# 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

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

   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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