A VERY IMPORTANT BUT VERY LESS SPOKEN TOPIC, THE TOPIC CAN BE STUDIED THROUGH , MECHANICS OF MATERIALS AND THEORY OF ELASTICITY , THE PRESENTATION SHOWS WHAT ARE TENSORS , RANK OF TENSOR, HOW TO DERIVE GENERALIZED HOOKS LAW , AND DERIVE FROM 81 CONSTANTS TO 2 ,DUE TO SYMMETRIES,

1. TENSORS &
GENERALIZED
HOOKS LAW
University of Tripoli / Faculty of Engineering
Department of Civil Engineering
Graduate Studies
PRESENTED BY: M.ABDUL AZEEM BAIG SUPERVISED BY :Prof. Lisaneldeen Galhoud
CE-601

2. ▸INTRODUCTION
2
Before starting our presentation
TENSORS
▸UNDERSTANDING THE BASICS OF TENSORS
▸GENERLIZED HOOKS
LAW
FROM BASCIS TO ADVANCED
▸REDUCTION OF CONSTANTS
HOW TO CONVERT 81 CONSTANTS INTO 1
▸APPLICATIONS OF HOOKS LAW
WHAT WE UNDERSTOOD
TABLE OF CONTENTS
▸1
▸2
▸3
4
▸5

3. 1.
TENSORS
Let’s start try to understand
3

4. “• Tensors arise when dealing with functions that take a
vector as input and produce a vector as output.
• For example, if a ball is thrown at the ground with a
certain velocity (which is a vector), then classical physics
principals can be use to come up with a formula for the
velocity vector after hitting the ground.
• In other words, there is presumably a function that takes
the initial velocity vector as input and produces the final
velocity vector as output
WHAT IS A TENSOR ?

5. WHAT IS TENSOR ?
TO UNDERSTAND TENSORS WE MUST UNDERSTAND:
• UNIT VECTORS
• CARTESIAN PLANE
▸TO EXPLAIN THIS IDEA, WE CAN USE AN ALTERNATIVE
TERMINOLOGY, RATHER THAN MATHEMATICAL APPROACH,WE WILL
USE :
5
▸ ARROW HEADS TO SHOW UNIT VECTORS IN 3D SPACE

6. “What is a vector?
A vector is a quantity , which has both magnitude and
direction, for e.g force,velocityand etc
Also , it will be alright to say that length of the vector
drawn is equal to the magnitude and orientation of
arrow gives its direction,

7. 7
TO UNDERSTAND VECTOR RESOLUTION
WE MUST UNDERSTAND UNIT VECTORS
VECTOR
RESOLUTION

8. VECTORS CAN ALSO REPRESENT AREA !
▸VECTORS CAN ONLY REPRESENT A SURFACE ONLY AND
ONLY IF IT IS PROJECTING PERPENDICULAR TO IT
▸IN THE GIVEN FIG THE YZ ,XZAND XY PLANES ARE
REPRESENTED BY GIVEN VECTORS
8

9. WHAT IS A BASIS VECTOR ?
Also known as unit vector,Unit means one ,
the unit vector defines the one of a given unit
in some direction, the unit vector is usally
written as
In other words,unit vector has a magnitude one
9

10. What ARE vector components?
▸it can be seen in given fig
that vector a has two
components.
▸a unit vector i and a unit
vector j ,
▸which means 4 unit vectors
at i direction and 3 unit
vectors ar j directions.
resulting in vector a
10

11. WHAT IS TENSOR RANK?
RANK OF A TENSOR
▸ represents a physical entity which may be characterized by
magnitude and multiple directions simultaneously.Therefore,
the number of simultaneous directions is denoted R and is
called the rank of the tensor
▸ IN n -dimensional space, it follows that a rank-0 tensor (i.e.,
a scalar) can be represented by 𝑁0 =1 number since scalars
represent quantities with magnitude and no direction
▸ a rank-1 tensor (i.e., a vector) in N -dimensional space can
be represented by 𝑁1 =N numbers and a general tensor
by 𝑁 𝑅 numbers.
11

12. WHAT IS TENSOR RANK?
RANK OF A TENSOR
▸ A rank-2 tensor (one that requires 𝑁2 numbers to describe)
is equivalent, mathematically, to an N X N matrix.
12
▸ A rank two tensor is what we typically
think of as a matrix, a rank one tensor is
a vector.
▸ For a rank two tensor you can access
any element with the syntax t[i, j].
▸ For a Rank three tensor you would need
to address an element with t[i, j, k].

