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Couple stress theory for polar solids

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					                Couple stress theory for polar solids
                          Ali R. Hadjesfandiari, Gary F. Dargush

                  Department of Mechanical and Aerospace Engineering
                   University at Buffalo, State University of New York
                                 Buffalo, NY 14260 USA
                         ah@buffalo.edu, gdargush@buffalo.edu


Abstract
The existing couple stress theory for polar media suffers from an indeterminacy of the
spherical part of the couple-stress tensor, which limits significantly its applicability in the
study of micro and nanoscale mechanics. Here we rely on concepts from virtual work,
along with some kinematical considerations, to establish a consistent polar theory for
solids that resolves all of the indeterminacies by recognizing the character of the couple
stress tensor. We then develop the corresponding theory of small deformations in elastic
bodies, including the energy and constitutive relations, displacement formulations, the
uniqueness theorem for the corresponding boundary value problem and the reciprocal
theorem for linear elasticity theory. Next, we consider the more restrictive case of
isotropic materials and present general solutions for two-dimensional problems based on
stress functions and for problems of anti-plane deformation. Finally, we examine several
additional elementary boundary value problems within this consistent theory of polar
elasticity.


1. Introduction
Classical first gradient approaches in continuum mechanics do not address the size-
dependency that is observed in smaller scales. Consequently, a number of theories that
include higher gradients of deformation have been proposed to capture, at least partially,
size-effects at the nano-scale. Additionally, consideration of the second gradient of
deformation leads naturally to the introduction of the concept of couple-stresses. Thus, in
the current form of these theories, the material continuum may respond to body and
surface couples, as well as spin inertia for dynamical problems.



                                              1
The existence of couple-stress in materials was originally postulated by Voigt (1887).
However, Cosserat and Cosserat (1909) were the first to develop a mathematical model to
analyze materials with couple stresses. The idea was revived and generalized much later
by Toupin (1962), Mindlin and Tiersten (1962), Mindlin (1964), Koiter (1964), Nowacki
(1986) and others. In these developments, the gradient of the rotation vector, as a
curvature tensor, has been recognized as the effect of the second gradient of deformation
in polar materials.     Unfortunately, there are some difficulties with the present
formulations. Perhaps the most disturbing troubles are the indeterminacy of the spherical
part of the couple-stress tensor and the appearance of the body couple in the constitutive
relation for the force-stress tensor (Mindlin and Tiersten, 1962).


Here we develop a consistent couple stress theory for polar media and organize the
current paper in the following manner. In Section 2, we present stresses, couple stresses
and the equilibrium equations per the usual definitions in the existing couple stress
literature. Based on purely kinematical considerations as provided in Section 3, we first
suggest the mean curvature tensor as the measure of deformation compatible with the
couple stress tensor for the infinitesimal theory.      Then, by using the virtual work
formulation of Section 4, we demonstrate that in couple stress materials, body couples
must be transformed to an equivalent body force system. More importantly, based on
resolving properly the boundary conditions, we show that the couple-stress tensor is
skew-symmetric and, thus, completely determinate.          This also confirms the mean
curvature tensor as the fundamental deformation measure, energetically conjugate to the
couple stress tensor. Afterwards, in Section 5, the general theory of small deformation
polar elasticity is developed. The constitutive and equilibrium equations for a linear
elastic material also are derived under the assumption of infinitesimal deformations in
Section 6, along with the uniqueness theorem for well-posed boundary value problems
and the reciprocal theorem. Section 7 provides the general solution based on stress
functions for two-dimensional infinitesimal linear polar elasticity, while the
corresponding anti-plane deformation problem is examined in Section 8. Section 9
presents solutions for several elementary problems in polar elasticity. Finally, Section 10
contains a summary and some general conclusions.



                                             2
2. Stresses and equilibrium
For a polar material, it is assumed that the transfer of the interaction in the current
configuration occurs between two particles of the body through a surface element dS
                                                       (n)
with unit normal vector ni by means of a force vector ti dS and a moment vector

mi(n )dS , where ti(n ) and mi(n ) are force and couple traction vectors. Surface forces and

couples are then represented by generally non-symmetric force-stress σ ji and couple-

stress μ ji tensors, where

                                                  ti( ) = σ ji n j
                                                     n
                                                                                         (1)

                                                 mi( ) = μ ji n j
                                                    n
                                                                                         (2)


Consider an arbitrary part of the material continuum occupying a volume V enclosed by
boundary surface S as the current configuration. Under quasistatic conditions, the linear
and angular balance equations for this part of the body are
                                                 ( n)
                                            ∫ti         dS + ∫ Fi dV = 0                 (3)
                                             S                 V


                                 (n)
                     ∫ ⎡ε ijk x jtk    + mi( ) ⎤ dS + ∫ ⎡ε ijk x j Fk +Ci ⎤ dV = 0
                                            n
                                                        ⎣                 ⎦              (4)
                       ⎢
                       ⎣                       ⎥
                                               ⎦
                     S                                     V

where Fi and Ci are the body force and the body couple per unit volume of the body,

respectively. Here ε ijk is the permutation tensor or Levi-Civita symbol.


By using the relations (1) and (2), along with the divergence theorem, and noticing the
arbitrariness of volume V, we finally obtain the differential form of the equilibrium
equations, for the usual couple stress theory, as
                                             σ ji , j + Fi = 0                          (5)

                                         μ ji , j + ε ijk σ jk + Ci = 0                  (6)

where the comma denotes differentiation with respect to the spatial variables.




                                                           3
3. Kinematics
Here we consider the kinematics of a polar continuum under the assumptions of
infinitesimal deformation.     In Cartesian coordinates, we define ui to represent the
displacement field of the continuum material. Consider the neighboring points P and Q
with position vectors xi and xi + dxi in the reference configuration.            The relative
displacement of point Q with respect to P is
                                             dui = ui , j dx j                            (7)

where ui , j is the displacement gradient tensor at point P. As we know, although this

tensor is important in analysis of deformation, it is not itself a suitable measure of
deformation. This tensor can be decomposed into symmetric and skew-symmetric parts
                                           ui , j = eij + ωij                             (8)

where

                                       eij = u(i , j ) =
                                                           1
                                                             (ui, j + u j ,i )            (9)
                                                           2

                                      ωij = u[i , j ] =
                                                           1
                                                             (ui, j − u j ,i )           (10)
                                                           2


Notice that here we have introduced parentheses surrounding a pair of indices to denote
the symmetric part of a second order tensor, whereas square brackets are associated with
the skew-symmetric part. Of course, in (9) and (10), the tensors eij and ωij are the small

deformation strain and rotation tensor, respectively. The rotation vector ωi dual to the

rotation tensor ωij is defined by

                                            1                  1
                                     ωi = ε ijk ω kj = ε ijk u k , j                    (11a)
                                            2                  2
which in vectorial form is written
                                              1
                                           ω = ∇×u                                     (11b)
                                              2


Alternatively, this rotation vector is related to the rotation tensor through
                                           ω ji = ε ijkωk                                (12)


                                                  4
which shows
                               ω1 = −ω 23 , ω 2 = ω13 , ω3 = −ω12                        (13)


Therefore, the relative displacement is decomposed into
                                       dui = dui(1) + dui(2 )                            (14)
where
                                          dui(1) = eij dx j                              (15)

                                         dui(2 ) = ωij dx j                              (16)


Then, ωij is seen to generate a rigid-like rotation of element dxi about point P, where

                                   dui(2 )dxi = ωij dxi dx j = 0                         (17)


Since ωij does not contribute to the elongation or contraction of element dxi , it cannot

appear in a tensor measuring material stretches. Therefore, as we know, the symmetric
strain tensor eij is the suitable measure of deformation in classical infinitesimal theories,

such as Cauchy elasticity.


In couple stress theory, we expect to have an additional tensor measuring the curvature of
the arbitrary fiber element dxi . To find this tensor, we consider the field of rotation

vector ωi . The relative rotation of two neighboring points P and Q is given by

                                          dωi = ωi , j dx j                              (18)

where the tensor ωi, j is the gradient of the rotation vector at point P. It is seen that the

components ω1,1 , ω 2, 2 and ω3,3 represent the torsion of the fibers along corresponding

coordinate directions x1 , x2 and x3 , respectively, at point P.           The off-diagonal
components represent the curvature of these fibers in planes parallel to coordinate planes.
For example, ω1, 2 is the curvature of a fiber element in the x2 direction in a plane parallel




                                                 5
to the x2 x3 plane, while ω 2,1 is the curvature of a fiber element in the x1 direction in a

plane parallel to the x1 x3 plane.


