vibrations by stariya

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									                                             MECH2200/2210 Dynamics and Orbital Mechanics
                                                      Department of Mechanical Engineering,
                                                              The University of Queensland




VIBRATIONS

Common engineering phenomenon concerned with linear displacement- angular
rotation- deflection (bending) of components.

Simplest system is 1 DOF involving 1 displacement.

                                                          In general, multi DOF will
                                                          apply- but 1 DOF is a
                                                          basic + useful analytical
                                                          tool




Single wheel on car ~ 1DOF
                                                MECH2200/2210 Dynamics and Orbital Mechanics
                                                         Department of Mechanical Engineering,
                                                                 The University of Queensland




Free Vibration
After displacement, system oscillates before returning to the equilibrium position in
absence of imposed external forces- eg. car over single bump


Forced Vibration
Motion continuously excited by disturbing force- external or due to system dynamics-
eg. car out of balance

Damping
In all engineering applications, some forces resist motion (mechanical + fluid friction)
 mechanism for energy losses


                                         Friction always opposes motion.

                                           Fdx is always -ve




May be deliberately induced- eg. shockers
                                     .
Damping is generally proportional to x, v
        .                                           .
FD  k x (viscous damping, convenient sign change- x goes –ve)




Linear and non-linear systems consider vibration when restoring force &
displacement.
                                                  MECH2200/2210 Dynamics and Orbital Mechanics
                                                           Department of Mechanical Engineering,
                                                                   The University of Queensland




Example of a linear system

- simple spring




This is applicable to many systems- our
analysis is restricted to linear – very useful.




Examples of non-linear systems

- Air spring

P V  R T  isothermal

     Po Vo
P         
      V




- Overload springs on a 4WD  only contact at high load – large displacement

        k
n 
        m
                                                    MECH2200/2210 Dynamics and Orbital Mechanics
                                                                  Department of Mechanical Engineering,
                                                                          The University of Queensland




Single degree of freedom – linear analysis




                                             .    ..
Newton’s Law          F        kx        cx  mx

                              opposes     opposes
                           displacement   motion


                       ..    .
        Fo cos t  m x  c x  k x                                                        (8)


     forcing term    inertia      damping      spring stiffness




FREE VIBRATION


Case #1 Undamped, free vibration ( c  0 , F  0 )

        ..
       mx k x  0

         .. k        ..
        x  x  0  x  n x
                          2
                                                                                            (9)
            m

              k
where  n       natural frequency
              m
                                                MECH2200/2210 Dynamics and Orbital Mechanics
                                                          Department of Mechanical Engineering,
                                                                  The University of Queensland


We would expect a simple harmonic motion (acceleration & displacement) to produce
an oscillatory solution.

        x  A sin  n t  B cos  n t
        .
        x  A n cos n t  B n sin n t
        ..
        x   A n sin n t  B n cos n t
                 2               2



 satisfies (9) choose A, B to match boundary conditions (starting values)

        . .
        x  x o cos  n t  xo  n sin  n t
                                                             .
                                          . .                xo
If it is known that at t  0 , x  xo and x  x o , then A     and B  xo .
                                                             n
 full solution for x is
             .
             xo
        x        sin  n t  xo cos  n t                                          (10)
             n

Examples:

1. vibration of a car




2. vibration of a bed
                                                         MECH2200/2210 Dynamics and Orbital Mechanics
                                                                     Department of Mechanical Engineering,
                                                                             The University of Queensland




FORCED VIBRATION

Undamped F  Fo cos  t (    n in general)
         ..
       m x  k x  Fo cos t

Solution has 2 parts     complementary (RHS = 0) - natural frequency
                         particular (response to forcing)


       ..         F
       x   n x  o cos  t
             2
                                                                                               (11)
                   m
The main purpose of this analysis is to get the long-term (steady) response to forcing
input. The SHM component will damp out anyway. Eventually, the system will
vibrate at   need to find amplitude

eg. vibration of engine on mounts

Assume form x p  x cos  t
       .
       x   x sin  t
       ..                                   (11)
       x   2 x cos t

