Download Anisotropy of Elastic Behavior

Elastic Properties and Anisotropy of
Elastic Behavior
Figure 7-1 Anisotropic materials: (a) rolled material,
(b) wood, (c) glass-fiber cloth in an epoxy matrix, and
(d) a crystal with cubic unit cell.
• Real materials are never perfectly isotropic. In some
cases (e.g. composite materials) the differences in
properties for different directions are so large that one
can not assume isotropic behavior - Anisotropic.
• There is need to discuss Hooke’s Law for anisotropic
cases in general. This can then be reduced to isotropic
cases - material property (e.g., elastic constant) is the
same in all directions.
• In the general 3-D case, there are six components of
stress and a corresponding six components of strain.
• In highly anisotropic materials, any one component of
stress can cause strain in all six components.
• For the generalized case, Hooke’s law may be expressed
as:
 i  Cij j
(7-1)
 i  Sij j
(7-2)
where,
C  Stiffness (or Elastic cons tan t )
S  Compliance
• Both Sijkl and Cijkl are fourth-rank tensor quantities.
• Expansion of either Eqs. 7-1 or 7-2 will produce nine
(9) equations, each with nine (9) terms, leading to 81
constants in all.
• It is important to note that both ij and  ij are symmetric
tensors.
• Symmetric tensor
Means that the off-diagonal
components are equal. For example, in case of stress:
13   31 , 12   21
 23   32
We can therefore write:
 11 12 13   11 12 13 

 

  21  22  23    12  22  23 
       
 31 32 33   13 23 33 
(7-3)
Similarly, the strain tensor can be written as:
 11 12 13   11 12 13 

 

  21  22  23    12  22  23 
       
 31 32 33   13 23 33 
(7-4)
• Symmetry effect leads to a significant simplification of
the stress-strain relationship of Eqs. 7-1 and 7-2.
We can write:
 ij  Sijkl kl or  ij  Sijlk  lk
and since
Sijkl kl  Sijlk  lk ;  kl   lk ; and Sijkl  Sijlk
We can also write:
 ij  Sijkl kl   ji  S jikl kl
Sijkl  S jikl
• The direct consequence of the symmetry in the
stress and strain tensors is that only 36
components of the compliance tensor are
independent and distinct terms.
• Similarly, only 36 components of the stiffness
tensor are independent and distinct terms.
Additional simplification of the stress-strain relationship
can be realized through simplifying the matrix notation
for stresses and strains.
We can replace the indices as follows:
11  1
23  4
22  2
13  5
33  3
12  6
 11 12 13


22 23



33
Notation I
1
=
5


2 4

3
6
Notation II
• The foregoing transformation is easy to remember: In
other to obtain notation II, one must proceed first along
the diagonal (1  2  3 ) and then back ( 4  5  6 ).
• Notation II method makes life very easy when
correlating the stresses and strains for general case, in
which the elastic properties of a material are dependent
on its orientations.
We now have the stress and strain, in general form, as
 1

 6

 5
6
2
4

 1
5 



 4  and  6
 2

3 
 5

 2
6
2
2
4
2
5 

2
4 
2

3 

It should be noted that
and  3   33, but
22
  2  ,11  1 ,
 4  2 23   23
 5  213   13
 6  212   12
(7-5)
In matrix format, the stress-strain relation showing the
36 (6 x 6) independent components of stiffness can be
represented as:
  1   C11
  
  2   C21
   C
 3    31
  4   C41
   C
 5   51
  6   C61
C12
C13
C14
C15
C22
C23 C24
C25
C32
C33
C34
C35
C42
C43 C44
C45
C52
C53
C54
C55
C62
C63
C64
C65
C16  1 
 
C26   2 
C36   3 
 
C46   4 
C56   5 
 
C66   6 
Or in short notation, we can write:
 i  Cij j and  i  Sij j
(7-6)
Further reductions in the number of independent
constants are possible by employing other symmetry
considerations to Eq. 7-6.
• Symmetry in Stiffness and Compliance matrices
requires that:
Cij  C ji and Sij  S ji
Of the 36 constants, there are six constants where i = j,
leaving 30 constants where i  j.
But only one-half of these are independent constants
since Cij = Cji
Therefore, for the general anisotropic linear elastic solid
there are: 30
2
 6  21
independent elastic constant.
• The 21 independent elastic constants can be reduced still
further by considering the symmetry conditions found in
different crystal structures.
• In Isotropic case, the elastic constants are reduced from
21 to 2.
• Different crystal systems can be characterized exclusively
by their symmetries. Table 7-1 presents the different
symmetry operations defining the seven crystal systems.
• The seven crystalline systems can be perfectly described
by their axes of rotation. For example, a threefold rotation
is a rotation of 120o (3 x 120o = 360o); after 120o the
crystal system comes to a position identical to the initial
one.
Table 7.1 Minimum Number of Symmetry Operations in
Various Systems
______________________________________________
System
Rotation
______________________________________________
Triclinic
None (or center of symmetry)
Monoclinic
1 twofold rotation
Orthorhombic
2 perpendicular twofold rotation
Tetragonal
1 fourfold rotation around [001]
Rhombohedral
1 threefold rotation around [111]
Hexagonal
1sixfold rotation around [0001]
Cubic
4 threefold rotations around <111>
abc
      90
o
abc
      90
o
abc
      90
o
The hexagonal system exhibits a sixfold rotation around the
[0001] - c axis; after 60 degrees, the structure superimposes
upon itself.
In terms of a matrix, we have the following:
Orthorhombic
Tetragonal
11 12 13 0 0 0  11 12 13 0 0 16 
 . 22 23 0 0 0   . 11 13 0 0  16



