zhao.pdf

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W. ZhaoFaculty of Engineering and Applied Science,

Memorial University of Newfoundland,

St. Johns, NF, Canada, A1B 3X5

R. SeshadriFaculty of Engineering and Applied Science,

Memorial University of Newfoundland,

St. Johns, NF, Canada, A1B 3X5

R. N. DubeyDepartment of Mechanical Engineering,

University of Waterloo,

Waterloo, ON, Canada, N2L 3G1

On Thick-Walled Cylinder UnderInternal Pressure

A technique for elastic-plastic analysis of a thick-walled elastic-plastic cylinder underinternal pressure is proposed. It involves two parametric functions and piecewise linear-ization of the stress-strain curve. A deformation type of relationship is combined with

Hookes law in such a way that stress-strain law has the same form in all linear segments,

but each segment involves different material parameters. Elastic values are used to de-scribe elastic part of deformation during loading and also during unloading. The tech-nique involves the use of deformed geometry to satisfy the boundary and other relevantconditions. The value of strain energy required for deformation is found to depend onwhether initial or final geometry is used to satisfy the boundary conditions. In the case oflow work-hardening solid, the difference is significant and cannot be ignored. As well, itis shown that the new formulation is appropriate for elastic-plastic fracture calculations. DOI: 10.1115/1.1593082

Introduction

Thick walled cylinders subjected to high internal pressure are

widely used in various industries. In general, vessels under high

pressure require a strict analysis for an optimum design for reli-able and secure operational performance. Efforts were continually

made to increase reliability. Solutions have been obtained either in

analytical form or with numerical implementations. Literature in-

cludes solutions of Hill 1 , Mendelson 2 , Durban 3 , Chen 4 ,and Chakrabarty 5 . Durban and Kubi 6 suggested an analyticalmethod of solving pressurized elastic-plastic tubes in plane strain.

Jahed and Dubey 7 proposed a numerical method for solutionfor elastic-plastic tubes using total deformation theory of plastic-

ity. Parker 8 implemented a numerical procedure to calculateautofrettage pressure and associated residual stress fields for open-

ended cylinder. Dubey, Seshadri, and Bedi 9 obtained solutionfor an elastic-plastic work hardening model using piecewise lin-

earization of constitutive law. Zhao, Dubey, and Seshadri 10presented a formulation of bilinear work hardening model to esti-

mate residual stress in pressurized cylinders 1117 .The proposed formulation provides a general method of solu-

tion for elastic-plastic cylinders. The work involves several fea-

tures. Solution of the problem is expressed in terms of two para-

metric functions. The relationship between them suggests that

Lames elastic solution and solution for perfectly plastic material

depend on special choices for these parameters. Both solutions use

linear material behavior. The proposed technique uses piecewise

linear approximation of the actual stress-strain curve to obtain a

general solution in terms of the two parametric functions. For

application, the material is divided into several finite segments. A

linear stress-strain curve is assumed for each segment, but the

slope of the curve differs from segment to segment. Accordingly

the cylinder is divided into sections or domains as shown in Fig.

1. Thus, several sets of parametric functions are selected that sat-

isfy the boundary conditions, continuity at the interface betweenthe segments and any other condition that formulation of the prob-

lem requires. One important restriction in the selection is that

stress-strain field associated with each set must be consistent with

the chosen stress-strain curve. The restriction is imposed to allowfor unloading and reloading, if necessary. This requirement leads

to another feature used in the analysis. It involves stress-strain law

in plastic deformation. A deformation type of relationship is com-bined with Hookes law in such a way that stress-strain law has

the same form in all segments, but each segment involves differ-ent material parameters. Elastic values are used in elastic defor-mation during loading and also when the material unloads. A thirdfeature involves the preference for the use of deformed geometry

in the analysis. Equations are developed that help evaluate thedeformed geometry as part of the solution. This feature has beensuccessfully employed in elastic-plastic crack analysis as well.Yet, another feature of the formulation is that it can account forBauschinger effect. The method again involves piecewise linear-ization of stress strain curve during unloading to the point wherereverse yielding begins.