                                               Fo
         2 x cos t   n x cos t 
                          2
                                                  cos  t
                                               m
                   
        x cos t  n   2
                    2
                                 
                       Fo                     Fo                Fo
        x                                                                                  (12)
               m   
                   n
                    2
                           2
                                              2
                                         m n 1  2 
                                            2 
                                                                2
                                                             K 1  2 
                                                                
                                               n                n 




Notes: Role of  n - natural frequency of undamped system
       As    n , x   ‘resonance’ occurs. When this happens, energy input is
       at all times adding to natural motion (swing). This will increase until
       something breaks or non-linearities occur (eg suspension hits stops). In
       practice, there will always be damping which stops  
                                                             MECH2200/2210 Dynamics and Orbital Mechanics
                                                                      Department of Mechanical Engineering,
                                                                              The University of Queensland


                                        Fo
Case #2 For    n , x 
                                K
i.e. the system follows a ‘static’ response, in phase with input (eg. car going up hill)


                                        Fo  n
                                             2         Fo
Case #3 For    n , x                        
                                        K 2          m 2

i.e. output is attenuated from static
                  Fo
deflection by          and inverted
                m 2
 out of phase



                         Fo
Case #4 If  st               static deflection, then
                         K

               x               1
                                     = magnification factor
               st            
                                    2
                         1    
                            n 
                               

 very important for engineering design.



eg. rockets


want to make sure  n of any vibration
modes << lowest exciting frequency
                                                      MECH2200/2210 Dynamics and Orbital Mechanics
                                                               Department of Mechanical Engineering,
                                                                       The University of Queensland




Resonance occurs when    n (forcing frequency = natural frequency)

         Free vibration will occur at natural frequency if device accelerates rapidly
          through  n
         Resonance may not have time to start- eg. wheel wobbles
         For connected network of rigid bodies, number of  n = no. DOF
         Non-rigid elastic system has  number of frequencies- eg. diving board
                                          .
         Energy x  x cos t  v  x   x  sin  t
                                   1
           Kinetic energy T  m x 2  2 sin2 t
                                   2
           more energy needed to excite higher  (high  vibration is less
              important)
           for given energy T, amplitudes at higher  will be less




DAMPED, FREE VIBRATION c  0, F=0


            ..    .
          m x  c x  kx  0

                                             u 
                      viscous damping  Y  
                                                
                                              y
                                                 


                       .              ..
let       x  A e t , x  A e  t , x  A2 e  t



                                                                 Expected solution


                                                                       Steady position
                                                      MECH2200/2210 Dynamics and Orbital Mechanics
                                                               Department of Mechanical Engineering,
                                                                       The University of Queensland




Want to find 

        m  2
                            
                  c  k e t A  0        ‘characteristic equation’

The two roots are:


                    c  c2  4 k m       c    c 
                                                       2
                                                        k
         1, 2                                                                     (13)
                             2m           2m    2m    m


                       c k
        2              0
                       m m


Response depends on roots of the characteristic equation.


Critical damping

                                                           2
                                                    c     k
Critical damping is defined as being when                   0. The characteristic
                                                    2m    m
equation has two equal, negative roots

                        cc
                                 ( cc  critical damping coefficient)
                        2m
                   2
         cc   k   2
and            n
         2m   m


        cc  2 m n  2 k m                                                             (14)

At this condition, two real roots are
                            2 km
        1  2                    n (undamped natural frequency)
                             2m

                             k
                       
                             m

                                    n t
        x  A e  t  xo e                                                              (15)
                                                    MECH2200/2210 Dynamics and Orbital Mechanics
                                                              Department of Mechanical Engineering,
                                                                      The University of Queensland




Response is non-vibratory.
Exponentially decays to zero in
minimum amount of time.


