3．Field Equations of Elasticity - Differential Formulation
This chapter will explore the field equations of elasticity viewed as differential
equations. The materials presented here have partial overlap with the general
formulation in continuum mechanics. The focus will be on the simplest case: an
elastic body under infinitesimal deformation. We start with the balance laws.
3.1
Balance Laws of Momentum and Moment
Field Variables
Consider a general three-dimensional body (in the shape of a potato) occupying a
domain V under linear elastic and infinitesimal deformation, as shown in the figure
below. The basic field variables include a displacement vector ui (3 components), a
symmetric second rank strain tensor  ij (6 components), and another symmetric
second rank stress tensor  ij (6 components). That gives a total number of field
variables of 15. In general, they are all functions of the spatial coordinates xi and
temporal variable t.
St
V
X3
X2
X1
Su
Figure 3.1
A three-dimensional body
Obviously, one needs to formulate 15 equations to solve these field variables. To
shorten the equations to be derived, the temporal and spatial derivatives are
abbreviated as:


     and     
t
xi
,i
in the sequel.
Balance Law of Linear Momentum
Consider the balance of the linear momentum first. From the previous knowledge in
continuum mechanics, one has
 ij, j  f i  ui
(3.1)
where  denotes the density (mass per unit volume) of the body, and f i the body
force per unit volume. For the special case of quasi-static motion where the particle
velocity is much less than the stress wave speeds that will be specified after (3.20),
the balance equation (3.1) reduces to the equilibrium one:
 ij, j  f i  0 .
(3.2)
Balance of Angular Momentum
Next consider the balance of angular momentum. If the effect of body couple can be
regarded as higher order infinitesimal, the equilibrium of an infinitesimal cube will
lead to the following reciprocal theorem of shear stress:
 ij   ji .
(3.3)
A basic assumption to derive (3.3) is that the body moment (  m  V ) is much less
than the moment   V introduce by the surface traction on the cube. I would like
to remind you that the assumption (3.3) on a symmetric stress tensor sometimes
breaks down. The case of strong magnetic field is one exception, along with the others.
The theory of Cossara brothers were proposed more than a century ago to deal with
the situation, where 9 independent components (instead of six) of stress tensor should
be engaged. In more recent literatures, the asymmetric shear stresses are termed the
“couple stresses”. Mindlin was largely responsible in developing the couple stress
concept in elasticity. The current developments of strain gradient theory (such as those
undertaken by K. C. Hwang and his collaborators) somewhat borrow the idea of
asymmetric stress under the framework of couple stress elasticity.
3.2
Compatibility Equation
Kinematics
Under infinitesimal deformation, the strain tensor can be easily derived as the
symmetric gradient of the displacement vector:
 ij 
1
ui, j  u j ,i .
2
(3.4)
The 6 symmetric components of the strain tensor can be expressed through 3
displacement components. Accordingly, the strain tensor cannot be arbitrarily
prescribed. Otherwise, gap or overlap would occur in the deformation manifold. The
conditions to avoid gap or overlap in deformation are called the compatibility
equations.
gap
overlap
compatibility
yyy
Figure 3.2 Compatibility of deformation
Compatibility Conditions
We examine the compatibility condition in the general three-dimensional case. From
(3.4), it is straightforward to verify that the symmetric double curl operation on the
strain tensor would vanish
  ε    emjk enil ij,kl  Lmn  0
(3.5)
where  denotes the gradient operator,  the vector product, and emjk the
alternation symbol. They are all introduced in the course of continuum mechanics.
From the first glance, equations (3.5) seems to represent 9 conditions for the
“incompatible tensor” Lmn . However, a large portion of equations in (3.5) are
redundant. Listed below are two useful properties for Lmn to remove the redundancy:
Property 1 Symmetry of Lmn
Lmn  Lnm
(3.6)
 mn  mi  ml
This property can be proved by means of the identity emjkenil   jn  ji  jl
 kn  ki  kl
between the alternation symbol emjk and the Kronecker delta  mj .
Property 2 Bianchi identity
Lmn, n  0
(3.7)
Namely, the “incompatible tensor” Lmn is divergence free. Conditions (3.6) and (3.7)
give 6 constraints on the incompatible tensor Lmn . Therefore, among 9 compatibility
conditions in (3.5), only three of them are independent.
For the special case of planar elasticity, the strain compatibility degenerates to a scalar
condition, written as
  ,     ,
(3.8)
where the Greek indices only have the range from 1 to 2.
Simply and Multiply Connected Region
Figure 3.3 Simply-connected regions
The compatibility conditions (3.5) or (3.8) are expressed in a local sense, namely in
the vicinity of a given point. The strain compatibility also requires global impact, in
terms of the single value condition of the displacement fields. To illustrate this point,
one has to distinguish a simply-connected region and a multiply-connected one. The
former is shown in Figure 3.3. Any closed curves in a simply-connected region can
continuously shrink to a point.
Several cases for multiply-connected regions are delineated in Fig. 3.4, where at least
some closed curves can not be shrunk onto a point.
Figure 3.4 Multiply-connected regions.
Global Compatibility
Consider two points, A and B, as shown in Fig. 3.5. One may ask the question: if the
displacement of point A is known, can we uniquely compute the displacement at point
B from the information of the strain field?
B
S
A
Figure 3.5 From the displacement at point A to the displacement at point B
There are infinitely many paths to joint point A and point B. Take an arbitrary path,
mark by S in Figure 3.5. If the displacement of point A is known, one can heuristically
compute the displacement at point B through a path integral accounting for the
variation of the displacement along the curve as:
u i
ds .
s
A
B
u iB  u iA  
(3.9)
Is the displacement that computed according to (3.9) unique? If so, (3.9) should be the
same for any curve S connecting A and B. Namely, the single-valued-ness of the
displace field requires that
ui
 s ds  0 , for any closed curves passing A and B
(3.10)
For a simply-connected region, the single-valued-ness condition (3.10) can be relaxed
to an infinitesimal closed loop:
ui
 s ds  0 , for any infinitesimal loops
(3.11)
It is a straightforward matter to show that (3.11) is equivalent to (3.5), assuming a
smooth strain field in the neighborhood.
We next consider the global strain compatibility in a multiply-connected (or
n-connected to be specific) region. Beside the local strain compatibility condition
(3.5), one also need to include n－1 conditions listed below to guarantee the
single-valued-ness condition of the displacement
ui
ds  0
i s

