rock mechanics.pdf

Rock Mechanics Self Learning Package Sugar Land Learning Center

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SUGAR LANDLEARNING CENTER

Rock MechanicsSELF LEARNING COURSE

USEFUL PRE-REQUISITES

Basic understanding of drilling terms and proceduresStuck Pipe Self Learning Package


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Table of Contents

OBJECTIVES………………………………………………………………………….. 3INTRODUCTION……………………………………………………………………... 4

SECTION I:FUNDAMENTALS OF ROCK MECHANICS………………..…………………….. 5

THE STRESS IN THE EARTH BEFORE WE DRILL A WELLBORE……………...…………6THE STRESS IN THE EARTH AFTER WE DRILL A WELLBORE….………………...……15ROCK FAILURE……………………………..……………………………………………...….20FRACTURING……...…….……………………………………………………………………..26SANDING…………………………………………………………………………………...…..27

SECTION II:ROCK MECHANICS FROM LOG IMAGES…………..………………………......28

THE GEOMETRY OF BOREHOLE SHEAR FAILURES ….…………………………………29THE GEOMETRY OF BOREHOLE TENSILE FAILURES ….…………………….…………39IDENTIFICATION OF ROCK MECHANICS FEATURES ON BOREHOLE IMAGES……..44

SECTION III:WELLBORE STABILITY PLANNING……………………………………...……...50

CALCULATING CONDITIONS FOR ROCK FAILURE……….…….……………………….51BOREHOLE STABILITY IN DEVIATED WELLBORES….…………………………………57THE MECHANICAL EARTH MODEL (MEM)………………….……………………………60

SECTION IV:PRACTICAL WELLBORE STABILITY MONITORING……….………...……...67

THE MEM AND PERFORM………………..………………………….……………………….69REAL TIME WELLBORE STABILITY MANAGEMENT....…………………………………70

BRAIN TEASERS……………………………………………………………………...81

APPENDIX A……………………………………………..…………………………... 83APPENDIX B……………………………………………………………………….….91APPENDIX C……………………………………………………………………...…...94APPENDIX D………………………………………………………………………......98

ANSWERS TO BRAIN TEASERS……………………...…………………....……....102

REFERENCES….………………………………………………………..…………... 105


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Objectives

Upon completion of this training module you should be able to:

� Describe the stresses in the earth before we drill a borehole� Describe the stresses in the earth after we drill a borehole� Describe how a rock fails when we drill it� Understand the features of an extended leak off test� Explain the geometry of borehole shear failures and how to detect them on lateralog images� Explain the geometry of borehole tensile failures and how to detect them on lateralog images� Understand how a mud weight window is calculated� Understand the differences in borehole stability in deviated wells from vertical wells� Describe the basic components of the Mechanical Earth Model and how the input data is

derived� Describe the different types of cavings that are found at the wellsite� Detect the 4 most common wellbore instability mechanisms from surface and downhole

signatures and offset well data� Propose remedial actions for the common instability mechanisms that occur at the wellsite


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Introduction

Rock Mechanics can be defined as:

“The theoretical and applied science of the mechanical behavior of rock to the forcefields of its physical environment.”

Rock Mechanics can be used in the following main applications:

� Wellbore Stability Prediction / Drilling Optimisation� Sand Production / Completion Design� Hydraulic Fracturing / Reservoir Stimulation Design

Wellbore stability is quoted as costing the industry between 0.6 and 1.0 billion dollars per year.Shell estimated in 1998 that it has annual wellbore instability costs of $200 million.

Drilling costs typically require 70% of field development capital. Therefore cutting costs cansignificantly reduce field development costs.

A example of cost savings from improved handling of wellbore stability is in the Cusiana field,Colombia. Well costs had been in the order of tens of millions of dollars. After geomechanicalsolutions had been made, great reductions were made in these costs.

This Self Learning Package focuses primarily on Wellbore Stability during drilling. However,some discussion of hydraulic fracturing mechanism and sanding is also included.

Section I Fundamentals of Rock Mechanics: teaches the basic concepts in Rock Mechanics.This includes an explanation of the stresses in the earth, together with the stresses atthe wellbore wall. The section ends with analysis of rock failure.

Section II Rock Mechanics from Log Images: examines the different conditions for borehole failure during drilling and gives examples of these in log images.

Section III Wellbore Stability Planning: discusses wellbore stability planning (how mud weight windows are calculated) and the MEM (Mechanical Earth Model).

Section IV Practical Wellbore Stability Monitoring: presents a hands on approach to Real Time Wellbore Stability and includes detection and remedial solutions in real time.


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SECTION I:Fundamentals of Rock Mechanics


6

The Stress in the Earth before we Drill a Borehole

Stress is the force per unit area acting at a point in a material.

It has the same units as pressure (psi) but differs from pressure:

When a fluid exerts a pressure we are only concerned with its magnitude (a column of liquidexerts a pressure that acts in all directions).When a force is exerted on a material, this causes a stress within the material. We are concernedwith the both magnitude and the direction of this stress.

Before we drill a borehole the rock in the earth is in a state of equilibrium.

In rock mechanics, we have 3 stresses in a material that are perpendicular to each other:

σ1 Maximum Principal Stressσ2 Intermediate Principal Stressσ3 Minimum Principal Stress

These can act in any orientation in 3 dimensions.

Also (and independently from the above 3 stresses) tfollows:

σv Principal stress acting in the vertical axisσh Principal stress acting in the horizontal axisσH Principal stress acting in the horizontal axis

σH is the maximum of the 2 horizontal stresses and (ie σH > σh )

The term “Principal Stress” is described in Appendix

σσσσ1

σσσσ3

σσσσ2

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he earth’s principal stresses are defined as

σh is the minimum.

A


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Slip Fault Regime

σH = σ1

σv = σ2

σh = σ3

Thrust (Reverse) FaultRegime

Gentle sloping

σH = σ1

σh = σ2

σv = σ3

Normal FaultRegime

Steep sloping

σv = σ1

σh = σ3σH = σ2

Figure 1.1: Tectonic dependence on earth stresses


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The earth’s stresses are related to a number of different variables including:

� Tectonic Setting� Depth� Pore pressure� Lithology� Temperature� Structure

The relationship between stress and the above variables is complicated due to local geographicaldifferences between basins and interdependence of the above variables.

However, it can be seen that:

Intrabasin stress variations are correlated with lithology and pore pressureInterbasin stress variations are correlated with tectonic setting and diagenesis (consolidation

and cementation).

Tectonic Setting (see Figure 1.1):

a) Normal fault regime, the vertical stress (σv) is the maximum principal stress (σ1). σv > σH > σh

b) Thrust (reverse) fault regime, the horizontal stress (σH) is the maximum principal stress (σ1).σH > σh > σv

c) Slip fault regime, the horizontal stress (σH) is the maximum principal stress (σ1).σH > σv > σh


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Depth:

In practice it is observed that at shallow depths the minimum principal stress is the verticalstress. Here a hydraulic fracture is most likely to occur in a horizontal plane.

At greater depths the principal stresses generally follow fault regimes described in Figure 1.1.

For example, in a normally pressured sedimentary basin, the minimum stress σ3 is most probablyin the horizontal plane at depths greater than 3300 ft (Plumb). This stress scenario is probablythe most common to be found in the oilfield.

Generally the vertical stress is depth dependent and expressed as follows:

=H

v gdHH0

)(ρσ

ρ(H) = density of rock at given depth H (from density log or core sample)g = gravity

Hence the value of this stress component is obtained from the integration of a density log.σv is known as the Overburden Gradient and varied from 0.8 psi/ft in young, shallowformations (eg Gulf Coast) to 1.25 psi/ft in high-density formations.Given the density of quartz (1.65 g/cc) the overburden gradient ranges between 1.0 and 1.1 psi/ftfor brine saturated sandstone with porosity ranging between 20% and 7% respectively.

If tectonic loading is present yielding horizontal compressive forces then we cannot use thisexpression for vertical stress.


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Pore Pressure:

Pore Pressure supports a portion of the total applied stress in a rock.

In general:

Total stress (in given direction) = Effective Stress of Rock Grains (given direction) + Pore Pressure

If a formation is “normally pressured” the pore pressure mechanism can be described asfollowing:

Sediment burial →→→→ full pore fluid escape →→→→ porosity decreases →→→→ effective rock stress increases→→→→ pore pressures are hydrostatic (normal)

A “Cam-Clay” law under these (normal) conditions shows how porosity evolves with depth3:

[ ])(ln11 0

wbgz ρρλφφ

φφ −−

−=

−

φ = Porosity at a given depthφ0 = Initial porosity of materialλ = Compressibility coefficient (constant)ρb = Density of rockρw = Density of waterz = Vertical Depth

(ie at surface z = 0 and ln0 = –∞, hence φφ−1

= +∞ and φ → ∞ or 100% porosity)


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If a formation is “over-pressured” the pressure in the formation is greater than the pressureexerted by a column of water at that same depth.

3 mechanisms have been proposed:

a) Loading mechanisms:Sediment burial →→→→ pore fluid escape fully restricted →→→→ porosity & effective stress are bothconstant →→→→ pore pressures increases at the same rate as the overburden (ie overpressure)

– Restricted fluid escape may occur in thick low-permeability shale sections.– Depends on: rate of sediment compaction Vs rate at which fluid is expelled (higher

this ratio the more likely that overpressure occurs).– Also depends on: pore fluid compressibility. The lower the compressibility of the pore fluid,

the worse the overpressure (for all else equal). This is because the overallrock is stiffer and this will lower the efficiency of pore fluid expulsion.

b) Unloading mechanisms7:

When unloading occurs, stress is relieved and sediments do not revert to their original state.

2 natural mechanisms occur that can cause ‘unloading’.

(i) Aquathermal expansion / hydrocarbon generation / mineral dehydration (smectite → illite) / osmosis → sealed formation → fluid-volume increase can result in rapid pore pressure increases that unload the rock grain matrix.

(ii) Uplift / Erosion → unloading rock grain matrix → sealed formation → formation has same pore pressure as before but due to closed system is abnormally pressured compared with neighbor formations at same depth.

c) Tectonic Stress7:

Horizontal tectonic stress → Very low permeability seal (eg non-communicating fault) → rate ofpore fluid escape cannot keep up with additional tectonic stress → system does not fracture /fault to relieve stress → pore pressure increases.


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Terzaghi proposed the following relationship:

σ’ = σ – p

σ’= Effective rock stress (in a given direction)σ = Total applied stress (in a given direction)p = Pore pressure

Biot proposed an equation to account for the following: any change in pore pressure isaccompanied by variation in pore volume – this change in pore volume affects the overallmechanical response of the rock:

σ’ = σ – αp (1.1)

α = Biot’s constant (varies between 0 and 1)

αααα describes the efficiency of the fluid pressure in counteracting the total applied stress.

If αααα = 1 this means that the pore fluid has maximum efficiency in counteracting the totaloverburden stress (σv) and therefore implies that the effective stress of the rock is lower, apessimistic condition for rock failure.

If αααα < 1 this means that the pore fluid is less efficient in counteracting the total overburdenstress (σv) and the effective stress of the rock is greater for all else equal.

αααα is close to 1 for stiff rocks and close to 0 for rocks with low stiffness

In Petroleum Rock Mechanics always effective rock stress σ’ (not total rock stress) is used incalculations.


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Lithology:

The effect of lithology on earth stresses is complicated.

To analyze the effect on horizontal stresses, we define Rh:

Rh = σh / σv

σh = horizontal stressσv = vertical stress

Rh is merely the normalized horizontal stress - we cannot just compare horizontal stresses for thevarious lithologies. We need to remove the dependence of vertical stress Sv.

For example if a lithology has a high σh , this could be due to there being a high σv . The high σvcould have the affect of “squeezing out sediment laterally”. Assuming there is a lateral force (atlarge distances) to counteract this, σh will then be higher.

Relaxed-state Basins:Rh Carbonates < Rh Sandstones < Rh Siltstones < Rh Mudstones < Rh Shales

(In the North Sea Rh shales is up to 15% higher than Rh for sandstones.).The poisson’s ratio ν for sands is generally lower than for shales. This implies that shales havegreater affinity for lateral expansion FOR ALL ELSE CONSTANT.(For an explanation of ν please refer to appendix B)

The lateral expansion encounters lateral forces of resistance causing the greater lateral stress inshales.

Compressed-state Basins:Rh Shales < Rh Sandstones

This is characterized by higher stress magnitudes in elastically stiffer rocks. Sandstones aregenerally stiffer than shales.

Pore pressure effects:Rh generally increases for overpressured formations (in relaxed-state basins) due to theformation becoming more elastic.


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Fractured rocks:Rh decreases (is generally lower than 0.7) in naturally fractured lithologies (eg carbonates).These rocks are more brittle and lateral stresses are released as microfractures.

Temperature:

A reduction of temperature (eg injection of cooler injection fluid) causes a reduction in the rockstress.The cooler temperature acts to contract the rock – if the rock is confined on all sides (which italways is) by a stress of some kind, the rock will contract leaving a ‘void space’ and weaknesswhich could favor a tensile failure as a reduction in the rock stress.

