Dynamic Responses - Mechanics of Structures: Variational and Computational Methods - page 639

The other Ritz vectors are defined by linear algebraic equations in which M is multiplied by
the previously calculated Ritz vector and the resulting vector is used as the external force for
a static analysis. Thus, for i = 2 , 3 , ... ,n, the equation

Kx i =

Mx i − 1 (10.139)
is solved for x i . Next, the vector x i is orthogonalized and normalized with respect to M .
These operations are carried out by calculating for j = 1 , 2 , ... ,i − 1 the scalars

x j Mx i

=

c j (10.140)

followed by

i − 1

x ∗∗
i

x i −

= c j x j (10.141)

j = 1

and finally

x ∗∗
i

x i = x ∗∗ T

i

(10.142)

Mx ∗∗
i

Since the Ritz vectors x i , 1 n, are orthogonalized and normalized with respect to

the mass matrix M , the condensed mass matrix M 1 given by Eq. (10.133) is diagonal. The
matrices K 1 and C 1 are, in general, full. Once the matrices M 1 and K 1 are calculated, it
is possible to find approximate values for the lowest n natural frequencies and the corre-
sponding mode shape vectors by solving the generalized eigenvalue problem stated in Eq.
(10.82) with M replaced with M 1 and K with K 1 .

≤ i ≤

EXAMPLE 10.12 Ritz Vector Method for a Frame
Return to the frame shown in Fig. 10.2. The system mass and stiffness matrices M and K
are given by Eq. (9) of Example 10.1 and Eq. (11) of Chapter 5, Example 5.5.
As in Example 10.11 the d a mping matrix C is assumed to be proportional to the mass ma-
trix, C

of Fig. 10.8 is applied at node b in the X direction. The system
governing equation is given by Eq. (10.130) with V

= 0 . 000625 M . Force P ( t )

c ] T ,

= [ U Xb U Zb U Xc U Zc

b

=

T

.

and, for the loading applied at node b in the X direction, F [100000 Also,

( )

the function g is expressed in Fig. 10.8.

We choose to use Ritz vectors to condense this six-DOF system to a three-DOF system.
From Eq. (10.137), the first Ritz vector x 1

t

is the solution of Kx 1 = F . This gives

x 1 = 10 − 7 10 − 7 10 − 7

[1 . 74034 × 0 . 97015 × − 0 . 02479 ×

10 − 7 10 − 7 10 − 7 ] T 1 . 71722 × 0 . 00789 × − 0 . 43951 × (1)

Normalize x 1 according to Eq. (10.138) to find the first Ritz vector.

10 − 1 10 − 1 10 − 1

x 1 = [0 . 77242 × 0 . 43058 × − 0 . 01100 ×

10 − 1 10 − 1 10 − 1 ] T

0 . 76216 × 0 . 00350 × − 0 . 19507 × (2)

Follow Eqs. (10.139) to (10.142) to find the next two Ritz vectors.

10 − 1 10 − 1 10 − 1 x 2 = [ − 0 . 90063 × 0 . 03594 × 1 . 79391 ×

10 − 1 10 − 1 10 − 1 ] T − 0 . 11120 × − 0 . 01446 × 2 . 12710 × (3)

10 − 1 10 − 1 10 − 1 x 3 = [0 . 51388 × − 0 . 24340 × − 2 . 07095 ×

10 − 1 10 − 1 10 − 1 ] T

− . × . × . ×

0 37934 0 06251 1 34613 (4)