Plane-Strain Problems for a Class of Gradient …Plane-Strain Problems for a Class of Gradient Elasticity Models 47 We consider the special case in which a 1 =a 3 =a 5 =0,a 2 = ν

J Elast (2011) 104:45–70DOI 10.1007/s10659-011-9308-7

Plane-Strain Problems for a Class of Gradient ElasticityModels—A Stress Function Approach

Nikolaos Aravas

Received: 11 October 2010 / Published online: 26 January 2011© Springer Science+Business Media B.V. 2011

Abstract The plane strain problem is analyzed in detail for a class of isotropic, compress-ible, linearly elastic materials with a strain energy density function that depends on both thestrain tensor ε and its spatial gradient ∇ε. The appropriate Airy stress-functions and double-stress-functions are identified and the corresponding boundary value problem is formulated.The problem of an annulus loaded by an internal and an external pressure is solved.

Keywords Gradient elasticity · Stress functions

Mathematics Subject Classification (2000) 74A35

1 Introduction

Theories with intrinsic- or material-length-scales find applications in the modeling of size-dependent phenomena. In elasticity, length scales enter the constitutive equations throughthe elastic strain energy function, which in this case depends not only on the strain tensorbut also on gradients of the rotation and strain tensors; in such cases we refer to “gradientelasticity” theories.

A first attempt to incorporate length scale effects in elasticity was made by Mindlin[34], Koiter [27, 28] and Toupin [42]. They solved also a number of problems and demon-strated the effects of the material length scales that enter the strain-gradient elasticity the-ories (Mindlin and Tiersten [37], Mindlin [34, 35], Koiter [28]). Several theoretical issuesrelated to strain-gradient elasticity were addressed later by Germain [23–25]. More recently,

The paper is dedicated to the memory of Professor Donald Carlson.

N. Aravas (�)Department of Mechanical Engineering, University of Thessaly, Volos 38334, Greecee-mail: [email protected]

N. AravasThe Mechatronics Institute, Center for Research and Technology—Thessaly (CERETETH),1st Industrial Area, Volos 38500, Greece

mailto:[email protected]

46 N. Aravas

a variety of “non-local” or “gradient-type” theories have been used in order to introduce ma-terial length scales into constitutive models (Aifantis [1], Pijaudier-Cabot and Bazant [39],Vardoulakis et al. [44–46], de Borst [9], Fleck et al. [16–18], Leblond et al. [32], Tvergaardand Needleman [43]). A common feature of all the aforementioned theories is the non-symmetry of the true stress tensor, and the existence of couple and higher order stresses.Some of the early developments in the theory are summarized in the Nowacki’s book [38]that was published in the 80s. Lazar and Maugin [29–31] have solved a series of problemsincluding dislocations, disclinations, and line forces in the context of gradient elasticity.A summary of the applicability of gradient elasticity to certain micro/nano problems hasbeen given recently by Aifantis [3].

In the present work, the plane strain problem is analyzed in detail for a class of isotropic,linearly elastic materials with a strain energy density function that depends on both the straintensor ε and its spatial gradient ∇ε. The appropriate Airy stress-functions and double-stress-functions are identified and the corresponding boundary value problem is formulated. Theproblem of an annulus loaded by an internal and an external pressure is solved and the effectsof strain gradients are examined. The special cases of a thin-walled tube and an infinite bodywith a cylindrical hole are also analyzed.

Standard notation is used throughout. Boldface symbols denote tensors the orders ofwhich are indicated by the context. The usual summation convention is used for repeatedLatin and Greek indices of tensor components with respect to a fixed Cartesian coordinatesystem with base vectors ei (i = 1,2,3). Latin indices take the values (1,2,3), whereasGreek indices range over the integers (1,2), unless indicated otherwise. A comma followedby a subscript, say i, denotes partial differentiation with respect to the spatial coordinatexi , i.e., f,i = ∂f/∂xi . Let a, b, c be vectors, A, B second-order tensors, and κ , μ third-order tensors; the following products are used in the text: (ab)ij = aibj , (abc)ijk = aibj ck ,(aA)ijk = aiAjk , (a · A)i = ajAji , (aA)ijk = aiAjk , A : B = AijBij , (a · μ)ij = akμkij , κ :μ = κijkμijk , (∇ · μμμ)ij = μkij,k ,1 (μμμ · ∇)ij = μijk,k , (a × A)i = eipqapAqj , (∇ × A)ij =eipqAqj,p , (A ×∇)ij = −ejpqAiq,p , and (∇ ×κκκ)ijk = eipqκqjk,p , where eijk is the alternatingsymbol.

2 The Constitutive Model

We consider the class of elastic materials in which the elastic strain energy density functionW depends on the infinitesimal strain tensor and its spatial gradient. Mindlin and Eshel [36]have shown that for an isotropic, compressible, linearly elastic material, the general form ofW is

W (εεε,κκκ) = G

(εij εij + ν

1 − 2νεiiεjj

)

+ a1κiikκkjj + a2κijj κikk + a3κiikκjjk + a4κijkκijk + a5κijkκkji , (1)

where εεε = 12 (u∇ + ∇u) is the infinitesimal strain tensor, u the displacement field, κκκ = ∇εεε

(κijk = κikj = εjk,i ) the strain gradient third-order tensor, G the elastic shear modulus, ν

Poisson’s ratio, and (a1, a2, a3, a4, a5) material constants.

1With respect to a Cartesian system xi with unit base vectors ei , the gradient operator is written in the form

∇ = ei∂

∂xi= ei ∂i .

Plane-Strain Problems for a Class of Gradient Elasticity Models 47

We consider the special case in which

a1 = a3 = a5 = 0, a2 = ν

1 − 2νG�2, a4 = G�2, (2)

where � is a material length, so that

W (εεε,κκκ) = G

[εij εij + ν

1 − 2νεiiεjj + �2

(κijkκijk + ν

1 − 2νκijj κikk

)]. (3)

We define the stress-like and double-stress-like quantities τττ and μ as follows:

τij = ∂W

∂εij

= 2G

(εij + ν

1 − 2νεkkδij

), (4)

and

μijk = ∂W

∂κijk

= 2G�2

(κijk + ν

1 − 2νκippδjk

). (5)

The above equation can be inverted to yield

εij = 1

2G

(τij − ν

1 + ντkkδij

)and κijk = 1

2G�2

(μijk − ν

1 + νμippδjk

). (6)

Equations (5) imply also that μijk = �2τjk,i and μijk,i = �2τjk,ii = �2∇2τjk , or

μμμ = �2∇τττ and ∇ ·μμμ = �2∇2τττ . (7)

The quantity τ − ∇ · μ that enters (20) below can be written as

τ − ∇ · μ = λεkkδδδ + 2με − �2∇2 (λεkkδδδ + 2με) . (8)

The right hand side of the above equation is formally similar to the expression used for thestress tensor by Aifantis [2] and Altan and Aifantis [4] in their version of a gradient elasticitytheory.

