Bottom Hole Assembly Problems Solved by Beam-Column Theory

Jiazhi, Bai

doi:10.2118/10561-MS

ABSTRACT

The bending portion of a multiple-stabilizer bottom hole assembly can be considered as a string of several simply-supported beams loaded by axial compressive forces together with transverse ones. By the beam-column theory, a system of "equations of three moments" can be obtained. The essential aim to design a bottom hole assembly with special requirements is to produce a desirable reactive side force on the bit from the formation. While there are numerous parameters that would influence the value of this force, the effect of each of them can be evaluated from the position it occupies in those equations. This paper is particularly focused on the analysis of pendulum assemblies with two stabilizers and affirms their obvious advantages over those with only one stabilizer or no stabilizer.

INTRODUCTION

When the angle of inclination or a borehole drilled in crooked regions is confined within very small value, simple pendulum assembly without any stabilizer may be used, but the weight on the bit should be confined to very small values, too. Many investigators had tried various procedures to promote the effectiveness of the pendulum by using heavy metal collars,⁽¹⁾ or by installing one or more stabilizers at proper places on the drill collar string.^(2,3) When the borehole is inclined and the bit weight is applied, the lower portion of a multi-stabilized bottom hole assembly will be bended in a special manner. Indeed, quite a \ill\ methods to solve this problem,^(4,5,6,7,8) had been published, still, it is presented here another one by applying the beam-column theory. An assembly is, then, cut at tne stabilized junctures into several simply-supported beams submitted to axial compression in company with lateral loads.

The parameters which would influence the force distribution and deflecting configuration of a multiple stabilizer bottom hole assembly include: (see fig. 2)

D_B = diameter or bit, cm.
P_B = weight on bit, kg.
α = angle of inclination, deg.
D_ci = outside diameter of i-th collar, cm.
W_i = effective unit weight of collar, kg/cm.
γ = specific gravity of drilling fluid, kg/cm³.
I_i = moment of inertia, cm⁴.
E = modulus of elasticity, kg/cm².
EI_i = flexural rigidity, kg/cm².
D_sj = diameter of j-th stabilizer, cm.
l_i = length of i-th drill collar section, cm. (in general, i = j+1)
e_j = $\frac{1}{2} (D_{B} {-D}_{sj})$ = radial clearance of stabilizer, cm.

It will be seen later in this paper, the various roles played by all of the above parameters can be exposed by examining the positions they seat in a system or "equations of three moments" derived by the beam-column theory. To do a specific drilling job successfully, a good designed bottom hole assembly should create an appropriate side force on the bit, so that, a resultant force including a component from the formation will be formed to drive the bit to the proper direction.

SIDE FORCE ON BIT

Fig. 1 show how a side force R_B is created at the bit on the lowest section l₁ of a multiple stabilizer bottom hole assembly of pendulum type. This side force is shown directing upward because it is a reactive force of a simply-supported beam l₁ with the bit as the lower end. The transverse loads acting on the beam include a bending moment M₁ at the upper end S₁, besides the uniformed distributed load q₁

Fig. 1

View large Download slide

q_{1} = w_{1} sin \propto

By the static equilibrium condition, the magnitude of R_B may be computed by the following equation:

R_{B} = \frac{1}{2} w_{1} l_{1} sin \propto + \frac{M_{1}}{l_{1}} - \frac{{Pe}_{1}}{l_{1}}

(1)

Where P is the average value of the axial compressive load and may be taken as:

P = P_{B} - \frac{1}{2} w_{1} l_{1}

(2)

Equation (1) may be regarded as a criterion to appraise the effectiveness of different pendulum designs. (see table 1) But, right now, the internal bending moment M₁ at stabilizer S₁ is still an unknown quantity. It is the purpose of this paper to derive the formulas to calculate M₁ and other moments at the supporting stabilizers first, and then to see how to choose those optional parameters to render R_B as large as possible.

TABLE 1*

Side forces of Example 4

P_B

3000

12000

20000

l

1742

1612

1520

P

1830

10900

19000

X

1.031

1.406

1.773

\frac{P_{s^{'}}}{1}

-1.9

-12.5

-23.1

R_B

38.8

25.2

12.4

* Numerical values computed by eqs. (18)(19) conform well with the charts commonly used⁽¹⁰⁾.

