Civil Engineering Reference
In-Depth Information
■
Solution
As the integrand is a polynomial of order 2, we have, for exact integration,
2
m
−
1
=
2
,
which results in the required number of sampling points as
m
=
3
/
2
. The calculated
number of sampling points must be rounded
up
to the nearest integer value, so in this
case, we must use two sampling points. Per Table 6.1, the sampling points are
r
i
=
±
0
.
5773503
and the weighting factors are
W
i
=
1
.
0,
i
=
1, 2
.
Therefore,
1
(
r
2
5773503)
2
−
3
r
+
7) d
r
=
(1)[(0
.
−
3(0
.
5773503)
+
7]
−
1
5773503)
2
+
(1)[(
−
0
.
−
3(
−
0
.
5773503)
+
7]
1
(
r
2
−
3
r
+
7) d
r
=
14
.
666667
−
1
The result is readily verified as, indeed, being exact by direct integration.
The Gaussian quadrature numerical integration procedure is by no means
limited to one dimension. In finite element analysis, integrals of the forms
1
1
=
I
f
(
r
,
s
)d
r
d
s
−
1
−
1
(6.103)
1
1
1
I
=
f
(
r
,
s
,
t
)d
r
d
s
d
t
−
1
−
1
−
1
are frequently encountered. Considering the first of Equation 6.103, we integrate
first with respect to
r
(using the Gaussian technique) to obtain
1
1
1
1
n
=
=
=
(6.104)
I
f
(
r
,
s
)d
r
d
s
[
W
i
f
(
r
i
,
s
)] d
s
g
(
s
)d
s
i
=
1
−
1
−
1
−
1
−
1
which, in turn, is integrated via quadrature to obtain
m
=
(6.105)
I
W
j
g
(
s
j
)
j
=
1
combining Equations 6.98 and 6.99, we find
1
1
m
n
=
=
(6.106)
I
f
(
r
,
s
)d
r
d
s
W
j
W
i
f
(
r
i
,
s
j
)
j
=
1
i
=
1
−
1
−
1