，i＝1，…，n－1,
(3.12)
where i are contours surrounding different cavities in the n-connected region.
Displacements via Strain Integration
A formal derivation for the displacement fields under prescribed strain fields is now in
order. From the kinematics of (3.4), one can start from the equation ui , j  u j ,i  2 ij
to get ui from the prescribed  ij values. Obviously, the integration of this relation
would be by no means unique. A rigid body motion in the form of uirigid  0ji x j  Ci
(corresponding to null strain field) can always be super-imposed on the derived
displacement field. The symbol ij0  0ji represents a skew-symmetric tensor of
rigid body rotation, it relates to a rotation vector through a relation Ω 0  x  w 0  x .
To derive an expression for the displacement gradient ui , j , one needs to represent the
rotation (the skew-symmetric part of the displacement gradient) ij 
1
ui, j  u j ,i 
2
in terms of the gradients of the strain tensor. That can be accomplished by
2ij, k  ui , jk  u j ,ik  ui , k  uk ,i , j  uk , j  u j , k ,i
 2 ik , j  2 jk ,i  2U ijk
.
The integration gives
x
 ij   ij 0    U ijk  x 'dx k'   ij 0   I ji  x  .
(a)
o
With the above expression of the rotation tensor, the displacement gradient can be cast
in the following form: ui , j   ij  ij . Further integration gives
ui  x   ui 0   ij 0x j    ij  I ij dx' j .
x
(b)
0
The double integration can be reduced to single integration through integration by
parts. The final result is
ui  x   ui 0   ij 0x j    ij ( y)dy j   x j  y j U ijk  y dy k
x
x
0
o
(3.13)
Examine the path independency conditions for integrations in (a) and (b) would shed
lights on the strain compatibility condition. The integration in (b) would be path
independent provided

ij
 I ij , k   ik  I ik , j . That can be interpreted as
 ij, k  U ijk   ik , j  U ikj , and the identity is validated by the definition of U ijk . On the
other hand, the integration in (a) would be path independent if U ijk ,l  U ijl, k , which
can
be
written
in
an
equivalent
form
of
 ki , jl   kj ,il   li, jk   lj,ik , or
emij ki , jl  emij li, jk , or emijenkl ki , jl  0 . One then reaches the sufficiency of condition
(3.5) for a single-valued displacement field (3.13).
3.3
Field Equations of Elastic Dynamics
Displacement Equations
For a linear elastic solid of general anisotropy, the elastic constitutive law in Chapter
2 describes a linear relation,  ij  Cijkl kl , between the stress tensor and the strain
tensor. Substituting the kinematics (3.4) and the Voigt symmetry of the stiffness tensor
Cijkl , one has the following relation between the stress tensor and the displacement
gradient:
 ij  Cijkluk ,l
(3.14)
Combining (3.14) with the balance law of linear momentum (3.1), one obtains the
displacement governing equation for elastic dynamics
C
u

ijkl k , l , j
 fi  ui
(3.15)
Energy Principle
That equation solves three displacement components via three partial differential
equations. An energetic interpretation of (3.15) can be stated as follows. Multiply both
sides of (3.15) by ui , and integrate over the domain V, one has
 u u  u C
i i
i
u

