2
D RILLING & P RODUCTION =A 0 ln r +B 0 r 2 +C 0 r 2 ln r + A 1 r 2 + B 1 r 4 + +D 1 cos 2 (1) r = + = + 2B 0 2A 1 + + cos 2 (2) = =– + 2B 0 + 2A 1 + 12B 1 r 2 + cos 2 (3) r =– = 2A 1 + 6B 1 r 2 sin 2 (4) where boundary conditions are: (5) ( r ) r=r 1 = 0; ( r ) r=r 1 = 0; ( r ) r=r 2 = –q 1 –q 2 cos2; ( r ) r=r 2 = –q 2 sin2 A 0 =q 1 r 2 1 (6) B 0 =– (7) C 0 =0 (8) A 1 = (9) B 1 = (10) C 1 = (11) D 1 =q 2 r 2 1 (12) Kr = r 1 / r 2 (13) r = 1.154 Y P (14) r r=r 1 = + 2B 0 2A 1 + + (15) r=r 1 =– + 2B 0 + 2A 1 + 12B 1 r 2 1 + (16) P c = (17) A 0 = (18) A 1 = (19) B 1 = (20) K r 4 (K r 2 –1) 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 1 + 3K r 4 – 4K r 6 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 1 2 1 1–K r 2 1.154Y P –(1 + K p )A 0 + (1 – K p ) (2A 1 + 6B 1 + 6C 1 + 2D 1 ) 6C 1 r 4 1 A 0 r 2 1 4D 1 r 2 1 6C 1 r 4 1 A 0 r 2 1 K r 6 –1 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 1–K r 4 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 q 2 r 4 1 2 K r 2 (K r 2 – 1) 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 q 2 r 2 2 1 + 3K r 4 – 4K r 6 4K r 2 (1 – K r 2 ) 2 – (1 – K r 4 ) 2 q 2 2 1 1–K r 2 q 1 2 1 1–K r 2 2D 1 r 2 6C 1 r 4 r r 6C 1 r 4 A 0 r 2 2 r 2 4D 1 r 2 6C 1 r 4 A 0 r 2 2 r 2 2 rr C 1 r 2 E QUATIONS =0 =0

Empirical Formulas for Collapse Resistance Under Nonuniform Loading

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Empirical Formulas for Collapse Resistance Under Nonuniform Loading

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Page 1: Empirical Formulas for Collapse Resistance Under Nonuniform Loading

nonuniform loading forces, as indicated inEquation 5.

Equations 2-4 evaluate radial, tangen-tial, and shear stresses at these boundaryconditions. Also, the stress function coeffi-cients A0, B0, A1, B1, C1, D1 (Equations 6-12) are determined, given the boundaryconditions of Equation 5.

There are two ways to determine thecasing collapse failure pressure—elasticinstability and plastic yielding.2 For plastic

yielding, engineers use the Von MisesCriterion in which the stress distributionis elastic up to the point of yielding. Forcasing sizes ranging from 5 to 95⁄8-in.,most diameter-to-WT (D/t) ratios liebetween 10 and 25.Thus,Von Mises plas-tic yielding shows a more representativeview and will be used in the followinganalysis (Equation 14).

Plastic yielding first occurs at the inter-nal wall of casing. Applying the boundaryconditions of r = r1 and � = 0 intoEquations 2 and 3 yields Equations 15 and

16.Then substituting Equations 15 and 16into Equation 14 yields Equation 17,where the required coefficients are givenin Equations 18-23.

Casing imperfectionsCalculation of collapse resistance pres-

sure from Equation 17 results in valuesthat are higher than test values. Casingdoes not form a perfectly round cylinderbut has many imperfections from themanufacturing process such as ovality,eccentricity, and residual stress.

Consideration of these imperfectionsrequires Equation 17 to be modified tothe form shown in Equation 24, beforeusing it to calculate collapse resistancepressure.

The term PAPI is the collapse resistance

D R I L L I N G & P R O D U C T I O N

46 Oil & Gas Journal / June 18, 2001

Casing sample no. 1 2 3 4 5 7 8 10 13 14

Grade N-80 N-80 N-80 N-80 N-80 N-80 N-80 J-55 J-55 J-55Outside diameter, mm 139.7 139.7 139.7 139.7 139.7 139.7 139.7 139.7 139.7 139.7Wall thickness, mm 10.54 10.54 10.54 9.17 9.17 7.72 7.72 7.72 6.99 6.20

CASING DATA Table 1

� = A0 ln r + B0 r2 + C0 r2 ln r + �A1 r2 + B1 r4 + + D1� cos 2� (1)

�r = + = + 2B0 – �2A1 + + � cos 2� (2)

�� = = – + 2B0 + �2A1 + 12B1 r2 + � cos 2� (3)

�r� = – � � = �2A1 + 6B1 r2 – – � sin 2� (4)

where boundary conditions are: (5)(�r)r=r1

= 0; (�r�)r=r1= 0;

(�r)r=r2= –q1 – q2 cos2�; (�r�)r=r2

= –q2 sin2�

A0 = q1 r21 � (6)

B0 = – � (7)

C0 = 0 (8)

A1 = � (9)

B1 = � (10)

C1 = � (11)

D1 = q2 r21 � (12)

Kr = r1 / r2 (13)

�� – �r = 1.154 YP (14)

�r r=r1= + 2B0 – �2A1 + + � (15)

�� r=r1= – + 2B0 + �2A1 + 12B1 r2

1 + � (16)

Pc = (17)

A�0 = (18)

A�1 = � (19)

B�1 = (20)Kr

4 (Kr2 –1)

