This is a revised version of manuscripts PO-0516-0001 (2015) and PO-0815-0022 (2016). These

manuscripts have been submitted to SPE Production & Operations. Manuscripts have been subjected to the peer

review process. In view of the reviewers' comments, these manuscripts have been declined for publication in

SPE Production & Operations.

Simultaneous Buckling Of Tubing And Rod Strings In

Pumping Wells

S.Vagapov and Sh. Vagapov, (Scientific and Production Enterprise Limited Liability Company

Burintekh, Ufa, Republic Bashkortostan, Russian Federation)

Abstract

In this paper, the phenomenon of simultaneous buckling of tubing and rod strings is explained. Rod

string is bent by the buckled tubing. Both strings are buckled during pump upstroke. This means that the buckled

sucker rod string moves inside a helically buckled tubing during pump upstroke. Interestingly, simultaneous

buckling of tubing and rod strings does not depend upon the direction of motion of the plunger but only upon the

value of the pressure drop across the plunger. In some instances, the buckled sucker rod string moves inside the

buckled tubing both upward and downward.

It is theoretically explained that the use of rod guides may entirely prevent buckling, because the energy

source of buckling disappears after the rod guides installation.

The experimental results from a study of simultaneous buckling of tubing and rod strings are given.

The annular production logging have long been used as a source of acquiring real time data in sucker

rod pumping wells. However, the simultaneous buckling of tubing and rod strings can prevent wireline tools

from passing through the annulus. As a result, the simultaneous buckling can pose extreme risk of loss to logging

tools, because the wireline twists and wraps itself around the tubing. Means for minimizing a risk of loss are

explained.

Introduction

It is well known that the lower part of freely suspended tubing buckles and wraps itself around the

sucker rod during pump upstroke (Lubinski and Blenkarn 1957). Extensive researches on this subject have been

conducted, both theoretically and experimentally. The most well known analysis of this problem have been done

by Lubinski (1957, 1962). Buckling of tubing, buckling of concentric pipes has been studied by Mitchell (1982,

2002, 2008, 2012); Paslay, P.R. and Bogy, D.B. (1964); Dawson, R. and Paslay, P.R. (1984); Miska, S. and

Cunha, J.C. (1995); Chen, Y. C., Lin, Y. H., and Cheatham, J. B. (1990); Gao, G. and Miska, S. (2008); Wenjun,

H., Deli, G., Shaolei, W. (2015) and others. In Russia (USSR), buckling of tubing, buckling of rod string has

been studied by Adonin, A.N. (1964); Virnovski, A.S. (1971); Peslyak Y.A. (1966); Muravyev, I.M. and

Mishchenko, I.T. (1967); Saroyan, A.E. (1985); Vagapov Y.G. and Sultanov, B.Z. (1980); Vagapov, S.Y. (2000)

and others.

However, in all cases these theories have always followed the same assumptions: during upstroke the

rod string remains straight in spite of the forces exerted upon it by the helically buckled tubing. The following

researches were focused more on the buckling behavior of the rod string during downstroke (Lukasiewicz, S.A.

and Knight, C. 2006; Nickens, H., Lea, J.F., Cox, J.C., et al. 2004; Zifeng, L. 1999) and others.

When consideration is given to the rod string during pump upstroke, the questions may be asked:

Why do researchers assume that the rod string remains straight?

Will the rod string be buckled during pump upstroke?

At first, it seemingly should not be buckled. The conventional wisdom says that the rod string remains

straight. In the past, theoretical works on buckling of rod string were confined to conditions for which such

buckling does not occur. However, this is not so. Both the rod and the tubing strings may be buckled during up

stroke. In this paper, the phenomenon of simultaneous buckling of tubing and rod strings is explained.

The buckling of the rod string is initiated when the energy being delivered to the rod string by the

buckled tubing is sufficient to overcome the resistance of the rod string. However, the rod string is under great

tension during upstroke. In spite of this, it is to be expected that the rod string is not capable of withstanding the

forces exerted upon it by helically buckled tubing. Why?

The rod string has much lower lateral stiffness than the lateral stiffness for tubing string. Ratio of lateral

stiffness is 50 to 215 (Table 1). Hence, the lateral stiffness for rod string may be neglected. It may be appropriate

to assume zero lateral stiffness. In addition, the radial clearance between the tubing and casing is more than the

radial clearance between the rod and tubing. This means that the simultaneous buckling of tubing and rod strings

is not limited by the casing.

Solution was obtained by use Lagrange-Dirichlet stability theorem, virtual work principle and

Timoshenko strain energy method (1955, 1956).

The investigation was based upon the following assumptions.

1. Rod and tube strings behave as linear elastic bodies.

2. Wellbore is assumed to be straight and vertical.

3. The tubing string is freely suspended in the well.

4. Friction force between the rod and the tubing strings may be neglected. Tubing and rod considered as

two separate physical entities.

5. Friction force between the buckled tubing and the constraining casing may be neglected.

6. Lateral stiffness for rod string may be neglected.

7. Inertial forces may be neglected.

8. Dynamic load may be neglected.

9. Elastic behavior induced by stretch may be neglected.

10. It is assumed that the tubing volume is kept constant during buckling.

11. Fluid level is near the pump intake.

12. No tail pipe below the pump.

13. Non tapered rod string.

Energy Source of Tubing Buckling In Pumping Well

In order to identify the root cause of buckling of tubing and rod string, the following question shall be

answered first:

What is the source of energy for buckling of the tubing in pumping wells?

Consider a pumping well as shown in Fig.1a. We assume that the lower part of freely suspended tubing

buckles during the upstroke portion of the pumping cycle. Fig.1b shows this situation when the sucker rod string

stops the tubing deflection.

