00056628

7/30/2019 00056628

1/16

Copyright 1999, Society of Petroleum Engineers Inc.

This paper was prepared for presentation at the 1999 SPE Annual Technical Conference andExhibition held in Houston, Texas, 36 October 1999.

This paper was selected for presentation by an SPE Program Committee following review ofinformation contained in an abstract submitted by the author(s). Contents of the paper, as

presented, have not been reviewed by the Society of Petroleum Engineers and are subject tocorrection by the author(s). The material, as presented, does not necessarily reflect anyposition of the Society of Petroleum Engineers, its officers, or members. Papers presented at

SPE meetings are subject to publication review by Editorial Committees of the Society ofPetroleum Engineers. Electronic reproduction, distribution, or storage of any part of this paperfor commercial purposes without the written consent of the Society of Petroleum Engineers is

prohibited. Permission to reproduce in print is restricted to an abstract of not more than 300

words; illustrations may not be copied. The abstract must contain conspicuousacknowledgment of where and by whom the paper was presented. Write Librarian, SPE, P.O.

Box 833836, Richardson, TX 75083-3836, U.S.A., fax 01-972-952-9435.

Abstract

The complexity of the wells have increased significantly inlater years. Reach has more than doubled, and highinclination and fully 3-D well paths are common. However,statistics shows that sidetracking the boreholes due to stuckpipe has also shown a significant increase, and is presently ahigh cost factor. The margins between success and failure arenow much smaller.

A larger study was initiated to understand the stuck pipesituation better, and to develop improved procedures. Amechanistic approach was chosen. The following elementswere analyzed: 1) the forces developed during differentialsticking, 2) pipe strength under combined loads; tension,torque and pressure, 3) effects of buoyancy under variousconditions like equal or different mud densities in drillpipeand annulus, 4) wellbore friction as related to torque and drag.

This paper presents new equations to determine depth to thestuck point in deviated wellbores, based on pulling tests andtorsion tests. In particular it is shown that bends in thewellbore leads to more friction, which with the new equations

results in a deeper stuck point in a deviated well compared to asimilar vertical well. Knowing all the forces involved in astuck pipe case, another analysis was performed to determinethe action that has largest impact to free the pipe. One of themain conclusions is that the most important element to freethe string is to keep the bottomhole pressure as low aspossible. The paper will present three methods to free thepipe, which where developed from the analysis: 1) maximummechanical force method, 2) minimum density method, and 3)maximum buoyancy method. A detailed field case from the

Yme field in the North Sea will demonstrate these methods,and show the effect on the stuck point using each method.

In addition to the field case, the paper will in the Appendicespresent the complete equations for pipe stress and strengthunder 3-dimensional loading, and define the effects ofbuoyancy in deviated wellbores. In particular is the buoyancyissue resolved, showing the differences and similarities

between the piston force approach and the law ofArchimedes, as applied to deviated wellbores.

Forces in the drill string

Axial weight and buoyancy

The calculation of buoyancy on tubulars has led toconsiderable confusion over the years. Instead of resolvingthe issue, one has typically used one of the followingapproaches; the Archimedes law, or the piston forceapproach. The first approach simply states that the buoyancyequals the weight of the displaced fluid, whereas the pistonforce is a uniaxial force balance applied to each geometricchange in the string.

In Appendices A and B of this paper this issues has beenresolved. It is shown that the two approaches mentionedabove gives identical results if correctly applied, and that theArchimedes principle can be used for all cases. We will herestate the conclusions, but refer the reader to Appendices A andB for details.

If the unit weight of a pipe is wdrill pipe, the weight whensubmerged into a fluid is:

w wdrillpipe= (1)

The buoyancy is given by the factor , which is defined asfollows:

If the pipe has fluid of a different density inside compared tothe outside, the buoyancy factor becomes (App. B4):

SPE 56628

Analysis of Stuck Pipe in Deviated BoreholesBernt S. Aadny/Stavanger University, Kenneth Larsen/Statoil and Per C. Berg/Statoil

= 1 2

fl

p ipe

( )

7/30/2019 00056628

2/16

2 B. S. AADNY, K. LARSEN AND P. C. BERG SPE 56628

=

1

2 2

2 2o o

pipe o i

r i r

r r

i

( )(3)

If a surface pressure is applied inside the string, outside thestring or on both sides of the string, the buoyancy remainsunchanged. However, for a surface pressure applied on the

outside or on both sides of the string a lifting force results,which equals the pressure multiplied with the drill string areabeneath the wellhead, or (App. B5):

For inclined boreholes, the buoyancy factors are identical tothose for vertical holes. However, the axial weight is equal tothe unit weight multiplied with the projected vertical depth, or(App. B6):

W wDaxial TVD= (5)

The application of these equations will be demonstrated inThe Yme Field Case. The simplicity of this equation is thatthe static pipe weight at surface is always the unit weightmultiplied with the projected depth, regardless of holeinclination.

Pipe strength

There are two types of pipe failure, a tensile failure and ashear failure. Which failure type that dominates depends onthe loading and on the material properties. Pure tension maylead to a tensile failure, whereas pure torsion will lead to ashear failure. Since we often have combined loading, theprincipal stresses should be computed. Below are some load

cases defined.

Uniaxial tensile pipe strength. This case only consider theaxial stretch of the drillpipe at the rig floor. It can beformulated as:

a yield (6)

Often we use the yield strength of the pipe material. However,the ultimate strength, which is higher, can also be used.

