Mechanical Vibrations
Fifth Edition in SI Units
Singiresu S. Rao
Chapter 2
Free Vibration of Single-Degree-of-Freedom Systems
3
© 2011 Mechanical Vibrations Fifth Edition in SI Units
2
Learning Objective
• Derive the equation of motion of SDOF –using
Newton’s second law, D’Alembert ‘s principle, Virtual
displacement and energy Conservation.
• Linearise the non linear equation of motion
• Solve a spring-mass-damper system for different type
of free vibration
• Compute the natural frequency of vibration system
• Determine whether the system stable or not
4
© 2011 Mechanical Vibrations Fifth Edition in SI Units
Chapter Outline
2.1
Introduction
2.2
Free Vibration of an Undamped Translational System
2.3
Free Vibration of an Undamped Torsional System
2.4
Response of First-Order Systems and Time Constant
2.5
Rayleigh’s Energy Method
2.6
Free Vibration with Viscous Damping
2.7
Graphical Representation of Characteristic Roots and Corresponding Solutions
2.8
5
Parameter Variations and Root Locus Representations
© 2011 Mechanical Vibrations Fifth Edition in SI Units
Chapter Outline
6
2.9
Free Vibration with Coulomb Damping
2.10
Free Vibration with Hysteretic Damping
2.11
Stability of Systems
© 2011 Mechanical Vibrations Fifth Edition in SI Units
2.1
Introduction
7
© 2011 Mechanical Vibrations Fifth Edition in SI Units
2.1
2.1 Introduction
• 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)
8
© 2011 Mechanical Vibrations Fifth Edition in SI Units
2.1 Introduction
• Several
mechanical and structural systems can be idealized as
single degree of freedom systems, for example, the mass and
stiffness of a system
9
© 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2
Free Vibration of an Undamped Translational System
10 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2
2.2 Free Vibration of an Undamped Translational
System
• Equation
of Motion Using Newton’s Second Law of Motion:

If mass m is displaced
a distance x (t ) when acted 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
2

d
x (t )


where x 
is the acceleration of the mass
2
dt
11 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
For a rigid body undergoing rotational motion, Newton’s Law gives


M (t )  J

where M

(2.2)

is the resultant moment acting on the body and  and

  d 2 (t ) / dt 2
are the resulting angular displacement and angular
acceleration, respectively.
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
12 © 2011 Mechanical Vibrations Fifth Edition in SI Units
( 2.3)
2.2 Free Vibration of an Undamped Translational
System
•
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
13 © 2011 Mechanical Vibrations Fifth Edition in SI Units
or
mx  kx  0
(2.3)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion Using Other Methods:
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,
the virtual work done by each force can be
computed as:
Virtual work done by the spring force  WS  (kx)x
Virtual work done by the inertia force  Wi  (mx)x
14 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion Using Other Methods:
2.
Principle of Virtual Displacements (Cont)
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
15 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.3)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion Using Other Methods:
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
16 © 2011 Mechanical Vibrations Fifth Edition in SI Units
d
(T  U )  0
dt
( 2.6)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion Using Other Methods:
3.
Principle of Conservation of Energy (Cont)
The kinetic and potential energies are given by:
1 2
T  mx
2
1 2
U  kx
2
(2.7)
(2.8)
Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6) yields the desired
equation
mx  kx  0
17 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.3)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion of a Spring-Mass System in Vertical Position:
Consider the configuration of the spring-mass system shown in the
figure.
18 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion of a Spring-Mass System in Vertical Position:
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
19 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.10)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion of a Spring-Mass System in Vertical Position:
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.
Hence, Eq. (2.3) can be expressed as
x(t )  C1e int  C2 e  int
(2.15)
where C1 and C2 are constants
By using the identities
x(t )  A1 cos nt  A2 sin nt
where A1 and A2 are new constants
20 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.16)
2.2 Free Vibration of an Undamped Translational
System
•
Equation of Motion of a Spring-Mass System in Vertical Position:
From Eq (2.16), we have
x(t  0)  A1  x0
x (t  0)  n A2  x 0
(2.17)
Hence, A1  x0 and A2  x 0 / n
Solution of Eq. (2.3) is subjected to the initial conditions of Eq.
(2.17) which is given by
x 0
x(t )  x0 cos nt 
sin nt
n
21 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.18)
2.2 Free Vibration of an Undamped Translational
System
•
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 )
where A0 and
respectively:
(2.23)
0 are new constants, amplitude and phase angle

 x 
A0  A   x   0 

 n 
2
0
 x0n 

0  tan 
 x 0 
1
22 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2 1/ 2



(2.24)
(2.25)
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
The nature of harmonic oscillation can be represented graphically as
shown in the figure.
23 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
1.
When the spring-mass system is in a vertical position
Circular natural frequency: n   k 
 m
Spring constant, k:
 g 

