Research ArticleCollapse Pressure Analysis of Transversely IsotropicThick-Walled Cylinder Using Lebesgue Strain Measure andTransition Theory

A K Aggarwal Richa Sharma and Sanjeev Sharma

Department of Mathematics Jaypee Institute of Information Technology A-10 Sector 62 Noida 201307 India

Correspondence should be addressed to Sanjeev Sharma sanjeevsharmajiitacin

Received 28 August 2013 Accepted 10 October 2013 Published 9 January 2014

Academic Editors F Berto A Di Schino Y-k Gao and F Peeters

Copyright copy 2014 A K Aggarwal et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The objective of this paper is to provide guidance for the design of the thick-walled cylinder made up of transversely isotropicmaterial so that collapse of cylinder due to influence of internal and external pressure can be avoided The concept of transitiontheory based on Lebesgue strain measure has been used to simplify the constitutive equations Results have been analyzedtheoretically and discussed numerically From this analysis it has been concluded that under the influence of internal and externalpressure circular cylinder made up of transversely isotropic material (beryl) is on the safer side of the design as compared to thecylinders made up of isotropic material (steel) This is because of the reason that percentage increase in effective pressure requiredfor initial yielding to become fully plastic is high for beryl as compared to steel which leads to the idea of ldquostress savingrdquo that reducesthe possibility of collapse of thick-walled cylinder due to internal and external pressure

1 Introduction

Lame in 1852 first studied the hollow circular cylinder underpressure which is widely used in structures aerospace andnuclear reactorsTheproblems of homogeneous and isotropiccircular cylinder under internal pressure have been foundin most of the standard elasticity and plasticity books [1 2]These days pressurized (internal and external) cylinders havebecome a point of interest of researchers due to their wideapplication in nuclear industry especially in advanced smalland medium-sized light water reactors A steam generatortube is an example of the problem of circular cylinder underinternal and external pressure in which primary coolantflows outside the tubes (external pressure) while secondarywater flows inside the tubes (internal pressure) Anotherexample is pipelines under seawater to transport gas oil andso forth In general vessels under high pressure require astrict analysis for an optimum design for reliable and secureoperational performance and thus efforts were continuallymade to increase reliability Solutions are obtained eitheranalytically or with the implementations of numerical meth-ods Rimrott [3] used the assumptions of constant density

zero axial strain and distortion-energy law to calculate creepstrain rate and stresses in a thick-walled closed-end hollowcylinder under internal pressure made of an isotropic andhomogeneous material under internal pressure A knowncreep strain rate versus stress relation is then used to solvethis specific problem Zhao et al [4] discussed elastic-plasticanalysis of a thick-walled cylinder under internal pressureThey involve two parametric functions and piecewise lin-earization of the stress-strain curve A deformation type ofrelationship is combined with Hookersquos law in such a way thatstress-strain law has the same form in all linear segmentsbut each segment involves different material parametersThis technique involves the use of deformed geometry tosatisfy the boundary and other relevant conditions Yoo etal [5] investigated the collapse pressure of cylinders withintermediate thickness subjected to external pressure basedon finite element (FE) analysis According to the concept ofthe partial safety factor the yield strength was concludedto be the most sensitive and the initial ovality of tube wasnot so effective in the proposed collapse pressure estimationmodel Perry and Aboudi [6] discussed the optimal design ofa modern gun barrel with two main objectives the first one

Hindawi Publishing Corporatione Scientific World JournalVolume 2014 Article ID 240954 10 pageshttpdxdoiorg1011552014240954

2 The Scientific World Journal

is to increase its strength-weight ratio and the second is toextend its fatigue life This can be carried out by generatinga residual stress field in the barrel wall A Von-Misesrsquoyield criterion isotropic strain hardening with the Prandtl-Reuss theory has been taken into the consideration withBauschinger effect and plane stress conditions The stressesare calculated incrementally by using the finite differencemethod Davidson et al [7] determine the residual-stressdistribution as a function of magnitude of overstrains anddiametric ratio and discuss the effects on the reyieldingcharacteristics of cylinders All these authors consideredyield criterion jump conditions and linear strain measureto calculate the stresses using the concept of infinitesimalstrain theory According to the approach of the above authorsthe spectrum of deformations is divided into two regionsthat is one is elastic region and another one is plastic regionwhich is physically not possible because transition from onestate into another state is a continuous phenomenon Attransition the fundamental structure undergoes a change andthe particles constituting the material rearrange themselvesand give rise to spin rotation vorticity and other nonlineareffects This suggests that at transition nonlinear terms arevery important and ignoring themmay not represent the realphysical phenomenon

Transition theory [8 9] does not require any of the aboveassumptions and thus solves a more general problem usingthe concept of generalized strain measure [10] which notonly gives the well-known strain measures but can also beused to find the stresses in plasticity and creep problemsby determining the asymptotic solution at the transitionpoints of the governing differential equations This theoryhas been applied to many problems for example Sharmaet al [11] analyzed thermal elastic-plastic stresses in trans-versely isotropic thick-walled rotating cylinder under inter-nal pressure Aggarwal et al [12] investigated safety factorsin thick-walled functionally graded cylinder under internaland external pressure and concluded that functionally gradedcylinder is a better choice for designers as compared tocylindersmade up of homogeneousmaterials Aggarwal et al[13] discussed safety factors in creep nonhomogeneous thick-walled circular cylinder under internal and external pressurewith steady state temperature and concluded that cylindermade up of nonhomogeneous material is better choice fordesigning as compared to cylinder made up of homogeneousmaterial

If a continuous phenomenon is represented by a spec-trum nonlinearity exhibits itself at its ends which corre-sponds to transition state Thus the classical measures ofdeformation are inadequate to deal with such transitions Inclassical theory such transition state requires semiempiricallaws to match the solutions and thus discontinuities haveto be introduced where they do not exists A continuumapproach necessarily means the introduction of nonlinearmeasures If in a very small interval the number of fluc-tuations is very large the concept of ordinary measurebased on Riemann integral fails and Lebesgue measures maybe used In like manner the generalized measures givenby weighted integral representations give very satisfactoryresults in problems of plasticity and creep Seth has defined

the generalized principal strain measure 120576119894119894by taking the

Lebesgue integral of the weighted function

120576119894119894= int

119890119860

119894119894

0

[(1 minus 2119890119860

119894119894)]

(1198992)minus1

119889119890119860

119894119894

=

1

119899

[1 minus (1 minus 2119890119860

119894119894)

(1198992)minus1

]

(1)

where 119899 is themeasure and 119890119860

119894119894is the principal Almansi strain

component For uniaxial case

119890 = [

1

119899

1 minus (

1198970

119897

)

119899

]

119898

(2)

where 119898 is the irreversibility index and 1198970and 119897 are the

undeformed and deformed lengths of the rod respectivelyFor 119899 = minus2minus1 0 1 2 in (2) gives theGreenHencky Swaingerand Almansi measure respectively where in all cases 119898 = 1The most important contribution made by the generalizedLebesgue measure is that they eliminate the use of empiricallaws and jump conditions

As we know the permissible stress of anymaterial is someproportion of the yield or ultimate stress of the materialand as such it incorporates a ldquosafety factorrdquo This safetyfactor provides a margin against the collapse condition of thecylinder that occurs due to high pressure Since the conditionof high pressure can cause failure of the cylinder therefore itis necessary to examine the state of the cylinder at collapseand to design the cylinder accordingly In order to provideguidance on a design and integrity evaluation of a cylinderunder pressure the failure characteristics of a cylinder shouldbe considered carefully In this paper it is our main aimto eliminate the need of assuming semiempirical laws yieldcondition creep-strain laws jump conditions and so forthto obtain the collapse pressure in transversely isotropic thick-walled circular cylinder under internal and external pressureusing generalized Lebesgue strain measure The stresses arecalculated in both transition and fully plastic state

2 Mathematical Formulation of Problem

We consider a thick-walled circular cylinder made up oftransversely isotropic material with internal and externalradii ldquoardquo and ldquobrdquo respectively subjected to internal pressure1199011and external pressure 119901

2as shown in the Figure 1 The

cylinder is taken so large that the plane sections remain planeduring the expansion and hence the longitudinal strain is thesame for all elements at each stage of the expansion

The components of the displacement in cylindrical polarcoordinates are given by

119906 = 119903 (1 minus 120573) V = 0 119908 = 119889119911 (3)

where120573 is a function of 119903 = radic1199092+ 1199102 only and119889 is a constant

The Scientific World Journal 3

p1 p1

p2

p2

a

b

Figure 1 Geometry of thick-walled transversely isotropic cylinderunder internal and external pressure

The generalized components of strain [10] are given asfollows

119890119903119903

=

1

119899

[1 minus (1199031205731015840+ 120573)

119899

] 119890120579120579

=

1

119899

[1 minus 120573119899]

119890119911119911

=

1

119899

[1 minus (1 minus 119889)119899] 119890

119903120579= 119890120579119911

= 119890119911119903

= 0

(4)

