1540 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBAghiba/lucrari/GhibaBulgariuD... · 2013-07-26 ·...

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Transcript of 1540 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBAghiba/lucrari/GhibaBulgariuD... · 2013-07-26 ·...

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1540 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

by prescribing the corresponding resultant force and resultant moment. Using themethod introduced by Toupin in classical elasticity, Batra and Yang [1] have provedthat these changes of the ends conditions produce negligible effect, except possiblynear the ends.

In this paper we study the deformation of right cylinders made of an inhomoge-neous anisotropic material with voids subjected to a prescribed thermal field. Thepresent results complete the results established by Ghiba [9] in the isothermal caseand show how the thermal field which is linear in the axial coordinate influencesthe deformation of the cylinder. The results obtained in [9] are established usingthe method described in the books [14, 21] and in the papers [3, 4]. This methodgives the possibility to reduce the Saint-Venant’s problem to some generalized planestrain problems. In fact, in the paper [9], two classes of semi-inverse solutions weredescribed in the set of solutions of Saint-Venant’s problem that may be expressedin terms of solutions of some generalized plane strain probleme. These classes arerelevant in obtaining a semi-inverse solution to the relaxed Saint-Venant’s problem.

The influence of the thermal field in the deformation of the cylinder filled withisotropic porous materials was studied by Iesan [19], while the case of orthotropicporous materials was considered in [20]. In these papers the thermal field wasconsidered to be independent of the axial coordinate.

Our purpose is to study the deformation of the cylinders made by anisotropicmaterials when the thermal field is linear in the axial coordinate. We show that thisthermal field produces extension, bending, torsion, flexure and a plane deformation.When the thermal field is independent of the axial coordinates we can see that thethermal field influences the deformation but produces only extension, bending andtorsion (see also [19, 20]).

We outline that a study of Saint-Venant’s problem for homogeneous and isotropicporous elastic cylinders was presented by Dell’Isola and Batra [7]. The semi-inversemethod used in the present paper, was employed in order to study the Saint-Venant’sproblem for different type of materials in the papers [4, 5, 8, 16, 18, 19, 23, 20, 25,22, 2].

2. Formulation of the problem. We consider a right cylinder of length L, occu-pied by an inhomogeneous anisotropic porous material. We denote by B the interiorof cylinder, by ∂B the boundary of B and by D ⊂ R2 the interior of the boundedcross section. The boundary ∂D of cross section is assumed sufficiently smooth toadmit application of the divergence theorem in the plane cross section. We choosea rectangular cartesian system Ox1x2x3 so that the Ox3 axis is parallel with thegenerator of the cylinder and one of the ends lies in the x1Ox2 plane. The lateralboundary of the cylinder is Π = ∂D × (0, L) and D0 and DL are, respectively, thecross section located at x3 = 0 and x3 = L.

The Latin subscripts and superscripts are understood to range over the integers1, 2, 3, unless we specify else, whereas Greek subscripts and superscripts are confinedto the range 1, 2; summation over repeated subscripts is implied and comma fol-lowed by a subscript to denote partial derivative with respect to the correspondingcartesian coordinate; where no confusion may occur, we suppress the dependenceupon the spatial variables.

Let u be the displacement field over B, θ describes the changes of the temperaturefrom the reference configuration and ψ is the volume distribution function. In thefollowing we consider that the thermal field is prescribed. We denote by U the four

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ON THE THERMAL STRESSES IN ANISOTROPIC POROUS CYLINDERS 1541

dimensional vector (ui, ψ). The linear strain measure eij is given by

eij(U) =1

2(ui,j + uj,i). (1)

The components of the stress tensor, the components of the equilibrated stressvector and the intrinsic equilibrated body force [6] are

tij(U) = Aijrsers +Dijrψ,r +Aijψ − βijθ,hi(U) = Drsiers +Dijψ,j + diψ − ξiθ,g(U) = −Arsers − diψ,i −mψ + ζθ,

(2)

where Aijrs, Aij , Dij , Dijk, di, βij ,m, ζ and ξi are the constitutive constants whichsatisfy the symmetry relations

Aijrs = Arsij = Ajirs, Aij = Aji, Dij = Dji, Dijr = Djir, βij = βji. (3)

In this paper we assume that the porous materials are cross-section inhomogeneous.Thus, we have

Aijrs = Aijrs(x1, x2), Dijr = Dijr(x1, x2), Aij = Aij(x1, x2),

Dij = Dij(x1, x2), di = di(x1, x2), m = m(x1, x2),

βij = βij(x1, x2), ξi = ξi(x1, x2), ζ = ζ(x1, x2).

