Ruhr-University Bochum
Faculty of Civil and Environmental Engineering
Institute of Mechanics
An Introduction to
Linear Continuum Mechanics
Klaus Hackl
Mehdi Goodarzi
2010
ii
Contents
1 Fundamentals
2
3
5
8
12
2 The Field Equations of Linear Continuum Mechanics
17
3 Displacement Functions
27
1.1 Stress and equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Balance equations: dierential form . . . . . . . . . . . . . . . . . .
1.2 Strain and compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Compatibility conditions . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Hooke's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Hooke's law for isotropic materials . . . . . . . . . . . . . . . . . . .
2.2.2 Alternative formulations of Hooke's law for isotropic materials . . .
2.2.3 Plane-strain and plane-stress problems . . . . . . . . . . . . . . . . .
2.3 Navier and Beltrami-Michell equations . . . . . . . . . . . . . . . . . . . . .
3.1
3.2
3.3
Scalar and vector potentials . . . . . . . . . . . . . . . . . . . . . . . . . . .
Galerkin vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Love's strain function . . . . . . . . . . . . . . . . . . . . . . . . . .
Displacement functions of Papkovich-Neuber . . . . . . . . . . . . . . . . .
iii
1
17
18
19
20
22
23
27
32
33
35
iv
CONTENTS
Chapter 1
Fundamentals
The subject matter of mechanics is the study of motion, in how a physical object changes
position with time and why. Here we shall conne ourselves within principles of classical
mechanics. Studying how a body moves is called kinematics which is basically a geometrical investigation of motion dealing with fundamental concepts like space, time and frame of
reference. In a second step Newton's laws of motion are introduced based on other fundamental concepts such as inertial frame of reference,mass and force to deal with the question
of why things move, and this is called kinetics. The alternative strategy is formulation of
mechanics based on conservation laws of energy, momentum and angular momentum which
turns out to have broader range of validity than Newton's formulation.
Scientic hypotheses try to explain physical observations based on a set of assumptions
and more often than not, neglecting certain eects. One of the most universal aspects of
observations is the scale, that is the range of values of quantities during an observation.
For instance, having a model proposed by an observation over the ordinary length scales
of laboratory, we can not draw the conclusion that the same model will hold over cosmic
length scales, or on sub-atomic length scales unless it is somehow tested or proved.
Considering length scales, mechanical model of a moving body can belong to one of the
following in order of sophistication
Particle is a model of a body of negligible dimensions so that it can be treated geometrically as a point, which means all the body mass is concentrated at its center of
mass. For example, in study of planetary motion dimensions of planets are negligible compared to orbital distances therefore they are often treated as point masses.
Kinematics of a particle deals with translation only. In this model, forces are free
vectors which means they can move freely without their mechanical eect being altered. Model equations for particle motion consist of balance of energy and linear
momentum.
Rigid body is an object of nite dimensions and negligible change in shape during motion.
That is where the distance between every two points on the body remains constant.
A rigid body is assumed to have a continuous mass distribution although in reality
matter is quantized at small scales. Kinematics of rigid body motion is expressed
in terms of translation and rotation. Forces are sliding vectors viz they can slide on
1
2
CHAPTER 1.
FUNDAMENTALS
their line of action without their mechanical eects being altered. Model equations
are balance of energy, linear momentum and angular momentum.
Deformable body is a mechanical object of continuous mass distribution whose relative
distance of points can change. This is ocially the starting point of continuum mechanics although a rigid body has a continuous mass distribution too. Continuum
mechanics deals with the length scales large enough to neglect all the molecular effects and, at the same time, small enough to observe the shape changes in the body.
Together with translations and rotations we have deformation in kinematics study
of deformable bodies. As the point of action of forces is important here, forces are
xed vectors. Balance equations of linear and angular momentum and energy are
sucient to model deformation. In addition we will need the so called constitutive
equations to have a closed model.
In the following section we work out the concept of forces in continuum media.
not
1.1
Stress and equilibrium
In the view of causality, forces are the cause for change in momentum and also for deformation. According Newton's third law, forces exist only in couples, to wit the presence of
any force is justiable only in the presence of its source. In this regard, we distinguish two
types of forces
Field forces which are applied from a distance through a eld such as electromagnetic or
gravitational eld.
Contact forces which are applied by direct contact of objects such as friction. Note
that these are also manifestation of eld forces that act at inter-molecular distances,
therefore observed as direct contact at large scales.
Distribution of eld forces is usually expressed as force per unit volume or per unit mass,
and distribution of contact forces is usually expressed as force per unit area. The exception
is the so called concentrated force which is an attempt to model the situation when a nite
amount of force is applied over a very small region of area or volume. In this specic case
the force distribution will not be nite-valued anymore. Therefore concentrated forces are
treated as individual force vectors rather than distributions, and presented by Dirac's delta
function which will be introduced later on. Let Ω be a body at
t
∂Ω
equilibrium with a nite and continuous distribution of solid
mass m throughout its simply connected volume V , and be
bounded by a smooth surface ∂Ω of area A. This is our typΩ
ical picture of a body under consideration in continuum solid
f (x)
mechanics. You will see potato-like illustrations (Fig. 1.1)!
Assume a distribution of eld forces presented by a volumetric force eld f (x) and contact force (also called traction )
distribution over ∂Ω denoted by t. Then equilibrium condi- Fig. 1.1: Solid body.
1.1.
3
STRESS AND EQUILIBRIUM
tions for the body read
(1.1)
X
(1.2)
M =0 :
This provides enough ground for analysis of the overall body motion which is a combination
of translation and rotation, as if the body was rigid. However, deformation is a local
phenomenon which depends on the state of forces at each body point.
X
Z
Z
F =0
t dA = 0
Z
x × t dA = 0
x × f dV +
f dV +
:
ZΩ
∂Ω
∂Ω
Ω
1.1.1 Stress tensor
In order to analyze the forces at each body point, we shall write the equilibrium equations
for an innitesimal volume element at the body point. For an innitesimal element at
X3
dX1
−e1
t3
dX3
f
t2
−e2
n
t1
X2
dX2
(a)
X1
−e3
(b)
Fig. 1.2: a) Innitesimal volume element, and b) inintesimal tetrahedron.
point x with volume dV the volumetric force is given by dF = f (x) dV . Also, there are
tractions at element faces (Fig. 1.2.a), because an area element within a body of mass is
an interface that experiences contact forces due to adhesion and/or cohesion. This traction
force t at each point depends on the location x of area element dA, and on its direction
denoted by unit normal vector n
t = t(x, n) .
(1.3)
This is Cauchy's assumption which is substantial in continuum mechanics, although it
might seem obvious. The functionality of t(x, n) is further specied by Cauchy's theorem
through which the concept of stress tensor is introduced.
V
Cauchy's theorem
Assume that we want to write the equation of motion for an innitesimal tetrahedron (Fig.
1.2.b) whose edges coincide with the three Cartesian axes. The four triangular faces have
unit normal vectors
−e1 , −e2 , −e3 , n .
We identify face triangles of the tetrahedron by their normal vectors i.e. 1, 2, 3 and n. Notice
that projection of the triangle n over coordinate planes are triangles 1, 2, 3. Therefore if
4
the center point of triangle n has the position
CHAPTER 1.
FUNDAMENTALS
(1.4)
then the center points of the other three triangles are projections of x onto coordinate
planes, given by
x =x e +x e , x =x e +x e , x =x e +x e .
(1.5)
Furthermore, if the triangle n has the area dA then the triangles 1, 2, 3 (which are its
projections) have the areas
x = x1 e 1 + x2 e 2 + x3 e 3 ,
c
1
2 2
c
2
3 3
1 1
c
3
3 3
1 1
2 2

 dA1 = n1 dA
dA2 = n2 dA
dAi = ei · n dA = ni dA :

dA3 = n3 dA
(1.6)
Also note that the tetrahedron has the mass ρdV and acceleration a. Finally, the equation
of motion is written as
ρ a dV = f dV + t(x, n) dA + t(x , −e ) dA + t(x , −e ) dA + t(x , −e ) dA . (1.7)
Substitution of dA 's according equation (1.6) gives
ρ a dV = f dV + t(x, n) dA + t(x , −e ) n dA + t(x , −e ) n dA + t(x , −e ) n dA (1.8)
and dividing both sides by dA
dV
dV
ρa
=f
+ t(x, n) + t(x , −e ) n + t(x , −e ) n + t(x , −e ) n .
(1.9)
dA
dA
However the fraction dV /dA tends to zero as the volume shrinks. This is because dV is
proportional to L and dA is proportional to L , where L is the characteristic length of the
element, hence
c
1
1
1
c
2
2
c
2
2
2
c
2
2
2
c
3
2
3
3
i
c
1
1
c
1
1
1
3
1
c
3
c
3
3
3
3
3
2
L→0
:
Therefore equation (1.9) yields
dV
L3
∝ 2 = L → 0.
dL
L
t(x, n) = −t(xc1 , −e1 ) n1 − t(xc2 , −e2 ) n2 − t(xc3 , −e3 ) n3 .
When the tetrahedron element shrinks, the center points of faces tend to each other
L→0
:
(1.10)
xc1 → xc2 → xc3 → x
and equation (1.10) gives
t(x, n) = −t(x, −e1 ) n1 − t(x, −e2 ) n2 − t(x, −e3 ) n3 .
(1.11)
1.1.
5
STRESS AND EQUILIBRIUM
Now writing components gives
(1.12)
(1.13)
(1.14)
t1 (x, n) = −t1 (x, −e1 ) n1 − t1 (x, −e2 ) n2 − t1 (x, −e3 ) n3
t2 (x, n) = −t2 (x, −e1 ) n1 − t2 (x, −e2 ) n2 − t2 (x, −e3 ) n3
t3 (x, n) = −t3 (x, −e1 ) n1 − t3 (x, −e2 ) n2 − t3 (x, −e3 ) n3
and in matrix
form 


