Tensors, Dyads

1
Tensors, Dyads

• 27-765, Advanced Characterization and
  Microstructural Analysis,
• Spring 2001, A. D. Rollett

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Objective

• The objective of this lecture is to introduce the
  student to the concept of tensors and to review
  some basic concepts relevant to tensors,
  including dyads.
• Many of the concepts reviewed in this lecture are
  useful or essential in discussions of elasticity
  and plasticity.

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Tensors

• Tensors are extremely useful for describing
  anisotropic properties in materials. They permit
  complicated behaviors to be described in a
  compact fashion that can be easily translated
  into numerical form (i.e. programming).

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Dyads 1

• We are familiar with constructing vectors as
  triples of coefficients multiplying the unit
  vectors we call these tensors of first order.
• In order to work with higher order tensors, it is
  very useful to construct dyads from the unit
  vectors.

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Dyads 2

• Define the dyadic product of two vectors. Note
  coordinate free. Properties are the following

scalar
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Dyads 3

• Transformation (l) of the dyadic product, from
  one coordinate system to another, leaves it
  invariant as can easily be seen from this
  construction

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Inner products from Dyadics

• The dyadic product is similar to the vector
  product it is not commutative.
• Inner products can be combined with the dyadic
  product

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Unit Dyads

• We can construct unit dyads from the unit
  vectorsFor now we will leave these as they
  are and not introduce any new symbols.

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Dyad example dislocation slip

• We commonly form a dyad for the strain, m,
  produced on a slip system (or twinning system) by
  combining unit vectors that represent slip (twin
  shear direction) direction, b, and slip plane
  normal (twin plane), n.

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Second Order Tensors

• Unit dyads form the basis for second order (rank)
  tensors, just as the unit vectors do for vectors,
  where the Tij are the (nine) coefficients of the
  tensor.

Example stress
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Second Order Tensors example strain from slip

• The dyad for crystallographic slip forms the
  basis for a second order (rank) strain tensor,
  where the magnitude of the tensor is given by the
  amount of shear strain on the given system.

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Unit (spherical) tensors

• The unit tensor is formed from the unit dyad
  thusNote that this tensor is invariant
  under transformations.

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Tensor transformations

• Transformation of tensors follows the rules set
  up for vectors and the unit vectorsthus

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Right, left inner products

• Right and left inner-products of the second-order
  tensor, T, with a vector left
  right

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Inner products of tensors

• Inner-product between two second-order tensors in
  the dyadic notation
• Notice that the inner-product involves a
  contraction of the inner indices,r s.

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Outer products of tensors

• Consider the outer product of a tensor of
  second-order with a vector to produce a tensor of
  third-order
• Fourth-order tensor is similar

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General Cartesian tensors

• More generally, Cartesian tensors of order n are
  defined by components by the
  expression
• The nth order polydyadics form a complete
  orthogonal basis for tensors of order n.

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nth order tensor transformations

• Changes in the coordinate frame change the
  components of the nth order tensor according to a
  simple extension

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Inner products of higher order tensors

• Inner-products on tensors of higher order are
  defined by contracting over one or more indices.
  For example, contracting the last n-p indices of
  tensor T (of order n) with the first n-p indices
  of a tensor U (of order m) gives a new tensor S
  (of order 2pm-n) according to the following.

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Higher order inner products
Here, 0ltpltn. From (2.30) it should be evident
that the order of each of the tensors S, T and U
(as specified by m, n and p) must be known in
order to correctly form the product. The order
of the contraction is n-p (sometimes denoted by
the number of dots between the symbols).
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Higher order outer products

• A natural generalization of the outer product to
  higher-order tensors is obvious. The outer
  product of two tensors T and U (of order n and m,
  respectively) is a new tensor S of order nm
  according to the expression

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Eigenanalysis of tensors

• It is very useful to perform eigenanalysis on
  tensors of all kinds, whether rotations, physical
  quantities or properties.
• We look for solutions to this equation, where µ
  is a scalar

or,
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Characteristic equation

• The necessary condition for the relation above to
  have non-trivial solutions is given byWhen
  the (cubic) characteristic equation is solved,
  three roots, µi, are obtained which are the
  eigenvalues of the tensor T. They are also
  called the principal values of the tensor.