Hence, n  
  st 
k
W mg

 st  st
1/ 2
24 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.28)
1/ 2
(2.26)
(2.27)
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
1.
When the spring-mass system is in a vertical position (Cont)
Natural frequency in cycles per second:
1
fn 
2
Natural period:
 g

  st

1/ 2


  st
1
n 
 2 
fn
 g
25 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.29)
1/ 2



(2.30)
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
2.
Velocity x (t ) and the acceleration x(t ) of the mass m at time t
can be obtained as:
dx

(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
x (t ) 
26 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.31)
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
3.
If initial displacement
 x0 
is zero,
x 0
x 0


x(t ) 
cos nt   
sin nt
n
2  n

If initial velocity  x 0  is zero,
x(t )  x0 cos n t
27 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.33)
(2.32)
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
4.
The response of a single degree of freedom system can be
represented by:
x (t )   An sin(nt   )
(2.34)
sin(nt   )  
x
y

An
A
(2.35)
By squaring and adding Eqs. (2.34) & (2.35)
cos 2 (nt   )  sin 2 (nt   )  1
x2 y 2
 2 1
2
A
A
28 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.36)
where
y  x / n
2.2 Free Vibration of an Undamped Translational
System
•
Harmonic Motion
Note the following aspects of spring-mass systems:
4.
Phase plane representation of an undamped system
29 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
Example 2.2
Free Vibration Response Due to Impact
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 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
Example 2.2
Free Vibration Response Due to Impact
Solution
Using the principle of conservation of momentum:
mvm  ( M  m) x 0

m 
m 

 vm  

 M m
 M m
x 0  
2 gh
(E.1)
The initial conditions of the problem can be stated:
mg
x0  
,
k
31 © 2011 Mechanical Vibrations Fifth Edition in SI Units

m 

 M m
x 0  
2 gh
(E.2)
2.2 Free Vibration of an Undamped Translational
System
Example 2.2
Free Vibration Response Due to Impact
Solution (Cont)
Thus the resulting free transverse vibration of the beam can be
expressed as
x(t )  A cos(nt   )
where

 x 0 

A   x  

 n 
2
0
2 1/ 2



,
x 0 

  tan 
 x0n 
32 © 2011 Mechanical Vibrations Fifth Edition in SI Units
1

, n 
k
3EI
 3
M m
l ( M  m)
2.2 Free Vibration of an Undamped Translational
System
Example 2.5
Natural Frequency of Pulley System
Determine the natural frequency of the system.
Assume the pulleys to be frictionless and of negligible mass.
33 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.2 Free Vibration of an Undamped Translational
System
Example 2.5
Natural Frequency of Pulley System
Solution
 2W 2W 
The total movement of the mass m (point O) is 2
 k  k 
2 
 1
The equivalent spring constant of the system is
Weight of the mass
 Net displacement of the mass
Equivalent spring constant
 1 1  4W (k1  k 2 )
W
 4W 
  
k eq
k1k 2
 k1 k 2 
k1k 2
k eq 
4(k1  k 2 )
34 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.1)
2.2 Free Vibration of an Undamped Translational
System
Example 2.5
Natural Frequency of Pulley System
Solution
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)
Natural frequency is given by
 keq 

n  
 m
1/ 2

k1k 2 

m
(
k

k
)
1
2 



1 
k1k 2 
fn  n 


2 4  m(k1  k 2 ) 
35 © 2011 Mechanical Vibrations Fifth Edition in SI Units
1/ 2
rad/sec
(E.3)
cycles/sec
(E.4)
1/ 2
2.3
Free Vibration of an Undamped Torsional System
36 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.3
2.3 Free Vibration of an Undamped Torsional System
•
From the theory of torsion of circular shafts, we have the relation:
Mt 
GI 0
l
(2.37)
where
Mt = torque that produces the twist θ,
G = shear modulus,
l = is the length of shaft,
I0 = polar moment of inertia of cross section of shaft
37 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.3 Free Vibration of an Undamped Torsional System
•
Polar Moment of Inertia:
d 4
I0 
32
•
(2.38)
Torsional Spring Constant:
M t GI 0 Gd 4
kt 