 
t1 (x, n)
−t1 (x, −e1 ) −t1 (x, −e2 ) −t1 (x, −e3 )
n1
 t2 (x, n)  =  −t2 (x, −e1 ) −t2 (x, −e2 ) −t2 (x, −e3 )   n2  .
t3 (x, n)
−t3 (x, −e1 ) −t3 (x, −e2 ) −t3 (x, −e3 )
n3
(1.15)
The 3 × 3 matrix in the latter equation is a second order tensor which is only a function of
x. It is called stress tensor and is usually denoted by σ . This equation can be written in
the more compact form
t(x, n) = σ(x) · n ,
(1.16)
which is called Cauchy's formula. Let's have a closer
X
σ
look at the matrix form in equation (1.15). It is easy
to show that −t (x, −e ) = t (x, e ). Then the comσ
σ
ponent at the i row and the j column is
σ
3
i
th
j
i
33
j
th
23
13
32
σ31
σij = ti (x, ej )
σ22
σ
σ
which is the i component of the traction force applied
σ
to the surface normal to the j axis (Fig. 1.3). ThereX
fore the second index j in the stress tensor σ shows X
the surface to which the traction is applied, and the
rst index i shows the direction of the traction compoFig. 1.3: Stress components.
nent.
The importance of equation (1.16) is that the functionality of traction on x and n
is separated as the stress tensor σ is a function of x only at the right-hand-side. If you
remember, when introducing tensors, we mentioned that second order tensors are linear
mappings from vectors to vectors. Cauchy's formula is a physical example of this, where
stress tensor maps directions denoted by n to tractions t exerted on an area element normal
to n.
th
21
th
11
ij
1.1.2 Balance equations: dierential form
12
2
1
As mentioned before we need the local state of forces in order to analyze deformation. The
integrals in equations (1.1) and (1.2) are written over the whole body Ω, while they are
equally true for any sub-domain ofZ the body,Z say Ω̃ ⊂ Ω which includes the body point x
f dV +
t dA = 0
(1.17)
Z
Z
x × f dV +
x × t dA = 0
(1.18)
Ω̃
Ω̃
∂ Ω̃
∂ Ω̃
6
Let's focus at the rst equation. If we substitute Cauchy's formula into (1.17) it yields
Z
Z
f dV +
σ · n dA = 0 .
(1.19)
Note that ∂Ω̃ is a closed surface, therefore the second integral according Gauss theorem
gives
Z
Z
σ · ∇ dV
(1.20)
σ · n dA =
Therefore equation (1.19) becomes
Z
Z
Z
f dV +
σ · ∇ dA =
(f + σ · ∇) dV = 0 .
(1.21)
This integral vanishes for any arbitrary sub-body Ω̃ only if the integrand equals zero
σ·∇+f =0
(1.22)
which is the dierential equation for balance of forces.
For the second balance equation (1.18), again we substitute Cauchy's formula to obtain
Z
Z
x × f dV +
x × (σ · n) dA = 0
(1.23)
however it holds that
x × (σ · n) = (x × σ) · n
(1.24)
therefore
Z
Z
x × f dV +
(x × σ) · n dA = 0
(1.25)
and once again Gauss theorem gives
Z
Z
(x × σ) · n dA =
(x × σ) · ∇ dV
(1.26)
which turns equation (1.25) into
Z
Z
Z
x × f dV +
(x × σ) · ∇ dV =
(x × f + (x × σ) · ∇) dV = 0 .
(1.27)
There is an identity that can be easily veried
(x × σ) · ∇ = : σ + x × (σ · ∇)
(1.28)
where is permutation tensor. Putting this into equation (1.27) gives
Z
x × (f + σ · ∇) + : σ dV = 0
(1.29)
CHAPTER 1.
Ω̃
∂ Ω̃
Ω̃
∂ Ω̃
Ω̃
Ω̃
Ω̃
Ω̃
∂ Ω̃
Ω̃
∂ Ω̃
∂ Ω̃
Ω̃
Ω̃
Ω̃
Ω̃
Ω̃
FUNDAMENTALS
1.1.
7
STRESS AND EQUILIBRIUM
however the term in parentheses vanishes due to (1.22), then
(1.30)
Z
: σ dV = 0
Ω̃
which holds for every sub-body Ω̃, therefore it must hold that
: σ = 0.
which can be true only if σ is a symmetric tensor
σ = σT .
which is the local condition for balance of moments.
Exercise 1. Show that : σ = 0 results in σ = σ .
Exercise 2. Write equilibrium equations and Cauchy's formula in components.
Exercise 3. Consider a body with some uid exposed to X
hydrostatic pressure (Fig. 1.4).
(1.31)
(1.32)
T
3


0
0 
 −p(x3 )
0
−p(x3 )
0
σ = −p(x3 ) I =


0
0
−p(x3 )
with body force: f = −ρge . Solve equilibrium equations X
X
for p(x ) with boundary condition p(0) = p .
Fig. 1.4: Body lled with uid.
Example 1.1. Write equilibrium equations in cylindrical coordinates.
Solution. The equilibrium equation is a vector equation or a set of three scalar equations
given by (1.22) as
3
1
3
0
σ · ∇ + f = 0.
The divergence operator in cylindrical coordinates is given by
σ·∇ =
1 ∂
1 ∂σ rϕ ∂σ rz
(rσ rr ) +
+
er +
r ∂r
r ∂ϕ
∂z
1 ∂
1 ∂σ ϕϕ ∂σ ϕz
(rσ ϕr ) +
+
eϕ +
r ∂r
r ∂ϕ
∂z
1 ∂
1 ∂σ zϕ ∂σ zz
(rσ zr ) +
+
ez
r ∂r
r ∂ϕ
∂z
and decomposition of body forces gives
f = fr er + fϕ eϕ + fz ez .
2
8
Substitution into equilibrium equation results in
CHAPTER 1.
FUNDAMENTALS
(1.33)
(1.34)
(1.35)
1 ∂
1 ∂σ rϕ ∂σ rz
(rσ rr ) +
+
+ fr = 0
r ∂r
r ∂ϕ
∂z
1 ∂
1 ∂σ ϕϕ ∂σ ϕz
(rσ ϕr ) +
+
+ fϕ = 0
r ∂r
r ∂ϕ
∂z
1 ∂
1 ∂σ zϕ ∂σ zz
(rσ zr ) +
+
+ fz = 0 .
r ∂r
r ∂ϕ
∂z
Using the transformation from Cartesian to cylindrical coordinates and the
nabla operator in Cartesian coordinates, show that in cylindrical coordinates the nabla
operator has the following form
Exercise 4.
1
∇ ≡ er ∂r + eϕ ∂ϕ + ez ∂z .
r
Calculate σ (r) for a tube loaded by torsion
(Fig. 1.5). There are no body forces and the boundary condition σ (r ) = t is given. Note that
Exercise 5.
rϕ
1
rϕ
1
σ = σrϕ (er eϕ + eϕ er ) .
1.2
r0
r1
Fig. 1.5: Tube under torsion.
Strain and compatibility
As pointed out before, the motion of a continuum medium is a combination of translation,
rotation and deformation. Deformation is the change in shape. Since the shape of an object
is characterized by the relative position of its points, in order to analyze deformation, we
focus on the change of innitesimal line elements that connect the neighboring points of
the body. Let us assume a typical image of a continuum body (Fig. 1.6) at initial/undeformed conguration whose points are denoted by the position vector X , and then in its
moved/deformed conguration whose points are denoted by the position vector x. The
basic assumption is that x is a one-to-one function of X . This guarantees that the material
points are neither created, nor are they annihilated. So, the starting point is a description
of motion by
x = x(X, t) .
(1.36)
It is also customary to dene displacement eld as
u = u(X, t) = x(X, t) − X .
(1.37)
1.2.
9
STRAIN AND COMPATIBILITY
dX
dx
u
X
x
Fig. 1.6: Undeformed and deformed congurations of a continuum body.
Excluding dynamic cases from our discussion, the time dependence is dropped. Moreover,
we assume that the displacement eld is innitesimal
kuk 1 .
(1.38)
Now imagine two neighboring material points in the undeformed conguration connected by the dierential line element dX , and in the deformation conguration connected
by dx. We would like to study the relation dX 7−→ dx. According denition of displacement eld we have
dx = dX + u(X + dX) − u(X) .
(1.39)
Then substitution of Taylor's expansion as
u(X + dX) = u(X) + u∇ · dX + · · ·
(1.40)
will result in
dx = dX + u∇ · dX = x∇ · dX .
(1.41)
Here u∇ is the displacement gradient and x∇ the deformation gradient which are related
clearly by
x∇ = I + u∇ .
(1.42)
The displacement gradient is a second order tensor and can be decomposed into symmetric
and antisymmetric parts designated by
1
ε = (u∇ + ∇u)
(1.43)
2
1
ω = (u∇ − ∇u)
(1.44)
2
the symmetric part ε is called strain tensor and the antisymmetric part ω is the rotation tensor. As their names suggest the symmetric part ε reects deformation and the
antisymmetric part ω carries information about rotation only. It is immediately clear that
ε = ε , ω = −ω .
(1.45)
T
T
10
To stay within linear continuum theory the fundamental assumption is
CHAPTER 1.
FUNDAMENTALS
(1.46)
Every antisymmetric tensor has three independent elements only, and therefore can be
specied by a vector. So the rotation tensor ω can be expressed by the so called rotation
vector
1
1
ω
= − : ω.
(1.47)
w = (w ) =
2
2
Exercise 6. Show that ω = w .
By expanding the equations (1.47) and (1.44), one can show that
1
w = ∇ × u.
(1.48)
2
kεk 1 ,
i
ij
ikj
kωk 1 .
jik jk
k
Physical meaning of the decomposition
Putting the decomposition u∇ = ε + ω into (1.41) we obtain
dx = dX + u∇ · dX = dX + ε · dX + ω · dX .
(1.49)
= ε · dX is the
We show that dx = ω · dX is the rotational part of deformation and dx
deformational part of deformation. Let us focus at dx rst
[ω · dX] = ω dX = w dX
(1.50)
which gives
dv
dx = w × dX .
(1.51)
This, according assumption (1.46), means that dx is obtained
by a small rotation of dX around w. Note that dx is normal
ϕ
dv
to dX , and since kwk 1 then the ratio kdx k / kdXk which
is equal to the angle of rotation (Fig. 1.7) is also small. In
short form notation one can state
w

