75

3. TUBING

Before considering tubular goods, we should have decided upon the tubing sizeneeded. As design input, we should know the range of pressures andtemperatures the tubing may be subjected to, and the fluid composition.

The tubing design principles are based on mechanical formula and laid downin industry policy, government regulation and company procedures. Belowthe design principles and basic formular are presented.

3.1 Steel

3.1.1 Metallurgy

Steel is made of iron and small amounts (0.2-1%) of carbon. The carbon isreacted with parts iron to form iron carbide, F’e3C. Thus steel is primarily analloy of iron and iron carbide, and much stronger than pure iron. The ironcarbide is distributed as small islands, plates, globes or other shapes.

The rnicrostructure or morphology of the steel is primarily dictated by themanner it is cooled and heating during manufacturing. The heating andcooling process is referred to as heat treatment. Its primary purpose is toalter the mechanical properties of the material, although it also influencesthe corrosion resistance.

Plain carbon steel are normally subjected to either of four heat treatments:annealing, normalizing. spheroclizing, or quenching and tempering (Q & T).

Annealing

Annealing involves heating steel to a temperature above its criticaltemperature of approximately 871 C (1600 F) and slow cooling in the furnace.Cooling in this manner may take as long as 8 to 10 hours. The resultingm1crostru’ture is pearlite with relatively coarse crystals. The Fe3C in anannealed rnicrostructure are distributed as platelets, have significantthickness and are rather widely spaced.

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Normalizin

A steel is normalized by heating it to a temperature above its critical

temperature, then removing it from the furnace and allowing it to cool in

still, ambient air. Under such conditions, a piece of steel will likely cool to

room temperature within 10 to 15 minutes. This more rapid cooling will

produce Fe3C platelets that are much thinner and are dispersed on a much

finer spacing than for an annealed carbon steel. A normalized

microstructure is usually described as “fine pearlite.

S p hero i ci iz in

Spheroidizing of carbon steel involves soaking the steel at a relatively high

subcritical temperature [approximately 650 C (1200 F)[ for 24 hours or more.

Under these conditions, the Fe3C phase has an opportunity to coalesce into

its most stable globular fonm

uenchin and Tempering

This is a two-step treatment in which the steel is heated to a temperature

above its critical temperature. rapidly cooled by immersing in water or oil

(quenched). and then heated at an elevated temperature below the critical

temperature for periods of approximately one hour (tempered). The F’e3C

precipitated during the tempering operation is in the form of very fine

particles.

High-strength oil field tubulars are often produced by quenching and

tempering. Such steel may have low resistance against several forms of

corrosion.

Normalizing is somewhat used for lower strength tubulars. By sacrificing

some strength and hardness, the corrosion and fatigue properties can he

improved.

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3.1.2 Mechanical strength

Subjected to mechanical loads, steel will bend, compress or stretch. As long

as the steel piece is not overloaded, ii. vill return to its original shape once the

load is relieved.

The load capacity obviously depends on the dimensions of the steel piece.

Thus, it is logical to divide the load by the cross-sectional area. This gives the

stress.

_Fc— (3-1)

stress is analogous to pressure, except that solid materials can take both

tensile and compressive stress. Pressure can only be compressive.

When subjected to tension, the steel \vill elongate proportionally to the stress,

and to the original length. Hooks law gives the relation between stress and

eastic elongation

a=E=EE (3-2)

Where:

AL : length change

L : original length

relative elongation

E : elasticity constant (Youngs modulus)

(forsteelE=2lOGPa=210 lO9Pa)

At sufficiently high stress, the steel will yield and elongate (flow) without any

further increase of the stress level, as illustrated by Fig. 3.1. This plastic

elongation is permanent and associated by changes in the microstructure

and metallurgical properties of the steel.

For construction purposes the yield point is obviously important, since steel

can take uniaxial stresses up to this level without breaking or being

permanently deformed or changed. The yield point dependents on the

microstructure of the steel and thus on the manufacturing processes.

However, the elasticity, E, is not affected by rnanufactu ring and is essentially

equal for all low-alloy steels. Fig. 3.1.

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The yield point is not easy to measure exactly. In the industry standard (API

5 series) most often used for well tubular goods, the yield point is defined as

the stress corresponding to 0.5% elongation. Such an elongation causes

permanent deformation.

Triaxial strentth

Well tubing are subjected to triaxial loads, usually decomposed in axial,

tangential and radial stress components. Obviously the strength considera

tion under triaxial loads can not be limited to any single stress components.

Triaxial stresses can be averaged into an equivalent uni-directional stress by

von-Mises equation

2a = (Ga -a)2 + (Or - Ga)2 + (a - Or)2 (3-3)

where

ae : equivalent unidirectional stress level (should not exceed the yield

point)

aa axial stress

at : tangential stress

ar radial stress

3.1.3 Elongation

Subjected to stress an elastic material will elongate. The axial elongation

due to axial stress can be estimated by inverting Hook’s law, Eq. 3-2. This

gives

= .L. = ! (3-4)

For triaxial stress, the elongation behaviour is slightly more complicated,

since each stress component also causes some elongation (or contraction)

perpendicular to its direction: transverse elongation.

Thus, for well Itibulars, differences between the inside and the outside

pressure will cause some length change. The ratio between transverse and

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directional elongation is a constant for elastic materials. Thus the

elongation clue to radial and tangential stresses is estimated as

Ea2 =- (G + Gr)

(35)

where:

Ea2 : axial elongation due to tangential and radial stress

v : Poisson’s number. For steel : v = 0.3.

3.1.4 Well tubulars

Standard carbon steel well tubing and casing are usually manufactured

according to API-specifications (API= American Petroleum Institute). These

specifications primarily ensures that pipes from different mills are

interchangeable. Thus, the API-specifications are precise on dimensions,

less precise on strength and unpresize on manufacturing and metallurgy.

Well tubular goods are covered by the API 5 series. These requirements are

generally less severe than for process pipe,. This has some logic, since the

consequence of failure is less severe for xvell pipe than for surface pipe. Also,

the safety factors are relatively modest for tubing design.