l
32l
38 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.39)
2.3 Free Vibration of an Undamped Torsional System
•
Equation of Motion:
Applying Newton’s Second Law of Motion,
J 0  kt  0
(2.40)
The natural circular frequency is
 kt 

n  
 J0 
1/ 2
(2.41)
The period and frequency of vibration in cycles per second are:
 J0 

 n  2 
 kt 
1/ 2
39 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.42) ,
1
fn 
2
 kt 


 J0 
1/ 2
(2.43)
2.3 Free Vibration of an Undamped Torsional System
•
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
hD 4 WD 4
J0 

32
8g
3)
where ρ = mass density
h = thickness
D = diameter
W = weight of the disc
An important application of a torsional pendulum is in a mechanical
clock
40 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
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.
41 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
Solution
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 is
J 0  Wd  0
(E.2)
The natural frequency of the1/compound
pendulum is
2
1/ 2
 Wd 

n  
 J0 
42 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 mgd 

 J0 
 
(E.3)
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
Solution
Comparing with natural frequency, the length of equivalent simple
pendulum is
J0
l
md
(E.4)
If J0 is replaced by mk02, where k0 is the radius of gyration of the body
1/ 2
about O,
 gd 
 k 02 
n  

 k0 


2
43 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.5) , l  
 d 



(E.6)
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
Solution
If kG denotes the radius of gyration of the body about G, we have:
k k d
2
0
2
G
2
 kG2

(E.7) and l  
 d 
 d

If the line OG is extended to point A such that
kG2
GA 
d
Eq.(E.8) becomes
l  GA  d  OA
44 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.9)
(E.10)
(E.8)
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
Solution
Hence, from Eq.(E.5), ωn is given by

g 
n   2

k
/
d
 0


1/ 2

 g
 
 l
1/ 2
 g 


 OA 
1/ 2
(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.
45 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.3 Free Vibration of an Undamped Torsional System
Example 2.6
Natural Frequency of Compound Pendulum
Solution
Applications of centre of percussion
46 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.4
Response of First-Order Systems and Time Constant
47 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.4
2.4 Response of First-Order Systems and Time
Constant
•
Consider a turbine rotor mounted in bearings as shown
48 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.4 Response of First-Order Systems and Time
Constant
•
The application of Newton’s second law of motion yields the
equation of motion of the rotor as
 2.47 
Jw  ct w  0
where w 
•
dw
dt
Assuming the trial solution as
w t   Ae st  2.48
where A and s are unknown constants
•
Using the initial condition, w t  0   w0 , Eq. (2.48) can be written as
w t   w0 e st
49 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.49
2.4 Response of First-Order Systems and Time
Constant
•
By substituting Eq. (2.49) into Eq. (2.47), we obtain
w0 e st  Js  ct   0
 2.50
Since w0  0 leads to “no motion” of the rotor, we assume w0  0 and
Eq. (2.50) can be satisfied only if
Js  ct  0
 2.51
Equation (2.51) is known as the characteristic equation which yields
ct
c
 Jt t
. Thus the solution, Eq. (2.49), becomes w t   w0 e
J
Because the exponent of Eq. (2.52) is known to be  ct , the time
J
constant will be equal to   J
 2.53
ct
s
50 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.4 Response of First-Order Systems and Time
Constant
•
For t  
w t   w0 e
•
c
 Jt 
 w0 e 1  0.368w0
 2.54
Thus the response reduces to 0.368 times its initial value at a time
equal to the time constant of the system.
51 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.5
Rayleigh’s Energy Method
52 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.5
2.5 Rayleigh’s Energy Method
•
The principle of conservation of energy, in the context of an
undamped vibrating system, can be restated as
T1  U1  T2  U 2
(2.55)
where subscripts 1 and 2 denote two different instants of time
•
If the system is undergoing harmonic motion, then
Tmax  U max
53 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.57)
2.5 Rayleigh’s Energy Method
Example 2.8
Effect of Mass on wn of a Spring
Determine the effect of the mass of the spring on the natural
frequency of the spring-mass system shown in the figure below.
54 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.5 Rayleigh’s Energy Method
Example 2.8
Effect of Mass on wn of a Spring
Solution
The kinetic energy of the spring element of length dy is
1 m
  yx 
dTs   s dy  

