Page 333 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
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320                      Introduction to the Finite Element Method   Chap. 10

                                       i/2             t/2
                                                                     Figure 10.5-2.  Two-element dis-
                                                 mi              ml  erete  mass  model  of a  uniform  ean-
                                                  2              4   etilever  beam.
                                  Thus,  the  original  4 x 4   equation  has  been  reduced  to  a  2  X  2  equation,  the  final
                                  form  being
                                              m^  0
                                                                    16  - 5 ‘
                                              0   m^ \ r /    1  7/-’  f  - 5  2
                                      An  acceptable  discrete  mass  distribution  is  one  in  which  the  mass  of  each
                                  element  is  divided  into  half at  each  end  of the  clement.  Thus,  if the  total  mass  of the
                                  uniform  beam  of length  /  is  ml,  the  mass  of each  element  is  m l / 2   and   =  2{ml/4)
                                  =  m l / 2   and  m^  =  m l / 4,  as  shown  in  Fig.  10.5-2.
                                      The  equation  of motion  and  solution  then  becomes
                                                     2   0     16   ~ 5 ] '

                                                  —A A     1
                                                     0   1    - 5    2j
                                  where
                                                        rml   11-    .  7  f ml^
                                                    A  =  -
                                                         4   48 T7    192  [  El
                                              A,  =   0 . 3 6 3 2    o;,  =  3.516   exact value  =  3.516
                                              A .   =   9 .6 3 7  ia.  =  22.033  exact value  =  22.034
                                                         0.327           1.5271
                                                    ^1  =  1.000  ^2  =  1.000
                              Example  10.5-3
                                  Determine  the  equation  of  free  vibration  of the  portal  frame  with  identical  elements.
                              Solution:  By  labeling  the  joints  as  shown  in  Fig.  10.5-3,  the  stiffness  and  mass  for  each
                                  element  arc  available  from  Eqs.  (10.4-7)  and  (10.4-8).  Because  joints  0  and  3  have
                                  zero  displacements,  we  write  only  the  terms  corresponding  to joints  1 and  2.

                                       Element  0-1,  a  =  90°,  c  =  0,  x  =  1:
                                                                 1    “   0   - 6 / “
                                                                 1    ()  - R   0
                                                                 1    6/  0   2 / “
                                            k   =  ^ ,3          1
                                            '^0-1
                                                                 1    12  0    6/  î'E
                                                                 1    0   R     0
                                                                 1    6 /  0  4 l \
                                                     -
                                                                         5 4   0    1 3 / '
                                                                          0          0
                                                  ml                    -   1 3 /  0  - 3 / “
                                                 420
                                                                         156   0    2 2 /
                                                                          0    0     0
                                                                         2 2 /  0   4 / “
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