IJAAMMInt. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 (ISSN: 2347-2529)

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International Journal of Advances in Applied Mathematics and Mechanics

Stress analysis in strain hardening two-layer composite tubes subjectto cyclic loading of internal pressure

Research Article

Ahmet N. Eraslana, Tolga Akisb, ∗

a Department of Engineering Sciences, Middle East Technical University, 06531, Ankara, Turkeyb Department of Civil Engineering, Atılım University, 06836, Ankara, Turkey

Received 13 August 2015; accepted (in revised version) 11 November 2015

Abstract: Cyclically loaded two-layer composite tubes in the elastic and elastic-plastic stress states are examined in this study.An analytical model based on Tresca’s yield criterion, its associated flow rule, and linear strain hardening is devel-oped to estimate the stresses in the tube assembly with axially constrained ends. Then a comprehensive analysis isperformed for tubes having different dimensions under one cycle of loading, unloading, and reloading of internalpressure. This cycle demonstrates an autofrettage process in which a residual stress field is formed inside the tubeassembly.

MSC: 74A10 • 74C05

Keywords: Elastoplasticity • Composite tubes • Strain hardening • Tresca criterion • Cyclic loading© 2015 The Author(s). This is an open access article under the CC BY-NC-ND license (https://creativecommons.org/licenses/by-nc-nd/3.0/).

1. Introduction

Thick-walled tubes are commonly used structures that have a wide range of applications in engineering. Chemi-cal and nuclear industry, power plants, pipelines, and military equipment are the major areas of usage of these assem-blies. Due to having wide application areas in different branches of industry, the analysis and design of thick walledtubes, most of which are subjected to internal pressure, are important.

The problem of a single-layered tube under internal pressure was treated in purely elastic stress state by severalresearchers (Timoshenko [1], Timoshenko and Goodier [2], Ugural and Fenster [3], Boresi, Schmidt and Sidebottom[4]). On the other hand, the fully plastic solution of the problem was reported by Mendelson [5], Nadai [6] and Boresi,Schmidt and Sidebottom [4]. In addition to these studies, Durban and Kubi [7], Lazzarin and Livieri [8], Parker [9],and Perry and Aboudi [10] investigated the internally pressurized tube problem in elastic-plastic stress state. Amongthem, Lazzarin and Livieri [8], Parker [9], and Perry and Aboudi [10] also studied the autofrettage process of thick-walled tubes. Autofrettage is defined as a process that increases the elastic strength of the tube assembly by initiallysubjecting the vessel to high internal pressure so that plastic deformations occur in the tube [4]. This process alsoincreases the fatigue lifetime by slowing down the development and propagation of cracks [10].

There also exist studies in which internally pressurized multilayer composite tubes in the elastic and partiallyplastic states were investigated [11–13]. In recent years, studies focusing on tubes made of functionally graded mate-rials (FGM) were performed by several researchers both in elastic and elastic-plastic stress states. For example, Fukuiand Yamanaka [14], Horgan and Chan [15], Tutuncu and Ozturk [16], and Tutuncu [17] treated the internally pressur-ized FGM tube problem in elastic stress state. On the other hand, Eraslan and Akis [18] investigated the elastoplasticresponse of such tubes and Jahromi et al. [19, 20] studied the autofrettage of FGM pressure vessels. Besides thesestudies, the mechanical and thermal stresses in FGM tubes were studied by Jabbari, Sohrabpour and Eslami [21] andEraslan [22].

