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Semi-inverse solution for Saint-Venant’s problem inthe theory of porous elastic materials
- Extended form of the paper -
Ionel-Dumitrel Ghiba
Department of Mathematics, University “A.I. Cuza” of Iasi, 700506 Iasi, Romania, Blvd. Carol I, no. 11,700506 Iasi, Romania &
Octav Mayer Institute of Mathematics, Romanian Academy, 700505 - Iasi,email: [email protected]
Abstract
The purpose of this research is to study the Saint-Venant’s problem for right cylinderswith general cross-section made of inhomogeneous anisotropic elastic materials with voids.We reformulate the equilibrium equations with the axial variable playing the role of aparameter. Two classes of semi-inverse solutions to Saint-Venant’s problem are describedin terms of five generalized plane strain problems. These classes are used in order toobtain a semi-inverse solution for the relaxed Saint-Venant’s problem. An applicationof this results in the study of extension, bending, torsion and flexure of right circularcylinders in the case of isotropic materials is presented.
Keywords: Saint-Venant’s problem; Semi-inverse solution; Anisotropic elastic materials withvoids
1
Contents
1 Introduction 2
2 Materials with voids 32.1 Equations of the nonlinear theory . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 The linear theory of materials with voids . . . . . . . . . . . . . . . . . . . . . . 7
3 Some mathematical results in the equilibrium theory 113.1 Reciprocity and uniqueness theorems . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Existence and uniqueness theorems . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 The generalized plane strain state 15
5 Saint-Venant’s problem 165.1 Analysis of Saint-Venant’s problem by plane section solutions . . . . . . . . . . 185.2 The relaxed Saint-Venant’s problem . . . . . . . . . . . . . . . . . . . . . . . . . 24
6 Extension, bending and torsion of isotropic cylinder 28
7 Flexure of isotropic porous elastic cylinder 33
8 Saint-Venant’s principle 35
1 Introduction
The theory of elastic materials with voids has been established by Nunziato and Cowin (1979).In this theory, the bulk density is written as the product of two fields, the matrix materialdensity field and the volume fraction field. The first investigations in the theory of thermoelasticmaterials with voids are due to Nunziato and Cowin (1979) and Iesan (1986). The intendedapplications of the theory are to geological materials and to manufactured porous materials. Apresentation of this theory can be found in (Iesan and Ciarletta, 1993) and in (Iesan, 2004).
Cowin and Nunziato (1983) solved the problem of pure bending of an isotropic and homoge-neous porous beam of rectangular cross section. In (Iesan and Ciarletta, 1993) is presented theproblem of extension and bending of right cylinders made of an isotropic elastic material withvoids and in a recent paper Iesan and Scalia (2007) investigates the problem of extension andbending of right cylinder made of inhomogeneous and orthotropic elastic materials with voids.A study of Saint-Venant’s problem for homogeneous and isotropic porous elastic cylinders hasbeen presented by Dell’Isola and Batra (1997). The Saint-Venant’s principle in the theory ofelastic materials with voids has been studied by Batra and Yang (1995).
In this paper we consider the Saint-Venant’s problem for right cylinders made of an inhomo-geneous anisotropic elastic materials with voids. For the treatment of this problem, we resortto the method used by Chirita (1992), (see also (Iesan, 1987)), for the study of Saint-Venant’sproblem in linear viscoelasticity. In the first part of the paper we formulate the Saint-Venant’sproblem for an anisotropic elastic cylinder with voids and we define the generalized plane strainfor the cross section of the considered cylinder. Then, we study the possibility to reduce the
2
problem to some generalized plane strain problems, which are more tractable. For this, we treatthe Saint-Venant’s problem by reformulating the equilibrium equations with the axial variableplaying the role of a parameter. Further, we point out two classes of semi-inverse solutionsin the set of solutions of Saint-Venant’s problem that may be expressed in terms of a state ofgeneralized plane strain. These classes are relevant in obtaining a semi-inverse solution to therelaxed Saint-Venant’s problem. We use the results obtained in the anisotropic case to solvethe extension, bending, torsion and flexure problem of isotropic porous elastic circular cylindersfor the relaxed Saint-Venant’s problem.
An analysis and important proprieties of the solutions of the relaxed Saint-Venant’s problemin the classical elasticity have been presented in (Iesan, 1987). Iesan (2007) study the extensionof an isotropic circular cylinder with a special kind of inhomogeneity. Follow the method usedin (Chirita, 1992), Gales (2000) studied the Saint-Venant’s problem in micropolar viscoelaticity.The deformation of isotropic microstretch elastic cylinders has been studied by Iesan and Nappa(1995) and by De Cicco and Nappa (1995) and the extension, bending and torsion of anisotropicmicrostretch elastic cylinders have been studied by Scalia (2000).
The present paper can be useful for the study of orthotropic porous material, for porousmaterial with different types of inhomogeneous and for functionally graded porous materials.
2 Materials with voids
2.1 Equations of the nonlinear theory
We consider a body made by porous thermoelastic materials, which at the time t0 occupiesthe region B of the three-dimensional Euclidian space E3. In the following, the configurationof the body at the initial time t0 is considered as the reference configuration.
We refer the motion of the body to a fixed system of rectangular axes OXK (K = 1, 2, 3).We denote by XK , (K = 1, 2, 3) the coordinates of a generic material point M0, of the domainB. We suppose that after the deformation process, at the time t, the body occupies a newdomain B which has the boundary ∂B, the material point M0 arriving in the position M . Wewill refer the configuration B of the body at the time t to a new fixed system of rectangularaxes oxi (i = 1, 2, 3). Thus, the coordinates of the position M are denoted xi, (i = 1, 2, 3). Inthe rest of this book, X denotes the vector with the components (X1, X2, X3), and x denotesthe vector (x1, x2, x3)
We assume that the body should not penetrate itself and hence there is a one–to–oneapplications between B and B. Let us consider a fixed time interval I = [t0, t1), where t0 ≥ 0,while t1 > 0 can be infinite.
The deformation of the body will be described by the relation
x = x(X, t), (X, t) ∈ B × I. (2.1)
These applications are of class C2 on B × I and we have
det
(∂xi
∂XK
)> 0. (2.2)
The coordinates XK will be called material coordonates, while the coordonates xi will becalled spatial coordinates.
3
Throughout this book (when we do not specify else) Latin subscripts take the values 1, 2, 3whereas the Greek indices are understood to range over the integers (1, 2), summation is carriedout over repeated indices. Typical conventions for differential operations are implied such as asuperposed dot or comma followed by a subscript to denote the partial derivative with respectto the time or to the corresponding cartesian coordinate.
We say that a function f defined on B × (t0, t1) is of class CM,N , where m and n are givenpositive number, if the functions
∂mf (n) ≡ ∂m
∂XP∂XQ· · ·∂XK
(∂nf
∂tn
), m ∈ {0, 1, ...M}, n ∈ {0, 1, ..., N},
m+ n ≤ max{M,N},
exist and are continuous on B × (t0, t1).In order to introduce the theory of elastic materials with voids, Nunziato and Cowin [216]
used the concept of a distributed body previously introduced by Goodman and Cowin [140] inthe formulations of a theory of granular materials.
According with the paper of Goodman and Cowin [140], a distributed body is a one-parameterfamily {Bt}, 0 < t <∞, of regions of Euclidean three space such that
(a) for any times t and t′, the region Bt is homeomorphic to the region Bt′ , and
(b) for each t, the region Bt is endowed with a structure given by two real valued set ofapplications mt and vt subjected to the following axioms:
(b1) mt and vt are non-negative measures defined for all Borel subsets Pt ⊂ Bt,
(b2) vt(Pt) ≤ v(Pt), for all Pt ⊂ Bt (v is the Lebesque volume measure),
(b3) mt is absolutely continuous with respect to vt.
It is easy to see that, Bt represents the configuration of the distributed body at the time t,Pt is subset of Bt, mt and vt are the distributed volume and distributed mass, respectively.
For the case of porous bodies, the volume is always less than or equal to the total volume ofany part of the body. The axiom (b2) takes into account this fact. Axiom (b3) implies that,in the theory of Goodman and Cowin [140] the void mass of a porous or granular material areneglected.
Because, from axiom (b2) it follows that the distributed volume measure vt is absolutelycontinuous with respect to the Lebesgue volume measure, by the Radon-Nikodym theorem,there exists a real valued Lebesgue integrable function ν(x, t) defined on Bt such that for anypart Pt ⊂ Bt
vt(Pt) =
∫Pt
νdv. (2.3)
The function ν will be called the volume distribution function (0 < ν ≤ 1) and represent ameasure of the change of the bulk materials which results from the presence of voids.
Similarly, from axiom (b3), it follows that there is an essentially bounded vt–integrablefunction, defined on Bt such that
mt(Pt) =
∫Pt
ρdvt for almost all Pt ⊂ Bt. (2.4)
4
We call the function ρ distributed mass density or simply the distributed density. From theabsolute continuity of distributed volume with respect to Lebesgue volume, we have
mt(Pt) =
∫Pt
ρνdv. (2.5)
This relation suggest the introduction of the function
ρ = νρ, (2.6)
which is called is called the bulk density of the distributed body.Clearly, we have this representation in the reference configuration B, that is
ρ0 = ν0ρ0, (2.7)
where ρ0, ρ0 and ν0 are the bulk density, the distributed density and the volume distributionfunction in the reference configuration.
The theory of Nunziato and Cowin [216] is based on this decomposition which introduceda new kinematic variable ν. In the rest of this paper, we will suppose that ν is of class C2 onB × I and ρ0 is a strictly positive continuous function on B × I.
In what follows, we present a short path in the deduction of the nonlinear equations whichdescribe the thermal bahavior of the materials with voids. For this we use the way describedby Iesan in [161, 173], using the procedure of Green and Rivlin [143].
Let us consider an arbitrary regular subset P of the body which after the deformationprocess becomes P .
We have the following relation which postulate the conservation of energy for the regularsubset P of B and for every time, in the form [173]
d
dt
∫P
ρ0
(e+
1
2x2 +
1
2κν2
)dV =
∫P
ρ0(f · x + `ν + S)dV
+
∫∂P
(T · x +Hν +Q)dA,
(2.8)
where dV and dA are element of volume and area in the reference configuration, e is the internalenergy per unit mass, f is the body force per unit mass, ` is the extrinsic equilibrated body forceper unit mass, S is the heat supply per unit mass, T is the stress vector associated with thesurface ∂P , but measured per unit area of the surface ∂P , H is the equilibrated stress with thesurface ∂P , but measured per unit area of the surface ∂P , Q is the heat flux across the surface∂P , but measured per unit area of the surface ∂P , κ is the equilibrated inertia. We supposethat f , ` and S are continuous functions B× (t0, t1), e is of class C0,1 on B× (t0, t1), T, H andQ are of class C1,0 on B × (t0, t1) and continuous on B × [t0, t1), and κ is a strictly positiveknown function on B.