Torsion, pressure and axial load. The most importantcombined loading scenario is if an axial load, a differential

pressure and a moment is applied simultaneously. Separately,these stresses are as follows for a thin-walled pipe(seeAppendix A):

Axial stress:

aF

A= (7)

F can be an external force and/or tension due to internaloverpressure (Eqn. A7).

Tangential stress caused by a pressure differenceinside/outside, assuming thin-walled pipe:

t Pr

t= (8)

Shear stress caused by applied moment, assuming thin-walled

pipe:

=M

r t2 2(9)

The two normal stresses above may have yield as an upperlimit, while the shear stress has yield as the upper limit. Oneway to handle this situation is to take the vector sum of thecapacities of each component. Two approaches will be shownin the following.

Principal stresses.The classical way to consider 3-dimensionaloading is to compute the principal stresses, and to compare

these to the failure condition. Having the three stresscomponents defined above, the principal stresses are:

[ ]

=

=

a

t

0

0, which solved becomes:

( )

1 2

21

2 22

, = +

+a t

a t(10)

Equating the material yield strength to the maximum principal

stress, the maximum permissible axial load is from thisequation:

a yield

t yield

+

2

(11)

Von Mises yield strength.This equation is often used forductile materials. It is based on the 2nd deviatoric invariantwhich looks as follows:

( ) ( ) ( )J x y x z y z

xy xz yz

2

2 2 2

2 2 2

1

2

1

2

1

2

3 3 3

= + +

+ + +

(12)

Computing the invariant for a uniaxial tensile test leading toyield, and inserting the loads defined above, the resultbecomes:

a t yield t + 1

2

3

432 2 2 (13)

F P ro o= 2 4( )

7/30/2019 00056628

3/16

SPE 56628 ANALYSIS OF STUCK PIPE IN DEVIATED BOREHOLES 3

Differential sticking.

The two most common reasons for stuck drillstring ismechanical sticking due to jamming in the hole anddifferential sticking. It is often difficult to determine whichtype that applies. However, we will here address thedifferential sticking case below.

For a deviated well, the force needed to pull the drillstring canbe expressed as the sum of the pipe weight, the drag force andthe diff. sticking force, or:

F = wh(cos + sin) + d hP (14)

where h is the length of the stuck interval. Fig. 1 illustrates adiff. sticking scenario.Here we assume that the stuck point is in a straight holesection. The force normal to the stuck point is the componentof the weight of the pipe plus the hydraulic force that causesdifferential sticking.The equation above says that the stuck force is equal to the

normal force multiplied by a coefficient of friction. Thepressure differential, P, is the difference between the outsidemud pressure and the pore pressure in the rock. The boreholepressure may be the static mud pressure plus the surfacepressure if the well is pressurized in the annulus.

Depth to the stuck point

Aadnoy and Andersen(1998) defines the equations needed tomodel torque and drag for any well geometry and for any caseof loading. We will use some of these equations, but point attwo differences between the ordinary drilling and the stuckpipe case.

1) During drilling the complete drillstring is pulled orrotated from top to bottom. During the stuck case,the top part is pulled/rotated whereas the bottom partis fixed. The average displacement for the stuck caseis just the strain of the drillstring, which is very small.

2) During drilling a considerable speed is used (e.g.rotating at 100 rpm), while in the stuck case the speedis considerably lower. Although speed is not aparameter in the basic torque/drag models, it can beargued that low speed combined with jerking loadminimizes the friction. It is well known that e.g.

vibrations lower the effective friction coefficient.

The analysis to follow will apply the friction models fromAadny and Andersen(1998), but take the two considerationsabove into account.

Pull test.

When a drillstring is stuck it is important to determine thedepth to the stuck point. Often the string has to be cut just

above the stuck point before a sidetrack operation is initiated.To estimate the depth to the stuck point, the string is pulledwith an additional force dF, and the elongation dL is measuredon the surface. Since steel behaves linearly elastic, Hookeslaw describes the relationship between force and elongation,dF = AEdL/L, where A is the crossectional area and Eis themodulus of elasticity.

The reference is the string hanging in its own weight. If thewell is vertical, and if friction is neglected, the depth to thestuck point becomes, if the drillstring is composed ofidifferent elements:

dL dLE

L

AdFn

n

in

nn

i

i= == =

0 0

1(15)

Assuming that the bottom-hole-assembly is negligible becauseit is much stiffer than the drillpipe, and that one drillpipe sizeis used, the depth to the stuck point becomes for a verticalwell:

L EAdL

dF= (16)

In the above example we assume that there is no friction in thewell. This means that when a pull force dF is applied atsurface, this force is reflected all the way to the stuck point.This mean that this force is acting directly on the stuck point.

If we introduce drag, some of the pulling force goes tofriction. However, usually the pull-rate is slow, nearly static,and it can be a reasonable assumption to neglect drag in avertical well.

As a general rule, the pull force is assumed unaffected offriction for vertical wells. For straight sections in deviatedwells, friction can either be neglected, or be taken care by thedrag forces measured before the drillstring became stuck.However, Aadnoy and Andersen(1998) showed that curvedsections of the borehole has the effect of amplifying thefriction. This is seen in Fig. 2, where the vertical well showsthe pull force all the way down to the stuck point in Fig. 2a.In Fig. 2b, the deviated well, the pull force is remainingconstant all the way to the build-up section. Through thebuild-up section some of the pull force is taken up as friction.Below the build-up section, a smaller pull force is seen.