The strain energy density function given in (3) can be written also in the form

W = 1

2τττ : εεε + 1

2μμμ : κκκ = 1

2τττ : εεε + �2

2∇τττ : ∇εεε. (9)

3 The Compatibility Conditions

The compatibility conditions for the strain tensor have the well know form

S ≡ ∇ × εεε × ∇ = −eipmejqnεpq,mneiej = 0. (10)

The above equations are the necessary and sufficient conditions for the existence of a dis-placement field u(x) such that the relation ε = (1/2)(u∇ + ∇u) is satisfied. If the regionof interest B is simply connected, then the ε-compatibility conditions (10) guarantee alsothat u(x) is single valued. An interesting discussion of the strain compatibility conditionsand their relationship to a Stokes theorem for second-order tensor fields has been given byFosdick and Royer-Carfagni [19].

48 N. Aravas

Fig. 1 Triply-connected region(N = 2)

The definition of the strain gradient κ = ∇ε implies the κ-compatibility conditions:

P ≡ ∇ × κκκ = eipqκqjk,peiej ek = 0. (11)

Conversely, satisfaction of the κ-compatibility conditions implies the existence of a tensorfield ε(x) such that κ = ∇ε. Furthermore, if the region is simply connected, ε(x) is single-valued (Courant and John [14]).

If ε and κ are related to τ and μ though (6), the above compatibility conditions can bewritten in the form

S = 1

2G

{∇ × τττ × ∇ + ν

1 + ν

[∇2τkkδδδ − ∇ (∇τkk)]}

= 1

2G

[−eipmejqnτpq,mn + ν

1 + ν

(∇2τkkδij − τkk,ij

)]eiej = 0, (12)

and

P = 1

2G�2

[∇ ×μμμ − ν

1 + ν(∇ ×μμμ)irr eiδδδ

]

= eipq

2G�2

(μqjk,p − ν

1 + νμqrr,pδjk

)eiej ek = 0. (13)

We conclude this section with a discussion of the compatibility conditions in multiply con-nected domains. Assume that the region B is (N +1)-tuply connected, i.e., it has N “holes”.Starting with B, we can form a simply-connected domain by inserting N interior surfacesSn (n = 1,2, . . . ,N) between the outside boundary and each hole (doted lines in Fig. 1).

For u(x) to be single valued, in addition to the ε-compatibility conditions (10), the fol-lowing 6N conditions are required (e.g., see Boley and Weiner [7] pp. 92–95 and 97–100,


Fung [21] pp. 104–107, Fraeijs de Veubeke [20] pp. 90–97):

Su ≡∮Cn

[ε + x × (∇ × ε)] · dx =∮Cn

(εij − xkerikerpqεpj,q

)dxj ei

=∮Cn

[εij − xk

(εij,k − εkj,i

)]dxj ei

= −∮Cn

[xk

(εik,j + εij,k − εkj,i

)]dxj ei = 0, n = 1,2, . . . ,N, (14)

Sω ≡∮Cn

(∇ × ε) · dx =∮Cn

eipqεpj,q dxj ei = 0, n = 1,2, . . . ,N, (15)

where Cn (n = 1,2, . . . ,N) are simple closed curves in B that surround only a single holeor alternatively, each cut only a single surface Sn as introduced above (see Fig. 1).

Equations (15) guarantee that the infinitesimal rotation field is single valued and, underthese conditions, (14) guarantee that the displacement filed u(x) is single valued.

Similarly, it can be shown readily that for ε(x) to be single valued, in addition to theκ-compatibility conditions (11), the following 6N conditions are required:

Pε ≡∮Cn

(dx · κ) =∮Cn

κkij dxkeiej = 0, n = 1,2, . . . ,N. (16)

If ε and κ are related to τ and μ though (6), the compatibility conditions on Cn (14)–(16)can be written in the form

Su = 1

2G

∮Cn

[τττ − ν

1 + ντkkδδδ + x ×

(∇ × τττ + ν

1 + νeijpτkk,peiej

)]· dx, (17)

sω = 1

2G

∮Cn

[x ×

(∇ × τττ + ν

1 + νeijpτkk,peiej

)]· dx, (18)

Pε = 1

2G�2

∮Cn

dx ·(μμμ − ν

1 + νμippeiδδδ

), (19)

where n = 1,2, . . . ,N in all three equations above.

4 The Boundary Value Problem

Consider a homogeneous elastic body that occupies a region B in a fixed reference configu-ration and obeys the constitutive equations (4) and (5). The field equations in B are

∇ · (τττ − ∇ ·μμμ) + b = 0, (20)

εεε = 1

2(u∇ + ∇u) , κκκ = ∇εεε, (21)

50 N. Aravas

τττ = 2G

(εεε + ν

1 − 2νεkkδδδ

), μμμ = 2G�2

(κκκ + ν

1 − 2νκippeiδδδ

)= �2∇τ , (22)

where b is the body force per unit volume. Let ∂B be the smooth boundary of B. Thecorresponding boundary conditions are (Mindlin [34], Mindlin and Eshel [36])

u = u on ∂Bu, (23)

n · (τττ − ∇ ·μμμ) − D · (n ·μμμ) + (D · n)n · (n ·μμμ) = P on ∂BP , (24)

Du = v on ∂Bv, (25)

n · (n ·μμμ) = R on ∂BR, (26)

where (u, v,P,R) are known functions, D = n · ∇ = ∂∂n

is the normal derivative to ∂B,D = ∇ − nD is the “surface gradient” on ∂B, ∂Bu ∪ ∂BP = ∂Bv ∪ ∂BR = ∂B, and ∂Bu ∩∂BP = ∂Bv ∩ ∂BR = ∅.

The boundary value problem defined by (20)–(26) determines u, ε, κ , τ , and μ. Mindlinand Eshel [36] have shown that the conditions G > 0 and −1 < ν < 1/2 are sufficient for aunique solution (see also Georgiadis et al. [22]).

If the boundary ∂B is not smooth, additional boundary conditions are required on theedges of ∂B [34, 36].

The quantities P and R in (24) and (26) are work conjugate to u and Du, and are referredto as the “generalized” or “auxiliary” or “mathematical” forces and double-forces respec-tively. It should be emphasized that τττ and μμμ are defined as work-conjugate quantities to εεε

and κκκ , i.e.,

τττ = ∂W

∂εεεand μμμ = ∂W

∂κκκ, (27)

and then (20) and the boundary conditions (23)–(26) are derived from Hamilton’s principle(Mindlin [34], Mindlin and Eshel [36]). The principles of conservation of linear and angularmomentum introduce the true stress σσσ and the true couple-stress ¯μμμ second-order tensors.The relation of σσσ and ¯μμμ to τττ and μμμ is discussed in Sect. 5. The true force and moment perunit area are defined by t = n · σ and m = n · ¯μμμ and not by P and R, which appear in (24)and (26) (see Mindlin and Eshel [36], p. 124).