API-pipe is referred to by steel grade and dimensions. The steel grade is

designated by letters and numbers (e.g. N-80). The letter refers to

requirements of the manufacturing process. The number refers to the

minimum yield strength in thousands of psi. Tab. 3.1. The minimum yield

strength is commonly used as load limit. Thus, N-SO and L-80 tubulars have

the same load capacity. The manufacturing process differs, resulting in

different corrosion resistance. Some manufacturers produce N-80 tubulars

approaching L-80 quality.

H-40, J-55 and K-55 are one low carbon, lower strength steels. They are

ductile and seems to be immune to sulfide stress cracking. The K-55 is

primarily a casing grade and available only in casing sizes (diameters above

4.5 inches). It has the same yield strength as J-55. but higher ultimate

strength. Tab. 3.1.

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N-SO is the lowest tubular grade considered to be high strength. II. is verycommon in North Sea wells. Chemically it is similar to automobile steel.

C-75, L-80 and C-95 are restricted yield steels, heat treated to avoid localtension and hardening. These grades were specifically developed for sourservice.

P- 105 arid P-i 10 are highest yield strength tubulars covered by the APIspecifications. The manufacturing process may according to the cificationAPI- 5AX be either quenched and tempered or normalized and tempered.Experience seems to indicate that no.nnalized and tempered pipes Is lesssuitable for these grades.

The API-pipe dimensions available are listed in Tables 3.2 and 3.3. The API-5 series also contains tolerances, testing procedures, some chemical andmanufacturing requirements and other information.

Since the APIspecificaUons are broad on manufacturing and metallurgy,many mills also manufacture proprietar pipe grades. Usually these exceedsthe generally API-requirements in some aspects. often in corrosionresistance. Most non-API pipe confirms to the API standard dimensions.

Table 3.1 API standard steel grades for well tubulars. (Ref. API Rp. 5A, 5AC.5AX).

API Mm. yield Max yield Ultimate strengthclassification psi MPa psi MPa psi MPa

H-40 40.000 276 80.000 552 60.000 414J-55 55.000 379 80.000 552 75.000 517K-55 55.000 379 80.000 552 95.0(X) 655C-75 75.000 517 90.000 620 95.000 655L-80 80.000 552 95.000 655 95.000 655N-SO 80.000 552 110.000 758 100.000 689C-95 95.000 110.000 758 105.000 724P-105 105.000 724 135000 931 120.000 827P-HO 110.000 758 140.000 965 125.000 862

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Table 3.2 Tubing list after API spec. 5A

1 2 3 4 5

Size NominalOutoide Weight, Wall

Diameter Threadu Grade Thiel,nesg Type o(

r“— and -‘-—--—- End5l

in. mm Coupling, in. mm

lb. per ft.

I)

ff1 050 26,7 1.14 H, J, N 0,113 2,87 Non-Upset

LOSO 26,7 1.20 H, J, N 0.113 2,87 Ext. Upset

1.315 33,4 1.70 H, J, N 0.133 3,38 Non-Upset

1.315 33,4 1.72 H, J, N 0.133 3,58 Integral Joint

1.315 33,4 1.80 H, 3, N 0.133 3,38 Ext. Upset

1.660 42,2 2.10 H, 3 0.125 3,18 Integral Joint

1.660 42,2 2.30 H, J, N 0.140 3,56 Non-Upset

1.660 42,2 2.33 H, J, N 0.140 3,56 Integral Joint

J.660 42,2 2.40 H, J, N 0.140 3,56 Ext. Upset

1.900 48,3 2.40 H, 3 0.125 3,18 Integral Joint

1.900 48,3 2.75 H, J, N 0.145 3,68 Non-Upset

1.900 4,3 2.76 H, J, N 0.145 3,63 Integral Joint

1.900 48,3 2.90 H, J, N 0.145 3,68 Ext. Upset

2.063 52,4 3.25 H, J, N 0.156 3,96 Integral Joint

2% 60,3 4.00 H, J, N 0.167 4,24 Non-Upset

2% 60,3 4.60 H, J, N 0.190 4,83 Non-Upset

2% 60,3 4.70 H, J, N 0.190 4,83 Ext. Upset

2% 60,3 5.80 N 0.254 6,45 Non-Upset

2% 60,3 5.95 N 0.254 6,45 Ext. Upset

2% 73,0 6.40 H, J, N 0.217 5,51 Non-Upset

2% 73,0 6.50 H, J, N 0.217 5,51 Ext. Upset

2% 73,0 7.80 N 0.276 7.01 Non-Upset

2% 73.0 7.90 N 0.276 7,01 Ext. Upset

2% 73,0 8.60 N 0.308 7,82 Non-Upset

2% 73,0 8.70 N 0.308 7,82 Ext. Upset

3½ 88,9 7.70 H, J, N 0.216 5,49 Non-Upset

3½ 88,9 9.20 H, 3, N 0.254 6,45 Non-Upset

3½ 88,9 9.30 H, J, N 0.254 6,45 Ext. Upset

3½ 88,9 10.20 H, J, N 0.289 7,34 Non-Upset

3½ 88,9 12.70 N 0.375 9,52 Non-Upset

3½ 88,9 12.95 N 0.375 9,52 Ext. Upset

4 101,6 9.50 H, J, N 0.226 5,74 Non-Upset

4 101,6 11.00 H, J, N 0.262 6,65 Ext. Upset

4½ 114,3 12.60 H J, N 0.271 6,88 Non-Upset

4½ 114,3 12.75 H, J, N 0.271 6,88 Ext. Upset

Non-ui,set tubing is acalIble with regularspecial bevel. ,,,‘ sj’ecil clearance cout’l ings.

ttFor inCormation purposes only.

c,,uidings ru special bevel c,,,,lrli,,gr, E,,ternal’ui’set tubing is a ail;,bl with

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Table 3.3 casing list after API spec. 5A

1 2 3 4 5

Slee: Nominal Type o ThreadOutside Weight. Wall

Diameter, Threads and Thickness t U U

,............., Coupling, C C

lb. Per ft. Grade ,_—--———-_. .0 0

U)in. mm in. mm o xD t

1 2 3 4

Site: Nominal Type of ThreadOutside Weight.