Assume that the well consist of a vertical section to the kick-off point, L1. A build-up section of radius R builds up to anangle . A sail section maintaining this inclination continuesto the stuck point. We assume that the pipe is pulled slowlywith a force dF a distance dL. Drag is neglected. This forcedF is reflected down to the kick-off point. Due to thecurvature, the force at the bottom of the build-up bend is

7/30/2019 00056628

4/16


(Aadnoy and Andersen, 1998):

dF dFe1 =

The elongation through the build-up bend is:

( ) ( )dL dF dF R

AE

R dF

AEe2 1

1

2 21= + = +

The elongation in the vertical section is:

dLdFL

AE1

1=

The elongation in the sail section down to the stuck point L is:

( )( )dL

dF L L R

AE

dFe

AE

L L R31 1

1=

=

The total elongation measured at surface is the sum of thethree contributions defined above, or:

( ) ( )dLdF

AEL

Re e L L R= + + +

1 12

1

The depth to the stuck point is from the equation above:

( )L AEedL

dFe L R= +

1

1

21(17)

The equation above is valid if one drillpipe size is used, andfor a well consisting of one build-up section and a constantsail angle in the section below. Assuming two drillpipe sizesthe equation can be modified. Referring to the top pipe sizewith index 1, and the bottom pipe size with index 2, the depthto the stuck point is:

( )L A EedL

dF

A

Ae L R L

A

A= +

2

2

11 2

2

11

1

21

(18)Torsion test.If we rotate the drill string at surface, an applied moment dMleads to an twist angle d. The corresponding expression forthe angle of rotation as a function of depth is similar to thepulling case:

dG

L

Jnd M

i

n

i

==

10

(19)

The shear modulus is:( )

GE

=+2 1

and the polar

moment: ( )J D dp =

324 4

Using an elastic modulus of 215 kN/mm2, a Poissons ratio of

0.25 for steel, Equation 19 can be expressed in degrees as:

( )( )

( )( )d x

L m

J mmdM kNm

n

pnn

i

==

6 66 105 40

.

If pure rotation is applied, the force amplification through thebuildup bend will be neglected. Again for small rotation thecumulative frictional torque will be neglected, and the totaltwist angle at surface is determined from the total length of thedrillstring.

If only one drillpipe type is used, the stuck point can beestimated from the following equation:

Again the equation will be expanded to the case of using twodrillpipe sizes. From Eqn. 20, the depth to the stuck point nowbecomes:

The equations above are valid for well paths consisting of avertical section to the kick-off point, a constant build sectionand a constant inclination sail section to the stuck point. For

more complex geometries similar equations can be derived,but these will become more cumbersome. A simple numericalroutine may be handy for these cases.

The Yme field case

During drilling of the 7854 m long exploration well 9/2-8S onYme, several stuck pipe incidents occurred. Two of thesestuck pipe incidents resulted in the bottom-hole-assembly(BHA) being left in the hole and the well had to be sidetrackedtwice in the 12-1/4 in section. The second incident whichresulted in the BHA being shot off will be analyzed, due to

better data from this incident. Fig.3 shows the well path forthis case.

From the drilling operation, the following friction data weredetermined:

On bottom torque at stuck point depth: 26 kNmOff bottom torque at stuck point depth: 23 kNmFriction factor with syntetic mud: 0.12

( )L

JE d

dM=

+2 120

( )

( )L

J E d

dML

J

J=

+

2 1

2

12 11 21

( )

7/30/2019 00056628

5/16


The friction factor above, which was measured during drilling,will be used in the analysis to follow. We assume that thesame coefficient of friction is valid also after the pipe wasstuck.

Description of case. After sidetracking the hole following the

first stuck pipe incident, a sand reservoir was entered at 5830mMD. At 6011 mMD the washpipe started leaking, and it wasdecided to pull out to above the reservoir for replacement.The first stand was pulled without excessive drag. From thetime rotation was stopped to set back the stand 7 minutespassed before the second stand was attempted pulled. It wasnot possible to pull the drillstring and 260 tons hookload wasused. The full string weight was also applied, without anyeffect. Combined torque and pull was also applied withoutsuccess. The jar did not go off in any of the attempts to freethe string.

Full circulation was maintained throughout these attempts,leading to the conclusion of differential sticking. The section(13-3/8 shoe at 2281m) was drilled with a mud weight of1.65 s.g. for borehole stability reasons, and the pore pressurewas determined to be 1.24 s.g.

First we will estimate the depth to the stuck point in the Ymewell. Pipe data are shown in Table 1. The followingadditional data are given from the stretch test on the rig:

dF= 392 kN L1 = 3548 mdL = 2.60 m L2 = 2298 mE= 215 kN/mm2 R= 840 mA1 = 3401 mm

2 = 68A2 = 4210 mm

2 = 0.12E= 215 kN/mm2.

Inserting these data into Eqn. 18, the depth to the stuck pointis as follows:

First,

= =01268

1800142. . and: e = 1153. and:

1

2

68

2 180840 498

R

xm= =

For comparison, the estimated depth for a vertical well is:5457m. This is found by setting =0 in the equation above.

The above calculation is repeated with different frictioncoefficients. The result is shown in Fig. 4 shows the depth tothe stuck point for various friction coefficients for theexample.

The torsion test also provide important information. Thedrillstring was rotated 14 turns, with an applied moment of 20kNm. Using data from Table 1, the estimated length of thedrillstring is:

It is seen that two independent tests can be performed todetermine the depth to the stuck point, both pull test andtorsion test. Clearly, the string is stuck in the bottom-hole-assembly.

The weight of the drillstring.

