Biomedical Engineering Reference
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and the Lagrangian strain tensor component
E
III
is related to
d
I
,
d
II
and to the
change in the right angle
f
between the filaments lying in the
I
and
II
directions by
ð
1
2
E
III
¼
þ d
I
Þð
þ d
II
Þ
f:
1
1
sin
(11.25)
are small so that the square of each may
be neglected, the traditional geometric interpretation for small strains,
Thus, unless the extensions
d
I
,
d
II
, and
f
E
III
2
;
E
II
d
I
;
(11.26)
namely that
E
II
is the extension in the
I
direction and
E
III
is one-half the change in
an angle that was originally a right angle, is not accurate. Eqs. (
11.23
) and (
11.24
)
representing
E
II
and
E
III
in terms of
show that the geometric
interpretation of the finite strain tensors in terms of extensions and changes in
right angles is possible, but is awkward and not very useful due to its nonlinear
nature.
Example 11.4.1
Compute the Lagrangian strain tensor
E
and the Eulerian strain tensor
e
for the
motion given by (2.12). Determine the range of values of
t
for which these two
strain measures coincide in this special motion.
d
I
,
d
II
, and
f
Solution
: The deformation gradient and inverse deformation gradient for the motion
(2.12) are given in Example 2.1.1,
2
4
3
5
and
F
1
2
4
3
5
:
1
þ
t
t
0
1
þ
t
t
0
1
F
¼
t
1
þ
t
0
¼
t
1
þ
t
0
1
þ
2
t
0
0
1
0
0
1
þ
2
t
These values for
F
and
F
1
can then be substituted into (
11.15
) for the Lagrang-
ian strain tensor
E
and the Eulerian strain tensor
e
, thus
2
4
3
5
and
e
2
4
3
5
:
110
110
000
110
110
000
t
ð
1
þ
t
Þ
E
¼
t
ð
1
þ
t
Þ
¼
2
ð
1
þ
2
t
Þ
These expressions are, of course, valid for large strains. If we restrict ourselves
to small strains, then the two strain tensors must coincide. Since each component of
E
is proportional to
t
t
2
and each component of
e
is proportional to
þ
t
ð
1
þ
t
Þ
3
t
2
8
t
3
2
¼
t
þ
ð
1
þ
2
t
Þ
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