Diameter. Threads and,.._. Coupling.

lb. per ft. GradeIn. mm

D

.-‘ 54

‘ r4½ 114,3 9.50 HJ, K 0.205 5,21 X4½ 114,3 10.50 J, K 0.224 5,69 X . . X4½ 114,3 11.60 J, K 0.250 6,35 X X X4½ 114,3 11.60 N 0.250 6,35 . . X X41/a 114,3 13.50 N 0.290 7,37 .. X X

5 127,0 11.50 J, K 0.220 5,59 X5 127,0 13.00 J, K 0.253 6,43 X X X5 127,0 15.00 J, K 0.296 7,52 X X X X5 127,0 15.00 N 0.296 7,52 .. X X X5 127,0 18.00 N 0.362 9,19 .. X X X5 127,0 21.40 N 0.437 11,10 . . X X5 127,0 23.20 N 0.478 12.14 .. X X5 127.0 24.10 N 0.500 12,70 .. X X

5½ 139,7 14.00 H, J, K 0.244 6,20 X5½ 139,7 16.50 J, K 0.275 6,98 X X X X5½ 139,7 17.00 J, K 0.304 7,72 X X X X5½ 139,7 17.00 N 0.304 7,72 .. X X X5½ 139,7 20.00 N 0.361 9,17 .. X X X5½ 139,7 23.00 N 0.416 10,54 .. X X X

6% 168,3 20.00 H 0.288 7,32 X6% 168,3 20.00 J, K 0.288 7,32 X X X6% 168,3 24.00 J, K 0.352 8,94 X X X X6% 168,3 24.00 N 0.352 8,94 .. X X X6% 168,3 28.00 N 0.417 10,59 .. X X X6% 168,3 32.00 N 0.475 12,06 .. X X X

7 177,8 17.00 H 0.231 5,87 X7 177,8 20.00 H, J, K 0.272 6,91 X7 177,8 23.00 J, K 0.317 8,05 X X X X7 177,8 23.00 N 0.317 8,05 .. X X X7 177,8 26.00 J, K 0.362 9,19 X X X X7 177,8 26.00 N 0.362 9,19 .. X X X7 177,8 29.00 N 0.408 10,36 .. X X X7 177,8 32.00 N 0.453 11,51 .. X X X7 177,8 35.00 N 0.498 12,65 .. X X X7 177,8 38.00 N 0.540 13,72 .. X X X

7% 193,7 24.00 H 0.300 7,62 X7% 193,7 26.40 J, K 0.328 8,33 X X X X7% 193,7 26.40 N 0.328 8,33 .. X X X7% 193,7 29.70 N 0.375 9,52 .. X X X7% 193,7 33.70 N 0.430 10,92 .. X X X7% 193,7 39.00 N 0.500 12,70 .. X X X7% 193,7 42.80 N 0.562 14,27 . . X X

7% 177,8 45.30 N 0.595 15,11 . . X X7% 193,7 47.10 N 0.625 15,86 . . X X

WallThickness

—--

in. mm

8% 219,1 24.00 J, K 0.264 6,71 X8% 219,1 28.00 H 0.304 7,72 X8% 219,1 32.00 H 0.352 8,94 X8% 219,1 32.00 J, K 0.352 8,94 X X X X8% 219,1 36.00 J, K 0.400 10,16 X X X x8% 219,1 36.00 N 0.400 10,16 .. X X X8% 219,1 40.00 N 0.450 11,43 .. X X X8% 219,1 44.00 N 0.500 12,70 . . X X X8% 219,1 49.00 N 0.557 14,15 .. X X X

9% 244,5 32.30 H 0.312 7,92 X9% 244,5 36.00 H 0.352 8,94 X9% 244,5 36.00 J, K 0.352 8,94 X X X9% 244,5 40.00 J, K 0.395 10,03 X X X X9% 244,5 40.00 N 0.395 10,03 .. X X X9% 244,5 43.50 N 0.435 11,05 . . X X X9% 244,5 47.00 N 0.472 11,99• . . X X X9% 244,5 53.50 N 0.545 13,84 .. X X X

10% 273,0 32.75 H 0.279 7,09 X10% 273,0 40.50 H 0.350 8,89 X10% 273,0 40.50 J, K 0.350 8,89 X .. X10% 273,0 45.50 J, K 0.400 10,16 X .. X x10% 273,0 51.00 J, K, N 0.450 11,43 X .. X x10% 273,0 55.50 N 0.495 12,57 X .. X x

11% 298,4 42.00 H 0.333 8,46 X11% 298,4 47.00 J, K 0.375 9,62 X .. X11% 298,4 54.00 J, K 0.435 11,05 X .. X

:11% 298,4 60.00 J, K, N 0.489 12,42 X .. X

J3% 339,7 48.00 H 0.330 8,38 I13% 339,7 54.50 J, K 0.380 9,65 X . . X13% 339,7 61.00 J, K 0.430 10,92 X .. X13% 339,7 68.00 J. K. N 0.480 12,19 X . . X13% 339,7 72.00 N 0.514 13,06 I .. X

16 406,4 65.00 H 0.375 9,52 I16 406,4 75.00 J, K 0.438 11,13 X .. X16 406,4 84.00 J, K 0.495 12,57 X .. X

18% 473,1 87.50 H, J, K 0.435 11,05 X18% 473,1 87.50 J, K 0.435 11,05 X .

20 508,0 94.00 H, J, K 0.438 11,13 X X ..

20 508,0 94.00 J, K 0.438 11,13 X20 508,0 106.50 J, K 0.500 12,70 X X X .

20 508,0 133.00 J, K 0.635 16,13 X X X .

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3.2 Tubing design

A tubing hanging in a vell will be subjected to the tension caused by its own

weight, initial tensioning, thermal expansion, pressure differences,

buoyancy, flow friction. These loads will result in stresses. The tubing

should be dimensioned such that the actual stresses stay some safety margin

below the yield point of the steel.