The axial weight of the string will be computed with Eqn. 5.The mud weight used was 1.65 s.g. resulting in a buoyancyfactor of(Eqn.2):

= =1165

7 85079

.

..

The effective weight of the drillstring is given by the unitweight multiplied with the projected depth. The 5 in. drillpipehas a vertical depth of 2060 m, whereas the 6 5/8 in. pipeextends to a vertical depth of 2700 m. The total weight,excluding the bottom-hole-assembly is:

( ){ }W kN mx m kN m

kN

= +

=

0 79 0 336 2060 0 406 2700 2060

752

. . / . /

This is the static hookload on the rig. During hoisting orlowering of the string, drag will be added to this load.

Maximum axial load of the pipe.The upper part of the drillstring consists of a 5 in. drill pipe.The highest load will occur in the upper part. Therefore, wewill compute the maximum load at this position. The pipedata from Table 1 applies.

Inserting the data into Eqns. 7, 8 and 9, the pipe stresses canbe expressed as:

Axial load: a barF N

( )( )

.=

340 1

Tangential stress caused by a pressure differenceinside/outside, assuming thin-walled pipe:

t bar P bar ( ) . ( )= 8 36

Shear stress caused by applied moment, assumingthin-walled pipe:

( ) . ( )bar M Nm= 0054

( )( )L x

m

= +

=

215 42102 60

3921153 1153 1 3548 498

22984210

3401 1 5756

.. .

L x x xx

m

=

=

15 10 27 05 1014 360

203548

2705

11881 56976 6. .

.

.

7/30/2019 00056628

6/16


The maximum axial load from Eqn. 11 is:

F NM

P( ) .

.

.= +

3401 7340

000287

836 7340

2

(22)

A few conditions of this equation is shown below:

Condition 1: P = 0 bar, M = 0 Nm F = 2496 kNCondition 2: P = 50 bar, M = 0 Nm F = 2496kNCondition 3: P = 0 bar, M = 10000 Nm F = 2487 kNCondition 4: P = 200 bar, M = 10000 Nm F = 2479 kN

The classical principal stress approach has a shortcoming asthe pressure only has an effect if there simultaneous is a torqueapplied. We will instead use the 3-dimensional approachgiven by the von Mises equation (Eqn. 13).

Inserting the equations above into Eqn. 13, the following

equation results:

{ }F N P P M( ) . . . .= + 340 1 418 7340 52 42 0 008752 2 2

A few conditions of this equation is shown below:Condition 1: P = 0 bar, M = 0 Nm F = 2496 kNCondition 2: P = 200 bar, M = 0 Nm F = 2732 kNCondition 3: P = -200 bar, M = 0 Nm F = 2163 kNCondition 4: P = 0 bar, M = 10000 Nm F = 2476 kN

Condition 1 defines pure axial pull. The yield strength is then2496 kN. Condition 3 shows the case of applying a pressureoutside the drillpipe only, resulting in an axial strengthreduction. Condition 4 shows the case of applyingsimultaneous pull and torque, again resulting in a reducedaxial strength.

Condition 2 shows the case of applying pressure on theinside of the pipe only. This is possible only if the inside isclosed e.g. by plugged bit nozzles. The result is that the axialstrength increases by 236 kN.

Increase in strength by combined loading.

It is observed in the examples above that the acceptable axialload in the drillpipe may be increased by applying a combined

load. Let us first rewrite Eqn. 13 for the case of zero torque:

a

yield

t

yield

t

yield

=

1

21

3

4

2

(23)

This form of the equation is well known. Actually it describes

an ellipse, with a ratio of the axis of 3 . The fourth quadrantis very much used in the petroleum industry to compute the

reduction in collapse resistance caused by axial tension. Forthat case an outside drillpipe overpressure is required.

The first quadrant is of interest to the present topic. If acombined tangential and axial tension is applied, the axialstrength may increase. Physically it means that bysimultaneous loading in two directions the deviatoric stress is

kept at a lower value. The following defines the maximumaxial loading from Eqn. 22:

a yield

t yield

=

=

1155

0577

.

.

The axial load to yielding can actually be increased by 15.5 %by applying a bi-axial loading.

Using the same data as before, we will compute the pipestrength under combined loading. Using Eqn. 22, themaximum axial load is:

F x x kN

Px

bar

= =

= =

340 1 1155 7340 28830577 7340

836506

. ..

.

To obtain maximum pipe strength at surface, the surfacedrillpipe pressure should be 506 bar. The mud pump on adrilling rig can at most supply about 300 bar pressure.Inserting this condition, Eqn. 22 modifies to the followingmaximum condition:

a yield

t yield

=

=

1126

0342

.

.

The pipe strength is now: F = 340.1x1.126x7340 = 2811 kNat P = 300 bar. Please observe that the numbers above arevalid only for the particular drillpipe used in the example.

So far we have concluded that the total axial load on thedrillpipe can be increased by a combined loading. However,as we pressure up the inside of the pipe the hydraulic pressureadds an axial component as the pipe is closed at both ends.The force is here the pressure multiplied with the inside pipearea or:

F = 300(bar)93(cm2)100=277 kN

The net external pull force the pipe can be subjected to atsurface is then:

2811 kN - 277 kN = 2534 kN

The pipe strength in simple tension has been determined to2679 kN.

The conclusion is that the maximum lifting force is nearly

7/30/2019 00056628

7/16


equal if a simple tension is applied, or if a combined loadingis applied. If pumping is applied to provide lift at the bottom,the maximum allowable lift force is not significantly reduced,and nearly the same pull can be applied.