kwk 1

Fig. 1.7: Rotational part of
r
s
r
ij
i
j
ikj
j
k
r
r
r
r
r
⇒ ϕ 1.
ϕ ≈ kdxr k / kdXk = kwk
displacement.

Now let us have a look at dx . Deformation is expressed in terms of change in lengths
and angles. The length of the line element in the deformed conguration is obtained by
s
dx · dx = (dX + w × dX + ε · dX) · (dX + w × dX + ε · dX)
(1.52)
where higher order terms, denoted by · · · , are neglected due to assumption (1.46). Also we
know that
= dX · dX + 2dX · (w × dX) + 2dX · ε · dX + · · ·
dX · (w × dX) = 0
1.2.
11
STRAIN AND COMPATIBILITY
because dX is orthogonal to (w × dX). Therefore
(1.53)
dx2 = dX 2 + 2dX · ε · dX
and then
(1.54)
where Taylor's expansion is used because kεk 1. The change of length is nally obtained
by
1
dX · ε · dX .
(1.55)
∆l = dx − dX =
dX
Remember that this result is based on the assumption (1.46) of small strain and rotation.
To study the changes in angle, assume two adjacent innitesimal vectors normal to
each other in the undeformed conguration denoted by dX and dX , and their deformed
counterparts denoted by dx and dx respectively. Inner multiplication of the two vectors
will give us information on the angle between them
dx · dx = (dX + w × dX + ε · dX ) · (dX + w × dX + ε · dX )
(1.56)
dx
dX · ε · dX 1/2
dX · ε · dX
= 1+2
≈1+
,
2
dX
dX
dX 2
1
1
1
2
1
2
2
1
1
2
2
2
= dX 1 · dX 2 + dX 1 · (w × dX 2 ) + (w × dX 1 ) · dX 2 + 2dX 1 · ε · dX 2 + · · ·
where the higher order terms are neglected. The rst term vanishes due to the assumption
of orthogonality dX · dX = 0, and the sum of second and third terms is also zero because
1
Therefore
On the other hand
2
dX 1 · (w × dX 2 ) = − (w × dX 1 ) · dX 2 .
dx1 · dx2 ≈ 2dX 1 · ε · dX 2 .
dx1 · dx2 = dx1 dx2 cos ϕ ≈ dX1 dX2 cos ϕ = dX1 dX2 sin(π/2 − ϕ)
(1.57)
(1.58)
where π/2 is the angle in the undeformed conguration. Finally comparing (1.58) and (1.57)
yields
dX · ε · dX
∆ϕ = 2
.
(1.59)
dX dX
Again, note that this formula is obtained under the assumption (1.46). As can be seen
from equations (1.55) and (1.59), deformation depends on the strain tensor ε only. Therefore our objective of interpreting the additive decomposition of displacement gradient into
symmetric (strain) and antisymmetric (rotation) parts is fullled.
Exercise 7. For the given displacement gradient
= dX1 dX2 sin(∆ϕ) ≈ dX1 dX2 ∆ϕ
1
2
1

u∇ = 10−3
2

1 −2 0
3 0 0
0 1 1
12
CHAPTER 1.
FUNDAMENTALS
1. calculate ε, ω, w.
 
