Page 36 - Introduction to Continuum Mechanics

Part B Dyadic Product of Two Vectors 21

7 7
[T ] = [T] "
ff>
i.e., the matrix of T is the transpose of the matrix of T.
We also note that by Eq. (2B6.1), we have
T T
7
a-T b = b-(T ) a
7 r
Thus, b-Ta = b- (T ) a for any a and b, so that

It can also be established that (see Prob. 2B13)

That is, the transpose of a product of the tensors is equal to the product of transposed tensors
in reverse order. More generally,

287 Dyadic Product of Two Vectors
The dyadic product of vectors a and b, denoted by ab, is defined to be the transformation
which transforms an arbitrary vector c according to the rule:

Now, for any c, d, a and/3, we have, from the above definition:

(ab)(ac+£d) = a(b-(ac+0d)) = a((ab-c)+(0b-d)) = a(ab)c+0(ab)d
Thus, ab is a tensor. Letting W=ab, then the components of W are:

i.e.,

In matrix notation, Eq. (2B7.2a) is

In particular, the components of the dyadic product of the base vectors e, are:
"l 0 0] [b 1 0
[e^i] =000 , [0^2] = 000,.. .
00 0 00 0
Thus, it is clear that any tensor T can be expressed as: