Elastic Properties and 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:

where,

• Both Sijkl and Cijkl are fourth-rank tensor quantities.

jiji

jiji

S

C

ComplianceS

tconsElasticorStiffnessC

)tan(

(7-2)

(7-1)

• 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:

322321123113 ,

We can therefore write:

332313

232212

131211

333231

232221

131211

(7-3)

Similarly, the strain tensor can be written as:

332313

232212

131211

333231

232221

131211

(7-4)

• Symmetry effect leads to a significant simplification of the stress-strain relationship of Eqs. 7-1 and 7-2.

We can write:

We can also write:

lkijlkijklijklij SorS and since

ijlkijkllkkllkijlkklijkl SSandSS ;;

jiklijkl

kljikljiklijklij

SS

SS

• 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:

333

222

111

612

513

423

11 12 13 1 6 5 22 23 2 4

33 3

=

Notation I Notation II

• The foregoing transformation is easy to remember: In other to obtain notation II, one must proceed first along the diagonal ( ) and then back ( ).

• 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.

321 654

We now have the stress and strain, in general form, as

345

42

6

561

345

426

561

22

22

22

and

It should be noted that , and , but

222111 , 333

12126

13135

23234

2

2

2

(7-5)

In matrix format, the stress-strain relation showing the 36 (6 x 6) independent components of stiffness can be represented as:

6

5

4

3

2

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

6

5

4

3

2

1

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

jijijiji SandC Or in short notation, we can write:

(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:

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:

independent elastic constant.

jiijjiij SSandCC

2162

30

• 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 inVarious Systems______________________________________________ System Rotation______________________________________________Triclinic None (or center of symmetry)Monoclinic 1 twofold rotationOrthorhombic 2 perpendicular twofold rotationTetragonal 1 fourfold rotation around [001]Rhombohedral 1 threefold rotation around [111]Hexagonal 1sixfold rotation around [0001]Cubic 4 threefold rotations around <111>

o

cba

90

o

cba

90

o

cba

90

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

66.....

044....

0044...

00033..

16001311.

1600131211

,

66.....

055....

0044...

00033..

0002322.

000131211

(7.7a)

x44....

0044...

00033..

0001311.

000131211

Hexagonal

where

)(21

),(2

1211

1211

CCx

ssx

or

(7.7b)

• 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.

The number of independent elastic constants in a cubic system is three (3). For isotropic materials ( most polycrystalline aggregates canbe treated as such) there are two (2) independent constants, b/c :

44....

044...

0044..

00011.

0001211.

000121211

2CC

C 121144

(7.8)

Cubic

(7.7c)

The stiffness matrix of an isotropic system is:

2.....

02

....

002

...

000..

000.

000

1211

1211

1211

11

1211

121211

CC

CC

CCC

CC

CCC

(7.9)

For cubic systems, Equation (7-8) does not apply, and wedefine an anisotropy ratio (also called the Zener anisotropy ratio, in honor of the scientist who introduced it):

• 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.

1211

442CC

CA

(7.10)

)(2.....

0)(2....

00)(2...

000..

000.

000

1211

1211

1211

11

1211

121211

SS

SS

SS

S

SS

SSS

(7.11)

Elastic compliances - for the isotropic case:

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

Rigidity or Shear modulus

11

1S

E (7.12)

41211

1)(2

1SSS

G

(7.13)

Compressibility (B) and bulk modulus (K):

Poisson’s ratio

Lame’s constants:

)(31

1

131211

332211

K

B

11

12

SS

v

12

44121144

1)(

21

C

GS

CCCu

(7.14)

(7.15)

(7.16)

(7.17)

• The equation to determine the compliance of isotropic materials can be written as (by using Eqs. 7-2 and 7-11):

6

5

4

3

2

1

1211

1211

1211

111212

121112

121211

6

5

4

3

2

1

200000

020000

002000

000

000

000

SS

SS

SS

SSS

SSS

SSS

(7.18)

The relationship of Eq. 7-18 can be expanded and equated to Eq. 6-9 to give:

2133112121123

3123122111122

3213122121111

1

1

1

ESSS

ESSS

ESSS

(7.19)

Also,

,1

2

,1

2

,1

2

6612116

5512115

4412114

GSS

GSS

GSS

(7.19)

• Expressing the strains as function of stresses, we have

6612116

5512115

4412114

3213112121123

3213122111122

3213122121111

212121

,2

,2

,2

CC

CC

CC

CCC

CCC

CCC

(7.20)

G:Note

• 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.

23

21

23

22

22

2144121144

23

21

23

22

22

2144121111

)(21

(41

)(21

(21

kikjjiijk

kikjjiijk

llllllSSSSG

llllllSSSSE

(7.21a)

(7.21b)

and are the Young’s and shear modulus, respectively, in the [ijk] direction; are the direction cosines of the direction [ijk]

ijkG321 ,, kji lll

ijkE

In a cubic material, the elastic moduli can be determined along any orientation, from the elastic constants, by the application of the following equations:

Table 7-2 Stiffness and compliance constants for cubic crystals___________________________________________________Metal ___________________________________________________Aluminum 10.82 6.13 2.85 1.57 -0.57 3.15Copper 16.84 12.14 7.54 1.49 -0.62 1.33Iron 23.70 14.10 11.60 0.80 -0.28 0.86Tungsten 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

11C 12C 44C 11S 12S 44S

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:

)](21

)[(21 222222

44121111 nlnmmlSSSSE

(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 Elasticconstants are applicable. This include:

ES

111 E

vS 12

GS

144

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 = Cijj

and

the applied hydrostatic stress is:

p = 1 = (50 + 40 + 40)(-3.33) 103 Pa

= -130 x 3.33 x 103 = -433 kPa

3122121111 CCC

Example: Determine the modulus of elasticity for tungsten and ironin the <111> and <100> directions. What conclusions can be drawnabout their elastic anisotropy? From Table 7.1____________________________

________________Fe: 0.80 -0.28 0.86W: 0.26 -0.07 0.66

44S12S11S

SOLUTIONThe direction cosines for the chief directions in a cubic lattice are:_______________________________________Directions _______________________________________<100> 1 0 0<110> 0<111>

2/1 2/13/1 3/1 3/1

1il 2jl 3kl

For iron:

91

91

91

}2/86.0)28.080.0{(280.01

111E

31

30.180.031

)43.008.1(280.01

111E

43.080.0

PaxE 11111 1070.2

37.01

80.0)0(30.180.01

100

E

PaxE 11100 1025.1

For tungsten:

31

266.0

)07.026.0(226.01

111E

26.031

33.033.0226.01

111

E

GPaPaxE 3851085.326.01 11

111

26.00266.0

)07.026.0(226.01

100

E

PaxE 11100 1085.3

26.01

Therefore, we see that tungsten is elastically isotropic while iron is elastically anisotropic.