1

a = 0  mm
1


−1
b =  1  mm
1
2. For the two orthogonal vectors
and
calculate the change
of their lengths and the angle between them.
Remark 1. Strain is dened as the symmetric part of displacement gradient. Sometimes
this is indicated by the short notation
ε=∇ u
(1.60)
called symmetric gradient of displacement.
s
1.2.1 Compatibility conditions
If the equation (1.43) is expanded as
ε11
ε22
ε33
∂u1
,
=
∂X1
∂u2
=
,
∂X2
∂u3
=
,
∂X3
ε12 = ε21
ε23 = ε32
ε31 = ε13
1 ∂u1
∂u2
=
+
,
2 ∂X2 ∂X1
1 ∂u2
∂u3
=
+
,
2 ∂X3 ∂X2
1 ∂u3
∂u1
=
+
.
2 ∂X1 ∂X3
(1.61)
we see that the six independent components of the strain tensor are obtained from the three
components of the displacement vector, assuming that u is dierentiable. Now suppose that
the strain components are given and the displacement components are to be determined
from these equations. In this case since we have six equations and three unknowns there
is not necessarily a unique acceptable solution for displacement eld. In other words the
set of equations (1.61) poses a restriction on the strain eld, i.e. we can not choose strain
components arbitrarily and they are somehow connected.
Denition 2. A compatible strain eld is the one for which a single-valued and continuous
displacement eld can be found by solving the equation ε = ∇ u.
Our objective then will be to nd the condition(s) under which a given strain eld
is compatible. One would naturally think of three additional equations to connect strain
components to each other. Applying the following identities from tensor calculus
∇ × ∇u = 0 and u∇ × ∇ = 0 .
(1.62)
applying the curl operator from the left-hand-side to equation (1.43) we obtain
1
1
1
∇×ε=∇×
(∇u + u∇) = (
∇
×
∇u + ∇ × u∇) = ∇ × u∇
(1.63)
2
2
2
s
1.2.
13
STRAIN AND COMPATIBILITY
and then from the right-hand-side
∇×ε×∇=
(1.64)
1
1
∇ × u∇ × ∇ = ∇ × (u∇
×∇)
2
2
which nally leaves us with
∇ × ε × ∇ = 0,
(1.65)
the so called compatibility condition. This condition is obtained from the denition of
strain, therefore it is a necessary condition for compatibility, viz if the displacement eld
is acceptable then the above condition should be fullled. In a next step we show that this
condition is also sucient, i.e. if the condition is fullled then the strain eld is compatible,
that is to say, a physically meaningful displacement eld exists such that ε = ∇ u.
Since the expression ∇ × ε × ∇ is to vanish for a compatible strain eld, one would
dene its value as a measure of incompatibility. Therefore, the incompatibility of the strain
is dened as
Inc(ε) ≡ ∇ × ε × ∇ ,
(1.66)
which makes sense as the compatibility condition can be restated as
Inc(ε) = 0 .
(1.67)
Exercise 8. Prove that the alternative form of compatibility condition
s
tr(I × ∇ × ε × ∇ × I) = 0
can also be expressed as
(1.68)
Now we intend to show that the condition (1.65) is sucient for the strain eld to be
compatible in a simply connected domain. For this we assume that the strain eld ε is
given in a simply connected body Ω, and it fullls (1.65), and then it has to be proven that
a continuous single-valued displacement u can be obtained from (1.43) if the displacement
is given at one point over the body.
We denote the point where the displacement is given by
X and the given value by u . From (1.41) we have
X
du = ε · dX + ω · dX
(1.69)
whose integration gives
Ω
X
Z
Z
u(X) = u(X ) +
ε X̄ · dX̄ +
ω X̄ · dX̄ (1.70)
Fig. 1.8: Integration path.
where the integral is taken over any continuous path C connecting X and X . Using index
notation
the lastZ term can be calculated
as
Z
Z
¯ · dX̄
ω · dX̄ =
ω · I · dX̄ =
ω · X̄ − X ∇
(1.71)
∆ε + ∇tr(ε) ∇ = ∇∇ · ε + ε · ∇∇ .
0
0
X
X
X0
X0
0
0
0
X
X0
X
X
X0
X
X0
Z
=
X0
Z
X
¯
ωij ∂k X̄j − Xj dX̄k = ωij X̄j − Xj −
X0
X
X0
∂¯k ωij X̄j − Xj dX̄k
14
here integration by parts is used. Then we have
CHAPTER 1.
FUNDAMENTALS
(1.72)
X
ωij X̄j − Xj = ω(X) · (X − X) − ω(X 0 ) · (X 0 − X) = ω(X 0 ) · (X 0 − X)
X0
and using the result in exercise 6
¯ × X̄ − X
∂¯k ωij X̄j − Xj = ∂¯k ilj wl X̄j − Xj = ∇w
1¯
1 ¯
¯
¯ × X̄ − X
=∇
×∇
∇ × u × X̄ − X = − ∇u
2
2
¯ × X̄ − X
= − ε×∇
which nally gives
(1.73)
the so called Michell-Cesaro formula. Now we have to show that the integral term in
the above formula is path-independent i.e. u(X) is a single-valued function. However, we
remember that path-independence of a line integral is fullled if and only if the integral
vanishes over every closed path, viz we should prove that
I
¯ × X̄ − X = 0 .
dX̄ · ε + ε × ∇
(1.74)
Using Stokes' theorem and the compatibility condition (1.65) we have
I
Z
¯
¯ × X̄ − X (1.75)
dX̄ · ε + ε × ∇ × X̄ − X =
dAn · ∇ × ε + ε × ∇
Z
=
dAn · (∇ × ε × ∇) × X̄ − X = 0 . (1.76)
Proof of the second step is left as exercise 10.
x
Exercise 9. Let the tensor A be given as
Z
X
u(X) = u(X 0 ) + ω(X 0 ) · (X − X 0 ) +
¯ × X̄ − X
dX̄ · ε + ε × ∇
X0
C
C
A
A
2
A = ax21 e1 e1 + bx1 x2 e2 e2 .
For which values of a and b holds A × ∇ = 0? Calculate
Z
X2
X3
X3
A · ds
u(X 3 ) =
0
for the two paths indicated in Fig. 1.9.
Exercise 10. Prove the following identity
O
Fig. 1.9:
tion.
X1
x1
Paths of integra-
¯ × ε+ ε×∇
¯ × (x̄ − x) = ∇
¯ ×ε×∇
¯ × (x̄ − x) .
∇
(1.77)
1.2.
15
STRAIN AND COMPATIBILITY
Let us have a look at the classical solution to
the cantilever beam problem, considering its compatibility.
The solution to the cantilever beam problem shown in the y
gure is of the form
ε
ε = −κ(x) y e e .
(1.78) x
x
where κ(x) is the curvature at each cross section. The beam is Fig. 1.10: Cantilever beam.
built-in at its left end x = 0, where u(0) = 0 and ω(0) = 0.
Check if ε is compatible. If yes, nd the displacement eld
components u (x, y) and u (x, y).
Solution. We start with the compatibility condition, applying the curl operator from the
right-hand-side to the strain
ε × ∇ = (−κ(x) ye e ) × (∂ e + ∂ e + ∂ e ) = −κ(x) e e
(1.79)
and then from the left
∇ × ε × ∇ = (∂ e + ∂ e + ∂ e ) × (−κ(x) e e ) = 0 . X
(1.80)
Now with the compatibility condition being fullled, we want to calculate the displacement
eld from Michell-Cesaro formula. Applying the boundary conditions at x = 0 we obtain
Z
¯ × (x̄ − x)
u(x) =
dx̄ · ε + ε × ∇
(1.81)
in which
Example 1.2.
xx
0
x x
x
y
x x
x x
x x
y y
y y
z z
z z
x z
x z
x
0
¯ × (x̄ − x) = (−κ(x̄) ex ez ) × ((x̄ − x) ex + (ȳ − y) ey + (z̄ − z) ez )
ε×∇
= −κ(x̄) (x̄ − x) ex ey − (ȳ − y) ex ex .
Substitution into (1.81) yields
Z
x
u(x) =
h
i
[ex dx̄ + ey dȳ] · κ(x̄) − ȳex ex − (x̄ − x) ex ey + (ȳ − y) ex ex
Z0x
Therefore
(1.83)
dx̄ κ(x̄) [−yex − (x̄ − x) ey ] .
=
(1.82)
0
κ(x̄) dx̄
(1.84)
κ(x̄) (x̄ − x) dx̄ .
(1.85)
x
Z
ux (x, y) = −y
and
0
Z
uy (x, y) = −
Then it is also clear that
x
0
u00y = κ(x)
and
ux = −y u0y .
(1.86)
16
Exercise 11.
CHAPTER 1.
For the given strain tensor and boundary conditions
ε = x22 e1 e1 + x1 x2 (e1 e2 + e2 e1 )
u(0) = 0 , ω(0) = ω0 (e1 e2 + e2 e1 )
calculate the displacement eld u(x).
FUNDAMENTALS
Chapter 2
The Field Equations of Linear
Continuum Mechanics
Until now we have developed the basic kinematical and dynamical notions concerning material bodies. As mentioned before, balance laws of linear momentum, angular momentum
and energy are not sucient to model mechanical behavior of deformable bodies, and we
additionally need the so called constitutive laws. Constitutive laws try to express material
deformation versus loading, based on specic thermo-mechanical assumptions. Depending
on the type of constitutive formulation one can classify materials to e.g. elastic, plastic,
viscoelastic etc.
In this chapter we briey address constitutive laws and governing equations of linear
elasticity. Derivation of these laws and the respective thermodynamics is postponed to the
second part of these lectures in nonlinear continuum mechanics where hyper-elasticity is
studied.
2.1
Elasticity
So far we have established two separate sets of equations, namely equilibrium equations
(1.22) and kinematic equations of strain-displacement (1.61), and now we want to study
stress-strain relations. As will be seen, this completes the set of equations, i.e. there will
be as many independent equations as unknowns.
We shall conne our study within the so called elastic range, that is the range of strain
for which the material behavior remains elastic. A body of material is called elastic if at
each body point the strain is a one-to-one function of stress at that point regardless of the
history of loading. In formal notation
σ(x) ←→ ε(x) .
(2.1)
This ensures that stress and strain are both single-valued functions of each other which
in turn means that the material follows the same stress-strain curve during loading and
unloading, and shows no hysteresis eect. This is more rigorously presented within the
framework of hyper-elasticity later on.
17
18
CHAPTER 2.
2.2
THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS
Hooke's law
Experimental data for most hard solids shows a linear relation between load and strain
within a specic interval during loading and unloading. In the most general representation
this linearity is of the form of generalized Hooke's law as
σ(x) = C(x) : ε(x)
(2.2)
which says that each of stress components at any point of a body is a linear function of the
strain components at that point. Here the so called stiness tensor C is introduced which
is a fourth-order tensor. This linearity could be also represented as
ε(x) = S(x) : σ(x)
(2.3)
where S is called compliance tensor which is a fourth-order tensor satisfying the relation
S : C = C : S = I.
(2.4)
and I is the fourth-order identity tensor
I=δ δ ee e e .
(2.5)
The functionality of position x is emphasized in equation (2.2), however we usually
write
σ = C : ε.
(2.6)
Specially for a homogeneous body when C does not depend on x, we have
ik jl
i j k l
∗
σ(x) = C : ε(x) .
To further clarify the above equations we can write them in index notation as
(2.7)
The stiness tensor, being of fourth-order, has 3 × 3 × 3 × 3 = 81 components. However,
these components are not independent. Looking into equation (2.2), since stress and strain
tensors are both symmetriceach having six independent componentsthe stiness tensor
can only have 6×6 = 36 independent components. Because the most general linear relation
between the six stress components and the six strain components can take the form of
σij = Cijkl εkl .
 K K K K K K 

11
12
13
14
15
16
σ11
ε11
 ε 
K21
···
 σ22  
  22 

 

 σ33  

  ε33 


 =  K31

 σ23  
  2ε23 


 

