Page 317 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
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304                      Introduction to the Finite Element Method   Chap. 10

                              we  find
                                                     d  dT    A  1  ••  1  ..
                                                          =  ml\
                                                    dt  dii|       +
                                                     d  dT    ,l  1  ••  1  ..
                                                          =  ml\^-^u^ +  3 U2
                                                    dt  dû 2
                              which  establishes  the  mass  matrix  for  the  axial  element  as
                                                          r ^ \ 2    1
                                                                                         (10.1-7)
                                                          “6” [ l   2
                              The  terms  of the  mass  matrix  are  also  available  from

                                                        m,  =  /  <P<P dm
                                                                  j
                                                                i
                              Example  10.1-1
                                  Determine  the  equation  of  motion  for  the  longitudinal  vibration  of  the  two-section
                                  bar shown  in  Fig.  10.1-6.
                              Solution:  Numbering the joints  as  1,  2,  and  3, we  have  two  axial  elements,  1-2,  and  2-3,
                                  with  displacements  Wj,  U2,  and  Uy  Although   is  zero,  we  at  first  allow  it  to  be
                                  unrestrained  and  later impose  its zero value.
                                       The  element  mass  and  stiffness  terms  from  Eqs.  (10.1-7)  and  (10.1-3)  are  as
                                  follows:
                                                                2    1    k   1  -1
                                                            6   1    2       -1    1
                                                           K    2    1    Ir  1  -T
                                               Element  b:
                                                            6   1    2       -1    1
                                  where   =  £/!„//„,  ki, = EAf,/l^,  M„  = mj^, and  A/j,  = mjf,.
                                       The  element  matrices  have  a  common  coordinate  U2,  and  by  superimposing
                                  them,  they can be  assembled  into  a 3  X  3  matrix as follows:
                                                             2M„  ! 1          0
                                              Mass matrix         2M,  +  2M,
                                                              0               2M,

                                                              k.  \          0
                                              Stiffness matrix
                                                             -
                                                               0

                                      We note  here that the stiffness matrix is singular and does not have  an  inverse.
                                  This is to be  expected because no limitations have been placed on the  displacements.
                                            t -
                                            1“
                                                                  3   Figure  10.1-6.
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