We consider next the traction boundary value problem in which ∂Bu = ∂Bv = ∅ and∂BP = ∂BR = ∂B. Let (u(0),ε(0),τττ (0)) be the solution of the classical elasticity problem(� = 0) that corresponds to loads b and P, i.e., the solution of the following boundary valueproblem

∇ · τττ (0) + b = 0, (28)

εεε(0) = 1

2

(u(0)∇ + ∇u(0)

), (29)

τττ (0) = 2G

(εεε(0) + ν

1 − 2νε

(0)kk δδδ

), and (30)

n · τττ (0) = P on ∂B. (31)

We introduce the “gradient correction” (u(1),ε(1),τττ (1),μμμ(1)) so that the solution of theboundary value problem defined by (20)–(26) is written as

u = u(0) + u(1), ε = ε(0) + ε(1), τττ = τττ (0) + τττ (1), μμμ =μμμ(0) +μμμ(1), (32)


where μμμ(0) = �2∇τττ (0). Then, the gradient correction is defined by the following boundaryvalue problem:

∇ · (τττ (1) − ∇ ·μμμ(1)) = ∇ ·μμμ(0), (33)

εεε(1) = 1

2

(u(1)∇ + ∇u(1)

), (34)

τττ (1) = 2G

(εεε(1) + ν

1 − 2νε

(1)kk δδδ

), μμμ(1) = �2∇2τττ (1), (35)

and

n · (τττ (1) − ∇ ·μμμ(1)) − D · (n ·μμμ(1)

) + (D · n)n(n ·μμμ(1)

)= n · (∇ ·μμμ(0)

) + D · (n ·μμμ(0)) − (D · n)n

(n ·μμμ(0)

)on ∂B, (36)

n · (n ·μμμ(1)) = R − n · (n ·μμμ(0)

)on ∂B. (37)

Equations (33)–(37) show that if R = 0, then the “loads” on the right hand side of (33),(36), and (37) are all proportional to μ(0) (which equals �2∇τ (0)). Therefore, when R = 0,if μ(0) is proportional to a parameter, then the gradient correction (u(1),ε(1),τττ (1),μμμ(1)) isproportional to the same parameter. An application of this property will be discussed inSect. 7.1, where the problem of an annulus loaded by internal and external pressure is solved.

5 The “True” Stress and the “True” Couple Stress

The quantities τττ and μμμ have dimensions of stress and double-stress respectively and aredefined by (4) and (5) as work-conjugate to ε and κ (Mindlin [34]). However, the physicalinterpretation of τττ and μμμ is not clear. Mindlin and Eshel [36] established the relationshipbetween τττ and μμμ and the true stress tensor σ and the true couple stress tensor ¯μμμ as describedbriefly in the following.

Mindlin and Eshel [36] consider the standard true stress vector t and true couple-stressvector m, defined the usual way as force and moment per unit area. Then, the standardargument of force and moment equilibrium of an infinitesimal tetrahedron leads to the wellknow relations

t = n · σ and m = n · ¯μ, (38)

where n is the unit vector normal to the infinitesimal area, and (σ , ¯μ) are the usual stressand couple-stress second-order tensors.

The principles of conservation of linear and angular momentum lead to the well knownequations

∇ · σ + b = 0 and ∇ · ¯μμμ + s + c = 0, (39)

where c is the body moment per unit volume and si = eijkσjk .Let �� be the double-force per unit volume. Mindlin and Eshel [36] identify the sym-

metric part of �� with the body double-force without moment per unit volume, and the anti-symmetric part of �� with the body moment per unit volume with its components defined as

52 N. Aravas

�[ij ] = 12 eijkck or ci = eijk�[jk]. Mindlin and Eshel [36] have shown that (see also Germain

[23–25], Aravas and Giannakopoulos [6])

σσσ = τττ − 2

3μμμ · ∇ − 1

3∇ ·μμμ −�� or σij = τij − 2

3μijk,k − 1

3μkij,k − �ij , (40)

and

¯μij = 2

3μpqiepqj . (41)

6 Plane Strain Problems

We consider the plane displacement field

u = uα (x1, x2) eα. (42)

The corresponding strain εεε and strain-gradient κκκ fields are of the form

εεε = εαβeαeβ and κκκ = ∇εεε = καβγ eαeβeγ , (43)

with

εαβ = 1

2

(uα,β + uβ,α

)and καβγ = εβγ,α = 1

2

(uβ,γ α + uγ,βα

). (44)

The plane strain conditions

ε33 = 1

2μ

(τ33 − ν

1 + ντkkδ33

)= 0 and κα33 = 1

2�2μ

(μα33 − ν

1 + νμαppδ33

)= 0,

(45)imply that

τ33 = νταα and μα33 = νμαββ, (46)

and τττ and μμμ are of the form

τττ = ταβeαeβ + ντααe3e3 and μμμ = μαβγ eαeβeγ + νμαββeαe3e3. (47)

The constitutive equations can be written now in the form

ταβ = 2G

(εαβ + ν

1 − 2νεγγ δαβ

), μαβγ = 2�2G

(καβγ + ν

1 − 2νκαδδδβγ

), (48)

with

ταα = 2G

1 − 2νεαα and μαββ = 2�2G

1 − 2νκαββ . (49)

The above equations can be inverted to yield

εαβ = 1

2G

(ταβ − ντγγ δαβ

), καβγ = 1

2�2G

(μαβγ − νμαδδδβγ

). (50)


6.1 The Compatibility Equations

The compatibility equations (12) and (13) now take the form

S = (2ε21,12 − ε11,22 − ε22,11

)e3e3

= 1

2G

(2τ21,12 − τ11,22 − τ22,11 − ν∇2ταα

)e3e3 = 0, (51)

and

P = (κ2βγ,1 − κ1βγ,2

)e3eβeγ

= 1

2�2G

[μ2βγ,1 − μ1βγ,2 − ν

(μ2αα,1 − μ1αα,2

)δβγ

]e3eβeγ = 0. (52)

The above compatibility equations can be written in the form

2τ21,12 − τ11,22 − τ22,11 − ν∇2ταα = 0, (53)

and

(1 − ν)(μ211,1 − μ111,2) − ν(μ222,1 − μ122,2) = 0, (54)

(1 − ν)(μ222,1 − μ122,2) − ν(μ211,1 − μ111,2) = 0, (55)

μ212,1 − μ112,2 = 0. (56)

For ν �= 1/2, the κ-compatibility equations (54)–(56) are equivalent to

μ211,1 − μ111,2 = 0, (57)

μ222,1 − μ122,2 = 0, (58)

μ212,1 − μ112,2 = 0. (59)

If the region of interest is (N + 1)-tuply connected, the following additional 6N conditionsshould be satisfied on the closed curves Cn (n = 1,2, . . . ,N) introduced in Sect. 3 above:

Suα =

∮Cn

(εaβ − xζ eαζ eγ δεγβ,δ

)dxβ

= 1

2G

∮Cn

[ταβ − ντγγ δαβ − xζ eαζ eγ δ

(τγβ,δ − ντηη,δδγβ

)] = 0, n = 1,2, . . . ,N,

(60)