Critical damping is the target
condition for many designs




                                       - eg. the car wheels




As critical damping is a common design objective, it is convenient to define a
‘damping ratio’ 

               c
                damping coefficient
              cc
          ..         .
eom      x  2  n x   n x  0
                           2




Roots can be expressed as:

         1, 2    n   n  2 1
                                                                                        (16)

                      1, 2 t
         x  Ae                                                                        (17)

Type of solution will depend on damping ratio  .
                                                   MECH2200/2210 Dynamics and Orbital Mechanics
                                                            Department of Mechanical Engineering,
                                                                    The University of Queensland




Over damped   1 , c  cc .

eq. (16) has 2 unequal, -ve real roots.

                     1 t          2t
         x  Ae            Be           with A, B depending on boundary conditions

        .         1 t      2t
        x  A 1 e  B 2 e


Not normally achieved (or desired) in passive mechanical systems.

Non-vibratory motion, decays to 0 as t  0


eg. car overdamped-
after step input stays
low.




Over damped car wheel example:
                                                     MECH2200/2210 Dynamics and Orbital Mechanics
                                                              Department of Mechanical Engineering,
                                                                      The University of Queensland


Critically damped   1 , c  cc .

From eq.(16)                     n  0   n        (2 equal, real roots)


Solution of this equation has the form:

                             n t
       x   A  Bt  e                                                                 (18)

Response is non-vibratory. Exponentially decays to zero in minimum amount of time.

               n t         t               t
                       n e n  A  B t   e n B   n  A  B t 
       x  Be
       .


       ..   e n t B    A  B t   B  e n t                               ( 19)
       x      n              n                  n

                    n t
           n e             n A  B t   2B

Car example:
                                                            MECH2200/2210 Dynamics and Orbital Mechanics
                                                                     Department of Mechanical Engineering,
                                                                             The University of Queensland


Under damped   1, c  cc .

        ..         .                                 k                     c
CE:     x  2  n x   n x  0 ,
                         2                  n        ,             
                                                     m                    cc


         1,2    n   n  2  1

For   1 , only imaginary roots exist (complex conjugate pairs)

         1,2    n   n  2  1 ,      with j   i


         1,2    n  j d                                                                  (20)
where

        d   n 1   2 (damped natural frequency)                                            (21)

This leads to attenuating oscillations at  d - most common passive system.

The solution is of the form


        x  A1 e
                   n  jd  t  A e  n  jd  t
                                       2


         x  e n t  A sin  d t  B cos  d t 
                                                  


                may be represented as [sin + phase shift]


         x(t )  Ce n t cos d t                                                       (22)


Two coefficients C,  are to be found from initial conditions and parameters  d ,  n
are to be found from known physical properties k , m, c .

Oscillations at  d , with exponential decay  d   n - viscous damping slows motion
down.

                                                                 . .
To solve for C,  , we need two conditions- eg. t  0 , x  xo , x  x o
or t  0 , x  xo , t   , x  x etc.
                                                    MECH2200/2210 Dynamics and Orbital Mechanics
                                                               Department of Mechanical Engineering,
                                                                       The University of Queensland


                n t
      x Ce              cos d t   

      . dx
      x      n C e n t cos d t     C d e n t sin d t   
         dt                                                                              (23)
         C e n t n  cos  d t      d sin  d t   




example: underdamped wheel
                                                          MECH2200/2210 Dynamics and Orbital Mechanics
                                                                    Department of Mechanical Engineering,
                                                                            The University of Queensland




Observation of the rate of decay of ‘peaks’ may be used to estimate damping ratio


From (22)        x(t )  Ce n t cos d t   

Peaks occur after a cycle so that
                           2
        d T  2 T             (period)
                           d
After n periods t  nT

        x(t  nT )  Ce   n (t  nT ) cos  d (t  nT )   
              x( t )
                      e  n nT    (as cos terms are equal)
           x(t  nT )
             x( t )                         2   2 n
        ln                n nT   n n                            from (21)
             x(t  nT )                     d   1  2


Given the magnitude of x at 2 matched points on different cycles ( n apart), damping
ratio may be found.