 σ31  
 2ε31 
2ε12
σ12
K61
···
K66

...
..
(2.8)
which is called contracted notation. Here the original stress and strain tensors are recast
in a vectorial form. The factor 2 in front of shear strains has technical reasons beyond our
scope, however it can be omitted in principle.
∗
A homogeneous body is the one whose physical properties are the same for all of its points.
2.2.
19
HOOKE'S LAW
3 Note that the symmetry of stress and strain tensors is fullled only if
Remark .
Cijkl = Cjikl = Cijlk
(2.9)
which is called minor symmtery of the stiness tensor.
We will see later on that the existence of strain energy potential requires the fourth
order tensor C to satisfy
C
=C
(2.10)
the so called major symmetry, which in turn requires the symmetry of the stiness matrix
K in equation (2.8). Therefore the matrix K has only 21 independent components.
The above arguments which prove the reduction of number of independent stiness
components from 81 to 21 can be summarized as follows
Summary 4. Since the stiness tensor has to fulll major and minor symmetries
C
=C
=C
=C
(2.11)
it has only 21 independent components for a linear elastic material.
ijkl
ijkl
klij
klij
jikl
ijlk
2.2.1 Hooke's law for isotropic materials
As yet, we have simplied the general Hooke's law based on symmetry properties (2.11).
Now we consider material symmetries based on which we can further reduce the number
of required stiness components.
Solids have crystal structures at the atomic scale. These crystal structures are classied
based on their symmetry properties. A detailed study of how these symmetries inuence
the stiness tensor goes beyond our agenda (see [4]), but regardless of crystallography a
typical solid material is composed of microscopic grains with random distributions and
orientations throughout the body. Although each grain has a specic crystal structure with
its respective symmetries, the overall behavior of the material at the macro scale seems
isotropic due to this randomness.
Denition 5. An isotropic material is the one that has identical thermo-mechanical properties in all directions at each material point.
This requires that the elastic constants be invariant under an arbitrary rotation of coordinate system
C
=C
(2.12)
and consequently we come up with two substantial results.
†
ijkl
0
ijkl
Theorem 6. For an isotropic material, the principal axes of the strain tensor coincide with
the principal axes of the stress tensor.
†
Components of stiness tensor are also called elastic constants.
20
CHAPTER 2.
THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS
Theorem 7. For an isotropic material, the stiness tensor in principal coordinates follows
the relationships
C1111 = C2222 = C3333 = C
C1122 = C2233 = C3311 = C2211 = C1133 = C3322 = λ
(2.13)
(2.14)
and the rest of components are zero.
So, there are only two independent stiness components for an isotropic material in
principal coordinates. But this holds in any other coordinates as well because the rotation
transformation from principal coordinates to any other coordinates leaves us with a stiness
tensor whose components are expressed in terms of the two components above. For an
arbitrary coordinate system one can show the following constitutive equation
σ = λ tr(ε) I + 2µ ε
(2.15)
where λ and µ are Lamé parameters. Where λ is the same as in (2.14), and µ is related to
the constant C in (2.13) by 2µ = C − λ. The above equation in index notation is given by
σ = λ ε δ + 2µ ε .
(2.16)
The inverse form of equation (2.15) is given by
1
λ
ε=
σ−
tr(σ) I .
(2.17)
2µ
2µ (3λ + 2µ)
In view of Hooke's law as given by (2.2) and (2.3) stiness and compliance tensors for an
isotropic material are given in terms of Lamé parameters as
λ
1
C = λI ⊗ I + 2µI and S = −
I ⊗I +
I,
(2.18)
2µ (3λ + 2µ)
2µ
where I and I are the second-order and fourth-order identity tensors respectively.
ij
kk ij
ij
2.2.2 Alternative formulations of Hooke's law for isotropic materials
We proceed with reformulations of Hooke's law which are suitable for dierent types of
problems.
Volumetric-deviatoric decomposition
The strain tensor can be additively decomposed into the so called volumetric and deviatoric
parts
1
1
(2.19)
ε = vol(ε) + dev(ε) : vol(ε) = tr(ε) I , dev(ε) = ε − tr(ε) I .
3
3
Kinematically, the volumetric part reects dilatation which is the relative change of volume
dv − dV
= tr(ε)
(2.20)
dV
21
and the deviatoric part reects distortion. Interesting is that if we decompose the stress in
the same fashion into deviatoric and volumetric parts as
1
1
σ = vol(σ) + dev(σ) : vol(σ) = tr(σ) I , dev(σ) = σ − tr(σ) I
(2.21)
3
3
then the Hooke's law is decomposed such that volumetric stress depends on volumetric
strain only, and deviatoric stress depends on deviatoric strain only by
σ = (3λ + 2µ) ε , σ
= 2µε
(2.22)
The so called bulk modulus or compression modulus of elasticity is introduce as
2
K =λ+ µ
(2.23)
3
which based on (2.22) yields
vol(σ) = 3K vol(ε)
(2.24)
Remark 8. The deviatoric parts of stress and strain tensors are traceless i.e. tr(dev(•)) = 0.
Maybe the most important aspect of the above decomposition is that deviatoric and
volumetric parts are orthogonal to each other in the sense that
vol(σ) : dev(σ) = 0 , vol(ε) : dev(ε) = 0 .
(2.25)
Then it follows that
σ : ε = vol(σ) : vol(ε) + dev(σ) : dev(ε)
(2.26)
which means the elastic energy is decomposed into the volumetric elastic energy and the
deviatoric elastic energy.
2.2.
HOOKE'S LAW
vol
vol
dev
dev
Young's modulus, shear modulus and Poisson's ratio
Hooke's law for isotropic solids is often expressed in terms of a set of material constants
which are more favorable in applied mechanics because they can be directly measured in
experiments. Those constants are Young's modulus
µ (2λ + 2µ)
E=
,
(2.27)
λ+µ
Poisson's ratio
ν=
λ
,
2 (λ + µ)
and usually renaming one of Lamé parameters to shear modulus
Then Hooke's law is reformulated as
σ=
and its inverse as
G = µ.
(2.28)
(2.29)
Eν
E
tr(σ) I +
ε
(1 + ν) (1 − 2ν)
1+ν
(2.30)
1+ν
ν
σ − tr(σ) I .
E
E
(2.31)
ε=
22
CHAPTER 2.
THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS
Exercise 12. During the tension test of a prismatic bar with square cross section having
the dimensions l = 500mm and a = 30mm, the changes in the dimensions of ∆l = 8mm and
∆d = −0.2mm are obtained after application of a force F = 2000N (Fig. 2.1). Calculate the
l + ∆l
a
F
F
a − ∆a
l
Fig. 2.1: Prismatic bar subjected to tension.
material constants: λ, µ, E, ν .
2.2.3 Plane-strain and plane-stress problems
Whenever a general theory is developed there are special cases that draw special attention,
sometimes due to their broad range of application and other times for the impact they
have in development of models. Two well know examples of such cases in elasticity are
plane-strain and plane-stress problems.
Plane-strain
As the name suggests, plane-strain is a type of deformation in a material body when given
