Transcript Lecture 2 Free Vibration of Single Degree of

ERT 452
VIBRATION
MUNIRA MOHAMED NAZARI
SCHOOL OF BIOPROCESS
ERT 452 SESION 2011/2012UNIVERSITI MALAYSIA PERLIS 1
CO 2
Ability to DEVELOP and PLAN the solutions to
vibration problems that contain free and
forced-vibration analysis of one degree of
freedoms system.
ERT 452 SESION 2011/2012
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 2.1
Introduction
 2.2
Free Vibration of an Undamped
Translational System
 2.3
Free Vibration of an Undamped Torsional
System
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


Free Vibration occurs when a system oscillates
only under an initial disturbance with no external
forces acting after the initial disturbance.
Undamped vibrations result when amplitude of
motion remains constant with time (e.g. in a
vacuum).
Damped vibrations occur when the amplitude of
free vibration diminishes gradually overtime, due
to resistance offered by the surrounding medium
(e.g. air).
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
A spring-mass system in horizontal position.
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
Several mechanical and structural systems can be
idealized as single degree of freedom systems, for
example, the mass and stiffness of a system.
Equivalent spring-mass
system for the cam follower
system.
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
Modeling of tall structure as spring-mass system.
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

Equation of Motion Using Newton’s Second Law of
Motion:
Procedure
◦ Select a suitable coordinate to describe the position of
the mass of rigid body in the system (linear or angular).
◦ Determine the static equilibrium configuration of the
system and measure the displacement of the mass or
rigid body from its static equilibrium position.
◦ Draw free body diagram.
◦ Apply Newton’s second law of motion.
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
x (t ) when acted
If mass m is displaced a distance

upon by a resultant force F (t ) in the same
direction,


d  dx (t ) 
F (t ) 
m

dt 
dt 
If mass m is constant, this equation reduces to
2


d x (t )

F (t )  m
 mx
(2.1)
2
dt

 d x (t )

where x 
is the acceleration of the mass.
2
dt
2
9

 d x (t )

is the acceleration of the mass.
where x 
2
dt
2
For a rigid body undergoing rotational motion,
Newton’s Law gives


M (t )  J
(2.2)

where M is
moment acting on the
 the resultant

body and  and   d 2 (t ) / dt 2 are the resulting
angular displacement and angular acceleration,
respectively.
10
For undamped single degree of freedom system,
the application of Eq. (2.1) to mass m yields the
equation of motion:
F (t )  kx  mx
or
mx  kx  0
(2.3)
11

Equation of Motion Using Other Methods:
1)D’Alembert’s Principle.
The equations of motion, Eqs. (2.1) & (2.2) can
be rewritten as



F (t )  mx  0


M (t )  J  0
(2.4a )
(2.4b)
The application of D’Alembert’s principle to
the system shown in Fig.(c) yields the equation
of motion:
 kx  mx  0
or
mx  kx  0
(2.3)
12
2)Principle of Virtual Displacements.
“If a system that is in equilibrium under the
action of a set of forces is subjected to a
virtual displacement, then the total virtual
work done by the forces will be zero.”
Consider spring-mass system as shown in
figure, the virtual work done by each force
can be computed as:
13
Virtual work done by the spring force  WS  (kx)x
Virtual work done by the inertia force  Wi  (mx)x
When the total virtual work done by all the forces
is set equal to zero, we obtain
 mxx  kxx  0
(2.5)
Since the virtual displacement can have an
arbitrary value,
x  0 , Eq.(2.5) gives the
equation of motion of the spring-mass system
as
mx  kx  0
(2.3)
14
3)Principle of Conservation of Energy.
A system is said to be conservative if no energy
is lost due to friction or energy-dissipating
nonelastic members.
If no work is done on the conservative system
by external forces, the total energy of the
system remains constant. Thus the principle
of conservation of energy can be expressed
as:
T  U  constant
or
d
(T  U )  0
dt
(2.6)
15
The kinetic and potential energies are given by:
1 2
T  mx
2
or
(2.7)
1 2
U  kx
2
(2.8)
Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6)
yields the desired equation
mx  kx  0
(2.3)
16

