Holpzaphel_-_Nonlinear-Solid-Mechanics-a-Contin

vectors relative to each other. In particular, Q rotates the basis vectors ei into ei, while QT rotates ei back to ei. Using eqs. ( 1.59), ( 1.53)2, ( 1.21) and rule ( 1.61 ), we find that

which is the analogue of eq. ( 1.63)4 •

By comparing the equations above we may extract the orthogonality condition of the cosines, characterized by QTQ = QQT = I. Equivalently, expressed in index or in matrix notation

where [Q] contains the collection of the components QiJ. It is an orthogonal matrix which is referred to as the transformation matrix. Note that [Q]T = [QT] (compare

with Exercise 3 on p.19). In order to maintain the right-handedness of the basis vectors we have admitted only rotations of the basis vectors, consequently det[Q] = +1.

Vectorial transformation law. We consider any vector u resolved along the two sets {ei} and {ei} of basis vectors, i.e.

We assume that the relation between the basis vectors ei and ej is known by the

cosines QiJ. Combining ( 1.185) with (1.182) 1 and using (1.183)1 and ( 1. 186), we obtain the vectorial transformation law for the Cartesian components of the vector u,

1.e.

(given here in both index and matrix notation). In an analogous manner, we find that

or [u] = [Q][ii] . (I. I 88)

These equations determine the relationship between the components of a vector associated with the (old) basis {ei} and the components of the same vector associated with another (new) basis {ei}·

It is important to emphasize that the equations [ii] = [Q]'I'[u]and [u] = [Q][ii] are not identical to ii = QTu and u = Qii, respectively. In ( 1.187)2 and ( 1. l 88h, [ii] and

[u] are column matrices characterizing components of the same vector in two different coordinate systems, whereas ii and u are different vectors.

Tensorial transformation law. To determine the transformation laws for the Cartesian components of any second-order tensor A, we describe its components along the

Combining (l.189) with (1.182) 1 and using (1.183) 1 and (l.190), then the (rectangular Cartesian) components Aii and Aii are related via the so-called tensorial trans-

formation law

Aij = ei. Aej = (Qkiek). A(Qmjem)

Transformation (A] = [Q]T[A](Q] relates the different matrices [A] and (A], which have the components of the same tensor A. In an analogous manner, we find that

[Q][A](Q]T differ from the tensor equations A = QTAQ and A = QAQT, relating two different tensors, namely A and A.

Finally, the 3n components Aiih···in of a tensor of order n (with n indices j 1, h,

... , in) transform as

(1.193)

This tensorial transformation law relates the different components Ai1i~...in (along the directions e1, e2 , e3 ) and Ai•h···jn (along the directions e1, e2 , e:i) of the same tensor of order n.

A tensor A is said to be isotropic if its components are the same under arbitrary rotations of the basis vectors. The requirement is deduced frdm

Note that all scalars as well as zero tensors and unit tensors of all orders are isotropic tensors.

EXAMPLE 1.6 Show that the unit tensor, as given in ( 1.65), is an isotropic tensor of order two.

Solution. In accord with manipulation (1.191 )4 we can show that the components of the unit tensor I are 8ii in any coordinate system. Namely, using ( 1.61) and ( 1.184)

we obtain

(1.195)

which means that the components §ii do not change upon a rotation or reflection of axes. •

By multiplying both sides of oii = 8ii with a scalar we can see that every scalar multiple of I is an isotropic tensor. Consequently, any spherical tensor in the form of

A == a:I or (1.196)

is an isotropic tensor, where a denotes a scalar.

The permutation tensor ( 1.143) is an isotropic tensor of order three with respect to rotation of coordinate axes (proper orthogonal transformations). With respect to

reflection it emerges that the permutation tensor is not isotropic.

It can be shown that the fourth-order unit tensors lI, Il and the tensor I ® I with components (I ® I)iikl = 8ii8k1 are also isotropic tensors. The most general isotropic tensor of order four is of the form

EXERCISES

1.Let a new right-handed Cartesian coordinate system be represented by the set {ei} of basis vectors with transformation law e2 = -sin0e1 + cos/Je2 , e3 = e3 •

The origin of the new coordinate system coincides with the old origin.

(a)Find e1 in terms of the old set {ei} of basis vectors.

(b)Find the orthogonal matrix [Q] and express the new coordinates in terms of the old one.

(c)Express the vector u = -Ge 1 - 3e2 + e:i in tem1s of the new set {ei} of basis vectors.

2.Considering eqs. ( 1.189) and ( 1.190), show that Aii = Aii.

3.Assume that A is an isotropic tensor of order two, for example, nl. Show that the deviatoric part of A is always the zero tensor of order two, i.e.

(In general, equations within exercises are not numbered. However, particular results may be given equation numbers, for later reference.)

1.6 General Bases

There are more general ways to introduce the notion of a vector and a tensor. A more general theory assumes the existence of a vector space and its so-called dual space.

Thereby, a vector and a tensor of order n are introduced through a linear and a mul-

tilinear jimction, respectively. Within this broader framework we must distinguish

between the scalar product (which is only defined between elements of a vector space and between elements of its dual) and the inner product. For a detailed explanation of the underlying theory the interested reader is referred to, for example, BOWEN and WANG [1976b] and OGDEN [1997, Section 1.4.3].