a specic plane, all the o-plane components of strain are zero or negligibly small. Assume
that the plane of strain is normal to X axis in Cartesian coordinates. Then we may express
the plane-strain condition by
ε =0
for i = 3 or j = 3 .
(2.32)
Then Hooke's law for an isotropic material becomes
σ = λ (ε + ε ) + 2µ ε
(2.33)
σ = λ (ε + ε ) + 2µ ε
(2.34)
σ = 2µ ε
(2.35)
σ = λ (ε + ε )
(2.36)
σ =σ =0
(2.37)
3
ij
11
11
22
11
22
11
22
22
12
12
33
11
13
Plane-stress
22
23
Now the meaning of plane-stress should be immediately clear by analogy. Plane stress is
a condition where for a given plane all the o-plane stress components are zero. If the
plane of stress is normal to X axis in Cartesian coordinates then we state the plane-stress
condition as
σ =0
for i = 3 or j = 3 .
(2.38)
3
ij
2.3.
23
NAVIER AND BELTRAMI-MICHELL EQUATIONS
With introduction of a new constant
λ̄ =
(2.39)
2λµ
λ + 2µ
the Hooke's law for isotropic materials becomes
(2.40)
(2.41)
(2.42)
σ11 = λ̄ (ε11 + ε22 ) + 2µ ε11
σ22 = λ̄ (ε11 + ε22 ) + 2µ ε22
σ12 = 2µ ε12 .
Also note that as a result of condition (2.38) we have
(2.43)
(2.44)
λ
(ε11 + ε22 )
λ + 2µ
= ε23 = 0 .
ε33 = −
ε13
Exercise 13.
Prove the following relationships
ν=
2.3
3K − E
,
6K
λ=
3K − 2µ
,
3
E=
λ (1 + ν) (1 − 2ν)
.
ν
Navier and Beltrami-Michell equations
Now we shall be able to write the governing equations of isotropic elasticity as Hooke's law
completes the set of equations. Let us restate all those equations together
1
1
ε = (∇u + u∇)
ε = (∂ u + ∂ u )
(2.45)
2
2
∇·σ+f =0
∂ σ +f =0
(2.46)
σ = λ tr(ε) I + 2µ ε
σ = λ ε δ + 2µ ε .
(2.47)
In the rst step we would like to have an equation in terms of displacement eld only.
For this we have to substitute ε from (2.45) into (2.47) which yields
ij
i j
i ij
j
ij
kk ij
j i
ij
1
1
σij = λ (∂k uk + ∂k uk ) δij + 2µ (∂i uj + ∂j ui )
2
2
= λ ∂k uk δij + µ (∂i uj + ∂j ui )
and then plug in this result into (2.46)
(2.48)
∂i (λ ∂k uk δij + µ (∂i uj + ∂j ui )) + fj = λ∂i ∂k uk δij + µ∂i ∂i uj + µ∂i ∂j ui + fj
= λ∂j ∂k uk + µ∂i ∂i uj + µ∂j ∂k uk + fj
= (λ + µ) ∂j ∂k uk + µ ∂i ∂i uj + fj = 0 .
This equation can be written in operator notation as
(2.49)
(2.50)
which is the so called Navier equation. Since u is a vector the equation is basically a system
of three partial dierential equations.
(λ + µ) ∇∇ · u + µ ∆u + f = 0 ,
24
CHAPTER 2.
Exercise 14.
THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS
Prove the alternative form of the Navier equation as
(λ + 2µ) ∇∇ · u − µ∇ × ∇ × u + f = 0 .
Another step would be to set up a system of dierential equations in terms of stress
tensor. Stress tensor has six independent components. The equation (2.46) provides three
partial dierential equations which is not sucient. On the other hand equations (2.45)
and (2.47) do not provide more equations in terms of stress. Therefore we need additional
equations. It turns out the additional equation is the alternative form of compatibility
condition (1.68) that we repeat here
∆ε + ∇tr(ε) ∇ = ∇∇ · ε + ε · ∇∇ .
(2.51)
From equation (2.22) we have
1 − 2ν
tr(ε) =
tr(σ) ,
(2.52)
E
and the inverse form of Hooke's law (2.31)
1+ν
ν
ε=
σ − tr(σ) I .
(2.53)
E
E
Now we put (2.53) and (2.52) into (2.51) to obtain
ν
1 − 2ν
1+ν
∆σ − ∆tr(σ) I +
∇tr(σ) ∇ =
E
E
E
ν
1+ν
(∇∇ · σ + σ · ∇∇) − (∇∇ · (tr(σ) I) + (tr(σ) I) · ∇∇) .
E
E
(2.54)
For the second term on the left-hand-side based on equilibrium equation (2.46) we have
∆tr(σ) = ∂ ∂ σ = ∇ · σ · ∇ = −∇ · f .
(2.55)
For the rst term on the right-hand-side again based on (2.46) we have
∇∇ · σ + σ · ∇∇ = − (∇f + f ∇) ,
(2.56)
because ∇ · σ = σ · ∇ = −f . Considering the second term on the right-hand-side
(∇∇ · (tr(σ) I)) = ∂ ∂ σ δ = ∂ ∂ σ
= (∇tr(σ) ∇) .
(2.57)
The same can be easily shown for (tr(σ) I) · ∇∇ therefore
∇∇ · (tr(σ) I) = (tr(σ) I) · ∇∇ = ∇ tr(σ) ∇ .
(2.58)
Substituting all these results into (2.54) yields
1
ν
∆σ +
∇ tr(σ) ∇ +
∇ · f I + ∇f + f ∇ = 0 , (ν 6= −1)
(2.59)
1+ν
1−ν
the so called Beltrami-Michell formulation. Note that Beltrami-Michell equation is in fact
compatibility equation in terms of stress which together with equilibrium equation ∇ · σ +
f = 0, establishes a complete system of partial dierential equations for the stress eld.
m m kk
ij
i k mm kj
i j mm
ij
2.3.
NAVIER AND BELTRAMI-MICHELL EQUATIONS
Exercise 15.
Assume f = const., using Beltrami-Michell equation show that
∆∆σ = 0 and ∆∆ε = 0 .
Exercise 16.
Assuming that
σ = σ(r)
and f = f (r) ,
derive Beltrami-Michell equation in cylindrical coordinates.
25
26
CHAPTER 2.
THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS
Chapter 3
Displacement Functions
An elastostatics problem is basically a boundary value problem, with boundary conditions
of the type Dirichlet, Neumann or mixed. In chapter 2 we derived the governing equations
of linear isotropic elasticity in terms of displacement eld (Navier equation) and in terms
of stress eld (Beltrami-Michell and equilibrium equations).
In this chapter we present some classical solutions to the displacement formulation in a
coherent way. All the methods are meant to
the solution of the original boundary
value problem. We shall often assume that the body forces are not present which is not
a serious restriction as in practical situations, nding the particular solution belonging to
body forces is not complicated . Throughout our derivations we frequently employ tensorial
and vectorial identities.
Our point of departure is the Navier equation with zero body force
(λ + µ) ∇∇ · u + µ ∆u = 0 ,
(3.1)
with prescribed boundary conditions
u(x) = u ,
x∈Γ
(3.2)
σ(x) · n = t , x ∈ Γ .
(3.3)
The idea is to substitute u with a combination of scalar or vector functions, called displacement functions, or their derivatives so that the governing equation in terms of these
functions takes the form of harmonic or biharmonic equations which are basically simpler
to solve than the Navier equation. However, the boundary conditions expressed in terms
of displacement functions become more complicated.
simplify
∗
∗
u
∗
3.1
σ
Scalar and vector potentials
The rst approach we are going to present is based on the so called Helmholtz theorem as
follows.
∗
We remember that the solution to an inhomogeneous dierential equation is the sum of the general
solution to the homogenized equation and the particular solution to the inhomogeneous equation.
27
28
CHAPTER 3.
DISPLACEMENT FUNCTIONS
Theorem 9. Every nite vector eld which is uniform, continuous and vanishing at innity may be decomposed into the sum of a curl-free (irrotational) and a divergence-free
(solenoidal) eld.
We know that a curl-free vector eld has the form of v = ∇φ, and a divergence-free
vector eld has the form of w = ∇ × ψ. Therefore if the displacement eld u fullls the
assumptions of Helmholtz theorem we can write
u = ∇φ + ∇ × ψ ,
(3.4)
where φ is a scalar and ψ is a vector eld. Now if u is substituted from (3.4) into (3.1) it
