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                                            Design at Resonance of Mechanical Microsystems

                              30   Chapter One

                                Example:  Study the coupling and/decoupling of the 3-DOF system shown in
                                Fig. 1.23, and find the transformed matrices which will decouple the system
                                in case it is coupled. Consider that  m 1  =  m 2  =  m 3  =  m and  k 1  =  k 2  =  k 3
                                = k 4  = k.
                                  As Eq. (1.79) shows, the system is statically coupled as the stiffness matrix
                                is not in diagonal form. The eigenvectors corresponding to the mechanical
                                system of Fig. 1.23 are assembled into the modal matrix of Eq. (1.93) as

                                                         ෹1   1   1
                                                    P =    0  2 ෹ 2                      (1.100)
                                                           1  1   1

                                As a result of the definition in Eq. (1.96), the transformed stiffness matrix is

                                                          1   0     0
                                                K   =4k 0 2 ෹ 2     0                    (1.101)
                                                  m
                                                          0   0   2+ 2
                                which is in diagonal form. The mass matrix [Eq. (1.78)] is already in diagonal
                                form, and there is no need to calculate a transformed mass matrix. The de-
                                coupled dynamic equation of motion will therefore be
                                                       ..
                                                    M {x}+ K   {x}= {0}                  (1.102)
                                                             m

                              1.3.3  Lagrange’s equations
                              Lagrange’s method and equations can be utilized to derive the equa-
                              tions of motion of a mechanical system with (generally) multiple de-
                              grees of freedom by utilizing the energy forms that define the system’s
                              motion. Presented next are Lagrange’s equations for conservative sys-
                              tems as well as for nonconservative ones.
                                In conservative systems there are  no  energy  gains  or losses,  as
                                                                         3
                                                      1
                                                                2
                              indicated by Timoshenko,  Thomson,  or Rao,  and Fig. 1.27 shows a
                              serial chain composed of only masses and springs. The system possesses
                              n degrees of freedom. The number n represents the minimum number
                              of parameters that can describe the state of the system at any given
                              time, and these parameters (depending on time) are called generalized
                              coordinates.
                                Lagrange’s equation for the ith mass (and degree of freedom) is
                                                 d ˜T    ˜T   ˜U
                                                      .  í  +     =0                    (1.103)
                                                 dt( ˜x ) ˜x  ˜x
                                                      i    i    i
                              where T is the kinetic energy of the whole system and is calculated as




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