ijl k ,l , j
dV   f u dV .
i i
V
V
Through integration by parts, one arrives at
 u u  C
i i
V
u u
ijkl k ,l i , j
dV   C
u u  j dS   f i ui dV
ijkl k ,l i
V
V
where  j is the unit outward normal of the boundary and V denotes the boundary
of V. The Gauss theorem is utilized to convert the volume integral to a surface integral.
By equation (3.14) and the Cauchy’s principle of stress, one can phrase this equality
in the following form:
Wt u   K t u    Tiui dS   fiui dV
V
(3.16)
V
where T denotes the traction along the boundary, Wt u 
1
Cijkl u i , j u k ,l dV
2 V
represents the total strain energy within the domain V, and Kt u  
1
uiui dV
2 V
labels the total kinetic energy in the body. Equation (3.16) states that the changing rate
of the field energy within the body equals to the power done by the boundary traction
and the body force.
Isotropic Materials and Navier Equation
Attention is now focused on the simple case of isotropic materials. The fourth-rank
stiffness tensor adopts the following representation
Cijkl   ij kl    ik  jl   il  jk 
(3.17)
by two Lamè constants  and  . Substituting (3.17) into (3.15), the following
Navier equation is derived:
   u j , ji   2ui  ui
(3.18)
Helmholtz Representation and Wave Equations
Though simpler than (3.15), the Navier equation for isotropic elastic dynamics
appears to be rather complicated. A reduction can be made via Helmholtz
representation proposed more than a century ago. A rigorous proof for its
completeness was given by Eli Sternberg (1960). The central idea of the Helmholtz
representation is to express the displacement field ui (3 components) by a scalar
potential  and a divergence free vector potential  k such that
ui  ,i  eijk k , j ,
 k ,k  0 .
(3.19)
This can also be written as u    rotψ , div ψ  0 by absolute notation.
Substituting
e  
ijk
k , j ,i
(3.19)
into
the
Navier
equation
(3.18),
and
noting
 eijk k , ji  0 , one may replace the Navier equation by the following wave
equations:
cl2 2   ,
where cl 
  2

cs2 2 k  k ,
 k ,k  0
denotes the longitudinal wave speed, and c s 
(3.20)


the
transverse (or shear) wave speed. Apparently, cl  cs , the longitudinal wave that
induces dilatational deformation is faster than the transverse wave that carries
deviatoric deformation. Equation (3.20) states that the solution of the Navier equation
can be composed of one dilatational wave and two transverse waves.
Plane Waves and Traveling Waves
Before the encountering on any boundary, the wave solution can be fairly described
by the following form:
  x, t    k  x  cl t  .
(3.21)
Combination of (3.20) and (3.21) gives cl2 ki ki   cl2  , that defines k as a phase
vector of unit amplitude. For one dimensional problem, the following traveling wave
solution is recovered:
 x, t    (1) x  cl t    ( 2) x  cl t  .
Similarly,  k   k n  x  cs t 
(3.22)
and n  1 . Since ψ is divergence free, one
concludes nk k  0 , or the vectorial wave potential ψ is unchanged along the
direction of the phase vector n.
Consider the special case of plane wave. The displacement vector can be expressed as
u x, t   A exp ik  x  ipt 
where A signifies the wave amplitude. Let k  kn 
(3.23)
p
n . At an arbitrary instant t,
c
all points along the plane of k  x  constant will maintain the same phase. The
plane sharing the same phase has a normal of n and a phase speed of c. Substitution of
the plane wave solution (3.23) into the Navier equation (3.18) gives rise of
 Cijkl Ak kl k j   ip  Ai . Denote M ik  Cijkl nl n j as the Christoffel acoustic tensor,
2
one has
M ik Ak  c 2 Ai .
(3.24)
From linear algebra, one understands that (a) A is the eigen-vector of the acoustic
tensor; (b) c 2 is the eigen-value; and (c) the positive definiteness of the acoustic
tensor and the positive nature of density  infer that c2 is non-negative and real.
3.4
Quasi-static Field Equations
Quasi-static Processes
Equations in (3.20) have spatial Laplacian terms on the left hand side and temporal
inertial terms on the right hand side. The so-called quasi-static case refers the situation
where cl2 2   and cs2 2 k  k . To put this statement in a specific form, we
denote t0 as a characteristic time for temporal variation, and l0 a characteristic length
for spatial variation. The condition to justify a quasi-static process can be phrased as
c s2
1
l
 2 , or cs  0 .
2
t0
l0
t0
(3.25)
The quasi-static solution will be challenged in the following cases: (1) a very small t0,
referring rapid temporal jubilation; and (2) a very large l0, referring an almost uniform
field.
Displacement Equations
In the quasi-static case, the elastodynamics equation (3.15) is reduced to:
Cijkluk ,lj  f i  0 .
(3.26)
For an isotropic material, it further reduces to the following quasi-static Navier
equation:
   u j , ji   2ui  f i
 0.
(3.27)
Green Functions
We conclude this section by providing a fundamental solution for (3.26) that can be
easily specialize to the isotropic case of (3.27). The fundamental solution refers to a
unit point force applied at x' along the direction of xm on an infinite body,
characterized by a body force of
f i   im  x  x' . The solution for such a
fundamental problem is termed the “Green function” in the literatures. The Green
function Gkm  x  x' is defined as the displacement u k at a point x due to a unit
force acting in xm direction at point x' , namely
C ijkl G km ,lj   im  x  x '
x, x ' R 3 .
(3.28)
The solution for such a problem is of fundamental importance. First, the solutions for
an infinite body under arbitrary body force can be obtained by integrating the Green
function as u k   Gkm  x  x ' f m  x 'dx ' . Second, for a finite body with arbitrary body
V
force, this solution can be used to reduced the problem to that without body force but
with known boundary traction. Third, it serves as the basis for the so-called
“boundary integral equation” and “boundary element method”. Fourth, it reduces to
the famous Kelvin solution for the special case of an isotropic material,.
Fourier Transform
The approach of using Fourier transform for the Green function in an infinite body
was pioneered by Fredholm (1900). We recapitulate its essence on solving (3.28) here.
The direct and inverse Fourier transform can be written as:

g k    G  x  exp(ik  x )dx
and

G x  
1
8 3

 g k exp( ik  x )dk
(3.29)

respectively. Let us perform the direct Fourier transform to (3.28). The property of the
Dirac delta facilitates the evaluation of the Fourier transform on the right hand side of

(3.28). So one arrives at
 Cijkl Gkm,lj exp[ ik   x  x ']d x  x '   im . Integration by

parts twice, and making use of the rapid decaying property of the solution away from
x' , one has
Cijkl k j k l g km k    im
(3.30)
where g km k  denotes the Fourier transform of the Green function. Introduce a unit
vector n parallel to k in the transform space, such that n 
k
， k  k . Recall our
k
previous definition of the Christoffel acoustic tensor M ik  Cijkl nl n j . The Fourier
transform g km k  can be solved as
g km k  
1
1
M km
.
2
k
(3.31)
Applying the inverse Fourier transform defined in the second expression of (3.29),
and after rather lengthy algebras, the Green function can be obtained as

1
1
1
Gkm  x  x '  3  2 M km
exp[ ik   x  x ']dk 
2
8
k
8 x  x '
1

2
M
1
km
0
 
 ,  d .
2 
(3.32)
The detailed derivation can be found in Yang and Lee (Mesoplasticity and Its
Application, Springer-Verlag, 1993, Appendix 4A). The definite integration in the last
expression of (3.32) is evaluated along a unit circle perpendicular to the line from x to
x' , as delineated in the figure below.
X

1
X'
Figure 3.6 Contour to evaluate the Green function according to (3.32)
We now outline the derivation for the last equality in (3.32). Define R  x  x' ,
T
x  x'
and λ  n  Rk with   kR , the triple inversion in (3.32) can be
R
written as Gkm 
1
8 3

 

M km1 n 
2 R
cosn  T dλ since the Green function is real.
Take a spherical coordinate system ( , , ) whose volume element can be written
as dλ  2 sin   d d d . The inverse of the acoustic tensor can be written as
1
n  M km1  ,   and the n T term equals to cos . That puts the Green
M km
function into the following form
Gkm 
1

8 3 R 0
2

0
0
1
 ,  cos cos d
d  d  sin  M km
To evaluate the integral with respect to  , one might quote the following formula



 cos cos d    2  .
from the integration table
If we further use the
0
property of the Dirac delta function, the last equality in (3.32) is recovered.
The expression of the Green function reveals its following properties:
Gkm  Gmk
and
Gkm  x  x '  Gkm  x ' x  .
(3.33)
The first relation reflects the symmetry of the Green function with respect to its
indices, namely the displacement in k direction by a unit force in m direction equals to
the displacement in m direction by a unit force in k direction. The second relation
reveals its symmetry with respect to the argument, namely the displacement at x by a
unit force at x' is identical to the displacement at x' by a unit force at x.
Kelvin Solution
We next consider the special case of an isotropic elastic solid, so that (3.17) applies.
For this case, it is a straightforward matter to show that the Christoffel acoustic tensor
and its inverse are reduced to
M ik  Cijkl n j nl     ni nk   ik and M ik1 

1
1
  ik 
ni nk  (3.34)

21  v 

respectively. Substituting the second expression in (3.34) into (3.32), one yields
Gkm 

1
 km  Tk Tm 
2 km 
8R 
21  v 

1
(3.35)
which is the famous Kelvin solution. An extensive discussion on the Kelvin solution,
through a less compact but insightful construction, will be given in Chapter 8.
Stress Formulation
The approach given above is centralized on (3.26), featuring the solution by
displacements. Beside an infinite body, the displacement approach works well for
all-around displacement boundary conditions. It would be less convenient if traction is
prescribed along the boundary. We give below an alternative approach that uses
stresses as the primitive field variables. In a stress approach, one has three equilibrium
equations in (3.2), three independent compatibility conditions in (3.5) with the
redundant ones excluded by (3.6) ( Lmn  Lnm ) and (3.7) ( Lmn, n  0 ). One may use the
six independent equations to solve for the six symmetric components of the stress
tensor. The stress approach is formulated by
emjkenil Sijpq pq, kl  0 ,
 ij, j  f i  0
(3.36)
where S  C 1 is the fourth rank compliance tensor such that  ij  Sijpq pq .
Isotropic Materials
Let us consider the special case of isotropic material under the generalized Hooke’s
law  ij   ij  kk  2 ij 