∗ Corresponding author.E-mail address: aeraslan@metu.edu.tr (Ahmet N. Eraslan), tolga.akis@atilim.edu.tr (Tolga Akis)

66 Stress analysis in strain hardening two-layer composite tubes subject to cyclic loading of internal pressure

Nomenclature

a,b,c Inner, interface, and outer radii of the tube assembly, respectivelyCi Integration constantsE Modulus of elasticityP Pressurer,θ, z Cylindrical coordinatesri Elastic-plastic border radiusu Radial displacementεi Strain componentsεEQ Equivalent plastic strain

η Hardening parameterν Poisson’s ratioσi Stress componentsσ0,σy Initial and subsequent yield stress

In the literature, there are also several studies in which thick-layered tubes under combined loading conditionsare analyzed. For example, Imaninejad and Subhash [23] investigated thick-walled cylinders subjected to internalpressure and proportional axial loading. Limam, Lee and Kyriakides [24] studied dented tubes under combined bend-ing and internal pressure. Eraslan and Apatay [25] investigated nonlinear strain hardening preheated tubes subjectedto internal pressure. Other examples of different types of loading including combined ones on the tubes and otherbasic structures can be found in [26–31].

In the present work, we extend the internally pressurized two-layer composite tube problem to a cyclicallyloaded two-layer tube in the elastic and elastic-plastic stress states. A detailed analysis is performed for these tubesunder one cycle of loading, unloading, and reloading of internal pressure. This loading cycle under considerationdemonstrates an autofrettage process in which a residual stress field is formed inside the tube assembly. In the liter-ature, there are few studies on the cyclic behavior of pressurized tubes. Mahbadi and Eslami [32] studied the cyclicloading behavior of single-layer thick-walled tubes under different types of loading including internal pressure andthey used a numerical iterative method in their study. Megahed and Abbas [33] studied the influence of reverse yield-ing on the residual stresses that developed after autofrettage of a single-layer tube. Darijani, Kalgarnovin and Naghd-abadi [34] obtained closed-form solutions of the internally pressurized elastoplastic tube problem in which forwardand reversed loading is considered. In contrast to those studies, in our work two-layer composite thick-walled tubesare analyzed in which the locations of the yielding along the tube assembly depend on the tube dimensions. The so-lutions obtained in this study may be used in the analysis and design of the two-layer tube assemblies under internalpressure. In addition, the results of this analytical work may be used as benchmark problems for numerical solutions.

As mentioned above, the geometry considered here consists of two tightly fitted concentric tubes. A long tube ofinner radius a and outer radius b is placed in a tube of the same length and of inner radius b and outer radius c. Theassembly is then constrained axially and at this stage it is stress-free. As the internal pressure is applied, stresses areformed in both of the tubes and increase in magnitude with increasing pressures. Depending on the tube dimensions,the yielding commences at the inner surface, at the interface, or at both locations at a critical value of pressure P = Pe .For P > Pe , the assembly will consist of plastic and elastic regions. When the pressure load is removed, the plasticstrains will be frozen and the assembly will have elastic behavior when it is reloaded. In the framework of smalldeformations, a state of plane strain and Tresca’s yield condition analytical solutions are obtained for the two-layertubes under the above-mentioned load cycle (loading-unloading-reloading).

2. Formulation

2.1. All elastic

Cylindrical polar coordinates (r,θ, z) are considered. A state of plane strain (εz = 0) and infinitesimal deforma-tions are presumed. Furthermore, the notation and basic equations of elasticity as given by Timoshenko and Goodier[2] are used. The strain-displacement relations: εr = du/dr , εθ = u/r , the equation of equilibrium in the radial direc-tion

dσr

dr+ σr −σθ

r= 0, (1)

Ahmet N. Eraslan, Tolga Akis / Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 67

and generalized Hooke’s law

εr = εpr + 1

E[σr −ν (σθ+σz )] , (2)

εθ = εpθ+ 1

E[σθ−ν (σr +σz )] , (3)

εz = εpz + 1

E[σz −ν (σr +σθ)] , (4)

form the basis for the present analysis and are valid in both elastic (with plastic strains εpi = 0) and plastic regions. In

the equations above, σ j represents the normal stress, r the radial coordinate, ε j the normal strain, E the modulus ofelasticity, ν the Poisson’s ratio, and u the radial component of the displacement vector. For purely elastic deformationsof a tube, the axial stress is obtained from Eq. (4) as

σz = ν (σr +σθ) . (5)