(4)

This type of anisotropy was used in the study of the deformation of functionallygraded porous elastic cylinders by Iesan and Scalia [18] and by Iesan [19, 20].

The surface force and the equilibrated surface force at a regular point of ∂B, aregiven by

ti(U) = tij(U)nj , h(U) = hj(U)nj , (5)

respectively, where nj are the components of the outward unit normal to ∂B.In the absence of the body force and the extrinsic equilibrated body force, the

equilibrium equations are

tji,j(U) = 0, hi,i(U) + g(U) = 0 in B. (6)

The cylinder is assumed to be free from lateral loading, so that the conditionson the lateral surface are

ti(U) = 0, h(U) = 0 on Π. (7)

We consider that the loading on the end D0 is statically equivalent to a force Rand a moment M. Then, for x3 = 0, we have the conditions∫

D

t3i(U)da=−Ri,∫D

εijkxjt3k(U)da=−Mi. (8)

From the existence results concerning the equilibrium problem [17], we must havesimilar conditions on the end DL.

The main aim of this paper is to find the solution of the following problems:(P1): extension–bending–torsion (Rα = 0), when the temperature distribution donot depends of the axial coordinate;(P2): extension–bending–torsion–flexure, when the temperature distribution is lin-ear in the axial coordinate.

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1542 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

The internal energy density associated to the solution of the boundary valueproblem defined by the equations (6) and the boundary condition (7) and (8) isdefined by

W =1

2Aijrseijers +

1

2Dijψ,iψ,j +

1

2mψ2+

+Dijreijψ,r +Aijeijψ + diψ,iψ.(9)

Throughout this paper we assume that the internal energy density is positive defi-nite.

Following Iesan [12, 14, 19, 20] we define the state of generalized thermoelasticplane strain for the interior of the cross section domain, D ⊂ R2, to be the statein which the displacement field w, the volume distribution ω and the temperaturefield τ depend only on x1 and x2

wi = wi(x1, x2), ω = ω(x1, x2), τ = τ(x1, x2), (x1, x2) ∈ D. (10)

For every state of generalized thermoelastic plane strain W=(w(x1, x2), ω(x1, x2))and temperature field τ(x1, x2), (x1, x2) ∈ D, we define the operators

Ti(W, τ) ≡[Aαiβsws,β +Dαiβω,β +Aαiω − βαiτ

],α,

P(W, τ) ≡[Dβsαws,β +Dαβω,β + dαω − ξατ

],α

−[Aβjwj,β + dβω,β +mω − ζτ

]in D,

Ai(W, τ) ≡[Aαiβsws,β +Dαiβω,β +Aαiω − βαiτ

]nα,

C(W, τ) ≡[Dβsαws,β +Dαβω,β + dαω − ξατ

]nα on ∂D.

(11)

In what follows, by using the semi-inverse method [12, 14, 3, 4], we construct thesolutions of the problems (P1) and (P2) defined above.

3. The solution of the problem (P1). The problem (P1) consists in finding asolution U0 = (u0, ψ0) of the problem defined by the equations (6) and the boundarycondition (7) and (8) when the temperature θ is given and it is independent of theaxial coordinate, i.e.

