Page 248 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 248

Sec. 8.1   Root Solving                                        235


                       8.1  ROOT SOLVING
                              Figure  8.1-1  shows  a  3-DOF  system  for  which  the  normal  modes  and  natural
                              frequencies are desired. The equation of motion for this system is
                                          "2                  3   -1    0 “  f-^i'  fo]
                                        m     1    1 ^2 /   ^  -1  2   -1   X2 »  =  0  .
                                                 1            0   -1    1         loj
                              or
                                             ‘2             3   -1    O'
                                          -A     1     +  -1     2   -1    -^9   =
                                                    1       0   -1    1
                              where  A =  oj^m/k.

                                  The eigenvalues of the system are found from the characteristic determinant
                              equated  to zero:
                                                 (3  -   2A)   -1     0
                                                   -1     ( 2 - A )  - 1  = 0
                                                     0      - 1   ( 1 - A )
                              This determinant  reduces  to  a third-degree  algebraic equation.  Using the  method
                              of  minors  (see  Appendix  C)  and  choosing  the  elements  of  the  first  column  as
                              pivots, we have
                                                 ( 2 - A )  - 1      - 1     0
                                         ( 3 - 2 A )             +  1             =  0
                                                   - 1    ( 1 - A )  - 1  ( 1 - A )
                              and  the characteristic equation becomes
                                                    A^  -   4.50A^  +  5A  -   1  =  0
                                  There is no simple equation to find the roots of this equation.  However,  it  is
                              a simple  matter to plot  it  as  a function  of  A and  find  its zero  crossing. There  are,

                              however, a number of computer programs that will solve for the roots (eigenvalues)
                              of the  polynomial  equation.  The  procedure  is  quite  straightforward  and  based  on
                              the following idea.
                                  By letting the  A-degree equation be  expressed  as
                                            /(A)  = A^  + CiA"-^  +   +  • • •  +c,  =  0
                                                    '

                              and  by  assuming  a  number  for  A and  substituting  it  into  this  equation,  a value  is
                                                k         k
                                          2mI 2m—AAAAr- m - v w v - m

                                                                 ^3   Figure 8.1-1.
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