Civil Engineering Reference
In-Depth Information
at the boundaries (the “reactions”) are time dependent and can be computed at each time
step, as will be explained. Hence, the gradient “force vector” is
q
1
A
0
0
0
−
q
5
A
{
F
g
}=
Having taken care of the boundary conditions, we now consider the initial conditions
and examine the totality of the conditions on the solution procedure. It should be
clear that, since we have the temperature of two nodes specified, the desired solution
should provide the temperatures of the other three nodes and, therefore, should be a
3
×
3
system. The reduction to the
3
×
3
system is accomplished via the following
observations:
1.
If
T
1
=
80
◦
C
=
constant, then
T
1
=
0
.
2.
If
T
5
=
30
◦
C
=
constant, then
T
5
=
0
.
The equations can be modified accordingly. In this example, the general equations
become
0
T
2
T
3
T
4
0
80
T
2
T
3
T
4
30
q
1
A
0
0
0
[
C
]
+
[
K
]
=
−
q
5
A
Consequently, the first and fifth equations become
1
.
1451
T
2
+
0
.
9408 (80)
−
0
.
9408
T
2
=
q
1
A
1
.
1451
T
4
−
0
.
9408
T
4
+
0
.
9408 (30)
=−
q
5
A
respectively. The three remaining equations are then written as
+
T
2
T
3
T
4
4
.
5804
1
.
1451
1
.
8816
−
0
.
9408
0
0
T
2
T
3
T
4
.
.
.
−
.
.
−
.
1
1451
4
5804
1
1451
0
9408
1
8816
0
9408
0
1
.
1451
4
.
5804
0
−
0
.
9408
1
.
8816
75
.
264
0
=
28
.
224
For this example, the capacitance matrix is inverted (using a spreadsheet program) to
obtain
0
.
2339
−
0
.
0624
0
.
0156
[
C
]
−
1
=
−
.
.
−
.
0
0624
0
2495
0
0624
0
.
0156
−
0
.
0624
0
.
2339
