Page 249 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

236 Computational Methods Chap. 8

obtained for /(A). If the procedure is repeated and /(A) is plotted as a function of
A, a change of sign in /(A) will indicate the proximity of a root. By using a straight
line or Newton’s interpolation, the roots are easily found.
For an estimate of the range covered by A and the interval A A for the
computer, the polynomial can be assumed to be factored as
/(A) = (A - A,)(A - A 2)(A - A , ) = 0
where A^ are the roots of the equation. Multiplying out the factored form of the
equation, one finds that the coefficient Cj for the next to the highest power of A is
always equal to the sum of the roots, regardless of N; i.e., for the third-degree
equation, we have
/(A) == A^ — (Aj + A2 A3)A^ + (A|A2 ■(" AjA^ H- A2A3)A —AjA2A3 = 0
'

The preceding procedure or modification of this procedure is used for most
computer programs. Because the computer can carry out thousands of calculations
in a few seconds, A A can be chosen very small, in which case, the interpolation can
be minimized or even eliminated for the accuracy required.

8.2 GAUSS ELIMINATION

In solving for the mode shapes, the eigenvalues are substituted, one at a time, into
the equation of motion. The Gauss method offers one way in which to solve for the
ratio of amplitudes. Essentially, the Gauss procedure reduces the matrix equation
to an upper triangular form that can be solved for the amplitudes starting from the
bottom of the matrix equation.
Applying the Gauss method to the previous problem, we start with the
equation of motion written in terms of A:
"(3 - 2A,) - 1 0 •^1 'O'
- 1 ( 2 -A, ) - 1 <X2 = <0
0 - 1 ( l - A , ) [oj
The eigenvalues solved for the problem were
^ [0.25536

A = w^-j- = 1.3554
i 2.8892
Substituting AI = 0.25536 into the preceding equation, we have
2.489 - 1 0 ¡0
1 i; 745 - 1 - 0
0 - 1 0.7446 (o

method. the first step is to eliminate the
column in the second and third rows. Because the first column of the third row is
already equal to zero, we need only to zero the first term of the second row. This is