3  2
 kk  ij  2 ij . The contraction of this relation
gives a relation  kk  3  2  kk between the hydrostatic stress and the volumetric
strain. Facilitated by this relation, the generalized Hooke’s law can be inverted as
 ij 

1 

  ij 
 kk  ij  .
2 
3  2

(3.37)
That leads to the following expression for the compliance tensor
S ijpq 

1 
2
  ip jq   iq jp 
 ij pq  .
4 
3  2

(3.38)
Beltrami-Michell Equations
Substituting (3.38) into the first set of equations in (3.36), one arrives at


2
 emqk enpl  empk enql 
emik enil pq  pq,kl  0 .
3  2


The symmetry of the stress tensor would reduce the above equation to
v

 kl mn   ml kn  pq  pq, kl  0 .
 emqk enpl 
1 v


Denote the trace of stress as  kk   and use the property of the Kronecker delta,
one has
emqk enpl pq, kl 


v
 2 mn  , mn  0 .
1 v
Substituting the equilibrium equation (3.2) into the first term on the left hand side of
the equation, one yields
 2 mn 
1
v
, mn 
 2 mn   f m, n  f n , m  .
1 v
1 v
Taking contraction of the above equation, one finds a governing equation for the
hydrostatic stress
 2  
1 v
f n, n .
1 v
(3.39)
Namely the hydrostatic stress satisfies a Poisson equation for the general body force
situation, and a Laplace equation if the body force is divergence free. Replacing the
 2  term above by (3.39), one finally arrives at the following stress compatibility
equation
 2 mn 
1
v
, mn  
f k , k mn   f m , n  f n, m  .
1 v
1 v
(3.40)
That relates 6 symmetric components of the stress tensor by 6 equations. Historically,
they are called Beltrami-Michell equations.
If the body force is ignored, equations (3.39) and (3.40) can be further simplified to
 2  0 ,  2 mn 
1
, mn  0 ,  2 2 mn  0 .
1 v
(3.41)
Namely  is a harmonic function, and any component of stress,  mn , is a
bi-harmonic function.
Stress Functions
The Beltrami-Michell formulation is based on individual stress components. For
many circumstances, a wiser choice can be made for the stress approach by using
stress functions. Consider the special case of zero body forces. The equilibrium
equation (3.2) is reduced to  ij , j  0 , or the stress tensor is divergence free. Recall
(3.2) and the Bianchi identity (3.7) that state the incompatibility tensor L    ε  
is also symmetric and divergence free. That leaves σ and L as good analogies.
Remember that   L  0 only relies on the symmetry of ε . One may similarly
construct a symmetric tensor Φ such that σ    Φ   , and conclude by analogy
that the stress tensor so constructed should be divergence free. Thus, the equilibrium
equation is automatically satisfied. These symmetric tensorial functions Φ are called
the stress functions. To satisfy the homogeneous Beltrami-Michell equations, the
stress functions Φ should obey the following equations
 2 einke jml mn, kl  
1
e pnke pml mn,kl ,ij  0
1 v
(3.42)
They are called the compatibility equations for the stress function.
Like stress tensor σ and incompatibility tensor L, the stress function tensor Φ is
symmetric and divergence free. That leaves at least 3 components in Φ as
independent. The literature in elasticity stated 17 possible choices of 3 independent
components. The most frequently used ones are:
1. Maxwell stress function (Edinburgh Roy. Soc. Trans., Vol. 26(1870))
The Maxwell stress function only take the diagonal part of the general stress function,
the only non-trivial components are: 11  1 , 22  2 , 33  3 . The resulted
formulation has been well described in the textbook by Lu and Luo.
2.
Morera stress function (Roma. Acc. Lincei
(ser.5 ), t.1 1892)
The Morera stress function instead takes the off-diagonal components of the general
0  3  2 
stress function, such as 
0  1  . The simplified expressions are also given in