Formulations

Piecewise Linearization of Stress Strain Law. GeneralizedHookes law for isotropic elastic solid is of the form

i j

1 i jkki j

E . (1)

Even though the stress strain behavior in plastic domain is non-linear, it is often linearized for analysis. The incremental theory isan example of a linear relationship between stress and strain in-crements. This paper aims to use piecewise linearization to de-scribe elastic plastic stress and strain behavior of kinematic hard-ening material. For convenience, the proposed constitutiveequation is expressed in a form that resembles generalizedHookes law. As an example, constitutive equation for a bilinearstress strain curve of Fig. 2 is written as

i j 12 i j2kki j

E2

12 i j2kki j

E2

rr1

11 i j1kki j

E1 rr1, (2)

where E1E and 1 in the elastic segment, E2 and 2 are thecorresponding modulus and Poissons ratio in the plastic segment.The stresses reach yield point at the boundary r1 between elasticand plastic segments. This interface moves outwards with increasein loading. The mechanical parameters of material, E2 and 2 , arerelated to plastic modulus Ep and p as follows:

1

E2

1

E

1

Ep, (3a)

Contributed by the Pressure Vessels and Piping Division for publication in the

JOURNAL OF PRESSURE VESSEL TECHNOLOGY. Manuscript received by the PVP

Division March 13, 2003; revision received May 6, 2003. Associate Editor: M. Perl.

Copyright 2003 by ASMEJournal of Pressure Vessel Technology AUGUST 2003, Vol. 125 267

Downloaded 15 Jan 2008 to 129.5.160.97. Redistribution subject to ASME license or copyright; see http://www.asme.org/terms/Terms_Use.cfm

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2

E2

E

p

Ep. (3b)

The second and third terms are included in Eq. 2 for correctionto the plastic strain, which is overestimated by the linearization of

stress strain curve in plastic domain. The overestimated strain, oe

shown in Fig. 2, has to be subtracted from the total value.

Formulation of General Elastic-Plastic Solution. For cylin-der in axisymmetric deformation, stress can be obtained by solv-ing the equilibrium equation

dr

dr

r

r0 (4)

along with the prescribed boundary conditions.For a general solution of the equation of equilibrium, choose

two parametric functions g and f such that

rgf, gf. (5)

The functions satisfy the equilibrium equation provided:

r2dg

dr

d

dr r2f . (6)

Note that rconstant is a solution of the equation of equi-librium. It represents a uniform stress field independent of r and

hence associated with the coordinate function 1. In the elasticor Lames solution, g is assumed constant, and its coordinate func-

tion is therefore 1. The coordinate function of f is 1/r2 and the

general form for stresses in the elastic solution is AB/r2. In thecase of plastic solid 1,5 , f is assumed constant and hence g2f ln(r) involves the coordinate function ln(r). This solution

does not use the coordinate function 1/r2. A general solution ofthe boundary value problem may involve all three coordinatefunctions. Therefore, solution for elastic plastic solids is obtainedby choosing

gA2Cln r,(7)

fCB

r2,

where A, B, and C are constants.The constants of the problem are determined by the boundary

and other relevant conditions. For internal pressure vessels, theboundary condition can be prescribed in the form

r0 at the external boundary rro ,(8)

rPn at the internal boundary rrn .

Suppose the solution is continuous across the elastic plastic inter-face, assumed located at rr1 . To maintain continuity of stressesacross the elastic plastic boundary, choose the plastic domain pa-rameters such that there is no jump in stresses across the interface.This condition leads to

gA12C2 ln r/r1 ,(9)

fC2A1ro

2C2r1

2

r2.

Choice 9 satisfies equilibrium as well as the continuity condi-tions. As a result, stresses in the plastic domain are

rA1

ro2

r

21

C2

r12

r

212 ln

r1

r

,

(10)

A 1ro

2

r21 C2

r12

r212 ln

r1

r.