May use any corresponding points but it is sensible to use peaks (most accurate)
     no phase to worry about
     max displacement
                                                MECH2200/2210 Dynamics and Orbital Mechanics
                                                         Department of Mechanical Engineering,
                                                                 The University of Queensland




For the special case when n  1 (2 successive peaks)

                 x( t )              2 
         ln               n T                                              (24)
               x(t  T )             1  2
where   logarithmic decrement
                     
         
                  4 2   2


For very lightly damped systems   1  2 ~ 0

          2                                                                   (25)

For car wheel example:  d =21.65 r/s,  n =25 r/s,  =0.5
                                                                  MECH2200/2210 Dynamics and Orbital Mechanics
                                                                            Department of Mechanical Engineering,
                                                                                    The University of Queensland




DAMPED, FORCED VIBRATION c  0, F  0

Need full equation of motion

                       ..    .
        Fo cos  t  m x  c x  k x                                            from (8)


  forcing input      inertia          damping            spring stiffness


        F  0, c  0

Solution consists of superimposed steady state +
transient solutions.

Steady state solution (i.e. what happens when
starting process dies down) is of most
engineering interest study that.

eg. accelerate through wheel wobble- essential
difference– will oscillate at  , not  n or  d .

Many techniques have been developed to solve
the equation eg. Laplace transforms -Math.

Output has the form:

        x  X cos( t   )

with

                               Fo
        X                                                     and
                                                
                                                  1/ 2
                   k  m   2 2
                                    c 
                                              2


                                                                                                      (26)

                      c 
          tan 1             
                      k  m 2 


i.e. system will oscillate at the driving frequency  , with displacement peak X and
phase shift  .
                                                                       MECH2200/2210 Dynamics and Orbital Mechanics
                                                                                  Department of Mechanical Engineering,
                                                                                          The University of Queensland


Critical parameters are:

                                                  c
damping ratio                        
                                                  cc

                                                   forcing frequency
and frequency ratio                                
                                                   n natural frequency

                                     c   c        c       
noting that                                 cc       2 km    2 
                                      k   cc    k   cc      k


                                                                           from (14)

gives

                            Fo k
        X                                                                  and
                                            
                                              1/ 2
                1     
                       2 2
                               2  
                                          2


                                                                                                            (27)
                    2  
          tan 1      2 
                    1  

phase response


Noting that the static displacement (  =0) has the value of X st  Fo k .

Ratio of dynamic to static deflection (amplitude response) is given by

        X       X                                  1
                     
        X st  Fo k 
                                                                 
                                                                   1/ 2
                                          
                                              2
                                1  2              2  
                                                               2

                                                                                                            (28)
                                              MECH2200/2210 Dynamics and Orbital Mechanics
                                                       Department of Mechanical Engineering,
                                                               The University of Queensland


Example

A platform of mass 150 kg has an item of rotating machinery mounted on it,
providing an out of balance force created by a 5kg mass, mounted 50 mm off the axis
of rotation, 600 rpm. The platform is supported on a spring-damper system. It is
required that the peak deflection of the table is to be 1mm. Find the spring and
damping parameters needed to achieve this for (i)   0 (ii)   0.25
                                               MECH2200/2210 Dynamics and Orbital Mechanics
                                                           Department of Mechanical Engineering,
                                                                   The University of Queensland


General comments


     Solution for a viscously damped, forced oscillating system for any value of
      damping (through  ) and frequency (through  )

     Includes the undamped analysis from earlier (eq. 12) with   0

     Most common engineering application is to reduce or limit the amplitude of
      vibration from an applied force. For X X st <1,  must be > 2 for an
      undamped system.

     For finite values of  , this occurs at lower 

     For   1, all frequencies are attenuated

     As a general design rule- design the system for a lower  n than the lowest
      frequency of the expected load.

     A system is said to be in ‘resonance’ if    n .

     At low  (  ),   0 . Characterised by low mass eg. spring on its own.

     At high  ,   180 o (out of phase). Mass dominates.

								
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