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ME375 Dynamic System Modeling and Control

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ME375 Dynamic System Modeling and Control Powered By Docstoc
					         MESB 374
 System Modeling and Analysis

Translational Mechanical System
Translational Mechanical Systems

•   Basic (Idealized) Modeling Elements
•   Interconnection Relationships -Physical Laws
•   Derive Equation of Motion (EOM) - SDOF
•   Energy Transfer
•   Series and Parallel Connections
•   Derive Equation of Motion (EOM) - MDOF
Key Concepts to Remember

• Three primary elements of interest
  –   Mass ( inertia ) M
  –   Stiffness ( spring ) K
  –   Dissipation ( damper ) B
  –   Usually we deal with “equivalent” M, K, B
       • Distributed mass  lumped mass
• Lumped parameters
  – Mass maintains motion            (Kinetic Energy)

  – Stiffness restores motion        (Potential Energy)
  – Damping eliminates motion        (Eliminate Energy ? )
                                            (Absorb Energy )
Variables
• x : displacement [m]                 d
                               vx x
• v : velocity [m/sec]                 dt
                                   dv d  d  d 2
• a : acceleration [m/sec2]    a    x  2 x  x
                                   dt dt  dt  dt
• f : force [N]
                                                d
• p : power [Nm/sec]          p  f v  f  x  w
                                                dt
• w : work ( energy ) [Nm]                      t1
                              w(t1 )  w(t0 )   p(t ) dt
   1 [Nm] = 1 [J] (Joule)                       t0
                                                 t1
                                     w(t0 )   ( f  x ) dt
                                                t0
Basic (Idealized) Modeling Elements
• Spring
   – Stiffness Element
                                              – Reality
                                     x2
                                                 • 1/3 of the spring mass may be
                                                   considered into the lumped model.
                  K                              • In large displacement operation
                                                   springs are nonlinear.
                                                                    Linear spring
                                                                     nonlinear spring
        – Idealization
                                                                     broken spring !!
             • Massless
             • No Damping
             • Linear
       – Stores Energy                        Hard Spring
              Potential Energy
                               1
                         U      k  x 
                                          2

                               2                   Soft Spring
Basic (Idealized) Modeling Elements
• Damper                                     • Mass
      – Friction Element                       – Inertia Element
                  x1               x2
                                                                    x
 fD                                     fD        f2
                           B                      f3        M            f1
              .        .
       f D  B x2  x1  v2  v1
      – Dissipate Energy
                  fD


                                               – Stores Kinetic Energy
                               .         .                         1
                               x2  x1                     T        M x2
                                                                   2
Interconnection Laws
• Newton’s Second Law                                       Lumped Model of a Flexible Beam



                            f
   d         ..
      Mv  M x                     EXTi                                           K,M
   dt                          i
• Newton’s Third Law
                                                                               x
    – Action & Reaction Forces                                                           x


                                    x                         x                      M
                0                                                          K

    fs                                  fs
                                             fs         M
                     K

            f s  K  x  0  Kx                 M x   fs
                                                  M x   Kx
            Massless spring
                                                  M x  Kx  0    E.O.M.
Modeling Steps
• Understand System Function, Define Problem, and
  Identify Input/Output Variables
• Draw Simplified Schematics Using Basic Elements
• Develop Mathematical Model (Diff. Eq.)
  – Identify reference point and positive direction.
  – Draw Free-Body-Diagram (FBD) for each basic element.
  – Write Elemental Equations as well as Interconnecting
    Equations by applying physical laws. (Check: # eq = # unk)
  – Combine Equations by eliminating intermediate variables.
• Validate Model by Comparing Simulation Results
  with Physical Measurements
        Vertical Single Degree of Freedom (SDOF) System
•       Define Problem      The motion of the object
                                                                                                                                      g
•       Input                 f
•       Output                x
•       Develop Mathematical Model (Diff. Eq.)
                                                                                                                             xs
         – Identify reference point and positive direction.                                                                 x
                                                                                                        B,K,M
                                                                                                                             f
                    xs
                                       x  xs  xd
                    xd x               Kxs  Mg