13. LETS START WITH RANK 1 TENSOR
▸ RANK ONE TENSOR 𝑁1 =N ,
Which is actually a normal vector,
13

14. LETS START WITH RANK 2 TENSOR
▸ RANK ONE TENSOR 𝑁2 =N ,
Which is actually a MATRIX FORM,𝟑 𝟐
=9
14

15. Tensor of rank 3
15
• RANK 3 TENSOR 𝑁3 ,
Which is actually for 3 dimensional
. space is 9 ,
• Before we had 3 components and 3 unit
vectors
• Now , we have 9 components and
9 sets of 3 unit vectors
𝟑 𝟑
=27 unit vectors

16. Tensor of rank 4
16
• RANK two TENSOR 𝑁4 ,
Which is actually for 3 dimensional
. space is 81 ,
• Before we had 9 components
and 3 unit vectors
• Now , we have 27 components or
27 sets of 3 unit vectors 𝟑 𝟒=81
unit vectors

17. Tensor of rank 4
17
The surface 1 has x as third index
and hence pertains to these 9 sets of
unit vector on yz plane
The surface 2 has y as third index
and hence pertains to these 9 sets of
unit vector on xz plane
The surface 3 has z as third index
and hence pertains to these 9 sets of
unit vector on xy plane

18. Tensor of rank 4
18
Transformation of cube to 9x9 matrix

19. 2.
GENERALIZED
HOOKS LAW
Let’s start try to understand
19

20. HOOKS LAW AND ITS GENERALIZATION
20
HOOKS LAW IS APPLICATION IS LIMITED TO LIMIT OF
PROPOTIONALITY ONLY, OF STRESS AND STRAINS,FOR
HOOKS LAW TO BE VALID ,
1- HOMOGENOUS MATERIAL
2- ISOTROPIC
3- ELASTIC
FOLLOWING ASSUMPTIONS SHOULD BE VALID:

21. HOOKS LAW AND ITS GENERALIZATION
21
NO OF CONSTANTS FOR DIFFERENT KIND OF MATERIAL
BASED ON THEIR STIFNESS MATRIX
• Isotopic material has two independent elastic constants.
• Anisotropic materials has 81 ,which boils down to 21(i.e
21 independent elastic constants) due to the symmetry of
stiffness matrix and strain energy function.
• orthotropic material :Applying the symmetry of
orthotropic material number of independent elastic
constants becomes 9.

22. HOOKS LAW AND ITS GENERALIZATION
22
• Homogeneous :Materials having uniform chemical Composition
and cannot separated into its constituent elements. Ex. metal
alloys.
• Isotropic :Materials whose properties like density, refractive index,
thermal and electric conductivity etc., do not vary with direction.
• Orthotropic : Materials whose properties like density, refractive
index, thermal and electric conductivity etc., vary only with three
perpendicular directions say X , Y & Z- direction.
• Anisotropic : Materials whose properties like density, refractive
index, thermal and electric conductivity etc., do not follow any
regular trend in terms of direction.

23. HOOKS LAW AND ITS MATHEMATICAL GENERALIZATION
23
• The Generalized Hook's Law of proportionality of stress and strain in
general form can be written as:
• We can write the general form of the law by the statement: Each of the
components of the state of stress at a point is a linear function of the
components of the state of strain at the point. mathematically this is
expressed by:
• Where the 𝐶 𝑘𝑙𝑚𝑛 are elasticity constants. there are 81 such constants
corresponding to the indices i,j,k,l . Taking values equal to 1,2 and 3.
𝐶𝑖𝑗𝑘𝑙relating the stress in the I,j directions to the strain in the k,l
directions
𝜎𝑖𝑗 = 𝐶𝑖𝑗𝑘𝑙 𝑒 𝑘𝑙 (1,2,3) (2)
𝜎 = 𝐶𝑒 1

24. HOOKS LAW AND ITS MATHEMATICAL GENERALIZATION
24

25. HOOKS LAW AND ITS MATHEMATICAL GENERALIZATION
25
• PLAIN STRESS • PLAIN STRAIN

26. EFFECT OF SYMMETRY ON MATRIX
26

27. EFFECT OF SYMMETRY ON MATRIX
27

28. HOOKS LAW AND ITS MATHEMATICAL GENERALIZATION
28

29. HOOKS LAW AND ITS GENERALIZATION
29
GENERAL MATRIX FORM OF STRESS-STRAIN TENSOR RANK 4 OF
STIFFNESS MATRIX

30. HOOKS LAW AND ITS GENERALIZATION
30
To reduce the constant from 𝟖𝟏 to 𝟑𝟔
Symmetry of stresses:
since the stress tensor is symmetric
then 𝛔𝒊𝒋 = 𝛔𝒋𝒊
𝛔𝒊𝒋 = 𝐂𝒊𝒋𝒌𝒍 𝐞 𝒌𝒍 ; 𝛔𝒊𝒋 = 𝐂𝒋𝒊𝒌𝒍 𝐞 𝒌𝒍
Therefore
𝐂𝒊𝒋𝒌𝒍 = 𝐂 𝒌𝒍𝒊𝒋

