Continuum mechanics and nonlinear elasticity

Stefano Giordano
Department of Physics - University of Cagliari
Cittadella Universitaria - 09042 Monserrato (Ca), Italy
E-mail: stefano.giordano@dsf.unica.it

Contents

1 Symbols 2

2 Lagrangian versus Eulerian formalism 2

2.1 Derivative of a volume integral . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 Derivative of a surface integral . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Strain 8

4 Stress 11

5 Continuity equation 14

6 Balance equations: Euler description 15

7 Balance equations: Lagrange description 17

7.1 Novozhilov formulation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

8 Nonlinear constitutive equations 20

9 The small-strain approximation 23

CONTENTS 2

1. Symbols

List of the most important tensor quantities used in the following sections
F̂ deformation gradient
Ĝ inverse deformation gradient
L̂ velocity gradient
J deformation Jacobian
B̂ and Ĉ left and right Cauchy tensors
Û and V̂ left and right stretching tensors
R̂ rotation tensor
η̂ Green-Lagrange tensor
ê Almansi-Eulero tensor
JˆL Lagrangian displacement gradient
JˆE Eulerian displacement gradient
D̂ rate of deformation tensor
Ŵ spin tensor
T̂ Cauchy stress tensor
1PK
T̂ first Piola-Kirchhoff stress tensor
2PK
T̂ second Piola-Kirchhoff stress tensor
Jˆ small-strain displacement gradient
ˆ small-strain tensor
Ω̂ local rotation tensor
Cˆ stiffness tensor

2. Lagrangian versus Eulerian formalism

The motion of a body is typically referred to a reference configuration Ω0 ⊂ 3 , which

is often chosen to be the undeformed configuration. After the deformation the body
occupies the current configuration Ωt ⊂ 3 . Thus, the current coordinates (x ∈ Ωt ) are
 ∈ Ω0 ):
expressed in terms of the reference coordinates (X


X → x = Ft X (2.1)
where Ft is the transformation function at any time t (see Fig. 1). More explicitely, it
means that
x1 = x1 (X1 , X2 , X3 , t)
x2 = x2 (X1 , X2 , X3 , t) (2.2)
x3 = x3 (X1 , X2 , X3 , t)
CONTENTS 3

Figure 1. Reference configuration and current configuration after a deformation.

We call the set (X and t) Lagrangian coordinates, named after Joseph Louis Lagrange
[1736-1813], or material coordinates, or reference coordinates. The application of these
coordinates is called Lagrangian description or reference description. We can obtain
also the inverse function of Eq.(2.1) in the form
 = F −1 (x)
x → X (2.3)
t

or, in components
X1 = X1 (x1 , x2 , x3 , t)
X2 = X2 (x1 , x2 , x3 , t) (2.4)
X3 = X3 (x1 , x2 , x3 , t)
The set (x and t) is called Eulerian coordinates, named after Leonhard Euler [1707-
1783], or space coordinates, and their application is said Eulerian description or spatial
description. The Lagrangian coordinates were introduced by Euler in 1762, while Jean
le Rond D’Alembert [1717-1783] was the first to use the Eulerian coordinates in 1752.
In general Continuum Mechanics Lagrangian coordinates and the reference description
are the most common. The same holds true in solid Mechanics. However, in Fluid
Mechanics, due to large displacements and complex deformations, it is usually necessary
and most practical to use Eulerian coordinates and spatial description.
One of the key quantities in deformation analysis is the deformation gradient of Ωt
relative to the reference configuration Ω0 , denoted F̂ , which gives the relationship of a
material line dX  before deformation to the line dx (consisting of the same material as
 after deformation. It is defined as
dX)





x = Ft X : F̂ X,  t =∇ 
Ft X
X
 ⇒ dx = F̂ X,  t dX  (2.5)
Its components are given by
∂xi
FiK = ∀(i, K) ∈ {1, 2, 3}2 (2.6)
∂XK
CONTENTS 4

As before, we can define a deformation gradient Ĝ of the inverse function relating Ω0 to

the current configuration Ωt
 = F −1 (x) : Ĝ (x, t) = ∇
X 
x F −1 (x) ⇒ dX
 = Ĝ (x, t) dx (2.7)
t t

In components, it assumes the form

∂Xk
GKi = ∀(i, K) ∈ {1, 2, 3}2 (2.8)
∂xi
Of course, the tensors F̂ and Ĝ are related by the relationships


 

Ĝ Ft X  , t = F̂ −1 X,  t (2.9)
 
F̂ Ft−1 (x) , t = Ĝ−1 (x, t) (2.10)
In fact, F̂ is a Lagrangian tensor while Ĝ is an Eulerian tensor. The velocity and
 (Lagrangian
acceleration fields, related to the trajectory of the particle starting at X
description) are given by



v X,  t = ∂x X,  t (2.11)
∂t

 ∂ 2 x


a X, t = 2 X, t  (2.12)
∂t
On the other hand, the velocity and acceleration fields in the Euler description are given
by
∂x  −1 
v (x, t) = Ft (x) , t (2.13)
∂t
∂ 2 x  −1 
a (x, t) = F (
x ) , t (2.14)
∂t2 t

Any time-dependent scalar, vector, or tensor field can be regarded as a function of

 t) (Lagrangian or material variables) or (x, t) (Eulerian or spatial variables)
(X,

whenever the motion x = Ft X  is given. For example, for a scalar field we can


write φ (x, t) = Φ X, t where



 

Φ X, t = φ Ft X  ,t (2.15)


The time derivative of the field Φ X,  t can be calculated as

∂Φ ∂φ ∂φ ∂x ∂φ ∂φ
= + · = + · v (2.16)
∂t ∂t ∂x ∂t ∂t ∂x
Instead of using different symbols for the quantities (i.e. φ and Φ) in the Lagrangian
and Eulerian descriptions, we can use the dot for the Lagrangian or material derivative
(φ̇) and the partial differentiation symbol ( ∂φ
∂t
) for the Eulerian or spatial derivative:
therefore, Eq.(2.16) assumes the simpler form
∂φ ∂φ
φ̇ = + · v (2.17)
∂t ∂x
CONTENTS 5

The Eulerian tensor

∂v
L̂ = (2.18)
∂x
with components
∂vi
Lij = (2.19)
∂xj
satisfies the important relation
˙
F̂ = L̂F̂ (2.20)
It can be proved as follows
∂ ∂


˙
F̂ = Ft X = ∂ ∂ Ft X  = ∂v = ∂v ∂x = L̂F̂ (2.21)

∂t ∂ X ∂X ∂t ∂X 
∂x ∂ X
It is also important an inverse relation given by
˙ −1
F̂ = −ĜL̂ (2.22)


˙ −1 ˙ ˙ −1
Since dtd F̂ −1 F̂ = 0 we have F̂ = −F̂ −1 F̂ F̂ −1 (where F̂ represents the Lagrangian
time derivative of the inverse of F̂ ) and, therefore, we obtain the proof of Eq.(2.22)
˙ −1 ˙
F̂ = −F̂ −1 F̂ F̂ −1 = −F̂ −1 L̂F̂ F̂ −1 = −F̂ −1 L̂ = −ĜL̂ (2.23)