Both elastic and plastic solutions are embedded in the above gen-eral solutions. As a result of the choice C20 suffix 1 refers tothe constants in domain 1 or elastic domain, while suffix 2 is fordomain 2 or plastic domain , elastic or Lames solution is recov-ered from the general form. The plastic solution discussed by Hill 1 and Chakrabarty 5 is for ideally plastic or nonwork harden-ing Tresca solid. It can be obtained from the general solution by

choosing B2A1ro2C2r1

20. Such a choice makes the boundary

value problem statically determinate.For work hardening material, however, it is necessary to find an

independent condition to evaluate the constants. For this purpose,consider first the constitutive equation.

Strains in the elastic domain are obtained from the generalizedHookes law,

rdu

dr

rz

E,

u

r

rz

E, (11)

zzr

E.

The compatibility condition and equilibrium equation yield

d

dru

r1

Er

1

rdu

dru

r 1

E

r

r . (12)

In view of the constitutive equation and the expression forstresses, the above equation leads to

d

dr

2A1z

E d

dr

2A 1 12

Ez 0. (13)

Since A1 is constant, this equation suggests that both axial stressand strain remain constant in the elastic domain. The constantvalue for axial strain suggests that plane sections normal to theaxis remain plane and normal. According to Hill 1 , this assump-tion is expected to hold away from the two ends.

Fig. 1 Cylinder geometry and load conditions: a two domainsmodel; b finite domains model

Fig. 2 Bilinear stress strain behavior

268 Vol. 125, AUGUST 2003 Transactions of the ASME


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Suppose z1 is the axial strain and r1 and 1 are the radial and

hoop strain at the elastic plastic interface rr1 . Use the follow-ing version of the linearized stress strain relation in plastic do-main:

rr22z

E2

r22z

E2

rr1

r1,

2r2z

E2

2r2z

E2

rr1

1, (14)

zz2r2

E2

z2r2

E2

rr1

z 1.

The compatibility condition and the equation of equilibrium yield

d

dr

u

r

12

E2r

1

r

du

dr

u

r

12

E2

r

r. (15)

In view of the constitutive relation, the above equation can berewritten as follows:

d

dr

u

r

12

E2r 2A1

1p

Ep

ro2

r12

1

r. (16)

The assumption that plane sections normal to the axis remain

plane and normal implies zconstant. To satisfy this condition,choose

C2A11p

2Ep

E2

122

ro2

r12

(17)

or in view of plastic incompressibility, p1/2,

C23A1

4Ep

E2

122

ro2

r12

. (18)

This provides the additional equation to solve for constants A1and C2 . Equation 18 , the boundary condition

PnA1ro

2rn

2

rn2C2

r1

2rn

2

rn22 ln

r1

rn, (19)

along with the yield criterion provides three equations to solve forthree constants. Tresca yield condition yields

A1r1

2

2ro2Y , (20)

where Y is the yield stress. For Mises solid,

3 A1 ro2

r12

2

zA 12Y . (21)

Besides, three end conditions that have been used are

z0, plane stress,

z2A1 , plane strain, (22)

zA1 , closed ends.

The corresponding values for axial strains are

z2A1

E, plane stress,

z0, plane strain, (23)

z12

EA1 , closed ends.

Unloading and Shakedown. If the internal pressure is re-moved after part of the cylinder has become plastic, a residualstress will remain in the wall. Assuming that during unloading thematerial follows Hookes law, the internal pressure is removed byapplying Pn on the inside surface. The residual stresses in thatcase can be obtained from

r,re sA 1ro

2

r21 C2

r12

r212 ln

r1

rD 2

ro2

r21 ,

(24)

, re s

A1ro

2

r2

1

C2r1

2

r2

1

2 ln

r1

r

D2ro

2

r2

1 ,where D2 is the constant during unloading. However, in determin-ing the residual stresses after removal of autofrettage pressure apurely elastic unloading is assumed. This would be a nonconser-vative assumption since the Bauschinger effect significantly re-duces the residual compressive bore hoop stress.