We assume that ρ0, e, f , `, T, H and ν are not affected when we superposed a given rigidmotion with constant translation velocity and with constant angular velocity. Following theprocedure of Green and Rivlin [143], we have that there is the tensor TAi called Piola-Kirchhoffstress tensor, the vector QK called heat flux vector and the vector HK called equilibrated stressvector such that
Ti = TKiNK , (H −HKNK)ν +Q−QKNK = 0, (2.9)
5
where NK is the unit outward normal vector to the surface ∂P .Also, in our hypothesis, from the relation (2.8) we have the equations of motion
TAi,A + ρ0fi = ρ0xi, (2.10)
the balance of equilibrated forces
HK,K + g + ρ0` = ρ0κν (2.11)
and the energy equation
ρ0e = TKivi,K +HK ν,K − gν +QK,K + ρ0S, (2.12)
on B× (t0, t1), where g is a dependent constitutive variable which is called intrinsic equilibratedbody force.
As in classical theory, we consider the Piola-Kirchhoff stress tensor TAB given by
TAi = xi,BTAB. (2.13)
and the Lagrangian strain tensor
EKL =1
2(xi,Kxi,L − δKL), (2.14)
where δKL is the Kronecker’s delta.Assuming the invariance of the quantities ρ0, e, TKL, HK , QK , S, g and ν to a superposed
uniform rigid body motion with a constant angular velocity, we have
TKL = TLK (2.15)
and hence, the energy equation can be written in the following form
ρ0e = TKLEKL +HK ν,K − gν +QK,K + ρ0S. (2.16)
We have not insisted here on the complete proof of the above results. More details aboutthe deduction of the above equations can be found in the books by Iesan [165, 173], whilethe physical significance of the quantities of the present theory can be found in the works ofGoodman and Cowin [140], Nunziato and Cowin [216], Cowin and Leslie [69].
According with the classical thermodynamics theory (see Truesdell and Noll [261]), for everysubset P of the body B and for every time, we must have∫
P
ρ0ηdV −∫
P
1
Tρ0SdV −
∫∂P
1
TQdA ≥ 0, (2.17)
where η is the entropy per unit mass and T is the absolute temperature. We consider that T isa positive function of class C2,1 and η is of class C0,1 on B × (t0, t1).
6
We say that a media is a thermoelastic materials with voids if the following constitutiveequations hold
ψ = ψ(EKL, T, T,B, ν, ν,M , XN),
TKL = TKL(EKL, T, T,B, ν, ν,M , XN),
η = η(EKL, T, T,B, ν, ν,M , XN),
QK = QK(EKL, T, T,B, ν, ν,M , XN),
HK = HK(EKL, T, T,B, ν, ν,M , XN),
g = g(EKL, T, T,B, ν, ν,M , XN),
Q = Q(EKL, T, T,B, ν, ν,M , XN),
H = H(EKL, T, T,B, ν, ν,M , XN),
(2.18)
whereψ = e− Tη (2.19)
is the Helmholtz free-energy. We assume that the response functions are of class C2 on theirdomain.
In view of the constitutive hypothesis we obtain
H = HKNK , Q = QKNK , (2.20)
and from the entropy inequality (2.17), as in classical theory, we have that the constitutiveequations of the thermoelastic materials with voids are
W = W(EKL, T, ν, ν,M , XN),
TKL =∂W∂EKL
, ρ0η = −∂W∂T
HK =∂W∂ν,K
, g = −∂W∂ν
,
QK = QK(EKL, T, T,M ν, ν,M , XN),
(2.21)
where W = ρ0ψ andQKT,K ≥ 0. (2.22)
In conclusion, the basic equations of the nonlinear theory of thermoelastic materials with voidsconsists of equations of motion (2.10), the balance of equilibrated forces (2.11), the energyequation (2.16), the constitutive equations (2.21) and the geometrical equations (2.14), onB × (t0, t1). We must adjoin to these equations boundary conditions and initial conditions.The boundary conditions can be of Dirichlet type, of Neumann type or we can have mixedboundary conditions.
2.2 The linear theory of materials with voids
In this section we give the equations of the linear theory of thermoelastic and of the linear
theory of elastic materials with voids. We use the following notations Xi = δiAXA, f,i =∂f
∂Xiand we introduce the displacement vector u defined by
ui = xi −Xi. (2.23)
7
We denote by T0 and ν0 the absolute temperature and the volume distribution function in thereference configuration and we suppose that are given positive constants. We also suppose thatin the natural state, the body is free from stresses and has zero intrinsic equilibrated body forceand entropy.
Let us consider now, the changes of the temperature and the variation of the volume dis-tribution function from the reference configuration by
ϕ = ν − ν0, θ = T − T0. (2.24)
In the linear theory, we can suppose that u = εu′, ϕ = εϕ′ and θ = εθ′ where ε is a constantsmall enough to have εn ' 0, for n ≥ 2.
As in classical theory, the Lagrangian strain tensor reduces to
eij =1
2(ui,j + uj,i). (2.25)
The linear theory of thermoelastic materials with voids
In the case of linear theory of thermoelastic materials with voids, the response functional canbe consider functions of eij, θ, θ,i, ϕ, ϕ,k and Xm. Using the notations
tji = δjKTKi, hi = δiKHK , qi = δiKQK ,
from the relations (2.21), we can write
tij =∂W∂eij
, ρ0η = −∂W∂θ
, hi =∂W∂ϕ,i
, g = −∂W∂ϕ
(2.26)
andqiθ,i ≥ 0. (2.27)
In the linear approach, we have
W =1
2(Cijrseijers − 2βijeijθ − aθ2 + Aijϕ,iϕ,j + 2Bijϕeij
+2Dijkeijϕ,k + 2diϕϕ,i + ξϕ2 − 2mθϕ− 2aiϕ,iθ),(2.28)
where the constitutive coefficients are functions on Xm and have the following symmetries
Cijrs = Crsij = Cjirs, βij = βji, Dijk = Djik,Aij = Aji, Bij = Bji.
(2.29)
Thus, the constitutive equations of the linear theory of anisotropic elastic materials withvoids are
tij = Cijrsers +Bijϕ+Dijkϕ,k − βijθ,hi = Aijϕ,j +Drsiers + diϕ− aiθ,g = −Bijeij − ξϕ− diϕ,i +mθ,ρ0η = βijeij + aθ +mϕ+ aiϕ,i
qi = kijθ,j
(2.30)
8
The inequality (2.27) implies that the conductivity tensor kij is positive semi-definite, i.e.
kijξiξj ≥ 0 for all ξs.
In view of the assumption of the linear theory, the equations which describe the bahaviorof the thermoelastic materials with voids are
tji,j + ρ0fi = ρui, (2.31)
hi,i + g + ρ0` = ρ0κϕ, (2.32)
ρ0T0η = qi,i + ρ0S, (2.33)
and the components of the surface traction t, the equilibrated surface traction h and the heatflux q at regular points of ∂B are given by
ti = tjinj, h = hini, q = qini, (2.34)
where nj = cos(nx, Oxj) and nx is the unit vector of the outward normal to ∂B at x.We consider that the constitutive coefficients are known and have the properties:
Cijrs, Bij, Dijk, βij, Aij, di, ai, and kij are function of class C1 on B and a, ξ, and m arecontinuous on B.
We recall that the unknowns are the displacement vector, the variation of the volume dis-tribution function (which will be called in the following the porosity function) and the variationof the temperature from the reference configuration. In terms of this unknowns, the equationsof the linear theory are
(Cijrsur,s +Bijϕ+Dijkϕ,k − βijθ),j + ρ0fi = ρ0ui,
(Aijϕ,j +Drsiur,s + diϕ− aiθ),i −Bijui,j − ξϕ− diϕ,i+
+mθ + ρ0` = ρ0κϕ,(kijθ,j),i − T0βijui,j − T0aθ − T0mϕ− T0aiϕ,i = −ρ0S,
(2.35)
on B × (0, t1).In the case of isotropic materials, we have only nine constitutive coefficients,
λ, µ, b, β, α, ξ, m, a, k, and the constitutive equations are
tij = λerrδij + 2µeij + bϕδij − βθδij,hi = αϕ,i,g = −berr − ξϕ+mθ,ρ0η = βerr + aθ +mϕ,qi = kθ,i
(2.36)
and the internal energy density W is positive defined quadratic form if and only if
µ > 0, α > 0, ξ > 0, 3λ+ 2µ > 0, (3λ+ 2µ)ξ > 3β2. (2.37)
It follows that the basic equations of homogeneous and isotropic materials are
µ∆ui + (λ+ µ)uj,ji + bϕ,i − βθ,i + ρ0fi = ρ0ui,
α∆ϕ− buj,j − ξϕ+mθ + ρ0` = ρ0κϕ,k∆θ − T0βuj,j − T0aθ − T0mϕ = −ρ0S,
(2.38)
9
where ∆ is the Laplace operator.We consider the following mixed boundary conditions
ui = ui on S1 × (t0, t1), tjinj = ti pe S2 × (t0, t1),
ϕ = ϕ on S3 × (t0, t1), hini = h on S4 × (t0, t1),
θ = θ on S5 × (t0, t1), qini = q on S6 × (t0, t1),
(2.39)
where Sr, (r = 1, 2, ..., 6), are subsets of ∂B so that S1∪S2 = S3∪S4 = S5∪S6 = ∂B, S1∩S2 =
S3 ∩ S4 = S5 ∩ S6 = ∅. The functions ui, ϕ and θ are prescribed continuous functions onS1 × (t0, t1), S3 × (t0, t1), S5 × (t0, t1), respectively, while ti, h and q are prescribed continuousfunctions on S2 × (t0, t1), S4 × (t0, t1), S6 × (t0, t1), respectively.
The initial conditions are
u(X, t0) = u0(X), u(X, t0) = v0(X), ϕ(X, t0) = ϕ0(X),ϕ(X, t0) = ζ0(X), θ(X, t0) = θ0(X), X ∈ B, (2.40)
with u0, v0, ϕ0, ζ0 and θ0 are prescribed continuous functions on B;For more details regarding the equations of the linear theory, the readers can use the book
of Iesan [173].
The linear theory of elastic materials with voids. Equilibrium equations
In the isothermal linear theory of elastic materials with voids (see [165]), the constitutivefunctionals depend only on eij, ϕ, ϕ,k and Xm. In consequence, we obtain
W =1
2(Cijrseijers + Aijϕ,iϕ,j
+2Bijϕeij + 2Dijkeijϕ,k + 2diϕϕ,i + ξϕ2),(2.41)
and the constitutive equations becomes
tij = Cijrsers +Bijϕ+Dijkϕ,k,hi = Aijϕ,j +Drsiers + diϕ,g = −Bijeij − ξϕ− diϕ,i.
(2.42)
The equations which describe the bahavior of the elastic materials are
tji,j + ρ0fi = ρ0ui (2.43)
andhi,i + g + ρ0` = ρ0κϕ. (2.44)
We must adjoin boundary conditions and initial conditions which are similar with (2.39)1−4
and (2.40)1−4.When we study the equilibrium of elastic materials, we assume that all the quantities of the
equations are time-independent quantities.
10
Thus, the above equations are reduced to the equilibrium equations
tji,j + Fi = 0,
hi,i + g + L = 0(2.45)
in B, where Fi = ρ0fi and L = ρ0` are continuous functions in B.The mixed boundary value problem of the equilibrium theory of elastic materials with void
consists in finding the functions ui and ϕ which satisfy the equations (2.43), (2.44), (2.25) and(2.42) and the boundary conditions
ui = ui on S1 × (t0, t1), tjinj = ti pe S2 × (t0, t1),
ϕ = ϕ on S3 × (t0, t1), hini = h on S4 × (t0, t1),(2.46)
where Sr, (r = 1, 2, ..., 6), are subsets of ∂B so that S1∪S2 = S3∪S4 = ∂B, S1∩S2 = S3∩S4 = ∅.The functions ui, ϕ and θ are prescribed continuous functions on S1, S3, respectively, while ti, hand q are prescribed continuous functions on S2, S4, respectively.
3 Some mathematical results in the equilibrium theory
3.1 Reciprocity and uniqueness theorems in the equilibrium theory
In this section we consider the equilibrium problem defined by the equations (2.45) and theboundary conditions (2.46). For this problem we present a uniqueness theorem and a reciprocitytheorem.
We consider two sets of external data I(α) = (f(α)i , `(α), u
(α)i , ϕ(α), t
(α)i , h(α)), (α = 1, 2). For
this external data, let us consider the corresponding solutions s(α) = (u(α)i , ϕ(α), e
(α)ij , t
(α)ij , h
(α)i ,
g(α)).We have the following result [165].
Theorem 3.1 (Reciprocity theorem) If an elastic material with void occupies the domain Band it is subjected to the following external data I(α), (α = 1, 2), then the corresponding solutionss(α) satisfy the following reciprocity relation∫
B
ρ0(f(1)i u
(2)i + `(1)ϕ(2))dv +
∫∂B
(t(1)i u
(2)i + h(1)ϕ(2))da
=
∫B
ρ0(f(2)i u
(1)i + `(2)ϕ(1))dv +
∫∂B
(t(2)i u
(1)i + h(2)ϕ(1))da,
(3.1)
wheret(α)i = t
(α)ji nj, h
(α) = h(α)i ni, (α = 1, 2).
Proof. Let us consider2Wαβ = t
(α)ij e
(β)ij + h
(α)i ϕ
(β),i − g(α)ϕ(β).
From the relation (2.42) we deduce that
2Wαβ = Cijrse(β)ij +Bij(e
(β)ij ϕ
(α) + e(α)ij ϕ
(β))
+Dijr(e(β)ij ϕ
(α),r + e
(α)ij ϕ
(β),r )
+di(ϕ(α)ϕ
(β),i + ϕ(β)ϕ
(α),i ) + Aijϕ
(α),i ϕ
(β),i + ξϕ(α)ϕ(β).
(3.2)
11
It is easy to see that, from the symmetries of the coefficients, we have
W12 = W21.
Using the relations (2.42), we deduce
2Wαβ = (t(α)ij u
(β)i ),j + (h
(α)i ϕ(β)),i − t
(α)ji,ju
(β)i − (h
(α)i,i + g(α))ϕ(β). (3.3)
By a direct integration of this relation, and taking into account the boundary conditions wededuce the desired result.
Now, let us consider I(1) = (fi, `, ui, ti, ϕ, h) and s(1) = (ui, ϕ, eij, tij, hi, g). Then, from (3.2)and (3.3) we obtain
2
∫B
W11dv =
∫B
(tijeij + hiϕ,i − gϕ)dv = 2
∫B
Wdv,
where W is given by (2.41). With the help of the previous relation, we deduce that
2
∫B
Wdv =
∫∂B
(tiui + hϕ)da+
∫B
ρ0(fiui + `ϕ)dv. (3.4)
A direct consequence of these relations is the following result:
Theorem 3.2 (Uniqueness theorem). If the internal energy W is a positive defined quadraticform in terms of eij, ϕ and ϕ,k for every X ∈ B, then two solutions of the mixed boundaryproblems defined by (2.45) and (2.46) are equals, excepting a rigid motion. Moreover, if S1 isnot empty, then the considered mixed boundary problem has at most one solution.
Proof. Let us consider two solutions, (u′i, ϕ′) and (u′′i , ϕ
′′), of the mixed boundary problem andu = (ui, ϕ) their difference.
In view of the linearity of the problem, (ui, ϕ) is solution of the problem defined by theequilibrium equations (2.45) and by null boundary conditions. Moreover, we have uiti + hϕ =0 on ∂B. From (3.4) we deduce that ∫
B
Wdv = 0,
where W is the internal energy corresponding to (ui, ϕ). Because W is a positive definedquadratic form, we have that eij = 0, ϕ = 0 and, hence
ui = ai + εijrbjxr, ϕ = 0,
where ai and bi are arbitrary constants. If S1 is not empty, then ai = bi = 0, which completethe proof.
12
3.2 Existence and uniqueness theorems in the equilibrium theory
We consider the equilibrium problem of elastic materials with voids. We assume that thedomain B is bounded by a Lipschitz surface. Let consider that ∂B = Su ∪St ∪C, where C is anull measurable subset of the boundary and Su and St are empty or open subsets. We considerthe boundary conditions
ui = ui, ϕ = ϕ on Su, tijnj = ti, hini = h on St, (3.5)
where ui, ϕ ∈ W 1,2(B) and ti, h ∈ L2(St). We define the space
W(B) = {U = (ui, ϕ); ui, ϕ ∈ W 1,2(B)}, (3.6)
and the following norm
||U||2W(B) = ||ϕ||2W 1,2(B) +3∑
i=1
||ui||2W 1,2(B). (3.7)
We consider the space V(B) defined as the subspace of W(B) for which
ui = 0, ϕ = 0 pe Su, (3.8)
in the trace sense. We also consider the bilinear form A(V,U) defined on W(B)×W(B) by
A(V,U) =
∫B
[Cijrseij(v)eij(u) +Bij(eij(v)ϕ+ eij(u)ψ)
+Dijr(eij(v)ϕ,r + eij(u)ψ,r)
+di(ϕψ,i + ϕ,iψ) + Aijϕ,iψ,j + ψϕψ]dv,
(3.9)
where U = (ui, ϕ), V = (vi, ψ), eij(u) =1
2(ui,j + uj,i).
In the following, the constitutive coefficients are considered measurable bounded functions inB which satisfy the symmetry relations (2.29). We have
A(U,V) = A(V,U), A(U,U) = 2
∫B
Wdv, (3.10)
where W is the internal energy associated to U.Let us define the functionals f, g : W(B) → R by
f(V) =
∫B
ρ0(fivi + `ψ)dv,
g(V) =
∫St
(tivi + hψ)da, V = (vi, ψ) ∈ W(B).(3.11)
We assume that ρ0, fi, ` ∈ L2(B). Let be U = (ui, ϕ) such that ui, ϕ on Su are restrictions onSu of some functions ui, ϕ ∈ W 1,2(B), in the trace sense.
13
We say that U ∈ W(B) is a solution of the boundary value problem defined by the relations
(2.45) and (3.5) if U− U ∈ V(B) and verifies
A(V,U) = f(V) + g(V), (3.12)
for every V ∈ V(B).We suppose that there is a positive constant α for which
A(U,U) ≥ α
∫B
[eij(u)eij(u) + ψ2 + ψ,iψ,i]dv, (3.13)
for all U ∈ W(B).We define the operators Ns : W(B) → L2(B), (s = 1, 2, ..., 10) by
N1U = u1,1, N2U =1
2(u1,2 + u2,1), N3U =
1
2(u1,3 + u3,1),
N4U = u2,2, N5U =1
2(u2,3 + u3,2), N6U = u3,3, N7U = ψ,1,
N8U = ψ,2, N9U = ψ,3, N10U = ψ,
(3.14)
for all U = (ui, ψ) ∈ W(B). In view of the relation (3.13) the following inequality occurs
A(U,U) ≥ 2α10∑
s=1
||NsU||2L2(B). (3.15)
Let prove that this operators form a coercive operator system on W(B), i.e. there is a constantk > 0 such that [165], [214]
||U||L2(B) +10∑
s=1
||NsU|2L2(B) ≥ k||U||2W(B), ∀ U ∈ W(B). (3.16)
Using the results derived by Necas [214], we can say that the above inequality is true if theoperator system Ns has the order 4. These can be prove if we take into account the definitionsgiven to the operators Ns.
We consider the subspace P of V(B) defined by
P = {V ∈ V(B);10∑
s=1
||NsV||2L2(B) = 0}. (3.17)
In view of the definitions of the operators Ns we deduce that
P = {V = (vi, ψ) ∈ V(B); vi = ai + εirsbrxs, ϕ = 0}, (3.18)
where ai and bi are arbitray constants, and let us consider the factor space W(B)/P of theclasses of equivalence U = {U + p; p ∈ P}, U ∈ W(B), endowed with the norm
||U||W(B)/P = infp∈P
||U + p||W(B). (3.19)
14
We remark that for every p ∈ P and U ∈ W(B) we have
A(U,p) = 0
and we can define the bilinear form A on the product space W(B)/P ×W(B)/P by
A(U, V) = A(U,V), U ∈ U, V ∈ V. (3.20)
If Su is not empty, then P = {0} and we have the following result
Theorem 3.3 Let consider that P = {0}. In this case the boundary value problem has auniqueness weak solution.