θ = T0(x1, x2) (12)

and Rα = 0.In view of the results established in [19, 20, 9], we search a solution of the problem

(P1) to be

u0α = −1

2aαx

23 − ε3αβa4xβx3 + wα(x1, x2),

u03 = (a1x1 + a2x2 + a3)x3 + w3(x1, x2),

ψ0 = ω(x1, x2),

(13)

where as, s = 1, 2, 3, 4, are for now arbitrary constants. The vector W = (w, ω) isdefined by

W =

4∑s=1

asW(s) (14)

where W(s) = (w(s), ω(s)), s = 1, 2, 3, 4, are solutions of the problems, denoted by(∆s) and characterized by the equations

Ti(W(s), T0) + f

(s)i = 0, P(W(s), T0) + h(s) = 0, (15)

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ON THE THERMAL STRESSES IN ANISOTROPIC POROUS CYLINDERS 1543

and the boundary conditions

Ai(W(s), T0) = T

(s)i , C(W(s), T0) = Π(s), (16)

with

f(γ)i = (Aαi33xγ),α, f

(3)i = (Aαi33),α,

f(4)i = (ε3βρAαi3ρxβ),α,

h(γ) = (D33αxγ),α −A33xγ , h(3) = (D33α),α −A33,

h(4) = (ε3βρD3ραxβ),α − ε3βαA3αxβ ,

T(γ)i = −Aαi33xγnα, T

(3)i = −Aαi33nα,

T(4)i = −ε3βρAαi3ρxβnα,

Π(γ) = −D33αxγnα, Π(3) = −D33αnα,

Π(4) = −ε3βρD3ραxβnα.

(17)

According to the existence results presented in [15], the above generalized ther-moelastic plane strain problem has a unique solution.

In the following we assume that the solutions of the generalized thermoelasticplane strain problems are known. It is easy to see that the solution U0 can bewritten in the form

U0 =

4∑s=1

asU(s), (18)

where U(s) = (u(s), ω(s)), with

u(γ)α = −1

2x23δαγ + w(γ)

α , u(3)α = w

(3)α , u

(4)α = −ε3αβxβx3 + w

(4)α ,

u(γ)3 = xγx3 + w

(γ)3 , u

(3)3 = x3 + w

(3)3 , u

(4)3 = w

(4)3 .

(19)

Similar as in [9], in order to satisfy the ends conditions, we find that the unknownconstants as, s = 1, 2, 3, 4, are solutions of the following algebraic system

−R3 =

4∑s=1

asD3s, ε3αβMβ =

4∑s=1

asDαs, −M3 =

4∑s=1

asD4s, (20)

where

D3s =

∫D

t33(U(s))da, Dβs =

∫D

xβt33(U(s))da,

D4s =

∫D

ε3αβxαt3β(U(s))da, s = 1, 2, 3, 4.

(21)

It follows from the positive definiteness of the internal energy (see [9]) that

det(Drs) 6= 0, (22)

so that the system (20) uniquely determines the constants as, s = 1, 2, 3, 4.We can verify that the solution given in this section corresponds to the loads for

which

Rα = 0. (23)

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1544 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

Remark 1. The solution of the problem (P1), constructed in this section, corre-sponds to the following equilibrated stress vector’s values on the ends of cylinder

h = −4∑s=1

ash3(U(s)) on D0

h =

4∑s=1

ash3(U(s)) on DL.

(24)

4. The solution of the problem (P2). The problem (P2) consists in the study ofthe deformation of a porous cylinder subjected to a prescribed thermal field whichis linear in the axial coordinate, namely

θ = T0(x1, x2) + x3T1(x1, x2), (x1, x2) ∈ D. (25)

So, we have to determine an equilibrium displacement field U = (ui, ψ) whichsatisfies the boundary lateral conditions

ti(U) = 0, h(U) = 0 on ∂D × (0, L), (26)

and ∫D

t3i(U)da=−Ri,∫D

εijkxjt3k(U)da=−Mi. (27)

where Ri and Mi are preassigned functions representing the resultant force and theresultant moment about the origin of the cartesian system of the traction acting onthe end for which x3 = 0. Similar conditions are assumed on the end located atx3 = L.

These conditions satisfy the necessary and sufficient conditions for the existenceof solution. In this moment we do not specify the form of the equilibrated vectoron the bases. The form of this vector will be specified later. We will see that theequilibrated vector will be given in every point and must have a specific form. Thisfact is a consequence of the considered mathematical model and it is in concordancewith the physical reality [24]. The punctual conditions are usually considered inthe framework of the theory of materials with voids (see [18, 19, 20]).

