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 ue1 ve 2 we 3 , e1, e2, e3 are unit basis vectors
Matrices – Nine components in three dimensions
stress matrix
x xy xz
[] yx y yz
zx zy z
Other – Variables with more than nine components
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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.
a1
a11 a12 a13
Index notation is a shorthand scheme whereby
a whole set of numbers or components can be a i a 2 , a ij a 21 a 22 a 23
represented by a single symbol with subscripts
a 3
a 31 a 32 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.
a1 b1
a11 b11 a12 b12
ai bi a2 b2 , aij bij a21 b21 a22 b22
a3 b3
a31 b31 a32 b32
a1
a11 a12
ai a2 , aij a21 a22
a3
a31 a32
Elasticity
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a13
a23
a33
a13 b13
a23 b23
a33 b33
a1b1 a1b2
ai b j a2 b1 a2 b2
a3b1 a3b2
a1b3
a2 b3
a3b3
Notation Rules and Definitions
Summation Convention - if a subscript appears twice in the same term,
summation over that subscript from one to three is implied; for example
3
aii aii a11 a22 a33
i 1
3
aij b j aij b j ai1b1 ai 2b2 ai 3b3
j 1
A symbol aij…m…n…k is said to be symmetric with respect to index pair mn if
aij... m... n... k aij... n... m... k
A symbol aij…m…n…k is said to be antisymmetric with respect to index pair mn if
aij... m... n... k aij... n... m... k
1
2
1
2
Useful Identity aij (aij a ji ) (aij a ji ) a( ij) a[ij]
a ( ij )
Elasticity
1
( aij a ji ) . . . symmetric
2
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M.H. Sadd , University of Rhode Island
a[ij ]
1
( aij a ji ) . . . antisymmetric
2
Example 1-1: Index Notation Examples
The matrix aij and vector bi are specified by
1 2 0
2
aij 0 4 3 , bi 4
2 1 2
0
Determine the following quantities: aii , aij aij , aij a jk , aij b j , aij bi b j , bi bi , bi b j , a(ij ) , a[ij ]
Indicate whether they are a scalar, vector or matrix.
Following the standard definitions given in section 1.2,
aii a11 a 22 a33 7 (scalar)
aij aij a11a11 a12 a12 a13 a13 a 21a 21 a 22 a 22 a 23 a 23 a31a31 a32 a32 a33 a33
1 4 0 0 16 9 4 1 4 39 (scalar)
1 10 6
aij a jk ai1 a1k ai 2 a 2 k ai 3 a3k 6 19 18 (matrix)
6 10 7
10
aij b j ai1b1 ai 2 b2 ai 3b3 16 (vector)
8
a (ij )
aij bi b j a11b1b1 a12 b1b2 a13b1b3 a 21b2 b1 84 (scalar)
bi bi b1b1 b2 b2 b3b3 4 16 0 20 (scalar)
4 8 0
bi b j 8 16 0 (matrix)
0 0 0
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a[ij ]
1
1
1
aij a ji 0
2
2
2
1
1
1
aij a ji 0
2
2
2
2 0
1
1
4 3 2
2
0
1 2
2 0
1
1
4 3 2
2
0
1 2
0 2 1 1 1
4 1 1 4 2
3 2 1 2 2
0 2 0 1
4 1 1 0
3 2 1 1
(matrix)
1
1 (matrix)
0
Special Index Symbols
Kronecker Delta
1 0 0
1 , if i j ( no sum )
ij
0 1 0
0 , if i j
0 0 1
ij ji
ii 3 , i i 1
Properties: ij a j ai , ij ai a j
ij a jk aik , jk aik aij
ij aij aii , ij ij 3
Alternating or Permutation Symbol ijk
1 , if ijk is an even permutation of 1,2,3
1 , if ijk is an odd permutation of 1,2,3
0 , otherwise
123 = 231 = 312 = 1, 321 = 132 = 213 = -1, 112 = 131 = 222 = . . . = 0
a11 a12 a13
Useful in evaluating determinants
det[ aij ] | aij | a21 a22 a23 ijk a1i a2 j a3k ijk ai1a j 2 ak 3
and vector cross-products
a31 a32
Elasticity
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a33
Coordinate Transformations
x3
x 3
v
e3
e3
x 2
e2
e2
e1
x2
e1
x1
To express elasticity variables in different
coordinate systems 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 Qij cos( xi, x j )
e1 Q11e1 Q12 e 2 Q13e3
e 2 Q21e1 Q22 e 2 Q23e3
e3 Q31e1 Q32 e 2 Q33e3
v v1e1 v2 e 2 v3e3 vi ei
v1e1 v2 e 2 v3 e3 viei
Elasticity
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ei Qij e j
ei Q ji e j
transformation laws
for Cartesian vector
vi Q ji v j
components
vi Qij v j
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)
ai Qip a p , first order (vector)
aij QipQ jq a pq , second order (matrix)
QipQ jqQkr a pqr , third order
aijk
QipQ jqQkr Qls a pqrs , fourth order
aijkl
... m QipQ jqQkr Qmt a pqr... t general order
aijk
Elasticity
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Example 1-2 Transformation Examples
The components of a first and second order tensor in a particular coordinate frame are given by
1
1 0 3
ai 4 , aij 0 2 2
2
3 2 4
x3
x 3
Determine the components of each tensor in a new coordinate system
found through a rotation of 60o (/6 radians) about the x3-axis.