The axial stressσz may be eliminated from the radial and circumferential strain expressions and results are substitutedin the strain-displacement relations to obtain stress-displacement relations. Substituting the stresses in the equationof equilibrium (1), one obtains a differential equation in radial displacement u with the solution

u(r ) = C1

r+C2r, (6)

where Ci represents an arbitrary integration constant. The stresses are then determined as

σr (r ) = E

1+ν

[−C1

r 2 + C2

1−2ν

], (7)

σθ(r ) = E

1+ν

[C1

r 2 + C2

1−2ν

], (8)

σz (r ) = 2νEC2

(1+ν)(1−2ν). (9)

Eqs. (6)-(9) are nothing but the elastic equations of a homogeneous tube with fixed ends. For a two-layer compositetube assembly, the same stress and displacement expressions are valid. However, these expressions contain four un-known integration constants to be determined: C I

1 and C I2 for the inner tube and C I I

1 and C I I2 for the outer tube. The

elastic solution of composite tubes subjected to internal pressure follows.

2.2. Onset of yield

Using subscript 1 to mark material properties (E , ν and σ0 (yield stress)) of the inner and 2 for the outer tubeand superscripts I and I I to mark inner and outer tubes, respectively, the boundary conditions for a tube assemblysubject to internal pressure and a traction-free outer surface can be stated as σI

r (a) =−P and σI Ir (c) = 0. In addition,

at the interface of the tubes radial stresses and displacements are equal and hence σIr (b) = σI I

r (b), u I (b) = u I I (b).Application of these four conditions results in

C I1E1 = a2b2P N1[b2(E1M2 −E2M1)+ c2(E1N2 +E2M1)]

E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1), (10)

C I2E1 = a2P M1[b2(E1M2 +E2N1)+ c2(E1N2 −E2N1)]

E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1), (11)

C I I1 = a2b2c2P N2(M1 +N1)

E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1), (12)

C I I2 = a2b2P M2(M1 +N1)

E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1), (13)

where

N1 = (1+ν1), M1 = N1(1−2ν1), N2 = (1+ν2), M2 = N2(1−2ν2). (14)

Like in homogeneous ones, the stress state in the two-layer tube assemblies under internal pressure satisfies σθ >σz >σr throughout. However, unlike the deformation behavior of a single tube, in a two-layer composite tube underinternal pressure different modes of plastic flow may take place. According to the Tresca’s yield criterion σθ−σr =σ0,plastic deformation may first begin in the inner tube at r = a, or in the outer tube at the interface r = b. These twodifferent modes imply the existence of a critical interface radius b = bC at which yielding commences simultaneouslyin both tubes at the inner surface (r = a) and at the interface (r = bC ). This critical radius may be determined from

σIθ

(a)−σIr (a)

σ01= σI I

θ(b)−σI I

r (b)

σ02. (15)

68 Stress analysis in strain hardening two-layer composite tubes subject to cyclic loading of internal pressure

After some algebraic manipulations, we get

bC =

c√σ02

[c2σ02(E1N2 +E2M1)2 +4a2E2σ01(E1M2 −E2M1) (M1 +N1)

]2σ02(E1M2 −E2M1)

+c2(E1N2 +E2M1)

2(E2M1 −E1M2)

}1/2

. (16)

The existence of the critical interface radius depends mainly on the material properties of the tubes. This subject isdiscussed comprehensively in [13]. If the critical interface radius exists, the plastic flow will start at the inner surfacefor the values of b ≥ bC and at the interface of the assembly for b < bC . The nondimensional elastic limit pressure thatcauses yielding at the inner surface r = a, which is determined from σI

θ(a)−σIr (a) =σ01, is obtained as

P e = Pe

σ01= E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1)

2b2[E1(b2M2 + c2N2)+E2M1(c2 −b2)]. (17)

If the interface radius is less than bC , yielding begins at the interface according to σI Iθ (b)−σI I

r (b) = σ02. The elasticlimit becomes

P e = σ02[E1(b2 −a2)(b2M2 + c2N2)+E2(c2 −b2)(a2N1 +b2M1)]

