analytical solution of wbs problem using camclay model charlez

8/3/2019 Analytical Solution of WBS Problem Using Camclay Model Charlez

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Oil & Gas Science and Technology Rev.IFP, Vol. 54 (1999), No. 5, pp. 551-563Copyright 1999, ditions Technip

A Fully Analytical Solutionof the Wellbore Stability Problem

under Undrained ConditionsUsing a Linearised Cam-Clay Model

Ph. A. Charlez1 and S. Roatesi2

1 TOTAL FINA, La Dfense - France

2 University of Bucharest, Bucharest - Romania

e-mail: [email protected] [email protected]

Rsum Solution analytique au problme de stabilit de puits en conditions non drainesutilisant un modle de Cam-Clay linaris Cet article prsente une version linarise du modle deCam-Clay intgr dans le cadre gnral de la thorie de la poroplasticit isotherme. La loi decomportement dveloppe partir du concept des contraintes effectives plastiques ne contient que deuxparamtres plastiques (module dcrouissage et pente de la droite critique). Le modle est valid sur deschemins de contrainte homognes (hydrostatique, triaxial drain et triaxial non drain) et compar desessais exprimentaux obtenus partir dessais conventionnels de laboratoire.

Ensuite, une version simplifie du modle est applique au problme du puits dans un champ decontrainte axisymtrique et en conditions non draines. Compte tenu de la linarit de la loi et de laconnaissance a priori de la rgion plastique, la solution (contraintes, dformations et pression de pore) estanalytique.

Les rsultats montrent que pour un matriau surconsolid (degr de consolidation infrieur 2) lacontrainte tangentielle est fortement relche dans la zone plastique. Plus la compressibilit du fluide estgrande, plus cette relaxation est importante. En termes de stabilit, plus le fluide est compressible, plus lepuits sera stable. Finalement, plus le degr de consolidation est grand, moins le puits sera stable.

Mots-cls : stabilit, puits, lasticit, non drain.

Abstract A Fully Analytical Solution of the Wellbore Stability Problem under Undrained

Conditions Using a Linearised Cam-Clay Model This paper presents a linearised version of the

Cam-Clay model fully integrated in the scope of the general theory of poroplasticity. The constitutive law

which is developed in the scope of the effective plastic stress concept, only contains two plastic

parameters (hardening modulus and slope of the critical state line). To be validated, the model is

integrated over homogeneous stress paths (hydrostatic, drained triaxial and undrained triaxal) then

compared with experimental data issued from conventional laboratory triaxial tests.

In the second part, a simplified version of the model is applied to the wellbore boundary problem

(vertical well) in an axisymmetric horizontal stress field and under undrained conditions. Given the

linearity of the constitutive law and the a priori knowledge of the shape of the plastic region, the solution

(stress, strain and pore pressure) is fully analytical.
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Oil & Gas Science and Technology Rev. IFP, Vol. 54 (1999), No. 5

NOMENCLATURE

P mean effective stress (MPa)

Q stress deviator (MPa)

p pore pressure

KB

drained bulk modulus (MPa)

Ku

undrained bulk modulus (MPa)

EB drained Young's modulus (MPa)E

uundrained Poisson's ratio (MPa)

B

drained Poisson's ratio

G shear modulus (MPa)

m mass of fluid per total volume (kg/m3)

A, V hardening forces, hardening flux variables

f yield locus

M slope of critical state line

H hardening modulus (MPa)

k plastic bulk modulus (MPa)

Pcon

consolidation pressure (MPa)

Pcr

critical pressure (MPa)

pR

virgin pore pressure (MPa)

v

vertical stress (MPa)

h

horizontal stress (MPa)

B Skempton coefficient

kk

p mean plastic strain

radial stress (MPa)

hoop stress (MPa)

R well radius (m)

Rp

plastic radius (m).

INTRODUCTION

The propping of underground cavities is one of the mostbasic geotechnical problem in mining engineering (stabilityof underground galleries) and in civil engineering(tunnelling). The objective of any stability calculation is tosize a propping agent. The oil well, which is the key elementin petroleum engineering, is no exception to the rule.

During the drilling phase, the well remains open hole andis propped by the drilling mud which is a complex mixture of

a fluid (water or oil base) and a solid (usually bentonite or

barite). Depending on the quantity of solid used, it is possibleto adjust the density of the mixture over a wide range, andconsequently the pressure (equal to the weight of the mudcolumn) exerted on the borehole wall. The required densitywill obviously be strongly connected to the type of formationbeing crossed and may vary considerably during a samedrilling operation.

The mud weight required to stabilise a given formation

mainly depends both on the mechanical state existing beforedrilling (in situ stresses and virgin pore pressure) but also onthe rheological behaviour of the rock concerned. From ageneral view point (Charlez, 1997), deep sedimentary rockscan be classified in damaging/brittle materials (cohesivesandstones and limestones) and poroplastic rocks (claystones,shales and loose sands). Due to their low shear strength, deepporoplastic materials are the most sensitive to stabilityproblems (Veeken et al., 1989; Charlez and Heugas, 1991;Ewy, 1991; Zhou et al., 1996). A proper understanding oftheir rheological behaviour is consequently a key issue inview of assessing stability of deep wells.