However, for the purpose we pursue in this text it is sufficiently general to work with a theory which assumes the existence of Euclidean vector spaces and which identifies one vector space with its dual so that the scalar product is identified with the inner

product.

Thus far we have worked with Cartesian tensors which are resolved along a set of orthonormal basis vectors. However, tensor quantities can be projected onto more general bases. The aim of this section is to present the fundamental rules and standard

results for these projections onto the basis of the aforementioned identification.

General basis vectors. We now resolve vectors and tensors along a fixed ~et {gJie{i,2,3 } of basis vectors. The elements g 1, g2 , g3 of this basis are called covariant basis vectors. They are assumed to be non-zero and non-parallel but they need not necessarily be orthogonal to each other or of unit length. The general basis vectors are linearly independent, requiring (g1 x g2 ) • g:1 =f 0 (the vectors do not lie in a plane).

We now introduce the so-called reciprocal (or dual) basis of {g; }, i.e. the fixed set {gi}ie{i,2 ,a} (note that for a more general theory which assumes the existence of a vector space and its dual the reciprocal and dual bases are in different spaces). The elements g 1, g2 , g:1 of the reciprocal basis are called contravariant basis vectors or reciprocal basis vectors and have superscripts rather than subscripts. The two bases

{g;} and {gi} satisfy the condition

Solution. We tentatively write

(l.203)

where aik are unknown quantities. Taking the dot product of (1.203) with gi and using relation ( 1.199), we obtain

( 1.204) This means that aik is only multiplied with the factor 1 if j = k otherwise it is multi-

Taking the dot product of gi with gk and using (1.202) we obtain gi ·gk = gii 9klgi ·g1.

The distinction between subscript and superscript is essential for arbitrary bases. It is important to note that afree (or live) index, which is not summed over, such as 1: and j in eq. (1.199), must be either a subscript or a superscript on both sides of an equati.Qn. In the following we will sometimes use the term 'index' to denote either a subscript or a superscript. Only if the basis is orthonormal may we write gi == gi and then no distinction between upper and lower indices is necessary. For this case we write all indices as subscripts, as in previous sections where we used a Cartesian basis.

Covariant and contravariant components of a vector. Any vector u may be

represented uniquely by a linear combination of the contravariant or covariant basis vectors. We write

(1.207)

where 'lli are the covariant components of u (with respect to the general basis {gi} ), while 'Ui are the contravariant components of u (with respect to the general basis

{gi}). It is only in this theory, in which the reciprocal basis is identified with the dual basis, that the vector u has both covariant components ui and contravariant components ui. We shall call a vector with covariant components a covariant vector and a vector with contravariant components a contravariant vector.

By analogy with eq. (1.22)2, the Kronecker delta o} has the useful (replacement) properties

in which the index ion u is replaced by .i. Note that summation can only take place over a dummy (or summation) index if one index is a subscript and the other is a superscript, such as index i in eq. (1.207). Hence, the .i-th covariant and contravariant components of any vector u are the projections of u onto the bases {gi} and {gi}, respectively. Taking the dot product of (I .207)1 with gi (and the dot product of ( l.207)2 with gi), we obtain, by means of properties (l.13), (1.199), (1.208) 1 (and (l.208h), the useful relations

which should be compared with eq. (I .23) for the case of a Cartesian basis.

The contravariant components of any vector u may also be expressed in tenns of its covariant components, and vice versa. Thus, taking the dot product of u = u1gi and u == uigi with gi and gi, respectively, on both sides gives with definitions (1.200) and eqs. (1.21 Oh and ( l .209)J, i.e.

which are properties analogous to those given in ( 1.202). Note that the new vector components farmed depend on the original one. As indicated in eqs. (1.211 ), the set of six quantities gii (or .<Jii) can be used as a useful operator to raise or lower an index on vector components. This process of raising or lowering indices can similarly be applied to tensor components (see Exercise 5 on p. 39).

Consider the two vectors u and v which may be represented by (1.207). Their dot product reads

above, any second-order tensor A may be represented with respect to a basis consisting

as defined in eqs. ( 1.200) and ( 1.199), are actually the components of the second-order unit tensor I. Verify the representation

where I = gi.igi 0 g1 is the unit tensor in the representation ( 1.224)1 and .r/i == (I)ii are identified as the six cmztravariant components of I with respect to the basis {gi}.

In an analogous manner, starting from u = u.;gi, we may obtain representation and the six covariant components 9ii = (I)ij of I with respect to the reciprocal basis {gi}. Since the metric coefficients .'lii and gii are components of I, we also call I

the Euclidean metric tensor.

In order to show the mixed component fonn we apply ui = 8jui to u = 11.ig;. Then, with eq.( l .21 Oh and rule ( 1.53 ), we obtain

( 1.226)

A generalization to higher-order tensors is a straightforward task. A general tensor of order n,, having n indices i 1, i'2, ... , in, may be represented in the form

are involved. Now, a general tensor of order n may have mixed components of various

kinds according to, for example,