Mechanical Vibrations
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 Introduction
1
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 Examples
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 Examples
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 Equation of motion
Equation of
motion
Newton's 2nd
law of
motion
Other
methods
D’Alembert’s
principle
Virtual
displacement
Energy
conservation
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 Newton's 2nd law of motion
Draw the
free-body
diagram
Apply
Newton s
second law
of motion
Determine the static
equilibrium configuration
of the system
Select a
suitable
coordinate
The rate of change of
momentum of a
mass is equal to the
force acting on it.
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 Newton's 2nd law of motion
..
d  d xt  
d 2 xt 
F t  
m
m
mx
2
dt 
dt 
dt
..
..
F t    kx  m x  m x  kx  0

Energy conservation
Kinetic energy
1 .2
T  mx
2
Potential energy
1
U  kx 2
2
T  U  const ant
d
T  U   0
dt
..
m x  kx  0
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 Vertical system
..
m x  k x   st   m g ; m g  k st
..
 m x kx  0
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Solution to the equation of motion:
mx  kx  0
(1), it a homogenuos2nd order
ordinarydifferental
i equat ion.Equat ion(1) can be writ t enas,
x  n2 x  0
where n 
k / m is t heknown as t henat ural frequency
solut ion o
t equat ion (1) may has t hesolut ion of t he t ype:
x(t )  Best where B is const antand S are t heroot sof solut ion.
subst it ut e t hesolut ionint oeqat ion(1) yields :
( s 2  n2 ) Best  0
.
or
s 2  n2  0
e st  0
i.e
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s   j n
T hesolut ion ot equation (1) can now be writtenas :
x(t )  B1e jnt  B2 e  jnt
Note that:
e  jnt  cosnt  j sin nt ,
Therefore
x(t )  ( B1  B2 ) cosnt  j ( B1  B2 ) sin nt
P ut A1  B1  B2 and A2  j ( B1  B2 )
T hen
x ( t )  A 1c o sω n t  A 2 s inω n t
(2)
T hesolut ion ot equat ion (1) can also be writ t enin t heform:
v
(3)
x ( t )  Ac o s ( ω n t  φ )
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W hereA 
  t an1
A12  A22 is t heamplit udeof mot ionand
A2
is t hephase angle.
A1
t free vibrat ionof t hesyst emconsist sof a harmonicmot ion
we can say t hat he
of an amplit ude A and frequancyn .
Note : T heconst ant sA1 and A2 (or A and  ) can be obt ainedfrom
applyinginitial conditionsof t hemot ion.
Inat ialcondit ions:
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Let at t  0 ; x(t )  X 0 and x(t )  X 0 . Sust it ut ing t hese
condit ionsint o t hegeneralsolut ion(2) or (3) yields :
.
A1  X 0 and A2 
X0
n
t hen t hegeneralsolut ionbecomes:
 . 
 X0 
x(t )  X 0 cosnt  
sin nt ........(4)

ω
 n


T heamplit udeA and t hephaseangle φ become:
2
.
 . 
X0
2
 X0 
1
A  X0  
and


t
an
X 0ωn
 ω n 


Equat ion(4) can be used t o obt ain t hevalue of displacement of t he
syst emunder consideration at any t imet.
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Graphical representation
x(t ) = A cos(ωnt – ϕ )
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Torsional vibration
A systemhas a shaft of torsionalstiffness K t and a disk
of momentof ineria J D as shown. Note that thestiffness
of theshaft is calculatedas;
 d4
Kt 
where I p 
is the polar moment of ineria.
L
32
ApplyingNewtons 2 nd law :
 M 0 J 
GI p
D
 K t  J D
J D  K t  0 is theequation of motion
and from which thenaturalfrequencycan be obtainedas
n 
.

Kt
rad/sec.
JD
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
Natural Frequency (ωn) :

It is a system property. It depends, mainly, on the stiffness and the mass
of vibrating system.