Sω3 =

∮Cn

eβγ εβα,γ dxα = 1

2G

∮Cn

eβγ

(ταβ,γ − ντδδ,γ δαβ

)dxα, n = 1,2, . . . ,N, (61)

P eαβ =

∮Cn

κγαβ dxγ = 1

2G

∮Cn


)dxγ , n = 1,2, . . . ,N. (62)

54 N. Aravas

6.2 The Plane-Strain Traction Boundary Value Problem in Terms of τττ and μ

For simplicity we assume for the rest of the paper that the body forces vanish (b = 0). Thenthe plane strain boundary value problem becomes

(ταβ − μγαβ,γ

),β

= 0, (63)

2τ21,12 − τ11,22 − τ22,11 − ν∇2ταα = 0, (64)

μγαβ = �2ταβ,γ , (65)

and

nβ

(τβα − μγβα,γ

) − Dβ

(nγ μγβα

) + (Dδnδ)nγ nβμγβα = Pα on ∂B, (66)

nγ nβμγβα = Rα on ∂B, (67)

where B is now a two-dimensional domain and ∂B its smooth boundary.Equations (63)–(65) form a linear second-order system of equations, which together

with the boundary conditions (66) and (67), define (τ11, τ22, τ12) and (μ111,μ122,μ112,μ211,

μ222,μ212).We can eliminate the double-stress μμμ from the above equations and restate the boundary

value problem in the following form:

(ταβ − �2∇2ταβ

),β

= 0, (68)

2τ21,12 − τ11,22 − τ22,11 − ν∇2ταα = 0 (69)

and

nβ


) − �2Dβ

(nγ ταβ,γ

) + �2 (Dδnδ)nγ nβταβ,γ = Pα on ∂B, (70)

�2nγ nβταβ,γ = Rα on ∂B. (71)

Equations (68) and (69) form a linear third-order system of equations, which together withthe boundary conditions (70) and (71), define the three unknowns (τ11, τ22, τ12).

If the region of interest is (N + 1)-tuply connected, the following additional conditionsshould be satisfied on the closed curves Cn (n = 1,2, . . . ,N) introduced in Sect. 3 above:

Suα = 1

2G

∮Cn

[ταβ − ντγγ δαβ − xζ eαζ eγ δ

(τγβ,δ − ντηη,δδγβ

)] = 0, (72)

Sω3 = 1

2G

∮Cn

eβγ


)dxα, (73)

P eαβ = 1

2G

∮Cn


)dxγ . (74)

6.3 Stress and Double-Stress Functions

Schaefer [40], Mindlin [33] and Carlson [10] derived the appropriate stress functionsfor plane problems with couple stresses. For the particular case of a homogeneous,


isotropic, linearly elastic constrained-Cosserat material, they used the equilibrium and the κ-compatibility conditions to show that the stress and couple-stress components can be writtenin terms of two stress functions (see also Boresi and Chong [8], pp. 390–398).

We follow a similar approach here, starting with the κ-compatibility conditions (57)–(59):

μ211,1 = μ111,2, μ222,1 = μ122,2, μ212,1 = μ112,2. (75)

According to the theory of total differentials (Courant and John [14]), the above equationsimply the existence of functions F11, F22, and F12 (single valued if B is simply connected)such that

μ111 = ∂F11

∂x1, μ211 = ∂F11

∂x2, (76)

μ122 = ∂F22

∂x1, μ222 = ∂F22

∂x2, (77)

μ112 = ∂F12

∂x1, μ212 = ∂F12

∂x2. (78)

The above equations can be written compactly in the form

μαβγ = ∂Fβγ

∂xα

or μμμ = ∇F, (79)

with F21 = F12. Comparison of (79) with the constitutive equation (7), i.e.,

μμμ = �2∇τττ (80)

leads to the conclusion that F can be identified with �2τ , i.e., the second order tensor τττ canbe viewed as a double-stress function in the present model.

We consider next equations (68):


),β

= 0, (81)

which imply the existence of the well-known Airy [5] stress function f (single valued if Bis simply connected) such that

τ11 − �2∇2τ11 = f,22, τ22 − �2∇2τ22 = f,11, τ12 − �2∇2τ12 = −f,12. (82)

The above equations can be written in compact form as

ταβ − �2∇2ταβ = ∇2f δαβ − f,αβ = eαγ eβδf,γ δ, (83)

where eαβ is the alternating symbol in two dimensions that takes the values e11 = e22 = 0and e12 = −e21 = 1.

It should be noted that (82) are a consequence of (81) and are independent of any con-stitutive equations τττ and μμμ may be required to satisfy. However, (79) are valid only for theparticular gradient elasticity model described in Sect. 2. Günther [26], Schaefer [40], andCarlson [11–13] presented stress functions for couple and dipolar stresses appropriate forall materials.

56 N. Aravas

The plane-strain boundary value problem can be written now in terms of τij and the stressfunction f as follows

τ11 − �2∇2τ11 = f,22, (84)

τ22 − �2∇2τ22 = f,11, (85)

τ12 − �2∇2τ12 = −f,12, (86)

2τ21,12 − τ11,22 − τ22,11 − ν∇2ταα = 0, (87)

and

nβ

(∇2f δαβ − f,αβ

) − �2Dβ

(nγ ταβ,γ

) + �2 (Dδnδ) nγ nβταβ,γ = Pα on ∂B, (88)

�2nγ nβταβ,γ = Rα on ∂B. (89)

Equations (84)–(87) form a linear second-order system of equations, which together withthe boundary conditions (88) and (89), define the four unknowns (f, τ11, τ22, τ12). The stressfunction f is defined to within a linear function of the form kαxα + c that causes no stress.

If the domain is multiply connected, conditions (72)–(74) are imposed as well.If polar coordinates are used, equations (84)–(87) have the form

τrr − �2

{∂2τrr

∂r2+ 1

r

∂τrr

∂r+ 1

r2

[∂2τrr

∂θ2− 4

∂τrθ

∂θ− 2 (τrr − τθθ )

]}= 1

r

∂f

∂r+ 1

r2

∂2f

∂θ2, (90)

τθθ − �2

{∂2τθθ

∂r2+ 1

r

∂τθθ

∂r+ 1

r2

[∂2τθθ

∂θ2+ 4

∂τrθ

∂θ+ 2 (τrr − τθθ )

]}= ∂2f

∂r2, (91)

τrθ − �2

{∂2τrθ

∂r2+ 1

r

∂τrθ

∂r+ 1

r2

[∂2τrθ

∂θ2+ 2

∂

∂θ(τrr − τθθ ) − 4τrθ

]}= − ∂

∂r

(1

r

∂f

∂θ

), (92)

−∂2τθθ

∂r2+ 2

r

∂2τrθ

∂r∂θ+ 1

r

∂

∂r(τrr − 2τθθ ) + 2

r2

∂τrθ

∂θ− 1

r2

∂2τrr

∂θ2+ ν∇2 (τrr + τθθ ) = 0. (93)

