Biomedical Engineering Reference
In-Depth Information
and
T
=−
.
(10.63)
For an interpretation of the symmetrical tensor
D
we depart from the relations that
have been derived in Section
10.3
:
F
·
e
0
=
λ
e
(10.64)
and
2
=
e
0
·
F
T
λ
·
F
·
e
0
.
(10.65)
2
can be elaborated as follows:
The material time derivative of the equation for
λ
·
F
F
T
λλ
=
e
0
·
·
F
+
F
T
2
·
e
0
I
I
·
F
T
F
T
·
F
=
e
0
·
·
F
+
·
·
e
0
F
T
F
·
F
T
F
−
T
F
T
·
F
F
−
1
=
e
0
·
·
·
F
+
·
·
·
e
0
F
−
T
+
F
·
F
−
1
·
F
T
=
e
0
·
F
T
·
·
F
·
e
0
2
=
λ
e
·
(2
D
)
·
e
,
(10.66)
eventually resulting in the simple relation:
λ
λ
=
e
·
D
·
e
(
=
ln(
λ
)
)
.
(10.67)
This equation shows that the deformation velocity tensor
D
completely determines
the current rate of (logarithmic) strain for an arbitrary line segment in the current
state with a direction specified by
e
. The analogous equation in component form
is written as
λ
λ
=
∼
T
D
∼
.
(10.68)
The terms on the diagonal of the matrix
D
represent the rate of strain in the direc-
tions of the
x
-,
y
- and
z
-coordinates. The off-diagonal terms represent the rate of
shear.
For the interpretation of the skew symmetric spin tensor Eq. (
10.61
) can be
used directly. After all, it is clear that the contribution
x
to
d
˙
·
d
x
is always
perpendicular to
d
x
, because for all
d
x
:
T
d
x
·
·
d
x
=
0 because
=−
,
(10.69)
meaning that the contribution
x
has to be considered as the effect of a rotation.
For the material time derivative of the volume change factor
J
·
d
=
det (
F
)itcan
be derived (without proof):
J
J
tr(
F
F
−
1
)
∇·
=
·
=
J
tr(
L
)
=
J
v
=
J
tr(
D
) .
(10.70)
