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Similar expressions apply for the sum, product, dyadic product, and trans-
13
2.2 VECTOR AND TENSOR ALGEBRA



pose of tensors as,




[S 1 + S 2 ] = [S 1 ] + [S 2 ] (2.40a)
[S 1 S 2 ] = [S 1 ][S 2 ] (2.40b)
[S T ] = [S]T (2.40c)
[u — v] = [u][v]T (2.40d)




The association between second-order tensors and 3 — 3 matrices is thus
identical to that between vectors and column matrices. Similarly, the dis-
tinction between the tensor and its components is only useful when more
than one basis is being considered at the same time. Under such circum-
stances, a single tensor S is expressed by two sets of components, [S] or Sij
in the base ei — ej and [S] or Sij in the base ei — ej as,




3 3
S= Sij ei — ej = Sij ei — ej (2.41)
±,β=1 ±,β=1




Introducing the relationship between the new and old bases given by the
angle cosines as shown in Equations (2.11a“b) and after simple algebra, a
relationship between both sets of tensor components emerges as,



3
T
[S] = [Q] [S][Q] or Sij = Qki Skl Qlj (2.42)
k,l=1




Using this equation and assuming some elementary knowledge of a two-
dimensional state of stresses, the invariant nature of the stress tensor is
shown in Example 2.4.
14 MATHEMATICAL PRELIMINARIES




EXAMPLE 2.4: Tensor invariance
0 t
x2 x2



0
x1
n
20 “
x1



2.232 3.079
[t] = ; [t] =
2.866 1.932
0.866 0.985
[n] = ; [n] =
0.500 0.174
21 2.878 1.409
[σ] = ; [σ] =
14 1.409 3.125


Consider a two-dimensional example of a stress tensor σ having Carte-
sian components [σ] and an associated unit normal vector n having
components [n]. The traction force t having components [t] on the
surface normal to n is,

2.232 21 0.866
[t] = [σ] [n] or =
2.866 14 0.500

Now rotate the Cartesian axes anticlockwise through an angle ± =
20—¦ in which the same stress tensor σ now has components [σ] and
the normal vector n components [n] . The traction vector t having
components [t] on the surface normal to n is,

3.079 2.878 1.409 0.985
[t] = [σ] [n] or =
1.932 1.409 3.125 0.174

The vectors n and t have remained unchanged even though their com-
ponents have changed. Likewise the stress tensor σ has remained un-
changed in that it is that which operates on n to give t, even though
its components have changed. Hence, the general expression t = σn
can justi¬ably be written.
15
2.2 VECTOR AND TENSOR ALGEBRA




EXAMPLE 2.4 (cont.)
Noting that from Equation (2.13b) [t] = [Q]T [t] and [n] = [Q]T [n] it
is easy to show that,
cos ± ’ sin ±
[σ] = [Q]T [σ] [Q] ; [Q] =
sin ± cos ±



2.2.3 VECTOR AND TENSOR INVARIANTS
The above sections have shown that when di¬erent bases are used, vectors
and tensors manifest themselves via di¬erent sets of components. An inter-
esting exception to this rule is the case of the identity tensor or its multiples
±I. Using Equation (2.42) it is easy to show that the components of these
tensors remain unchanged by the rotation of the axes as,

[±I] = [Q]T [±I][Q]
= ±[QT Q]
= [±I] (2.43)

Tensors that satisfy this property are said to be isotropic and are used in
continuum mechanics to describe materials that exhibit identical properties
in all directions.
In general, however, the components of vectors and second-order ten-
sors will change when the axes are rotated. Nevertheless, certain intrinsic
magnitudes associated with them will remain invariant under such trans-
formations. For example, the scalar product between two vectors u and v,
de¬ned in Section 2.2.1 as,
3
ui vi = [u]T [v]
u·v = (2.44)
±=1

remains invariant when the components change as a result of a rotation Q
of the axes. This is easily proved with the help of Equation (2.13a) and the
orthogonality of Q as,

u · v = [u]T [v]
= ([Q][u] )T [Q][v]
= ([u] )T [QT Q][v]
= ([u] )T [v] (2.45)
16 MATHEMATICAL PRELIMINARIES