ASSEMBLY WITH TWO STABILIZERS

As shown in Fig. 2, a pendulum assembly with two stabilizers is taken as a sample to illustrate how the beam-column theory can be utilized to solve the bottom hole assembly problems. A pendulum assembly is characterized by its long heavy collar section l₁ above the bit. After cutting off at the stabilizers S₁ and S₂, the bending portion of the assembly would be divided into three spans, l₁l₂l₃, with two internal bending moments M₁, M₂ appeared as end couples. Each span may be considered as a simply-supported beam. Two lengths of them, l₁ and l₂, can be arbitrarily selected and therefore are taken as know quantities in the derivation, the third one l₃, its upper end being the point of tangency T is, however, an unknown, which, following and M₁ and M₂, represents the third unknown to be determined.

Fig. 2

View large Download slide

Fig. 3 shows the acting forces and angular deflections at the ends of three beams, or rather beam-columns, putting horizontally. In elementary mechanics of materials, we usually apply the method of "Three Moments Equations" to solve a continuous beam problem. Now, each of these beams in Fig 3 is loaded axially besides the uniformly distributed forces q_i = w_isinα: (i = 1, 2, 3.) the same technique can, still, be utilized to establish the corresponding equations by satisfying the following conditions:

Fig. 3

View large Download slide

$\begin{array}{l} Condition of continuity at S_{1} & θ_{1}^{'} = - θ_{1}^{"} \end{array}$
(3)
$\begin{array}{l} Condition of continuity at S_{2} & θ_{2}^{'} = - θ_{2}^{"} \end{array}$
(4)
$Zero slope at point of tangency T θ_{τ} = 0$
(5)

The formulas expressing the above angles of rotation can be found from Timoshenkos book⁽⁹⁾ as:

{θ'}_{1} = \frac{w_{1} \sin \propto l_{1}^{3}}{{24E}_{1} I_{1}} X_{1} + \frac{M_{1} l_{1}}{3 E_{1} I_{1}} V_{1} - \frac{e_{1}}{l_{1}}

(6)

(7)

θ_{2}^{'} = \frac{w_{2} \sin \propto l_{2}^{3}}{24 E_{2} I_{2}} X_{2} + \frac{M_{1} l_{2}}{6 E_{2} I_{2}} W_{2} + \frac{M_{1} l_{2}}{3 E_{2} I_{2}} V_{2} + \frac{e_{1} - e_{2}}{l_{2}}

(8)

θ_{2}^{"} = \frac{w_{3} \sin \propto l_{3}^{3}}{24 E_{3} I_{3}} X_{3} + \frac{M_{2} l_{3}}{3 E_{3} I_{3}} V_{3} + \frac{e_{3} - e_{2}}{l_{3}}

(9)

θ_{T} = \frac{w_{3} \sin \propto l_{3}^{3}}{24 E_{3} I_{3}} X_{3} + \frac{M_{2} l_{3}}{6 E_{3} I_{3}} W_{3} - \frac{e_{3} - e_{2}}{l_{3}}

(10)

It should be noted, here, that three trancendental functions. X_i, V_i, W_i, representing the effects of axial compressive forces on the transverse bending deflections, do appear in the above five equations and are expressed by the following formulas:

\begin{array}{l} X_{i} = \frac{3 (tg μ_{i} - μ_{i})}{μ_{i}} & (i = 1, 2, 3,) \end{array}

(11)

\begin{array}{l} V_{i} = \frac{3}{2 μ_{i}} & (\frac{2}{2 μ_{i}} - \frac{1}{t g 2 μ_{i}}) & (i = 1, 2, 3,) \end{array}

(12)

\begin{array}{l} W_{i} = \frac{3}{μ_{i}} & (\frac{1}{S i n_{2} μ_{i}} - \frac{1}{2 μ_{2}}) & (i = 1, 2, 3,) \end{array}

(13)

Where

\begin{array}{l} μ_{i} = \frac{l_{i}}{2} \sqrt{\frac{P_{i}}{E_{i} I_{i}}} & (i = 1, 2, 3,) \end{array}

(14)

and P_i = average axial compressive force on i-th section. Substituting eqs. (6), (7), (8), (9), (10), into eqs. (3), (4), (5), and simplifying, then the following important equations to determine the unknown values of M₁, M₂, and M₃ are obtained:

(Assume E₁ = E₂ = E₃ = E = 2.1 × 10⁶ kg/cm²)

\begin{array}{l} {2M}_{1} (V_{1} + \frac{l_{2} I_{1}}{l_{1} I_{2}} V_{2}) + M_{2} \frac{l_{2} I_{1}}{l_{1} I_{2}} W_{2} \\ = - \frac{q_{1} l_{1}^{2}}{4} X - \frac{q_{2} l_{2}^{2}}{4} \cdot \frac{l_{2} I_{1}}{l_{1} I_{2}} X_{2} + \frac{6 {EI}_{1} e_{1}}{l_{1}} + \frac{6 {EI}_{1} (e_{1} {-e}_{2})}{l_{1} l_{2}} \end{array}

(15)

\begin{array}{l} M_{1} W_{2} + 2 M_{2} (V_{2} + \frac{l_{3} I_{2}}{l_{2} I_{3}} V_{3}) \\ = - \frac{q_{2} l_{2}^{2}}{4} X_{2} - \frac{q_{3} l_{3}^{2}}{4} \cdot \frac{l_{3} I_{2}}{l_{2} I_{3}} X_{3} - \frac{6 {EI}_{2} (e_{1} {-e}_{2})}{l_{2}^{2}} - \frac{6 {EI}_{2} (e_{3} {-e}_{2})}{l_{2} l_{3}} \end{array}

(16)

L_{3}^{4} + \frac{{4M}_{2}}{q_{3}} (\frac{W_{3}}{X3}) L_{3}^{2} = \frac{24 {EI}_{3} (e_{3} - e_{2})}{q_{3} X_{3}}

(17)

In solving the above simultaneous equations, as the unknown length l₃ is entangled in the trancendental functions X_i, V_i, and W_i, a reasonable value of l₃ is assume at the beginning, and then use the iteration method to get the final answers.

Now, it is the time to examine eqs. (15), (16), (17) to fix up the related parameters in order to design a most effective pendulun assembly with two stabillizers.

In general, at first, a long and heavy enough section, is chosen for l₁, then the problem is, with known values of l₁, W₁, I₁, α, P_B, D_B, how to choose those parameters such as l₁, D_c2, e₁, e₂, etc, so as to make the value of R_B in E_q(1) as large as possible. E_q(1) indicates that a large positive value of M₁, is profitable. Careful examination will reveal that for a larger M₁, the following choices are preferable:

smaller value of l₂
thinner section of L₂
larger value of e₁
e₂ as small as possible.

Among these preferences, the third one is harmful, as it will produce a larger negative part in E_q(1) ( i.e $\frac{Pe}{l_{1}}$ ⁠. If both a large positive M₁ and a small e₁ are required, assemblies with three stabilizers may be used.

Two illustrative examples are given below, shotting that there are two options, one is featured with large value of e₁ the other with very slender section of l₂. The solutions give the main numerical results only.

Example 1:

solution: The main numerical results are obtained as:

P₁ = 15000 − $\frac{1}{2}$ 1.84 × 1800 = 13344kg
e₁ = 1st stabilizer radial clearance = $\frac{1}{2}$ (24.4−21.4)=1.5cm
e₂ = 2nd stabilizer radial clearance = $\frac{1}{2}$ (24.4−24.4)=0
e₃ = radial clearance at
T = $\frac{1}{2}$ (24.4-17.8) = 3.3cm
M₁ = 21100kg-cm
M₂ = -66400kg-cm
l₃ = 2300cm
R_B = $\frac{1}{2}$ x 1.84 x sin3° − $\frac{21100}{1800}$ − $\frac{13344 x 1.5}{1800}$ =87.3kg

Example 2: Given l₁ = 1850cm,

W₁=W₂=1.34kg/cm (7" collar in mud with γ=1.2),

EI₁=EI₂=10 x 10⁹kg/cm²,

l₂=1000cm,

W₂=0.225kg/cm (5" pipe),

EI₂=1.25 x 10⁹kg-cm²,

D_B=21.5cm(8 $\frac{1}{2}$ ⁠" bit),

D_c1D_c3=17.8cm,

D_c2=12.7cm,

D_s1=D_s2=21.3cm,

α = 2°

P_B = 12000 kg

Solution:The main numerical results are obtained as:

P₁ = 12000 − $\frac{1}{2}$ 1.34 × 1850 = 10760 kg
e₁ = e₂ = 0.1 cm
e₃ = 1.85 cm
M₁ = 1200 kg - cm
M₂ = - 3000kg-cm
l₃ = 1650 cm
R_B = $\frac{1}{2}$ x 1.34 x sin2° 1850− $\frac{1200}{1850}$ − $\frac{10760 x 0.1}{1850}$ =\ill\

PENDULUM ASSEMBLY WITHOUT ANY STABILIZER

Without any stabilizer, it is the simplest bottom hole assembly which is naturally welcomed by all drillers and under certain conditions is reliable too. But in crooked area this will result an unfavorable conditon as only very light weight can be exerted on the bit.

As shown in Fig 4 there exists only one unknown quantity which is the length l between the bit B and the point of \ill\ T and can be easily determined by the following equation:

Fig. 4

View large Download slide

I^{4} = \frac{24 EIe}{w \sin \propto X}

(18)

Where e = radial clearance of collar = $\frac{1}{2} (D_{B} {-D}_{c})$

\begin{array}{l} X & = & \frac{3 (t g μ - μ)}{μ^{3}} μ = \frac{1}{2} / \frac{P}{EI} \\ P & = & P_{B} - \frac{1}{2} w_{1} \end{array}

The reactive side force R_B, in the present case, then consists of two items as:

R_{B} = \frac{1}{2} w \sin \propto 1 - \frac{Pe}{1}

(19)

From the above relations, it can be seen that as the value of P_B is small, X is nearly equal to one, l is large and consequently R_B will be fairly large. So this is a good deviation correction tool. But when P_B is large, X will grow to a value much greater than one, and l will become short and at the same time, the negative part of eq (19) grows larger at even greater rate and consequently, R_B will be reduces to a very small value. Such a situation will be illustrated apparently in Example 4.

Example 3

\begin{array}{c} Given D_{B} = 24.4 (9 {\frac{5}{8}}^{"}), D_{c} = 20.3 cm (8 ") \\ w = 1.84 kg / cm, \propto = 3 °, \\ P_{B} = 15000 kg, EI = 17.2 \times 10^{9} kg - {cm}^{2} \end{array}

Solution: The main numerical results are obtained as:

\begin{array}{l} l & = & 1628 cm \\ e & = & \frac{1}{2} (24.4 - 20.3) = 2.05 cm \\ P & = & 15000 - \frac{1}{2} \times 1.84 \times 1628 = 13500 kg \\ R_{B} & = & \frac{1}{2} \times 1.84 \times \sin 3^{°} \times 1628 - \frac{13500 \times 2.05}{1628} = 61.4 Kg \end{array}

Example 4:

\begin{array}{l} Given D_{B} = 21.5 cm (8 \frac{1 "}{2}) & D_{c} = 17.8 cm (7 ") \\ W = 1.34 kg / cm, & \propto = 2 ° \\ EI = 10 \times 10^{9} kg - {cm}^{2} & P_{B} = 4000 kg, 12000 kg, 20000 kg. \end{array}

Solution: Table 1 gives the main numerical results.

PENDUIUM ASSEMBLY WITH ONE STABLIZER

Woods and Lubinski pointed out in their paper (2) that more weight may be carried on the bit by using one stabilizer properly placed on the assembly. As shown in Fig 5, the pendulum length l₁. can be fixed to value much longer than 1 in Fig 4. Besides, the negative item in Eq (19) may be materially eliminated. Some profits are really obtained, but, on the other hand, a negative M₁ of large value is induced at S₁ which will nullify the profits to a certain extent.

