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1 Tensors
1
Tensors
1.1 Introduction
As seen previously in the introductory chapter, the goal of continuum mechanics is to
establish a set of equations that governs a physical problem from a macroscopic
perspective. The physical variables featuring in a problem are represented by tensor fields,
in other words, physical phenomena can be shown mathematically by means of tensors
whereas tensor fields indicate how tensor values vary in space and time. In these equations
one main condition for these physical quantities is they must be independent of the
reference system, i.e. they must be the same for different observers. However, for matters
of convenience, when solving problems, we need to express the tensor in a given
coordinate system, hence we have the concept of tensor components, but while tensors are
independent of the coordinate system, their components are not and change as the system
change.
In this chapter we will learn the language of TENSORS to help us interpret physical phenomena.
These tensors can be classified according to the following order:
Zeroth-Order Tensors (Scalars): Among some of the quantities that have magnitude but
not direction are e.g.: mass density, temperature, and pressure.
First-Order Tensors (Vectors): Quantities that have both magnitude and direction, e.g.:
velocity, force. The first-order tensor is symbolized with a boldface letter and by an arrow
r
at the top part of the vector, i.e.: • .
Second-Order Tensors: Quantities that have magnitude and two directions, e.g. stress and
strain. The second-order and higher-order tensors are symbolized with a boldface letter.
In the first part of this chapter we will study several tools to manage tensors (scalars,
vectors, second-order tensors, and higher-order tensors) without heeding their dependence
NOTES ON CONTINUUM MECHANICS
10
on space and time. At the end of the chapter we will introduce tensor fields and some field
operators which can be used to interpret these fields.
In this textbook we will work indiscriminately with the following notations: tensorial,
indicial, and matricial. Additionally, when the tensors are symmetrical, it is also possible to
represent their components using the Voigt notation.
1.2 Algebraic Operations with Vectors
There now follows a brief review of vectors, in the Euclidean vector space (E ) , so that we
may become acquainted with the nomenclature used in this textbook.
r
r
Addition: Let a , b be arbitrary vectors, we can show the sum of adding them, (see Figure
r
1.1 (a)), with a new vector ( c ) thus defined as:
r r r r r
c =a+b=b+ a
r
a
(1.1)
r
d
r
c
r
a
r
−b
r
b
a)
r
c
r
b
b)
Figure 1.1: Addition and subtraction of vectors.
r r
Subtraction: The subtraction between two arbitrary vectors ( a , b ), (see Figure 1.1 (b)), is given
as follows:
r r r
d=a−b
r
r r
Considering three vectors a , b and c the following properties are satisfied:
r r r r r
r r
r r
(a + b) + c = a + (b + c ) = a + b + c
(1.2)
(1.3)
r
r
Scalar multiplication: Let a be a vector, we can define the scalar multiplication with λa .
r
The product of this operation is another vector with the same direction of a , and whose
length and orientation is defined with the scalar λ as shown in Figure 1.2.
λ =1
r
a
λ >1
r
a
λ<0
r
λa
0 < λ <1
r
a
r
λa
r
a
r
λa
Figure 1.2: Scalar multiplication.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
11
Scalar Product: The Scalar Product (also known as the dot product or inner product) of two
r r
r r
vectors a , b , denoted by a ⋅ b , is defined as follows:
r r
r r
γ = a ⋅ b = a b cos θ
(1.4)
where θ is the angle between the two vectors, (see Figure 1.3(a)), and • represents the
Euclidean norm (or magnitude) of • . The result of the operation (1.4) is a scalar.
r r r r
r r
Moreover, we can conclude that a ⋅ b = b ⋅ a . The expression (1.4) is also true when a = b ,
therefore:
r r r r
r r r r
r
=0 º
a ⋅ a = a a cos θ θ
→ a ⋅ a = a a ⇒ a
r
2
r r
= a⋅a
(1.5)
r r
Hence, the norm of a vector is a = a ⋅ a .
r
Unit Vector: A unit vector, associated with the a -direction, is shown with a â , which has
r
the same direction and orientation of a . In this textbook, the hat symbol ( •ˆ ) denotes a
r
unit vector. Thus, the unit vector, â , codirectional with a , is defined as:
r
ˆa = a
r
a
(1.6)
r
r
where a represents the norm (magnitude) of a . If â is the unit vector, then the
following must be true:
aˆ = 1
(1.7)
Zero Vector (or Null Vector): The zero vector is represented by a:
r
0
(1.8)
r
r
Projection Vector: The projection vector of a onto b , (see Figure 1.3(b)), is defined as:
r
r
r
r
proj br a = proj br a bˆ Projection vector of a onto b
r
r
(1.9)
r
where proj br a is the projection of a onto b , and b̂ is the unit vector associated with the
r
r
b -direction. The magnitude of proj br a is obtained by means of the scalar product:
r r
r
r
proj br a = a ⋅ bˆ Projection of a onto b
(1.10)
So, taking into account the definition of the unit vector, we obtain:
r r
r
⋅b
a
projbr a = r
b
(1.11)
r
Then, the projection vector, proj br a , can be calculated by:
r r
r
a
⋅b
proj br a = r bˆ =
b
r r
a⋅b
r
b
r
r r
b a⋅b r
r = r 2 b
b
b
{
(1.12)
scalar
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
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0≤θ≤π
r
a
r
a
θ
r
b
θ
b̂
r r r r
a ⋅ b = a b cos θ
a) Scalar product
b̂
.
r
projbr a
r
b
b) Projection vector
Figure 1.3: Scalar product and projection vector.
r
r
Orthogonality between vectors: Two vectors a and b are orthogonal if the scalar product
between them is zero, i.e.:
r r
a⋅b = 0
(1.13)
r r
Vector Product (or Cross Product): The vector product of two vectors, a , b , results in
r
another vector c , which is perpendicular to the plane defined by the two input vectors,
(see Figure 1.4). The vector product has the following characteristics:
ƒ
ƒ
ƒ
Representation:
r r
r r r
c = a ∧ b = −b ∧ a
(1.14)
r
r
r
The vector c is orthogonal to the vectors a and b , thus:
r r r r
a⋅c = b ⋅c = 0
r
The magnitude of c is defined by the formula:
r
r r
c = a b sin θ
(1.15)
(1.16)
r
r
where θ measures the smallest angle between a and b , (see Figure 1.4).
r
r
The magnitude of the vector product a ∧ b is geometrically expressed as the area of the
parallelogram defined by the two vectors, (see Figure 1.4):
r r
A= a∧b
(1.17)
Therefore, the triangle area defined by the points OCD , (see Figure 1.4 (a)), is:
1 r r
a∧b
(1.18)
2
r
r
r
r
If a and b are linearly dependent, i.e. a = αb with α denoting a scalar, the vector product
r r r
r r
of two linearly dependent vectors becomes a zero vector, a ∧ b = αb ∧ b = 0 .
r r r
Scalar Triple Product (or Mixed Product): Let a , b , c be arbitrary vectors, we can
AT =
define the scalar triple product as:
(
)
( )
(
)
r r r r r r r r r
a ⋅ b ∧ c = b ⋅ (c ∧ a) = c ⋅ a ∧ b = V
r r r
r r r
r r r
V = −a ⋅ c ∧ b = −b ⋅ (a ∧ c ) = −c ⋅ b ∧ a
(
)
(1.19)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
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r r r
where the scalar V represents the volume of the parallelepiped defined by a, b, c , (see
Figure 1.5).
r r r
c = a∧b
r
b
θ
..
D
A
O
r
a
O
r
b
r
c
C
a)
AT
θ
..
D
r
a
C
r r r
−c =b∧a
b)
Figure 1.4: Vector product.
If two vectors are linearly dependent then, the scalar triple product is zero, i.e.:
(
)
r r r r
a⋅ b ∧ a = 0
(1.20)
r r r r
Let a , b , c , d , be vectors and α , β be scalars, the following property is satisfied:
r r r
r r r
r
r r r
(1.21)
(α a + β b) ⋅ (c ∧ d) = α a ⋅ (c ∧ d) + β b ⋅ (c ∧ d)
r
r
r r r
r
NOTE: Some authors represent the scalar triple product as, [a, b, c ] ≡ a ⋅ b ∧ c ,
r r r r r r
r r r r r r
[b, c, a] ≡ b ⋅ (c ∧ a) , [c, a, b] ≡ c ⋅ a ∧ b . ■
(
(
)
)
V ≡ Scalar triple product
r
r
a∧b
r
c
..
V
r
b
θ
r
a
Figure 1.5: Scalar triple product.
r
r
r
Vector Triple Product: Let a , b , c be vectors, we can define the vector triple product as
r r r r
w = a ∧ b ∧ c . Then, we can demonstrate that the following relationships to be true:
(
)
(
r r r r
w =a∧ b∧c
)
(
)
(
r r r r r r
= −c ∧ a ∧ b = c ∧ b ∧ a
r r r r r r
= (a ⋅ c ) b − a ⋅ b c
( )
)
(1.22)
r
whereby it is clear that the result of the vector triple product is another vector w , belonging to
r
r
the plane Π1 formed by the vectors b and c , (see Figure 1.6).
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
14
r r
Π1 - plane defined by b , c
Π1
Π2
r r r
Π 2 - plane defined by a , b ∧ c
r
c
r
b
r
w belonging to the plane Π1
r
a
r r
b∧c
r
w
Figure 1.6: Vector triple product.
r
r
Problem 1.1: Let a and b be arbitrary vectors. Prove that the following relationship is
true:
(ar ∧ br )⋅ (ar ∧ br ) = (ar ⋅ ar )(br ⋅ br ) − (ar ⋅ br )
2
Solution:
(ar ∧ br )⋅ (ar ∧ br )
r r 2
= a∧b
2
r r
= a b sin θ
r 2 r 2
= a b sin 2 θ
r 2 r 2
= a b 1 − cos 2 θ
r 2 r 2
r 2 r 2
= a b − a b cos 2 θ
2
r 2 r 2
r r
= a b − a b cos θ
r 2 r 2 r r 2
= a b − a⋅b
r r r r
r r 2
= (a ⋅ a) b ⋅ b − a ⋅ b
)
(
(
)
(
( )
( ) ( )
)
Linear Transformation
r
r
Let u and v be arbitrary vectors, and α be a scalar, we can state F is a linear
transformation if the following is true:
ƒ
ƒ
r r
r
r
F (u + v ) = F (u) + F ( v )
r
r
F (αu) = αF (u)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
15
1
2
Problem 1.2: Given the following functions σ(ε) = Eε and ψ(ε) = Eε 2 , demonstrate
whether these functions show a linear transformation or not.
Solution:
σ(ε 1 + ε 2 ) = E [ε1 + ε 2 ] = Eε 1 + Eε 2 = σ(ε1 ) + σ(ε 2 ) (linear transformation)
σ( ε)
σ (ε 1 + ε 2 ) = σ (ε 1 ) + σ ( ε 2 )
σ (ε 2 )
σ (ε 1 )
ε1
ε2
ε
ε1 + ε 2
1
2
ψ (ε1 + ε 2 ) = ψ (ε1 ) + ψ (ε 2 ) has not been satisfied:
1
1
1
1
1
ψ(ε1 + ε 2 ) = E [ε1 + ε 2 ]2 = E ε12 + 2ε1ε 2 + ε 22 = Eε12 + Eε 22 + E 2ε1ε 2
2
2
2
2
2
= ψ ( ε 1 ) + ψ ( ε 2 ) + Eε 1 ε 2 ≠ ψ ( ε 1 ) + ψ ( ε 2 )
The function ψ(ε) = Eε 2 does not show a linear transformation because the condition
[
]
ψ ( ε)
ψ (ε1 + ε 2 )
ψ (ε1 ) + ψ (ε 2 )
ψ (ε 2 )
ψ (ε1 )
ε1
ε2
ε1 + ε 2
ε
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
16
1.3 Coordinate Systems
A tensor, which has physical meanings, must be independent of the adopted coordinate
system. Sometimes for reasons of convenience, we need to represent a tensor in a specific
coordinate system, hence, we have the concept of tensor components, (see Figure 1.7).
TENSORS
Mathematical representation of the physical
quantities
(Independent of the coordinate system)
COMPONENTS
Tensor Representation in a
Coordinate System
Figure 1.7: Tensor components.
r
Let a be a first-order tensor (vector) as shown in Figure 1.8 (a), the tensor representation
in a general coordinate system, defined as ξ1 , ξ 2 , ξ 3 , is made up of its components
( a1 , a 2 , a 3 ), (see Figure 1.8 (b)). Some examples of coordinate system are: the Cartesian
coordinate system; the cylindrical coordinate system; and the spherical coordinate system.
ξ3
ξ2
r
a ( a1 , a 2 , a 3 )
r
a
r
a
a)
b)
ξ1
Figure 1.8: Vector representation in a general coordinate system.
1.3.1
Cartesian Coordinate System
The Cartesian coordinate system is defined by three unit vectors: î , ĵ , k̂ , denoted by the
Cartesian basis, which make up an orthonormal basis. The orthonormal basis has the
following properties:
1. The vectors that make up this basis are unit vectors:
ˆi = ˆj = kˆ = 1
(1.23)
ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1
(1.24)
or:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
17
2. The unit vectors ( î , ĵ , k̂ ) are mutually orthogonal, i.e.:
ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0
(1.25)
3. The vector product between the vectors ( î , ĵ , k̂ ) is the following:
ˆi ∧ ˆj = kˆ
ˆj ∧ kˆ = ˆi
;
;
kˆ ∧ ˆi = ˆj
(1.26)
The direction and orientation of the orthonormal basis can be obtained using the righthand rule as shown in Figure 1.9.
ˆj ∧ kˆ = ˆi
ˆi ∧ ˆj = kˆ
ĵ
kˆ ∧ ˆi = ˆj
ĵ
ĵ
î
î
î
k̂
k̂
k̂
(1.27)
Figure 1.9: The right-hand rule.
1.3.2
Vector Representation in the Cartesian Coordinate
System
r
The vector a , (see Figure 1.10), in the Cartesian coordinate system, is represented by its
different components ( a x , a y , a z ) and by the Cartesian bases ( î , ĵ , k̂ ) as:
r
a = a x ˆi + a y ˆj + a z kˆ
(1.28)
y
ay
ĵ
k̂
r
a
î
ax
x
az
z
Figure 1.10: Cartesian coordinate system.
r
r
r
Let a , b , c be arbitrary vectors, we can describe some vector operations in the Cartesian
coordinate system, as follows:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
18
ƒ
r r
The scalar product a ⋅ b becomes a scalar, which is defined in the Cartesian
system as:
r r
a ⋅ b = (a x ˆi + a y ˆj + a z kˆ ) ⋅ (b x ˆi + b y ˆj + b z kˆ ) = (a x b x + a y b y + a z b z )
r r
r
(1.29)
Thus, it is true that a ⋅ a = a x a x + a y a y + a z a z = a 2x + a 2y + a 2z = a .
2
NOTE: The projection of a vector onto a given direction was established in the
equation (1.10), thus defining the component concept. For example, if we want to
know the vector component along the y -direction, all we need to do is calculate:
ƒ
ƒ
r
a ⋅ ˆj = (a x ˆi + a y ˆj + a z kˆ ) ⋅ (ˆj) = a y .■
r
The norm of a is:
r
a = a 2x + a 2y + a 2z
r
Then, the unit vector codirectional with a is:
r
ax
a
ˆi +
aˆ = r =
2
2
2
a
ax + a y + az
ƒ
ƒ
(1.30)
ay
a 2x + a 2y + a 2z
ˆj +
az
a 2x + a 2y + a 2z
kˆ
(1.31)
The zero vector is:
r
0 = 0 ˆi + 0 ˆj + 0 kˆ
r
r
Addition: The vector sum of a and b is represented by:
(1.32)
r r
a + b = (a x ˆi + a y ˆj + a z kˆ ) + (b x ˆi + b y ˆj + b z kˆ ) = (a x + b x ) ˆi + a y + b y
r
r
ƒ Subtraction: The difference between a and b is:
r r
a − b = (a x ˆi + a y ˆj + a z kˆ ) − (b x ˆi + b y ˆj + b z kˆ ) = (a x − b x ) ˆi + a y − b y
r
ƒ Scalar multiplication: The resulting vector defined by λa
r
λa = λa x ˆi + λa y ˆj + λa z kˆ
r r
ƒ The vector product ( a ∧ b ) is evaluated as:
ˆi
r r r
c = a ∧ b = ax
(
) ˆj + (a
z
+ b z ) kˆ
(1.33)
(
) ˆj + (a
z
− b z ) kˆ
(1.34)
ˆj
ay
is:
kˆ
a y az
a ay
ˆi − a x a z ˆj + x
kˆ
az =
by bz
bx b y
bx bz
bx b y bz
= (a y b z − a z b y )ˆi − (a xb z − a z b x )ˆj + (a xb y − a y b x )kˆ
(1.35)
(1.36)
where the symbol • ≡ det (•) denotes the matrix determinant.
ƒ
r r r
The scalar triple product [a, b, c] is the determinant of the 3 by 3 matrix, defined
as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
19
ax
r r r
r r r r r r r r r
V (a, b, c ) = a ⋅ b ∧ c = b ⋅ (c ∧ a) = c ⋅ a ∧ b = b x
cx
by bz
bx by
bx bz
= ax
− ay
+ az
cy cz
cx cy
cx cz
(
)
(
(
)
)
ay
az
by
cy
bz
cz
(1.37)
(
)
= a x b y c z − b z c y − a y (b x c z − b z c x ) + a z b x c y − b y c x
r r r
The vector triple product made up of the vectors ( a, b, c ) is obtained, in the
ƒ
Cartesian coordinate system, as:
(
r r r
a∧ b∧c
)
( )
r r r r r r
= (a ⋅ c ) b − a ⋅ b c
(1.38)
= (λ 1b x − λ 2 c x ) ˆi + λ 1b y − λ 2 c y ˆj + (λ 1b z − λ 2 c z ) kˆ
r r
r r
where λ 1 = a ⋅ c = a x c x + a y c y + a z c z , and λ 2 = a ⋅ b = a x b x + a y b y + a z b z .
(
)
Problem 1.3: Consider the points: A(1,3,1) , B (2,−1,1) , C (0,1,3) and D(1,2,4 ) , defined in the
Cartesian coordinate system.
→
→
1) Find the parallelogram area defined by AB and AC ; 2) Find the volume of the
→
→
→
→
parallelepiped defined by AB , AC and AD ; 3) Find the projection vector of AB onto
→
BC .
Solution:
→
→
1) Firstly we calculate the vectors AB and AC :
→
→
→
r
a = AB = OB − OA =
r
→
→
→
b = AC = OC − OA =
(2ˆi − 1ˆj + 1kˆ ) − (1ˆi + 3ˆj + 1kˆ ) = 1ˆi − 4ˆj + 0kˆ
(0ˆi + 1ˆj + 3kˆ )− (1ˆi + 3ˆj + 1kˆ ) = −1ˆi − 2ˆj + 2kˆ
With reference to the equation (1.36) we can evaluate the vector product as follows:
ˆi
r r
a∧b= 1
ˆj kˆ
− 4 0 = ( −8)ˆi − 2ˆj + ( −6)kˆ
−1 − 2
2
Then, the parallelogram area can be obtained using definition (1.19), thus:
r r
A = a ∧ b = (−8) 2 + (−2) 2 + ( −6) 2 = 104
→
2) Next, we can evaluate the vector AD as:
(
) (
)
→
→
→
r
c = AD = OD − OA = 1ˆi + 2ˆj + 4kˆ − 1ˆi + 3ˆj + 1kˆ = 0ˆi − 1ˆj + 3kˆ
and using the equation (1.37) we can obtain the volume of the parallelepiped:
(
r r r
r r r
V (a, b, c ) = c ⋅ a ∧ b
)
(
= 0ˆi − 1ˆj + 3kˆ
)⋅ (− 8ˆi − 2ˆj − 6kˆ )
= 0 + 2 − 18 = 16
→
3) The BC vector can be calculated as:
(
) (
)
→
→
→
BC = OC − OB = 0ˆi + 1ˆj + 3kˆ − 2ˆi − 1ˆj + 1kˆ = −2ˆi + 2ˆj + 2kˆ
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
20
→
→
Hence, it is possible to evaluate the projection vector of AB onto BC , (see equation
(1.12)), as:
→
→
proj BC→ AB =
→
BC ⋅ AB
→
→
BC
BC
⋅4
1
42
3
→
BC
→
BC
=
2
=
1.3.3
(− 2ˆi + 2ˆj + 2kˆ )⋅ (1ˆi − 4ˆj + 0kˆ ) (− 2ˆi + 2ˆj + 2kˆ )
(− 2ˆi + 2ˆj + 2kˆ )⋅ (− 2ˆi + 2ˆj + 2kˆ )
(− 2 − 8 + 0 ) (− 2ˆi + 2ˆj + 2kˆ ) = 5 ˆi − 5 ˆj − 5 kˆ
(4 + 4 + 4 )
3
3
3
Einstein Summation Convention (Einstein Notation)
r
As we saw in equation (1.28) a in the Cartesian coordinate system was defined as:
r
a = a x ˆi + a y ˆj + a z kˆ
(1.39)
Said expression can be rewritten as:
r
a = a1eˆ 1 + a 2 eˆ 2 + a 3 eˆ 3
(1.40)
where we have considered that: a1 ≡ a x , a 2 ≡ a y , a 3 ≡ a z , eˆ 1 ≡ î , eˆ 2 ≡ ĵ , eˆ 3 ≡ k̂ , (see
Figure 1.11). In this way we can express equation (1.40) by means of the summation
symbol as:
r
a = a1eˆ 1 + a 2 eˆ 2 + a 3 eˆ 3 =
3
∑ a eˆ
i
i
(1.41)
i =1
Then, we introduce the summation convention, according to which the “repeated indices”
indicate summation. So, equation (1.41) can be represented as follows:
r
a = a1eˆ 1 + a 2 eˆ 2 + a 3 eˆ 3 = a i eˆ i
(i = 1,2,3)
r
a = a i eˆ i
(i = 1,2,3)
(1.42)
NOTE: The summation notation was introduced by Albert Einstein in 1916, which led to
the indicial notation. ■
1.4 Indicial Notation
Using indicial notation, the three axes of the coordinate system are designated by the letter
x with a subscript. So, xi is not a single value but i values, i.e. x1 , x 2 , x3 (if i = 1,2,3 )
where these values x1 , x 2 , x3 correspond to the axes x , y , z , respectively.
r
Let a be a vector represented in the Cartesian coordinate system as:
r
a = a1eˆ 1 + a 2 eˆ 2 + a 3 eˆ 3
(1.43)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
21
where the orthonormal basis is represented by ê1 , ê 2 , ê 3 , (see Figure 1.11), and a1 , a 2 ,
a 3 are the vector components. In indicial notation the vector components are represented
by a i . If the range of the subscript is not indicated, we assume that 1,2,3 show these
values. Therefore, the vector components are represented as:
 a1 
r
(a) i = a i = a 2 
 a 3 
(1.44)
y ≡ x2
a y ≡ a2
r
a
ˆj ≡ eˆ
2
ˆi ≡ eˆ
1
kˆ ≡ eˆ 3
a z ≡ a3
a x ≡ a1
x ≡ x1
z ≡ x3
Figure 1.11: Vector representation in the Cartesian coordinate system.
r
Unit vector components: Let a be a vector, the normalized vector â is defined as:
r
a
aˆ = r
a
aˆ = 1
with
(1.45)
whose components are:
aˆ i =
ai
a12
+
a 22
+
a 32
=
ai
a ja j
=
ai
ak ak
(i, j , k = 1,2,3)
(1.46)
In light of the previous equation we can emphasize two types of indices:
The free index (live index) is that which only appears once in a term of the expression. In the
above equation the free index is the ( i ). The number of the free index indicates
the tensor order.
The dummy index (summation index) is that which is repeated only twice in a term of the
expression, and indicates summation. In the above equation (1.46) the
dummy index is the ( j ), or the ( k ) index.
OBS.: An index in a term of an expression can only appear once or twice. If it
appears more times, then a large error has occurred.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
22
Scalar product: Using definitions (1.4) and (1.29), we can express the scalar product
r r
( a ⋅ b ) as follows:
r r
r r
γ = a ⋅ b = a b cos θ = a1b1 + a 2 b 2 + a 3b 3 = a i b i = a j b j
(i, j = 1,2,3)
(1.47)
Problem 1.4: Rewrite the following equations using indicial notation:
1) a1 x1 x 3 + a 2 x 2 x 3 + a 3 x 3 x 3
Solution:
a i xi x 3
(i = 1,2,3)
2) x1 x1 + x2 x2
Solution:
xi x i
(i = 1,2)
 a11 x + a12 y + a13 z = b x

3) a 21 x + a 22 y + a 23 z = b y

 a 31 x + a 32 y + a 33 z = b z
Solution:
 a11 x1 + a12 x 2 + a13 x 3 = b1

a 21 x1 + a 22 x 2 + a 23 x 3 = b2
a x + a x + a x = b
32 2
33 3
3
 31 1
 a1 j x j = b1

→ a 2 j x j = b2

 a 3 j x j = b3
dummy
index j
free
index
i →
a ij x j = bi
As we can appreciate in this problem, the use of the indicial notation means that the
equation becomes very concise. In many cases, if algebraic operation do not use indicial or
tensorial notation they become almost impossible to deal with due to the large number of
terms involved.
Problem 1.5: Expand the equation: Aij x i x j
(i, j = 1,2,3)
Solution: The indices i, j are dummy indices, and indicate index summation and there is no
free index in the expression Aij x i x j , therefore the result is a scalar. So, we expand first the
dummy index i and later the index j to obtain:
expanding j

i → A1 j x1 x j + A2 j x 2 x j + A3 j x 3 x j
Aij x i x j expanding
1
424
3 1
424
3 1
424
3
A11 x1 x1 A21 x 2 x1 A31 x 3 x1
+
+
+
A12 x1 x 2
A22 x 2 x 2
A32 x 3 x 2
+
+
+
A13 x1 x 3
A23 x 2 x 3
A33 x 3 x 3
Rearranging the terms we obtain:
Aij x i x j = A11 x1 x1 + A12 x1 x 2 + A13 x1 x 3 + A21 x 2 x1 + A22 x 2 x 2 +
A23 x 2 x 3 + A31 x 3 x1 + A32 x 3 x 2 + A33 x 3 x 3
1.4.1
Some Operators
1.4.1.1
Kronecker Delta
The Kronecker delta δ ij is defined as follows:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
 1 iff

δ ij = 
 0 iff

23
i= j
(1.48)
i≠ j
Also note that the scalar product of the orthonormal basis eˆ i ⋅ eˆ j is equal to 1 if i = j and
equal to 0 if i ≠ j . Hence, eˆ i ⋅ eˆ j can be expressed in matrix form as:
 eˆ 1 ⋅ eˆ 1

eˆ i ⋅ eˆ j = eˆ 2 ⋅ eˆ 1
eˆ 3 ⋅ eˆ 1

eˆ 1 ⋅ eˆ 2
eˆ 2 ⋅ eˆ 2
eˆ 3 ⋅ eˆ 2
eˆ 1 ⋅ eˆ 3  1 0 0

eˆ 2 ⋅ eˆ 3  = 0 1 0 = δ ij
eˆ 3 ⋅ eˆ 3  0 0 1
(1.49)
An interesting property of the Kronecker delta is shown in the following example. Let Vi
r
be the components of the vector V , therefore:
δ ij Vi = δ 1 jV1 + δ 2 jV2 + δ 3 jV3
(1.50)
As ( j = 1,2,3) is a free index, we have three values to be calculated, namely:
j = 1 ⇒ δ ij Vi = δ 11V1 + δ 21V2 + δ 31V3 = V1 

j = 2 ⇒ δ ij Vi = δ 12V1 + δ 22V2 + δ 32V3 = V 2  ⇒ δ ij Vi = V j

j = 3 ⇒ δ ijVi = δ 13V1 + δ 23V 2 + δ 33V3 = V3 
(1.51)
That is, in the presence of the Kronecker delta symbol we replace the repeated index as
follows:
δi
j
V
i
=V j
(1.52)
For this reason, the Kronecker delta is often called the substitution operator.
Other examples using the Kronecker delta are presented below:
δ ij Aik = A jk , δ ij δ ji = δ ii = δ jj = δ 11 + δ 22 + δ 33 = 3 , δ ji a ji = a ii = a11 + a 22 + a 33
(1.53)
r
To obtain the components of the vector a in the coordinate system represented by ê i , it
r
r
is sufficient to obtain the scalar product with a and ê i , i.e. a ⋅ eˆ i = a p eˆ p ⋅ eˆ i = a p δ pi = a i .
With that, it is also possible to represent the vector as:
r
r
a = a i eˆ i = (a ⋅ eˆ i )eˆ i
(1.54)
Problem 1.6: Solve the following equations:
1) δ ii δ jj
Solution:
δ ii δ jj = (δ 11 + δ 22 + δ 33 )(δ 11 + δ 22 + δ 33 ) = 3 × 3 = 9
2) δ α1δ αγ δ γ1
Solution:
δ α1δ αγ δ γ1 = δ γ1δ γ1 = δ 11 = 1
NOTE: Note that the following algebraic operation is incorrect δ γ1δ γ1 ≠ δ γγ = 3 ≠ δ 11 = 1 ,
since what must be replaced is the repeated index, not the number ■
1.4.1.2
Permutation Symbol
The permutation symbol  ijk (also known as Levi-Civita symbol or alternating symbol) is defined as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
24
 ijk
+ 1 if (i, j , k ) ∈ {(1,2,3), (2,3,1), (3,1,2)}

= − 1 if (i, j , k ) ∈ {(1,3,2), (3,2,1), (2,1,3)}
0 for the remaining cases i.e. : if (i = j ) or ( j = k ) or (i = k )

(1.55)
NOTE:  ijk are the components of the Levi-Civita pseudo-tensor, which will be introduced
later on. ■
The values of  ijk can be easily memorized using the mnemonic device shown in Figure
1.12(a), in which if the index values are arranged in a clockwise direction, the value of  ijk
is equal to 1 , if not it has the value of − 1 . In the same way we can use this mnemonic
device to switch indices, (see Figure 1.12(b)).
ijk = 1
3
1
i
ijk = −1
ijk = −ikj
ijk =  jki = kij
k
2
 ijk = − ikj
j
= − kji
= − jik
b)
a)
Figure 1.12: Mnemonic device for the permutation symbol.
Another way to express the permutation symbol is by means of its indices:
1
2
 ijk = (i − j )( j − k )(k − i )
(1.56)
Using both the definition seen in (1.55) and Figure 1.12 (b), it is possible to verify that the
following relations are valid:
 ijk =  jki =  kij
(1.57)
 ijk = − ikj = − jik = − kji
Using the Kronecker delta property, we can state that:
 ijk
=  lmn δ li δ mj δ nk
= δ 1i δ 2 j δ 3k − δ 1i δ 3 j δ 2 k − δ 2i δ 1 j δ 3k + δ 3i δ 1 j δ 2 k + δ 2i δ 3 j δ 1k − δ 3i δ 2 j δ 1k
= δ 1i δ 2 j δ 3k − δ 3 j δ 2 k − δ 1 j (δ 2i δ 3k − δ 3i δ 2 k ) + δ 1k δ 2i δ 3 j − δ 3i δ 2 j
(
)
(
)
(1.58)
The above equation can be represented by means of the following determinant:
 ijk
δ 1i δ 1 j δ 1k δ 1i δ 2i δ 3i
= δ 2i δ 2 j δ 2 k = δ 1 j δ 2 j δ 3 j
δ 3i δ 3 j δ 3k δ 1k δ 2 k δ 3k
(1.59)
After which, the term  ijk  pqr can be evaluated as follows:
 ijk  pqr
δ 1i
= δ1 j
δ 1k
δ 2i δ 3i δ 1 p δ 1q δ 1r
δ 2 j δ 3 j δ 2 p δ 2q δ 2r
δ 2 k δ 3k δ 3 p δ 3q δ 3r
(1.60)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
25
Taking into account that det (AB ) = det (A )det (B ) , where det (•) ≡ • is the determinant
of the matrix • , the equation (1.60) can be rewritten as:
 ijk  pqr
 δ 1i
= δ 1 j
δ 1k
δ 3i   δ 1 p δ 1q δ 1r 