3.2.1 Initial stress

Consider the tubing before production start. It will be subject to different

mechanical loads. Below we will consider these loads and the resulting stress

components. In the end, these separate stress components will be added

(superposed) to give the actual stress level.

a) Tubing weight:

This will cause tensional load, largest at the tubing hanger where the first

pipe section carries all sections below. The axial tension due to weight

alone is expressed as

= - z (3-6)

where:

Gaw : axial stress due to the steel weight

average steel weight per meter of tube, including tool joints

g : acceleration of gravity

cross sectional pipe wall area

z : distance from the lower end of the pipe

b) Tensioning

The tubing is often pulled in tension initially, to avoid later compression.

Thus, the axial stress is increased over the whole tubing. The tension

force effect is readily added to (he tubing weight stress Eq. (36) to give

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F-w g z

0a2 =(37)

where:

F : packer force, negative downward. (Thus pulling the tubingcauses a negative packer force). N

aa2 axial stress due to steel weight and initial packer forces.

c) Pressure differences

Pressure dilferenccs between the tubing and the annulus will cause radialand tangential stresses .i.n the pipe walls, The tangential sli-ess will be thelargest of these. For stress calculations the tangential stress is usuallyestimated by formula for thin-walled pipe.

(P0 -P)D(3-8)

where:

P0 : outer (annulus) pressure

Pj : inner (tubing) pressure

pipe wall thickness

D : (using the outer diameter of a thick-walled pipe provides a safetymargin)

More precise estimates for radial and tangential stresses are obtained byLamé’s equations for thick-walled pipe.

— p r?-

p r p-p r r+

——

-r2..r2 r2. r2C e d

—p r?

-

p r p-p r r?

.2 .2 2 r2C C d

kiipc1 u n /1 ‘rod .1 I /333 /27 0 9 IT IAA/ 9

85

where:

re external radius of the tube

inner radius of the tuber : radial distance to the point considered (rj r re)Pc : external pressure (in the annulus)

Pi inner tubing pressure

radial stress

tangential stress

It may be noted that Lamés tangential stress equation (3-10) approachesthe thin-walled stress equation (3-8), when the pipe wall thickness goes tozero. (The proof is suggested as an exercise). It may also be noted that withno external pressure, the radial stress at the inner pipewall equals theinternal pressure. (Exercise: What is the radial stress at the outerpipewall with 100 bar internal pressure and 1 bar external).

d) Buoyancy

Buoanc is the force resulung from the pressure distribution around asubmerged object. Often buoyancy is taken as the weight of the displacedliquid volume. However, this assumes static pressure equilibrium in thesurrounding liquid. This assumption notably breaks down for the case ofpressure differences across a production packer. (It also breaks down forthe case of gas release in water, where one may erroneously predict that aship will sink due to aeration of the water).

Due to geometry, the resulting buoyancy of a vertically suspended pipeequals the pressure force at the lower pipe end. Thus the axial stress dueto buoyancy simply equals the well pressure.

F,0a,h =

A,(3-11)

Thus, buoyancy is not constant. The buoyancy stress component isusually relatively small. However, since it is directed upward at the lowerend of the pipe, ft may cause compression and buckling.

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3.2.2 Bi-axial loads

The tubing is subject to axial, tangential and radial stresses. The equivalent

unidirectional stress is determined by von-Mises criterium, Eq. (3-3). The

radial stresses Eq. (3-9) is small compared to the axial and tangential

stresses and is commonly neglected.

Doing so, von-Mises criterium can be displayed graphically by the ellipse of

biaxial yield stress, Fig. 3.2. This generally implies that moderate tension

increases the burst capacity.

3.2.3 Stabffity

Intuitively a freely hanging pipe will always hangs straightly. This intuition

is correct, provided that the pipe material is heavier than the surrounding

fluid.

Intuitively it is also easy to accept that a long slender pipe can not withstand

much compressive force without bending or buckling. Thus, as a first

approximation, a production tubing may be expected to remain straight only

as long as it is kept in axial tension.

However, a production tubing is also subject to differences between internal

and external pressures. These causes additional stresses in the pipe walls,

which also affect the tendency to remain straight, or to buckle.

It is useful to approach buckling as a stability problem. Compression stores

elastic energy in the tubing walls, which is released if the tubing buckles.

Consequently, a straight, compressed tubing represents a higher energy level

than a buckled. Thus, a straight tubing under compression is unstable.

Extending this basically simple idea, Klinkenberg /1951/ and Woods /1951/

showed that a straight tubing will be stable provided that (he axial stress is

less than the average of the radial and tangential stresses.

0r + 0tO <(3-12)

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Lamé’s equation (3-9) (3-10), with Woods criterium (3-12) gives the followingcommonly used stability criterium.

-) 2pi r-

l:)c r — pA, - pAOa<2 2 A (3-13)s

It may be noted that for equal inner and outer pressure (pi Pe) and thin-walled tubing, the right hand side of Eq. 3-13 vanishes. Thus, the stabilitycriterium becomes: 0a < 0. which brings us back to the simple intuitivecriterium that a tubing is unstable under compression.

3.2.4 Length changes during production

During production the tenWerature and pressure and flow rate will vary.These variations causes elongation and contraction of the steel and affect thetotal length of the tubing.

Length changes should be accounted for in the design. For surface pipelinesthis is usually done by turns or loops. If you visit a refinery you may look forthese turns that may appear unmotivated at first glance.

For well tubing the limited space inside a casing does not allow anything likea loop. Length changes may be designed for by providing an expansion Jointwhere the tubing is allowed to slide inside an overshooting tube unit (sealunit/tubing seal receptacle). This requires dynamic, axial, elastomer seals.In a chemically unfriendly environment, seals have not been problem-free.

Alternatively, a fixed connection may be used between the tubing and thepacker (tubing anchor). This prevents axial tubing movement. The restrictedelongation causes elastic stress that must be designed for.

The anticipated length changes must be known so that they can he designedfor. Below the most important causes of tubing elongation are treated.

a) Thermal expansion

When hot reservoir fluids are produced, the tubing will expand due totemperature changes. The thermal expansion is proportional to thechange in temperature and can be estimated by

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ALT = L cxAT (3-14)

where:

L initial length

AT : temperature change

a : thermal expansion coefficient (for steel, a = 11.5 10-6 K-1)

b) Ballooning

If the pressure inside the tubing increases, the tubing length will decrease

slightly. The analogy to inflating a balloon is obvious and is the reason

for the terminology. When the well is put into production, the reservoir

pressure makes the tubing inside pressure increase. The setting of a

hydraulic packer requires the tubing pressure to be pumped up.

