The Structure of the Elastic Tensor

1
The Structure of the Elastic Tensor

• A study of the possibilities opened up by Kelvin
  150 years ago

Klaus Helbig, Hannover
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The customary official representation
The tensor cijkl connects the symmetric stress
tensor ?ij symmetrically with the symmetric
strain tensor ekl The tensor cijkl has
3481components
The two symmetries of stress and strain mean?
?ij ?ji , ?kl ?lk , thus cijkl cjikl
cijlk The symmetrical connection means cijkl
cklij
Thus only 21 of the 81 components are significant!
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Hookes Law
There are still 36 terms, but, e.g., c2313c1323
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The Voigt mapping
With only 21 significant components, the elastic
tensor can obviously be mapped on a symmetric 6?6
matrix
Such mapping should preserve the elastic energy
density
2E ?ij ?ij ?p?p
The Voigt mapping achieves this by the mapping
rules
p i ?ij (1?ij)(9ij ) , q k ?kl
(1?kl)(9kl )
?p ?ij , cpq cijkl , ?q (2 ?kl )?kl
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The Voigt mapping, visual
?11 ?12 ?13
?12 ?22 ?23
?13 ?23 ?33
?11 ?12 ?13
?12 ?22 ?23
?13 ?23 ?33
??????11
??????22
??????33
??????23
??????13
??????12
?11
?22
?33
?23
?13
?12
??????11
??????22
??????33
???23
???13
??12
Mapping for the stress tensor
Mapping for the strain tensor
But this would not keep the scalar product s.?!
But this does.
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Properties of the Voigt mapping
Advantages
The Voigt mapping preserves the elastic energy
density
The Voigt mapping preserves the elastic
stiffnesses
Disadvantages
Stress and strain are treated differently
The norms of the three tensors are not preserved
The entries in all three Voigt arrays are not
tensor- or vector components, thus we loose all
advantages of tensor algebra
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Lost Advantages of Tensor Algebra
There is no invariant representation
Representation in another coordinate system by
the simple rule ?kl rki rlj ?ij is not
possible
For rotation of the coordinate system, one has to
use the Bond relations, Mohrs Circle, and other
constructs.
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The Kelvin mapping
2E ?ij ?ij ?p?p
The Kelvin mapping preserves E by the mapping
rules
p i ?ij(1-?ij)(9ij ) , q k ?kl
(1?kl)(9kl )
?p (?ijv2(1?ij )) ?ij , ?q
(?klv2(1?kl)) ?kl , Cpq (?ijv2(1 ?ij ))
(?klv2(1 ?kl)) cijkl ,
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Properties of the Kelvin mapping
Advantages
The Kelvin mapping preserves the elastic energy
density
The norms of the three tensors are preserved
Stress and strain are treated identically
The maps of stress, strain, and stiffness have
all properties of tensors of 1st respectively 2nd
rank in 6D-space, thus we keep all advantages of
tensor algebra
Only disadvantage the values of the stiffness
components are changed
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A tensor is a tensor by any name!
It is important to realize that a tensor is a
physical entity that does not depend on the way
we describe it.
For instance if in a mapping (a change of
description) the norm of the tensor is not
defined (as in the Voigt mapping), it does not
mean that the norm is lost, only that we cannot
access it easily as long as we use this
description.
The description we choose is not a question of
ideology, but of scientific economy for many
problems the Voigt mapping is the natural choice,
but there are problems that are much more easily
solved in the Kelvin form.
For some problems the 4th-rank tensor is the most
convenient notation.
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Which notation? Each has is place!
The Voigt notation is the de facto standard in
the outside world in the entire literature,
elastic parameters are listed in this notation,
and users expect that results are listed in
this form. For this reason, algorithms to deal
with tensors in Voigt notation are useful.
The strength of the Kelvin notation is the
possibility to reduce an elastic tensor to its
invariant (coordinate free) representation, and
conversely to construct tensors with given
invariants. It should be used in the analysis of
tensors.
The 4-subscript tensor notation is convenient for
operations as the change of coordinate systems
the very definition of a tensor is based on this
operation.
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Practical aspects of multiple notations
Even ten years ago, a conversion from one
notation to another was not a trivial matter, and
the 6561 multiplications needed for a
transformation of the coordinate system might
have taken up to a minute.
Today a scientist is hardly ever without access
to a computer, and the relevant routines can be
freely exchanged. Most of the important
operations give results instantly, i.e., with
response times below two seconds.
