### Chapter 1

```Common Variable Types in Elasticity
Elasticity theory is a mathematical model of material deformation. Using principles of
continuum mechanics, it is formulated in terms of many different types of field
variables specified at spatial points in the body under study. Some examples include:
Scalars - Single magnitude
mass density , temperature T, modulus of elasticity E, . . .
Vectors – Three components in three dimensions
displacement vector u  u e 1  v e 2  w e 3 , e1, e2, e3 are unit basis vectors
Matrices – Nine components in three dimensions
stress matrix
x

[  ]    yx

 zx
 xy
y
 zy
 xz 

 yz 
 z 
Other – Variables with more than nine components
Index/Tensor Notation
With the wide variety of variables, elasticity formulation makes use of a tensor
formalism using index notation. This enables efficient representation of all
variables and governing equations using a single standardized method.
Index notation is a shorthand scheme whereby
a whole set of numbers or components can be
represented by a single symbol with subscripts
a1 
 a 11
 

a i  a 2 , a ij  a 21
 

 a 3 
 a 31
a 12
a 22
a 32
a 13 

a 23

a 33 
In general a symbol aij…k with N distinct indices represents 3N distinct numbers
Addition, subtraction, multiplication and equality of index symbols are defined in
the normal fashion; e.g.
 a 1  b1 
 a 11  b11



a i  b i  a 2  b 2 , a ij  b ij  a 21  b 21



 a 3  b 3 
 a 31  b 31
  a1 
  a 11



 a i   a 2 ,  a ij   a 21



  a 3 
  a 31
 a 12
 a 22
 a 32
 a 13 

 a 23

 a 33 
a 12  b12
a 22  b 22
a 32  b 32
a 13  b13 

a 23  b 23

a 33  b 33 
 a 1 b1

a i b j  a 2 b1

 a 3 b1
a 1b2
a 2 b2
a 3b2
a 1b3 

a 2 b3

a 3 b 3 
Notation Rules and Definitions
Summation Convention - if a subscript appears twice in the same term,
then summation over that subscript from one to three is implied; for
example
3
a ii 
a
 a 11  a 22  a 33
ii
i 1
3
a ij b j 
a
ij
b j  a i 1 b1  a i 2 b 2  a i 3 b 3
i 1
A symbol aij…m…n…k is said to be symmetric with respect to index pair mn if
a ij ... m ... n ... k  a ij ... n ... m ... k
A symbol aij…m…n…k is said to be antisymmetric with respect to index pair mn if
a ij ... m ... n ... k   a ij ... n ... m ... k
Useful Identity
a ( ij ) 
1
2
( a ij  a ji )
a ij 
1
2
. . . symmetric
( a ij  a ji ) 
a [ ij ] 
1
2
1
2
( a ij  a ji )  a ( ij )  a [ ij ]
( a ij  a ji )
. . . antisymmetric
Example 1-1: Index Notation Examples
The matrix aij and vector bi are specified by
a ij
1

 0

 2
2
4
1
0
2

 
3 , bi  4

 
 0 
2 
Determine the following quantities: a ii , a ij a ij , a ij a jk , a ij b j , a ij b i b j , b i b i , b i b j , a ( ij ) , a [ ij ]
Indicate whether they are a scalar, vector or matrix.
Following the standard definitions given in section 1.2,
a ii  a 11  a 22  a 33  7 (scalar)
a ij a ij  a 11 a 11  a 12 a 12  a 13 a 13  a 21 a 21  a 22 a 22  a 23 a 23  a 31 a 31  a 32 a 32  a 33 a 33
 1  4  0  0  16  9  4  1  4  39 (scalar)
a ij a jk  a i1 a 1 k  a i 2 a 2 k  a i 3 a 3 k
1

 6

 6
10
19
10
6

18 (matrix)

7 
10 
 
a ij b j  a i1 b1  a i 2 b 2  a i 3 b 3  16 (vector)
 
 8 
a ( ij )
a ij b i b j  a 11 b1 b1  a 12 b1 b 2  a 13 b1 b 3  a 21 b 2 b1    84 (scalar)
b i b i  b1 b1  b 2 b 2  b 3 b 3  4  16  0  20 (scalar)
4

bi b j  8

 0
8
16
0
0

0 (matrix)

0 
a [ ij ]
1
1
 a ij  a ji  
0
2
2
 2
1
1
1
 a ij  a ji  
0
2
2
 2
1
2
4
1
2
4
1
0
1
 1
3 
2
 2
 0
2 
0
0
1
 1
3 
2
 2
 0
2 
0
4
3
4
3
2  1
 
1  1
 
2  1
2  0
 
1  1
 
2   1
1

2 (matrix)

2 
1
4
2
1
0
1
 1

1 (matrix)

0 
Special Indexed Symbols
Kronecker Delta
1
