Biomedical Engineering Reference
In-Depth Information
e
w
·
fd
.
N
el
N
el
(
∇
w
)
T
:
σ
d
=
e
w
·
(
σ
·
n
)
d
+
(18.24)
e
e
=
1
e
=
1
Clearly, the above
e
(.)
d
denotes the integral along those element boundaries
that coincide with the domain boundary
). As an example of how
to proceed based on Eq. (
18.24
)a
plane strain
problem is considered.
(
e
=
e
∩
Step 1
The integrand of the integral in the left-hand side of Eq. (
18.24
)is
elaborated first. If the double inner product of a symmetric tensor
A
and a
skew-symmetric tensor
B
is taken, then
A
:
B
=
0if
A
T
=
A
and
B
T
=−
B
.
(18.25)
This can easily be proved. Using the properties of the double inner product it
follows that
A
:
B
=
A
T
:
B
T
=−
A
:
B
.
(18.26)
This equality can only hold if
A
:
B
=
0.
The dyadic product (
∇
w
)
T
appearing in (
∇
w
)
T
:
σ
may be split into a
symmetric and a skew-symmetric part:
2
(
∇
w
)
T
2
(
∇
w
)
T
.
1
1
(
∇
w
)
T
(
∇
(
∇
=
w
)
+
−
w
)
−
(18.27)
Because of the symmetry of the Cauchy stress tensor it follows that
(
∇
w
)
T
:
1
2
(
∇
w
)
T
:
(
∇
. (18.28)
Notice that the expression
2
(
∇
w
)
+
(
∇
w
)
T
has a similar form to the infinites-
imal strain tensor
ε
defined as
σ
=
w
)
+
σ
(
∇
u
)
T
,
1
2
(
∇
ε
=
u
)
+
(18.29)
where
u
denotes the displacement field. This motivates the abbreviation:
2
(
∇
w
)
+
(
∇
w
)
T
.
1
w
ε
=
(18.30)
Consequently, the symmetry of the Cauchy stress tensor allows the double inner
product (
∇
w
)
T
:
σ
to be rewritten as
(
∇
w
)
T
:
w
:
σ
=
ε
σ
.
(18.31)
Step 2
To elaborate further, it is convenient to introduce a vector basis. Here, a
Cartesian vector basis
{
e
x
,
e
y
,
e
z
}
is chosen. In the plane strain case it is assumed
that
ε
zz
=
ε
xz
=
ε
yz
=
0 while the displacement field is written as
u
=
u
x
(
x
,
y
)
e
x
+
u
y
(
x
,
y
)
e
y
.
(18.32)