We search a solution, U∗ = (u∗, ψ∗), of the problem (P2) in the class of functionwith the expressions

u∗α = −1

6bαx

33 −

1

2cαx

23 −

1

2ε3αβb4xβx

23 − ε3αβc4xβx3

+

4∑s=1

(cs + bsx3)w(s)α + wα(x1, x2) + x3wα(x1, x2),

u∗3 =1

2(b1x1 + b2x2 + b3)x23 + (c1x1 + c2x2 + c3)x3

+

4∑s=1

(cs + bsx3)w(s)3 + w3(x1, x2) + x3w3(x1, x2),

ψ∗ =

4∑s=1

(cs + bsx3)ω(s) + ω(x1, x2) + x3ω(x1, x2),

(28)

where W(s) = (w(s), ω(s)) are solutions of the generalized plane strain problemdefined in the previous section, wi, wi, ω and ω are unknown functions solutions ofsome generalized thermoelastic plane strain problems and bs and cs, (s = 1, 2, 3, 4)are unknown constants.

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ON THE THERMAL STRESSES IN ANISOTROPIC POROUS CYLINDERS 1545

From the geometrical relations we have that eij , corresponding to the state of

generalized thermoelastic plane strain defined by the field W = (wi, ω) and T0, are

eαj =1

2(wj,α + wα,j), e33 = 0. (29)

Hence, the corresponding stress tensor, the equilibrated stress vector and the in-trinsic equilibrated force are given by

tij = Aijαseαs +Aij3αe3α +Dijαω,α +Aijω − βijT0,

hi = Dαsieαs +D3αie3α +Diαω,α + diω − ξT0,−g = Aαj eαj +A3αe3α + dαω,α +mω − ζT0.

(30)

Similarly, the geometrical relations, corresponding to the state of generalized

thermoelastic plane strain defined by the field W = (wi, ω) and T1, are

eαj =1

2(wj,α + wα,j), e33 = 0, (31)

and

tij = Aijαseαs +Aij3αe3α +Dijαω,α +Aijω − βijT1,

hi = Dαsieαs +D3αie3α +Diαω,α + diω − ξT1,−g = Aαj eαj +A3αe3α + dαω,α +mω − ζT1.

(32)

Using the relations (28), (29) and (31), we have

eαi =

4∑s=1

(cs + bsx3)e(s)αi + eαi + x3eαi,

e3α =

4∑s=1

(cs + bsx3)e(s)3α + e3α + x3e3α − ε3αβ(c4 + b4x3)xβ + wα +

4∑s=1

bsw(s)α ,

e33 = c1x1 + c2x2 + c3 + (b1x1 + b2x2 + b3)x3 + w3 +

4∑s=1

bsw(s)3 ,

ψ∗,α =

4∑s=1

(cs + bsx3)ω(s),α + ω,α + x3ω,α, ψ∗

,3 =

4∑s=1

bsω(s) + ω,

(33)

where e(s)ij , (s = 1, 2, 3, 4), are the strain corresponding to the functions w

(s)i and

ω(s) from the previous section.Hence, taking into account the relations (33), the constitutive equations (2) be-

come

tij =

4∑s=1

(cs + bsx3)t(s)ij + tij + x3tij +

4∑s=1

bs(Aij3rw

(s)r +Dij3ω

(s))

+Aij3rwr +Dij3ω − ε3αβAij3α(c4 + b4x3)xβ

+Aij33(c1x1 + c2x2 + c3) +Aij33(b1x1 + b2x2 + b3)x3,

(34)

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1546 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

hi =

4∑s=1

(cs + bsx3)h(s)i + hi + x3hi +

4∑s=1

bs(D3riw

(s)r +Di3ω

(s))

