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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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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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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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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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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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
top related