Standards poroplastic constitutive laws are very attractivemodelling tools both for the simplicity of their formulationbut also for their theoretical robustness. However, does thenormality hypothesis complies with experimental reality?

Poroplastic rocks are generally governed by two differentelastoplastic mechanisms: a collapse mechanism purelycontractant and associated with the spherical component of thestress tensor and a deviatoric mechanism governed by internalfriction successively contractant, at constant plastic volume,then possibly dilatant just before macroscopic failure (Charlezand Shao, 1993). If described by a Mohr-Coulomb straight linethe deviatoric mechanism (contractant or at constant plastic

volume) appears incompatible with respect to the normalityconcept which imposes the volumetric plastic strain incrementto be dilatant. The advantage of the Cambridge model residesin the successful marriage between these two concepts.

More widely known under the name Cam-Clay it wasdeveloped in the mid 1960's by Burland and Roscoe (1968).Although it was initially written for normally consolidated claymaterials from surface, the very realistic physics it integratesallows adapting it to deeper materials. In the classical Cam-Clay model, the state equations, the yield locus and the strainhardening law are non linear. We propose below a linearisedversion of the Cam-Clay which can be fully integrated in the

case of the wellbore boundary problem.

552

The solution shows that for an overconsolidated material (overconsolidation degree less than 2) the hoop

stress is strongly relaxed in the plastic zone. The higher the compressibility of the saturating fluid, the

larger the relaxation of the hoop stress.

In terms of stability, the more compressible the fluid saturating the porous medium is, the more stable the

well will be. Finally the larger the overconsolidation ratio is, the less stable the well will be.Keywords: wellbore stability, undrained, plasticity.


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Ph A Charlez and S Roatesi / A Fully Analytical Solution of the Wellbore Stability Problem under Undrained Conditions

1 LINEARISATION OF THE CAM-CLAY MODEL:BILINEAR MODEL

Poroplastic rocks (Biot, 1973; Lade, 1977; Coussy,1989;

Charlez, 1991; Coussy, 1995) are materials likely to undergoirreversible instantaneous deformations (no delayed effects)of both their skeleton, and their fluid mass content. Undersmall perturbation hypothesis, the partitioning rule ofclassical plastic materials (decomposition of the total straininto an elastic and a plastic part) is extended to fluid massincrements, i.e.:

(1)

dm, dme and dmp being respectively total, elastic (recoverable)and plastic fluid mass increments per unit of initial volume.

1.1 Main Hypothesis

Two fundamental hypothesis are considered. On one hand, theplastic flow rule is considered as standard (merging of theyield locus and the plastic potential) on the other, the matrix(i.e. the grains) is assumed elastically and plasticallyincompressible. As shown by Coussy (1989) these hypothesesallow writing the various equations directly in terms ofeffective stress. Assuming a single strain hardening mecha-nism (A, V), the plastic flow rule is written:

(2)

where f is the yield locus and F the plastic potentialassociated with the strain hardening mechanism (the materialis not considered as standard generalised). In equations (2), is the plastic multiplier.

1.2 Basic Equations

In the P, Q space (mean effective stress and second invariant

of the stress tensor), the various equations governing themodel are the following.

1.2.1 State Law

(3)

where Ku, K

Band G are respectively the undrained bulk

modulus (i.e. for m = 0), the bulk drained modulus and the

shear modulus (similar to that of continuous media). B is the

Skempton's coefficient and 0 the fluid density in thereference state.

1.2.2 Yield Locus

The yield locusfis written as following:

(4)

In (4), Pcr

is the critical pressure and M a materialconstant. Their meaning will be discussed later.

1.2.3 Strain Hardening Law

The strain hardening mechanism is defined by a flowvariable Vand a strain hardening forceA such that:

(5)

where har

is the locked up energy per unit volume. Thecombination of the various plastic equations leads to theincremental law (Loret, 1987):

(6)

with:

(7)

and being respectively the elastoplastic and thedrained elastic matrix. In (7), H is the strain hardeningmodulus such that:

(8)

The proposed linear strain hardening law of the model iswritten:

(9)

In (9), k is a second material constants to be experi-mentally measured.

