magma-051102

The Structure of Tensors 1

The Structure of the Elastic Tensor

A study of the possibilities opened up by

Kelvin 150 years ago

Klaus Helbig, Hannover


The customary “official” representation

The tensor cijkl connects the symmetric stress tensor ij

symmetrically with the symmetric strain tensor kl

The tensor cijkl has 34=81components

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!


Hooke’s Law

There are still 36 terms, but, e.g., c2313=c1323


The Voigt mapping

With only 21 significant components, the elastic tensor can

obviously be mapped on a symmetric 66 matrix

Such “mapping” should preserve the elastic energy density

2E = ij ij = pp

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 , q = (2 – kl )kl


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

1

1

22

33

23

13

12

Mapping

for the

stress tensor

Mapping

for the

strain tensor

But this would not keep

the scalar product .! But this does.


Properties of the Voigt mapping Advantages

The Voigt mapping preserves the elastic energy density

The entries in all three Voigt arrays are not tensor-

or vector components, thus

we loose all advantages of tensor algebra

The Voigt mapping preserves the elastic stiffnesses

Disadvantages

Stress and strain are treated differently

The norms of the three tensors are not preserved


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, Mohr’s Circle, and other

constructs.


The Kelvin mapping

2E = ij ij = pp

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 ,

q = (kl+√2(1–kl)) kl ,

Cpq = (ij+√2(1– ij )) (kl+√2(1– kl)) cijkl ,


Properties of the Kelvin mapping Advantages

The Kelvin mapping preserves the elastic energy density

The norms of the three tensors are preserved

Only disadvantage: the values of the stiffness

components are changed

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


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.

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 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.

For some problems the 4th-rank tensor is the most convenient

notation.


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 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.

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.


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.

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.

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.


A simple tool for conversion For conversion between the Voigt- and Kelvin notation, only

one array is needed:

For the stiffnesses, one uses the “outer product” of

1 1 1 √2 √2 √2 =

K = V V = K/K = V / V = K

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

√2 √2 √2 2 2 2

CK = CV


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


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 66 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 66 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”.


Eigensystem of isotropic media


Nomenclature

I have called the eigenvalues eigenstiffnesses, and the

eigenvectors eigenstrains. Why these special terms?

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 33 tensors

and brought into invariant (canonical) form. The elements of these

representations are called eigenvalues and eigenvectors.

A similar eigenvalue problem exists for the inverse of Hooke’s Law.

The resulting items are called eigencompliances and eigenstresses.


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 =>(cijkl – ikjl ) kl = 0

Meaningful problem, solution has to be programmed

Voigt notation: the problem cannot be formulated easily, since

stresses and strains are expressed in terms of different bases.

Kelvin notation: p = Cpq q = p => (cpq - pq) q= 0

Standard eigenvalue/eigenvector problem, routines for

solution in every math-package


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

Conjecture: All geometric parameters must be real.

Only the eigenstiffnesses can assume complex values.

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


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”.


Eigenvalue composition of a tensor

If E is a 66 orthonormal matrix and

L is a diagonal matrix with 6 positive elements, then

C = E .L.ET

is a symmetric, positive definite, 66 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.


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.


Example: a dissipative orthotropic medium

The following complex-valued stiffness tensor C was used:

The real part was “invented” in 1991, the imaginary

part came from a different source.

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

Are these data consistent with our assumptions?

Kelvin form


The eigensystem of the tensor C

All eigenstiffnesses have a small imaginary part, but so have

two components of each of three eigenstrains.

What does that mean?

.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

Moreover, this matrix is not orthonormal !

14.29-.37 i 5.80-.015 i 4.68-.99 i 4.36-.011i 4.00-.011 i 3.20-0.11 i


The improved dissipative tensor C’ The real part of the eigenstrains together with the

complex eigenstiffnesses result in:

The changes are moderate, but now everything is consistent.

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

Note: the shear stiffnesses have not changed at all.

The “shear”eigenstiffnesses are equal to the corresponding

shear stiffnesses (in Kelvin form)

Kelvin form


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”


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.


“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.


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


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


All strains that are “wave compatible”in

a given coordinate direction

Note that each

shear strain

could travel in

two directions!

Six mutually orthogonal unit strains


All strains (as 6D-vectors) that are “wave

compatible”in a given coordinate direction

Note that each shear strain could travel in two

directions!

Six mutually orthogonal unit strains

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.


Symmetry planes and shear eigenstrains

Consider a set of six orthonormal eigenstrains, two of them plane shear

strains in the 23- and 13- planes. It has the form

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.

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.


Symmetry planes and shear waves

Two shear eigenstrains that share

an axis thus define a symmetry

plane perpendicular to this axis

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

This holds for any two shear

eigenstrains


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


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).


Effect of one single shear eigenstrain

A single shear eigenstrain has no effect on symmetry,

but a strong effect on the stiffness matrix.

The black disks represent real numbers

The A, B,…F are arbitrary positive numbers

This orthonormal

eigensystem leads to

this Voigt stiffness

matrix with 16

independent stiffnesses

The tensor is stable “by design”


Two shear eigenstrains -> monoclinic Two shear eigenstrains generate a plane of symmetry, thus

the tensor has “monoclinic” symmetry.

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.

This orthonormal



matrix with twelve



Three shear eigenstrains –> orthotropic Three shear eigenstrains generate three planes of symmetry,

thus the tensor has “orthotropic” symmetry.

This orthonormal



matrix with nine


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.


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.


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 2c55 2c66 2c66

11

22

33

23

13

12

0

0

0

0

1

0

0

0

0

1

0

0

0

0

0

0

0

1

–1

1

0

0

0

0


The eigensystem of TI: #5 and #6 p = c11 + (c33 - c66)/2, q = √[(p-c33)

2 + 2c132]/2

Estff p – q p + q

11

22

33

23

13

12

A

A

C

0

0

0

B

B

D

0

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


Overview of Eigensystems I

Eigensystems of monoclinic, orthotropic, and trigonal

symmetry

Capital letters: E

plane shear;

asterisk: d*

isochoric strain


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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