2.1. Derivative of a volume integral

We consider a subset Pt ⊂ Ωt which is the time deformed version of P0 ⊂ Ω0 . We search
a property giving the time derivative of an arbitrary volume integral. In this context,
the symbol d/dt can be used when it is applied to a quantity depending only on the
time t. In fact, in this case, there is no ambuguity. As before we consider a scalar field
φ and, through a change of variables between Eulerian and Lagrangian coordinates, we
obtain
 
d d 
φdx = φJdX (2.24)
dt Pt dt P0
where J is the determinant of the deformation gradient
∂x
J = det = det F̂ (2.25)
∂X
Then, the time derivation can enter the integral written in the reference configuration
  

d d  =
φdx = (φJ) dX φ̇J + φJ˙ dX  (2.26)
dt Pt P0 dt P0

The derivative of a determinant follows the rule

d


˙ −1
det F̂ = det F̂ tr F̂ F̂ (2.27)
dt
˙
From Eq.(2.20) we obtain F̂ F̂ −1 = L̂ and, therefore,we have


J˙ = Jtr L̂ = J ∇ 
x · v (2.28)
CONTENTS 6

So
 
 

d
φdx = φ̇ + φ∇
x · v JdX =
  φ̇ + φ∇
x · v dx
 (2.29)
dt Pt P0 Pt

Since φ̇ = ∂φ
∂t
+ · v we obtain
∂φ
∂

x
   
d ∂φ 
φdx = + ∇
x φ · v + φ∇

x · v dx (2.30)
dt Pt Pt ∂t
or, finally
  
d ∂φ 
φdx = + ∇
x · (φv ) dx (2.31)
dt Pt Pt ∂t
This property has been called Reynolds theorem or transport theorem. It is the most
important result used to obtain the balance equations for continuum materials. If φ = 1
we obtain
 
d
dx = ∇

x · v dx (2.32)
dt Pt Pt
which represent the rate of variation of the volume of the region Pt .

2.2. Derivative of a surface integral

We begin by describing the deformation of a given surface moving from the reference to
the current configuration. We therefore consider a surface X  =X  (α, β) in the reference
configuration described in parametric form by two parameters

αand β. The deformed
surface in the current configuration is given by x = Ft X  (α, β) . We define N  dS and
nds as the unit normal vector multiplied by the area element in the reference and in the
current configuration, respectively. From standard differential geometry we have
∂X ∂X

NdS = ∧ dαdβ (2.33)
∂α ∂β
The deformed version can be straightforwardly obtained as



∂x ∂x 
∂x ∂ X 
∂x ∂ X
nds = ∧ dαdβ = ∧ dαdβ
∂α ∂β ∂X  ∂α ∂X  ∂β



∂X 
∂X
= F̂ ∧ F̂ dαdβ (2.34)
∂α ∂β
The last expression can be written component by component
∂Xs ∂Xt
ni ds = εijk Fjs Fkt dαdβ
∂α ∂β
and it can be multiplied by Fir on both sides
∂Xs ∂Xt
Fir ni ds = εijk Fir Fjs Fkt dαdβ
∂α ∂β
Since εijk Fir Fjs Fkt = det F̂ εrst we obtain
∂Xs ∂Xt
Fir ni ds = Jεrst dαdβ
∂α ∂β
CONTENTS 7

or
∂X 
∂X
∧
F̂ T nds = J 
dαdβ = J NdS (2.35)
∂α ∂β

and finally we have obtained the relationship between NdS and nds
nds = J F̂ −T NdS
 (2.36)
This property has been called Nanson theorem. Now, it is interesting to evaluate the
time derivative of the surface integral of a vector field a. It can be brought back to the
reference configuration as
 
d d
a · nds = a · J F̂ −T N
 dS
dt St dt S0
 
˙ −T −T ˙ −T 
= a · J F̂ + a · J F̂ + a · J F̂
˙ N dS (2.37)
S0

 −T

x · v and F̂˙ ˙T
Now J˙ = Jtr L̂ = J ∇ = −F̂ −T F̂ F̂ −T and therefore
   T

d ˙ −T  −T −T ˙ −T 
a · nds = a · J F̂ + a · J ∇
x · v F̂ − a · J F̂ F̂ F̂ NdS
dt St
S0 
= a˙ + a∇
x · v − L̂a · J F̂ −T NdS
 (2.38)
S0
˙
where the relation F̂ = L̂F̂ has been used. Finally, coming back to the current
configuration we obtain
  
d
a · nds = a˙ + a∇

x · v − L̂a · nds (2.39)
dt St St

Since the material derivative is given by a˙ = ∂
a ∂
a

+ ∂
· v, we obtain
   ∂t x

d ∂a ∂a
a · nds = + · v + a∇
x · v − L̂a · nds (2.40)
dt St St ∂t ∂x
It is simple to verify that ∇ 
x ∧ (a ∧ v ) + v∇ 
x · a = ∂
a · v + a∇
∂

x

x · v − L̂a and therefore
we can write   
d ∂a  
a · nds = + ∇
x ∧ (a ∧ v ) + v ∇
x · a · nds (2.41)
dtSt St ∂t

The Nanson relation nds = J F̂ −T NdS  can be also applied in order to obtain the so-
called Piola identity. To this aim we use the standard divergence theorem
 
∂Ψ
Ψni ds = dx (2.42)
Pt ∂xi
∂Pt

if Ψ = 1 identically, we obtain ∂Pt ni ds = 0 and, therefore
  

−T  
· J F̂ −1 dX
nds = J F̂ N dS = ∇ X
 =0 (2.43)
∂Pt ∂P0 P0
which means  

 ∂ ∂X
−1 j
∇

· J F̂
X =0 ⇒ J =0 (2.44)
∂Xj ∂xi
This relation will be useful to obtain the balance equations of the continuum mechanics
in the Lagrangian description.
CONTENTS 8

Figure 2. Infinitesimal vector dX

in Ω0 and its deformed version d
x in Ωt .

3. Strain

The measure of the deformation between the reference and the current configuration is
an important topic in continuum mechanics and it can be performed in several ways.
The starting quantity is the deformation gradient F̂ (X)  (in the Lagrangian formalism)
or its inverse Ĝ(x) (in the Eulerian formalism). We consider two infinitesimal vectors
dX and dY in Ω0 and their deformed versions dx and dy in Ωt (see Fig. 2 for the
deformation of dX).  The changes of lengths and angles are controlled by the scalar
product of the vectors and, therefore, we define the right and the left Cauchy tensors Ĉ
and B̂ in order to obtain dx · dy = dX  · dY = dx · B̂ −1 dy (see Table 1). The
 · ĈdY or dX
variations of the scalar product (moving from the reference to the current configuration)
are described by the Green-Lagrange tensor η̂ and by the Almansi-Eulero tensor ê as
summarized in Table 1.
Moreover, the gradients of the displacements field u(X)  and u(x) are defined by
JˆL = ∂∂
Xu
and JˆE = ∂

∂

u
x
in the Lagrangian and Eulerian vision, respectively. In Table 2
we can find: (i) the effective variation of length for the vector dX  = N
d X
 deformed
into dx = n
dx
; (ii) the variation of the right angle between the unit vectors N  and T
(N · T = 0 in Ω0 ) deformed into n and t (in Ωt ): θnt is the angle in Ωt and, therefore,
γN T = π2 − θnt is the angle variation (with opposite sign); (iii) the variation of the right
angle between the unit vectors n and t (n · t = 0 in Ωt ) originally placed at N
 and T (in
Ω0 ): θN T is the angle in Ω0 and, therefore, γnt = θN T − 2 is the angle variation (with
π
CONTENTS 9

Table 1. Strains definitions and properties in Lagrangian and Eulerian formalisms.

