Page 233 - Intro to Tensor Calculus

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0
For Q a neighboring point of P which moves to Q when the material is in a state of strain, we have
from the figure 2.3-14 the following vectors:

position of P : y i , i =1, 2, 3
0
position of P : y i + u i (y 1 ,y 2 ,y 3 ), i =1, 2, 3
(2.3.52)
position of Q : y i +∆y i , i =1, 2, 3
0
position of Q : y i +∆y i + u i (y 1 +∆y 1 ,y 2 +∆y 2 ,y 3 +∆y 3 ), i =1, 2, 3
Employing our earlier one dimensional definition of strain, we define the strain associated with the point P
L − L 0
in the direction PQ as e = , where L 0 = PQ and L = P Q . To calculate the strain we need to first
0
0
L 0
2
2
calculate the distances L 0 and L. The quantities L and L are easily calculated by considering dot products
0
2
of vectors. For example, we have L =∆y i ∆y i , and the distance L = P Q is the magnitude of the vector
0
0
0
y i +∆y i + u i (y 1 +∆y 1 ,y 2 +∆y 2 ,y 3 +∆y 3 ) − (y i + u i (y 1 ,y 2 ,y 3 )), i =1, 2, 3.
Expanding the quantity u i (y 1 +∆y 1,y 2 +∆y 2 ,y 3 +∆y 3 ) in a Taylor series about the point P and neglecting
higher order terms of the expansion we find that
2
L =(∆y i + ∂u i ∆y m )(∆y i + ∂u i ∆y n ).
∂y m ∂y n
Expanding the terms in this expression produces the equation
2
L =∆y i ∆y i + ∂u i ∆y i ∆y n + ∂u i ∆y m ∆y i + ∂u i ∂u i ∆y m ∆y n .
∂y n ∂y m ∂y m ∂y n
2
2
Note that L and L 0 are very small and so we express the difference L − L in terms of the strain e. We can
0
write
2
2
2
L − L =(L + L 0)(L − L 0 )= (L − L 0 +2L 0)(L − L 0 )= (e +2)eL .
0 0
2
Now for e very small, and e negligible, the above equation produces the approximation
2
L − L 2
2 0 1 ∂u m ∂u n ∂u r ∂u r
eL ≈ = + + ∆y m ∆y n .
0
2 2 ∂y n ∂y m ∂y m ∂y n
The quantities

1 ∂u m ∂u n ∂u r ∂u r
e mn = + + (2.3.53)
2 ∂y n ∂y m ∂y m ∂y n
is called the Green strain tensor or Lagrangian strain tensor. To show that e ij is indeed a tensor, we consider
the transformation y i = ` ij y +b i , where ` ji ` ki = δ jk = ` ij ` ik . Note that from the derivative relation ∂y i = ` ij
j
∂y j
and the transformation equations u i = ` ij u j ,i =1, 2, 3 we can express the strain in the barred system of
coordinates. Performing the necessary calculations produces

1 ∂u i ∂u j ∂u r ∂u r
e ij = + +
2 ∂y ∂y ∂y ∂y
j i i j

1 ∂ ∂y n ∂ ∂y m ∂ ∂y k ∂ ∂y t
= (` ik u k ) + (` jk u k ) + (` rs u s ) (` rmu m )
2 ∂y n ∂y ∂y m ∂y ∂y k ∂y ∂y t ∂y
j i i j

1 ∂u m ∂u k ∂u s ∂u p
= ` im ` nj + ` jk ` mi + ` rs` rp ` ki ` tj
2 ∂y n ∂y m ∂y k ∂y t

1 ∂u m ∂u n ∂u s ∂u s
= + + ` im ` nj
2 ∂y n ∂y m ∂y m ∂y n
or e ij = e mn ` im ` nj .Consequently, the strain e ij transforms like a second order Cartesian tensor.