The suitable measure of curvature must be a tensor measuring pure curvature of an
arbitrary element dxi . Therefore, in this tensor, the components ω1,1 , ω 2, 2 and ω3,3

cannot appear. However, simply deleting these components from the tensor ωi, j does not

produce a tensor.      Consequently, we expect that the required tensor is the skew-
symmetric part of ωi , j . By decomposing the tensor ωi , j into symmetric and skew-

symmetric parts, we obtain
                                             ωi , j = χ ij + κ ij                      (19)

where

                                     χ ij = ω(i , j ) =
                                                          1
                                                            (ωi, j + ω j ,i )          (20)
                                                          2

                                     κ ij = ω[i , j ] =
                                                          1
                                                            (ωi, j − ω j ,i )          (21)
                                                          2


The symmetric tensor χ ij results from applying the strain operator to the rotation vector,

while the tensor κ ij is the rotation of the rotation vector at point P. From (20),

                                χ11 = ω1,1 , χ 22 = ω 2, 2 , χ 33 = ω3,3               (22)

and
                                                          1
                                      χ12 = χ 21 =          (ω1,2 + ω2,1 )            (23a)
                                                          2

                                     χ 23 = χ 32 =
                                                          1
                                                            (ω2,3 + ω3,2 )            (23b)
                                                          2

                                      χ13 = χ 31 =
                                                          1
                                                            (ω1,3 + ω3,1 )            (23c)
                                                          2


The diagonal elements χ11 , χ 22 and χ 33 defined in (22) represent pure torsion of fibers

along the x1 , x2 and x3 directions, respectively, as mentioned above. On the other hand,



                                                    6
from careful examination of (23), we find that χ12 , χ 23 and χ13 measure the deviation

from sphericity (Hamilton, 1866) of deforming planes parallel to x1 x2 , x2 x3 and x1 x3 ,

respectively. Furthermore, we may recognize that this symmetric χij tensor must have

real principal values, representing the pure twists along the principal directions. Thus,
we refer to χij as the torsion tensor and we expect that this tensor will not contribute as a

fundamental measure of deformation in a polar material. Instead, we anticipate that the
fundamental curvature tensor is the skew-symmetric rotation of rotation tensor κ ij . This

will be confirmed in the next section through consideration of couple stresses and virtual
work.


We also may arrive at this outcome by noticing that only the part of dωi that is normal to

element dxi produces pure curvature. Therefore, by decomposing dωi into

                                       dωi = dωi(1) + dωi(2 )                           (24)
where
                                         dωi(1) = χ ij dx j                             (25)

                                         dωi(2 ) = κ ij dx j                            (26)

we notice
                                   dωi(2 )dxi = κ ij dxi dx j = 0                       (27)

This shows that dωi(2 ) is the component of dωi normal to dxi . Therefore, the tensor κ ij

seems to be the suitable curvature tensor, which is represented by
                                             ⎡ 0           κ 12     κ 13 ⎤
                                     [ ]
                                      κ ij = ⎢− κ 12
                                             ⎢                 0    κ 23 ⎥
                                                                         ⎥              (28)
                                             ⎢ − κ 13
                                             ⎣            − κ 23     0 ⎥
                                                                       ⎦
where the non-zero components of this tensor are

                                     κ12 = −κ 21 =
                                                        1
                                                          (ω1,2 − ω2,1 )               (29a)
                                                        2
                                                        1
                                     κ 23 = −κ 32 =       (ω2,3 − ω3,2 )               (29b)
                                                        2



                                                7
                                                           1
                                        κ13 = −κ 31 =        (ω1,3 − ω3,1 )              (29c)
                                                           2


Now we may recognize that κ12 , κ 23 , and κ13 are the mean curvatures of planes parallel

to the x1 x2 , x2 x3 , x3 x1 planes, respectively, at point P after deformation. Therefore, the

skew-symmetric tensor κ ij will be referred to as the mean curvature tensor or simply the

curvature tensor. The curvature vector κ i dual to this tensor is defined by

                                          1                1
                                    κ i = ε ijk ω k , j = ε ijk κ kj                      (30)
                                          2                2


Thus, this axial vector is related to the mean curvature tensor through
                                              κ ji = ε ijkκ k                             (31)

which shows
                                 κ1 = −κ 23 , κ 2 = κ13 , κ 3 = −κ12                      (32)


It is seen that the mean curvature vector can be expressed as
                                                 1
                                              κ = ∇×ω                                     (33)
                                                 2


This shows that κ is the rotation of the rotation vector, which can also be expressed as
                             1              1           1
                          κ = ∇ × (∇ × u ) = ∇(∇ • u ) − ∇ 2 u                           (34a)
                             4              4           4
                                1           1          1           1
                           κ i = u k ,ki − ui ,kk = u k ,ki − ∇ 2 ui                     (34b)
                                4           4          4           4


What we have presented here is applicable to small deformation polar theory, which
requires the components of the strain tensor and mean curvature vector to be
infinitesimal. These conditions can be written as
                                                  eij << 1                                (35)




                                                   8
                                                                             1
                                                                    κ i <<                      (36)
                                                                             lS

where, lS is the smallest characteristic length in the body.


While analogous measures of strain and curvature can be obtained for finite deformation
polar theory, this would take us beyond the scope of the present work, which is directed
toward the infinitesimal linear couple stress theory.


4. Virtual work formulation and its consequences for polar media
Consider now a polar material continuum occupying a volume V bounded by a surface S
as the current configuration. The standard form of the equilibrium equations for this
medium were given in (5) and (6).


Let us multiply equation (5) by a virtual displacement δui and integrate over the volume

and also multiply equation (6) by the corresponding virtual rotation δωi , where

                                                                        1
                                                       δωi = ε ijk δu k , j                     (37)
                                                                        2
and integrate this over the volume as well. Therefore, we have

                                           ∫ (σ
                                           V
                                                      ji , j   + Fi )δui dV = 0                 (38)


                                   ∫ (μ
                                   V
                                            ji , j   + ε ijk σ jk + Ci )δωi dV = 0              (39)



By noticing the relation
                                       σ ji , j δui = (σ jiδui ), j − σ jiδui , j               (40)

and using the divergence theorem, the relation (38) becomes

                               ∫σ   ji   δui , j dV = ∫ ti(n )δui dS + ∫ Fiδui dV               (41)
                               V                                S                 V




Similarly, by using the relation
                     μ ji , j δωi + ε ijk σ jk δωi = (μ jiδωi ), j − μ jiδωi , j − σ jk δω jk   (42)


                                                                    9
equation (39) becomes

                              ∫μ   ji   δωi , j dV − ∫ σ jiδωij dV = ∫ mi(n )δωi dS + ∫ Ciδωi dV                       (43)
                              V                       V                         S               V




Then, by adding (41) and (43), we obtain

                                                      V
                                                               (
                              ∫ μ jiδωi , j dV + ∫ σ ji δ ui , j − δωij dV =
                              V
                                                                                        )
                                                                                                                       (44)
                                            ( n)                         (n)
                                         ∫ ti δ ui dS + ∫ Fiδ ui dV + ∫ mi δωi dS + ∫ Ciδωi dV
                                         S                 V                        S               V




However, by noticing the relation
                                                          δeij = δui , j − δωij                                        (45)

for compatible virtual displacement, we obtain the virtual work theorem as

      ∫σ   ji   δeij dV + ∫ μ jiδωi , j dV = ∫ t i(n )δui dS + ∫ Fiδui dV + ∫ mi(n )δωi dS + ∫ Ciδωi dV                (46)
      V                   V                       S                        V                S              V




Now, by using this virtual work formulation, we investigate the fundamental character of
the body couple and couple stress in a material continuum.


It is seen that the term

                                                               ∫ C δω dV
                                                               V
                                                                   i        i                                          (47)

in (46) is the only term in the volume that involves δωi .                                              However, δωi is not

independent of δui in the volume, because we have the relation

                                                                       1
                                                          δωi = ε ijk δu k , j                                         (48)
                                                                       2


Therefore, by using (48), we find

                                  Ciδωi = Ci ε ijk δu k , j = (ε ijk Ciδu k ), j − ε ijk Ci , j δu k
                                         1                   1                    1
                                                                                                                       (49)
                                         2                   2                    2
and, after applying the divergence theorem, the body couple virtual work in (47) becomes




                                                                       10
                                                        1                          1
                               ∫ C δω dV = ∫ 2 ε
                               V
                                       i   i
                                                    V
                                                            ijk   Ck , jδui dV + ∫ ε ijk C j nk δui dS
                                                                                 S
                                                                                   2
                                                                                                                (50)

                                                                                                         1
which means that the body couple C i transforms into an equivalent body force                              ε ijk Ck , j
                                                                                                         2
                                                                1
in the volume and a force traction vector                         ε ijk C j nk on the bounding surface. This shows
                                                                2
that in polar materials, the body couple is not distinguishable from the body force.
Therefore, in the couple stress theory for polar media, we must only consider body
forces. This is analogous to the impossibility of distinguishing a distributed moment load
in Euler–Bernoulli beam theory, in which the moment load must be replaced by the
equivalent distributed force load and end concentrated loads. Therefore, for a proper
couple stress theory, the equilibrium equations become
                                                        σ ji , j + Fi = 0                                       (51)

                                                     μ ji , j + ε ijk σ jk = 0                                  (52)

where
                                                  1
                                               F + ∇ ×C → F                       in V                         (53a)
                                                  2
and
                                                     1
                                               t( ) + C × n → t( )
                                                 n              n
                                                                                  on S                        (53b)
                                                     2
and the virtual work theorem reduces to

             ∫σ   ji   δeij dV + ∫ μ jiδωi , j dV = ∫ ti(n )δui dS + ∫ Fiδui dV + ∫ mi(n )δωi dS                (54)
             V                     V                        S                 V              S




Next, we investigate the fundamental character of the couple stress tensor based on
boundary conditions.


The prescribed boundary conditions on the surface of the body can be either vectors ui

and ωi , or ti(n ) and mi(n ) , which makes a total number of six boundary values for either
case. However, this is in contrast to the number of geometric boundary conditions that
can be imposed (Koiter, 1964). In particular, if components of ui are specified on the


                                                                   11
boundary surface, then the normal component of the rotation ωi corresponding to
twisting
                                          ωi(n ) = ω (nn )ni = ωk nk ni               (55)
where
                                                 ω (nn ) = ωk nk                      (56)

cannot be prescribed independently. However, the tangential component of rotation ωi
corresponding to bending, that is,
                                    ωi(ns ) = ωi − ω (nn )ni = ωi − ωk nk ni          (57)
may be specified in addition, and the number of geometric or essential boundary
conditions that can be specified is therefore five.


Next, we let m( nn ) and mi( ns ) represent the normal and tangential components of the

surface couple vector mi(n ) , respectively, where

                                        m (nn ) = mk n ) nk = μ ji ni n j
                                                   (
                                                                                      (58)

causes twisting, while
                                           mi(ns ) = mi(n ) − m (nn )ni               (59)
is responsible for bending.


From kinematics, since ω ( nn ) is not an independent generalized degree of freedom, its
apparent corresponding generalized force must be zero. Thus, for the normal component
of the surface couple vector mi(n ) , we must enforce the condition

                                m( ) = mkn ) nk = μ ji ni n j = 0 on S
                                  nn    (
                                                                                      (60)


Furthermore, the boundary couple surface virtual work in (54) becomes
                              (n)
                         ∫ mi    δωi dS = ∫ mi( ns )δωi dS = ∫ mi( ns )δωi( ns ) dS   (61)
                         S                   S                     S

This shows that a polar material in couple stress theory does not support independent
distributions of normal surface couple m (nn ) , and the number of mechanical boundary



                                                      12
conditions also is five. In practice, it might seem that a given m (nn ) has to be replaced by
an equivalent shear stress and force system. Koiter (1964) gives the detail analogous to
the Kirchhoff bending theory of plates. However, we should realize that there is a
difference between couple stress theory and the Kirchhoff bending theory of plates. Plate
theory is an approximation for elasticity, which is a continuum mechanics theory.
However, couple stress theory is a continuum mechanics theory itself without any
approximation.


From the above discussion, we should realize that on the surface of the body, a normal
couple m (nn ) cannot be applied. By continuing this line of reasoning, we may reveal the
subtle character of the couple stress-tensor. First, we notice that the virtual work theorem
can be written for every arbitrary volume with arbitrary surface within the body.
Therefore, for any point on any arbitrary surface with unit normal ni , we must have

                                  m (nn ) = μ ji ni n j = 0 in V                         (62)


Since ni n j is symmetric and arbitrary in (62), μ ji must be skew-symmetric. Thus,

                                      μ ji = − μij in V                                  (63)


This is the fundamental property of the couple-stress tensor in polar continuum
mechanics, which has not been recognized previously. Here we can see the crucial role
of the virtual work theorem in this result.


In terms of components, the couple stress tensor now can be written as
                                      ⎡ 0           μ12    μ13 ⎤
                               [ ]
                               μ ij = ⎢− μ12
                                      ⎢              0     μ 23 ⎥
                                                                ⎥                        (64)
                                      ⎢ − μ13
                                      ⎣         − μ 23      0 ⎥
                                                              ⎦
and one can realize that the couple stress actually can be considered as an axial vector.
This couple stress vector μ i dual to the tensor μ ij can be defined by

                                                1
                                         μi = ε ijk μ kj                                 (65)
                                                2


                                                13
where we also have
                                               ε ijk μ k = μ ji                                    (66)


These relations simply show
                                   μ1 = − μ 23 , μ 2 = μ13 , μ 3 = − μ12                           (67)


It is seen that the surface couple vector can be expressed as

                                       mi( ) = μ ji n j = ε ijk n j μk
                                          n
                                                                                                   (68)

which can be written in vectorial form
                                               m (n ) = n × μ                                      (69)

This obviously shows that the surface couple vector m (n ) is tangent to the surface.


Interestingly, the angular equilibrium equation (52) can be expressed as
                                        ε ijk (μ k , j + σ jk ) = 0                                (70)

which indicates that μ k , j + σ jk is symmetric.                    Therefore, its skew-symmetric part

vanishes and
                                            σ [ ji ] = − μ[i , j ]                                 (71)

which produces the skew-symmetric part of the force stress tensor in terms of the couple
stress vector. This result could have been expected on the grounds that the skew-
symmetric stress tensor σ [ ji ] is actually an axial vector and should depend on the axial

couple stress vector μ i . Therefore, it is seen that the sole duty of the angular equilibrium
equation (52) is to produce the skew-symmetric part of the force stress tensor. This
relation can be elaborated if we consider the axial vector si dual to the skew-symmetric

part of the force-stress tensor σ [ij ] , where

                                                1
                                            si = ε ijk σ [kj ]                                    (72a)
                                                2
which also satisfies
                                            ε ijk sk = σ [ ji ]                                   (72b)



                                                      14
or simply
                                      s1 = −σ [23] , s2 = σ [13] , s3 = −σ [12 ]                      (73)

By using (71) and (72a), we obtain
                                                1                 1
                                          si = − ε ijk μ[ j ,k ] = ε ijk μ k , j                     (74a)
                                                2                 2
which can be written in vectorial form
                                                        1
                                                   s=     ∇×μ                                        (74b)
                                                        2
This simply shows that half of the curl of the couple stress vector μ produces the skew-
symmetric part of the force-stress tensor through s. Interestingly, it is seen that
                                                    ∇•s = 0                                           (75)


Returning to the virtual work theorem, we notice since μ ji is skew-symmetric

                                               μ jiδωi , j = μ jiδκ ij                                (76)

which shows that the skew-symmetric mean curvature tensor κ ij is energetically

conjugate to the skew-symmetric couple-stress tensor μ ji . This confirms our speculation

of κ ij as a suitable curvature tensor in Section 2. Furthermore, the virtual work theorem

(54) becomes

            ∫σ jiδ eij dV + ∫ μ jiδκij dV = ∫ti     δ ui dS + ∫ Fiδ ui dV + ∫ mi( ns )δωi( ns ) dS
                                                  (n)
                                                                                                      (77)
            V                V                S                 V                  S




Interestingly, by using the dual vectors of these tensors, we have
                μ jiδκ ij = ε ijp μ pε jiqδκ q = −ε ijpε ijq μ pδκ q = −2δ pq μ pδκ q = −2μiδκ i      (78)

which shows the conjugate relation between twice the mean curvature vector − 2κ i and

the couple-stress vector μ i .


Since δeij is symmetric, we also have

                                            σ jiδeij = σ ( ji )δeij                                   (79)

where


                                                        15
                                                   σ ( ji ) =
                                                                1
                                                                  (σ ji + σ ij )                         (80)
                                                                2
is the symmetric part of the force-stress tensor. Thus, the principle of virtual work can be
written

                ∫σ (       δeij dV + ∫ μ jiδκ ij dV = ∫ ti(n )δui dS + ∫ Fiδui dV + ∫ mi(n )δωi dS
                        ji )                                                                             (81)
                V                           V                    S                 V          S




Therefore, it is seen that the symmetric small deformation strain tensor eij is the

kinematical tensor energetically conjugate to the symmetric part of the force-stress tensor
σ ( ji ) . Finally, the virtual work theorem (81) can be rewritten as

                    ∫ (σ
                    V
                               ji   δeij − 2μ iδκ i )dV = ∫ ti(n )δui dS + ∫ Fiδui dV + ∫ mi(n )δωi dS
                                                            S                  V          S
                                                                                                         (82)



What we have presented so far is a continuum mechanical theory of couple stress polar
materials, independent of the material properties. In the following section, we specialize
the theory for elastic materials.