Lagrangian vision Eulerian vision

Right Cauchy tensor Left Cauchy tensor
Ĉ = F̂ T F̂ B̂ = F̂ F̂ T
Ĉ −1 = ĜĜT B̂ −1 = ĜT Ĝ
dx · dy = dX
 · ĈdY dX · dY = dx · B̂ −1 dy

Green-Lagrange

 tensor Almansi-Eulero

tensor
1 1 ˆ
η̂ = 2
Ĉ − Iˆ ê = 2
I − B̂ −1
 · dY = 2dX
dx · dy − dX  · η̂dY  · dY = 2dx · êdy
dx · dy − dX
Lagrange displacement gradient Eulero displacement gradient
JˆL = ∂∂
Xu
JˆE = ∂
u
∂

x
F̂ = Iˆ + JˆL F̂ −1 = Iˆ − JˆE
Ĉ = Iˆ
+ JˆL + JˆLT + JˆLT JˆL B̂ −1 =
Iˆ − JˆE − JˆET + JˆE
T ˆ
JE
η̂ = 12 JˆL + JˆLT + JˆLT JˆL ê = 1 JˆE + JˆT − JˆT JˆE
2 E E

opposite sign); (iv) the variations of volume and surface measures.

Any non singular tensor (describing a deformation) can be decomposed in two
different ways
F̂ = R̂ Û = V̂ R̂ (3.1)
where R̂ is a rotation matrix (R̂R̂T = R̂T R̂ = I) ˆ while Û and V̂ are symmetric
and positive definite tensors. In order to prove this polar decomposition theorem
due to Cauchy, we use the right Cauchy tensor Ĉ = F̂ T F̂ : it is symmetric since

T
T
F̂ F̂ = F̂ T F̂ T T = F̂ T F̂ and it is positive definite as proved by the following relation

T

 T F̂ T F̂ w
w  = F̂ w   =
F̂ w

F̂ w  ≥0 ∀ w  (3.2)

If F̂ T F̂ is symmetric and positive definite then it can be diagonalized in the field of real
numbers. Therefore, we can write F̂ T F̂ = Q̂−1  ˆ Q̂ where Q̂ is non singular and  ˆ is
diagonal. We define
 
T
Û = F̂ F̂ = Ĉ (3.3)
The square root of the tensor can be defined (and calculated) as follows
  
ˆ −1 ˆ Q̂
Û = F̂ F̂ = Q̂ Q̂ = Q̂
T −1  (3.4)
in fact
  2    
−1 ˆ −1 ˆ −1 ˆ −1 ˆ ˆ Q̂ = Q̂−1 
ˆ Q̂ (3.5)
Q̂ Q̂ = Q̂ Q̂Q̂ Q̂ = Q̂ 
CONTENTS 10

Table 2. Variations measure in Lagrangian and Eulerian formalisms.

Lagrangian vision Eulerian vision

Lagrangian length variation Eulerian length variation
N = dX
d

x
n =
d

x


dX



 
N N =
d
x
d

−
dX



nn =
d
x
−
d


= N · Ĉ N
 −1 X

= 1 − n · B̂ −1n


X

d

x

N N + 1 2 = N
2 NN
 · η̂ N
 nn − 12 2nn = n · ên
Lagrangian angle variation Eulerian angle variation
N · T = 0 n · t = 0
γN T = π2 − θnt γnt = θN T − π2
sin(γN T ) = √ 2N ·η̂√T



sin(γnt ) = √ 2
n·ê√t



·Ĉ N
N
T
·Ĉ T
n·B̂ −1


n
t·B̂ −1
t

Lagrangian volume variation Eulerian volume variation

J = det(F̂ ) J −1 = det(Ĝ)
ΘV = dv−dV
dV
=J −1 Θv = dv−dV
dv
= 1 − J1
Lagrangian surface variation Eulerian surface variation
 dS = J −1 F̂ T nds
N nds = J F̂ −T N
 dS



nds
−
N
dS



nds
−
N
dS

ΘN = Θn =



N
dS
nds


ΘN = J  · Ĉ −1 N
N  −1 Θn = 1 − J −1 n · B̂n


√
where  ˆ = diag(λi ) (the symbol diag explicitely indicates the entries
ˆ = diag( λi ) if 
of a diagonal matrix). Finally, we define R̂ = F̂ Û −1 and we verify its ortogonality

T
T
−1 −1 −1
T
R̂ R̂ = Û T
F̂ F̂ Û = Û Û 2 Û −1 = Û −1 Û Û Û −1 = Iˆ (3.6)
This concludes the proof of the first polar decomposition. We have to prove the unicity
of the right decomposition F̂ = R̂ Û . We can suppose the two different decompositions
F̂ = R̂ Û = R̂∗ Û ∗ exist. It follows that F̂ T F̂ = Û 2 = Û ∗2 from which Û = Û ∗ and,
therefore, R̂ = R̂∗ . It proves the unicity of the  right decomposition.
 Similarly we can
T
obtain the left decomposition by defining V̂ = F̂ F̂ = B̂: it is possible to prove
that it is symmetric and positive definite and we define R̂ = V̂ −1 F̂ , which is orthogonal.

T
To conclude we must verify that R̂ = R̂. Since R̂ R̂ = Iˆ we have F̂ = V̂ R̂ =

T
R̂ R̂ V̂ R̂ . The unicity of the right decomposition (F̂ = R̂ Û ) allows us to affirm
that R̂ = R̂ and that Û = R̂T V̂ R̂. This completes the proof of the polar decomposition
Cauchy theorem.
This decomposition implies that the deformation of a line element dX  in the
undeformed configuration onto dx in the deformed configuration, i.e. dx = F̂ dX  may
  followed by
be obtained either by first stretching the element by Û i.e. dx = ÛdX,
CONTENTS 11

Figure 3. Polar decomposition applied to a given deformation.

a rotation R̂, i.e. dx = R̂dx or, equivalently, by applying a rigid rotation R̂ first, i.e.
dx = R̂dX
 followed later by a stretching V̂ , i.e. dx = V̂ dx (seeFig. 3).

4. Stress

In continuum mechanics we must consider two systems of forces acting on a given region
of a material body:
• the body forces. They are dependent on the external fields acting on the elastic
body and they are described by the vector field b(x) representing their density on
the volume in the current configuration. The physical meaning of such a density of
forces can be summed up stating that the total force dFv applied to a small volume
dx centered on the point x is given by dFv = b(x)dx. A typical example is given by
the gravitational forces proportional to the mass of the region under consideration.
In this case we can write dFv = g dm where g is the gravitational acceleration and
dm is the mass of the volume dx. If we define ρ = dm d

x
as the density of the body,
we simply obtain b(x) = ρg .
• the surface forces. In continuum mechanics we are additionally concerned with the
interaction between neighbouring portions of the interiors of deformable bodies. In
reality such an interaction consists of complex interatomic forces, but we make the
simplifying assumption that the effect of all such forces across any given surface
may be adequately represented by a single vector field defined over the surface. It
is important to observe that the nature of the forces exerted between two bodies
in contact is identical to the nature of the actions applied between two portions of
CONTENTS 12
x3
C
dA1
n = (n1 , n2 , n3 )

dA2 dAn
P
B x2
dA3
A

x1

Figure 4. Cauchy tetrahedron on a generic point P.

the same body, separated by an ideal surface.

