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.

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)

• DB = diameter or bit, cm.

• PB = weight on bit, kg.

• α = angle of inclination, deg.

• Dci = outside diameter of i-th collar, cm.

• Wi = effective unit weight of collar, kg/cm.

• γ = specific gravity of drilling fluid, kg/cm3.

• Ii = moment of inertia, cm4.

• E = modulus of elasticity, kg/cm2.

• EIi = flexural rigidity, kg/cm2.

• Dsj = diameter of j-th stabilizer, cm.

• li = length of i-th drill collar section, cm. (in general, i = j+1)

• ej = $12DB-Dsj$ = 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.

Fig. 1 show how a side force RB 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

$q1=w1sin∝$

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

$RB=12w1l1sin∝+M1l1−Pe1l1$
(1)

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

$P=PB−12w1l1$
(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 RB as large as possible.

TABLE 1*

Side forces of Example 4

 PB 3000 12000 20000 l 1742 1612 1520 P 1830 10900 19000 X 1.031 1.406 1.773 $Ps′1$ -1.9 -12.5 -23.1 RB 38.8 25.2 12.4
 PB 3000 12000 20000 l 1742 1612 1520 P 1830 10900 19000 X 1.031 1.406 1.773 $Ps′1$ -1.9 -12.5 -23.1 RB 38.8 25.2 12.4

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

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. 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 qi = wisinα: (i = 1, 2, 3.) the same technique can, still, be utilized to establish the corresponding equations by satisfying the following conditions:

1. $Condition of continuity at S1θ1'=−θ1"$
(3)
2. $Condition of continuity at S2θ2'=−θ2"$
(4)
3. $Zero slope at point of tangency T θτ=0$
(5)

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

$θ'1=w1sin∝l1324E1I1X1+M1l13E1I1V1−e1l1$
(6)
(7)
$θ2'=w2sin∝l2324E2I2X2+M1l26E2I2W2+M1l23E2I2V2+e1−e2l2$
(8)
$θ2"=w3sin∝l3324E3I3X3+M2l33E3I3V3+e3−e2l3$
(9)
$θT=w3sin∝l3324E3I3X3+M2l36E3I3W3−e3−e2l3$
(10)

It should be noted, here, that three trancendental functions. Xi, Vi, Wi, 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:

$Xi=3(tg μi−μi)μi(i=1,2,3,)$
(11)
$Vi=32 μi(22μi−1tg2μi)(i=1,2,3,)$
(12)
$Wi=3μi(1Sin2μi−12μ2)(i=1,2,3,)$
(13)

Where

$μi=li2PiEiIi(i=1,2,3,)$
(14)

and Pi = 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)

$2M1(V1+l2I1l1I2V2)+M2l2I1l1I2W2=−q1l124X−q2l224⋅l2I1l1I2X2+6EI1e1l1+6EI1(e1-e2)l1l2$
(15)
$M1W2+2M2(V2+l3I2l2I3V3)=−q2l224X2−q3l324⋅l3I2l2I3X3−6EI2(e1-e2)l22−6EI2(e3-e2)l2l3$
(16)
$L34+4M2q3(W3X3)L32=24EI3(e3−e2)q3X3$
(17)

In solving the above simultaneous equations, as the unknown length l3 is entangled in the trancendental functions Xi, Vi, and Wi, 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, PB, DB, how to choose those parameters such as l1, Dc2, e1, e2, etc, so as to make the value of RB in Eq(1) as large as possible. Eq(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 Eq(1) ( i.e $Pel1$. 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 − $12$ 1.84 × 1800 = 13344kg

• e1 = 1st stabilizer radial clearance = $12$(24.4−21.4)=1.5cm

• e2 = 2nd stabilizer radial clearance = $12$(24.4−24.4)=0

• e3 = radial clearance at

• T = $12$ (24.4-17.8) = 3.3cm

• M1 = 21100kg-cm

• M2 = -66400kg-cm

• l3 = 2300cm

• RB = $12$ x 1.84 x sin3° − $211001800$$13344x1.51800$=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,

