Mechanical Vibrations
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Single degree of freedom system with viscous damping:
System free body diagram
C is the damping
constant or
coefficient
v Mathematical model (governing equation of motion):
.. .
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m x  c x  kx  0
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c x  Fd is thedampingforce.
a
It is proportion
al to velocity
t
i Solution:
o Assume x(t) = B est, substitute into the equation of motion:
2
n

c

c
s1, 2 
s
m s2  cs  k  0
 4m k
2m
.
s1, 2  
c

2m
2
k
 c 



m
 2m 
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DefineC  2mn , then therootsof thesolutioncan be writtenas :


s1, 2      2  1 n , where is known as thedam ping ratio.
C
  , where Cc is thecritical dam ping coe
fficient
Cc
Cc  2mn . In theface of the value of thesquare rootshown above
threepossible solutionsare exist.T hesesolutionsare as follows:
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Solution # 1: under-damped vibration:
When ζ<1
The roots S1,2 of the characteristic equation can now be written
as: s     i 1   2 
1, 2
n
T hesolut ionfor t hiscase becomes:




xt   e  nt B1 cos 1   2 nt  B2 sin 1   2 nt
OR
.


 
xt   Xent cos 1   2 nt  


Defined  1   2 nt as t hedam ped nat
ural frequency, t hen
xt   Ae nt cosd t   
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The solution (above) for the under-damped vibration case (<1
consists of a harmonic motion of frequency d and an amplitude
 n t
Ae
Note that:
 B2 
A  B  B and   t an  
 B1 
is t hephase angle. A and  can be det erminedfrom t heinat ial
2
1
2
2
1
condit ionsof t hemotion.
.
T heamplit defor t hiscase is exponant ially decaying wit h
t imeas shown.
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Frequency of damped vibration:
d  1   2 n
Under-damped vibration
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Critical damping (cc)
The critical damping cc is defined as the value of the damping constant c
for which the radical in s-equation becomes zero:
2
k
k
 cc 
 0  cc  2m
 2mn

 
m
m
 2m 
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Solution # 2: critically damped vibration (ζ=1):
For this case the roots of the characteristic equation become:
s1, 2   n
T herefor he
t solutioncan be writtenas :
xt   e nt B1  B2t
B1 and B2 are constantsthatcan be determinedfrominatial
conditions. T hemotionis no longer harmonicas shown in the
figure.
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-The system returns to the equilibrium position in short time
-The shape of the curve depends on initial conditions as shown
-The moving parts of many electrical meters and instruments are critically
damped to avoid overshoot and oscillations .
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Solution # 3: over-damped vibration (ζ>1)
For this case the roots of the characteristic equation become:

s1, 2    

 2  1 n
T herefore,thesolutioncan be writtenas,
xt   B1e
    2 1  n t


 B2 e
    2 1  n t


B1 and B2 are constantsto be determinedfrom knowing
theinitialconditionsof themotion.
.
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Graphical representation of the motions of the damped systems
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Logarithmic decrement
The logarithmic decrement represents the rate at which the
amplitude of a free-damped vibration decreases. It is defined as
the natural logarithm of the ratio of any two successive
amplitudes.
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Logarithmic decrement:
x1 (t ) Aent1 cosd t1   
 nt2
x2 (t ) Ae
cosd t2   
So
t2  t1   d   d 
But
x1 (t )
ent1
 n t1  d   en d
x2 (t ) e
d
cosd t2     cos2  d t1   
 cosd t1   
Assume: δ is the logarithmic decrement
x1 (t )
2
2
  ln
  n d   n

2
x2 (t )
1   n
1  2
Logarithmic decrement: is dimensionless
2
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Logarithmic decrement
 

2 2   2
For small damping ; ζ <<1
  2
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Generally, when the amplitude after a number of cycles “n” is
known, logarithmic decrement can be obtained as follows:
1
n
  ln
x1 (t )
2

xn (t )
1  2
where x1 (t)is thefirst known amplitude,xn (t)is theotherknown
amplitudeaftern cyclesof thedecayedmotion.
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Example 2.6
An under-damped shock absorber is to be designed for a motorcycle
of mass 200 kg (Fig.(a)).When the shock absorber is subjected to an
initial vertical velocity due to a road bump, the resulting
displacement-time curve is to be as indicated in Fig.(b).
Requirements:
1. Find the necessary stiffness and damping constants of the shock
absorber if the damped period of vibration is to be 2s and the
amplitude x1 is to be reduced to one-fourth in one half cycle (i.e. x1.5 =
x1/4 ).
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Example 2.6
Note that this system is under-damped system
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Solution:
Finding k and c
Here, n  0.5 and
1
n
  ln
x1
1
2

ln4   2.7726
x n 0 .5
1  2
   0.4037
d  2 
.
x1
 4, t hen t helogarit hmic decrementis :
xn
2
d

2
n 1  
2
 n 
2
n 1  0.4037
2
 3.4338rad / s
v The critical damping can be found as:
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cc  2mn  22003.4338  1373.54 N.sec/m
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t Damping coefficient C and stiffness K can be found as:
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o c   c  0.40371373.54  554.4981N.sec/m.......(Ans.)
c
n
s
k  mn2  2003.4338  2358.2652N/m........(Ans.)
2
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End of chapter