Fig. 5

View large Download slide

It is very easy, now, to derive the required equations to determine the unknown quantities M₁ and l₃ as:

(20)

l_{2}^{4} + \frac{4 M_{1}}{q_{2}} \frac{W_{2}}{X_{2}} l_{2}^{2} = \frac{24 {EI}_{2} (e_{2} {-e}_{1})}{q_{2} X_{2}}

(21)

Example 5: Given D_B = 24.4cm, D_c1 = 20.3cm,

D_c2 = 17.8cm, D_s1 = 24.4cm,

W₁ = 1.84 kg/cm (γ = 1.2),

W₂ = 1.34 kg/cm l₁ = 1800cm

α=3° P_B=15000

Solution: The main numerical results are obtained as:

\begin{array}{l} M_{1} & = & 56400 kg - cm I_{2} = 2180 cm \\ R_{B} & = & \frac{1}{2} \times 1.84 \times \sin 3^{°} \times 1800 - \frac{56400}{1800} = 55.3 kg \end{array}

The above calculated value of R_B is even smaller than that of Example 3. It indicates that the pre-selected I₁ is \ill\ short, and also explains the harmful effect of M₁,

Example 6: Given D_B = 21.5cm, D_c1=D_c2=17.8cm

W₁ = W₁ = 1.34 kg/cm (γ = 1.2),

D_s1 = 21.3 cm, l₁ = 1850cm,

α=2° P_B=12000kg

Solution: The main numerical results are obtained as:

COMPARISON OF SIDE FORCES

It is stated that the magnitude of reactive side force at the bit represents the effectiveness to prevent crookedness of different designs of pendulum assemblies. The numerical results obtained in the above six illustrative examples are compilled in Table 2. The influencing parameters have not been optimized, and there are still rooms for improvement, however, the advantages of those assemblies with two stabilizers can be observed clearly. It should be pointed out here once more that there exist two options of combination one with small D_s1 and stiff l₂ and the other with large D_s2 and flexible l₂.

Table 2

Comparison of Side Forces

FIELD VERIFICATIONS

Field testings had been performed on eight wells with depths ranging from 1200m to 1800m in crooked regions of different classes of severity. All of the experimental results show very close agreement with the theoretical predictions and may be explained by the following typical examples:

Example 1. (A) Well number: No. Tong 13

\begin{array}{l} BHA ϕ 244 - 3 A (i,e, 9 \frac{5^{"}}{8} tricone bit) + 8 " colour \times 17. 5m \\ + ϕ 220 mm (lst Stabilizer) + 7 " collar \times 6 m \\ + ϕ 242 mm (2 nd Stabilizer) + 7 " collar \times 105 m \\ + 5 " drill pipes . \end{array}

Exp. Results:

\begin{array}{l} Depth H 100 m to 800 & \propto < 1^{°} \\ 800 m to 1025 m & 1^{°} < \propto < {3.5}^{°} \\ 1025 m to 1300 m & {3.5}^{°} < \propto < 5^{°} \end{array}

Weight on bit: 16000kg to 18000kg.

(B) Offset well for reference. No. Tong 10 used 7" collar pendulum without any stabilizer. Max. Weight on bit, 12000kg reached 15° at depth 1300m

Example 2. (A) Well number: No, Jing 113

\begin{array}{l} BHA : ϕ 215 - 3 A (i .e,8 \frac{1}{2} " Tricone bit) + 7 " \times 18.2 m \\ + ϕ 188 mm (1st stabilizer) + 7 " \times 8m \\ + ϕ 210 mm (2nd stabilizer) \end{array}

Exp. results:

\begin{array}{l} H< 1100 m 0 .5 ° < \propto <2 .0 ° \\ 1100 m < H< 1300 m 2 ° < \propto <3 .8 ° \\ 1300 m < H<1650m 3 .8< \propto <4 ° \end{array}

Weight on bit 12000kg to 14000kg

(B) offset wells: average of several wells. $8 \frac{1}{2}$ bit use 7" pendulum (no stabilizer). Weight on bit, \ill\ to 14000kg. α₁ doubled as much at same depths.

Example 3. (A) well number: Ko. Wangdong 12-7

Exp. results

\begin{array}{l} H < 1100 m & \propto < 1^{°} 45' \\ 1100 m <H<1388m, & 1^{°} 45' < \propto < 3^{°} \end{array}

(B) offset well: No. Wangdong 9-7. Use 8" pendulum without any stabilizer. Wt on lit, 3000kg about the same as (A).