The boundary conditions (88) and (89) can be stated in polar coordinated, if we take intoaccount that

∇ = er

∂

∂r+ eθ

1

r

∂

∂θ, ∇2 = ∂2

∂r2+ 1

r

∂

∂r+ 1

r2

∂2

∂θ2, (94)

D = ernθ

(nθ

∂

∂r− nr

1

r

∂

∂θ

)+ eθnr

(nr

r

∂

∂θ− nθ

∂

∂r

), (95)

and

∇τττ = ∂τrr

∂rererer + ∂τθθ

∂rereθ eθ + ∂τrθ

∂rer (ereθ + eθ er )

+ 1

r

(∂τrr

∂θ− 2τrθ

)eθ erer + 1

r

(∂τθθ

∂θ+ 2τrθ

)eθ eθ eθ

+ 1

r

(∂τrθ

∂θ+ τrr − τθθ

)eθ (ereθ + eθ er ) . (96)


In particular, on a circular arc of radius r , n = ±er (nr = ±1, nθ = 0) and

D = eθ

1

r

∂

∂θ, D · n = Dαnα = ±1

r. (97)

The polar components of (72)–(74), required in multiply connected domains, are given inAppendix A.

6.4 Axisymmetric Problems

We consider plane-strain axisymmetric problems, in which the solution is independent of θ

and

uθ = 0, τrθ = 0. (98)

Equations (90)–(93) reduce now to

−�2 d2τrr

dr2− �2

r

dτrr

dr+

(1 + 2

�2

r2

)τrr − 2

�2

r2τθθ = 1

r

df

dr, (99)

−�2 d2τθθ

dr2− �2

r

dτθθ

dr+

(1 + 2

�2

r2

)τθθ − 2

�2

r2τrr = d2f

dr2, (100)

νd2τrr

dr2− (1 − ν)

d2τθθ

dr2+ 1 + ν

r

dτrr

dr− 2 − ν

r

dτθθ

dr= 0. (101)

On a circular boundary of radius r with outward unit normal n = ±er , the boundary condi-tions (88) and (89) become

±1

r

(df

dr+ �2 dτθθ

dr

)= Pr and �2 dτrr

dr= Rr on ∂B. (102)

Equations (99)–(102) form a second-order system of linear ordinary differential equationswhich defines the Airy stress function f and the “double stress functions” τrr and τθθ .

If the domain is multiply connected, conditions (72)–(74) should be imposed as well;the form of these conditions on circular boundaries for axisymmetric problems is given inAppendix A.

In the following we present the general solution of the aforementioned system of equa-tions. The form of (99) and (100) suggests that the problem can be simplified if we add andsubtract them. In fact, if we define

S = τrr + τθθ , D = τrr − τθθ , and F = 1

r

df

dr, (103)

so that

τrr = S + D

2, τθθ = S − D

2, and f =

∫rF dr, (104)

addition and subtraction of (99) and (100) yield

−�2 d2S

dr2− �2

r

dS

dr+ S = 1

r

d

dr

(r2F

)(105)

58 N. Aravas

and

�2 d2D

dr2− �2

r

dD

dr+

(1 + 4

�2

r2

)D = −r

dF

dr, (106)

whereas (101) becomes

(1 − 2ν)d

dr

(rdS

dr

)= 1

r2

d

dr

(r3 dD

dr

). (107)

Elimination of D and F from (105)–(107) leads to the following fourth-order ordinary dif-ferential equation for S

�2 d4S

dr4+ 2�2

r

d3S

dr3−

(1 + �2

r2

)d2S

dr2− 1

r

(1 − �2

r2

)dS

dr= 0, (108)

and rest of the solution is defined by

dD

dr= 1

r3

[(1 − 2ν)

∫r2 d

dr

(rdS

dr

)dr + c5

], (109)

D = 1

2

[2�2 (1 − ν)

(rd3S

dr3+ d2S

dr2− 1

r

dS

dr

)− r

(dS

dr+ dD

dr

)], (110)

τrr = S + D

2, τθθ = S − D

2, (111)

F = −�2 dτ 2rr

dr2− 2

�2

r

dτrr

dr+ τrr + 2

�2

r2(τrr − τθθ ), (112)

f =∫

rF dr. (113)

where c5 is an arbitrary constant.The general solution of (108) is

S(r) = c1 + c2K0

( r

�

)+ c3I0

( r

�

)+ c4 ln

r

�, (114)

where c1, c2, c3, and c4 are arbitrary constants, and (In,Kn) are modified Bessel functionsof the first and second kind. The rest of the solution is determined by (109)–(113):

τrr (r) = 1

2

(c1 − c6

r2

)+ c2

2

[K0

( r

�

)+ (1 − 2ν)K1

( r

�

)]

+ c3

2

[I0

( r

�

)+ (1 − 2ν) I1

( r

�

)]+ c4

2

(ln

r

�− 1

2

), (115)

τθθ (r) = 1

2

(c1 + c6

r2

)+ c2

2

[K0

( r

�

)− (1 − 2ν)K1

( r

�

)]

+ c3

2

[I0

( r

�

)− (1 − 2ν) I1

( r

�

)]+ c4

2

(ln

r

�+ 1

2

), (116)

f (r) = c1

4r2 − c6

2ln r − c4�

2

[ln r + r2

4�2

(1 − ln

r

�

)], (117)

where c6 = c5 + 4 (1 − 2ν) c2�2.


The terms that involve c4 in the above solution are not consistent with the assumption ofan axisymmetric solution;2 therefore we set c4 = 0.

The corresponding displacement is

ur(r) = 1

2G

{1 − 2ν

2c1r + c6

2r− (1 − 2ν) �

[c2K1

( r

�

)− c3I1

( r

�

)]}. (119)

Finally, the boundary conditions (102) on a circular boundary with outward unit normaln = ±er become

Pr(r) = ±{

1

2

(c1 − c6

r2

)− c2

�

r

[νK1

( r

�

)− (1 − 2ν)K2

( r

�

)]

+ c3�

r

[νI1

( r

�

)+ (1 − 2ν) I2

( r

�

)]− c6

2

�2

r4

}, (120)

and

Rr(r) = −c2�[(1 − ν)K1

( r

�

)+ (1 − 2ν)K2

( r

�

)]

+ c3�[(1 − ν) I1

( r

�

)− (1 − 2ν) I2

( r

�

)]+ c6

�2

r3. (121)

Assuming that the double-force per unit volume vanishes, i.e. �� = 0, and using (40) and(41), we conclude that the only non-zero in-plane components the double-stress μμμ, the truestress σσσ , and true couple stress ¯μμμ are as follows (see also Appendix A in Aravas and Gian-nakopoulos [6])

μrrr = �2 dτrr

dr, μrθθ = �2 dτθθ

dr, μθrθ = μθθr = �2 τrr − τθθ

r, (122)

σrr = τrr − dμrrr

dr− 1

r

(μrrr − 2

3μrθθ − 4

3μθrθ

), (123)

σθθ = τθθ − 1

3

dμrθθ

dr− 2

3

dμθrθ

dr− μrθθ + 2μθrθ

r, (124)

and

¯μθ3 = − ¯μ3θ = 2

3(μrθθ − μθrθ ) . (125)

2The strains corresponding to the c4-terms are

εrr = − c4

4G[1 − 2 (1 − 2ν) ln r] and εθθ = c4

4G[1 + 2 (1 − 2ν) ln r] . (118)

In axisymmetric problems with uθ = 0, the radial and hoop strains are εrr = dur/dr and εθθ = ur/r . There-fore, if we evaluate ur = rεθθ and substitute it into εrr = dur/dr , we conclude that c4 = 0. The c4-termsare known to correspond to uθ displacements that are not single-valued and cannot be used in multiply con-nected domains (such as an annulus) expect in those special problems involving “dislocations” (see followingSect. 7.1 and Soutas-Little [41], p. 159).