+D3riwr +Di3ω − ε3αβD3αi(c4 + b4x3)xβ

+D33i(c1x1 + c2x2 + c3) +D33i(b1x1 + b2x2 + b3)x3,

−g =

4∑s=1

(cs + bsx3)g(s) + g + x3g +

4∑s=1

bs(A3rw

(s)r + d3ω

(s))

+A3rwr + d3ω − ε3αβA3α(c4 + b4x3)xβ

+A33(c1x1 + c2x2 + c3) +A33(b1x1 + b2x2 + b3)x3,

where t(s)ij , h

(s)i and g(s) are given by

t(s)ij = Aijαre

(s)αr +Aij3αe

(s)3α +Dijαγ

(s)α +Aijω

(s),

h(s)i = Dαrie

(s)αr +D3αie

(s)3α +Diαγ

(s)α + diω

(s),

−g(s) = Aαre(s)αr +A3αe

(s)3α + dαγ

(s)α +mω(s).

(35)

Now, we are able to say that the unknown functions wi, wi, ω and ω are solutionsof the following equations

tβi,β + fi = 0, hα,α + g + h = 0 in D (36)

andtβi,β = 0, hα,α + g = 0 in D. (37)

Moreover, the solutions of the above equations must verify the following boundaryconditions

tβinβ = Ti, παnα = P on ∂D (38)

andtβinβ = 0, παnα = 0 on ∂D. (39)

In relations (36) to (39), we have used the notations

fi = t3i +

4∑s=1

bs[t(s)3i +

(Aαi3rw

(s)r +Dαi3ω

(s)),α

]+(Aαi3rwr +Dαi3ω

),α− ε3αβA3i3αb4xβ

+A3i33(b1x1 + b2x2 + b3),

h = h3 +

4∑s=1

bs[h(s)3 +

(D3rαw

(s)r +Dα3ω

(s)),α

−A3rw(s)r − d3ω(s)

]+(D3rαwr +Dα3ω

),α−A3rwr

−d3ω − ε3αβD3α3b4xβ +D333(b1x1 + b2x2 + b3),

Ti = −[ 4∑s=1

bs(Aαi3rw

(s)r +Dαi3ω

(s))

+Aαi3rwr +Dαi3ω

]nα,

P = −[ 4∑s=1

bs(D3rαw

(s)r +Dα3ω

(s))

+D3rαwr +Dα3ω

]nα.

(40)

We denote by (Π0) the generalized thermoelastic plane strain problem definedby the equations (29), (30) and (36) and the boundary conditions (38), while (Π1)

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ON THE THERMAL STRESSES IN ANISOTROPIC POROUS CYLINDERS 1547

denotes the generalized thermoelastic plane strain problem defined by the equations(31), (32) and (37) and the boundary conditions (39).

According to the existence result, we always have a solution of the problem (Π1),while the problem (Π0) has a solution if and only if the following conditions aresatisfied ∫

D

fida+

∫∂D

Tids = 0,∫D

ε3αβxαfβda+

∫∂D

ε3αβxαTβds = 0.

(41)

But, in view of the relations (40) and (34) we deduce that∫D

fαda+

∫∂D

Tαds =

∫D

t3α,3da (42)

and moreover ∫D

f3da+

∫∂D

T3ds =

4∑s=1

bsD3s +

∫D

t33da,

∫D

ε3αβxαfβda+

∫∂D

ε3αβxαTβds =

4∑s=1

bsD4s +

∫D

ε3αβxαt3βda.

(43)

Using the equilibrium equations (6), we remark that

t3α = (xαtρ3),ρ + xαt33,3,

t3α,3 = (xαtρ3),ρ3 + xαt33,33.(44)

Thus, the divergence theorem leads to∫D

t3αda =

∫D

xαt33,3da,∫D

t3α,3da =

∫D

xαt33,33da.

(45)

We use the form (34) of the constitutive equations to deduce that∫D

t3αda =

4∑s=1

bsDαs +

∫D

xαt33da,

∫D

t3α,3da = 0. (46)

Having in mind the above identities, we can say that the existence conditions (41)are satisfied if and only if the unknown constants bs are solutions of the algebraicsystem

4∑s=1

bsDαs = −Rα −R∗α,

4∑s=1

bsD3s = K1,

4∑s=1

bsD4s = K2, (47)

where

R∗α =

∫D

xαt33da, K1 = −∫D

t33da,

K2 = −∫D

ε3αβxαt3βda.