The stress-strain relations for transversely isotropic materialare

119879119903119903

= 11986211119890119903119903

+ (11986211

minus 211986266) 119890120579120579

+ 11986213119890119911119911

119879120579120579

= (11986211

minus 211986266) 119890119903119903

+ 11986211119890120579120579

+ 11986213119890119911119911

119879119911119911

= 11986213119890119903119903

+ 11986213119890120579120579

+ 11986233119890119911119911

119879119911119903

= 119879120579119911

= 119879119903120579

= 0

(5)

Using (4) in (5) we get

119879119903119903

= (

11986211

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ [

(11986211

minus 211986266)

119899

] [1 minus 120573119899] + 11986213119890119911119911

119879120579120579

= [

(11986211

minus 211986266)

119899

] [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986211

119899

) [1 minus 120573119899] + 11986213119890119911119911

119879119911119911

= (

11986213

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986213

119899

) [1 minus 120573119899] + 11986233119890119911119911

119879119903120579

= 119879120579119911

= 119879119903119911

= 0

(6)

Equations of equilibrium are all satisfied except

119889

119889119903

(119879119903119903) + (

119879119903119903

minus 119879120579120579

119903

) = 0 (7)

3 Identification of Transition Point

We know that as the point in the material has yielded thematerial at the neighbouring points is on its way to yieldrather than remain in its complete elastic state or fully plasticstate Thus we can assume that there exists some state in-between elastic and plastic which is called transition state Soat transition the differential system defining the elastic stateshould attain some criticalityThe differential equation whichcomes out to be nonlinear at transition state is obtained bysubstituting equations (6) in (7)

11989911987511986211120573119899+1

(1 + 119875)119899minus1 119889119875

119889120573

= minus11989911987511986211120573119899(1 + 119875)

119899

minus (11986211

minus 211986266) 119899119875120573119899

+ 211986266

[1 minus 120573119899(1 + 119875)

119899]

minus 211986266

(1 minus 120573119899)

(8)

where 1199031205731015840= 120573119875

The critical points or transitional points of (8) are 119875 rarr

minus1 and 119875 rarr plusmninfinThe boundary conditions are given by

119879119903119903

= minus1199011

at 119903 = 119886

119879119903119903

= minus1199012

at 119903 = 119887

(9)

The resultant axial force is given by

2120587int

119887

119886

119903119879119911119911

119889119903 = 1205871198862(1199011minus 1199012) (10)

4 Mathematical Approach

The material from elastic state goes into plastic state as119875 rarr plusmninfin or to creep state as 119875 rarr minus1 under internal andexternal pressure It has been shown [8 9 11ndash13] that theasymptotic solution through the principal stress leads fromelastic to plastic state at the transition point 119875 rarr plusmninfin Forfinding the plastic stress at the transition point 119875 rarr plusmninfin wedefine the transition function 119877 in terms of 119879

119903119903as

119877 = 2 (11986211

minus 11986266) + 119899119862

13119890119911119911

minus 119899119879119903119903

= 120573119899[11986211

minus 211986266

+ 11986211(

1 + 119875)119899]

(11)

4 The Scientific World Journal

Taking the logarithmic differentiation of (11) with respect toldquorrdquo with asymptotic value as 119875 rarr plusmninfin on integration yields

119877 = 1198601119903minus1198621

(12)

where 1198601is a constant of integration and 119862

1= 21198626611986211

Using (11) and (12) we get

119879119903119903

= 1198623minus (

1198601

119899

) 119903minus1198621

(13)

Using boundary condition (9) in the above equation we get

1198601= 1198991198871198621

[

119901

(119887119886)1198621

minus 1

] 1198623= [

119901

(119887119886)1198621

minus 1

] (14)

Substituting the value of 1198601and 119862

3in (13) we get

119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1199012 (15)

Using (15) in (7) we have

119879120579120579

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1199012 (16)

119879119911119911

=

11986213

2 (11986211

minus 11986266)

[(

1199011minus 1199012

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862(1199011minus 1199012)

1198872minus 1198862

minus

119886211986213

(1199011minus 1199012)

(11986211

minus 11986266) (1198872minus 1198862)

(17)

From (15) and (16) we get

119879120579120579

minus 119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119903

)

1198621

(18)

From the above equation it is found that the value of |119879120579120579

minus119879119903119903|

is maximum at 119903 = 119886 which means yielding of the cylinderwill take place at the internal surface Therefore we have

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119886

=

100381610038161003816100381610038161003816100381610038161003816

[

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119886

)

1198621 100381610038161003816100381610038161003816100381610038161003816

equiv 119884 (say) (19)

Let 1199011minus 1199012= 119901

The pressure required for initial yielding is given by

1003816100381610038161003816119875119894

1003816100381610038161003816=

1003816100381610038161003816100381610038161003816

119901

119884

1003816100381610038161003816100381610038161003816

=

[(119887119886)1198621

minus 1]

1198621(119887119886)1198621

(20)

where (1199011119884) minus (119901

2119884) = 119875

1198941minus 1198751198942

= 119875119894

Using (20) in (15) (16) and (17) we get transitionalstresses as

119879119903119903

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1198751198942

119879120579120579

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1198751198942

119879119911119911

119884

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862119875119894

1198872minus 1198862minus

119886211986213119875119894

(11986211

minus 11986266) (1198872minus 1198862)

(21)

Equation (21) give elastic-plastic transitional stresses in thick-walled cylinder under internal and external pressure

For fully plastic state (1198621

rarr 0) (20) becomes

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119887

=

10038161003816100381610038161003816100381610038161003816

119901

log (119887119886)

10038161003816100381610038161003816100381610038161003816

equiv 1198841(say) (22)

where 119875119891

= (1199011minus 1199012)1198841= 1198751198911

minus 1198751198912

The stresses for fully plastic state are

119879119903119903

1198841

= log(

119903

119887

) minus 1198751198912

119879120579120579

1198841

= (1 + log(

119903

119887

)) minus 1198751198912

119879119911119911

1198841

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log(

119903

119887

))

+

1198862 log (119887119886)

1198872minus 1198862

minus

119886211986213log (119887119886)

(11986211

minus 11986266) (1198872minus 1198862)

(23)

Now we introduce the following nondimensional compo-nents as

119877 = (

119903

119887

) 1198770= (

119886

119887

) 120590119903119905

= [

119879119903119903

119884

]

120590120579119905

= [

119879120579120579

119884

] 120590119911119905

= [

119879119911119911

119884

] 120590119903119891

= [

119879119903119903

1198841

]

120590120579119891

= [

119879120579120579

1198841

] 120590119911119891

= [

119879119911119911

1198841

]

(24)

The necessary effective pressure required for initial yieldingis given by (20) in nondimensional form as

1003816100381610038161003816119875119894

1003816100381610038161003816=

[(1198770)1198621

minus 1]

1198621(1198770)minus1198621

(25)

The Scientific World Journal 5

The transitional stresses given by (21) become

120590119903119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (119877)minus1198621

] minus 1198751198942

120590120579119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (1 minus 1198621) (119877)minus1198621

] minus 1198751198942

120590119911119905

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(1198770)minus1198621

minus 1

)

times (2 minus (2 minus 1198621) (119877)minus1198621

) ]

+

119875119894

(119877minus2

0minus 1)

minus

11986213119875119894

(11986211

minus 11986266) (119877minus2

0minus 1)

(26)

The effective pressure required for full plasticity is given by

119875119891

= log(

1

1198770

) (27)

Now stresses for full plasticity are obtained by taking 1198621

rarr

0 we have

120590119903119891

= minus (1198751198911

minus 1198751198911

)

log (119877)

log (1198770)

minus 1198751198912

120590120579119891

= minus (1198751198911

minus 1198751198912

)

[1 + log (119877)]

log (1198770)

minus 1198751198912

120590119911119891

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log (119877))

+

log (1198770)

(1 minus 1198770

minus2)

minus

11986213log (119877

0)

(11986211

minus 11986266) (1 minus 119877

0

minus2)

(28)

5 Numerical Discussion and Conclusion

To observe the effect of pressure required for initial yieldingand fully plastic state against various radii ratios Tables 2ndash4and Figures 2 and 3 using Table 1 have been drawn

From Table 2 it can be seen that in case of isotropicmaterial (steel) effective pressure required for initial yieldingand fully plastic state is high for the cylinder whose radii ratiois 02 as compared to the cylinder with other radii ratios thatis 03 04 and so forth It has also been noted that percentageincrease in effective pressure required for initial yielding tobecome fully plastic is high for the cylinder with radii ratio02 as compared to cylinder with other radii ratios It hasbeen noticed from Tables 3 and 4 that in case of transverselyisotropic materials (beryl and magnesium) effective pressurerequired for initial yielding to fully plastic state is again highfor cylinder with radii ratio 02 as compared to the cylinderwith other radii ratios It has also been observed from Tables2ndash4 that percentage increase in effective pressure required forinitial yielding to become fully plastic is high for cylindermade up of beryl material as compared to magnesium and