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
cos 60 cos 30 cos 90 1 / 2
Qij cos 150 cos 60 cos 90 3 / 2
cos 90 cos 90 cos 0 0
3 / 2 0
1 / 2 0
0
1
60
x2
o
x 1
x1
The transformation for the vector quantity follows from equation (1.5.1)2
1/ 2
3 / 2 0 1 1 / 2 2 3
ai Qij a j 3 / 2 1 / 2 0 4 2 3 / 2
0
2
0
1
2
and the second order tensor (matrix) transforms according to (1.5.1)3
aij Qip Q jq a pq
Elasticity
1/ 2
3 / 2
0
3 / 2 0 1 0 3 1 / 2
1 / 2 0 0 2 2 3 / 2
0
1 3 2 4 0
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T
7/4
3 / 2 0
3/4
3/ 2 3
1 / 2 0 3 / 4
5/ 4
1 3 3 / 2
3 / 2 3 1 3 3 / 2
0
1
4
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
aij n j ni
( aij ij )n j 0
Relation 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[ aij ij ] 3 I a 2 II a III a 0
scalars Ia, IIa and IIIa are called the fundamental invariants of the tensor aij
I a aii a11 a 22 a33
II a
a
a12 a 22
1
( aii a jj aij aij ) 11
a 21 a 22 a32
2
III a det[ aij ]
Elasticity
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M.H. Sadd , University of Rhode Island
a 23
a33
a11
a13
a31 a33
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
x3
a11
aij a21
a31
a12
a22
a32
a13
a23
a33
1 0
aij 0 2
0 0
n(3)
0
0
3
n(2)
x2
n(1)
x1
x1
Original Given Axes
Elasticity
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Principal Axes
x2
Example 1-3 Principal Value Problem
Determine the invariants, and principal values and directions of
2 0 0
aij 0 3 4
0 4 3
First determine the principal invariants
2 0
3 4
2 0
I a aii 2 3 3 2 , II a
6 25 6 25
0 3
4 3
0 3
2 0
III a 0 3
0
4 2(9 16) 50
0 4 3
The characteristic equation then becomes
det[aij ij ] 3 22 25 50 0 ( 2)(2 25) 0
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
2n 2(1) 4n3(1) 0
4n 2(1) 8n3(1) 0
1
( 2e 2 e 3 )
which gives a normalized solution n(1)
5
1
In similar fashion the other two principal directions are found to be n( 2 ) e1 n( 3)
( e 2 2e 3 )
5
It is easily verified that these directions are mutually orthogonal.
Note for this case, the transformation matrix Qij defined by (1.4.1) becomes
0 2 / 5 1 / 5
Qij 1
0
0
0 1 / 5 2 / 5
Elasticity
5 0 0
aij 0 2 0
0 0 5
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Vector, Matrix and Tensor Algebra
Scalar or Dot Product
a b a1b1 a2 b2 a3b3 ai bi
e1
Vector or Cross Product
e2
a b a1 a2
b1 b2
e3
a3 ijk a j bk ei
b3
Common Matrix Products
Aa [ A]{a} Aij a j a j Aij
aT A {a}T [ A] ai Aij Aij ai
AB [ A][ B] Aij B jk
ABT Aij Bkj
AT B A ji B jk
tr( AB) Aij B ji
tr( ABT ) tr( AT B) Aij Bij
Elasticity
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Second Order
Transformation Law
aij Qip Q jq a pq
a QaQ T
Calculus of Cartesian Tensors
Field concept for tensor components a a ( x1 , x2 , x3 ) a ( xi ) a( x )
ai ai ( x1 , x2 , x3 ) ai ( xi ) ai ( x )
aij aij ( x1 , x2 , x3 ) aij ( xi ) aij ( x )
Comma notation for partial differentiation a,i
a , ai , j
ai , aij ,k
aij ,
xi
x j
xk
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
ai , j
Elasticity
a1
x
1
a
2
x1
a
3
x1
Theory, Applications and Numerics
M.H. Sadd , University of Rhode Island
a1
x 2
a 2
x 2
a 3
x 2
a1
x3
a 2
x3
a 3
x3
Vector Differential Operations
Directional Derivative of Scalar Field
df f dx f dy f dz
n f
ds x ds y ds z ds
n unit normal vector in direction of s
vector differential operator e1
dx
dy
dz
e1 e 2 e 3
ds
ds
ds
e2
e3
x
y
z
f grad f gradient of scalar function f e1
f
f
f
e2
e3
x
y
z
Common Differential Operations
Gradient of a Scalar ,i ei
Gradient of a Vector
u u i , j e i e j
Laplacian of a Scalar 2 ,ii
Divergence of a Vector u ui ,i
Curl of a Vector
Laplacian of a Vector
Elasticity
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u ijk uk , j ei
2 u ui ,kk ei
Example 1-4: Scalar/Vector Field Example
Scalar and vector field functions are given by x 2 y 2 u 2 xe1 3 yze 2 xye 3
Calculate the following expressions, , 2, ∙ u, u, u.