1 , if i  j ( no sum ) 
 ij  
 0

0
,
if
i

j

 0
 ij   ji
0
1
0
0

0

1 
 ii  3 ,  i i  1
Properties:
 ij a j  a i ,  ij a i  a j
 ij a jk  a ik ,  jk a ik  a ij
 ij a ij  a ii ,  ij  ij  3
Alternating or Permutation Symbol
 ijk
  1 , if ijk is an even permutatio n of 1, 2 , 3

   1 , if ijk is an odd permutatio n of 1, 2 , 3
 0 , otherwise

123 = 231 = 312 = 1, 321 = 132 = 213 = -1, 112 = 131 = 222 = . . . = 0
Useful in evaluating determinants
and vector cross-products
a 11
a 12
a 13
det[ a ij ]  | a ij |  a 21
a 22
a 23   ijk a 1 i a 2 j a 3 k   ijk a i 1 a j 2 a k 3
a 31
a 32
a 33
Coordinate Transformations
x3
x 3
v
e3
e3
x 2
e2
e2
e1
x2
e1
x1
We wish to express elasticity variables in
different coordinate systems. This requires
development of transformation rules for
scalar, vector, matrix and higher order
variables – a concept connected with basic
definitions of tensor variables. The two
Cartesian frames (x1,x2,x3) and ( x 1 , x 2 , x 3 )
differ only by orientation
x 1
Using Rotation Matrix
Q ij  cos( x i , x j )
e 1  Q 11 e 1  Q 12 e 2  Q 13 e 3
e 2  Q 21 e 1  Q 22 e 2  Q 23 e 3
e 3  Q 31 e 1  Q 32 e 2  Q 33 e 3
v  v1 e1  v 2 e 2  v 3 e 3  v i e i
 v 1e 1  v 2 e 2  v 3 e 3  v ie i
e i  Q ij e j
e i  Q ji e j
v i  Q ij v j
v i  Q ji v j
transformation laws
for Cartesian vector
components
Cartesian Tensors
General Transformation Laws
Scalars, vectors, matrices, and higher order quantities can be represented by an
index notational scheme, and thus all quantities may then be referred to as
tensors of different orders. The transformation properties of a vector can be
used to establish the general transformation properties of these tensors.
Restricting the transformations to those only between Cartesian coordinate
systems, the general set of transformation relations for various orders are:
a   a , zero order (scalar)
a i  Q ip a p , first order (vector)
a ij  Q ip Q jq a pq , second order (matrix)
  Q ip Q jq Q kr a pqr , third order
a ijk
  Q ip Q jq Q kr Q ls a pqrs , fourth order
a ijkl

 ... m  Q ip Q jq Q kr    Q mt a pqr ... t
a ijk
general order
Example 1-2 Transformation Examples
The components of a first and second order tensor in a particular coordinate frame are given by
1 
1
 

a i  4 , a ij  0
 

 2 
 3
0
2
2
3

2

4 
x3
x 3
Determine the components of each tensor in a new coordinate system
Choose a counterclockwise rotation when viewing down the negative
x3-axis, see Figure 1-2.
x 2
The original and primed coordinate systems are shown in Figure 1-2.
The solution starts by determining the rotation matrix for this case
Q ij
 cos 60 

 cos 150 

 cos 90 
cos 30 
cos 60 
cos 90 
cos 90    1 / 2
 
cos 90     3 / 2

cos 0   
0

3/2
1/2
0
0

0
1

60
x2
o
x 1
x1
The transformation for the vector quantity follows from equation (1.5.1)2
 1/2

a i  Q ij a j    3 / 2

0

3/2
1/2
0
0

0
1

 1  1 / 2  2 3 

  
4  2 3/2
 


2
 2  

and the second order tensor (matrix) transforms according to (1.5.1)3
a ij  Q ip Q jq a pq
 1/ 2

  3 / 2

0

3/2
1/ 2
0
0  1

0 0

1   3

0
2
2
3  1 / 2

2  3 / 2

4  
0

3/2
1/ 2
0
0

0
1

T
 7/4

 
3/4
3 / 2  3

3/4
5/4
1 3 3 / 2
3

1  3 3 / 2

4

3/2
Principal Values and Directions for
Symmetric Second Order Tensors
The direction determined by unit vector n is said to be a principal direction or
eigenvector of the symmetric second order tensor aij if there exists a parameter
 (principal value or eigenvalue) such that
( a ij   ij ) n j  0
a ij n j   n i
which is a homogeneous system of three linear algebraic equations in the
