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                                                                  A O3  > A O2  > A O1

                                          Cylinder Speed  A O1  A O2  A O3












                                                                            F max
                                                           External Load

                       FIGURE 10.4  Hydraulic cylinder load-speed relationship under the same system pressure.

                         Equation (10.27) indicates that the increase in speed stiffness can be achieved either by increasing the
                       system pressure or the cylinder size, or by decreasing the speed.

                       System Dynamic Characteristics
                       To analyze the dynamic characteristics of this hydraulic cylinder actuation system, one can use  flow
                       continuity and system momentum equations to model the cylinder motion. Neglecting system leakage,
                       friction, and line loss, the following are the governing equations for the hydraulic system:

                                                q =  kx P P –  P 1 =  A 1 dy  V 1 dP 1          (10.28)
                                                                  ----- +
                                                                       ----- --------
                                                                   dt  b dt
                                                                 2
                                                                d y
                                                       P 1 A 1 =  m -------- +  F               (10.29)
                                                                dt 2
                         To perform dynamic analysis on this hydraulic system, it is essential to derive its transfer function
                       based on the above nonlinear equation, which can be obtained by taking the Laplace transform on the
                       linearized form of the above equations (Watton, 1989).
                                                                     1 
                                                     ----------dis() –   
                                                               V 1
                                                     k 1 K i   ---------s + ------------- dFs()
                                                                      2
                                                                2
                                                      A 1      A 1 b  A 1 k 3 R o
                                              dvs() =  ----------------------------------------------------------------------------  (10.30)
                                                                   1
                                                              2
                                                         ---------ms +  -------------ms +  1
                                                         V 1
                                                                  2
                                                          2
                                                         A 1 b   A 1 k 2 R o
                         Making
                                                2
                                               A 1 b
                                                                mb
                                                           1
                                        w n =  ----------,  z =  ------------- ------------,  and  K s =  k 1 K i
                                                                                 ----------
                                               V 1 m     2k 2 R o V 1 A 1 2      A 1
                       Equation (10.30) can be represented as
                                                                         1 
                                                                  
                                                                ---- -----s +  -------- d Fs()
                                                                 1 V 1
                                                      K s dis()  A 1 2   b  k 3 R o 
                                            dvs() =  -------------------------------- –  -----------------------------------------------  (10.31)
                                                                   1
                                                                        ------s +
                                                    1
                                                         ------s +
                                                   ------s +  2z  1  ------s +  2z  1
                                                      2
                                                                     2
                                                     2              2
                                                   w n   w n       w n  w n
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