2 l
 l 
2
(E.1)
where ms is the mass of the spring
The total kinetic energy of the system can be expressed as
T  kinetic energy of mass (Tm )  kinetic energy of spring (Ts )
1 2 l 1  ms   y 2 x 2  1 2 1 ms 2
  mx 
 mx  y 0 
dy  
x
2 
2
2 l
2 3
 l  2
55 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.2)
2.5 Rayleigh’s Energy Method
Example 2.8
Effect of Mass on wn of a Spring
Solution
The total potential energy of the system is given by
U
1 2
kx
2
(E.3)
By assuming a harmonic motion
x(t )  X cos n t
(E.4)
The maximum kinetic and potential energies can be expressed as
Tmax
1
ms  2 2
  m
 X n
2
3 
56 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.5) and U max
1 2
 kX
2
(E.6)
2.5 Rayleigh’s Energy Method
Example 2.8
Effect of Mass on wn of a Spring
Solution
By equating Tmax and Umax, we obtain the expression for the natural
frequency:


k


1/ 2
n  
 m  ms 


3 


(E.7)
Thus the effect of the mass of spring can be accounted for by adding
one-third of its mass to the main mass.
57 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6
Free Vibration with Viscous Damping
58 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6
2.6 Free Vibration with Viscous Damping
•
Equation of Motion:
F  cx
(2.58)
where c = damping
•
From the figure, Newton’s law yields
that the equation of motion is
mx  cx  kx
mx  cx  kx  0
59 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.59)
2.6 Free Vibration with Viscous Damping
•
We assume a solution in the form
x(t )  Ce st
(2.60)
where C and s are undetermined constants
•
The characteristic equation is
ms 2  cs  k  0
•
(2.61)
The roots and solutions are
s1, 2 
2
 c  c  4mk
c
k
 c 

 


2m
2m
2
m
m


2
x1 (t )  C1e s1t and x2 (t )  C2 e s2t
60 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.63)
(2.62)
2.6 Free Vibration with Viscous Damping
•
Thus the general solution is
x(t )  C1e s1t  C2 e s2t

c
k 
 c 
 
   t
 
2
m
2
m
m





 C1e

2
c
k 
 c 
 
   t
 
2
m
2
m
m





 C2 e
2
(2.64)
where C1 and C2 are arbitrary constants to be determined from the initial conditions of the system
61 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
The critical damping cc is defined as the value of the damping
constant c for which the radical in Eq.(2.62) becomes zero:
2
k
k
 cc 
 2 km  2mn

   0  cc  2m
m
m
 2m 
The damping ratio ζ is defined as:
  c / cc
62 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.66)
(2.65)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Thus the general solution for Eq.(2.64) is
x(t )  C1e
    2 1   t

 n
 C2 e
    2 1   t

 n
(2.69)
Assuming that ζ ≠ 0, consider the following 3 cases:
Case 1: Underdamped system (  1 or c  cc or c/ 2m  k / m )
For this condition, (ζ2-1) is negative and the roots are

    i
s1     i 1  
2
1 
2
s2
63 © 2011 Mechanical Vibrations Fifth Edition in SI Units


n
n
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system (  1 or c  cc or c/ 2m  k / m )
The solution can be written in different forms:
x(t )  C1e
e
   i 1 2   t

 n
 n t
C e

1
 C2 e
i 1 2  n t
  i 1 2   t

 n
 C2 e
i 1 2  n t

 e  nt C1 cos 1   2 nt  C2 sin 1   2 nt

 Xe  nt sin 1   2 nt  


 X 0 e  nt cos 1   2 nt  0
64 © 2011 Mechanical Vibrations Fifth Edition in SI Units

(2.70)

where (C’1,C’2), (X,Φ),
and (X0, Φ0) are arbitrary constants
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system (  1 or c  cc or c/ 2m  k / m )
For the initial conditions at t = 0,
C1  x0 and C2 
x 0   n x0
(2.71)
1   n
2
and hence the solution becomes
x(t )  e

 n t
 x0 cos 1   nt 
2

65 © 2011 Mechanical Vibrations Fifth Edition in SI Units
x 0   n x0
1   n
2
sin 1   nt 

