CONTINUUM MECHANICS for ENGINEERS

which becomes (after dividing both sides by scalar dX)
ni Λ = xi,A NA or ˆnΛ = F ⋅ (4.10-8)
ˆ N
If we take the material derivative of this equation (using the symbolic notation
for convenience),
so that
By forming the inner product of this equation with ˆn we obtain
But nˆ⋅nˆ= 1 and so ˆ = 0, resulting in
˙ n⋅nˆ (4.10-9)
or Λ˙ / Λ = v n n
(4.10-10)
which represents the rate of stretching per unit stretch of the element that
originated in the direction of ˆN , and is in the direction of ˆn of the current
configuration. Note further that Eq 4.10-10 may be simplified since W is
skew-symmetric, which means that
and so
ˆ˙ nΛ+ nˆΛ˙= F˙⋅ Nˆ = L⋅F⋅ Nˆ = L⋅nˆΛ ˆ˙ n+ nˆΛ˙ / Λ = L⋅nˆ ˆ ˆ˙ n⋅ n+ nˆ⋅ nˆΛ˙ / Λ = nˆ⋅L⋅nˆ Λ˙ / Λ = nLn ˆ⋅ ⋅ˆ
i,j i j
L ijn in j = (D ij + W ij)n in j = D ijn in j
˙ Λ/ Λ = nˆ⋅D⋅nˆ or (4.10-11)
˙ Λ/ Λ = Dnn
ij i j
For example, for the element in the x1 direction, nˆ eˆ
and
⎡D
D D
Λ˙ ⎢
/ Λ =[ 100 , , ] ⎢
D D D
⎣
⎢D
D D
11 12 13
12 22 23
13 23 33
Likewise, for nˆ = eˆ
, and for , . Thus the
2 Λ˙ / Λ = D22 nˆ = eˆ3Λ˙
/ Λ = D33
diagonal elements of the rate of deformation tensor represent rates of extension,
or rates of stretching in the coordinate (spatial) directions.
= 1
⎤ ⎡1⎤
⎥ ⎢ ⎥
⎥ ⎢
0
⎥
= D
⎦
⎥
⎣
⎢0⎦
⎥
11