δ 3 j  δ 2 p δ 2 q δ 2 r  ⇒
δ 3k  δ 3 p δ 3q δ 3r 
δ 2i
δ2j
δ 2k
 ijk  pqr
δ ip δ iq δ ir
= δ jp δ jq δ jr
δ kp δ kq δ kr
(1.61)
The term δ ip was obtained by means of the operation δ 1i δ 1 p + δ 2i δ 2 p + δ 3i δ 3 p = δ mi δ mp
and δ mi δ mp = δ ip , the other terms were obtained in a similar fashion.
For the special exception when r = k , the equation (1.61) is reduced to:
 ijk  pqr
δ ip δ iq δ ik
= δ jp δ jq δ jk
δ kp δ kq 3
⇒  ijk  pqk = δ ip δ jq − δ iq δ jp
i, j , k , p, q = 1,2,3
(1.62)
Problem 1.7: a) Prove the following is true  ijk  pjk = 2δ ip and  ijk  ijk = 6 . b) Obtain the
numerical value of  ijk δ 2 j δ 3k δ 1i .
Solution: a) Using the equation in (1.62), i.e.  ijk  pqk = δ ip δ jq − δ iq δ jp , and by substituting q
for j , we obtain:
 ijk  pjk = δ ip δ jj − δ ij δ jp = δ ip 3 − δ ip = 2δ ip
Based on the above result, it is straight forward to check that:
 ijk  ijk = 2δ ii = 6
b)  ijk δ 2 j δ 3k δ 1i = 123 = 1
ƒ
r
r
r
The vector product of two vectors ( a ∧ b ) leads to a new vector c , defined in
r
(1.36), and the components of c , in Cartesian system, are given by:
eˆ 1 eˆ 2 eˆ 3
r r r
c = a ∧ b = a1 a 2 a 3
b1 b 2 b 3
= (a 2 b 3 − a 3b 2 )eˆ 1 + (a 3b1 − a1b 3 )eˆ 2 + (a1b 2 − a 2 b1 )eˆ 3
142
4 43
4
14243
14243
c1
c2
(1.63)
c3
Using the definition of the permutation symbol  ijk , defined in (1.55), we can express the
r
components of c as follows:
c 1 = 123 a 2 b 3 + 132 a 3b 2 = 1 jk a j b k 

c 2 =  231a 3b1 +  213 a1b 3 =  2 jk a j b k  ⇒ c i =  ijk a j b k
(1.64)

c 3 =  312 a1b 2 +  321a 2 b1 =  3 jk a j b k 
r r
Then, the vector product ( a ∧ b ) can be represented by means of the permutation symbol
as:
r r
a ∧ b =  ijk a j b k eˆ i
a j eˆ j ∧ b k eˆ k = a j b k  ijk eˆ i
(1.65)
a j b k (eˆ j ∧ eˆ k ) = a j b k  ijk eˆ i = a j b k  jki eˆ i
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
26
Therefore, we can also conclude that the following relationship is valid:
(eˆ j ∧ eˆ k ) =  ijk eˆ i
(1.66)
The permutation symbol and the orthonormal basis can be interrelated using the triple
scalar product as follows:
(eˆ
)⋅ eˆ
(
)
=  ijm eˆ m ⋅ eˆ k =  ijm δ mk ⇒ eˆ i ∧ eˆ j ⋅ eˆ k =  ijk
r r r
ƒ The triple scalar product made up of the vectors a, b, c is expressed by:
r r r
λ = a ⋅ b ∧ c = a i eˆ i ⋅ b j eˆ j ∧ c k eˆ k = a i b j c k eˆ i ⋅ eˆ j ∧ eˆ k =  ijk a i b j c k
r r r
λ = a ⋅ b ∧ c =  ijk a i b j c k
(i, j , k = 1,2,3)
eˆ i ∧ eˆ j =  ijm eˆ m
⇒
(
)
i
(
(
∧ eˆ j
k
)
)
(
)
(1.67)
(1.68)
(1.69)
or
a1 a 2
r r r r r r r r r
λ = a ⋅ b ∧ c = b ⋅ (c ∧ a) = c ⋅ a ∧ b = b1 b 2
(
)
(
)
c1
c2
a3
b3
(1.70)
c3
Starting from the equation (1.69) we can prove the following are true:
(
)
(
)
r r r r r r r r r
a ⋅ b ∧ c = b ⋅ (c ∧ a) = c ⋅ a ∧ b :
r r r r r r
[a, b, c] ≡ a ⋅ b ∧ c = ijk aib j c k
(
)
r r r
r r r
=  jki aib j c k = b ⋅ (c ∧ a) ≡ [b, c, a]
r r r
r r r
=  kij aib j c k = c ⋅ a ∧ b ≡ [c, a, b]
r r r
r r r
= −ikj aib j c k = −a ⋅ c ∧ b ≡ −[a, c, b]
r r r
r r r
= − jik aib j c k = −b ⋅ (a ∧ c ) ≡ −[b, a, c ]
r r r
r r r
= − kji aib j c k = −c ⋅ b ∧ a ≡ −[c, b, a]
(
(
)
(
)
(1.71)
)
where we take into account the property of the permutation symbol as given in (1.57).
(r r ) (r r )
Problem 1.8: Rewrite the expression a ∧ b ⋅ c ∧ d without using the vector product
symbol.
r r
Solution: The vector product a ∧ b can be expressed as
(
(
)
)
(
)
(
)
r r
r r
a ∧ b = a j eˆ j ∧ b k eˆ k =  ijk a j b k eˆ i . Likewise, it is possible to express c ∧ d as
r r
c ∧ d =  nlm c l d m eˆ n , thus:
r r r r
a ∧ b ⋅ c ∧ d =  ijk a j b k eˆ i ) ⋅ ( nlm c l d m eˆ n ) =  ijk  nlm a j b k c l d m eˆ i ⋅ eˆ n
=  ijk  nlm a j b k c l d m δ in =  ijk  ilm a j b k c l d m
(
)(
)
Taking into account that  ijk  ilm =  jki  lmi (see equation (1.57)) and by applying the
equation (1.62), i.e.:  jki  lmi = δ jl δ km − δ jm δ kl =  jki  ilm , we obtain:
 ijk  ilm a j b k c l d m = (δ jl δ km − δ jm δ kl ) a j b k c l d m = a l b m c l d m − a m b l c l d m
r r
r r
(
)
r r r r
(a ∧ b)⋅ (c ∧ d) = (arr ⋅ cr )r(br ⋅ dr ) − (ar ⋅ dr )(br ⋅ cr )
Since a l c l = (a ⋅ c ) and b m d m = b ⋅ d holds true, we can conclude that:
r
r
Therefore, it is also valid when a = c and b = d , thus:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
(ar ∧ br )⋅ (ar ∧ br ) = ar ∧ br
2
27
( ) ( )( )
r r r r
r r r r
r
= (a ⋅ a ) b ⋅ b − a ⋅ b b ⋅ a = a
2
r
b
2
( )
r r
− a⋅b
2
which is the same equation obtained in Problem 1.1.
(r r ) (r r ) r [r
r
r
] r [r
r
r
]
Problem 1.9: Prove that a ∧ b ∧ c ∧ d = c d ⋅ (a ∧ b) − d c ⋅ (a ∧ b)
Solution: Expressing the correct equality term in indicial notation we obtain:
[
] [
]
r r r
rr r r
cr d ⋅ (a
∧ b) − d c ⋅ (a ∧ b)  = c p d i  ijk a j b k − d p c i  ijk a j b k


p
⇒  ijk a j b k c p d i −  ijk a j b k c i d p
⇒
 ijk a j b k c p d i − c i d p
[ (
)]
[ (
(
)]
)
Using the Kronecker delta the above equation becomes:
( ijk a j b k ) c m d n (δ pm δ ni − δ im δ np )
⇒  ijk a j b k (δ pm c m d n δ ni − δ im c m d n δ np ) ⇒
and by applying the equation δ pm δ ni − δ im δ np =  pil  mnl , (see eq. (1.62)), the above equation
can be rewritten as follows:
⇒ ( ijk a j b k ) c m d n ( pil  mnl ) ⇒
 pil [( ijk a j b k ) ( mnl c m d n )]
Since  ijk a j b k and  mnl c m d n represent the components of
respectively, we can conclude that:
(ar ∧ br )
and
(cr ∧ dr ) ,
[(r r ) (r r )]
 pil [( ijk a j b k ) ( mnl c m d n )] = a ∧ b ∧ c ∧ d
r
r
p
r
r
Problem 1.10: Let a , b , c be linearly independent vectors, and v be a vector,
demonstrate that:
r
r
r r
r
v = αa + βb + γ c ≠ 0
where the scalars α , β , γ are given by:
 ijk v i b j c k
 ijk a i v j c k
 ijk a i b j v k
α=
; β=
; γ=
 pqr a p b q c r
 pqr a p b q c r
 pqr a p b q c r
r
r
r
Solution: The scalar product made up of v and ( b ∧ c ) becomes:
r r r
r r r
r r r
r r r
v ⋅ (b ∧ c ) = αa ⋅ (b ∧ c ) + β b ⋅ (b ∧ c ) + γ c ⋅ (b ∧ c )
14243
14243
=0
⇒
=0
r r r
v ⋅ (b ∧ c )
α= r r r
a ⋅ (b ∧ c )
which is the same as:
α=
v1
b1
v2
b2
v3
b3
c1
c2
c3
a1
b1
a2
b2
a3
b3
c1
c2
c3
=
v1
v2
b1
b2
c1
c2
v3
b3
c3
a1
a2
b1
b2
c1
c2
a3
b3
c3
=
 ijk v i b j c k
 pqr a p b q c r
One can obtain the parameters β and γ in a similar fashion.
Problem 1.11: Prove the relationship given in (1.38) is valid, i.e.:
r r r
r r r r r r
a ∧ b ∧ c = (a ⋅ c ) b − a ⋅ b c .
(r ) r
( )
r
r r
Solution: Taking into account that (d) = (b ∧ c ) =  b c and that (a ∧ d)
i
i
ijk
j
k
q
=  qjk b j c k , we
obtain:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
28
[ar ∧ (br ∧ cr )]
q
=  qsi a s ( ijk b j c k ) =  qsi  ijk a s b j c k =  qsi  jki a s b j c k
= (δ qj δ sk − δ qk δ sj ) a s b j c k = δ qj δ sk a s b j c k − δ qk δ sj a s b j c k
( )
( )]
r r
r r
= a k b q c k − a j b j c q = b q (a ⋅ c ) − c q a ⋅ b
r r r rr r
r r r
⇒ a ∧ b ∧ c q = b(a ⋅ c ) − c a ⋅ b q
[ (
)] [
1.5 Algebraic Operations with Tensors
1.5.1
Dyadic
r
r
The tensor product, made up of two vectors v and u , becomes a dyad, which is a particular
case of a second-order tensor. The dyad is represented by:
rr r r
uv ≡ u ⊗ v = A
(1.72)
where the operator ⊗ denotes the tensor product. Then, we define a dyadic as a linear
combination of dyads. Furthermore, as we will see later, any tensor can be represented by
means of a linear combination of dyads, (see Holzapfel (2000)).
The tensor product has the following properties:
1.
2.
3.
r r r r r r
r
r r
(u ⊗ v ) ⋅ x = u( v ⋅ x ) ≡ u ⊗ ( v ⋅ x )
r
r
r
r r
r
r
u ⊗ (αv + βw ) = αu ⊗ v + βu ⊗ w
r r
r r r
r r r
r r r
(αv ⊗ u + βw ⊗ r ) ⋅ x = α ( v ⊗ u) ⋅ x + β ( w ⊗ r ) ⋅ x
r
r r
r
r r
= α[v ⊗ (u ⋅ x )] + β [w ⊗ (r ⋅ x )]
(1.73)
(1.74)
(1.75)
where α and β are scalars. By definition, the dyad does not contain the commutative
r r r r
property, i.e., u ⊗ v ≠ v ⊗ u .
The equation (1.72) can also be expressed in the Cartesian system as:
r r
A = u ⊗ v = (u i eˆ i ) ⊗ ( v j eˆ j )
= u i v j (eˆ i ⊗ eˆ j )
(i, j = 1,2,3)
(1.76)
(i, j = 1,2,3)
(1.77)
= A ij (eˆ i ⊗ eˆ j )
A = A ij eˆ i ⊗ eˆ j
{
{ 1
424
3
Tensor
components
basis
In this textbook, the components of a second-order tensor can be represented in different
ways, namely:
r r
A4
=2
u4
⊗3
v
1
↓
components
↓
(1.78)
r r
( A ) ij = (u ⊗ v ) ij = u i v j = A ij
These components are explicitly expressed in matrix form as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
 A 11
( A ) ij = A ij = A =  A 21
 A 31
A 12
A 22
A 32
29
A 13 
A 23 
A 33 
(1.79)
It is easy to identify the tensor order by the number of free indices in the tensor
components, i.e.:
Second-order tensor U = U ij eˆ i ⊗ eˆ j
Third-order tensor
Fourth-order tensor
T = Tijk eˆ i ⊗ eˆ j ⊗ eˆ k
(i, j , k , l = 1,2,3)
(1.80)
I = I ijkl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
OBS.: The tensor order is given by the number of free indices in its components.
OBS.: The number of tensor components is given by a n , where the base a is the
maximum value in the index range, and the exponent n is the number of the free
index.
Problem 1.12: Define the order of the tensors represented by their Cartesian components:
v i , Φ ijk , Fijj , ε ij , C ijkl , σ ij . Determine the number of components in tensor C .
Solution: The order of the tensor is given by the number of free indices, so it follows that:
r r
First-order tensor (vector): v , F ; Second-order tensor: ε , σ ; Third-order tensor: Φ ;
Fourth-order tensor: C
The number of tensor components is given by the maximum index range value, i.e.
i, j , k , l = 1,2,3 , to the power of the number of free indices which is equal to 4 in the case of
C ijkl . Thus, the number of independent components in C is given by:
3 4 = (i = 3) × ( j = 3) × (k = 3) × (l = 3) = 81
The fourth-order tensor C ijkl has 81 components.
Let A and B be second-order tensors, we can then define some algebraic operations
including:
ƒ
Addition: The sum of two tensors of same order is a new tensor defined as follows:
C = A +B =B + A
(1.81)
The components of C are represented by:
(C) ij = ( A + B) ij
or
C ij = A ij + B ij
(1.82)
or, in matrix notation as:
ƒ
C =A+B
(1.83)
D = λA incomponents
 →(D) ij = λ( A ) ij
(1.84)
Multiplication of a tensor by a scalar: The multiplication of a second-order tensor
( A ) by a scalar ( λ ) is defined by a new tensor D , so that:
or, in matrix form:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
30
 A 11
A = A 21
 A 31
A 13 
 λA 11

A 23  
→ λA = λA 21
λA 31
A 33 
A 12
A 22
A 32
λA 12
λA 22
λA 32
λA 13 
λA 23 
λA 33 
(1.85)
It is also true that:
r
r
(λ A ) ⋅ v = λ ( A ⋅ v )
(1.86)
r
for any vector v .
ƒ
Scalar Product (or Dot Product): The scalar product (also known as single
r
contraction) between a second-order tensor A and a vector x is another vector
r
(first-order tensor) y , defined as:
δ kl
r
r
y = A⋅x
= ( A jk eˆ j ⊗ eˆ k ) ⋅ ( x l eˆ l )
= A jk x l δ kl eˆ j
= A jk x k eˆ j
123
(1.87)
yj
= y j eˆ j
The scalar product between two second-order tensors A and B is another second-order
tensor, that verifies: A ⋅ B ≠ B ⋅ A :
δ jk
δ jk
C = A ⋅ B = ( A ij eˆ i ⊗ eˆ j ) ⋅ (B kl eˆ k ⊗ eˆ l )
= A ij B kl δ jk eˆ i ⊗ eˆ l
= A ik B kl eˆ i ⊗ eˆ l
123
AB
D = B ⋅ A = (B ij eˆ i ⊗ eˆ j ) ⋅ ( A kl eˆ k ⊗ eˆ l )
= B ij A kl δ jk eˆ i ⊗ eˆ l
= B ik A kl eˆ i ⊗ eˆ l
123
(1.88)
BA
= C il eˆ i ⊗ eˆ l
= D il eˆ i ⊗ eˆ l
It also satisfies the following properties:
A ⋅ (B + C) = A ⋅ B + A ⋅ C
ƒ
A ⋅ (B ⋅ C) = ( A ⋅ B) ⋅ C
;
(1.89)
The powers of second-order tensors
The scalar product allows us to define the power of second-order tensors, as seen below:
A 0 = 1 ; A1 = A ; A 2 = A ⋅ A
A 3 = A ⋅ A ⋅ A , and so on,
;
(1.90)
where 1 is the second-order unit tensor (also called the identity tensor).
ƒ
Double Scalar Product (or Double contraction)
r
r
r
r
Consider two dyads, A = c ⊗ d and B = u ⊗ v . The double contraction between them is
defined in different ways, namely: A : B and A ⋅ ⋅B .
Double contraction (⋅ ⋅ ) :
(cr ⊗ dr ) ⋅ ⋅ (ur ⊗ vr ) = (cr ⋅ vr )(dr ⋅ ur )
(1.91)
In components
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
31
δ il
δ jk
A ⋅ ⋅ B = ( A ij eˆ i ⊗ eˆ j )⋅ ⋅ (B kl eˆ k ⊗ eˆ l )
= A ij B kl δ jk δ il
(1.92)
= A ij B ji
=γ
( scalar )
The double contraction (⋅ ⋅ ) is commutative, i.e. A ⋅ ⋅B = B ⋅ ⋅ A .
Double contraction ( : ):
(
)
( )
r r r r
r r r r
A : B = c ⊗ d : (u ⊗ v ) = (c ⋅ u) d ⋅ v
The double contraction ( : ) is commutative, so:
(
)
( )
(1.93)
( )
r r r r
r r r r
r r r r
B : A = (u ⊗ v ) : c ⊗ d = (u ⋅ c ) v ⋅ d = (c ⋅ u) d ⋅ v = A : B
(1.94)
The breakdown into its components appears like this:
δ ik
A : B = ( A ij eˆ i ⊗ eˆ j ) : (B kl eˆ k ⊗ eˆ l )
δ jl
= A ij B kl δ ik δ jl
= A ij B ij
=λ
(1.95)
( scalar )
In general, A : B ≠ A ⋅ ⋅B , however, they are equal if at least one of them is symmetric, i.e.
A sym : B = A sym ⋅ ⋅ B or A : B sym = A ⋅ ⋅ B sym , so A sym : B sym = A sym ⋅ ⋅ B sym .
The double contraction with a third-order tensor ( S ) and a second-order tensor ( B )
becomes:
(
)
( )cr
( )ar
r r r r r
r r r r
S : B = c ⊗ d ⊗ a : (u ⊗ v ) = (a ⋅ v ) d ⋅ u
r r r r r
r r r r
B : S = (u ⊗ v ) : c ⊗ d ⊗ a = (u ⋅ c ) v ⋅ d
(
)
(1.96)
As we can verify the result is a vector. In symbolic notation, the double contraction ( B : S )
is represented by:
S ijk eˆ i ⊗ eˆ j ⊗ eˆ k : B pq eˆ p ⊗ eˆ q = S ijk B pq δ jp δ kq eˆ i = S ijk B jk eˆ i
(1.97)
The double contraction of a fourth-order tensor ( C ) with a second-order tensor ( ε ) is
defined as:
C ijkl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l : ε pq eˆ p ⊗ eˆ q = C ijkl ε pq δ kp δ lq eˆ i ⊗ eˆ j
= C ijkl ε kl eˆ i ⊗ eˆ j
= σ ij eˆ i ⊗ eˆ j
(1.98)
where σ ij are the components of σ = C : ε .
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
32
Next, we express some properties of the double contraction ( : ):
a) A : B = B : A
b) A : (B + C ) = A : B + A : C
c) λ(A : B ) = (λA ) : B = A : (λB )
(1.99)
where A , B, C are second-order tensors, and λ is a scalar.
Via the definition of the double scalar product, it is possible to obtain the components of
the second-order tensor A in the Cartesian system, i.e.:
( A ) ij = ( A kl eˆ k ⊗ eˆ l ) : (eˆ i ⊗ eˆ j ) = eˆ i ⋅ ( A kl eˆ k ⊗ eˆ l ) ⋅ eˆ j = A kl δ ki δ lj = A ij
(1.100)
r r
If we consider any two vectors a , b , and an arbitrary second-order tensor, A , we can
demonstrate that:
r
r
a ⋅ A ⋅ b = a p eˆ p ⋅ A ij eˆ i ⊗ eˆ j ⋅ b r eˆ r = a p A ij b r δ pi δ jr = a i A ij b j = A ij (a i b j )
r r
= A : ( a ⊗ b)
ƒ
(1.101)
Vector product
r
The vector product between a second-order tensor A and a vector x is a second-order
tensor given by:
r
A ∧ x = ( A ij eˆ i ⊗ eˆ j ) ∧ ( x k eˆ k ) =  ljk A ij x k eˆ i ⊗ eˆ l
(1.102)
where we have used the definition (1.67), i.e. eˆ j ∧ eˆ k =  ljk eˆ l . In Problem 1.11, we have
r
(r r )
(r r ) r
r r r
shown that the relation a ∧ b ∧ c = (a ⋅ c ) b − a ⋅ b c holds, which is also represented by
means of dyads as:
[ar ∧ (br ∧ cr )]
[(
) ]
r r r r r
= (a k c k )b j − (a k b k )c j = (b j c k − c j b k )a k = b ⊗ c − c ⊗ b ⋅ a
r r
In the particular case when a = c we obtain:
r r r
a ∧ b ∧ a j = (a k a k )b j − (a k b k )a j = (a k a k )b p δ jp − (a k b p δ kp )a j
[ (
j
)]
[
]
[
]
= (a k a k )δ jp − (a k δ kp )a j b p = (a k a k )δ jp − a p a j b p
r r
r r r
= [(a ⋅ a)1 − a ⊗ a] ⋅ b j
{
}
j
(1.103)
(1.104)
Thus, the following relationships are valid:
( ) (
)
r r
r
r r r r r r r r r r r
a ∧ (b ∧ c ) = (a ⋅ c ) b − a ⋅ b c = b ⊗ c − c ⊗ b ⋅ a
r r
r
r r
r r r
a ∧ (b ∧ a) = [(a ⋅ a)1 − a ⊗ a]⋅ b
1.5.1.1
(1.105)
Component Representation of a Second-Order Tensor in the
Cartesian Basis
As seen before, a vector which has 3 independent components is represented in a
Cartesian space as shown in Figure 1.11. An arbitrary second-order tensor has 9
independent components, so we would need a hyperspace to represent all its components.
Afterwards, a device is introduced to represent the second-order tensor components in the
Cartesian basis.
An arbitrary second-order tensor T is represented in the Cartesian basis by:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
33
T = Tij eˆ i ⊗ eˆ j = Ti1eˆ i ⊗ eˆ 1 + Ti 2 eˆ i ⊗ eˆ 2 + Ti 3 eˆ i ⊗ eˆ 3
= T11eˆ 1 ⊗ eˆ 1 + T12 eˆ 1 ⊗ eˆ 2 + T13 eˆ 1 ⊗ eˆ 3 +
+ T21eˆ 2 ⊗ eˆ 1 + T22 eˆ 2 ⊗ eˆ 2 + T23 eˆ 2 ⊗ eˆ 3 +
(1.106)
+ T31eˆ 3 ⊗ eˆ 1 + T32 eˆ 3 ⊗ eˆ 2 + T33 eˆ 3 ⊗ eˆ 3
Next, we calculate the projection of T onto ê k :
T ⋅ eˆ k = Tij eˆ i ⊗ eˆ j ⋅ eˆ k = Tij eˆ i δ jk = Tik eˆ i = T1k eˆ 1 + T2 k eˆ 2 + T3k eˆ 3
(1.107)
thereby defining three vectors, namely:
r
k = 1 ⇒ Ti1eˆ i = T11eˆ 1 + T21eˆ 2 + T31eˆ 3 = t ( eˆ 1 )

r ( eˆ )
⇒
T ⋅ eˆ k = Tik eˆ i
(1.108)
k = 2 ⇒ Ti 2 eˆ i = T12 eˆ 1 + T22 eˆ 2 + T32 eˆ 3 = t 2
r (eˆ )

3
k = 3 ⇒ Ti 3 eˆ i = T13 eˆ 1 + T23 eˆ 2 + T33 eˆ 3 = t
r ˆ r ˆ
r ˆ
Graphical representation of these three vectors t ( e1 ) , t (e 2 ) , t (e 3 ) , in the Cartesian basis, is
r ˆ
shown in Figure 1.13. Note also that t ( e1 ) is the projection of T onto ê1 , nˆ (i1) = [1,0,0] ,
which can be verified by:
 T11
( T ⋅ nˆ ) i =  T21
 T31
T12
T22
T13  1  T11 
ˆ
T23  0 =  T21  = t i( e1 )
T33  0  T31 
T32
x3
r ˆ
t (e3 )
ê 3
r ˆ
t ( e1 )
(1.109)
r ˆ
t (e 2 )
ê 2
x2
ê 1
x1
Figure 1.13: The projection of T in the Cartesian basis.
The same result obtained in (1.109) could have been evaluated by the scalar product of T ,
given in (1.106), with the basis ê1 , i.e.:
T ⋅ eˆ 1 = [ T11eˆ 1 ⊗ eˆ 1 + T12 eˆ 1 ⊗ eˆ 2 + T13 eˆ 1 ⊗ eˆ 3 +
+ T21eˆ 2 ⊗ eˆ 1 + T22 eˆ 2 ⊗ eˆ 2 + T23 eˆ 2 ⊗ eˆ 3 +
+ T31eˆ 3 ⊗ eˆ 1 + T32 eˆ 3 ⊗ eˆ 2 + T33 eˆ 3 ⊗ eˆ 3 ] ⋅ eˆ 1
r ˆ
= T11eˆ 1 + T21eˆ 2 + T31eˆ 3 = t ( e1 )
(1.110)
where we have used the orthogonality property of the basis, i.e. eˆ 1 ⋅ eˆ 1 = 1 , eˆ 2 ⋅ eˆ 1 = 0 ,
eˆ 3 ⋅ eˆ 1 = 0 . Taking into account the components are represented in matrix form, (see
Figure 1.14), we can establish that, the diagonal terms ( T11 , T22 , T33 ) are normal to the
plane defined by the unit vectors ( ê1 , ê 2 , ê 3 ), hence they will be referred to as normal
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
34
components. The components displayed tangentially to the plane are called tangential
components, and correspond to the off-diagonal terms of Tij .
x3
 T11
Tij =  T21
 T31
T12
T22
T32
rˆ
t (e 3 )
T33 ê 3
T13 
T23 
T33 
rˆ
t (e 2 )
T23 ê 2
T13 ê1
T32 ê 3
T31ê 3
r ˆ
t ( e1 )
T22 ê 2
T21ê 2
T12 ê1
x2
T11ê1
x1
Figure 1.14: Representation of the second-order tensor components in the Cartesian
coordinate system.
NOTE: Throughout the textbook, we will use the following notations:
Tensorial notation
Symbolic notation
A ⋅ B = ( A ij eˆ i ⊗ eˆ j ) ⋅ (B kl eˆ k ⊗ eˆ l )
(1.111)
= A ij B kl δ jk (eˆ i ⊗ eˆ l )
= A ij B jl (eˆ i ⊗ eˆ l )
Cartesian basis
Indicial notation
Note that the index is not repeated more than twice either in symbolic notation or in
indicial notation. Also note that the indicial notation is equivalent to the tensor notation
r r
only when dealing with scalars, e.g. A : B = A ijB ij = λ , or a ⋅ b = a i b i . ■
1.5.2
1.5.2.1
Properties of Tensors
Tensor Transpose
Let A be a second-order tensor, the transpose of A is defined as:
A T = A ji (eˆ i ⊗ eˆ j ) = A ij (eˆ j ⊗ eˆ i )
(1.112)
If A ij are the components of A , it follows that the components of the transpose are:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
35
(A T ) ij = A ji
r
(1.113)
r
r
r
If A = u ⊗ v , the transpose of the dyad A is given by A T = v ⊗ u :
AT
r r T
= (u ⊗ v )
= u i eˆ i ⊗ v j eˆ j
= u i v j eˆ i ⊗ eˆ j
(
(
= (A
ˆ ⊗ eˆ j
ij e i
)
)
)
T
T
T
r r
= v ⊗u
= v j eˆ j ⊗ u i eˆ i
= u i v j eˆ j ⊗ eˆ i
= v i eˆ i ⊗ u j eˆ j
= u j v i eˆ i ⊗ eˆ j
= A ij eˆ j ⊗ eˆ i
= A ji eˆ i ⊗ eˆ j
(1.114)
Let A and B be second-order tensors and α , β be scalars, and the following
relationships are valid:
; (αB + βA ) T = αB T + βA T
(A T )T = A
(
: B = (A
)
⊗ eˆ ) : (B
; (B ⋅ A ) T = A T ⋅ B T
(1.115)
A : B T = A ij eˆ i ⊗ eˆ j : (B kl eˆ l ⊗ eˆ k ) = A ij B kl δ il δ jk = A ij B ji = A ⋅ ⋅ B
A
T
ˆ
ij e j
i
(1.116)
ˆ ⊗ eˆ l ) = A ij B kl δ jk δ il = A ij B ji = A ⋅ ⋅ B
kl e k
The transpose of the matrix A is formed by changing rows for columns and vice versa,
i.e.:
 A 11
A = A 21
 A 31
A 12
A 22
A 32
A 13 
 A 11

transpose
T
A 23   → A =  A 21
 A 31
A 33 
T
A 12
A 22
A 13 
 A 11

A 23  =  A 12
 A 13
A 33 
A 32
A 21
A 22
A 23
A 31 
A 32 
A 33 
(1.117)
Problem 1.13: Let A , B and C be arbitrary second-order tensors. Demonstrate that:
(
)
(
)
A : (B ⋅ C ) = B T ⋅ A : C = A ⋅ C T : B
Solution: Expressing the term A : (B ⋅ C ) in indicial notation we obtain:
A : (B ⋅ C ) = A ij eˆ i ⊗ e j : (B lk eˆ l ⊗ e k ⋅ C pq eˆ p ⊗ eˆ q )
= A ij B lk C pq eˆ i ⊗ e j : (δ kp eˆ l ⊗ eˆ q )
= A ij B lk C pq δ kp δ il δ jq = A ij B ik C kj
Note that, when we are dealing with indicial notation the position of the terms does not
matter, i.e.:
A ij B ik C kj = B ik A ij C kj = A ij C kj B ik
We can now observe that the algebraic operation B ik A ij is equivalent to the components of
the second-order tensor (B T ⋅ A ) kj , thus,
(
)
B ik A ij C kj = (B T ⋅ A ) kj C kj = B T ⋅ A : C .
Likewise, we can state that A ij C kj B ik = (A ⋅ C ): B .
T
r
r
Problem 1.14: Let u , v be vectors and A be a second-order tensor. Show that the
following relationship holds:
Solution:
r
r r
r
u⋅ AT ⋅ v = v ⋅ A ⋅u
r
r
u ⋅ AT ⋅ v
u i eˆ i ⋅ A jl eˆ l ⊗ eˆ j ⋅ v k eˆ k
r
r
= v ⋅ A ⋅u
= v k eˆ k ⋅ A jl eˆ j ⊗ eˆ l ⋅ u i eˆ i
u i A jl δ il v k δ jk
u l A jl v j
= v k δ kj A jl u i δ il
= v j A jl u l
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
36
1.5.2.2
Symmetry and Antisymmetry
1.5.2.2.1
Symmetric tensor
A second-order tensor A is symmetric, i.e.: A ≡ A sym , if the tensor is equal to its transpose:
A = A T incomponents
 → A ij = A ji
if
⇔
A is symmetric
(1.118)
A 12
A 22
A 13 
A 23 
A 33 
(1.119)
in matrix form:
T
→ A ≡ A
A=A 
sym
 A 11
=  A 12
 A 13
A 23
From the above it is clear that a symmetric second-order tensor has 6 independent
components, namely: A 11 , A 22 , A 33 , A 12 , A 23 , A 13 .
According to equation (1.118), a symmetric tensor can be represented by:
A ij = A ji
A ij + A ij = A ij + A ji
(1.120)
2 A ij = A ij + A ji
A ij =
1
( A ij + A ji )
2
⇒
A=
1
(A + A T )
2
A fourth-order tensor C , whose components are C ijkl , may have the following types of
symmetries:
Minor symmetry:
C ijkl = C jikl = C ijlk = C jilk
(1.121)
C ijkl = C klij
(1.122)
Major symmetry:
A fourth-order tensor that does not exhibit any kind of symmetry has 81 independent
components. If the tensor C has only minor symmetry, i.e. symmetry in ij = ji (6) , and
symmetry in kl = lk (6) , the tensor features 36 independent components. If besides
presenting minor symmetry it also provides major symmetry, the tensor features 21
independent components.
1.5.2.2.2
Antisymmetric tensor
A tensor A is antisymmetric (also called skew-symmetric tensor or skew tensor), i.e.: A ≡ A skew :
if
A = − A T incomponents
 → A ij = − A ji
⇔
A is antisymmetric
(1.123)
which broken down into its components is the same as:
T
→ A
A = −A 
skew
 0
= − A 12
 − A 13
A 12
0
− A 23
A 13 
A 23 
0 
(1.124)
Therefore, an antisymmetric second-order tensor has 3 independent components, namely:
A 12 , A 23 , A 13 .
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
37
Under the conditions expressed in (1.123), an antisymmetric tensor can be represented by:
A ij + A ij = A ij − A ji
2 A ij = A ij − A ji
A ij =
(1.125)
1
( A ij − A ji )
2
⇒
A=
1
(A − A T )
2
Let us consider an antisymmetric second-order tensor denoted by W , then satisfy the
above relationship (1.125):
Wij =
1
1
1
(Wij − W ji ) = (Wkl δ ik δ jl − Wkl δ jk δ il ) = Wkl (δ ik δ jl − δ jk δ il )
2
2
2
(1.126)
Using the relation between the Kronecker delta and the permutation symbol given by
(1.62), i.e. δ ik δ jl − δ jk δ il = − ijr  lkr , the equation (1.126) is rewritten as:
1
Wij = − Wkl  ijr  lkr
2
(1.127)
Expanding the term Wkl  lkr , for the dummy indices ( k , l ), we can obtain the following
nonzero terms:
Wkl  lkr = W12  21r + W13  31r + W2112 r + W23  32 r + W3113r + W32  23r
(1.128)
Wkl  lkr = − W23 + W32 = −2W23 = 2 w1 

Wkl  lkr = W13 − W31 = 2W13 = 2w2  ⇒ Wkl  lkr = 2wr

Wkl  lkr = − W12 + W21 = −2W12 = 2w3 
(1.129)
thus,
r =1
⇒
r=2
⇒
r =3
⇒
In which we assume the following variables have changed:
− w3 w2 
W13   0
W12
W13   0




0
0
− w1 
W23  =  − W12
W23  =  w3
(1.130)
0   − W13 − W23
0   − w2
0 
W32
w1
r
Hence, we introduce the axial vector w associated with the antisymmetric tensor, W , as:
r
w = w1eˆ 1 + w2 eˆ 2 + w3 eˆ 3
(1.131)
r
The magnitude of the axial vector w is given by:
r 2 r r
(1.132)
ω 2 = w = w ⋅ w = w12 + w22 + w32 = W232 + W132 + W122
 0
Wij =  W21
 W31
W12
0
Substituting (1.129) into (1.127) and by considering that  ijr =  rij we obtain:
Wij = − wr  rij
(1.133)
Multiplying both sides of the equation (1.133) by  kij we can obtain:
 kij Wij = − wr  rij  kij = −2 wr δ rk = −2 wk
(1.134)
where we have applied the relation  rij  kij = 2δ rk , which was evaluated in Problem 1.7,
thus we can conclude that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
38
1
wk = −  kij Wij
2
(1.135)
Graphical representation of the antisymmetric tensor components and its corresponding
axial vector, in the Cartesian system, is shown in Figure 1.15.
x3
r
w = w1eˆ 1 + w2 eˆ 2 + w3 eˆ 3
w3 = − W12
 0
Wij =  − W12
 − W13
W12
0
W13 
W23 
0 
− W23
W13
W23
W12
W12
w1 = − W23
x2
W23
w2 = W13
W13
x1
Figure 1.15: Antisymmetric tensor components and the axial vector.
r
r
Let a and b be arbitrary vectors and W be an antisymmetric tensor, it follows that:
r
r
r
r
r r
r
b ⋅ W ⋅ a = a ⋅ W T ⋅ b = −a ⋅ W ⋅ b
r
when a = b , it holds that:
r
(1.136)
r
r
r r
a ⋅ W ⋅ a = W : (a ⊗ a) = 0
(1.137)
r
NOTE: Note that (a ⊗ a) is a symmetric second-order tensor. Later on we will show that
the result of the double contraction between a symmetric tensor and an antisymmetric
tensor equals zero. ■
r
Let us consider an antisymmetric tensor W and an arbitrary vector a . The components of
r
the scalar product W ⋅ a are given by:
Wij a j = Wi1a1 + Wi 2 a 2 + Wi 3 a 3
i = 1 ⇒ W11a1 + W12 a 2 + W13 a 3
i = 2 ⇒ W21a1 + W22 a 2 + W23 a 3
i =3⇒
(1.138)
W31a1 + W32 a 2 + W33 a 3
Bearing in mind that the normal components are equal to zero for an antisymmetric tensor,
i.e., W11 = 0 , W22 = 0 , W33 = 0 , the scalar product (1.138) becomes:
i = 1 ⇒ W12 a 2 + W13 a 3
r
(W ⋅ a)i ⇒ i = 2 ⇒ W21a1 + W23 a 3
i = 3 ⇒ W a + W a
31 1
32 2

