Holpzaphel_-_Nonlinear-Solid-Mechanics-a-Contin
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1.2 Algebra of Tensors |
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Every tensor A can be decomposed into its so-called spherical part and its deviatoric part, i.e.
A= al+ devA |
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rl• = -trA =-(I: A)= -4-· |
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u • |
(1.131)
(l.132)
Any tensor of the fonn al, with a denoting a scalar, is known as a spherical tensor, while elevA is known as a deviator of A, or a deviatoric tensor. The deviatoric op-
erator dev( •) is denoted by the short-hand notation, i.e. dev( •) (1/3)tr( •)I, or dev( • )iJ = (• )ij - (1/3)( • )kk8iJ .
Computing the trace of ( 1.131) we deduce with ( 1.92), ( 1.87)5 , ( 1.22)1 and ( l. I32h
that Aii = 3(Aii/3) + dev Aii. This relation yields the important property |
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tr(devA) = 0 , |
(l.133) |
which means that the trace of the deviator of A is always zero. |
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EXERCISES
1.Let u be any vector and e any unit vector. Use eqs. ( 1.125), ( 1.126) and the triple vector product ( 1.47) to show that u can be resolved into components parallel, i.e. 011'and perpendicular, i.e. Uj_, to unit vector e, according to
u = u11 + u.l = (u · e)e +ex (u x e) .
2.Prove eqs. ( 1.57) and ( 1.58) and show that
3.Let A be a tensor with matrix [A] and AT the transpose of A with matrix [A]T. Show that [AT) = [A]T.
4.Show by means of representation (1.59) and properties ( 1.98) and ( J.61) that the
double contraction of the two tensors A and B yields the component form Aii Bii.
5.Starting from eq. (l.100), verify that detA = EijkAi 1Ai2 Ak3 and show property
(1.101).
6. Consider eq. (1.93) 1 and relations (l.103), (l.88) in order to obtain the property
A -·r:A=3 . |
(1.134) |
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1 Introduction to Vectors and Tensors |
7.Given that S is a symmetric tensor and W is antisymmetric, prove property
(l.117).
8.Two tensors A and B are given in the form of their matrix representations
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cos(} |
sinO |
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[A) = -sinO |
cosO |
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Verify that these matrices are orthogonal. Show that [A] describes a rotation and
[B] a reflection. Give a geometrical interpretation.
9. Find the axial vector of the skew tensor W = 1/2{u ® v - v 0 u).
10. Let A be a tensor whose matrix is
[A]= [ t t tl
Find the spherical and the deviatoric parts of A.
1.3 Higher-order Tensors
In the following we discuss higher-order tensors. Any tensor of order (or rank) n may be expressed in the form
Ai 1 i2 ••• in ei 1 © ei2 0 · · · ® ein . |
(1.135) |
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As (1.135) shows, a tensor of order n has 3n components Ai1 i 2 ••• i,., |
provided with n |
indices i1 , i 2 , ••• , in. In particular, a tensor of order zero has 3° = 1 component with 0 (no) index, which simply is a scalar a (i.e. a single real number). A tensor of order one has 31 = 3 components characterizing any vector v, i.e. 'Vi as indicated in ( 1.23 ). A tensor of order two, for example, A, has 32 = 9 components, i.e. Aij as indicated in ( 1.60).
Tensor of order three. A tensor of order three we distinguish by the notation A,
B, C, ... According to (1.135), A may be expressed as
(l.136)
where Aiik characterizes the 33 = 27 components of A.
1.3 Higher-order Tensors |
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A particular example for a third-order tensor is the so-called triadic product of
three vectors u, v, w, denoted by u ® v ® w, with the properties |
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(u®v)®w=u®v®w, |
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(1.137) |
(u ® v 0 w)x = (w · x)u 0 v |
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(1.138) |
(u ® v ® w) : (x 0 y) = (v · x)(w · y)u |
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(l.139) |
(u®v®w):l=(v·w)u. |
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(1.140) |
Hence, by analogy with the procedure which led to eq. (1.62) we may determine the components of A, i.e.