02 04 06 08 1Radii ratio

025

05

075

1

125

15

Effec

tive p

ress

ure

SteelBerylMagnesium

Figure 2 Effective pressure required for initial yielding for isotropic(steel) and transversely isotropic material (beryl and steel)

steel material It has also been observed from Table 2 thatfor isotropic material (steel) external pressure required forinitial yielding and fully plastic state when internal pressureis given (say 119875

1= 10) is high for the cylinder with radii

ratio 05 as compared to cylinders with less radii ratio whilepercentage increase in external pressure required for initialyielding to become fully plastic is high for the cylinder withlesser radii ratio as compared to cylinder with higher radiiratio From Tables 3 and 4 it has been noticed that fortransversely isotropic material external pressure requiredfor initial yielding and fully plastic state is again high forcylinder with high radii ratio as compared to cylinder withless radii ratio while percentage increase in external pressurerequired for initial yielding to become fully plastic is high forcylinder with lesser radii ratio as compared to higher radiiratio cylinder From Tables 2ndash4 it can be seen that withthe increase in internal pressure external pressure requiredfor initial yielding and fully plastic state also increases Alsoit has been noted that this percentage increase in externalpressure required for initial yielding to become fully plasticis high for cylinder made up of beryl as compared to cylindermade up of steel and magnesium

From Figure 2 it is noticed that effective pressurerequired for initial yielding is maximum at internal surfaceand this effective pressure is more for magnesium as com-pared to beryl as well as steel From Figure 3 it can be seenthat external pressure required for initial yielding for thecylinder with internal pressure (=10) is maximum at externalsurface Also it has been observed that external pressurerequired for initial yielding is high for beryl as comparedto steel and magnesium As internal pressure increasesexternal pressure required for initial yielding also increasesaccordingly

To see the effect of pressure on stresses Figures 4ndash6 aredrawn for transitional stresses while Figures 7ndash9 are for fullyplastic stresses

From Table 5 and Figure 4 it can be seen that stressesare maximum at internal surface Also it has been observed

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

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Nano

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Journal ofNanomaterials

2 The Scientific World Journal

is to increase its strength-weight ratio and the second is toextend its fatigue life This can be carried out by generatinga residual stress field in the barrel wall A Von-Misesrsquoyield criterion isotropic strain hardening with the Prandtl-Reuss theory has been taken into the consideration withBauschinger effect and plane stress conditions The stressesare calculated incrementally by using the finite differencemethod Davidson et al [7] determine the residual-stressdistribution as a function of magnitude of overstrains anddiametric ratio and discuss the effects on the reyieldingcharacteristics of cylinders All these authors consideredyield criterion jump conditions and linear strain measureto calculate the stresses using the concept of infinitesimalstrain theory According to the approach of the above authorsthe spectrum of deformations is divided into two regionsthat is one is elastic region and another one is plastic regionwhich is physically not possible because transition from onestate into another state is a continuous phenomenon Attransition the fundamental structure undergoes a change andthe particles constituting the material rearrange themselvesand give rise to spin rotation vorticity and other nonlineareffects This suggests that at transition nonlinear terms arevery important and ignoring themmay not represent the realphysical phenomenon

Transition theory [8 9] does not require any of the aboveassumptions and thus solves a more general problem usingthe concept of generalized strain measure [10] which notonly gives the well-known strain measures but can also beused to find the stresses in plasticity and creep problemsby determining the asymptotic solution at the transitionpoints of the governing differential equations This theoryhas been applied to many problems for example Sharmaet al [11] analyzed thermal elastic-plastic stresses in trans-versely isotropic thick-walled rotating cylinder under inter-nal pressure Aggarwal et al [12] investigated safety factorsin thick-walled functionally graded cylinder under internaland external pressure and concluded that functionally gradedcylinder is a better choice for designers as compared tocylindersmade up of homogeneousmaterials Aggarwal et al[13] discussed safety factors in creep nonhomogeneous thick-walled circular cylinder under internal and external pressurewith steady state temperature and concluded that cylindermade up of nonhomogeneous material is better choice fordesigning as compared to cylinder made up of homogeneousmaterial

If a continuous phenomenon is represented by a spec-trum nonlinearity exhibits itself at its ends which corre-sponds to transition state Thus the classical measures ofdeformation are inadequate to deal with such transitions Inclassical theory such transition state requires semiempiricallaws to match the solutions and thus discontinuities haveto be introduced where they do not exists A continuumapproach necessarily means the introduction of nonlinearmeasures If in a very small interval the number of fluc-tuations is very large the concept of ordinary measurebased on Riemann integral fails and Lebesgue measures maybe used In like manner the generalized measures givenby weighted integral representations give very satisfactoryresults in problems of plasticity and creep Seth has defined

the generalized principal strain measure 120576119894119894by taking the

Lebesgue integral of the weighted function

120576119894119894= int

119890119860

119894119894

0

[(1 minus 2119890119860

119894119894)]

(1198992)minus1

119889119890119860

119894119894

=

1

119899

[1 minus (1 minus 2119890119860

119894119894)

(1198992)minus1

]

(1)

where 119899 is themeasure and 119890119860

119894119894is the principal Almansi strain

component For uniaxial case

119890 = [

1

119899

1 minus (

1198970

119897

)

119899

]

119898

(2)

where 119898 is the irreversibility index and 1198970and 119897 are the

undeformed and deformed lengths of the rod respectivelyFor 119899 = minus2minus1 0 1 2 in (2) gives theGreenHencky Swaingerand Almansi measure respectively where in all cases 119898 = 1The most important contribution made by the generalizedLebesgue measure is that they eliminate the use of empiricallaws and jump conditions

As we know the permissible stress of anymaterial is someproportion of the yield or ultimate stress of the materialand as such it incorporates a ldquosafety factorrdquo This safetyfactor provides a margin against the collapse condition of thecylinder that occurs due to high pressure Since the conditionof high pressure can cause failure of the cylinder therefore itis necessary to examine the state of the cylinder at collapseand to design the cylinder accordingly In order to provideguidance on a design and integrity evaluation of a cylinderunder pressure the failure characteristics of a cylinder shouldbe considered carefully In this paper it is our main aimto eliminate the need of assuming semiempirical laws yieldcondition creep-strain laws jump conditions and so forthto obtain the collapse pressure in transversely isotropic thick-walled circular cylinder under internal and external pressureusing generalized Lebesgue strain measure The stresses arecalculated in both transition and fully plastic state

2 Mathematical Formulation of Problem

We consider a thick-walled circular cylinder made up oftransversely isotropic material with internal and externalradii ldquoardquo and ldquobrdquo respectively subjected to internal pressure1199011and external pressure 119901

2as shown in the Figure 1 The

cylinder is taken so large that the plane sections remain planeduring the expansion and hence the longitudinal strain is thesame for all elements at each stage of the expansion

The components of the displacement in cylindrical polarcoordinates are given by

119906 = 119903 (1 minus 120573) V = 0 119908 = 119889119911 (3)

where120573 is a function of 119903 = radic1199092+ 1199102 only and119889 is a constant

The Scientific World Journal 3

p1 p1

p2

p2

a

b

Figure 1 Geometry of thick-walled transversely isotropic cylinderunder internal and external pressure

The generalized components of strain [10] are given asfollows

119890119903119903

=

1

119899

[1 minus (1199031205731015840+ 120573)

119899

] 119890120579120579

=

1

119899

[1 minus 120573119899]

119890119911119911

=

1

119899

[1 minus (1 minus 119889)119899] 119890

119903120579= 119890120579119911

= 119890119911119903

= 0

(4)

The stress-strain relations for transversely isotropic materialare

119879119903119903

= 11986211119890119903119903

+ (11986211

minus 211986266) 119890120579120579

+ 11986213119890119911119911

119879120579120579

= (11986211

minus 211986266) 119890119903119903

+ 11986211119890120579120579

+ 11986213119890119911119911

119879119911119911

= 11986213119890119903119903

+ 11986213119890120579120579

+ 11986233119890119911119911

119879119911119903

= 119879120579119911

= 119879119903120579

= 0

(5)

Using (4) in (5) we get

119879119903119903

= (

11986211

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ [

(11986211

minus 211986266)

119899

] [1 minus 120573119899] + 11986213119890119911119911

119879120579120579

= [

(11986211

minus 211986266)

119899

] [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986211

119899

) [1 minus 120573119899] + 11986213119890119911119911

119879119911119911

= (

11986213

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986213

119899

) [1 minus 120573119899] + 11986233119890119911119911

119879119903120579

= 119879120579119911

= 119879119903119911

= 0

(6)

Equations of equilibrium are all satisfied except

119889

119889119903

(119879119903119903) + (

119879119903119903

minus 119879120579120579

119903

) = 0 (7)

3 Identification of Transition Point

We know that as the point in the material has yielded thematerial at the neighbouring points is on its way to yieldrather than remain in its complete elastic state or fully plasticstate Thus we can assume that there exists some state in-between elastic and plastic which is called transition state Soat transition the differential system defining the elastic stateshould attain some criticalityThe differential equation whichcomes out to be nonlinear at transition state is obtained bysubstituting equations (6) in (7)

11989911987511986211120573119899+1

(1 + 119875)119899minus1 119889119875

119889120573

= minus11989911987511986211120573119899(1 + 119875)