Structure:

In a typical anticline structure the lowest stress magnitudes are found at the crest of the structurewhile greater stresses are found along the flanks8.

The crest is usually under a greater tensile state whereas the flanks are under a greatercompressive state.


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The Stress in the Earth after we Drill a Borehole

Before a wellbore is drilled the rock is in a state of equilibrium. This state is called the “InitialState”.

The stresses in the earth under this condition are known as the Far Field Stresses (σσσσh , σσσσH , σσσσv )or in-situ stresses.

When a well is drilled it introduces a perturbation in the initial stress field. The perturbationcauses a ‘new’ set of stresses known as wellbore stresses that act on the formation at the mud-formation interface.

The far field stresses have therefore been altered near the wellbore due to the removal of rockand substitution of drilling fluid during the creation of the borehole6.


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Figure 1.2:The 3 Wellbore Stresses

Figure 1.3: Variation of Wellbore Stresses away from the wellbore wall


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The wellbore stresses are described as follows:

σσσσr = Radial Stressσσσσt = Tangential Stress (or Hoop Stress)σσσσa = Axial Stress

Figure 1.2 shows the 3 wellbore stresses.

Figure 1.3 shows the how these wellbore stresses vary with distance away from the boreholewall.This shows that the wellbore stresses diminish rapidly from the borehole wall converting to farfield stresses. This makes sense because away from the wellbore the rock is in an unperturbedstate (with stresses σh , σH , σv ).σr is seen to converge to σh and σt is seen to converge to σH at very large distances from thewellbore. This is not always the case and is described in appendix C.

Appendix C describes these wellbore stresses in more detail.Appendix C shows that at the wellbore wall itself :

σr = pw (1.2)

( ) ( ) whHhHt pcos2 θσσ2σσσ −−−+= (1.3)

( ) cos2 θσσ2σσ νhH −−= va (1.4)

Where:

σσσσr = radial stressσσσσt = tangential stressσσσσa = axial stressσσσσv = vertical stressσσσσh = minimum horizontal stressσσσσH = maximum horizontal stresspw = mud weightθθθθ = Angle between a point on the circumference of the wellbore and the direction ofmaximum Horizontal Stress (σσσσH)

Arbitrary point on the circumferenceof the wellbore wall

θ

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σH


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Thus:

• The radial stress σr depends on the mud weight (pw)It acts in all directions perpendicular to the wellbore wall as shown in figure 1.2

• When the mud weight is high, then at the wellbore wall σr is large and σt is small andvice versa. This can be seen from examination of figure 1.3 and equations 1.2 and 1.3

• The tangential (hoop) stress σt circles the borehole. Its magnitude depends on:- The far field stresses (σh and σH)- The mud weight (pw)- The azimuthal position around the wellbore (θ)

Hence σt is a direction dependent variable.

When θ = 0°

(σσσσt) θθθθ = 0 = 3σσσσh – σσσσH – pw (1.5)

When θ = 90°

(σσσσt) θθθθ =90°°°° = 3σσσσH – σσσσh – pw (1.6)

The maximum tangential stress occurs in a position that is 90° from the maximum horizontalstress σH (equation 1.6), ie in the orientation of the minimum horizontal stress σh


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eg.Wellbore pressure = 3000 psi (in equilibrium with the pore pressure of the reservoir) = pw

σh = 3500 psi, σH = 5000 psi

The equations lead to values for the effective tangential stress (σt – p) of:

σt = (3x3500) – 5000 – 3000 – p = -500 psi (θ = 0°)σt = (3x5000) – 3500 – 3000 – p = 5500 psi (θ = 90°)

(assuming pore pressure p = 3000 psi)

The tangential stress is in compression at θ = 90° and in tension at θ = 0.The latter result indicates the possibility for the occurrence of tensile failure in a directionperpendicular to the minimum stress, solely as a result of drilling the borehole.

The wellbore stresses (given by equations 1.2, 1.3, 1.4) are extremely important because it isat the borehole wall where problems occur that affect the drilling, not far from the wellbore.These wellbore stresses form the basis for wellbore shear failure (described in the next section)

The mud weight (pw) is the major parameter that the Drilling Engineer has in order to influencethese wellbore stresses.


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Rock Failure

For the purpose of this account, rocks behave as elastic materials. If a stress is applied to a rock,its stress/strain relationship can be described as follows:

- Elastic Region: If the stress is relieved, the specimen will return to its original state- Yield Point: The point beyond which permanent changes will occur. The sample

will not return to its original state upon stress relief- Ductile Region: A region in which the sample undergoes permanent deformation

without losing the ability to support load.- Brittle Region: A region in which the specimens ability to withstand stress

decreases rapidly as deformation is increased

Shear Failure:

A shear stress is caused by 2 perpendicular stresses acting in the same plane (see Appendix A)When this shear stress exceeds a threshold for the material, shear failure occurs in the material.

Shear failure will occur when :

σ1’ ≥ C0 + σ3’ tan2β (1.7)

Where:

C0 = 2So tanθθθθ = Unconfined Compressive Strengthσσσσ1’ = Maximum Effective Stressσσσσ3’ = Minimum Effective Stressββββ = Angle between the normal stress σσσσn and the maximum effective stress σσσσ1’

ββββ = 45°°°° + φφφφ/2 (see Appendix D for derivation)φφφφ = Friction angle (Where µµµµ = Tan φφφφ and µµµµ is the coefficient of friction)

σσσσ1’ and σσσσ3’ are any 2 wellbore stresses (both different in magnitude, 1 will be max the other min)

If the above inequality is satisfied then the rock will fail through shear failure. Equation 1.7 is the formalcriterion for shear failure, known as Mohr-Coulomb. (SeeAppendix D for full derivation of this)

From equation 1.7

• Pure shear failure depends only on the minimum and maximum principalstresses σσσσ1 and σσσσ3 and not on the intermediate stress σσσσ2.

• The failure condition is approached as the difference between the twoprincipal stresses σσσσ1 and σσσσ3 becomes larger (the diameter of the hemispherein figure 1.4 increases).


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At the point of rock failure the maximum effective stress σ1’ is also termed the Peak Strength.This is the maximum strength, above which the rock fails. In this case a constant lateral stress(confining stress) σ3’ is present.

If there is no confining stress acting on the rock then σ3’ = 0In this case the peak strength σ1’ is known as the unconfined compressive strength Co

Graphically, Co is seen to be the diameter of the largest possible Mohr circle for when σ3’ = 0(see figure 1.4)

If Co is known (through rock tests), and we know the values of σσσσ1 , σσσσ3 and θθθθ then we candetermine whether the rock will fail through shear failure.

The direction of the shear fracture is always inclined at an angle (0°°°° – 45°°°°) to the directionof maximum stress, σ1.

(Refer to Appendix D for discussion)

This makes perfect sense because when:σ1 = σv Normal fault regime Fault is steep sloping (wrt horizontal plane)σ1 = σH Thrust fault regime Fault is gentle sloping (wrt horizontal plane)


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The criterion for shear failure is when the Mohrcircle touches the failure line

Failure line

σ3’=0

Figure 1.4: The graphical representation of shear failure

Mohr Circle

σn

τ

σ1’
σ3’
2β
φφφφ
Co

So


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The Mohr-Coulomb criterion for the condition of shear failure can also be expressed as:

τ = S0 + (σn – p) Tan φ

wherep = pore pressureφ = Friction angle (Tan φ = µ µ = Coefficient of friction, see appendix D for explanation)

This is a convenient expression for shear failure as it can be represented graphically in the samemanner as previously.The Mohr circle merely defines the shear stress of a material for a given normal stress σn.

Figure 1.5 shows the graphical representation for this expression – the y-axis intercept being S0

and the gradient being Tanφ.

The graph shows that:

• The smaller the friction angle φ the more likely the shear failure criterion is reachedShales have a low coefficient of friction µµµµ and hence low friction angle φφφφ.Quartz has a higher friction angle (φ = 50°)

• The greater the difference between the two effective stresses (σ1’ and σ3’) the larger theMohr circle produced and the more likely the shear failure (as the surface of the circle iscloser to the failure line.)

• An increase in pore pressure decreases the effective stresses, causing the Mohr Circle to beshifted to the left and increasing the likelihood of shear failure

• If we subject the rock to a higher confinement σ3’ we can apply a much greater stress σ1’before shear failure is initiated.This is a peculiar characteristic to many porous rocks: they effectively become strongerthe more they are confined.(Some very soft rocks, clays, and some strong and tight igneous rocks do not behave likethis and their corresponding failure envelopes appear near horizontal.11)The low φ of clay → rock strength does not increase much with increased rock confinementHigher φ of quartz → rock strength increases with increased rock confinement.

• Granular rocks such as sandstones derive a great part of their strength from internal friction– they become stronger the more they are confined. In contrast clays exhibit little internalfriction but derive their strength almost entirely from cohesion So between grains and theirfailure lines appear as near horizontal lines.


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Tensile Failure:

When a formation is subject to a tensile stress, the grains are pulled apart in the direction of thetensile stress.A crack perpendicular to the tensile stress is created, and the formation fails in tension6.Ie the orientation of the crack is parallel to the direction of maximum horizontal stress and willopen in the direction of the minimum horizontal stress.

The criterion for tensile failure is given by:

σ3’ ≤ –T0 (1.8)

where:

TO = Tensile Strength of the materialσ3’ = Minimum Effective Stress

The minimum effective stress (σ3’) in this case is any one of the 3 wellbore stresses (ie σ3’ canbe radial stress, tangential stress or axial stress – see “Geometry of Tensile Failures” in the nextsection.)

Shear and tensile failure are completely independent of one another.

Most rocks are exceptionally weak in tension. All are far weaker in tension than they are incompression (i.e. similar to concrete in buildings, which must be reinforced with steel bars if it isto carry any tensile loads).


Formation Breakdown Pressure pbdpw

Tensile Strength To

Figur

Leak off Pressure

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Fracture Closure Pressure = σh

e 1.5: Extended Leak off Test (Min

Pumping stops

ifra

Fracture opening pressure

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Time

c)


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Fracturing

Fracturing is related to tensile failure.

The condition for tensile failure was given by equation 1.8.Hydraulically induced fractures will develop in directions perpendicular to the least in situprincipal stress.

The fracture is typically held open by the fluid pressure acting against the formation stresses; ifthe well pressure is reduced the fracture will close. In reservoir stimulation proppants arepumped into the fracture so that fracture will close on the proppants and acts as a highpermeability flow channel to the well.

Figure 1.5 shows a typical extended leak-off test or minifrac.When fluid is pumped into the formation the wellbore pressure pw builds up linearly. When mudbegins to leak off, pressure increases less quickly – the pressure departs from a linear increase.The point where this change occurs is the leak off pressure.

More fluid is pumped and the rock is broken down at a formation breakdown pressure pbd.

As the fracture forms pressure usually drops. After the fracture is extended for a while pumpingis stopped and the fracture closes as the pressure levels off. The pressure at that point is thefracture closure pressure and equals the minimum horizontal stress σσσσh.

The inflexion point in the curve (the fracture closure pressure) physically represents thetransition from linear to radial flow as the fracture closes.

In a 2nd cycle fluid can be pumped again to reopen the fracture. The difference between pbd andthe pressure required to reopen gives the tensile strength of the formation To.10

In hydraulic fracturing, fluid is injected above the far field stress. If the formation is highlypermeable the fracture fluid will have a greater tendency to leak off into the formation reducingthe fracture growth. For this reason fracture fluids often contain polymers/gels for viscosificationthus reducing leak off.

The fracture will propagate when the stress intensity factor KI at the fracture tip reaches thecritical stress intensity factor KIc

KIc is given by:

KI = (pf - σσσσ3) √√√√ππππL

KI = stress intensity factorPf = Pressure in the crackσ3 = Minimum Horizonatal StressL = Half Length of the crack


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Sanding

Sand production is the production of small or large amounts of solids together with the reservoirfluid.Sand production may lead to 3 types of problems:

a) Wear / erosion of the production equipment downhole and on the platforms.

b) Problems with the stability of the wellbore and production cavities leading to sand up,abandonment of a well. Casing collapse may also result from excessive sand production

c) Handling significant amounts of sand at the rig (environmental problems with dirty sand)

There are 2 mechanisms for sand production:

a) Shear Failure: Related to too low well pressure (ie too high pressure drawdown)

A plane in the near wellbore region is subjected to a higher shear stress than it cansustain

b) Tensile Failure: Related to too high production rate

If the production rate is too high, this will increase the friction due to the drag forces onthe grains of the formation. This will lead to tensile failure.Fines migration (such as clays) can reduce permeability in the near wellbore regionwhich can also increase fluid drag forces.

In practice the 2 mechanisms will interact – a formation altered by shear failure may be muchmore susceptible to fluid drag (hence tensile failure).Shear failure is the mechanism that generally gives the catastrophic amounts of sand; sandproduction by tensile failure is less malignant. This is because as the cavity grows, the fluidgradient becomes smaller and production rate tends to decrease.Tensile sand production therefore has a self-stabilizing effect.