60 N. Aravas

Fig. 2 Annulus loaded byinternal and external pressure

7 Examples

7.1 An Annulus Subjected to Internal and External Pressure

We consider the problem of an annulus loaded by an internal pressure pi and an externalpressure po under conditions of plane strain (Fig. 2).

The domain is doubly connected; therefore, the conditions (72)–(74) should be imposedon the inner and outer circular boundaries of the annulus. These conditions together withthe strain compatibility equation (101) guarantee that the corresponding displacement fieldis single valued in the doubly connected domain. As discussed in Appendix A, conditions(72) and (74) are satisfied identically, whereas (73) requires that the following condition besatisfied [

−νdτrr

dr+ (1 − ν)

dτθθ

dr− τrr − τθθ

r

]r=a

= 0. (126)

Using (115) and (116) for τrr and τθθ in the above equation, we find

c41 − ν

a= 0, (127)

i.e., it is required that c4 = 0 for the displacement field to be single valued. This finding isconsistent with the remark made immediately after (116) concerning the vanishing of c4.

In the case of the classical “local” linear isotropic elasticity (� = 0), the solution is of theform

f (0) = 1

2Ar2 + B ln r, σ (0)

rr = τ (0)rr = A + B

r2, σ

(0)θθ = τ

(0)θθ = A − B

r2, (128)

and

u(0)r = 1

2G

[(1 − 2ν)Ar − B

r

], (129)

where

A = pia2 − pob

2

b2 − a2, B = (po − pi)

a2b2

b2 − a2, (130)

and (pi,po) are the inner and outer applied pressure loads as shown in Fig. 2.


In order to make connection with the aforementioned “local” solution, we set

c1 = c7 + 2A, c6 = c8 − 2B, (131)

so that the gradient elasticity solution is written in the form

f (r) = f (0)(r) + c7

4r2 − c8

2ln r, (132)

τrr (r) = τ (0)rr (r) + 1

2

(c7 − c8

r2

)+ c2

2

[K0

( r

�

)+ (1 − 2ν)K1

( r

�

)]

+ c3

2

[I0

( r

�

)+ (1 − 2ν) I1

( r

�

)], (133)

τθθ (r) = τ(0)θθ (r) + 1

2

(c7 + c8

r2

)+ c2

2

[K0

( r

�

)− (1 − 2ν)K1

( r

�

)]

+ c3

2

[I0

( r

�

)− (1 − 2ν) I1

( r

�

)], (134)

and

εrr (r) = ε(0)rr (r) + 1

2G

{1 − 2ν

2c7 − c8

2r2

+ (1 − 2ν)

{c2

[K0

( r

�

)+ �

rK1

( r

�

)]+ c3

[I0

( r

�

)− �

rI1

( r

�

)]}}, (135)

εθθ (r) = ε(0)θθ (r) + 1

2G

{1 − 2ν

2c7 + c8

2r2

− (1 − 2ν)�

r

[c2K1

( r

�

)− c3I1

( r

�

)]}, (136)

ur(r) = rεθθ = u(0)r (r) + 1

2G

{1 − 2ν

2c7r + c8

2r

− (1 − 2ν) �[c2K1

( r

�

)− c3I1

( r

�

)]}. (137)

The only non-zero in-plane components of the true stress σσσ and the true couple-stress¯μμμ are

σrr (r) = σ (0)rr (r) + 1

2

(c7 − c8

r2

)− 2ν

3

�

r

[c2K1

( r

�

)− c3I1

( r

�

)], (138)

σθθ (r) = σ(0)θθ (r) + 1

2

(c7 + c8

r2

)+ 2ν

3

{c2

[K0

( r

�

)+ �

rK1

( r

�

)]

+ c3

[I0

( r

�

)− �

rI1

( r

�

)]}, (139)

and

¯μ3θ (r) = − ¯μθ3(r) = 2ν

3�[c2K1

( r

�

)− c3I1

( r

�

)]. (140)

62 N. Aravas

Fig. 3 Spatial variation ofnormalized displacement(|ur |/a)/(pi/G) for ν = 0.3,�/a = 0.2, b = 2a, and po = 2pi

The terms that involve c2, c3, c7, and c8 define now the “gradient correction”, where theconstants c2, c3, c7, and c8 are determined from the following boundary conditions:

on r = a (n = −er ) : Pr(a) = pi, Rr(a) = 0, (141)

on r = b (n = er ) : Pr(b) = −po, Rr(b) = 0. (142)

The boundary conditions (141) and (142) define a system of four algebraic linear equationsthat is solved for c2, c3, c7 and c8. The resulting expressions for c2, c3, c7 and c8 are lengthyand are listed in Appendix B.

We note that the solution is of the form

f (r) = f (0)(r) + (po − pi)Q

(r

a, ν,

b

a,�

a

), (143)

τrr (r) = τ (0)rr (r) + (po − pi)Trr

(r

a, ν,

b

a,�

a

), (144)

τθθ (r) = τ(0)θθ (r) + (po − pi)Tθθ

(r

a, ν,

b

a,�

a

), (145)

ur(r) = u(0)r (r) + po − pi

G/aU

(r

a, ν,

b

a,�

a

), (146)

where (Q,Trr , Tθ,θ ,U) are dimensionless functions defined in Appendix B.It is interesting to note that, whereas the classical solution depends on the individual val-

ues of pi and po though A and B , the “gradient correction” is proportional to the differencepo − pi (see (143)–(146)). An explanation of this is given at the end of Sect. 4, where isnoted that, if R = 0, the “gradient correction” is proportional to the magnitude of μμμ(0). Inthe present problem R = 0, and (128) and (130) show that

μμμ(0) = �2∇τττ (0) ∼ B ∼ p0 − pi. (147)

Therefore, the “gradient correction” is proportional to p0 − pi , i.e.,

(f (1),u(1),τττ (1)

) ∼ p0 − pi. (148)

Figure 3 shows the radial variation of the normalized radial displacement ur . In Fig. 3and in all following figures of this section ν = 0.3, �/a = 0.3, b/a = 2, and po/pi = 2.


