SINTEF Petroleum Research The strength of fractured rock Erling Fjær SINTEF Petroleum Research 1

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SINTEF Petroleum Research

The strength of fractured rock

Erling Fjær


1

SINTEF Petroleum Research 2

Porosity, Density, Sonic, . . . .

Challenge: Estimation of rock strength from log data

Strength

Available Wanted

Traditional approach:

correlations

bPUCS a t



Challenge: Estimation of rock strength from log data

Strength

Available WantedBrandås et al. (2012)


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


0

20

40

60

80

100

120

140

160

0.005 0.010 0.015 0.020 0.025

Str

ess

(MP

a)

Peak stress

Axial

Radial

static moduli vs dynamic moduli

Rock mechanical testincluding acoustic measurementson a dry sandstone

static moduli dynamic moduli

The differences changes with stress and strain

Dry, weak sandstone

-5

0

5

10

15

20

0.005 0.010 0.015 0.020 0.025

Axial strain

You

ng's

mod

ulus

(G

Pa)

Dynamic

Static


0

20

40

60

80

100

120

140

160

0.005 0.010 0.015 0.020 0.025

Str

ess

(MP

a)

Peak stress

Axial

Radial

static moduli vs dynamic moduli

Rock mechanical testincluding acoustic measurementson a dry sandstone

static moduli dynamic moduli

The differences changes with stress and strain

Dry, weak sandstone

-5

0

5

10

15

20

0.005 0.010 0.015 0.020 0.025

Axial strain

You

ng's

mod

ulus

(G

Pa)

Dynamic

Static

We are seeking mathematical relations between the static and the dynamic moduli


We introduce a parameter P, defined as:

P is a measure of the inelastic part of the deformation caused by a compressive

hydrostatic stress increment.

Building relations

,

3v v eP

,v e

eK

v - total volumetric strain

Hydrostatic test

- elastic strain


We introduce a parameter P, defined as:

P is a measure of the inelastic part of the deformation caused by a compressive

hydrostatic stress increment.

Building relations

,

3v v eP

,v e

eK

v - total volumetric strain

Hydrostatic test

- elastic strain

K = Static bulk modulusKe = Dynamic bulk modulusK

K

PKe

e

1 3


Observations

0

5

10

15

20

25

0 5 10 15 20 25 30

Hydrastatic stress [MPa]

1/P

[G

Pa-1

]Hydrostatic test


Observations

0

5

10

15

20

25

0 5 10 15 20 25 30

Hydrastatic stress [MPa]

1/P

[G

Pa-1

]Hydrostatic test

gPT


We introduce a parameter F, defined as:

F is a measure of the inelastic part of the deformation caused by a shear stress

increment.

Building relations

, ,z z e z p

z

F

,z eeE

,z p z zP

z - total axial strain

Uniaxial loading test

- elastic strain


We introduce a parameter F, defined as:

F is a measure of the inelastic part of the deformation caused by a shear stress

increment.

Building relations

, ,z z e z p

z

F

,z eeE

,z p z zP

z - total axial strain


- elastic strain

E = Static Young’s modulusEe = Dynamic Young’s modulus 1

1e

z e

EE F

P E


Observations

0.00

0.01

0.02

0.03

0.04

0.000 0.005 0.010 0.015

Shear strain

F*

* z rF F S



Observations

0.00

0.01

0.02

0.03

0.04

0.000 0.005 0.010 0.015

Shear strain

F*

* z rF F S

z r o

z r

F AS



Discussion: the F - parameter

Since E (1 - F)

when F =1 then E = 0

peak stress

11

e

z e

EE F

P E

0

20

40

60

80

100

120

140

160

0.005 0.010 0.015 0.020 0.025

Str

ess

(MP

a)

Peak stress

Axial

Radial

Note:

F = 1 rock strength


Griffith’s failure criterion:

If we can assume that: (1 - 3) (1 - 3)

then we could state that

F = 1 Fulfilment of the Griffith criterion

Our model:

2

1 3

1 3

18 oT

2

1 32 2

1 3

oA FS



(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) !



0.00

0.10

0.20

0.30

0.40

0.000 0.005 0.010 0.015 0.020 0.025

Shear strain

F*

Peak stress

The development of F can be seen as a gradual fulfillment of the Griffith criterion

May be associated with local failure at various places in the rock,

triggered at different stress levels due to variable local strength



We have a set of equations……

These represent a constitutive model for the rock

We may use it to predict rock behavior, and thereby derive mechanical properties for the rock

gPT

11

e

z e

EE F

P E

KK

PKe

e

1 3

z r o

z r

F AS



Constitutive model

Application for logging purposes

Simulates rock mechanical test on fictitious core

0

2

4

6

8

10

12

-5 0 5 10 15

Strain (mStrain)

Str

ess

(MP

a)

Strength


0

50

100

150

200

0 25 50 75 100 125

Strength (MPa) @ 2MPa

Dep

th (m

- fr

om

a r

efer

ence

po

int)

Courtesy of Statoil

Prediction from logs

Core measurements

… an example:


In the lab

2 = 3

1 2 3

in general

In the field

Challenge: What is the impact of the intermediate principal stress on rock strength?


