Tensor Product

45 Since a tensor primary objective is to operate on vectors, the tensor product of two vectors provides a fundamental building block of second-order tensors and will be examined next.

46 The Tensor Product of two vectors u and v is a second order tensor u ® v which in turn operates on an arbitrary vector w as follows:

In other words when the tensor product u ® v operates on w (left hand side), the result (right hand side) is a vector that points along the direction of u, and has length equal to (v-w)||u||, or the original length of u times the dot (scalar) product of v and w.

47 Of particular interest is the tensor product of the base vectors ej ® ej. With three base vectors, we have a set of nine second order tensors which provide a suitable basis for expressing the components of a tensor. Again, we started with base vectors which themselves provide a basis for expressing any vector, and now the tensor product of base vectors in turn provides a formalism to express the components of a tensor.

48 The second order tensor T can be expressed in terms of its components Tij relative to the base tensors ei ® ej as follows:

Thus Tik is the ith component of Tek. We can thus define the tensor component as follows

49 Now we can see how the second order tensor T operates on any vector v by examining the components of the resulting vector Tv:

vk=i which when combined with Eq. 1.60-c yields

which is clearly a vector. The ¿th component of the vector Tv being

Lij vj

The identity tensor I leaves the vector unchanged Iv = v and is equal to

51 A simple example of a tensor and its operation on vectors is the projection tensor P which generates the projection of a vector v on the plane characterized by a normal n:

the action of P on v gives Pv = v — (v-n)n. To convince ourselves that the vector Pv lies on the plane, its dot product with n must be zero, accordingly Pv-n = v-n — (v-n)(n-n) = 0^/. Product of Two Second-Order Tensors

52 The product of two tensors is defined as

in any rectangular system. 53 The following axioms hold

Note again that some authors omit the dot. Finally, the operation is not commutative

0 -1

Post a comment