Civil Engineering Reference
In-Depth Information
Table 2.16.
(
Continued
)
0.1912 −0.3291
0.3319
1.0843
A
-1
=
0.2084
0.4844 −0.2871 −0.3651
−0.0791
0.1394
0.0891
0.3720
−0.0754
0.0866
0.1588 −0.3483
Now the third and fourth rows are swapped in coefficient matrix and
the solution is found as [
x
]=[
A
]
-1
[
C
] in Table 2.17:
Table 2.17.
Example 2.13 Matrix inversion method
1.65
1.65
1
C
=
13.18
C
=
13.18
x
=
2
-
0.67
17.32
3
17.32
4
−
0.67
2.12
gAuSS-SEiDEL itERAtiOn MEtHOD
Some methods such as the Gaussian elimination are not appropriate
when a
sparse matrix
exists. A matrix normally can be considered sparse
if approximately two-thirds or more of the entries in a matrix are zero.
The Gauss-Seidel iteration method was developed for such systems. The
method is named for Carl Friedrich Gauss and Philipp Ludwig von Seidel
(Gauss 1903). This method is an iteration method in which the last cal-
culated values are used to determine a more accurate solution. Typically,
all unknown
x
values are assumed to be zero to begin the iteration. This
method mainly works best with a diagonal system in which the largest
values lie on the diagonal. The elastic stiffness matrix used to analyze
structures is a typical example of a diagonal system and will be presented
in Chapter 4. A diagonal system is sufficient, but not necessary to provide
convergence. During the process, each row is used to find a better approx-
imation of the variable corresponding to the row using all other variables
as known.
Example 2.14
Gauss-Seidel lteration method
Determine the solution to the following set of equations using the
Gauss-Seidel iteration method with
e
= 0.01 and assume
x
1
= x
2
= x
3
= x
4
=
0.