The length changes due to ballooning can be derived from basic material

properties. The elongation due to radial and tangential stresses has

already been given by Eq. (3-5), rewritten below

E2

The radial and tangential stresses are related to inside and outside

pressures by Lames equations (3-9) and (3-10). Substituting Eqs. (3-9) and

(3-10) into Eq. (3-5) gives

— —‘

pr-prEa2 — ER

— E 2 - 7 (3-15)C

where:

elongation due to ballooning

This accounts for the absolute elongation (negative elongation =

contraction) due to different inside-outside pressures. referenced to

surface conditions. However, in most cases we are more interested in the

elongation referenced to the tubing length at setting conditions. The

resulting relative elongation formula is usually written as

AL’ — L — 2v Ap - R2Ap0— ER — -

E R2 -

L (3-16)

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where:

R : diameter ratio,

d r

change in internal pressure from setting conditions: Ap1 = P -

change in annular pressure from setting conditions; Ap = Pc - Pc°

During completion, the tubing and the annulus are often both filled withthe same fluid. When setting a hydraulic packer, pressure is applied to thetubing. Since the fluid density are the same inside and outside (he tubing.the inside-outside pressure thfferences will be the same all along thetubing. Thus, the ballooning effect can be estimated directly from Eq. (3-16).

During production, the tubing contains oil and/or gas (hopefully),whereas the annulus still contains completion fluid. Since these fluiddensity usually differs, the inside-outside pressur.e difference will changealong the tubing.

In the literature, more complicated ballooning formula are sometimespresented. e.g. Hammerlindl/ 1979/. These may account for pressuredifferences at the wellhead, density differences and pressure loss due toflow friction. Since the ballooning is linear with respect to pressure,density differences and constant friction gradients may be accounted formuch easier by using aritmetic averaged pressures and Eq. (3-16) directly.

c) Buckling

A theoretical expression of length changes due buckling has been given byLubinski /1962/. This results substantial length changes.

d) Other effects

All changes in axial forces during production may also cause lengthchanges. Thus. flow friction will have some effect on the tubing length. asvill also variations in the pressure below the packer. Such lengthchanges may he estimated by Hooks law, when the force is known.Usually these length change effects are small.

koIj3cl1liI/I’IJd.Ul/331/27.C) .4/j (j’/>

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e) Total length change

The total length change may be estimated simply by adding U all the

contributions by the independent mechanisms considered above.

AL = ALT + ALB + (3-17)

If the tubing is designed with an expansion joint, the stroke length of this

must be at least as large as the maximum length change.

3.2.5 Pre-tensioning to avoid later compression

An expansion joint involves dynamic axial seals that may become future

sources of leaks. An alternative design is to use a fixed connection between

the tubing and the packer (tubing anchor). This requires initial tensioning of

the tubing, to avoid compressive loads and buckling later.

Consider an expected length change AL, computed by Eq. (3-17) above. Such

an anticipated length change may picked up initially by tensioning the

tubing after landing. This leaves the tubing with an additional axial stress

initially. When the later elongation occurs, it will release this tension, and

not cause compressive loads.

The relationship between a tension force, -F, and the corresponding elastic

elongation, AL, is obtained by combining Eqs. (3-1) and (3-2).

-F = AE (3-17)L

where:

steel wall cross sectional area of the tubing

E : elasticity constant (210. i09 Pa)

At the tubing hanger, the tensioning to avoid later compression adds to the

tension due to the weight of the tubing. Thus the relative increase may be

n ode rate.

At the packer, the tensioning causes an upward-directed packer force equal to

the force computed by Eq. (3-17). This may be the dominating packer force.

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91

3.2.6 Design factors

In most cases we have reasons to select a somewhat stronger pipe thanabsolutely required to withstand the design loads. This extra margin is

commonly expressed as a design factor, DF, defined as the ratio between the

actual strength and the minimum strength required, or equivalently

DF= actual material yield strength

design stress

Minimum design factors are usually provided as company policy. There alsoexists some industry standards like API-recommendations. Governmentalagencies may set minimum requiremenis. Special circumstances mayfavour the choice of more generous design factors than these minima.

Some considerations that go into the choice of design factors are:

a) Ma nu fact u ring tolerances:

The actual pipewall thickness may vary above and below the nominalthickness. API accepts a minimal wall thickness of 0.875 of the nominal.Thus, considering the most thin-wafled pipe, we may use a design factor:DF= 1/0.875 1.14.

b) Material inhomogenities:

The steel properties vill vary within the same pipe grade. Also, differentsteel mills achieves different variances of the material properties. This isonly partly accounted for by designing based on the minimum yieldstrength.

c) Corrosion:

During production, the pipe wall thickness may be reduced due tocorrosion. If uniform corresion is expected, it may be cost efficient to useplain steel and allow for wall thickness reduction.

d) Dynamic loads:

Pipe design is based on static loads. During setting and pulling, the pipe Issubjected to substantial dynamic loads. These usually dictate a somewhatgenerous tension design factor.

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e) Safety considerations:

Unforeseen circumstances may weaken the pipe. If the design loads are

uncertain, a higher design factor should be used. Likewise, if the

consequence of a rupture is severe.

Examples of minimum tubing design factors listed in Table 3.4.

The engineer may find that special circumstances of the design considered

justifies higher design factors than such minimum requirements.

Table 3.4

Collapse: DF = 1.13

Burst: DF1 = 1.13

Tension: DF = 1.33

3.3 Corrosion

In nature, most metals occur as ore. Common iron ores are oxides (Fe203) and

sulfides (FeS). Iron oxide is equivalent to common rust.

Iron ore is converted to iron metal by adding energy and a reducing agent

(commonly carbon monoxide). The energy stored in the iron during this

conversion is released when the iron corrodes. This energy supplies the

driving force for corrosion.