The yield locus of the bilinear model is studied inFigure 1. For P > P

cr, the volumetric behaviour is contractant

and the yield locus moves parallel to itself such that point Aremains on a line of slopeM(called critical straight line). ForP < P

cr, the yield locus is parallel to the P axis and the

material plastifies at constant plastic volume. We should notethat the yield locus cut the P axis for a value P

con

(consolidation pressure) which is equal to twice the critical

pressure. The plastic formalism can be written indifferently

P P kcr cr kkp kkp= + ( )0 2 0

Hf

A V

F

A

har=

2

2

Bep

= +

ep B- -

H

f f1 1 1

d d = ep:

AV

har=

P P f P Q P Q MP

P P f P Q P Q+MP MP

cr cr cr

cr cr cr

( ) =

> ( ) =

, ,

, , 2

P P K G i j

p p BK

B kk kk

p

ij ij ij

p

R u kk

= + ( ) = ( ) ( )

= + +

0

0

2

m

d and d

with

p f V F

A

Ip

= =

=

d d d d d d = + = +p e p em m m

553


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Figure 1

Bilinear model: yield locus.

using Pcr

or Pcon

. In the hardening law, Pcr

appears as ahardening force and as a hardening flux variable. Byintegrating Equation (5), we calculate the locked-up energy:

(10)

which is a quadratic function of the plastic volumetric strain.Identification of plastic flow rules (2) allows determining theplastic potential F associated with the strain hardeningprocess (the model is not standard generalised), then through

Equation (8), the strain hardening modulusH:

(11)

which, in both cases, is a constant. This is a direct consequenceof the linear choice of the different constitutive equations. Thestrain hardening modulus is positive in the contractant zone(i.e., for P > P

cr). For P = P

cr, the strain hardening modulus is

nil and the material behaves as an ideal plastic one. Thelocked-up energy corresponds to grains contact energy. Inthe hardening phase, the grains are brought close together andthe number of contacts increases, as does the locked-up

energy; the material is plastically contractant and the strainhardening modulus is positive. When the representativeloading point reaches the critical state line, the locked-upenergy and the number of contacts remain stationary; thebehaviour is then ideal plastic. For a mean stress value higherthan the critical pressure, calculation of the plastic matrix iscarried out by applying Equation (7) and leads to the followinglinear incremental law:

(12)

dp being the deviatoric strain. Unlike a purely elastic model,the deviatoric stress induces volumetric deformations duringthe strain hardening phase. Apart from the elastic constants

EB

and B

, the bilinear model contains 3 plastic parameters

H,Mand Pcr(or Pcon).

1.3 Integration on Homogeneous Stress Paths

For a homogeneous triaxial stress path ('1, '2 = '3), theincremental elastoplastic law (12) becomes:

(13)

For a hydrostatic compression test (d'1 = d'3 = dP), thevolumetric deformation evolves with the mean stressaccording to:

(14)

The constant k(introduced in the strain hardening law (9))plays the role of a plastic bulk modulus.

The drained triaxial test, which consists of keeping theconfining pressure constant (d'3 = 0) under drainedconditions, corresponds in the P-Q diagram to a straight lineof slope +3. For such a stress path, the incremental law (13)is written:

(15)

During the strain-hardening phase, the axial stress evolveslinearly versus the axial strain with a slope such that:

(16)

As shown on Figure 2, for normal consolidationconditions, the material exhibits a strain hardening/idealplastic behaviour whereas, for an overconsolidated material(degree of overconsolidation less than 2), an elastic/strain

hardening/ideal plastic behaviour will be observed. These

=+ + +

1

1 1

3 32 3

2

E H

MM

B

d d

d d

1

2

1

3

2

1

1 1

3 32 3

1

2

2

9 3

1

= + + +

= + +

E H

MM

E H

M M

B

B

B

d dkkBK k

P= +

1 1

d d

d

d d

1

2

1

2

3

3

2

1

1 1

3 32 3

2 1

3

2

33

1

2

2

9 31

1

= + + +

+

+ +

= + +

+

E H

MM

E H

MM

E H

M M

B

B

B

B

B

BBE H

M M+ +

1

2

4

9

4

31

2

3d

P P

M

HP

M

HQ

M

H

P

H

Qcr

kk

p

p

>= +

= +

d d d

d d d

2

1

H kM P P

H P P

cr

cr

= >=

2

0

if

if

har cr kk p

har kk

p

kk

p= P

k = + ( ) 0

2

4 0

kkp

Q

Q = MPcr A

Pcr

Q =MP

Pcon

Q+MP_MPcon

P

554


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Figure 2

Behaviour of the bilinear model under drained triaxial

loading.

results also demonstrate the lower rupture limit of an over-consolidated material which fails at a lower Q value(compare Q atB and C) than a normally consolidated one.

A calibration example of the bilinear model on drainedtriaxial tests is presented on Figure 3 (shallow sand fromOrenico Delta, Venezuela). The best fitting gives thefollowing set of parameters: E

B=125MPa,H= 1000 MPa,

B

= 0.33 andM= 1.16. However, given the linearity of theconstitutive equations, the model is not able to take into

account the strong non linearities experienced during strainhardening.

Figure 3

Validation of the bilinear model on drained tests (loosesand/Orenico, Venezuela).

In the case of an incompressible fluid, the undrained pathcorresponds to an isochore transformation:

(17)

In the P-Q diagram, the undrained triaxial test follows aslope b such that:

(18)

The undrained path is therefore plastic contractant for anormally consolidated material, elastic (this time the slope isequal to the undrained Young's modulus), then contractantplastic in the case of an overconsolidated material. Formaterials with an overconsolidation ratio equal to 2, thematerial reaches the critical state by following a purelyelastic path (Fig. 4). In the Q- diagram, the strain hardeningphase follows a slope ' such that:

(19)

Figure 4

Behaviour of the bilinear model under undrained triaxialloading.