How can sand production be controlled ?

a) Natural Completion of the well: where there are perforations through the casing. Sandproduction is controlled by the control of production parameters such as well pressure(drawdown) and production rate.

b) Active Sand Control: including gravel packing, sand screens, chemical consolidation. Activesand control methods are expensive and are thought to decrease the productivity of the well.


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SECTION II:Rock Mechanics from Log Images


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The Geometry of Borehole Shear Failures

Section I showed that shear failure will occur when 2 perpendicular stresses of differentmagnitudes are acting on each other in a plane.

Assuming the rocks that we drill are linear elastic, the largest stress differences will occur at theborehole wall.

Therefore we would expect rock failure to initiate at the borehole wall.

At the borehole wall, there are 3 wellbore stresses (σr , σt , σa)

Their stress magnitudes can be ordered in 6 different ways:

a) σa > σt > σr

b) σt > σa > σr

c) σa > σr > σt

d) σr > σa > σt

e) σr > σt > σa

f) σt > σr > σa

All 3 wellbore stresses are orthogonal.

In each of these 6 different scenarios we have a maximum and minimum wellbore stress.For example in a) the maximum stress is σa and the minimum stress is σr.

The difference between these 2 stresses will govern the magnitude of the shear stress. (Seeequation A4 Appendix A, the equation for shear stress).Pure shear failure does not depend on the intermediate stress σt (see Section I)

If this shear stress is large enough to satisfy the failure condition (see equation 1.8), then rockfailure will occur in the wall of the borehole.

The most common scenarios are conditions a) b) and c).

It should be noted that post failure behavior is very difficult to model.


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Figure 2.1: Wellbore Failure in RAB images


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a) Shear Failure Shallow Knockout (ssko):

σa > σt > σr

This mode of failure occurs when the axial stress (σa) is maximum and the radial stress (σr) isminimum.

Geometry and Orientation of ssko Failure:

The failure will occur in the radial / axial plane because the maximum (σa) and minimum (σr)stresses are oriented in this plane (a vertical plane).

The thickness of the failure (ie the circumferential coverage) will be small as this plane (theplane tangent to the circumference of the borehole) contains the tangential stress σt which doesnot affect the ssko failure mechanism.

The orientation of ssko failures will always be in the direction of minimum horizontal stress.

This is because the failure will occur preferentially in the direction where the axial stress isgreatest and the radial stress is smallest. This condition gives a maximum shear stress.

The radial stress is constant ( = pw) for all azimuthal orientations (see equation 1.11),Reference to equation 1.13 shows that σa is maximum when θ = 90° (→ cos2θ = -1)θ is defined as the direction from the maximum horizontal stress. Hence when θ = 90° this is inthe direction of minimum horizontal stress.

The image in Figure 2.1 shows an example of ssko. The failure is highlighted and is shown tocover a small circumferential area. Also we can see that the direction of minimum horizontalstress is NE-SW.The ssko failure can be confused with a vertical fracture due to the small circumferential area.

σr

σa

Shear failure shallow knockout occurring in theplane of σa / σr (ie in a plane perpendicular to theborehole wall).The failure region is given by the light blue color.


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Figure 2.2: Appearance and geometry of Shear Failure Wide Breakout on RAB images


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b) Shear Failure Wide Breakout (swbo):

σt > σa > σr

This mode of failure occurs when the tangential stress (σt) is maximum and the radial stress (σr)is minimum.

Geometry and Orientation of swbo Failure:

The failure will occur in the radial / tangential plane because the maximum (σt) and minimum(σr) stresses are oriented in this plane (the horizontal plane).

This failure is called a breakout. It is generally wide because the failure covers a large arc, from30° to 90°.

The orientation of swbo failures will always be in the direction of minimum horizontal stress.

This is because the failure will occur preferentially in the direction where the tangential stress isgreatest and the radial stress is smallest. This condition gives a maximum shear stress.

The radial stress is constant ( = pw) for all azimuthal orientations (see equation 1.11),Reference to equation 1.12 shows that σt is maximum when θ = 90° (→ cos2θ = -1)When θ = 90° this is in the direction of minimum horizontal stress.

Figure 2.2 shows an example of swbo. The failure is highlighted and is shown to cover a largecircumferential area (30° to 90°). The direction of minimum horizontal stress can be determinedfrom the location of the breakout on the image.

=

σr

σt

Shear failure wide breakout occurring in the planeof σt / σr (ie in a horizontal plane). The failureregion is given by the light blue color.


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Figure 2.3: Shear Failure High Angle Echelon failures on RAB images


35

c) Shear Failure High-Angle Echelon (shae):

σa > σr > σt

This mode of failure occurs when the axial stress (σa) is maximum and the tangential stress (σt)is minimum.

Geometry and Orientation of shae Failure:

The failure will occur in the axial / tangential arc because the maximum (σa) and minimum (σt)stresses are oriented in this arc (the arc of the borehole wall).

This failure forms high-angle fractures that cover up to a quarter of the borehole circumference.

In determining the orientation of shae failures, σa is maximum when θ = 90° (see equation 1.13)and σt is minimum when θ = 45° (see equation 1.12).

Figure 2.3 shows an example of shae. The failure originates in the orientation of maximumhorizontal stress and extends away at high angle (60°).

It is very difficult to predict the post failure behavior the shae failures.

=

σt

σa
Shear failure high angle echelon occurring in thearc of σa / σt (ie in a the arc of the borehole wall).The failure originates at the azimuth of themaximum horizontal stress and extends away athigh angle (60°).
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d) Shear Failure Narrow Breakout (snbo):

σr > σa > σt

This mode of failure occurs when the radial stress (σr) is maximum and the tangential stress (σt)is minimum.

Geometry and Orientation of snbo Failure:

The failure will occur in the radial / tangential plane because the maximum (σr) and minimum(σt) stresses are oriented in this plane (the horizontal plane).

This failure is called a breakout. It is generally narrow because the failure covers an arc< 30°.

The orientation of snbo failures will always be in the direction of maximum horizontal stress.

This is because the failure will occur preferentially in the direction where the tangential stress isgreatest and the radial stress is smallest. This condition gives a maximum shear stress.

The radial stress is constant ( = pw) for all azimuthal orientations (see equation 1.11),Reference to equation 1.12 shows that σt is minimum when θ = 0° (→ cos0 = 1)When θ = 0° this is in the direction of maximum horizontal stress.

Figure shows an example of swbo. The failure is highlighted and is shown to cover a largecircumferential area (30° to 90°). The direction of minimum horizontal stress can be determinedfrom the location of the breakout on the image.

=

σr

σt

Shear failure narrow breakout occurring in theplane of σt / σr (ie in a horizontal plane). Thefailure region is given by the light blue color.


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e) Shear Failure Deep Knockout (sdko):

σr > σt > σa

This mode of failure occurs when the radial stress (σr) is maximum and the axial stress (σa) isminimum.

Geometry and Orientation of sdko Failure:

The failure will occur in the radial / axial plane because the maximum (σr) and minimum (σa)stresses are oriented in this plane (a vertical plane).

The thickness of the failure (ie the circumferential coverage) will be small as this plane (theplane tangent to the circumference of the borehole) contains the tangential stress σt which doesnot affect the sdko failure mechanism.

The orientation of sdko failures will always be in the direction of maximum horizontal stress.

This is because the failure will occur preferentially in the direction where the radial stress isgreatest and the axial stress is smallest. This condition gives a maximum shear stress.

The radial stress is constant (=pw) for all azimuthal orientations (see equation 1.11),Reference to equation 1.13 shows that σa is minimum when θ = 0° (→ cos0 = 1)θ is defined as the direction from the maximum horizontal stress. Hence when θ = 0° this is inthe direction of maximum horizontal stress.

The sdko failure can be confused with a vertical fracture due to the small circumferential area.

σr

σa

Shear failure deep knockout occurring in the planeof σa / σr (ie in a plane perpendicular to theborehole wall).The failure region is given by the light blue color.


38

f) Shear Failure Low-Angle Echelon (slae):

σt > σr > σa

This mode of failure occurs when the tangential stress (σt) is maximum and the axial stress (σa)is minimum.

Geometry and Orientation of slae Failure:

The failure will occur in the axial / tangential arc because the maximum (σt) and minimum (σa)stresses are oriented in this arc (the arc of the borehole wall).

This failure forms low-angle fractures.

In determining the orientation of slae failures, σt is maximum when θ = 90° (see equation 1.12)and σa is minimum when θ = 0° (see equation 1.13). This could be the reason why the high-angle echelon failures are spread over a larger circumferential area than knockouts or breakouts.

The slae failure originates in the orientation of minimum horizontal stress and extends away atlow angle.

=

σt

σa
Shear failure low angle echelon occurring in the arcof σa / σt (ie in a the arc of the borehole wall).The failure originates at the azimuth of theminimum horizontal stress and extends away at lowangle.
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The Geometry of Borehole Tensile Failures

When a formation is subject to a tensile stress, the constituent grains are pulled apart in thedirection of the tensile stress.

A crack perpendicular to the tensile stress is created and the formation fails in tension.

There are 3 different tensile stresses that can exceed the tensile strength of the formation:

a) Radial stress σr

b) Axial stress σa

c) Tangential stress σt

The tensile strength To of the formation is usually rather low – generally < 10 % of theunconfined compressive strength Co.


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a) Tensile Failure Cylindrical (tcyl):

σr ≤ –To

This mode of failure occurs when the radial stress (σr) is lower than the negative of the tensilestrength of the formation (To)

Geometry and Orientation of tcyl Failure:

This failure is concentric with the borehole wall and cannot be seen on images.A low mud weight would favor the tcyl failure due to the magnitude of σr being lower.

σσσσr


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b) Tensile Failure Horizontal (thor):

σa ≤ –To

This mode of failure occurs when the axial stress (σa) is lower than the negative of the tensilestrength of the formation (To)

Geometry and Orientation of thor Failure:

This failure creates horizontal fractures.On an image the horizontal fracture would appear as a thin black horizontal line throughout allazimuthal orientations.

σσσσa


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Figure 2.4: Tensile Failure Vertical in RAB image as labelled above


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c) Tensile Failure Vertical (tver):

σt ≤ –To

This mode of failure occurs when the tangential stress (σt) is lower than the negative of thetensile strength of the formation (To)

Geometry and Orientation of tver Failure:

This failure creates a vertical fracture parallel with the maximum horizontal stress direction.

This is because in this orientation the tangential stress has to overcome the smallest formationtensile strength (that consisting of a minimum horizontal stress, σh).

Figure 2.4 shows an example of tver. The thin vertical line represents the vertical fracture. Thefracture is oriented parallel to the direction of maximum horizontal stress (in figure 2.4 SW-NE).

If the mud weight increases then σt decreases (becomes more negative) (see equation 1.9) untilthe condition for the tver failure described above. This is the opposite scenario for the tensilefailure cylindrical where a low mud weight decreases the magnitude of σr .

σσσσt


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Identification of Rock Mechanics Features on Borehole Images

We can summarize the various failure mechanisms by using a truth table:

FailureType

Failure name Stress Involved Appearance onImage

Orientation

Shallow Knockout(ssko) Max = σa Min = σr

Dark vertical featurewith narrow width(around 20°)

In axis of σh

Deep Knockout(sdko) Max = σr Min = σa


In axis of σH

ShearFailure Wide Breakout

(swbo)Max = σt Min = σr

Dark vertical featurewith wide width (30°- 90°)

In axis of σh

Narrow Breakout(snbo)

Max = σr Min = σt


In axis of σH

High Angle Echelon(shae)

Max = σa Min = σt

Dark feature inclinedat high angle (>50°)

Originates atσH , extendingaway at angle

Low Angle Echelon(slae)

Max = σt Min = σa

Dark feature inclinedat low angle (<40°)

Originates atσh , extendingaway at angle

Cylindrical(tcyl)

σr Not observed onimages

N/A

TensileFailure Horizontal (thor) σa

Dark narrowhorizontal featurecovering all azimuths

All azimuthalorientations - ahorizontal line

Vertical (tver) σt

Dark vertical featurewith very width (<20°)

In axis of σH

Table 2.1: Table summarizing the various borehole failure mechanisms


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Table 2.1 can help us determine the mode of borehole failure.RAB or FMI images are used to identify drilling induced borehole failures.ADN images are generally not able to pick out borehole failure features.

High and low angle echelon features are discernable from all others with the exception offractures that are non vertical.

If we already know the orientation of the minimum horizontal stress σh then a feature occurringin the direction of σh could either be a shear failure shallow knockout or a shear failure widebreakout. These two possibilities can be distinguished by the relative width of the feature. Thewide breakout would have wider borehole coverage than the shallow knockout.

If the feature occurs in the direction of σH then it could be one of 3 failure modes: shear failuredeep knockout, shear failure narrow breakout and vertical tensile failure.All 3 failures occur because σr is too large. Hence the mud weight is too high.