If Su empty then:
Theorem 3.4 Necessary and sufficient conditions for the existence of a weak solution u ∈ Wof the traction problem are ∫
B
ρ0fidv +
∫∂B
tida = 0,∫B
ρ0εijrxjfrdv +
∫∂B
εijrxj trda = 0.(3.21)
From the above theorem, we can conclude that to have a solution of the traction problem,the forces must have vanishing resultant and torque. This results are established using themethod established in the paper by Hlavacek si Necas [150] (see also Fichera [107]) for ellipticequations.
4 The generalized plane strain state
Throughout this section, B denotes the interior of a right cylinder of length L, occupied by aporous elastic material. We choose a rectangular cartesian system Ox1x2x3 so that the Ox3 axisis parallel with the generator of the cylinder and one of the ends lies in the x1Ox2 plane. Wedenote by ∂B the boundary of B and by D(x3) ⊂ R2 the interior of the bounded cross sectionat distance x3 from the x1Ox2 plane. The boundary, ∂D, of each cross section is assumedsufficiently smooth to admit application of the divergence theorem in the plane cross section.The lateral boundary of the cylinder is π = ∂D × [0, L].
Moreover, we assume that the porous materials which made the cylinder are cross-sectioninhomogeneous. Thus, we have
Cijrs = Cijrs(x1, x2), Aij = Aij(x1, x2), Bij = Bij(x1, x2),Dijk = Dijk(x1, x2), di = di(x1, x2), ξ = ξ(x1, x2).
(4.1)
Following Iesan [160] and Chirita [34] we define the state of generalized plane strain for theinterior of the cross section domain, D ⊂ R2, of the considered cylinder to be the state in whichthe displacement field w and the volume distribution ψ depend only on x1 and x2
wi = wi(x1, x2), ψ = ψ(x1, x2), (x1, x2) ∈ D. (4.2)
15
In this case, the components of the stress tensor, the components of the equilibrated stressvector and the intrinsic equilibrated body force are function of x1 and x2, i.e. Tij = Tij(x1, x2),Hi = Hi(x1, x2) si G = G(x1, x2). Moreover, we have
Tij(W) = Cijkαwk,α +Bijψ +Dijαψ,α,Hi(W) = Aiαψ,α +Drαiwr,α + diψ,G(W) = −Biαwi,α − ξψ − dαψ,α,
(4.3)
where W = (wi, ψ).Given the body force, the tensions and the extrinsic equilibrated force, the plane strain
problem for D ∪ ∂D consists in finding a solution w, ψ of the boundary value problem Sdefined by the equations
Tαi,α(W) + Fi = 0, Hα,α(W) +G(W) + L = 0 in D, (4.4)
and by the boundary conditions
Tαi(W)nα = Ti, Hα(W)nα = H on ∂D. (4.5)
We substitute the relation (4.3) into (4.4) and (4.5) and we write the boundary value problemS in the following form
[Ciαkβwk,β +Biαψ +Diαβψ,β],α = −Fi,[Aαβψ,β +Drβαwr,β + dαψ],α −Biαwi,α − ξψ − dαψ,α = −L in D,
(4.6)
and[Ciαkβwk,β +Biαψ +Diαβψ,β]nα = Ti,
[Aαβψ,β +Drβαwr,β + dαψ]nα = H on ∂D.(4.7)
The generalized plane strain state was studied in various papers (see for example the bookby Iesan [162] and by Leknitski [199]).
We assume that Fi, L ∈ C∞(D) and Ti, H ∈ C∞(∂D). Following an appropriate method asthat used in [162] and in view of the Theorem 3.4, we have that the necessary and sufficientconditions for the existence of a solution of the generalized plane problem [165] are given by∫
D
Fi da+
∫∂D
Ti ds = 0,
∫D
ε3αβxαFβ da+
∫∂D
ε3αβxαTβ ds = 0, (4.8)
where εijk is the alternating symbol. The assumptions of regularity imply that the solutionbelong to C∞(D) (cf. [107]).
5 Saint-Venant’s problem
As in the previous section, we will consider that the domain B is the interior of a right cylinderof length L. All the notations and assumptions introduced in previous section will be used inthis section.
16
We recall that the component of the tension vector and the equilibrated tension in a regularpoint of the frontier ∂B are given by
ti(U) = tij(U)nj, h(U) = hj(U)nj, (5.1)
where nj nj are the components of the outward unit normal to ∂B and U = (u, ϕ).The equations of equilibrium of porous continua, in the absence of body loads, are given by
tji,j(U) = 0, hj,j(U) + g(U) = 0. (5.2)
Saint-Venant’s problem for B consists in the determination of the displacement field u andthe volume distribution function ϕ on B, subjected to the requirements
ti(U) = 0, h(U) = 0 pe π,
ti(U) = t(1)i , h(U) = h(1) on D(0),
ti(U) = t(2)i , h(U) = h(2) on D(L),
(5.3)
where t(α)i and h(α) are functions preassigned on D(0) and D(L), respectively.
Following the results presenting in the Section 3.2, necessary and sufficient conditions forthe existence of a solution to this problem are given by∫
D(0)
t(1)i da+
∫D(L)
t(2)i da = 0,∫
D(0)
εijkxjt(1)k da+
∫D(L)
εijkxjt(2)k da = 0.
(5.4)
This equations imply that, for the equilibrium of cylinder, the tractions at the bases havevanishing resultant and torque.
The displacement field u and the volume distribution function ϕ satisfy the boundary valueproblem defined by the equations
Ti(U) ≡ [Cijrsur,s +Bijϕ+Dijkϕ,k],j = 0,H(U) ≡ [Aijϕ,j +Drsiur,s + diϕ],i −Bijui,j − ξϕ− diϕ,i = 0,
(5.5)
on B = D × (0, L), the lateral boundary conditions
Bi(U) ≡ [Ciαrsur,s +Biαϕ+Diαkϕ,k]nα = 0,G(U) ≡ [Aαjϕ,j +Drsαur,s + dαϕ]nα = 0,
(5.6)
on ∂D × (0, L), and the end boundary conditions
t3i(U) = t(1)i , h3(U) = h(1) on D(0),
t3i(U) = t(2)i , h3(U) = h(2) on D(L).
(5.7)
In the study of Saint–Venant’s problem [127, 130], we assume that the internal energy densityassociated to the solution of the boundary value problem T , defined by the equations (5.5) andthe boundary conditions (5.6)–(5.7), is positive defined.
17
5.1 Analysis of Saint-Venant’s problem by plane section solutions
In this subsections, following the semi-inverse method used by Iesan [160, 163] and [34, 36], westudy the possibility to reduce the system (5.5) and the lateral boundary conditions (5.6) to ageneralized plane problem.
For this aim, we consider the system (5.5) and the boundary condition (5.6) on the crosssection D ∪ ∂D. More exactly, we consider the plane boundary value problem
Ti(U) = 0, H(U) = 0 ın D (5.8)
andBi(U) = 0, G(U) = 0 pe ∂D, (5.9)
considered x3 ∈ (0, L) as parameter. We formulate the following problem: when the solutionof the Saint-Venant’s problem described above, by the relations (5.5)–(5.7), can be reduced tothe case of generalized plane strain problem associated to the cross section of the cylinder?
The components of the resultant force and of the resultant moment of the traction aboutthe origin of the rectangular cartesian system Ox1x2x3, acting on the cross-section D of thecylinder are defined by
Ri(U) =
∫D
t3i(U) da, Mi(U) =
∫D
εijkxjt3k(U) da. (5.10)
We will have a answer to the above questions, in terms of the vector-valued linear functionalsR and Mi. We remark that
Mα(U) = ε3αβ
∫D
xβt33(U) da− x3ε3αβRβ(U),
M3(U) = ε3αβ
∫D
xαt3β(U) da.
(5.11)
We write the plane boundary value problem (5.8) and (5.9) in the form
[Ciαkβuk,β +Biαϕ+Diαβϕ,β],α
+[Ciαk3uk,3 +Diα3ϕ,3],α + t3i,3(U) = 0,[Aαβϕ,β +Drβαur,β + dαϕ],α
−Biαui,α − ξϕ− dαϕ,α −Bi3ui,3 − d3ϕ,3++[Aα3ϕ,3 +Dr3αur,3],α + h3,3(U) = 0 in D,
(5.12)
and[Ciαkβuk,β +Biαϕ+Diαβϕ,β]nα = −[Ciαk3uk,3 +Diα3ϕ,3]nα,
[Aαβϕ,β +Drβαur,β + dαϕ]nα = −[Aα3ϕ,3 +Dr3αur,3]nα
(5.13)
on ∂D, and we see the boundary value problem defined by (5.12) and (5.13) as a generalizedplane strain boundary value problem, with
Fi = [Ciαk3uk,3 +Diα3ϕ,3],α + t3i,3(U),L = [Aα3ϕ,3 +Dr3αur,3],α −Bi3ui,3 − d3ϕ,3 + h3,3(U),
Ti = −[Ciαk3uk,3 +Diα3ϕ,3]nα,
H = −[Aα3ϕ,3 +Dr3αur,3]nα.
(5.14)
18
The necessary and sufficient conditions (4.8), show us that we must have∫D
t3i,3(U) da = 0,
∫D
ε3αβxαt3β,3(U) da = 0. (5.15)
From (4.1) and (2.42) we can write
t3i,3(U) = t3i(U,3), (5.16)
so that the relations (5.15) take the form∫D
t3i(U,3) da = 0, (5.17)
∫D
ε3αβxαt3β(U,3) da = 0. (5.18)
We observe, in view of relations (5.15), (5.10) si (5.11), that a sufficient condition forexpressing a solution of Saint-Venant’s problem in terms of a generalized plane strain is thefact that Ri(U) and M3(U) are independent of x3.
We are thus led to the following result.
Proposition 5.1 Let U be a solution to Saint-Venant’s problem. If Ri(U) and M3(U) areindependent of x3 then U can be expressed in term of a state of generalized plane strain.
Corollary 5.1 Let U be a solution to Saint-Venant’s problem for which Ri(U) and M3(U)are independent of x3. Then Mα(U) are independent of x3.