The fluid which contributes directly to elastic deformationunder undrained conditions (E

uis different, depending on

whether the fluid is compressible or not) plays no role in theplastic strain hardening. This is perfectly consistent with thephysical aspects of strain hardening phenomena (nohardening energy is locked up in the fluid).

An example of validation (remoulded samples fromVilleperdue marl) of the bilinear model on triaxial undrained

tests is presented on Figure 5. The best set of parameters

Overconsolidated

Axial deformation

slope

Hardening plasticity:

Linear elasticity: slope Eu

A

B

Q Q

Q = MP

Normally consolidated

=+ + +

1

1 1

3 32 3

2

E H

MM

u

= +

MH

MKu

d d dkkuK

M

H

PM

H

Q= +

+ =

10

2

20

18

16

14

12

10

8

6

4

0

2

-15 -5 5 15Strain (%)

2 MPa

9 MPa

4.5 MPa

Bilinear model

Q(MPa)

+3

PconP

Hardening plasticity: slope

Linear elasticity: slope EB

Axial deformation

Overconsolidated

Normallyconsolidated

Q

C

B

A

Initial yield locus

Q

555


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obtained from these undrained tests is the following: Eu

=2800 MPa,

u= 0.32, H= 1150MPa,M= 0.71. The elastic

constants are in good agreement with the pore fluidcompressibility modulus (water in this case).

In the P-Q plane the model fits quite well the experimentalresults. However, in the stress/strain diagram due to thestrong experimental nonlinearities the bilinear model doesnot accurately predict the experimental curves.

2 THE PLASTIC WELLBORE PROBLEM

Analytical solutions of plastic boundary problems are notcommon, as the incremental equations can only be integratedfor simplified geometries. First of all the stress state beingnot calculated using the same equations in the plastic and theelastic zones, for an explicit solution to be found, the shapeof the potential plastic zone must a priori be known. For thewellbore problem, only a circular geometry (isotropichorizontal stress field) allows obtaining analytical solutions.By contrast, if an anisotropic stress field prevails, the plasticzone being not a priori known, an implicit numerical method

(finite elements) shall be used.Although limited in their applications, these analytical

solutions do have the advantage of revealing howelastoplastic parameters act on stability.

2.1 Definition of the Problem

Let us consider an infinite circular domain initially loaded(Fig. 6) at infinity by an isotropic horizontal geostatic stress

hand a vertical geostatic stress

v. Before drilling, the pore

pressure in the whole domain is uniform and equal topR

.At time t= 0, an infinite cylinder of radius R is instanta-

neously drilled parallel toz axis. It is filled with a fluid of the

same nature as that saturating the porous medium but at adifferent pressure p

w. Since the borehole is assumed to be

infinitely long parallel to its axis, calculation of the perturba-tion can be performed under plane strain hypothesis (

zz= 0).

The boundary conditions can be summarised as follows:

(20)

The resolution of the undrained problem is presented inAppendix. We discuss below the main results.

2.2 Discussion of the Results

The analytical solution is studied below using the followingset of parameters. For simplicity, calculations are performedconsidering that the initial pore pressurep

R

is equal to 0.

TABLE 1

Set of parameters used to study the wellbore boundary problem

EB

MPa 10 000

B

- 0.3

- 0.1

Rp

- 5

h MPa 30

v

MPa 40

pR

MPa 0

P0c MPa 40

t p p

t R, t p , t

p R, t p p , t p

R

w h

w R

zz

=

> =( ) = = ( ) =

=( ) = = ( ) =

=

0

0

0

( )

556

Figure 5

Validation of the bilinear model on undrained tests (remoul-

ded specimens/Villeperdue field).

Figure 6

Well axisymmetric plane problem.

zz= 0

h

Rp

R

PwR

0 2 4 6 8 10 12 10 20 30 40

22

18

14

6

2

10

Q(MPa)

45 MPa

35 MPa

25 MPa

15 MPa

M= 0.71

Axial deformation (%) P(MPa)

Bilinear model


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As pointed out on Figure 7 a substantial relaxation of thehoop stress occurs in the strain hardened zone with respect tothe elastic solution (dashed lines). However, the radialcomponent is only slightly affected by plasticity. That resultis caused by the fact that the radial stress is imposed on bothboundaries (well pressure for the internal boundary, geostaticstress at infinity).

The effect of the Skempton's coefficient on hoop stressrelaxation is discussed on Figure 8. The lower theSkempton's coefficient, the more the hoop stress is relaxed in

the well vicinity. In other words, the higher the fluidcompressibility (B close to 0), the higher the hoop stressrelaxation.

For the same reasons as previously, Skempton's coeffi-cient has a marginal effect on the radial component. In fact,

for infinitely compressible fluids, the undrained solutionconverges towards the drained solution. Drained and undrai-ned solutions characterises respectively instantaneous andvery long-term behaviours of the structure.