119899

minus (11986211

minus 211986266) 119899119875120573119899

+ 211986266

[1 minus 120573119899(1 + 119875)

119899]

minus 211986266

(1 minus 120573119899)

(8)

where 1199031205731015840= 120573119875

The critical points or transitional points of (8) are 119875 rarr

minus1 and 119875 rarr plusmninfinThe boundary conditions are given by

119879119903119903

= minus1199011

at 119903 = 119886

119879119903119903

= minus1199012

at 119903 = 119887

(9)

The resultant axial force is given by

2120587int

119887

119886

119903119879119911119911

119889119903 = 1205871198862(1199011minus 1199012) (10)

4 Mathematical Approach

The material from elastic state goes into plastic state as119875 rarr plusmninfin or to creep state as 119875 rarr minus1 under internal andexternal pressure It has been shown [8 9 11ndash13] that theasymptotic solution through the principal stress leads fromelastic to plastic state at the transition point 119875 rarr plusmninfin Forfinding the plastic stress at the transition point 119875 rarr plusmninfin wedefine the transition function 119877 in terms of 119879

119903119903as

119877 = 2 (11986211

minus 11986266) + 119899119862

13119890119911119911

minus 119899119879119903119903

= 120573119899[11986211

minus 211986266

+ 11986211(

1 + 119875)119899]

(11)

4 The Scientific World Journal

Taking the logarithmic differentiation of (11) with respect toldquorrdquo with asymptotic value as 119875 rarr plusmninfin on integration yields

119877 = 1198601119903minus1198621

(12)

where 1198601is a constant of integration and 119862

1= 21198626611986211

Using (11) and (12) we get

119879119903119903

= 1198623minus (

1198601

119899

) 119903minus1198621

(13)

Using boundary condition (9) in the above equation we get

1198601= 1198991198871198621

[

119901

(119887119886)1198621

minus 1

] 1198623= [

119901

(119887119886)1198621

minus 1

] (14)

Substituting the value of 1198601and 119862

3in (13) we get

119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1199012 (15)

Using (15) in (7) we have

119879120579120579

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1199012 (16)

119879119911119911

=

11986213

2 (11986211

minus 11986266)

[(

1199011minus 1199012

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862(1199011minus 1199012)

1198872minus 1198862

minus

119886211986213

(1199011minus 1199012)

(11986211

minus 11986266) (1198872minus 1198862)

(17)

From (15) and (16) we get

119879120579120579

minus 119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119903

)

1198621

(18)

From the above equation it is found that the value of |119879120579120579

minus119879119903119903|

is maximum at 119903 = 119886 which means yielding of the cylinderwill take place at the internal surface Therefore we have

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119886

=

100381610038161003816100381610038161003816100381610038161003816

[

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119886

)

1198621 100381610038161003816100381610038161003816100381610038161003816

equiv 119884 (say) (19)

Let 1199011minus 1199012= 119901

The pressure required for initial yielding is given by

1003816100381610038161003816119875119894

1003816100381610038161003816=

1003816100381610038161003816100381610038161003816

119901

119884

1003816100381610038161003816100381610038161003816

=

[(119887119886)1198621

minus 1]

1198621(119887119886)1198621

(20)

where (1199011119884) minus (119901

2119884) = 119875

1198941minus 1198751198942

= 119875119894

Using (20) in (15) (16) and (17) we get transitionalstresses as

119879119903119903

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1198751198942

119879120579120579

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1198751198942

119879119911119911

119884

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862119875119894

1198872minus 1198862minus

119886211986213119875119894

(11986211

minus 11986266) (1198872minus 1198862)

(21)

Equation (21) give elastic-plastic transitional stresses in thick-walled cylinder under internal and external pressure

For fully plastic state (1198621

rarr 0) (20) becomes

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119887

=

10038161003816100381610038161003816100381610038161003816

119901

log (119887119886)

10038161003816100381610038161003816100381610038161003816

equiv 1198841(say) (22)

where 119875119891

= (1199011minus 1199012)1198841= 1198751198911

minus 1198751198912

The stresses for fully plastic state are

119879119903119903

1198841

= log(

119903

119887

) minus 1198751198912

119879120579120579

1198841

= (1 + log(

119903

119887

)) minus 1198751198912

119879119911119911

1198841

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log(

119903

119887

))

+

1198862 log (119887119886)

1198872minus 1198862

minus

119886211986213log (119887119886)

(11986211

minus 11986266) (1198872minus 1198862)

(23)

Now we introduce the following nondimensional compo-nents as

119877 = (

119903

119887

) 1198770= (

119886

119887

) 120590119903119905

= [

119879119903119903

119884

]

120590120579119905

= [

119879120579120579

119884

] 120590119911119905

= [

119879119911119911

119884

] 120590119903119891

= [

119879119903119903

1198841

]

120590120579119891

= [

119879120579120579

1198841

] 120590119911119891

= [

119879119911119911

1198841

]

(24)

The necessary effective pressure required for initial yieldingis given by (20) in nondimensional form as

1003816100381610038161003816119875119894

1003816100381610038161003816=

[(1198770)1198621

minus 1]

1198621(1198770)minus1198621

(25)

The Scientific World Journal 5

The transitional stresses given by (21) become

120590119903119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (119877)minus1198621

] minus 1198751198942

120590120579119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (1 minus 1198621) (119877)minus1198621

] minus 1198751198942

120590119911119905

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(1198770)minus1198621

minus 1

)

times (2 minus (2 minus 1198621) (119877)minus1198621

) ]

+

119875119894

(119877minus2

0minus 1)

minus

11986213119875119894

(11986211

minus 11986266) (119877minus2

0minus 1)

(26)

The effective pressure required for full plasticity is given by

119875119891

= log(

1

1198770

) (27)

Now stresses for full plasticity are obtained by taking 1198621

rarr

0 we have

120590119903119891

= minus (1198751198911

minus 1198751198911

)

log (119877)

log (1198770)

minus 1198751198912

120590120579119891

= minus (1198751198911

minus 1198751198912

)

[1 + log (119877)]

log (1198770)

minus 1198751198912

120590119911119891

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log (119877))

+

log (1198770)

(1 minus 1198770

minus2)

minus

11986213log (119877

0)

(11986211

minus 11986266) (1 minus 119877

0

minus2)

(28)

5 Numerical Discussion and Conclusion

To observe the effect of pressure required for initial yieldingand fully plastic state against various radii ratios Tables 2ndash4and Figures 2 and 3 using Table 1 have been drawn

From Table 2 it can be seen that in case of isotropicmaterial (steel) effective pressure required for initial yieldingand fully plastic state is high for the cylinder whose radii ratiois 02 as compared to the cylinder with other radii ratios thatis 03 04 and so forth It has also been noted that percentageincrease in effective pressure required for initial yielding tobecome fully plastic is high for the cylinder with radii ratio02 as compared to cylinder with other radii ratios It hasbeen noticed from Tables 3 and 4 that in case of transverselyisotropic materials (beryl and magnesium) effective pressurerequired for initial yielding to fully plastic state is again highfor cylinder with radii ratio 02 as compared to the cylinderwith other radii ratios It has also been observed from Tables2ndash4 that percentage increase in effective pressure required forinitial yielding to become fully plastic is high for cylindermade up of beryl material as compared to magnesium and

02 04 06 08 1Radii ratio

025

05

075

1

125

15

Effec

tive p

ress

ure

SteelBerylMagnesium

Figure 2 Effective pressure required for initial yielding for isotropic(steel) and transversely isotropic material (beryl and steel)

steel material It has also been observed from Table 2 thatfor isotropic material (steel) external pressure required forinitial yielding and fully plastic state when internal pressureis given (say 119875

1= 10) is high for the cylinder with radii

ratio 05 as compared to cylinders with less radii ratio whilepercentage increase in external pressure required for initialyielding to become fully plastic is high for the cylinder withlesser radii ratio as compared to cylinder with higher radiiratio From Tables 3 and 4 it has been noticed that fortransversely isotropic material external pressure requiredfor initial yielding and fully plastic state is again high forcylinder with high radii ratio as compared to cylinder withless radii ratio while percentage increase in external pressurerequired for initial yielding to become fully plastic is high forcylinder with lesser radii ratio as compared to higher radiiratio cylinder From Tables 2ndash4 it can be seen that withthe increase in internal pressure external pressure requiredfor initial yielding and fully plastic state also increases Alsoit has been noted that this percentage increase in externalpressure required for initial yielding to become fully plasticis high for cylinder made up of beryl as compared to cylindermade up of steel and magnesium

From Figure 2 it is noticed that effective pressurerequired for initial yielding is maximum at internal surfaceand this effective pressure is more for magnesium as com-pared to beryl as well as steel From Figure 3 it can be seenthat external pressure required for initial yielding for thecylinder with internal pressure (=10) is maximum at externalsurface Also it has been observed that external pressurerequired for initial yielding is high for beryl as comparedto steel and magnesium As internal pressure increasesexternal pressure required for initial yielding also increasesaccordingly

To see the effect of pressure on stresses Figures 4ndash6 aredrawn for transitional stresses while Figures 7ndash9 are for fullyplastic stresses