. 33 0 0 0   .
. 33 0 0
0 
.
,
.


.
. 44 0 0
.
.
. 44 0
0



.
.
. 55 0   .
.
.
. 44 0 
.



.
.
.
. 66  .
.
.
.
.
66 
.
(7.7a)
Hexagonal
11 12 13 0 0
 . 11 13 0 0

. 33 0 0
.
.
.
. 44 0

.
.
. 44
.


0
0

0

0

x


(7.7b)
where
x  2( s11  s12 ),
or
1
x  (C11  C12 )
2
• Laminated composites made by the consolidation of
prepregged sheets, with individual piles having
different fiber orientations, have orthotropic symmetry
with nine independent elastic constant.
• This is analogous to orthorhombic symmetry, and
possess symmetry about three orthogonal (oriented 90o
to each other) planes. The elastic constants along the
axes of these three planes are different.
Cubic
11 12 12 0 0 0 
 . 11 12 0 0 0 


. 11 0 0 0 



.
. 44 0 0


.
.
. 44 0 



.
.
.
. 44

(7.7c)
The number of independent elastic constants in a cubic system
is three (3).
For isotropic materials ( most polycrystalline aggregates can
be treated as such) there are two (2) independent constants, b/c :
C44
C11  C12

2
(7.8)
The stiffness matrix of an isotropic system is:
C11 C12 C12
 . C
C12
11

. C11
 .

.
.
 .

 .
.
.

 .
.
.

0
0
0
0
0
C11  C12
2
0
0
.
C11  C12
2
.
.


0

0


0



0

C11  C12 

2
0
(7.9)
For cubic systems, Equation (7-8) does not apply, and we
define an anisotropy ratio (also called the Zener anisotropy
ratio, in honor of the scientist who introduced it):
2C44
A
C11  C12
(7.10)
• Several metals have high “A” anisotropy ratio.
• Aluminum and tungsten, have values of A very close
to 1. Single crystals of tungsten are almost isotropic.
Elastic compliances - for the isotropic case:
0
0
0
 S11 S12 S12

 . S

S
0
0
0
11
12


. S11
0
0
0
 .

 .

.
. 2( S11  S12 )
0
0


.
.
.
2( S11  S12 )
0
 .



.
.
.
.
.
2
(
S

S
)

11
12 
(7.11)
Similarly, the 81 components of elastic compliance for the
cubic system have been reduced to three (3) independent
ones while for the isotropic case, only two (2) independent
elastic constants are needed.
The elastic constants for an isotropic material are given
by:
Young’s modulus
1
E
S11
(7.12)
Rigidity or Shear modulus
1
1
G 

2( S11  S12 )
S4
(7.13)
Compressibility (B) and bulk modulus (K):
1
11   22   33
B 
K  1 (     )
11
12
13
3
(7.14)
Poisson’s ratio
S12
v
S11
(7.15)
1
1
u  C44  (C11  C12 ) 
G
2
S44
(7.16)
  C12
(7.17)
Lame’s constants:
• The equation to determine the compliance of isotropic
materials can be written as (by using Eqs. 7-2 and 7-11):
 1   S11
  
  2   S12
   S
 3    12
4   0
   0
 5  
6   0
S12
S12
0
0
0
S11
S12
0
0
0
S12
S11
0
0
0
0
0
2S11  S12 
0
0
0
0
0
2S11  S12 
0
0
0
0
0
2S11 
  1 
 