5. Infinitesimal polar elasticity
Now, we develop the theory of small deformation for elastic polar materials. In a polar
elastic material, there is a strain energy function W , where for arbitrary virtual
deformations about the equilibrium position, we have
                                        δW = σ jiδeij + μ jiδκ ij = σ ( ji )δeij − 2μ iδκ i              (83)

Therefore, W is a positive definite function of the symmetric strain tensor eij and the

mean curvature vector κ i . Thus,

                                                 W = W (e, κ ) = W (eij , κ i )                          (84)


However, for a variational analysis the relation (83) should be written as
                                                  δW = σ ( ji )δui , j − 2μ iδκ i                        (85)

where the all components of δui , j and δκ i can be taken independent of each other. From

the relations (84) and (85), we obtain


                                                                     16
                                                             ∂W
                                               σ ( ji ) =                                       (86)
                                                             ∂ui , j

                                                              ∂W
                                               2μ i = −                                         (87)
                                                              ∂κ i


However, it is seen that
                                            ∂W ∂W ∂ekl
                                                   =                                            (88)
                                            ∂ui , j ∂ekl ∂u i , j

By noticing

                                            ekl =
                                                    1
                                                      (uk ,l + ul ,k )                          (89)
                                                    2
we obtain
                                        ∂ekl 1
                                               = (δ kiδ lj + δ liδ kj )                         (90)
                                        ∂ui , j 2


Therefore
                                     ∂W 1 ∂W
                                            =       (δ kiδ lj + δ liδ kj )                      (91)
                                     ∂ui , j 2 ∂ekl

which shows

                                       ∂W 1 ⎛ ∂W ∂W                      ⎞
                                              = ⎜      +                 ⎟                      (92)
                                       ∂ui , j 2 ⎜ ∂eij ∂e ji
                                                 ⎝
                                                                         ⎟
                                                                         ⎠


Then
                                                    ⎛
                                                1 ∂W ∂W ⎟              ⎞
                                      σ ( ji ) = ⎜
                                                 ⎜
                                                    +
                                                        ⎟
                                                                                                (93)
                                                 2 ⎝ ∂eij         ∂e ji ⎠

                                                        1 ∂W
                                             μi = −                                             (94)
                                                        2 ∂κ i
It is also seen that
                                                           1 ⎡⎛ ∂W      ⎞       ⎛ ∂W   ⎞ ⎤
                       μ [i , j ] = (μ i , j − μ j ,i ) = − ⎢⎜
                                   1
                                                                        ⎟      −⎜      ⎟ ⎥      (95)
                                   2                       4 ⎢⎜ ∂κ i    ⎟       ⎜ ∂κ   ⎟ ⎥
                                                             ⎣⎝         ⎠, j    ⎝ j    ⎠ ,i ⎦

Therefore, for the skew-symmetric part of the force-stress tensor, we have


                                                        17
                                                         1 ⎡⎛ ∂W     ⎞     ⎛ ∂W     ⎞ ⎤
                             σ [ ji ] = − μ [i , j ] =     ⎢⎜        ⎟ −⎜           ⎟ ⎥          (96)
                                                         4 ⎢⎜ ∂κ i   ⎟     ⎜
                                                                     ⎠ , j ⎝ ∂κ j
                                                                                    ⎟ ⎥
                                                           ⎣⎝                       ⎠ ,i ⎦


Finally, we obtain the constitutive relations as

                              1 ⎛ ∂W ∂W                  ⎞ 1 ⎡⎛ ∂W      ⎞ ⎛ ∂W          ⎞ ⎤
                        σ ji = ⎜      +                  ⎟ + ⎢⎜
                                                         ⎟ 4 ⎢⎜ ∂κ
                                                                        ⎟ −⎜
                                                                        ⎟ ⎜ ∂κ
                                                                                        ⎟ ⎥      (97)
                              2 ⎜ ∂eij ∂e ji                                            ⎟ ⎥
                                ⎝                        ⎠   ⎣⎝ i       ⎠, j ⎝ j        ⎠ ,i ⎦

                                                          1 ∂W
                                              μi = −                                             (98)
                                                          2 ∂κ i


The total potential energy functional for polar elastic body is defined as
                       Π {u} = ∫ WdV − ∫ Fi ui dV − ∫ ti(n )ui dS − ∫ mi(n )ωi dS                (99)
                                V              V                S               S




It can be easily shown that this functional attains its absolute minimum when the
displacement field corresponds to the elastic solution that satisfies the equilibrium
equations. The kinematics of deformation and variation of (99) reveal an important
character of the strain energy function W . We know there are two sets of equilibrium
equations (51) and (52) corresponding to linear and angular equilibrium of an
infinitesimal element of material. Therefore, the geometrical boundary conditions are the
displacement ui and rotation ωi as we discussed previously. As we showed in Section 4,

polar continuum mechanics supports the geometrical boundary conditions u i and ω i(ns ) ,

and their corresponding energy conjugate mechanical boundary conditions t i(n ) and mi(ns ) .
Consequently, there is no other possible type of boundary condition in polar continuum
mechanics. Therefore, in the variation of the total potential energy Π in (99), the strain
energy function W at most can be in the form (84). This means at most the strain energy
function W is a function of the second derivative of deformation in the form of the mean
curvature vector κ i . In other words, the continuum mechanics strain energy function

W cannot depend on third and higher order derivative of deformations.




                                                         18
6. Infinitesimal linear polar elasticity


Strain energy and constitutive relations
For a linear elastic material, based on our development, the quadratic positive definite
strain energy must be in the form
                                             1                1
                              W (e, κ ) =      Aijkl eij ekl + Bij κ iκ j             (100)
                                             2                2
The tensors Aijkl and Bij contain the elastic constitutive coefficients. It is seen that the

tensor Aijkl is actually equivalent to its corresponding tensor in Cauchy elasticity. The

symmetry relations
                                      Aijkl = Aklij = A jikl                          (101)

                                            Bij = B ji                                (102)

are trivial. These symmetry relations show that for the most general case the number of
distinct components for Aijkl and Bij are 21 and 6, respectively. It is seen that the couple

stress vector and symmetric part of stress tensor can be found as
                                                  1
                                        μ i = − Bij κ j                               (103)
                                                  2
                                          σ ( ji ) = Aijkl ekl                        (104)

Additionally, we find that
                                                    1
                                        μ i , j = − Bik κ k , j                       (105)
                                                    2


The skew-symmetric part of this tensor is
                                                      1               1
                             μ [i , j ] = −σ [ ji ] = − Bik κ k , j + B jk κ k ,i     (106)
                                                      4               4


Therefore, for the force stresses, we find
                                                  1               1
                             σ ji = Aijkl ekl + Bik κ k , j − B jk κ k ,i             (107)
                                                  4               4




                                                      19
For an isotropic polar material, the symmetry relations require
                               Aijkl = λδ ijδ kl + μδ ikδ jl + μδ ilδ jk                   (108)

                                             Bij = 16ηδ ij                                 (109)


The moduli λ and μ have the same meaning as the Lamé constants for an isotropic
material in Cauchy elasticity. It is seen that only one extra material constant η accounts
for couple-stress effects in an isotropic polar material and the strain energy becomes
                                        1
                             W (e, κ ) = λ (ekk ) + μeij eij + 8ηκ iκ i
                                                 2
                                                                                           (110)
                                        2
with the following restrictions on elastic constants for positive definite strain energy
                                3λ + 2μ > 0,            μ > 0,         η >0                (111)


Then, the constitutive relations can be written
                                               μ i = −8ηκ i                                (112)

                                        σ ( ji ) = λekk δ ij + 2μeij                       (113)

Interestingly, it is seen that for an isotropic material
                                          ∇ • μ = μ i ,i = 0                               (114)

By using the relation
                                                  1          1
                                        κ i = u k ,ki − ui ,kk                             (115)
                                                  4          4
we obtain
                                      μ i = 2η (∇ 2 ui − u k ,ki )                     (116a)

or in vectorial form
                                              [
                                    μ = 2η ∇ 2 u − ∇(∇ • u )           ]               (116b)
Additionally,
                                     μ i , j = 2η (∇ 2 ui , j − u k ,kij )                 (117)

Therefore,
                                       μ[i , j ] = η∇ 2 (ui , j − u j ,i )                 (118)

or


                                                      20
                                          μ [i , j ] = 2η∇ 2ωij                           (119)

and we obtain
                                    σ [ ji ] = − μ[i , j ] = 2η∇ 2ω ji                    (120)

or by exchanging indices
                                           σ [ij ] = 2η∇ 2ωij                             (121)


Recall that the axial vector si is dual to σ [ij ] , as shown in (72). Then, from (74a) and

(112), si can be written in terms of the curvature vector as

                                          si = −4ηε ijkκ k , j                            (122)

Therefore, the constitutive relation for vector s is
                                           s = −4η∇ × κ                               (123a)
which can be written alternatively as
                                 s = −2η∇ × ∇ × ω = 2η∇ 2 ω                           (123b)
or
                                     s = −η∇ × ∇ × ∇ × u                              (123c)


This remarkable result shows that in an isotropic polar material the vector s ,
corresponding to skew-symmetric part of stress tensor, is proportional to the curl of curl
of curl of the displacement vector u .


