Transcript Undamped Single Degree of Freedom System

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
Potential energy
1 .2
T  mx
2
U
T  U  constant
d
T  U   0
dt
1 2
kx
2
..
m x  kx  0
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 Vertical system
..
m x  k x   st   mg ; mg  k st
..
 m x  kx  0
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Solution to the equation of motion:
mx  kx  0
(1), it a homogenuos 2nd order
ordinary differenti al equation. Equation (1) can be written as,
x   n2 x  0
where  n 
k / m is the known as the natural frequency
solution t o equation (1) may has the solution of the type :
x (t )  Be st where B is constant and S are the roots of solution.
substitute the solution into eqation (1) yields :
( s 2   n2 ) Be st  0
.
or
s 2   n2  0
e st  0
s   j n
i.e
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The solution t o equation (1) can now be written as :
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
Put
.
A1  B1  B2 and A2  j ( B1  B2 )
Then
x(t)  A 1cosω n t  A 2 sinω n t
(2)
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The solution t o equation (1) can also be written in the form :
x(t)  Acos( ω n t  φ)
Where A 
  tan
1
(3)
A12  A22 is the amplitude of motion and
A2
is the phase angle.
A1
we can say that t he free vibration of the system consists of a harmonic motion
of an amplitude A and frequancy n .
Note : The constants A1 and A2 (or A and  ) can be obtained from
applying initial co nditions of the motion.
Inatial conditions :
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Let at t  0 ; x(t )  X 0 and x(t )  X 0 . Sustitutin g these
conditions into the general solution (2) or (3) yields :
.
A1  X 0 and A2 
X0
n
then the general solution becomes :
 . 
 X0 
x(t )  X 0 cos nt  
sin n t ........(4)

ω
 n


The amplitude A and the phase angle φ become :
2
.
 . 
X0
2
 X0 
1
A  X0  
and


tan
X 0ωn
 ω n 


Equation (4) can be used to obtain the value of displaceme nt of the
system under considerat ion at any time t.
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Graphical representation
x(t ) = A cos(ωnt – ϕ )
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Torsional vibration
A system has a shaft of torsional stiffness
of moment of ineria J D as shown. Note that the stiffness
of the shaft is calculated as;
 d4
Kt 
where I p 
is the polar moment of ineria.
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32
Applying Newtons 2 nd law :
 M 0 J D
 K   J 

J   K   0 is the equation of motion
GI p
t
D
D
t
and from which the natural frequency can be obtained as
n 
.
K t and a disk
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
3x30 x109 x 2.3475
 289,812 N / m
So: k 
3
90
2. Natural frequency : n 
k
289,812

 0.9829 rad / 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 maximum when 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 maximum when 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 the displaceme nt of the centres of pulleys 1 and 2
are x1 and x 2 respective ly. The free body diagrams of the pulleys
show that :
2w  k1 x1
and 2w  k 2 x 2 where w  mg. Note that
 2w 2w 

x  2

k2 
 k1
The system is equivalent to a mass spring system of mass m
x  2x1  2x 2 or
and stiffness k eq such that
w  k eq x, then by substituti ng
for x the equivalent stiffness of the system becomes :
k eq 
1  k1k 2 

, the equation of motion is then
4  k1  k 2 
mx  k eq x  0, and the natural frequency is n 
k eq
m
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Simple
pendulum
Governing equation:
..
 M O  J O  J O 
..
J O   mgl sin    0
Assume θ is very small
sin    
..
J O   mgl   0
Natural frequency (ωn)
JO
mgl
n 
   2
JO
mgl
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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
mgd
 

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