2a2c2E2σ01(M1 +N1). (18)

3. Elastic-Plastic solution

As stated before, for composite tubes under internal pressure, the stresses satisfy σθ > σz > σr and the yieldcondition is σy = σθ −σr . The associated flow rule and the equivalence of increment of plastic work give ε

pθ= −εp

r ,

εpz = 0 and ε

pθ= εEQ with εEQ , being the equivalent plastic strain. On the other hand, for a linear strain material the

hardening law can be expressed as

σy =σ0(1+ηεEQ ), (19)

where η represents the hardening parameter. Manipulating strain-displacement relations, Hooke’s law, and plasticstrain relations, one may obtain the following stress-displacement relations:

σr =− σ0

2+H(1+ν)+ E

[2+H(1+ν)](1+ν)(1−2ν)

{[1+Hν(1+ν)]u

r+ [1+H(1−ν2)]

du

dr

}, (20)

σθ =σ0

2+H(1+ν)+ E

[2+H(1+ν)](1+ν)(1−2ν)

{[1+H(1−ν2)]u

r+ [1+Hν(1+ν)]

du

dr

}, (21)

where H = ησ0/E represents the nondimensional hardening parameter. The governing differential equation for thisplastic region is obtained by combining Eqs. (20) and (21) and the equation of equilibrium (1). The result is

r 2 d 2u

dr 2 + rdu

dr−u = 2σ0(1+ν)(1−2ν)

E [1+H(1−ν2)]r. (22)

The general solution is

u(r ) = C3

r+C4r + σ0(1+ν)(1−2ν)(2lnr −1)r

2E [1+H(1−ν2)], (23)

and as a result

σr (r ) =− E HC3

[2+H(1+ν)]r 2 + EC4

(1+ν)(1−2ν)+ (2lnr −1)σ0

2[1+H(1−ν2)], (24)

σθ(r ) = E HC3

[2+H(1+ν)]r 2 + EC4

(1+ν)(1−2ν)+ (2lnr +1)σ0

2[1+H(1−ν2)], (25)

σz (r ) = 2EνC4

(1+ν)(1−2ν)+ 2ν lnrσ0

1+H(1−ν2), (26)

εpθ=−εp

r = 2C3

[2+H(1+ν)]r 2 − (1−ν2)σ0

E [1+H(1−ν2)]. (27)

Ahmet N. Eraslan, Tolga Akis / Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 69

4. Elastic reloading

As soon as the tube assembly is completely unloaded, the plastic strains are frozen and the assembly behavespurely elastic again. Since the plastic strain distribution is known, the stress distributions upon reloading can becalculated considering the plastic strains on the right-hand side as permanent plastic strains.

The elastic strains are related to the stresses via Hooke’s law,

εr = 1

E[σr −ν (σθ+σz )]+ε

pr , (28)

εθ =1

E[σθ−ν (σr +σz )]+ε

pθ

, (29)

εz = 0 = 1

E[σz −ν (σr +σθ)]+ε

pz , (30)

where εpi represents the permanent plastic strains after unloading of the internal pressure. Since εp

z = 0, the axial stressis the same as that given by Eq. (5). The plastic strain expressions in Eq. (27) can be written as

εpr =− A

r 2 +B ; εpθ= A

r 2 −B , (31)

where

A = 2C3

2+H(1+ν), (32)

B = (1−ν2)σ0

E [1+H(1−ν2)]. (33)

The axial stressσz may be eliminated from the radial and circumferential strain expressions given in Eqs. (28) and (29)and results, and the plastic strains given in Eq. (31) are substituted in the strain-displacement relations εr = du/dr ,εθ = u/r to obtain the following stress displacement relations:

σr = E

(1+ν)(1−2ν)

[νu

r+ (1−ν)

du

dr

]+ E

1+ν

[A

r 2 −B

], (34)

σθ =E

(1+ν)(1−2ν)

[(1−ν)u

r+ν

du

dr

]− E

(1+ν)

[A

r 2 −B

]. (35)

