CONTINUUM MECHANICS for ENGINEERS

Solution
By definition of symmetric tensor A ij and skew-symmetric tensor S ij, we have
S ij A ij = –S ji A ji = –S mn A mn = –S ij A ij
where the last two steps are the result of all indices being dummy indices.
Therefore, 2S ijA ij = 0, or S ij A ij = 0.
One of the most important advantages of the indicial notation is the compactness
it provides in expressing equations in three dimensions. A brief
listing of typical equations of continuum mechanics is presented below to
illustrate this feature.
1. φ = SijTij – SiiTjj (1 equation, 18 terms on RHS)
2. ti = Qijnj (3 equations, 3 terms on RHS of each)
3. Tij = λδ E + 2µ
E (9 equations, 4 terms on RHS of each)
Example 2.3-2
By direct expansion of the expression vi = εijkWjk determine the components
of the vector vi in terms of the components of the tensor Wjk. Solution
By summing first on j and then on k and then omitting the zero terms, we
find that
Therefore,
ij kk ij
v = ε W + ε W + ε W
i i1k 1k i2k 2k i3k 3k
= ε W + ε W + ε W + ε W + ε W + ε W
i12 12 i13 13 i21 21 i23 23 i31 31 i32
32
v = ε W + ε W = W −W
1 123 23 132 32 23 32
v = ε W + ε W = W −W
2 213 13 231 31 31 13
v = ε W + ε
W = W −W
3 312 12 321 21 12 21
Note that if the tensor W jk were symmetric, the vector v i would be a null
(zero) vector.