The concept of shakedown provides another avenue that can beused for design of thick walled cylinders. Essentially, a structuredevelops residual state of stress and strain under certain loadingand unloading conditions. The structure is said to shakedown if itsbehavior is elastic during subsequent cycle of same loading andunloading. Under these conditions, stresses during the subsequentcycle can be obtained from

rr,re sA ro

2

r21 , (25)

, re sAro

2

r21 , (26)

where r,re s and , re s are the residual radial and hoop stresses atr. The radial stress must vanish at the external boundary rro .Here the case of plane stress is considered. At this point, therefore,the only nonzero stress in plane stress is

,re s2A. (27)

When there is no internal pressure, residual radial stresses at theexternal and internal boundaries must vanish. Therefore, stressesat rrn during subsequent loading to inside pressure PS are

rPSAro

2

rn21 , (28)

rrn, re s rrn

PSro

2rn

2

ro2rn

2. (29)

At the external boundary,

r0, (30)

rro,re s rro

2 PSrn

2

ro2rn

2. (31)

Whatever the value of the residual stress, the conditions at the

external boundary yield

PSYro

2rn

2

rn2

. (32)

If the residual stress at the internal boundary is zero, the shake-down pressure is

PS2Yro

2rn

2

3ro4rn

4. (33)

For infinitely large value of residual stress at the inside boundary,the shakedown pressure increases to an asymptotic value of

Journal of Pressure Vessel Technology AUGUST 2003, Vol. 125 269


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PS4Yro

2rn

2

3ro2rn

2. (34)

Since infinitely large residual stress at the inside surface is dueto infinite applied pressure, consequent stress is too high for thecylinder to sustain during loading cycle. This state ought to beruled out for design purposes. Therefore, the value of shakedown

pressure at no residual stress is a safe criterion for design. Shake-down pressure is higher if it is evaluated on the basis that therewould be residual stress once the cylinder yields during the firstloading cycle.

Formulation of Finite Domains and Analysis Based on De-formed Geometry. Linearized stress strain curve implies con-stant value for material constants in each linear segment. It ispossible to divide the cylinder in sections or domains Fig. 1 suchthat the stress strain law of one segment describes the materialbehavior in that domain. If the segments are sufficiently small, asshown in Fig. 1b then the piecewise linearization closely approxi-mates the true stress strain law. To derive the generalized formu-lation of k domains in elastic plastic solid, first consider the caseof three domains, as shown in Fig. 3. It shows that strains indomains 2 and 3 are over estimated if it is based entirely on the

respective linear stress strain curve. To find the true strain, it isnecessary to apply a correction for each crossing from one domainto the next. For example, linear stress strain law of domain 2

overestimates the true strain by oe 1. This value must be sub-tracted from the strain of the constitutive equation to obtain actualstrain in this domain. Similarly, a crossing from domain 2 to do-

main 3 would require a correction of oe 2. In this figure, theplastic region consists of two domains. The concept can be con-veniently expanded to k domains.

Suppose the stress strain curve is divided into a finite number ofdomains. As stress point moves from k1th domain to kth do-main, as shown in Fig. 4, it crosses the common boundary at rk1as illustrated in Fig. 1b. From Eq. 9 ,

gAk12Ck1 ln r, (35)

fCk1Bk1

r2,

and

gAk2Ck ln r,(36)

fCkBk

r2.

Continuity of stresses across rrk1 requires that

Ak2Ck ln rk1Ak12Ck1 ln rk1 (37)

and

CkBk

rk12Ck1

Bk1

rk12

. (38)

Therefore, stresses in domain k can be obtained as follows:

rAkCk2Ck ln rBk

r2

, (39)

AkCk2Ck ln rBk

r2. (40)

The axial stress can be determined once the end condition isgiven, as shown for the case of two domains. The values used inthe previous domain help evaluate the constants of the followingdomains. For kth domain, with the assumption that plane sectionnormal to the axis remains plane and normal, the following rela-tion could be derived in the way it was done for the case involvingthe two domains. Thus,

4Ck1k

2

Ek3

m1

k

1Epm

1

Epm1 Cm1Bm1

rm12 . (41)

Strain obtained from

i j 1k i jkpp i j

Ek(42)

overestimates the value in kth domain by an amount

i jk oe 1k

Ek

1k1

Ek1 i j k

Ek

k1

Ek1pp i j

rrk1

.

