Page 126 - Introduction to Continuum Mechanics

The Spin Tensor and the Angular Velocity Vector 111

and

(b) The material element chr'is currently in the ej-direction and therefore its rate of extension
r t le
=
is equal to DU = 0. Similarly, the rate of extension of dr ^ is equal to Z>22 ^- P° *
element rfx= (<&)n, where n = rrr (ej+2e2)
D
\ J

(c) From the characteristic equation

we determine the eigenvalues of the tensor D as K - 0, ± k/2, therefore, k/2 is the maximum
/vT\
and -k/2 is the minimum rate of extension. The eigenvectors HI = — (61+62) and
/vTi ve tne
«2 = -r- (61-62) g^ directions of the elements having the maximum and the minimum
i- '
stretching respectively.

3.14 The Spin Tensor and the Angular Velocity Vector
In section 2B.16 of Chapter 2, it was shown that an antisymmetric tensor W is equivalent to
a vector <o in the sense that for any vector a

The vector m is called the dual vector or axial vector of the tensor W and is related to the three
nonzero components of W by the relation:

Now, since the spin tensor W is an antisymmetric tensor (by definition, the antisymmetric
part of Vv), therefore