unknowns n1, n2, n3. The system possesses nontrivial solution if and only if
determinant of coefficient matrix vanishes
det[ a ij   ij ]     I a   II a   III
3
2
a
 0
scalars Ia, IIa and IIIa are called the fundamental invariants of the tensor aij
I a  a ii  a 11  a 22  a 33
II a 
III
a
1
2
( a ii a jj  a ij a ij ) 
 det[ a ij ]
a 11
a 12
a 21
a 22

a 22
a 23
a 32
a 33

a 11
a 13
a 31
a 33
Principal Axes of Second Order Tensors
It is always possible to identify a right-handed Cartesian coordinate system such
that each axes lie along principal directions of any given symmetric second order
tensor. Such axes are called the principal axes of the tensor, and the basis
vectors are the principal directions {n(1), n(2) , n(3)}
x3
x3
 a 11

a ij  a 21

 a 31
a 12
a 22
a 32
a 13 

a 23

a 33 
 1

a ij  0

 0
n(3)
0
2
0
0 

0

 3 
n(2)
x2
n(1)
x1
x1
Original Given Axes
Principal Axes
x2
Example 1-3 Principal Value Problem
Determine the invariants, and principal values and directions of
2

a ij  0

 0
0
3
4
0 

4

 3 
First determine the principal invariants
I a  a ii  2  3  3  2 , II a 
III
a
2
0
0
 0
3
4
0
4
3
2
0
0
3

3
4
4
3

2
0
0
3
 6  25  6   25
 2 (  9  16 )   50
The characteristic equation then becomes
det[ a ij   ij ]     2   25   50  0  (   2 )(   25 )  0
3
2
2
 1  5 ,  2  2 ,  3  5
Thus for this case all principal values are distinct
For the 1 = 5 root, equation (1.6.1) gives the system
 3n1
(1 )
 0
 2n2  4n3
(1 )
(1 )
4 n 2  8n 3
(1 )
(1 )
 0
 0
which gives a normalized solution n (1 )  
1
5
(2e2  e3 )
In similar fashion the other two principal directions are found to be n ( 2 )   e 1 n ( 3 )  
It is easily verified that these directions are mutually orthogonal.
Note for this case, the transformation matrix Qij defined by (1.4.1) becomes
Q ij
0

 1
0

2/
5
0
1/
5
5 

0
 
 2 / 5

1/
5

a ij  0

 0
0
2
0
0 

0

 5 
1
5
(e2  2e3 )
Vector, Matrix and Tensor Algebra
Scalar or Dot Product
Vector or Cross Product
a  b  a 1 b1  a 2 b 2  a 3 b3  a i bi
e1
e2
e3
a  b  a1
a2
a 3   ijk a j b k e i
b1
b2
b3
Common Matrix Products
Aa  [ A ]{ a }  Aij a j  a j Aij
a A  { a } [ A ]  a i Aij  Aij a i
T
T
AB  [ A ][ B ]  Aij B jk
AB
T
 Aij B kj
Transforma
tion Law
a ij  Q ip Q jq a pq 
A B  A ji B jk
T
a   QaQ
tr ( AB )  Aij B ji
tr ( AB )  tr ( A B )  Aij B ij
T
Second Order
T
T
Calculus of Cartesian Tensors
Field concept for tensor components
a  a ( x1 , x 2 , x 3 )  a ( x i )  a ( x )
a i  a i ( x1 , x 2 , x 3 )  a i ( x i )  a i ( x )
a ij  a ij ( x 1 , x 2 , x 3 )  a ij ( x i )  a ij ( x )

Comma notation for partial differentiation
a ,i 

xi
a , a i, j 

x j
a i , a ij , k 

x k
a ij , 
If differentiation index is distinct, order of the tensor will be increased by one;
e.g. derivative operation on a vector produces a second order tensor or matrix
a i, j
 a1

x
 1
a
  2
  x1
 a
 3
  x 1
a1
x 2
a 2
x 2
a 3
x 2
a1 

x3

a 2 
x3 
a 3 

 x 3 
Vector Differential Operations
df
Directional Derivative of Scalar Field
n  unit normal vector in direction

ds
of s 
 f dx
 x ds
dx
ds
  vector differenti al operator
 e1
of scalar function

x
e1 
 e2
f  e1
 f dy

dy
ds
 y ds
e2 

y
f
x
 e3
 e2
Common Differential Operations
  ie i
 u  ui, j e i e j
Laplacian
         ,ii
Divergence
of a Scalar
of a Vector
2
  u  u i ,i
Curl of a Vector
  u   ijk u k , j e i
Laplacian
 u  u i , kk e i
of a Vector
2

dz
ds
 f dz
 z ds
e3

z
f
y
 e3
f
z
 n f
Example 1-4: Scalar/Vector Field Example
Scalar and vector field functions are given by
 x  y
2
2
u  2 x e 1  3 yz e 2  xy e 3
Calculate the following expressions, , 2,  ∙ u, u,   u.