(43)

The material constants of this domain can be expressed in termsof plastic modulus because

1k

Ek

1

E

1p

Epk,

(44)1k1

Ek1

1

E

1p

Epk1.

When the material is incompressible in plastic deformation, p1/2, and strain overestimated due to crossing of the boundaryrrk1 is

i jk oe

3

2 1

Epk

1

Epk1i j

1

2 1

Epk

1

Epk1pp i j

rrk1

(45)

or

Fig. 3 Three domains piecewise linearization of stress strainlaw

Fig. 4 Generalization of stress strain law for finite domains



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i jk oe

3

2 1

Epk

1

Epk1i j rrk1

. (46)

Therefore, the total overestimated strain including all the previousdomains is

i joe

3

2 m1

k

1Epm

1

Epm1i j

rrm1

. (47)

The correct value of strain in kth segment is therefore given by

i j 1k i jkpp i j

Ek

3

2 m1

k

1Epm

1

Epm1i j

rrm1

. (48)

Deformation due to external load causes the geometry tochange. As a result, the load applied to a surface moves it from itsundeformed to deformed position. In the classical theory, dis-placement is assumed small and hence it is permissible to replacethe deformed position of a particle by its initial undeformed posi-tion. This argument can be reversed in favor of deformed geom-etry. That is, it is permissible to use the deformed position of aparticle in place of its undeformed position because of small dis-placement. Further, the consistency of analysis requires the use ofdeformed geometry if the boundary value problem is formulatedin terms of true stress and true traction. Hence, the deformedgeometry is used in this research to predict elastic plastic behav-ior. Thus, the traction boundary conditions are satisfied on thedeformed radii.

Suppose the radial displacement

uRr, (49)

where R and r are respectively the deformed and initial radii. It ispossible to express deformed radius in terms of initial radius andhoop strain:

R 1 r. (50)

This equation relates the deformed body geometry to its initial

configuration.

Energy Work Principle. Energy required for deformationper unit volume of the body, or energy density of deformation is

U drd rzdz . (51)The energy of deformation is given by

Urdrd. (52)In elastic segment of a bilinear solid, the elastic strain energydensity is

Uel1

E A12

1

E

ro4

r4A12 in plane stress, (53)

Uel 122

EA1

2

1

E

ro4

r4A1

2 in plane strain,

(54)

Uel 12

EA1

2

1

E

ro4

r4A1

2in closed end. (55)

Therefore, the energy required for elastic deformation of the ma-terial can be obtained from

el2 A12ro

2

E

A12 r1

4r1

4ro

4r0

4

2Er12 in plane stress, (56)

el2 A12

ro2 1

E

A1

2 r1

4r

1

42r

1

4

2r

o

4r

o

4

2Er12 in plane strain,

(57)

el2 3A12ro

2

2E

A12 r1

42r1

4ro

4ro

4

2Er12 in closed end. (58)

Alternatively, a general form of strain energy density for differ-ent end conditions can be derived using the principle of virtualwork:

Uel

1 i j2kk

2

2E . (59)

Therefore, the general form of strain energy density required forplastic deformation can be obtained by using the concept of piece-wise linearization of the stress strain curve:

Upl 12 i j

22kk

2

2E2

12 i j22kk

2

2E2

rr1

1 i j2kk

2

2E

rr1

. (60)

The energy required for deformation in plastic zone is therefore

pl2

rn

r1

Upl rd r. (61)

The work done by the external traction is

WP n Rn

2rn

2

2. (62)

The principle of work energy balance or W can be used toderive the condition

P n Rn2rn

2

22

r1

ro

Uel rd rrn

r1

Upl rd r . (63)

Equation 63 provides the condition to solve boundary valueproblems of deformed geometry in an iterative manner.