The current project will be completed with a
Tensor Toolbox written as a Mathematica notebook.
In Reader format it can be used on any computer
without the program.
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A simple tool for conversion
For conversion between the Voigt- and Kelvin
notation, onlyone array is needed
?
1 1 1 v2 v2 v2
?K ?V ??? ?V ?K/??? ?K ?V / ??? ?V ?K ??
For the stiffnesses, one uses the outer product
of ??
1 1 1 v2 v2 v2
1 1 1 v2 v2 v2
1 1 1 v2 v2 v2
v2 v2 v2 2 2 2
v2 v2 v2 2 2 2
v2 v2 v2 2 2 2
CK
CV
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How did Kelvin come to his description?
Not as a mapping of a 4th-rank 3D tensor on a
2nd-rank 6D tensor, because neither did exist
then.
The line of thought can be described like this
1. A strain can be described by 6 linearly
independent base strains.
2. A stress can be described by 6 linearly
independent base stresses.
3. It is convenient to use bases of the same
type for stress and strain.
4. It is convenient to use a set orthogonal
types for the common basis.
5. The weight of the different components
should be such that theproduct of aparallel
pair of stress and strain is preserved under
coordinate transformations from one to another
orthonormal base (hence the v2).
In this way Kelvin had defined a 6D Cartesian
vector space
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Kelvin went on to the invariant description
Hookes law can be thought of as a linear mapping
of the strain space on the stress space
The map is described by the 6?6 stiffness matrix
6. The ideal base strain generates a parallel
base stress (of the same type). Kelvin called
such strains principle strains, we call them
eigenstrains. As examples he gave for isotropic
media hydrostatic pressure -gt uniform volume
compression and shear strain -gt shear stress.
7. In a base consisting of (orthogonal)
eigenstrains, the 6?6 representation of the
stiffness tensor is diagonal, with the
eigenstiffnesses the only non-zero components.
Kelvin thus had produced the eigenvalue
decomposition or the canonical representation
of the stiffness tensor by its invariants.
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Eigensystem of isotropic media
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Nomenclature
I have called the eigenvalues eigenstiffnesses,
and the eigenvectors eigenstrains. Why these
special terms?
A similar eigenvalue problem exists for the
inverse of Hookes Law.The resulting items are
called eigencompliances and eigenstresses.
The nature of the 6 eigenstrains is of great
importance for many tasks, but not easily seen in
an arbitrarily oriented coordinate system. If
these details are important, the eigenstrain
vectors are treated as 3?3 tensors and brought
into invariant (canonical) form. The elements of
these representations are called eigenvalues and
eigenvectors.
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How does eigenvalue decomposition work in the
different notations?
The eigenvalue decomposition can be formulated
as find a stress that is generated by a parallel
strain
Four-subscripts ?ij cijkl ?kl??ij gt(cijkl
?ik?jl ?) ?kl 0Meaningful problem, solution
has to be programmed
Kelvin notation ?p Cpq ?q ??p gt (cpq -
?pq?) ?q 0 Standard eigenvalue/eigenvector
problem, routines for solution in every
math-package
Voigt notation the problem cannot be formulated
easily, since stresses and strains are expressed
in terms of different bases.
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What does the invariant description mean?
Of the 21 parameters of a general elastic tensor,
only the six eigenstiffnesses are genuinely
elastic
The remaining 15 parameters are geometric the 15
free parameters that describe a set of six
mutually perpendicular 6D unit vectors
Of these 15, three are extraneous to the problem
they are the Euler angles that describe the
orientation of the material with respect to the
global coordinate system
Conjecture All geometric parameters must be
real. Only the eigenstiffnesses can assume
complex values.
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How does all this affect our work?
The invariant description is possible for every
medium. Obviously the medium is stable if all six
eigenstiffnesses are positive.
In modeling a dissipative medium one cannot
freely assign imaginary parts to the stiffnesses.
If we would do that, we might end up with complex
eigenstrains, in violation of the concept of a
strain type.
How do we assign imaginary parts to the 21
stiffnesses? We need a method to construct
tensors from their eigensystem, i.e., a
eigenvalue composition.
