Civil Engineering Reference
In-Depth Information
and these expressions simplify to
4
a
(1
s
)
T
(
e
)
2
T
(
e
1
+
s
)
T
(
e
)
3
T
(
e
4
k
x
q
(
e
)
x
=−
−
−
(1
+
−
(7.50)
4
b
(1
r
)
T
(
e
)
4
T
(
e
1
+
r
)
T
(
e
)
3
T
(
e
2
k
y
q
(
e
)
y
=−
−
−
(1
+
−
The flux components, therefore the temperature gradients, vary linearly in a four-
node rectangular element. However, recall that, for a
C
0
formulation, the gradi-
ents are not, in general, continuous across element boundaries. Consequently, the
element flux components associated with an individual element are customarily
taken to be the values calculated at the centroid of the element. For the rectangu-
lar element, the centroid is located at
(
r
,
s
)
=
(0, 0)
, so the centroidal values are
simply
4
a
T
(
e
)
T
(
e
4
k
x
q
(
e
)
x
T
(
e
)
3
T
(
e
)
1
=−
+
−
−
2
(7.51)
4
b
T
(
e
)
T
(
e
2
k
y
q
(
e
)
y
T
(
e
)
4
T
(
e
)
1
=−
+
−
−
3
The centroidal values calculated per Equation 7.51, in general, are quite accurate
for a fine mesh of elements. Some finite element software packages compute the
values at the integration points (the Gauss points) and average those values for
an element value to be applied at the element centroid. In either case, the com-
puted values are needed to determine solution convergence and should be
checked at every stage of a finite element analysis.
EXAMPLE 7.6
Calculate the centroidal heat flux components for elements 2 and 3 of Example 7.5.
■
Solution
From Example 7.4, we have
a
=
b
=
0
.
5in
.,
k
x
=
20 Btu/(hr-ft-
◦
F)
, and from
=
k
y
Example 7.5, the nodal temperature vector is
T
1
T
2
T
3
T
4
T
5
T
6
T
7
T
8
T
9
180
180
180
106
.
507
111
.
982
106
.
507
89
.
041
90
.
966
89
.
041
◦
F
{
T
}=
=
For element 2, the element-global nodal correspondence relation can be written as
T
(2)
1
T
(2
4
=
[
T
4
T
(2)
2
T
(2)
3
T
7
T
8
T
5
]
=
[106
.
507
89
.
041
90
.
966
111
.
982]