(1.139)
r
r
The above components are the same as the result of the algebraic operation w ∧ a :
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
eˆ 1
r r
w ∧ a = w1
a1
eˆ 2
eˆ 3
w2
a2
w3
a3
39
= (− w3 a 2 + w2 a 3 ) eˆ 1 + (w3 a1 − w1a 3 ) eˆ 2 + (− w2 a1 + w1a 2 ) eˆ 3
= (W12 a 2 + W13 a 3 )eˆ 1 + (W21a1 + W23 a 3 ) eˆ 2 + (W31a1 + W32 a 2 ) eˆ 3
(1.140)
where w1 = −W23 = W32 , w2 = W13 = −W31 , w3 = −W12 = W21 . Then, given an antisymmetric
r
tensor W and the axial vector w associated with W , it holds that:
r r r
W⋅a = w ∧ a
(1.141)
r
for any vector a . The property (1.141) could easily have been obtained by taking into
account the components of W given by (1.133), i.e.:
r
r
(W ⋅ a)i = Wik a k = −w j  jik a k =  ijk w j a k = (wr ∧ a)i
(1.142)
r
r
The vector w can be represented by its magnitude, w = ω , and by the unit vector
r
r
codirectional with w , i.e. w = ωê1* . Then, the equation (1.141) can still be expressed as:
r r r
r
W ⋅ a = w ∧ a = ωeˆ 1* ∧ a
(1.143)
Additionally, we can choose two unit vectors ê *2 , ê *3 , which make up an orthonormal basis
with the unit vector ê1* , (see Figure 1.16), so that:
eˆ 1* = eˆ *2 ∧ eˆ *3
eˆ *2 = eˆ *3 ∧ eˆ 1*
;
ê*2
eˆ *3 = eˆ 1* ∧ eˆ *2
;
(1.144)
r
w = ωê1*
ê3
ê 2
ê1*
ê1
ê*3
Figure 1.16: Orthonormal basis defined by the axial vector.
r
r
By representing the vector a in this new basis, a = a1* eˆ 1* + a*2 eˆ *2 + a*3 eˆ *3 , the relationship
shown in (1.143) obtains the form below:
r
r
W ⋅ a = ωeˆ 1* ∧ a = ωeˆ 1* ∧ (a1* eˆ 1* + a *2 eˆ *2 + a *3 eˆ *3 )
= ω (a1* eˆ 1* ∧ eˆ 1* + a *2 eˆ 1* ∧ eˆ *2 + a *3 eˆ 1* ∧ eˆ *3 ) = ω (a *2 eˆ *3 − a *3 eˆ *2 )
1
424
3
1
424
3
1
424
3
[
=0ˆ
=eˆ *3
]
r
= ω (eˆ *3 ⊗ eˆ *2 − eˆ *2 ⊗ eˆ *3 ) ⋅ a
=− eˆ *2
(1.145)
Thus, an antisymmetric tensor can be represented, in the space defined by the axial vector,
as follows:
W = ω (eˆ *3 ⊗ eˆ *2 − eˆ *2 ⊗ eˆ *3 )
(1.146)
Note that by using the antisymmetric tensor representation shown in (1.146), the
projections of the tensor W according to directions ê1* , ê *2 and ê *3 are respectively:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
40
r
W ⋅ eˆ 1* = 0
;
W ⋅ eˆ *2 = ωeˆ *3
;
W ⋅ eˆ *3 = −ωeˆ *2
[
⋅ [ω (eˆ
] ⋅ eˆ
⊗ eˆ )] ⋅ eˆ
(1.147)
We can also verify that:
eˆ *3 ⋅ W ⋅ eˆ *2 = eˆ *3 ⋅ ω (eˆ *3 ⊗ eˆ *2 − eˆ *2 ⊗ eˆ *3 )
*
2
=ω
eˆ *2 ⋅ W ⋅ eˆ *3 = eˆ *2
*
3
= −ω
*
3
⊗ eˆ *2 − eˆ *2
*
3
(1.148)
Then, the tensor components of W in the basis formed by the orthonormal basis ê1* , ê *2 ,
ê *3 , are given by:
Wij*
0 
0 0

= 0 0 − ω 
0 ω 0 
(1.149)
In Figure 1.17 we can see these components and the axial vector representation. Note that
if we take any basis that is formed just by rotation along the ê1* -axis, the components of
W in this new basis will be the same as those provided in (1.149).
x3
x2
r
w = ωê1*
ê*2
Wij*
0 
0 0

= 0 0 − ω 
0 ω 0 
ê1*
ω
x1
ω
ê*3
Figure 1.17: Antisymmetric tensor components in the space defined by the axial vector.
1.5.2.2.3
Additive decomposition. Symmetric and antisymmetric part
Any arbitrary second-order tensor A can be split additively into a symmetric and an
antisymmetric part:
A=
1
1
( A + A T ) + ( A − A T ) = A sym + A skew
2 4243 1
2 4243
1
A
sym
A
(1.150)
skew
or, into its components:
A ijsym =
1
( A ij + A ji ) and
2
A ijskew =
1
( A ij − A ji )
2
(1.151)
If A and B are arbitrary second-order tensors, it holds that:
(A
T
⋅B ⋅ A)
sym
(
) (
)
[
T
1 T
1
A ⋅ B ⋅ A + A T ⋅ B ⋅ A  = A T ⋅ B ⋅ A + A T ⋅ BT ⋅ A




2
2
1
= A T ⋅ B + B T ⋅ A = A T ⋅ B sym ⋅ A
2
=
[
]
]
(1.152)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
41
Problem 1.15: Show that σ : W = 0 is always true when σ is a symmetric second-order
tensor and W is an antisymmetric second-order tensor.
Solution:
σ : W = σ ij (eˆ i ⊗ eˆ j ) : Wlk (eˆ l ⊗ eˆ k ) = σ ij Wlk δ il δ jk = σ ij Wij (scalar)
Thus,
σ ij Wij = σ1 j W1 j + σ 2 j W2 j + σ 3 j W3 j
123
1
424
3
1
424
3
σ11W11
σ21W21
σ31W31
+
+
+
σ12 W12
σ 22 W22
σ32 W32
+
+
+
σ13W13
σ23W23
σ33W33
Taking into account the characteristics of a symmetric and an antisymmetric tensor, i.e.
σ12 = σ 21 , σ 31 = σ13 , σ 32 = σ 23 , and W11 = W22 = W33 = 0 , W21 = − W12 , W31 = − W13 ,
W32 = − W23 , the equation above becomes:
σ :W =0
r
r
r
r
Problem 1.16: Show that a) M ⋅ Q ⋅ M = M ⋅ Q sym ⋅ M ; b) A : B = A sym : B sym + A skew : B skew
r
where M is a vector, and Q , A , B are arbitrary second-order tensors.
Solution:
r
r r
r r
r r
r
a) M ⋅ Q ⋅ M = M ⋅ (Q sym + Q skew )⋅ M = M ⋅ Q sym ⋅ M + M ⋅ Q skew ⋅ M
r
(r
r
r
)
Since the relation M ⋅ Q skew ⋅ M = Q skew : 1
M ⊗ M = 0 holds, it follows that:
424
3
symmetric tensor
r
r r
r
M ⋅ Q ⋅ M = M ⋅ Q sym ⋅ M
b)
= ( A sym + A skew ) : (B sym + B skew )
A :B
skew
sym
skew
= A sym : B sym + 1
A sym
: B43
+1
A skew
: B skew
42
42: B
43 + A
=0
= A sym : B sym + A skew : B skew
=0
Then, it is also valid that:
A : B sym = A sym : B sym
;
A : B skew = A skew : B skew
r
Problem 1.17: Let T be an arbitrary second-order tensor, and n be a vector. Check if the
r
r
relationship n ⋅ T = T ⋅ n is valid.
Solution:
r
n ⋅ T = n i eˆ i ⋅ Tkl (eˆ k ⊗ eˆ l )
= n i Tkl δ ik eˆ l
= n k Tkl eˆ l
and
r
T ⋅ n = Tlk (eˆ l ⊗ eˆ k ) ⋅ n i eˆ i
= n i Tlk δ ki eˆ l
= n k Tlk eˆ l
= (n1 T1l + n 2 T2 l + n 3 T3l )eˆ l
= (n1 Tl1 + n 2 Tl 2 + n 3 Tl 3 )eˆ l
With the above we can prove that n k Tkl ≠ n k Tlk , then:
r
r
n⋅ T ≠ T ⋅n
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
42
r
r
If T is a symmetric tensor, it follows that the relationship n ⋅ T sym = T sym ⋅ n holds.
r
Problem 1.18: Obtain the axial vector w associated with the antisymmetric tensor
r r
( x ⊗ a ) skew .
r
Solution: Let z be an arbitrary vector, it then holds that:
r r
r r r
( x ⊗ a ) skew ⋅ z = w ∧ z
r
r r
where w is the axial vector associated with ( x ⊗ a ) skew . Using the definition of an
antisymmetric tensor:
[
]
r r
r r
1 r r r r
1 r r
( x ⊗ a ) skew = ( x ⊗ a ) − ( x ⊗ a ) T = [ x ⊗ a − a ⊗ x ]
2
2
r r skew r r r
and by replacing it with ( x ⊗ a ) ⋅ z = w ∧ z , we obtain:
1 r r r r r r r
[x ⊗ a − a ⊗ x ] ⋅ z = w ∧ z ⇒ [xr ⊗ ar − ar ⊗ xr ] ⋅ zr = 2wr ∧ zr
2
r r r r r r
r r
By using the equation [x ⊗ a − a ⊗ x ] ⋅ z = z ∧ ( x ∧ a ) , (see Eq. (1.105)), the above equation
becomes:
[xr ⊗ ar − ar ⊗ xr ] ⋅ zr = zr ∧ ( xr ∧ ar ) = (ar ∧ xr ) ∧ zr = 2wr ∧ zr
with the above we can conclude that:
r 1 r r
r r
w = (a ∧ x ) is the axial vector associated with ( x ⊗ a ) skew
2
1.5.2.3
Cofactor Tensor. Adjugate of a Tensor
r
r
Let A be a second-order tensor and a , b be arbitrary vectors then there is then a unique
tensor cof(A ) , known as the cofactor of A , as we can see below:
r
r r
r
cof( A ) ⋅ (a ∧ b) = ( A ⋅ a) ∧ ( A ⋅ b)
(1.153)
We can also define the adjugate of A as:
adj( A ) = [cof (A )]
T
(1.154)
which satisfies the following condition:
[adj(A)]T
= adj( A T )
(1.155)
The components of cof(A ) are obtained by expressing the equation (1.153) in terms of its
components, i.e.:
[cof(A)]it  tpr a p b r =  ijk A jp a p A kr b r
⇒
[cof(A )]it  tpr =  ijk A jp A kr
(1.156)
By multiplying both sides of the equation by  qpr and by also considering that
 tpr  qpr = 2δ tq , we can conclude that:
[cof(A)]it  tpr =  ijk A jp A kr
⇒
[cof(A)]it  tpr  qpr =  ijk  qpr A jp A kr
1
424
3
= 2δ tq
⇒ [cof( A )]iq
1.5.2.4
1
=  ijk  qpr A jp A kr
2
(1.157)
Tensor Trace
Let’s start by defining the trace of the basis (eˆ i ⊗ eˆ j ) :
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
43
Tr (eˆ i ⊗ eˆ j ) = eˆ i ⋅ eˆ j = δ ij
(1.158)
Thus, we can define the trace of a second-order tensor A as follows:
Tr ( A ) = Tr ( A ij eˆ i ⊗ eˆ j ) = A ij Tr (eˆ i ⊗ eˆ j ) = A ij (eˆ i ⋅ eˆ j ) = A ij δ ij = A ii
= A 11 + A 22 + A 33
r r
And, the trace of the dyad (u ⊗ v ) can be evaluated as:
r r
r r
Tr (u ⊗ v ) = Tr (u ⊗ v ) = u i v j Tr (eˆ i ⊗ eˆ j ) = u i v j (eˆ i ⋅ eˆ j ) = u i v j δ ij = u i v i
r r
= u1 v 1 + u 2 v 2 + u 3 v 3 = u ⋅ v
(1.159)
(1.160)
NOTE: As we will show later, the tensor trace is an invariant, i.e. it is independent of the
coordinate system. ■
Let A , B be arbitrary tensors, then:
ƒ
The transposed tensor trace is equal to the tensor trace:
( )
Tr A T = Tr (A )
ƒ
(1.161)
The trace of (A + B ) is the sum of traces:
Tr(A + B ) = Tr (A ) + Tr (B )
(1.162)
Tr(A + B ) = Tr (A ) + Tr (B )
[(A 11 + B 11 ) + (A 22 + B 22 ) + (A 33 + B 33 )] = (A 11 + A 22 + A 33 ) + (B 11 + B 22 + B 33 )
(1.163)
Proving this is very simple:
ƒ
The scalar product trace ( A ⋅ B) becomes:
[
Tr(A ⋅ B ) = Tr ( A ij eˆ i ⊗ eˆ j ) ⋅ (B lm eˆ l ⊗ eˆ m )
= A ij B lm δ jl Tr (eˆ i ⊗ eˆ m )
144244
3
[
]
]
δ im
(1.164)
= A il B li = A ⋅ ⋅B = Tr ( B ⋅ A )
and, the double scalar product ( : ) can be expressed in trace terms as:
A : B = A ij B ij
= A kj B lj δ ik δ il
= A kj B lj δ kl
123
( A⋅BT )
(
kl
= A ⋅ BT
)
kk
= Tr ( A ⋅ B )
T
= A ik B il δ jk δ jl
= A ik B il δ kl
123
( AT ⋅B )
(
kl
= AT ⋅B
= Tr ( A
T
(1.165)
)
kk
⋅ B)
Similarly, it is possible to show that:
Tr(A ⋅ B ⋅ C ) = Tr (B ⋅ C ⋅ A ) = Tr (C ⋅ A ⋅ B ) = A ij B jk C ki
(1.166)
Tr (A ) = A ii
[Tr (A )]2 = Tr (A ) Tr(A ) = A ii A jj
Tr(A ⋅ A ) = Tr (A 2 ) = A il A li ; Tr(A ⋅ A ⋅ A ) = Tr (A 3 ) = A ij A jk A ki
(1.167)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
44
Problem 1.19: Let T be a second-order tensor. Show that:
(T ) = (T )
m T
Solution:
(T )
T m
( )
Tr T T
and
( )
= Tr T m .
( )
= (T ⋅ T L T ) = T T ⋅ T T L T T = T T
m T
m
T
m
For the second demonstration we can use the trace property Tr (T T ) = Tr (T ) , thus:
( )
Tr T T
m
( )
= Tr T m
T
( )
= Tr T m
1.5.2.5
Particular Tensors
1.5.2.5.1
Unit Tensors
ƒ
The second-order unit tensor, also called the identity tensor, is defined as:
1 = δ ij eˆ i ⊗ eˆ j = eˆ i ⊗ eˆ i = 1 eˆ i ⊗ eˆ j
(1.168)
where 1 is the matrix with the components of tensor 1 . δ ij is the Kronecker delta symbol
defined in (1.48).
ƒ
Fourth-order unit tensors can be defined as follows:
I = 1⊗1 = δ ik δ jl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = I ijkl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
(1.169)
I = 1⊗1 = δ il δ jk eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = I ijkl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
(1.170)
I = 1 ⊗ 1 = δ ij δ kl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = I ijkl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
(1.171)
Taking into account the fourth-order unit tensors defined above, it holds that:
(
)(
)
(
I : A = δ ik δ jl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l : A pq eˆ p ⊗ eˆ q = δ ik δ jl A pq δ kp δ lq eˆ i ⊗ eˆ j
= δ ik δ jl A kl eˆ i ⊗ eˆ j = A ij eˆ i ⊗ eˆ j
(
)
(
)
)
(1.172)
=A
and
(
)(
)
(
I : A = δ il δ jk eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l : A pq eˆ p ⊗ eˆ q = δ il δ jk A pq δ kp δ lq eˆ i ⊗ eˆ j
= δ il δ jk A kl eˆ i ⊗ eˆ j = A ji eˆ i ⊗ eˆ j
=A
(
)
(
)
)
(1.173)
T
and
(
)(
(
)
(
I : A = δ ij δ kl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l : A pq eˆ p ⊗ eˆ q = δ ij δ kl A pq δ kp δ lq eˆ i ⊗ eˆ j
= δ ij δ kl A kl eˆ i ⊗ eˆ j = A kk δ ij eˆ i ⊗ eˆ j
(
)
)
)
(1.174)
= Tr ( A )1
The symmetric part of the fourth-order unit tensor I is defined as:
I sym ≡ I =
(
)
(
1
1
 → I ijkl = δ ik δ jl + δ il δ jk
1⊗1 + 1⊗1 incomponents
2
2
)
(1.175)
The property of the tensor product ⊗ is presented below. Consider a second-order unit
tensor, 1 = δ ij eˆ i ⊗ eˆ j . Then, the tensor product ⊗ can be defined as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
(
)
45
(
1⊗1 = δ ij eˆ i ⊗ eˆ j ⊗(δ kl eˆ k ⊗ eˆ l ) = δ ij δ kl eˆ i ⊗ eˆ k ⊗ eˆ j ⊗ eˆ l
)
(1.176)
which is the same as:
(
I = 1⊗1 = δ ik δ jl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
)
(1.177)
and the tensor product ⊗ is defined as:
(
)
(
1⊗1 = δ ij eˆ i ⊗ eˆ j ⊗(δ kl eˆ k ⊗ eˆ l ) = δ ij δ kl eˆ i ⊗ eˆ l ⊗ eˆ k ⊗ eˆ j
or
(
I = 1⊗1 = δ il δ jk eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l
)
(1.178)
)
(1.179)
The antisymmetric part of I is defined as:
I skew =
(
1
1⊗1 − 1⊗1
2
)
incomponents
 →
skew
I ijk
l =
r
(
1
δ ik δ jl − δ il δ jk
2
)
(1.180)
With a second-order tensor A and a vector b , the following relationships are valid:
r
r
b ⋅1 = b
I:A = A
I sym : A = A sym
;
A : 1 = Tr (A ) = A ii
( )
: 1 = Tr (A ) = Tr (A ⋅ A ⋅ A ) = A
(1.181)
A 2 : 1 = Tr A 2 = Tr (A ⋅ A ) = A il A li
A3
3
ij A jk A ki
Problem 1.20: Show that T : 1 = Tr (T ) , where T is an arbitrary second-order tensor.
Solution:
T : 1 = Tij eˆ i ⊗ eˆ j : δ kl eˆ k ⊗ eˆ l
= Tij δ kl δ ik δ jl
= Tij δ ij = Tii = T jj
= Tr ( T )
1.5.2.5.2
Levi-Civita Pseudo-Tensor
The Levi-Civita Pseudo-Tensor, also known as the Permutation Tensor, is a third-order pseudotensor and is denoted by:
 =  ijk eˆ i ⊗ eˆ j ⊗ eˆ k
(1.182)
which is not a “true” tensor in the strict meaning of the word, and whose components  ijk
were defined in (1.55), the permutation symbol.
1.5.2.6
Determinant of a Tensor
The determinant of a tensor is a scalar and is expressed as:
det ( A ) ≡ A =  ijk A 1i A 2 j A 3k =  ijk A i1 A j 2 A k 3
144244
3
AT
(1.183)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
46
It is also an invariant (independent of the adopted system). Demonstrating the equation
above (1.183) can be done starting directly from the determinant:
A 11
A 12
A 13
det ( A ) = A = A 21 A 22 A 23
A 31 A 32 A 33
= A 11 ( A 22 A 33 − A 23 A 32 ) − A 21 ( A 12 A 33 − A 13 A 32 ) + A 31 ( A 12 A 23 − A 13 A 22 )
= A 11 (1 jk A j 2 A k 3 ) − A 21 (− 2 jk A j 2 A k 3 ) + A 31 ( 3 jk A j 2 A k 3 )
= 1 jk A 11 A j 2 A k 3 +  2 jk A 21 A j 2 A k 3 +  3 jk A 31 A j 2 A k 3
=  ijk A i1 A j 2 A k 3
(1.184)
Some determinant properties with second-order tensors are described below:
det (1) = 1
ƒ
(1.185)
We can conclude from (1.183) that:
det ( A T ) = det ( A )
ƒ
(1.186)
We can also show that:
det ( A ⋅ B) = det ( A )det (B) , det (αA ) = α 3 det ( A )
where α is a scalar
(1.187)
ƒ
A tensor (A ) is said to be singular if det ( A ) = 0 .
ƒ
If you exchange two rows or columns, the determinant sign changes.
ƒ
If all elements of a row or column equal zero, the determinant is also zero.
ƒ
If you multiply all the elements of a row or column by a constant c (scalar), the
determinant is c A .
ƒ
If two or more rows (or column) are linearly dependent, the determinant is zero.
Problem 1.21: Show that A  tpq =  rjk A rt A jp A kq .
Solution:
We start with the following definition:
A =  rjk A r1 A j 2 A k 3
⇒
A  tpq =  rjk  tpq A r1 A j 2 A k 3
and also taking into account that the term  rjk  tpq can be replaced by (1.61):
 rjk  tpq
δ rt
= δ jt
δ kt
δ rp
δ jp
δ kp
δ rq
δ jq
δ kq
(1.188)
(1.189)
= δ rt δ jp δ kq + δ rp δ jq δ kt + δ rq δ jt δ kp − δ rq δ jp δ kt − δ jq δ kp δ rt − δ kq δ jt δ rp
Then, by substituting (1.189) into (1.188) we can obtain:
A  tpq = A t1 A p 2 A q 3 + A p1 A q 2 A t 3 + A q1 A t 2 A p 3 − A q1 A p 2 A t 3 − A t1 A q 2 A p 3 − A p1 A t 2 A q 3
= A t1 ( 1 jk A pj A qk ) + A t 2 ( 2 jk A pj A qk ) + A t 3 ( 3 jk A pj A qk )
=  rjk A rt A jp A kq =  rjk A tr A pj A qk
1
6
Problem 1.22: Show that A =  rjk  tpq A rt A jp A kq .
Solution:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
47
Starting with the definition A  tpq =  rjk A rt A jp A kq , (see Problem 1.21), and by multiplying
both sides of the equation by  tpq , we obtain:
A  tpq  tpq =  rjk  tpq A rt A jp A kq
(1.190)
Using the property defined in expression (1.62), we obtain
 tpq  tpq = δ tt δ pp − δ tp δ tp = δ tt δ pp − δ tt = 6 . Then, the relationship (1.190) becomes:
A =
1
 rjk  tpq A rt A jp A kq
6
r
r
Problem 1.23: Let a , b be arbitrary vectors and α be a scalar. Show that:
(
)
r r
r r
det µ1 + αa ⊗ b = µ 3 + µ 2 α a ⋅ b
(1.191)
Solution: The determinant of A is given by A =  ijk A i1 A j 2 A k 3 . If we denote by
A ij µδ ij + αa i b j , thus, A i1 = µδ i1 + αa i b1 , A k 3 = µδ k 3 + αa k b 3 , A k 3 = δ k 3 + αa k b 3 , then
the equation in (1.191) can be rewritten as:
r r
det µ1 + αa ⊗ b =  ijk (µδ i1 + αa i b1 )(µδ j 2 + αa j b 2 )(µδ k 3 + αa k b 3 )
(1.192)
By developing the equation (1.192), we obtain:
r r
det µ1 + αa ⊗ b =  ijk [µ 3 δ i1δ j 2 δ k 3 + µ 2 αa k b 3 δ i1δ j 2 + µ 2 αa j b 2 δ i1δ k 3 + µ 2 αa i b 1δ j 2 δ k 3 +
(
(
)
)
+ µα 2 a j b 2 a k b 3 δ i1 + µα 2 a i a k b 1b 3 δ j 2 + µα 2 a i a j b 1b 2 δ k 3 + α 3 a i a j a k b 1b 2 b 3
]
Note that: µ 3  ijk δ i1δ j 2 δ k 3 = µ 3 123 = µ 3 ,
µ 2 α ( ijk a k b 3 δ i1δ j 2 +  ijk a j b 2 δ i1δ k 3 +  ijk a i b1δ j 2 δ k 3 ) =
r r
µ 2 α (12 k a k b 3 +  1 j 3 a j b 2 +  i 23 a i b1 ) = µ 2 α (a 3b 3 + a 2 b 2 + a1b1 ) = µ 2 α (a ⋅ b)
 ijk a i a k b1b 3 δ j 2 =  i 2 k a i a k b1b 3 = a1a 3b1b 3 − a 3 a1b1b 3 = 0
 ijk a i a j b1b 2 δ k 3 =  ij 3 a i a j b1b 2 =  123 a1 a 2 b1b 2 −  213 a 2 a1b1b 2 = 0
 ijk a i a j a k b1b 2 b 3 = 0
Notice that, there was no need to expand the terms  ijk a i a k b1b 3 δ j 2 ,  ijk a i a j b1b 2 δ k 3 , and
 ijk a i a j a k b1b 2 b 3 to realize that these terms equal zero, since
r
r
 ijk a i a k b1b 3 δ j 2 = (a ∧ a) j b1b 3 δ j 2 = 0 , similarly for other terms.
Taking into account the above considerations we can prove that:
r r
r r
det µ1 + αa ⊗ b = µ 3 + µ 2 α a ⋅ b
For the particular case when µ = 1 the above equation becomes:
(
)
(
(
)
r r
r r
det 1 + αa ⊗ b = 1 + α a ⋅ b
r r
Then, it is simple to prove that det αa ⊗ b = 0 , since
r r r
r r
det αa ⊗ b = α 3  ijk a i a j a k b1b 2 b 3 = α 3b1b 2 b 3 [a ⋅ (a ∧ a)] = 0
(
)
The following relations are satisfied:
[
]
)
[
]
r r
r r
r r
r r
r r
r r r r
(1.193)
det 1 + α (a ⊗ b) + β (b ⊗ a) = 1 + α (a ⋅ b) + β (a ⋅ b) + αβ (a ⋅ b) 2 − (a ⋅ a)(b ⋅ b)
r r
r r
where α , β are scalars. If β = 0 we can regain the equation det 1 + αa ⊗ b = 1 + αa ⋅ b ,
(see Problem 1.23). If α = β we obtain:
(
)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
48
(
)
( ) ( ) ( )
( ) ( )
r r
r r
r r
r r
r r
det 1 + α a ⊗ b + α b ⊗ a = 1 + α a ⋅ b + α a ⋅ b + α 2  a ⋅ b

r
r
2
r
r
= 1 + α 2 a ⋅ b − α a ∧ b 


(r r )
( )
r r r r
− (a ⋅ a) b ⋅ b 

2
(1.194)
( ) (r r )
r r r r
where we have used the property a ⋅ b − (a ⋅ a) b ⋅ b = − a ∧ b , (see Problem 1.1).
2
2
It is also true that:
(1.195)
det (α A + β B ) = α 3 det ( A ) + α 2 β Tr[B ⋅ adj( A )] + αβ 2 Tr[A ⋅ adj(B)] + β 3 det (B)
r r
Moreover, in the particular case when α = 1 , A = 1 , B = a ⊗ b , and bearing in mind that
r r
r r
det a ⊗ b = 0 , and cof a ⊗ b = 0 , we can conclude that:
r r
r r
r r
(1.196)
det 1 + βa ⊗ b = det (1) + βTr a ⊗ b ⋅ 1 = 1 + βTr a i b j = 1 + βa ⋅ b
(
)
(
)
(
)
[
]
[
]
which has already been demonstrated in Problem 1.23.
We next show that the following property is valid:
[
]
[
r
r
r
r r r
( A ⋅ a) ⋅ ( A ⋅ b) ∧ ( A ⋅ c ) = det ( A ) a ⋅ (b ∧ c)
]
(1.197)
To achieve this goal we start with the definition of the scalar triple product given in (1.69),
r r r
i.e. a ⋅ b ∧ c =  ijk a i b j c k , and by multiplying both sides of this equation by the
determinant of A we obtain:
(
)
(
)
r r r
a ⋅ b ∧ c A =  ijk a i b j c k A
(1.198)
It was proven in Problem 1.21 that A  ijk =  pqr A pi A qj A rk , thus:
(
)
r r r
a ⋅ b ∧ c A =  ijk a i b j c k A =  pqr A pi A qj A rk a i b j c k =  pqr ( A pi a i )( A qj b j )( A rk c k )
r
r
r
r
r
r
= ( A ⋅ a) ⋅ ( A ⋅ b) ∧ ( A ⋅ c ) = ( A ⋅ b) ∧ ( A ⋅ c) ⋅ ( A ⋅ a)
1.5.2.7
[
] [
]
(1.199)
Inverse of a Tensor
We use the notation A −1 to denote the inverse of A , which is defined as follows:
if A ≠ 0 ⇔ ∃ A −1 A ⋅ A −1 = A −1 ⋅ A = 1
(1.200)
Or in indicial notation:
if A ≠ 0 ⇔ ∃ A ij−1 A ik A kj−1 = A ik−1 A kj = δ ij
(1.201)
To obtain the inverse of a tensor we start from the definition of the adjugate tensor given
r
r r
r
in (1.153), i.e. adj( A T ) ⋅ (a ∧ b) = ( A ⋅ a) ∧ ( A ⋅ b) . Then by applying the dot product
r
between an arbitrary vector d and this equation we obtain:
{[adj(A)]
T
⋅ (a ∧ b)}⋅ d = [( A ⋅ a) ∧ ( A ⋅ b)]⋅ d = [( A ⋅ a) ∧ (A ⋅ b)]⋅ 1 ⋅ d
r
r
r
r
r
[
r
r
]
r
r
(1.202)
r
r
r
−1
= ( A ⋅ a) ∧ ( A ⋅ b) ⋅ A ⋅ A
⋅
d
12r3
(
)
=c
[
]
r
r r r
r
r
In equation (1.199) we demonstrated that c ⋅ a ∧ b A = ( A ⋅ a) ∧ ( A ⋅ b) ⋅ ( A ⋅ c) thus,
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
49
⋅ (a ∧ b)}⋅ d = [( A ⋅ a) ∧ ( A ⋅ b)]⋅ A ⋅ A −1 ⋅ d =
{[adj(A)]
r
r
T
r
r
r
r
r
(
)
r
r r
A a ∧ b ⋅ A −1 ⋅ d
(1.203)
r
r
Denoted by p = (a ∧ b) , the above equation (1.203) can be rearranged as follows:
{[adj(A)]ki p k }di = A p k A ki−1di
⇒ [adj( A )]ki p k d i = A A ki−1p k d i
r
r
r r
r r
⇒ [adj( A )] : [(a ∧ b) ⊗ d] = A A −1 : [(a ∧ b) ⊗ d]
(1.204)
Thus, we can conclude that:
[adj(A)] = A A −1
ƒ
⇒
A −1 =
1
[adj(A)] = 1 [cof(A)]T
A
A
(1.205)
Let A and B be invertible tensors, the following properties hold:
( A ⋅ B) −1 = B −1 ⋅ A −1
(A )
−1 −1
=A
(1.206)
(βA )−1 = 1 A −1
β
det ( A ) = [det ( A )]
−1
−1
ƒ
The following notation will be used to represent the inverse transpose:
A −T ≡ ( A −1 ) T ≡ ( A T ) −1
(1.207)
Next, we prove the relation adj( A ⋅ B) = adj(B) ⋅ adj( A ) holds. To do this, we take the
definition of the inverse of a tensor given in (1.205) as a starting point:
B −1 ⋅ A −1 =
[adj(B)] ⋅ [adj(A)] ⇒ A B B −1 ⋅ A −1 = [adj(B)] ⋅ [adj(A)]
B
A
⇒ A B (A ⋅ B ) = [adj(B)] ⋅ [adj( A )] ⇒ A B
−1
⇒ adj( A ⋅ B) = [adj(B)] ⋅ [adj( A )]
[adj(A ⋅ B)] = [adj(B)] ⋅ [adj(A)]
A ⋅B
(1.208)
where we have used the property A ⋅ B = A B . Similarly, it is possible to show that
cof( A ⋅ B) = [cof( A )] ⋅ [cof(B)] .
Procedure for obtaining the inverse of the matrix A
1) Obtain the cofactor matrix: cof (A ) as follows:
Consider the matrix A as:
 A 11
A = A 21
 A 31
A 12
A 22
A 32
A 13 
A 23 
A 33 
(1.209)
We define the matrix M , where the component Mij is the determinant of the resulting
matrix by removing the i th row and the j th column, i.e.:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
50
 A 22

 A 32
A
M =  12
 A 32
 A 12

 A 22
A 23
A 33
A 13
A 21
A 31
A 11
A 23
A 33
A 13
A 21
A 31
A 11
A 22
A 32
A 12
A 33
A 13
A 31
A 11
A 33
A 13
A 23
A 21
A 23
A 31
A 11
A 21
A 32
A 12
A 22









