Elastic Properties and Anisotropy of Elastic Behavior

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Elastic Properties and Anisotropy of Elastic
Behavior
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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.
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• 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 Hookes 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.

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• 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, Hookes law may be
  expressed as
• where,
• Both Sijkl and Cijkl are fourth-rank tensor
  quantities.

(7-1)
(7-2)
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• 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

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We can therefore write
(7-3)
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Similarly, the strain tensor can be written as
(7-4)
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• 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

and since
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• 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.

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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 12 13 1 6 5
22 23 2 4 33
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Notation II
Notation I
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• 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.

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We now have the stress and strain, in general
form, as
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It should be noted that
, and , but
(7-5)
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In matrix format, the stress-strain relation
showing the 36 (6 x 6) independent components of
stiffness can be represented as
Or in short notation, we can write
(7-6)
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• 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.

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• 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.

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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 lt111gt
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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
(7.7a)
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Hexagonal
(7.7b)
where
or
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• 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.

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Cubic
(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
(7.8)
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The stiffness matrix of an isotropic system is
(7.9)
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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)
(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.

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Elastic compliances - for the isotropic case
(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.
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The elastic constants for an isotropic material
are given by Youngs modulus
(7.12)
Rigidity or Shear modulus
(7.13)
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Compressibility (B) and bulk modulus (K)
(7.14)
Poissons ratio
(7.15)
Lames constants
(7.16)
(7.17)
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• The equation to determine the compliance of
  isotropic materials can be written as (by using
  Eqs. 7-2 and 7-11)

(7.18)
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• The relationship of Eq. 7-18 can be expanded and
  equated to Eq. 6-9 to give

(7.19)
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• Also,

(7.19)
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• Expressing the strains as function of stresses,
  we have

(7.20)
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• 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.

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In a cubic material, the elastic moduli can be
determined along any orientation, from the
elastic constants, by the application of the
following equations
(7.21a)
(7.21b)
and are the Youngs and shear
modulus, respectively, in the ijk direction
are the direction cosines of the
direction ijk
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Table 7-2 Stiffness and compliance constants for
cubic crystals ___________________________________
________________ Metal
_________________________________________________
__ 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
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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
(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
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• 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 Hookes
  law for the material, and compute the applied
  hydrostatic stress.

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• 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 Hookes law,
• ?i Cij?j
• and
• the applied hydrostatic stress is
• ?p ?1 (50 40 40)(-3.33) 103 Pa
• -130 x 3.33 x 103 -433 kPa

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Example Determine the modulus of elasticity for
tungsten and iron in the lt111gt and lt100gt
directions. What conclusions can be drawn about
their elastic anisotropy? From Table
7.1 ____________________________
________________ Fe 0.80 -0.28 0.86 W 0
.26 -0.07 0.66
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SOLUTION The direction cosines for the chief
directions in a cubic lattice are _______________
________________________ Directions
_______________________________________ lt100gt 1
0 0 lt110gt 0 lt111gt
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For iron
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For tungsten
44
Therefore, we see that tungsten is elastically
isotropic while iron is elastically anisotropic.