As pointed on Figure 9a, for similar boundary conditions(same well pressure), a low Hvalue tends on one hand toincrease the extension of the plastic zone (larger plasticradius), but also to further relax the hoop stress. It is thuspossible, depending on the value of the strain hardeningmodulus, to obtain either a decreasing hoop stress (high

values ofH) or a peak opposite the plastic radius (low Hvalues). Finally, M (slope of critical straight line) has asimilar effect on the hoop stress: a low value ofMtends toincrease the extension of the plastic zone and further relaxesthe hoop stress (Fig. 9 b).

557

Figure 7

Stress state around the borehole. Comparison with the elastic

solution (M= 0.7,H= 100 MPa,B = 0).

Figure 8

Effect of Skempton's coefficient on hoop stress relaxation

(M= 0.7,H= 100 MPa).

30

26

1 2 3 4 5 6 7 8

32

33

31

29

28

27

Dimensionless radius

(MPa)

0.96

0.7

0.19

0

40

20

10

30

1 2 3 4 5 6 7 8 9

50


Elastic solution

Rp

(MPa

)

27

1 2 3 4 5 6 7 8

31

35

29


(M

Pa)

331.25

0.7

30

26

1 2 3 4 5 6 7 8

32

33

31

29

28

27


(MP

a)

1000 MPa

100 MPa

Figure 9a

Influence ofH (M= 0.7) on hoop stress relaxation (B = 0).

Figure 9b

Influence ofM (H= 100 MPa) on hoop stress relaxation (B = 0).


8/13


With respect to Equation (A9), the pressure response isthe sum of two terms respectively connected to the meantotal stress variation (Skempton's effect) and the plasticvolumetric strain.

In fact, these two effects play in opposite sense (Fig. 10),as the mean stress and the plastic volumetric strain arerespectively minimum and maximum at the borehole wall. Inmost cases, the term connected to the plastic volumetricstrain is the largest. Consequently, a maximum pore pressureat the borehole wall will prevail followed by a

Figure 10

Decomposition of the undrained pressure response.

Figure 11

Incompatibility between boundary conditions at the boreholewall and the undrained response.

decrease which is nullified opposite the plastic radius. In thecase of low compressible fluids, however, the term related tothe mean stress can, at a certain distance away from theborehole, become predominant. The pressure curve then

passes through a slightly marked minimum.Nevertheless (Detournay and Cheng, 1988), the undrainedpressure response does not properly respect the boundarycondition imposed at the borehole wall. Physically twosituations are possible. If the borehole wall is impermeable(oil-base mud, perfect cake) or the pressure in the borehole isnot applied directly by the fluid (but through a rubbermembrane such as a packer for instance), the downholepressure will not necessarily be equal to the pore pressure atthe wall. On the contrary, in the case of a permeable borehole,the undrained solution at the borehole wall is no longer valid.Instantaneously, the pore pressure achieves equilibrium with

the condition imposed at the boundary and the pressureexhibits a peak lying inside the block (Fig. 11).

2.3 Critical Instantaneous Pressure

The critical pressure corresponds to the well pressure forwhich the critical state is reached that is:

(21)

For a given Mvalue, the lower the Skempton coefficient,the lower the critical instantaneous pressure (Fig. 12a). In otherwords, the more compressible the fluid saturating the porousmedium is, the more stable the well will be. A zero Skemptoncoefficient corresponding to the perfectly drained case theundrained solution is less stable than the drained solution. Formore complex constitutive laws (nonlinear Cam-Clay,Laderock) and depending on the rock permeability, drainage(that is time) have sometimes a negative effect on stability.

The critical pressure decreases when M increases (morestable well) but, for a same Mvalue, the critical pressureincreases with the hardening modulusH(Fig. 12b). If critical

pressures are compared with those issued from a purelyelastic solution, the latter is more pessimistic. With respect tothe hardening modulus H, the elastic solution appears as alimit of the elastoplastic solution for which the strainhardening modulus is infinite. Generally speaking, plastichardening plays a favourable role on stability.

2.4 Impact of the Consolidation Degree

For a normally consolidated material the initial consolidationpressure is equal to the mean effective stress, i.e.:

(22)

Pconh0 2

3

=+

2 1

3

= + + +

M

B h Bp

( )( )

1

2 ( )

27

25

23

21

19

17

151 1.2 1.4 1.6 1.8 2.2 2.42

(M

Pa)


P

-10

10

5

0

-5

0 1 2 3 4 5 6 7 8


a

Bkk/3

P

BKB P

kk

558


9/13


In such a material, the plastic radius moves out to infinity.In other words, all the points are initially plasticallyadmissible and the material hardens as soon as the wellpressure drops below the horizontal stress

h.

As pointed out on Figure 13 (the vertical stress is used as asingle parameter), the critical pressure is an increasingfunction of the over consolidation ratio. In other words, thelarger the over consolidation ratio is the less stable the well

will be.Finally, the plastic radius (corresponding to the critical

pressure) strongly decreases versus the consolidation ratio,the normally consolidated material being the case for whichthe plastic radius is infinite.