From this follows that iron, and most other metals, are energetically

unstable and will lend to convert back to ores. Corrosion control/prevention

means that we Liy to temporarily stop or slow down the rate of this

conversion

3.3.1 The common weight-loss corrosion

Corrosion as occurring in petroleum equipment is an electrochemical

reaction involving an anode and a cathode, covered by an electrolyte. The

anode and cathode may be made up by different areas of the same metal

surface, or by different metal components. The electrolyte is usually water

with dissolved salts or acids.

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93

a) Anode reaction:

At the anode, the metal is oxidized (stripped of electrons). The anodic

reaction can be written as

Fe — Fe2 + 2e(3-18)

iron iron+ free

atom ions electrons

The iron ions goes into solution in he electrolyte. Thus, the anodic area

is that portion of the metal that is corroded. The electrons are left behind

in the metal and must be consumed at the cathode for the reaction to

proceed.

b) Cathode reaction:

AL the cathodic area, the electrons are consumed by reaction with an

oxidizing agent present in the water. The chemical reaction differs

depending on the oxtdizing agent. Exposed to the atmosphere. water vili

absorb some oxygen, as well as other gases. Oxygen, even in minor

amounts, is a very potent oxidizing agent. The cathodic reactions with

oxygen present may be written as

in acidic solutions:

02 + 4l-l + 4e — 2H20(3-19)

in neutral or alkaline solutions:

02 + 2H20 + 4e — 40H (3-20)

Oxygen corrosion may occur in open tanks, or in injection wells if oxygen

is not adequately removed from the injected fluids. Oxygen is normally

not present in reservoir fluids. In fact most sub-surface environment are

reducing and will consume oxygen over time. Inside petroleum

equipment, the oxidizing agent is commonly weak acids. This gives the

following cathodic half-reaction.

2H + 2e —> -l2 (3-2 1)

konipendftim/l’rodAIl/333/27-O I -93/I IAA/alb

94

hydroTen hydrogen+ electrons —

ions gas

c) Electrolyte:

The anodic and cathodic reactions listed above are half-reactions. Each

reaction produces or consumes electrons. Isolated, such reactions would

build up or deplete electrical changes, and stop as soon as a limiting

electrical potential has been reached. To proceed, the anode and cathode

must be electrically connected to allow flow of electrons, and covered by a

electrolyte to allow movement of ions to match the electrons.

Figure 3.3 illustrates acidic corrosion of iron. The anode reaction

described by Eq. (3-18) and the cathode reaction by Eq. (3-2 1). The overall

reaction adding Eqs. (3-18) and (3-2 1) becomes:

Fe + 2H — Fe + H, (3-22)

iron +hydrogen iron

+hydrogen

atOmS ions ions gas

The corresion rate depends on the conductivity of electrolyte. Pure water

at neutral pH is a poor electrical conductor. However, the conductivity

increases rapidly with increasing concentration of salts or acids.

d) Electromotive force:

As described above, at the anode the metal releases electrons and

dissolves. At the cathode, the metal takes no apparent part in the overall

reaction.

The relative tendency for different metal to act as an anode or a cathode

may be measured directly by submerging the metal pieces in the same

electrolyte and measuring the electromotive force (measured as voltage at

zero current flow) that developes between them. However, it is more

practical to measure the electric potential between each metal and a

standard reference electrode.

Typical potential values for different metals in neutral soil containing

oxygen or water, measured with respect to a copper/copper-sulfate

koni 1,cnd ti ii / ‘mcI .111 / 333/27 () I 94/1 IAA/ aib

95

reference electrode are presented in Table 3.5. A high negative potential

here implies a high tendency to release electrons. Thus, if two metals

with different potentials are connected, the one with the highest negative

potential will preferably release electrons and act as the anode. The one

with the less negative (or more positive) potential will act as a cathode

and thus not con-ode.

From Table 3.5 it may be noted that e.g. mill scale on steel (EMF = -0.2) is

less negative than the steel itself (EMF -0.5). Thus, the mill scale will act

as a cathode; whereas the steel around the mill scale will anodic and

corrode locally.

Table 3.5 - Practical Galvanic Series

Metal Volts(’)

Commercially pure magnesium -1.75

Magnesium alloy (6% Al, 3% Zn. 0. 15% Mn) -1.6

Zinc -1.1

Aluminum alloy (5% zinc) -1.05

Commercially pure aluminum -0.8

Mild steel (clean and shiny) -0.5 to -0.8

Mild steel (rusted) -0.2 to -0.5

Cast iron (not graphitized) -0.5

Lead -0.5

Mild Steel in concrete -0.2

Copper. brass, bronze -0.2

High silicon cast iron -0.2

Mill scale on steel -0.2

Carbon, graphite, coke +0.3(1) Typical potential normally observed in neutral soils and

water, measured with respect to copper sulfate reference

electrode

knnipendium/T’rodW/333/27.O 1 -94/I lAA/lb

96

3.3.2 Secondary reactions

The iron ions by react with different components in the electrolyte to form

solid precipitates like

iron sullide:

Fe2 + S2 — FeS (3-23)

iron carbonate:

Fe2 + CO — Fe CO3 (3-24)

Solid precipitates on the anodic areas will usually slow down the diffusion of

ions. Thus, the corrosion rate may be reduced, often significantly.

Conversly, if the precipitates are dissolved by acids or eroded by flow

turbulence or droplet impingement, the corrosion rate increase

dramatically.

3.3.3 Other forms of corrosion

The effect of the common corrosion reactions discussed above is to weaken

the pipe by dissolving the metal. There are also other forms of corrosion that

weakens the pipe by altering the rnicrostructure of the metal, without any

appreciable metal loss. The most feared of these are

a) Sulfide stress corrosion (SSC).

This is the traditional problem of all fields containing H2S. When H2S

dissolves in water, it causes acidic corrosion with an overall reaction as

described by Eq. (3-22). However, the sulfide ions catalyse the diffusion of

hydrogen atoms into the metal, rather than creation of hydrogen gas at

the metal surface. When the hydrogen atoms later re-combine inside the

metal matrix, it may cause blistering in low strength steel.