In order to begin the mathematical descriptions of the forces, it is useful to introduce
the following notation for the surface force dFs applied to the area element ds (with unit
normal vector n) of the deformed configuration
dFs = f (x, n, t) ds (4.1)
where f assumes the meaning of a density of forces distributed over the surface. By
definition, the force dFs is applied by the region where the unit vector n is directed to
the other region beyond the ideal surface (or interface). We can now recall the Cauchy
theorem on the existence of the stress tensor describing the distribution of the surface
forces in a given elastic body. More precisely, we can say that a tensor T̂ exists such
that
f (x, n, t) = T̂ (x, t)n (4.2)
where n is the external normal unit vector to the surface delimiting the portion of body
subjected to the force field f. The quantity T̂ has been called Cauchy stress tensor or
simply stress tensor. This very important result has been firstly published by Cauchy in
1827 in the text “Exercices de mathématique”. The forces applied to the area element
can be therefore written in the following form
dFs = T̂ (x)nds (4.3)
dF
or, considering the different components dss,i = Tij nj . So, we may identify the stress
tensor T̂ with a sort of vector pressure. Its physical unit is therefore the Pa (typical
values in solid mechanics range from MPa to GPa). The proof of the Cauchy theorem
can be performed as follows.
We consider a generic point P in the deformed configuration and a small tetrahedron
as described in Fig. 4. The oblique plane is defined by a unit vector n and by the distance
CONTENTS 13

dh from P. The faces of the tetrahedron have areas dA1 , dA2 , dA3 and dAn and the
outgoing normal unit vectors are −E  1 , −E
 2 , −E
 3 and n (where the vectors E
 i belong
to the reference base). We define f1 , f2 , f3 and fn as the surface forces acting on each
face and b as the body force distributed over the volume. The motion equation is
fn dAn + f1 dA1 + f2 dA2 + f3 dA3 + b dv = ρadv (4.4)
where a is the acceleration of the tetrahedron

with mass ρdv. From Eq.(4.1) we
can identify fn = f (n, x, t) and fk = f −E
 k , x, t , ∀ k = 1, 2, 3. Moreover,
dAi = ni dAn , ∀ i = 1, 2, 3 and dv = 13 dAn dh, so we can write Eq. (4.4) as follows
(sum over j)

 1 1
 
f (n, x, t) + f −Ej , x, t nj + b dh = ρ a dh
 (4.5)
3 3
In the limit of dh → 0 we obtain (sum over j)


f (n, x, t) = −f −E  j , x, t nj (4.6)

Now we can use the previous result with n = E  i (for any i = 1, 2, 3), by obtaining



f E i , x, t = −f −E
 i , x, t (4.7)
This is a sort of third law of the dynamics written in term of surface forces. Now,
Eq.(4.6) can be simply rewritten as (sum over j)


f (n, x, t) = f E j , x, t nj (4.8)

This result shows that the surface force f on a given plane is determined by the three
surface forces on the three coordinate planes; in components


fi (n, x, t) = f (n, x, t) · E
 i = f E j , x, t · E  i nj = Tij nj (4.9)


where the Cauchy stress T̂ is represented by Tij = f E  j , x, t · E
 i . To better understand
the physical meaning of the stress tensor we consider the cubic element of volume shown
in Fig.5, corresponding to an infinitesimal portion dV = (dl)3 taken in an arbitrary solid
body. The six faces of the cube have been numbered as shown in Fig.5. We suppose
that a stress T̂ is applied to that region: the Tij component represents the pressure
applied on the j-th face along the i-th direction.
The Cauchy stress tensor is the most natural and physical measure of the state
of stress at a point in the deformed configuration and measured per unit area of the
deformed configuration. It is the quantity most commonly used in spatial or Eulerian
description of problems in continuum mechanics. Some other stress measures must
be introduced in order to describe continuum mechanics in the Lagrangian formalism.
From Cauchy formula, we have dFs = T̂ nds, where T̂ is the Cauchy stress tensor. In a
similar fashion, we introduce a stress tensor T̂ 1PK , called the first Piola-Kirchhoff stress
tensor, such that dFs = T̂ 1PK NdS.  By using the Nanson formula nds = J F̂ −T N  dS we
obtain
dFs = T̂ J F̂ −T N
 dS = T̂ 1PK N
 dS (4.10)
CONTENTS 14
x3

T33
3 T13 T23
2
T31 T32
T11 T12 T22
T21
x1 1 x2

Figure 5. Geometrical representation of the stress tensor T̂ : the Tij component

represents the pressure applied on the j-th face of the cubic volume along the i-th
direction.

and therefore
T̂ 1PK = J T̂ F̂ −T (4.11)
Sometimes it is useful to introduce another state of stress T̂ 2PK , called the second Piola-
Kirchhoff stress tensor, defined as F̂ −1 dFs = T̂ 2PK N
 dS. We simply obtain

F̂ −1 dFs = F̂ −1 T̂ J F̂ −T N
 dS = T̂ 2PK N
 dS (4.12)
and therefore
T̂ 2PK = J F̂ −1 T̂ F̂ −T = F̂ −1 T̂ 1PK (4.13)
The stress tensors T̂ 1PK and T̂ 2PK will be very useful for the finite elasticity theory
described within the Lagrangian formalism.

5. Continuity equation

The first balance equation of the continuum mechanics concerns the mass distribution.
 in the Lagrangian formalism and ρ (x, t)
We define the mass density: we will use ρ0 (X)
in the Eulerian description. The total mass of the region Pt is given by

m (Pt ) = ρ(x, t)dx (5.1)
Pt

The consevation of the mass gives

  
 X  or d
ρ(x, t)dx = ρ0 (X)d ρ(x, t)dx = 0 (5.2)
Pt P0 dt Pt
The first equality in Eq.(5.2) can be also written
 

ρJdX = 
ρ0 dX (5.3)
P0 P0
CONTENTS 15

and we simply obtain

ρJ = ρ0 (5.4)
On the other hand, from the second equality in Eq.(5.2) we have

  
∂ρ
ρ̇ + ρ∇

x · v dx = +∇

x · (ρv ) dx = 0 (5.5)
Pt Pt ∂t

and therefore we obtain two forms of the continuity equation

ρ̇ + ρ∇ 
x · v = 0 (5.6)
∂ρ 
+ ∇
x · (ρv ) = 0 (5.7)
∂t
It is important for the following applications to evaluate expressions of this kind:
d

dt Pt
ρ(x, t)Ψ(x, t)dx; to this aim we use the Reynolds theorem with φ = ρΨ
 
 
d
ρΨdx = 
x · v dx =
ρ̇Ψ + ρΨ̇ + ρΨ∇ ρΨ̇dx (5.8)
dt Pt Pt Pt

It means that, when there is the density in the integrand, the time derivative must be
applied directly to the function Ψ.