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Eigenvalue composition of a tensor
If E is a 6?6 orthonormal matrix andL is a
diagonal matrix with 6 positive elements, thenC
E .L.ETis a symmetric, positive definite, 6?6
matrix with the elements of L as eigenvalues and
the column vectors of E as eigenvectors
If we choose the intended eigenstrains as column
vectors of E and place the intended
eigenstiffnesses on the diagonal of L, we can
generate a stiffness tensor C with an arbitrarily
chosen eigensystem.
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How should the eigensystem be chosen?
Obviously, the symmetry of the stiffness tensor
and some other properties like the order of
elastic waves is controlled by the chosen
eigensystem. An educated choice requires that
we know more about the nature of the eigenstrains
and the way they influence (together with the
eigenstiffnesses) the stiffness tensor.
Such a study is not necessary if we just want to
make an existing tensor dissipative we just
have to determine the eigensystem, add imaginary
parts (of the correct sign) to the
eigenstiffnesses, and re-compose.
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Example a dissipative orthotropic medium
The following complex-valued stiffness tensor C
was used
9.00-.03i 3.60-.02i 2.25-.16i
9.84-.03i 2.40-.16i
5.94-1.3i
4.00-.01i
3.20-.01i
4.36-.01i
Kelvin form
The real part was invented in 1991, the
imaginary part came from a different source.
Are these data consistent with our assumptions?
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The eigensystem of the tensor C
14.29-.37 i 5.80-.015 i 4.68-.99 i
4.36-.011i 4.00-.011 i 3.20-0.11 i
.622-.002i .737 -.26.073i 0 0 0
.693 -.67.024i -.240.02i 0 0 0
.359-.066i .033-.037i .93 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 1 0 0
All eigenstiffnesses have a small imaginary part,
but so have two components of each of three
eigenstrains. What does that mean?
Moreover, this matrix is not orthonormal !
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The improved dissipative tensor C
The real part of the eigenstrains together with
the complex eigenstiffnesses result in
Kelvin form
9.00-.22i 3.58-.22i 2.17-.16i
9.78-.24i 2.37-.13i
5.90-.90i
4.00-.01i
3.20-.01i
4.36-.01i
The changes are moderate, but now everything is
consistent.
Note the shear stiffnesses have not changed at
all.The sheareigenstiffnesses are equal to the
corresponding shear stiffnesses (in Kelvin form)
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Anatomy of the strain tensor
Since the appearance of the eigenstrains depends
on their orientation with respect to the global
coordinate system (three Euler angles!), we have
to look at the invariant representation of
strains.
Any (unit) strain tensor can be part of the
eigensystem of a stable stiffness tensor
Two invariants of a strain tensor are the
traceand the determinant
Strains with vanishing trace are called
isochoric
Strains with vanishing determinant are called
wave-compatible
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Wave compatibility?
The term wave compatible sounds oddly out of
place in a discussion that so far concerned only
static aspects.
Consider a homogeneous strain in an unbounded
medium. Even an infinitesimal strain can lead to
very large displacements at large distance from
the reference point.
Homogeneous strain in unbounded media cannot
exist.
Some homogeneous strains can exist along a plane.
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Wave compatibility explained
Ultimately we are interested in plane waves. Any
displacement can be attached to a plane wave, but
only those strains that can exist along a plane
without creating infinite displacements.
Such strains are compatible with wave
propagation perpendicular to the plane. Hence
their name.
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Strains that are not wave compatible
These two strains could not travel as a plane
wave in 3-direction with distance from the
center the displacement would grow without limit,
and a shear strain in the 13- and 23-planes would
be enforced
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Three strains that are wave compatible
These three strains could travel as a plane wave
in 3-direction without generating locally large
displacements
The plane shear strain (lowest example) could
travel not only in 3-direction, but also in
1-direction
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All strains that are wave compatiblein a given
coordinate direction
Six mutually orthogonal unit strains
Note that each shear strain could travel in two
directions!
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All strains (as 6D-vectors) that are wave
compatiblein a given coordinate direction
Six mutually orthogonal unit strains
Note that each shear strain could travel in two
directions!
Of the two shear strains that can travel in
3-direction,
one can also travel in 1-direction, the other
also in 2-direction, etc.
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Symmetry planes and shear eigenstrains
The symmetry of a medium is controlled by its
symmetry planes.
Proposition Two shear eigenstrains that share an
axis define a symmetry plane perpendicular to
this axis.