gives
(λ + µ) ∇∇ · u + µ ∆u = (λ + µ) ∇∇ · [∇φ + ∇ × ψ] + µ ∆ [∇φ + ∇ × ψ]
((
= (λ + ν) [∇∇ · ∇φ + ∇(
∇(
· (∇
×(ψ)] + µ ∆∇φ + µ ∆∇ × ψ
(
= (λ + µ) ∇∆φ + µ∇∆φ + µ∇ × ∆ψ = 0 ,
(3.5)
which nally gives
(λ + 2µ) ∇∆φ + µ∇ × ∆ψ = 0 .
(3.6)
Any set of φ and ψ that satises this equation will also satisfy (3.1) when u = ∇φ + ∇ × ψ,
and for any solution u to (3.1) there exits at least one set of φ and ψ that satises (3.6),
which is not unique because u is expressed in terms of rst derivatives of φ and φ. However
this is not important because we are interested in nding the solution in terms of u.
One particular solution of (3.6) is the solution to
∆φ = const. and ∆ψ = const.
(3.7)
and as long as this solution is capable of satisfying the boundary conditions our objective
is fullled. The two above equations are called Laplace's equations or harmonic equations.
Remark 10. The reader may wonder why we reduced the original problem to the particular
case of harmonic functions. The answer is: simplication. However it might be the case
that the problem is over-simplied, but we do not know until we try the harmonic solution
and see if the boundary conditions are fullled. The good news is that harmonic solutions
are rich enough to reect the boundary conditions in many engineering problems we face.
Example 3.1. Wave propagation. The dynamic case of wave propagation can be studied
with Helmholtz decomposition. Instead of equilibrium equation we have the balance of
momentum
∇ · σ + f = ρü ,
(3.8)
where ρ is density and ü is acceleration. Consequently Navier equation becomes
(λ + µ) ∇∇ · u + µ ∆u + f = ρü .
(3.9)
After substitution of Helmholtz decomposition and derivations we get the particular case
of
(λ + 2µ) ∆φ = ρφ̈ , µ ∆ψ = ρψ̈ ,
(3.10)
3.1.
29
SCALAR AND VECTOR POTENTIALS
which are called wave equations. The solution will be of the form
(3.11)
with k being a unit vector called wave-number vector or propagation vector and c and c
are compression wave speed and shear wave speed. In order to obtain the wave speed we
have
(
∆φ = ∂ ∂ φ = k k φ̄ = φ̄
φ = φ̄(k x − c t) =⇒
(3.12)
φ̈ = ∂ ∂ φ = c φ̄
which by substitution into the wave equation for φ yields
(λ − 2µ) φ̄ = c ρφ̄ ,
(3.13)
therefore
s
λ + 2µ
.
(3.14)
c =
ρ
In the same way we can show that the shear wave speed is
r
2µ
.
(3.15)
c =
ρ
φ = φ̄(k · x − cp t) ,
ψ = ψ̄(k · x − cs t) ,
p
i i
j j
i i
2 00
p
p
t t
00
2
p
00
s
00
00
p
s
z
p(r)
h
Fig. 3.1: Axially symmetric slab.
r
A two-dimensionally innite slab is loaded at the top with a given axially
symmetric distribution p(r) in cylindrical coordinates, and constrained with a frictionless
surface at the bottom. We want to nd the solution in terms of Helmholtz functions.
Solution. To understand the problem we have to express boundary conditions. At the top
we have a vertical force distribution therefore
z=h :
σ = −p(r) , σ = 0
(3.16)
and at the bottom there is no vertical movement and tangential force (no friction!)
z=0 :
u = 0 , σ = 0.
(3.17)
Example 3.2.
zz
rz
z
rz
30
Axial symmetry means that u, ε and σ are independent of ϕ. Back to equations (3.7) we
assume the special case
ψ=0
(3.18)
∆φ = 0 ,
(3.19)
and we have to see if the solution can fulll all the boundary conditions. If it does not then
we will assume a more sophisticated case. Now we need to express the boundary conditions
in terms of φ. Starting from the above assumption
1
ε = (∇u + u∇) = ∇φ∇ ,
(3.20)
2
therefore based on Hooke's law (2.15)
I + 2µ∇φ∇ = 2µ∇φ∇ .
σ = λ tr(ε) I + 2µ ε = λ ∆φ
(3.21)
Then in cylindrical coordinates
u =∂ φ,
u =∂ φ
(3.22)
1
(3.23)
σ = 2µ∂ φ , σ = 2µ ∂ φ
r
σ = 2µ∂ φ , σ = 2µ∂ ∂ φ ,
(3.24)
and the boundary conditions (3.16) and (3.17) in terms of φ are obtained as
z = h : 2µ∂ φ = −p(r) , 2µ∂ ∂ φ = 0
(3.25)
z = 0 : ∂ φ = 0 , 2µ∂ ∂ φ = 0
(3.26)
The Laplace's equation (3.19) in cylindrical coordinates reads
1
∆φ = ∂ φ + ∂ φ + ∂ φ = 0 .
(3.27)
r
For separation of variables we put
φ(r, z) = f (r) g(z) ,
(3.28)
which results in
1
f g + f g + fg = 0
(3.29)
r
after factorization
f + 1/r f
g
= − = ±k ,
(3.30)
f
g
where ±k reects the possibility of a positive or negative constant. Since the rst expression in the above equation is a function of r only and the second is a function of z only,
therefore the only possible way for them to be equal is to be constant.
CHAPTER 3.
r
r
z
DISPLACEMENT FUNCTIONS
z
rr
2
r
ϕϕ
zz
2
z
rz
r
r z
2
z
r z
z
r z
2
r
2
z
r
†
00
00
0
00
00
2
2
†
Separation of variables is a standard method to reduce PDE's to ODE's.
31
The negative choice −k leads to contradiction with boundary values. Therefore we
take the positive constant +k which yields two ordinary dierential equations
1
f + f − k f = 0 , g + k g = 0.
(3.31)
r
The rst equation is the so called Bessel's equation of zeroth order which has the general
solution
f (r) = A J (kr) + B H (kr) ,
(3.32)
where J is Bessel function of zeroth order and H is Hankel function of zeroth order.
The second equation has the well know real solution
g(z) = C sin(kz) + D cos(kz) .
(3.33)
Since
H (x) → ∞ for x → 0
(3.34)
in order to have a nite solution in the neighborhood of r = 0 it must hold that B = 0,
then the general solution becomes
φ = J (kr) (C sin(kz) + D cos(kz)) .
(3.35)
Applying the boundary conditions gives
u (0) = ∂ φ(0) = J (kr) k C = 0
⇒
C=0
(3.36)
σ (0) = 2µ∂ ∂ φ(0) = 2µ k J (kr) k C = 0 X
(3.37)
3.1.
SCALAR AND VECTOR POTENTIALS
2
2
00
0
00
2
0
2
0
0
0
0
0
z
rz
z
0
0
0
r z
σrz (h) = sµk J00 (kr) (−D k sin(kh)) = 0 ⇒ sin(kh) = 0 ⇒ kn =
nπ
, n = 0, 1, . . .
h
(3.38)
As you can see there are innitely many possible values for k and therefore there are
innitely many solutions of the form
φ (r, z) = J (k r) cos(k z) ,
(3.39)
which are called normal modes. The general solution is the linear combination of all possible
solutions as
X
φ=
a φ .
(3.40)
This situation is a bit tricky because there are innitely many constants a 's to determine
but there is only one remaining boundary condition to be fullled as
n
0
n
n
∞
n
n
n=0
n
σzz (h) = 2µ∂z2 φ = −2µ
X
kn a2n J0 (kn r) cos(kn h) =
n
− 2µ
X
n
(−1)n an kn2 J0 (kn r) = −p(r) .
(3.41)
32
The key idea is the orthogonality
property of Bessel functions as
Z
J (k r) J (k r) r dr = 0
for i 6= j .
(3.42)
Therefore if we multiply both sides of (3.1) with J (k r) r and then integrate over [0, ∞]
we obtain
Z
Z
p(r) J (k r) r dr
(3.43)
(J (k r)) r dr =
2µ(−1) a k
because all the terms except the i term vanish
due to (3.42), and a is obtained as
R
p(r) J (k r) r dr
1
a =
(3.44)
R
2µ(−1) k
(J (k r)) r dr
which nally yields
R
X
p(r) J (k r) r dr
1
J (k r) cos(k z) .
(3.45)
φ=
R
2µ(−1) k
(J (k r)) r dr
CHAPTER 3.
DISPLACEMENT FUNCTIONS
∞
0
i
0
j
0
0
∞
i
2
i i
0
i
∞
2
i
0
th
i
Exercise 17.
potentials
i
i 2
i
∞
i 2
n
n=0
i
0
0
∞
0
∞
0
∞
0
∞
0
0
0
n
0
0
i
i
2
n
0
2
n
n
Using Helmholtz theorem calculate u, ε and σ for the given scalar and vector