DB=21.5cm(8$12$" bit),

Dc1Dc3=17.8cm,

Dc2=12.7cm,

Ds1=Ds2=21.3cm,

α = 2°

PB = 12000 kg

Solution:The main numerical results are obtained as:

• P1 = 12000 − $12$ 1.34 × 1850 = 10760 kg

• e1 = e2 = 0.1 cm

• e3 = 1.85 cm

• M1 = 1200 kg - cm

• M2 = - 3000kg-cm

• l3 = 1650 cm

• RB = $12$ x 1.34 x sin2° 1850−$12001850$$10760x0.11850$=\ill\

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:

$I4=24 EIew sin∝X$
(18)

Where e = radial clearance of collar = $12DB-Dc$

$X=3(tgμ−μ)μ3 μ=12/PEIP=PB−12w1$

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

$RB=12wsin∝1−Pe1$
(19)

From the above relations, it can be seen that as the value of PB is small, X is nearly equal to one, l is large and consequently RB will be fairly large. So this is a good deviation correction tool. But when PB 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, RB will be reduces to a very small value. Such a situation will be illustrated apparently in Example 4.

Example 3

$Given DB=24.4(958"),Dc=20.3cm(8")w=1.84kg/cm,∝=3°,PB=15000kg,EI=17.2×109kg−cm2$

Solution: The main numerical results are obtained as:

$l=1628 cme=12(24.4−20.3)=2.05 cmP=15000−12×1.84×1628=13500 kgRB=12×1.84×sin3°×1628−13500×2.051628=61.4Kg$

Example 4:

$GivenDB=21.5cm(81"2)Dc=17.8cm(7")W=1.34kg/cm, ∝=2°EI=10×109kg−cm2 PB=4000kg,12000kg,20000kg.$

Solution: Table 1 gives the main numerical results.

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.

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

(20)
$l24+4M1q2W2X2l22=24EI2(e2-e1)q2X2$
(21)

Example 5: Given DB = 24.4cm,    Dc1 = 20.3cm,

Dc2 = 17.8cm,    Ds1 = 24.4cm,

W1 = 1.84 kg/cm (γ = 1.2),

W2 = 1.34 kg/cm    l1 = 1800cm

α=3°    PB=15000

Solution: The main numerical results are obtained as:

$M1=56400kg−cm I2=2180 cmRB=12×1.84×sin3°×1800−564001800=55.3kg$

The above calculated value of RB 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 DB = 21.5cm,    Dc1=Dc2=17.8cm

W1 = W1 = 1.34 kg/cm (γ = 1.2),

Ds1 = 21.3 cm,    l1 = 1850cm,

α=2°    PB=12000kg

Solution: The main numerical results are obtained as:

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 Ds1 and stiff l2 and the other with large Ds2 and flexible l2.

Table 2

Comparison of Side Forces

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

$BHA ϕ244−3A(i,e,95"8tricone bit)+8" colour×17.5m +ϕ220mm(lst Stabilizer )+7" collar×6m +ϕ242mm(2nd Stabilizer)+7"collar×105m +5" drill pipes.$

Exp. Results:

$Depth H 100m to 800∝<1°800m to 1025m1°<∝<3.5°1025m to 1300m3.5°<∝<5°$

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

$BHA:ϕ215−3A(i.e,812" Tricone bit)+7"×18.2m +ϕ188mm(1st stabilizer)+7"×8m +ϕ210mm(2nd stabilizer)$

Exp. results:

$H<1100m 0.5°<∝<2.0°1100m

Weight on bit 12000kg to 14000kg

(B) offset wells: average of several wells. $812$ 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

$H<1100m∝<1°45'1100m

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

1. 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.

2. 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.

3. 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.

4. The theory proposed in this paper is also applicable to assemblies in building-up, maintaining, or dropping the angle of inclination.

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