Forces during pipe release.

The main objective of this paper is to determine the most

efficient way to free a stuck drillstring. Now we will derivethe equations required to perform this analysis. In theprevious chapter on determining the depth to the stuck point,we neglected friction because we assumed negligible motionas we applied a pull force. However, if this pull force isincreased such that the string starts to move and eventually isfreed, one may assume that axial friction is present. Theanalysis of this paper is based on these assumptions.

Aadnoy and Andersen(1998) presented all equations requiredto determine the string weight and the well friction for all wellgeometries. We will use the equations that are required toanalyze the present case, but refer the reader to the abovereference for other well geometries.

Applying pull force. Figure 3 shows the well geometry for theYme well. The pipe is stuck in the bottom-hole-assembly.Above this point the well path consists of a long sail section, abuild-up section and a vertical section to surface. A studyconcluded that the width of the stuck point was d= 0.05m,along a permeable interval of h = 22 m. Inserting the well datainto Eqn. 14, the pull force of the bottom-hole assembly is:

( )F x x x x P

F kN P kN m

1

12

1344 155 68 012 68 012 005 22

101 2 0132

= + +

= +

. cos . sin . .

( ) . . ( / )

The axial load of the pipe in the sail section again consists ofthe axial weight component and the drag force, or:

( )( )F x x2 0406 1848 0 336 2298 68 012 68 740= + + = . . cos . sin

The axial load through the build-up section depends on thetotal forces below and is (from Aadny and Andersen, 1998):

( ) ( ) ( )F F F e wR e3 1 2 2 12 1 2 1= +

sin sin

Inserting the data for the well, the following equation results:

( )F F F3 1 21153 302= + +.

Adding the effective weight of the vertical pipe, the total pullon top equals:

F F x F4 3 30 336 700 235= + = +.

Inserting the respective equations into the last equation, the

pull force as surface can be expressed as:

F kN P kN m421507 0152( ) . ( / )= + (24)

Please observe the multiplicative effect of the build-up bend.

Applying torsion.One may also apply torsion to the string to

attemt to free it. Aadny and Andersen(1998) defines allequations required to determine the torque when the drillstring is rotated. The torque is defined as the normal forcemultiplied with the friction coefficient and the radius to thecontact point. We use an average radius of the bottom-holeassembly of 0.1m. Since the drillpipe contact is on theconnections, we assume the same radiushere .

The torque of the bottom-hole-assembly becomes:

( )T x x x P P 1 01 012 1344 155 0 05 22 25 0 0132= + = +. . . . . .

The torque at the top of the sail section is:

( )T T x x x

T

2 1

1

01 012 0406 1848 0336 2298 68

16 9

= + +

= +

. . . . sin

.

The torque at the top of the build-up section equals the torqueat surface, since the upper drill string is vertical. The axialweight at the end of the build-up section is, neglecting thebottom-hole-assembly, which is stuck:

( ) ( )( )F kN1 0 336 2060 1450 0406 2700 2060 465= + = . . ( )

The torque at the top of the well is:

( )( )

( )

T T r F wR

wR

3 2 1 1 2 1

2 12

= + +

+

sin

cos cos

Inserting the data for the well, the resulting torque equationbecomes:

T T3 2 1296= + .

Combining the equations above, the total torque can beexpressed as:

( ) ( )T kNm P kN m32

32 4 0 0132= +. . / (25)

Combined pulling and rotation.Often one attempt to free thedrillstring by simultaneously pulling and rotating. Aadnyand Andersen(1998) have analyzed this scenario and haveshown that by combined motion the axial drag is reduced.This effect is also seen during backreaming. The pipe has atotal frictional capacity given by the weight and the friction

7/30/2019 00056628

8/16


coefficient. If a combined motion is applied, the vector sumof rotation and pulling equals this same frictional capacity.

For the present study we assume that the string is rotated 100rpm for a short period of time. The tangential speed of thebottom-hole-assembly with a radius of 0.1 m is 1.05 m/s.Next, we assume that the pipe is simultaneously hoisted with

the same speed. According to Aadny and Andersen(1998)the total frictional capacity is now equally split between thetorque and the axial drag, or:

T

rF

Fdrag

= =2

22

2(26)

Methods to free the pipe

In the following an analysis will be performed using some ofthe equations derived. This analysis will demonstrate theeffects of the various ways a stuck pipe problem can behandled.

Maximum mechanical force

This simplest approach is to pull or rotate the drillstringtowards the strength limit of the drillpipe. For the presentfield case, the mud density used was 1.65 s.g., and the porepressure in the permeable interval was estimated to 1.24 s.g.The projected area at the stuck position was estimated to 0.05m times 22 m. The differential pressure and the buoyancy is:

( )P bar kN m= = =

= =

0 098 165 124 2936 118 11800

1165

7 850 79

2. . . ) /

.

..

Inserting these data into Eqn. 23, the force required to free thedrillstring is 2984 kN. However, from Table 1 the axialstrength of the drillpipe is 2495 kN. Therefore, this is notsufficient to free the pipe.

Inserting the same data into Eqn. 24, the torque required is181.3 kNm, which also far exceeds the torque limit of thedrillpipe of 78.8 kNm. The conclusion is that by applyingmaximum mechanical force only, the string will remain stuck,and a parted drillpipe may result.