2

(2.72)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 1: Underdamped system (  1 or c  cc or c/ 2m  k / m )
Eq.(2.72) describes a damped harmonic motion. Its amplitude
decreases exponentially with time, as shown in the figure below.
The frequency of damped vibration is: d  1   2 n
66 © 2011 Mechanical Vibrations Fifth Edition in SI Units
( 2.76)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system (  1 or c  cc or c/ 2m  k / m )
In this case, the two roots are:
cc
s1  s2  
 n (2.77)
2m
Due to repeated roots, the solution of Eq.(2.59) is given by
x(t )  (C1  C2t )e nt
67 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.78)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system (  1 or c  cc or c/ 2m  k / m )
Application of initial conditions gives:
C1  x0 and C2  x 0  n x0
(2.79)
Thus the solution becomes:
x(t )   x0   x 0  n x0  t  e nt
68 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.80)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 2: Critically damped system (  1 or c  cc or c/ 2m  k / m )
It can be seen that the motion represented by Eq.(2.80) is a
periodic (i.e., non-periodic).
Since e  t  0 as t   , the motion will eventually diminish to zero,
as indicated in the figure below.
n
Comparison of motions with different types of damping
69 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 3: Overdamped system (  1 or c  cc or c/ 2m  k / m )
The roots are real and distinct and are given by:

   

 1 
s1      2  1 n  0
s2
2
n
0
In this case, the solution Eq.(2.69) is given by:
x(t )  C1e
    2 1   t

 n
70 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 C2 e
    2 1   t

 n
(2.81)
2.6 Free Vibration with Viscous Damping
•
Critical Damping Constant and Damping Ratio:
Case 3: Overdamped system (  1 or c  cc or c/ 2m  k / m )
For the initial conditions at t = 0,
C1 
C2 


x0n    2  1  x 0
2n  2  1


 x0n    2  1  x 0
71 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2n   1
2
(2.82)
2.6 Free Vibration with Viscous Damping
•
Logarithmic Decrement:
Using Eq.(2.70),
x1 X 0 e  nt1 cos(d t1  0 )

x2 X 0 e  nt 2 cos(d t 2  0 )

e  nt1
e
 n  t1  d 
 e n d
(2.83)
(2.84)
The logarithmic decrement can be obtained from Eq.(2.84):
x1
2
2 c
  ln   n d   n


(2.85)
2
x2
d 2m
1 
72 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
•
Logarithmic Decrement:
For small damping,
Hence,
  2

 
or
 
Thus

  1
if
 2 
2


2
1  x1 

ln
m  xm 1 
73 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2
(2.86)
(2.87)
(2.88)
(2.92)
where m is an integer
2.6 Free Vibration with Viscous Damping
•
Energy dissipated in Viscous Damping:
In a viscously damped system, the rate of change of energy with
time is given by:
dW
 dx 
2
 force  velocity  Fv  cv  c

dt
 dt 
2
(2.93)
The energy dissipated in a complete cycle is:
( 2 /  d )
W  t 0
2
 dx 
2
2
2
2
dt


0 cX d cos d t  d (d t )  cd X
 dt 
c
74 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.94)
2.6 Free Vibration with Viscous Damping
•
Energy dissipated in Viscous Damping:
Consider the system shown in the figure.
The total force resisting the motion is
F  kx  cv  kx  cx
(2.95)
If we assume simple harmonic motion is
x(t )  X sin d t
(2.96)
Eq.(2.95) becomes F   kX sin  d t  cd X cos d t
75 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.97)
2.6 Free Vibration with Viscous Damping
•
Energy dissipated in Viscous Damping:
The energy dissipated in a complete cycle will be
W  
2 /  d
t 0

2 /  d
t 0

2 /  d
t 0
Fvdt
kX 2d sin d t  cos d t  d (d t )
cd X 2 cos 2 d t  d (d t )  cd X 2
76 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.98)
2.6 Free Vibration with Viscous Damping
•
Energy dissipated in Viscous Damping:
Computing the fraction of the total energy of the vibrating system
that is dissipated in each cycle of motion,
 2 
W
cd X 2


 2
1
W
 d 
m d2 X 2
2
 c 

  2  4  constant
 2m 
(2.99)
where W is either the max potential energy or the max kinetic energy
The loss coefficient is defined as
(W / 2 ) W
loss coefficient 