x21 − x22
.
0
ψ = l
0

2
φ = k x21 − x2 ;
A circular plate with radius R and thickness h is constrained at the top and
bottom by frictionless surfaces (Fig. 3.2). A load distribution p(z) is applied on its vertical
boundary. Using Helmholtz theorem derive solution for axially symmetric problem.
Exercise 18.
z
R
p (z)
h
r
Fig. 3.2: Circular plate under radial loading.
Hint: If the ansatz φ = J (kr) [C sin(kz) + D cos(kz)] is used, the solution in the form
φ = J (k r) cos k z will be obtained. Then Fourier expansion in z should be used.
0
n
3.2
0
n
n
Galerkin vector
Helmholtz decomposition gives displacement eld in terms of rst derivatives of a scalar
and a vector potential. It is also possible to express u in terms of second derivatives of a
vector eld. Let us introduce the Galerkin vector V which is related to u by
u = k ∆V − ∇∇ · V
(3.46)
where k is a constant.
3.2.
33
GALERKIN VECTOR
Theorem 11. For any vector eld
satises the relation (3.46).
u
and constant k there exists a vector eld V that
Therefore the relation (3.46) provides a completely general solution if V is determined
somehow. If the Galerkin vector is plugged into Navier equation (2.50) it gives
(λ + µ) k∇∇ · ∆V − (λ + µ) ∇∇ · ∇∇ · V + µk ∆∆V − µ ∆∇∇ · V + f = 0
(3.47)
and then
[(λ + µ) k − (λ − 2µ)] ∇∇ · ∆V + µk∆∆V + f = 0 ,
(3.48)
as we are allowed to chose the constant k, the best choice is the one that simplies the
above equation the most, which turns out to be
λ + 2µ
k=
= 2 (1 − ν) ,
(3.49)
λ+µ
and this gives
2 (1 − ν) µ ∆∆V + f = 0 ,
(3.50)
which is a vectorial biharmonic equation. Biharmonic equations are well studied in analysis
and nding the answer to the above equation is basically simpler than the original Navier
equation. Then we obviously need the relation (3.46) to transform the answer in terms of
V to u.
Remark 12. Galerkin vector is related to Helmholtz potentials by
φ = −∇ · V , ∇ × ψ = 2 (1 − ν) ∆V .
(3.51)
3.2.1 Love's strain function
There is a special case of Galerkin vector which can be used in axially symmetric problems.
Assuming that the only non-vanishing component of Galerkin vector is the X component
we have
V =V =0, V =L
(3.52)
where L is Love's strain function. Then (3.50) simplies to
2 (1 − ν) µ ∆∆L + f = 0 .
(3.53)
Note that f = f = 0 which is the symmetry requirement. The relation (3.46) in terms of
Love's function reads
u = −∂ ∂ L , u = −∂ ∂ L , u = 2 (1 − ν) ∆L − ∂ L .
(3.54)
Example 3.3. In the case of spherical symmetry L is a function of r only
√
(3.55)
L = L(r) :
r = x x = kxk
3
1
2
3
3
1
2
1
1 3
2
2 3
2
3
3
i i
34
and Laplacian in spherical coordinates is given by
CHAPTER 3.
DISPLACEMENT FUNCTIONS
(3.56)
2
∆ ≡ ∂r2 + ∂r .
r
Therefore in the absence of body forces equation (3.53) becomes
∆∆L =
2
+ ∂r
r
∂r2
(3.57)
2
4
2
∂r + ∂r L = ∂r4 L + ∂r3 L = 0 .
r
r
The general solution to this equation has the form
(3.58)
with C 's being integration constants to be determined by application of boundary conditions.
L = C1 r−1 + C2 + C3 r + C4 r2 ,
i
Kelvin's problem
The classical problem of Kelvin deals with a concentrated
force in an innite medium. This is an important case whose
solution is used to solve more complicated problems. Let us
assume the concentrated force at the origin of coordinate sysf e
tem and directed towards X axis as
f = f δ(x) e
(3.59)
where δ(x) is Dirac's delta function in 3-dimensional space.
Due to the axial symmetry we use the Love's formulation of
Fig. 3.3: Kelvin's problem.
Navier equation with body force
2 (1 − ν) µ∆∆L = −f δ(x) .
(3.60)
Integrating both sides within a sphere of radius R surrounding the origin yields
Z
Z
2 (1 − ν) µ
∆∆L dV = −f
δ(x) dV = −f ,
(3.61)
where we have used the fundamental property of Dirac's delta R δ(x) dV = 1. On the
other hand we know that ∆∆L = ∇ · ∇∆L, and using Gauss theorem
Z
Z
Z
Z
∆∆L dV =
∇ · ∇∆L dV =
n · ∇∆L dS =
∂ ∆L dS .
(3.62)
The general solution of the form (3.58)
0 3
3
0
3
0
0
r≤R
0
r≤R
r≤R
r
r≤R
is simplied to
r≤R
r=R
r=R
L = C1 r−1 + C2 + C3 r + C4 r2
L = C1 r−1 + C2
(3.63)
3.3.
DISPLACEMENT FUNCTIONS OF PAPKOVICH-NEUBER
because otherwise
35
(3.64)
which is not acceptable as the solution has to stay nite. Then plugging (3.63) into the
surface integral in (3.62) gives
Z
1
∂ ∆L dS = 4πR −2C
= −8πC
(3.65)
R
then putting back into (3.61) results in
16π (1 − ν) µC = f
(3.66)
and nally the answer is
f
L=
r.
(3.67)
16π (1 − ν) µ
Exercise 19. For the Love strain function in (3.67) calculate u and σ.
x→∞
r
L→∞
:
2
3
r=R
3
2
3
0
0
3.3
Displacement functions of Papkovich-Neuber
So far we have managed to replace the Navier equation by the fourth-order biharmonic
equation (3.50) and the third-order equation (3.6). Note that the harmonic solutions to
(3.6) are special cases that satisfy the original third-order equation, not the general solution
to it. Now it would be nice if we could have an equivalent formulation to Navier equation
which is of the second-order i.e. one order lower with respect to (3.6) and two orders lower
with respect to (3.50).
Starting from Galerkin vector (3.46), if the vector eld A and the scalar eld b are
introduced such that
1
A = ∆V , b = ∇ · V
(3.68)
2
then we can write displacement eld in terms of A and b
u = 4 (1 − ν) A − ∇b .
(3.69)
Therefore
1
∆A = ∆∆V = 0 X , see (3.50)
(3.70)
2
∆b = ∆∇ · V = ∇ · ∆V = 2∇ · A .
(3.71)
Exercise 20. For the Papkovich-Neuber solution u = 4 (1 − ν) A − ∇b show that
∆ (A · X) = 2∇ · A .
(3.72)
36
Using the result in the recent exercise the general form of b is given by
CHAPTER 3.
b=A·x+a,
DISPLACEMENT FUNCTIONS
∆a = 0 .
Note that a can be any harmonic function in general. Then this all comes down to
(3.73)
(3.74)
∆A = 0 , ∆a = 0 .
(3.75)
This is called Papkovich-Neuber solution which is equivalent to Galerkin's solution, and
since Galerkin's solution is equivalent to Navier equation, then the above formulation is
completely general and equivalent to our original formulation of Navier.
Example 3.4. Problem of Leon. Find the displacement eld in a uni-axially loaded
innite medium with a spherical cavity (Fig. 3.4).
σ
Solution. Laplace's equation in spherical coordinates (independent of ϕ due to axial symmetry of our problem), reads
∂ f
2 ∂f
cos θ ∂f
1 ∂ f
∆f =
+
+
+
=0
(3.76)
∂r
r ∂r
r sin θ ∂θ
r ∂θ
R
and we have to solve (3.75)
∆A = 0 , ∆a = 0 .
(3.77)
Because of axial symmetry around the axis of loading the only
nonzero component of A is A , which means A = A = 0
and we are left with
∆A = 0 , ∆a = 0 .
(3.78) Fig. 3.4: Leon's problem.
By separation of variables we seek solutions of the form
f (r, θ) = R(r) Θ(θ)
(3.79)
that ends up in
1 d
dR
1
d
dΘ
r
=−
sin θ
= l (l + 1) , l ∈ {0, 1, 2, . . .} .
(3.80)
R dr
dr
sin θ Θ dθ
dθ
The special form of the constant l (l + 1), for l being a non-negative integer, guarantees a
convergent solution of Legendre equation that appears here in terms of η = cos θ
d Θ
dΘ
− 2η
+ l (l + 1) Θ = 0 .
(3.81)
1−η
dη
dη
u = 4 (1 − ν) A − ∇ [A · x + a]
◦
2
2
2
2
2
2
z
x
y
z
2
‡
2
2
2
‡
.
To derive (3.80) into (3.81) we need to know from the chain rule that
d
d
= − sin θ
dθ
dη
⇒
d
1 d
=−
dη
sin θ dθ
3.3.
DISPLACEMENT FUNCTIONS OF PAPKOVICH-NEUBER
37
The general solution is given by
(3.82)
where P (x) is Legendre polynomial of degree l. Then the general solution to (3.78) is
obtained
X
1
P (cos θ)
(3.83)
a=
cr +d
r
1
l
f = Ar + B l+1 Pl (cos θ) ,
r
l
∞
l
l
Az =
l=0
∞ X
l
l l+1
Cl r + D l
l=0
1
l
Pl (cos θ) .
rl+1
The boundary conditions are given as
(3.84)
(3.85)
r = R : σ = σ = 0.
(3.86)
In order to homogenize the boundary condition the following re-parameterization is introduced
σ̄ = σ − σ , σ = σe e .
(3.87)
Then according transformation rules for second-order tensors
σ = σ cos θ , σ = σ sin θ , σ = − σ sin θ cos θ .
(3.88)
And in terms of σ̄ the boundary conditions are
r → ∞ : σ̄ → 0 ⇒ c = 0 , C = 0
(3.89)
(3.90)
r = R : σ̄ = −σ , σ̄ = − σ .
Considering the number of boundary conditions to be fullled we can start from a limited
number of terms in (3.83) and (3.84), say l = 0, 1, 2 to see if the boundary conditions are
fullled. If we need more terms we can add them later on, so let us assume
1
1
1
a=d +d
cos θ + d
3 cos θ − 1
(3.91)
r
r
r
1
1
1
cos θ + D
3 cos θ − 1 .
(3.92)
A =D +D
r
r
r
(3.93)
To express the boundary conditions in terms of a and A we use the following result whose
proof is left as an exercise
u = 4 (1 − ν) A cos θ − ∂ (A r cos θ + a)
(3.94)
1
u = −4 (1 − ν) A sin θ − ∂ (A r cos θ + a) .
(3.95)
r
◦
r→∞ :
σzz = σ
rr
◦
rθ
◦
◦
z z
◦
◦
◦
2
rr
◦
◦
2
θθ
◦
rθ
l
l
◦
◦
rr
rr
z
0
1 2
0
1 2
rθ
rθ
2
2 3
2
2 3
z
r
θ
z
z
r
z
θ
z
38
We also need the following formulas
CHAPTER 3.
DISPLACEMENT FUNCTIONS
(3.96)
(3.97)
(3.98)
It is left to the reader as an important exercise to apply the boundary conditions and obtain
the following results
1 6 − 5ν
d =−
R σ,
D = 0,
(3.99a)
4µ 7 − 5ν
5
1
d = 0,
D =
R σ,
(3.99b)
4µ 7 − 5ν
1
1
R σ,
D = 0.
(3.99c)
d =−
4µ 7 − 5ν
1
1
εrr = ∂r ur , εθθ = ∂θ uθ + ur
r
r
1 1
1
εrθ =
∂θ ur + ∂r uθ − uθ
2 r
r
1
1
εϕϕ = ur + cot θ uθ .
r
r
3◦
0
0
1
1
5◦
2
Exercise 21.
3◦
2
Using Papkovich-Neuber solution
u = 4 (1 − ν) A − ∇b ,
b=A·X +a
for the ansatz
∞ X
1
n
a=
cn r + dn n+1 pn (cos θ)
r
n=0
A1 = A2 = 0
A3 =
∞ X
n
Cn r + D n
n=0
1
rn+1
pn (cos θ)
with p (cos θ) being Legendre polynomial, show that
n
ur = 4 (1 − ν) A3 cos θ − ∂r (A3 r cos θ + a)
1
uθ = −4 (1 − ν) A3 sin θ − ∂θ (A3 r cos θ + a) .
r
Using Papkovch-Neuber solution for a sphere with radius R and surface
displacement u (R, θ) = 0 and u (R, θ) = u sin θ nd σ(r, θ).
Hint: use ansatz of the previous assignment but only up to n = 1 for a and A .
Exercise 22.
r
θ
0
3
Bibliography
[1] A.I. Borisenko and I.E. Tarapov. Vector and Tensor Analysis with Applications. Dover
Pub. Inc, 1968.
[2] P.C. Chou and N.J. Pagano. Elasticity: Tensor, Dyadic, and Engineering Approaches.
Dover Pub. Inc, 1992.
[3] G.A. Holzapfel. Nonlinear Solid Mechanics: A Continuum Approach for Engineering.
Wiley, 2000.
[4] A.E. Love. Treatise on the Mathematical Theory of Elasticity. Dover Pub. Inc, 4
edition, 1927.
[5] R.W. Ogden. Non-Linear Elastic Deformations. Dover Pub. Inc, 1997.
[6] M.R. Spiegel. Mathematical Handbook of Formulas and Tables. McGraw-Hill (Schaum's
outline), 1979.
[7] T.C.T. Ting. Anisotropic Elasticity: Theory and Applications. Oxford University Press,
1996.
th
39