(1.210)
Thus, we define the cofactor matrix as:
cof (A ) = (−1) i + j Mij
(1.211)
2) Obtain the adjugate matrix, adj(A ) , as follows:
adj(A ) = [cof (A )]
T
3) Obtain the inverse matrix:
A −1 =
(1.212)
adj(A )
(1.213)
A
So, the relation A [adj(A )] = A 1 holds, where 1 is the identity matrix.
Taking into account (1.64), we can express the components of the first, second, and third
row of the cofactor matrix, (1.211), as: M1i =  ijk A 2 j A 3k , M 2i =  ijk A 1 j A 3k , M3i =  ijk A 1 j A 2 k ,
respectively.
Problem 1.24: Let A be an arbitrary second-order tensor. Show that there is a nonzero
r r
r r
vector n ≠ 0 so that A ⋅ n = 0 if and only if det ( A ) = 0 , Chadwick (1976).
r
r
Solution: Firstly, we show that, if det ( A ) ≡ A = 0 ⇒ n ≠ 0 . Secondly, we show that, if
r r
n ≠ 0 ⇒ det ( A ) ≡ A = 0 .
r r r
We assume that det ( A ) ≡ A = 0 , and we choose an arbitrary basis {f , g, h} (linearly
independent). We apply these vectors in the definition seen in (1.197):
(
[
)
r r r
r
r
r
f ⋅ g ∧ h A = ( A ⋅ f ) ⋅ ( A ⋅ g) ∧ ( A ⋅ h)
]
Due to the fact that det ( A ) ≡ A = 0 , the implication is that:
[
]
r
r
r
( A ⋅ f ) ⋅ ( A ⋅ g) ∧ ( A ⋅ h) = 0
r
r
r
Thus, we can conclude that the vectors ( A ⋅ f ) , ( A ⋅ g) , ( A ⋅ h) , are linearly dependent.
This implies that there are nonzero scalars α , β , γ so that:
r
r
r
r
r
(r
r
r
)
r
r
r
α ( A ⋅ f ) + β ( A ⋅ g) + γ ( A ⋅ h) = 0 ⇒ A ⋅ αf + βg + γh = 0 ⇒ A ⋅ n = 0
r
r
r
r r r
r
where n = αf + βg + γh ≠ 0 since {f , g, h} is linearly independent, (see Problem 1.10).
r
r
r
Now we choose two vectors k , m , which are linearly independent to n . We apply
definition (1.199) once more:
r r r
r
r
r
k ⋅ (m ∧ n) A = ( A ⋅ k ) ⋅ [( A ⋅ m) ∧ ( A ⋅ n)]
r r r
r r r
r r
Considering that A ⋅ n = 0 , and k ⋅ (m ∧ n) ≠ 0 owing to the fact that k , m , n are linearly
independent, we can conclude that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
r r r
k ⋅ (m ∧ n) A = 0
14243
⇒
51
A =0
≠0
1.5.2.8
Orthogonal Tensors
Orthogonal tensors play an important role in continuum mechanics. A second-order tensor
( Q ) is said to be orthogonal when the transpose ( Q T ) is equal to the inverse ( Q −1 ), i.e.
Q T = Q −1 . Then, it follows that:
Q ⋅ QT = QT ⋅ Q = 1
Q ik Q jk = Q ki Q kj = δ ij
;
(1.214)
A proper orthogonal tensor has the following properties:
ƒ
The inverse of Q is equal to the transpose (orthogonality):
Q −1 = Q T
ƒ
(1.215)
The tensor Q is a proper, rotation tensor, if:
det (Q) ≡ Q = +1
(1.216)
If Q = −1 , the orthogonal tensor is said to be improper (a reflection tensor).
If A and B are orthogonal tensors, the resulting tensor A ⋅ B = C is also an orthogonal
tensor, i.e.:
(1.217)
C −1 = ( A ⋅ B) −1 = B −1 ⋅ A −1 = B T ⋅ A T = ( A ⋅ B) T = C T
r
r
Consider two arbitrary vectors a and b . An orthogonal transformation applied to these
vectors becomes:
r
r
~
b = Q ⋅b
r
r
~
And the dot product between these new vectors ( ~a ) and ( b ) is given by:
r r
r
r r r
r
r
~ ~
a ⋅ b = (Q ⋅ a) ⋅ (Q ⋅ b) = a ⋅ Q T ⋅ Q ⋅ b = a ⋅ b
123
r
r
~
a = Q⋅a
;
=1
~~
ai b i = (Q ik a k )(Q ij b j ) = a k Q ik Q ij b j = a k b k
123
(1.218)
(1.219)
δ kj
r r r 2 r r r 2
r ~r
~ ~
So, it is also true when ~a = b , thus ~a ⋅ a
= a = a ⋅ a = a . Therefore, we can conclude
that in an orthogonal transformation, the magnitude vectors and the angle between them
are maintained, (see Figure 1.18).
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
52
Q
r
r
~
a = a
r
b
θ
r
~
b
θ
r
a
r
~
a
r
r
~
b = b
r r
~ ~ r r
a⋅b = a⋅b
Figure 1.18: Orthogonal transformation applied to vectors.
1.5.2.9
Positive Definite Tensor, Negative Definite Tensor and SemiDefinite Tensors
A tensor is said to be positive definite when the following notations hold:
Tensorial notation
Indicial notation
Matrix notation
(1.220)
x i Tij x j > 0
xˆ ⋅ T ⋅ xˆ > 0
xT T x > 0
r
for all vectors xˆ ≠ 0 . Conversely, a tensor is said to be negative definite when these notations
hold:
Tensorial notation
Indicial notation
Matrix notation
x i Tij x j < 0
xT T x < 0
xˆ ⋅ T ⋅ xˆ < 0
r r
for all vectors x ≠ 0 .
(1.221)
r
A tensor is said to be semi-positive definite if xˆ ⋅ T ⋅ xˆ ≥ 0 for all vectors xˆ ≠ 0 . Similarly, we
define a semi-negative definite tensor when the following holds: xˆ ⋅ T ⋅ xˆ ≤ 0 .
r
If α = xˆ ⋅ T ⋅ xˆ = T : ( xˆ ⊗ xˆ ) = Tij x i x j , then the derivative of α with respect to x is given
by:
∂x j
∂x i
∂α
x j + Tij x i
= Tij δ ik x j + Tij x i δ jk = Tkj x j + Tik x i = (Tki + Tik ) x i
= Tij
∂x k
∂x k
∂x k
(1.222)
Thus, we can conclude that:
∂α
= 2 T sym ⋅ xˆ
∂xˆ
⇒
∂ 2α
= 2 T sym
∂xˆ ⊗ ∂xˆ
(1.223)
Remember that it is also true that xˆ ⋅ T ⋅ xˆ = xˆ ⋅ T sym ⋅ xˆ , therefore if the symmetric part of
a tensor is positive definite the tensor is too.
NOTE: As we will demonstrate later, the eigenvalues must be positive for T to be
positive definite. The proof is in the subsection “Spectral Representation of Tensors”. ■
Problem 1.25: Let F be an arbitrary second-order tensor. Show that the resulting tensors
C = F T ⋅ F and b = F ⋅ F T are symmetric tensors and semi-positive definite tensors. Also check in
what condition are C and b positive definite tensors.
Solution: Symmetry:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
53
C T = (F T ⋅ F )T = F T ⋅ (F T )T = F T ⋅ F = C
b T = (F ⋅ F T ) T = (F T )T ⋅ F T = F ⋅ F T = b
Thus, we have shown that C = F T ⋅ F and b = F ⋅ F T are symmetric tensors.
To prove that the tensors C = F T ⋅ F and b = F ⋅ F T are semi-positive definite tensors, we
start with the definition of a semi-positive definite tensor, i.e., a tensor A is semi-positive
r
definite if xˆ ⋅ A ⋅ xˆ ≥ 0 holds, for all xˆ ≠ 0 . Thus:
xˆ ⋅ ( F T ⋅ F ) ⋅ xˆ = F ⋅ xˆ ⋅ F ⋅ xˆ
= ( F ⋅ xˆ ) ⋅ ( F ⋅ xˆ )
= F ⋅ xˆ
Or in indicial notation:
2
xˆ ⋅ ( F ⋅ F T ) ⋅ xˆ = xˆ ⋅ F ⋅ F T ⋅ xˆ
= ( F T ⋅ xˆ ) ⋅ ( F T ⋅ xˆ )
2
= F T ⋅ xˆ ≥ 0
≥0
= x i ( Fki Fkj ) x j
= ( Fki x i )( Fkj x j )
x i C ij x j
= Fki x i
2
x i bij x j
≥0
= x i ( Fik F jk ) x j
= ( Fik x i )( F jk x j )
= Fik x i
2
≥0
Thus, we proved that C = F T ⋅ F and b = F ⋅ F T are semi-positive definite tensors. Note
that xˆ ⋅ C ⋅ xˆ = F ⋅ xˆ
2
r
r
equals zero, when xˆ ≠ 0 , if F ⋅ xˆ = 0 . Furthermore, by definition
r
r
F ⋅ xˆ = 0 with xˆ ≠ 0 if and only if det ( F ) = 0 , (see Problem 1.24). Then, the tensors
C = F T ⋅ F and b = F ⋅ F T are positive definite if and only if det ( F ) ≠ 0 .
1.5.2.10 Additive Decomposition of Tensors
Given two arbitrary tensors S , T ≠ 0 , and a scalar α , we can represent S by means of the
following additive decomposition of tensors:
S = αT + U
where
U = S − αT
(1.224)
Note that, depending on the value of α , we have an infinite number of possibilities for
representing S . But, if Tr ( T ⋅ UT ) = Tr (U ⋅ T T ) = 0 , the additive decomposition is unique.
From the relationship in (1.224), we can evaluate the value of α as follows:
S ⋅ T T = αT ⋅ T T + U ⋅ T T ⇒ Tr (S ⋅ T T ) = αTr ( T ⋅ T T ) + Tr (U ⋅ T T ) = αTr ( T ⋅ T T )
14243
⇒
α=
Tr (S ⋅ T T )
Tr ( T ⋅ T T )
=0
(1.225)
(1.226)
For example, let us suppose that T = 1 . In this case α is evaluated as follows:
α=
Tr (S ⋅ T T ) Tr (S ⋅ 1) Tr (S ) Tr (S )
=
=
=
3
Tr ( T ⋅ T T ) Tr (1 ⋅ 1) Tr (1)
(1.227)
We can then define U as:
Tr (S )
1 ≡ S dev
3
(1.228)
Tr (S )
1 + S dev = S sph + S dev
3
(1.229)
U = S − αT = S −
Thus:
S=
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
54
NOTE: S sph =
Tr (S )
Tr (S )
1 is the spherical part of the tensor S , and S dev = S −
1 is
3
3
known as a deviatoric tensor. ■
1
2
Now suppose that T is given by T = (S + S T ) then α can be evaluated as follows:
[
]
1
Tr S ⋅ (S + S T ) T
Tr (S ⋅ T T )
2
=
α=
=1
Tr ( T ⋅ T T ) 1 Tr (S + S T ) ⋅ (S + S T ) T
4
[
(1.230)
]
1
2
1
2
We can then define U as U = S − αT = S − (S + S T ) = (S − S T ) . Then we obtain S
represented by the additive decomposition as follows:
S=
1
1
(S + S T ) + (S − S T ) = S sym + S skew
2
2
(1.231)
which is the same as the equation obtained in (1.150) in which we split the tensor into
symmetric and antisymmetric parts.
Problem 1.26: Find a fourth-order tensor P so that P : A = A dev , where A is a secondorder tensor.
Solution: Taking into account the additive decomposition into spherical and deviatoric parts,
we obtain:
A = A sph + A dev =
Tr ( A )
1 + A dev
3
⇒
A dev = A −
Tr ( A )
1
3
Referring to the definition of fourth-order unit tensors seen in (1.172), and (1.174), where
the relations I : A = Tr ( A )1 and I : A = A hold, we can now state:
A dev = A −
Tr ( A )
1
1 
1



1 = I : A − I : A =  I − I  : A =  I − 1 ⊗ 1 : A
3
3
3 
3



Therefore, we can conclude that:
1
P = I − 1 ⊗1
3
The tensor P is known as a fourth-order projection tensor, Holzapfel(2000).
1.5.3
Transformation Law of the Tensor Components
The tensor components depend on the coordinate system, so, if the coordinate system is
changed due to a rotation so do the tensor components. The tensor components between
these coordinate systems are interrelated to each other by the component transformation
law, which is defined below, (see Figure 1.19).
Consider a Cartesian coordinate system (x1 , x 2 , x3 ) formed by the orthogonal basis
(eˆ 1 , eˆ 2 , eˆ 3 ) , (see Figure 1.20). In this system, an arbitrary vector vr is represented by its
components as follows:
r
v = v i eˆ i = v 1 eˆ 1 + v 2 eˆ 2 + v 3 eˆ 3
(1.232)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
55
TENSORS
Mathematical interpretation of physical concepts
(Independent of the coordinate system)
COMPONENTS
Representation of a tensor in
a coordinate system
COMPONENT
TRANSFORMATION
LAW
COORDINATE SYSTEM
I
COORDINATE SYSTEM
II
Figure 1.19: Transformation law of the tensor components.
x3
x 2′
γ1
x3′
x1′
ê′2
ê′3
ê3
ê1′
β1
x2
ê 2
ê1
α1
x1
Figure 1.20: Rotation of the Cartesian system.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
56
The components, v i , are represented in matrix form as:
 v1 
r
( v ) i = v i = v =  v 2 
 v 3 
(1.233)
Now consider a new coordinate system (x1′ , x 2′ , x3′ ) represented by the orthogonal basis
(eˆ 1′ , eˆ ′2 , eˆ ′3 ) , (see Figure 1.20). In this new system, the vector vr is represented by v ′j eˆ ′j . As
mentioned before, a tensor is independent of the adopted system, so:
r
v = v ′k eˆ ′k = v j eˆ j
(1.234)
To obtain the components of a tensor in a given system one need only make the dot
product between the tensor and the system basis:
v ′k eˆ ′k ⋅ eˆ ′i = ( v j eˆ j ) ⋅ eˆ ′i
v ′k δ ki = ( v j eˆ j ) ⋅ eˆ ′i
(1.235)
v ′i = ( v 1 eˆ 1 + v 2 eˆ 2 + v 3 eˆ 3 ) ⋅ eˆ ′i
Or in matrix form:
 v 1′   ( v 1eˆ 1 + v 2 eˆ 2 + v 3 eˆ 3 ) ⋅ eˆ 1′ 
 v ′  = ( v eˆ + v eˆ + v eˆ ) ⋅ eˆ ′ 
2 2
3 3
2
 2  1 1
 v ′3   ( v 1 eˆ 1 + v 2 eˆ 2 + v 3 eˆ 3 ) ⋅ eˆ ′3 
(1.236)
After restructuring, the previous equation looks like:
 v 1′   eˆ 1 ⋅ eˆ 1′
 v ′  = eˆ ⋅ eˆ ′
 2  1 2
 v ′3   eˆ 1 ⋅ eˆ ′3
eˆ 2 ⋅ eˆ 1′
eˆ 2 ⋅ eˆ ′2
eˆ 2 ⋅ eˆ ′3
eˆ 3 ⋅ eˆ 1′   v 1 

eˆ 3 ⋅ eˆ ′2   v 2 
eˆ 3 ⋅ eˆ ′3   v 3 
∴
a ij = eˆ j ⋅ eˆ ′i = eˆ ′i ⋅ eˆ j
(1.237)
Or in indicial notation:
v ′i = a ij v j
(1.238)
where we have introduced the transformation matrix A ≡ aij as:
 eˆ 1 ⋅ eˆ 1′

A ≡ a ij = eˆ 1 ⋅ eˆ ′2
eˆ 1 ⋅ eˆ ′3

eˆ 2 ⋅ eˆ 1′
eˆ 2 ⋅ eˆ ′2
eˆ 2 ⋅ eˆ ′3
eˆ 3 ⋅ eˆ 1′   eˆ 1′ ⋅ eˆ 1
 
eˆ 3 ⋅ eˆ ′2  = eˆ ′2 ⋅ eˆ 1
eˆ 3 ⋅ eˆ ′3   eˆ ′3 ⋅ eˆ 1
 a11
a ij ≡ A =  a 21
 a 31
a12
a 22
a 32
a13 
a 23 
a33 
eˆ 1′ ⋅ eˆ 2
eˆ ′2 ⋅ eˆ 2
eˆ ′3 ⋅ eˆ 2
eˆ 1′ ⋅ eˆ 3 

eˆ ′2 ⋅ eˆ 3 
eˆ ′3 ⋅ eˆ 3 
(1.239)
The matrix ( A ) is not symmetric, i.e. A ≠ A T . With reference to the scalar product
eˆ ′i ⋅ eˆ j = eˆ ′i eˆ j cos(xi′ , x j ) = cos(xi′ , x j ) , (see equation (1.4)), the relationship in (1.237) is
expressed by means of the direction cosines as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
57
 v 1′   cos( x1′ , x1 ) cos( x1′ , x 2 ) cos( x1′ , x3 )  v 1 
 v ′  = cos(x ′ , x ) cos(x ′ , x ) cos(x ′ , x )  v 
2
1
2
2
2
3   2
 2 
 v ′3   cos(x 3′ , x1 ) cos(x3′ , x 2 ) cos( x3′ , x 3 )  v 3 
{ 1444444424444444
3{
v′
(1.240)
v
A
v ′ = Av
The direction cosines of a vector are those of the angles between the vector and the three
coordinate axes. According to Figure 1.20 we can verify that cos α 1 = cos(x1′ , x1 ) ,
cos β 1 = cos( x1′ , x 2 ) and cos γ 1 = cos( x1′ , x3 ) .
In the equation (1.235) we have projected the vector onto eˆ ′i . Now, we can project the
vector onto ê i :
v k eˆ k ⋅ eˆ i = v ′j eˆ ′j ⋅ eˆ i
v k δ ki = v ′j a ji
(1.241)
v i = v ′j a ji
v = A T v′
Therefore, it is also true that:
eˆ i = a ji eˆ ′j
(1.242)
The inverse relationship of equation (1.240) is obtained as follows:
A −1 v ′ = A −1 A v
⇒
v = A −1 v ′
(1.243)
and by comparing the equations (1.243) with (1.241) we can conclude that the matrix A is
an orthogonal matrix, i.e.:
A −1 = A T
⇒
Indicial
A T A = 1 notation

→ a ki a kj = δ ij
(1.244)
Second-order tensor
Consider a coordinate system formed by the orthogonal basis ê i then, how the basis
changes from the ê i system to a new one represented by the orthogonal basis eˆ ′i . This is
illustrated in transformation law as eˆ k = a ik eˆ ′i , which allow us to represent a second-order
tensor T as follows:
T = Tkl eˆ k ⊗ eˆ l
= Tkl a ik eˆ ′i ⊗ a jl eˆ ′j
= Tkl a ik a jl eˆ ′i ⊗ eˆ ′j
= Tij′ eˆ ′i ⊗ eˆ ′j
(1.245)
Then, the transformation law of the components between systems for a second-order
tensor is given by:
Tij′ = Tkl a ik a jl = a ik Tkl a jl
Matrix
 form

→
T ′ = A T AT
(1.246)
Third-order tensor
A third-order tensor ( S ) can be shown in two systems represented by orthogonal bases ê i
and eˆ ′i as follows:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
58
S = S lmn eˆ l ⊗ eˆ m ⊗ eˆ n
= S lmn ail eˆ ′i ⊗ a jm eˆ ′j ⊗ a kn eˆ ′k
= S lmn ail a jm a kn eˆ ′i ⊗ eˆ ′j ⊗ eˆ ′k
= S ′ijk eˆ ′i ⊗ eˆ ′j ⊗ eˆ ′k
(1.247)
In conclusion the components of the third-order tensor in the new basis ( eˆ ′i ) are:
S ′ijk = S lmn a il a jm a kn
(1.248)
The following table summarizes the transformation law of the components according to
the tensor rank:
rank
to
from (x1 , x 2 , x 3 ) →
(x1′ , x 2′ , x3′ )
to
from (x1′ , x 2′ , x 3′ ) →
(x1 , x 2 , x3 )
0 (scalar)
λ′ = λ
λ = λ′
1 (vector)
S i′ = a ij S j
S i = a ji S ′j
2
S ij′ = a ik a jl S kl
S ij = a ki a lj S kl′
3
′ = a il a jm a kn S lmn
S ijk
′
S ijk = a li a mj a nk S lmn
4
′ = a im a jn a kp a lq S mnpq
S ijkl
′
S ijkl = a mi a nj a pk a ql S mnpq
(1.249)
Problem 1.27: Obtain the components of T ′ , given by the transformation:
T′ = A ⋅ T ⋅ AT
where the components of T and A are shown, respectively, as Tij and a ij . Afterwards,
given that a ij are the components of the transformation matrix, represent graphically the
components of the tensors T and T ′ on both systems.
Solution: The expression T ′ = A ⋅ T ⋅ A T in symbolic notation is given by:
′ (eˆ a ⊗ eˆ b ) = a rs (eˆ r ⊗ eˆ s ) ⋅ T pq (eˆ p ⊗ eˆ q ) ⋅ a kl (eˆ l ⊗ eˆ k )
Tab
= a rs T pq a kl δ sp δ ql (eˆ r ⊗ eˆ k )
= a rp T pq a kq (eˆ r ⊗ eˆ k )
To obtain the components of T ′ one only need make the double scalar product with the
basis (eˆ i ⊗ eˆ j ) , the result of which is:
′ (eˆ a ⊗ eˆ b ) : (eˆ i ⊗ eˆ j ) = a rp T pq a kq (eˆ r ⊗ eˆ k ) : (eˆ i ⊗ eˆ j )
Tab
′ δ ai δ bj = a rp T pq a kq δ ri δ kj
Tab
Tij′ = a ip T pq a jq
The above eqaution is shown in matrix notation as:
T ′ = A T A T inverse
→ T = A −1 T ′ A −T
Since A is an orthogonal matrix, it holds that A T = A −1 . Thus, T = A T T ′ A . The
graphical representation of the tensor components in both systems can be seen in Figure
1.21.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
59
x3′
T ′ = A T AT
′
T33
′
T23
x3
x3′
T13′
T33
′
T31
T13
T23
x2′
T32
x1
T12′
′
T22
x2′
′
T21
T11′
T22
T31
T11
′
T32
T21
T12
x1′
x2
T = AT T ′ A
Figure 1.21: Transformation law of the second-order tensor components.
x1′
Problem 1.28: Let T be a symmetric second-order tensor and I T , II T , III T be scalars,
where:
I T = Tr ( T ) = Tii
;
II T =
{
}
1
2
I T − Tr ( T 2 )
2
;
III T = det ( T )
Show that I T , II T , III T are invariant with a change of basis.
Solution:
a) Taking into account the transformation law for the second-order tensor components
given in (1.249), i.e. Tij′ = a ik a jl Tkl or in matrix form T ′ = A T A T . Then, Tii′ is:
Tii′ = a ik a il Tkl = δ kl Tkl = Tkk = I T
Hence we have proved that I T is independent of the adopted system.
b) To prove that II T is an invariant, one only need show that Tr ( T 2 ) is one also, since I T 2
is already an invariant.
Tr ( T ′ 2 ) = Tr ( T ′ ⋅ T ′) = T ′ : T ′ = Tij′ Tij′
= ( a ik a jl Tkl )( a ip a jq T pq )
= a ik a ip a jl a jq Tkl T pq
123 123
δ kp
δ lq
= T pl T pl
= T : T = Tr ( T ⋅ T ) = Tr ( T 2 )
c)
det ( T ′) = det ( T ′) = det (A T A T ) = det (A )det ( T )det (A T ) = det ( T )
1
424
3
1424
3
=1
=1
Consider now four sets of coordinate systems, represented by (x1 , x 2 , x3 ) , (x1′ , x 2′ , x3′ ) ,
(x1′′, x 2′′ , x3′′ ) and (x1′′′, x 2′′′, x3′′′) , (see Figure 1.22), and consider also the following
transformation matrices:
A : Transformation matrix from (x1 , x 2 , x3 ) to (x1′ , x 2′ , x3′ ) ;
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
60
B : Transformation matrix from (x1′ , x 2′ , x3′ ) to (x1′′, x 2′′ , x3′′ ) ;
C : Transformation matrix from (x1′′, x 2′′ , x3′′ ) to (x1′′′, x 2′′′, x3′′′) .
X′
A
−1
=A
B −1 = B T
T
A
B
BA
X ′′
X
A T B T = (BA ) T
CBA
(CBA ) T = A T B T C T
C
C −1 = C T
X ′′′
Figure 1.22: Transformations matrices between several systems.
r
If we consider a v column matrix made up of components of v in the coordinate system
(x1 , x 2 , x3 ) , the components of this vector in the system (x1′ , x 2′ , x3′ ) are given by:
v ′ = Av
(1.250)
v = AT v′
(1.251)
v ′′ = Bv ′
(1.252)
v ′ = B T v ′′
(1.253)
v ′′ = BA v
(1.254)
v = A T B T v ′′
(1.255)
and the inverse transformation of relation (1.250) is:
Now, starting with the system (x1′ , x ′2 , x3′ ) , the components of the vector in the system
(x1′′, x ′2′ , x3′′ ) are given by:
and the inverse transformation is:
By substituting the equation (1.250) into (1.252) we obtain:
The resulting matrix BA is also an orthogonal matrix, and shows the transformation
matrix from (x1 , x 2 , x3 ) to (x1′′, x ′2′ , x3′′ ) , (see Figure 1.22). The inverse form of (1.254) is
evaluated by substituting (1.253) into (1.251), the result of which is:
This equation could have been obtained by using equation (1.254), i.e.:
(BA )−1 v ′′ = (BA )−1 (BA )v
⇒
v = (BA ) v ′′ = A −1B −1 v ′′ = A T B T v ′′
−1
(1.256)
Then, it is easy to find the components of the vector in the coordinate system (x1′′′, x 2′′′, x3′′′) ,
(see Figure 1.22):
v ′′′ = CBA v
inverse
 form

→
v = A T B T C T v ′′′
(1.257)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
1.5.3.1
61
Component Transformation Law in Two Dimensions (2D)
Now, consider two sets of coordinate systems, shown in Figure 1.23.
y
α y ′y = α
x′
y′
α x′y
α y ′y
α y ′x
π
−α
2
π
= +α
2
α x′y =
α y ′x
α x′x = α
x
Figure 1.23: Transformation of a coordinate system in 2D.
The transformation matrix from ( x − y ) to ( x ′ − y ′ ) is given by direction cosines, (see
Figure 1.23), as:
 a11
A = a 21
 0
a12
a 22
0
0 cos(α x′x ) cos(α x′y ) 0
0 = cos(α y ′x ) cos(α y ′y ) 0
1  
0
0
1
(1.258)
By using trigonometric identities we can deduce that:
π

− α  = sin(α ) ,
2

α x′x = α y′y ⇒ cos(α x′x ) = cos(α y′y ) = cos(α ) , cos(α x′y ) = cos
π

cos(α y′x ) = cos + α  = − sin(α )
2

(1.259)
Thus, the transformation matrix in 2D is dependent on a single parameter, α , i.e.:
 cos(α) sin(α ) 

− sin(α ) cos(α )
A=
(1.260)
Another way to prove (1.260) is by considering the vector position of the point P in both
systems, (see Figure 1.24).
Moreover, in view of Figure 1.24, said coordinates are interrelated as shown below:

π

 x ′P = x P cos(α ) + y P cos(β )
 x ′P = x P cos(α ) + y P cos 2 − α 





π
 ⇒
′
 y ′ = − x cos π − α  + y cos(α )
 y P = − x P cos(β ) + y P cos 2 − β 


P
P

 P
2

(1.261)
 x ′P = x P cos(α ) + y P sin(α )
⇒
 y ′P = − x P sin(α) + y P cos(α )
Or in matrix form:
Inverse
 x   cos(α ) sin(α ) 
 x ′P   cos(α) sin(α )   x P  transforma


 tion
→ P  = 
=
 y ′  − sin(α ) cos(α )  y 

  P
 y P   − sin(α ) cos(α )
 P 
−1
 x ′P 
 y′ 
 P
(1.262)
Since A −1 = A T , it is true that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
62
 x P  cos(α ) − sin(α )  x ′P 
 y  =  sin(α ) cos(α )   y ′ 
  P
 P 
(1.263)
yP′
y
x′
x ′P
α
r
r
β
y ′P
α
yP
xP
xP
α)
s(
co
y′
)
(α
sin
yP
xP
P
yP
)
(α
n
si
α)
s(
o
c
x
Figure 1.24: Transformation of a coordinate system in 2D.
Problem 1.29: Find the transformation matrix between the systems: x, y , z and x′′′, y ′′′, z′′′ .
These systems are represented in Figure 1.25.
z = z′
z ′′ = z ′′′
β
y ′′′
y ′ = y ′′
γ
α
x
α
y
x′
γ
x ′′′
x ′′
Figure 1.25: Rotation.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
63
Solution: The coordinate system x′′′, y ′′′, z′′′ can be obtained by different combinations of
rotations as follows:
♦
Rotation along the z -axis
z = z′
from x, y , z to x′, y ′, z′
 cos α sin α 0
A = − sin α cos α 0
 0
0
1 
y′
α
y
α
with 0 ≤ α ≤ 360 º
x′
x
♦
Rotation along the y′ -axis
from x′, y ′, z′ to x′′, y ′′, z′′
z = z′
z ′′
y ′ = y ′′
β
α
α
x
cos β
B =  0
 sin β
with 0 ≤ β ≤ 180 º
y
z = z′
x′
z ′′
x′
β
x ′′
♦
0 − sin β 
1
0 
0 cos β 
Rotation along the z ′′ -axis
x ′′
z = z′
z ′′ = z ′′′
from x′′, y ′′, z′′ to x′′′, y ′′′, z′′′
β
y ′′′
y ′ = y ′′
γ
α
x
α
y
x′
 cos γ
C = − sin γ
 0
sin γ
cos γ
0
0
0 
1 
with 0 ≤ γ ≤ 360º
γ
x ′′′
x ′′
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
64
NOTES ON CONTINUUM MECHANICS
The transformation matrix from ( x, y , z ) to ( x′′′, y ′′′, z′′′ ), (see Figure 1.22), is given by:
D = CBA
After multiplying the matrices, we obtain:
(sin α cos β cos γ + cos α sin γ ) − sin β cos γ 
 (cos α cos β cos γ − sin α sin γ )

D = ( − cos α cos β sin γ − sin α cos γ ) ( − sin α cos β sin γ + cos α cos γ ) sin β sin γ 


cos α sin β
sin α sin β
cos β
The angles α, β , γ are known as Euler angles and were introduced by Leonhard Euler to
describe the orientation of a rigid body motion.
Problem 1.30: Let T be a second-order tensor whose components in the Cartesian system
(x1 , x 2 , x3 ) are given by:
 3 − 1 0
= Tij = T =  − 1 3 0 
 0
0 1 
(T )ij
Given that the transformation matrix between two systems, (x1 , x 2 , x 3 ) - (x1′ , x 2′ , x 3′ ) , is:

 0
 2
A=
 2

2
−
 2
0
2
2
2
2

1

0


0

Obtain the tensor components Tij in the new coordinate system (x1′ , x 2′ , x 3′ ) .
Solution: As defined in equation (1.249), the transformation law for second-order tensor
components is:
Tij′ = aik a jl Tkl
To enable the previous calculation to be carried out in matrix form we use:
[ ]
Tij′ = [a i k ] [Tk l ] a l j
T
Thus
 0

 2
T ′= 
 2

2
−
 2
T ′ = A T AT
0
2
2
2
2

1
0

3
1
0
−







0  − 1 3 0  0




0
0
1


1
0


2
2
2
2
0
−
2

2 
2 
2 

0 

On carrying out the operation of the previous matrices we now have:
1 0 0 
T ′ = 0 2 0 
0 0 4 
NOTE: As we can verify in the above example, the components of the tensor T , in the
new basis, have one particular feature, i.e. the off-diagonal terms are equal to zero. The
question now is: Given an arbitrary tensor T , is there a transformation which results in the
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
65
off-diagonal terms being zero? This type of problem is called the eigenvalue and eigenvector
problem. ■
1.5.4
Eigenvalue and Eigenvector Problem
As we have seen, the scalar product between a second-order tensor T and a vector (or unit
vector nˆ ′ ) leads to a vector. In other words, projecting a second-order tensor onto a
certain direction results in a vector that does not necessarily have the same direction as nˆ ′ ,
(see Figure 1.26(a)).
The aim of the eigenvalue and eigenvector problem is to find a direction n̂ , in such a way
r ˆ
that the resulting vector, t (n) = T ⋅ nˆ , coincides with it, (see Figure 1.26 (b)).
b) Principal direction.
a) Projection of T onto an
arbitrary plane.
r ˆ
t (n′) = T ⋅ nˆ ′
r ˆ
t (n) = T ⋅ nˆ = λnˆ
n̂
nˆ ′
n̂ - principal direction of T
x3
λ - eigenvalue of T associated
with the direction n̂ .
x2
x1
Figure 1.26: Projecting a tensor onto a direction.
Let T be a second-order tensor. A vector n̂ is said to be eigenvector of T if there is a scalar
λ , called the eigenvalue, so that:
T ⋅ nˆ = λnˆ
(1.264)
The equation (1.264) can be rearranged in indicial notation as:
Tij nˆ j = λnˆ i
⇒ Tij nˆ j − λnˆ i = 0 i
(
)
⇒ Tij − λδ ij nˆ j = 0 i
(1.265)
r
 → (T − λ1) ⋅ nˆ = 0
Tensorial
notation
r
The previous set of homogeneous equations only have nontrivial solution, i.e. nˆ ≠ 0 , if and
only if:
det ( T − λ1) = 0 ;
Tij − λδ ij = 0
(1.266)
The determinant (1.266) is called the characteristic determinant of the tensor T , explicitly given
by:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
66
T11 − λ
T21
T12
T22 − λ
T13
T23
T31
T32
T33 − λ
=0
(1.267)
Developing this determinant, we obtain the characteristic polynomial, which is shown by a
cubic equation in λ :
λ3 − λ2 I T + λ II T − III T = 0
(1.268)
where I T , II T , III T are the principal invariants of T , and are defined in components terms
as:
I T = Tr (T ) = Tii
II T
=
=
=
=
=
δ jk
1
( TrT ) 2 − Tr ( T 2 )
2
1
Tr ( Tij eˆ i ⊗ eˆ j ) Tr ( Tkl eˆ k ⊗ eˆ l ) − Tr Tij eˆ i ⊗ eˆ j ⋅ (Tkl eˆ k ⊗ eˆ l )
2
1
Tij δ ij Tkl δ kl − Tij Tkl δ jk Tr (eˆ i ⊗ eˆ l )
2
1
Tii Tkk − Tij Tkl δ jk δ il
2
1
Tii Tkk − Tij T ji = Mii = Tr[cof( T )]
2
[
{
{
{
{
]
[
}
}
[(
]}
)
]}
(1.269)
III T = det ( T ) = Tij =  ijk Ti1 T j 2 Tk 3
where Mii is the matrix trace defined in equation (1.210), Mii = M11 + M 22 + M33 . More
explicitly the invariants are given by:
IT
II T
III T
= T11 + T22 + T33
T22 T23
T11 T13
T
T12
=
+
+ 11
T32 T33 T31 T33 T21 T22
= T22 T33 − T23 T32 + T11 T33 − T13 T31 + T11 T22 − T12 T21
= T11 ( T22 T33 − T32 T23 ) − T12 (T21 T33 − T31 T23 ) + T13 ( T21 T32 − T31 T22 )
(1.270)
If T is a symmetric tensor, the principal invariants are summarized as follows:
IT
= T11 + T22 + T33
II T
III T
2
= T11 T22 + T11 T33 + T22 T33 − T122 − T132 − T23
2
= T11 T22 T33 + T12 T13 T23 + T13 T12 T23 − T122 T33 − T23
T11 − T132 T22
(1.271)
The eigenvalues, λ 1 , λ 2 , λ 3 , are found by solving the characteristic polynomial (1.268).
Once the eigenvalues are evaluated, the eigenvectors are found by applying equation
(1.265), i.e. ( Tij − λ 1δ ij ) nˆ (j1) = 0 i , ( Tij − λ 2 δ ij ) nˆ (j2) = 0 i , ( Tij − λ 3δ ij ) nˆ (j3) = 0 i . These
eigenvectors constitute a new space denoted as the principal space.
If T is a symmetric tensor, the principal space is defined by an orthonormal basis and all
eigenvalues are real numbers. If the three eigenvalues are different, λ 1 ≠ λ 2 ≠ λ 3 , the three
principal directions are unique. If two of them are equal, e.g. λ 1 = λ 2 ≠ λ 3 , we can state that
the principal direction, n̂ (3) , associated with the eigenvalue λ 3 , is unique, and, any
direction defined in the plane normal to n̂ (3) is a principal direction, and othorgonality is
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
67
the only constraint to determining n̂ (1) and n̂ ( 2) . If λ 1 = λ 2 = λ 3 , any direction is principal.
A tensor that has three equal eigenvalues is called a Spherical Tensor, (see Appendix A-The
Tensor ellipsoid).
The T -components in the principal space are only made up of normal components, i.e.:
λ 1
Tij′ =  0
 0
0
λ2
0
  T1
=0
 
λ 2   0
0
0
0
T2
0
0
0 
T3 
(1.272)
Therefore, the principal invariants can also be evaluated by:
I T = T1 + T2 + T3 , II T = T1 T2 + T2 T3 + T1 T3 , III T = T1 T2 T3
(1.273)
whose values must match the values obtained in (1.270), since they are invariant with a
change of basis.
If T is a spherical tensor, i.e. T1 = T2 = T3 = T , it holds that I T2 = 3 II T , III T = T 3 .
Let W be an antisymmetric tensor. The principal invariants of W are given by:
IW
II W
= Tr (W ) = 0
− Tr (W 2 )
1
= ( TrW ) 2 − Tr (W 2 ) =
2
2
0
0
W23
W13
0
=
+
+
− W23
0
− W13
0
− W12
[
]
W12
0
(1.274)
= W23 W23 + W13 W13 + W12 W12
III W
= ω2
=0
r
r r
where ω 2 = w = w ⋅ w = W232 + W132 + W122 as defined in (1.132). Then, the characteristic
equation for an antisymmetric tensor is reduced to:
2
λ3 − λ2 I W + λ II W − III W = 0
⇒
λ3 + ω 2 λ = 0
⇒
(
)
λ λ2 + ω 2 = 0
(1.275)
In this case, one eigenvalue is real and equal to zero and the others are imaginary roots:
λ2 + ω 2 = 0
1.5.4.1
λ2 = −ω 2 = 0
⇒
⇒
λ (1, 2 ) = ±ω − 1 = ±ω i
(1.276)
The Orthogonality of the Eigenvectors
Consider a symmetric second-order tensor T . By the definition of eigenvalues, given in
(1.264), if λ 1 , λ 2 , λ 3 are the eigenvalues of T , then it follows that:
T ⋅ nˆ (1) = λ 1nˆ (1)
;
T ⋅ nˆ ( 2 ) = λ 2 nˆ ( 2)
T ⋅ nˆ (3) = λ 3 nˆ (3)
(1.277)
Applying the dot product between n̂ ( 2) and T ⋅ nˆ (1) = λ 1nˆ (1) , and the dot product between
n̂ (1) and T ⋅ nˆ ( 2 ) = λ 2 nˆ ( 2 ) we obtain:
nˆ ( 2 ) ⋅ T ⋅ nˆ (1) = λ 1nˆ ( 2 ) ⋅ nˆ (1)
nˆ (1) ⋅ T ⋅ nˆ ( 2 ) = λ 2 nˆ (1) ⋅ nˆ ( 2 )
(1.278)
Since T is symmetric, it holds that nˆ ( 2) ⋅ T ⋅ nˆ (1) = nˆ (1) ⋅ T ⋅ nˆ ( 2) , so:
λ 1nˆ ( 2) ⋅ nˆ (1) = λ 2 nˆ (1) ⋅ nˆ ( 2 ) = λ 2 nˆ ( 2 ) ⋅ nˆ (1)
(1.279)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
68
⇒ (λ 1 − λ 2 ) nˆ (1) ⋅ nˆ ( 2 ) = 0
(1.280)
To satisfy the equation (1.280), with λ1 ≠ λ 2 ≠ 0 , the following must be true:
nˆ (1) ⋅ nˆ ( 2 ) = 0
(1.281)
Similarly, it is possible to show that nˆ (1) ⋅ nˆ (3) = 0 and nˆ ( 2) ⋅ nˆ (3) = 0 and then we can
conclude that the eigenvectors are mutually orthogonal, and constitute an orthogonal basis,
(see Figure 1.27), where the transformation matrix between systems is:
 nˆ (1)   nˆ 1(1)