                                                                                                          K             B
                          M             f s  Kx                  f d  Bx                                                        g
          –   Draw Free-Body-Diagram (FBD)
                                                       M
                                                                                                                    M        x
                                                   Mg         f
                                                                                                                f
          –    Write Elemental Equations           Mx   Bx  Kx  Mg  f
                                                   Mx  Bx  Kx  Mg  f
                              Mxu  Bxu  Kxu  Mg  f                   From the undeformed position
                              Mxd  Bxd  Kxd  f                From the deformed (static equilibrium) position
    •     Validate Model by Comparing Simulation Results with Physical Measurement
Energy Distribution
• EOM of a simple Mass-Spring-Damper System
                                                                                                                     x
                ..          .                                                                 K

     M x  B x  K x  f (t )                                                                                 M             f
                                                                                              B
   We want to look at the energy distribution of the system. How should we start ?

• Multiply the above equation by the velocity term v :                                      What have we done ?

      M x  x  B x  x  K x  x  f (t )  x
• Integrate the second equation w.r.t. time:                                    What are we doing now ?


                                                                                                           f  t   v dt
          t1                            t1                           t1                               t1
       t0
               M x  x dt          t0
                                             B x  x dt          t0
                                                                          K x  x dt             t0

                KE                   t1
                                     t0 B x dt  0
                                            2                              PE                                W
                                                                                               Total work done by the
                                                                                             applied force f ( t ) from

     Change of kinetic energy
                                                                                                    time t0 to t1
                                    Energy dissipated by       Change of potential energy
                                          damper
Example -- SDOF Suspension (Example)
• Suspension System
                                         – Simplified Schematic (neglecting tire
  Minimize the effect of the surface
  roughness of the road on the drivers
                                           model)
                                                                                  From the “absolute zero”
  comfort.
                                               Mx  B  x  x p   K  x  x p    Mg
                                                                                               From the path

                                               Mxop  Bxop  Kxop   Mg  Mx p
                                                                                                xop  x  x p
                                              Mxon  Bxon  Kxon   Mx p                  From nominal position

                                                                      g                        xon  x  xn
                                                                                           K  xn  x p    Mg
                                                           M          x


                                                     K           B


                                          x
                                                                       xp
Series Connection
• Springs in Series
                 x1                              x2                                        x1           x2

  fS                                                  fS            fS                                      fS
                      K1              K2                                                        KEQ

                  x1             xj               x2

   fS                                              fS
                                                                                  K1K 2
                       K1             K2                                  fs              x2  x1 
                                                                                 K1  K 2
     K1  x j  x1   K2  x2  x j                                               K eq
               1
    xj                K2 x2  K1 x1 
           K1  K 2
                                                                
                                1
                                                                
   f s  K1  x j  x1   K1             K2 x2  K1 x1   x1 
                                                                 
                                K1  K 2                        
                               
                                            xj                  
                                                                 
Series Connection
• Dampers in Series
           x1              x2                                     x1         x2

  fD                            fD                   fD                          fD
                B1    B2                                               BEQ




                                             B1B2
                                     fd             x2  x1 
                                            B1  B2
                                              Beq
Parallel Connection
• Springs in Parallel
                          x1              x2
                                                             x1         x2

                                                                             fS
      fS
                               K1
                                               fS      fS
                                                                  KEQ

                               K2


     f s  K1  x2  x1   K2  x2  x1 



           f s   K1  K2  x2  x1 
                    Keq
Parallel Connection
• Dampers in Parallel
                x1              x2
                                                         x1         x2

      fD             B1              fD            fD                   fD
                                                              BEQ
                     B2



                 f s   B1  B2  x2  x1 
                          Keq
Horizontal Two Degree of Freedom (TDOF) System
•   DOF = 2
                                                   x1                             x2
                                  B1                              B2