Proof. In view of equation (5.2) and the boundary condition (5.3), we get∫D
xαt33(U,3) da =
∫D
xαt33,3(U) da = −∫
D
xαtβ3,β(U) da
= −∫
D
(xαtβ3(U)),β da+
∫D
tα3(U) da
= −∫
∂D
xαtβ3(U)nβ ds+Rα(U) = Rα(U).
(5.19)
On the other hand, by means of relation (5.11), we have
Mα,3(U) = ε3αβ
(∫D
xβt33(U)da
),3
− ε3αβRβ(U)− x3ε3αβRβ,3(U) (5.20)
and the proof is complete.
Remark 5.1 Relation (5.19) combined with the fact that Rα(U) is independent of x3, yields∫D
xαt33(U,33) da = 0. (5.21)
19
In what follows, we describe the two classes of semi-inverse solutions to Saint-Venant’sproblem, solutions which will correspond to some particular types of tensions. This classes canbe expressed in term of a state of generalized plane strain.
The class CI. We consider, inspired by (5.17) and (5.18), the class CI of solutions of Saint-Venant’s problem for which the expression of u,3 is the same of that of a rigid displacementand ϕ is independent of x3.
In view of relations (5.10), (5.11), (5.17) and (5.18), we deduce that for U0 = (u0, ϕ0) ∈ CI ,we have (
Rα(U0))
,3= 0,
(R3(U
0))
,3= 0,
(Mi(U
0))
,3= 0. (5.22)
On the other hand, by a direct integration of the expression of a rigid displacement, wededuce that if U0 ∈ CI , then
u0α = −1
2aαx
23 − ε3αβa4xβx3 + wα(x1, x2),
u03 = (a1x1 + a2x2 + a3)x3 + w3(x1, x2),ϕ = ψ(x1, x2),
(5.23)
except for an additive rigid motion, where as, s = 1, 2, 3, 4, are arbitrary constants. To put it inother words, a solutions from class CI is written in terms of four constants and four functionsindependent of x3.
In this case, we deduce that the components of the stress tensor and the components of theequilibrated stress vector are the followings
tij(U0) = Cij33(a1x1 + a2x2 + a3)− a4Cijα3ε3αβxβ + Tij(W),
hi(U0) = D33i(a1x1 + a2x2 + a3)− a4Dα3iε3αβxβ +Hi(W),
g(U0) = −B33aαxα −B33a3 + a4Bρ3ερβ3xβ +G(W),
(5.24)
where Tij(W) and Hi(W) are given by the relations (4.3)1,2.To simplify the expressions, we introduce the following notations
Si(W) = [Ciαkβwk,β +Biαψ +Diαβψ,β],α,
C(W) = [Aαβψ,β +Drβαwr,β + dαψ],α −Biαwi,α − ξψ − dαψ,α
Pi(W) = [Ciαkβwk,β +Biαψ +Diαβψ,β]nα,
D(W) = [Aαβψ,β +Drβαwr,β + dαψ]nα.
(5.25)
The boundary value problem defined by (5.8) and (5.9) becomes
Ti(U0) = Si(W) + (Ciα33aβxβ + Ciα33a3 − a4ε3ρβCiαρ3xβ),α = 0,
H(U0) = C(W) + (D33αaβxβ +D33αa3 − a4D3ραε3ρβxβ),α−−B33aβxβ −B33a3 + a4Bρ3ερβ3xβ = 0 in D
(5.26)
andBi(U
0) = Pi(W) + (Ciα33aβxβ + Ciα33a3 − a4ε3ρβCiαρ3xβ)nα = 0G(U0) = D(W) + (D33αa3 − a4D3ραε3ρβxβ)nα = 0 on ∂D.
(5.27)
We remark, from (4.6), (5.26) and (5.27), that for any as, s = 1, 2, 3, 4 the necessary andsufficient conditions to have the solution W = (wi, ψ) of the above plane boundary valueproblem are satisfied.
20
We denote by W(j) = (w(j), ψ(j)) a solution of the boundary value problem (5.26) and (5.27)when ai = δij, a4 = 0 and by W(4) = (w(4), ψ(4)) a solution of the boundary value problem(5.26) and (5.27) when ai = 0 and a4 = 1. Therefore, these solutions, W(s), s = 1, 2, 3, 4, arecharacterized by the equations
Si(W(s)) + F
(s)i = 0, C(W(s)) + L(s) = 0 in D (5.28)
and the boundary conditions
Pi(W(s)) = T
(s)i , D(W(s)) = H(s) on ∂D, (5.29)
whereF
(γ)i = (Ciα33xγ),α , F
(3)i = (Ciα33),α , F
(4)i = − (ε3ρβCiαρ3xβ),α ,
L(γ) = (D33αxγ),α −B33xγ, L(3) = (D33α),α −B33,
L(4) = − (ε3ρβD3ραxβ),α +Bρ3ερβ3xβ,
T(γ)i = − (Ciα33xγ)nα, T
(3)i = − (Ciα33)nα,
T(4)i = (ε3ρβCiαρ3xβ)nα,
H(γ) = − (D33αxγ)nα, H(3) = − (D33α)nα,
H(4) = (ε3ρβD3ραxβ)nα.
(5.30)
From the linearity of the problems, we have
W =4∑
s=1
asW(s). (5.31)
In what follows we assume that the solutions W(s) of the above problems are known. Then thesolution U0 ∈ CI can be written in the form
U0 =4∑
s=1
asU(s), (5.32)
where
u(β)α = −1
2x2
3δαβ + w(β)α , u
(β)3 = xβx3 + w
(β)3 ,
u(3)α = w
(3)α , u
(3)3 = x3 + w
(3)3 ,
u(4)α = ε3βαxβx3 + w
(4)α , u
(4)3 = w
(4)3 .
(5.33)
It follows from (5.24) and (5.32) that
tij(U0) =
4∑s=1
astij(U(s)), hi(U
0) =4∑
s=1
ashi(U(s)),
g(U0) =4∑
s=1
asg(U(s)),
(5.34)
wheretij(U
(α)) = Tij(W(α)) + Cij33xα, tij(U
(3)) = Tij(W(3)) + Cij33,
tij(U(4)) = Tij(W
(4))− ε3αβCijα3xβ,hi(U
(α)) = Hi(W(α)) +D33ixα, hi(U
(3)) = Hi(W(3)) +D33i,
hi(U(4)) = Hi(W
(4))− ε3αβDα3ixβ,g(U(α)) = G(W(α))−B33xα, g(U
(3)) = G(W(3))−B33,g(U(4)) = G(W(4)) + εαβ3Bα3xβ.
(5.35)
21
Obviously, relations (5.28), (5.29) and (5.33) give
tαi,α(U(s)) = 0, hα,α(U(s)) + g(U(s)) = 0 ın D (5.36)
andtαi(U
(s))nα = 0, hα(U(s))nα = 0 pe ∂D. (5.37)
This relations imply
Rα(U(0)) =
∫D
4∑s=1
ast3α(U(s)) da =
=
∫D
4∑s=1
as
(xαtρ3(U
(s)))
,ρda = 0.
(5.38)
Finally, we note that for (u0, ϕ0) ∈ CI we have
R3(U(0)) =
4∑s=1
asD3s, Mα(U(0)) =4∑
s=1
ε3αβasDβs,
M3(U(0)) =
4∑s=1
asD4s,
(5.39)
where
D3s =
∫D
t33(U(s)) da, Dβs =
∫D
xβt33(U(s)) da,
D4s =
∫D
ε3αβxαt3β(U(s)) da, s = 1, 2, 3, 4.
(5.40)
Let write u0{a} and ϕ0{a} to indicate the dependence of the vector u0 and ϕ0, defined by therelation (5.32), on the four-dimensional vectora = (a1, a2, a3, a4).
In view of Remark 5.1 we are led to introduce the following class of semi-inverse solutionsfor the Saint-Venant’s problem.
The class CII. We denote by CII the class of semi-inverse solutions of Saint-Venant’sproblem for which the conditions (5.17) and (5.18) hold true, and, moreover the expression ofu,33 is the same of that of a rigid displacement and ϕ,3 is independent of x3.
For (u∗, ϕ∗) ∈ CII it follows that (u∗,3, ϕ∗,3) ∈ CI and, by means of the above discussion, we
haveu∗,3 = u0{b}, ϕ∗,3 = ϕ0{b} (5.41)
and hence
u∗ =
∫ x3
0
u0{b} dx3 + u0{c}+ w∗(x1, x2),
ϕ∗ =
∫ x3
0
ϕ0{b} dx3 + ϕ0{c}+ ψ∗(x1, x2),
(5.42)
where b and c are arbitrary four-dimensional vectors, w∗ and ψ∗ are independent functions ofx3.
22
The components of the stress tensor, the components of the equilibrated stress vector and theintrinsic equilibrated body force corresponding toU∗ = (u∗i , ϕ
∗) defined by (5.42), have the form
tij(U∗) =
4∑s=1
(cs + x3bs)tij(U(s)) + kij(W
(s)) + Tij(W∗),
hi(U∗) =
4∑s=1
(cs + x3bs)hi(U(s)) + si(W
(s)) +Hi(W∗),
g(U∗) =4∑
s=1
(cs + x3bs)g(U(s)) + q(W(s)) +G(W∗),
(5.43)
where
kij(W(s)) =
4∑s=1
(Cijk3bsw(s)k +Dij3bsψ
(s)),
si(W(s)) =
4∑s=1
(D3kibsw(s)k + Ai3bsψ
(s)),
q(W(s)) = −4∑
s=1
(B3kbsw(s)k + d3bsψ
(s)),
(5.44)
and we denote W∗ = (w∗i , ψ
∗).In view of relations (5.38)–(5.40) and Corollary 5.1, we get
4∑s=1
bsD3s = 0 and4∑
s=1
bsD4s = 0. (5.45)
On the basis of relations (5.8) and (5.9) we deduce that W∗ is a solution of the followingboundary value problem
Si(W∗) + kαi,α(W(s)) +
4∑s=1
bst3i(U(s)) = 0,
C(W∗) + sα,α(W(s)) +4∑
s=1
bsh3(U(s)) + q(W(s)) = 0 in D,
(5.46)
andPi(W
∗) + kαi(W(s))nα = 0, D(W∗) + sα(W(s))nα = 0 on ∂D. (5.47)
The necessary and sufficient conditions for the existence of a solution of this problem aresatisfied on the basis of relations (5.9) and (5.45).