Fig. 4 Spatial variation of normalized radial and hoop stresses for ν = 0.3, �/a = 0.2, b = 2a, and po = 2pi .All stress components shown in the figure are compressive

Fig. 5 Spatial variation of normalized strains εrr /(pi/G) and |εθθ |/(pi/G) for ν = 0.3, �/a = 0.2, b = 2a,and po = 2pi

As shown in Fig. 3 |ugrr (r)| < |ucl

r (r)| ∀r , where the superscripts “gr” and “cl” denote thegradient and classical solution respectively; i.e., the gradient elastic material appears to bestiffer. Also, |ugr

r (r)| varies monotonically with r , whereas the classical solution is such that|ucl

r (r)| has a minimum at r 1.2a.Figure 4 shows the spatial variation of the normalized radial and hoop stress components

of τττ , the true stress σσσ , and the classical solution σσσ (0). All stress components shown in Fig. 4are compressive. The gradient solution predicts a higher compressive radial stress and alower compressive hoop stress. It is interesting to note that τrr is substantially different fromthe true stress σrr , whereas τθθ σθθ .

Figure 5 shows the spatial variation of the normalized strain components εrr/(pi/G)

and |εθθ |/(pi/G) for both the classical and the gradient solution. The radial strain εrr in thegradient solution remains compressive everywhere in the annulus, whereas both tensile andcompressive radial strains appear in the classical solution, reflecting the fact that |ucl

r (r)| hasa minimum in the range a ≤ r ≤ b (see Fig. 3).

We consider next the case of a thin-walled annulus. Let t be the thickness and R the meanradius of the annulus, i.e.,

t = b − a, R = a + b

2, so that a = R − t

2, b = R + t

2, (149)

64 N. Aravas

where

t

R≡ ε � 1. (150)

If we set a = R − t/2, b = R + t/2, evaluate (133), (134), (139), and (140) at r = R, andexpand the solution in ε = t/R, we find that

τrr (R) = − (po − pi)

3−2ν1−ν

( �R)2

1 + 2 3−2ν1−ν

( �R)2

R

t− po + pi

2+ O (εp) , (151)

τθθ (R) = − (po − pi)1 + 3−2ν

1−ν( �

R)2

1 + 2 3−2ν

1 − ν( �

R)2

R

t− po + pi

2+ O (εp) , (152)

and

σrr (R) = − (po − pi)

3−2ν3(1−ν)

( �R)2

1 + 2 3−2ν1−ν

( �R)2

R

t− po + pi

2+ O (εp) , (153)

σθθ (R) = − (po − pi)1 + (3−2ν)(4−5ν)

3(1−ν)2 ( �R)2

1 + 2 3−2ν1−ν

( �R)2

R

t− po + pi

2+ O (εp) , (154)

where p is a typical pressure of order po or pi .The corresponding thin-wall solution of the classical theory (� = 0) is

σ (0)rr (R) = τ (0)

rr = −po + pi

2+ O (εp) , (155)

σ(0)θθ (R) = τ

(0)θθ = − (po − pi)

R

t− po + pi

2+ O (εp) . (156)

It is interesting to note that the gradient theory compared to the classical local theory predictshigher values for the radial stress and lower values for the hoop stress. In particular, whereasthe classical theory predicts a radial stress of order p, the gradient theory predicts a radialstress of order p

ε( �

R)2.

7.2 Infinite Body with Cylindrical Hole

We consider the problem of a infinite body with a cylindrical hole of radius a (Fig. 6). Thehole is loaded by an internal pressure pi and a pressure po is applied at infinity (Fig. 6). Thesolution to this problem can be found from the solution developed in Sect. 7.1 by consideringthe limit b → ∞.

In the limit as b → ∞, the constants that enter the solution take the values

A = −po, B = (po − pi)a2, c3 = c7 = 0, (157)

and

c2 = −po − pi

c, c8 = 2

po − pi

ca�K1

(a

�

), (158)


Fig. 6 Infinite body withcircular hole

with

c = 1 − 2ν

2K0

(a

�

)+ 1 − ν

2

(a

�+ 4

�

a

)K1

(a

�

)(159)

and the solution becomes

τrr (r) = −po + (po − pi)a2

r2

− po − pi

c

{a�

r2K1

(a

�

)+ (1 − ν)K0

( r

�

)+ (1 − 2ν)

�

rK1

( r

�

)}, (160)

τθθ (r) = −po − (po − pi)a2

r2

+ po − pi

c

[a�

r2K1

(a

�

)− νK0

( r

�

)+ (1 − 2ν)

�

rK1

( r

�

)], (161)

f (r) = −1

2por

2 + (po − pi)a2 ln r − po − pi

ca�K1

(a

�

)ln r. (162)

The corresponding displacement field is

ur(r) = − (1 − 2ν)po

2Gr − po − pi

2G

a2

r+ po − pi

2Gc�[a

rK1

(a

�

)+ (1 − 2ν)K1

( r

�

)].

(163)

This solution was communicated to the author by Prof. Exadaktylos [15] in 2001.

66 N. Aravas

The corresponding non-zero in-plane true stresses and true couple-stresses are

σrr = −p0 + (p0 − pi)a2

r2− p0 − pi

c

[a�

r2K1

( r

�

)− 2ν

3

�

rK1

( r

�

)], (164)

σθθ = −p0 − (p0 − pi)a2

r2+ p0 − pi

c

{a�

r2K1

(a

�

)− 2ν

3

[K0

( r

�

)+ �

rK1

( r

�

)]}, (165)

and

¯μ3θ = − ¯μθ3 = −2ν

3

p0 − pi

c�K1

( r

�

). (166)

Acknowledgements Fruitful discussions with Prof. A.E. Giannakopoulos of the University of Thessalyare gratefully acknowledged. The author would like also to thank Prof. D. Panagiotounakos of the NationalTechnical University of Athens for helpful discussions and material.

Appendix A: Plane Strain Compatibility Equations in Polar Coordinatesfor Multiply-Connected Regions

In multiply connected domains the additional compatibility conditions are

Su =∮Cn

[εεε + x × (∇ × εεε)] · dx = 0, (167)

Sω =∮Cn

(∇ × εεε) · dx = 0, (168)

Pε =∮Cn

(dx · κκκ) = 0. (169)

The polar coordinates of the quantities that appear in gradient elasticity theories can befound in Appendix A of Aravas and Giannakopoulos [6]. Here, we present the form of theabove compatibility equations in polar coordinates.