Minimum density methodIt is very obvious from Eqn. 23 and 24 that the dominatingparameter is the differential pressure that is the main cause ofdifferential sticking. This can be reduced by displacing thewell to a lighter mud. We will repeat the analysis assumingthat the well has been displaced to a 1.3 s.g. mud. Thedifferential pressure and the buoyancy now becomes:

( )P bar kN m= = =

= =

0 098 130 124 2936 17 36 1730

1130

7 85083

. . . ) . /

.

..

Inserting these data into Eqn. 23 leads to a surface force of1514 kN which is below the pipe strength of 2495 kN. Eqn.

24 gives a torque of 49.7 kNm, which is below the torquestrength of 78.8 kNm. In other words, displacing the well to alighter mud is the best measure to free the drillstring. Thebuoyancy decreases, but this effect is negligible to the effectof reducing the bottomhole pressure.

Maximum buoyancy method.

The last method analyzed is to displace the inside of thedrillpipe to seawater to increase buoyancy. This is permissibleproviding that well control is maintained as the seawater pilllater is displaced up the annulus. For this case we assume theinitial mud of 1.65 s.g. in the annulus which results in theinitial differential pressure of 118 bar. The buoyancyincreases and is given by Eqn. 3:

( ) =

=1

165 127 103 108 6

7 85 127 108 6057

2 2

2 2

. . .

. ..

x x

Again inserting these numbers into Eqn. 23 leads to a requiredpull force of 2652 kN and Eqn. 24 leads to a required torqueof 174 kNm.

Below are the results summarized.

Method: Reqd. pull force (kN): Reqd. torque(kNm)

Max. mech. force: 2984 181.3Min. density: 1514 49.7Max. buoyancy: 2652 174

Pipe strength: 2495 78.8

This example demonstrates that there is some potential inincreasing buoyancy by displacing the inside of the drillstringto seawater. The most important measure, however, is toreduce the annulus pressure by displacing the well to a lightermud.

Summary

This paper present a mechanistic analysis of differentiallystuck pipe in a deviated well. The following mechanisms areanalyzed; buoyancy effects, pipe strength, differential stickingand well friction.

Equations are derived to estimate the depth to the stuck pointin deviated wellbores based on pull and rotation tests. Due to

7/30/2019 00056628

9/16


friction in bends, the stuck point appears deeper in a deviatedwell compared to a vertical well.

Equations for pipe strength under combined loading aredefined. In particular, it is shown that the drillpipe can beloaded towards uniaxial pipe strength even if a full pumppressure is applied on the inside.

The differential pressure across the stuck interval is thedominating factor. The most important remedy to free thepipe is to reduce the bottom hole pressure. This can be doneby displacing the well with a lighter mud. One can alsoincrease buoyancy by displacing the inside of the drillstringwith seawater.

Nomenclature

P = pressureA = areaF = force

r = pipe radiusd= width of mudcakeR = radius of build-up sectionw = unit pipe weightW = total weightD = well depthh = length of stuck intervalE = Youngs modulusG = shear modulusJ = polar moment of inertiaL = pipe length = buoyancy factor = density

= strain = coefficient of friction = hole inclination= rotation of the drillstring = hole azimuth

= Poissons ratio

Subscripts

a = axialr = radialt = tangential/wall thicknessi = inner/initial

o = outerf = frictionMD = measured depthTVD = projected vertical depthfl = fluid

References

Aadnoy, B.S. and Andersen, K.: Friction analysis for long-reachwells. Paper IADC/SPE 39391 presented at the 1998 IADC/SPEDrilling Conference, Dallas, TX, Mar. 3-6.

Adams,A.J., Parfitt,S.H.L., Reeves,T.B. and Thorogood,J.L.: Casingsystem risk analysis using structural reliability. Paper SPE/IADC25693 presented at the 1993 SPE/IADC Drilling Conference,Amsterdam, Feb. 22-25.

Chesney,A.J. and Garcia,J., 1969: Load and stability analysis oftubular strings. Paper 69-PET-15, ASME Petroleum Mechanical

Engineering conference, Tulsa, OK

Dellinger,T.B., 1973: Preventing instability in partially-cementedintermediate casing strings. Paper presented at the 48th Annual FallMeeting of the SPE, Las Vegas, Nov.

Goins, W.C., 1980: Better understanding prevent tubular bucklingproblems. Part 1- Buckling tendency, causes and resulting problemsdescribed. World Oil, January, pp.101-105.

Goins, W.C., 1980: Better understanding prevent tubular bucklingproblems. Part 2 Graphic solutions are presented for fieldsituations. World Oil, February, pp. 35-40.

Hawkins,M.F. and Lamont,N, 1949: The analysis of stresses in drill

pipe. Drilling and production practice, American Petroleum Institutep. 358.

Hammerlindl,D.J., 1978: Basic fluid and pressure forces on oil welltubulars. Paper presented at the 55th Annual Fall Meeting of theSociety of Petroleum Engineers, Houston, TX.

Kastor,R.L., 1986: Triaxial casing design for burst, Paper IADC/SPE14727 presented at the 1986 IADC/SPE Drilling Conference, Dallas,TX, Feb. 10-12.

Klementich,E.F. and Jellison,M.J., 1986: A service-life model forcasing strings. SPE Drilling Engineering, April 1986, pp. 141-152.

Klinkenberg,A., 1951: The neutral zones in drill pipe and casing andtheir significance in relation to buckling and collapse. Drilling andProduction Practice, American Petroleum Institute, pp. 64-76.

Kyllingstad, , 1995: Buckling of tubular strings in curved wells.Journal of Petroleum science and Engineering 12 (1995), pp. 209-218.