0 
the textbook by Lu and Luo (Sections 4.9 and 4.10).
Detailed discussions for various stress functions can be found from the historic paper
of Michell J. H. in London Math Soc. Proc. Vol.32 (1901).
3.5
Constraints
Internal Constraints
A special class of problems involve the internal deformation constraints within the
body, and they impose an additional issue in the formulation of elasticity. We listed
below a few representatives under infinitesimal deformation.
One frequently encountered constraint is the almost incompressibility of a certain
class of materials, such as rubbers. The famous Bridgeman test for the volumetric
deformation of metal (which contributed to his Nobel prize) also indicated that the
plastic deformation of metals is nearly incompressible. The deformation of this kind is
also termed isochoric in the literatures. Inextensibility is another type of deformation
constraints. Textile products typically have large elastic moduli in the knitting
directions but significantly small shear modulus. You can convince yourself by testing
on a handkerchief or a piece of cloth. Fiber-reinforced materials are another type of
materials that exhibit inextensibility along the reinforced fiber directions. A new
material, multiwalled carbon nano-tubes (MWCNT), has in-plane moduli much larger
(3 orders of magnitudes) than the out-of-plane (tension and shear) moduli.
Constraint Equations
We now give mathematical representations for constraints of different natures. The
incompressibility constraint can be described by  kk  0 , and the inextensible
constraint by  ijlil j  0 , with l being the fiber directions. In general, a constraint can
be described mathematically as
 ε   constant .
(3.43)
For the even more general case of multiple constraints, one has
 ε   constant ,   1,    , n .
(3.44)
Constraining Stress
The internal constraints induce un-determined stresses, termed the constraining stress
N. It can only be determined through equilibrium. The constraining stress cannot
deliver any work, namely
N ijij  0 .
(3.45)
Combining the constraint equation (3.43) or (3.44) and the null-work expression
(3.45), one can determine the constraining stress N up to a scalar undetermined
constant. Take the time derivative of the constraint equation (3.43), one has
 

ij  0 .
 ij
Comparison of this expression with (3.45) naturally concludes that
Nij  

.
 ij
(3.46)
The expression in (3.46) prescribes the structure of the constraining stress tensor, and
leaves an amplitude scalar  to be determined by equilibrium.
Similar expression can also be derived for the case of multiple constraints. Combining
(3.44) and (3.45), one has
n
N ij   
 1

.
 ij
(3.47)
One unknown constant   arises from one family of internal constraint.
Examples
We discuss two examples before the conclusion of this section. The first example
concerns with incompressible materials, in which the internal constraint function is
written as  ε    kk . From (3.46), one easily calculates the form of the constraining
stress as Nij  
 kk
 ij   p ij , with p being the hydrostatic pressure, a quantity
 ij
of vital importance in fluid mechanics. For the case of isotropic materials, the
generalized Hooke’s law for an incompressible solid can be revised as
 ij   p ij  2 eij
(3.48)
The second example refers to a material reinforced by two families of inextensible
fibers, as shown in Fig. 3.7, where 1 and  2 label the fiber alignment angles.
X2
2
1
X3
X1
Figure 3.7 Material reinforced by two families of inextensible fibers.
The constraint equations can be written in the following form
cos  
  l   ε  l  ,  ＝1, 2 where l    sin    .
 0 
Guided by (3.47) we take the partial derivative of the constraint as


lk kqlq   lil j . That leads to the following expression for the

 ij  ij
constraining stress
N  T1l1  l1  T2 l 2  l 2
(3.49)
where the undetermined scalars T1 and T2 have the physical significance of the
tensions experienced by two families of fibers.
3.6
Boundary Conditions
Boundary Composition
St
V
X3
X2
X1
Su
Figure 3.8 Boundary conditions for a three-dimensional body.
After a rather thorough examination on the basic field equations of elasticity, we
direct our attention to boundary conditions and the related issues. Consider the
boundary loading exemplified in Figure 3.8. The volumetric domain V is enclosed by
its boundary V  S . We assume the entirety of S is composed of non-overlapped
traction boundary S t and displacement boundary S u such that S  St  Su and
St  Su   .
The displacement boundary condition is given by
ui  ui , on Su.
(3.50)
To study the deformation rather than the movement of a mechanism, one has to
eliminate the rigid body motion from the displacement of an elastic body.
The traction boundary condition is given by
 jiv j  ti , on St.
(3.51)
All quantities in (3.50) and (3.51) with superimposed bars have prescribed values
along the boundary. For the general case of n-dimensional problem, a number of n
independent boundary conditions should be prescribed.
Types of Boundary Conditions
Different combinations of boundary conditions promote different solution strategies.
Listed below are typical boundary conditions.
(1) All-around displacement boundary condition. Solution can be approach via
Navier equation under a displacement formulation.
(2) All around traction boundary condition. Solution can be approached via stress
functions. Cares should be exercised for this case since
a) rigid body motion cannot be eliminated;
b) the all around traction should satisfy the global equilibrium of the body to
prevent accelerated rigid body motion.
(3) Mixed boundary condition. Part of the boundary is prescribed by displacement,
and part by traction. Singularity arises across the line where the boundary
condition changes. The solution for this type of problems is typically intricate.
(4) Hybrid boundary condition. Two or more types of the boundary conditions are
prescribed in the any segment of the boundary. For examples, the displacement
boundary condition in one component, while the others are of the traction type.
Figure 3.9 gives an example for roller-type boundary condition, where the
vertical boundary condition is null displacement, but the horizontal boundary
condition is vanishing traction.