It has the units rad./sec, or cycles/sec. (Hz)

It is related to the natural period of oscillation (τn) such that, τn = 2π/ωn
and
ωn = 2 π fn where fn is the natural frequency in Hz.
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Example
2.1
The column of the water tank shown in Fig.
is 90m high and is made of reinforced
concrete with a tubular cross section of
inner diameter 2.4m and outer diameter
3m. The tank mass equal 3 x 105 kg when
filled with water. By neglecting the mass of
the column and assuming the Young’s
modulus of reinforced concrete as 30 Gpa.
determine the following:
the natural frequency and the natural
period of transverse vibration of the water
tank
the vibration response of the water tank
due to an initial transverse displacement of
tank of 0.3 m and zero initial velocity.
the maximum values of the velocity and
acceleration experienced by the tank.
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Example 2.1 solution:
Initial assumptions:
1.
the water tank is a point mass
2.
the column has a uniform cross section
3.
the mass of the column is negligible
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Example 2.1 solution:
a. Calculation of natural frequency:
1. Stiffness: k  3EI
, But: I 
3
l


d
64
4
o

 d i4 


3
64
4

 2.4 4  2.3475 m 4
3x30x109 x 2.3475
 289,812 N / m
So: k 
3
90
2. Natural frequency : n 
k
289,812

 0.9829rad / s
5
m
3x10
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Example 2.1 solution:
b. Finding the response:
1. x(t ) = A cos (ωn t - ϕ )
2
 . 
 Xo 
A  X o2  
 X o  0.3m

 n 


So,
.
 . 
0 
1  X o 
1 

  0
  tan 
 tan 

 X on 
 X on 


x(t ) = 0.3 cos (0.9829 t )
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Example 2.1 solution:
c. Finding the max. velocity:
.
xt   0.30.9829sin 0.9829t 
x (t ) is maximumwhen 0.9829t /2
then t /(2 * 0.9829)
.
x max  0.30.9829  0.2949m / s
Finding the max. acceleration :
..
2
xt   0.30.9829 cos0.9829t 
..
xt  is maximumwhen 0.9829t  
then t  / 0.9829
..
2
x max  0.30.9829  0.2898m / s 2
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Example 2.2 :
Free vibration of single degree of freedom system
Example 2.2 : solution
Let t hedisplacement of t hecent resof pulleys1 and 2
are x1 and x 2 respect ively.T hefree body diagrams of t hepulleys
show t hat:
2w  k1 x1
and 2w  k 2 x 2 where w  mg. Not e t hat
 2w 2w 

x 2

k2 
 k1
T hesyst emis equivalent t o a mass springsyst emof mass m
x  2x1  2x2 or
and st iffnessk eq such t hat
w  k eq x, t henby subst it ut ing
for x t heequivalentst iffnessof t hesyst embecomes:
k eq 
1  k1k 2 

, t heequat ion of mot ionis t hen
4  k1  k 2 
mx  k eq x  0, and t henat uralfrequencyis n 
k eq
m
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Simple
pendulum
Governing equation:
M
..
O
 J O  J O 
..
J O   m glsin    0
Assume θ is very small
sin    
..
J O   m gl  0
Natural frequency (ωn)
JO
m gl
n 
   2
JO
m gl
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
Solution
 t   A1 cosnt   A2 sinnt 
 t  0  A1   o
.
 t  0
.
o
 A2n   o ,  A2 
n
.
.
o
  t    o cosnt  
sin nt 
n
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Example
2.3
Any rigid body pivoted at a
point other than its center of
mass will oscillate about the
pivot point under its own
gravitational force. Such a
system is known as a compound
pendulum (as shown). Find the
natural frequency of such a
system.
Solution
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the governing equation is found as:
b
..
r
J O   Wd sin   0
a Assume small angle of vibration:
t
..
i
J O   Wd   0
o
n So:
s
Wd
m gd
 

n 
.

JO

JO
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Example
2.5: Q2.45
Draw the free-body diagram and derive the equation of motion
using Newton s second law of motion for each of the systems
shown in Fig