Figure 2.4 shows the swbo failure oriented NE and the tver failure oriented SE, hence they are90° apart. This is consistent with the theory that vertical tensile failures are oriented in thedirection of σh and shear wide breakouts oriented in are the direction of σH.

Figures 2.5 and 2.6 show the example of the 90° phase shift in azimuthal orientation between thetensile failure and the shear failure wide breakout.Figure 2.7 shows an example of finely bedded shaly features affected by borehole breakouts.The breakouts are seen to occur only in the shalier beds and is due to the value of σH being largeenough in the shaly beds to satisfy the shear failure criterion. In the surrounding beds σH is toosmall to satisfy this criterion.

Figure 2.8 shows breakouts oriented NW-SE. The well is deviated so the breakout direction doesnot correspond exactly to σh.


Figure 2.5: FMI image; Tens

Tver

ile

swbo

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failure vertical and Shear failure Wide Breakout


Figure 2.6: FMI image;

Tver

swbo

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Tensile failure vertical and Shear failure Wide Breakout


Figure 2.7: FMI image; B

s
Breakout
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orehole Breakouts in shaly intervals


Figure 2.8: FMI image; Breakouts

Breakouts

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oriented NW-SE in deviated well


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SECTION III:Wellbore Stability Planning


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The Mechanical Earth Model (MEM)

Wellbore stability planning is used in the well design of most high profile wells.

Wellbore instability is undesirable because it can lead to Non Productive Time (NPT) such as:

• Pack offs (formation failure leading to excess of cuttings)• Excessive trip and reaming time• Mud losses• Stuck Pipe and BHAs → Loss of equipment / Fishing / Sidetracks• Inability to land casing, casing collapse• Poor logging and cementing conditions

One process used to reduce these events is the Integrated Geomechanics Process and inparticular the Mechanical Earth Model.

The Mechanical Earth Model (MEM) is a numerical representation of allgeomechanical knowledge available for a field or basin.

The ultimate objective of the MEM is to reduce NPT due to wellbore instability andgeomechanical issues. As wells become more complex an MEM becomes an increasinglyimportant part in reducing NPT.

Even relatively minor wellbore stability problems in tectonically passive settings can beextremely expensive ($100,000 to $250,000 per day offshore).Planning and drilling high-risk wells without the aid of an earth model can lead to numeroussurprises which could cost the well:

• Wellbore instability develops unexpectedly, what action is required ? Raise or drop mudweight, change drilling fluid etc.

• High angle extended reach wells can be drilled NE but not SW direction. Why ?• Pore pressure increases faster than expected. Can the well reach TD with the remaining

casing strings ?

raltman@

slb.com52

(b) 3D Fram

ework M

odel

Figure 3.3: The 2 components of the M

echanical Earth Model:

1D

Mechanical Stratigraphy and 3D

Framew

ork Model

→→→ →+

Rock M

echanics Self Learning PackageSugar Land Learning C

enter

(c) 3D M

echanical Earth M

odel

(c) 1D M

echanical Stratigraphy


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Figure 3.3 shows that the MEM (c) is constructed by combination of the following 2 elements:

(a) 1D Mechanical Stratigraphy:

1D Mechanical Stratigraphy consists of all relevant geomechanical parameters Vs depthtogether with offset drilling data Vs depth (ie it is 1 Dimensional)

The geomechanical variables include: Poisson’s Ratio (ν), Young’s Modulus (E), Unconfined Compressive Strength (UCS), Friction Angle (φ), Pore Pressure (p), Minimum Horizontal Stress (σh), Maximum Horizontal Stress (σH), Vertical Stress (σv) and the Direction of the Horizontal Stress Axes

The above parameters are known as hard data and are used for the mud window calculations (next topic), which are made within the framework of classical rock mechanics.

The common modes of wellbore instability that are not amenable to this classical approach (e.g. fractured shales, fault reactivation) are analysed using soft data. Soft data consists of qualitative information such as certain offset drilling records.

The instability mechanism at a given offset depth is categorized as either:

breakouts, sloughing, natural fractures/weak planes, drilling induced fractures, faulting, undergauge hole, interbedded sequence, overpressured formation, unconsolidated formation, mobile formation, permeable formation, chemical activity.

(b) 3D Framework Model12:

The framework model consists of 2D-surfaces, folding, formation tops and faults, interpretedfrom seismic data, guided by log data and the geologist’s lithostratigraphic model.

(c) MEM:

In its most complete form (from the combination of the above 2 elements), the MEM willconsist of a complete 3D description of pore pressures, stresses and mechanical properties.


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(a) 1D Mechanical Stratigraphy: this is the most involved part of the MEM. Itcalculates all the variables that go into wellbore stability forecasting. These are also usedfor completions optimization and reservoir stimulation.

(i) Poisson’s ratio νννν / Young’s Modulus E

- Sonic data DSI or ISONIC (p and s wave)

1

121

2

2

−�

��

�

�

∆

∆

−��

��

�

�

∆

∆

=

c

s

c

s

tt

tt

ν)(

)43(22

222

s

s

vvvvrv

p

ppdynE

−

−=

Where:∆ts = shear slowness∆tc = compressional slownessvs = rock shear velocityvp = rock compressional velocityν = Poisson’s RatioEdyn = Dynamic Young’s Modulus

- Core Analysis can be used to measure E and ν in laboratory conditions. The Young’sModulus measured in this way is called the “Static” Young’s Modulus and is merely thestress/strain relationship of the rock in the linear elastic region.

Dynamic moduli (from sonic logs) are generally larger than static moduli measured in thelaboratory.

(ii) Vertical Stress σσσσv

- From ρρρρb data (integration of bulk density to surface) TLD or ADN- Core Analysis (To determine ρb and then upscale)- Sonic data p wave velocity profile, sonic log (DSI, ISONIC), VSP, checkshot- Cuttings density

The integration of density to surface to find σv is applicable when a uniaxial strain model can beapplied and σv > σH


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(iii) Minimum Horizontal Stress σσσσh

- Extended leak off test (mini-frac). This is a riskier exercise than the usual leak-off testbecause a full fracture is created.When pumping is stopped the fracture closes. This closure pressure = σhAs a rule of thumb, breakdown pressure = 1.1σh

- Calculated from Poisson’s ratio νννν (DSI, ISONIC).The coefficient of earth’s stress ko is given by:

νν−

=1ok

Then:σh = koσv + (1– ko)p

Where σv and p (pore pressure) are already known

If a rock has a large poisson’s ratio this means that it will have large lateral expansionfor a given axial contraction. Assuming the far field is resisting this lateral expansion,then the rock will be under greater lateral stress.

- Core Analysis (Differential Strain Analysis)

(iv) Maximum Horizontal Stress σσσσH

There is no direct method for determining σH.However we can infer it from stress ratios where:

σσσσH ≈≈≈≈ 1.2σσσσh

We can also infer it from an extended leak off test (pb is the breakdown pressure)

If σh and pore pressure p are known and T is determined from core tests, then σH can beinferred:

Pb = 3σσσσh– σσσσH – p + T

σH can also be inferred as follows: if we know the point on the wellbore wall where thefailure starts, from 2 neighboring (offset) wells we can calculate a unique set of stresses.This is called inversion of wellbore data to estimate stress tensor.


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(v) Pore Pressure p

- Eaton’s Method / PERT

- Sonic / Resistivity profiles ARC, RAB, AIT, HALS, DSI, ISONIC

- 3D Seismic / Tomographic velocity Schlumberger Seismic

- Formation Sampling MDT, DST

- Mud Weight from Offset Exploration Wells

(vi) Unconfined Compressive Strength Co / Friction Angle φφφφ

- Core Analysis (Rock Testing)

- Well logs Resistivity, Density, Sonic p&s waves (these can infer rock strength)

Co can be seen to be inferred empirically using a relationship of the form:

Co = K + AE

Where:

Co = Unconfined Compressive Strength,K, A = constantsE = Young’s Modulus

If the Young’s Modulus E is derived from sonic logs (ie a dynamic modulus) this will bedifferent from the static Young’s modulus (from laboratory tests) → see (i).This needs to be taken into account when inferring rock strength from sonic logs (using therelationship shown above.)

(vii) Tensile Strength To

- Extended Leak off Test. (Difference between opening pressure & breakdown pressure)

- Core Analysis (Rock Testing)


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(viii) Stress Direction ϕϕϕϕ

- Structural Maps / Seismic data. Extensional faulting: σH is parallel to fault strike` (note: geological structure does not necessarily indicate

the in-situ state of stress)

- Identify zones of breakout (can be misleading so care required) Dual Caliper,VISION ADN caliper

- Breakout features oriented in direction of σσσσH UBI/FMI/RAB images

(b) 3D Framework Model: provides the framework for the MEM and involvesbuilding a 3D geological representation of the field or basin:

→ Characterization of formation tops, faults, regional tectonics, basin analysis (folding etc)Also miscellaneous data (non quantifiable) such as swelling shales, fracture density etc.

Schlumberger measurements: 3D/2D Seismic, Borehole Seismic, Dipmeter Data (OBDT), Borehole image data (RAB, FMI, FMS)

(c) MEM Outputs: once the 3D model is established, wellbore stresses can becalculated for given wellbore trajectories and wellbore stability optimization proceeds.The following can then be carried out:

(i) Wellbore Stability Prediction / Failure Mechanisms (via wellbore stresses) →→→→Suitable Mud Weight Program

(ii) Wellbore Trajectory Design

(iii) Sanding Prediction and Completion Design

(iv) Hydraulic Fracture / Stimulation Design


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Calculating conditions for Rock Failure

The focal point for wellbore stability planning begins with the criteria for rock failure:

For shear failure to occur:

0 ≥ C0 + σ3’ tan2β – σ1’ (3.1)

For tensile failure to occur:

0 ≥ T0 + σ3’ (3.2)

These equations were explained in section I (see equations 1.7 and 1.8)

The RHS (right hand side) of equations 3.1 and 3.2 is known as the “Delta Stability”.

Therefore when Delta Stability is negative for eq 3.1 we have shear failure.When Delta Stability is negative for eq. 3.2 we have tensile failure.

To analyze the shear failure condition we take σ1’ and σ3’ as being any two out of the 3 wellborestresses σr , σt , σa

To analyze the tensile failure condition we take σ3’ as being any one out of the 3 wellborestresses σr , σt , σa


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For example:

Drilling a vertical well at 10000 ft

σh = 15 ppgσH = 16.5 ppg (in azimuth of 210°)σv = 19.4 ppgCo = 1500 psiφ = 30° β = 45 + 30/2 = 60°To = 150 psip = 9.5 ppg (pore pressure)

a) What is the minimum mud weight required to prevent shear failureb) What is the maximum mud weight required to prevent tensile failure

a) For shear failure:

We need to take all three wellbore stresses and present their maximum and minimum values:Assume that σa approximates to σv

Radial stress: σr = pw – p = pw – 4940

Tangential stress: σtmax = 3σH – σh – pw – p = 13000 – pw (occurring at 90° to σH) σtmin = 3σh – σH – pw – p = 9880 – pw (occurring at 0° to σH)

Axial stress: σa = σv – p = 5148 psi

We do not want to use a mud weight lower than the pore pressure for fear of a kick occurring.

The lowest mud weight we would reasonably use is 9.9 ppg (allowing for 200 psi overbalance)

Under this condition:

σr = 208 psiσtmax = 7852 psi (@ θ = 90°)σtmin = 4732 psi (@ θ = 0°)σa = 5148 psi

The most obvious shear condition occurs when the tangential stress is maximum (ie 7852 psi)and the radial stress is minimum (ie 208 psi). This condition will give a wide breakout (swbo).


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Testing for wide breakout failure:

σ1’ = σtmax = 7852 psi (or maximum stress to be used in the failure criterion)σ3’ = σr = 208 psi (or minimum stress to be used in the failure criterion)

Using equation 3.1:

RHS = 500 + 208*(Tan 60)2 – 7852 = – 6728 psi

Hence a wide breakout failure would occur.

The minimum mud weight we must use to prevent swbo failure is found as follows:substituting into equation 3.1 and solving for unknown pw.

At the limit:

0 = C0 + σ3’ tan2β – σ1’ from equation 3.1

ie:0 = 500 + 3(pw – 4940) – (13000 – pw) (tan2β = 3)

→ pw = 13.1 ppg

b) For Tensile failure:

We need to consider the smallest minimum effective stress σ3’ from one of the following:

Axial stress σa = 5148 psiRadial stress σr = pw – 4940Tangential stress (minimum) σtmin = 9880 – pw

If the minimum effective stress σ3’ is the tangential stress, then we have a tver failure where themud weight must not exceed a certain maximum value:

At the limit:

0 = T0 + σ3’ from equation 3.2

ie:0 = 150 + (9880 – pw)

→ pw = 19.3 ppg (mud weight must not exceed this value otherwise tver failure will occur)


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If the minimum effective stress σ3’ is the radial stress, then we have a tcyl failure where the mudweight must exceed a minimum value:

ie:0 = 150 – (pw – 4940)

pw = 9.8 ppg

Therefore to ensure that tensile failure does not occur the mud weight should be between11.4 ppg and 20.9 ppg.