                                                M1                           M2
                                 K1                               K2
                                                f1                            f2
•   Absolute coordinates                          x1                           x2
•   FBD
                        B1 x1                           B2  x2  x1 
                                       M1                                    M2
                        K1 x1                           K2  x2  x1 
                                       f1                                    f2
•   Newton’s law
                   M 1 x1   B1 x1  K1 x1  B2  x2  x1   K 2  x2  x1   f1  t 
                   M 2 x2   B2  x2  x1   K 2  x2  x1   f 2  t 
Horizontal Two Degree of Freedom (TDOF) System
                                               x1                    x2
                           B1                          B2

                                         M1                        M2
                           K1             f1          K2            f2
                                                                                          Static coupling
•   Absolute coordinates
                                    M1 x1   B1  B2  x1   K1  K2  x1  B2 x2  K2 x2  f1 t 
                                                     M 2 x2  B2 x2  K2 x2  B2 x1  K2 x1  f 2 t 


•   Relative coordinates



                                          M1 x1  B1 x1  K1 x1  B2 x21  K2 x21  f1 t 
                                M 2 x1  M 2 x21  B2 x21  K2 x21  f 2  t 

                     Dynamic coupling
Two DOF System – Matrix Form of EOM
                           B1                       B2

                                        M1                     M2
                           K1            f1         K2          f2
                                                                                              Input vector
•   Absolute coordinates
                                                                     Output vector
                       Mass matrix            Damping matrix

                 M 1 0   x1   B1  B2         B2   x1   K1  K 2    K 2   x1   f1  t  
                 0 M   x    B                     x    K                x    f t 
                      2 2          2          B2   2          2      K2   2   2 

•   Relative coordinates                                                                   SYMMETRIC
                                                                Stiffness matrix


                                 M1   0   x1   B1  B2   x1   K1  K 2   x1   f1  t  
                                M                                                        
                                 2    M 2   x21   0 B2   x21   0 K 2   x21   f 2  t  
                                                                            

                                                NON-SYMMETRIC
MDOF Suspension
• Suspension System   – Simplified Schematic (with tire model)


                           M2
                                     x2

                      K2        B2

                                           g
                           M1
                                     x1
                                               TRY THIS

                      K1        B1


                                      xp
MDOF Suspension
                      – Simplified Schematic (with tire model)
• Suspension System             Assume ref. is when springs are
                                Deflected by weights

                            Car body
                                         M2        x2

                      Suspension K 2          B2

                             Wheel       M1
                                                   x1

                              Tire K1         B1

                               Road
                                              xp
                        Reference
Example -- MDOF Suspension
  – Draw FBD
                      x2                – Apply Interconnection Laws
         M2

 FS2           FD2                              M 2 x2   FS 2  FD 2
  FS2           FD2
                                                FS 2  K 2  x2  x1 
                                                FD 2  B2  x2  x1 
               FD2                              M 1 x1  FS 2  FD 2  FS 1  FD1
FS2
                                                FS 1  K1  x1  x p 
FS2            FD2
        M1
                           x1
                                                FD1  B1  x1  x p 

FS1            FD1                  M 2 x2  B2 x2  B2 x1  K 2 x2  K 2 x1  0
                                
FS2      FD2                        M 1 x1  B2 x2   B2  B1  x1  K2 x2   K 2  K1  x1  B1x p  K1x p



               FD1
  FS1
Example -- MDOF Suspension
  – Matrix Form
    M 2 x2  B2 x2  B2 x1  K 2 x2  K 2 x1  0
    M 1 x1  B2 x2   B2  B1  x1  K 2 x2   K 2  K1  x1  B1 x p  K1 x p


                   x2 
Define vector x   
                   x1 
   M2    0   x2   B2          B2   x2   K2           K2   x2            0        
    0         x   B
          M1   1   2                   x   K
                                 B2  B1   1   2
                                                                                            
                                                             K2  K1   x1   B1 x p  K1 x p 
                                                                    


  Mass matrix              Damping matrix              Stiffness matrix            Input Vector

				
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posted:8/24/2011
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