Equation of Motion of a Spring-Mass System
in Vertical Position:
17
For static equilibrium,
W  mg  k st
(2.9)
where W = weight of mass m,
 st = static deflection
g = acceleration due to gravity
The application of Newton’s second law of motion
to mass m gives
mx  k ( x   st )  W
and since k st  W , we obtain
mx  kx  0
(2.10)
18
Notice that Eqs. (2.3) and (2.10) are identical.
This indicates that when a mass moves in a vertical
direction, we can ignore its weight, provided we
measure x from its static equilibrium position.
19
The solution of Eq. (2.3) can be found by assuming
x(t )  Ce
st
(2.11)
Where C and s are constants to be determined. Substitution
of Eq. (2.11) into Eq. (2.3) gives
C (ms2  k )
(2.12)
Since C ≠ 0, we have
ms2  k  0
(2.13)
20
And hence,
s  (
k 1/ 2
)   i n
m
( 2.13)
Roots of characteristic equation or known as eigenvalues of the problem.
Where i = (-1) 1/2 and
k 1/ 2
n  ( )
m
( 2.14)
21
Hence, the general solution of Eq. (2.3) can be
expressed as
x(t )  C1eint  C2eint
(2.15)
where C1 and C2 are constants. By using the
identities
 it
e  cos t  i sin t
x(t )  A1 cos nt  A2 sin nt
(2.16)
where A1 and A2 are new constants.
22
x(t  0)  A1  x0
x (t  0)  n A2  x0
(2.17)
Hence, A1  x0 and A2  x0 / n . Thus the solution
of Eq. (2.3) subject to the initial conditions of
Eq. (2.17) is given by
x(t )  x0 cos nt 
x0
n
sin nt
(2.18)
23

Harmonic Motion:
Eqs.(2.15),(2.16) & (2.18) are harmonic functions of
time. Eq. (2.16) can also be expressed as:
x(t )  A0 sin( nt  0 )
(2.23)
where A0 and 0 are new constants, amplitude
and phase angle respectively:

 x0 
2
A0  A   x0   

 n 
2 1/ 2



(2.24)
 x0n 

0  tan 
 x0 
1
and
(2.25)
24
Note the following aspects of spring-mass systems:
1) Circular natural frequency:
1/ 2
k

m
n  
(2.26)
Spring constant, k:
k
W
 st

mg
 st
(2.27)
Hence,
1/ 2
 g 
n   
  st 
(2.28)
25
Hence, natural frequency in cycles per
second:
1/ 2
1  g 
 
fn 
(2.29)
2   st 
and, the natural period:
1/ 2
  st 
1
 n   2  
fn
 g 
(2.30)
26
2) Velocity x (t ) and the acceleration x(t ) of the
mass m at time t can be obtained as:
dx

x (t )  (t )  n A sin( nt   )  n A cos(nt    )
dt
2
d 2x
x(t )  2 (t )  n2 A cos(nt   )   n2 A cos(nt     )
dt
3) If initial displacement
x0 
(2.31)
is zero,
  x0

x(t ) 
cos nt   
sin nt
n
2  n

x0
(2.32)
If initial velocity x0  is zero,
x(t )  x0 cos nt
(2.33)
27
4) The response of a single degree of freedom
system can be represented in the state space
or phase plane:
x
cos(nt   ) 
(2.34)
A
Where,
x
y
sin( nt   )  

(2.35)
y  x / n
An
A
By squaring and adding Eqs. (2.34) & (2.35)
cos 2 (nt   )  sin 2 (nt   )  1
x2 y2
 2 1
2
A
A
(2.36)
28
Phase plane representation of an undamped
system
29
A cantilever beam carries a mass M at the free end as
shown in the figure. A mass m falls from a height h on to
the mass M and adheres to it without rebounding.
Determine the resulting transverse vibration of the beam.
30
Using the principle of conservation of momentum:
or
mvm  ( M  m) x0
 m 
 m 
x0  
v m  
 2 gh
M m
M m
(E.1)
The initial conditions of the problem can be stated:
mg
x0  
,
k
 m 
x0  
 2 gh
M m
(E.2)
Thus the resulting free transverse vibration of the
beam can be expressed as:
x(t )  A cos(nt   )
31
where

 x0 
2
A   x0   

 n 
2 1/ 2



 x0 

  tan 
 x0n 
1
n 
k
3EI
 3
M m
l ( M  m)
32
Determine the natural frequency of the system shown
in the figure. Assume the pulleys to be frictionless and
of negligible mass.
33
The total movement of the mass m (point O) is:
 2W 2W 