By using the relations (113) and (120), the total force-stress tensor can be written as
                               σ ji = λekkδ ij + 2μeij + 2η∇ 2ω ji                        (124)

We also notice that
                                     μ ji = −8ηκ ji
                                                                                          (125)
                                           = 4η (ωi , j − ω j ,i )

which is more useful than μi in practice.




                                                  21
It is seen that these relations are similar to those in the indeterminate couple stress theory
(Mindlin and Tiersten, 1962), when η ′ = −η . Here we have derived the couple stress
theory for polar materials in which all former troubles with indeterminacy disappear.
There is no spherical indeterminacy and the second couple stress coefficient η ′ depends
on η , such that the couple stress tensor becomes skew-symmetric.


Interestingly, the ratio
                                                             η 2
                                                               =l                                           (126)
                                                             μ
specifies a characteristic material length l , which is absent in Cauchy elasticity, but is
fundamental to small deformation couple-stress polar elasticity. We realize that this is
the characteristic length in an elastic material and that lS → l in (36).                               Thus, the
requirements for small deformation polar elasticity are
                                                          eij << 1                                         (127a)

                                                                     1
                                                          κ i <<                                           (127b)
                                                                     l


Displacement formulations
When the force-stress tensor (107) is written in terms of displacements, as follows
                                           1                1
                    σ ji = Aijkl ekl + Bikκ k , j − B jkκ k ,i
                                           4                4
                                                                                                            (128)
                    = Aijkl uk ,l
                                      1
                                     16
                                               (                 1
                                                                16
                                                                         )           (
                                    + Bik um , mkj − ∇ 2uk , j − B jk um, mki − ∇ 2uk ,i            )

and is carried into the linear equilibrium equation, we obtain

                                  Bik (∇ 2um,mk − ∇ 2∇ 2uk ) − B jk (um ,mkij − ∇ 2uk ,ij ) = 0
                                1                              1
              Aijkl uk ,lj +                                                                                (129)
                               16                             16


For an isotropic material, the force-stress tensor becomes
                           σ ji = λekk δ ij + 2μeij − 2η∇ 2ωij
                                                                                                            (130)
                                  = λu k ,k δ ij + μ (ui , j + u j ,i ) − η∇ 2 (ui , j − u j ,i )



                                                                22
and for the linear equilibrium equation, we have

                        ( λ + μ + η∇ ) u
                                     2
                                           k ,ki   + ( μ − η∇ 2 )∇ 2ui + Fi = 0               (131a)

which can be written in the vectorial form

                        ( λ + μ + η∇ ) ∇ (∇ • u ) + (μ −η∇ )∇ u + F = 0
                                     2                               2   2
                                                                                              (131b)


This relation can also be written as

                         ( λ + 2μ ) ∇ ( ∇ • u ) − ( μ − η∇ 2 )∇ × ∇ × u + F = 0                (132)

which was derived previously by Mindlin and Tiersten (1962) within the context of the
indeterminate couple stress theory.                    However, recall that the Mindlin-Tiersten
formulation involved two couple stress parameters η and η ′ . In hindsight, the fact that
η ′ does not appear in (132) should have been an indication that this coefficient is not
independent of η . We now know that η ′ = −η .


The general solution for the displacement in isotropic polar elasticity also has been
derived by Mindlin and Tiersten (1962) as

                      u = G − l 2 ∇∇ • G −
                                                      1
                                                   4(1 − ν )
                                                              [
                                                             ∇ r • (1 − l 2 ∇ 2 )G + G0   ]    (133)

where the vector function G and scalar function G0 satisfy the relations

                                     μ (1 − l 2 ∇ 2 )∇ 2 G = −F                               (134a)

                                         μ∇ 2 G0 = r • F                                      (134b)
These functions reduce to the Papkovich functions in the classical theory, when l = 0 . It
is easily seen that
                               λ + 2μ
                                      ∇ • u = (1 − l 2 ∇ 2 )∇ • G                             (135a)
                                 μ
                                           ∇×u = ∇×G                                          (135b)


Uniqueness theorem for boundary value problems
Now we investigate the uniqueness of the linear polar elasticity boundary value problem.
The proof follows from the concept of strain energy, similar to the approach for Cauchy


                                                         23
elasticity. By replacing the virtual deformation with the actual deformation in the virtual
work theorem (82) and accounting for the symmetry of eij , we obtain

                            (                        ) ( n)                     (n)
                        ∫ σ ( ji )eij − 2μiκ i dV = ∫ ti ui dS + ∫ Fiui dV + ∫ mi ωi dS
                        V                                      S                   V                 S
                                                                                                                              (136)



Using the constitutive relations (103) and (104), we have
                                       σ ( ji )eij − 2μiκ i = Aijkl eij ekl + Bijκ iκ j = 2W (e, κ )                          (137)


Therefore, (136) can be written as

                        ∫ (A
                        V
                                       e e + Bij κ iκ j )dV = ∫ t i(n )ui dS + ∫ Fi ui dV + ∫ mi(n )ωi dS
                                ijkl ij kl
                                                                   S                   V                 S
                                                                                                                              (138)

This relation gives twice of total strain energy in terms of the work of external body
forces and surface tractions.


Now, we consider the general boundary value problem.                                                     The prescribed boundary
conditions on the surface of the body can be any well-posed combination of vectors ui

and ωi , ti(n ) and mi(n ) as discussed on Section 4. Assume that there exist two different

solutions {u i(1) , eij1) , κ i(1) , σ (ji ) , μ i(1) } and {u i(2 ) , eij2 ) , κ i(2 ) , σ (ji2 ) , μ i(2 ) } to the same problem with
                     (                   1                              (


identical body forces and boundary conditions. Thus, we have the equilibrium equations
                                                               α
                                                           σ (ji , )j + Fi = 0                                                (139)

                                                           σ [(α ]) = − μ[(iαj)]
                                                               ji           ,                                                 (140)

where
                                                                       1
                                                         μi(α ) = − Bijκ (jα )                                               (141a)
                                                                       2
                                                                          (α
                                                         σ ((α ) = Aijkl ekl )
                                                             ji
                                                                )
                                                                                                                             (141b)
                                (α )                                        (1)            (2)
and the superscript                     references the solutions                   and           .


Let us now define the difference solution
                                                           ui′ = ui(2 ) − ui(1)                                              (142a)



                                                                       24
                                                              (        (
                                                       eij = eij2 ) − eij1)
                                                        ′                                                                      (142b)

                                                        κ i′ = κ i(2 ) − κ i(1)                                                (142c)

                                                     σ ′ji = σ (ji2 ) − σ (ji )
                                                                            1
                                                                                                                               (142d)

                                                       μ i′ = μ i(2 ) − μ i(1)                                                 (142e)


Since the solutions {u i(1) , eij1) , κ i(1) , σ (ji ) , μ i(1) } and {u i(2 ) , eij2 ) , κ i(2 ) , σ (ji2 ) , μ i(2 ) } correspond to the
                               (                   1                              (


same body forces and boundary conditions, the difference solution must satisfy the
equilibrium equations
                                                            σ ′ji , j = 0                                                        (143)

                                                        σ [′ji ] = − μ[′i , j ]                                                  (144)

with zero corresponding boundary conditions.                                          Consequently, twice the total strain
energy (137) for the difference solution is

                                   V
                                       (                            )
                                    ∫ Aijkl eij ekl + Bijκ i′κ ′j dV = ∫ 2W ′ dV = 0
                                             ′ ′
                                                                                  V
                                                                                                                                 (145)

Since the strain energy density W ′ is non-negative, this relation requires
                                           2W ′ = Aijkl eij ekl + Bijκ i′κ ′j = 0 in V
                                                         ′ ′                                                                     (146)


However, the tensors Aijkl and Bij are positive definite. Therefore the strain, curvature

and associted stresses for difference solution must vanish
                                           eij = 0, κ i′ = 0, σ ij = 0, μ i′ = 0
                                            ′                   ′                                                           (147a-d)


These require that the difference displacement ui′ can be at most a rigid body motion.
However, if displacement is specified on part of the boundary such that rigid body
motion is prevented, then the difference displacement vanishes everywhere and we have
                                                               ui(1) = ui(2 )                                                  (148a)
                                                                (       (
                                                               eij1) = eij2 )                                                  (148b)

                                                               κ i(1) = κ i(2 )                                                (148c)

                                                             σ (ji ) = σ (ji2 )
                                                                 1
                                                                                                                               (148d)


                                                                    25
                                                               μi(1) = μi( 2 )                                                     (148e)


Therefore, the solution to the boundary value problem is unique. On the other hand, if
only force and couple tractions are specified over the entire boundary, then the
displacement is not unique and is determined only up to an arbitrary rigid body motion.