Proposition 5.2 If U∗ ∈ CII , it has the form (5.42), where b satisfies the conditions (5.45).Moreover, W∗ can be obtained from the generalized plane strain problem defined by the relations(5.46) and (5.47).
23
Remark 5.2 Let U∗ ∈ CII . Then, from relations (5.19), (5.38) and (5.43), we have
Rα(U∗) =4∑
s=1
bsDαs, (5.48)
R3(U∗) =
4∑s=1
csD3s +
∫D
(k33(W
(s)) + T33(W∗)
)da,
Mα(U∗) = ε3αβ
[4∑
s=1
csDβs +
∫D
xβ
(k33(W
(s)) + T33(W∗)
)da
],
M3(U∗) =
4∑s=1
csD4s +
∫D
ε3αβxα
(k3β(W(s)) + T3β(W∗)
)da,
where the components of the vector fields b satisfy the conditions (5.45).
Remark 5.3 The solutions of Saint-Venant’s problem residing within CI correspond to theend loads
t(1)i = −
4∑s=1
ast3i(U(s)), t
(2)i =
4∑s=1
ast3i(U(s)),
h(1) = −4∑
s=1
ash3(U(s)), h(2) =
4∑s=1
ash3(U(s)),
(5.49)
while, the solutions of Saint-Venant’s problem residing within CII correspond to the end loads
t(1)i = −
4∑s=1
cst3i(U(s))− ki3(W
(s))− T3i(W∗),
t(2)i =
4∑s=1
(cs + Lbs)t3i(U(s)) + ki3(W
(s)) + T3i(W∗),
h(1) = −4∑
s=1
csh3(U(s))− s3(W
(s))−H3(W∗),
h(2) =4∑
s=1
(cs + Lbs)h3(U(s)) + s3(W
(s)) +H3(W∗).
(5.50)
5.2 The relaxed Saint-Venant’s problem
The relaxed Saint-Venant’s problem for the porous elastic cylinder B consists in the deter-mination of a equilibrium displacement field U = (ui, ϕ) that satisfies the boundary lateralconditions
ti(U) = 0, h(U) = 0 on ∂D × (0, L), (5.51)
andRi(U) = −Ri, Mi(U) = −Mi on x3 = 0, (5.52)
where Ri and Mi are preassigned functions representing the resultant force and the resultantmoment about the origin of the cartesian system of the traction acting on the end for whichx3 = 0. Similar conditions are assumed on the end located at x3 = L.
24
These conditions satisfy the necessary and sufficient conditions for the existence of solution.In this moment we do not specify the form of the equilibrated vector on the bases. The formof this vector will be specified later. We will see that the equilibrated vector will be not givenglobal and will be given in every point and must have a specific form. This fact is a consequenceof the considered mathematical model and is in concordance with the physical reality [216]. Thepunctual conditions are usually used in the framework of the theory considered in this chapter(see [175, 177])).
In the relaxed Saint-Venant’s problem we do not specify the tension in every point of theends, we give the resultant force and the resultant moment about the origin of the cartesiansystem of the traction acting on the ends. In classical elasticity, Saint-Venant said that thedeformations are the same without in a neighborhood of the loaded end. This hypothesis inknown as the Saint-Venant’s principle and was mathematical formulated and proof by Toupin[258] (see also [147], [162]). In the theory of porous materials, the Saint-Venant’s principle hasbeen sudied by Batra and Yang [6] and will be presented in the Section 8.
In what follows we proceed to determine a semi-inverse solution of the above problemwhich corresponds to a class of pointwise value of the equilibrated stress vector on the ends ofcylinder. In view of the discussion from the previous Section (see also [36, 162, 177]), we usethe decomposition of the relaxed problem into problems (P1) and (P2) characterized by(P1) (extension–bending–torsion): Rα = 0,(P2) (flexure): R3 = Mi = 0.
The solution of the problem (P1). In view of the results of the previous Section, asolution of the problem (P1) has the form
UI = U0 =4∑
s=1
asU(s). (5.53)
Relations (5.39) and (5.52) give for determination of the unknown constants as, s = 1, 2, 3, 4,the following system
4∑s=1
asDαs = ε3αβMβ,
4∑s=1
asD3s = −R3,
4∑s=1
asD4s = −M3. (5.54)
Let us prove that this system can be solved. For this, we follow the method used in [162],[169] and [238]. We consider two solution U = (ui, ϕ) and V = (vi, φ) of the equilibriumequations (5.2), corresponding to two external loads. The strain energy U(U) on B is
U(U) =
∫B
W(U) dv. (5.55)
The functional U(·) generates the bilinear symmetric functional A(U,V) given by (3.9).We can remark that
U(U) =1
2A(U,U), (5.56)
and from the reciprocal theorem presented in Section 3.1, we have
A(U,V) =
∫∂B
(ti(U)vi + h(U)φ) da. (5.57)
25
In view of (5.53), (5.55) si (5.56), we get
U(U0) =1
2
4∑r, s=1
A(U(r),U(s))aras. (5.58)
Because W(U0) is positive definite and u0 is not a rigid motion, it follows that
det(A(U(r),U(s))) 6= 0. (5.59)
ButA(U(r),U(s)) = LDsr, (5.60)
and using (3.1) and (5.60) it follows that Drs = Dsr, where s, r = 1, 2, 3, 4 and, thus
det(Drs) 6= 0, (5.61)
so that the system (5.54) uniquely determines the constants as, s = 1, 2, 3, 4.Concluding, the solution of the problem (P1) has the form (5.53), where U(s), s = 1, 2, 3, 4
are solutions of the problems defined by (5.33), (5.28) and (5.29), and as, s = 1, 2, 3, 4 aresolutions of the algebraic system (5.54).
The solution UI corresponds to the pointwise values of the equilibrated stress vector on theends of cylinder who have the forms
h(U) =
−
4∑s=1
ash3(U(s)), x3 = 0
4∑s=1
ash3(U(s)), x3 = L.
(5.62)
In the case of homogeneous isotropic materials these quantities vanish and in consequencethe solution corresponds to the null equilibrated stress vector’s values h(1) and h(2) on theends of the cylinder. Moreover, this observation holds true for orthotropic elastic materials.This hypothesis has been discussed in connection with the study of extension and bendingof isotropic elastic cylinders with voids presented by Iesan and Ciarletta (1993) and with thestudy of extension and bending of orthotropic elastic cylinders with void presented by Iesanand Scalia [175].
The solution of the problem (P2). On the basis of the results from the previous Section,we seek a solution of the problem (P2) in the class CII . Therefore, we seek in the form (5.42),i.e.
UII = U∗ = (u∗i , ϕ∗), (5.63)
26
where
u∗α = −1
6bαx
33 −
1
2cαx
23 −
1
2b4ε3αβxβx
23 − c4ε3αβxβx3
+4∑
s=1
(cs + x3bs)w(s)α + w∗
α,
u∗3 =1
2(bρxρ + b3)x
23 + (cρxρ + c3)x3
+4∑
s=1
(cs + x3bs)w(s)3 + w∗
3,
ϕ∗ =4∑
s=1
(cs + x3bs)ϕ(s) + ψ∗,
(5.64)
and the unknown constants bs satisfy the conditions
4∑s=1
bsD3s = 0 si4∑
s=1
bsD4s = 0. (5.65)
From the relations (5.48) and (5.52), we get
4∑s=1
bsDαs = −Rα, (5.66)
so that from the system (5.65) and (5.66) we can uniquely determine the constants we canuniquely determine the constants bs, s = 1, 2, 3, 4. In what follows, we assume that the constantsbs are known. Then w∗, ϕ∗ can be determined from the generalized plane strain problemdefined by relations (5.46) and (5.47). Further, from the relations (5.48) and (5.52), for thedetermination of the unknown constants cs, s = 1, 2, 3, 4, we get the following system
4∑s=1
csDβs = −∫
D
[xβ (k33 + T33(W∗))] da,
4∑s=1
csD3s = −∫
D
(k33 + T33(W∗)) da,
4∑s=1
csD4s = −∫
D
[ε3αβxα (k3β + T3β(W∗))] da.
(5.67)
We can conclude that a solution of the problem (P2) has the form (5.64), where the unknownconstants bs and cs are determined from the system defined by relations (5.65)–(5.67), whilew∗, ψ∗ are determined as a solution of the generalized plane strain problem defined by therelations (5.46) si (5.47). The solution UII corresponds to pointwise values of the equilibratedstress vector on the ends of cylinder who have the forms
h(U) =
−
4∑s=1
csh3(U(s))− s3(W
(s))−H3(W∗), x3 = 0,
4∑s=1
(cs + Lbs)h3(U(s)) + s3(W
(s)) +H3(W∗), x3 = L.
(5.68)
27
In the case of isotropic and for transversal isotropic materials with voids and for circular cylin-ders, this solution correspond for equilibrated vectors stress which have null resultants of theends.
Finally, we note that the relaxed Saint-Venant’s problem has a solution of the form
U = UI + UII , (5.69)
where UI and UII are defined by the relations (5.53), (5.63) and (5.64).This solution was constructed in the paper [127]. In the following two sections we illustrate
the above results on an isotropic porous elastic circular cylinder with radius a. Firstly, wecarry up the solution of the extension-bending-torsion problem which corresponds to a null fluxof porosity on the ends of cylinder. Then, we treat the flexure problem when the resultantflux of porosity vanishes on the ends of cylinder. For this we follow the way indicated inthe previous section; and to find the solutions of the corresponding generalized plane strainproblems defined by (5.28) and (5.29), we use the results obtained by Iesan and Nappa [167] inthe study of extension and bending of microstretch elastic circular cylinders.
6 Extension, bending and torsion of isotropic porous
elastic circular cylinder
In this section, we denote by (P∗1 ), the problem consisting in finding the solution U = (ui, ϕ)
of the equilibrium equations which verifies the boundary conditions
ti(U) = 0, h(U) = 0 on ∂D × (0, L), (6.1)
and the ends conditions ∫D
t3α(U) da = 0
∫D
t33(U) da = −R3,∫D
εijkxjt3k(U) da = −Mi, h(U) = 0 on x3 = 0,
(6.2)
where R3 and Mi are given functions [127].For an elastic homogeneous isotropic porous material the constitutive coefficients has the
following formCijkl = λδijδkl + µ(δikδjl + δilδjk),Bij = bδij, Aij = αδijDijk = di = 0,
(6.3)
and in consequence, we have
t(s)βi = λw
(s)ρ,ρδiβ + µ(w
(s)i,β + w
(s)β,i) + bψ(s)δiβ,
t(s)33 = λw
(s)ρ,ρ + bψ(s),
h(s)α = αψ
(s),α ,
g(s) = −bw(s)α,α − ξψ(s), s = 1, 2, 3, 4.
