Page 240 - Mechanical design of microresonators _ modeling and applications
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                                                 Resonant Micromechanical Systems

                                                             Resonant Micromechanical Systems  239
                                                      k I            Ew
                                               I       b,e      t      1
                                              Ȧ b,e  =  I  =  2                           (5.27)
                                                      m b,e    l 1  ȡw l l
                                                                      2 1 2
                              As a reminder, the assumptions are made in Eqs. (5.25), (5.26), and
                              (5.27) that the three segments have identical thicknesses t and that
                              their widths are different: w 1  for the end segments and w 2  for the middle
                              one (w 2  > w 1 ).
                                The torsional stiffness associated with the midpoint of the paddle
                              microbridge is again a combination of the two beam-springs, namely,
                                                        2GI t1  2Gw t 3
                                                                    1
                                                  k I  =      =                          (5.28)
                                                   t,e    l 1     3l 1
                              and the torsional  moment  of  inertia of the paddle microbridge is
                              identical to that of the middle segment according to the assumptions of
                              model I, namely,

                                                                 2
                                                                     2
                                                         ȡl w t(w + t )
                                                          2 2
                                                                 2
                                                    I
                                                   J t,e  =   12                         (5.29)
                              The torsional resonant frequency can now be calculated as
                                                  k  I             Gw
                                           Ȧ I  =  t,e  =2 2t         1                  (5.30)
                                            t,e     I                  2  2
                                                  J t,e       ȡw l l (w + t )
                                                                 2 1 2
                                                                      2
                              Model II.  Model I ignored the inertia contributions of the end compliant
                              segments, and this is acceptable as long as the mass of the middle seg-
                              ment is (considerably) larger than the  masses of the  end segments.
                              However, in cases where the two different segments comprising the
                              paddle bridge are comparable, the compliant end segments also con-
                              tribute to the total mass, and this additional mass can be calculated by
                              Rayleigh’s principle, which states that the deformation (displacement)
                              distribution of a beam in this case is identical to the velocity distribution
                              of the same member. By equating the kinetic energy of the real, dis-
                              tributed-parameter beam to that of the equivalent lumped-parameter
                              system, the additional mass is calculated as
                                                             l 1
                                                                2
                                                 m   = ȡtw 1ฒ  f (x) dx                  (5.31)
                                                   1,e          b
                                                            0







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