31. HOOKS LAW AND ITS GENERALIZATION
31
• Due to the symmetries of the stress tensors There are six
independent ways to express i and j taken together and still nine
• independent ways to express k and l taken together. Thus, with
this symmetry the number
• of independent elastic constants reduces to ( 6×9 = ) 54 from 81.

32. HOOKS LAW AND ITS GENERALIZATION
32
• Due to the symmetries of the strain tensors (, 𝛔𝒊𝒋 = 𝐂𝒊𝒋𝒌𝒍 𝐞 𝒌𝒍 , 𝛔𝒋𝒊 =
𝐂𝒋𝒊𝒌𝒍 𝐞 𝒌𝒍 ), the expression above can be simplified by removing the
last three columns , WE GET A 6X6 STIFNESS MATRIX=36

33. To reduce the constant from 𝟑𝟔 to 𝟐𝟏:
33
• USING STRAIN ENERGY
 ijijklklkl
ij
ij
ij
ij
kl
kl
ijij
ij
ij
kl
kl
ij
klijijklij
Ce
d
W
de
e
W
de
d
W
d
W
ijdedW
eeCijeW





















2
1
2
1
2
1
ij= Cijklekl = Cijklekl
Cijkl = Cklij

kl
dekl
dW
e
FIG 1

34. HOOKS LAW AND ITS GENERALIZATION
34
To reduce the constant from 𝟑𝟔 to 𝟐𝟏:
THE UPPER TRIANGLE EQUALS THE LOWER TRIANGLE , NOW
WE CAN REDUCE THE COEFFICIENTS TO 21 SUCH THAT
36-(N0 OF ELEMENT IN LOWER TRIANGLE(15))=21
=>

35. To reduce the constant from 𝟐𝟏 to 𝟏𝟑:35
1) Symmetry with Respect to One Plane :
(OX1, OX2)
Cabcd = akblcmdnCklmn
11, 12, 13 = (1, 0, 0)
21, 22, 23 = (0, 1, 0)
31, 32, 33 = (0, 0, -1)
C
1123 = 112233 C1123
= -C1123
C
1111 = C1111
11 = C1123 e23 , 
11 = C
1123 e
23
for symmetry : C
1123 = C1123 = -C1123C1123 = 0
also C1333 = 0

36. HOOKS LAW AND ITS GENERALIZATION
36
• ALL ODD NUMBERS OF
SUBSCRIPT (3) EQUALS TO ZERO
𝐂 𝟏𝟏𝟏𝟏
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟏𝟏𝟑𝟑
𝐂 𝟏𝟏𝟏𝟐
𝟎
𝟎
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟐𝟐𝟐𝟐
𝐂 𝟑𝟑𝟐𝟐
𝐂 𝟐𝟐𝟏𝟐
𝟎
𝟎
𝐂 𝟏𝟏𝟑𝟑
𝐂 𝟐𝟐𝟑𝟑
𝐂 𝟑𝟑𝟑𝟑
𝐂 𝟑𝟑𝟏𝟐
𝟎
𝟎
𝐂 𝟏𝟏𝟏𝟐
𝐂 𝟐𝟐𝟏𝟐
𝐂 𝟑𝟑𝟏𝟐
𝐂 𝟏𝟐𝟏𝟐
𝟎
𝟎
𝟎
𝟎
𝟎
𝟎
𝐂 𝟏𝟑𝟏𝟑
𝐂 𝟏𝟑𝟐𝟑
𝟎
𝟎
𝟎
𝟎
𝐂 𝟏𝟑𝟐𝟑
𝐂 𝟐𝟑𝟐𝟑

37. To reduce the constant from 𝟏𝟑 to 𝟗:37
• Symmetry with Respect to Two
Orthogonal Plane :
Let the two planes be the 𝐎𝐗 𝟏, 𝐎𝐗 𝟐 plane and
the 𝐎𝐗/
𝟏 ,𝐎𝐗/
𝟐 plane
11, 12, 13 = (-1, 0, 0)
21, 22, 23 = (1, 0, 0)
31, 32, 33 = (0, 0, -1)
C1323 = C1112 = C2221 = C2223 = 0
FIG.