Figure 13

Influence of over consolidation ratio on critical pressure andplastic radius (B = 0).

CONCLUSION

In a range of confining pressures of several tens of MPa, softdeep materials (particularly shales and unconsolidated sands)clearly exhibit a ductile/plastic behaviour. Modelisation ofsuch materials is therefore of a strategic importance to predictstability of oil wells. Complex elastoplastic models includingseveral hardening mechanisms generally require heavynumerical tools (finite elements). Consequently, simplified

constitutive laws can be very usefull in day to day engineering.Such simplified laws have to be developped in the scope

of the Terzaghi's effective stress concept which allowssolving the poroplastic problem as an equivalent dryproblem. This concept is only valid for an incompressiblerock matrix (both elastically and plastically) and anassociated plastic flow rule.

For any boundary problem to be solved analytically, theshape of the plastic zone has first of all to be a priori known. Inthe case of the wellbore boundary problem, it will only bepossible for plane strain axisymmetric cases (vertical well andhorizontal isotropic stress) for which the plastic zone is a

circular ring. Secondly, to be analytically integrated, theconstitutive relations will be linear (linear relationship betweenstress and strain, linear yield locus, linear hardening law).

A linearised Cam-Clay has been proposed in the scope ofthis paper. It has been validated on homogeneous drained andundrained triaxial tests then a simplified version has beenintegrated in the case of the undrained boundary wellproblem. The main results can be summarised as follows: compared to a purely elastic solution, a strong relaxation of

the hoop stress is observed in the plastic zone; by contrast, the radial stress (for which the value is

imposed on the two boundaries) is only slightly affected by

plasticity;

20

10

1.32 1.28 1.24 1.2 1.16 1.12 1.08 1.04

Overconsolidation ratio

35 40 45 50 5530

Vertical stress (MPa)

Dimensionless plastic radius

Critical well pressure (MPa)

P = 40 MPa h= 30 MPa

0c

559

M

14

12

10

8

6

4

2

0.6 0.7 0.8 0.9 1

Criticalwellpres

sure(MPa)

0.96

0.19

0

M

2

4

6

8

10

12

14

16

18

0.6 0.7 0.8 0.9 1

Criticalwellpressure(MPa)

1000 MPa100 MPa

Elastic

Figure 12a

Critical instantaneous pressure versus M; effect of Skempton

coefficient (H= 1000).

Figure 12b

Critical instantaneous pressure versus M; effect of Skempton

coefficientH (B = 1000).


10/13


plastic hardening plays in favour of stability: the higher thehardening modulus, the less the hoop stress relaxation isand the less the well will be stable;

the elastic solution can be considered as the limiting case

of an elastoplastic model for which the hardening modulusis infinite. For this reason, the elastic solution will alwaysbe the more pessimistic and will always overestimate thecritical well pressure;

a low fluid compressibility decreases the relaxation of thehoop stress in the well vicinity and plays a negative role onstability;

the critical pressure is an increasing function of the overconsolidation ratio. In other words, the larger the overconsolidation ratio is, the less stable the well will be.

ACKNOWLEDGMENTSThis work has been performed in the scope of the Star Artepproject TOTAL FINA. The authors thank TOTAL for allowingto publish this paper.

REFERENCES

1 Charlez, P.A. (1997) Impact of Constitutive Laws onWellbore Stability. SPE Drill. and Compl .Jour.

2 Veeken, C., Walters, J.V, Enter, C.J. and Davies, D. (1989)Use of Plasticity Models for Predicting Borehole Stability.Symposium Rock at Great Depth, Pau 1989, A.A. Balkema.

3 Charlez, P.A. and Heugas, O. (1991) Evaluation of OptimalMud Weight in Soft Shale Levels. Proceedings of the 32ndUS Symposium Rock Mechanics as a MultidisciplinaryScience, Norman, A.A. Balkema.

4 Ewy, R.T. (1991) 3D Stress Effects in Elastoplastic WellboreFailure Models. Proceedings of the 32nd US SymposiumRock Mechanics as a Multidisciplinary Science, Norman,A.A. Balkema.

5 Zhou, S., Hillis, R. and Sandiford, M. (1996) A Supplementto a Study of the Design of Inclined Wellbores with Regardto Both Mechanical Stability and Fracture Insertion. Jour.Appl. Geophy., 36, 145-147.

6 Biot, M. (1973) Nonlinear and Semilinear Rheology of

Porous Solids.Jour. Geophy. Res., 78, 23.7 Lade, P.V. (1977) Prediction of Undrained Behaviour of

Sand.Jour. of the Geot. Eng. Div.