6. Balance equations: Euler description

The other two important balance equations can be derived by the principles of linear
and angular momentum. When dealing with a system of particles, we can deduce from
Newton’s laws of motion that the resultant of the external forces is equal to the rate of
change of the total linear momentum of the system. By taking moments about a fixed
point, we can also show that the resultant moment of the external forces is equal to the
rate of change of the total moment of momentum. Here we define the linear and angular
momentum density for a continuum and we introduce balance laws for these quantities.
We consider a portion Pt in a material body and we define P as its linear momentum,
F as the resultant of the applied forces, L  as the total angular momentum and, finally,
M as the resultant moment of the applied forces. The standard principles for a system
of particles can be written as follows
dP dL
= F =M  (6.1)
dt dt
We start with the first principle, applied to the portion of body contained to the region
Pt , limited by the closed surface ∂Pt
  
d bdx
ρv dx = T̂ nds + (6.2)
dt Pt ∂Pt Pt

where we have utilized the decomposition of the forces (body forces and surface forces)
as described in the previous section. The previous equation can be simplified by means
of Eq.(5.8) and the divergence theorem, by obtaining
  
˙
ρv dx = ∇
x · T̂ dx +
 bdx (6.3)
Pt Pt Pt
CONTENTS 16

Since the volume Pt is arbitrary, we easily obtain the first balance equation for the
elasticty theory (Eulerian description)

x · T̂ + b = ρv˙
∇ (6.4)
This is the basic linear momentum equation of continuum mechanics. We remark that
the divergence of a tensor is applied on the second index; in fact, in components, we
simply obtain
∂Tji
+ bj = ρv˙j (6.5)
∂xi
Further, we observe that
∂v ∂v ∂v 1 

v˙ = + · v = + ∇
x (v · v ) + ∇ 
x ∧ v ∧ v (6.6)
∂t ∂x ∂t 2
and, therefore Eq.(6.4) is equivalent to

 ∂v ∂v
∇
x · T̂ + b = ρ
 + · v (6.7)
∂t ∂x
or


 ∂v 1 
∇
x · T̂ + b = ρ
 + ∇
x (v · v ) + ∇
x ∧ v ∧ v
 (6.8)
∂t 2
Now, we consider the principle of the angular momentum. For the region Pt such
a balance equation can be written in the following form
 
 
d
x ∧ ρv dx = x ∧ T̂ n ds + x ∧ b dx (6.9)
dt Pt ∂Pt Pt

As before, the surface integral can be simplified with the application of the divergence
theorem, by obtaining, after some straightforward calculations

  
∂Tkp
x × T̂ n ds = Tkh + xh ηhkj ej dx (6.10)
∂Pt Pt ∂xp
So, the second balance equation assumes the form
   
∂Tkp
xh ρv̇k − − bk − Tkh ηhkj ej dx = 0 (6.11)
Pt ∂xp
The term in bracket is zero because of the first balance equation. Therefore, we obtain

T η e dx = 0 or, equivalently, Tkh ηhkj = 0. Finally, the second principle leads to
Pt kh hkj j

Tij = Tji (6.12)

In other words, we may state that the principle of the angular momentum assures the
symmetry of the Cauchy stress tensor.
CONTENTS 17

7. Balance equations: Lagrange description

In finite elasticity theory the Lagrangian description is the most important

 point of
view since it allows to determine the exact transformation x = Ft X between the

reference and the actual configurations. In the case of finite deformations (arbitrarily
large), the Piola-Kirchhoff stress tensors above defined are used to express the stress
relative to the reference configuration. This is in contrast to the Cauchy stress tensor
which expresses the stress relative to the current configuration. In order to obtain the
Lagrangian equations of motion it is useful to introduce the so-called Piola transform
W (X,
 t) (which is a Lagrangian vector field) of a given Eulerian vector field w(
 x, t)


 x, t) ⇒ W
w(  t) = J F̂ −1 w(F
 (X,  t X  , t) (7.1)
An important relation gives the relationship between the divergence of the two fields: of
course, the divergence of W (X,
 t) is calculated with respect to the Lagrangian variables
 while that of w(
X  x, t) is calculated with respect to the Eulerian variables x
 
∂Wi ∂ ∂Xi
∇X
· W (X, t) =
   = J ws
∂Xi ∂Xi ∂xs
 
∂ ∂Xi ∂Xi ∂ws
= J ws + J (7.2)
∂Xi ∂xs ∂xs ∂Xi
The first term is zero for the Piola identity given in Eq.(2.44), and therefore

∇
·W  t) = J ∂Xi ∂ws = J ∂ws

 (X, (7.3)
X
∂xs ∂Xi ∂xs
It means that we have obtained the important relation
∇  (X,

·W  t) = J ∇

x · w(
 x, t) (7.4)
X

We can also make a Piola transformation on a given index of a tensor. For example,
if Tji the Cauchy stress tensor, we may use the above tranformation on the last index.
We apply this procedure to transform the motion equation from the Eulerian to the
Lagrangian coordinates
 
∂Tji 1 ∂ ∂Xi
+ bj = ρv˙j ⇒ J Tjs + bj = ρv˙j (7.5)
∂xi J ∂Xi ∂xs
or, identifying the deformation gradient
∂  −1

J(F̂ )is Tjs + Jbj = ρJ v˙j (7.6)
∂Xi
By using the relation ρ0 = Jρ we obtain
∂  ρ
0
JTjs (F̂ −T )si + bj = ρ0 v˙j (7.7)
∂Xi ρ
Since we have defined the first Piola-Kirchhoff stress tensor as T̂ 1PK = J T̂ F̂ −T we obtain

· T̂ 1PK + ρ0 b = ρ0v˙
∇ (7.8)
X
ρ
CONTENTS 18
1 1PK T
Now, we consider the principle of the angular momentum: since T̂ = J
T̂ F̂ and
T̂ = T̂ T we obtain
T̂ 1PK F̂ T = F̂ T̂ 1PKT (7.9)
These two important results can be also expressed in terms of the second Piola-Kirchhoff
stress tensor T̂ 2PK = F̂ −1 T̂ 1PK . We simply obtain the linear momentum balance

 ρ
2PK 0
∇X
· F̂ T̂
 + b = ρ0v˙ (7.10)
ρ
and the angular momentum balance
T̂ 2PK = T̂ 2PKT (7.11)
Of course, Eqs.(7.10) and (7.11) must be completed by the constitutive equations and
by the boundary conditions.

7.1. Novozhilov formulation.

We consider the standard base of unit vectors E  2 and E
 1, E  3 in the point X  of
the reference

 configuration. Since the motion is controlled by the tranformation
x = Ft X  , the unit vectors ei in the deformed configuration are given by the direction
of the deformed coordinate lines

)
∂Ft (X
∂Xi
i
F̂ E
ei = = (7.12)

)
∂Ft (X
F̂ E i

∂Xi

We remark that they do not form an orthogonal base. First of all, we simply obtain the
i
norm of F̂ E


  
 

F̂ E
 i
=  i · F̂ E
F̂ E  i = Fki Fki = F̂ T F̂ = Cii (7.13)
ii

where Ĉ is the right Cauchy tensor. We define the unit vectors n1 , n2 and n3
perpendicular to the planes (e2 , e3 ), (e1 , e3 ) and (e1 , e2 ). It means that we can write




F̂ Ei ∧ F̂ Ej 
1 ei ∧ ej 1
nk = ηkij = ηkij

 (7.14)
2
ei ∧ ej
2
F̂ Ei ∧ F̂ Ej

 



 i ∧ F̂ E
Now, we start with the calculation of
F̂ E j


 

F̂ E
 i ∧ F̂ E j
= ηkst Fsi Ftj ηkab Fai Fbj

= (δsa δtb − δsb δta ) Fsi Ftj Fai Fbj

= Cii Cjj − Cij2 (7.15)
We can also write
dsk 
= Cii Cjj − Cij2 (7.16)
dSk
CONTENTS 19

where the indices i and j are complementary to k and dSk and dsk are the surface
elements


the reference and current configuration having unit normal vector nk . Since
 in
  j = ηqst Fsi Ftj E
F̂ Ei ∧ F̂ E  q , we therefore obtain