From Table 5 and Figure 4 it can be seen that stressesare maximum at internal surface Also it has been observed

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

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Nano

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Journal ofNanomaterials

The Scientific World Journal 3

p1 p1

p2

p2

a

b

Figure 1 Geometry of thick-walled transversely isotropic cylinderunder internal and external pressure

The generalized components of strain [10] are given asfollows

119890119903119903

=

1

119899

[1 minus (1199031205731015840+ 120573)

119899

] 119890120579120579

=

1

119899

[1 minus 120573119899]

119890119911119911

=

1

119899

[1 minus (1 minus 119889)119899] 119890

119903120579= 119890120579119911

= 119890119911119903

= 0

(4)

The stress-strain relations for transversely isotropic materialare

119879119903119903

= 11986211119890119903119903

+ (11986211

minus 211986266) 119890120579120579

+ 11986213119890119911119911

119879120579120579

= (11986211

minus 211986266) 119890119903119903

+ 11986211119890120579120579

+ 11986213119890119911119911

119879119911119911

= 11986213119890119903119903

+ 11986213119890120579120579

+ 11986233119890119911119911

119879119911119903

= 119879120579119911

= 119879119903120579

= 0

(5)

Using (4) in (5) we get

119879119903119903

= (

11986211

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ [

(11986211

minus 211986266)

119899

] [1 minus 120573119899] + 11986213119890119911119911

119879120579120579

= [

(11986211

minus 211986266)

119899

] [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986211

119899

) [1 minus 120573119899] + 11986213119890119911119911

119879119911119911

= (

11986213

119899

) [1 minus (120573 + 1199031205731015840)

119899

]

+ (

11986213

119899

) [1 minus 120573119899] + 11986233119890119911119911

119879119903120579

= 119879120579119911

= 119879119903119911

= 0

(6)

Equations of equilibrium are all satisfied except

119889

119889119903

(119879119903119903) + (

119879119903119903

minus 119879120579120579

119903

) = 0 (7)

3 Identification of Transition Point

We know that as the point in the material has yielded thematerial at the neighbouring points is on its way to yieldrather than remain in its complete elastic state or fully plasticstate Thus we can assume that there exists some state in-between elastic and plastic which is called transition state Soat transition the differential system defining the elastic stateshould attain some criticalityThe differential equation whichcomes out to be nonlinear at transition state is obtained bysubstituting equations (6) in (7)

11989911987511986211120573119899+1

(1 + 119875)119899minus1 119889119875

119889120573

= minus11989911987511986211120573119899(1 + 119875)

119899

minus (11986211

minus 211986266) 119899119875120573119899

+ 211986266

[1 minus 120573119899(1 + 119875)

119899]

minus 211986266

(1 minus 120573119899)

(8)

where 1199031205731015840= 120573119875

The critical points or transitional points of (8) are 119875 rarr

minus1 and 119875 rarr plusmninfinThe boundary conditions are given by

119879119903119903

= minus1199011

at 119903 = 119886

119879119903119903

= minus1199012

at 119903 = 119887

(9)

The resultant axial force is given by

2120587int

119887

119886

119903119879119911119911

119889119903 = 1205871198862(1199011minus 1199012) (10)

4 Mathematical Approach

The material from elastic state goes into plastic state as119875 rarr plusmninfin or to creep state as 119875 rarr minus1 under internal andexternal pressure It has been shown [8 9 11ndash13] that theasymptotic solution through the principal stress leads fromelastic to plastic state at the transition point 119875 rarr plusmninfin Forfinding the plastic stress at the transition point 119875 rarr plusmninfin wedefine the transition function 119877 in terms of 119879

119903119903as

119877 = 2 (11986211

minus 11986266) + 119899119862

13119890119911119911

minus 119899119879119903119903

= 120573119899[11986211

minus 211986266

+ 11986211(

1 + 119875)119899]

(11)

4 The Scientific World Journal

Taking the logarithmic differentiation of (11) with respect toldquorrdquo with asymptotic value as 119875 rarr plusmninfin on integration yields

119877 = 1198601119903minus1198621

(12)

where 1198601is a constant of integration and 119862

1= 21198626611986211

Using (11) and (12) we get

119879119903119903

= 1198623minus (

1198601

119899

) 119903minus1198621

(13)

Using boundary condition (9) in the above equation we get

1198601= 1198991198871198621

[

119901

(119887119886)1198621

minus 1

] 1198623= [

119901

(119887119886)1198621

minus 1

] (14)

Substituting the value of 1198601and 119862

3in (13) we get

119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1199012 (15)

Using (15) in (7) we have

119879120579120579

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1199012 (16)

119879119911119911

=

11986213

2 (11986211

minus 11986266)

[(

1199011minus 1199012

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862(1199011minus 1199012)

1198872minus 1198862

minus

119886211986213

(1199011minus 1199012)

(11986211

minus 11986266) (1198872minus 1198862)

(17)

From (15) and (16) we get

119879120579120579

minus 119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119903

)

1198621

(18)

From the above equation it is found that the value of |119879120579120579

minus119879119903119903|

is maximum at 119903 = 119886 which means yielding of the cylinderwill take place at the internal surface Therefore we have

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119886

=

100381610038161003816100381610038161003816100381610038161003816

[

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119886

)

1198621 100381610038161003816100381610038161003816100381610038161003816

equiv 119884 (say) (19)

Let 1199011minus 1199012= 119901

The pressure required for initial yielding is given by

1003816100381610038161003816119875119894

1003816100381610038161003816=

1003816100381610038161003816100381610038161003816

119901

119884

1003816100381610038161003816100381610038161003816

=

[(119887119886)1198621

minus 1]

1198621(119887119886)1198621

(20)

where (1199011119884) minus (119901

2119884) = 119875

1198941minus 1198751198942

= 119875119894

Using (20) in (15) (16) and (17) we get transitionalstresses as

119879119903119903

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1198751198942

119879120579120579

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1198751198942

119879119911119911

119884

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862119875119894

1198872minus 1198862minus

119886211986213119875119894

(11986211

minus 11986266) (1198872minus 1198862)

(21)

Equation (21) give elastic-plastic transitional stresses in thick-walled cylinder under internal and external pressure

For fully plastic state (1198621

rarr 0) (20) becomes

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119887

=

10038161003816100381610038161003816100381610038161003816

119901

log (119887119886)

10038161003816100381610038161003816100381610038161003816

equiv 1198841(say) (22)

where 119875119891

= (1199011minus 1199012)1198841= 1198751198911

minus 1198751198912

The stresses for fully plastic state are

119879119903119903

1198841

= log(

119903

119887

) minus 1198751198912

119879120579120579

1198841

= (1 + log(

119903

119887

)) minus 1198751198912

119879119911119911

1198841

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log(

119903

119887

))

+

1198862 log (119887119886)

1198872minus 1198862

minus

119886211986213log (119887119886)

(11986211

minus 11986266) (1198872minus 1198862)

(23)

Now we introduce the following nondimensional compo-nents as

119877 = (

119903

119887

) 1198770= (

119886

119887

) 120590119903119905

= [

119879119903119903

119884

]

120590120579119905

= [

119879120579120579

119884

] 120590119911119905

= [

119879119911119911

119884

] 120590119903119891

= [

119879119903119903

1198841

]

120590120579119891

= [

119879120579120579

1198841

] 120590119911119891

= [

119879119911119911

1198841

]

(24)

The necessary effective pressure required for initial yieldingis given by (20) in nondimensional form as

1003816100381610038161003816119875119894

1003816100381610038161003816=

[(1198770)1198621

minus 1]

1198621(1198770)minus1198621

(25)

The Scientific World Journal 5

The transitional stresses given by (21) become

120590119903119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (119877)minus1198621

] minus 1198751198942

120590120579119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (1 minus 1198621) (119877)minus1198621

] minus 1198751198942

120590119911119905

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(1198770)minus1198621

minus 1

)

times (2 minus (2 minus 1198621) (119877)minus1198621

) ]

+

119875119894

(119877minus2

0minus 1)

minus

11986213119875119894

(11986211

minus 11986266) (119877minus2

0minus 1)

(26)

The effective pressure required for full plasticity is given by

119875119891

= log(

1

1198770

) (27)

Now stresses for full plasticity are obtained by taking 1198621

rarr

0 we have

120590119903119891

= minus (1198751198911

minus 1198751198911

)

log (119877)

log (1198770)

minus 1198751198912

120590120579119891

= minus (1198751198911

minus 1198751198912

)

[1 + log (119877)]

log (1198770)

minus 1198751198912

120590119911119891

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log (119877))

+

log (1198770)

(1 minus 1198770

minus2)

minus

11986213log (119877

0)

(11986211

minus 11986266) (1 minus 119877

0

minus2)

(28)

5 Numerical Discussion and Conclusion

To observe the effect of pressure required for initial yieldingand fully plastic state against various radii ratios Tables 2ndash4and Figures 2 and 3 using Table 1 have been drawn