Let the following designations be made.

δ1- displacement of the lower end of tubing due to buckling

δ2- displacement of the center of gravity of buckled tubing

h1 - height of the liquid column inside tubing above the pump

h2 - height of the liquid column outside tubing above the pump

L - length of the of the buckled tubing (measured along its axis)

i - density of liquid in the tubing

o - density of liquid in the annulus

- density of steel

A1 - area corresponding to tubing outside diameter (OD)

A2 - area corresponding to tubing inside diameter (ID)

A3 - plunger cross-section area

A4- area corresponding to rod OD (rod area)

Pi - pressure above fluid surface inside the tubing

Po- pressure above fluid surface outside the tubing

P1- pressure inside the pump chamber below the traveling valve

∆P - pressure differential across the plunger

∆t - deflection of tubing

∆t max - maximum deflection of tubing

R - radial clearance between the tubing and the rod

Rt- tubing inside radius

Rr- rod outside radius

l - half wave length

E - modulus of elasticity

I - moment of inertia of tubing cross-section with respect to its diameter

qr - the rod string weight per unit distance in fluid

qt - the tubing weight per unit distance in fluid

q - tubing weight with fluid per unit distance in fluid.

λt - displacement the end of tubing string

λr - displacement the end of rod string

δt , δt - small virtual displacement (for tubing)

∆r- deflection of rod string

δr , δλr - small virtual displacement (for rod)

p - pitch of a helix

U, U2, U4, U5 – potential energy of bending

U1 - work done by fictitious load

U3 - work done by the tubing weight and rod weight

Consider equilibrium of the system between the cross sections m1 n1 and m2 n2 (Fig.1). Let the tubing

will buckle and moves from position Fig.1a to position Fig.1b. Between the conditions of Fig.1a and Fig.1b, the

tubing shortens due to buckling. It is evident that center of gravity of tubing is raised. Therefore, the point of

application of the tubing weight undergoes displacement δ2. The work done by the tubing weight during the

displacement δ2 is

−δ2 (A1 − A2)L ρ ……………………………………………………………………………………(1)

The point of application of the tubing weight moves in upward direction and does negative work during

this motion.

Note that the pump barrel is rising but the plunger is stationary with respect to ground coordinate

system. As the pump barrel continues to move upward, the volume in the working chamber above the plunger

increases (Fig.1b).

Because of this, the fluid is flowing down from tubing into working chamber. This means that the center

of gravity of the fluid column inside tubing is lowered. Hence, the point of application of the weight of fluid

column inside tubing undergoes displacement δ1. The work done by the weight of the fluid column inside tubing

during the displacement δ1 is

δ1 A3Pi + δ1A3h1ρi ...…………………………………………………………………………………..(2)

The point of application of the weight of fluid column inside tubing moves in downward direction and

positive work will be done during this motion.

To determine the work done by the weight of the fluid column inside tubing during displacement δ2 , we

can break buckling process down into two steps that are more easily solved. It is does not matter that the virtual

displacements (steps) look unrealistic, because the work of force acting on a body that moves from a start to an

end position is determined only by these two positions, and does not depend on the trajectory of the body.

Consider the following hypothetical scenarios of buckling.

Step 1. It is assumed that both strings are buckled (Fig.1c). Let the rod string axis coincides with the

tube axis. It is evident that the point of application of the weight of the fluid column inside tubing undergoes

displacement δ 2. Hence, the work done by the weight of fluid column inside tubing during displacement δ2 is

−δ2(A2 − A4) L ρi ………………………………………………………………………………………(3)

The point of application of the weight of fluid column inside tubing moves in upward direction and

negative work will be done during this motion.

Step 2. The rod string becomes straight and moves to an end position (Fig.1b). This means that the rod

string displaces an amount of fluid in upward direction. Because of this, the center of gravity of the fluid column

inside tubing is raised to reflect the new volume of displaced fluid. Hence, the point of application of the weight

of the fluid column inside tubing undergoes displacement δ 2. The work done by the weight of fluid column

inside tubing during displacement δ2 is

−δ2A4Lρi ………………………………………………………………………………………………(4)

The point of application of the weight of fluid column inside tubing moves in upward direction and

negative work will be done during this motion.

Thus, the work done by the weight of fluid column inside tubing during displacement δ2 will be found

by using Eqs.3 and 4

−δ2(A2 − A4)Lρi − δ2A4Lρi = −δ2A2Lρi …………………………………………………………...(5)

To determine the work done by the weight of fluid column outside tubing during displacement δ2, we

can break buckling process down into two hypothetical scenarios of buckling.

Step 1. It is assumed that the plunger is rising but stationary with respect to the pump barrel (Fig 1d).

This means that the tubing string displaces an amount of fluid outside tubing in downward direction. Because of

this, the center of gravity of the fluid column outside tubing is lowered to reflect the new volume of displaced

fluid. Therefore, the point of application of the weight of fluid column outside tubing undergoes displacement δ2.

The work done by the weight of fluid column inside tubing during displacement δ2 is

δ2A1Lρ0 ………………………………………………………………………………………………..(6)

The point of application of the weight of fluid column outside tubing moves in downward direction and

positive work will be done during this motion.

Step 2. The plunger moves to an end position in downward direction (Fig.1b). This means that the

plunger displaces fluid into the annulus. The work done by the pressure (pressure-volume work) in the working

chamber will be

−δ1A3P1 ………………………………………………………………………………………………..(7)

where P1 = (h2 o + Po) - pressure inside the pump chamber below the traveling valve.

The point of application of the weight of fluid column outside tubing moves in upward direction and

negative work will be done during this motion.

Thus, the work done by the weight of fluid column outside tubing during displacement δ2 will be found

by using Eqs.6 and 7.