1 q
ηqst Fsi Ftj E
nk = ηkij  (7.17)
2 C C − C2
ii jj ij

Since ηqst Fsi Ftj Fqa = Jηaij we can simply write ηqst Fsi Ftj = Jηaij (F̂ −1 )aq ; this result
can be used in Eq.(7.17) to yield
1 Jηaij (F̂ −1 )aq E
q
nk = ηkij  (7.18)
2 C C − C2
ii jj ij

When k is fixed the indices i and j can assume two couples of values [if k =1 we have
(i, j)=(2,3) or (3,2), if k =2 we have (i, j)=(1,3) or (3,2) and if k =3 we have (i, j)=(2,1)
or (1,2)] and the index a must assume the value k. At the end we evetually obtain
J(F̂ −1 )kq E
q dSk
nk =  = J(F̂ −1 )kq E
q (7.19)
Cii Cjj − Cij 2 ds k

where the indices i and j are complementary to k (there is not the sum on k). We may
consider the forces acting on the three deformed coordinate planes (e2 , e3 ), (e1 , e3 ) and
(e1 , e2 ) (having normal unit vectors n1 , n2 and n3 , respectively) through the expressions
J(F̂ −1 )kq T E
q dSk
T nk =  = J(F̂ −1 )kq T E
q (7.20)
Cii Cjj − Cij 2 ds k

 i and ei as follows

These vectors can be represented on both the base E
E 
T nk = σsk Es (7.21)

T nk = σsk
e
es (7.22)
where, since E  2 and E
 1, E  3 is an orthonormal base, we have
E
σsk  s = dSk J(F̂ −1 )kq T E
= T nk · E  s = dSk J(F̂ −1 )kq Tsq
q · E (7.23)
dsk dsk
E e
Moreover, we have the following relation between σsk and σsk
j · E
s
E
σsk = T nk · E  s = σ e F̂
 s = σ e ej · E E 1
=  Fsj σjk
e
(7.24)
jk jk
Cjj Cjj
E e
The representations σsk and σsk have been introduced by Novozhilov in his pioneering
book on nonlinear elasticity. The Lagrangian equation of motion can be written as (see
Eq.(7.6))
∂ 
J(F̂ −1 )kq Tsq + Jbs = ρJ v˙s (7.25)
∂Xk
CONTENTS 20
E
and then it can be expressed in terms of σsk

∂ dsk E
σ + Jbs = ρJ v˙s (7.26)
∂Xk dSk sk
e
or in terms of σsk
 
∂ dsk 1
 Fsj σjk e
+ Jbs = ρJ v˙s (7.27)
∂Xk dSk Cjj

Finally, since it is evident that Cjj = dlj /dLj , we can state the Lagrangian equations
of motion in the Novozhilov form
 ds 
k
∂ dSk e
Fsj σjk + Jbs = ρJ v˙s (7.28)
∂Xk dlj
dLj

8. Nonlinear constitutive equations

The constitutive equations represent the relation between the stress and the strain and,
therefore, they depend on the material under consideration. Here we prove that there
is a strong conceptual connection between the constitutive equations and the energy
balance for a continuum body. We start from the motion equation in the Eulerian
formalism and we multiply both sides to the velocity component vj
∂Tji
vj + vj bj = ρvj v˙j (8.1)
∂xi
This expression can also be written as
∂ (vj Tji ) ∂vj
− Tji + vj bj = ρvj v˙j (8.2)
∂xi ∂xi
∂v
The Eulerian velocity gradient Lji = ∂xji can be decomposed in the symmetric and
skew-symmetric parts
   
∂vj 1 ∂vj ∂vi 1 ∂vj ∂vi
Lji = = + + − = Dji + Wji (8.3)
∂xi 2 ∂xi ∂xj 2 ∂xi ∂xj
     
symmetric skew−symmetric

where D̂ is the rate of deformation tensor and Ŵ is the spin tensor. Therefore, the
energy balance equation assumes the local form form
∂ (vj Tji )
− Tji Dji + vj bj = ρvj v˙j (8.4)
∂xi
By using the property in Eq. (5.8) we also obtain the global version on the region Pt
   
d 1
ρvj vj dx + TjiDji dx = Tjini vj dx + vj bj dx (8.5)
dt Pt 2 Pt ∂Pt Pt

The second side of this balance represents the power input (product between force and
velocity) consisting of the rate of work done by external surface tractions Tji ni per unit
area and body forces bj per unit volume of the region Pt bounded by ∂Pt . Since the
time-rate of change of the total energy is equal to the the rate of work done by the
CONTENTS 21

external forces (first principle of thermodynamics without thermal effects), we identify

the first side as dE/dt where E is the total energy contained in Pt . Moreover, the total
energy can be written as E = K + U where K is the kinetic energy and U is the potential

energy. Since K = Pt 12 ρvj vj dx is the standard kinetic energy, we identify

dU
= Tji Djidx (8.6)
dt Pt

We define the energy density U per unit volume in the reference configuration and
therefore ρρ0 U is the energy density per unit volume in the current configuration. We
obtain

ρ
U= Udx (8.7)
Pt ρ0

By drawing a comparison between Eqs.(8.6) and (8.7) we obtain

 
d ρ
Tji Djidx = Udx (8.8)
Pt dt Pt ρ0
By using the property in Eq. (5.8) we obtain
ρ
U̇ = Tji Dji (8.9)
ρ0
We introduce now a general statement affirming that the strain energy function U
depends upon the deformation gradient F̂ : therefore, we have U = U(F̂ ). This relation
can be simplified by means of the principle of material objectivity (or material frame
indifference), which says that the energy (and the stress) in the body should be the same
regardless of the reference frame from which it is measured. If we consider a motion
x = Ft (X)
 we obtain a corresponding deformation gradient F̂ ; on the other hand, if we
 + c(t) (where Q̂(t) is an orthogonal
consider a roto-translated motion x = Q̂(t)Ft (X)
matrix and c(t) is an arbitrary vector), then the deformation gradient is Q̂F̂ . In both
cases we must have the same energy and therefore
U(F̂ ) = U(Q̂F̂ ) ∀Q̂ : Q̂Q̂T = Iˆ (8.10)
Now, the deformation gradient F̂ can be decomposed through F̂ = R̂Û by obtaining
U(F̂ ) = U(Q̂R̂Û ) ∀Q̂ : Q̂Q̂T = Iˆ (8.11)
By imposing Q̂ = R̂T we have U(F̂ ) = U(Û ) and, since Û 2 = Ĉ, we finally obtain the
dependance
U(F̂ ) = U(Ĉ) (8.12)
where Ĉ is the right Cauchy tensor. The choice of Ĉ as an independent variable
is convenient because, from its definition, Ĉ = F̂ T F̂ is a rational function of the
deformation gradient F̂ . Now we can calculate U̇ as follows
∂U ∂U

U̇ = Ċij = FkiḞkj + Ḟki Fkj (8.13)
∂Cij ∂Cij
CONTENTS 22

We remember that Ḟkj = Lks Fsj (see Eq.(2.20)) and we obtain

∂U
U̇ = (Fki Lks Fsj + Lks Fsi Fkj )
∂Cij
 
∂U T ∂U T T ∂U T
= tr F̂ L̂F̂ + F̂ L̂ F̂ = tr 2 F̂ D̂ F̂ (8.14)
∂ Ĉ ∂ Ĉ ∂ Ĉ
where D̂ is the rate of deformation tensor defined as the symmetric part of the velocity
gradient L̂. Through the comparison of Eqs.(8.9) and (8.14) we obtain
 