Most convenient description:

-plane cross sections(planes normal to the

hydrostatic axis)

-plane Hydrostatic axis

Projections of the principal axes

Cross section of the failure surface


Failure criteria (-plane):

Assumption: Rotational symmetry in -plane (No physical argument)

No impact of the intermediate stress

Empirical


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

123

The intermediate principal stress (2) has no impact

Stress symmetry is not important


Experimental observations:

0

50

100

150

200

250

300

350

0 50 100 150 200 250 300 350

Intermediate principal stress 2 [MPa]

La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1

2 = 3

No impact of intermediate stress



0

50

100

150

200

250

300

350

0 50 100 150 200 250 300 350


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1

2 = 3

Takahashi & Koide (1989)


Numerical simulations:

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1

2 = 3

Fjær & Ruistuen (2002)



0

50

100

150

200

250

300

350

0 50 100 150 200 250 300 350


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a] 2 = 1

2 = 3

-plane

Mohr-Coulomb

Drucker-Prager


Question:

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1

2 = 3

What is similar when

2 = 3

and

2 = 1

but different when

1 > 2 > 3

?

It’s the stress symmetry!

Tetragonal

Tetragonal

Orthorhombic


How can stress symmetry affect the strength?

- It’s because it affects the probability for failure!

123


Classical picture

123

Probability for failure

0

1

m


Classical picture

123


0

1

m


Classical picture

123


0

1

m


Classical picture

123


0

1

m


Classical picture

123


0

1

m


Classical picture

123


0

1

m


Classical picture

123


0

1

m


Classical picture


0

1

Classical picture:

Failure occurs if the shear stress

across any plane in the rock sample

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.

The probability for failure

increases when So +

So +


Classical picture

12

3

All planes oriented at an angle relative to the 1 axis

2

Many potential failure planes in a critical state

High probability for failure


Classical picture

123

Only planes oriented at an angle relative to the 1 axis,

and parallel to the 2 axis

2

Few potential failure planes in a critical state

Low probability for failure


Classical picture

3 2

1

All planes oriented at an angle /2 -

relative to the 3 axis

2

Many potential failure planes in a critical state

High probability for failure


Mathematical model

Probability for failure of a plane with orientation specified by (,):

(n classical Mohr-Coulomb)

,n

f nno

pS

Overall probability for failure:

failureall ,

1 1 ,N

TfP p

Expected strength of the material:failure

exp 1 11

dP


Mathematical model

Probability for failure of a plane with orientation specified by (,):

(n classical Mohr-Coulomb)

,n

f nno

pS

Overall probability for failure:

failureall ,

1 1 ,N

TfP p

Expected strength of the material:

failureexp 1 1

1

dP

0

0.005

0.01

0.015

0.75 1 1.25 1.5

1/ M-C

Pro

bab

ility

dis

trib

uti

on

f

(1,

1+

1)

2 =

2 = 1

2 = 0.5( 1 + 3)


Mathematical model

failure1 1 1 1

1

,Pf

0

0.005

0.01

0.015

0.75 1 1.25 1.5

1/ M-C

Pro

bab

ility

dis

trib

uti

on

f

(1,

1+

1)

2 =

2 = 1

2 = 0.5( 1 + 3)


Mathematical model

0

2

4

6

8

0 2 4 6 8

2

exp

2 = 1

2 = 3

n = 30n = 175n = 1000Mohr-Coulomb

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


Comparing model and observations

0

50

100

150

200

250

300

350

0 50 100 150 200 250 300 350


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1

Takahashi and Koide, 1989

n = 30


Comparing model and observations

Numerical model

n = 25

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3


La

rge

st

pri

nc

ipa

l s

tre

ss

1 [

MP

a]

2 = 1


Outcrop from a Marcellus shale

formation

Han, 2011

Fractures are planes with

largely reduced or no strength


Borehole breakouts in a non-fractured rock


Shear failure planes




Shear failure planes


Simple example

1500

1505

1510

1515

1520

1525

1530

1 1.2 1.4 1.6 1.8Mudweight [sg]

No fractures


Simple example

1500

1505

1510

1515

1520

1525

1530

1 1.2 1.4 1.6 1.8Mudweight [sg]

No fractures

Sealed fractures|| borehole


Simple example

1500

1505

1510

1515

1520

1525

1530

1 1.2 1.4 1.6 1.8Mudweight [sg]

No fractures

Sealed fractures|| borehole

Open fractures|| borehole


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.


Other possible failure modes – bedding plane splitting














Økland and Cook 1998


Økland and Cook 1998

To avoid the problem:

The “angle of attack” between the well and the bedding plane should be at least

20.

20

Well


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)


Geologocal Strength Index - GSI


Rocks are heterogeneous –

treating them as homogeneous comes at a price…..


Hoek & Brown (1980)

The strength of a homogeneous material is size invariant.

Rocks, on the other hand, -


Current work: Relate the failure probability model to Hoek-Brown


Data from Hoek; Kaiser (2008)

Challenge: Match with observations

0

10

20

30

40

50

60

0 5 10 15 20 1

3

Failure probability model


Kaiser (2008)

Consideravble scatter in measured strength


0

10

20

30

40

50

60

0 5 10 15 20 1

3

Failure probability model

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0 10 20 30 40 50

Prob

abili

ty d

istr

ibuti

on

1


Conclusions:

Physics helps us to make better tools for rock mechanics applications

There is still room for more physics in rock mechanics

Documents

SINTEF Petroleum Research The strength of fractured rock Erling Fjær SINTEF Petroleum Research 1