Page 252 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 252

Sec. 8.3   Matrix Iteration                                    239












                                                                     Figure 8.3-1.
                              Solution:  The  mass and  the  flexibility matrices for the  system  are
                                                       ■4  0  o'           1  1  r
                                                m]  =  m 0  2  0  W - 5 f  1  4  4
                                                       0  0  1             1  4  7
                                  and  substituting into Eq. (8.3-1), we  have
                                                 1  1  r ■4  0  0-
                                                 1  4  4  0  2  0
                                                 1  4  7  0  0  1
                                  or




                                       To start  the  iteration, we arbitrarily assume
                                                                    / 0.2
                                                                     0.6
                                                         ^1  =  " 4   =
                                                             W      \l.O
                                                             0.2 )  / 3.0      ^0.238 \

                                             AX.  =          0.6  =  9.6   12.6(  0.762
                                                             1.0 /  1 12.6     1.000 j
                                  By using the  new  normalized column  for  X2, the  second  iteration yields
                                                          0.238]    3.476]       (0.247]
                                                          0.762    11.048   =  14.048  0.786
                                                          1.000 j  14.048)       \ 1.000)

                                  In  a similar manner,  the third iteration  gives
                                                          0.247     3.560]       (0.249]
                                           AX,  =         0.786  )  =   11.276  =  14.276  0.790
                                                          1.000    14.276)       ( 1.000)
                                  By  repeating  this  procedure  a  few  more  times,  the  iteration  procedure  converges  to
                                                     '0.250^     M   i  3k  \  '0.250^
                                                14.324(  0.790  } = A{ ^2  =  —     0.790

                                                      1.000J    h            1.000
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