Civil Engineering Reference
In-Depth Information
of
r
and
s
. Therefore, the integrals can be evaluated exactly by using two Gauss points
in
r
and
s
. Per Table 6.1, the required Gauss points and weighting factors are
r
i
,
s
j
=
±
0
.
57735
and
W
i
,
W
j
=
1
.
0
,
i
,
j
=
1, 2
. Using the numerical procedure for
k
11
, we write
1
1
1
1
=
k
x
t
b
a
1
16
(
s
−
d
r
d
s
+
k
y
t
b
a
1
16
(
r
−
1)
2
1)
2
k
11
d
r
d
s
−
1
−
1
−
1
−
1
1
1
16
(
r
−
1)
2
(
s
−
1)
2
+
2
hab
d
r
d
s
−
1
2
2
2
2
=
k
x
t
b
a
1
16
W
i
W
j
(
s
j
−
+
k
y
t
a
b
1
16
W
i
W
j
(
r
i
−
1)
2
1)
2
i
=
j
=
i
=
j
=
1
1
1
1
2
2
1
16
W
i
W
j
(1
−
r
i
)
2
(1
−
s
j
)
2
+
2
hab
i
=
j
=
1
1
and, using the specified integration points and weighting factors, this evaluates to
1
3
1
3
2
hab
4
9
=
k
x
t
b
a
+
k
y
t
a
b
k
11
+
It is extremely important to note that the result expressed in the preceding equation is the
correct value of
k
11
for
any
rectangular element used for the two-dimensional heat con-
duction analysis discussed in this section. The integrations need not be repeated for each
element; only the geometric quantities and the conductance values need be substituted to
obtain the value. Indeed, if we substitute the values for this example, we obtain
=
0
.
6327
Btu/(hr-
◦
F
)
k
11
as per the analytical integration procedure.
Proceeding with the Gaussian integration procedure (calculation of some of these
terms are to be evaluated as end-of-chapter problems), we find
=
0
.
6327
Btu/(hr-
◦
F
)
k
11
=
k
22
=
k
33
=
k
44
Why are these values equal?
The off-diagonal terms (again using the numerical integration procedure) are calcu-
lated as
k
12
=−
0
.
1003
k
13
=−
0
.
2585
k
14
=−
0
.
1003
k
23
=−
0
.
1003
k
24
=−
0
.
2585
k
34
=−
0
.
1003
