Biomedical Engineering Reference
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where
p
þ
p
1
2
2
p ;
1
2
2
p :
cos 15
¼
sin 15
¼
A
¼
B
¼
The fact that
U
and
V
given above are the square roots of
U
2
and
V
2
, respec-
tively, many be verified simply by squaring
U
and
V
. The orthogonal tensor
R
may
be computed in several ways,
2
3
AB
0
4
5
:
V
1
U
1
R
¼
F
¼
F
¼
BA
0
001
The polar right and left decompositions of the given deformation gradient tensor
are then given by
2
4
3
5
2
4
p
A
p
B
3
5
¼
2
4
3
p
10
020
001
3
5
;
AB
0
0
p
B
2
A
F ¼ R U ¼
BA
0
001
þ
B
0
0
0
1
and
2
4
3
5
2
4
3
5
¼
2
4
3
p
10
020
001
3
5
;
2
A
2
B
0
2
B
2
A
0
001
AB
0
F
¼
V
R
¼
BA
0
001
respectively.
Example Problem 11.3.2
Develop a geometric interpretation of the deformation gradient tensor
F
of Example
Problem 11.3.1 by considering it as representing a homogeneous deformation
x
¼
F·X
acting on a unit square with vertices (0, 0), (1, 0), (1, 1), (0, 1).
Solution
: The scalar equations equivalent to the homogeneous deformation
x
¼
F·X
where
F
is given in the statement of Example 11.3.1 are
x
1
¼
p
3
X
I
þ
X
II
;
x
2
¼
2
X
II
;
x
3
¼
X
III
:
This unit square is deformed by
F
into a parallelogramwith corners at the points (0,
0), (
√
3, 0), (1
þ √
3, 2), and (1, 2) as illustrated in Fig.
11.10
. Consider the left
decomposition,
F
R
, first. In this decomposition the rotation
R
is applied first,
then the left deformation or left stretch,
V
. The effect of
R
on the unit square is a
clockwise rotation of 15
; this is illustrated in Fig.
11.11
. Following this rotation of the
unit square, there is a left stretch
V
that carries the rotated square into the deformed
shape illustrated in Fig.
11.9
. The other decomposition choice
F
¼
V
¼
R
U
reverses the
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