stress (MPa)
Axial
Peak
stress
100
80
60
40
Radial
20
static moduli  dynamic moduli
The differences changes with stress
and strain
0
20
0.005
Young's modulus (GPa)
We are seeking
mathematical relations between
the static and the dynamic moduli
Dynamic
0.010
0.015
0.020
0.025
0.015
0.020
0.025
15
10
5
Static
0
-5
0.005
0.010
Axial strain
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Hydrostatic test
Building relations
We introduce a parameter P, defined as:
P
v - total volumetric strain
 v   v ,e
3
 v ,e
P is a measure of the inelastic part of the
deformation caused by a compressive
hydrostatic stress increment.

- elastic

K e strain
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Hydrostatic test
Building relations
We introduce a parameter P, defined as:
P
v - total volumetric strain
 v   v ,e
3
 v ,e
P is a measure of the inelastic part of the
deformation caused by a compressive
hydrostatic stress increment.

- elastic

K e strain

Ke
K
1  3 PKe
K = Static bulk modulus
Ke = Dynamic bulk modulus
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Hydrostatic test
Observations
25
1/P [GPa-1]
20
15
10
5
0
0
5
10
15
20
25
30
Hydrastatic stress [MPa]
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Hydrostatic test
Observations
25
1/P [GPa-1]
20
15
P
10
g
 T
5
0
0
5
10
15
20
25
30
Hydrastatic stress [MPa]
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Uniaxial loading test
Building relations
We introduce a parameter F, defined as:
F
z - total axial strain
 z   z ,e   z , p
 z
F is a measure of the inelastic part of the
deformation caused by a shear stress
increment.
 z ,e

- elastic

Ee strain
 z , p  Pz  z
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Uniaxial loading test
Building relations
We introduce a parameter F, defined as:
F
z - total axial strain
 z   z ,e   z , p
 z
F is a measure of the inelastic part of the
deformation caused by a shear stress
increment.
 z ,e

- elastic

Ee strain
 z , p  Pz  z

Ee
E
1  F 
1  Pz Ee
E = Static Young’s modulus
Ee = Dynamic Young’s modulus
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Uniaxial loading test
Observations
F*  F  z   r  S
0.04
F*
0.03
0.02
0.01
0.00
0.000
0.005
0.010
0.015
Shear strain
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Uniaxial loading test
Observations
F*  F  z   r  S
0.04
F*
0.03
 z  r  o
FA
z r  S
0.02
0.01
0.00
0.000
0.005
0.010
0.015
Shear strain
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Discussion: the F - parameter
Ee
E
1  F 
1  Pz Ee
160
140
120
Note:
Since E  (1 - F)
 when F =1 then E = 0
Stress (MPa)
Axial
Peak
stress
100
80
60
40
Radial
20
 peak stress
0
0.005
0.010
0.015
0.020
0.025
F = 1  rock strength
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Discussion: the F - parameter
 1   3   1
8To  1   3 
2
Griffith’s failure criterion:
 1   3   o 
 1   3  S 
2
Our model:
A
2
If we can assume that:
 F2
(  1 -  3)  (  1 -  3 )
then we could state that
F = 1  Fulfilment of the Griffith criterion
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Discussion: the F - parameter
(  1 -  3)  (  1 -  3) ?
OK for a purely elastic material
Also OK at the intact parts of the material
even after local failure has occurred
elsewhere
Local (1 -
3)  Global (1 - 3) !
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Discussion: the F - parameter
The development of F can be
seen as a gradual fulfillment of
the Griffith criterion
0.40
F*
0.30
0.20
May be associated with
local failure at various
places in the rock,
0.10
triggered at different
stress levels due to
variable local strength
Peak
stress
0.00
0.000
0.005
0.010
0.015
0.020
0.025
Shear strain
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We have a set of equations……
Ke
K
1  3 PKe
These represent a constitutive model
for the rock
E
We may use it to predict rock behavior,
and thereby derive mechanical properties
for the rock
P
Ee
1  F 
1  Pz Ee
g
 T
 z  r  o
FA
z r  S
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Application for logging purposes
Porosity,
Density, Sonic, .
...
Strength
Constitutive model
12
10
Stress (MPa)
8
6
4
2
0
-5
0
5
10
15
Strain (mStrain)
Simulates rock mechanical
test on fictitious core
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… an example:
Strength (MPa) @ 2MPa
0
25
50
75
100
125
Depth (m - from a reference point)
0
Prediction from logs
Courtesy
of Statoil
50
100
150
200
Core measurements
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21
Challenge:
What is the impact of the intermediate principal stress
on rock strength?
In the field
  