  2 
  
 3 
  4 
  
 5 
S12   6 
(7.18)
The relationship of Eq. 7-18 can be expanded and equated to
Eq. 6-9 to give:
1
1  S11 1  S12 2  S12 3   1    2   3 
E
1
 2  S12 1  S11 2  S12 3   2    1   3 
E
1
 3  S12 1  S12 2  S11 3   3    1   2 
E
(7.19)
Also,
 4  2S11
 5  2S11
 6  2S11
1
 S12  4   4 ,
G
1
 S12  5   5 ,
G
1
 S12  6   6 ,
G
(7.19)
• Expressing the strains as function of stresses, we have
 1  C111  C12 2  C12 3  2    1   2   3 ,
 2  C121  C11 2  C12 3  1  2     2   3 ,
 3  C121  C12 2  C11 3  1   2  2     3 ,
1
 4  C11  C12  4   4
2
1
 5  C11  C12  5   5
2
1
 6  C11  C12  6   6
2
(7.20)
Note :   G
• A great number of materials can be treated as isotropic,
although they are not microscopically so.
• Individual grains exhibit the crystalline anisotropy and
symmetry, but when they form a poly-crystalline aggregate
and are randomly oriented, the material is microscopically
isotropic.
• If the grains forming the poly-crystalline aggregate have
preferred orientation, the material is microscopically
anisotropic.
• Often, material is not completely isotropic; if the
elastic modulus E is different along three perpendicular
directions, the material is Orthotropic; composites are a
typical case.
In a cubic material, the elastic moduli can be determined
along any orientation, from the elastic constants, by the
application of the following equations:
1
1
 S11  2( S11  S12  S44 )(li21l 2j 2  l 2j 2lk23  li21lk23
Eijk
2
(7.21a)
1
1
 S44  4( S11  S12  S44 )(li21l 2j 2  l 2j 2lk23  li21lk23
Gijk
2
(7.21b)
Eijk and Gijk are the Young’s and shear modulus, respectively, in
the [ijk] direction; li1 , l j 2 , lk 3 are the direction cosines of the
direction [ijk]
Table 7-2 Stiffness and compliance constants for cubic crystals
___________________________________________________
Metal
C11
C12
S11
S44
C44
S12
___________________________________________________
Aluminum
10.82
6.13
2.85
1.57
-0.57
3.15
Copper
16.84 12.14
7.54
1.49
-0.62
1.33
Iron
23.70 14.10 11.60 0.80
-0.28 0.86
Tungsten
50.10 19.80 15.14 0.26
-0.07
0.66
___________________________________________________
Stiffness constants in units of 10-10 Pa.
Compliance in units of 10-11 Pa
Using the direction cosines l, m, n (as described in the text book)
the equation for determining the Elastic Moduli along any direction
is given by:
1
1
 S11  2[(S11  S12 )  S44 ](l 2 m 2  m 2 n 2  l 2 n 2 )
E
2
(7.22)
Typical values of elastic constants for cubic metals are given in
Table 7.2.
All the relations described in Eqs. 7-12 to 7-20 for obtaining Elastic
constants are applicable. This include:
1
S11 
E
v
S12  
E
1
S44 
G
Example
A hydrostatic compressive stress applied to a material
with cubic symmetry results in a dilation of -10-5. The
three independent elastic constants of the material are
C11 = 50 GPa, C12 = 40 GPa and C44 = 32 GPa. Write an
expression for the generalized Hooke’s law for the
material, and compute the applied hydrostatic stress.
SOLUTION
Dilation is the sum of the principal strain components:
 = 1 + 2 + 3 = -10-5
Cubic symmetry implies that 1 = 2 = 3 = -3.33 x10-5
and
4 = 5 = 6 = 0
From Hooke’s law,
i = Cijj
1  C111  C12 2  C12 3
and
the applied hydrostatic stress is:
p = 1 = (50 + 40 + 40)(-3.33) 103 Pa
= -130 x 3.33 x 103 = -433 kPa
Example: Determine the modulus of elasticity for tungsten and iron
in the <111> and <100> directions. What conclusions can be drawn
about their elastic anisotropy? From Table 7.1
____________________________
Fe:
W:
S12 S44
S11
________________
0.80 -0.28 0.86
0.26 -0.07 0.66
SOLUTION
The direction cosines for the chief directions in a cubic lattice are:
_______________________________________
Directions
li1
l j2
lk 3
_______________________________________
<100>
1
0
0
<110>
0
1/ 2
1/ 2
1/ 3
<111>
1/ 3
1/ 3
For iron:
1
1 1 1

 0.80  2{(0.80  0.28)  0.86 / 2}   
E111
9 9 9
1
1
1


 0.80  2(1.08  0.43)   0.80  1.30 
E111
 3
 3
 0.80  0.43
1
E111 
 2.70 x1011 Pa
0.37
1
 0.80  1.30(0)  0.80
E100
E100  1.25x1011 Pa
For tungsten:
1
0.66  1 

 0.26  2(0.26  0.07) 
 
E111
2  3 

1
1

 0.26  20.33  0.33   0.26
E111
 3
1
E111 
 3.85x1011 Pa  385GPa
0.26
1
0.66 

 0.26  2(0.26  0.07) 
0   0.26
E100
2 

E100
1

 3.85x1011 Pa
0.26
Therefore, we see that tungsten is elastically isotropic while iron
is elastically anisotropic.