From Table 2 it can be seen that in case of isotropicmaterial (steel) effective pressure required for initial yieldingand fully plastic state is high for the cylinder whose radii ratiois 02 as compared to the cylinder with other radii ratios thatis 03 04 and so forth It has also been noted that percentageincrease in effective pressure required for initial yielding tobecome fully plastic is high for the cylinder with radii ratio02 as compared to cylinder with other radii ratios It hasbeen noticed from Tables 3 and 4 that in case of transverselyisotropic materials (beryl and magnesium) effective pressurerequired for initial yielding to fully plastic state is again highfor cylinder with radii ratio 02 as compared to the cylinderwith other radii ratios It has also been observed from Tables2ndash4 that percentage increase in effective pressure required forinitial yielding to become fully plastic is high for cylindermade up of beryl material as compared to magnesium and

02 04 06 08 1Radii ratio

025

05

075

1

125

15

Effec

tive p

ress

ure

SteelBerylMagnesium

Figure 2 Effective pressure required for initial yielding for isotropic(steel) and transversely isotropic material (beryl and steel)

steel material It has also been observed from Table 2 thatfor isotropic material (steel) external pressure required forinitial yielding and fully plastic state when internal pressureis given (say 119875

1= 10) is high for the cylinder with radii

ratio 05 as compared to cylinders with less radii ratio whilepercentage increase in external pressure required for initialyielding to become fully plastic is high for the cylinder withlesser radii ratio as compared to cylinder with higher radiiratio From Tables 3 and 4 it has been noticed that fortransversely isotropic material external pressure requiredfor initial yielding and fully plastic state is again high forcylinder with high radii ratio as compared to cylinder withless radii ratio while percentage increase in external pressurerequired for initial yielding to become fully plastic is high forcylinder with lesser radii ratio as compared to higher radiiratio cylinder From Tables 2ndash4 it can be seen that withthe increase in internal pressure external pressure requiredfor initial yielding and fully plastic state also increases Alsoit has been noted that this percentage increase in externalpressure required for initial yielding to become fully plasticis high for cylinder made up of beryl as compared to cylindermade up of steel and magnesium

From Figure 2 it is noticed that effective pressurerequired for initial yielding is maximum at internal surfaceand this effective pressure is more for magnesium as com-pared to beryl as well as steel From Figure 3 it can be seenthat external pressure required for initial yielding for thecylinder with internal pressure (=10) is maximum at externalsurface Also it has been observed that external pressurerequired for initial yielding is high for beryl as comparedto steel and magnesium As internal pressure increasesexternal pressure required for initial yielding also increasesaccordingly

To see the effect of pressure on stresses Figures 4ndash6 aredrawn for transitional stresses while Figures 7ndash9 are for fullyplastic stresses

From Table 5 and Figure 4 it can be seen that stressesare maximum at internal surface Also it has been observed

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

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Journal ofNanomaterials

4 The Scientific World Journal

Taking the logarithmic differentiation of (11) with respect toldquorrdquo with asymptotic value as 119875 rarr plusmninfin on integration yields

119877 = 1198601119903minus1198621

(12)

where 1198601is a constant of integration and 119862

1= 21198626611986211

Using (11) and (12) we get

119879119903119903

= 1198623minus (

1198601

119899

) 119903minus1198621

(13)

Using boundary condition (9) in the above equation we get

1198601= 1198991198871198621

[

119901

(119887119886)1198621

minus 1

] 1198623= [

119901

(119887119886)1198621

minus 1

] (14)

Substituting the value of 1198601and 119862

3in (13) we get

119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1199012 (15)

Using (15) in (7) we have

119879120579120579

= [

1199011minus 1199012

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1199012 (16)

119879119911119911

=

11986213

2 (11986211

minus 11986266)

[(

1199011minus 1199012

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862(1199011minus 1199012)

1198872minus 1198862

minus

119886211986213

(1199011minus 1199012)

(11986211

minus 11986266) (1198872minus 1198862)

(17)

From (15) and (16) we get

119879120579120579

minus 119879119903119903

= [

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119903

)

1198621

(18)

From the above equation it is found that the value of |119879120579120579

minus119879119903119903|

is maximum at 119903 = 119886 which means yielding of the cylinderwill take place at the internal surface Therefore we have

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119886

=

100381610038161003816100381610038161003816100381610038161003816

[

1199011minus 1199012

(119887119886)1198621

minus 1

]1198621(

119887

119886

)

1198621 100381610038161003816100381610038161003816100381610038161003816

equiv 119884 (say) (19)

Let 1199011minus 1199012= 119901

The pressure required for initial yielding is given by

1003816100381610038161003816119875119894

1003816100381610038161003816=

1003816100381610038161003816100381610038161003816

119901

119884

1003816100381610038161003816100381610038161003816

=

[(119887119886)1198621

minus 1]

1198621(119887119886)1198621

(20)

where (1199011119884) minus (119901

2119884) = 119875

1198941minus 1198751198942

= 119875119894

Using (20) in (15) (16) and (17) we get transitionalstresses as

119879119903119903

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (

119887

119903

)

1198621

] minus 1198751198942

119879120579120579

119884

= [

119875119894

(119887119886)1198621

minus 1

][1 minus (1 minus 1198621) (

119887

119903

)

1198621

] minus 1198751198942

119879119911119911

119884

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(119887119886)1198621

minus 1

)

times (2 minus (2 minus 1198621) (

119887

119903

)

1198621

)]

+

1198862119875119894

1198872minus 1198862minus

119886211986213119875119894

(11986211

minus 11986266) (1198872minus 1198862)

(21)

Equation (21) give elastic-plastic transitional stresses in thick-walled cylinder under internal and external pressure

For fully plastic state (1198621

rarr 0) (20) becomes

1003816100381610038161003816119879120579120579

minus 119879119903119903

1003816100381610038161003816119903=119887

=

10038161003816100381610038161003816100381610038161003816

119901

log (119887119886)

10038161003816100381610038161003816100381610038161003816

equiv 1198841(say) (22)

where 119875119891

= (1199011minus 1199012)1198841= 1198751198911

minus 1198751198912

The stresses for fully plastic state are

119879119903119903

1198841

= log(

119903

119887

) minus 1198751198912

119879120579120579

1198841

= (1 + log(

119903

119887

)) minus 1198751198912

119879119911119911

1198841

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log(

119903

119887

))

+

1198862 log (119887119886)

1198872minus 1198862

minus

119886211986213log (119887119886)

(11986211

minus 11986266) (1198872minus 1198862)

(23)

Now we introduce the following nondimensional compo-nents as

119877 = (

119903

119887

) 1198770= (

119886

119887

) 120590119903119905

= [

119879119903119903

119884

]

120590120579119905

= [

119879120579120579

119884

] 120590119911119905

= [

119879119911119911

119884

] 120590119903119891

= [

119879119903119903

1198841

]

120590120579119891

= [

119879120579120579

1198841

] 120590119911119891

= [

119879119911119911

1198841

]

(24)

The necessary effective pressure required for initial yieldingis given by (20) in nondimensional form as

1003816100381610038161003816119875119894

1003816100381610038161003816=

[(1198770)1198621

minus 1]

1198621(1198770)minus1198621

(25)

The Scientific World Journal 5

The transitional stresses given by (21) become

120590119903119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (119877)minus1198621

] minus 1198751198942

120590120579119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (1 minus 1198621) (119877)minus1198621

] minus 1198751198942

120590119911119905

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(1198770)minus1198621

minus 1

)

times (2 minus (2 minus 1198621) (119877)minus1198621

) ]

+

119875119894

(119877minus2

0minus 1)

minus

11986213119875119894

(11986211

minus 11986266) (119877minus2

0minus 1)

(26)

The effective pressure required for full plasticity is given by

119875119891

= log(

1

1198770

) (27)

Now stresses for full plasticity are obtained by taking 1198621

rarr

0 we have

120590119903119891

= minus (1198751198911

minus 1198751198911

)

log (119877)

log (1198770)

minus 1198751198912

120590120579119891

= minus (1198751198911

minus 1198751198912

)

[1 + log (119877)]

log (1198770)

minus 1198751198912

120590119911119891

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log (119877))

+

log (1198770)

(1 minus 1198770

minus2)

minus

11986213log (119877

0)

(11986211

minus 11986266) (1 minus 119877

0

minus2)

(28)

5 Numerical Discussion and Conclusion

To observe the effect of pressure required for initial yieldingand fully plastic state against various radii ratios Tables 2ndash4and Figures 2 and 3 using Table 1 have been drawn

From Table 2 it can be seen that in case of isotropicmaterial (steel) effective pressure required for initial yieldingand fully plastic state is high for the cylinder whose radii ratiois 02 as compared to the cylinder with other radii ratios thatis 03 04 and so forth It has also been noted that percentageincrease in effective pressure required for initial yielding tobecome fully plastic is high for the cylinder with radii ratio02 as compared to cylinder with other radii ratios It hasbeen noticed from Tables 3 and 4 that in case of transverselyisotropic materials (beryl and magnesium) effective pressurerequired for initial yielding to fully plastic state is again highfor cylinder with radii ratio 02 as compared to the cylinderwith other radii ratios It has also been observed from Tables2ndash4 that percentage increase in effective pressure required forinitial yielding to become fully plastic is high for cylindermade up of beryl material as compared to magnesium and

