Dynamic Responses - Mechanics of Structures: Variational and Computational Methods - page 613

A simple iterative solution of this problem appears to be effective in converging rapidly to
a precise eigenvalue solution.

φ 0

2 M 0 0 0

[ K ( 0 ) − ω ( 0 ) ] φ = 0 ⇒ ω = ω

1 ,

ω

2 ,

ω

3 ,

... φ =

K ω

1 − ω

2 M ω

1 φ

0 0 1

1 ,

1
2 ,

1
3 ,

φ 1 = 0 ⇒ ω = ω ω ω ... φ =

··· (10.50)

K ω

1 − ω

2 M ω

1 φ

j − 1 j − 1 j

1 ,

j
2 ,

j
3 ,

φ j = 0 ⇒ ω = ω ω ω ... φ =

where the superscript j indicates the eigensolution for the j th iteration. The frequencies
ω

0 0 0

are the same as would be found using a consistent mass matrix and the usual
static stiffness matrix.
This approach, although significantly less efficient than solving the problem of Eq. (10.34),
is more efficient than the determinant search required to solve Eq. (10.42). This iterative
technique of Eq. (10.50) has the advantage of being able to utilize highly reliable, stan-
dard eigenvalue solvers that should result in an accurate set of frequencies and mode
shapes. Moreover, the frequency-dependent mass and stiffness matrices permit a model to
be employed with fewer (larger) elements than is possible with consistent or lumped mass
matrices. Sometimes, more accurate higher eigenvalues are obtained from Eq. (10.50) if K
and M are evaluated at

1 ,

ω

2 ,

ω

3 ,

...

1
1 .
The iterative scheme of Eq. (10.50) can be replaced by a higher order, e.g., quadratic,
generalized eigenvalue problem, using matrices expanded in series [Fergusson and Pilkey,
1992]. Expand the mass and stiffness matrices in Taylor series to obtain

j −

−

ω

j 1

, where n > 1, rather than at the lowest natural frequency ω

n

∞ ∞ ∞

m i

2 n

m i

2 n

k i

2 n

= ω =

0

ω = ω

m n or m n k n (10.51)

n = 0 n = n = 0

The dynamic stiffness matrix would be expanded similarly, giving

∞

k i dyn =

2 n

0 (

k dyn ) n ω (10.52)

n =

It follows from Eq. (10.48) that the dynamic stiffness matrix expansion terms would be
defined as

( k dyn ) n = k n − m n − 1 , n ≥ 1 (10.53)

A simple relationship between the terms m 0 , m 1 , , etc., is useful in that

only a few terms in these frequency expansions need be determined. From Fergusson and
Pilkey (1992),

... and k 1 , k 2 , ...

( n + 1 ) k n + 1 = n m n = n ( n + 1 ) m n =− n ( n + 1 )( k dyn ) n + 1 , n ≥ 1 (10.54)

Also,

( ) =

k dyn k 0 (the traditional stiffness matrix)

0

( k dyn ) =− m 0 (the traditional consistent mass matrix) (10.55)

1

k 1 = 0

There is a sizeable literature on the solution of higher order eigenvalue problems.