In high strength steels the result is the much more damaging

embrittlement, whereby the steel loses its strength and ductility.

Especially for steel under tension, this may result in spontaneous brittle

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97

failure. The suspectibility to SSC is primarily determined by thefollowing variables.

1. Strength: Carbon steels with yield strength below 620 MPa (90 000psi) (corresponding to hardness below Rc 22) are generally consideredimmune to SSC. Above this, increased strength usually impliesincreased susceptibility.

2. Stress level: The time to failure decreases with increased stress level.In most cases failure occurs due to axial tension or applied pressure.1-lowever, residual stress crealed by bending, wrench marks and otherforms of incidental or intended cold working may iniUate SSC.

3. HS concentration: The time to failure generally decreases withincreased concentration of HnS. As a rule of thumb. f12 S partialpressures above 0.001 atmosphere or 100 Pa (0.0 14 psi) may causeSSC, although the time to failure may be very long at the lowestconcentrations.

4. pH-level, or competing reactions: pH indicates the amount of H ionsin a solution and thus the tendency of H2S to dissociate. The presenceof other ions may consume H ions (e.g. C0321 thus prevent the acidcorrosion reaction. Thus the cracking tendency decreases withincreasing pH, and is drastically reduced if the pH can be maintainedabove 9.

5. Temperature: SSC mainly occurs at temperatures below 70 C (150 F).Thus it basically concerns the upper parts of the wells surfaceequipment and drilling operations

b) Stress corrosion cracking (SCC)

This phenomena affects most alloys subjected to tensile stress, saltsolutions and high temperatures. In the oil field, SCC is mostlyexperienced in very hot wells containing clorides (Cli and H2S. Stresscorrosion cracking is known to affect plain carbon steel as well as veryexpensive stainless alloys, only pure metals seems to be immune. Thetendency to SCC increases by increasing tensile stress.

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3.3.4 Principles of corrosion control

Generally, any action that retards the slowest reaction step in the corrosion

process, will also retard the overall rate. Actions that may retard the

corrosion rate are:

a) Covering the metal by an impervious layer: to break the electric circuit.

b) Replacing the electrolyte by a non-conductive fluid: to break the electric

circuit.

c) Removing the oxidizing agent, to prevent the reaction.

d) Allowing the formation of reaction precipitates, to slow down the

dissolution of metal.

e) Replacing the steel by a more resistant metal or alloy.

f) Seeking uniform metal properties: to avoid concentrated corrosion

attack around local anodes.

g) Inlemen1ing sacrificial anodes or imposed current, to make the whole

steel surface cathodic. Thus, the corrosion is removed to the sacrificial

anode.

3.3.5 StaInless steels

The use of stainless steels in the petroleum industry is increasing. In the

North Sea area it has been successful in controlling C02 corrosion. Stainless

steels is not a single alloy, but a group of at least 4 different classes of alloys.

It is not stainless either, but resistant to some forms of corrosion whereas

susceptible to others.

The corrosion resistance of stainless steels is associated with the ability of

chromium to passivate the surface of the alloy. Chromium is a reactive

element, but ii and its alloys passivate in oxidizing environments. The

passivation process is not completely understood but is believed to be

korn pciniiu in / ‘roil. F / 333/27-0 1 o)4 /1 FAA/OIl)

99

associated with the formation of an electrically insulating monolayer

chromium-oxide film on the surface of the alloy.

Based on their microstructure and properties, the stainless steels can be

classified as follows.

Martensitic stainless steels

The martensitic stainless steels contain chromium as their principal

alloying element. As their name implies, the rnicrostructure of these steels

consists of martensile. The most common types contain approximately 13%

chromium: alihough for the different grades, the chromium may be as high

as 18%. The carbon content ranges from 0.08% to 1.10% and other elements

such as nickel, columbium, molybdenum, selenium SliCOfl, and sulfur are

added in small amounts for special purposes in certain grades. The most

important characteristic that distinguishes these steels from other grades is

their response to heat treatment. The martensitie stainless steels can be

hardened by he same heat treatmeins used to harden carbon and aLloy steel.

The martencitic stainless steels comprise part of the ‘400’ series of stainless

steels. The most common of the niartensitic stainless steels is AISI Type 410.

All of the martensitic stainless steels are strongly magnetic under all

conditions of heat treatment.

Ferritic stainless steels

The second class of stainless steels, the ferritic stainless steels, are similar to

the martensitic stainless steels in that they have only chromium as the

principal alloying element. The microstructure of the ferritic stainless steels

consists of ferrite. Ferrite is simply body-centered cubic iron or an alloy

based on this structure. The chromium contents of the ferritic stainless

steels are, however, normally higher than those of the marlensitic stainless

steels, and the carbon contents are generally lower. The chromium content

of the ferritic stainless steels ranges from about 13% to about 27%. They are

not hardenable by heat treatment and are used principally for their good

corrosion resistance and high temperature properties.

100

The ferritic stainless steels are also part of the 400” series, the principaltypes being Type 405. 430 and 436. They are also strongly magnetic.

Austenitic stainless steels

The austenitic stainless steels have two principal alloying elements:chromium and nickel. Their microstructure consists essentially ofaustenite. Austenite is face-centered cubic iron or an iron alloy based on thisstructure. They contain a minimum of 18% chromium and 8% nickel, withother elements added for special purposes, and range up to as high as 25%chromium and 20% nickel. The austenitic stainless steels have the highestgeneral corrosion resistance of any of the stainless steels, but their strengthis lower than that of the martensitic and ferritic stainless steels. They arenot hardenable by heat treatment, although they are hardenable to someextent by cold work.

The austenitic stainless steels constitute the ‘300” series, the most commonbeing Type 304. Others commonly used are Type 303 (free machining), Type

316 (high Cr and Ni which may include Mo), and Type 347 (stabilized forwelding and corrosion resistance). They are generally nonmagnetic.

Duplex stainless steels

The duplex (dual phase/precipitation hardening) stainless steels usuallycontain around 22% chromium and 5% nickel and 3% molybdcnium. Themanufacturing process involve rolling and heat treating at moderatetemperature to induce precipitation of dual crystal structures of martensitedifferent from the feritic matrix. By controlling (he precipitation process,the steel can be hardened to various strength levels. Duplex steels maycombine (lie high strength of martensitic stainless steels with the corrosionresistance of the austenitic ones.

