Biomedical Engineering Reference
In-Depth Information
•
I
1
=
0
(6.32)
Proof:
I
1
=
(
I
−
1
3
T
)
1
=
I1
−
1
3
T
11
11
1
=
1
−
1
=
0
3
•
I
I
=
I
or in general
I
n
=
I
for any
n
∈
IN
(6.33)
Proof:
1
3
1
⊗
1
1
3
1
⊗
1
I
I
=
(
I
−
)(
I
−
)
T
11
T
T
1
3
1
3
1
9
T
=
II
−
I11
−
11
I
+
11
1
1
I
11
3
T
T
2
3
T
1
3
T
=
I
−
11
+
11
1
3
T
=
I
−
11
This relationship expresses the fact that the deviatoric part of the stress cannot be
changed by any other deviatoric projection and that the spherical part is equal to
zero. A corresponding projection tensor can be derived for the spherical part of the
actual state of stress:
1
3
1
⊗
1
=
1
3
11
o
=
I
−
I
=
T
I
(6.34)
Thus, we can rewrite Eq. (6.29) based on the transformation tensors as
o
o
σ
+
I
σ
σ
=
σ
+
s
=
I
(6.35)
In the following, the strain energy per unit volume
w
will be derived in its canonical
form of two decoupled contributions, that is, its spherical and deviatoric parts:
2
σ
ε
1
2
σε
=
1
T
T
w
=
1
ε
1
+
s
(6.36)
m
m
By using Eq. (6.29), we can substitute
σ
m
and
s
and express the strain energy as
2
3
K
ε
m
1
e
=
2
9
K
ε
e
1
1
T
ε
m
1
+
2
G
L
−
1
T
2
m
T
L
−
1
w
=
e
+
2
G
e
(6.37)
In Eq. (6.37), the positive strain energy argument delimits the range of possible
values of Poisson's ratio to
5 (cf. Table 6.3).
In the last part of this section, we will provide relationships between the strain
vector and the stress vector. Equation (6.16) can be written in the matrix form as
−
1
≤
ν
≤
0
.
ε
=
D
σ
(6.38)
where the elastic compliance matrix
D
is given by the inverse of the elasticity
matrix
:
D
=
C
−
1
C
(6.39)
