Civil Engineering Reference
In-Depth Information
EXAMPLE 6.9
Use Gaussian quadrature to obtain an exact value of the integral
1
1
1
r
2
(
s
2
1)(
t
4
I
=
−
−
2) d
r
d
s
d
t
−
1
−
1
−
1
■
Solution
In this case, we have a q
ua
dratic polynomial in
r
, so two sampling points are required, with
r
i
=±
√
3
=±
0
.
5773503
/
3
and
W
i
=
1
per Ta
b
le 6.1. The quadratic in
s
similarly requires
=±
√
3
two sampling points,
s
j
0
. For
the quartic function in
t
, three sampling points are required for exactness and the values
and weighting factors per Table 6.1 are
=±
0
.
5773503
/
3
, with weighting factors
W
j
=
1
.
t
1
=
0
.
0
W
1
=
0
.
8888889
t
2
=
7745967
t
3
=−
0
.
W
2
=
0
.
5555556
0
.
7745967
W
3
=
0
.
5555556
For an exact solution, we then have
3
2
2
I
=
W
k
W
j
W
i
f
(
r
i
,
s
j
,
t
k
)
k
=
1
j
=
1
i
=
1
so a total of 12 terms is required. The required calculations are summarized in Table 6.2.
So we obtain
1
1
1
r
2
(
s
2
1)(
t
4
I
=
−
−
2) d
r
d
s
d
t
=
3
.
2
−
1
−
1
−
1
and this result is exact.
Table 6.2
Sampling Points, Weighting Factors, and Calculations for Example 6.9
Cumulative
Point
r
i
s
j
t
k
W
i
W
j
W
k
f
(
r
i
,
s
j
,
t
k
)
Sum
1
0.57735
0.57735
0
1
1
0.88888889
0.395062
0.395062
2
0.57735
0.57735
0.774597
1
1
0.55555556
0.202469
0.597531
3
0.57735
0.57735
−
0.774597
1
1
0.55555556
0.202469
0.8
4
0.57735
−
0.57735
0
1
1
0.88888889
0.395062
1.195062
5
0.57735
−
0.57735
0.774597
1
1
0.55555556
0.202469
1.397531
6
0.57735
−
0.57735
−
0.774597
1
1
0.55555556
0.202469
1.6
7
−
0.57735
0.57735
0
1
1
0.88888889
0.395062
1.995062
8
−
0.57735
0.57735
0.774597
1
1
0.55555556
0.202469
2.197531
9
−
0.57735
0.57735
−
0.774597
1
1
0.55555556
0.202469
2.4
10
−
0.57735
−
0.57735
0
1
1
0.88888889
0.395062
2.795062
11
−
0.57735
−
0.57735
0.774597
1
1
0.55555556
0.202469
2.997531
12
−
0.57735
−
0.57735
−
0.774597
1
1
0.55555556
0.202469
3.2
