Page 160 - Mechanical design of microresonators _ modeling and applications
P. 160

0-07-145538-8_CH03_159_08/30/05



                                   Microhinges and Microcantilevers: Lumped-Parameter Modeling and Design

                               Microhinges and Microcantilevers: Lumped-Parameter Modeling and Design  159
                              which is the known relationship for a one-component microrod. By com-
                              bining Eqs. (3.161) and (3.163), the resonant frequency corresponding
                              to free axial vibrations of the patched microcantilever becomes
                                            k a,e
                                    Ȧ   =
                                      a,e   m
                                             a,e
                                                                                        (3.165)
                                                           p p /
                                           1.73l E t (E t + E t ) [E t l + E t (l í l )]
                                                                1 1
                                                                       p p
                                                     1 1
                                                                             p
                                                 1 1
                                         =
                                                                        2
                                              3
                                             l t ȡ + l [3(l í l ) 2 í 3(l í l )l + l ] t ȡ
                                               1 1  p      1       1 p  p  p p
                                In torsion, similar to the axial problem, the stiffness of the patched
                              microcantilever can be found by  considering the series connection
                              between the three portions above-mentioned. The stiffness equation is
                                                       3
                                                            3
                                                                   3
                                                    G t (G t + G t )w
                                                      1 1
                                                           1 1
                                                                 p p
                                             k   =                                      (3.166)
                                              t,e        3            3
                                                   3 G lt + G (l í l )t p
                                                                   p
                                                      1 1
                                                              p
                              For l p ĺ 0 and t p ĺ 0, Eq. (3.166) reduces to
                                                          G wt 1 3
                                                           1
                                                     k =                                (3.167)
                                                      t     3l
                              which is the stiffness of a homogeneous fixed-free bar.
                                The equivalent torsional mechanical moment of inertia is determined
                              again by means of Rayleigh’s principle, and its equation is
                                                                               2
                                                                           2
                                        3
                                                        2
                                                                                    2
                                              2
                                                   2
                                     w{l ȡ t (t + w ) +3l l t 3ȡ t t + ȡ (3t + t + w )
                                          1 1 1           p p   1 1 p  p   1   p
                                                                          2
                                                                  2
                                                                     2
                                     ෹3ll t (2l + l ) 3ȡ t t + ȡ (3t + t + w )
                                         p p  1   p   1 1 p   p  1   p
                                                                                        (3.168)
                                            2
                                                                      2
                                     +l t (3l +3l l + l ) 3ȡ t t + ȡ (3t + t + w ) }
                                                       2
                                                                          2
                                                                               2
                                       p p
                                                                      1
                                                       p
                                                           1 1 p
                                            1
                                                                          p
                                                 1 p
                                                                   p
                              J   =
                               t,e                            2
                                                           36l
                              When l p ĺ 0 and t p ĺ 0, Eq. (3.168) simplifies to
                                                               2
                                                                   2
                                                     1  ȡ lwt (w + t )
                                                            1
                                                                   1
                                                        1
                                                J =  3       12                         (3.169)
                                                  t
                              which is the equation expressing the equivalent mechanical moment of
                              inertia in torsion for a homogeneous fixed-free bar.
                                The torsional resonant frequency of the bar is  calculated by
                              Eqs. (3.166) and (3.168), and its equation is
                           Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)
                                      Copyright © 2004 The McGraw-Hill Companies. All rights reserved.
                                        Any use is subject to the Terms of Use as given at the website.
   155   156   157   158   159   160   161   162   163   164   165