Page 345 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 345

332                      Introduction to the Finite Element Method   Chap. 10


                                       The  equation of motion  can  now be written  as
                                                      (o~ml  .,   El
                                                                   V =   - m i l ' l H V
                                  By  multiplying  and  dividing  the  right  hand  term  by  El/P  and  transferring  it  to  the
                                  left  side, we  obtain
                                                    (jo^ml ,   El
                                                                K + ' ^ H   v = o
                                                                      t l


                                                       M -   A          =  0
                                                                  t l
                                  where
                                                                420 £ /

                                                             A =
                                       With  the  boundary conditions  [  j  =   =  0, we need only to maintain the  lower
                                  right quadrant of the matrices. Also by remembering that the   inside the matrices all
                                  cancel  in  the  solution  for  the  eigenvalues  and  eigenvectors,  we  can  let  I =  1.0.  The
                                  final  equation  of motion  is  then

                                      156  - 2 2 '   12  - 6 '  ~r  '    0.4286  -0.06429   ^’2
                                                              1  ,
                                      - 2 2  4   a(  - 6   4   i   E l   j  -0.06429  -0.02381    )J
                                                 7                                      ^  J  V^2
                                           0
                                           0

                                  This  equation  can  be  solved  for  the  eigenvalue  A by  assuming  a  number  for  the
                                  rotation  parameter.  If we  choose  il^ml^/EI =  1.0, we  obtain
                                                          _
                                               156  - 2 2 ‘  -  A  12.43  -6.064
                                              _-22    4      -6.064   4.024

                                                     (156  -   12.43A)   -(2 2   -   6.064A)
                                                                                =  0
                                                     -(22  -   6.064A)   (4  -   4.024A)
                                  The  eigenvalues and  natural  frequencies  from  the  determinant  are

                                                     A,  =  30.65


                                                     A,  =  0.345
                                  and  the  associated  eigenvectors  and  mod
                                                          0.545                 0.0807
                                                          0.749                 0.615
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