( 1.141)
The double contraction of a third-order tensor A with a second-order tensor B produces a vector. It is denoted by A : B and may be found with representations (l.136), (1.59) and properties (1.139), (1.21) and (l.61) as
A: B = AiikB1m(ei 0 ei ® ek) : (e1 0 em)
=
=
AijkBtm(ej ·eL)(ek ·em)ei = AijkBtm8j18kmei
AijkBjkei . |
(1.142) |
EXAMPLE 1.4 Let E denote the third-order permutation tensor, which may be expressed as
(1.143)
Here, Cijk = (ei x ei) ·ek are the 33 components of E which we have introduced as the permutation symbol (recall eqs. (1.33) and (1.35)).
Show that the double contraction of E with the dyad of any vectors u and v, i.e. u ® v, gives the cross product of v and u.
Solution. With expressions (E)ijk = €ijk and (u 0 v)1m = 'U(Vm we may write
(1.144)
Using properties (1.139), (1.21) and (1.22)2 , we find that
(l.145)
which is a vector with components €ijk"UfVk. Finally, by means of the permutation
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1 |
Introduction to Vectors and Tensors |
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property €ijk |
= -€kji and expressions (1.43h, ( 1.26) we obtain |
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e : (u ® v) = v x u ' |
(1.146) |
which is the desired result. |
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Tensor of order four. |
Any tensor of order four, denoted by A, JR, C, ... , has |
34 = 81 components, i.e. |
Aiikl, Biikl, Ciikl, .. ., respectively. We may express any |
tensor A in terms of the three Cartesian basis vectors, we write
(l.147)
in which Aijkl are the components of A.
One example for a fourth-order tensor is the product of the four vectors u, v, w, x,
denoted by u ® v 0 w ® |
x. We have the useful properties: |
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(u 0 v) 0 (w 0 x) = u ® v ® w ® x |
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(I. 148) |
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(u 0 v ® |
w ® x) : (y 0 z) = (w . y)(x . z)(u ® |
v) . |
(1.149) |
With these properties we can show that the components Aiikl of A can be expressed as
(1.150)
Hence, the double contraction of A with a second-order tensor B is a second-order tensor, denoted by A : B. With representations (I. I47), (1.59), and properties (1.149), ( 1.21) and (1.61) we find that
A: B = AijklBmn(ei ® ei ® |
ek ® e,) : (em® en) |
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= Aijk1Bmn8krn81nei ® |
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= Aijk1Bk1ei ® ei . |
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(1.151) |
Following the definitions above and by analogy with (1.53), (1.56) and (1.57), note the important rules
{A® B) : C = A(B : C) = |
(B : C)A |
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(1.152) |
A : (B ® C) = {A : B)C = C(A : B) |
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(1.153) |
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(A® B) : (C ® D) = (B: C){A ® |
D) = (A 0 D)(B: C) , |
(1.154) |
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and the property |
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A(B ® C)D = (AB) 0 (CD) . |
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(1.155) |
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1.3 Higher-order Tensors |
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EXAMPLE 1.5 Show eq. (l.152) which is the analogue of eq. (l.53).
Solution. Express tensor A, B, C in terms of their basis vectors according to ( 1.59).
With properties (1.148), (1.149), (1.21) and (1.61) we find that
(A® B): C = AijBktCmn(ei ® ej) ® (ek ® e,): (em® en)
=AijBklCmn(ek · em)(e, · en)(ei ® ej)
=AiiBktCmn8km81nei ® ei
= AiiBk1Ck,ei ® ei . |
( 1.156) |
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By using representation ( 1.59) |
and definition (1.93) we obtain AiiBk,Cklei ® ei |
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A{B : C) = (B : C)A, which is |
the desired result. |
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A further example for a fourth-order tensor is the tensor product D = A ® B, while
A and Bare second-order tensors with components Aii and Bii, respectively. In index notation we may write Dijkl = AiiBkt·
The unique transpose of a fourth-order tensor A denoted by AT is governed, by
analogy with (I .81 ), by the identity |
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B : AT : C = C : A : B = {A : B) : C |
(1.157) |
for all second-order tensors B and C. The properties of a fourth-order tensor |
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(A?)T =A , |
( 1.158) |
{A®B)T = B®A |
( 1.159) |
hold. From the identity ( 1.157) we deduce the index relation, namely (AT) iJkl = AkliJ.