The governing differential equation is obtained by substitution of these stress expressions in the equation of equilib-rium (Eq. (1)). The result is

r 2 d 2u

dr 2 + rdu

dr−u = 2B(1−2ν)r

1−ν. (36)

The general solution is derived as

u(r ) = C5

r+C6r − B(1−2ν)(1−2lnr )r

2(1−ν). (37)

and as a result

σr = E

1+ν

[−C5

r 2 + C6

1−2ν

]+ E

1+ν

[A

r 2 − B(1−2lnr )

2(1−ν)

], (38)

σθ =E

1+ν

[C5

r 2 + C6

1−2ν

]− E

1+ν

[A

r 2 − B(1+2lnr )

2(1−ν)

], (39)

σz = 2EνC6

(1+ν)(1−2ν)+ 2EνB lnr

1−ν2 . (40)

5. Numerical results

A composite system consisting of steel inner (E = 200 GPa, ν= 0.3, σ0 = 430 MPa, H = 0.5) and aluminum outer(E = 70 GPa, ν = 0.35, σ0 = 100 MPa, H = 0.25) tubes is considered. In addition, for the presentation of numericalresults the following nondimensional and normalized quantities are used:

r = r

c; σ j =

σ j

σ01; u = uE1

σ01c; εp

j =ε

pj E1

σ01; P = P

σ01. (41)

70 Stress analysis in strain hardening two-layer composite tubes subject to cyclic loading of internal pressure

0.0

0.2

0.4

0.6

0.8

1.0

1.2

str

esses a

nd d

ispla

cem

ent

rσσ

θ−

rσσ

θ−

θσ

θσ

zσ

zσ

u

ST AL

rσ

-0.4

-0.2

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

70.0=a 886638.0=Cb

Fig. 1. Elastic response of steel-aluminum tube assembly of inner radius a = 0.7 subjected to the elastic limit internal pressureP = P e = 0.213214 for b = bC = 0.886638.

Here we also define the following dimensionless integration constants:

C 1 =C1/c2 ; C 2 =C2 ; C 3 =C3/c2 ;

C 4 =C4 ; C 5 =C5/c2 ; C 6 =C6. (42)

In order to obtain the unknown integration constants, the following boundary and interface conditions are valid:Boundary conditions: 1. σr (a) =−P ; 2. σr (c) = 0.Interface conditions: 1. σI

r (b) =σI Ir (b); 2. u I (b) = u I I (b).

In addition, for a tube having elastic and plastic deformations, the following conditions should be taken into consid-eration:

1. σpr (ri ) =σe

r (ri ); 2. σpθ

(ri ) =σeθ(ri ); 3. up (ri ) = ue (ri ).

Here, ri is defined as the plastic-elastic border radius at the inner (i = 1) or outer tube (i = 2) and the superscriptse and p represent the plastic and elastic regions, respectively. For example, if the inner tube has plastic and elastic

regions, the conditions that should be used are: 1. σpIr (r1) =σeI

r (r1); 2. σpIθ

(r1) =σeIθ (r1); and 3. upI (r1) = ueI (r1). The

superscripts eI and pI are used to indicate the plastic and elastic regions in the inner tube. For the determination ofthe integration constants of the plastically predeformed tube assembly subjected to the reloading of the internal pres-sure, the general boundary and interface conditions given above are valid. Additionally, at the plastic-elastic interfaceof the tube, the following conditions should be used: 1. σr

r (ri ) =σer (ri ); 2. ur (ri ) = ue (ri ). Here, the superscript r rep-

resents the reloading region. Finally, the solution data for which the figures are drawn are given in the Appendix. Thevalues used in the solution and the parameters that are evaluated during calculations are presented for each figure inthis part.