We consider the plane-strain problem of Sect. 6 and introduce polar coordinates (r, θ).The position vector x, its differential dx, and the gradient operator on the plane are

x = rer , dx = dr er + r dθ eθ , ∇ = ∂

rer + 1

r

∂

∂θeθ , (170)

where (er , eθ ) are the unit vectors of the polar coordinate system.The strain tensor ε and the strain gradient tensor κ can be written as

εεε = εrr er er + εθθ eθ eθ + εrθ (er eθ + eθ er ) , (171)

κκκ = κrrr er er er + κrθθ er eθ eθ + κrrθ er (er eθ + eθ er )

+ κθrr eθ er er + κθθθ eθ eθ eθ + κθrθ eθ (er eθ + eθ er ) . (172)


We can evaluate the quantities that appear in (167)–(169) as follows:

[εεε + x × (∇ × εεε)] · dx

=[εrr er −

(r∂εrθ

∂r− ∂εrr

∂θ+ εrθ

)eθ

]dr

+ r

[εrθ er −

(r∂εθθ

∂r− ∂εrθ

∂θ+ 2εθθ − εrr

)eθ

]dθ, (173)

(∇ × εεε) · dx

=[(

∂εrθ

∂r− 1

r

∂εrr

∂θ+ 2εrθ

r

)dr +

(r∂εθθ

∂r− ∂εrθ

∂θ+ εθθ − εrr

)dθ

]e3, (174)

dx · κκκ = [κrrrerer + κrθθ eθ eθ + κrrθ (ereθ + eθ er )]dr

+ [κθrrerer + κθθθ eθ eθ + κθrθ (ereθ + eθ er )] r dθ. (175)

In the above equations the unit vectors (er , eθ ) are functions of θ :

er (θ) = cos θ e1 + sin θ e2, eθ (θ) = − sin θ e1 + cos θ e2, (176)

where (e1, e2) are the base vectors of a fixed Cartesian coordinate system in the plane wherethe polar system is defined.

On a circular contour of radius r , dr = 0 and the conditions (167)–(169) can be writtenas follows:

Su = r

2π∫0

[εrθ er −

(r∂εθθ

∂r− ∂εrθ

∂θ+ 2εθθ − εrr

)eθ

]dθ = 0, (177)

Sω = r

2π∫0

(∂εθθ

∂r− 1

r

∂εrθ

∂θ− εrr − εθθ

r

)dθ e3 = 0, (178)

Pε = r

2π∫0

[κθrr er er + κθθθ eθ eθ + κθrθ (er eθ + eθ er )]dθ = 0, (179)

where er (θ) and eθ (θ) are defined by (176).All the above equations can be written in terms of the components of τττ if we use the

constitutive equations

εαβ = 1

2G

(ταβ − ντγγ δαβ

), καβγ = 1

2G�2

[(∇τττ)αβγ − ν (∇τδδ)α δβγ

], (180)

where the polar components of ∇τττ are defined by (96) and

∇τδδ = ∂(τrr + τθθ )

∂rer + 1

r

∂(τrr + τθθ )

∂θeθ . (181)

In axisymmetric problems we have that

εrθ = 0, κrrθ = κrθr = κθrr = κθθθ = 0,∂

∂θ= 0. (182)

68 N. Aravas

In that case, if we take into account (176) and that the axisymmetric solution is independentof θ , we conclude that (177) and (179) are satisfied identically, i.e.,

Su = Pε = 0, (183)

whereas (178) takes the form

Sω = 2π

Gr

[−ν

dτrr

dr+ (1 − ν)

dτθθ

dr− τrr − τθθ

r

]e3 = 0. (184)

Appendix B: Constants in the Solution of the Annulus Problem

The constants c2, c3, c7, and c8 that enter (132)–(140) of the solution of the annulus problemtake the values

c22(po−pi )

�a�

b2

= 1 − ν

1 − 2ν

[a3

b3I1

(a

�

)− I1

(b

�

)]− �

b

[a2

b2I2

(a

�

)− I2

(b

�

)], (185)

c32(po−pi )

�a�

b2

= 1 − ν

1 − 2ν

[a3

b3K1

(a

�

)− K1

(b

�

)]+ �

b

[a2

b2K2

(a

�

)− K2

(b

�

)], (186)

c7

4(po−pi )

�a �2

b(b2−a2)

=(

1 + a2

b2

)�2

b2− a

b

{�

b

[I2

(b

�

)K1

(a

�

)+ a

bI2

(a

�

)K1

(b

�

)]

−[

1 − ν

1 − 2ν

(1 − a2

b2

)K1

(a

�

)− a�

b2K2

(a

�

)]I1

(b

�

)

+[

1 − ν

1 − 2ν

(1 − a2

b2

)K1

(b

�

)+ �

bK2

(b

�

)]I1

(a

�

)}, (187)

c8

4(po−pi )

�a2 �2

b2−a2

= 2a �2

b3− �

b

[I2

(b

�

)K1

(a

�

)+ a3

b3I2

(a

�

)K1

(b

�

)]

+[

1 − ν

1 − 2ν

(1 − a4

b4

)K1

(a

�

)− a3�

b4K2

(a

�

)]I1

(b

�

)

−[

1 − ν

1 − 2ν

(1 − a4

b4

)K1

(b

�

)+ �

bK2

(b

�

)]I1

(a

�

), (188)

where

� = 4a�4

b5+ a�

b2

[(1 − ν) − a2

b2

(1 − ν + 2

�2

b2

)]I2

(a

�

)K1

(b

�

)

− �

b

{[2�2

b2+ (1 − ν)

a2

b2

(1 − a2

b2

)]K1

(a

�

)

+ (1 − 2ν)a�

b2

(1 − a2

b2

)K2

(a

�

)}I2

(b

�

)

+{

1 − ν

1 − 2ν

(1 − a2

b2

)[2�2

b2+ a2

b2

(1 − ν + 2

�2

b2

)]K1

(a

�

)


+ a�

b2

[1 − ν − a2

b2

(1 − ν + 2

�2

b2

)]K2

(a

�

)}I1

(b

�

)

+ (1 − 2ν)a �2

b3

(1 − a2

b2

)I2

(a

�

)K2

(b

�

)

−{

1 − ν

1 − 2ν

(1 − a2

b2

)[2�2

b2+ a2

b2

(1 − ν + 2

�2

b2

)]K1

(b

�

)

+ �

b

[2�2

b2+ (1 − ν)

a2

b2

(1 − a2

b2

)]K2

(b

�

)}I1

(a

�

). (189)

In the limiting case of an incompressible material (ν → 1/2), the constants take the follow-ing values

c2 → (po − pi)8 �

b[ b3

a3 I1(b�) − I1(

a�)]

( b2

a2 − 1)[1 + 4 �2

b2 ( b2

a2 + 1)][I1(a�)K1(

b�) − I1(

b�)K1(

a�)] , (190)

c3 → (po − pi)4 �

b[ b3

a3 K1(b�) − K1(

a�)]

( b2

a2 − 1)[1 + 4 �2

b2 ( b2

a2 + 1)][I1(a�)K1(

b�) − I1(

b�)K1(

a�)] , (191)

c7 → (po − pi)8 �2 b2

a4

( b2

a2 − 1)[ b2

a2 + 4 �2

a2 ( b2

a2 + 1)] , (192)

c8 → (po − pi)8�2 b2

a2 ( b2

a2 + 1)

( b2

a2 − 1)[ b2

a2 + 4 �2

a2 ( b2

a2 + 1)] . (193)

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