Following the energy method (Timoshenko 1955), the strain energy to the work done by the load during

deflection, may be obtain by assuming the tubing is only compressed by a centrally applied load F.

Consequently, the work done by the load F during displacement δ1 will be transformed into potential energy of

strain. Hence, strain energy of bending stored in the tubing during upstroke is

−Fδ1 …………………………………………………………………………………………................(8)

The point of application of the load moves in upward direction and negative work will be done during

displacement δ1.

The principle of virtual displacement states that the virtual work done by the actual forces is zero if and

only if the body is in equilibrium.

Thus, the total work done by the forces within on a system is

−δ2(A1 − A2)Lρ + δ1A3Pi + δ1A3h1ρi − δ2A2Lρi + δ2A1Lρ0 − δ1A3P1 − Fδ1 = 0 ………………(9)

Using the notation ∆P = (h1ρi + Pi) − (h2ρ0 + P0) and q = (A1 − A2)ρ + A2ρi − A1ρ0 this equation

may be transformed into the significantly simpler form

δ1A3∆P − σ2Lq − Fδ1 = 0 …………………………………………………………………………………. (10)

It should be noted, that P is the pressure differential across the plunger, q is the tubing weight with

fluid per unit distance in fluid.

Let us consider equation in more detail.

The first term (δ1A3 P) is the positive work done by the force (A3 P) during displacement δ1. The sign

of the two terms (- δ2Lq – Fδ1) are negative.

This equation shows that the source of energy for buckling of tubing in pumping wells is the pressure

differential across the plunger during upstroke. So one can conclude that the positive work (δ1A3 P) necessary

to buckle tubing is the work done by the weight of fluid column inside tubing during displacement δ1.

It is well known that condition of equilibrium is obtained by minimizing the total potential energy of

the system (Lagrange-Dirichlet stability theorem). It is to be remembered that following the energy method,

buckling occurs when the work done by the weight of fluid column inside tubing (δ1A3 P) exceeds to the sum

of the potential energy of bending (F δ1) and the work done by the tubing weight with fluid (δ2 Lq) during

upstroke.

At the beginning of the upstroke, the tubing is buckled while the rod string is straight. The pump barrel

is rising but the plunger is stationary with respect to ground coordinate system. As the pump barrel continues to

move upward, the plunger undergoes displacement with respect to the working chamber and the volume in the

working chamber above the plunger increases. Because of this, the fluid column inside tubing moves in

downward direction and does the positive work δ1A3 P necessary to induce buckling. This means that the center

of gravity and potential energy of system is lowered.

On the other hand, Lubinski A. and Blenkarn K.A. (1957) proposed so-called "fictitious force" in the

analysis of tubing buckling in pumping wells. During pump upstroke the tubing will buckle as if subjected to an

upward fictitious load and equal to

F = A3∆P …………………………………………………………………………………………….(11)

where P is the pressure differential across the plunger.

Comparing this with the Eq.10 we conclude, that the positive work necessary to buckle tubing is equal

to the work done by the fictitious load during displacement δ1.

It should be noted, that the tubing buckles being under great tension. In fact, the compressive force does

not exist; tubing buckling is due to the bending moment. Based on the terminology of Lagrange mechanics the

fictitious load F can be considered as a generalized force and the displacement δ1 can be considered as a

generalized displacement.

Simultaneous Buckling of Tubing and Rod Strings

Lagrange-Dirichlet stability theorem states that the condition of equilibrium is obtained by minimizing

the total potential energy of the system. Let us determine the deflection of the tube with the rod corresponding to

the minimum total potential energy of the system.

Lateral Configuration. The tubing is assumed to buckle into a lateral shape. The value of deflection

may be found by using principle of virtual displacement. Taking the case of the buckled tubing string shown in

Fig.2, deflection curve may be represented by the simple sinusoidal curve (Timoshenko 1955)

уt = ∆t sin πx

l ……………………………………………………….………….………………………………(12)

The displacement the end of tubing string is (Timoshenko 1955)

λt =1

2∫ (

dyt

dx)

2

dx ………………………………………………………….………………………….(13)

The derivative of y, from (12), is

dуt

dx= π

∆t

lcos π

x

l ……………………..…………………………………..………..….…………….(14)

It is well known, that

cos2α =1

2+

1

2cos 2α………………………………………………………..…………………………(15)

Then by using Eq.15 and substituting Eq.14 into Eq.13, the general solution is

λt = (π∆t

2l)

2

(x −l

2πsin 2π

x

l) + C ………………………………………………..………….……….(16)

where C is constant of integration which must be adjusted to satisfy the conditions x = 0, = 0. These

conditions are fulfilled, if C = 0.

Integrating this between the limits 0 and 2l, we find

λt =π2∆t

2

2l ……………………………………………………………………….…………….……...(17)

The displacement of the end of tubing string may be calculated from this expression.

We assume that the tubing string undergoes a small virtual displacement δt as shown in Fig.2. It is

evident that the end of the tubing string undergoes a small virtual displacement δt during a small virtual

displacement δt.

Substituting into Eq.17, we have

λt + δλt =π2

2l(∆t + δ∆t)

2 ………………………………………………………...…………………(18)

Using the notation π2

2l= K and λt = t

2 K (see Eq.17), this equation may be transformed into the form

δλt = K(∆t2 + 2∆tδ∆t + δ∆t

2 − ∆t2) ………………………………………………………………(19)

Let's make the following assumption: the value of deflection t is large compared a small virtual

displacement δ∆t , therefore the value δt 2 may be neglected.

Equation then becomes

δλt =π2

l∆tδ∆t ………………………………………………………………………………………...(20)

Equation gives the displacement of the end of tubing string as a function of the deflection.

