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)