Reciprocal theorem
We derive now the general reciprocal theorem for the equilibrium states of a linear polar
elastic material under different applied loads. Consider two sets of equilibrium states of
                                                {                                        }         {
compatible elastic solutions ui(1) , ωi(1) , t i(n )(1) , mi(n )(1) , Fi (1) and ui(2 ) , ωi(2 ) , ti(n )(2 ) , mi(n )(2 ) , Fi (2 ) . }
Let us apply the virtual work theorem (82) in the forms

            ∫ (σ         eij − 2μ i(1)κ i(2 ) )dV = ∫ t i(n )(1)ui(2 )dS + ∫ Fi (1)ui(2 )dV + ∫ mi(n )(1)ωi(2 )dS
                    (1) ( 2 )
                    ji                                                                                                              (149)
            V                                         S                      V                      S



             ∫ (σ        eij − 2μi(2 )κ i(1) )dV = ∫ ti(n )(2 )ui(1)dS + ∫ Fi (2 )ui(1)dV + ∫ mi(n )(2 )ωi(1)dS
                    ( 2 ) (1)
                    ji                                                                                                              (150)
            V                                         S                      V                      S


By using the general constitutive relations
                                                                   1                 1
                                                            (1
                                           σ (ji ) = Aijkl ekl ) + Bik κ k(1,)j − B jk κ k(1,i)
                                               1
                                                                                                                                    (151)
                                                                   4                 4
                                                                    1
                                                       μi(1) = − Bijκ (j1)                                                          (152)
                                                                    2
                                                                   1                1
                                                            (
                                          σ (ji2 ) = Aijkl ekl2 ) + Bik κ k(2j) − B jk κ k(2i)
                                                                            ,              ,                                        (153)
                                                                   4                4
                                                                    1
                                                      μ i(2 ) = − Bijκ (j2 )                                                        (154)
                                                                    2
it is seen that
                                                                    1                        1
                    1 (                                (1 (                       (                     (
                σ (ji )eij2 ) − 2μi(1)κ i(2 ) = Aijkl ekl )eij2 ) + Bik κ k(1,)j eij2 ) − B jk κ k(1,i)eij2 ) + Bijκ (j1)κ i(2 )    (155)
                                                                    4                        4


However, the symmetry relation (102) shows
                                                 1                      1
                                                                 (                     (
                                                   Bik κ k(1,)j eij2 ) − B jk κ k(1,i)eij2 ) = 0                                    (156)
                                                 4                      4
Therefore



                                                                        26
                                     1 (                                (1 (
                                 σ (ji )eij2 ) − 2μi(1)κ i(2 ) = Aijkl ekl )eij2 ) + Bijκ (j1)κ i(2 )       (157)

Similarly,
                                          (                             ( (
                                 σ (ji2 )eij1) − 2μi(2 )κ i(1) = Aijkl ekl2 )eij1) + Bijκ (j2 )κ i(1)       (158)


Now we see that the symmetry relations (101) and (102) require
                                             1 (                                  (
                                         σ (ji )eij2 ) − 2μi(1)κ i(2 ) = σ (ji2 )eij1) − 2μi(2 )κ i(1)      (159)

which shows

                          ∫ (σ           eij − 2μi(1)κ i(2 ) )dV = ∫ (σ (ji2 )eij1) − 2μi(2 )κ i(1) )dV
                                    (1) ( 2 )                                  (
                                    ji                                                                      (160)
                          V                                              V




Therefore, the general reciprocal theorem for these two elastic solutions is
                       ( n )(1) ( 2 )
                               ui dS + ∫ Fi( )ui( ) dV + ∫ mi( )( )ωi( ) dS
                                            1    2            n 1     2
                   ∫ ti
                   S                               V                    S
                                                                                                            (161)
                                         ( n )( 2 ) (1)           ( 2 ) (1)              ( n )( 2 ) (1)
                              = ∫ ti             ui dS + ∫ Fi ui dV + ∫ mi                      ωi dS
                                S                           V                       S




7. Two-dimensional infinitesimal linear isotropic polar elasticity theory
In this section, we reconsider the two-dimensional infinitesimal linear isotropic polar
elasticity developed by Mindlin (1963). We start this development by assuming that the
displacement components are two-dimensional, where
                                              u1 = u ( x, y ) , u 2 = v( x, y ) , u3 = 0                  (162a-c)


This is exactly the conditions for plane strain theory in Cauchy elasticity. The non-zero
components of strains are
                                                  ∂u        ∂v         1 ⎛ ∂u ∂v ⎞
                                         ex =        , ey =    , e xy = ⎜ + ⎟                             (163a-c)
                                                  ∂x        ∂x         2 ⎜ ∂y ∂x ⎟
                                                                         ⎝       ⎠
and the only non-zero rotation component is
                                                                       1 ⎛ ∂v       ∂u ⎞
                                                       ω z = ω yx = ⎜ − ⎟                                   (164)
                                                                   2 ⎜ ∂x ∂y ⎟
                                                                     ⎝       ⎠


Therefore, the components of the mean curvature vector are


                                                                  27
                                           1 ∂ω z                             1 ∂ω z
                           κ x = −κ yz =          ,         κ y = κ xz = −             (165a,b)
                                           2 ∂y                               2 ∂x


It is seen that the compatibility equations for this case are
                                       ∂ 2 ex ∂ e y ∂ 2 e xy
                                               2

                                             + 2 =2                                     (166a)
                                       ∂y 2   ∂x    ∂x∂y
                                               ∂κ yz        ∂κ xz
                                                        =                               (166b)
                                                ∂x           ∂y

                                            ∂ω z ∂e xy ∂e x
                                                =     −                                 (166c)
                                             ∂x   ∂x    ∂y

                                            ∂ω z ∂e y ∂e xy
                                                =    −                                  (166d)
                                             ∂y   ∂x   ∂y


Then, the corresponding couple stress and force stress components can be written
                                               ∂ω z                    ∂ω z
                                  μ x = −4η         ,       μ y = 4η                   (167a,b)
                                                ∂y                      ∂x

                                   σ [xy ] = 2η∇ 2ω xy = −2η∇ 2ω z                      (168a)

                                    σ [ yx ] = 2η∇ 2ω yx = 2η∇ 2ω z                     (168b)


                                      σ ( xx ) = (λ + 2 μ )ex + λe y                    (169a)

                                       σ ( yy ) = λex + (λ + 2 μ )e y                   (169b)

                                               σ ( xy ) = 2μexy                         (169c)

It is also found that
                                                              ∂ω z
                                               μ xz = 4η                                (170a)
                                                               ∂x
                                                              ∂ω z
                                               μ yz = 4η                                (170b)
                                                               ∂y
Finally, it is seen that
                                       σ x = (λ + 2μ )ex + λe y                         (171a)




                                                      28
                                    σ y = λex + (λ + 2 μ )e y                    (171b)

                                    σ xy = 2 μe xy − 2η∇ 2ω z                    (171c)

                                    σ yx = 2μe xy + 2η∇ 2ω z                     (171d)

where
                                      σ xy + σ yx = 4μexy                        (171e)

It is also seen that
                                       σ z = ν (σ x + σ y )                      (171f)

similarly to plane strain Cauchy elasticity, while for the couple stresses
                                                               ∂ω z
                                    μ zx = − μ xz = −4η                          (171g)
                                                                ∂x
                                                               ∂ω z
                                    μ zy = − μ yz = −4η                          (171h)
                                                                ∂y


When there is no body force, these stresses satisfy the equilibrium equations
                                          ∂σ x ∂σ yx
                                              +      =0                          (172a)
                                           ∂x   ∂y
                                       ∂σ xy           ∂σ y
                                                   +          =0                 (172b)
                                          ∂x            ∂y

                                 ∂μ xz ∂μ yz
                                      +      + σ xy − σ yx = 0                   (172c)
                                  ∂x    ∂y


To solve for stresses, we need to derive compatibility equations in terms of stresses as
follows. It is seen that

                                   ex =
                                           1
                                          2μ
                                               [
                                             (1 −ν )σ x −νσ y         ]          (173a)


                                   ey =
                                           1
                                          2μ
                                               [
                                             (1 −ν )σ y −νσ x         ]          (173b)


                                     2e xy =
                                                1
                                                  (σ xy + σ yx )                 (173c)
                                               2μ