Duplex steel alloys are generally proprialory. Thus they are designated bycompany names. They have found application in corrosion environmentstoo severe for Cr 13 martensiltic steels, as an alternative to much more

expensive nickel-based alloys. Table 3.6 lists (lie composition and relative

price for same duplex steels and nickel alloys. Payne-Hurst / 1986/.

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101

Table 3.6 - Partial list of duplex steels and nickel alloys for tubulars

Composition (Wt%)

Price* Fe Ni Cr Mo N C Si MnSAF 2205 1.0 BaP* 5.5 22 3 0.14 0.03 0.8 2SM 22 1.0 Bal 6 22 3 - 0.03 - -

VS22 1.0 Bal 5.5 22 3 - 0.03 1 1Ferralium 255 - Bal 5.5 25.5 3.5 0.24 0.03 0.5 0.8AF’22 - Bal 5.5 22 3 0.16 0.018 0.41* 1.57Sanicro 28 1.5 Bal 31 27 3.5 - 0.02 1 2SM 2035 1.4 Ba] 35 20 5 - 0.01 0.44 0.52NIC 32 1.7 Bal 32 22 4.5 - 0.02 0.5 1‘s’S 28 1.5 Bal 30 27 3.5 0.4 0.02 1 2Hasteiley 535 1.8 35 37 22 3 - 0.05 1 2.5Incoloy 825 1.4 30 41.8 21.5 3 - 0.03 0.5 1Incoloy 925 0.8 32 42 21 3 - 0.02 - -

SM 2550 2.5 - 50 25 6 0.01 0.018 0.31 0,59Hastelloy G3 2.2 19.5 Bal 22 7 - 0.015 0.4 0.8Hastelloy C276 4.9 5 Ba] 15.5 16 - 0.01 0.08 1Inconel 625 1.6 5 Ba] 21.5 9 - 0.1 0.5 0.5

Quoted in Dcc. 1983 for plain-end 2 7/8-in., 1 1,65-lbljft (0.440-in wall) with 130-

ksi yield strength (0.2% offset method) for 20.000-fl order; price relative In 22Cr

duplex.

Bal balance

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102

EXERCISES

Exercise 3.1

Strength and elongation

Two different steel pipes are available. Both have an outside diameter of

114.3 mm, wall thickness of 6.88 mm and will have an installed length of

1000 m. One pipe is made of H-40 steel with a minimum yield of 276 MPa

(276 106 Pa), the other N-80 steel with a minimum yield of 552 MPa.

a) How many tonnes of tension can be applied to each of these pipes, without

exceeding the yield point.

b) What will be the elongation under maximum load for each pipe.

koiii tint / ‘rot] III / 333/27-0 94 / I IAA/aIIt

103

Exercise 32

Tubing of grade 1-1-40 (ref. Tables 3.1, 3.2) is considered for a well.

a) What is the maximum suspended length of a 4 1/2” non-upset-tubing

string, considering only weight and strength.

h) Will the maximum length be different for 1.90 tubing.

.9/f,\A/Ib

104

Exercise 3.3

A vertical well is drilled down to a gas reservoir. The following reservoir

data are known:

Depth (datum) : 3000 m

Reservoir pressure at datum : 250 bar

Reservoir temperature : 90 °C

Surface temperature : 10 °C

Gas gravity : 0.6

The well is planned completed as follows:

Packer setting depth: 3000 m

Tubing: 4 1/2, non-upset

Production casing: 7”, 29 lb rn/ft

Annular fluid is

salt water, with density: 1025 kg/rn3

a) Make a diagram showing the pressure in the annulus and the shut in

pressure in the tubing, as function of depth. (Hint: the shut-in gas

pressure profile may be estimated as shown by oppgave 2.4’ in the

textbook lbr Production I).

b) At what depth will the annular pressure equal the pressure inside the

tubing.

C) What is the magnitude and direction of the force due to the pressure

difference across the packer.

konjcdiun/l’rod.I/33/27-() I .C)4 /1 Ii\i\/iIh

105

Exercise 3.4

The well described in exercise 3.3 is proposed completed with an expansion

joint above the packer.

a) Prepare a diagram showing the axial stress as function of depth. (Assume

no transfer of forces between the tubing and the packer).

b) How many meters of tubing above the packer will be in axial compression

due to the pressure below the packer (buoyancy effect).

c) ‘Will any section of the tubing be unstable under iurrent conditions.

d) What API-grades of tubing steel will be sufficient for the current

conditions, using the design factors of tab. 3.4.

km rthurn/rrudIH/333/27-fl I •94/IIAA/alb

06

Exercise 3.5

During production, the well is expected to heat up to a welihead temperature

of 60 C. For simplicity, let us assume linear temperature variation between

the reservoir temperature (90 C) and this wellhead temperature.

a) Estimate the tubing length change due to this temperature change.

b) Estimate the tubing length change due to ballooning, based on the

pressure profiles computed in Exercise 3.3. (We assume that the tubing was

set full of completion fluid, which during production displaced by

reservoir gas).

c) What length would you recommended for the expansion joint.

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107

Exercise 3.6

The well described in example 3.3 may later be needed to re-inject producedwater. The density of the water produced is 1025 kg/m3. The injectionpressure needed is estimated to 50 bar (at the weliheacl). The weilboreaveraged temperature during water injection is estimated to 30 C. The flowfriction in the weilbore may be neglected.

a) Investigate the stability of the tube during injection

N Estimate the length changes due to ballooning and due to the temperaturechange.

c) How much tensioning would be needed (to avoid later buckling) if the wellis completed by a tubing anchor. Consider both production, computed inExercise 3.5, and injection.

d) What tubing grade will be needed if a tubing anchor is used.

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108

Figures chapter 3

Figure 3.1 Strength and elasticity, carbon steels

Figure 3.2 Ellipse of biaxial yield stress

Figure 3.3 Basic reactions of CO2 corrosion

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