Consider now rod string movement due to buckling.

Let us determine displacement of the end of rod string corresponding to its deflection.

Taking the case of the buckled rod string shown in Fig.2, deflection curve may be represented by the

simple sinusoidal curve (Timoshenko 1955). It should be noted (see Fig.2) that the half wave length of the

buckled tubing equals the half wave length of the buckled rod string.

уr = ∆r sin πx

l ………………………………………………………………..………………………(21)

where r = t - (Rt – Rr), where Rt is inside radius of the tubing and Rr is outside radius of the rod (see

Fig.2).

Based on the results of previous discussion (see Eq.20), the displacement of the end of rod string may

be found

δλr =π2

l∆rδ∆t ……………………………………………………………………………………….(22)

where δt = δr , (see Fig.2).

Consider now this equation and Eq.20.

The end of tubing undergoes displacement δt and the end of rod string undergoes displacement δr

during simultaneous buckling of rod and tubing. This means that the plunger undergoes displacement δt - δr

with respect to working chamber.

It has been explained previously that the positive work necessary to buckle tubing is equal to the work

done by fictitious load during displacement δ1 (see Eqs.10 and 11). Hence, the positive work U1 necessary to

buckle tubing and rod is equal to the work done by fictitious load during displacement δt - δr.

U1 = A3∆P(δλt − δλr) ……………………………………………………………………………….(23)

where A3 is the plunger cross-section area,

P is the pressure differential across the plunger.

Substituting Eqs.20 and 22 and observing that t - r = Rt – Rr (see Fig.2) we obtain

U1 = A3∆Pπ2 δ∆t

l(Rt − Rr) ……….……………………………………………………………….....(24)

The simultaneous buckling of rod and tubing occurs when the positive work U1 exceeds to the sum of

the potential energy of bending U2 and work U3 done by the tubing weight and rod weight during displacement

δt - δr .

U1 = U2 + U3 …………………………………………………………………………………………(25)

Let us estimate the maximum deflection for buckled columns.

The tubing will buckle more severely in the vicinity of the pump. We assume that the work done by the

weight of tubing and rod may be neglected (there is no tail pipe below the pump), U3 = 0.

The strain energy of bending is (Timoshenko 1955)

U =EI

2∫ (

d2y

dx2)2

dx …………………………………………………………………………………...(26)

where E is the modulus of elasticity,

I is the moment of inertia of tubing cross-section with respect to its diameter (the lateral stiffness for rod

string may be neglected (see assumptions)).

Since the assumption is that the lateral stiffness for rod string may be neglected, the strain energy of

bending in the rod string need not be considered.

The second derivative of y, from (12), is

d2уt

dx2 = −∆t (π

l)

2

sinπ

lx …………………...………………………………………………………….(27)

The general solution is

EI∆t2 (

π

l)

3

(2πx

l− sin 2π

x

l) + C ……………………………………………………………………(28)

In which C is constant of integration which must be adjusted to satisfy the conditions x = 0, U = 0.

These conditions are fulfilled if C = 0.

Integrating this between the limits 0 and 2l, we find

U = ∆t2π4 EI

2l3 ………………………………………………………………………………………….(29)

The strain energy stored in the tubing during deflection δt is

U2 = U4 − U5 ....……………………………………………………………………………………...(30)

where U4 is the strain energy stored in the tubing during deflection t + δt ,

U5 is the strain energy stored in the tubing during deflection t .

Therefore

U2 = π4 EI

2l3(∆t + δ∆t)2 − π4 EI

2l3 ∆t2 …………………………………………………………………(31)

As it has been mentioned above, the value δt 2 may be neglected. Hence, this equation may be

transformed into the significantly simpler form

U2 = ∆tδ∆tπ4 EI

l3 …..………………………………………………………………………………….(32)

Substituting Eqs. 24 and 32 into Eq.25 and observing U3 = 0, we obtain

A3∆Pπ2 δ∆t

l(Rt − Rr) = ∆tδ∆tπ

4 EI

l3 …………………………………………………………………..(33)

Then

∆t max = A3∆P(Rt − Rr)l2

π2EI ………………………………………………………….……………...(34)

This equation gives the maximum deflection for the laterally buckled columns corresponding to the

minimum total potential energy of the system.

Consider the numerical example from a paper Lubinski and Blenkarn (1957)

A3 = 2.4 in2. (1 ¾ in. plunger).

P = 2,500 psi (working fluid level 5,000 ft., gradient 0.5 psi/ft, i.e., fluid specific gravity 1.154).

Pump depth 6,000 ft.

Tubing size, 2 7/8 (OD = 2.875, ID = 2.441) in., q t = 6.4 lbm/ft.

E = 30 х 106 psi.

Rod size ¾ (0.75) in., q r = 1.634 lbm/ft., sucker rod coupling OD = 1.625 in.

Casing size, 5 ½ (OD = 5.5, ID = 5.012) in.

Determine the amount of deflection during pump upstroke from Eq.34.

Solution. The moment of inertia of the tubing is

I =π 2.8754

64(1 −

2.4414

2.8754) = 1.61 in4

The half wave length is (Timoshenko 1955)

l = π√EI

A3∆P= 3.14√

30×106×1.61

2.4×2,500= 282 in.

The wave length is 282 x 2 = 564 in.

Substituting these into Eq.34, we have

∆t= 2.4 × 2,500 (2.441

2−

0.75

2)

5642

3.142×30×106×1.61= 3.39 in.

Thus, for the above example the maximum deflection for the tube due to the lateral buckling is 3.39 in.

However, the radial clearance between the rod and the tubing is 2.441−0.75

2= 0.845 in. This means that the rod

string is bent by the buckled tubing during pump upstroke. On the other hand, buckling is limited by the casing.