1
2
in general
In the lab
3
 =
2
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22
Most convenient
description:
-plane
Hydrostatic
axis
-plane cross sections
(planes normal to the
hydrostatic axis)
Projections of the
principal axes
Cross section of
the failure surface
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Failure criteria (-plane):
No impact of the intermediate stress
Empirical
Assumption: Rotational symmetry
in -plane (No physical argument)
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Basic theory on shear failure:

Shear failure occurs when the shear stress over some plane
within the rock exceeds the shear strength of the rock
The intermediate principal
stress (2) has no impact
 
3
2

Stress symmetry is not
important
1
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Experimental observations:
2 = 3
Largest principal stress  1 [MPa]
350
2 = 1
300
250
200
150
No impact of intermediate stress
100
50
0
0
50
100
150
200
250
300
350
Intermediate principal stress  2 [MPa]
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Experimental observations:
2 = 3
Largest principal stress  1 [MPa]
350
2 = 1
300
250
200
150
100
50
0
0
50
100
150
200
250
300
350
Intermediate principal stress  2 [MPa]
Takahashi & Koide (1989)
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Numerical simulations:
2 = 3
Largest principal stress  1 [MPa]
3
2 = 1
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
Intermediate principal stress  2 [MPa]
Fjær & Ruistuen (2002)
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Experimental observations:
2 = 3
Largest principal stress  1 [MPa]
350
2 = 1
300
MohrCoulomb
250
DruckerPrager
200
150
100
50
0
0
50
100
150
200
250
300
350
Intermediate principal stress  2 [MPa]
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-plane
29
Question:
2 = 3
Largest principal stress  1 [MPa]
3
2 = 1
What is similar when
2.5
2 = 3
Tetragonal
2
and
1.5
2 = 1
Tetragonal
but different when
1
1 > 2 > 3
Orthorhombic
0.5
?
0
0
0.5
1
1.5
2
2.5
3
Intermediate principal stress  2 [MPa]
It’s the
stress symmetry!
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How can stress symmetry affect the strength?
- It’s because it affects the probability for failure!

 
3
2

1
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Classical picture
Probability
for failure

1

 
3
2

0
1

m
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
32
Classical picture
Probability
for failure


 
3
2
1

0
1

m
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
33
Classical picture
Probability
for failure


 
3
2
1

0
1

m
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
34
Classical picture
Probability
for failure


 
3
2
1

0
1

m
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
35
Classical picture
Probability
for failure


1

3

2

0
1

m
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
36
Classical picture
Probability
for failure


1

3

2

0
1

m
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
37
Classical picture
Probability
for failure


1

3

2

0
1

m
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
38
Classical picture
Classical picture:
Failure occurs if the shear stress
across any plane in the rock sample
Probability
for failure
exceeds So +  – otherwise not.
Introducing fluctuations:
The shear strength varies
from plane to plane.
The rock fails when  exceeds the
shear strength for one of them.
1
0
So
+ 

The probability for failure
increases when 
 So + 
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Classical picture
All planes oriented at an
angle  relative to the 1 axis


Many potential failure planes
in a critical state
2
 
3
2
 High probability for failure

1
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Classical picture
Only planes oriented at an
angle  relative to the 1 axis,
and parallel to the 2 axis


Few potential failure planes
in a critical state
2
 
3
2
 Low probability for failure

1
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Classical picture
All planes oriented at an
angle /2 - 
relative to the 3 axis


