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The Structure of Tensors1 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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Presentation on theme: "The Structure of Tensors1 The Structure of the Elastic Tensor A study of the possibilities opened up by Kelvin 150 years ago Klaus Helbig, Hannover."— Presentation transcript:

1 The Structure of Tensors1 The Structure of the Elastic Tensor A study of the possibilities opened up by Kelvin 150 years ago Klaus Helbig, Hannover

2 The Structure of Tensors2 The customary official representation The tensor c ijkl connects the symmetric stress tensor ij symmetrically with the symmetric strain tensor kl The tensor c ijkl has 3 4 =81components The two symmetries of stress and strain mean ij = ji, kl = l k, thus c ijkl = c jikl = c ijlk The symmetrical connection means c ijkl = c klij Thus only 21 of the 81 components are significant!

3 The Structure of Tensors3 Hookes Law There are still 36 terms, but, e.g., c 2313 =c 1323

4 The Structure of Tensors4 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 )(9–i–j ), q = k kl + (1– kl )(9–k–l ) p = ij, c pq = c ijkl, q = (2 – kl ) kl

5 The Structure of Tensors5 The Voigt mapping, visual Mapping for the stress tensor Mapping for the strain tensor But this would not keep the scalar product. ! But this does.

6 The Structure of Tensors6 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

7 The Structure of Tensors7 Lost Advantages of Tensor Algebra There is no invariant representation Representation in another coordinate system by the simple rule kl = r ki r lj ij is not possible For rotation of the coordinate system, one has to use the Bond relations, Mohrs Circle, and other constructs.

8 The Structure of Tensors8 The Kelvin mapping 2E = ij ij = p p 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, C pq = ( ij +2(1– ij )) ( kl +2(1– kl )) c ijkl,

9 The Structure of Tensors9 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

10 The Structure of Tensors10 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.

11 The Structure of Tensors11 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.

12 The Structure of Tensors12 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.

13 The Structure of Tensors13 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 = K = V V = K / K = V / V = K C K = CVCV

14 The Structure of Tensors14 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 aparallel 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

15 The Structure of Tensors15 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 -> 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.

16 The Structure of Tensors16 Eigensystem of isotropic media

17 The Structure of Tensors17 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 3 3 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 Hookes Law. The resulting items are called eigencompliances and eigenstresses.

18 The Structure of Tensors18 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 = c ijkl kl = ij =>(c ijkl – ik jl ) 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 = C pq q = p => (c pq - pq ) q = 0 Standard eigenvalue/eigenvector problem, routines for solution in every math-package

19 The Structure of Tensors19 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

20 The Structure of Tensors20 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.

21 The Structure of Tensors21 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. E T 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.

22 The Structure of Tensors22 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.

23 The Structure of Tensors23 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 i i i i i i i i i Are these data consistent with our assumptions? Kelvin form

24 The Structure of Tensors24 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? i i i i i i Moreover, this matrix is not orthonormal ! i i i i i i

25 The Structure of Tensors25 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 i i i i i i i i i Note: the shear stiffnesses have not changed at all. The sheareigenstiffnesses are equal to the corresponding shear stiffnesses (in Kelvin form) Kelvin form

26 The Structure of Tensors26 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

27 The Structure of Tensors27 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.

28 The Structure of Tensors28 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.

29 The Structure of Tensors29 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

30 The Structure of Tensors30 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

31 The Structure of Tensors31 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

32 The Structure of Tensors32 All strains (as 6D-vectors) that are wave compatiblein 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.

33 The Structure of Tensors33 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.

34 The Structure of Tensors34 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

35 The Structure of Tensors35 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

36 The Structure of Tensors36 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).

37 The Structure of Tensors37 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

38 The Structure of Tensors38 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 eigensystem leads to this Voigt stiffness matrix with twelve independent stiffnesses

39 The Structure of Tensors39 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.

40 The Structure of Tensors40 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.

41 The Structure of Tensors41 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 2c 55 2c 55 2c 66 2c –

42 The Structure of Tensors42 The eigensystem of TI: #5 and #6 p = c 11 + (c 33 - c 66 )/2, q = [(p-c 33 ) 2 + 2c 13 2 ]/2 Estff p – q p + q AAC000AAC000 BBD000BBD000 Each of the two tensors is orthogonal to the other four. To be orthogonal to each other, we need C. D = –2 A. B

43 The Structure of Tensors43 Overview of Eigensystems I Eigensystems of monoclinic, orthotropic, and trigonal symmetry Capital letters: E plane shear; asterisk: d* isochoric strain

44 The Structure of Tensors44 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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