Civil Engineering Reference
In-Depth Information
Solving the equations by Gaussian elimination (Appendix C), the nodal tempera-
tures are
T
1
=
95
.
11
◦
C
T
2
=
90
.
14
◦
C
T
3
=
85
.
14
◦
C
T
4
=
82
.
57
◦
C
and the heat flux at node 5 is calculated using the fifth equation
−
Aq
5
=
0
.
7332
T
3
−
5
.
8660
T
4
+
5
.
1327 (80)
to obtain
9
W/m
2
q
5
=
4001
.
which is observed to be in quite reasonable numerical agreement with the heat input at
node 1.
The solution for the nodal temperatures in this example is identical for both
the linear and quadratic interpolation functions. In fact, the solution we obtained
is the exact solution (Problem 7.1) represented by a linear temperature distribu-
tion in each half of the bar. It can be shown [1] that, if an exact solution exists
and the interpolation functions used in the finite element formulation include the
terms appearing in the exact solution, then the finite element solution corre-
sponds to the exact solution. In this example, the quadratic interpolation func-
tions include the linear terms in addition to the quadratic terms and thus capture
the exact, linear solution. The following example illustrates this feature in terms
of the field variable representation.
EXAMPLE 7.2
For the quadratic field variable representation
(
x
)
=
a
0
+
a
1
x
+
a
2
x
2
determine the explicit form of the coefficients
a
0
,
a
1
,
a
2
in terms of the nodal variables if
the three nodes are equally spaced. Then use the results of Example 7.1 to show
a
2
=
0
for that example.
■
Solution
Using the interpolation functions from Example 7.1, we can write the field variable
representation in terms of the dimensionless variable
s
as
(
s
)
=
(2
s
2
−
3
s
+
1)
1
+
4(
s
−
s
2
)
2
+
(2
s
2
−
s
)
3
Collecting coefficients of similar powers of
s
,
(
s
)
=
1
+
(4
2
−
3
1
−
3
)
s
+
(2
1
−
4
2
+
2
3
)
s
2