Therefore, taking into account 3 criteria –Shear failureTensile failurePore pressure

The mud weight should be:13.1 ppg – 20.9 ppg

Before we use this window of stability, we need to consider the minimum horizontal stress σh.

In a pessimistic condition, if the rock we drill has a network of natural fractures near thewellbore, these will preferentially open against the minimum horizontal stress σh .

If when we are drilling these fractures are closed then it is only necessary to have our mudweight exceed σh to open the fractures again and cause a lost circulation problem.

We must not have our mud weight exceed the minimum horizontal stress (15 ppg).

Therefore the final mud weight window to be used is:

13.1 ppg – 15 ppg

In this case:

swbo < mud weight < σσσσh


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The diagram below summarises in graphical form the results we have found for safe mud weightwindow.

If the mud weight is in the yellow region then swbo can occur and the cavings from the wellborewall fall into the wellbore leaving a washed out zone (in the direction of σh) as shown.

If the mud weight is above σh then there is a risk of lost circulation if natural fractures exist. Ifthe mud weight is in the red region there tver can occur with a vertical fracture as shown causinglost circulation.

If the mud weight is very high then snbo and sdko failures could occur but in practice shae ortver would occur before this. Similarly, at low mud weights slae, ssko and tcyl could occur.

Swbo = Shear failure Wide BreakoutSnbo = Shear failure Narrow BreakoutShae = Shear failure High Angle EchelonSlae = Shear failure Low Angle EchelonSsko = Shear failure Shallow KnockoutSdko = Shear failure Deep Knockouttver = Tensile failure VerticalTcyl = Tensile failure Cylindrical

Safe Mud WeightWindow

σσσσhPorePressure

snbo

swbo

shae

sdko

slae ssko tcyl

tver

Increasing MudWeight


Stable region

Collapse

Fracture Initiation

WIn

Vertica

Figure 3.1: Deviated borehole in an anisotropic stress field, relaxed basin (σv is max)

135 125 115 105 95 85 75 65 55 45

Well Azimuth Sh

MudWeight(g/cc)

ellc.

l well

S

Horizontalwell

Increase the mudweight or increasethe risk of shearfailure

Figure 3.2: Trajectory Analysis for anisotropic stress field, relaxed basin (σv is max)
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Borehole Stability in Deviated Wellbores

In deviated wells whilst we still use the 3 wellbore stresses (radial, axial and tangential) to testfor the shear failure criterion, the solution for these wellbore stresses is complex (Appendix C).

The problem can be simplified by examining horizontal wells that are drilled in the orientationof one of the horizontal stresses.

There are 4 separate cases:

1) In a Relaxed basin:a) drilling parallel to σHb) drilling parallel to σh

2) In a Tectonically stressed basin:a) drilling parallel to σHb) drilling parallel to σh

In all cases the radial stress σr is given by pw – p (as for a vertical well)For shear failure:

0 ≥ C0 + σ3’ tan2β – σ1’

For wide breakouts, the minimum stress (σ3’) is σr and the maximum stress (σ1’) is σtmax (themaximum tangential stress)To compare these 4 cases for shear failure we only need to examine how the tangential stresschanges for each case.

0 ≥ T0 + σ3’

Where σ3’ is σtmin (the minimum tangential stress)To compare these 4 cases for tensile failure we only need to examine how the tangential stresschanges for each case.

In each case we will be comparing with the case for a vertical well where

σσσσtmax = 3σσσσH – σσσσh – pw – p (for shear failure swbo, ie in direction of σσσσh)σσσσtmin = 3σσσσh – σσσσH – pw – p (for tensile failure tver, ie in direction of σσσσH)


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1) In relaxed basins (σσσσv > σσσσH > σσσσh):

a) Drilling parallel to σH

σtmax = 3σv – σh – pw – p (greater than the σtmax for a vertical well, shear failure more likely)σtmin = 3σh – σv – pw – p (less than the σtmin for a vertical well, tensile failure more likely)

To prevent shear failure need greater mud weight.To prevent tensile failure need lower mud weight.

MUD WEIGHT WINDOW DECREASES IN GOING FROM VERTICAL TO HORIZONTAL

Figure 3.1 shows that the condition can become acute in wells where the deviation exceeds 70°.There is no stable mud weight window. A lower mud weight to prevent tensile failure is oftenmore desirable: a lower ROP would better handle the additional cuttings caused by too low anECD rather than the time and safety consequences resulting from a lost circulation event.

b) Drilling parallel to σh

σtmax = 3σv – σH – pw – p (greater than the σtmax for a vertical well, shear failure more likely)σtmin = 3σH – σv – pw – p (less than the σtmin for a vertical well, tensile failure less likely)

Comparison of case a) and b) shows that:

IT IS SAFER TO DRILL PARALLEL TO THE MINIMUM HORIZONTAL STRESS WHENDRILLING A HORIZONTAL WELL IN A RELAXED BASIN(Figure 3.2 shows this where the most dangerous place to drill is parallel to σH and at 90° inc.)

This is because σtmax (case a) > σtmax (case b) case a is more likely to have shear failure σtmin (case a) < σtmin (case b) case a is more likely to have tensile failure

σσσσH

σσσσhσσσσvσσσσt

σσσσh

σσσσHσσσσvσσσσt


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2) In tectonically stressed basins (σσσσH > σσσσh > σσσσv):

a) Drilling parallel to σH

σtmax = 3σh – σv – pw – p (probably less than σtmax for vertical well, shear failure less likely)σtmin = 3σv – σh – pw – p (probably lower than σtmin for vertical well, tensile failure morelikely)

To prevent shear failure don’t need such a high mud weight.To prevent tensile failure need a lower mud weight.

b) Drilling parallel to σh

σtmax = 3σH – σv – pw – p (greater than the σtmax for a vertical well, shear failure more likely)σtmin = 3σv – σH – pw – p (less than the σtmin for a vertical well, tensile failure more likely)

To prevent shear failure need greater mud weight.To prevent tensile failure need lower mud weight.

MUD WEIGHT WINDOW DECREASES IN GOING FROM VERTICAL TO HORIZONTAL

Comparison of case 2a) and 2b) shows that:

IT IS SAFER TO DRILL PARALLEL TO THE MAXIMUM HORIZONTAL STRESS WHENDRILLING A HORIZONTAL WELL IN A TECTONICALLY STRESSED BASIN

This is because σtmax (case b) > σtmax (case a) case b is more likely to have shear failure σtmin (case b) < σtmin (case a) case b is more likely to have tensile failure

The situation is different if in a tectonically active basi,n σH > σv > σh. YOUSHOULD ALWAYS MAKE CALCULATIONS TO DETERMINE THEMOST FAVOURABLE DIRECTION IN WHICH TO DRILL

σσσσH

σσσσhσσσσvσσσσt

σσσσh

σσσσHσσσσvσσσσt


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SECTION IV:Practical Wellbore Stability Monitoring


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DRILL COMPAREWITH

FORECASTPERFORM / HRT

MONITORDRILLING

PERFORM

Client /PERFORM

(Operations)

REVISE MEM

HRT

Earth

HRT

GeoMechanics TeamHRT (Office)

HRT

STABILITYFORECAST

NEWINFORMATION

PERFORM / HRT

DIAGNOSE

TREATHRT / PERFORM / CLIENT

CONTINUE

NEWDATA

YES

N

AGREE?

Figure 4.1: The Geomechanics Solutions Approach

HRT = Holditch Reservoir Technologies


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The MEM and PERFORM

Wellbore Stability management is a dynamic discipline:

Plan → Execute

Evaluate

The planning stage consists of the construction of the initial MEM as outlined in section III.The wellbore stability forecast is for the well to be drilled and the corresponding drillingprocedures are made for zones of potential instability.

The execution stage occurs while drilling the well: the PERFORM engineer monitors drillingconditions both surface and downhole and the appropriate drilling practices are used to keep thewell stable.If the drilling conditions are different from those predicted by the MEM then appropriate actionis taken at the wellsite and the MEM / stability forecast is further refined.

The evaluation stage occurs at the end of the well: all the information acquired during and afterdrilling is used to further refine the MEM.

This dynamic approach to drilling optimization is known as The Solutions Approach (seefigure 4.1)The solutions approach collocates information from various sources in the different stages ofdrilling from pre-spud to completion.This requires an integrated team of drilling engineers, geologists, geomechanics engineers andwellsite (PERFORM) engineers. Their inputs keep the MEM dynamic and enhance drillingoptimization.

In basins where wellbore instability is notorious (ie Cusiana Field, Colombia) the mostdesirable solution is not necessarily to prevent it from occurring but instead toacknowledge its occurrence and manage the problem in real time.The borehole may be unstable but the well could remain stable if correct precautions aretaken.


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Real Time Wellbore Stability Management

The open hole section of a wellbore must be maintained in a condition that is good enough toallow drilling and casing to be run. This does not mean that it is necessary to eliminate allformation failure.

Indeed the wellbore can remain stable even after a period of prolonged formation failure.

An example of this is the Cuisiana field, Colombia where the wellbore has remained stablebecause the cavings from the breakouts can be cleaned out of the hole.In this example the wellbore instability was managed (or contained) rather than prevented.

Often the problem occurs in highly deviated wells where the conditions for tensile and shearfailure become similar and the stable mud weight window becomes hard to define (see previoussection).

In these cases it becomes difficult to find a solution that will completely prevent the instabilityfrom occurring in the first place and wellbore stability management is required: for example,loss circulation might be avoided at all costs, and techniques to manage the shear failure areimplemented such as good hole cleaning practices.

Real time Wellbore stability management (control) is a twofold process involving:

a) Continuous monitoring – ie downhole/surface signatures to diagnose onset of a problem.

b) Remedial actions – ie drilling parameters to fix a failed or failing wellbore.


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a) Continuous monitoring

Real Time Wellbore Stability Control relies on an integration of all data available at theWellsite:

Surface signatures:Cavings analysis – Wellbore Failure,Cuttings volume – Hole Cleaning,Pit volumes – Gains (overpressured zone), losses,Surface Drilling Parameters

MWD data:Downhole Drilling ParametersDWOB, DTORQ – Friction / DragECD behaviour – Hole Cleaning, pack off

LWD data:Gamma Ray, Resistivity – Identify zones of potential instability from MEMSonic – Pore pressure prediction while drillingCaliper measurements – if pattern is forming in some intervals, can identify unstable formations

A reliable diagnosis of the instability mechanism requires use of all available data.

If tabular cavings due to natural fracturing are observed then the resistivity log should bechecked for evidence of mud invasion into fractures and the mud records require examining forlosses.

Similarly, if splintered cavings due to over-pressured formations are seen then high gas levels,kicks or mud gains may also be present.

The observation of angular cavings due to breakouts requires the debris levels in the hole to bediscerned. In all cases, the cavings volume should be compared to the ECD and the degrees oftight hole and restricted circulation to discern the effectiveness of the hole cleaning and theseverity of instability.

(see cavings analysis on the following pages)


Figure 4.2: Tabular Cavings

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Cavings Analysis:

An analysis of cavings can provide a signal that the borehole is failing and indicates both thenature of the instability and the troublesome formations.

Cavings dimensions range from a few millimetres to 10 cm or more, with larger examples risingto the surface while lodged in the BHA.

There are four main types of caving:

Tabular,Angular,SplinteredThose which cannot be characterized.

Tabular cavings are the result of natural fractures or weak planes.In the case of natural fractures, the fluid pressure in the annulus exceeds the minimum horizontalstress, resulting in mud invasion of fracture networks surrounding the wellbore.

This can result in severe destabilization of the near wellbore region, due to the movement ofblocks of rock, leading rapidly to high cavings rates, lost returns and stuck pipe.

The blocks of rock are bounded by natural fractures planes and, therefore, have flat, parallel,faces.Figure 4.2 shows examples of tabular cavings due to natural fractures.The other characteristic is that bedding, if any, will not be parallel to the faces of the caving.

In the case of weak planes, the combination of low mud weight and a borehole axis that is withinapproximately 15 degrees of the bedding direction can induce massive failure along the planes ofweakness, leading to the symptoms described above.Cavings that are the result of weak planes are characterized by having flat, parallel, faces. Thebedding direction is also parallel to the faces.

Angular cavings are a consequence of breakouts. These cavings are characterized by curvedfaces with a rough surface structure. The surfaces intersect at acute angles (much less than 90degrees). Figure 4.3 shows Angular Cavings.

Splintered cavings have two nearly-parallel faces with plume structures. This type of caving isdue to tensile failure (tcyl) occurring parallel to the borehole wall and commonly occurs inoverpressured zones drilled with a small overbalance. The effective radial stress is now evenlower due to the overpressured formation pore pressure. Figure 4.4 shows Splintered Cavings.