Many potential failure planes
in a critical state
2

3
 High probability for failure
 
1
2
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Mathematical model
Probability for failure of a plane
with orientation specified by (,):
p f  ,   
n
 n   So   
n
(n    classical Mohr-Coulomb)
Overall probability for failure:
Expected strength of the material:
Pfailure  1 
1  p  ,  


all ,

f
 Pfailure
  1
 exp    1 
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N
T

 d 1

43
Mathematical model
Probability for failure of a plane
with orientation specified by (,):
(n    classical Mohr-Coulomb)
p f  ,   
f ( 1 ,  1+  1 )
Probability distribution
0.015
 n   So   
n
 2 =  
0.01
2 = 1
Overall probability for failure:
 2 = 0.5( 1 +  3)
0.005
0
0.75
n
1
1.25
 1/ M-C
1.5
Pfailure  1 
1  p  ,  


all ,

N
T
f
Expected strength of the material:
 Pfailure
  1
 exp    1 
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
 d 1

44
Mathematical model
Pfailure
f 1 , 1  1  
1
1
f ( 1 ,  1+  1 )
Probability distribution
0.015
 2 =  
0.01
2 = 1
 2 = 0.5( 1 +  3)
0.005
0
0.75
1
1.25
1.5
 1/ M-C
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Mathematical model
8
2 = 3
2 = 1
6
 exp 4
n = 30
n = 175
n = 1000
Mohr-Coulomb
2
The impact of the
intermediate principal
stress is directly linked
to the non-sharpness of
the failure criterion
(represented by 1/n)
i.e. to the rock
heterogeneity
0
0
2
4
6
8
2
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Comparing model and observations
2 = 1
Largest principal stress  1 [MPa]
350
300
Takahashi and Koide,
1989
250
200
150
n = 30
100
50
0
0
50
100
150
200
250
300
350
Intermediate principal stress  2 [MPa]
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Comparing model and observations
2 = 1
Largest principal stress  1 [MPa]
3
2.5
Numerical model
2
1.5
1
n = 25
0.5
0
0
0.5
1
1.5
2
2.5
3
Intermediate principal stress  2 [MPa]
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Fractures are
planes with
largely reduced
or no strength
Outcrop from a
Marcellus shale
formation
Han, 2011
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Borehole breakouts in a non-fractured rock
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Borehole breakouts in a non-fractured rock
Shear failure planes
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Borehole breakouts in a non-fractured rock
Shear failure planes
Failure would be more easily triggered by a set
of suitably oriented
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Simple example
Mudweight [sg]
1
1.2
1.4
1.6
1.8
1500
No fractures
1505
1510
1515
1520
1525
1530
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Simple example
Mudweight [sg]
1
1.2
1.4
1.6
1.8
1500
No fractures
1505
1510
Sealed fractures
1515
|| borehole
1520
1525
1530
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Simple example
Mudweight [sg]
1
1.2
1.4
1.6
1.8
1500
No fractures
1505
1510
Sealed fractures
1515
|| borehole
1520
Open fractures
|| borehole
1525
1530
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Several fracture sets complicates the situation.
Blocks may become detached at washed away by the circulating
mud.

More fractures will be exposed to the drilling fluid.
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Other possible failure modes – bedding plane splitting
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Økland and Cook 1998
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 20
To avoid the problem:
Økland and Cook 1998
The “angle of attack” between the well
and the bedding plane should be at least
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20.
Challenge:
What is the strength of a fractured rock
(if we consider it as homogeneous)?
Available alternative:
Hoek-Brown
Purely empirical criterion
Hoek & Brown (1980)
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Geologocal Strength Index GSI
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Rocks are heterogeneous –
treating them as homogeneous comes at a price…..
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The strength of a homogeneous
material is size invariant.
Rocks, on the other hand, -
Hoek & Brown (1980)
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Current work:
Relate the failure probability model to Hoek-Brown
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Challenge: Match with observations
60
50
1
40
30
20
10
0
0
5
10
15
20
3
Data from Hoek; Kaiser (2008)
Failure probability model
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Consideravble scatter in
measured strength
Kaiser (2008)
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60
50
1
40
Probability distribution
0.16
30
0.14
20
0.12
0.1
10
0.08
0.06
0
0.04
0
5
10
15
20
3
0.02
0
0
10
20
30
1
40
50
Failure probability model
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Conclusions:
Physics helps us to make better tools for rock mechanics applications
There is still room for more physics in rock mechanics
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