The radial clearance between the tubing and the casing is 5.012−2.875

2= 1.078 in. Until the clearance between the

tubing and the casing is large enough, the tubing will still buckle the rod string. However, if the lateral

displacements are constrained (i.e. a casing constraint), then contact of the buckled tubing with the constraint

will induce a rearrangement to a helix form. For this reason, the buckled sucker rods move inside a helically

buckled tubing during pump upstroke.

Helical Configuration. The tubing is assumed to buckle into a helical shape. It has been mentioned

above that the equilibrium position of a conservative mechanical system is stable if the potential energy U

assumes a minimum at this equilibrium. Let us determine the maximum deflection for buckled columns shown

in Fig.3 by using this theorem.

We have

dU

d∆t= 0 ………………………………………………………………………………………………. (35)

where U is the strain energy of bending,

t is the radius of the helix.

Length of the helix corresponding to one pitch (measured along its axis) is (Lubinski et al 1962)

L = √(p2 + 4π2∆t2) …………………………………………………………………………………(36)

where p is the pitch of a helix.

It is evident that

L = p + λt …………………………………………………………………………………………...(37)

where t - the displacement the end of tubing string.

Substituting Eqs.37 into 36 and observing that p = 2 l , we obtain

4l2 + 4λtl + λt2 = 4l2 + 4π2∆t

2 ….……………………………….…………………………………(38)

Let's make the following assumption: the value l and t is large compared λt therefore the value λt2

may be neglected (error arising from this approximation is only 0,1 per cent). Thus the displacement of the end

of tubing string due to the helical buckling is

λt = π2 ∆t2

l …………………………………………………………………………………………….(39)

In the same manner for rod string

λr = π2 ∆r2

l ……………………………………………………………………………………………(40)

The end of tubing undergoes displacement t and the end of rod string undergoes displacement r

during simultaneous helical buckling of rod and tubing. This means that the plunger undergoes displacement t -

r with respect to working chamber.

The positive work U1 necessary to buckle tubing is equal to the work done by the fictitious load during

displacement t - r (the lateral stiffness for rod string may be neglected (see assumptions)).

U1 = A3∆P(λt − λr)……………………………...……………………………………………………(41)

where A3 is the plunger cross-section area,

P is the pressure differential across the plunger.

Let (see Fig.3)

Rt − Rr = R …………………………………………………………………………………………(42)

Substituting Eqs.39 and 40 and observing that ∆t − ∆r= R and ∆r= ∆t − R (see Fig 3), we obtain

U1 = A3∆Pπ2 (2R∆t−R2)

l ……………………………………………………………………………...(43)

Let us estimate the maximum deflection for buckled columns. The tubing will buckle more severely in

the vicinity of the pump. The simultaneous buckling of the rod and tubing occurs when the positive work U1

exceeds to the sum of the potential energy of bending U2 stored in the tubing and work U3 done by the tubing

weight during displacement t and the rod weight during displacement r . (There is no tail pipe below the

pump).

U1 = U2 + U3 ……………………………………………………………………………………….(44)

On the other hand the strain energy of bending is (Timoshenko 1955)

U =EI

2∫ (

d2y

dx2)2

dx ……………………………………………………………………………………(45)

where E is the modulus of elasticity,

I is the moment of inertia of the tubing (lateral stiffness for rod string may be neglected (see

assumptions)).

For the helical buckled tubing this expression becomes (Lubinski et al 1962)

U2 =8π4∆t

2EIL

(p2+4π2∆t2)

2 ……………………………………………………………………………………..(46)

where L = 2l + λt (see Eqs.37, 38).

Let's make the following assumption: the value 2l is large compared t in pumping wells, therefore this

equation may be rewritten L = 2l (error arising from this approximation is only 0,1 per cent). Substituting this

equation and observing that p = 2l, we obtain

U2 =8π4∆t

2EI2L

(4l2+4π2∆t2)

2 …………………………………………………………………………………….(47)

Let us determine the deflection of tube taking into account the weight of tubing and rod. Due to the

helical buckling, the point of application of the weight of tubing undergoes displacement λt = π2 ∆t2

l (see

Eq.39). Hence the work done by the weight of tubing corresponding to one pitch is

qtπ2∆t2 ……………………………………………………………………………………………...(48)

where q t is the tubing weight per unit distance in fluid.

In the same manner the work done by the weight of rod string corresponding to one pitch may be found

qrπ2∆r2 ………………………………………………………………………………………………(49)

where q r is the rod string weight per unit distance in fluid.

It must be noted, however, that r= t - R (see Fig.3). Equation becomes

qrπ2(∆t − R)2 ………………………………………………………………………………………..(50)

Substituting Eqs.43, 46, 48 and 49 into Eq.44 we obtain

A3∆Pπ2 (2R∆t−R2)

l=

8π4∆t2EIL

(p2+4π2∆t2)

2 + qtπ2∆t2 + qrπ2∆r

2 ……………………………………………..(51)

Let's make the following assumption: the value 4l2 is large compared 4 2t 2 in pumping wells,

therefore this equation may be rewritten (4 l 2+ 4 2t2) = 4l2 (error arising from this approximation is only 0,1

percent). Substituting this equation, we obtain

U = U1 − U2 − U3 = A3∆Pπ2 (2R∆t−R2)

l−

π4∆t2EI

l3 − qtπ2∆t2 − qrπ2(∆t − R)2 ……………………(52)

The derivative of this expression with respect to t is

dU

d∆t= 2A3∆Pπ2 R

l−

2π4∆tEI

l3 − 2π2(qt∆t + qr∆t − qrR) ……………………………………………..(53)

The deflection of the tube with rod corresponding to the minimum total potential energy of the system

from the condition dU

d∆t= 0 is

∆t max =R(

A3∆P

l+qr)

π2EI

l3 +qt+qr

…………………………………………………..………...……………………(54)

This equation gives the maximum deflection for a helically buckled columns corresponding to the

minimum total potential energy of the system.