02 04 06 08 1Radii ratio

025

05

075

1

125

15

Effec

tive p

ress

ure

SteelBerylMagnesium

Figure 2 Effective pressure required for initial yielding for isotropic(steel) and transversely isotropic material (beryl and steel)

steel material It has also been observed from Table 2 thatfor isotropic material (steel) external pressure required forinitial yielding and fully plastic state when internal pressureis given (say 119875

1= 10) is high for the cylinder with radii

ratio 05 as compared to cylinders with less radii ratio whilepercentage increase in external pressure required for initialyielding to become fully plastic is high for the cylinder withlesser radii ratio as compared to cylinder with higher radiiratio From Tables 3 and 4 it has been noticed that fortransversely isotropic material external pressure requiredfor initial yielding and fully plastic state is again high forcylinder with high radii ratio as compared to cylinder withless radii ratio while percentage increase in external pressurerequired for initial yielding to become fully plastic is high forcylinder with lesser radii ratio as compared to higher radiiratio cylinder From Tables 2ndash4 it can be seen that withthe increase in internal pressure external pressure requiredfor initial yielding and fully plastic state also increases Alsoit has been noted that this percentage increase in externalpressure required for initial yielding to become fully plasticis high for cylinder made up of beryl as compared to cylindermade up of steel and magnesium

From Figure 2 it is noticed that effective pressurerequired for initial yielding is maximum at internal surfaceand this effective pressure is more for magnesium as com-pared to beryl as well as steel From Figure 3 it can be seenthat external pressure required for initial yielding for thecylinder with internal pressure (=10) is maximum at externalsurface Also it has been observed that external pressurerequired for initial yielding is high for beryl as comparedto steel and magnesium As internal pressure increasesexternal pressure required for initial yielding also increasesaccordingly

To see the effect of pressure on stresses Figures 4ndash6 aredrawn for transitional stresses while Figures 7ndash9 are for fullyplastic stresses

From Table 5 and Figure 4 it can be seen that stressesare maximum at internal surface Also it has been observed

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

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Nano

materials

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Journal ofNanomaterials

The Scientific World Journal 5

The transitional stresses given by (21) become

120590119903119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (119877)minus1198621

] minus 1198751198942

120590120579119905

= [

119875119894

(1198770)minus1198621

minus 1

] [1 minus (1 minus 1198621) (119877)minus1198621

] minus 1198751198942

120590119911119905

=

11986213

2 (11986211

minus 11986266)

[(

119875119894

(1198770)minus1198621

minus 1

)

times (2 minus (2 minus 1198621) (119877)minus1198621

) ]

+

119875119894

(119877minus2

0minus 1)

minus

11986213119875119894

(11986211

minus 11986266) (119877minus2

0minus 1)

(26)

The effective pressure required for full plasticity is given by

119875119891

= log(

1

1198770

) (27)

Now stresses for full plasticity are obtained by taking 1198621

rarr

0 we have

120590119903119891

= minus (1198751198911

minus 1198751198911

)

log (119877)

log (1198770)

minus 1198751198912

120590120579119891

= minus (1198751198911

minus 1198751198912

)

[1 + log (119877)]

log (1198770)

minus 1198751198912

120590119911119891

=

11986213

2 (11986211

minus 11986266)

(1 + 2 log (119877))

+

log (1198770)

(1 minus 1198770

minus2)

minus

11986213log (119877

0)

(11986211

minus 11986266) (1 minus 119877

0

minus2)

(28)

5 Numerical Discussion and Conclusion

To observe the effect of pressure required for initial yieldingand fully plastic state against various radii ratios Tables 2ndash4and Figures 2 and 3 using Table 1 have been drawn

From Table 2 it can be seen that in case of isotropicmaterial (steel) effective pressure required for initial yieldingand fully plastic state is high for the cylinder whose radii ratiois 02 as compared to the cylinder with other radii ratios thatis 03 04 and so forth It has also been noted that percentageincrease in effective pressure required for initial yielding tobecome fully plastic is high for the cylinder with radii ratio02 as compared to cylinder with other radii ratios It hasbeen noticed from Tables 3 and 4 that in case of transverselyisotropic materials (beryl and magnesium) effective pressurerequired for initial yielding to fully plastic state is again highfor cylinder with radii ratio 02 as compared to the cylinderwith other radii ratios It has also been observed from Tables2ndash4 that percentage increase in effective pressure required forinitial yielding to become fully plastic is high for cylindermade up of beryl material as compared to magnesium and

02 04 06 08 1Radii ratio

025

05

075

1

125

15

Effec

tive p

ress

ure

SteelBerylMagnesium

Figure 2 Effective pressure required for initial yielding for isotropic(steel) and transversely isotropic material (beryl and steel)

steel material It has also been observed from Table 2 thatfor isotropic material (steel) external pressure required forinitial yielding and fully plastic state when internal pressureis given (say 119875

1= 10) is high for the cylinder with radii

ratio 05 as compared to cylinders with less radii ratio whilepercentage increase in external pressure required for initialyielding to become fully plastic is high for the cylinder withlesser radii ratio as compared to cylinder with higher radiiratio From Tables 3 and 4 it has been noticed that fortransversely isotropic material external pressure requiredfor initial yielding and fully plastic state is again high forcylinder with high radii ratio as compared to cylinder withless radii ratio while percentage increase in external pressurerequired for initial yielding to become fully plastic is high forcylinder with lesser radii ratio as compared to higher radiiratio cylinder From Tables 2ndash4 it can be seen that withthe increase in internal pressure external pressure requiredfor initial yielding and fully plastic state also increases Alsoit has been noted that this percentage increase in externalpressure required for initial yielding to become fully plasticis high for cylinder made up of beryl as compared to cylindermade up of steel and magnesium

From Figure 2 it is noticed that effective pressurerequired for initial yielding is maximum at internal surfaceand this effective pressure is more for magnesium as com-pared to beryl as well as steel From Figure 3 it can be seenthat external pressure required for initial yielding for thecylinder with internal pressure (=10) is maximum at externalsurface Also it has been observed that external pressurerequired for initial yielding is high for beryl as comparedto steel and magnesium As internal pressure increasesexternal pressure required for initial yielding also increasesaccordingly

To see the effect of pressure on stresses Figures 4ndash6 aredrawn for transitional stresses while Figures 7ndash9 are for fullyplastic stresses

From Table 5 and Figure 4 it can be seen that stressesare maximum at internal surface Also it has been observed

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

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Nano

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Journal ofNanomaterials

6 The Scientific World Journal

02 04 06 08 1Radii ratio

85

875

925

95

975

10

Exte

rnal

pre

ssur

eP1 = 10

SteelBerylMagnesium

(a)

02 04 06 08 1Radii ratio

485

4875

4925

495

4975

50

Exte

rnal

pre

ssur

e

P1 = 50

SteelBerylMagnesium

(b)

Figure 3 External pressure required for initial yielding when internal pressure 1198751

= 10 50 for isotropic (steel) and transversely isotropicmaterial (beryl and steel)

02 04 06 08 1R

100

200

300

400

500

Stre

sses

minus100

P1 = 10 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

P1 = 50 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 4 Transitional stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

20

40

60

80

100

120

Stre

sses

minus20

P1 = 0 P2 = 10

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

100

200

300

400

500

600

Stre

sses

minus100

P1 = 0 P2 = 50

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 5 Transitional stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials

The Scientific World Journal 7

02 04 06 08 1R

Stre

sses

minus20

minus40

minus60

minus80

minus100

minus120

minus140

P1 = 10 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(a)

02 04 06 08 1R

Stre

sses

minus100

minus200

minus300

minus400

minus500

minus600

minus700

P1 = 50 P2 = 0

120590r (beryl)120590120579 (beryl)120590r (magnesium)

120590120579 (magnesium)120590r (steel)120590120579 (steel)

(b)

Figure 6 Transitional stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

02 04 06 08 1R

50

100

150

Stre

sses

minus50

minus100

P1 = 10 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

ss

minus50

minus100

minus150

minus200

P1 = 50 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 7 Fully plastic stresses when internal pressure = 10 external pressure = 50 and internal pressure = 50 external pressure = 10respectively

02 04 06 08 1R

10

20

30

40

Stre

sses

minus10

minus20

P1 = 0 P2 = 10

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1

R

50

100

150

200

Stre

sses

minus50

minus100

P1 = 0 P2 = 50

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 8 Fully plastic stresses when external pressure 1198752= 10 and external pressure 119875

2= 50 (without internal pressure)

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials

8 The Scientific World Journal

02 04 06 08 1R

10

Stre

sses

minus10

minus20

minus30

minus40

minus50

P1 = 10 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(a)

02 04 06 08 1R

50

Stre

sses

minus50

minus100

minus150

minus200

minus250

P1 = 50 P2 = 0

120590r (steel)120590120579 (steel)

120590r (beryl and magnesium)120590120579 (beryl and magnesium)

(b)

