Civil Engineering Reference
In-Depth Information
With the exception of the mass transport term, Equation 7.65 is identical to
Equation 7.4. Consequently, if we apply Galerkin's finite element method, the
procedure and results are identical to those of Section 7.3, except for additional
stiffness matrix terms arising from mass transport. Rather than repeat the deriva-
tion of known terms, we develop only the additional terms. If Equation 7.65
is substituted into the residual equations for a two-node linear element (Equa-
tion 7.6), the additional terms are
x
2
mc
d
T
d
x
˙
N
i
d
x
i
=
1, 2
(7.66)
x
1
Substituting for
T
via Equation 7.5, this becomes
x
2
mc
d
N
1
d
x
T
2
N
i
d
x
d
N
2
d
x
˙
T
1
+
i
=
1, 2
(7.67)
x
1
Therefore, the additional stiffness matrix resulting from mass transport is
d
N
1
d
x
d
N
2
d
x
x
2
N
1
N
1
[
k
m
]
=˙
mc
d
x
(7.68)
N
2
d
N
1
d
x
N
2
d
N
2
d
x
x
1
EXAMPLE 7.8
Explicitly evaluate the stiffness matrix given by Equation 7.68 for the two-node element.
■
Solution
The interpolation functions are
x
L
N
1
=
1
−
x
L
N
2
=
and the required derivatives are
d
N
1
d
x
1
L
=−
1
L
Utilizing the change of variable
s
=
x
/
L
, Equation 7.68 becomes
d
N
2
d
x
=
=
1
2
1
2
−
(1
−
s
)(1
−
s
)
−
s
L
d
s
=
mc
−
11
−
11
−
1
mc
L
mc
2
[
k
m
]
=
s
1
2
1
2
−
0
and note that the matrix is not symmetric.