Using the basic relations
  2 x e 1  2 y e 2
2  2  2  0 - (satisfies Laplace equation)
 ∙ u  2  3z  0  2  3z
u  u i , j
2

 0

 y
e1
0
3z
x
e2
  u   / x  / y
2x
3 yz
0 

3y

0 
e3
 / z  ( x  3 y )e 1  ye 2
xy
Contours =constant and vector distributions of 
Note vector field  is orthogonal to -contours,
in general
for all scalar fields
Vector Distribution
10
8
6
4
y
2
0
x
-2
-4
-6
-8
-10
-10
-5
0
5
10
Vector/Tensor Integral Calculus
Divergence Theorem
Stokes Theorem

 u  dr
C

u  n dS 
S



  u dV
V

(   u )  n dS
S
C
S
a ij ... k n k dS 
a ij ... k dx t 

S

V
a ij ... k , k dV
 rst a ij ... k , s n r dS
Green’s Theorem in the Plane
 g f 
 S   x   y dxdy 



Zero-Value Theorem
( fdx  gdy )
C

V

g
S
x
dxdy 

C
gn x ds ,
f ij ... k dV  0  f ij ... k  0  V

f
S
y
dxdy 

C
fn y ds
Orthogonal Curvilinear Coordinate Systems
x3
x3
z
eˆ z
eˆ r
eˆ 
e1

r
Cylindrical Coordinate System (r,,z)
x 1  r cos  , x 2  sin  , x 3  z
x 1  x 2 ,   tan
2
2
1
x2
x1
, z  x3
eˆ 
e3
x2
e2
x1
r 

eˆ r
e3
eˆ 
e1
R
x2
e2

x1
Spherical Coordinate System (R,,)
x1  R cos  sin  , x 2  R sin  sin  , x 3  R cos 
R 
  cos
x1  x 2  x 3
2
2
2
x3
1
x1  x 2  x 3
2
  tan
1
2
x2
x1
,
2
General Curvilinear Coordinate Systems
Common Differential Forms
x3
3

eˆ 3
 f  eˆ 1
eˆ 1
e3
m
1
h1 h 2 h 3
x2
e2
1
 
2
h1 h 2 h 3
u 
x1

h1  
  (x , x , x ) , x
m
1
2
3
m
 x ( ,  ,  )
m
1
2
( ds )  ( h1 d  )  ( h 2 d  )  ( h3 d  )
2
1
2
2
2
j
3
3
2
u 

i
1
j

h2 
2
1 f
 eˆ 2
h2 
2
 eˆ 3
 eˆ 3
h3 
h3 




  h1 h 2 h 3  

i 
2
i
 ( hi ) 




 
k
 ijk

h j hk 
j
 

 
  j

 eˆ
i
3

 eˆ
i
1
i

hi 
i
1 f
i
hi 
i
( u  k  h k )eˆ i
 eˆ j
eˆ i   u  j 

eˆ j  u  j 
i
i

hi  


eˆ 
2
 u    i
i
 i hi 
3
1 f
  h1 h 2 h 3

u i
i 
 hi
i
i

1
 

i
1
 eˆ 2
1
1 f
 u 
1
e1
h1  
2
eˆ 2

1
  eˆ 1

k




 eˆ j  
eˆ k   u  j 

ˆ
e

u

j

j

k
k 
hk  
   
Example 1-5: Polar Coordinates
From relations (1.9.5) or simply using the geometry shown in Figure
eˆ r  cos  e 1  sin  e 2
 eˆ r
eˆ    sin  e 1  cos  e 2

 eˆ  ,
 eˆ 

  eˆ r ,
 eˆ r
r

 eˆ 
r
x2
 0
eˆ 
eˆ r
The basic vector differential operations then follow to be
  eˆ r
   eˆ r
 u 

r
 eˆ 

r
1 
r r
1 
 eˆ 
2

1 
r 
( ru r ) 
u r
r
eˆ r eˆ r 
u 
( ds )  ( dr )  ( rd  )
2
1 u 
r 
1    
1  
r
 2
2
r r  r  r 
r
eˆ r eˆ  
1  u r
1  u 


 u   eˆ  eˆ r  
 u r  eˆ  eˆ 

r  
r  


u 
u 
2 u 
2 u r
 2
 2
2
 u   ur  2
 2r  eˆ r    u   2
 2  eˆ 
r 
r 
r 
r 


where
x1
e1
1 u r 
1 
u 
( ru  ) 
 eˆ z
r  
 r r
u 
r
r 
2
 
e2
u  u r eˆ r  u  eˆ θ , eˆ z  eˆ r  eˆ θ
2
2
 h1  1 , h 2  r
```