 
A = nˆ ( 2)  = nˆ 1( 2)
nˆ (3)  nˆ (3)

  1
nˆ (21)
nˆ (22 )
nˆ (23)
nˆ 3(1) 

nˆ 3( 2 ) 
nˆ (33) 
(1.282)
diagonalization
T ′ = A T AT
x3
T3
T33
x3′
x3′
T23
T13
T2
n̂( 2 )
x 2′
x 2′
T23
T13
n̂(1)
T22
T12
T12
T1
x2
T11
x1
n̂
( 3)
x1′
x1′
T = AT T ′ A
Principal Space
Figure 1.27: Diagonalization.
Problem 1.31: Show that the following relations are invariants:
C12 + C 22 + C 32
C14 + C 24 + C 34
where C1 , C 2 , C 3 are the eigenvalues of the second-order tensor C .
;
C13 + C 23 + C 33
;
Solution: Any combination of invariants is also an invariant, so, on this basis, we can try to
express the above expressions in terms of their principal invariants.
(
)
I C2 = (C1 + C 2 + C 3 ) = C12 + C 22 + C 32 + 2 C1 C 2 + C1 C 3 + C 2 C 3 ⇒ C12 + C 22 + C 32 = I C2 − 2 II C
1444424444
3
2
II C
So, we have proved that C12 + C 22 + C 32 is an invariant. Similarly, we can obtain:
C13 + C 23 + C 33 = I C3 − 3 II C I C + 3 III C
C14 + C 24 + C 34 = I C4 − 4 II C I C2 + 4 III C I C + 2 II C2
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
69
Problem 1.32: Let Q be a proper orthogonal tensor, and E be an arbitrary second-order
tensor. Show that the eigenvalues of E do not change with the following orthogonal
transformation:
E* = Q ⋅ E ⋅ QT
Solution: We can prove this as follows:
(
)
= det (Q ⋅ E ⋅ Q − λ1)
= det (Q ⋅ E ⋅ Q − Q ⋅ λ1 ⋅ Q )
= det [Q ⋅ (E − λ1 ) ⋅ Q ]
(3) det (E − λ1 ) det
(Q3)
= det
12Q
1
424
0 = det E * − λ1
T
T
T
T
T
1
1
= det (E − λ1 )
(
= det (Q
= det (Q
0 = det E *ij − λδ ij
)
)
ik E kp Q jp
− λδ ij
ik E kp Q jp
− λQ ik Q jp δ kp
[ (
)
) ]
= det (Q )det (E − λδ ) det (Q )
= det (E − λδ )
= det Q ik E kp − λδ kp Q jp
ik
kp
kp
jp
kp
kp
*
Thus, we have proved that E and E have the same eigenvalues.
1.5.4.2
Solution of the Cubic Equation
Let T be a symmetric second-order tensor. The roots of the characteristic equation
( λ3 − λ2 I T + λ II T − III T = 0 ) are all real numbers, and are expressed as:
  α  I
λ 1 = 2 S cos   + T
  3  3
  α 2π  
λ 2 = 2 S cos +
 +
3 
 3
  α 4π  
λ 3 = 2S cos +
 +
3  
 3
IT
3
(1.283)
IT
3
where
R=
I T2 − 3 II T
;
3
S=
R
;
3
Q=
I T II T
2I 3
− III T − T ;
3
27
T=
R3
;
27
 Q 

 2T 
α = arccos −
where α is in radians.
(1.284)
By restructuring the solution (1.283) in matrix form, we obtain:
λ 1
0

 0
0
λ2
0
 α

0
0
cos 3 

0
1 0 0
  

IT 
 α 2π 




0=
0 1 0 + 2 S
0
cos +
0



3 
3
3


0 0 1


λ 3 
 α 4π  
144244
3
 0
+
0
cos


Spherical part

 344343

14
4444444244444
(1.285)
Deviatoric part
where we clearly distinguish the spherical and the deviatoric part of the tensor in the
principal space. Note that, if T is a spherical tensor the following relationship holds
I T2 = 3 II T , then S = 0 .
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
70
Problem 1.33: Find the principal values and directions of the second-order tensor T ,
where the Cartesian components of T are:
(T )ij
 3 − 1 0
= Tij = T =  − 1 3 0 
 0
0 1 
Solution: We need to find nontrivial solutions for (Tij − λδ ij ) nˆ j = 0 i , which are constrained
by nˆ j nˆ j = 1 (unit vector). As we have seen, the nontrivial solution requires that:
Tij − λδ ij = 0
Explicitly, the above equation is:
T11 − λ
T21
T12
T22 − λ
T13
T23
3 − λ −1
= −1 3 − λ
T31
T32
T33 − λ
0
0
0
=0
1− λ
0
Developing the above determinant, we can obtain the cubic equation:
[
]
(1 − λ ) (3 − λ ) 2 − 1 = 0
3
2
λ − 7 λ + 14λ − 8 = 0
We could have obtained the characteristic equation directly in terms of invariants:
I T = Tr ( Tij ) = Tii = T11 + T22 + T33 = 7
II T =
T
1
Tii T jj − Tij Tij = 22
T32
2
(
)
T23
T33
+
T11
T13
T31
T33
+
T11
T12
T21
T22
= 14
III T = Tij =  ijk Ti1 T j 2 Tk 3 = 8
Then, using the equation in (1.268), the characteristic equation is:
λ3 − λ2 I T + λ II T − III T = 0
→
λ3 − 7λ2 + 14λ − 8 = 0
On solving the cubic equation we obtain three real roots, namely:
λ 1 = 1;
λ 2 = 2;
λ3 = 4
We can also verify that:
I T = λ1 + λ 2 + λ 3 = 1 + 2 + 4 = 7 ✓
II T = λ 1 λ 2 + λ 2 λ 3 + λ 3 λ 1 = 1 × 2 + 2 × 4 + 4 × 1 = 14 ✓
III T = λ 1 λ 2 λ 3 = 8 ✓
Thus, we can see that the invariants are the same as those evaluated previously.
Principal directions:
Each eigenvalue, λ i , is associated with a corresponding eigenvector, nˆ (i ) . We can use the
equation in (1.265), i.e. ( Tij − λδ ij ) nˆ j = 0 i , to obtain the principal directions.
Œ λ1 = 1
3 − λ 1
 −1

 0
−1
3 − λ1
0
0   n 1  3 − 1 − 1
0   n1   0 




0  n 2  =  − 1 3 − 1 0  n 2  = 0 
1 − λ 1  n 3   0
0 1 − 1 n 3  0 
These become the following system of equations:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
71
2n1 − n 2 = 0 

 ⇒ n1 = n 2 = 0
 − n1 + 2n 2 = 0
0n = 0
 3
n i n i = n12 + n 22 + n 32 = 1
Then we can conclude that: λ1 = 1 ⇒ nˆ i(1) = [0 0 ± 1] .
NOTE: This solution could have been directly determined by the specific features of the
T matrix. As the terms T13 = T23 = T31 = T32 = 0 imply that T33 = 1 is already a principal
value, then, consequently, the original direction is a principal direction. ■
λ2 = 2
−1
0   n 1  3 − 2 − 1
0   n1  0 
3 − λ 2
 −1




3 − λ2
0  n 2  =  − 1 3 − 2
0  n 2  = 0 

 0
0
1 − λ 2  n 3   0
0
1 − 2  n 3  0 
n1 − n 2 = 0 ⇒ n1 = n 2

 − n1 + n 2 = 0
− n = 0
 3
The first two equations are linearly dependent, after which we need an additional equation:
n i n i = n12 + n 22 + n 32 = 1 ⇒ 2n12 = 1 ⇒ n1 = ±
1
2
Thus:
λ2 = 2
⇒

1
nˆ i( 2 ) =  ±
2


0

1
2
±
λ3 = 4
−1
0   n 1  3 − 4 − 1
0   n1   0 
3 − λ 3
 −1




3 − λ3
0  n 2  =  − 1 3 − 4
0  n 2  = 0 

 0
0
1 − λ 3  n 3   0
0
1 − 4  n 3  0 
 − n1 − n 2 = 0 

 ⇒ n1 = −n 2
 − n1 − n 2 = 0 
− 3n = 0
3

n i n i = n12 + n 22 + n 32 = 1 ⇒ 2n 22 = 1 ⇒ n 2 = ±
1
2
Then:
λ3 = 4
⇒

nˆ i(3) =  m

1
2
±
1
2

0

Afterwards, we summarize the eigenvalues and eigenvectors of T :
λ1 = 1
⇒ nˆ i(1)
= [0 0 ± 1]
λ2 = 2
⇒ nˆ i( 2)

= ±

1
2
±
1
2

0

λ3 = 4
⇒ nˆ i(3)

= m

1
2
±
1
2

0

Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
72
NOTE: The tensor components of this problem are the same as those used in Problem
1.30. Additionally, we can verify that the eigenvectors make up the transformation matrix,
A , between the original system, (x1 , x 2 , x 3 ) , and the principal space, ( x1′ , x 2′ , x 3′ ) , (see
Problem 1.30). ■
1.5.5
Spectral Representation of Tensors
Based on the solution of the equation in (1.268), if T is a symmetric second-order tensor
there are three real eigenvalues: T1 , T2 , T3 each of which is associated with an
eigenvector, i.e.:
[
= [ nˆ
= [ nˆ
T1
⇒ nˆ (i1)
= nˆ 1(1)
T2
⇒ nˆ (i 2 )
T3
⇒ nˆ i(3)
]
nˆ (21)
nˆ 3(1)
( 2)
1
nˆ (22)
nˆ 3( 2 )
( 3)
1
nˆ (23)
nˆ (33)
]
]
(1.286)
The principal space is formed by the orthogonal basis nˆ (1) , nˆ ( 2) , nˆ (3) , and the tensor
components are represented by their eigenvalues as:
 T1
Tij′ = T ′ =  0
 0
0
T2
0
0
0 
T3 
(1.287)
With reference to the fact that eigenvectors form a transformation matrix, A , so that:
T ′ = A T AT
(1.288)
T = AT T ′ A
(1.289)
Since A −1 = A T , the inverse form is:
where
 nˆ (1)   nˆ 1(1)

 
A = nˆ ( 2)  = nˆ 1( 2)
nˆ (3)  nˆ (3)

  1
nˆ (21)
nˆ (22 )
nˆ (23)
nˆ 3(1) 

nˆ 3( 2 ) 
nˆ (33) 
(1.290)
Explicitly, the relation in (1.289) is given by:
 T11
T
 12
 T13
T12
T22
T23
nˆ 1(1) nˆ 1( 2 )
T13 

T23  = nˆ (21) nˆ (22 )
nˆ (1) nˆ ( 2 )
T33 
3
 3
 T1 0
= A T  0 T2
 0 0
nˆ 1(3)   T1

nˆ (23)   0
nˆ 3(3)   0
0
0  A
T3 
0
T2
0
0   nˆ 1(1)

0  nˆ 1( 2)
T3  nˆ 1(3)
nˆ (21)
nˆ (22 )
nˆ (23)
nˆ (31) 

nˆ 3( 2) 
nˆ 3(3) 
(1.291)
1 0 0
0 0 0 
0 0 0 



T 
T 
= T1A 0 0 0 A + T2 A 0 1 0A + T3 A 0 0 0 A
0 0 0
0 0 0
0 0 1
T
Whereas:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
nˆ 1(1) nˆ 1(1)
1 0 0

A T 0 0 0A = nˆ (21) nˆ 1(1)
nˆ (1) nˆ (1)
0 0 0
 3 1
nˆ 1(1) nˆ (21)
nˆ (21) nˆ (21)
nˆ 3(1) nˆ (21)
nˆ 1( 2 ) nˆ 1( 2 )
0 0 0 

A T 0 1 0A = nˆ (22) nˆ 1( 2)
nˆ ( 2 ) nˆ ( 2 )
0 0 0
 3 1
nˆ 1( 2 ) nˆ (22 )
nˆ (22 ) nˆ (22 )
nˆ 3( 2 ) nˆ (22 )
nˆ 1(3) nˆ 1(3)
0 0 0 

A T 0 0 0A = nˆ (23) nˆ 1(3)
nˆ (3) nˆ (3)
0 0 1
 3 1
nˆ 1(3) nˆ (23)
nˆ (23) nˆ (23)
nˆ (33) nˆ (23)
73
nˆ 1(1) nˆ 3(1) 

nˆ (21) nˆ 3(1)  = nˆ i(1) nˆ (j1)
nˆ 3(1) nˆ 3(1) 
nˆ 1( 2 ) nˆ (32 ) 

nˆ (22 ) nˆ (32 )  = nˆ i( 2 ) nˆ (j2 )
nˆ (32 ) nˆ (32 ) 
nˆ 1(3) nˆ 3(3) 

nˆ (23) nˆ 3(3)  = nˆ i(3) nˆ (j3)
nˆ 3(3) nˆ 3(3) 
(1.292)
Then, it is possible to represent the components of a second-order tensor in function of
their eigenvalues and eigenvectors (spectral representation) as:
Tij = T1 nˆ i(1) nˆ (j1) + T2 nˆ (i 2 ) nˆ (j2 ) + T3 nˆ (i 3) nˆ (j3)
(1.293)
As we can see, the tensor is represented as a linear combination of dyads and the above
representation in tensorial notation becomes:
T = T1 nˆ (1) ⊗ nˆ (1) + T2 nˆ ( 2 ) ⊗ nˆ ( 2 ) + T3 nˆ (3) ⊗ nˆ (3)
(1.294)
or:
T=
3
∑T
a
nˆ ( a ) ⊗ nˆ ( a )
a =1
Spectral representation of a
second-order tensor
(1.295)
which is the spectral representation of the tensor. Note that, in the above equation we have to
resort to the summation symbol, because the dummy index appears thrice in the
expression.
NOTE: The spectral representation in (1.295) could easily have been obtained from the
definition of the second-order unit tensor, given in (1.168), i.e. 1 = nˆ i ⊗ nˆ i , which can also
3
be represented by means of the summation symbol as 1 = ∑ nˆ ( a ) ⊗ nˆ ( a ) . Then, it follows
a =1
that:
 3

T = T ⋅ 1 = T ⋅  nˆ ( a ) ⊗ nˆ ( a )  =
 a =1

∑
3
∑
T ⋅ nˆ ( a ) ⊗ nˆ ( a ) =
a =1
3
∑T
a
nˆ ( a ) ⊗ nˆ ( a )
(1.296)
a =1
where we have used the definition of eigenvalue and eigenvector T ⋅ nˆ ( a ) = Ta nˆ ( a ) . ■
We now consider the orthogonal tensor R . The orthogonal transformation applied to the
ˆ . Therefore, it is also possible to
unit vector N̂ leads to the unit vector n̂ , i.e. nˆ = R ⋅ N
represent the orthogonal tensor R as follows:
 3 ˆ (a) ˆ (a) 
⊗ N  =
R = R ⋅ 1 = R ⋅  N
 a =1

∑
3
∑ R ⋅ Nˆ
a =1
(a)
ˆ (a ) =
⊗N
3
∑ nˆ
(a )
ˆ (a)
⊗N
(1.297)
a =1
The spectral representation is very useful for making algebraic operations with tensors. For
example, tensor power in the principal space can be expressed as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
74
 T1n

= 0
0

(T )
n
ij
0
T2n
0
0

0
T3n 
(1.298)
So, the spectral representation of T n is given by:
3
∑T
Tn =
n
a
nˆ ( a ) ⊗ nˆ ( a )
(1.299)
a =1
T , this can easily be obtained from the
Now, if we need the square root of the tensor,
spectral representation as:
T=
3
∑
Ta nˆ ( a ) ⊗ nˆ ( a )
(1.300)
a =1
Next, we can show that a positive definite tensor has positive eigenvalues. For this
purpose, we can consider a semi-positive definite tensor, T , by which the condition
r
xˆ ⋅ T ⋅ xˆ ≥ 0 holds for all xˆ ≠ 0 . Replacing the tensor by its spectral representation, we
obtain:
xˆ ⋅ T ⋅ xˆ ≥ 0
 3

⇒ xˆ ⋅  Ta nˆ ( a ) ⊗ nˆ ( a )  ⋅ xˆ ≥ 0
 a =1

∑
⇒
3
∑T
a
(1.301)
xˆ ⋅ nˆ ( a ) ⊗ nˆ ( a ) ⋅ xˆ ≥ 0
a =1
Note that the result of the operation ( xˆ ⋅ nˆ (a ) ) is a scalar, thus:
3
∑
Ta xˆ ⋅ nˆ ( a ) ⊗ nˆ ( a ) ⋅ xˆ ≥ 0
a =1
⇒
( xˆ ⋅ nˆ ) ] ≥ 0
∑ T [1
4243
3
(a ) 2
a
a =1
>0
(1.302)
⇒ T1 ( xˆ ⋅ nˆ (1) ) 2 + T2 ( xˆ ⋅ nˆ ( 2 ) ) 2 + T3 ( xˆ ⋅ nˆ (3) ) 2 ≥ 0
r
The above expression must hold for all xˆ ≠ 0 . If we take xˆ = nˆ (1) , the above equation is
reduced to T1 (nˆ (1) ⋅ nˆ (1) ) 2 = T1 ≥ 0 . The same is true for T2 and T3 . Thus, we have
demonstrated that if a tensor is semi-positive definite, its eigenvalues are greater than or
equal to zero, i.e. T 1 ≥ 0 , T 2 ≥ 0 , T 3 ≥ 0 . Therefore we can conclude that a tensor is positive
definite, i.e. xˆ ⋅ T ⋅ xˆ > 0 , if and only if its eigenvalues are positive and nonzero, i.e. T 1 > 0 ,
T 2 > 0 , T 3 > 0 . Consequently, the positive definite tensor trace is greater than zero. If the
positive definite tensor trace is zero, this implies that the tensor is the zero tensor.
The spectral representation of the fourth-order unit tensor, I , can be obtained starting
from the definition in (1.169), i.e.:
I = δ ik δ jl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = eˆ i ⊗ eˆ j ⊗ eˆ i ⊗ eˆ j =
3
3
∑∑ eˆ
a
⊗ eˆ b ⊗ eˆ a ⊗ eˆ b
(1.303)
a =1 b =1
As I is an isotropic tensor, (see 1.5.8 Isotropic and Anisotropic tensors), then the
representation in (1.303) is also valid in any orthonormal basis, nˆ ( a ) , so:
I=
3
3
∑∑ nˆ
(a)
⊗ nˆ (b ) ⊗ nˆ ( a ) ⊗ nˆ (b )
(1.304)
a =1 b =1
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
75
Similarly, we obtain the spectral representation for I and I as:
I = δ il δ jk eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = eˆ i ⊗ eˆ j ⊗ eˆ j ⊗ eˆ i
I=
3
∑ nˆ
(1.305)
⊗ nˆ (b ) ⊗ nˆ (b ) ⊗ nˆ ( a )
(a)
(1.306)
a ,b =1
and
I = δ ij δ kl eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l = eˆ i ⊗ eˆ i ⊗ eˆ k ⊗ eˆ k
I=
3
∑ nˆ
(1.307)
⊗ nˆ ( a ) ⊗ nˆ (b ) ⊗ nˆ (b )
(a)
(1.308)
a ,b =1
Problem 1.34: Let w be an antisymmetric second-order tensor and V be a positive
definite symmetric tensor whose spectral representation is given by:
V=
3
∑λ
a
nˆ ( a ) ⊗ nˆ ( a )
a =1
w can be represented by:
3
w = ∑ w ab nˆ (a ) ⊗ nˆ (b)
Show that the antisymmetric tensor
a ,b =1
a ≠b
Demonstrate also that:
3
w ⋅ V − V ⋅ w = ∑ w ab (λ b − λ a ) nˆ ( a ) ⊗ nˆ (b)
a ,b =1
a ≠b
Solution:
It is true that


3
3

a =1
a =1
w ⋅ 1 = w ⋅  ∑ nˆ ( a ) ⊗ nˆ ( a)  = ∑ w ⋅ nˆ (a ) ⊗ nˆ (a ) = ∑ (wr ∧ nˆ ( a) ) ⊗ nˆ ( a)
3
 a =1
∑ w (nˆ
3
=
b
(b )
)
∧ nˆ ( a ) ⊗ nˆ ( a )
a ,b =1
r
r
where we have applied an antisymmetric tensor property w ⋅ nˆ = w ∧ nˆ , where w is the
axial vector associated with w . Expanding the above equation, we obtain:
w = wb (nˆ (b) ∧ nˆ (1) ) ⊗ nˆ (1) + wb (nˆ (b) ∧ nˆ ( 2) ) ⊗ nˆ ( 2) + wb (nˆ (b) ∧ nˆ (3) ) ⊗ nˆ (3) =
(
+ w (nˆ
+ w (nˆ
)
(
)
(
)
= w1 nˆ (1) ∧ nˆ (1) ⊗ nˆ (1) + w2 nˆ ( 2) ∧ nˆ (1) ⊗ nˆ (1) + w3 nˆ (3) ∧ nˆ (1) ⊗ nˆ (1) +
1
1
(1)
∧ nˆ
(2)
(1)
∧ nˆ
( 3)
) ⊗ nˆ
) ⊗ nˆ
(2)
( 3)
(
(nˆ
+ w2 nˆ
+ w2
( 2)
∧ nˆ
( 2)
∧ nˆ
On simplifying the above expression we obtain:
( 2)
( 3)
) ⊗ nˆ
) ⊗ nˆ
( 2)
( 3)
(
(nˆ
+ w3 nˆ
+ w3
( 3)
∧ nˆ
( 3)
∧ nˆ
) ⊗ nˆ
) ⊗ nˆ
( 2)
( 3)
( 2)
+
( 3)
w = −w2 (nˆ (3) ) ⊗ nˆ (1) + w3 (nˆ ( 2) ) ⊗ nˆ (1) +
( )
(nˆ ) ⊗ nˆ
( )
(nˆ ) ⊗ nˆ
+ w1 nˆ (3) ⊗ nˆ ( 2 ) − w3 nˆ (1) ⊗ nˆ ( 2 ) +
( 2)
( 3)
( 3)
− w1
+ w2 (1)
Taking into account that w1 = −w 23 = w 32 , w2 = w13 = −w 31 , w3 = −w12 = w 21 , the
above equation becomes:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
76
w = w 31 nˆ (3) ⊗ nˆ (1) + w 21 nˆ ( 2) ⊗ nˆ (1) +
+ w 32 nˆ (3) ⊗ nˆ ( 2 ) + w12 nˆ (1) ⊗ nˆ ( 2 ) +
+ w 23 nˆ ( 2 ) ⊗ nˆ (3) + w13 nˆ (1) ⊗ nˆ (3)
which is the same as:
3
w = ∑ w ab nˆ (a ) ⊗ nˆ (b)
The terms
w⋅V
a ,b =1
a ≠b
and V ⋅ w can be expressed as follows:

 3
  3


(
)
(
)
a
b
w ⋅ V =  w ab nˆ ⊗ nˆ  ⋅  λ b nˆ (b) ⊗ nˆ (b) 
  b =1
 a ,b =1


 a≠b
∑
=
∑
3
∑λ w
b
ab
3
∑λ w
nˆ ( a ) ⊗ nˆ (b ) ⋅ nˆ (b ) ⊗ nˆ (b ) =
a ,b =1
a ≠b
b
ab
nˆ ( a ) ⊗ nˆ (b )
a ,b =1
a≠b
and

 3
3


 3
(a )
(b )
(a )
(a) 
ˆ
ˆ
ˆ
ˆ
V ⋅ w =  λ a n ⊗ n  ⋅ 
w ab n ⊗ n  = λ a w ab nˆ ( a) ⊗ nˆ (b)
 a ,b =1
 a =1
  a ,b =1
 a≠b
 a≠b
∑
∑
∑
Then,

  3
 3

 

(
)
(
)
(
)
(
)
a
b
a
b
w ⋅ V − V ⋅ w =  λ b w ab nˆ ⊗ nˆ  −  λ a w ab nˆ ⊗ nˆ 

  a ,b =1
 a ,b =1

  a≠b
 a ≠b
∑
=
3
∑w
∑
ab (λ b
− λ a ) nˆ ( a ) ⊗ nˆ (b )
a ,b =1
a ≠b
Similarly, it is possible to show that:
3
w ⋅ V 2 − V 2 ⋅ w = ∑ w ab (λ2b − λ2a ) nˆ ( a ) ⊗ nˆ (b)
a ,b =1
a ≠b
1.5.6
Cayley-Hamilton Theorem
The Cayley-Hamilton theorem states that any tensor, T , satisfies its own characteristic
equation, i.e. if the eigenvalues of T satisfy the equation λ3 − λ2 I T + λ II T − III T = 0 , so
does the tensor T :
T 3 − T 2 I T + T II T − III T 1 = 0
(1.309)
One of the applications of the Cayley-Hamilton theorem is to express the power of tensor,
T n , as a combination of T n −1 , T n − 2 , T n −3 . For example, T 4 is obtained as:
T 3 ⋅ T − T 2 ⋅ TI T + T ⋅ T II T − III T 1 ⋅ T = 0
⇒
T 4 = T 3 I T − T 2 II T + III T T
(1.310)
Using the Cayley-Hamilton theorem, it is possible to express the third invariant as a
function of traces. According to the Cayley-Hamilton theorem, the expression
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
77
T 3 − I T T 2 + II T T − III T 1 = 0 remains valid. Additionally, by applying the double scalar
product with the second-order unit tensor, 1 , we obtain:
T 3 : 1 − I T T 2 : 1 + II T T : 1 − III T 1 : 1 = 0 : 1
(1.311)
Taking into consideration T 3 : 1 = Tr ( T 3 ) , T 2 : 1 = Tr ( T 2 ) , T : 1 = Tr (T ) , 1 : 1 = Tr (1) = 3 ,
0 : 1 = Tr (0) = 0 in the equation (1.311) we obtain:
Tr ( T 3 ) − I T Tr ( T 2 ) + II T Tr ( T ) − III T Tr (1) = 0
123
=3
[
1
⇒ III T = Tr ( T 3 ) − I T Tr ( T 2 ) + II T Tr ( T )
3
]
(1.312)
Replacing the values of the invariants, I T , II T , given by equation (1.269), we obtain:
1
3
1
3
III T =  Tr ( T 3 ) − Tr ( T 2 ) Tr ( T ) + [Tr ( T )] 
3
2
2

(1.313)
or in indicial notation
1
3
1

III T =  Tij T jk Tki − Tij T ji Tkk + Tii T jj Tkk 
3
2
2

(1.314)
Problem 1.35: Based on the Cayley-Hamilton theorem, find the inverse of a tensor T in
terms of tensor power.
Solution: The Cayley-Hamilton theorem states that:
T 3 − T 2 I T + T II T − III T 1 = 0
Carrying out the dot product between the previous equation and the tensor T −1 , we
obtain:
T 3 ⋅ T −1 − T 2 ⋅ T −1 I T + T ⋅ T −1 II T − III T 1 ⋅ T −1 = 0 ⋅ T −1
T 2 − TI T + 1 II T − III T T −1 = 0
⇒ T −1 =
(
1
T 2 − I T T + II T 1
III T
)
The Cayley-Hamilton theorem also applies to square matrices of order n . Let An×n be a
square n by n matrix. The characteristic determinant is given by:
λ1n×n − A = 0
(1.315)
where 1n×n is the identity n by n matrix. Developing the determinant (1.315) we ontain:
λn − I 1 λn −1 + I 2 λn − 2 − L (−1) n I n = 0
(1.316)
where I 1 , I 2 , L , I n are the invariants of A . In the particular case when n = 3 , the
invariants are the same obtained for a second-order tensor, i.e.: I 1 = I A , I 2 = II A , I 3 = III A .
Applying the Cayley-Hamilton theorem it is true that:
A n − I 1A n −1 + I 2 A n − 2 − L + (−1) n I n 1 = 0
(1.317)
By means of the relationship (1.317), we can obtain the inverse of the matrix An×n by
multiplying all the terms by the inverse, A −1 , i.e.:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
78
A n A −1 − I 1A n −1A −1 + I 2 A n − 2 A −1 − L + (−1) n I n 1A −1 = 0
then
⇒ A n −1 − I 1 A n − 2 + I 2 A n −3 − L (−1) n −1 I n −11 + (−1) n I n A −1 = 0
A −1 =
(
(−1) n −1
A n −1 − I 1A n − 2 + I 2 A n−3 − L (−1) n −1 I n −11
In
)
(1.318)
(1.319)
I n is the determinant of An×n . Then, the inverse exists if I n = det (A ) ≠ 0 .
Problem 1.36: Check the Cayley-Hamilton theorem by using a second-order tensor whose
Cartesian components are given by:
5 0 0 
T = 0 2 0
0 0 1 
Solution:
The Cayley-Hamilton theorem states that:
T 3 − T 2 I T + T II T − III T 1 = 0
where I T = 5 + 2 + 1 = 8 , II T = 10 + 2 + 5 = 17 , III T = 10 , and
T
3
5 3

=0
0

0
23
0
0  125 0 0 

0  =  0 8 0 
1   0 0 1 
; T
2
5 2

=0
0

0
22
0
0  25 0 0 

0  =  0 4 0 
1   0 0 1 
By applying the Cayley-Hamilton theorem, we can verify that it is true:
125 0 0
 25 0 0
5 0 0 
1 0 0   0 0 0 
 0 8 0 − 8  0 4 0 + 17 0 2 0  − 10 0 1 0 = 0 0 0 







 