38. To reduce the constant from 𝟏𝟐 to 𝟗:38
The stiffness matrix is written as
:
𝐂 𝟏𝟏𝟏𝟏
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟏𝟏𝟑𝟑
𝟎
𝟎
𝟎
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟐𝟐𝟐𝟐
𝐂 𝟑𝟑𝟐𝟐
𝟎
𝟎
𝟎
𝐂 𝟏𝟏𝟑𝟑
𝐂 𝟐𝟐𝟑𝟑
𝐂 𝟑𝟑𝟑𝟑
𝟎
𝟎
𝟎
𝟎
𝟎
𝟎
𝐂 𝟏𝟐𝟏𝟐
𝟎
𝟎
𝟎
𝟎
𝟎
0
𝐂 𝟏𝟑𝟏𝟑
𝟎
𝟎
𝟎
𝟎
𝟎
𝟎
𝐂 𝟐𝟑𝟐𝟑
FIG.
If material possesses three mutually perpendicular
planes of elastic symmetry, the material is called
orthotropic and its elastic matrix is of the form having
12 independent constants or 9 :

39. To reduce the constant from 𝟗to 𝟓:39
• Symmetry of rotation with respect
to one axis (transversely isotropic)
• The symmetry is expressed by the requirement that
the elastic constants are unaltered in any rotation
𝜽around the axis of symmetry:
11, 12, 13 = (cos , sin (90-), cos 90)
21, 22, 23 = (cos (90+), cos , cos 90)
31, 32, 33 = (cos 90, cos , cos 0)
In the OX1, OX2, OX3 system, the elastic stress-strain
relations are written :
• kl = Cklmnemn
• And in the OX
1, OX
2, OX
3 system, the elastic
stress-strain relations are written :
• 
kl = kmlnmn
FIG.

40. To reduce the constant from 𝟗to 𝟓:40
FIG.
𝐂 𝟏𝟏𝟏𝟏
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟏𝟏𝟑𝟑
𝟎
𝟎
𝟎
𝐂 𝟏𝟏𝟐𝟐
𝐂 𝟏𝟏𝟏𝟏
𝐂1133
0
0
0
C1133
C2233
C3333
0
0
0
0
0
0
1
2
C1111 − C1122
0
0
0
0
0
0
C1313
0
0
0
0
0
0
C1313
The stiffness matrix is written as

41. To reduce the constant from 𝟓 to 𝟐:
41
FIG.
Isotropy:
An isotropic material possesses elastic
properties which are independent of the
orientation of the axes.
𝐶1313=
1
2
𝐶1111 − 𝐶1122 , 𝐶3333 = 𝐶1111,
𝐶1133 = 𝐶1122
So that in fact we only have two independent
constants. the stiffness matrix is written
𝐶1111
𝐶1122
𝐶1122
0
0
0
𝐶1122
𝐶1111
𝐶1122
0
0
0
𝐶1122
𝐶1122
𝐶1111
0
0
0
0
0
0
1
2
𝐶1111 − 𝐶1122
0
0
0
0
0
0
1
2
𝐶1111 − 𝐶1122
0
0
0
0
0
0
1
2
𝐶1111 − 𝐶1122

42. To reduce the constant from 𝟓 to 𝟐:
42
Elastic stress−strain relations for isotropic media:
Let𝐶1122 =  ,
𝐶1212 =
1
2
𝐶1111 − 𝐶1122 =  , 𝐶1111 =  + 2
The pair of constants  and  are called Lame′s constant and  is referred to as
the shear modulus (also called G).
The stress−strain relations for an isotropic material are new written as follows:
𝜎11
𝜎22
𝜎33
𝜎12
𝜎13
𝜎23
=
 + 2


0
0
0

 + 2

0
0
0


 + 2
0
0
0
0
0
0

0
0
0
0
0
0

0
0
0
0
0
0

𝑒11
𝑒22
𝑒33
2𝑒12
2𝑒13
2𝑒23
∴ 𝜎𝑖𝑗= 2𝑒𝑖𝑗 + 𝛿𝑖𝑗 𝑒 𝑛𝑛
𝑒𝑖𝑗=
−𝛿𝑖𝑗
23  + 2
𝜎 𝑛𝑛 + 1
2

𝜎𝑖𝑗

43. 43
THANKS!
Any questions?
CONTACT ME AT:fb.com/azeem.baig.33
azeembaig94@gmail.com
“THINGS SHOULD BE DESCRIBED AS SIMPLY AS POSSIBLE
BUT NO SIMPLER”
ALBERT EINSTEIN