Consequently, the modulus or magnitude of a vector, de¬ned as,

||u|| = u · u (2.46)
is an invariant, that is, an intrinsic physical property of the vector.
Insofar as tensors can be expressed in terms of dyadic products of base
vectors it is reasonable to enquire if invariant quantities can be associated
with tensors. This is indeed the case, and the ¬rst of these magnitudes,
denoted as IS , is given by the trace which is de¬ned as the sum of the
diagonal components of the tensor S as,
3
IS = trS = Sii (2.47)
±=1

The invariance of this magnitude can be easily checked using Equation (2.41)
in an indicial form and recalling the orthogonality of Q given by Equa-
tion (2.26), to give,
3 3
Sii = Qki Skl Qli
±=1 i,k,l=1
3
= Skl δkl
k,l=1
3
= Skk (2.48)
k=1

Some useful and simple-to-prove properties of the trace are,
tr(u — v) = u · v (2.49a)
trS T = trS (2.49b)
trS 1 S 2 = trS 2 S 1 (2.49c)

EXAMPLE 2.5: Proof of Equation (2.49a)
Property (2.49a) follows from the components of the dyadic product
discussed in Example 2.3 and the de¬nition of the scalar product as,
3
tr(u — v) = (u — v)ii
±=1
3
= ui vi
±=1
= u·v
17
2.2 VECTOR AND TENSOR ALGEBRA



Analogous to the scalar product of vectors, the double product or double
contraction of two tensors A and B is an invariant magnitude de¬ned in
terms of the trace as,
A : B = tr(AT B) (2.50)
which recalling the properties of the trace can be variously written as,
3
T T T T
A : B = tr(A B) = tr(BA ) = tr(B A) = tr(AB ) = Aij Bij
i,j=1
(2.51)
Further useful properties of the double contraction are,
trS = I : S (2.52a)
S : (u — v) = u · Sv (2.52b)
(u — v) : (x — y) = (u · x)(v · y) (2.52c)
ST = S and W T = ’W
S : W = 0 if (2.52d)
A second independent invariant of a tensor S can now be de¬ned as* ,
IIS = S : S (2.53)
A third and ¬nal invariant of a second-order tensor is provided by its
determinant, which is simply de¬ned as the determinant of the matrix of
components as,
IIIS = det S = det[S] (2.54)
Proof of the invariance of the determinant follows easily from Equation (2.42)
and the orthogonality of the transformation tensor Q as,
det [S] = det([Q]T [S][Q])
= det[Q]T det[S] det[Q]
= det[QT Q] det[S]
= det[S] (2.55)
An alternative way in which invariant magnitudes of second-order ten-
sors can be explored is by studying the existence of eigenvectors and eigen-
values. For a given tensor S, a vector n is said to be an eigenvector with

* In other texts the following alternative de¬nition is frequently used,
…2 ć
IIS = 1 IS ’ S : S
2
18 MATHEMATICAL PRELIMINARIES



an associated eigenvalue » if,
Sn = »n (2.56)
Linear algebra theory shows that the eigenvalues » that satisfy the above
equation are the roots of a third-degree polynomial obtained as,
det(S ’ »I) = 0 (2.57)
In general, however, the roots of such an equation can be imaginary and
hence of little practical use. An important exception is the case of sym-
metric tensors, for which it is possible to prove not only that there exist
three real roots of Equation (2.57), »1 , »2 , and »3 , but also that the three
corresponding eigenvectors n1 , n2 , and n3 are orthogonal, that is,
Sni = »i ni ; i = 1, 2, 3 (2.58a)
ni · nj = δij ; i, j = 1, 2, 3 (2.58b)
Consequently, the above unit eigenvectors can be used as an alternative
Cartesian base in which Equations (2.33) and (2.58a) show that the o¬-
diagonal components of S vanish whereas the three diagonal components
are precisely the eigenvalues »i . Accordingly, the symmetric tensor S can
be conveniently written as,
3
S= »i ni — ni (2.59)
i=1

The eigenvalues of a symmetric tensor are independent of the Cartesian axes
chosen and therefore constitute an alternative set of invariants. Finally, rela-
tionships between the invariants IS , IIS , and IIIS and the eigenvalues can be
derived by applying De¬nitions (2.47), (2.53), and (2.54) to Equation (2.59)
to give,
IS = »1 + »2 + »3 (2.60a)

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