First, the inner radius is taken as a = 0.7. By virtue of Eqs. (16) and (17), the critical interface radius and thecorresponding critical elastic limit internal pressure are calculated as bC = bC /c = 0.886638 and P e = 0.213214, re-spectively. If these values are substituted in Eqs. (10)-(13), the unknown integration constants can be obtained. (The

calculated integration constants CI1, C

I2, C

I I1 ,and C

I I2 are given in the Appendix-Fig. 1). Fig. 1 shows the consequent

stresses and displacement. As seen in this figure, σθ(a)−σr (a) = 1 and σθ(bC )−σr (bC ) = 1, which implies the forma-tion of plastic regions at the inner surface and at the interface of the tube assembly. It should also be noted for thisfigure that the radial stress and displacement are continuous at the interface satisfying interface conditions.

For the values of pressure P > P e the composite tube becomes partially plastic, consisting of an inner plasticregion in a ≤ r ≤ r1, an inner elastic region in r1 ≤ r ≤ bC , an outer plastic region in bC ≤ r ≤ r2, and an outer elasticregion in r2 ≤ r ≤ c. The unknowns to be determined for the solution are: C I

3 , C I4 (inner plastic), C I

1 , C I2 (inner elastic),

C I I3 , C I I

4 (outer plastic), C I I1 , C I I

2 (outer elastic), and the unknown plastic elastic border radii r1 and r2. The evolution

Ahmet N. Eraslan, Tolga Akis / Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 71

of the plastic regions is shown in Fig. 2(a). As seen in this figure, both plastic regions expand with increasing P , and atP = P 1 = 0.257815 the outer tube becomes fully plastic since r2 = c. The stresses in partially plastic state for this tubeassembly may be observed by considering an internal pressure P in Pe < P < P1.

For the tube assembly considered (a = 0.7), assigning b = 0.90 > bC and using Eq. (17) the elastic limit internalpressure is obtained as P e = 0.218906. In Fig. 2(b), the expansion of plastic regions in the pressurized tube is given.Plastic deformation commences at the inner surface as soon as P = P e = 0.218906. As this plastic region spreads intothe assembly as P increases, another critical limit, P = P 1 = 0.226060, is reached at which the outer tube begins toyield at r = b based on σθ −σr =σ0 as well. Thereafter, both plastic regions expand with increasing internal pressureand at P = P 2 = 0.266289 the outer tube becomes fully plastic while the inner tube has plastic and elastic regions.

Assigning the interface radius b = 0.80 < bC and using Eq. (18), the elastic limit pressure is obtained as P e =0.135444. Fig. 2(c) shows the expansion of plastic regions in the tube assembly. The yielding begins at the interface assoon as P = P e = 0.135444 is reached. As this plastic region spreads into the assembly as P increases, another criticallimit, P = P 1 = 0.166937, is reached at which the inner tube begins to yield at r = a based on σθ −σr = σ0 as well.Thereafter, both plastic regions expand with increasing internal pressure and at P = P 2 = 0.191958 the outer tubebecomes fully plastic while the inner tube has plastic and elastic regions.

In order to show the distributions of stresses, plastic strains, and displacement, the same assembly (a = 0.7)with b = bC = 0.886638 is considered. For P = 0.24 < P 1 the distributions of stress and plastic strain components, andradial displacement are plotted in Fig. 3(a). The corresponding residual stresses and displacement upon completeremoval of the load P 1 = 0.24 are displayed in Fig. 3(b). In order to determine the elastic response of the plasticallypredeformed tube assembly subjected to the reloading of the internal pressure, four regions should be considered: aninner reloading region in a ≤ r ≤ r1, an inner elastic region in r1 ≤ r ≤ bC , an outer reloading region in bC ≤ r ≤ r2, andan outer elastic region in r2 ≤ r ≤ c. Eight unknown integration constants are determined: C I

5 and C I6 (inner reloading

region), C I1 and C I

2 (inner elastic region), C I I5 and C I I

6 (outer reloading region), and C I I1 and C I I

2 (outer elastic region).For this case, the elastic limit reloading pressure is calculated as P er = 0.230293 and the corresponding stresses anddeformation are plotted in Fig. 3(c). As seen in this figure, plastic deformation caused by the reloading of internalpressure commences at the inner tube r = a according to the same yield condition σy =σθ−σr .