Determine the amount of deflection by using the data of the foregoing example.

Solution.

The pitch of a helix is (Lubinski et al 1957)

P = π√8EI

A3∆P= π√

8×30×106×1.61

2.4×2,500= 797 in.

Noting that l =P

2=

797

2= 399 in., q r = 1.634 lbm/ft = 0.136 lbm/in., q t = 6.4 lbm/ft = 0.533 lbm/in.,

𝑅 =2.441−0.75

2= 0.85 in., we obtain, from Eq.54

∆t=0.85(

2.4×2,500

399+0.136)

π2×30×106×1.61

3993 +0.533+0.136= 1.58 in.

Thus, the maximum deflection for the tubing due to the helical buckling is 1.58 in. However, the radial

clearance between the rod and the tubing is 0.85 in. This means that the rod string is bent by the buckled tubing

during pump upstroke. So one can conclude that both strings will buckle during pump upstroke.

Experimental Facility

In the middle 1970s, Vagapov Y.G. (1980) have carried out some experiments with a specially

developed apparatus in order to investigate the tubing and rod buckling in pumping well. The test facility was

designed such that it allows human observation for characterization of modes of buckling and general behavior

of the tubing, rod string and logging tool. Fig.4 shows a general schematic of this facility. The casing string was

oriented perpendicular to the ground (i.e., in the vertical orientation). The tubing string was freely suspended

inside the casing. The rod string is attached to the plunger. The rod string passes around the pulley. A pulley is

attached to a supporting structure and has an axle mounted in bearings. The other end of the rod string is attached

to the weight 1. A weight 1 simulates a force applied to the plunger. A weight 2 creates a force to balance the

friction force between the plunger and O-ring during pump upstroke. During pump upstroke, a mercury is

flowing down through a rubber tubing.

The operational parameters are given in Table 2.

The features are:

- To increase the weight of the column of fluid in the tubing, a liquid mercury was chosen to simulate

the fluid (density of mercury 0.490 lb/in3).

- Plexiglas tube was chosen to simulate the casing string.

- Tubing string was modeled by using a rubber hose (rubber tube with synthetic cord). Modulus of

elasticity was experimentally determined from the test, E = 940 psi. The moment of inertia of the

tubing is

I =π 0.9254

64(1 −

0.4884

0.9254) = 0.033 in4

- Rod string was modeled using a mild steel wire.

- Plunger cross-section area A3 = 0.171 in2 (0.467 in. plunger), A3 = 0.108 in2 (0.370 in. plunger).

- Pressure differential across the plunger ∆P = 88.2 psi (working fluid level 14.8 ft), ∆P = 81 psi

(working fluid level 13.6 ft).

The fictitious force F and critical force Fcrt was selected as a scale factor. The scale factor is based on

the relation F /Fcrt. For the experimental facility:

- Fictitious force F = 88.2x0.171 = 15.1 lb (0.467 in. plunger, working fluid level 14.8 ft), F =

81x0.108 = 8.8 lb (0.370 in. plunger, working fluid level 13.6 ft).

- Critical force (Eq.55) Fcrt = 1.94 √940x0.033x0.10423 = 1.35 lb.

- Scale factor F /Fcrt = (8.8 ÷ 15.1)/1.35 = 6.5 ÷ 11.2.

Determine the working fluid level of the real prototype by using the data of the foregoing example (the

numerical example from a paper Lubinski and Blenkarn 1957).

- Critical force (see Table 3) Fcrt = 423 lb. Tubing size, 2 7/8 (OD = 2.875, ID = 2.441).

- Fictitious force F = 423x6.5 = 2750 lb (scale factor 6.5), F = 423x11.2 = 4738 lb (scale factor

11.2).

- Working fluid level 2750

2.4x0.5= 2292 ft (scale factor 6.5),

4738

2.4x0.5= 3948 ft (scale factor 11.2). A3 =

2.4 in2. (1 ¾ in. plunger), gradient 0.5 psi/ft.

The following observations have been made:

1. Results indicate that the rod string is bent by the buckled tubing. Both strings are buckled during

pump upstroke (Fig.5).

2. The buckling is initiated near the pump and progresses upward during upstroke. The tubing buckles

into a helix, however, the lateral buckling is observed close to the neutral point (Fig.5).

Fig.6 is a photograph of the typical form of buckled tubing which contact the casing.

Sucker Rod Guides

The presence of rod guides affect the buckling significantly (Lubinski et al. 1957). Lubinski have

written that the rod guides cannot prevent buckling but only minimize its detrimental effects. Let us consider

Eqs.34 and 54 in more detail.

Noting that Rt - Rr = R (see Fig.3), the magnitude of deflection of the tube and rod corresponding to

the minimum total potential energy of the system is becoming equal to zero when Rt = Rr . On the other hand Rt

- Rr = R is the radial clearance between the tubing and the rod. When setting rod guides, the radial clearance R

becomes minimum. This means that the magnitude of deflection ∆t becomes minimum.