ρ0 ∂U T
tr T̂ D̂ = tr 2 F̂ D̂ F̂ (8.15)
ρ ∂ Ĉ
Further, from the commutation rule tr(ÂB̂) = tr(B̂ Â) of the trace operation we arrive
at the following relationships, which must be satisfied for any possible D̂
 
ρ0 ∂U T
tr T̂ D̂ = tr 2F̂ F̂ D̂ (8.16)
ρ ∂ Ĉ
Therefore, we obtain the formal connection between the constitutive equation (giving
the Cauchy stress tensor) and the strain energy function in the form
ρ ∂U T
T̂ = 2 F̂ F̂ (8.17)
ρ0 ∂ Ĉ
Similarly for the first Piola-Kirchhoff stress tensor we obtain
∂U
T̂ 1PK = J T̂ F̂ −T = 2F̂ (8.18)
∂ Ĉ
and finally for the second Piola-Kirchhoff stress tensor
∂U
T̂ 2PK = F̂ −1 T̂ 1PK = 2 (8.19)
∂ Ĉ
We have proved that an arbitrarily nonlinear constitutive equation can be always written
by means of derivations of the strain energy function: it means that the strain energy
function contains the complete information about the nonlinear elastic response of a
given material. For the particular case of nonlinear isotropic material the strain energy
function U must depend only upon the invariants of the right Cauchy tensor Ĉ. We
observe that they are defined as

IC = tr Ĉ (8.20)


1
2
IIC = trĈ − tr Ĉ 2 (8.21)
2
IIIC = det Ĉ (8.22)
and therefore we have U = U(IC , IIC , IIIC ). We remember that the three invariants
define the characteristic polynomial of the tensor Ĉ


det Ĉ − λIˆ = −λ3 + λ2 IC − λIIC + IIIC (8.23)
and satisfy the so-called Cayley-Hamilton theorem
0̂ = −Ĉ 3 + IC Ĉ 2 − IIC Ĉ + IIIC Iˆ (8.24)
CONTENTS 23

It is possible to prove that

∂IC ˆ ∂IIC = IC Iˆ − Ĉ; ∂IIIC = IIIC Ĉ −1 ;
= I; (8.25)
∂ Ĉ ∂ Ĉ ∂ Ĉ
and therefore we obtain
∂U(IC , IIC , IIIC ) ∂U ∂IC ∂U ∂IIC ∂U ∂IIIC
= + +
∂ Ĉ ∂IC ∂ Ĉ ∂IIC ∂ Ĉ ∂IIIC ∂ Ĉ
∂U ˆ ∂U
ˆ  ∂U
= I+ IC I − Ĉ + IIIC Ĉ −1 (8.26)
∂IC ∂IIC ∂IIIC
This expression can be used in the Cauchy and Piola-Kirchhoff tensors given in
Eqs.(8.17), (8.18) and (8.19) in order to obtain their final form in terms of the invariants
of the right Cauchy tensor Ĉ. Sometime the
stress tensors can also be expressed in term
of the Green-Lagrange strain tensor η̂ = 12 Ĉ − Iˆ ; since 2dη̂ = dĈ, we have
ρ ∂U T ∂U ∂U
T̂ = F̂ F̂ ; T̂ 1PK = F̂ ; T̂ 2PK = (8.27)
ρ0 ∂ η̂ ∂ η̂ ∂ η̂
In this case the strain energy function U (for unit volume of the reference configuration)
may be developed in power series with respect to the components of η̂. This leads to
the expression
1 L 1 L
U(η̂) = Cijkh ηij ηkh + Cijkhnm ηij ηkh ηnm + ... (8.28)
2 6
L L
Here the Cijkh and the Cijkhnm denote the second order elastic constants (SEOC) and the
third order elastic constants (TOEC), respectively (within the Lagrangian formalism).

9. The small-strain approximation

In the infinitesimal elasticity theory the extent of the deformations is assumed small.
While this notion is rather intuitive, it can be formalized by imposing that for small
deformations F̂ is very similar to Iˆ or, equivalently, that Ĝ is very very similar to I.
ˆ
It means that both JˆL and JˆE are very small. Therefore, we adopt as an operative
definition of small deformation the relations
Tr(JˆL JˆLT )  1 and Tr(JˆE JˆET )  1 (9.1)
i.e., a deformation will be hereafter regarded to as small provided that the trace of the
product JˆL JˆLT or JˆE JˆET is negligible. It means that we can assume JˆL = JˆE = Jˆ and
that we can interchange the Eulerian and the Lagrangian variables without problems.
Here, we write all the equations with the Eulerian variables x. We observe that Jˆ can
be written as the sum of a symmetric and a skew-symmetric (antisymmetric) part as
follows
   
1 ∂ui ∂uj 1 ∂ui ∂uj
Jij = + + − = ij + Ωij (9.2)
2 ∂xj ∂xi 2 ∂xj ∂xi
     
symmetric skew−symmetric
CONTENTS 24

Figure 6. Two-dimensional geometric deformation of an infinitesimal material

element.

The meaning of the displacement gradient can be found in Fig. 6 for a two-dimensional
configuration. Accordingly, we define the (symmetric) infinitesimal strain tensor (or
small strain tensor ) as
 
1 ∂ui ∂uj
ij = + (9.3)
2 ∂xj ∂xi
and the (antisymmetric) local rotation tensor as
 
1 ∂ui ∂uj
Ωij = − (9.4)
2 ∂xj ∂xi
Such a decomposition is useful to obtain three very important properties of the small
strain tensor, which is the key quantity to determine the state of deformation of an
elastic body:
• for a pure local rotation (a volume element is rotated, but not changed in shape
and size) we have Jˆ = Ω̂ and therefore ˆ = 0. This means that the small strain
tensor does not take into account any local rotation, but only the changes of shape
and size (dilatations or compression) of that element of volume.
Let us clarify this fundamental result. We consider a point x inside a volume
element which is transformed to x + u(x) in the current configuration. Under a
pure local rotation we have: x + u(x) = R̂x, where R̂ is a given orthogonal rotation
ˆ We simply obtain u(x) = (R̂ − I)
matrix (satisfying R̂R̂T = I). ˆ x or, equivalently,
Jˆ = R̂ − I.ˆ Since the applied deformation (i.e., the local rotation) is small by
hypothesis, we observe that the difference R̂ − Iˆ is very small too. The product
JˆJˆT will be therefore negligible, leading to the following expression



∼
0 = J J = R̂ − I R̂ − I = R̂R̂T − R̂ − R̂T + Iˆ
ˆ ˆT ˆ T ˆ
CONTENTS 25

= Iˆ − R̂ − R̂T + Iˆ = −Jˆ − JˆT (9.5)