Using the basic relations:
Contours =constant and vector distributions of
vector field is orthogonal to -contours (ture in general )
2 xe1 2 ye 2
2 2 2 0 - (satisfies Laplace equation)
∙ u 2 3z 0 2 3z
2 0 0
u ui , j 0 3 z 3 y
y x 0
e1
e2
e3
u / x / y / z ( x 3 y )e1 ye 2
2x
3 yz
xy
Gradient Vector Distribution
10
8
6
4
2
y
0
-2
x
-4
-6
-8
-10
Elasticity
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M.H. Sadd , University of Rhode Island
-10
-5
0
5
10
Vector/Tensor Integral Calculus
Divergence Theorem
Stokes Theorem
S
u n dS u dV
u dr
C
V
S
( u) n dS
C
S
aij... k nk dS aij... k ,k dV
V
aij... k dxt rst aij... k ,s nr dS
S
Green’s Theorem in the Plane
g f
S x y dxdy C ( fdx gdy)
Zero-Value Theorem
Elasticity
V
g
S x dxdy C gn x ds ,
f ij... k dV 0 f ij... k 0 V
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f
S y dxdy C fny ds
Orthogonal Curvilinear Coordinate Systems
x3
x3
z
ê z
ê R
ê
e1
êr
e3
x1
e3
x2
e2
r
Cylindrical Coordinate System (r,,z)
x1 r cos , x2 r sin , x3 z
r x12 x22 , tan 1
x2
, z x3
x1
e1
R
Theory, Applications and Numerics
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ê
e2
x2
x1
Spherical Coordinate System (R,,)
x1 R cos sin , x2 R sin sin , x3 R cos
R x12 x22 x32
cos 1
x3
x12 x22 x32
tan 1
Elasticity
ê
x2
,
x1
General Curvilinear Coordinate Systems
Common Differential Forms
x3
3
eˆ1
1
1
1
1
eˆ 2
eˆ3
eˆ i
1
2
3
h1
h2
h3
hi i
i
f eˆ1
1 f
1 f
1 f
1 f
eˆ 2
eˆ3
eˆ i
1
2
3
h1
h2
h3
hi i
i
2
ê3
ê2
ê1
e3
1
x2
e2
e1
2
1
h1h2 h3
i
( x , x , x ) , x x ( , , )
m
i
i
u
x1
m
1
2
3
m
m
1
2
3
(ds ) 2 (h1d1 ) 2 (h2 d 2 ) 2 (h3 d3 ) 2
h1h2 h3
u i
i
hi
1
h1h2 h3
u
u
i
j
h1h2 h3
i
2
i
(
h
)
i
ijk
(u k hk )eˆ i
h j hk j
j
k
eˆ i
hi
eˆ j
u j
eˆ j u j i
i
eˆ j
eˆ
eˆ u
k j eˆ j u j
2 u i
i
k
k
h
h
i
j
k
i
k
Elasticity
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Example 1-5: Polar Coordinates
From relations (1.9.5) or simply using the geometry shown in Figure
eˆ r cos e1 sin e 2
eˆ sin e1 cos e 2
x2
eˆ
eˆ
eˆ r
eˆ
eˆ , eˆ r , r 0
r
r
ê
êr
The basic vector differential operations then follow to be
1
eˆ
r
r
1
eˆ r
eˆ
r
r
1
1 u
u
(ru r )
r r
r
1 1 2
2
r
r r r r 2 2
eˆ r
1 u r
1
u
(ru )
eˆ z
r
r r
u
u
1 u
1 u
u r eˆ r eˆ r eˆ r eˆ r u eˆ eˆ r u r eˆ eˆ
r
r
r
r
u
u
2 u
2 u
2 u 2 u r 2 2r eˆ r 2 u 2 r 2 eˆ
r r
r r
where u ur eˆ r ueˆ θ , eˆ z eˆ r eˆ θ
Elasticity
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e2
r
e1
x1
(ds ) 2 (dr ) 2 (rd) 2 h1 1 , h2 r
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