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Eigenvectors

• Assume that the three eigenvalues are distinct.
  The ith eigenvalue can be reintroduced into the
  previous relation in order to solve for the
  eigenvectors, v(i)

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Real, Symmetric Tensors

• Consider the special case where the components of
  T are real and symmetric, e.g. stress, strain
  tensors. Now lets evaluate the effect on the
  eigenvalues andeigenvectors,which the
  symmetric nature of the tensor allows it to be
  re-written as

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Eigenvalues of real, symmetric tensors

• Now take the complex conjugate of the components
  of each element in the above, keeping in mind
  that T is real
• Next, take the left inner product of the previous
  relation with and subtract it from the
  right inner product of the above relation with
  

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real eigenvalues

• Given this consequence of non-trivial solutions
  for the eigenvectors, we see that the eigenvalues
  must be value in order for the previous relation
  to be satisfied

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eigenvectors

• Next, take the left inner product of the previous
  relation, with and subtract it from the
  right inner product of with

If the eigenvectors are distinct, the inner
product of the associated eigenvectors must be
zero.
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Eigenvectors are orthogonal

• If inner (scalar) products of the eigenvectors
  are zero, then they are orthogonal.
• The eigenvectors of a real-symmetric tensor,
  associated with distinct eigenvalues, are
  orthogonal.
• In general the eigenvectors can be normalized by
  an appropriate selection of scalar multiplier to
  have unit length.

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Orthonormal eigenvectors

• Convenient to select the set of eigenvectors in a
  right-handed manner such that
• The axes of the coordinate system defined by this
  orthonormal set of eigenvectors are often called
  the principal axes of tensor T, and their
  directions are called principal directions.

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Diagonalizing the tensor

• Consider the right and left inner product of
  tensor T with the eigenvectors according
  toThe left hand side of this relation can be
  expressed in the dyadic notation as

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Transformation to Diagonal form

• are the direction cosines linking the
  orthonormal set of eigenvectors to the original
  coordinate system for T. Combining the equations
  above, we get the following, where superscript
  d denotes the diagonal form of the tensor

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Principal values, diagonal matrix

• are components of the real-symmetric
  tensor T in the coordinate frame of its
  eigenvectors. It is evident that the matrix of
  components of is diagonal, with the
  eigenvalues appearing along the diagonal of the
  matrix. (The superscript d highlights the
  diagonal nature of the components in the frame
  of the eigenvectors.)

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Invariants of 2nd order tensor

• The product of eigenvalues, µ1µ2µ3, is equal to
  the determinant of tensor T.

see slide 28, and recall that a transformation
has unit det.
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Invariants 2

• Other combinations of components which form
  (three) invariants of second-order tensors
  include, where T2TT (inner product) I3
  det T

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Deviatoric tensors

• Another very useful concept in elasticity and
  plasticity problems is that of deviatoric
  tensors. A A - 1/3I trA
• The tensor A has the property that its trace is
  zero. If A is symmetric then A is also
  symmetric with only five independent components
  (e.g. the strain tensor, e).

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Deviatoric tensors 2

• Frequently we decompose a tensor into its
  deviatoric and spherical parts (e.g. stress) A
  A 1/3I trA e.g. s s 1/3I trs s sm
  
• Non-zero invariants of A I2-1/2(tr A)2-
  tr A2I3 det A 1/3 tr A3
• Re-arrange I2-1/3I12I2.
  I3I3-(I1I2)/32/27I13

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Positive definite tensors

• The tensor T is said to be positive definite if
  the above relation holds for any non-zero values
  of the vector u. A necessary and sufficient
  condition for T to be positive definite is that
  the eigenvalues of T are all positive.

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Polar Decomposition

• Polar decomposition is defined as the unique
  representation of an arbitrary second-order
  tensor, T, as the product of an orthogonal
  tensor, R, and a positive-definite symmetric
  tensor, either U or V, according to

Why do this? For finite deformations, this
allows us to separate the rotation from the
stretch expressed as a positive definite matrix.
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Polar Decomposition 2

• Define a new second-order tensor, A T-1T. A is
  clearly symmetric, and that it is positive
  definite is clear from considering the
  followingThe right-hand side of (2.67) is
  positive for any non-zero vector v, and hence vAv
  is positive for all non-zero v.

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Polar Decomposition 3

• Having shown that A is symmetric,
  positive-definite, we are assured that A has
  positive eigenvalues. We shall denote these by
  µ12, µ22, µ32, where, without loss of generality,
  µ1, µ2, µ3, are taken to be positive. It is
  easily verified that the same eigenvectors which
  are obtained for T are also eigenvectors for A
  thus

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Polar Decomposition 4

• Next we define a new tensor, U, with a diagonal
  (principle values) matrix, D, and a rotation, R,
  according to

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Polar Decomposition 5

• Thus, D is a diagonal tensor whose elements are
  the eigenvalues of T, and R is the rotation that
  takes the base vectors into the eigenvectors
  associated with T. U is symmetric and positive
  definite, and since R is orthogonal

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Polar Decomposition 6

• The (rotation) tensor R associated with the
  decomposition is defined byThat R has the
  required orthogonality is clear from the
  following

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Polar Decomposition 7

• Thus the (right) U-decomposition of tensor T is
  defined by relations (2.66) and (2.69). If the
  (left) V-decomposition is preferred then the
  following applies