8 Coussy, O. (1989) General Theory of Thermoporo-elastoplasticity. Transport in Porous Media.

9 Coussy, O. (1995)Mechanics of Porous Media, John Wiley& Sons.

10 Charlez, P.A. (1991) Rock Mechanics Vol. I TheoreticalFundamentals, ditions Technip, Paris.

11 Loret, B. (1987) lastoplasticit simple potentiel. Manuelde rhologie des gomatriaux, ENPC Press.

12 Charlez, P.A. and Shao, J.F. (1993) Mechanical Behaviour ofSoft Deep Rocks. Geotechnical Engineering of Hard Soils -Soft Rocks, Athens 20-23 Sept. 1993, A.A. Balkema.

13 Burland, I.B. and Roscoe, K.H. (1968) On the GeneralisedStress Strain Behaviour of Wet Clay.Engineering Plasticity,Cambridge Heyman-Leckie.

14 Addis, M.A. and Wu, B. (1993) The Role of IntermediatePrincipal Stress in Wellbore Stability Studies: Evidence fromHollow Cylinder Tests. Int. Jour. Rock Mech. Min. Sc. &Geomech. Abstract, 30.

15 Guenot, A. (1987) Contraintes et ruptures autour des foragesptroliers. Proc. of the 6th ISRM Congr., Montreal, A.A.Balkema.

16 Jaeger, J.C. and Cook, N.G.W. (1979) Fundamentals of RockMechanics. McGraw Hill.

17 Graziani, A. and Ribacchi, R. (1993) Critical Conditions fora Tunnel in a Strain Softening Rock. Assessment andPrevention of Failure Phenomena in Rock Engineering ,Istanbul. A.A. Balkema.

18 Detournay, E. and Cheng, A.H-D (1988) PoroelasticResponse of a Borehole in a Nonhydrostatic Stress Field.Int.Jour. Rock Mech. Min. Sc.& Geomech. Abstract, 25 , 3,171-182.

Final manuscript received in July 1999

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APPENDIX

Resolution of the Wellbore Boundary Problem

Before drilling the whole domain is supposed to be in anelastic state (over consolidated material). Depending on thevalues of the well pressure, the points located around the holebecome plastically admissible. Given the axisymmetricalcharacter of the problem, the plastic zone is a circular ring ofradius R

p. Considering the partitioning rule and the plane

strain hypothesis, it can be assumed that both axial elasticand plastic deformations are null i.e.:

(A1)

On one hand, the condition (A1) on the elastic

deformation allows calculating the axial stress required toensure plane strain condition i.e.:

(A2)

Consequently, the mean effective stress is written:

(A3)

On the other hand, condition (A1) on the plastic axialdeformation implies that the yield locus is independent of

'zz. In the general case, Q is written:

(A4)

However, borehole stability only slightly depends on theintermediate stress (Addis and Wu, 1993). As in many casesthis is confirmed by the shape of the cuttings (Guenot, 1987)'

zzis the intermediate stress, (A4) can be rewritten:

(A5)

Considering (A3) and (A5), the yield locus (4) can besimplified (assuming a degree of consolidation less than 2):

(A6)

The strain hardening law (9) which depends only on thevolumetric strain remains unchanged. Expression (A6)

allows recalculating the incremental law (7) (in its inverse

form) which according to the incompressibility of the matrixcan be directly expressed in terms of effective plasticstresses, i.e.:

(A7)

with:

(A8)

Undrained Response

The undrained poroplastic response is the sum of theSkempton's effect and a contribution related to the plasticvolumetric strain. In the case of an elastically and plasticallyincompressible matrix the variation in pore pressure iswritten (Charlez, 1997):

(A9)

After elimination of dp between (A7) and (A9), oneobtains:

(A10)

where the new coefficients are such that:

(A11)

Ck B k B

(k k

Ck B k B

(k k

Ck B k B

(k k

Ck B k B

(k k

111 11 2 12

1

2

2

2

121 12 2 22

1

2

2

2

212 11 1 21

1

2

2

2

222 12 1 22

1

2

2

2

=+

=+

=+

=+

( )det

)

( ) det

)

( )det

)

( ) det

)

A

A

A

A

d d d

d d d

= +[ ]

= +[ ]

1

1

11 12

21 22

det

det

A

A

C C

C C

dd d

d dpB

BKu

B=+ +

+ +( ) ( )

( )1

3

AE H

M

A AE H

M

AE H

M

EE

B

B

B

B

B

B

B

11

2

12 21

2 2

22

2

2

1 1 1

3

1

2

1 1

9

1

2

1 1 1

3

1

2

1 1

=

++

+

= =

+

=

+ +

=

=

( )

( )

( )

'

with and

d d d d

d d d d

= +[ ]

= +[ ]

p A A

p A A

1

1

11 12

21 22

det

det

A

A

f P

M

MP

con

v B h B

con

( , , ) ( )

( )( )

=

+ + + + [ ]

1

2

32 1

Q = 1

2( )

Q zz zz= + + 1

2

2 2 2( ) ( ) ( )

P v B h B= + + + [ ]1

32 1 ( )( )

v B h B= + + 2 ( )

zz zz v zz= = + 0

zz zze

zz

p

zz

e

zz

p= = + = =0 0

561


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with:

(A12)

In the Bij

coefficients, the first term characterises theincremental skeleton stiffness (elastic and plastic), whereasthe second only depends on the fluid (via the Skempton'scoefficient B). Note that unlike matrix A, matrix C is notsymmetrical.