W
2W
77 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.100)
2.6 Free Vibration with Viscous Damping
•
Torsional systems with Viscous Damping:
Consider a single degree of freedom torsional system with a viscous
damper as shown in figure.
78 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
•
Torsional systems with Viscous Damping:
The viscous damping torque is given by
T  ct
(2.101)
The equation of motion can be derived as:
J 0  ct  kt  0
(2.102)
where J0 = mass moment of inertia of disc
kt = spring constant of system
θ = angular displacement of disc
79 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
•
Torsional systems with Viscous Damping:
In the underdamped case, the frequency of damped vibration is
given by
d  1   2 n
(2.103)
where
n 
and
 
kt
J0
(2.104)
ct
ct
ct


ctc 2 J 0n 2 kt J 0
80 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.105)
ctc = critical torsional damping constant
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a motorcycle of
mass 200kg (shown in 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). Find the
necessary stiffness and damping constants of the shock absorber if
the damped period of vibration is to be 2 s and the amplitude x1 is to
be reduced to one-fourth in one half cycle (i.e., x1.5 = x1/4). Also find
the minimum initial velocity that leads to a maximum displacement of
250 mm.
81 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
82 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
Since x1.5
becomes
 x1 / 4, x2  x1.5 / 4  x1 / 16 , the logarithmic decrement
 x1 
2


  ln
 ln 16   2.7726 

1  2
 x2 
83 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.1)
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
From which ζ can be found as 0.4037 and the damped period of
vibration is given by 2 s. Hence,
2
2
2 d 

d n 1   2
n 
2
2 1  (0.4037)
84 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2
 3.4338 rad/s
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
The critical damping constant can be obtained as
cc  2mn  2(200)(3.4338)  1.373.54 N - s/m
Thus the damping constant is
c  cc  (0.4037)(1373.54)  554.4981 N - s/m
The stiffness is
k  mn2  (200)(3.4338) 2  2358.2652 N/m
85 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
The displacement of the mass will attain its max value at time t 1 is
sin d t1  1   2
sin d t1  sin t1  1  (0.4037) 2  0.9149
sin 1 (0.9149)
t1 
 0.3678 sec

86 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
The envelope passing through the max points is x  1   2 Xe  nt
Since x = 250mm, 0.25  1  (0.4037) 2 Xe  ( 0.4037 )(3.4338)( 0.3678)  X  0.4550 m
The velocity of mass can be obtained by
x(t )  Xe  nt sin d t
x (t )  Xe  nt ( n sin d t  d cos d t )
87 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.3)
(E.2)
2.6 Free Vibration with Viscous Damping
Example 2.11
Shock Absorber for a Motorcycle
Solution
When t = 0,
x (t  0)  x 0  Xd  Xn 1   2
 (0.4550)(3.4338) 1  (0.4037) 2
 1.4294 m/s
88 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.7
Graphical Representation of Characteristic Roots and
Corresponding Solutions
89 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.7
2.7 Graphical Representation of Characteristic Roots
and Corresponding Solutions
•
Roots of the Characteristic Equation
The free vibration of a single-degree-of-freedom spring-massviscous-damper system is governed by Eq. (2.59):
mx  cx  kx  0
 2.106
whose characteristic equation can be expressed as (Eq. (2.61)):
ms 2  cs  k  0
s 2  2wn s  wn2  0
90 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.108
2.7 Graphical Representation of Characteristic Roots
and Corresponding Solutions
•
Roots of the Characteristic Equation
The roots of Eq. (2.107) or (2.108) are given by (see Eqs. (2.62)
and (2.68)):
 c  c 2  4mk
s1 , s2 
2m
s1 , s2  wn  iwn 1   2
91 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.110 
2.7 Graphical Representation of Characteristic Roots
and Corresponding Solutions
•
Graphical Representation of Roots and Corresponding Solutions
The response of the system is given by
x t   C1e s1t  C2 e s2t
 2.111
Following observations can be made by examining Eqs. (2.110) and
(2.111):
1. The roots lying farther to the left in the s-plane indicate that the
corresponding responses decay faster than those associated with
roots closer to the imaginary axis.
2. If the roots have positive real values of s—that is, the roots lie in
the right half of the s-plane—the corresponding response grows
exponentially and hence will be unstable.
92 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.7 Graphical Representation of Characteristic Roots
and Corresponding Solutions
•
Graphical Representation of Roots and Corresponding
Solutions
3.
4.
5.
6.
7.
If the roots lie on the imaginary axis (with zero real value), the
corresponding response will be naturally stable.
If the roots have a zero imaginary part, the corresponding response
will not oscillate.
The response of the system will exhibit an oscillatory behavior only
when the roots have nonzero imaginary parts.
The farther the roots lie to the left of the s-plane, the faster the
corresponding response decreases.
The larger the imaginary part of the roots, the higher the frequency of
oscillation of the corresponding response of the system.
93 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.7 Graphical Representation of Characteristic Roots
and Corresponding Solutions
•
Graphical Representation of Roots and Corresponding
Solutions
94 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8
Parameter Variations and Root Locus Representations
95 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8
2.8 Parameter Variations and Root Locus
Representations
•
Interpretations of wn , wd , 
and  in the s-plane
The angle made by the line OA with the imaginary axis is given by
wn
sin  

