Page 258 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 258

Sec. 8.5   Convergence to Higher Modes                         245


                                  which  can  be  expressed  by  the  matrix  equation
                                                             0   0   0.25
                                                             0   0   -0 .7 9
                                                             0   0   1.00
                                  This  matrix  is  devoid  of the  first  two  modes  and  can  be  used  as  a  sweeping  matrix  for
                                  the  third  mode.  Applying  this  to  the  original  equation,  we  obtain
                                              ■ 4  2  1 ■ '0  0  0.25'
                                               4  8  4  0  0  -0 .7 9
                                               4  8  7  0  0   1.00

                                  This  equation  results  immediately  in  the  third  mode,  which  is
                                                           0.25 \
                                                      1.68  -0.79  I  =   3k
                                                           1.00 I         ^3

                                  The  natural  frequency  of the  third  mode  is  then  found  to  be

                                                             ^     =  1 34,/ A
                                                          V  IT .68w   V  m
                                      These  natural  frequencies were  checked  by  solving the  stiffness  equation,  which

                                               ”4  0  0“          4  - 1


                                             m  0  2  0  < ^2  > +   A  -1  2     =  0
                                                        V ^ j
                                                0  0  1           0  -1

                                  With  A  =  mco^/k,  the  determinant  of this  equation  set  equal  to  zero  gives
                                             8(1  -   A)’  -   5(1  -   A)  =  (1  -   A )[8(l  -   A)-  -   5 ]  = 0
                                  Its  solutions  are
                                                                              0.250
                                            A,  =  0.2094  ca,  =  0.4576  0 ,  =  {  0.791
                                                                              1.000

                                            Ao  =  1.0000   =  1.0000-1/  —
                                                                             ; i i
                                                                             1 [   0.250
                                            A ^=  1.7906   ia.  =  1.3381-i/  —  -0.791
                                                                             1 1   1.000
                                      Figure  8.5-1  is  a  printout  from  the  computer program  ITERA TE  that  solves for
                                  A  =  k/m o)^  instead  of  A  =  mco^/k.  Thus,  for  mode  1,  we  have  for  comparison,
                                  A  =  1 /A  =  1 /4.775  =  0.2094.  The  mode  shapes,  however,  are  not  altered.
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