                                                   29
                                          ∇ 2ω z =
                                                      1
                                                        (σ yx − σ xy )                (173d)
                                                     4η


By inserting these in (166), we obtain the compatibility equations in terms of the force
and couple stress tensors. Thus,
                    ∂ 2σ x ∂ σ y
                             2
                                                       ∂2
                          +      −ν∇ 2 (σ x + σ y ) =      (σ yx + σ xy )             (174a)
                     ∂y 2   ∂x 2                      ∂x∂y

                                              ∂μ xz ∂μ yz
                                                   =                                  (174b)
                                               ∂y    ∂x

                     μ xz = l 2
                                  ∂
                                  ∂x
                                                                [              ]
                                     (σ yx + σ xy ) − 2l 2 ∂ σ x −ν (σ x + σ y )
                                                           ∂y
                                                                                      (174c)


                    μ yz = 2l 2
                                   ∂
                                   ∂x
                                      [                     ]  ∂
                                      σ x −ν (σ x + σ y ) − l 2 (σ yx + σ xy )
                                                               ∂y
                                                                                      (174d)



By combining these with the equilibrium equations, we obtain the set of equations
                                             ∂σ x ∂σ yx
                                                 +      =0                            (175a)
                                              ∂x   ∂y
                                             ∂σ xy       ∂σ y
                                                     +          =0                    (175b)
                                               ∂x         ∂y

                                      ∂μ xz ∂μ yz
                                           +      + σ xy − σ yx = 0                   (175c)
                                       ∂x    ∂y

                                             ∇ 2 (σ x + σ y ) = 0                     (175d)

                                                ∂μ xz ∂μ yz
                                                     =                                (175e)
                                                 ∂y    ∂x


By introducing the stress functions Φ = Φ( x, y ) and Ψ = Ψ ( x, y ) , we may write the force
stresses and couple stresses as follows:
                                               ∂ 2Φ ∂ 2Ψ
                                           σx = 2 −                                   (176a)
                                               ∂y   ∂x∂y

                                                    ∂ 2Φ ∂ 2 Ψ
                                           σy =         +                             (176b)
                                                    ∂x 2 ∂x∂y


                                                     30
                                                ∂ 2Φ ∂ 2 Ψ
                                     σ xy = −       −                              (176c)
                                                ∂x∂y ∂y 2

                                                ∂ 2Φ ∂ 2 Ψ
                                     σ yx = −       +                              (176d)
                                                ∂x∂y ∂x 2
                                                      ∂Ψ
                                             μ xz =                                (177a)
                                                      ∂x
                                                      ∂Ψ
                                             μ yz =                                (177b)
                                                      ∂y


Equilibrium equations satisfy and compatibility equations give
                                         ∇ 2∇ 2 Φ = 0                               (178)
                         ∂                              ∂
                         ∂x
                            (            )        (
                            Ψ − l 2∇ 2Ψ = −2 1 −ν 2 l 2
                                                        ∂y
                                                           ∇ 2Φ)       (       )   (179a)

                         ∂                             ∂
                         ∂y
                            (            ) (
                            Ψ − l 2∇ 2Ψ = 2 1 −ν 2 l 2
                                                       ∂x
                                                           )
                                                          ∇ 2Φ     (       )       (179b)



Combining (179a) and (179b) by eliminating Φ gives
                                      ∇ 2 Ψ − l 2∇ 4 Ψ = 0                          (180)


All these relations are exactly the equations derived by Mindlin (1963). This shows that
the solutions for two-dimensional cases based on Mindlin’s development, such as stress
concentration relations for a plate with a circular hole, still can be used. However, we
should notice that the couple stresses μ zx and μ zy are

                                              η′            ∂ω
                                    μ zx =       μ xz = 4η ′ z                     (181a)
                                              η              ∂x
                                              η′            ∂ω
                                     μ zy =      μ yz = 4η ′ z                     (181b)
                                              η              ∂y
in Mindlin’s development. These relations become identical to those in the present
theory, when we take η ′ = −η . Thus, we may solve the boundary value problem in an
identical manner to Mindlin (1963), but then evaluate the couple stresses through a
postprocessing operation.


                                                 31
More specifically, by comparing the relations (170) and (177), we can see
                                                     ∂ω z ∂Ψ
                                         μ xz = 4η       =                           (182a)
                                                      ∂x   ∂x
                                                     ∂ω z ∂Ψ
                                         μ yz = 4η       =                           (182b)
                                                      ∂y   ∂y


Therefore, we can take
                                                 4ηω z = Ψ                            (183)


If Ψ is zero, there are no couple stress tensor components, and the relations for the force-
stress tensor reduce to the relations in classical elasticity, where Φ is the Airy stress
function.


For force and couple traction vectors, we have
                                           t xn ) = σ xx n x + σ yx n y
                                             (
                                                                                     (184a)

                                            t yn ) = σ xy nx + σ yy n y
                                              (
                                                                                     (184b)

                                         m = mz n ) = μ xz nx + μ yz n y
                                              (
                                                                                     (184c)

which can be written in terms of stress functions as
                                     ⎛ ∂ Φ ∂2 Ψ ⎞     ⎛ ∂2 Φ ∂2 Ψ ⎞
                          t   (n)
                                     ⎜ 2−
                                    =⎜                ⎜ ∂x∂y + ∂x 2 ⎟n y
                                                ⎟nx + ⎜ −                            (185a)
                                           ∂x∂y ⎟                   ⎟
                              x
                                     ⎝ ∂y       ⎠     ⎝             ⎠

                                   ⎛ ∂2 Φ ∂2 Ψ ⎞   ⎛ ∂2 Φ ∂2 Ψ ⎞
                          t yn ) = ⎜ −
                            (
                                         − 2 ⎟nx + ⎜ 2 +
                                   ⎜ ∂x∂y ∂y ⎟     ⎜ ∂x        ⎟n y                  (185b)
                                   ⎝           ⎠   ⎝      ∂x∂y ⎟
                                                               ⎠
                                                   ∂Ψ      ∂Ψ
                                             m=       nx +    ny                     (185c)
                                                   ∂x      ∂y
If the location on the surface is specified by the coordinate s along the boundary in a
positive sense, we have
                                                       dy
                                                nx =                                 (186a)
                                                       ds




                                                      32
                                                          dx
                                                 ny = −                             (186b)
                                                          ds
Therefore
                                                 d ⎛∂Φ ∂Ψ⎞
                                      t xn ) =
                                        (
                                                    ⎜    −    ⎟                     (187a)
                                                 ds ⎜ ∂y
                                                    ⎝      ∂x ⎟
                                                              ⎠
                                                  d ⎛∂Φ ∂Ψ⎞
                                     t yn ) = −
                                       (
                                                     ⎜    +    ⎟                    (187b)
                                                  ds ⎜ ∂x
                                                     ⎝      ∂y ⎟
                                                               ⎠
                                                 ∂Ψ      ∂ω
                                        m=          = 4η                            (187c)
                                                 ∂n      ∂n


8. Anti-plane deformation infinitesimal linear isotropic polar elasticity theory
We assume the displacement components are
                                  u1 = 0 , u 2 = 0 , u3 = w(x, y )                 (188a-c)


These are exactly the conditions for anti-plane deformation in Cauchy elasticity. The
non-zero components of strains are
                                                 1 ∂w          1 ∂w
                                       ezx =          , e zy =      ,              (189a,b)
                                                 2 ∂x          2 ∂y
and the non-zero rotation components are
                                            1 ∂w                        1 ∂w
                                    ωx =         ,             ωy = −                (190)
                                            2 ∂y                        2 ∂x


Therefore, the only non-zero component of the mean curvature vector is

                                            1 ⎛ ∂ω ∂ω ⎞   1
                                κ z = κ yx = ⎜ y − x ⎟ = − ∇ 2 w
                                              ⎜ ∂x     ⎟                             (191)
                                            2⎝      ∂y ⎠  4

Then, the corresponding couple stress and force stress components can be written
                                          μ z = μ yx = 2η∇ 2 w                       (192)

                                                    ∂w                ∂w
                                     σ ( zx ) = μ      , σ ( zy ) = μ              (193a,b)
                                                    ∂x                ∂y

                             1 ∂μ xy     ∂                  1 ∂μ yx    ∂
                σ [ zx ] =           = −η ∇ 2 w , σ [zy ] =         = η ∇2w        (193c,d)
                             2 ∂x        ∂x                 2 ∂y       ∂y


                                                       33
Therefore
                                 ∂w    ∂                 ∂w   ∂
                      σ zx = μ      − η ∇ 2 w , σ xz = μ    +η ∇2w                  (194a,b)
                                 ∂x    ∂x                ∂x   ∂x
                                  ∂w    ∂                 ∂w   ∂
                       σ zy = μ      + η ∇ 2 w , σ yz = μ    −η ∇2w                 (194c,d)
                                  ∂y    ∂y                ∂y   ∂y


When there is no body force, these stresses satisfy the equilibrium equation
                                            ∂σ xz ∂σ yx
                                                 +      =0                             (195)
                                             ∂x    ∂y
which in terms of displacement gives the single fourth order equation
                                         μ∇ 2 w − η∇ 2∇ 2 w = 0                        (196)


If couple stresses are zero on the boundary, the solution reduces to the classical Cauchy
elasticity solution
                                                ∇2w = 0                                (197)
with
                                             μ z = μ yx = 0                           (198a)

                                                              ∂w
                                            σ zx = σ xz = μ                           (198b)
                                                              ∂x
                                                              ∂w
                                            σ yz = σ zy = μ                           (198c)
                                                              ∂y
everywhere in the domain. However, geometrical boundary conditions, such as
                                                        1 ∂w
                                     w = 0, ω (ns ) =        =0    on S             (199a,b)
                                                        2 ∂n
create couple stresses in polar media. In that case, the classical solution cannot be used.