CONCLUSIONS

According to beam-column theory, a multiple stability bottom hole assembly can be treated similarly as a continuous beam as in elementary mechanies of materials. A systems of equations of three moments can be obtained to find simple and accurate solution.
The problem of designing an effective bottom hole assembly in general is essentially to produce a desirable side force on the bit by proper selections of related influencing parameters. As the role played by such parameters and be judged by the positions they occupied in those equation of three moments, the designer can manage his job straightforward to get the optimum results without much troubles.
Particular attention is focused on the analysis of pendulum assemblies with two stabilizer. Theoretical analysis shows that with certain special arrangement, they are much more effective to resist crookedness than those with only one stabilizer or no stabilizer. The theoretical prediction:, confirmed by field testings. Pendulum assemblies may be further improved by using of three stabilizers.
The theory proposed in this paper is also applicable to assemblies in building-up, maintaining, or dropping the angle of inclination.

References

Bradley

,

W. B.

,

Murphey

C.E.

,

Mclamore

,

R.T.

, and

Dickson

,

L.L.

, "

Advantages of Heavy Metal Collars in Directional Drilling and Deviation Control

"

J. of pet. Tech.

,

May

,

1976

.

Google Scholar

Woods

H. B.

, and

Lubinski

,

A.

, "

Use of stabilisers in Drill-Collar String

",

OGJ

.

Apr

.

4

,

1955

.

Google Scholar

Callas

,

N.p.

, and

Callas

,

R.L.

, "

stabilisers \ill\1, Drill-string Analysis Shows Forces Overcome by Stabilisation

,"

OGJ

,

Nov

.

24

,

1980

.

Google Scholar

Millheim

,

K.

, "

Behavior of Multiple Stabilizer Bottom Hole Assemblies

",

OGJ

,

Jan

.

1

,

1979

.

Google Scholar

Callas

N.P.

, and

Callas

,

R.L.

, "

Stabilizer placement-3 Boundary Value problem is Solved

".

OGJ

.

Dec

.

15

,

1980

Google Scholar

Walker

,

B.H.

, "

Some Technical and Economic Aspec to of Stabilizer Placement

,"

J.of Pet. Tech.

,

June

,

1973

.

Google Scholar

Walker

,

B.H.

, and

Friedman

,

M.B.

, "

Three-Dimensional Force and Deflection Analysis of a Variable Cross Section Drill string

",

Journal of Pressure Vessel Technology

,

May

1977

.

Google Scholar

Hoch

,

R.S.

, "

A Review of the Crooked-Hole Problem and An Analysis of Packed Bottom hole Drill Collar Assemblies

"

Drilling and Production Practices

,

Apr

.,

1963

.

Google Scholar

Timoshenko

,

S.P.

, and

Gere

,

J.M.

, <

Theory of Elastic Stability

>,

McGraw Hill Book Company, Inc

,

1961

.

Google Scholar

Woods

,

H.B.

, and

Lubinski

,

A.

, "

Charts to Solve Hole Deviation Problems

",

OGJ

.

May

31

1954

.

Google Scholar

Bottom Hole Assembly Problems Solved by Beam-Column Theory

ABSTRACT

INTRODUCTION

SIDE FORCE ON BIT

ASSEMBLY WITH TWO STABILIZERS

PENDULUM ASSEMBLY WITHOUT ANY STABILIZER

PENDUIUM ASSEMBLY WITH ONE STABLIZER

COMPARISON OF SIDE FORCES

FIELD VERIFICATIONS

CONCLUSIONS

References

Email Alerts

Explore

Connect

Resources

Engage

Bottom Hole Assembly Problems Solved by Beam-Column Theory Free

ABSTRACT

INTRODUCTION

SIDE FORCE ON BIT

ASSEMBLY WITH TWO STABILIZERS

PENDULUM ASSEMBLY WITHOUT ANY STABILIZER

PENDUIUM ASSEMBLY WITH ONE STABLIZER

COMPARISON OF SIDE FORCES

FIELD VERIFICATIONS

CONCLUSIONS

References

Email Alerts

Suggested Reading

Explore

Connect

Resources

Engage

Sign In or Register to Continue

Bottom Hole Assembly Problems Solved by Beam-Column Theory