Due to the buckling, the fluid level in tubing drops down and the fluid column inside tubing does the

positive work necessary to induce buckling. Let us imagine the hypothetical scenarios of buckling in which the

rod string axis coincides with the tube axis during buckling. In this case, the pump barrel is rising but the plunger

is stationary with respect to barrel, therefore the fluid level in tubing does not drop down and the fluid column

does not the positive work necessary to induce buckling. Thus, when setting rod guides, the energy source of

buckling disappears. For this reason, we assume that the rod guides may entirely prevent buckling, because the

energy source of buckling disappears after the rod guides installation. It should be noted that the contribution of

friction forces between the rod guides and tubing string to the buckling process would be substantial. In general,

an increase a fictitious force causes an increase a friction forces. Therefore, this assumption is valid for shallow

to mid-range depth rod pumping applications. It should be noted that according to Lubinski and Blenkarn (1957)

the field trials with sucker rod guides conducted in a North Louisiana field resulted in reducing the average

occurrence of failures due to rod on tubing wear from one in 6.3 months to one in 11.2 months.

In order to effective prevent buckling effect, the rod guides must be properly spaced on the sucker rod

string. The rod guides must be set so that the axis of the sucker rod string coincides with the axis of the tubing

string during buckling. The rod guides must be molded directly onto the rod body and must not move on the rods.

The outside diameter of rod guides should be as close to the inside diameter of tubing as possible. It should be

noted, that the rod guides would prevent buckling but not breathing.

Annulus Production Logging

The annulus production logging have long been used as a source of acquiring real time data in sucker

rod pumping wells (Hammack, G.W. et al.1976). The logging is done through the annulus between the tubing

and casing. Before logging, the well must be prepared by removing the tubing anchor. For freely suspended

tubing, the rod string is bent by the buckled tubing and the buckled tubing will contact the wall of the casing. For

this reason, the simultaneous buckling of tubing and rod strings can prevent wireline tools from passing through

the annulus. Therefore, the simultaneous buckling of tubing and rod strings can pose extreme risk of loss to

logging tools, because the wireline twists and wraps itself around the tubing.

In practice, the walking beam stops when the horsehead reaches the uppermost position to gain access

to the wellhead equipment (Fig.7). Typically, the walking beam stops during upstroke, when the rod string is

under the great tension and the tubing is buckled. That is not correct. To avoid the tool loss, several procedures

should be followed before and during logging operations in sucker rod pumping well.

Procedure 1. Well is under static conditions.

1. Shut down pump jack and lock in place. To provide access to the wellhead equipment, the walking

beam must be stopped in the upper position, however, during downstroke (Fig.7, right). The standing valve must

be closed and the traveling valve must be opened. Because of this, the fluid load is transferred to the tubing

string as a result the tubing string is straight and is under the great tension. However, one should keep in mind

that the tubing buckling does not depend upon the direction of motion of the plunger but only upon the value of

the pressure drop across the plunger. The tubing may be buckled (traveling valve is closed) even during

downstroke.

The fictitious force is much greater than the critical force in the great majority of pumping wells

(Appendix A). For this reason, the buckling period is approximately equal to the upstroke period during pumping

cycle. Fig.8 and Fig.9 show areas indicating periods where buckling occurs (1-2-3-4). In some cases, the tubing

may be buckled both upstroke and downstroke (Fig.9, gas interference situation). Therefore, in some instances,

the rod string is bent by the buckled tubing and moves both upward and downward. This means that additional

information must be observed on a pump card during logging. The load on the polished rod must be monitored to

determine whether or not the travelling valve is opened during pump downstroke.

2. During logging operations, the load on the polished rod must be monitored to predict simultaneous

buckling. If the standing valve is leaking, the traveling valve ball will seat. As a result, the dynamometer will

measure a load increase on the polished rod. This means that the tubing begins to bend the rod string. For this

reason, the well logging must be stopped and the pumping unit must be restarted to repeat the cycle. Once the

traveling valve is opened, the well logging may be restarted.

Procedure 2. Well is under dynamic, producing conditions.

1. The tubing buckles into a helix, which contacts the casing below the neutral point. Determine the

distance from pump setting depth to neutral point. The distance n from the bottom of the tubing to the neutral

point is (Lubinski et al. 1957)

𝑛 =F

qt

Where F is the fictitious load (see Eq.11),

qt is the tubing weight per unit distance in fluid.

The fictitious load F can be determined from a pump card.

2. Shut down pump jack and lock in place. To provide access to the wellhead equipment, the walking

beam must be stopped in the uppermost position during upstroke (Fig.7, left). The pump must be restarted after

the tool string has been lowered into the annulus.

3. Once the neutral point is reached, the unit must be shut down immediately. The walking beam must

be stopped during downstroke (Fig.7, right). The standing valve must be closed and the traveling valve must be

opened. The load on the polished rod must be monitored to determine whether or not the travelling valve is

opened during pump downstroke (see step 1 from Procedure 1).

4. Run the logging tool from the neutral point to the pump. During logging, the load on the polished rod

must be monitored to predict simultaneous buckling (see step 2 from Procedure 1).

5. Release and activate pump jack when the logging tool is below the pump.

6. The logging tool is then run from the pump down through zones of interest.

7. Shut down pump jack prior to retrieving the tool. The walking beam must be stopped during

downstroke. The standing valve must be closed and the traveling valve must be opened. The load on the polished

rod must be monitored to determine whether or not the travelling valve is opened during pump downstroke (see

step 1 from Procedure 1).

8. Pull the logging tool out of the hole. The load on the polished rod must be monitored to predict

simultaneous buckling (see step 2 from Procedure 1).

Procedure 3. Well is under dynamic, producing conditions.

1. Determine the distance from pump setting depth to neutral point (see step 1 from Procedure 2).

2. Tubing anchor must be removed. Guides must be spaced on the rod string between the pump and the

neutral point to prevent buckling.

3. Shut down pump jack and lock in place. The walking beam stops when the horsehead reaches the

uppermost position to gain access to the wellhead equipment during upstroke. The pump must be restarted after

the tool string has been lowered into the annulus.