Figure 3.9 Roller contact boundary conditions.
(5) Spring-like boundary condition. The spring-like boundary condition refers to a
linear relationship between the boundary traction and the boundary displacement
ti  K iju j .
(3.52)
Imagine that you are sitting on a sofa, the boundary condition between you and
the base of sofa is spring-like, with spring stiffness of K ij . The same situation holds
for the case shown in Fig. 3.10 when an elastic inclusion is embedded in a rigid
surrounding via an elastic interlayer.
Figure 3.10 An elastic inclusion is embedded in a rigid matrix via an elastic interlayer.
Symmetry Conditions
Symmetry conditions are frequently encountered in elasticity. Symmetry represents
simplicity and beauty. Their recognition is not an art, but can be put into a systematic
and scientific way. The symmetry in geometric shape is easy to identify. Suppose the
geometry, such as the one depicted in Fig. 3.11 is symmetric with respect to the plane
of x  0 , and the material is isotropic or symmetric with respect to the same plane.
By symmetry splitting
2ti , ui 
s ,a
 t i x, y, z , ui x, y, z   t i  x, y, z , ui  x, y, z 
(3.53)
where the superscript “s” denotes the symmetric field, and superscript “a” the
anti-symmetric field. One can split any prescribed boundary loadings (either
kinematical or traction-like) to the sum of symmetric and anti-symmetric loading.
Y
Z
X
Figure 3.11 Symmetry condition.
Under symmetric loading t i , ui  , one has the following symmetry conditions along
s
the plane of x  0 .
(1) All anti-symmetric field variables ( u x ;  xy ,  xz ; xz , xy ) with respect to the plane
of x  0 , as well as their derivatives with respect to x of the even orders should
vanish.
(2) The derivatives with respect to x of the odd orders of all symmetric field variables
( u y , u z ;  xx ,  yy ,  zz ,  yz ; xx , yy , zz , yz ) should vanish.
Under anti-symmetric loading t i , ui  , the following symmetry conditions can be
a
obtained along the plane of x  0 as the counterparts of the previous ones.
(1) All symmetric field variables ( u y , u z ;  xx ,  yy ,  zz ,  yz ; xx , yy , zz , yz ) with respect
to the plane of x  0 , as well as their derivatives with respect to x of the even
orders should vanish.
(2) The derivatives with respect to x of the odd orders of all anti-symmetric field
variables ( u x ;  xy ,  xz ; xz , xy ) should vanish.
Several practical examples are given below to grasp the idea.
Y
X
Y
X
Figure 3.12 Beams under symmetric (above) and anti-symmetric (below) loadings.
For the case delineated in Fig. 3.12, the plane x＝0 is the plane of geometric
symmetry. For the top graph, the loading is symmetric, thus one has
x＝0:  xy   xz  0 , u x  0 .
For the bottom graph, the loading is anti-symmetric, that leads to an opposite set of
symmetry conditions
x＝0:  xx  0 ; u y  u z  0 .
Conditions for Single Valued Displacement Fields
For multiply-connected domain such as shown in Fig. 3.13, the determination of the
displacement fields needs the imposition of the single-valued-ness conditions
ui
 s ds  0
for any contours that can not shrink continuously into a point.


Figure 3.13 Contour integral to impose single value-ness of displacement fields for a
multiply-connected domain.
Initial Conditions
Initial conditions are required to solve the elastic dynamics problems. Since the
governing equations, such as the dynamic Navier equations (3.15), encounter the time
derivatives of the second order, both the initial position and the initial velocity have to
be prescribed, namely
ui  ui0 ， ui  ui0 at
t＝0.
(3.54)
Saint Venant Principle
Named after the famous French scientist Saint Venant, the principle claims that a
self-equilibrium traction distribution (namely the one giving neither resultant force
nor resultant couple) over a local area of the boundary of an elastic body would only
cause negligible disturbance at a distance far exceeding the size of that local boundary
area. The Saint Venant princple is a gateway to reason equivalent boundary conditions
to suit for the purpose on solving an elasticity problem. Its proof, as well as its precise
mathematical statement, are by no means trivial, and are still a subject of current
research. As a first course in solid mechanics, we are in no position to get a rigorous
feeling of the Saint Venant principle. Instead, a heuristic approach is taken. Consider
the situation shown in Fig. 3.14. Were a point force applied on the boundary, the stress
and strain fields would decay like 1 / r 2 . Consider two point forces of opposite
directions acting on the boundary, giving a zero resultant force but non-zero resultant
moment. Consider the limit of two forces become infinitely large and infinitely close,
while maintaining the resultant moment at a fixed value. The solution obtained is just
the spatial derivative of point force solution with respect to the boundary tangent. The
stress and strain fields decay like 1 / r 3 . If one puts two nearby couples together, and
conducts the similar process, the solution will again be the spatial derivative of the
point couple solution with respect to the boundary tangent. With null resultant force
and null resultant couple, the stress and strain fields decay like 1 / r 4 . The last case
results no effect in average sense when far away from the dipole.
1
 ij ,  ij  2
r
 ij ,  ij 
1
r3
 ij ,  ij 
1
r4
Figure 3.14 Point force, point couple and diple applied along the boundary.
y
Figure 3.15 Sinusoidal traction distribution along the boundary.
We substantiate our reasoning on the Saint-Venant principle by another method.
Consider the Fourier series expansion on the traction exerted along the boundary.
Take a particular term of period L, as demonstrated in Fig. 3.15. The sinusoidal
distribution gives neither long-range resultant force, nor long-range resultant moment.
Straightforward calculation indicates that the strain and stress fields decay in the
direction y away from the boundary like  ij , ij  e