We now define fourth-order unit tensors Il and Il so that
A=Il:A, AT=Il:A' ( 1.160)
for any second-order tensor A. These transformations assign to any A the tensor A itself, and its transpose, respectively. The fourth-order unit tensors may be represented by
n= 8ik8jlei |
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ek ® |
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= ei ® |
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(1.161) |
Il = c5ilc5jkei |
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ej 0 |
ek ® |
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= ei 0 |
ej 0 |
Cj 0 |
ei |
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(l.162) |
where (Il)iikl = 8ik8i1 and (Il)ijkt |
= 8itc5jk |
define the Cartesian components of Il and Il, |
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respectively. Note that ni= nT. |
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1 Introduction to Vectors and Tensors |
The deviatoric part of a second-order tensor A may be described by means of a fourth-order tensor. From relations ( 1.13 I), ( 1.132h and transformations ( 1. I 60) 1,
(1.152) we conclude that
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~I} : A , |
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devA =A - 3(1: A)I = {Il - |
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(1.163) |
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or with the definition of a fourth-order projection tensor P
devA = P: A , |
(1.164) |
The components of devA and A are related through the expression elev Aii = Pijkl Akt, with Pijkl
EXERCISE |
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I. Two fourth-order tensors §, W are given in the form |
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w = -(Il - |
JI) |
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(observe the similar structure to eq. ( 1.1I2)). Show that
1 {A +A·r ) =§:A , |
~(A - AT) = w : A , |
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which assign to any second-order tensor A its symmetric and skew parts, respectively.
1.4 Eigenvalues, Eigenvectors of Tensors
The scalars Ai characterize eigenvalues (or principal values) of a tensor A if there exist corresponding nonzero normalized eigenvectors Di (or principal directions or principal axes) of A, so that
(i = 1, 2, 3; no summation) . |
(1.165) |
To identify the eigenvectors of a tensor, we use subsequently a hat on the vector quantity concerned, for example, ft.
Thus, a set of homogeneous algebraic equations for the unknown eigenvalues ,.\,
i = 1, 2, 3, and the unknown eigenvectors ni, 'i = 1, 2, 3, is |
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(A |
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,.\·l)n· - |
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(i = 1, 2, 3; no summation) . |
(1.166) |
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1.4 Eigenvalues, Eigenvectors of Tensors |
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Eigenvalues characterize the physical nature of a tensor. They do not depend on coordinates. For a positive definite symmetric tensor A, all eigenvalues ,.\i are (real and) positive since, using (1.165), we have,.\ =Di·Ani > 0, i = 1, 2, 3. Moreover, the set of eigenvectors of a symmetric tensor A form a mutually orthogonal (orthonormal) basis {ni} (the proof of this statement is omitted).
Principal scalar invariants. For the system ( 1.166) to have solutions ni # o the determinant of the system must vanish. Thus,
(1.167)
where
( 1.168)
This requires that we solve a cubic equation in ,.\, usually written as
( 1.169)
called the characteristic polynomial (or equation) for A, the solutions of which are the eigenvalues Ai, i = 1, 2, 3.
Here, Ii(A), i = 1, 2, 3, are the so-called principal scalar invariants of A. In
terms of A and its principal values,.\, i |
= 1, 2, 3, these are given by |
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11 (A) = Aii = trA = .A1 + ,.\2 |
+ ,.\3 |
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(I. I70) |
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! 2 (A) |
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(trA) |
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tr(A |
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= trA |
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detA |
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= -(AiiAji - AiiAii) = - |
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= ,.\ 1.A2 + .A1.A3 + A2A3 |
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(1.171) |
/3(A) |
= EijkA1iA2jAak = detA = A1A2A3 . |
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(l.172) |
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Sometimes we shall use the alternative notation |
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l3(A) = |
IIIA |
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(1.173) |
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for the three principal invariants interchangeably. |
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A repeated application of tensor A to eq. ( 1.165) yields A0 ni = |
.Arni, i = 1, 2, 3, |
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for any positive integer a. Using this relation and ( 1.169) multiplied by Di, we obtain the well-known
(1.174)
It states that every (second-order) tensor A satisfies its own characteristic equation.
Spectral decomposition of a tensor. Any symmetric tensor A may be represented by its eigenvalues ,.\i, i = 1, 2, 3, and the corresponding eigenvectors of A forming an
26 1 Introduction to Vectors and Tensors
orthonormal basis {Di}· Using the unittensor, by analogy with ( l.65h, i.e. I= Di 0Di, and relations (1.58), (1.165) and (l.55) we obtain an expression which is known as the spectral decomposition (or spectral representation) of A, i.e.