(6.4)
28
In the following, we give the solution of the generalized plane strain problem defined bythe relations (5.28) and (5.29). With the help of these solutions we will find the unknownconstants as from (5.54) and in consequence we will have a explicit solution of the first problemconsidered in the previous section.
We denote by (Πγ) the first two problem, for the isotropic materials with voids, defined bythe relations (5.28) and (5.29). These problem are
t(γ)βα,β = − (λxγ),α , t
(γ)β3,β = 0,
h(γ)α,α + g(γ) = bxγ,
(6.5)
with the boundary conditions
t(γ)βαnβ = −λxγnα, t
(γ)β3 nβ = 0,
h(γ)α nα = 0.
(6.6)
It is easy to remark that w(γ)3 = 0 is solution of the problem (6.5)2 and (6.6)2. So, we will
exclude w(γ)3 when we talk about the problems (Πγ).
The conditions for the positive defined of the internal energy are equivalents for an isotropicmaterial are (2.37).
We introduce the functions (v(γ)α , φ(γ)) on D by
w(1)1 = v
(1)1 − 1
2ν1(x
21 − x2
2),
w(1)2 = v
(1)2 − ν1x1x2,
ψ(1) = φ(1) − ν2x1,
w(2)1 = v
(2)1 − ν1x1x2,
w(1)2 = v
(2)2 − 1
2ν1(x
21 − x2
2),
ψ(2) = φ(2) − ν2x2,
(6.7)
where
ν1 =λξ − b2
2[(λ+ µ)− b2], ν2 =
bµ
[(λ+ µ)ξ − b2]. (6.8)
Then (v(γ)α , φ(γ)) are solutions of the plane problems (Pγ) given by the homogeneous equations
t(γ)βα,β = 0, h
(γ)α,α + g(γ) = 0 in D (6.9)
and the boundary conditions
t(γ)βαnβ = 0, h
(γ)α nα = αν2nγ on ∂D, (6.10)
where tβα are quantities corresponding to a plane strain state.It is easy to remark that, in the case of homogeneous materials, the solution of the third
problem, defined by (6.5) and (6.6) is
w(3)1 = ν1x1, w
(3)2 = −ν1x2, w
(3)3 = 0, ψ(3) = −ν2, (6.11)
29
and for the fourth problem we have
w(4)1 = w
(4)2 = w
(4)3 = ψ(4) = 0. (6.12)
For a circular cylinder we have
D = {(x1, x2) ∈ R2; x21 + x2
2 < a}. (6.13)
In the following we study the plane problem (P1) using the polar coordinates (r, θ). In polar
coordinates, the displacement v(1)γ and the porosity function φ(1) become
v(1)r = u(r, θ), v
(1)θ = v(r, θ), φ(1) = ψ(r, θ) (6.14)
and the equilibrium equations are
∂τrr
∂r+
1
r
∂τrθ
∂θ+
1
r(τrr − τθθ) = 0,
∂τrθ
∂r+
1
r
∂τθθ
∂θ+
2
rτrθ = 0,
1
r
∂
∂r(rhr) +
1
r
∂hθ
∂θ+ g = 0,
(6.15)
whereτrr = (λ+ 2µ)err + λeθθ + bψ,
τθθ = λeθθ + (λ+ 2µ)err + bψ,
τrθ = 2µerθ, hr = α∂ψ
∂r, hθ =
1
rα∂ψ
∂r,
g = −β(err + eθθ)− ξψ,
(6.16)
and
err =∂u
∂r, eθθ =
1
r
(∂v
∂θ+ u
), erθ =
1
2
(1
r
∂u
∂θ+∂v
∂r− 1
rv
). (6.17)
The conditions (6.6) lead to the boundary conditions
τrr = 0, τrθ = 0, hr =1
aαν2 cos θ. (6.18)
In [167], Iesan and Nappa give the solution of such a system in the form
u(r, θ) = U (1) cos θ, v(r, θ) = V (1) sin θ, ψ = Ψ(1) cos θ, (6.19)
and thus, we obtain the differential equations
r2d2U (1)
dr2+ r
dU (1)
dr− (1 + c1)U
(1) + r(1− c1)dV (1)
dr
−(1 + c1)V(1) + c2r
2dΨ(1)
dr=0,
c1
(r2d
2V (1)
dr2+ r
dV (1)
dr
)− (1 + c1)V
(1) − r(1− c1)dU (1)
dr
−(1 + c1)U(1) − c2rΨ
(1) =0,
(6.20)
30
α
[1
r
d
dr
(rdΨ(1)
dr
)− 1
r2Ψ(1)
]− b
[1
r
d
dr(rU (1)) +
1
rV (1)
]− ξΨ(1) =0,
where
c1 =µ
λ+ 2µ, c2 =
b
λ+ 2µ. (6.21)
Using an adaptation of the results obtained in [167], we can say that
U (1) = A1 + A2r2 − c2
2
(∫Ψ(1)(x) dx+ r−2
∫x2Ψ(1)(x) dx
), (6.22)
V (1) = −A1 −3− c11− 3c1
A2r2 +
c22
(∫Ψ(1)(x) dx− r−2
∫x2Ψ(1)(x) dx
),
where As, s = 1, 2 are arbitrary constants.Replacing U, V given by (6.22) in (6.20)3, we obtain the differential equation
r2d2Ψ(1)
dr2+ r
dΨ(1)
dr− (1 + p2r2)Ψ(1) = − 8bc1r
3
α(1− 3c1), (6.23)
where
p2 =ξ
α− b2
(λ+ 2µ)α. (6.24)
We have that
Ψ(1) = A3I1(pr) +8bc1
α(1− 3c1)p2A2r, (6.25)
where In is the modified Bessel functions of order n and A1 an arbitrary constant, and inconsequence, in view of (6.22) we obtain
U (1) = A1 +Q1A2r2 − c2
2pA3[I0(pr) + I2(pr)],
V (1) = −A1 −Q2A2r2 +
c22pA3[I0(pr)− I2(pr)],
(6.26)
where
Q1 =1
1− 3c1
(1− 3c1 −
3c1c2b
αp2
),
Q2 =1
1− 3c1
(3− c1 −
c1c2b
αp2
).
(6.27)
It follows from (6.16)–(6.19) that
τrr =
(W1A2r +
2µc2pr
A3I2(pr)
)cos θ,
τrθ =
(−W2A2r +
2µc2pr
A3I2(pr)
)sin θ,
hr =
(αpA3I
′(pr) +8bc1
p2(1− 3c1)A2
)cos θ,
(6.28)
31
where
W1 = (3λ+ 4µ)Q1 − λQ2 +8b2c1
αp2(1− 3c1), W2 = −µ(Q1 +Q2). (6.29)
We see that W1 = −W2. The conditions (6.18) hold if and only if
A2 = − 2
aΓµc2I2(pa)ν2p
2α(1− 3c1)
A3 =1
apI ′1(pa)
(ν2 −
8bc1a
αp2(1− 3c1)A2
),
(6.30)
whereΓ = W1p
4a2α(1− 3c1)I′1(pa)− 16µbc1c2I2(pa). (6.31)
From this, we can conclude that the solution of the problem (P1) is given by
v(1)1 = U (1) cos2 θ − V (1) sin2 θ,
v(1)2 = (U (1) + V (1)) sin θ cos θ,ψ(1) = Ψ(1) cos θ.
(6.32)
A similar way, gives us the solution of the problem (P2) to be
v(2)1 = (U (2) − V (2)) sin θ cos θ,
v(2)2 = U (2) sin2 θ + V (2) cos2 θ,ψ(2) = Ψ(2) sin θ,
(6.33)
whereU (2) = U (1), Ψ(2) = Ψ(1), V (2) = −V (1). (6.34)
We have that the divergence of the displacement corresponding to the problems (Πγ) are
divv(1) = [(3Q1 −Q2)A2r − c2A3I′1(pr)] cos θ,
divv(2) = [(3Q1 −Q2)A2r − c2A3I′1(pr)] sin θ.
(6.35)
From (5.40), (6.4), (6.7), (6.11) si (6.35) we have the components of the matrix (Dij)4×4 tobe
D11 = D22 = J, D33 = Eπa2, D44 = πa4,D12 = D21 = Dβ3 = D3β = Dβ4 = D4β = D43 = D34 = 0,
(6.36)
whereJ =
π
4a4Q+
πa
p2(b− λc2)A3 (apI0(pa)− 2I1(pa)) ,
Q = E + λ(3Q1 −Q2)A2 +8b3c1
αp2(1− 3c1)A2,
E = (λ+ 2µ)− 2λν1 − bν2.
(6.37)
It follows from (6.6) and (6.36) that
a1 =M2
J, a2 = −M1
J, a3 = − R3
Eπa2, a4 = −M3
πa4, (6.38)
and from (6.16) we have
b1 = −R1
J, b2 = −R2
J, b3 = b4 = 0. (6.39)
With this, the solution of the extension-bending-torsion problem is constructed. We can saythat this solution corresponds to the null equilibrated stress’s values h(1) and h(2) on the endsof cylinder.
32
7 Flexure of isotropic porous elastic circular cylinder
In this section, we consider the flexure of isotropic porous elastic circular cylinder [127]. Theaim is to find the solution U = (ui, ϕ) of the equilibrium equations which verify the boundaryconditions
ti(U) = 0, h(U) = 0 on ∂D × (0, L), (7.1)
and the conditions ∫D
t3α(U) da = −Rα,
∫D
t33(U) da = 0,∫D
εijkxjt3k(U) da = 0, on x3 = 0, L,
h(U) =
α
J(φ(r)− ν2r)(R1 cos θ +R2 sin θ), x3 = 0
−αJ
(φ(r)− ν2r)(R1 cos θ +R2 sin θ), x3 = L.
(7.2)
where Rα are prescribed functions and J, φ(r) and ν2 are quantities defined in the previoussubsection.
The solution of this problem, denoted by (P∗2 ), belongs in the second class of inverse solutions
presented in the Subsection 5.1.The constants bs was determined by the relations (5.65) and (5.66). Then, we have to solve
the problem defined by the relations (5.46) and (5.47). The flexure problem will be completedetermined after the solving of the algebraic system (5.67).
