Transcript The Structure of the Elastic Tensor

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 34=81components
The two “symmetries” of stress and strain mean
ij = ji , ekl = elk , thus cijkl = cjikl = cijlk
The “symmetrical” connection means cijkl = cklij
Thus only 21 of the 81 components are
significant!
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Hooke’s Law
There are still 36 terms, but, e.g., c2313=c1323
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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 eij = pep
The Voigt mapping achieves this by the mapping rules
p = i ij+ (1–ij)(9–i–j ) , q = k kl+ (1–kl)(9–k–l )
p = ij , cpq = cijkl , eq = (2 – kl )ekl
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The Voigt mapping, visual
11 12 13
11
e11 e12 e13

e
12 22 23
22
e12 e22 e23
11
13 23 33
33
e13 e23 e33
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Mapping
for the
stress tensor
23
13
12

e

e
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Mapping
for the
strain tensor

e
23
23
e
13

e12

But this would not
keep
13
But this does. 12
the scalar product .e!
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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 tensoror 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 e’kl = rki rlj eij is not possible
For rotation of the coordinate system, one has to
use the Bond relations, Mohr’s Circle, and other
constructs.
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The Kelvin mapping
2E = ij eij = pep
The Kelvin mapping preserves E by the mapping rules
p = i ij+(1-ij)(9–i–j ) , q = k kl+ (1–kl)(9–k–l )
p = (ij+√2(1–ij )) ij ,
eq = (kl+√2(1–kl)) ekl ,
Cpq = (ij+√2(1– ij )) (kl+√2(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, only
one array is needed:
= 1
1
1
√2
√2
√2
K = V  V = K/ eK = eV /  eV = eK 
For the stiffnesses, one uses the “outer product” of 
CK =
1
1
1
√2
√2
√2
1
1
1
√2
√2
√2
1
1
1
√2
√2
√2
√2
√2
√2
2
2
2
√2
√2
√2
2
2
2
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√2
√2
√2
2
2
2
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 the
product of a“parallel” pair of stress and strain is preserved under coordinate
transformations from one to another orthonormal base (hence the √2).
In this way Kelvin had defined a 6D Cartesian vector space
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Kelvin went on to the invariant description
Hooke’s 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” -> “uniform
volume compression” and “shear strain” -> “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 Hooke’s 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 ekl=eij =>(cijkl – ikjl ) ekl = 0
Meaningful problem, solution has to be programmed
Kelvin notation: p = Cpq eq = ep => (cpq - pq) eq= 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 and
L is a diagonal matrix with 6 positive elements, then
C = E .L.ET
is 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
Kelvin form
5.94-1.3i
4.00-.01i
3.20-.01i
4.36-.01i
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
-.24+0.02i
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:
9.00-.22i
3.58-.22i
2.17-.16i
9.78-.24i
2.37-.13i
Kelvin form
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 “shear”eigenstiffnesses 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 trace
and 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 compatible”in
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
compatible”in 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 3axis, 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., N–E–down)
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 (mono:12->13, trigo & tetra-> 6->7).
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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 -> 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 –> 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 1–4
Estff 2c55
11
0
22
0
33
0
23
0
13
1
12
0
2c55
0
0
0
1
0
0
2c66
0
0
0
0
0
1
2c66
–1
1
0
0
0
0
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The eigensystem of TI: #5 and #6
p = c11 + (c33 - c66)/2, q = √[(p-c33)2 + 2c132]/2
Estff p – q p + q
11
A
B
22
A
B
33
C
D
23
0
0
13
0
0
12
0
0
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 shear;
asterisk: d*
isochoric strain
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Overview of Eigensystems II
Eigensystems of tetragonal, TI, cubic, and isotropic
symmetry
Capital letters: E
plane shear;
asterisk: d*
isochoric strain
Underlined:
coplanar shear tensors
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