Consider a set of six orthonormal eigenstrains,
two of them plane shear strains in the 23- and
13- planes. It has the form
The 12-plane is a symmetry plane if the set is
invariant under a change of sign of the 3-axis,
i.e. for elements with an even number of
subscripts 3, i.e., for all but the two shear
strains.
are arbitrary entries
The two shear eigenstrains change sign, but for
an eigenstrain this is irrelevant.
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Symmetry planes and shear waves
The symmetry of a medium is controlled by its
symmetry planes
A symmetry plane supports shear wave with
cross-plane polar- ization in all direction.
Conversely, a plane that does support these waves
is a symmetry plane
Two shear eigenstrains that share an axis thus
define a symmetry plane perpendicular to this axis
This holds for any two shear eigenstrains
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The equivalence set
We had found that three of the 21 parameters
describing an elastic tensor are extraneous
they describe not the tensor, but its orientation
with respect to the default system (e.g.,
NEdown)
For a tensors without any symmetry plane that
means that there is a three-parametric manifold
of equivalent tensors that differ only in
orientation. These are the elements of the
equivalence set.
If we construct a tensor with symmetry planes, we
can choose the shear eigenstrains to let the
symmetry planes coincide with the coordinate
planes
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Equivalence for a single symmetry plane
Two shear eigentensors define a single plane of
symmetry. Let the normal to this plane be the
3-axis. The 1- and 2-axis can have still any
orientation in the symmetry plane, thus there is
a 1-parametric manifold of equivalent tensors.
This problem occurs with monoclinic and trigonal
tensors, which have only one symmetry plane (but
also with tetragonal tensors, where the plane
orthogonal to the fourfold axis is regarded as
the symmetry plane).
We choose the orientation which gives the lowest
number of stiffnesses. As far as elasticity is
concerned, this is always possible. In a
different (crystallographic) coordinate system,
the number is higher (mono12-gt13, trigo
tetra-gt 6-gt7).
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Effect of one single shear eigenstrain
A single shear eigenstrain has no effect on
symmetry, but a strong effect on the stiffness
matrix.
This orthonormal eigensystem leads to this Voigt
stiffness matrix with 16 independent stiffnesses
The A, B,F are arbitrary positive numbers
The black disks represent real numbers
The tensor is stable by design
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Two shear eigenstrains -gt monoclinic
Two shear eigenstrains generate a plane of
symmetry, thus the tensor has monoclinic
symmetry.
This orthonormal eigensystem leads to this Voigt
stiffness matrix with twelve independent
stiffnesses
The matrix on the right is automatically in the
coordinate system by the two shear planes and the
symmetry planes and therefore has twelve not
thirteen stiffnesses.
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Three shear eigenstrains gt orthotropic
Three shear eigenstrains generate three planes of
symmetry, thus the tensor has orthotropic
symmetry.
This orthonormal eigensystem leads to this Voigt
stiffness matrix with nine independent stiffnesses
Orthotropy marks an important point up to now we
had just to add new planes to increase the
symmetry. Also, the shape of the matrix has
reached its final form. From now on, co-planar
shear tensors and identical eigenstiffnesses will
be needed to increase symmetry.
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Example Transverse Isotropy
TI is the anisotropy best known. On the other
hand, the symmetry is high enough to give a good
example of how additional symmetry planes depend
on the eigensystem.
The equatorial plane must be a symmetry plane.
The eigenstiffnesses control the velocities of
the waves with the appropriate in-plane shears
along the three red lines. Thus the
eigenstiffnesses must be identical.
This makes the 3-axis at least a 4-fold axis. In
addition we certainly have a shear eigenstrain in
the 12-plane, since the two other shear planes
are symmetry planes too.
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The eigensystem of Transverse Isotropy
To simplify the discussion, I determined with
Mathematica the eigensystem of a general TI
medium. I obtained for 14
Estff 2c55 2c55 2c66 2c66
112233231312
000010
000100
000001
110000
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The eigensystem of TI 5 and 6
p c11 (c33 - c66)/2, q v(p-c33)2 2c132/2
Estff p q p q
112233231312
AAC000
BBD000
Each of the two tensors is orthogonal to the
other four. To be orthogonal to each other, we
need C.D 2 A.B
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Overview of Eigensystems I
Eigensystems of monoclinic, orthotropic, and
trigonal symmetry
Capital letters E plane shearasterisk
disochoric strain
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Overview of Eigensystems II
Eigensystems of tetragonal, TI, cubic, and
isotropic symmetry
Capital letters E plane shearasterisk
disochoric strain
Underlinedcoplanar shear tensors