(48)

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1548 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

In view of (22), the system (47) gives us the constants bs, s = 1, 2, 3, 4. With thehelp of the relation (34), the conditions (27) become

4∑s=1

csε3αβDβs = −Mα −M∗α,

4∑s=1

csD3s = −R3 −R∗3,

4∑s=1

csD4s = −M3 −M∗3 ,

(49)

where

M∗α =

∫D

{ε3αβxβ

[t33 +

4∑s=1

bs(A333rw

(s)r +D333ω

(s))

+A333rwr +D333ω

]}da,

R∗3 =

∫D

[t33 +

4∑s=1

bs(A333rw

(s)r +D333ω

(s))

+A333rwr +D333ω

]da,

M∗3 =

∫D

{ε3αβxα

[t3β +

4∑s=1

bs(A3β3rw

(s)r +D3β3ω

(s))

+A3β3rwrνr +D3β3ω

]}da.

(50)

After solving this algebraic system the solution of the problem (P2) is completelysolved.

Remark 2. The solution of the problem (P2), constructed in this section, corre-sponds to the following equilibrated stress vector’s values on the ends of cylinder

h = −{ 4∑s=1

csh(s)3 + h3 +

4∑s=1

bs(D3r3w

(s)r +D33ω

(s))

+D3r3wr +D33ω − ε3αβD3α3c4xβ +D333(c1x1 + c2x2 + c3)

}on D0,

h =

4∑s=1

(cs + bsL)h(s)3 + h3 + Lh3 +

4∑s=1

bs(D3r3w

(s)r +D33ω

(s))

+D3r3wr +D33ω − ε3αβD3α3(c4 + b4L)xβ

+D333(c1x1 + c2x2 + c3) +D333(b1x1 + b2x2 + b3)L on DL.

(51)

5. Conclusions. In this paper we study the deformation of the porous right cylin-ders on the presence of a thermal field. First, we assume that the thermal field isindependent of x3 (the problem (P1)). Then, we consider that the thermal field islinear in x3 (the problem (P2)). In fact, the problems are reduced to some general-ized isothermal plane strain problems and to some algebraic systems.

The solving of the problem (P2) is sketched below:

- we determine the solutions of the generalized isothermal plane strain problems(∆s), s = 1, 2, 3, 4,

- with the help of these solutions we determine the components of the matrixDrs,

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ON THE THERMAL STRESSES IN ANISOTROPIC POROUS CYLINDERS 1549

- we determine the solutions of the algebraic system (20),- we solve the generalized isothermal plane strain problem (Π1),- from the system (47) we determine the constants bs, s = 1, 2, 3, 4,

- we deduce the form of the loads fi, h, Ti, P with the help of the relations (40),- we solve the generalized isothermal plane strain problem (Π0),- from the algebraic system (49) we determine the constants cs, s = 1, 2, 3, 4,- we substitute all the quantities in (28) and the solution is determined.

Remark 3. From (28), (47) and (49) we can see that a thermal field which islinear in the axial coordinate produces an extension characterized by the force R∗

3,a bending by the moments M∗

α, a torsion by the moment M∗3 , a flexure by the forces

R∗α and a plane deformation parallel to the end D(0). If T1 = 0, then from (31),

(32), (37) and (39) we find that for our solution tij , hi and g are zero and thus,from (47), we have R∗

α = 0 and Kα = 0, α = 1, 2. From the solution of the problem(P1) we can remark that the thermal field independent by x3 produces extension,bending and torsion.

We can see how the thermal field influences the deformation if we compare thepresent results with the results obtained in the isothermal case [9, 10].

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1550 EMILIAN BULGARIU AND IONEL-DUMITREL GHIBA

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Received June 2012; revised September 2012.

E-mail address: [email protected]

E-mail address: [email protected]

E-mail address: ghiba [email protected]