At the critical pressure P = P 1 = 0.257815, the outer tube becomes fully plastic and the assembly contains aninner plastic, an inner elastic, and an outer plastic region. The corresponding distributions of stress, displacement,and plastic strains are shown in Fig. 4(a). Furthermore, the residual stresses and displacement upon complete removalof the fully plastic load P 1 = 0.257815 is displayed in Fig. 4(b). For the plastically predeformed tube assembly the elasticlimit reloading pressure is calculated as P er = 0.239450 and the corresponding stresses and deformation are plottedin Fig. 4(c). Similar to the previous reloading case, yielding first begins at the inner surface of the tube assembly.

72 Stress analysis in strain hardening two-layer composite tubes subject to cyclic loading of internal pressure

0.22

0.23

0.24

0.25

0.26

0.27

pre

ssure

ST AL

1r

2r

257815.01=P

213214.0=P

0.20

0.21

0.70 0.75 0.80 0.85 0.90 0.95 1.00

border radius

70.0=a

213214.0=eP

886638.0=Cb

0.22

0.23

0.24

0.25

0.26

0.27

0.28

pre

ssure

ST AL

218906.0=eP

226060.01=P

266289.02=P

0.20

0.21

0.70 0.75 0.80 0.85 0.90 0.95 1.00

border radius

90.0=b70.0=a

(a) (b)

0.14

0.15

0.16

0.17

0.18

0.19

0.20

0.21

pre

ssure

ST AL

166937.01=P

191958.02=P

1r

2r

0.12

0.13

0.70 0.75 0.80 0.85 0.90 0.95 1.00

border radius

135444.0=eP

70.0=a 80.0=b

(c)

Fig. 2. Expansion of plastic regions with increasing pressure in the steel-aluminum tube assembly (H1 = 0.5, H2 = 0.25) of innerradius a = 0.7 subjected to internal pressure for (a) b = bC = 0.886638, (b) b = 0.9, (c) b = 0.8.

Ahmet N. Eraslan, Tolga Akis / Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 73

0.0

0.2

0.4

0.6

0.8

1.0

1.2

str

esses, pla

stic s

train

s a

nd d

ispla

cem

ent rσσ

θ− rσσ

θ−

θσ

θσ

zσ

zσ

u

ST AL

rσ

1r

2r

p

θε

p

rεp

ε

p

θε

-0.4

-0.2

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

p

rε

70.0=a 886638.0=Cb

-0.06

-0.04

-0.02

0.00

0.02

0.04

resid

ual str

esses a

nd d

ispla

cem

ent

0

θσ

0

θσ

0

zσ 0

zσ

0

rσ

0u

ST AL

0

rσ

0u

-0.10

-0.08

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

70.0=a 886638.0=Cb

(a) (b)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

str

esses a

nd d

ispla

cem

ent

rσσ

θ− r

σσθ−

θσ

θσ

zσ

zσ

u

ST AL

rσ

-0.4

-0.2

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

70.0=a 886638.0=Cb

(c)

Fig. 3. (a) Partially plastic response of the steel-aluminum tube assembly of inner radius a = 0.7 subjected to the internal pressureP = 0.24 for b = bC = 0.886638, H1 = 0.5, and H2 = 0.25, (b) residual stresses and displacement upon complete unloading, (c)elastic response of the plastically predeformed assembly subjected to internal pressure under the reloading elastic limitP = P er = 0.230293.

74 Stress analysis in strain hardening two-layer composite tubes subject to cyclic loading of internal pressure

0.0

0.2

0.4

0.6

0.8

1.0

1.2

str

esses, pla

stic s

train

s a

nd d

ispla

cem

ent

rσσθ−

rσσθ−

θσ

θσ

zσ

zσ

u

ST AL

rσ

p

θε

p

θε

p

rε

pε

1r

-0.4

-0.2

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

p

rε

70.0=a 886638.0=Cb

-0.12

-0.08

-0.04

0.00

0.04

0.08

resid

ual str

esses a

nd d

ispla

cem

ent

0

θσ

0

θσ

0

zσ 0

zσ

0

rσ

0u

ST AL

0

rσ

0u

-0.16

-0.12

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

70.0=a 886638.0=Cb

(a) (b)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

str

esses a

nd d

ispla

cem

ent

rσσ

θ−

rσσ

θ−

θσ

θσ

zσ

zσ

u

ST AL

rσ

-0.4

-0.2

0.70 0.75 0.80 0.85 0.90 0.95 1.00

radial coordinate

70.0=a 886638.0=Cb

(c)