Therefore Jˆ = −JˆT or, equivalently, Jˆ is a skew-symmetric tensor. It follows that
Jˆ = Ω̂ and ˆ = 0. We have verified that a pure rotation corresponds to zero strain.
In addition, we remark that the local rotation of a volume element within a body
cannot be correlated with any arbitrary force exerted in that region (the forces are
correlated with ˆ and not with Ω̂): for this reason the infinitesimal strain tensor is
the only relevant object for the analysis of the deformation due to applied loads in
elasticity theory.
• the infinitesimal strain tensor allows for the determination of the length variation
of any vector from the reference to the current configuration. By defining nn as
the relative length variation in direction n, we have from Table 2
nn = n · ˆn (9.6)
If n is actually any unit vector of the reference frame, it is straightforward to
attribute a geometrical meaning to the components 11 , 22 , 33 of the strain tensor.
Since nn = ei · (ˆ
 ei ) = ii , they describe the relative length variations along the
three axes of the reference frame.
• the infinitesimal strain tensor allows for the determination of the angle variation
between any two vectors from the reference to the current configuration. The
variation of the angle defined by the two orthogonal directions n and t can be
obtained from Table 2
γnt = 2n · ˆt (9.7)
The present result is also useful for giving a direct geometrical interpretation of the
components 12 , 23 and 13 of the infinitesimal strain tensor. As an example, we
take into consideration the component 12 and we assume that n = e1 and t = e2 .
The quantity γnt represents the variation of a right angle lying on the plane (x1 , x2 ).
Since 12 = e1 · (ˆ
 e2 ), we easily obtain the relationship γnt = 212 = ∂u
∂x2
1
+ ∂u
∂x1
2
. In
other words, 12 is half the variation of the right angle formed by the axis x1 and
x2 . Of course, the same interpretation is valid for the other components 23 and
13 .
The result of the application of the small strain approximation on the main
quantities of the continuum mechanics is summarized in Table 3.
Knowing the ˆ tensor field within a strained (i.e., deformed) elastic body allows us

to calculate the volume change ∆V of a given region. We get ∆V = V Tr(ˆ )dx, where
V is the volume of the unstrained region.
The above discussion states that, given a displacement field u(x), the components of
the infinitesimal strain tensor are easily calculated by direct differentiation. The inverse
problem is much more complicated. Given an arbitrary infinitesimal strain tensor ˆ(x)
we could search for that displacement field u(x) generating the imposed deformation. In
general, such a displacement field may not exist. There are, however, suitable conditions
CONTENTS 26

Table 3. The small strain approximation.

Lagrangian vision Eulerian vision

JˆL = Jˆ JˆE = Jˆ
F̂ = Ĝ−1 = Iˆ + Jˆ F̂ −1 = Ĝ = Iˆ − Jˆ
η̂ = ˆ ê = ˆ
Ĉ = B̂ = Iˆ + 2ˆ  Ĉ −1 = B̂ −1 = Iˆ − 2ˆ

−1 −1
Û = V̂ = Iˆ + ˆ Û = V̂ = Iˆ − ˆ
R̂ = Iˆ + Ω̂ R̂−1 = Iˆ − Ω̂
T̂ 1PK = T̂ 2PK = ∂U∂ˆ

T̂ = ∂U∂ˆ


under which the solution of this inverse problem is actually found. These conditions are
written in the very compact form
∂ 2 ij
ηqki ηphj =0 (9.8)
∂xk ∂xh
where η’s are the Levi-Civita permutation symbols. Eqs.(9.8) are known as infinitesimal
strain compatibility equations or Beltrami Saint-Venant equations. The balance
equations assume the standard form
∂Tji ∂ 2 uj
+ bj = ρ 2 (9.9)
∂xi ∂t
Tij = Tji (9.10)
The principles of linear and angular momentum, the definition of strain and its
compatibility conditions need to be supplemented by a further set of equations, known
as constitutive equations, which characterize the constitution of the elastic solid body.
In the case of small deformation we can write
∂U
T̂ = T̂ 1PK = T̂ 2PK = (9.11)
∂ˆ
where the strain energy function is expressed as U = U(ˆ ). Such a strain energy function
U may be developed in power series with respect to the components of ˆ. This leads to
the expression
1 1
U(η̂) = Cijkh ij kh + Cijkhnmij kh nm + ... (9.12)
2 6
Here the Cijkh and the Cijkhnm denote the second order elastic constant (SEOC) and
the third order elastic constant (TOEC), respectively, with reference to the small strain
tensor. We can determine the relations with the elastic constants defined in Eq.(8.28):
to this aim, we consider an homogeneous deformation with F̂ = Iˆ + ˆ (i.e. with Ω = 0
or Jˆ = ˆ) and we obtain η̂ = ˆ + 12 ˆ2 ; so, by imposing U(ˆ
) = U(η̂) we eventually obtain
L
Cijkh = Cijkh (9.13)
L 3 L 3 L
Cijkhnm = Cijkhnm + Cimkh δjn + Cijkm δhn (9.14)
2 2
CONTENTS 27

The linear law for the relation between stress and strain is called the generalized Hooke’s
law. The general form of writing Hooke’s law is as follows
Tij = Cijkh kh (9.15)
where Cijkh are constants (for homogeneous materials). Eq.(9.15) is of general validity,
including all the possible crystalline symmetry or, in other words, any kind of anisotropy.
The tensor of the elastic constants satisfies the following symmetry rules: 1) symmetry
in the first pair of indices: since Tij = Tji we have Cijkh = Cjikh : 2) symmetry in the
last pair of indices: since kh = hk we may take Cijkh = Cijhk ; 3) symmetry between the
first pair and the last pair of indices: energetic considerations leads to Cijkh = Ckhij . At
the end Cijkh has at most 21 independent components rather than the 34 = 81 which,
as a general fourth-rank tensor, it might have had. In the case of a linear and isotropic
material we have
E νE
T̂ = ˆ + Iˆ Tr (ˆ
) (9.16)
1+ν (1 + ν)(1 − 2ν)
where E and ν are the Young modulus and the Poisson ratio, respectively. We can also
introduce the Lamé coefficients µ and λ as follows
E νE
µ= λ= (9.17)
2(1 + ν) (1 + ν)(1 − 2ν)
Therefore, Eq.(9.16) assumes the standard form
ˆ )
T̂ = 2µˆ + λITr(ˆ (9.18)
When we are dealing with a linear, isotropic and homogeneous material the governing
equations of the elasticity theory can be summed up as follows

 2
(λ + µ) ∇ ∇  · u + µ∇  2u + b = ρ ∂ u (9.19)
∂t2
This is an equation of motion where the displacement field is the single unknown, which
have been called Lamé or Navier equation. Such a motion equation for a isotropic
elastic

bodycan be
also written in a different form by utilizing the general property
∇× ∇
  × u = ∇ ∇  · u − ∇  2 u, which holds for the differential operators. The result
is

 2
(λ + µ) ∇ × ∇  × u + (λ + 2µ)∇  2u + b = ρ ∂ u (9.20)
∂t2
Both Eq. (9.19) and Eq. (9.20) are linear partial differential equations of the second
order with a vector field u (r) as unknown. In order to find a solution of Eq. (9.19)
or Eq. (9.20) we must impose some boundary conditions depending on the physical
problem under consideration. If we consider a body with an external surface S, a first
type of boundary condition fixes the values of the displacement field on this surface at
any time. It means that u = u(x, t) for any x ∈ S and for any t in a given interval. When
the entire external surface is described by these conditions we say that we are solving
an elastic problem of the first kind (Dirichlet). A second kind of boundary conditions
fixes the stress applied on the external surface. It means that Tij nj = fi (x, t) for any
CONTENTS 28

x ∈ S and for any t in a given interval. When the entire external surface is described
by these conditions we say that we are solving an elastic problem of the second kind
(Neumann). Finally, a third case can be defined by dividing the surface S in two parts
and by applying the Dirichlet conditions to the first part and the Neumann conditions
to the second part. In this case we say that we are solving an elastic problem of the
third kind, subjected to mixed boundary conditions.