 0 0 1 
 0 0 1 
0 0 1 
0 0 1  0 0 0 
1.5.7
Norms of Tensors
The magnitude (module) of a tensor, also known as the Frobenius norm, is given below:
r r
r
v = v ⋅ v = v i vi (vector)
(1.320)
T = T : T = Tij Tij (second-order tensor)
(1.321)
A = A : ⋅A = A ijk A ijk (third-order tensor)
(1.322)
C = C :: C = C ijkl C ijkl (fourth-order tensor)
(1.323)
Interpreting the Frobenius norm of T is done by considering the principal space of T
where T1 , T2 , T3 are the eigenvalues of T . In this space, it follows that:
T = T : T = Tij Tij = T12 + T22 + T32 = I T2 − 2 II T
(1.324)
As we can verify T is an invariant, and T represents a measurement of distance as
shown in Figure 1.28.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
79
x 2′
T = T : T = I T2 − 2 II T
T2
T
T1
x3′
x1′
T3
Figure 1.28: Norm of a second-order tensor.
1.5.8
Isotropic and Anisotropic Tensor
A tensor is called isotropic when its components are the same in any coordinate system,
otherwise the tensor is said to be anisotropic.
Let T and T ′ represent the tensor components T in the systems ê i and eˆ ′i ,
respectively, so, the tensor is isotropic if T = T ′ on any arbitrary basis.
Isotropic first-order tensor
r
Let v be a vector that is represented by its components, v 1 , v 2 , v 3 , in the coordinate
system x1 , x 2 , x3 . The representation of these components in a new coordinate system,
x1′ , x 2′ , x3′ , are given by v 1′ , v ′2 , v ′3 , so the transformation law for these components is:
r
v = v i eˆ i = v ′j eˆ ′j
⇒
v ′i = a ij v j
(1.325)
r
By definition, v is an isotropic tensor if it holds that v i = v ′i , and this is only possible if
eˆ i = eˆ ′j , i.e. there is no change of system, or if the tensor is the zero vector, i.e.
r
v i = v ′i = 0 i . Then, the unique isotropic first-order tensor is the zero vector 0 .
Isotropic second-order tensor
An example of a second-order isotropic tensor is the unit tensor, 1 , whose components
are represented by δ kl (Kronecker delta). In the demonstration, we use the transformation
law for a second-order tensor components, obtained in (1.248), thus:
δ ′ij = a ik a jl δ kl = a ik a jk = δ ij
123
AA T =1
(1.326)
An immediate observation of the isotropy of unit tensor 1 is that any spherical tensor
( α1 ) is also an isotropic tensor. So, if a second-order tensor is isotropic it is spherical and
vice versa.
Isotropic third-order tensor
An example of a third-order isotropic tensor is the Levi-Civita pseudo-tensor, defined in
(1.182), which is not a “real” tensor in the strict meaning of the word. With reference to
the transformation law for the third-order tensor components, (see equation (1.248)), we
can conclude that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
80
′ijk = a il a jm a kn  lmn = A  ijk =  ijk (see Problem 1.21)
(1.327)
{
1
Isotropic fourth-order tensor
With reference to the transformation law for fourth-order tensor components, (see
equation (1.249)), it is possible to demonstrate that the following tensors are isotropic:
I ijkl = δ ij δ kl
I ijkl = δ ik δ jl
;
I ijkl = δ il δ jk
;
(1.328)
Therefore, any fourth-order isotropic tensor can be represented by a linear combination of
the three tensors given in (1.328), e.g.:
D = a 0 I + a1 I + a 2 I
D = a 0 1 ⊗ 1 + a1 1⊗1 + a 2 1⊗1
(1.329)
D ijkl = a 0 δ ij δ kl + a1δ ik δ jl + a 2 δ il δ jk
Problem 1.37: Let C be a fourth-order tensor, whose components are given by:
C ijkl = λδ ij δ kl + µ (δ ik δ jl + δ il δ jk )
where λ , µ are constant real numbers. Show that C is an isotropic tensor.
Solution:
Applying the transformation law for fourth-order tensor components:
C ′ijkl = a im a jn a kp a lq C mnpq
and by replacing the relation C mnpq = λδ mn δ pq + µ (δ mp δ nq + δ mq δ np ) in the above equation,
we obtain:
C ′ijkl = a im a jn a kp a lq [λδ mn δ pq + µ (δ mp δ nq + δ mq δ np )]
(
= λa im a jn a kp a lq δ mn δ pq + µ a im a jn a kp a lq δ mp δ nq + a im a jn a kp a lq δ mq δ np
(
= λa in a jn a kq a lq + µ a ip a jq a kp a lq + a iq a jn a kn a lq
(
= λδ ij δ kl + µ δ ik δ jl + δ il δ jk
)
)
)
= C ijkl
which is proof that C is an isotropic tensor.
1.5.9
Coaxial Tensors
Two arbitrary second-order tensors, T and S , are coaxial tensors if they have the same
eigenvectors. It is easy to show that if two tensors are coaxial, this means the dot product
between them is commutative, and vice versa, i.e.:
if
T ⋅S = S ⋅ T
⇔
S, T are coaxial
(1.330)
If T and S are coaxial as well as symmetric tensors, the spectral representations of these
tensors are given by:
T=
3
∑
a =1
Ta nˆ ( a ) ⊗ nˆ ( a )
;
S=
3
∑S
a
nˆ ( a ) ⊗ nˆ ( a )
(1.331)
a =1
An immediate result of (1.330) is that the tensor S and its inverse S −1 are coaxial tensors:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
81
S −1 ⋅ S = S ⋅ S −1 = 1
S=
3
∑
S a nˆ ( a ) ⊗ nˆ ( a )
S −1 =
;
a =1
where S a ,
3
a =1
(1.332)
1 ˆ (a ) ˆ (a )
n ⊗n
∑S
a
1
, are the eigenvalues of S and S −1 , respectively.
Sa
If S and T are coaxial symmetric tensors, the resulting tensor ( S ⋅ T ) becomes another
symmetric tensor. To prove this we start from the definition of coaxial tensors:
T ⋅S = S ⋅ T
⇒
T ⋅S − S ⋅ T = 0
⇒
T ⋅ S − ( T ⋅ S ) T = 0 ⇒ 2( T ⋅ S ) skew = 0 (1.333)
Then, if the antisymmetric part of a tensor is a zero tensor, it follows that this tensor is
symmetric:
( T ⋅ S ) skew = 0
⇒
( T ⋅ S ) ≡ ( T ⋅ S ) sym
(1.334)
1.5.10 Polar Decomposition
Let F
be an arbitrary nonsingular second-order tensor, i.e. ( det ( F ) ≠ 0 ⇒ ∃F −1 ).
r
r
r
ˆ = f (N) = f (N) nˆ = λ nˆ ≠ 0 ,
Additionally, as previously seen, it satisfies the condition F ⋅ N
(nˆ )
ˆ
ˆ
ˆ ( a ) , we can obtain:
since det ( F ) ≠ 0 . After that, given an orthonormal basis N
F −1 ⋅ F = 1 =
3
∑ Nˆ
(a)
ˆ (a)
⊗N
a =1
⇒ F = F ⋅1 = F ⋅
3
∑
ˆ (a) ⊗ N
ˆ (a ) =
N
a =1
⇒F =
3
∑ λ nˆ
a
(a)
3
∑ F ⋅ Nˆ
(a)
ˆ (a)
⊗N
(1.335)
a =1
ˆ (a )
⊗N
a =1
NOTE: The representation of F , given in (1.335), is not the spectral representation of F
ˆ (a)
in the strict sense of the word, i.e., λ a are not eigenvalues of F , and neither nˆ ( a ) nor N
are eigenvectors of F . ■
r
r
ˆ (1) = f (Nˆ (1) ) = f (Nˆ (1) ) nˆ (1) = λ nˆ (1)
F ⋅N
1
r ˆ (1 )
ˆ (1)
f (N ) = F ⋅ N
r ˆ (2)
ˆ ( 2)
f (N ) = F ⋅ N
n̂(1)
n̂( 2 )
r
r
ˆ ( 2) = f (Nˆ ( 2 ) ) = f (Nˆ ( 2 ) ) nˆ ( 2 ) = λ nˆ ( 2 )
F ⋅N
2
r
r
ˆ (3) = f (Nˆ ( 3) ) = f (Nˆ ( 3) ) nˆ (3) = λ nˆ (3)
F ⋅N
3
N̂(3)
N̂( 2 )
n̂(3)
N̂(1)
r ˆ (3)
ˆ ( 3)
f (N ) = F ⋅ N
ˆ (a) .
Figure 1.29: Projecting F onto N
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
82
ˆ ( a ) , the new basis nˆ ( a ) will not necessarily
Note that for the arbitrary orthonormal basis N
ˆ ( a ) so that the new basis nˆ ( a ) is orthonormal,
be orthonormal. We seek to find a basis N
r
r
r
r
r
r
(see Figure 1.29), i.e. f (N ) ⋅ f (N ) = 0 , f (N ) ⋅ f (N ) = 0 , f (N ) ⋅ f (N ) = 0 . Then we
look for a space in accordance with the following orthogonal transformation
ˆ ( a ) , which ensures nˆ ( a ) orthonormality since an orthogonal transformation
nˆ ( a ) = R ⋅ N
changes neither angles between vectors nor their magnitudes.
ˆ (1 )
ˆ (2)
ˆ (2)
ˆ (3)
ˆ ( 3)
ˆ ( 1)
ˆ ( a ) to nˆ ( a ) , which is given by the
Now, consider that there is a transformation from N
following orthogonal transformation nˆ ( a ) = R ⋅ Nˆ ( a ) , then we can state that:
F=
3
∑ λ nˆ
a
(a )
a =1
ˆ (a) =
⊗N
3
∑ λ R ⋅ Nˆ
a
(a)
ˆ (a) = R ⋅
⊗N
a =1
F = R ⋅U
3
∑ λ Nˆ
a
(a)
ˆ (a) = R ⋅ U
⊗N
a =1
U=R
⇒
T
(1.336)
⋅F
3
ˆ (a ) ⊗ N
ˆ ( a ) . Note that U is a symmetric
where we have defined the tensor U = ∑ λ a N
a =1
ˆ (a) ⊗ N
ˆ ( a ) is also
tensor, i.e. U = U . This condition is easily verified by the fact that N
ˆ (a) ⇒ N
ˆ ( a ) = R T ⋅ nˆ ( a ) = nˆ ( a ) ⋅ R , we obtain:
symmetric. Now considering that nˆ ( a ) = R ⋅ N
T
F=
3
∑
ˆ (a ) =
λ a nˆ ( a ) ⊗ N
a =1
3
∑
λ a nˆ ( a ) ⊗ nˆ ( a ) ⋅ R =
a =1
F = V ⋅R
3
∑λ
ˆ ( a ) ⊗ nˆ ( a ) ⋅ R = V ⋅ R
an
a =1
⇒
V = F ⋅R
(1.337)
T
3
where we have defined the symmetric second-order tensor V = ∑ λ a nˆ ( a ) ⊗ nˆ ( a ) . By
a =1
comparing the spectral representation of U with V , we can conclude that they have the
same eigenvalues but different eigenvectors, and they are related by nˆ ( a ) = R ⋅ Nˆ ( a ) .
With reference to the above considerations, we can define the polar decomposition:
F = R ⋅U = V ⋅R
Polar Decomposition
(1.338)
Carrying out the dot product between F T and F = R ⋅ U , we obtain:
T
F
F = F T ⋅ R ⋅ U = (R T ⋅ F ) T ⋅ U = UT ⋅ U = U 2
12⋅3
⇒ U= ± FT ⋅F = ± C
C
(1.339)
Moreover, by carrying out the dot product between F = V ⋅ R and F T , we obtain:
F2
⋅ F3T = V ⋅ R ⋅ F T = V ⋅ (F ⋅ R T ) T = V ⋅ V T = V 2
1
b
⇒ V =± F ⋅FT =± b
(1.340)
Since det ( F ) ≠ 0 , the tensors C and b are positive definite symmetric tensors, (see
Problem 1.25), which implies that the eigenvalues of C and b are all real and positive.
However, up to now, det ( F ) ≠ 0 is the only restriction imposed on the tensor F .
Therefore, we have the following possibilities:
ƒ
If det ( F ) > 0
In this scenario, we have det ( F ) = det (R )det(U) = det ( V )det(R ) > 0 , which results in the
following cases:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
83
R − Proper ortogonal tensor
R − Improper orthogonal tensor
or 

U, V − Positive definite tensors
U, V − Negative definite tensors
ƒ
If det ( F ) < 0
In this situation, we have det ( F ) = det (R )det (U) = det ( V )det (R ) < 0 , which give us the
following cases:
R − Proper orthogonal tensor
R − Improper orthogonal tensor
or 

U, V − Negative definite tensors
U, V − Positive definite tensors
NOTE: In Chapter 2 we will work with some special tensors where F is a nonsingular
tensor, det ( F ) ≠ 0 , and det ( F ) > 0 . U and V are positive definite tensors and R is a
rotation tensor, i.e. a proper orthogonal tensor. ■
1.5.11 Partial Derivative with Tensors
The first derivative of a tensor with respect to itself is defined as:
∂A ij
∂A
≡ A, A =
(eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l ) = δ ik δ jl (eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l ) = I
∂A
∂A kl
(1.341)
The derivative of a tensor trace with respect to a tensor:
∂A kk
∂[Tr ( A )]
≡ [Tr ( A )],A =
(eˆ i ⊗ eˆ j ) = δ ki δ kj (eˆ i ⊗ eˆ j ) = δ ij (eˆ i ⊗ eˆ j ) = 1
∂A
∂A ij
(1.342)
The derivative of the tensor trace squared with respect to the tensor is given by:
∂[Tr ( A )]
∂[Tr ( A )]
= 2 Tr ( A )1
= 2 Tr ( A )
∂A
∂A
2
(1.343)
And, the derivative of the trace of the tensor squared with respect to tensor is given by:
[
∂ Tr ( A 2 )
∂A
]
=

∂ ( A sr A rs )
∂ ( A sr )
∂( A rs ) 
(eˆ i ⊗ eˆ j ) =  A rs
+ A sr
 (eˆ i ⊗ eˆ j )
∂A ij
∂A ij
∂A ij 

[
]
[
]
= A rs δ si δ rj + A sr δ ri δ sj (eˆ i ⊗ eˆ j ) = A ji + A ji (eˆ i ⊗ eˆ j )
(1.344)
= 2 A ji (eˆ i ⊗ eˆ j ) = 2A T
We leave the reader with the following demonstration:
[
∂ Tr ( A 3 )
∂A
]
(1.345)
= 3( A 2 ) T
Then, if we are considering a symmetric second-order tensor, C , it is true that
[
∂ Tr (C 2 )
∂[Tr (C)]
∂[Tr (C)]
= 2 Tr (C)1 ,
=1,
∂C
∂C
∂C
2
]
= 2C T = 2C ,
[
∂ Tr (C 3 )
∂C
]
= 3(C 2 ) T = 3C 2 .
Moreover, we can say that the derivative of the Frobenius norm of C is given by:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
84
∂ Tr (C ⋅ C T ) 
∂ Tr (C 2 ) 
∂ C:C


 = 1 Tr (C 2 )
=
=
= 
∂C
∂C
∂C
∂C
2
−1
1
= Tr (C 2 ) 2 2C
2
(
∂C
)
[
[
] [Tr(C )],
−1
2
2
]
C
(1.346)
or:
∂C
C
=
∂C
=
2
Tr (C )
C
C
(1.347)
Another interesting derivative is presented below:
(
)
∂ n i C ij n j
=
∂n k
∂n j
∂n i
= δ ik C ij n j + n i C ij δ jk = C kj n j + n i C ik
C ij n j + n i C ij
∂n k
∂n k
= C kj n j + C jk n j = (C kj + C jk ) n j =
2C kjsym n j
(1.348)
= 2C kj n j
where we have assumed that C is symmetric, i.e., C kj = C jk .
Let C be a symmetric second-order tensor. The partial derivative of C −1 with respect to
the tensor C is obtained by using the following relationship:
(
)
∂1 ∂ C −1 ⋅ C
=O
=
∂C
∂C
(1.349)
where O is the fourth-order zero tensor and the above equation in indicial notation
becomes:
(
∂ C iq−1C qj
∂C kl
( )C
) = ∂(C ) C
−1
iq
∂C kl
∂ C iq−1
∂C kl
⇒
qj
= −C iq−1
( )δ
∂ C iq−1
∂C kl
whereas C qj =
( )δ
( )
∂ C qj
(
∂C kl
⇒
∂C kl
= −C iq−1
( )
∂ C qj
( )
∂ C qj
∂C kl
= O ikjl
( )C
∂ C iq−1
∂C kl
−1
qj C jr
= −C iq−1
( )
∂ C qj
∂C kl
C −jr1
C −jr1
)
qr
= −C iq−1
( )
(
1  ∂ C qj + C jq

∂C kl
2
)
−1
 C jr

[
∂ C ir−1
1
1
= −C iq−1 δ qk δ jl + δ jk δ ql C −jr1 = − C iq−1δ qk δ jl C −jr1 + C iq−1δ jk δ ql C −jr1
2
2
∂C kl
( )
(1.350)
1
C qj + C jq , so we can conclude that:
2
∂ C iq−1
∂C kl
qr
+ C iq−1
qj
[
[
]
∂ C ir−1
1
= − C ik−1C lr−1 + C il−1 C −kr1
2
∂C kl
]
(1.351)
]
Or in tensorial notation:
[
1
∂C −1
= − C −1 ⊗C −1 + C −1 ⊗C −1
2
∂C
]
(1.352)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
85
NOTE: Note that, if we had not replaced the symmetric part of C qj in (1.351), we would
have found that
( )δ
∂ C iq−1
∂C kl
qr
= −C iq−1
( )
∂ C qj
∂C kl
C −jr1 = −C iq−1δ qk δ jl C −jr1 = −C ik−1C lr−1 , which is a non-
symmetric tensor. ■
1.5.11.1 Partial Derivative of Invariants
Let T be a second-order tensor. The partial derivative of I T with respect to T , (see
equation (1.342)), is:
∂[I T ] ∂[Tr ( T )]
= [Tr ( T )], T = 1
=
∂T
∂T
(1.353)
The partial derivative of II T with respect to T , (see equation (1.342)), is:
[
2
∂[ II T ] ∂  1
∂ Tr ( T 2 )
1  ∂[Tr ( T )]
2
2 
[
]
Tr
Tr
T
T
−
=
=
(
)
−
(
)



∂T
∂T
∂T
∂T  2
2 

1
= 2( TrT )1 − 2 T T
2
= Tr ( T )1 − T T
[
]
[
]
]

(1.354)
Next, we apply the Cayley-Hamilton theorem so as to represent T as:
T 3 : T −2 − I T T 2 : T −2 + II T T : T −2 − III T 1 : T −2 = 0
T − I T 1 + II T T −1 − III T T − 2 = 0
(1.355)
⇒ T = I T 1 − II T T −1 + III T T − 2
By substituting (1.355) into the equation in (1.354), we obtain:
∂[ II T ]
= Tr ( T )1 − T T = Tr ( T )1 − I T 1 − II T T −1 + III T T − 2
∂T
(
) = ( II
T
TT
−1
− III T T − 2
)
T
(1.356)
To find the partial derivative of the third invariant, we can start with the definition given in
(1.313), so:
∂[ III T ]
3
∂ 1
1
1
3
2
=
 Tr ( T ) − Tr ( T ) Tr ( T ) + [Tr ( T )] 
∂T
∂T  3
2
6

[
]
2
∂[Tr ( T )] 3
1
1 ∂ Tr ( T 2 )
1
= 3( T 2 ) T −
+ [Tr ( T )] 1
Tr ( T ) − Tr ( T 2 )
∂T
∂T
3
2
2
6
= ( T 2 ) T − Tr ( T ) T T −
2
1
1
Tr ( T 2 )1 + [Tr ( T )] 1
2
2
= ( T 2 ) T − Tr ( T ) T T +
2
1
[Tr(T )] − Tr( T 2 ) 1
2
[
(1.357)
]
= ( T 2 ) T − I T T T + II T 1
Once again using the Cayley-Hamilton theorem we obtain:
T 3 ⋅ T −1 − I T T 2 ⋅ T −1 + II T T ⋅ T −1 − III T 1 ⋅ T −1 = 0
T 2 − I T T + II T 1 − III T T −1 = 0
⇒ III T T
−1
(1.358)
2
= T − I T T + II T 1
and the transpose:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
86
( III
TT
) = (T
−1 T
2
− I T T + II T 1
) = (T )
T
2 T
− I T T T + II T 1
(1.359)
By comparing (1.357) with (1.359) we find another way to express the derivative of III T
with respect to T , i.e.:
∂[ III T ]
= III T T −1
∂T
(
)
T
= III T T −T
(1.360)
1.5.11.2 Time Derivative of Tensors
Let us assume a second-order tensor depends on the time, t , i.e. T = T(t ) . Then, we define
the first time derivative and the second time derivative of the tensor T , respectively, as:
D
T = T&
Dt
D2
&&
T=T
Dt 2
;
(1.361)
The time derivative of a tensor determinant is defined as:
DT
D
[det(T )] = ij cof Tij
Dt
Dt
( )
(1.362)
where cof (Tij ) is the cofactor of Tij and defined as [cof (Tij )]T = det(T ) (T −1 )ij .
1
1
Problem 1.38: Consider that J = [det (b )] 2 = ( III b ) 2 , where b is a symmetric second-order
tensor, i.e. b = b T . Obtain the partial derivatives of J and ln(J ) with respect to b .
Solution:
⇒
⇒
1


∂ ( III b ) 2 
∂J

= 
∂b
∂b
−1 ∂ III
−1
1
1
b
= ( III b ) 2
= ( III b ) 2 III b b −T
2
2
∂b
1
1
1
= ( III b ) 2 b −1 = J b −1
2
2
1 
 

∂ ln III b 2 
1 ∂ III b 1 −1
∂[ln( J )]


= 
= b
=
∂b
2 III b ∂b
2
∂b
1.5.12 Spherical and Deviatoric Tensors
Any tensor can be decomposed into a spherical and a deviatoric part, so, for a given
second-order tensor T , this decomposition is represented by:
T = T sph + T dev =
I
Tr ( T )
1 + T dev = T 1 + T dev = Tm 1 + T dev
3
3
(1.363)
The deviatoric part of the tensor T is defined as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
T dev = T −
87
Tr ( T )
1 = T − Tm 1
3
(1.364)
For the following operations, we consider that T is a symmetric tensor, T = T T , then
under this condition the deviatoric tensor components, Tijdev , become:
Tijdev
 T11dev

=  T12dev
 T dev
 13
T12dev
dev
T22
dev
T23
T13dev   T11 − Tm

dev 
T23
 =  T12
T33dev   T13
 13 ( 2 T11 − T22 − T33 )

T12
=

T13

T12
T22 − Tm
T23
T12
1
3



T33 − Tm 
T13
T23


T23

1
(2 T33 − T11 − T22 )
3
T13
(2 T22 − T11 − T33 )
T23
(1.365)
Graphical representations of the Cartesian components of the spherical and deviatoric
parts are shown in Figure 1.30.
In the following subsections we obtain the deviatoric tensor invariants in terms of the
principal invariants of T .
1.5.12.1 First Invariant of the Deviatoric Tensor
I
T dev
Tr ( T ) 
Tr ( T )

= Tr ( T dev ) = Tr  T −
1 = Tr ( T ) −
Tr (1) = 0
3
3 123


δ =3
(1.366)
ii
Thus, we can conclude that the trace of any deviatoric tensor is equal to zero.
1.5.12.2 Second Invariant of the Deviatoric Tensor
For simplicity we can use the principal space to obtain the second and third invariant of the
deviatoric tensor. In the principal space the components of T are given by:
 T1
Tij =  0
 0
0
T2
0
0
0 
T3 
(1.367)
The principal invariants of T : I T = T1 + T2 + T3 , II T = T1 T2 + T2 T3 + T3 T1 , III T = T1 T2 T3 .
The deviatoric components, T dev = T − Tm 1 , in the principal space are:
Tijdev
 T1 − Tm
=  0
 0
0
T2 − Tm
0



T3 − Tm 
0
0
(1.368)
So, the second invariant of deviatoric tensor T dev is evaluated as follows:
II T dev
= ( T1 − Tm )( T2 − Tm ) + ( T1 − Tm )( T3 − Tm ) + ( T2 − Tm )( T3 − Tm )
= ( T1 T2 + T1 T3 + T2 T3 ) − 2 Tm ( T1 + T2 + T3 ) + 3Tm2
2I T
I T2
= II T −
(I T ) +
3
3
1
2
= 3 II T − I T
3
(
(1.369)
)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
88
We could also have obtained the above result, by directly starting from the definition of the
second invariant of a tensor given in (1.269), i.e.:
1
2
1
=
2
1
=
2
1
=
2
II T dev =
{(I T
dev
]}
[
) 2 − Tr ( T dev ) 2 = −
{− Tr[(T − T
{− Tr[(T
[− Tr(T
2
]}
− 2 Tm T ⋅ 1 + Tm2 1)
2
2
m 1)
{ [
1
Tr ( T dev ) 2
2
]}
) + 2 Tm Tr ( T ) − Tm2 Tr (1)
=
IT
I T2 
1
2
(
)
2
3
Tr
T
I
−
+
−

T
2
3
9 
=
I T2 
1
2
(
)
Tr
T
−
+


2
3
]}
]
(1.370)
Observing that Tr ( T 2 ) = T12 + T22 + T32 = I T2 − 2 II T , (see Problem 1.31), the equation
(1.370) becomes:
II T dev =
I T2  1 
2 I T2  1
1 2
2
−
I
+
2
I
I
+
=
2
I
I
−
 T


 = 3 II T − I T
T
2
3  2 T
3  3
(
x3
)
(1.371)
T 33
T23
T13
T13
T23
T12
T22
T12
x2
T11
x
14414444442444444443
x3
x3
Tm
T33dev
T23
T13
Tm
Tm
x1
T13
+
x2
T23
T12
T11dev
dev
T22
T12
x2
x1
Figure 1.30: Spherical and deviatoric part.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
89
Another equation for II T dev is presented in terms of deviatoric tensor components. To
calculate this, we can apply the equation (1.370):
II
T dev
=−
[
]
[
]
1
1
1
Tr ( T dev ) 2 = − Tr ( T dev ⋅ T dev ) = − T dev
2
2
2
⋅ ⋅T dev = − 1 Tijdev T jidev
2
(1.372)
Expanding the previous equation we obtain:
II
T dev
=−
[
1
dev 2
dev 2
( T11dev ) 2 + ( T22
) + ( T33dev ) 2 + 2( T12dev ) 2 + 2( T13dev ) 2 + 2( T23
)
2
]
(1.373)
Additionally, in the space of the principal directions we obtain:
II
T dev
=−
[
1 dev dev
1
Tij T ji = − ( T1dev ) 2 + ( T2dev ) 2 + ( T3dev ) 2
2
2
]
(1.374)
Another way to express the second invariant is shown below:
II T dev =
dev
T22
dev
T23
dev
T23
T33dev
+
T11dev
T13dev
T13dev
T33dev
+
T11dev
T12dev
T12dev
dev
T22
[
]
1
dev dev
dev
dev 2
) − ( T13dev ) 2
= − − 2 T22
T33 − 2 T11dev T33dev − 2 T11dev T22
− ( T12dev ) 2 − ( T23
2
(1.375)
or
( ) − 2 T T + (T ) + (T ) − 2 T T + (T )
) − 2 T T + (T )  + (T ) + (T ) + (T )
1
dev
II T dev = −  T22
2 
(T
dev 2
11
2
dev
11
dev
22
dev
22
dev 2
33
dev
33
dev 2
22
dev 2
11
dev 2
11
dev
11
dev 2
22
dev
33
dev 2
33
dev 2
33
+
(1.376)
dev 2
) − ( T13dev ) 2
− ( T12dev ) 2 − ( T23
Note that, from equation (1.373), we can state that:
dev 2
( T11dev ) 2 + ( T22
) + ( T33dev ) 2 = −2 II
T dev
dev 2
− 2( T12dev ) 2 − 2( T13dev ) 2 − 2( T23
)
(1.377)
Substituting (1.377) into (1.376), we find:
II
T dev
=−
[
]
1
dev
dev 2
dev 2
( T22
− T33dev ) 2 + ( T11dev − T33dev ) 2 + ( T11dev − T22
) − ( T12dev ) 2 − ( T23
) − ( T13dev ) 2
6
(1.378)
Moreover, if we consider the principal space we obtain:
II
T dev
=−
[
1
( T2dev − T3dev ) 2 + ( T1dev − T3dev ) 2 + ( T1dev − T2dev ) 2
6
]
(1.379)
1.5.12.3 Third Invariant of Deviatoric Tensor
The third invariant of the deviatoric tensor is given by:
III T dev
= ( T1 − Tm )( T2 − Tm )( T3 − Tm )
= T1 T2 T3 − Tm ( T1 T2 + T1 T3 + T2 T3 ) + Tm2 ( T1 + T2 + T3 ) − Tm3
I
I2
I3
= III T − T II T + T I T − T
3
9
27
I T II T 2 I T3
= III T −
+
3
27
1
3
=
2 I T − 9 I T II T + 27 III T
27
(
(1.380)
)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
90
Another way of expressing the third invariant is:
III
= T1dev T2dev T3dev =
T dev
1 dev dev dev
Tij T jk Tki
3
(1.381)
Problem 1.39: Let σ be a symmetric second-order tensor, and s ≡ σ dev be a deviatoric
tensor. Prove that s :
∂s
= s . Also show that σ and σ dev are coaxial tensors.
∂σ
Solution: First, we make use of the definition of a deviatoric tensor:
σ = σ sph + σ dev = σ sph + s =
Iσ
1+s
3
⇒
s=σ −
Iσ
1.
3
Afterwards we calculate:
I


∂ σ − σ 1 
3  ∂[σ ] 1 ∂[I σ ]
∂s
= 
=
−
1
3 ∂σ
∂σ
∂σ
∂σ
which in indicial notation is:
∂s ij
∂σ kl
Therefore
s ij
∂s ij
∂σ kl
=
∂σ ij
∂σ kl
−
1 ∂ [I σ ]
1
δ ij = δ ik δ jl − δ kl δ ij
3 ∂σ kl
3
1
1
1


= s ij  δ ik δ jl − δ kl δ ij  = s ij δ ik δ jl − s ij δ kl δ ij = s kl − δ kl s ii
{
3
3
3


=0
= s kl
s:
∂s
=s
∂σ
To show that two tensors are coaxial, we must prove that σ dev ⋅ σ = σ ⋅ σ dev :
σ ⋅ σ dev = σ ⋅ (σ − σ sph ) = σ ⋅ σ − σ ⋅ σ sph = σ ⋅ σ − σ ⋅
Iσ
1
3
Iσ
I
1 = σ ⋅σ − σ 1⋅σ
3
3
I


=  σ − σ 1  ⋅ σ = σ dev ⋅ σ
3 

Therefore, we have shown that σ and σ dev are coaxial tensors. In other words, they have
= σ ⋅σ − σ ⋅
the same principal directions.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
91
1.6 The Tensor-Valued Tensor Function
Tensor-valued tensor function can be of the types: scalar, vector, or higher-order tensors.
As examples of scalar-valued tensor functions we can list:
Ψ = Ψ ( T ) = det (T )
Ψ = Ψ (T , S) = T : S
(1.382)
where T and S are second-order tensors. Additionally, as an example of a second-ordervalued tensor function we have:
Π = Π (T ) = α1 + βT
(1.383)
where α and β are scalars.
1.6.1
The
The Tensor Series
function
∞
f (x)
can
be
approximated
by
the
Taylor
series
as
n
1 ∂ f (a)
( x − a ) n , where n! denotes the factorial of n , and f (a) is the value of
n
∂x
n =0
the function at the application point x = a . We can extrapolate that definition for use on
f ( x) =
∑ n!
tensors. For example, let us suppose we have a scalar-valued tensor function ψ in terms of
a second-order tensor, E , then we can approximate ψ (E ) as:
1
1 ∂ψ ( E 0 )
1 ∂ 2 ψ( E 0 )
( E ij − E 0 ij ) +
( E ij − E ij 0 )( E kl − E kl 0 ) + L
ψ( E ) ≈ ψ ( E 0 ) +
0!
1! ∂E ij
2! ∂E ij ∂E kl
∂ψ ( E 0 )
∂ 2 ψ( E 0 )
1
≈ ψ0 +
: (E − E0 ) + (E − E0 ) :
: (E − E0 ) + L
∂E
∂E ⊗ ∂E
2
(1.384)
A second-order-valued tensor function, S (E ) , can be approximated as:
∂ 2 S( E 0 )
1
1 ∂S ( E 0 )
1
: (E − E0 ) + (E − E0 ) :
: (E − E0 ) + L
S( E 0 ) +
0!
1! ∂E
2!
∂E ⊗ ∂E
∂S ( E 0 )
∂ 2S( E 0 )
1
≈ S0 +
: (E − E0 ) + (E − E0 ) :
: (E − E0 ) + L
∂E
2
∂E ⊗ ∂E
S( E ) ≈
(1.385)
Other tensor algebraic expressions can be represented by series, e.g.:
1 2 1 3
S + S +L
2!
3!
1
1
ln(1 + S ) = S − S 2 + S 3 − L
2
3
1 3 1 5
sin(S ) = S − S + S − L
3!
5!
exp S = 1 + S +
(1.386)
With reference to the spectral representation of a symmetric second-order tensor, S , it is
also true that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
92
exp S =
3

∑ 1 + λ
a =1 
a
+

λ2a λ3a
+
+ L nˆ ( a ) ⊗ nˆ ( a ) =
2!
3!

3
∑ exp λ
1
1


ln(1 + S ) =  λ a − λ2a + λ3a + L nˆ ( a ) ⊗ nˆ ( a ) =
2
3

a =1 
∑
where λ a and nˆ
1.6.2
nˆ ( a ) ⊗ nˆ ( a )
a =1
3
(a )
a
(1.387)
3
∑ ln(1 + λ
ˆ
a)n
(a)
⊗ nˆ
(a)
a =1
are the eigenvalues and eigenvectors, respectively, of the tensor S .
The Tensor-Valued Isotropic Tensor Function
A second-order-valued tensor function, Π = Π (T ) , is isotropic if after an orthogonal
transformation the following condition is satisfied:
(
)
Π * (T ) = Q ⋅ Π (T ) ⋅ Q T = Π Q ⋅ T ⋅ Q T
14
4244
3
( )
Π T*
(1.388)
We can show that Π (T ) has the same principal directions of T , i.e. Π (T ) and T are
coaxial tensors. To demonstrate this we can regard the components of T in the principal
space as:
(T )ij
λ 1
=  0
 0
0
λ2
0
0
0 
λ 3 
(1.389)
Then the tensor function is given in terms of the principal values of T : Π = Π (λ 1 , λ 2 , λ 3 )
and, the transformation of T is given by:
T * = Q ⋅ T ⋅ QT
(1.390)
Likewise, for the tensor function Π :
Π * (T ) = Q ⋅ Π (T ) ⋅ Q T
(1.391)
If we take as the orthogonal tensor components:
(Q)ij
0
1 0

= 0 − 1 0 
0 0 − 1
(1.392)
After having done the calculation for the matrices (1.391), we obtain:
 Π 11 − Π 12 − Π 13   Π 11
Π = − Π 12 Π 22 − Π 23  = Π 12
 − Π 13 − Π 23 Π 33  Π 13
0
0 
Π 11

*
0 
⇒ Π =  0 Π 22
 0
0
Π 33 
*
Π 12
Π 22
Π 23
Π 13 
Π 23  = Π
Π 33 
(1.393)
To satisfy that Π * = Π (isotropy), we conclude that Π 12 = Π 13 = Π 23 = 0 . Therefore, Π (T )
and T have the same principal directions.
Once again we observe, a tensor function Π ( T ) . This tensor function is isotropic if and
only if it can be represented by the following linear transformation, Truesdell & Noll
(1965):
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
Π = Π(T) = Φ 0 1 + Φ1T + Φ 2 T 2
93
(1.394)
where Φ 0 , Φ 1 , Φ 2 are functions of the tensor T invariants or functions of the T
eigenvalues.
A brief demonstration follows. We can now consider the spectral representations of T and
Π , respectively:
3
∑λ
T=
Π=
a =1
3
∑ω
ˆ ( a ) ⊗ nˆ ( a ) = λ 1nˆ (1) ⊗ nˆ (1) + λ 2 nˆ ( 2) ⊗ nˆ ( 2 ) + λ 3nˆ (3) ⊗ nˆ (3)
(1.395)
nˆ ( a ) ⊗ nˆ ( a ) = ω1nˆ (1) ⊗ nˆ (1) + ω 2 nˆ ( 2 ) ⊗ nˆ ( 2 ) + ω 3nˆ (3) ⊗ nˆ (3)
(1.396)
an
a
a =1
Note that T and Π have the same principal directions nˆ (i ) . Then, we can put the
following set of equations together:
1 = nˆ (1) ⊗ nˆ (1) + nˆ ( 2) ⊗ nˆ ( 2) + nˆ (3) ⊗ nˆ (3)

(1)
(1)
(2)
(2)
( 3)
( 3)
T = λ 1nˆ ⊗ nˆ + λ 2 nˆ ⊗ nˆ + λ 3nˆ ⊗ nˆ
 2
2 ˆ (1)
2 ˆ (2)
2 ˆ ( 3)
ˆ (1)
ˆ ( 2)
ˆ ( 3)
T = λ 1n ⊗ n + λ 3n ⊗ n + λ 3n ⊗ n
(1.397)
Solving the set above, we obtain nˆ ( a ) ⊗ nˆ ( a ) ≡ M ( a ) as a function of the tensor T , and we
obtain:
M (1) =
(λ 2 + λ 3 )
λ 2λ3
T2
1−
T+
(λ 1 − λ 3 )(λ 1 − λ 2 )
(λ 1 − λ 3 )(λ 1 − λ 2 )
(λ 1 − λ 3 )(λ 1 − λ 2 )
M ( 2) =
M ( 3) =
(λ 1 + λ 3 )
λ 1λ 3
T2
1−
T+
(λ 2 − λ 1 )(λ 2 − λ 3 )
(λ 2 − λ 1 )(λ 2 − λ 3 )
(λ 2 − λ 1 )(λ 2 − λ 3 )
(1.398)
(λ 1 + λ 2 )
λ 1λ 2
T2
1−
T+
(λ 3 − λ 1 )(λ 3 − λ 2 )
(λ 3 − λ 1 )(λ 3 − λ 2 )
(λ 3 − λ 1 )(λ 3 − λ 2 )
It is evident that, if we substitute the values of nˆ ( a ) ⊗ nˆ ( a ) ≡ M ( a ) in equation (1.395) we
obtain: T = T . Now, if we substitute the values of nˆ ( a ) ⊗ nˆ ( a ) ≡ M ( a ) in equation (1.396),
we obtain:
Π = Π(T ) = Φ 0 1 + Φ1T + Φ 2 T 2
(1.399)
where the coefficients Φ 0 , Φ 1 , and Φ 2 are functions of the eigenvalues of T ,
(λ 1 ≠ λ 2 ≠ λ 3 ) , and given by:
Φ0 =
ω1 λ 2 λ 3
ω 2 λ 1λ 3
ω 3 λ 1λ 2
+
+
(λ 1 − λ 3 )(λ 1 − λ 2 ) (λ 2 − λ 1 )(λ 2 − λ 3 ) (λ 3 − λ 1 )(λ 3 − λ 2 )
Φ1 = −
Φ2 =
ω1 (λ 2 + λ 3 )
ω 2 (λ 1 + λ 3 )
ω3 (λ 1 + λ 2 )
−
−
(λ 1 − λ 3 )(λ 1 − λ 2 ) (λ 2 − λ 1 )(λ 2 − λ 3 ) (λ 3 − λ 1 )(λ 3 − λ 2 )
(1.400)
ω3
ω1
ω2
+
+
(λ 1 − λ 3 )(λ 1 − λ 2 ) (λ 2 − λ 1 )(λ 2 − λ 3 ) (λ 3 − λ 1 )(λ 3 − λ 2 )
We can now show that if a tensor function Π ( T ) is given in (1.399), this tensor function is
isotropic:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
94
(
)
Π * (T ) = Q ⋅ Π (T ) ⋅ Q T = Q ⋅ Φ 0 1 + Φ 1 T + Φ 2 T 2 ⋅ Q T
= Φ 0 Q ⋅ 1 ⋅ Q T + Φ 1Q ⋅ T ⋅ Q T + Φ 2 Q ⋅ T 2 ⋅ Q T = Φ 0 1 + Φ 1 T * + Φ 2 T *
2
(1.401)
= Π(T * )
1.6.3
The Derivative of the Tensor-Valued Tensor Function
Firstly, we refer to a scalar-valued tensor function:
Π = Π (A )
(1.402)
The partial derivative of Π (A ) with respect to A is defined as:
∂Π
∂Π ˆ
≡ Π, A =
(e i ⊗ eˆ j )
∂A
∂A ij
(1.403)
where the comma denotes a partial derivative.
Then the second derivative of Π (A ) becomes a fourth-order tensor:
∂ 2Π
∂ 2Π
= Π, AA =
(eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l ) = D ijkl (eˆ i ⊗ eˆ j ⊗ eˆ k ⊗ eˆ l )
∂A ⊗ ∂A
∂A ij ∂A kl
(1.404)
Let C and b be positive definite symmetric second-order tensors defined as:
C = FT ⋅F
;
b = F ⋅FT
(1.405)
where F is an arbitrary second-order tensor with the restriction det ( F ) > 0 imposed on it.
We must also bear in mind that there is a scalar-valued isotropic tensor function,
Ψ = Ψ (I C , II C , III C ) , expressed in terms of the principal invariants of C , where I C = I b ,
II C = II b , III C = III b . Next, we can find the partial derivative of Ψ with respect to C , and
with respect to b . We must also verify that the following relation holds:
F ⋅ Ψ ,C ⋅F T = Ψ , b ⋅b
(1.406)
By applying the chain rule for derivative we obtain:
Ψ ,C =
∂Ψ (I C , II C , III C ) ∂Ψ ∂I C
∂Ψ ∂ II C
∂Ψ ∂ III C
=
+
+
∂C
∂I C ∂C ∂ II C ∂C ∂ III C ∂C
(1.407)
Considering the partial derivatives of the principal invariants, we can state that:
∂I C
=1
∂C
∂ II C
= I C 1 − C T = I C 1 − C = II C C −1 − III C C − 2
∂C
(1.408)
∂ III C
= III C C −T = III C C −1 = C 2 − I C C + II C 1
∂C
Now, by substituting the following values
into the equation in (1.407), we obtain:
Ψ ,C =
∂ II C
∂ III C
∂I C
=1,
= I C 1 − C and
= III C C −1
∂C
∂C
∂C
∂Ψ
∂Ψ
(I C 1 − C ) + ∂Ψ III C C −1
1+
∂I C
∂ II C
∂ III C
(1.409)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
 ∂Ψ