Fig. 4. (a) Partially plastic response of the steel-aluminum tube assembly of inner radius a = 0.7 subjected to the internal pressureP = P 1 = 0.257815 for b = bC = 0.886638, H1 = 0.5, and H2 = 0.25, (b) residual stresses and displacement upon completeunloading, (c) elastic response of the plastically predeformed assembly subjected to internal pressure under the reloading elasticlimit P = P er = 0.239450.

Ahmet N. Eraslan, Tolga Akis / Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 65 – 76 75

6. Concluding remarks

An analytical model, based on Tresca’s yield criterion, its associated flow rule, and linear hardening, is devel-oped to estimate the elastic, elastic-plastic, and elastic reloading behavior of two-layer concentric tubes with axiallyconstrained ends. Analytical expressions for the critical values of pressure leading to onset of yield for both elastic andreloading elastic cases are obtained for different tube dimensions. It is found that the elastic limit for the plasticallypredeformed tube assembly is higher than that for the undeformed tube assembly.

References

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Appendix

Solution Data :Fig. 1: a = 0.7, b = 0.886638, P = 0.213214, C

I1 = 6.84775× 10−4, C

I2 = 3.20626× 10−4, C

I I1 = 7.58051× 10−4, C

I I2 =

2.27415×10−4.

Fig. 2(a): a = 0.7, b = 0.886638, H1 = 0.5, H2 = 0.25.

Fig. 2(b): a = 0.7, b = 0.9, H1 = 0.5, H2 = 0.25.

Fig. 2(c): a = 0.7, b = 0.8, H1 = 0.5, H2 = 0.25.

Fig. 3 a = 0.7, b = 0.886638, H1 = 0.5, H2 = 0.25, P = 0.24, (a) CI3 = 1.00014×10−3, C

I4 = 5.90196×10−4, r 1 = 0.749230,

CI1 = 7.84480×10−4, C

I2 = 3.68356×10−4, C

I I3 = 1.08112×10−3, C

I I4 = 2.85347×10−4, r 2 = 0.948570, C

I I1 = 8.67650×10−4,

CI I2 = 2.60295×10−4, (b) C

I1 = 7.70802×10−4, C

I2 = 3.60906×10−4, C

I I1 = 8.53283×10−4, C

I I2 = 2.55985×10−4, (c) C

I5 =

9.68969×10−4, CI6 = 5.75599×10−4, C

I1 = 7.53305×10−4, C

I2 = 3.53759×10−4, C

I I5 = 1.04661×10−3, C

I I6 = 2.74994×10−4,

CI I1 = 8.33139×10−4, C

I I2 = 2.49942×10−4.

Fig. 4 : a = 0.7, b = 0.886638, H1 = 0.5, H2 = 0.25, P = 0.257815, (a) CI3 = 1.11350× 10−3,C

I4 = 5.92977× 10−4, r 1 =

0.790550, CI1 = 8.73395×10−4, C

I2 = 4.12386×10−4, C

I I3 = 1.20155×10−3, C

I I4 = 2.89287×10−4, (b) C

I1 = 8.28018×10−4,

CI2 = 3.87696× 10−4, C

I I1 = 9.16621× 10−4, C

I I2 = 2.74986× 10−4, (c) C

I5 = 1.05452× 10−3, C

I6 = 5.65360× 10−4, C

I1 =

8.14413×10−4, CI2 = 3.84769×10−4, C

I I5 = 1.13625×10−3, C

I I6 = 2.69698×10−4.

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