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2

1 + 3Kr4 – 4Kr

6

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)21�2

1�1 – Kr

2

1.154YP������–(1 + Kp) A�0 + (1 – Kp) (2A�1 + 6B�1 + 6C�1 + 2D�1)

6C1�

r41

A0�r21

4D1�

r21

6C1�

r41

A0�r21

Kr6 – 1

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2

1 – Kr4

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2q2 r4

1�

2

Kr2 (Kr

2 – 1)���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2q2�r22

1 + 3Kr4 – 4Kr

6

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2q2�2

1�1 – Kr

2

q1�2

1�1 – Kr

2

2D1�

r2

6C1�

r4

∂��r∂�

∂�∂r

6C1�

r4

A0�r2

∂2��

∂r2

4D1�

r2

6C1�

r4

A0�r2

∂2��r2∂�2

∂��r∂r

C1�r2

EQUATIONS

�=0

�=0

Page 2: Empirical Formulas for Collapse Resistance Under Nonuniform Loading

pressure of casing under uniform loadingwhen casing imperfections are taken intoaccount. Equations 25-37 illustrate how todetermine appropriate values for PAPI.

3

If casing imperfections from the manu-facturing process are known specifically,however, values for PAPI are calculated withthe equations explained in the previousarticle (OGJ, Jan. 22, 2001, p. 44-47).

The nonuniformity coefficient of load-ing, Kp, is the ratio of minimum earthstress to maximum earth stress. Equation24 calculates the collapse resistance of cas-ing for all nonuniform loading scenarios.

The value of Kp is varied in the intervalfrom 0 to 1.Thus, Equation 24 becomesapplicable for both uniform and nonuni-form loading.

Fig. 2 is a casing cross section that

shows ovality and eccentricity from man-ufacturing imperfections. Under uniformloading, the casing tends to collapse in thedirection of maximum radius and direc-tion of eccentricity.

Collapse resistance pressure, Pc,degrades further under nonuniform load-ing, if the maximum load, q2, is at rightangles to the maximum radius and eccen-tricity.

Test dataThis study employed actual 51⁄2-in. cas-

ing data to evaluate the validity of theequations generated.Table 1 shows thecasing grades, outside diameter, and WTof casing samples used in the study.4

The distribution of elliptical loadingalong the outer wall of casing appears inTable 2.4 Regression analysis of the data

generated maximum and minimum radiusand the ratio of the elliptical loading.

Table 3 presents the collapse resistanceof casing under nonuniform loading, aspredicted by Equation 24, which is inexcellent agreement with the actual testdata.

Nonuniform loadingThe nonuniform loading coefficient,

Kp, ranges from 1to 0 as the maximumand minimum directional earth stresscomponents, q1 and q2, change. For uni-form loading, q2 = 0 and Kp = 1. Forextreme nonuniform loading, q2 becomesvery large relative to q1, and Kp = 0.

Casing collapse resistance drops asnonuniform loading becomes moresevere, or as the nonuniform loading coef-

Oil & Gas Journal / June 18, 2001 47

C�1 = � (21)

D�1 = (22)

Kp = (23)

Pc = (24)

Z = (25)

A = 2.8762 + 0.10679 x 10–5 Z + 0.21301 x 10–10 Z2 –0.53132 x 10–16 Z3 (26)

B = 0.026233 + 0.50609 x 10–6 Z (27)

C = –465.93 + 0.030867 Z – 0.10483 x 10–7 Z2 + 0.36989 x 10–13 Z3 (28)

46.95 x 106 � �3

F = ———————————–—————— (29)

Z � – � �1 – �2

G = F � B/A (30)

(D/t)YP = (31)

(D/t)PT = (32)

(D/t)TE = (33)

If D/t (D/t)YP, then PAPI = 2YP � � (34)

If (D/t)YP D/t (D/t)PT, then PAPI = YP � – B� – 6.894757 x 10–3 C (35)

If (D/t)PT D/t (D/t)TE, then PAPI = YP � – G� (36)

If D/t (D/t)TE, then PAPI = (37)

NomenclatureD = Nominal outside diameter, mmD/t = Ratio of outside diameter to WT(D/t)YP = D/t intersection between yield strength collapse and plastic col-

lapse(D/t)PT = D/t intersection between plastic collapse and transition collapse(D/t)TE = D/t intersection between transition collapse and elastic collapseKp = Non-uniform loading coefficientKr = Ratio of inside of casing to outside of casingPAPI = Collapse resistance pressure under uniform loading, MPaPc = Calculated collapse resistance pressure, Mpaq1 = Uniform or minimum casing load, MPaq2 = Non-uniform or maximum casing load, Mpar = Radial distance within pipe, mmr1 = Inside radius, mmr2 = Outside radius, mmt = Nominal WT, mmYP = Minimum yield strength of the pipe, MPa� = Stress function� = Angle from reference axis�r = Radial stress, MPa�� = Tangential stress, MPa�r� = Shear stress, MPa

323.71 x 103

��(D/t) ((D/t) – 1)2

F�(D/t)

A�(D/t)

(D/t) – 1��

(D/t)2

2 + B/A��

3 B/A

Z (A – F)��C + Z (B – G)

�(A – 2�)2 + 8� (B +� C/Z)� + (A – 2)�����

2 (B + C/Z)

3B/A��2 + B/A

B�A

3B/A��2 + B/A

3B/A��2 + B/A

YP��6.894757 x 10–3

1������

– �(1

2P+

AP

K

I

p)� + �(11.1

–54

KY

p

P

)� (2A�

1 + 6B�1 + 6C�

1 + 2D�1)

q1�q1 + q2

Kr6 – 1

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)2

1 – Kr4

���4Kr

2 (1 – Kr2)2 – (1 – Kr

4)21�2