In this moment, we can say which is the form of the problem defined by the relations (5.46)and (5.47). From (5.35), (5.44), (6.3), (6.7), (6.11) and (6.12) we obtain that
t3α(u(γ), ψ(γ)) = h3(u(γ), ψ(γ)) = 0,
kαβ = k33 = sα = r = 0, kα3 = µ∑
γ=1,2
bγw(γ)α ,
t33(u(γ), ψ(γ)) = λw
(γ)β,β + bψ(γ) + (λ+ 2µ)xγ,
(7.3)
and in consequence the equations (5.46) become
Tβα,α(w∗, ψ∗) = 0,Hα,α(w∗, ψ∗) = 0,
Tα3,α(w∗, ψ∗) = −µ∑γ=1,2
bγw(γ)α −
∑γ=1,2
bγt33(u(γ), ψ(γ)),
(7.4)
and the boundary conditions (5.47) are
Tαβ(w∗, ψ∗)nα = 0,Hα(w∗, ψ∗)nα = 0,
Tα3(w∗, ψ∗)nα = −µ
∑γ=1,2
bγw(γ)α nα.
(7.5)
The problem defined by (7.4)1,2 and (7.5)1,2 has the solution
w∗α = 0, ψ∗ = 0, (7.6)
33
and the problem defined by (7.4)3 and (7.5)3 can be written in the form
∆w∗3 = − 1
µ
∑γ=1,2
bγ[(λ+ µ)w(γ)
α,α + bψ(γ) + (λ+ 2µ)xγ
],
w∗3,αnα = −
∑γ=1,2
bγw(γ)α nα.
(7.7)
In polar coordinates, with the help of the relations (6.7), (6.14), (6.15), (6.26) and (6.34), thisproblem can be written in the following form
∆w∗3 = (Mr +NI1(pr))(b1 cos θ + b2 sin θ),
∂w∗3
∂r= −(U (1)(a)− 1
2ν1a
2)(b1 cos θ + b2 sin θ),(7.8)
where
M = − 1
µ{(λ+ µ)[(3Q1 −Q2)A2 − 2ν1] + b(TA3 − ν2) + λ+ 2µ} ,
N = −c2A3, T =8bc1
αp2(1− 3c1).
(7.9)
We search the solution of the above problem in the form
w∗3 = W (b1 cos θ + b2 sin θ), (7.10)
with W solution of the equation
d2W
dr2+
1
r
dW
dr− 1
r2W = Mr +NI1(pr). (7.11)
More exactly, we will find
W = Mr3
8+N
I1(pr)
p2+ A4
r
2+ A5
1
r, (7.12)
with A4 and A5 constants. Because we want that W to be finite when r = 0, we must imposethat A5 = 0.From (7.10) and (7.8) we have the conditions
dW
dr
∣∣∣∣r=a
= −(U (1)(a)− 1
2ν1a
2), (7.13)
and thus, the constant A4 will be
A4 = −3
4Ma2 − 2N
I ′1(pr)
p2− 2(U (1)(a)− 1
2ν1a
2), (7.14)
and the solution w∗3 is completely known.
With the help of relations (5.44), (5.67), (6.7), (6.33), (6.35), (7.3), (7.7) and (7.10) weobtain, in the case of isotropic materials with voids, that the algebraic system (5.67) is homo-geneous, and thus, we have cs = 0, s = 1, 2, 3, 4.
34
The solution of the flexure problem is given by (5.64), (6.6), (6.9) and corresponds to thefollowing equilibrated stress vector on the ends of the cylinder
h(1) = −h(2) =α
J(φ− ν2r)(R1 cos θ +R2 sin θ). (7.15)
We see that this solution is in the class of solutions for which the resultant of the equilibratedstress vector vanish on the ends of the cylinder.
With this, the relaxed Saint-Venant’s problem for a circular cylinder made by a porouselastic isotropic materials is completed solved.
To view the mechanical significations of the quantities defined by the relations (6.8) and(6.37), let us consider the particular case for which
b = 0. (7.16)
In this case the equilibrium equations are not coupled (see the Section ??). We find the equi-librium equations from classical elasticity and a equations for the porosity and the restrictionsimposed upon the constitutive coefficients are the usual conditions from classical elasticity.
Thus, from (6.8), (6.37) and (2.37) we have
U = −V = A1, φ = ν2 = A2 = A3 = c2 = N = T = 0,
ν1 =λ
2(λ+ µ), Q1 = p = 1, Q2 =
3− c11− 3c1
,
Q = E =µ(3λ+ 2µ)
λ+ µ, J =
πa4E
4, M = −2,
A4 =3
2a2 − 2A1 + ν1a
2.
(7.17)
In this case ν1 is the Poisson ratio and E is the Young modulus. Moreover, by this particular-ization we obtain the solution from classical elasticity, as is presented in [162], [172].
8 Saint-Venant’s principle
As we mention in the Subsection 5.2, an important problem in the study of deformation ofcylinders is the Saint-Venant’s principle.
We consider two solutions of the relaxed Saint-Venant’s problem which correspond to twodifferent loaded which have the properties that the resultant and the torque of the tensionsacting on the ends are equals. We denote by U = (u, ϕ) the difference of these two solutions.This difference is solution of the problem defined by the equilibrium equations
tji,j = 0, hj,j + g = 0. (8.1)
and the boundary condition
ti = 0, h = 0 pe π ∪D(L),ti = ti, h = 0 on D(0),
(3.9.2)
35
where ti are prescribed functions on D(0) which have the property∫D(0)
tida = 0,
∫D(0)
εijkxk tjda = 0. (8.2)
We denote byD(s, z) = {x |x ∈ B, s ≤ x3 ≤ s+ z}, (8.3)
the region of the cylinder situated between the planes x3 = s and x3 = s+ z, and we define [6]
U(s) =
∫x3≥s
W dv. (8.4)
We consider the free vibration problem
−Ti(U) = λui, −H = λϕ in D(s, z),
Bi = 0, G = 0 on ∂D(s, z),(8.5)
where the operators Ti, H, B, G are those from the Section 5. We denote by λ0(z) the lowereigenvalues of the above problem.
Using a similar way with those used in [38], [42] or [162] we can prove that there is aconstant aM > 0, the largest eigenvalue of the symmetric and positive defined tensor given bythe internal energy W , such that
tij(U)tij(U) + hi(U)hi(U) + g2(U) ≤ 2aMW (U). (8.6)
Theorem 8.1 Suppose that the internal energy is positive defined. Let U = (u, ϕ) be solutionof the problem given by the relations (8.1)–(8.2). Then, we have
U(s) ≤ U(0)e−(s−z)/s(z), (s > z) (8.7)
wheres(z) = 2(aM/λ0(z))
1/2.
Proof. We introduce the vector u defined by
ui = ui + ai + εijkbjxk, (3.9.9)
where ai and bi are arbitrary constants. We denote by U = (u, ϕ).As we prove in the Section 3.1, the uniqueness result and the reciprocity theorem, we deduce
that
U(s) =1
2
∫x3≥s
(tij(U)eij(u) + hi(U)ϕ,i − g(U)ϕ) dv (8.8)
The method derived by Toupin [258] gives that
U(s) = −1
2
∫D
(ti3(U)ui + h3(U)ϕ) da (8.9)
Using the inequality
2
∫V
fg ≤ ε
∫V
f 2dv +1
ε
∫V
g2dv, (8.10)
36
for every ε > 0, arithmetic–geometric means inequality and the Schwarz’s inequality, we obtain
−1
2
∫D
t3i(U)uida ≤1
4
(ε1
∫D
t3i(U)t3i(U) da+1
ε1
∫D
uiui da
). (8.11)
Similarly, we prove that
−1
2
∫D
h3(U)ϕda ≤ 1
4
(ε2
∫D
h3(U)h3(U) da+1
ε2
∫D
ϕ2 da
), (8.12)
and in consequence, choosing ε1 = ε2 = ε, we deduce
U(s) ≤ 1
4
[ε
∫D
(tij(U)tij(U) + hi(U)hi(U) + g2(U))da
+1
ε
∫D
(uiui + ϕ2)da
].
(8.13)
From the inequalities (8.6) it follows that
U(s) ≤ 1
4
[ε
∫D
2aMW(U)da+1
ε
∫D
(uiui + ϕ2)da
]. (8.14)
By a direct integration from x3 = s to x3 = s+ z, z > 0, and using the following notation
1
z
∫ s+z
s
U(x3)dx3 = Q(s, z), (8.15)
we have
Q(s, z) ≤ εaM
2z
∫D(s,z)
W(U)da+1
4εz
∫D(s,z)
(uiui + ϕ2)da. (8.16)
In view of the minimum principle for the free vibrations problem defined by the equations (8.5)(see [147] and [162]), we have that for every admissible vector V = (v, φ) which verifies∫
D(s,z)
(vivi + φ2)dv 6= 0,
∫D(s,z)
vidv = 0,
∫D(s,z)
εijkxjvkdv = 0, (8.17)
it follows that
0 < λ0(z) ≤2
∫D(s,z)
W(V)dv∫D(s,z)
(vivi + φ2)dv. (8.18)
We can assume that U verifies the conditions of the minimum principle (if not we can chooseappropriate values of the constants ai and bi to reduce the problem to this case), and thus,from (8.16) we deduce
Q(s, z) ≤ s(z)
z
∫D(s,z)
W(U)dv, (8.19)
where
s(z) =1
2εaM +
2
λ0(z)ε. (8.20)
37
We choose
ε =2
(aMλ0(z))1/2(8.21)
and in consequence
s(z) = 2
(aM
λ0(z)
)1/2
. (8.22)
From the relation (8.15) we have
dQ
ds=
1
z[U(s+ z)− U(s)] = −1
z
∫D(s,z)
W(U)dv. (8.23)
Using this equality in (8.19) we obtain
s(z)dQ
ds+Q ≤ 0. (8.24)
From (8.23) we deduce that U(s) is a decreasing function, and from (8.15) se deduce that
U(s+ z) ≤ Q(s, z) ≤ U(s), (8.25)
By integration of the relation (8.24) and using the previous inequality, we have
U(s2 + z)
U(s2)≤ e−(s2−s1)/s(z). (8.26)
We can choose s1 = 0, s2 = s− z and we obtain the inequality (8.7).
38
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