2

k2 
 k1
The equivalent spring constant of the system:
Weight of the mass
 Net displaceme nt of the mass
Equivalent spring constant
 1 1  4W (k1  k 2 )
W
 4W    
keq
k1k 2
 k1 k 2 
k1k 2
keq 
4(k1  k 2 )
(E.1)
34
By displacing mass m from the static equilibrium
position by x, the equation of motion of the mass can
be written as
mx  keq x  0
(E.2)
Hence, the natural frequency is given by:
1/ 2
 keq 
n   
 m
1/ 2
 k1k 2 

 rad/sec
 m(k1  k 2 ) 
(E.3)
1/ 2
n
1  k1k 2 
fn 


 cycles/sec
2 4  m(k1  k 2 ) 
(E.4)
35
36
From the theory of torsion of circular shafts, we
have the relation:
Shear modulus
GI 0
Mt 
l
Torque
Polar moment of
inertia of cross
section of shaft
Length shaft
37
Polar Moment of Inertia:
I0 
d 4
32
(2.38)
Torsional Spring Constant:
GI0 Gd 4
kt 



l
32l
Mt
(2.39)
38

Equation of Motion:
Applying Newton’s Second Law of Motion,
J 0  kt  0
(2.40)
Thus, the natural circular frequency:
1/ 2
 kt 
n   
 J0 
(2.41)
The period and frequency of vibration in cycles per
second are:
1/ 2
 J0 
 n  2  
 kt 
1
fn 
2
(2.42)
1/ 2
 kt 
 
 J0 
(2.43)
39

Note the following aspects of this system:
1)If the cross section of the shaft supporting the disc
is not circular, an appropriate torsional spring
constant is to be used.
2)The polar mass moment of inertia of a disc is given
by:
4
4
J0 
hD
32
WD

8g
where ρ is the mass density
h is the thickness
D is the diameter
W is the weight of the disc
3)An important application: in a mechanical clock
40

General solution of Eq. (2.40) can be obtained:
 (t )  A1 cos nt  A2 sin nt
(2.44)
where ωn is given by Eq. (2.41) and A1 and A2 can
be determined from the initial conditions. If
d

 (t  0)   0 and  (t  0) 
(t  0)  0
dt
(2.45)
The constants A1 and A2 can be found:
A1   0
A2  0 / n
(2.46)
Eq. (2.44) can also represent a simple harmonic motion. 41
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 (shown in Figure). Find the
natural frequency of such a system.
42
For a displacement θ, the restoring torque (due to
the weight of the body W ) is (Wd sin θ ) and the
equation of motion is
J 0  Wd sin   0
(E.1)
Hence, approximated by linear equation:
J 0  Wd  0
(E.2)
The natural frequency of the compound pendulum:
1/ 2
 Wd 

n  
 J0 
1/ 2
 mgd 

 
 J0 
(E.3)
43
Comparing with natural frequency, the length of
equivalent simple pendulum:
J0
l
md
(E.4)
If J0 is replaced by mk02, where k0 is the radius of
gyration of the body about O,
1/ 2
 gd 
n   2 
 k0 
 
 k 02 
l  
d 
 
(E.5)
(E.6)
44
If kG denotes the radius of gyration of the body
about G, we have:
k02  kG2  d 2
and
 kG2

l    d 
 d

(E.7)
(E.8)
If the line OG is extended to point A such that
kG2
GA 
d
(E.9)
l  GA  d  OA
(E.10)
Eq.(E.8) becomes
45
Hence, from Eq.(E.5), ωn is given by
1/ 2
 g 
n   2

 k0 / d 


1/ 2
g
 
l
1/ 2
 g 


 OA 
(E.11)
This equation shows that, no matter whether the
body is pivoted from O or A, its natural frequency is
the same. The point A is called the center of
percussion.
46
Let’s try!!!
47

A simple pendulum is set into oscillation from
its rest position by giving it an angular velocity
of 1 rad/s. It is found to oscillate with an
amplitude of 0.5 rad. Find the natural
frequency and length of the pendulum.
ERT 452 SESION 2011/2012
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
Derive an expression for the natural frequency
of the simple pendulum shown in Fig. 1.10.
Determine the period of oscillation of a simple
pendulum having a mass, m = 5 kg and a
length l = 0.5 m.
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
A uniform circular disc is pivoted at point O, as
shown in Fig. 2.99. Find the natural frequency
of the system. Also find the maximum
frequency of the system by varying the value of
b.
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
Derive the equation of motion of the system
shown in Fig. 2.100, by using Newton’s second
law of motion method.
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