If the material is assumed over consolidated, the wholedomain is initially in elastic state. In that case, the initialconsolidation pressure is greater than the initial meanstress, i.e.:

(A13)

The analysis will be restricted to over consolidation ratioof less than 2. Prior to the appearance of the plastic zone, thestress state is purely elastic that is (Jaeger and Cook, 1979):

(A14)

K2 being a constant to be determined from boundaryconditions. It is interesting to note that the elastic stresses donot induce any variation in mean stress and consequently noundrained response (for a purely elastic path, the porepressure remains equal to the virgin value p

R). The structure

is loaded by decreasing gradually the well pressure (whilekeeping the geostatic stress

hconstant) until a plastic zone

of radiusRp

appears and extends. For =Rp

, the stress fieldis both elastically the stresses are compatible withEquation (A14) and plastically admissible (these samestresses verify the yield locus (A6)) i.e.:

(A15)

The overlying stresses are called initial effective plasticstresses, as they are associated with the current plasticextension. They are independent of the plastic radius. Let usnote that the radial stress increasing versus radial distance,the appearance of a plastic ring is subjected to the condition:

(A16)

otherwise the whole domain remains in a purely elastic state.In the plastic ring, the total stresses will be such that:

(A17)

where the increments are consistent with (A10). As forstresses, the initial plastic strains can be calculated fromHooke's law, i.e.1:

(A18)

whereas in the plastic ring they are expressed as follows:

(A19)

Calculation of Stresses and Strains

Replacing (A10), (A17), (A18) and (A19) in the equilibriumequation:

(A20)

and taking into account the compatibility equations:

(A21)

we get the following differential equation:

(A22)

The general solution to Equation (A22) is such that

(Graziani and Ribacchi, 1993):

(A23)

1 Equations (A14) and (A17) can be indifferently written using totalstresses (for a purely elastic path, pore pressure is unaffected under

undrained conditions).

u =C C m

E C

C C C C C C

C

h

11

22

11 1 2

1 212 21 12 21

211 22

11

1 1

4

2

+

= +

( )

( )( )

( ) ( )

,with

Cu u

Cu m

E

m E2E

C C C C

h

11

2

2 22 2

12 21 11 22

1 1

21

+

=

= ++

+

( )

det'( )with A

= =u u

+

= 0

= += +

d

d

= +

=+

1

1

E

E

h

h

( )

( )

= +

= +

d

d

p pM

Pw w hh

con< < + +

2

2

3

0

=

= + +

2

2

2

3

0

h

hh

con

MP

= +

=

h

h

KR

KR

2

2

2

2

2

2

P3

conh0 2> +

Pcon0

k B

kB

BA

BK B B

u u

ij

ij

B ij ji

1 2

3 1

3

1

3=

+=

+

= + =

( ) ( )

det

A

562


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The two constants C1 and C2 are determined by expressingcontinuity of radial stress and radial displacement (theyshould be derivable) across the plastic boundary. Consideringthe linearity of the incremental law, these conditions are

equivalent to:

(A24)

Taking into account the compatibility equations (A21) oneobtains:

(A25)

By introducing the dimensionless variable x = /Rp, thedeformations take the form:

(A26)

with:

(A27)

Stresses in the plastic ring can now be easily calculated by

replacing (A26) in (A10), i.e.:

(A28)

with:

(A29)

Finally, the plastic radius is calculated by expressing thewell boundary condition, i.e.:

(A30)

In the elastic zone (which extends from the plastic radiusto infinity), constant K2 (Eq. (A14)) is determined byexpressing the continuity of the radial stress through theplastic boundary, i.e.:

(A31)

= ==

+

= +

R

R

R

R

Rp

p

h

p

p

h

p

1

1

2

2

2

2

2

2

2

2

( )x p xR

Rw w w

p

= =with

d

d

= = +[ ]

= = [ ]

1 1 2 2

1 2

1

1

F x F x

F x F x

( ) ( )

( ) ( )

= + C C

= + C C

1

1

11 12

21 22

det

det

A

A

d d

d d

+[ ]

+[ ]

F x xm x

C

F x xm x

C

1

2 1

2

11

11 1

2

2 1

1

11

11 2

11

1

1 1

11

1

1 1

1

1

2

2

( ) ( )( )

( )( )

( ) ( )( )

( )( )

=

+

+

=

+

+

( ) ( ) ( )

( ) ( ) ( )

x F x F x

x F x F x

= [ ]

= [ ]

1 1 2 2

1 2

C =R

m

C

C =

R

m

C

p

p

1 1

2 1

2

11 1

2 1

2 1

1

11 2

1

2

11 1

11 1

+

+

+ +

( )( )

( )( )

( )( )

( )( )

== =

= = =

R

u up

e

d 0

563

analytical solution of wbs problem using camclay model charlez

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