cy
L
, where the constant c has an
order of unity, so that the decay length is measured by the period L.
The works of J.K. Knowles and the late Eli Sternberg contributed a lot on the modern
understanding of the Saint Venant principle. The field disturbance is measure by its
energy norm. They attempted to prove under what circumstances, the energy norm
decays under self-equilibrium local traction, as well as an estimate for the
characteristic decay length.
3.7
Formulation of Elasticity
Positive Definiteness of Strain Energy
The thermodynamics requirement for an elastic material is the positive definiteness of
its strain energy, stated as
W
 0
1
Cijkl ij  kl 
2
 0
  ij  0
.
if  ij  0
(3.55)
Alternatively, one can use the matrix notation of Voigt to cast (3.55) into the
following matrix form:
C11    C16  11 
11,  22 ,  , 12 
       0

C66  12 
11  12   0 .
Since the stiffness matrix is real and symmetric, our knowledge from linear algebra
 
informs us that (3.55) is satisfied if and only if all eigen-values of the Cij
matrix
are positive. The positive-definiteness of all eigen-values infers the ellipticity of an
elastostatics problem.
Consider the special case of elastic isotropy, where the elastic strain energy is reduced
to
W
K 2
 kk  eij eij .
2
(3.56)
Since both  kk2 and eij eij are positive definite, and they are independent of each
other, the positive definiteness of the strain energy implies that both the volumetric
modulous K and the shear modulus G should be strictly positive. By relating these
two moduli with the Young’s modulus E and the Poisson’s ratio v , one concludes
that
E>0，  1  v 
1
.
2
(3.57)
The situation of a negative Poisson’s ratio is admissible in thermodynamics. Such a
material can be design by carefully choosing its microscopic cell, such as those
conducted by Yin and Yang (“Optimality criteria method for topology optimization
under multiple constraints”, Computers and Structures, 79(2001), 1839-1850).
Positive Definiteness of Elasticity Formulation
We recall the derivation toward (3.16) for the internal energy U  Wt u  K t u  due
to elastic deformation, where the elastic strain energy Wt u 
positive definite by (3.55) and the kinetic energy K t u  
1
Cijkl u i , j u k ,l dV is
2 V
1
u i u i dV is obviously
2 V
positive definite for a positive material density. We reach the conclusion that the total
energy U and consequently the elasticity formulation are positive definite.
Principle of Superposition
The principle of superposition applies for all linear problems. We have used it in some
of our previous derivations.
Uniqueness Theorem
Kirchhoff (1887) established the uniqueness theorem of elastic dynamics. We
recapitulate his argument by the modern terminology.
The proof is constructed by contradiction. Assume there exist two sets of solutions
σ , ε, u(1)
and σ , ε, u  , both satisfying all the field equations of elastodynamics,
( 2)
along with all the boundary and initial conditions. Accordingly, the difference
between
σ , ε, u(1)
and
σ , ε, u( 2) ,
denoted as
σ, ε, u ,
should satisfy the
homogeneous field equations, boundary and initial conditions of elastodynamics.
Following the integral relation (3.15) W t u  K t u    Ti u i dS   f i u i dV derived
V
V
from the Navier equation for elastodynamics and the homogeneous body forces and
boundary conditions, one arrives at
W t u   K t u   0 .
Integrating the above expression with respect to time t from the homogeneous initial
conditions for the difference field, one yields
Wt u  K t u   0 .
The positive definiteness of the total internal energy claims that the above equality
can be observed only if
u  0 ， u  0
Consequently the two sets of solutions,
σ , ε, u(1)
and
σ , ε, u( 2) ,
should be
identical, namely the solution of an elastodynamics problem is unique.
Summary of Chapter 3
Differential Formulation
Differential equations (15)
Boundary conditions (3)
equilibrium (3), kinematics (6), constitutive (6)
(or replaced by symmetry conditions)
Initial conditions (6)
Constraint conditions and expressions for constraining stress (n constraints lead to n
unknown constraining parameters determined by equilibrium).
Simplified Formulations
Displacement formulation - 3 Navier equations and 3 displacement boundary
conditions.
Stress formulation - 6 Beltrami－Michell equations or 6  2 2 mn  0 equations or
stress function equations and 3 traction boundary conditions.
Basic Principles
Principle of superposition
Positive definite strain energy (ellipticity)
Uniqueness theorem
Saint Venant principle.
Basic Solutions
Traveling wave solution
Plane wave solution
Point force solution (Green function).