Calculations and Conclusions

The proposed finite domain method has been applied on severalsample cylinders of prescribed dimensions. The cylinders are sub-

ject to pressure on the internal surface, while the outside surface istraction free. When the internal pressure is lower than the initialyielding pressure, the results obtained are found to be exactly theLames solution. If the internal pressure is increased to valuesgreater than the initial yielding pressure, plastic deformation ini-tiates at the inner surface and proceeds outward through the cyl-inder wall. The interface between the plastically deformed mate-rial and the elastic material will eventually reach the outer surfaceat a value of pressure that is known as the full plastic flow



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pressure. The internal pressure that causes yielding throughoutthe wall thickness can be determined as shown in Fig. 5. Succes-sive distributions of hoop, radial, and axial stresses in the elasticplastic expansion are obtained, as shown in Fig. 6. Distributionsof residual hoop, radial, and axial stresses with different overstrainpercentage in the elastic plastic expansion are shown in Fig. 7.Equivalent residual stress distributions of different overstrain per-centages are illustrated in Fig. 8. For the purpose of verifying thenew formulation, limit loads are calculated by the proposedmethod. The results are compared with classical method, as shownin Fig. 9. The comparison indicates that the results from the pro-posed method matches well with those of lower bound method.

Computational results of strain energy required for deformationare listed in Table 1. It shows the influence of deformed geometry.The results indicate that the difference of energy required for de-formation and the elastic plastic interface based on deformed andundeformed geometry are not negligible, when E2 , the tangent

modulus of plastic domain, is small. Thus, the effect of geometricchange can be significant and must be taken into account espe-cially if the internal pressure is high. In the case of elastic plasticfracture mechanics, clearly the evaluation of J integral will beaffected for small E2 /E ratios. Therefore, the new formulation isappropriate for elastic-plastic fracture calculation, and the deter-mination of energy release rate based on undeformed and de-formed geometry will make a considerable difference.

The proposed method provides a general elastic-plastic solutionto thick wall cylinders, which accounts for the effect of deformedgeometry due to high internal pressure. All the input calculationpoints in this formulation can be taken from the actual uniaxialstress strain curve by using piecewise linear segments. If the seg-ments are sufficiently small, a close approximation to the actualstress strain law can be achieved. As a closed form solution ofelastic-plastic cylinder problems, this formulation can provide aneconomical complement to nonlinear finite element analysis sinceit requires less extensive computer resources.

Fig. 5 Internal pressure versus elastic-plastic interface

Fig. 6 Hoop, radial, and axial stress elastic-plastic tubes withboth ends closed

Fig. 7 Residual stress distributions of different overstrain per-centage closed end

Fig. 8 Equivalent residual stress distribution of different over-strain percentage closed end

Fig. 9 Limit load versus wall thickness ratio comparison withlower bound method

Table 1 Strain energy and elastic-plastic interface, rn4 in. 0.10 m ro20 in. 0.50 m, Pn50000 psi 344.74MPa

Modulusratio

Strain energy in. lb

Elastic-plastic interface in.

Undeformedgeometry

Deformedgeometry

Undeformedgeometry

Deformedgeometry

E2 /E1 6,060.31 6,091.39 7.02 7.04E2 /E10 17,702.21 18,014.35 9.60 9.68E2 /E100 32,303.11 33,746.14 11.88 12.13E2 /E1000 37,076.23 39,298.53 12.52 12.86



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Nomenclature

E Youngs modulusEk Tangent modulus in domain kEp Plastic modulus

f, g Parametric functions defined in Eq. 5Pn Internal pressurePS Shakedown pressure

r Radius of initial geometryR Radius of deformed geometryr1 Radius of elastic plastic interface

rk1 Radius of interface of domains k and k1

rn Inside radius of cylinderro Outside radius of cylinderU Strain energy densityW Work done by external force Uniaxial strain

i j Total strain tensor

i jk oe

Tensor of overestimated strain in kth domain

i joe Total overestimated strain tensor

Poissons ratiok Poissons ratio of domain kp Poissons ratio of plastic region Uniaxial stress

i j Stress tensor

i j Deviatoric stress tensor

r,re s Residual radial stress

,re s Residual hoop stressY Yield stress Strain energy

References

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