Figure 9 Fully plastic stresses when internal pressure 1198751= 10 and internal pressure 119875

1= 50 (without external pressure)

Table 1 Elastic constants 119862119894119895used (in units of 1010 Nm2)

Materials 11986211

11986212

11986213

11986233

11986244

Steel (isotropic material) 2908 127 127 2908 0819Magnesium (transverselyisotropic material) 597 262 217 617 164

Beryl (transverselyisotropic material) 2746 098 067 469 0883

Table 2 The pressure required for initial yielding and fully plastic state for isotropic material (steel)

Steel (isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100604 084032 069384 0560183 599777 432757 320616 237358119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899396 915968 930616 943982 670895 397012 239035 140861119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

48994 491597 496062 494398 123158 073983 045126 026881119875119891

483906 48796 490837 493069

Table 3 The pressure required for initial yielding and fully plastic state for transversely isotropic material (beryl)

Beryl (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

100261 083806 069237 0559264 60525 436614 323412 239391119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

899739 916194 930763 944074 674451 39938 240577 141822119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489974 491619 493076 494407 123843 074428 045409 027063119875119891

483906 48796 490837 493069

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials

The Scientific World Journal 9

Table 4 The pressure required for initial yielding and fully plastic state for transversely isotropic material (magnesium)

Magnesium (transversely isotropic material)

Pressure 119875119894119875119891 119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

((119875119891minus 119875119894)119875119891) lowast 100

119886119887 = 02 119886119887 = 03 119886119887 = 04 119886119887 = 05

Effective pressure 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 0 119875119894

10598 087527 07164 0574248 518626 375544 279018 207052119875119891

160944 120397 091629 0693147

Internal pressure = 10 119875119894

89402 912473 92836 942575 614796 36023 215315 126144119875119891

839056 879603 908371 930685

Internal pressure = 50 119875119894

489402 491247 492836 494258 1123 066911 040561 024056119875119891

483906 48796 490837 493069Where 119875119894 and 119875119891 are pressures required for initial yielding fully plastic state while 119886 and 119887 are internal and external radius of thick-walled circular cylinder

Table 5 Circumferential stresses for different pressures for isotropic (steel) and transversely isotropic material (beryl and steel)

1198770

Transitional circumferential stresses Fully plastic circumferential stressesSteel Beryl Magnesium Steel Beryl Magnesium

1198751= 10

1198752= 50

0 193946 193358 197498 175754 118046 11804605 minus814052 minus815226 minus796563 minus10 minus677078 minus6770781 minus957177 minus956331 minus970019 minus4942 minus107128 minus107128

1198751= 50

1198752= 10

0 minus253946 minus253358 minus257498 minus235754 minus178046 minus17804605 214052 215226 196563 minus50 77078 770781 357177 356331 370019 minus1058 471278 471278

1198751= 0

1198752= 10

0 509865 508395 518744 464386 320116 32011605 minus178513 minus178807 minus174141 0 minus14427 minus144271 minus214294 minus214083 minus217505 minus9855 minus24282 minus24282

Where 1198751 and 1198752 are internal and external pressures

that the cylinder in which internal pressure is less (comparedto external pressure) circumferential stress is less for berylthan that of magnesium and steel It has been seen fromFigure 5 that in case of cylinder having internal pressuremorethan that of external pressure circumferential stresses arecompressible in nature and maximum at internal surfaceAlso it has been noted that for such type of cylinderscircumferential stresses are less for beryl as compared tomagnesium and steel It can be observed from Figure 6 thattransitional stresses are maximum at internal surface for thecylinder under external pressure only

From Table 5 and Figure 7 it can be seen that fullyplastic circumferential stresses are maximum at internalsurface for isotropic material as well as for transverselyisotropic material Also these stresses are high for steel ascompared to beryl and magnesium for the cylinder havinghigh external pressure as compared to internal pressure It hasalso been noted that stresses are approaching to compressivefrom tensile for the above cases Circumferential stressesare compressive in nature for the cylinder whose internalpressure is more than that of external pressure and thesestresses approaching towards tensile from compressive as canbe seen from Figure 7 It has been observed from Table 5and Figures 8 and 9 that fully plastic circumferential stressesare again maximum at the internal surface and approaching

towards compressive from tensile Also these stresses arehigh for steel as compared to beryl and magnesium It hasalso been observed that fully plastic stresses are less for thecylinder under external pressure only as compared to othercases of full plasticity

6 Conclusion

From the above analysis it can be concluded that circularcylinder under internal and external pressure made up oftransversely isotropic material (beryl) is on the safer side ofthe design as compared to the cylinder made up of isotropicmaterial (steel) as well as of transversely isotropic material(magnesium)Themain reason is that the percentage increasein effective pressure required for initial yielding to becomefully plastic is high for beryl as compared to steel andmagnesium which leads to the idea of ldquostress savingrdquo thatreduces the possibility of collapse of thick-walled cylinderdue to internal and external pressure

Nomenclature

119886 and 119887 Internal and external radii of cylinder119889 Constant119909 119910 119911 Cartesian coordinates

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials

10 The Scientific World Journal

119903 120579 119911 Cylindrical polar coordinates119890119894119895and 119879

119894119895 Strain and stress tensor

119862119894119895 Material constants

120590119903= (11987911990311990311986266) Radial stress

120590120579= (119879120579120579

11986266) Circumferential stress

119906 V 119908 Displacement components119877 Radial distance119890119894119894 First strain invariant

120573 Function of 119903 only119875 Function of 120573 only120582 and 120583 Lamersquos constants

119877 = (119903119887) 1198770= (119886119887)

120590119911= (119879119911119911

11986266) Axial stress

Conflict of Interests

The authors declare that there is no conflict of interest

References

[1] R B Hetnarksi and J IgnaczakMathematicalTheory of Elastic-ity Taylor amp Francis London UK 2004

[2] J Chakrabarty Applied Plasticity Springer Berlin Germany2000

[3] F P J Rimrott ldquoCreep of thickmdashwalled tube under inter-nal pressure considering large strainsrdquo Journal of AppliedMechanicsmdashTransactions ASME vol 26 pp 271ndash274 1959

[4] W Zhao R Seshadri and R N Dubey ldquoOn thick-walledcylinder under internal pressurerdquo Journal of Pressure VesselTechnologymdashTransactions of the ASME vol 125 no 3 pp 267ndash273 2003

[5] Y-S YooN-SHuh S Choi T-WKim and J-I Kim ldquoCollapsepressure estimates and the application of a partial safety factorto cylinders subjected to external pressurerdquoNuclear Engineeringand Technology vol 42 no 4 pp 450ndash459 2010

[6] J Perry and J Aboudi ldquoElasto-plastic stresses in thick walledcylindersrdquo Journal of Pressure Vessel TechnologymdashTransactionsof the ASME vol 125 no 3 pp 248ndash252 2003

[7] T E Davidson D P Kendall and A N Reiner ldquoResidualstresses in thick-walled cylinders resulting from mechanicallyinduced overstrainmdashPurpose of investigation is to determinethe residual-stress distribution as a function of magnitude ofoverstrain and diameter ratio and how it affects the reyieldingcharacteristics of cylinders autofrettage by sliding-wedged tech-niquerdquo ExperimentalMechanics vol 3 no 11 pp 253ndash262 1963

[8] B R Seth ldquoTransition conditions The yield conditionrdquo Inter-national Journal of Non-Linear Mechanics vol 5 no 2 pp 279ndash285 1970

[9] BN Borah ldquoThermo elasticmdashplastic transitionrdquoContemporaryMathematics vol 379 pp 93ndash111 2005

[10] B R Seth ldquoMeasure-concept in mechanicsrdquo InternationalJournal of Non-Linear Mechanics vol 1 no 1 pp 35ndash40 1966

[11] S Sharma M Sahni and R Kumar ldquoThermo elasticmdashplastictransition of transversely isotropic thickmdashwalled rotating cylin-der under internal pressurerdquo Advances in Theoretical andApplied Mechanics vol 2 no 3 pp 113ndash122 2009

[12] A K Aggarwal R Sharma and S Sharma ldquoSafety Analysisusing Lebesgue Strain Measure of Thick-Walled Cylinder forFunctionally Graded Material under Internal and External

Pressurerdquo The Scientific World Journal vol 2013 Article ID676190 10 pages 2013

[13] A K Aggarwal R Sharma and S Sharma ldquoSafety analysis ofthermal creep non-homogeneous thick-walled circular cylinderunder internal and external pressure using Lebesgue strainmeasurerdquoMultidiscipline Modelling in Materials and Structuresvol 9 no 4 pp 499ndash513 2013

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials

Submit your manuscripts athttpwwwhindawicom

ScientificaHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CorrosionInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Polymer ScienceInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CeramicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CompositesJournal of

NanoparticlesJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Biomaterials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

NanoscienceJournal of

TextilesHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Journal of

NanotechnologyHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

CrystallographyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CoatingsJournal of

Advances in

Materials Science and EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Smart Materials Research

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MetallurgyJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

BioMed Research International

MaterialsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Nano

materials

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal ofNanomaterials