74

The higher the cavings rate the more severe the failure for a given hole cleaninThe dominant caving should be noted not the proportion of different cavings.

The cavings rate is measured by the time required to fill a bucket placed undernThe cavings volume is then proportional to the amount of cavings in the bucketCARE MUST BE TAKEN – IF HOLE CLEANING IS POOR THERE WILL BE FEW

Figure 4.3:Angular Cavings

Figure 4.4:Splintered Cavings

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g efficiency.

eath the shakers..ER CUTTINGS


Chemical Wellbore Instability:

Wellbore instability can be classified as either mechanical (for example, failure of the rockaround the hole because of high stresses, low rock strength, or inappropriate drilling practice) orchemical.

Chemical Wellbore Instability arises from damaging interactions between the rock, generallyshale, and the drilling fluid.The integration of understandings of chemical and mechanical damage remains problematical.

In wellbore stability monitoring, it is important to determine whether a particular drillingproblem is mechanical or chemical in origin.

Figure 4.5 describes how to diagnose the 4 most important wellbore stability mechanisms.3 of these are mechanical and 1 of these is chemical in origin. The 3 tables that follow showexamples of wellbore stability from surface, downhole and miscellaneous signatures.

Figure 4.5: Diagnosing the 4 most common wellbore instability mechanisms

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raltman@

slb.com76

Figure4.6:W

ellboreinstability

–M

iscellaneoussignatures

Rock M


enter

Mechanism

Lost T

ime

Wellbore

Trajectory

In-situstresses

Formation

StengthPorePressure

Geology

Permeable

formation

Stuck pipeLowcom

pared tom

ud pressureInterbeddedsoft/strong rocks

Stuck pipeTortuous

Frequentchanges

Thick sections collapse more

Fault slip/activation

Stuck pipe,excessive ream

ingH

igh stressdeviation

Faults present

SloughingH

ole fill after tripsW

eakProxim

ity to salt dome or faults,

tectonically activeO

verpressuredform

ationH

ole fill after trips H

ighR

ecently crossed faultU

ndergauge holeExcessive slackoff w

hile trippingH

igh mean

stressLow

yieldstrength

Unconsolidated

formation

Restricted pipe

movem

entLarge sand or fractured section

Mobile form

ationH

ole fill after tripsH

ighoverburden

Proximity to salt dom

e, evaporatesequence

Chem

ical activityProblem

s worsen

with tim

e, slightflow

Low

Breakouts

Stuck PipeH

ighstress/stengthratio

Drilling induced

fracturesC

losely spacednatural fracs /w

eak planes

Stuck Pipe, holefill after trips

Planes ofw

eaknessM

udpressure>porepressure

raltman@

slb.com77

Mechanism

RO

PD

WO

BD

TO

RC

aliperγγγ γ-ray

Resistivity

UB

I

Permeable

formation

Decreases

Thick filter cakeG

API ≤

60Interbeddedsoft/strong rocks

Frequent & rapid

changesFrequent &

rapidchanges

GA

PI>60,&

GA

PI ≤60 often

Frequent &rapid changes

Frequent well

diameter changes


Decrease

Local boreholeelongation

Detected.

Rotation of

breakoutsSloughing

Decreases

LowB

oreholeenlargem

entG

API >

60H

igh dip (>60°)

Borehole

enlargement

Overpressured

formation

High, given rock

strengthLow

Borehole

enlargement

Borehole

enlargement

Undergauge hole

LowLow

Diam

eter lessthan gauge

Diam

eter lessthan gauge

Unconsolidated

formation

High

Decreases

Borehole

enlargement

GA

PI <60

Borehole

enlargement

Mobile form

ationD

ecreases

Chem

ical activityD

ecreasesD

ecreasesH

ole tightensw

ith time, or

dissolves

GA

PI >60

Swelling detected

Breakouts

Decreases

LowB

oreholeenlargem

entO

rientation &span detected

Drilling induced

fracturesLow

GA

PI >60

Diam

etricallyopposed &

longPossibledetection

Close spaced

fracs/weak planes

Decreases

LowB

oreholeenlargem

entG

API >

60Fracture &bedding planeorientation

Borehole

geometry

Figure4.7:W

ellboreinstability

–M

WD

,LW

D&

Wireline

Rock M


enter

raltman@

slb.com78

Mechanism

Pump pressure

Circulation

Mud

Cuttings and

cavingsH

ookloadSurfaceT

orqueD

rillstring

Permeable

formation

Gradual decrease

Flowdecreases

Water loss,

high solidsIncreases

Higher

Interbeddedsoft/strong rocks

SpikesFlow

erraticV

olume rate

changesfrequently

High

ErraticPacked off


SpikesFlow

erraticLoss

High

IncreaseD

iametrical w

earSloughing

IncreaseFlowdecreases

Large & flat

High w

henpum

ps offH

ighPacked off

Overpressured

formation

IncreasePit levelincrease

Background

gas highLarge, brittle,fissile, concave

Large overpullat connections

Increase

Undergauge hole

SpikesFlow

erraticA

brasive &hard

High

High &

erraticU

ndergaugeB

HA

Unconsolidated

formation

IncreaseFlowdecreases

Unconsolidated

& uncem

entedLarge overpullat connections

Erratic

Mobile form

ationIncrease

Flowdecreases

Salt present,rise in Cl

Salt grainsLarge overpullat connections

ErraticPacked off

Chem

ical activityIncrease

Flowdecreases

MW

& solids

increaseSoft,w

atersoluble.G

umbo

Large overpullat connections

Increases

Breakouts

SpikesFlow

erraticA

pparent lossH

igh volume

High

High

Packed offD

rilling inducedfractures

Decrease

Flowdecreases

Loss

Close spaced

fracs/weak planes

Decrease

Flowdecreases

Loss at similar

weights across

field

Squarish, highvolum

eH

ighH

igh

Figure4.8:W

ellboreinstability

–surface

signatures

Rock M


enter

raltman@

slb.com79

A "0" indicates that the action has no influence on instability or

makes it w

orse.

Rock M


enter

FrequentW

iperTrips

InfrequentW

iperTrips

Decrease

RO

PIncreaseM

ud Gel

Strength

IncreaseM

udC

irculationR

ate

Limit

OD

size /D

rillC

ollars

IncreaseM

udW

eight

Decrease

Mud

Weight

Use

Minim

umO

verbalance(200 psi)

EnsureO

verbalanceexceeds 200psi

Add

FluidLossA

gents

Use

InhibitiveM

ud

Minim

izesw

ab andsurgeaffects

Breakouts

10

11

10

10

00

00

1

Sloughing0

10

10

00

00

00

00

Natural

Fractures /W

eak Planes

01

11

00

00

10

10

1

Drilling

InducedFractures

00

01

00

01

00

10

1

FaultA

ctivation0

00

00

00

01

01

01

Undergauge

Hole

10

00

01

10

00

00

1

InterbeddedSequence

10

01

10

10

00

00

1

Overpressured

Formation

00

00

00

00

01

00

1

Unconsolidated

Formation

01

01

10

10

00

00

1

Mobile

Formation

01

00

10

10

00

00

1

Permeable

Formation

01

00

00

00

10

10

1

Chem

icalA

ctivity0

10

00

00

00

00

10

Figure 4.9: Actions inhibiting the instability m

echanisms. A

"1" indicates that the action suppresses the instability.


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b) Remedial Actions

If wellbore instability becomes severe as detected from a) continuous monitoring, and holecleaning cannot remove cavings from the wellbore then the wellbore would be unstable.

The ability to deal effectively with the consequences of the unstable wellbore depend on theinstability mechanism and its severity.Remedial action generally involves the control of surface parameters (e.g. ROP, RPM, flow rate,mud weight/rheology).

For example, if mud losses are currently occurring, but a mud weight decrease is not possibledue to conditions that will be encountered while drilling through formations below the currenthole bottom (cavings generation), then decreasing the ROP will reduce cuttings loading andtherefore the ECD. This may be sufficient to eliminate mud losses and also reduce cuttingsloading in deeper intervals

The emphasis when considering remedial actions, which either suppress instabilities or minimizetheir consequences, should be the entire open hole interval, rather than focusing on problemfixing at the bit.

The ROP and hole cleaning efficiency form the key links between wellbore instability andoperations. Rock debris in the annulus, resulting from drilling and/or wall failure, will increase ifhole cleaning is inadequate, raising the risk of pack-offs and stuck pipe. The ability to clean thehole is also related to the ROP.

Figure 4.9 outlines the various actions that are recommended for various given wellbore stabilitymechanisms. It can be seen that minimizing swabbing and surging affects helps to surpress moreinstability mechanisms than any other drilling practice.

Also it can be seen that drilling practices such as wiper trips that are often considered as routineare sometimes detrimental to wellbore stability. Minimising wiper trips can help surpress actionsthat are sensitive to mechanical agitation of the formation such as mobile formations / sloughingshales, weak planes.

Increasing mud weight is not necessarily the answer to wellbore stability problems. Whilst thispractice can help surpress breakouts, it can cause drilling induced fractures or activate naturalfracture networks by drilling above the minimum horizontal stress. However, whereoverpressure occurs, it is desirable to drill with an overbalance that exceeds 200 psi. In all cases,calculations are required prior to drilling to determine optimal parameters.

This problem becomes amplified in deviated and especially horizontal wells where the mudwindow between shear and tensile failure becomes so small that sometimes there is no stablemud weight window.

Good drilling practices such as Circulation, Rotation, Reciprocation, of the drillstring toremove excess cuttings in highly deviated wells, and close examination of shale shakers toexamine volume of cuttings and their geometry is desirable to manage (surpress) unstablewellbores.


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Brain Teasers

1) Describe the stresses in the earth ina) A tectonically active basin b) A tectonically relaxed basin

2) Describe 3 common mechanisms that cause a formation to be overpressured

3) Describe:(a) Shear Failure Wide Breakout and how it would look on a log image(b) Tensile Failure Vertical and how it would look on a log image

4) You are responsible for the well design of Well X in the Rocket Field. The Rocket Field isnotorious for its wellbore stability problems. The plan is to drill vertical until 11000 ft, thenbuild angle to horizontal.The horizontal entry point is at 16000 ft MD (14183 ft TVD). The horizontal drainhole will be2000 ft, so the final depth of the well will be 18000 ft MD (14183 ft TVD).The formation at 11000 ft is believed to be a low porosity sandstone.

You have the following data available:

An ISONIC log from an offset well:At the offset depth, ∆tshear = 120µsec/ft ∆tcompressional = 65 µsec/ftThe vertical stress σv in the field is 22 ppg. The horizontal stress σH is approximated as 20%greater than the minimum horizontal stress σhIn an extended leak off test performed in the offset well in an equivalent formation as thatpredicted at 11000 ft in our well, the formation breakdown pressure was 10152 psi and thefracture opening pressure was 9952 psi.A core was taken in the offset well in an equivalent formation. The friction angle φ determinedfrom rock tests was 35° and the unconfined compressive strength Co was 1800 psi.The maximum stress σH for the field is known to be oriented at an azimuth of 245°The pore pressure at 11000 ft in well X is predicted to be 11 ppg.

a) Calculate what the ECD window should be at 11000 ft

b) What azimuth should the horizontal section be drilled to minimize wellbore instability?Calculate what the ECD window should be at total depth (18000 ft MD) given this well azimuth? Assume the same properties in the horizontal section (14183 ft TVD)

c) You are at the wellsite and are drilling the horizontal section of the well with a mud weightthat you had previously planned.You notice a trend of increasing excess torque and friction during each connection whenreaming the stand. A torque and drag analysis using DWOB/ DTORQ shows that this trend is inexcess of the trend established by the well profile. You notice angular cavings on the shaleshakers. The company man wants to increase the mud weight. What would you do ?


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d) A decision is made to increase the mud weight. You notice after a while that the trend isgetting worse and now there are intermittent pump pressure spikes. The company man wants toincrease the mud weight further and perform a wiper trip. What decision would you make?

e) The mud weight is increased further and a wiper trip is performed. A great deal of drag wasexperienced and there was difficulty in getting back down to bottom again. The mud pits havedecreased a little suggesting lost circulation. What would you do ?

5) Below are listed the 4 most common wellbore stability mechanisms:

a) Breakoutsb) Natural Fractures and Weak Planesc) Drilling Induced Fracturesd) Sloughing Shales

For each one state the most important remedial actions that should be made to suppress theactivity ?


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Appendix A: Stress

Force ∆F

OArea ∆A

σn

τ

x

Figure A1: Force ∆F acting at a point O on a plane ∆A. The componentsof the force perpendicular to ∆A are σn and τ


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Stress is the force per unit area acting at a point.Although it has the same units as pressure (psi), stress differs from pressure:

Pressure is a scalar (a property that is only magnitude dependent) - the pressure at the bottom of a fluid column acts in all directions, and is governed only by its magnitude.

Stress is a tensor (stresses acting in different directions act independently. They cannot be added together or resolved to obtain a single equivalent value as one can do with vector quantities).