9. Some elementary linear isotropic polar elasticity problems
In this section, we reconsider several elementary problems in Cauchy elasticity within the
framework of the present infinitesimal linear polar theory. Koiter (1964) considered all




                                                  34
of these problems, but within the context of the indeterminate couple stress theory.
Consequently, some differences appear.


Twist of a cylindrical bar
Consider the x3 -axis of our coordinate system along the axis of a cylindrical bar with
constant cross section. We assume the displacement components are in the form as in the
classical theory and examine the corresponding stress field in the couple stress theory.
The assumed displacement components are
                          u1 = −θ x2 x3 ,       u2 = θ x1x3 ,          u3 = 0            (200a-c)

where θ is the constant angle of twist per unit length. The non-zero components of the
strain tensor and rotation vector are
                                        1                    1
                                 e13 = − θ x2 ,         e23 = θ x1                       (201a,b)
                                        2                    2
                          1                             1
          ω1 = ω32 = − θ x1,           ω2 = ω13 = − θ x2 ,             ω3 = ω21 = θ x3   (202a-c)
                          2                             2
Interestingly, it is seen that the curvature vector vanishes
                                               1
                                            κ = ∇×ω = 0                                    (203)
                                               2
Therefore, the force-stress distribution is the classical result
                                 σ13 = − μθ x2 ,        σ 23 = μθ x1                     (204a,b)
and the twist of a cylindrical bar does not generate couple stresses. This is in contrast
with the Koiter (1964) result, in which couple stresses appear.


Cylindrical bending of a flat plate
Consider a flat polar material plate of thickness h bent into a cylindrical shell with
generators parallel to the x3 -axis. Let R denote the radius of curvature of the middle

plane x1 x3 in the deformed configuration. We assume the displacement components are
similar to those in Cauchy elasticity. Thus,
                          1                   11 2 1 ν 1 2
                 u1 = −     x1 x2 ,    u2 =      x1 +          x2 ,        u3 = 0        (205a-c)
                          R                   2R      2 1 −ν R




                                                   35
The non-zero components of the strain tensor, rotation vector and mean curvature vector
are
                                         1                               ν   1
                              e11 = −      x2 ,              e22 =             x2                              (206a,b)
                                         R                              1 −ν R
                                                                   x1
                                           ω3 = ω21 =                                                            (207)
                                                                   R
                                                                1
                                          κ 31 = −κ 2 =                                                          (208)
                                                               2R
Therefore, the non-zero force and couple stresses are written as
                                          2μ x2                               2μν x2
                              σ 11 = −          ,              σ 33 = −                                        (209a.b)
                                         1 −ν R                               1 −ν R
                                                                          η
                                    μ 2 = μ13 = − μ31 = 4                                                        (210)
                                                                             R
Notice that unlike the previous example of twisting deformation, bending does produce
couple stresses. This is due to the existence of non-zero mean curvature.


Pure bending of a bar with rectangular cross-section
We take the x1 -axis to coincide with the centerline of the rectangular beam and the other
axes parallel to the sides of the cross section of the beam. Let R denote the radius of
curvature of the central axis of the beam after bending in the x1 x3 -plane . We assume the
displacement components are the same as in the classical Cauchy elasticity theory as
follows:
                                            ν                                    ν
           u1 =
                  1
                  R
                    x1 x3 ,       u2 = −
                                            R
                                                  x 2 x3 ,               u3 =
                                                                                 2R
                                                                                    (x    2
                                                                                          2      )
                                                                                              − x3 −
                                                                                                 2      1 2
                                                                                                       2R
                                                                                                          x1   (211a-c)



Then, the strains, rotations and mean curvatures can be written
                                                x3                                   ν x3
                                     e11 =         ,           e22 = e33 = −                                   (212a,b)
                                                R                                     R
                                                       νx 2                             x1
                                   ω1 = ω32 =                  ,        ω 2 = ω13 =                            (213a,b)
                                                        R                               R

                                     κ1 = κ 32 =
                                                         1
                                                           (ω3,2 − ω2,3 ) = 0                                   (214a)
                                                         2



                                                              36
                                               1
                                κ 2 = κ13 =      (ω1,3 − ω3,1 ) = 0                        (214b)
                                               2

                              κ 3 = κ 21 =
                                             1
                                               (ω2,1 − ω1,2 ) = 1 −ν                       (214c)
                                             2                   2R
As a result, the non-zero force and couple stresses take the form
                                                           x3
                                      σ 11 = 2μ (1 + ν )                                   (215a)
                                                           R
                                                            1 −ν
                               μ 3 = μ 21 = − μ12 = −4η                                    (215b)
                                                              R


Again, for this problem, we find non-zero mean curvature and couple stresses.


10. Conclusions
By considering further the consequences of the kinematics of a continuum and the
principle of virtual work, we find that the couple stress theory for polar media can be
formulated as a practical theory without any ambiguity. In the resulting quasistatic
theory, independent body couples cannot be specified in the volume and surface couples
can only exist in the tangent plane at each boundary point. (Although not specifically
addressed here, in the corresponding dynamical theory, it should be clear that spin inertia
does not appear.) As a consequence, the couple stress tensor is found to be skew-
symmetric and energetically conjugate to the mean curvature tensor, which also is skew-
symmetrical. Then, for infinitesimal or small deformation linear polar elasticity, we can
write constitutive relations for all of the components of the force stress and couple stress
tensors. The most general anisotropic polar material is described by 27 independent
constitutive coefficients, including six coefficients relating mean curvatures to couple
stresses. At the other extreme, for isotropic materials, the two Lamé parameters and one
length scale completely characterize the behavior. In addition, strain energy relations and
uniqueness and reciprocal theorems have been developed for linear polar elasticity.
General formulations for two-dimensional and anti-plane problems are also elucidated for
the isotropic case.    The former employs a pair of stress functions, as introduced
previously by Mindlin for the indeterminate theory.                    Finally, several additional




                                                37
elementary problems are examined within the context of small deformation polar
elasticity.


By resolving the indeterminacy of the previous couple stress theory, the present polar
theory provides a fundamental basis for the development of scale-dependent material
response. Additional aspects of the linear theory, including fundamental solutions and
computational mechanics formulations, will be addressed in forthcoming work. Beyond
this, the present polar theory should be useful for the development of nonlinear elastic,
elastoplastic, viscoplastic and damage mechanics formulations that may govern the
behavior of solid continua at the smallest scales.


References
Cosserat, E., Cosserat, F., 1909. Théorie des corps déformables (Theory of deformable
bodies). A. Hermann et Fils, Paris.

Hamilton, W. R., 1866. Elements of quaternions. Longmans, Green, & Co., London.

Koiter, W. T., 1964. Couple stresses in the theory of elasticity, I and II. Proc. Ned. Akad.
Wet. (B) 67, 17-44.

Mindlin, R. D., 1963. Influence of couple-stresses on stress concentrations. Exp. Mech. 3,
1–7.

Mindlin, R. D., 1964. Micro-structure in linear elasticity, Arch. Rational Mech. Anal. 16,
51–78.

Mindlin, R. D., Tiersten, H. F., 1962. Effects of couple-stresses in linear elasticity, Arch.
Rational Mech. Anal. 11, 415–488.

Nowacki, W., 1986. Theory of asymmetric elasticity. Pergamon Press, Oxford.

Toupin, R. A., 1962. Elastic materials with couple-stresses. Arch. Rational Mech. Anal.
11, 385–414.

Voigt, W., 1887. Theoretische Studien fiber die Elastizitatsverhiltnisse der Kristalle
(Theoretical studies on the elasticity relationships of crystals). Abh. Gesch.
Wissenschaften 34.


                                             38