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A =AI= (Aili) 0 Di = L ,.\iDi ®Di . |
(1.175) |
i=l
The components Aii of tensor A relative to a basis of principal directions follow with ( 1.59) by replacing ei with the three orthonormal basis vectors Di. With
eqs. (l.165) and (1.21) we obtain
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(j = 1, 2, 3; no summation) , |
( 1.176) |
which produces a diagonal matrix [A] in the form |
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[A)= |
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(l.177) |
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where the diagonal elements are the eigenvalues of A. This result may be obtained directly from the spectral decomposition (1.175) of A.
For a pair of equal roots, i.e. ,.\ 1 = ,.\2 # ,.\3 ( ,.\ 1 = ,.\2 == ,.\), with one unique eigenvector D3 associated with ,.\3 , we deduce from (1.175) that
(l.178)
where P~3, Pt3 denote projection tensors introduced in (1.125) and (1.126). In (1.178)
the eigenvector D |
is perpendicular to a plane spanned by the mutually orthogonal |
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3d~ |
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eigenvectors n1 an |
n2. |
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The third case to distinguish, namely for equal roots (,.\ 1 = |
,.\2 = ;\3 = ,.\), is |
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(l.179) |
where every direction is a principal direction and every set of mutually orthogonal basis denotes principal axes.
EXERCISES
1.Show that the quantities in eqs. ( 1.170)-( 1.172), i.e. / 1, / 2 , / 3 , are indeed principal invariants of a tensor A under an orthogonal transformation of coordinates.
1.4 Eigenvalues, Eigenvectors of Tensors |
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2. Let a tensor A be given by |
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A=cr(I-e1®e1)+f3(e1®e2+e2®e1) |
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where er, {j are scalars and e 17 e2 are orthogonal unit vectors. |
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(a) Show that the eigenvalues Ai, i = 1, 2, 3, of A are |
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,,\3 = |
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4 + /3 |
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(b)Derive the associated normalized eigenvectors Di, i = 1, 2, 3, which in matrix representation read
3.Let a non-symmetric tensor A be given by the tensor product e1 ® e:h with the property e 1 • e2 = 0. Determine the eigenvalues and eigenvectors of tensor A.
4.Let A be a given second-order tensor. Show that the invariants of its deviatoric part, according to ( 1.163), are
fdevA = 0 , |
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(l.180) |
/IdcvA=-?tr(devA) |
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IIIdcvA = det.(devA) = |
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3tr(devA · . |
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5. Use the relations (I. I 70)-( I. I 72), ( 1. I 80) |
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(I. I 63) 1 to show |
that I Ic1evA' |
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I I IdevA may be expressed in terms of IA, I IA and I I IA as |
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JldevA = JJA - |
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(1.181) |
l/lc1cvA =///A - - /A//A +~IA · |
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Hint: To derive eq. ( 1.181) apply the Cayley-Hamilto11 equation in the form of
(devA) 3 - lc1evA (devA) 2 + IldevA devA - Illc1evA I= 0.
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1 Introduction to Vectors and Tensors |
1.5Transformation Laws for Basis Vectors and Components
Consider two sets of mutually orthogonal basis vectors which share a common origin. They correspond to a 'new' and an 'old' (original) Cartesian coordinate system which we assume to be right-handed characterized by the two sets {ei} and {ei} of basis vectors, respectively. Hence, the new coordinate system may be obtained from the original one by a rotation of the basis vectors ei about their origin (see Figure 1.7). We find from ( 1.10) that the cosine of the angle between ei and ei is represented by Qii = cosO(ei, ei) = ei · ei. Note that the first index on Qii indicates the 'old' components whereas the second index holds for the 'new' components.
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O(c2, c:!)'\
'~
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'~ ' -
~C2
Figure 1.7 Rotation of mutually orthogonal basis vectors ei into ei.
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It is worthwhile to mention that vectors and tensors themselves remain invariant upon a change of basis - they are said to be independent of any coordinate system. However, their respective components do depend upon the coordinate system introduced, which is arbitrary. The components change their magnitudes by a rotation
(and/or reflection) of the basis vectors, but are independent of any translation (for the
characterization of a translation, see p. 62).
We now set up the transformation laws for various components of vectors and
tensors under a change of basis. Following the considerations above we have
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'-1'J3 |
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and |
ei= Q ei ' |
(1.182) |
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i. - |
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where Q denotes an orthogonal tensor (seep. 16), with components QiJ which are the same in either basis. The components describe the orientation of the two sets of basis