Lubinsky, A., Althouse,W.S. and Logan,J.L., 1968: Helical bucklingof tubular goods. Journ. of Petroleum Technology, June, pp. 655-70

Patillo,P.D. and Randall, B.V., 1980: Two unresolved problems inhydrostatics. Paper presented at the 1980 IADC/SPE DrillingTechnology Conference, Houston, TX,March 18-20.

Stephens,D.R. and McConnell,D.P., 1985: Pipeline designcodescompared graphically. Oil and Gas journal, July 29, pp. 139-144.

Thulin,L.G., 1971: Mechanics of materials II. Course compendium,University of Colorado.

7/30/2019 00056628

10/16


Appendix A: The Lame solution

Most of the components in a petroleum well has a cylindricalshape. Therefore equations to determine stresses in cylinders

are most important to perform analysis of drill strings. If the

geometry of a particular component deviates from a cylinder,other models must be used.

The equations to follow are obtained from classicalmechanics. They are valid for linearly elastic materials.

Figure A.1 shows the stresses acting in a thick-walledcylinder. The so called Lame equations defines the magnitudeof these stresses. The radial and tangential stresses are givenby (Thulin, 1971):

r

o o i i

i o

o i

o i

p r p rr r

rp p

r r

=

2 22 2

2

2 2

( )

(A1)

t

o o i i

i o

o i

o i

p r p rr r

rp p

r r=

+

2 22 2

2

2 2

( )

(A2)

We observe that the two expressions are very similar. If theyare added, the sum is constant across the cross section. Thisproperty will be used to assess the net axial load.

( )( )1

2

2 2

2 2 r to o i i

o i

P r Pr

r r+ =

(A3)

The axial stress depends on the end condition. If the tube isopen ended, the pressures causes zero axial stress, whereas fora closed end pipe, the pressure causes the fol lowing axialstress:

ao o i i

o i

p r p r

r r=

2 2

2 2 (A4)

Often one uses a thin-walled solution for slim tubing.Introducing the wall thickness t, and assuming constant stressacross the wall, the above equations can be approximated by:

r o ip p= +12( ) (A5)

( ) t o i o ir

tp p p p= + +( )

1

2(A6)

( )a o i o ir

tp p p p= + +

2

1

2( ) (A7)

For the special case of zero external pressure, the thin-walledtubing equations can be formulated as:

r ip= (A8)

t ir

tp= (A9)

a ir

tp=

2(A10)

Note the sign convention to make these equations consistent.If the pipe is pressurized, the following strains results:

{ } rE

r t a= +1

( ) (A11)

{ } tE t

r a= +1

( ) (A12)

aE

a=

1 2(A13)

for the case of closed ends

For applications in petroleum wells, we also need an equationfor the strain caused by the weight of the string. Thiscomponent can be superimposed on the previous equationsand is:

a

o i

W

E r r=

( )2 2(A14)

Appendix B: Buoyancy effects

B1: The principle of Archimedes

This principle is widely used in nearly all applicationsinvolving buoyancy effects. In its simplest form it states: Thebuoyant force acting on a submerged body equals the forcegiven by the weight of the displaced fluid. Consider thefollowing:

Weight of body: w = pipeVWeight of displaced fluid: wfl = flVWeight suspended in fluid: w-wfl =(pipe -fl)V

The ratio suspended weight/weight in air is defined as the

buoyancy factor, or

= 1 1fl

pipe

B( )

7/30/2019 00056628

11/16


The buoyed weight of the drillpipe is the weight in airmultiplied with the buoyancy factor.

B2: The piston force consept

The oil industry also uses the piston force consept. It is a

simple force balance, where the forces are computedthroughout the drillstring. At each change in geometry ahydraulic force acts, which magnitude is equal to the pressuretimes the projected area at the point of interest. An examplewill demonstrate the application. Fig. B1 shows a hangingdrillstring with drillcollars at bottom. The forces are shown onthe figure, and the net buoyed weight at surface is defined.Also shown is the buoyed weight according to Archimedesprinciple. Not surprisingly, these two principles shows thesame buoyed weight at surface. However, the forces at thedepth varies significantly. This problem has createdsignificant confusion in the oil industry.

According to the Archimedes principle, the entire string is intension, whereas the piston force defines the drillcollars asbeing in compression. The external loading in one dimensionis defined by the piston force principle. However, this is notidentical to the net axial load as shown in the following.

B3: Net axial stresses

Actually the stress state at any point is three-dimensional, notone-dimensional as suggested by the piston force approach.Figure A1 shows the three principal stresses (excludingtorsion); the radial, the tangential and the axial stresses.

Consider the hypothetical example that the drillstring densitywas equal to the mud density. For that case, the net weightwould be zero, and the drillstring would remain floating.However, at any point inside the string stresses would act inall three directions. The radial, the tangential and the axialstress would be identical, creating a hydrostatic stress state.The difference between the stresses is zero. This illustratesthat the net weight of the drillstring is equal to a differencebetween the total axial load and the stresses acting on anorthonormal plane.

From classical mechanics we may therefore split the total axialstress into the following:

Total load equals the hydrostatic load plus the

deviatoric load.

The deviatoric component actually is identical to the buoyedweight. Also, we know that it is the magnitude of thedeviatoric component that governs failure in the steel pipe, notthe total stress state.