 ∂Ψ
∂Ψ
Ψ , C = 
I C 1 − 
+
 ∂ II C

 ∂I C ∂ II C

 ∂Ψ

C + 
III C C −1

 ∂ III C

Another way to express the relation (1.410) is by substituting
and
95
∂ III C
= C 2 − I C C + II C 1 into the equation in (1.407), thus:
∂C
∂ II C
∂I C
=1,
= IC 1 − C
∂C
∂C
 ∂Ψ

 ∂Ψ

 ∂Ψ
∂Ψ
∂Ψ
∂Ψ
Ψ , C = 
+
IC +
II C 1 − 
+
I C C + 
∂ III C
 ∂I C ∂ II C

 ∂ II C ∂ III C

 ∂ III C
If we now consider the values
equation in (1.407), we obtain:
 ∂Ψ
Ψ , C = 
 ∂I C
(1.410)
 2
C

(1.411)
∂ II C
∂ III C
∂I C
=1,
= II C C −1 − III C C − 2 ,
= III C C −1 , in the
∂C
∂C
∂C

 ∂Ψ

 ∂Ψ

∂Ψ
1 + 
II C +
III C C −1 − 
III C C − 2
∂ III C

 ∂ II C

 ∂ II C

(1.412)
If we now observe both I C = I b , II C = II b , III C = III b , and the equation in (1.410), we can
draw the conclusion that:
 ∂Ψ
Ψ , b = 
 ∂I b
+

∂Ψ
∂Ψ
∂Ψ
b+
I b 1 −
III b b −1
∂ II b 
∂ II b
∂ III b
(1.413)
Using the equation in (1.410), the equation F ⋅ Ψ ,C ⋅ F T becomes:
 ∂Ψ

∂Ψ
∂Ψ
∂Ψ
+
F ⋅ Ψ , C ⋅F T = 
I C  F ⋅ 1 ⋅ F T −
F ⋅C ⋅ F T +
III C F ⋅ C −1 ⋅ F T
∂ II C
∂ III C
 ∂I C ∂ II C

(1.414)
Then, if we observe that:
⇒ F ⋅1 ⋅ F T = F ⋅ F T = b
C = FT ⋅F
(1.415)
⇒ F ⋅C ⋅ F T = F ⋅ F T ⋅ F ⋅ F T = b ⋅ b = b2
C −1 = F −1 ⋅ b −1 ⋅ F
⇒ F ⋅ C −1 ⋅ F T = F ⋅ F −1 ⋅ b −1 ⋅ F ⋅ F T = b −1 ⋅ b
(1.416)
The equation (1.414) can be rewritten as:
 ∂Ψ

∂Ψ
∂Ψ 2
∂Ψ
+
F ⋅ Ψ , C ⋅F T = 
b +
I C b −
III C b −1 ⋅ b
∂
∂
∂
∂
I
I
I
I
I
I
I
I
C
C
C
 C

 ∂Ψ


∂Ψ
∂Ψ
∂Ψ
= 
+
b+
I C 1 −
III C b −1  ⋅ b
∂ II C
∂ III C


 ∂I C ∂ II C
(1.417)
In light of the equation in (1.413) and (1.417), we can draw the conclusion that:
 ∂Ψ


∂Ψ
∂Ψ
∂Ψ
F ⋅ Ψ , C ⋅F T = 
b+
I b 1 −
III b b −1  ⋅ b
+
∂ II b
∂ III b
 ∂I b ∂ II b 

= Ψ ,b ⋅ b = b ⋅ Ψ ,b
(1.418)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
96
which indicates that Ψ , b and b are coaxial tensors.
Once again, we can observe C given by the equation in (1.405). Next, we can evaluate the
derivative of the scalar-valued tensor function, Ψ = Ψ (C ) , with respect to the tensor F :
Ψ,F =
∂Ψ (C ) ∂Ψ ∂C
=
:
∂F
∂C ∂F
notation
indicial
 

→
(Ψ , F )kl
=
∂Ψ ∂C ij
∂C ij ∂Fkl
(1.419)
The derivative of tensor C with respect to F is evaluated as follows:
∂C ij
∂Fkl
=
=
(
∂ Fqi Fqj
)
∂Fkl
( )
∂ Fqi
∂Fkl
Fqj + Fqi
( )
∂ Fqj
(1.420)
∂Fkl
= δ qk δ il Fqj + δ qk δ jl Fqi
= δ il Fkj + δ jl Fki
Then, by substituting (1.420) into (1.419), we obtain:
(Ψ , F )kl
=
∂Ψ
δ il Fkj + δ jl Fki
∂C ij
(
)
∂Ψ
∂Ψ
= Fkj
+ Fki
∂C lj
∂C il
(1.421)
Due to the symmetry of C , i.e. C lj = C jl , we can draw the conclusion that:
(Ψ , F )kl
=2
∂Ψ
∂Ψ
Fkj = 2
Fkj
∂C lj
∂C jl
⇒
Ψ , F = 2Ψ , C ⋅ F T = 2 F ⋅ Ψ , C
(1.422)
Now, suppose that C is given by the equation C = UT ⋅ U = U ⋅ U = U 2 , where U is a
symmetric second-order tensor. To find Ψ (C ) ,U we can use the same equation as in
(1.422), i.e.:
Ψ , U = 2Ψ , C ⋅ U = 2U ⋅ Ψ , C
(1.423)
Therefore, we can draw the conclusion that Ψ , C and U are coaxial tensors.
Let A be a symmetric second-order tensor, and Ψ = Ψ (A ) be a scalar-valued tensor
function. The following relationships hold:
Ψ , b = 2b ⋅ Ψ , A for A = b T ⋅ b
Ψ , b = 2Ψ , A ⋅b for A = b ⋅ b T
Ψ , b = 2b ⋅ Ψ , A = 2Ψ , A ⋅b
= b ⋅ Ψ , A + Ψ , A ⋅b
for A = b ⋅ b and b = b T
(1.424)
1.7 The Voigt Notation
When dealing with symmetric tensors, it may be advantageous to just work with the
independent components. For example, a symmetric second-order tensor has 6
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
97
independent components, so, it is possible to represent these components by a column
matrix as follows:
 T11


Tij =  T12


 T13

T12
T22
T23
 T11 
T13 
T 

 22 

 T33 
→{T } =  
T23  Voigt

 T12 

 T23 
 
T33 
 T13 
(1.425)
This representation is called the Voigt Notation. It is also possible to represent a secondorder tensor as:
E11


E ij = E12


E13

E12
E 22
E 23
 E11 
E13 
E 

 22 

 E 33 

→{E } = 
E 23  Voigt


 2E12 

 2E 23 


E 33 
 2E13 
(1.426)
As we have seen before, a fourth-order tensor, C , that presents minor symmetry, i.e.
C ijkl = C jikl = C ijlk = C jilk , has 6 × 6 = 36 independent components. Note that, due to the
symmetry of (ij ) we have 6 independent components, and due to the symmetry of (kl )
we have 6 independent components. In Voigt Notation we can represent these
components in a 6 -by- 6 matrix as:
 C 1111
C
 2211
C
[C ] =  3311
 C 1211
C 2311

 C 1311
C1122
C1133
C 1112
C1123
C 2222
C 3322
C 2233
C 3333
C 2212
C 3312
C 2223
C 3323
C1222
C 2322
C1233
C 2333
C 1212
C 2312
C1223
C 2323
C1322
C1333
C 1312
C1323
C1113 
C 2213 
C 3313 

C1213 
C 2313 

C1313 
(1.427)
In addition to minor symmetry the tensor also has major symmetry, i.e. C ijkl = C klij , and the
number of independent components have reduced to 21 . One can easily memorize the
order of the components in the matrix [C ] if we consider the order of the second-order
tensor in Voigt Notation, i.e.:
 (11) 
(22)


 (33) 

 [(11) (22) (33) (12) (23) (13)]
 (12) 
(23) 


 (13) 
1.7.1
(1.428)
The Unit Tensors in Voigt Notation
The second-order unit tensor is represented in the Voigt notation as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
98
1
1
 
1
0
0


1


Voigt
δ ij ≡ 1 = 0 1 0 →{δ} =  
0 
0 0 1
0 
 
0
(1.429)
In the subsection 1.5.2.5.1 Unit Tensors we have defined three fourth-order unit tensors,
namely, I ijkl = δ ik δ jl , I ijkl = δ il δ jk and I ijkl = δ ij δ kl , among which only I ijkl = δ ij δ kl is a
symmetric tensor. The representation of I ijkl = δ ij δ kl in Voigt notation can be evaluated by
observing how a symmetric fourth-order tensor is represented in (1.427), thus:
1
1

1
Voigt

→ I = 
0
0

0
[]
I ijkl = δ ij δ kl
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0

0 0 0 0 0
0 0 0 0 0

0 0 0 0 0
(1.430)
where I1111 = δ 11δ 11 = 1 , I1122 = δ 11δ 22 = 1 , and so on.
The
components
I ijkl =
1
δ ik δ jl + δ il δ jk , which in Voigt notation becomes:
2
I ijkl
(
of
)
 I1111
I
 2211
I 3311
Voigt

→[I ] = 
 I1211
I 2311

 I1311
a
fourth-order
unit
I1122
I 2222
I1133
I 2233
I1112
I 2212
I1123
I 2223
I 3322
I1222
I 3333
I1233
I 3312
I1212
I 3323
I1223
I 2322
I1322
I 2333
I1333
I 2312
I1312
I 2323
I1323
tensor,
I sym ,
I1113  1

I 2213  0
I 3313  0
=
I1213  0
I 2313  0
 
I1313  0
are
represented
0 0 0 0 0
1 0 0 0 0 
0 1 0 0 0

0 0 12 0 0 
0 0 0 12 0 

0 0 0 0 12 
by
(1.431)
and the inverse of the equation in (1.431) becomes:
[I ]−1
1.7.2
1
0

0
=
0
0

0
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0

0 0 2 0 0
0 0 0 2 0

0 0 0 0 2
(1.432)
The Scalar Product in Voigt Notation
r
The dot product between a symmetric second-order tensor, T , and a vector n , is given by
r
r
r
b = T ⋅ n where the components of b can be evaluated as follows:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
 b1   T11
b  =  T
 2   12
b 3   T13
T12
T22
T23
99
T13   n1   b1 = T11n1 + T12 n 2 + T13n 3

T23  n 2  ⇒ b 2 = T12 n1 + T22 n 2 + T23n 3
T33  n 3  b 3 = T13n1 + T23n 2 + T33n 3
(1.433)
By observing how a second-order tensor is presented in Voigt notation, as in (1.425), the
scalar product (1.433) can be represented in the Voigt notation as:
 T11 
T 
22
 b1  n1 0 0 n 2 0 n 3   

T
b  =  0 n
 33 
0
0
n
n
2
2
1
3
  
 T 
b 3   0 0 n 3 0 n 2 n1   12 
144444244444
3  T23 
 
[N ]T
 T13 
1.7.3
⇒
{b} = [N ] {T }
T
(1.434)
The Component Transformation Law in Voigt Notation
The component transformation law for a second-order tensor is defined as:
Tij′ = Tkl a ik a jl
(1.435)
or in matrix form:
 T11′
T ′
 12
 T13′
T12′
′
T22
′
T23
T13′   a11
′  =  a 21
T23
 
′   a 31
T33
a12
a 22
a 32
a13   T11
a 23   T12
a 33   T13
T12
T22
T23
T13   a11
T23  a 21
T33   a 31
a12
a 22
a 32
a13 
a 23 
a 33 
T
(1.436)
By multiplying the matrices and by rearranging the result in Voigt notation we obtain:
{T ′} = [M] {T }
(1.437)
where:
 T11′ 
T′ 
 22 
T′ 
{T ′} =  33′  ;
 T12 
 T23
′ 
 
 T13′ 
 T11 
T 
 22 
T 
{T } =  33 
 T12 
 T23 
 
 T13 
(1.438)
and [M] is the transformation matrix for the second-order tensor components in Voigt
Notation. The matrix [M] is given by:
 a11 2

2
 a 21
2

[M] =  a 31
 a 21 a11
a a
 31 21
 a 31 a11
a12 2
a 22 2
a 32 2
a 22 a12
a 32 a 22
a 32 a12
a13 2
a 23 2
a 33 2
a13 a 23
a 33 a 23
a 33 a13
2a11 a12
2a 21 a 22
2a 31 a 32
(a11 a 22 + a12 a 21 )
(a 31 a 22 + a 32 a 21 )
(a 31 a12 + a 32 a11 )
2a12 a13
2a 22 a 23
2a 32 a 33
(a13 a 22 + a12 a 23 )
(a 33 a 22 + a 32 a 23 )
(a 33 a12 + a 32 a13 )

2a11 a13

2a 21 a 23


2a 31 a 33
 (1.439)
(a13 a 21 + a11 a 23 )
(a 33 a 21 + a 31 a 23 )
(a 33 a11 + a 31 a13 )
If the representation of tensor components is shown in (1.426), equation (1.436) in Voigt
Notation becomes:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
100
{E ′} = [N ]{E }
(1.440)
where
 a11 2

2
 a 21
2

[N ] =  a31
 2a 21 a11
 2a a
 31 21
 2a 31 a11
a12 2
a13 2
a11 a12
a12 a13
a 22
2
a 23
2
a 21 a 22
a 22 a 23
a 32
2
a 33
2
a 31 a 32
a 32 a 33
2a 22 a12
2a 32 a 22
2a13 a 23
2a 33 a 23
2a 32 a12
2a 33 a13


a 21 a 23


a 31 a 33

(a13 a 21 + a11 a 23 )
(a 33 a 21 + a 31 a 23 )
(a 33 a11 + a 31 a13 )
a11 a13
(a11 a 22 + a12 a 21 ) (a13 a 22 + a12 a 23 )
(a 31 a 22 + a 32 a 21 ) (a 33 a 22 + a 32 a 23 )
(a 31 a12 + a 32 a11 ) (a 33 a12 + a 32 a13 )
(1.441)
The matrices (1.439) and (1.441) are not orthogonal matrices, i.e. [M]−1 ≠ [M]T and
[N ]−1 ≠ [N ]T . However, it is possible to show that [M]−1 = [N ]T .
1.7.4
Spectral Representation in Voigt Notation
Regarding the spectral representation of a symmetric tensor T :
T=
3
∑T
a
nˆ ( a ) ⊗ nˆ ( a )
Matricial
  form
→
a =1
T = AT T ′ A
(1.442)
where A is the transformation matrix between the original set and the principal space,
made up of the eigenvectors nˆ ( a ) . The above equation can be rewritten in terms of
components as follows:
 T11
T
 12
 T13
T12
T22
T23
T13 
 T1

T 
T23  = A  0
 0
T33 
0 0
0 0

T 
0 0  A + A 0 T 2
0 0
0 0
0
0 0 0 

T 
0 A + A 0 0 0  A
0 0 T3 
0
(1.443)
or
 T11
T
 12
 T13
T12
T22
T23
2
 a11
T13 

T23  = T1 a11 a12
a a
T33 
 11 13


+ T3  a 31 a 32
a a
 31 33
2
a 31
a11 a12
2
a12
a12 a13
a 31 a 32
2
a 32
a 32 a 33
2
 a 21
a11 a13 


a12 a13  + T2 a 21 a 22
2 
a a
a13

 21 23
a 21 a 22
2
a 22
a 22 a 23
a 31 a 33 

a 32 a 33 
2

a 33

a 21 a 23 

a 22 a 23 
2

a 23

(1.444)
By regarding how second-order tensors are presented in Voigt Notation as in (1.438), the
spectral representation of a second-order tensor in Voigt notation becomes:
2
2
2
 a11

 a 21

 a 31

 T11 






T 
2
2
2
 a12 
 a 22 
 a 32 
 22 
 2 
 2 
 2 
T 
{T } =  33  = T1  a13  + T2  a 23  + T3  a33 
 a11 a12 
 a 21 a 22 
 a 31 a 32 
 T12 




a a 
 T23 
a a
a a
 12 13 
 22 23 
 32 33 
 
T
 13 
 a11 a13 
 a 21 a 23 
 a 31 a 33 
(1.445)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
1.7.5
101
Deviatoric Tensor Components in Voigt Notation
Observing the components of the deviatoric tensor:
Tijdev
 13 ( 2 T11 − T22 − T33 )

=
T12

T13

T12
1
3


T23

1
(2 T33 − T11 − T22 )
3
T13
(2 T22 − T11 − T33 )
T23
(1.446)
Tijdev in Voigt notation is given by:
 T11dev 
 2 − 1 − 1 0 0 0  T11 
 dev 
 − 1 2 − 1 0 0 0  T 
 T22 
  22 

 T33dev  1 − 1 − 1 2 0 0 0  T33 
 dev  = 
(1.447)
 
0
0 3 0 0  T12 
 T12  3  0
 T dev 
0
0
0 0 3 0  T23 

 23
 

dev
0
0 0 0 3  T13 
 0
 T13 
r
Problem 1.40: Let T ( x , t ) be a symmetric second-order tensor, which is expressed in
r
terms of the position ( x ) and time (t ) . Also, bear in mind that the tensor components,
along direction x3 , are equal to zero, i.e. T13 = T23 = T33 = 0 .
r
NOTE: In the next section we will define T ( x , t ) as a field tensor, i.e. the value of T
depends on position and time. As we will see later, if the tensor is independent of any one
r
r
direction at all points ( x ) , e.g. if T ( x , t ) is independent of the x3 -direction, (see Figure
1.31), the problem becomes a two-dimensional problem (plane state) so that the problem is
greatly simplified. ■
2D
x2
x2
T22
T22
T12
T12
T12
T11
T11
T11
x1
T12
x3
T22
x1
Figure 1.31: A two-dimensional problem (2D).
′ , T12′ in the new reference system ( x1′ − x 2′ ) defined in Figure 1.32.
a) Obtain T11′ , T22
b) Obtain the value of θ so that θ corresponds to the principal direction of T , and also
find an equation for the principal values of T .
c) Evaluate the values of Tij′ , (i, j = 1,2) , when T11 = 1 , T22 = 2 , T12 = −4 and θ = 45º .
Also, obtain the principal values and principal directions.
′ and
d) Draw a graph that shows the relationship between θ and components T11′ , T22
T12′ , and in which the angle varies from 0º to 360º .
Hint: Use the Voigt Notation, and express the results in terms of 2θ .
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
102
x2
x1′
x 2′
θ
x1
Figure 1.32: A two-dimensional problem (2D).
Solution:
a) Here we can apply the transformation law obtained in (1.437), which after removing
rows and columns associated with the x3 -direction becomes:
 T11′   a11
T′  =  a 2
 22   21
 T12′   a 21 a11

2
a12
2
a 22
2
a 22 a12
  T11 
 
2a 21 a 22
  T22 
a11 a 22 + a12 a 21   T12 

2a11 a12
(1.448)
T ′ = A T AT
x2
′
T22
x 2′
T22
T12′
T11′
T12
T11
T11
P
T11′
T12
T12′
x1
T22
x1′
P
θ
′
T22
x1
T = AT T ′ A
Figure 1.33: Transformation law for (2D) tensor components.
The transformation matrix, a ij , in the plane, can be evaluated in terms of a single
parameter, θ :
 a11 a12 a13   cos θ sin θ 0
a ij = a 21 a 22 a 23  =  − sin θ cos θ 0
(1.449)
 a 31
a 32
a 33 

0
1
0
By substituting the matrix components a ij given in (1.449) into (1.448) we obtain:
2
sin 2 θ
2 cos θ sin θ   T11 
 T11′   cos θ


T′  =
2
2
cos θ
− 2 sin θ cos θ   T22 
 22   sin θ
 T12′   − sin θ cos θ cos θ sin θ cos 2 θ − sin 2 θ   T12 


(1.450)
trigonometric identities, 2 cos θ sin θ = sin 2θ ,
1 − cos 2θ
1 + cos 2θ
cos 2 θ − sin 2 θ = cos 2θ , sin 2 θ =
, cos 2 θ =
, (1.450) becomes:
Making
use
of
the
following
2
2
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
 1 + cos 2θ 


2


′
T
 11  
 T ′  =  1 − cos 2θ 

 22  
2

 T12′  
2
sin
θ


 −

 
2 
103

 1 − cos 2θ 

 sin 2θ 
2


  T11 
 1 + cos 2θ 
 

 − sin 2θ   T22 
2


  T12 
 sin 2θ 
cos 2θ 



 2 
Explicitly, the above components are given by:

 1 + cos 2θ 
 1 − cos 2θ 
 T11 + 
 T22 + T12 sin 2θ
 T11′ = 
2
2





 1 − cos 2θ 
 1 + cos 2θ 

′ =
 T22 − T12 sin 2θ
 T11 + 
 T22
2
2






 sin 2θ 
 sin 2θ 
 T12′ =  −
 T11 + 
 T22 + T12 cos 2θ

2 

 2 
Reordering the previous equation, we obtain:

 T11 + T22   T11 − T22 
+
 cos 2θ + T12 sin 2θ
 T11′ = 
2
2

 



 T + T22   T11 − T22 
′ =  11
−
 cos 2θ − T12 sin 2θ
 T22
2
2

 



 T − T22 
 T12′ = − 11
 sin 2θ + T12 cos 2θ

2


(1.451)
b) Recalling that the principal directions are characterized by the lack of any tangential
components, i.e. Tij = 0 if i ≠ j , in order to find the principal directions in the plane, we let
T12′ = 0 , hence:
 T − T22 
 T − T22 
T12′ = − 11
 sin 2θ = T12 cos 2θ
 sin 2θ + T12 cos 2θ = 0 ⇒  11
2
2




2
2
T
T
sin 2θ
12
12
=
⇒
⇒ tg(2θ ) =
cos 2θ T11 − T22
T11 − T22
Then, the angle corresponding to the principal direction is:

2 T12
 T11 − T22
1
2
θ = arctg



(1.452)
To find the principal values (eigenvalues) we must solve the following characteristic
equation:
T11 − T
T12
T12
=0
T22 − T
⇒
(
And by evaluating the quadratic equation we obtain:
T(1, 2 )
=
=
− [− ( T11 + T22 )] ±
[− (T11 + T22 )]2
2(1)
T11 + T22
±
2
)
T 2 − T ( T11 + T22 ) + T11 T22 − T122 = 0
[(T11 + T22 )]2
(
(
− 4(1) T11 T22 − T122
− 4 T11 T22 − T122
4
)
)
By rearranging the above equation we obtain the principal values for the two-dimensional
case as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
104
T(1, 2 ) =
T11 + T22
 T − T22 
±  11

2
2


2
(1.453)
+ T122
c) We directly apply equation (1.451) to evaluate the values of the components Tij′ ,
(i, j = 1,2) , where T11 = 1 , T22 = 2 , T12 = −4 and θ = 45º , i.e.:

1 + 2  1 − 2 
 T11′ =  2  +  2  cos 90º −4 sin 90º = −2.5

 


1 + 2  1 − 2 

′ =
−
 cos 90º +4 sin 90º = 5.5
 T22
 2   2 


1 − 2 
 T12′ = −
 sin 90º −4 cos 90º = 0.5

 2 
And the angle corresponding to the principal direction is:
1
2

2 T12
 T11 − T22
θ = arctg
r
 2 × (−4)
 =
1− 2

(θ = 41.4375º )
⇒
The principal values of T ( x , t ) can be evaluated as follows:
T(1, 2 )
T + T22
 T − T22 
= 11
±  11

2
2


2
+ T122
⇒
 T1 = 5.5311

 T2 = −2.5311
d) By referring to equation in (1.451) and by varying θ from 0º to 360º , we can obtain
′ , T12′ , which are illustrated in the following graph:
different values of T11′ , T22
x1′
T1
θ = 41.437 º
T2
x1′
Components
θ = 131.437º
σ2
8
′
T22
T1 = 5.5311
6
4
T22
2
T12′
T11
0
0
50
100
-2
45º
T12
-4
T2 = −2.5311
150
200
250
300
350
θ
T11′
x1′
θ = 86.437º
-6
TS max = 4.0311
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
105
1.8 Tensor Fields
r
r
A tensor field indicates how the tensor, T ( x , t ) , varies in space ( x ) and time ( t ). In this
section, we regard the tensor field as a differentiable function of position and time. For
more information about it, we need to define some operators, e.g. gradient, divergence, curl,
which we can use as indicators of how these fields vary in space.
A tensor field which is independent of time is called a stationary or steady-state tensor
r
field, i.e. T = T ( x ) . However, if the field is only dependent on t then it is said to be
r
homogeneous or uniform. That is, T (t ) has the same value at every x position.
Tensor fields can be classified according to their order as: scalar, vector, second-order
r
tensor fields, etc. As an example of a scalar field we can quote temperature T ( x , t ) and in
Figure 1.34(a) we can see temperature distribution over time t = t1 . Then, as an example of
r r
a vector field we can quote velocity v ( x , t ) and Figure 1.34(b) shows velocity distribution,
r
in which each point is associated with a vector v over time t = t1 .
t = t1
x3
T5
T8
t = t1
x3
r r
v ( x , t1 )
r
T4 ( x ( 4) , t1 )
T6
T3
T7
T1
x2
x2
T2
x1
a) Scalar field
x1
b) Vector field
Figure 1.34: Examples of tensor fields.
Scalar Field
r
φ = φ( x, t )
(1.454)
r r r
v = v ( x, t )
r
v i = v i ( x, t )
(1.455)
r
T = T ( x, t )
r
Tij = Tij ( x , t )
(1.456)
Vector Field
Tensorial notation
Indicial notation
Second-Order Tensor Field
Tensorial notation
Indicial notation
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
106
1.8.1
Scalar Fields
r
The next analysis is carry out with reference to a stationary scalar field, i.e. φ = φ( x ) , with
continuous values of ∂φ / ∂x1 , ∂φ / ∂x 2 and ∂φ / ∂x 3 . Then, observe that the value of the
r
r
r
r
scalar function at point ( x ) is φ ( x ) , and if we observe a second point located at ( x + dx ) ,
the total derivative (differential) of the function φ is defined as:
r
r
r
φ ( x + dx ) − φ ( x ) ≡ dφ
φ ( x1 + dx1 , x 2 + dx 2 , x3 + dx3 ) − φ ( x1 , x 2 , x3 ) ≡ dφ
(1.457)
For any continuous function φ ( x1 , x 2 , x3 ) , dφ is linearly related to dx1 , dx 2 , dx3 . This
linear relationship can be evaluated by the chain rule of differentiation as:
∂φ
∂φ
∂φ
dx1 +
dx 2 +
dx3
∂x1
∂x 2
∂x 3
dφ =
(1.458)
dφ = φ, i dxi
The differentiation of the components of a tensor, with respect to coordinates xi , is
expressed by the differential operator:
∂•
≡ •, i
∂xi
1.8.2
(1.459)
Gradient
The gradient of a scalar field
The gradient ∇ xr φ or gradφ is defined as:
r
∇ xr φ 
→ dφ = ∇ xr φ ⋅ dx
(1.460)
where the operator ∇ xr is known as the Nabla symbol. Expressing the equation (1.460) in
the Cartesian basis we obtain:
∂φ
∂φ
∂φ
dx1 +
dx 2 +
dx3 =
∂x1
∂x 2
∂x3
[
= (∇ xr φ) x eˆ 1 + (∇ xr φ) x eˆ 2 + (∇ xr φ) x eˆ 3
1
2
3
]⋅ [(dx )eˆ
1
1
+ (dx 2 )eˆ 2 + (dx3 )eˆ 3
]
(1.461)
Evaluating the above scalar product we find:
∂φ
∂φ
∂φ
dx1 +
dx 2 +
dx3 = (∇ xr φ) x1 dx1 + (∇ xr φ) x2 dx 2 + (∇ xr φ) x3 dx3
∂x1
∂x 2
∂x 3
(1.462)
Therefore, we can draw the conclusion that the ∇ xr φ components in the Cartesian basis
are:
(∇ xr φ )1 ≡
∂φ
∂x1
;
(∇ xr φ) 2 ≡
∂φ
∂x 2
;
(∇ xr φ) 3 ≡
∂φ
∂x3
(1.463)
Hence, the gradient in terms of components is defined as such:
∇ xr φ =
∂φ ˆ
∂φ ˆ
∂φ ˆ
e1 +
e2 +
e3
∂x 2
∂x3
∂x1
(1.464)
The Nabla symbol ∇ xr is defined as:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
∇ xr =
107
∂ ˆ
e i ≡ ∂ , i eˆ i Nabla symbol
∂xi
(1.465)
The geometric meaning of ∇ xr φ
ƒ The direction of
∇ xr φ is normal to the equiscalar surface, i.e. it is perpendicular to
the isosurface φ = const . The direction of ∇ xr φ points to the direction where φ is
increasing the most, (see Figure 1.35).
ƒ The magnitude of
∇ xr φ is the rate of change of φ , i.e. the gradient of φ .
The normal vector to this surface is obtained as follows:
n̂ =
∇ xr φ
∇ xr φ
(1.466)
The surface φ = const , called the surface level, or isosurface or equiscalar surface, is the
surface formed by points which all have the same value of φ , so, if we move along the
level surface the values of the function do not change.
r r
The gradient of a vector field v ( x ) :
r
r
grad( v ) ≡ ∇ xr v
(1.467)
Using the definition of ∇ xr , given in (1.465), the gradient of the vector field becomes:
r ∂ ( v i eˆ i )
∇ xr v =
⊗ eˆ j = ( v i eˆ i ) , j ⊗ eˆ j = v i , j eˆ i ⊗ eˆ j
∂x j
n̂
(1.468)
∇ xr φ
φ = const = c1
φ = const = c 2
c1 > c 2 > c3
φ = const = c 3
Figure 1.35: Gradient of φ .
r
Therefore, we can define the gradient of a tensor field (•( x , t )) in the Cartesian basis as:
∇ xr (•) =
∂(•)
⊗ ê j Gradient of a tensor field in the
∂x j
Cartesian basis
(1.469)
As noted, the gradient of a vector field becomes a second-order tensor field, whose
components are:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
108
v i, j
 ∂v 1

 ∂x1
∂v i  ∂v 2
≡
=
∂x j  ∂x1

 ∂v 3
 ∂x1
∂v 1 

∂x3 
∂v 2 
∂x3 

∂v 3 
∂x3 
∂v 1
∂x 2
∂v 2
∂x 2
∂v 3
∂x 2
(1.470)
r
The gradient of a second-order tensor field T ( x ) :
∇ xr T =
∂ (Tij eˆ i ⊗ eˆ j )
∂x k
⊗ eˆ k = Tij ,k eˆ i ⊗ eˆ j ⊗ eˆ k
(1.471)
and its components are represented by:
(∇ xr T )ijk
≡ Tij ,k
(1.472)
Problem 1.41: Find the gradient of the function f ( x1 , x 2 ) = cos( x1 ) + exp x1x2 at the point
( x1 = 0, x 2 = 1) .
Solution: By definition, the gradient of a scalar function is given by:
∂f ˆ
∂f ˆ
e1 +
e2
∂x1
∂x 2
∂f
;
= x1 exp x1x2
∂x 2
∇ xr f =
where:
∂f
= − sin( x1 ) + x 2 exp x1x2
∂x1
[
]
[
]
∇ xr f ( x1 , x 2 ) = − sin( x1 ) + x 2 exp x1x2 eˆ 1 + x1 exp x1x2 eˆ 2 ⇒ ∇ xr f (0,1) = [1] eˆ 1 + [0] eˆ 2 = 2eˆ 1
r r
Problem 1.42: Let u( x ) be a stationary vector field. a) Obtain the components of the
r r
r
differential du . b) Now, consider that u( x ) represents a displacement field, and is
independent of x3 . With these conditions, graphically illustrate the displacement field in
the differential area element dx1 dx 2 .
Solution: According to the differential and gradient definitions, it holds that:
r r
u( x )
r r r
r r r
du ≡ u( x + dx ) − u( x )
r
r r
du = ∇ xr u ⋅ dx
x2
r
x
r
dx
r r
r
u( x + dx )
r
r
x + dx
x1
x3
Thus, the components are defined as:
du i =
∂u i
dx j
∂x j
⇒
 ∂u1

 du1   ∂x1
du  =  ∂u 2
 2   ∂x
 du 3   1
 ∂u 3
 ∂x1
∂u1
∂x 2
∂u 2
∂x 2
∂u 3
∂x 2
∂u1 

∂x3   dx 
1
∂u 2  

dx
2
∂x3  


∂u 3   dx3 
∂x3 
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
109
or:

∂u1
∂u
∂u
dx1 + 1 dx 2 + 1 dx3
du1 =
∂x1
∂x 2
∂x3


∂u 2
∂u
∂u
dx1 + 2 dx 2 + 2 dx3
 du 2 =
∂x1
∂x 2
∂x3


∂u
∂u
∂u
du 3 = 3 dx1 + 3 dx 2 + 3 dx 3
∂x1
∂x 2
∂x3

with
du1 = u1 ( x1 + dx1 , x 2 + dx 2 , x3 + dx3 ) − u1 ( x1 , x 2 , x3 )

du 2 = u 2 ( x1 + dx1 , x 2 + dx 2 , x3 + dx3 ) − u 2 ( x1 , x 2 , x3 )
du = u ( x + dx , x + dx , x + dx ) − u ( x , x , x )
3
1
1
2
2
3
3
3
1
2
3
 3
As the field is independent of x3 , the displacement field in the differential area element is
defined as:
∂u1
∂u1

du1 = u1 ( x1 + dx1 , x 2 + dx 2 ) − u1 ( x1 , x 2 ) = ∂x dx1 + ∂x dx 2

1
2

u
u
∂
∂
du = u ( x + dx , x + dx ) − u ( x , x ) = 2 dx + 2 dx
2
1
1
2
2
2
1
2
1
2
 2
∂x1
∂x 2
or:
∂u1
∂u1

u1 ( x1 + dx1 , x 2 + dx 2 ) = u1 ( x1 , x 2 ) + ∂x dx1 + ∂x dx 2

2
1

u ( x + dx , x + dx ) = u ( x , x ) + ∂u 2 dx + ∂u 2 dx
2
1
1
2
2
2
1
2
 2 1
∂x1
∂x 2
Note that the above equation is equivalent to the Taylor series expansion taking into
account only up to linear terms. The representation of the displacement field in the
differential area element is shown in Figure 1.36.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
110
u2 +
∂u 2
dx 2
∂x 2
u2 +
( x1 + dx1 , x 2 + dx 2 )
( x1 , x 2 + dx 2 )
u1 +
∂u
∂u 2
dx1 + 2 dx 2
∂x 2
∂x1
∂u1
dx 2
∂x 2
u1 +
r
du
dx 2
∂u
∂u1
dx1 + 1 dx 2
∂x 2
∂x1
u2 +
(u 2 )
( x1 + dx1 , x 2 )
( x1 , x 2 )
x2
∂u 2
dx1
∂x1
u1 +
(u1 )
∂u1
dx1
∂x1
dx1
x1
144444444444444444424444444444444444443
=
644444444444444444474444444444444444448
x 2 ,u 2
u2 +
∂u1
dx2
∂x2
∂u 2
dx2
∂x2
B′
B
B
dx 2
A′
O′
u2
A
O
dx1
u1
u1 +
+
B′
dx 2
A′
O′
A
∂u 2
dx1
∂x1
dx1
∂u1
dx1
∂x1
x1 ,u1
Figure 1.36: Displacement field in the differential area element.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
1.8.3
111
Divergence
r r
The divergence of a vector field, v ( x ) , is denoted as follows:
r
r
div ( v ) ≡ ∇ xr ⋅ v
(1.473)
r
r
r
r
div ( v ) ≡ ∇ xr ⋅ v = ∇ xr v : 1 = Tr (∇ xr v )
(1.474)
which by definition is:
Then:
[
][
r
r
∇ xr ⋅ v = ∇ xr v : 1 = v i , j eˆ i ⊗ eˆ j : δ kl eˆ k ⊗ eˆ l
∂v
∂v
∂v
= 1 + 2 + 3
∂x1 ∂x 2 ∂x3
]
= v i , j δ kl δ ik δ jl = v k ,k
(1.475)
or
[
][
= [v δ δ eˆ ]⋅ eˆ
= [v eˆ ]⋅ eˆ
∂[v eˆ ]
⋅ eˆ
=
r
r
∇ xr ⋅ v = ∇ xr v : 1 = v i , j eˆ i ⊗ eˆ j : δ kl eˆ k ⊗ eˆ l
i, j
kl
i ,k
i
i
i
lj
i
]
k
(1.476)
k
k
∂x k
Which we can use to insert the following operator into the Cartesian basis:
∇ xr ⋅ (•) =
∂ (•)
⋅ ê k Divergence of (•) in Cartesian
∂x k
basis
(1.477)
We can also verify that, when divergence is applied to a tensor field its rank decreases by
one order.
r
Divergence of a second-order tensor field T ( x )
The divergence of a second-order tensor field T is denoted by ∇ xr ⋅ T = ∇ xr T : 1 , which
becomes a vector:
∇ xr ⋅ T ≡ divT
=
=
∂(Tij eˆ i ⊗ eˆ j )
∂Tij
∂x k
δ jk eˆ i
⋅ eˆ k
(1.478)
∂x k
= Tik ,k eˆ i
r
NOTE: In this text book, when dealing with gradient or divergence of a tensor field, e.g. ∇ xr v
(the gradient of the vector field), ∇ xr T (the gradient of a second-order tensor field), ∇ xr ⋅ T
(divergence of a second-order tensor field), this does not indicate that we are making a
r
r
r
r
tensor operation between a vector and a tensor, i.e. ∇ xr v ≠ (∇ xr ) ⊗ ( v ) , ∇ xr T ≠ (∇ xr ) ⊗ ( T )
r
and ∇ xr ⋅ T ≠ (∇ xr ) ⋅ ( T ) and so on. In this textbook, ∇ xr is an operator which must be
applied to the entire tensor field, so, the tensor must be inside the operator, (see equations
r
(1.477) and (1.469)). Nevertheless, it is possible to relate ∇ xr v , ∇ xr T or ∇ xr ⋅ T to tensor
operations between tensors, and it is easy to show that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
112
r
r
r
∇ xr v = ( v ) ⊗ (∇ xr )
r
∇ xr T = ( T ) ⊗ (∇ xr )
r
r
∇ xr ⋅ T = ( T ) ⋅ (∇ xr ) = (∇ xr ) ⋅ ( T T ) ■
(1.479)
Once the Nabla symbol is defined we introduce the Laplacian operator ∇ 2 as:
 ∂
2
∇ xr = ∇ xr ⋅ ∇ xr = 
 ∂xi

 ∂ ∂
 ∂

∂2

=
eˆ j  ⋅ eˆ i  =
δ
 ∂x j

 ∂xi ∂x j ij ∂xi ∂xi



∂2
∂2
∂2
∇ xr ≡ 2 + 2 + 2 = ∂, k ∂, k = ∂, kk
∂x1 ∂x 2 ∂x3
r r
Then, the vector Laplacian of a vector field, v ( x ) , is given by:
(1.480)
2
[
]
r
r
2r
2r
∇ xr v = ∇ xr ⋅ (∇ xr v ) components
  → ∇ xr v = [∇ xr ⋅ (∇ xr v )] = v i ,kk
i
i
(1.481)
r
r
and b be vectors. Show that the following identity
r
r r
r
∇ xr ⋅ (a + b) = ∇ xr ⋅ a + ∇ xr ⋅ b holds.
Solution:
Problem 1.43: Let a
r r
∂
, we can express ∇ xr ⋅ (a + b) as:
∂x i
r
r
Observing that a = a j eˆ j , b = b k eˆ k , ∇ xr = ê i
∂ (a j eˆ j + b k eˆ k )
∂x i
⋅ eˆ i
=
∂a j
∂x i
eˆ j ⋅ eˆ i +
r
r
∂b k
∂a
∂b
eˆ k ⋅ eˆ i = i + i = ∇ xr ⋅ a + ∇ xr ⋅ b
∂x i
∂x i ∂x i
Working directly with indicial notation we obtain:
r
r r
r
∇ xr ⋅ (a + b) = (a i + b i ), i = a i , i +b i , i = ∇ xr ⋅ a + ∇ xr ⋅ b
r
r
Problem 1.44: Find the components of (∇ xr a) ⋅ b .
r
r
Solution: Bearing in mind that a = a j eˆ j , b = b k eˆ k , ∇ xr = ê i
∂
( i = 1,2,3 ), the following is
∂x i
true:

∂a j
∂a j
r r  ∂ (a j eˆ j )
 ∂a j

⊗ eˆ i  ⋅ (b k eˆ k ) = 
(∇ xr a) ⋅ b = 
eˆ j ⊗ eˆ i  ⋅ (b k eˆ k ) = b k δ ik
eˆ j = b k
eˆ j
 ∂x i

∂x i
∂x k
 ∂x i



Expanding the dummy index k , we obtain:
∂a j
∂a j
∂a j
∂a j
bk
= b1
+ b2
+ b3
∂x k
∂x1
∂x 2
∂x 3
Thus,
j =1
⇒ b1
∂a1
∂a
∂a
+ b2 1 + b3 1
∂x1
∂x 2
∂x 3
j = 2 ⇒ b1
∂a 2
∂a
∂a
+ b2 2 + b3 2
∂x1
∂x 2
∂x 3
j = 3 ⇒ b1
∂a 3
∂a
∂a
+ b 2 3 + b3 3
∂x 2
∂x 3
∂x1
Problem 1.45: Prove that the following relationship is valid:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
113
r
r
q 1
1 r
∇ xr ⋅   = ∇ xr ⋅ q − 2 q ⋅ ∇ xr T
T
T  T
r r
r
where q( x , t ) is an arbitrary vector field, and T ( x , t ) is a scalar field.
Solution:
1.8.4
r
q
∂  qi   qi 
1
1
∇ xr ⋅   =
  ≡   = q i ,i − 2 q i T,i
∂
T
x
T
T
T
T
   ,i
 
i 
r
1
1 r
= ∇ xr ⋅ q − 2 q ⋅ ∇ xr T (scalar)
T
T
The Curl
The curl of a vector field
r r
r
r
r
r
The curl (or rotor) of a vector field, v ( x ) is denoted by curl( v ) ≡ rot ( v ) ≡ ∇ xr ∧ v , and is
defined in the Cartesian basis as:
r
∂ ˆ
e j ∧ (•) The curl (rotor) of a tensor field in
∇ xr ∧ (•) =
∂x j
the Cartesian basis
(1.482)
Note that the curl is already a tensor operator between two vectors. Using the definition of
the vector product we obtain the curl of a vector field as:
r
r
r
∂v
∂v
∂ ˆ
e j ∧ ( v k eˆ k ) = k eˆ j ∧ eˆ k = k  ijk eˆ i =  ijk v k , j eˆ i
rot ( v ) = ∇ xr ∧ v =
∂x j
∂x j
∂x j
(1.483)
where  ijk is the permutation symbol defined in (1.55). Moreover, we have applied the
definition eˆ j ∧ eˆ k =  ijk eˆ i and we can also note that:
eˆ 1
r
r
r
∂
rot ( v ) = ∇ xr ∧ v =
∂x1
v1
 ∂v
=  3
 ∂x 2
eˆ 2
∂
∂x 2
v2
∂v
− 2
∂x3
eˆ 3
∂
=  ijk v k , j eˆ i
∂x3
v3
 ∂v
 ∂v

∂v 
∂v 
eˆ 1 +  1 − 3 eˆ 2 +  2 − 1 eˆ 3
 ∂x1 ∂x 2 

 ∂x3 ∂x1 
(1.484)
We can verify that the antisymmetric part of a vector field gradient, which is illustrated by
r
(∇ xr v ) skew ≡ W , has as components:
[(∇
r
]
skew
r
x v)
ij
≡ v iskew
,j
 0
= W21
 W31

0


 1  ∂v
∂v 
=   2 − 1 
 2  ∂x1 ∂x 2 
 1  ∂v
∂v 
  3 − 1 
 2  ∂x1 ∂x3 
W12
0
W32
1  ∂v 1 ∂v 2

−
2  ∂x 2 ∂x1



0
1  ∂v 3 ∂v 2

−
2  ∂x 2 ∂x3
W13   0
W23  =  − W12
0   − W13
W12
0
− W23



1  ∂v 1 ∂v 3  


−
2  ∂x 3 ∂x1  
1  ∂v 2 ∂v 3  


−
2  ∂x 3 ∂x 2  


0

W13   0
W23  =  w3
0   − w2
− w3
0
w1
(1.485)
w2 
− w1 
0 
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
114
r
where w1 , w2 , w3 are the components of the axial vector w associated with W , (see
subsection: 1.5.2.2.2. Antisymmetric Tensor).
With reference to the definition of the curl in (1.484) and the relationship in (1.485), we can
conclude that:
r
r
r
 ∂v
 ∂v
 ∂v
∂v 
∂v 
∂v 
rot ( v ) ≡ ∇ xr ∧ v =  3 − 2 eˆ 1 +  1 − 3 eˆ 2 +  2 − 1 eˆ 3
 ∂x1 ∂x 2 
 ∂x3 ∂x1 
 ∂x 2 ∂x3 
ˆ
ˆ
ˆ
= 2W32 e1 + 2W13 e 2 + 2W21e 3
= 2(w1 eˆ 1 + w2 eˆ 2 + w3 eˆ 3 )
r
= 2w
(1.486)
And, if we use the identity in (1.141), we obtain:
(
)
r
r
r r r 1 r
W ⋅ v = w ∧ v = ∇ xr ∧ v ∧ v
2
(1.487)
It could be interesting to note that the equation in (1.486) can be obtained by means of
Problem 1.18, in which we showed that
r
r
1 r r
(a ∧ x ) is the axial vector associated with the
2
antisymmetric tensor ( x ⊗ a ) skew . Therefore, the axial vector associated with the
[r
r
r
antisymmetric tensor W = (∇ xr v ) skew = ( v ) ⊗ (∇ )
]
skew
is the vector
(
)
1 rr r
∇x ∧ v .
2
As we can see, the curl describes the rotational tendency of the vector field.
Summary
•
Divergence
Gradient
div (•) ≡ ∇ xr ⋅ •
grad(•) ≡ ∇ xr •
Scalar
Curl
r
rot (•) ≡ ∇ xr ∧ •
vector
Vector
Scalar
Second-order tensor
Vector
Second-order tensor
Vector
Third-order tensor
Second-order tensor
We can now present some equations:
ƒ
r
r
r
r
r
r
rot (λa) = ∇ xr ∧ (λa) = λ (∇ xr ∧ a) + (∇ xr λ ∧ a)
r
r
The result of the algebraic operation ∇ xr ∧ (λa) is a vector, whose components are
given by:
[∇r
r
x
]
r
∧ (λa) i
=  ijk (λa k ) , j
=  ijk (λ , j a k + λa k , j )
=  ijk λa k , j  ijk λ , j a k
r
= λ (∇ xr ∧ a) i  ijk (∇ xr λ ) j a k
r
r
= λ (∇ xr ∧ a) i  (∇ xr λ ∧ a) i
(1.488)
We can use the above equation to check that the relationship
r
r
r
r
r
r
rot (λa) = ∇ xr ∧ (λa) = λ (∇ xr ∧ a) + (∇ xr λ ∧ a) holds.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
115
r r
r r
r
r r
r r
r r
(1.489)
∇ xr ∧ (a ∧ b) = (∇ xr ⋅ b)a − (∇ xr ⋅ a)b + (∇ xr a) ⋅ b − (∇ xr b) ⋅ a
r r
r r
The components of the vector product (a ∧ b) are given by (a ∧ b) k =  kij a i b j , thus:
ƒ
[∇r
]
r r
∧ (a ∧ b) l =  lpk ( kij a i b j ) , p =  kij  lpk (a i , p b j + a i b j , p )
r
x
(1.490)
Regarding that  kij =  ijk and  ijk  lpk = δ il δ jp − δ ip δ jl , the above equation becomes:
[∇r
r
x
]
r r
∧ (a ∧ b) l =  kij  lpk (a i , p b j + a i b j , p ) = (δ il δ jp − δ ip δ jl )(a i , p b j + a i b j , p )
(1.491)
= δ il δ jp a i , p b j − δ ip δ jl a i , p b j + δ il δ jp a i b j , p − δ ip δ jl a i b j , p
= al , p b p − a p, p b l + al b p, p − a p b l, p
[
r
r
]
[
r r
]
[
r r
]
We can also verify that (∇ xr a) ⋅ b l = a l , p b p , (∇ xr ⋅ a)b l = a p , p b l , (∇ xr ⋅ b)a l = a l b p , p ,
[
]
r r
(∇ xr b) ⋅ a l = a p b l , p .
r
r
r
r
2r
ƒ
∇ xr ∧ (∇ xr ∧ a) = ∇ xr (∇ xr ⋅ a) − ∇ xr a
r
r
r
r
The components of (∇ xr ∧ a) are given by (∇ xr ∧ a) i =  ijk a k , j , thus:
123
(1.492)
ci
[∇r
r
x
]
r
r
∧ (∇ xr ∧ a) q =  qli c i ,l =  qli ( ijk a k , j ) ,l =  qli  ijk a k , jl
(1.493)
Once again considering that  qli  ijk =  qli  jki = δ qj δ lk − δ qk δ lj , the above equation becomes:
[∇r
r
x
]
r
r
∧ (∇ xr ∧ a) q =  qli  ijk a k , jl = (δ qj δ lk − δ qk δ lj )a k , jl = δ qj δ lk a k , jl − δ qk δ lj a k , jl
= a k ,kq − a q ,ll
[
(1.494)
]
where [∇ xr (∇ xr ⋅ a)]q = a k , kq and ∇ xr 2 a q = a q ,ll .
r
ƒ
r
∇ xr ⋅ (ψ∇ xr φ ) = ψ∇ xr φ + (∇ xr ψ ) ⋅ (∇ xr φ)
2
(1.495)
∇ xr ⋅ (φ∇ xr ψ ) = (φψ ,i ) ,i = φψ ,ii + φ ,i ψ ,i = φ∇ xr ψ + (∇ xr φ ) ⋅ (∇ xr ψ )
2
(1.496)
where φ and ψ are scalar fields. Other interesting equations derived from the above are:
∇ xr ⋅ (φ∇ xr ψ ) = φ∇ xr ψ + (∇ xr φ ) ⋅ (∇ xr ψ )
2
(1.497)
∇ xr ⋅ (ψ∇ xr φ ) = ψ∇ xr φ + (∇ xr ψ ) ⋅ (∇ xr φ)
2
After subtracting the above two identities we obtain:
∇ xr ⋅ (φ∇ xr ψ ) − ∇ xr ⋅ (ψ∇ xr φ) = φ∇ xr ψ − ψ∇ xr φ
2
2
⇒ ∇ xr ⋅ (φ∇ xr ψ − ψ∇ xr φ ) = φ∇ xr ψ − ψ∇ xr φ
2
1.8.5
2
(1.498)
The Conservative Field
r r
A vector field, b( x , t ) , is said to be conservative if there exists a differentiable scalar field,
r
φ ( x , t ) , so that:
r
b = ∇ xr φ
(1.499)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
116
r r
If the function φ satisfies the relation (1.499), then φ is a potential function of b( x , t ) .
r r
r
r
r
A necessary but insufficient condition for b( x , t ) to be conservative is that ∇ xr ∧ b = 0 . In
r
r
other words, given a conservative field, the curl ∇ xr ∧ b equals zero. However, if the curl
of a vector field equals zero, this does not necessarily mean that the field is conservative.
r
Problem 1.46: Let φ be a scalar field, and u be a vector field. a) Show that
⋅ (∇ xr
)
r
r
r
∧ v = 0 and ∇ xr ∧ (∇ xr φ ) = 0 .
r
r
r
r
r r
r r
r
r r
r
r
b) Show that ∇ xr ∧ ∇ xr ∧ v ∧ v = (∇ xr ⋅ v )(∇ xr ∧ v ) + ∇ xr (∇ xr ∧ v ) ⋅ v − (∇ xr v ) ⋅ (∇ xr ∧ v ) ;
r
r
r
r
r
r
r
c) Referring ω = ∇ xr ∧ v , show that ∇ xr ∧ (∇ xr 2 v ) = ∇ xr 2 (∇ xr ∧ v ) = ∇ xr 2 ω .
∇ xr
r
[(
)
]
[
]
Solution:
r
r
Regarding that: ∇ xr ∧ v =  ijk v k , j ê i
⋅ (∇ xr
r
)
r
∂
∂
∂
∧v =
 ijk v k , j eˆ i ⋅ eˆ l =  ijk
v k , j δ il =  ijk
v k , j =  ijk v k , ji
∂x l
∂x l
∂x i
r
The second derivative of v is symmetrical with ij , i.e. v k , ji = v k ,ij , while  ijk is
∇ xr
(
)
( )
( )
antisymmetric with ij , i.e.,  ijk = − jik , thus:
 ijk v k , ji =  ij1v1, ji +  ij 2 v 2, ji +  ij 3 v3, ji = 0
We can observe that  ij1v1, ji equals the double scalar product by using a symmetric and an
antisymmetric tensor, so  ij1v1, ji = 0 .
Likewise, we can show that:
r
r
∇ xr ∧ (∇ xr φ ) =  ijk φ , kj eˆ i = 0 i eˆ i = 0
r
r
r
b) Denoting by ω = ∇ xr ∧ v we obtain:
[(
) ]
r
r
r r
∇ xr ∧ ∇ xr ∧ v ∧ v
r
r r
= ∇ xr ∧ (ω ∧ v )
Observing the equation in (1.489), it holds that:
r
r r
r r
r r
r r
r r
∇ xr ∧ (ω ∧ v ) = (∇ xr ⋅ v ) ω − (∇ xr ⋅ ω)v + (∇ xr ω) ⋅ v − (∇ xr v ) ⋅ ω
r
r
r
Note that ∇ xr ⋅ ω = ∇ xr ⋅ (∇ xr ∧ v ) = 0 . Then, we can draw the conclusion that:
r
r r
r r
r r
r r
∇ xr ∧ (ω ∧ v ) = (∇ xr ⋅ v )ω + (∇ xr ω) ⋅ v − (∇ xr v ) ⋅ ω
r
r
r r
r
r r
r
r
= (∇ xr ⋅ v )(∇ xr ∧ v ) + ∇ xr (∇ xr ∧ v ) ⋅ v − (∇ xr v ) ⋅ (∇ xr ∧ v )
[
c) Observing the equation in (1.492) we obtain:
]
r
r
r
r
2r
∇ xr v = ∇ xr (∇ xr ⋅ v ) − ∇ xr ∧ (∇ xr ∧ v )
r
r
r
= ∇ xr (∇ xr ⋅ v ) − ∇ xr ∧ ω
Applying the curl to the above equation we obtain:
r
r
r
r
r
r
2r
∇ xr ∧ (∇ xr v ) = ∇ xr ∧ [∇ xr (∇ xr ⋅ v )] − ∇ xr ∧ (∇ xr ∧ ω)
14442r 444
3
=0
r
r
r
Referring once again to the equation in (1.492) to express the term ∇ xr ∧ (∇ xr ∧ ω) :
[
]
r
r
r
r
r
r
r
2r
2r
2r
∇ xr ∧ (∇ xr v ) = −∇ xr ∧ (∇ xr ∧ ω) = −∇ xr (∇ xr ⋅ ω) + ∇ xr ω = −∇ xr ∇ xr ⋅ (∇ xr ∧ v ) + ∇ xr ω
144244
3
=0
r
r
2
r
r
= ∇ x (∇ x ∧ v )
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
117
1.9 Theorems Involving Integrals
1.9.1
Integration by Parts
Integration by parts states that:
b
∫
b
b
u ( x)v ′( x)dx = u ( x)v( x)
−
a
a
where v ′( x) =
1.9.2
∫ v( x)u ′( x)dx
(1.500)
a
dv
, and the functions u (x) , v(x) are differentiable in a ≤ x ≤ b .
dx
The Divergence Theorem
Given a domain B with a volume V , and bounded by the surface S , (see Figure 1.37), the
divergence theorem, also called the Gauss’ theorem, applied to the vector field states that:
∫
r
∇ xr ⋅ v dV =
V
∫v
∫
r
v ⋅ nˆ dS =
S
i ,i
∫
r r
v ⋅ dS
S
∫
(1.501)
∫
dV = v i nˆ i dS = v i dS i
V
S
S
where n̂ is the outward unit normal to surface S .
S
n̂
x2
r
dS
dS
B
r
x
x1
x3
Figure 1.37.
Let T be a second-order tensor field defined in the domain B . The divergence theorem
applied to this field is defined as:
∫∇
V
r
x
⋅T
dV =
r
∫ T ⋅ nˆ dS = ∫ T ⋅ dS
S
S
∫T
ij , j
V
∫
∫
dV = Tij nˆ j dS = Tij dS j
S
(1.502)
S
By using the divergence theorem we can also demonstrate that:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
118
∫ (x
k
∫
∫
= δ ik x i nˆ j dS
= (δ ik xi ), j dV
), j dV
V
V
=
S
∫ [δ
]
+ δ ik xi , j dV
ik , j x i
∫
= x k nˆ j dS
V
(1.503)
S
∫
∫
= x k nˆ j dS
= x k , j dV
V
S
in which we have assumed that δ ik , j = 0 ikj . Additionally, by observing that x k , j = δ kj , we
can obtain:
δ kj ∫ dV = ∫ x k nˆ j dS
V
∫
Vδ kj = x k nˆ j dS
⇒
S
S
r
V 1 = x ⊗ nˆ dS
(1.504)
∫
S
Given a second-order tensor σ defined in the domain B , the following is valid:
∫ (x σ
i
jk
∫
∫
), k dV = ( xi σ jk ), k dV = xi σ jk nˆ k dS
V
V
=
S
∫ [x
i , k σ jk
∫ [δ
ik σ jk
]
∫
+ xi σ jk ,k dV = xi σ jk nˆ k dS
V
=
(1.505)
S
]
∫
+ xi σ jk ,k dV = xi σ jk nˆ k dS
V
S
Hence proving that:
∫x σ
i
jk , k
∫
V
∫
∫
dV = xi σ jk nˆ k dS − σ ji dV
S
V
r
r
x ⊗ ∇ xr ⋅ σ dV = x ⊗ (σ ⋅ nˆ ) dS − σ T dV
∫
V
(1.506)
∫
S
V
or
∫
r
x ⊗ ∇ xr ⋅ σ dV =
V
r
r
T
∫ ( x ⊗ σ ) ⋅ dS − ∫ σ
S
dV
(1.507)
V
Likewise, one can prove that:
∫ (∇
V
r
x
r
r
⋅ σ ) ⊗ x dV = ∫ (σ ⋅ nˆ ) ⊗ x
S
∫
dS − σ dV
(1.508)
V
Problem 1.47: Let Ω be a domain bounded by Γ as shown in Figure 1.38. Further
consider that m is a second-order tensor field and ω is a scalar field. Show that the
following relationship holds:
∫ [m : ∇
Ω
r
r
x (∇ x ω )
]dΩ = ∫ [(∇ xr ω ) ⋅ m] ⋅ nˆ dΓ − ∫ [(∇ xr ⋅ m) ⋅ ∇ xr ω ]dΩ
Γ
Ω
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
∫Ω [m
ij ω , ij
] dΩ = ∫ ( ω , m
i
119
∫Ω [m
ˆ dΓ −
ij )n j
Γ
ij , j ω , i
] dΩ
n̂
Ω
x2
x1
Γ
Figure 1.38
Solution: We could directly apply the definition of integration by parts to demonstrate the
above relationship. But, here we will start with the definition of the divergence theorem.
r
That is, given a tensor field v , it is true that:
∫Ω ∇
r
x
r
⋅v
dΩ =
r
r
∫Γ v ⋅ nˆ dΓ → Ω∫ v
indicial
j, j
∫
dΩ = v j nˆ j dΓ
Γ
r
Observing that the tensor v is the result of the algebraic operation v = ∇ xr ω ⋅ m and the
equivalent in indicial notation to v j = ω , i m ij , and by substituting it in the above equation
we obtain:
∫v
Ω
j, j
∫
Γ
dΩ = v j nˆ j dΓ
⇒
⇒
⇒
∫ [ω,
Ω
i
∫Ω [ω,
ij
∫ [ω,
ij
m ij
]
,j
∫
dΩ = ω , i m ij nˆ j dΓ
Γ
]
∫
m ij + ω , i m ij , j dΩ = ω , i m ij nˆ j dΓ
]
Γ
∫
m ij dΩ = ω , i m ij nˆ j dΓ −
Ω
Γ
∫ [ω,
i
]
m ij , j dΩ
Ω
The above equation in tensorial notation becomes:
∫ [m : ∇
Ω
r
r
x (∇ x ω )
]dΩ = ∫ [(∇ xr ω ) ⋅ m] ⋅ nˆ dΓ − ∫ [∇ xr ω ⋅ (∇ xr ⋅ m)]dΩ
Γ
Ω
NOTE: Consider now the domain defined by the volume V , which is bounded by the
r
surface S with the outward unit normal to the surface n̂ . If N is a vector field and T is a
scalar field, it is also true that:
∫ N T,
i
V
ij
∫
∫
dV = N i T , i nˆ j dS − N i , j T , i dV
S
V
r
r
r
⇒ N ⋅ ∇ xr (∇ xr T )dV = (∇ xr T ⋅ N ) ⊗ nˆ dS − ∇ xr T ⋅ ∇ xr N dV
∫
V
∫
S
∫
V
where we have directly applied the definition of integration by parts.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
120
1.9.3
Independence of Path
A curve which connects two points A and B is denoted by the path from A to B , (see
Figure 1.39). We can then establish the condition by which a line integral is independent of
path, (see Figure 1.39).
C1
B
r
dr
r
b
If
∫
r r
r r
b ⋅ dr = b ⋅ dr
C1
A
x3
∫
C2
r
b - Conservative field
C2
x2
x1
Figure 1.39: Path independence.
r r
Let b( x ) be a continuous vector fields, then the integral
∫
r r
b ⋅ dr is independent of the path if
C
r
r
and only if b is a conservative field. This means that there is a scalar field φ so that b = ∇ xr φ .
Regarding the above, we can draw the conclusion that:
B
∫
r r B
r
b ⋅ dr = ∇ xr φ ⋅ dr
∫
A
A
B
B
 ∂φ
r
∂φ ˆ  r
∂φ ˆ
(b1 eˆ 1 + b 2 eˆ 2 + b 3 eˆ 3 ) ⋅ dr = 
eˆ 1 +
e2 +
e 3  ⋅ dr
∂x3
∂x 2
∂x1

A
A
∫
(1.509)
∫
Thus
b1 =
∂φ
∂x1
; b2 =
∂φ
∂x 2
; b3 =
∂φ
∂x3
(1.510)
r
As the field is conservative, the curl of b is the zero vector:
r r
r
∇ xr ∧ b = 0
⇒
eˆ 1
∂
∂x1
b1
eˆ 2
∂
∂x 2
b2
eˆ 3
∂
= 0i
∂x3
b3
(1.511)
We can therefore conclude that:
 ∂b 3 ∂b 2
−
=0

 ∂x 2 ∂x3
 ∂b1 ∂b 3
−
=0

 ∂x3 ∂x1
 ∂b 2 ∂b1
−
=0

 ∂x1 ∂x 2
 ∂b 3 ∂b 2
=

 ∂x 2 ∂x 3
 ∂b
∂b
⇒ 1 = 3
 ∂x3 ∂x1
 ∂b 2 ∂b1
=

 ∂x1 ∂x 2
(1.512)
Therefore, if the above condition is not satisfied, the field is not conservative.
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
1.9.4
121
The Kelvin-Stokes’ Theorem
r r
Let S be a regular surface, (see Figure 1.40), and F( x , t ) be a vector field. According to the
Kelvin-Stokes’ Theorem:
r r
r
r
r
r
r
r ∧ F ) ⋅ dS = (∇ r ∧ F ) ⋅ n
ˆ dS
F
⋅
d
Γ
=
(
∇
x
x
(1.513)
∫
∫
∫
Γ
Ω
Ω
If p̂ denotes the unit vector tangent to the boundary Γ , the Stokes’ theorem becomes:
r
r
∫Γ F ⋅ pˆ dΓ = Ω∫ (∇
r
x
r
r
r
r
∧ F) ⋅ dS = (∇ xr ∧ F ) ⋅ nˆ dS
∫
(1.514)
Ω
r
With reference to the vector representation in the Cartesian basis: F = F1eˆ 1 + F2 eˆ 2 + F3 eˆ 3 ,
r
r
r
dS = dS1eˆ 1 + dS 2 eˆ 2 + dS 3 eˆ 3 , dΓ = dx1 eˆ 1 + dx 2 eˆ 2 + dx3 eˆ 3 , the components of the curl of F
are given by:
eˆ 1
r
r
∂
∇ xr ∧ F i =
∂x1
F1
(
)
eˆ 2
∂
∂x 2
F2
eˆ 3
 ∂F
∂F
∂
=  3 − 2
∂x3  ∂x 2 ∂x3
F3

 ∂F
∂F
eˆ 1 +  1 − 3

 ∂x3 ∂x1

 ∂F
∂F
eˆ 2 +  2 − 1
 ∂x1 ∂x 2


eˆ 3

(1.515)

dS 3

(1.516)
Next, the Stokes’ theorem expressed in terms of components becomes:
∫ F dx
Γ
1
1
 ∂F
∂F
+ F2 dx 2 + F3 dx3 =  3 − 2
∂x
∂x3
Ω 2
∫
 ∂F

∂F
 dS1 +  1 − 3
 ∂x3 ∂x1


 ∂F
∂F
dS 2 +  2 − 1
 ∂x1 ∂x 2

n̂
x3
S
Ω
p̂
Γ
x2
x1
Figure 1.40: Stokes’ theorem.
In the special case when the surface S coincides with the plane Ω , (see Figure 1.41), the
equation (1.516) remains valid. Then, if the domain Ω coincides with the plane x1 − x 2 ,
the equation (1.513) becomes:
r r
r
r
r ∧ F) ⋅ e
ˆ 3 dS
F
(
⋅
d
Γ
=
∇
(1.517)
∫
∫ x
Γ
Ω
which is known as the Stokes’ theorem in the plane or Green’s theorem, which is expressed in
terms of components as:
∫ F dx
Γ
1
1
 ∂F
∂F
+ F2 dx 2 =  2 − 1
∂x
∂x 2
Ω 1
∫

dS 3

(1.518)
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
NOTES ON CONTINUUM MECHANICS
122
n̂
x3
Γ
Ω
x2
x1
Figure 1.41.
r
dS = dSê 3
x3
x2
Γ
ê3
Ω
x1
Figure 1.42: Green’s theorem.
1.9.5
Green’s Identities
r
Let F be a vector field, and by applying the divergence theorem we obtain:
∫
r
∇ xr ⋅ F dV =
V
∫
r
F ⋅ nˆ dS
(1.519)
S
With reference to the equations (1.496) and (1.498), i.e.:
∇ xr ⋅ (φ∇ xr ψ ) = φ∇ xr ψ + (∇ xr φ ) ⋅ (∇ xr ψ )
(1.520)
∇ xr ⋅ (φ∇ xr ψ − ψ∇ xr φ) = φ∇ xr ψ − ψ∇ xr φ
(1.521)
2
2
2
r
and also regarding that F = φ∇ xr ψ , and by substituting (1.520) into (1.519) we obtain:
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
1 TENSORS
∫ φ∇
r
x
2
V
123
ψ + (∇ xr φ) ⋅ (∇ xr ψ) dV = ∫ φ∇ xr ψ ⋅ nˆ dS
S
⇒ (∇ xr φ) ⋅ (∇ xr ψ ) dV =
∫
∫ φ∇
V
r
x
S
(1.522)
ψ ⋅ nˆ dS − ∫ φ∇ xr ψ dV
2
V
which is known as Green’s first identity.
Now, if we substituting (1.521) into (1.519) we obtain:
∫ φ∇
r
x
2
V
ψ − ψ∇ xr 2 φ dV = ∫ (φ∇ xr ψ − ψ∇ xr φ) ⋅ nˆ dS
(1.523)
S
which is known as Green’s second identity.
r
r
r
r
Problem 1.48: Let b be a vector field, which is defined as b = ∇ xr ∧ v . Show that:
∫ λb nˆ
i
i
∫
d S = λ, i b i dV
S
r
V
where λ = λ( x ) .
r
r
r
Solution: The Cartesian components of b = ∇ xr ∧ v are b i =  ijk v k , j and by substituting
them in the above surface integral we obtain:
∫ λb nˆ
i
i
∫
dS = λ ijk v k , j nˆ i dS
S
S
Applying the divergence theorem we obtain:
∫ λb nˆ
i
S
i
∫
dS = λ ijk v k , j nˆ i dS
S
∫
= ( ijk λv k , j ), i dV
V
∫
= ( ijk λ, i v k , j +  ijk λv k , ji ) dV
V
∫
∫
= (λ, i  ijk v k , j + λ  ijk v k , ji ) dV = λ, i b i dV
1
424
3
1
424
3
V
bi
0
V
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves
124
NOTES ON CONTINUUM MECHANICS
Notes on Continuum Mechanics(Springer/CIMNE) - (Chapter 01) - By: Eduardo W.V. Chaves