Consider a random plane of area ∆A. (Figure A1)If a resultant force ∆F acts on a point O that is centered on that plane, the stress vector P at pointO is defined as:

��

�

∆∆=

→∆ AFP

Alim

0(A1)

If a material has a positive stress, by convention it is in a state of compression.If a material has a negative stress, by convention it is in a state of tension.

The resultant stress P acting at O can be decomposed into 2 components

A normal stress component: σn (acting normal to the plane ∆A)A shear stress component: τ (acting in the plane ∆A)


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O

B

A

P

x

y

σx

σy

τ

σn

τy

τx

AB = a

Also assume a unit thicknessof triangle OAB

Figure A2: 2D decomposition of σn and τ stresses in the planeperpendicular to ∆A

ββββ

ββββ


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It is obvious that an infinite amount of planes can be drawn through a given point (ie O) in 3Dspace. For a given force ∆F, the values (magnitudes) of σn and τ will change depending on theorientation of the plane.

Figure A2 is a 2D decomposition of the normal and shear components of a stress P.Note that in this diagram the x-y plane is representing the plane perpendicular to ∆A (containingτ and σn).β is the angle between the x-axis and the normal component, σn

Objective 1: Derive expressions for shear stress ττττ and normal stress σσσσnReferring to figure A2:

Resolving the forces (ie stress multiplied by area = force) along the x and y directionsrespectively we get:

a) a.px = σx acosβ + τx asinβ (resolving in x-axis)b) a.py = σy asinβ + τy acosβ (resolving in y-axis)

Also:σn = px cos β + py sin β = σx cos2β + τx sinβcosβ + σy sin2β + τy cosβsinβ [using a) and b)]

However τx = τy

This is because the system is in equilibrium and if τx ≠ τy then there would be a tendency forrotation (ie non-equilibrium).

Therefore we have the final expression for the normal component σn :

σn = σxcos2ββββ + 2τycosββββsinββββ + σysin2ββββ (A2)

Also:τ = py cos β - px sin β = σy sinβcosβ + τy cos2β - σx cosβsinβ - τx sin2β = (σy - σx)sinβcosβ + τy(cos2β - sin2β)

Hence we have the final expression for shear component τ :

τ = ½ (σy – σx)sin2ββββ+ τycos2ββββ (A3)

(since 2cosββββsinββββ≡≡≡≡sin2ββββ and cos2ββββ - sin2ββββ≡≡≡≡cos2ββββ)


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Objective 2: Derive expressions for shear stress ττττ and normal stress σσσσn

In practice the plane ∆A could be oriented in many different ways.

If we keep the force ∆F constant but change the orientation of the plane ∆A we could make theshear stress component τ vanish.

This would occur in two orientations:a) When the surface of the plane ∆A is directly perpendicular to the force ∆F.b) When the surface of the plane ∆A at point x is parallel to the force ∆F.

In either case the shear stress vanishes, and these planes are known as the principal planes.The 2 normal stresses (σn) associated with these 2 planes are referred to as the principalstresses.The principal stresses are known as σσσσ1 and σσσσ2.

It should be evident that these 2 orientations are at 90°°°° to one another, and the 2 normal stresses(principal stresses) associated with these 2 planes are therefore at 90°°°° to one another.

The 2 angles β (associated with these 2 orientations) will be at 90°°°° to one another.

We can calculate these 2 angles β either by letting τ = 0 in equation 1.3 (and solving for β) or bytaking the derivative of equation 1.2 wrt β, setting this to zero and finding the correspondingvalues of β for which this derivative is zero. (We can take the derivative of σn with respect to βbecause the resulting expression will yield the maximum and minimum principal stresses.)

Either case obtains the following expression for the value of β for which the shear stressvanishes1:

��

�

−=

yx

y

σσ2τ

arctan21β

Note that there are 2 solutions of the above equation for β (within a range of 0 - 2π radians).These 2 solutions will be 90° to each other.

If we substitute these 2 solutions of β into equation 1.2 we will obtain 2 solutions of σn.σ1 and σ2 are the two solutions of σn from equation 1.2:


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σ1 : This is the maximum principal stress: σ2 : This is the minimum principal stress:

( )1/2

2yx

2yyx1 σσ

41τ)σ(σ

21σ �

��

� −+++= (a)

( )1/2

2yx

2yyx2 σσ

41τ)σ(σ

21σ �

��

� −+−+= (b)

Note: in 3 dimensions, there are 3 principal stress components σ1 ,σ2 ,σ3 for which the shearcomponents vanish.

If we remove the τy shear stress component from these equations (due to the shear stressvanishing), equation A3 becomes:

τ = -½ (σ1 – σ2)sin2ββββ (A4)

and we can solve equations (a) and (b) above to express σx in terms of σ1 and σ2 and σy interms of σ1 and σ2Equation A2 then becomes:

σn = ½ (σ1 + σ2) + ½ (σ1 – σ2)cos2ββββ (A5)


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σ1σ2

Figure A3: The Mohr Circle 2D relationship between shear stress τ andnormal stress σn

σn

τ

τ
M
2ββββ

σn


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Mathematically, equations A4 and A5 represent the equation of a circle in a (σn, τ) plane, withits center located on the axis at σ2 + ½(σ1 – σ2) or ½(σ1 + σ2) and of diameter (σ1 – σ2).

This circle is known as the “Mohr Circle”. Figure A3 shows this Mohr Circle.

This contains all the information necessary to determine the two-dimensional stress state at anyorientation in the sample1.

From figure A3 it can be seen that when the value of σn is either σ1 or σ2 the circle touches thex-axis and the shear stress τ is therefore zero.Under this condition, σ1 and σ2 are both perpendicular to each other (ie β is either 0 or 90degrees). This is the condition for principal stress (σ1 and σ2 are principal stresses).

For all other values of σn , where σn ≠ σ1 and σn ≠ σ2 the shear stress is non-zero.

For these values of σn there exists a shear stress in the material τ which is given by theintersection of the σn value with the Mohr Circle (ie point M).

Alternatively τ can be found using equation A5 assuming the principal stresses (σ1 and σ2)and β are already known.


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Appendix B: Strain

O

Q

P

P

O

P*O*

l

ψ

O*

P*

l*→

→

Q*

DeformedNormal

Normal

Deformed

Axial strain a)

Shear strain b)

Figure B1: Axial strain (a) and Shear strain (b)


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If a material is subjected to a stress it deforms.

If the new positions of the material cannot be made to correspond to the old positions bytranslation or rotation, the material is strained.

If the stress acts in an arbitrary direction, then the strain can be decomposed into twocomponents (Figure B1):

��

� −=→∆ l

lll

*lim

0ε Axial Strain (B1)

γγγγ = Tan ψψψψ Shear Strain (B2)

l*= Final length of material (after axial strain)l = Original length of material (before axial strain)ψ = Change of angle between two directions that were perpendicular before straining

Therefore strain is dimensionless

Stresses are taken positive in compression so:

+ve longitudinal strain → decrease in length-ve longitudinal strain → increase in length

+ve shear strain → increase in angle-ve shear strain → decrease in angle

Various relationships exist between stress and strain.

The simplest and most important one used in Petroleum Rock Mechanics isthe theory of linear elasticity.


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This is essentially Hooke’s Law and assumes a linear and unique relationship between stressand strain of a rock, and the deformation of the rock (strain) recovers when the stress isremoved (provided the elastic limit is not exceeded).

ie:

11εσ

=E (B3)

E = Young’s modulus

Poisson’s ratio ν defines the ratio of lateral expansion (ε2) to longitudinal contraction (ε1). Thenegative sign ensures ν is positive.

1

2

εεν −= (B4)

Other constants:

)1(2 ν+= EG (Shear Modulus)

Bulk modulus K (ratio of mean stress σm and volumetric strain εv), for isotropic loading (σx =σy = σz , εx = εy = εz)

)21(3 ν−= EK

The Poisson’s ratio ν Shear Modulus G, and Bulk Modulus K are used as inputs into RockMechanics models (see Section III).


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Appendix C: The Wellbore Stresses

y’

x’

z’

σh

σH

σv

z

x

y

Figure C1: Deviated borehole in an anisotropic stress field

i

a

i = borehole inclinationa = ‘borehole azimuth’ (relative to direction of σH)


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Figure C1 shows a deviated wellbore in an anisotropic stress field (ie where σH ≠ σh ).

Objective: Describe the various wellbore stresses acting at the wellbore wall.

With reference to figure C1:

A coordinate system is used to so that:x’ is parallel with σHy’ is parallel with σhz’ is parallel with σv

x is parallel to the low side of the holey is perpendicular to x (and perpendicular to the axis of the borehole)z is parallel to the axis of the borehole

A transform from (x’ y’ z’) to (x, y, z) can be performed by a rotation “a” around the z’-axis anda rotation “i” around the y-axis.

The 6 complete stress solutions are given below. (Refer to reference 4 for full derivations.)It is assumed that the rock remains linear elastic, and the wellbore fluid pressure pw does notpenetrate the rock (here we are assuming the presence of mudcake).

( ) ( ) 2

2

2

2

4

4

2

2

4

4

2

2

2sin4312cos43

1211

21

rr

prr

rr

rr

rr

rr w

www

xyww

yxw

yxr +�

��

�

�−++

�

��

��

�

−+−+�

��

��

�−+= θτθσσσσσ

( ) ( ) 2

2

4

4

4

4

2

2

2sin312cos3

1211

21

rr

prr

rr

rr w

ww

xyw

yxw

yxt −�

��

�

�+−

�

��

��

�+−−

�

��

��

�++= θτθσσσσσ

( ) �

��

�+−−= θτθσσνσσ 2sin42cos2 2

2

2

2

rr

rr w

xyw

yxza

θτθσστ 2cos2312sin231)(21

2

2

4

4

2

2

4

4 �

��

�

�+−+�

��

��

�

�+−−=

rr

rr

rr

rr ww

xyww

yxrt

��

�

�

��

�

�++−= 2

2

1)cossin(rrw

yzxzta θτθττ

��

�

�

��

�

�−+= 2

2

1)sincos(rrw

yzxzra θτθττ


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σr = Radial Stressσt = Tangential Stress (or Hoop Stress)σa = Axial Stressσh = Minimum Horizontal StressσH = Maximum Horizontal Stressσv = Vertical Stressrw = Wellbore Radiusr = Distance from the wellboreθ = Angle between a point on the circumference of the wellbore and the direction of MaximumHorizontal Stress (σH)pw = Wellbore fluid pressureτrt / τta / τra = Shear stress in radial/tangential, tangential/axial, radial/axial planes respectively

In the case of a vertical well in an anisotropic stress field (ie σh ≠ σH) the problem becomes farsimpler.In a vertical well, the shear stresses are zero and therefore the normal stresses become principalstresses and are σH and σh.The above equations simplify to the following 3 equations:

2

2

2

2

4

4

2

2

2cos431)(211)(

21

rr

prr

rr

rr w

www

hHw

hHr +�

��

�

�−+−+��

�

��

�

�−+= θσσσσσ

2

2

4

4

2

2

2cos31)(211)(

21

rr

pwrr

rr ww

hHw

hHt −�

��

�

�+−−��

�

��

�

�++= θσσσσσ

θνσσσσ 2cos4

)(21

2

2 �

��

�

�−−=

rrw

hHva

These equations merely define the value of the wellbore stress (σr , σt or σa) as a function ofthe distance r from the wellbore and the azimuthal position of the wellbore stress θ.

The variables are σσσσr , σσσσt , σσσσa , σσσσh , σσσσH , r , θθθθ

The given knowns (constants) are rw , pw(The mud weight pw and the wellbore radius rw can be varied before the rock is drilled to see theeffect).


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a) At the wellbore wall itself r = rw and 1=rrw

The wellbore stresses simplify to:

σr = pw (C1)( ) ( ) whHhHt pcos2 θσσ2σσσ −−−+= (C2)

( ) cos2 θσσ2σσ νhH −−= va (C3)

These wellbore stresses are used in rock failure analysis.

The radial stress depends only on the mud weight while the tangential stress depends on theazimuthal position around the wellbore θ as well as the mud weight.

Note that from equation C3 the axial stress σa also depends on the value of θ. The reason is thatthe axial stress is affected by the horizontal stresses (which themselves vary with θ), just as thehorizontal stresses can be increased (forced laterally) by applying force axially.

b) Far from the wellbore wall r → ∞ and rw / r → 0

The stresses simplify to:

( ) ( ) θσσσσσ 2cos21

21

hHhHr −++= (C4)

( ) ( ) θσσσσσ 2cos21

21

hHhHt −−+= (C5)

va σσ = (C6)

This shows that the wellbore stresses diminish rapidly from the borehole wall converting to farfield stresses. This makes sense because at large distances from the wellbore the rock is in anunperturbed state.

For example, when θ = 0 (cos2θ = 1)

σr = σHσt = σhσa = σv

(When θ = 90° σr = σh σt = σH etc)


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Appendix D: Rock FailureShear Failure

Figure D1: The 3 principal stresses σ1 , σ2 , σ3 and the Mohr Circles connecting th

Safe region

Failure region

σn

τ

σ1σ1
σ2
σ3
σn
2θ

σ2

Failure curve

em


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Shear failure occurs when the shear stress along a plane is too large.