Eqn. A3 shows that the average stress across a plane in the

pipe is constant regardless of the magnitude of the externaland the internal pressures. We will define this stress as aaverage hydrostatic stress and subtract from the total stress.Reverting to the example of Appendix B2, the deviatoric loadat bottom equals the piston force minus the average horizontalstress on the plane, or:

-mudgDA2 +mudgDA2 = 0

At the top of the drill collars we subtract again the averagehydrostatic force from the total force calculated in theprevious example:

-mudgDA2 + steelgD2A2 + mudgD1A2 , or:(steel - mud)gD2A2

At the bottom of the drillpipe, the net deviatoric forcebecomes:

-mudgDA2 + steelgD2A2 + mudgD1 (A2 - A1) +

mudgD1A1 , or:

(steel - mud)gD2A2

At this point we observe several points. Firstly, the deviatoricforce at the bottom of the drillpipe equals the deviatoric forceat the top of the drillcollars. According to Newtons 2nd lawthere must be a force balance here. Secondly, we observe thatthe deviatoric force above is equal to Archimedes principle.

The force at surface is the load at the bottom of the drillstringminus the weight of the string minus the hydrostatic force.The surface weight is:

(steel - mud)g(A1D1 + A2D2)

The conclusion of this example is that the piston forces merelydetermine the external loading. The internal axial stressevaluated in a three-dimensional context is defined by theArchimedes principle.

B4: Different fluid densities.

Sometimes the inside of the drillstring are displaced to a fluidof different density than the fluid on the outside. The general

solution to this problem is similar to Appendix B3. However,recognizing that the net axial stress is defined by theArchimedes principle, we will in the following derive thebuoyancy factor in a simple way.Consider the following case shown in Fig. B2a. A tubing isfilled with mud of density i, and the outside annulus is filledwith mud of density o. In the bottom we assume a seal.According to Archimedes law:

Weight of pipe: w = pipeV = pipe(ro2 ri2)D

7/30/2019 00056628

12/16


Weight of pipe plus mud inside:wpm = pipe(ro2 ri2)D + iri2D

The outside mud provides buoyancy for the total volume ofpipe plus mud inside.

Weight of displaced fluid: wfl = oro2D

The buoyed weight equals:wpm wfl = D(pipe(ro2 ri

2) + ir2- oro

2)

The buoyancy factor, or the ratio buoyed pipe weight/ pipeweight equals (wpm w fl /w):

=

1

2 2

2 2or i r

r r

o i

pipe o i( )(B2)

It is observed that for equal fluid densities inside and outsidethe pipe, the equation reduced to Eqn. B1. Fig. B2b showsthe general case considering a composite drillstring consisting

of various pipe sizes. It can be shown that the effectivebuoyancy at any depth level is given by the followingequation:

=

=

=

1 1

1

2 2

2 2

D r rik

n

D r rik

n

k o ok i k

pipe k ok k

( )

( )(B3)

B5: Surface pressures.

During drilling there is usually a considerable pressure insidethe drillstring. Some well operations also require pressure in

the annulus. In the following will we investigate these effectsas related to buoyancy and axial stress.

If an inside overpressure is applied, an axial tension is addedin the pipe, assuming both ends closed. This additional stressis:

However, the average stress on the crossection of the pipechanges an amount from Eqn. A3:

The net axial stress from Appendix B3 is the total stress minusthe average hydrostatic stress. Subtracting the two equationsabove shows that the net stress is zero. In other words,applying a surface pressure inside the drillstring has no effecton the buoyancy of the pipe. Furthermore, there is no changein effective axial stress in the pipe.

For an outside overpressure a similar argument can be used.For this case no change in net axial stress results as well.

However, a lifting force will act on the bottom of the string.This results in a reduction of the hook-load on the rig. Thislifting force is equal to:

F P r= 2 (B4)

For a vertical well, this reduction in hook-load is computed forthe pipe size that goes through the well head. For larger pipesizes in the bottom hole assembly, the additional forces willcancel.

The case of a s imultaneous overpressure both inside theannulus and the drillpipe, again will have no effect on thebuoyancy. Again, the reduction in hook load is given by Eqn.4.

B6: Deviated boreholes.

Most boreholes drilled today are deviated. Therefore we willin the following discuss buoyancy in deviated holes.

Figure B3a shows a length of an inclined pipe. Assuming themeasured length is given by DMD, the projected vertical depthis:

D DTVD MD= cos

Figure B3b shows the unit weight (kg/m) of the pipedecomposed into an axial and a normal component. The axialcomponent is:

Eliminating the inclination from the two equation above, theaxial weight can be expressed as:

W wDaxial TVD= (B5)

The buoyant weight for this case is again given by the totalaxial stress minus the average stress according to Appendix

B3. There is no effect on orientation for this case and we canconclude as follows:

For deviated holes the buoyancy factor is identical to that of avertical hole. The axial load of a string is equal to the unitweight multiplied with the projected depth.

a io i

i

r

r rP=

2

2 2

( ) 12

2

2 2 r t

o i

ir

r rP+

=

w waxial = cos

7/30/2019 00056628

13/16


Table 1: Drill String Data.

Figure A.1: Stresses acting in a thick-walled cylinder.

Figure 1: Differential sticking.

Figure 2: Pull force at stuck point for vertical and deviated well.

7/30/2019 00056628

14/16


Figure 3: Well path for the Yme well.

Figure 4: Depth to stuck point versis friction coefficient.

7/30/2019 00056628

15/16


Figure B1: Forces in the drill string from Archimedes principle and piston forces.

Figure B2: Different densities and various pipe size scenaria.

7/30/2019 00056628

16/16


Figure B3: Forces for inclined pipe.

Documents

00056628