CONTINUUM MECHANICS for ENGINEERS

In a similar fashion, from dX (1) ⋅ dX (2) = dX (1) dX (2) cos Θ, where Θ is the angle
between dX (1) and dX (2) , we obtain from Eqs 4.8-6 and 4.8-7,
cos Θ =
= λ λ nˆ c nˆ
(4.8-15)
( nˆ ) ( nˆ)
1⋅ ⋅ 2
which gives the original angle between elements in the directions and
of the current configuration.
ˆn 2
Example 4.8-1
A homogeneous deformation is given by the mapping equations, x1 = X1 –
X2 + X3, x2 = X 2 – X3 + X1, and x3 = X3 – X1 + X2 . Determine (a) the stretch
ratio in the direction of Nˆ ˆ ˆ , and (b) the angle θ12 in the
1 = ( I1+ I2)
/ 2
deformed configuration between elements that were originally in the directions
of and .
ˆ N ˆ ˆ
1 N2 = I2
Solution
For the given deformation (as the student should verify),
(a) Therefore, from Eq 4.8-4,
and
[ FiA]= 2
Λ
( Nˆ 1)
Λ ( ˆ )
N 1
( 1)
( 2)
dX
dX
nˆ c nˆ
1⋅ ⋅ 2
⋅ =
( 1)
( 2)
dX dX nˆ ⋅c⋅nˆ nˆ ⋅c⋅nˆ 1 2
⎡ 1 −1
1⎤
⎢
⎥
⎢
1 1 −1
⎥
⎣
⎢−1
1 1⎦
⎥
( )
and
1 1 2 2
[ CAB]= ⎡ 3 −1 −1⎤
⎢
⎥
⎢
−1 3 −1
⎥
⎣
⎢−1
−1
3⎦
⎥
⎡ 3 −1 −1⎤⎡1/
2⎤
⎢
⎥ ⎢ ⎥
= [ 1/ 2, 1/ 2, 0]
⎢
−1 3 −1
⎥ ⎢1/
2⎥=
2
⎣
⎢−1
−1
3⎦
⎥ ⎢
⎣
0 ⎥
⎦
=
2
(b) For ˆ ˆ 2
N2 = I2
, Λ so that from Eq 4.8-13, using the
( ˆ = Iˆ ˆ
I ) 2⋅C⋅ I2 = C22
=
3
2
result in part (a),
ˆn 1