Mohr assumed the failure could be described by the following function:

| τ | = f(σ) (D1)

This equation describes a curve (function) that separates a safe region from a “failure” region5.

Figure D1 shows 3 principal stresses and the Mohr circles connecting them. Note the position ofthe failure curve. The failure curve is obtained experimentally and corresponds to failure under avariety of conditions.

If we increase σ1, the circle connecting σ1 and σ3 will expand and eventually touch the failurecurve causing a failure.The value of the intermediate principal stress σ2 does not affect this process – if σ2 is increasedup to a maximum value of σ1 (by definition cannot exceed σ1) it does not affect the outermostcircle and therefore does not affect the failure.

Hence:

• Pure shear failure depends only on the minimum and maximum principalstresses σσσσ1 and σσσσ3 and not on the intermediate stress σσσσ2.

• The failure condition is approached as the difference between the twoprincipal stresses σσσσ1 and σσσσ3 becomes larger (the diameter of the hemispherein figure 1.4 increases)

The most important criterion for rock failure was introduced by Coulomb (1773).

A rock will undergo shear failure across a plane if the shear stress ττττ generated in the planehas a magnitude that exceeds the inherent shear strength of the rock (S0) plus the opposingfriction force µµµµσσσσn :

| τ | = S0 + µσn (D2)

S0 = Cohesion of the material (can be regarded as the shear strength of the material)µµµµ = Static coefficient of internal friction of the material||||ττττ|||| = Shear stress magnitude generated across planeσσσσn = Normal stress

Since the sign of τ only affects the direction of the sliding friction, only the magnitude of τ is inquestion.


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The significance of µ is as follows:Imagine a shear stress acting on a material. Not only does the stress has to exceed the naturalshear strength of the material, but it also has to exceed a friction force that occurs between 2planes that are slipping past one another (shearing).This friction force is given by µN where N is a reaction force normal to the slip plane and µ thecoefficient of friction (this is a basic mechanics principle).

Once shear failure has been initiated in a rock, and assuming that sufficient stress ismaintained, the rock will continue to undergo irrecoverable (i.e. plastic) strain11.

Therefore for shear failure to occur:

| τ | ≥ S0 + µσn

or | τ | - µσn ≥ S0

τ and σn have been defined previously (equations ) and the LHS of the above equation can nowbe expressed:

| τ | – µσn = ½ (σ1 – σ2)sin2β – µ{½ (σ1 + σ2) + ½ (σ1 – σ2)cos2β} (1.14a)

We can find the maximum value of | τ | – µσn by differentiating with respect to β andsetting to zero.

Solving for β in this (maximum) condition gives:

Tan2β = – 1/µ (1.14b)

Therefore the value 2β lies between 90o and 180o.

And β = 90o – ½ Tan-1(1/µ)

ieµ = 0 (ie no friction) → β = 45o

µ = 1 β = 67.5o

µ → ∞ β = 90o

Hence for shear failure the value of β lies between 45° and 90°

β is the angle between the normal σσσσn and the max. stress σ1 as described in Appendix A

(The normal σn is oriented 90° from the shear stress. The shear failure plane is always in theshear stress plane. Therefore the normal is 90° from the shear failure plane.)


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Therefore the direction of the shear fracture is always inclined at an angle (0°°°° – 45°°°°) to thedirection of maximum stress, σ1.

This makes perfect sense because when:σ1 = σv Normal fault regime Fault is steep sloping (wrt horizontal plane)σ1 = σH Thrust fault regime Fault is gentle sloping (wrt horizontal plane)

Substituting 1.14b into equation 1.14a , we get the following maximum condition:

| τ | – µσn (max) = ½(σ1 – σ2)(µ2 + 1)1/2 – ½µ(σ1 + σ2) (1.14c)

If this maximum stress value is less than S0, then failure will not occur (physically impossible).However, if the value is equal (or greater than S0) then failure will occur.

The criterion for failure is therefore:

σ1{(µ2 + 1)1/2 – µ} – σ2{(µ2 + 1)1/2 + µ} ≥ 2S0 (1.14d)

(Re-arranging eq 1.14c & setting LHS to So)

The coefficient of friction µ is related to the angle of internal friction φφφφ as:

µ = Tan φ

(generally φ is around 30°)Furthermore, it can be shown that:

β = 45° + 2φ

Hence solving for σ1 (from equation 1.14d) we get:

σ1 ≥ 2So tanβ + σ3 tan2β

and Co = 2Sotanββββ where Co = unconfined compressive strength

The well-known shear failure criterion is therefore given by:

σ1’ ≥ Co + σ3’ tan2β (1.15)

The ’ are included in the maximum and minimum stresses to signify effective stress (ie totalstress – pore pressure). We need to be working in effective stress.


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Answers to Brain Teasers

1a) σH > σh > σv b) σv > σH > σh

2) a) Loading mechanisms – ie for a buried sediment: Rate of sediment compaction > Rate at which pore fluid is expelled

Unloading mechanisms – stress is relieved as sediment is unloaded causing productionof fluids which cannot escape due to sealed formation. Uplift / erosion of formationcauses pore pressure to be higher than neighbouring formations at same depth (assumingsealed formation)

Tectonic Stress – high horizontal tectonic stress system does not fracture to relieve stress. Porepressure increases if very low permeability seal exists.

3) a)Shear Failure Wide Breakout – when tangential stress > radial stress. Will get a breakout ofthe borehole some 30 – 90° wide and in direction of. On image will be a thick vertical feature.b) Tensile Failure Vertical – when sum of minimum tangential stress and tensile strength is lessthan zero. Will get a thin (sharp) feature on log image in direction of σH.

4)Variables that are unknown:To and σh

To = Formation Breakdown Pressure – Fracture Opening Pressure = 10152 – 9952psi = 200 psi

From the ISONIC log we can determine Poisson’s ratio (see page 55 for formula)

ν = (0.5(1202/652) – 1) / (1202/652) – 1) = 0.292

ko = ν / 1 – ν = 0.413

The minimum horizontal stress σh can be calculated from the relation:

σh = koσv + (1– ko)p (see page 56)

σh = (0.413 * 22) + (1 – 0.413)*11 = 15.55 ppgHence σH = 1.2*σh = 18.65 ppg

Also if φ = 35° then β = 45° + 35/2 = 62.5° (see equation 1.8)Converting Co into ppg:Co = 1800psi = 11000*EMW*0.052→ Co = 3.15 ppgTo = 0.35 ppg


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a) First we need to calculate the wellbore stresses:pw = ECD (what we are trying to find), p = pore pressure

σr = pw – p = pw – 11 ppgσt max= 3σH – σh – pw – p = 29.4 – pw (in the direction of the minimum horizontal stress)σt min= 3σh – σH – pw – p = 17 ppg – pw (in the direction of the maximum horizontal stress)σa = 22 – p = 11 ppg

Testing for shear failure breakouts:

Will occur when tangential stress is maximum stress (σ1’), radial stress is minimum stress (σ3’)Tangential stress is maximum in the direction of minimum horizontal stress.

Shear failure criterion limit:

0 = C0 + σ3’ tan2β – σ1’ (originates from equation 1.8, here in the limiting case)

0 = 3.15 + (pw – 11)*tan262.5° – (29.4 – pw)

pw = 14.25 ppg

Testing for tensile failure:

0 = σ3’ + To

0 = 17 – pw + 0.35

pw = 17.35 ppg

However, the desirable mud weight window = 14.25 to 15.55 ppg

This is because if there are a network of natural fractures, which is likely due to the low porositysandstone, mud weight should not exceed this (ie should be less than 15.55 ppg)

b) This is a relaxed basin as σv > σH Therefore the most stable direction to drill the well willbe in an azimuth of 335° or 155°. This is the orientation of the minimum horizontal stress.

In this orientation the tangential stress is maximum in the orientation of σH as it will be equal to3σv – σHIf the well were drilled at an azimuth of 245° or 65° (ie in the direction of the maximumhorizontal stress) then the tangential stress is maximum in the orientation of σh as it will beequal to 3σv – σhThe radial stress is the same and as always equal to the mud weight.As 3σv – σh > 3σv – σH shear failures occur more readily when you are drilling in thedirection of maximum horizontal stress.


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The tensile stress will be minimum when the tangential stress is oriented to σv (ie verticallyupwards). The tangential stress will then be 3σh – σv (if you are drilling parallel to themaximum horizontal stress) and 3σH – σv (if you are drilling parallel to the minimumhorizontal stress).Therefore the tangential stress is lower when you are drilling parallel to maximum horizontalstress. (The lower the tangential stress the nearer the condition for tensile failure.)From tensile failure viewpoint, it is also better to drill parallel to the minimum horizontal stress.

ECD window:

Shear failure: σ3’ = σr = pw – pσ1’ = σtmax = 3σv – σH – pw – p

pw = 15.73 ppg

Tensile failure: σ3’ = σtmin = 3σH – σv – pw – p = 3(18.65) – 22 – pw – 11 = 22.95 – pw

pw = 23.30 ppg

The best approach would be to use as high an ECD as possible without exceeding 15.55 ppg.The shear failures that result should be managed with appropriate hole cleaning practices.If we increase the ECD higher than 15.73 ppg (to avoid shear failure) this could haveundesirable consequences for borehole stability causing lost circulation (which would be severein these circumstances) and excess of tabular cavings.

c) We are using a mud weight that will be causing shear failure so we need to contain it the bestway we can. Angular cavings confirm the shear failure and excess torque could be due to theshear breakouts loading up the annulus. Best practices to use are a controlled (slower) ROP andcirculation, reciprocation (slowly to prevent swab/surge) while rotating the drillstring at eachconnection to optimize hole cleaning. Plenty of time should be done for this operation. A wipertrip could also be beneficial. You should not increase the mud weight as this could cause naturalfractures to open and cavings to break out into the wellbore due to weak planes making thesituation worse. Do not use a high flow rate for fear of increasing ECD and losing circulation.

d) Increase the mud weight and you have now probably opened up natural fractures (as ECD isprobably greater than σh). You should decrease the mud weight as you want to prevent lostcirculation – that would be very serious under the circumstances. You should not perform awiper trip.

e) This is a very serious situation. Full measures should be taken to reduce or stop the losses.

5) a) Breakouts – Wiper trips, decrease ROP, increase mud gel strength, increase mud weight(depends on your upper mud weight window).b) Natural Fractures/Weak Planes – Add fluid loss agents, increase mud gel strength, don’tperform wiper trips.c) Drilling induced fractures – Increase mud gel strength, decrease mud weight, add fluid lossagents.d) Sloughing Shales – Don’t perform wiper trips, increase mud gel strength


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References

1) Thiercelin, M. C. and Roegiers, J. C., Formation Characterisation: Rock Mechanics

2) Rock Mechanics (Volume 1) Theoretical Fundamentals Ph A. Charlez

3) Rock Mechanics (Volume 2) Petroleum Applications Ph A. Charlez

4) Jaeger, JC and NGW Cook (1979), Fundamentals of Rock Mechanics, 3rd Edition, Chapman and Hall, London

5) Petroleum-related Rock Mechanics Fjaer, Holt, Horsrud, Raaen, Risnes. Elsevier: Developments in Petroleum Science 33

6) Logging-While-Drilling Images for Geomechanical, Geological and Petrophysical Interpretations – Bratton et al.

7) Pore Pressure Prediction / Evaluation Methods – Qiuguo, Chang (Schlumberger)

8) Stress Magnitudes in Sedimentary Basins: Relationship to Lithology, Pore Pressure and Geologic Setting – R. A. Plumb

9) A Wellbore Stability Planner – T Bratton

10) The Quest for Borehole Stability in the Cusiana Field, Colombia – D. Plumb

11) An Introduction to Near-Wellbore Rock Mechanics – R. Marsden

12) The Mechanical Earth Model Concept and its Application to High-Risk Well Construction Projects – R. Plumb et al.

Other useful references:• http://www.bgc.beijing.oilfield.slb.com• http://www.cambridge.scr.slb.com/SCR/Dept/IGM/geomechanics/• http://www.hub.slb.com/index.cfm?id=id554• http://www.cambridge.scr.slb.com/SCR/Dept/IGM/geomechanics/GM_home_pa

ge_.html• http://www.cambridge.scr.slb.com/OFSR/WCN/GM/WS/frontpage.html

http://www.bgc.beijing.oilfield.slb.com/

http://www.cambridge.scr.slb.com/SCR/Dept/IGM/geomechanics/

http://www.hub.slb.com/index.cfm?id=id554

http://www.cambridge.scr.slb.com/SCR/Dept/IGM/geomechanics/GM_home_page_.html

http://www.cambridge.scr.slb.com/SCR/Dept/IGM/geomechanics/GM_home_page_.html

http://www.cambridge.scr.slb.com/OFSR/WCN/GM/WS/frontpage.html

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