Vagapov Y.G. (1980) has studied running the logging tool through the annulus experimentally. It was

found that the wireline twists and wraps itself around the buckled tubing during logging (up to 5 times). The

photographs (Fig.6 and Fig.10) clearly illustrate the manner in which the wireline wraps itself around the

buckled tubing.

Conclusions

1. Energy source of tubing buckling in pumping wells is the pressure differential across the plunger

during upstroke.

2. Tubing buckling is accompanied by a lowering of the point of application of the weight of fluid

column inside tubing during pump upstroke. This means that the center of gravity and potential energy of system

is lowered.

3. During buckling the fluid column inside tubing moves in downward direction and does the positive

work necessary to induce buckling.

4. Positive work necessary to buckle tubing is equal to the work done by the fictitious load or

generalized force during generalized displacement.

5. Rod string is bent by the buckled tubing. Both strings are buckled during pump upstroke. This means

that the buckled sucker rod string moves inside a helically buckled tubing during pump upstroke.

6. Simultaneous buckling of tubing and rod string does not depend upon the direction of motion of the

plunger but only upon the value of the pressure drop across the plunger. In some instances, the buckled sucker

rod string moves inside the buckled tubing both upward and downward.

7. It is theoretically explained, that the rod guides may entirely prevent buckling, because the energy

source of buckling disappears after the rod guides installation. An assumption is perfectly valid for shallow to

mid-range depth rod pumping applications.

8. Simultaneous buckling of tubing and rod string can pose extreme risk of loss to logging tools during

annular production logging. To mitigate the risk of the tool loss, the walking beam must be stopped during

downstroke. During logging operation, the load on the polished rod must be monitored to predict simultaneous

buckling. Risk of the tool loss may also be minimized by sucker rod guides spaced as described in this paper.

Acknowledgment

This research article is based on the materials of our father and grandfather U.Vagapov. The authors are

grateful him for his support of this project. The authors also would like to thank the colleagues from Burintekh

company.

References

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Appendix A

The Critical Force For The Tubing String

According to Lubinski the critical force for the tubing is (see Table 3)

Fcrt = 1.94 √E I qt23 ………………………...………………………………..………………..……..(55)

Where E = 30 x 106 psi,

qt = 4.089 lb/ft = 0.341 lb/in, tubing size, 2 3/8, fluid specific gravity 1.154,

qt = 5.545 lb/ft = 0.462 lb/in, tubing size, 2 7/8, fluid specific gravity 1.154.

Tubing

Rod

Ratio Of

Lateral

Stiffness

Radial Clearance

Between Tubing

and Rod (in)

Radial Clearance

Between Tubing

and Coupling (in)

Outside

diameter

(in)

Inside

Diameter

(in)

Moment

Of

Inertia

I t

(in4)

Outside

Diameter

(Dr )

(in)

Coupling

Outside

Diameter

(Dc)

(in)

Moment

Of

Inertia

I r,

(in4)

𝐼 2 7/8

𝐼 𝑟

𝐼 2 3/8

𝐼 𝑟

2.441

2

−𝐷𝑟

2

1.995

2

−𝐷𝑟

2

2.441

2

−𝐷𝑐

2

1.995

2

−𝐷𝑐

2

2 7/8 2.441 1.61 0.625 1.5 0.0075 215 104 0.91 0.69 0.47 0.25

2 3/8 1.995 0.78 0.75 1.625 0.0155 104 50 0.85 0.62 0.41 0.19

Table 1 – Lateral stiffness data of the rod and tubing

Tubing

Rod

Outside

Diameter

(in)

(mm)

Casing

Inside

Diameter

(in)

(mm)

Pump

Depth

(ft)

(mm)

Pump

Bore

(in)

(mm)

Maximum

Radial

Clearance

Between

Tubing

and Rod

(in)

(mm)

Maximum

Radial

Clearance

Between

Tubing and

Casing

(in)

(mm)

Tubing

Weight

With Fluid

Per Unit

Distance

(lb/in)

(kg/m)

Logging tool

Outside

Diameter

(in)

(mm)

Inside

Diameter

(in)

(mm)

Outside

Diameter

(in)

(mm)

Length

(in)

(mm)

0.925

23.50

0.488

12.40

0.047

1.20

1.417

36.0

13.6

4138

0.467

11.85

0.441

11.2

0.492

12.5

0.104

1.860

0.315

8.0

4.330

110.0

14.8

4520

0.370

9.40

0.370

9.4

2.874

73.0

Table 2 – The operational parameters of research facility

Outside Diameter (in) 2 3/8 2 7/8

Inside Diameter (in) 1.995 2.441

Critical Force (lb) 272 423

Table 3 – Critical force for the tubing

Fig. 1 – Real (b) and hypothetical (c, d) buckling scenarios

Fig. 2 - Simultaneous buckling of tubing and rod strings. Lateral configuration

Fig. 3 - Simultaneous buckling of tubing and rod strings. Helical configuration

Fig.4 – General schematic of experimental facility

Tubing

Pump

100 centimeter

Lateral buckling

150 centimeter

Helical buckling 200 centimeter

Fig.5 – Simultaneous buckling of tubing and rod strings (no casing). View of portion of the buckled tubing

above the pump (left). View from above of the tubing (right)

Fig.6 - Typical form of buckled tubing which contacts the casing

Wireline

Casing

Logging tool Pump

Fig.7 – Annulus production logging. Walking beam stops during downstroke (right) and upstroke (left)

Fig.8 –Duration of buckling (right) and pump dynamometer card (left) for non- anchored tubing

Fig.9 - Gas interference situation. Duration of buckling (right) and pump dynamometer card (left) for non -

anchored tubing

Wireline

Fig.10 – Wireline wraps itself around the tubing during logging. View from above of the tubing