wn
  sin 1 
 2.113
The radial lines pass through the origin correspond to different
damping ratios
1
The time constant of the system is defined as  
wn
96 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Interpretations of wn , wd , 
97 © 2011 Mechanical Vibrations Fifth Edition in SI Units
and  in the s-plane
2.8 Parameter Variations and Root Locus
Representations
•
Interpretations of wn , wd , 
98 © 2011 Mechanical Vibrations Fifth Edition in SI Units
and  in the s-plane
2.8 Parameter Variations and Root Locus
Representations
•
Interpretations of wn , wd , 
and  in the s-plane
Different lines parallel to the imaginary axis denote reciprocals of
different time constants
99 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
A plot or graph that shows how changes in one of the parameters
of the system will modify the roots of the characteristic equation of
the system is known as the root locus plot.
Variation of the damping ratio:
We vary the damping constant from zero to infinity and study the
migration of the characteristic roots in the s-plane.
From Eq. (2.109) when c = 0,
s1, 2  
 4mk
k

 iwn
2m
m
100 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.115 
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the damping ratio:
Noting that the real and imaginary parts of the roots in Eq. (2.109)
can be expressed as
c
  
 wn
2m
For
and
4mk  c 2
 wn 1   2  wd
2m
0    1 , we have  2  wd2  wn2
101 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.117 
 2.116 
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the damping ratio:
The radius vector will make an angle with the positive imaginary
axis with
wd
 wn
sin  
  , cos  


wn
wn
wn
with   1   2
The two roots trace loci or paths in the form of circular arcs as the
damping ratio is increased from zero to unity as shown
102 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the damping ratio:
103 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
Example 2.13
Study of Roots with Variation of c
Plot the root locus diagram of the system governed by the equation by
varying the value of c >0
3s 2  c  27  0
104 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
Example 2.13
Study of Roots with Variation of c
Solution
The roots of equation are given by
s1, 2
 c  c 2  324

6
 E.2
We start with a value of C = 0 and the roots is as shown in the figure.
Eq. (E.2) gives the roots as indicated in the Table.
105 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
Example 2.13
Study of Roots with Variation of c
Solution
106 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the spring constant:
Since the spring constant does not appear explicitly in Eq. (2.108),
we consider a specific form of the characteristic equation (2.107)
as:
2
s  16 s  k  0  2.121
The roots of Eq. (2.121) are given by
s1, 2 
 16  256  4k
 8  64  k
2
107 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.122
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the mass:
To find the migration of the roots with a variation of the mass m,
we consider a specific form of the characteristic equation, Eq.
(2.107), as
ms 2  14 s  20  0
 2.123
whose roots are given by
s1, 2
 14  196  80m

2
108 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 2.124
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the mass:
Some values of m and the corresponding roots given by Eq.
(2.124) are shown in Table.
109 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the mass:
110 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.8 Parameter Variations and Root Locus
Representations
•
Root Locus and Parameter Variations
Variation of the mass:
111 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9
Free Vibration with Coulomb Damping
112 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9
2.9 Free Vibration with Coulomb Damping
•
Coulomb’s law of dry friction states that, when two bodies are in
contact, the force required to produce sliding is proportional to the
normal force acting in the plane of contact. Thus, the friction force
F is given by:
F  N  W  mg
(2.125)
where N is normal force,
μ is the coefficient of sliding or kinetic friction
μ is 0.1 for lubricated metal, 0.3 for non-lubricated metal on metal, 1.0 for rubber on metal
•
Coulomb damping is sometimes called constant damping
113 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Equation of Motion:
Consider a single degree of freedom system with dry friction as
shown in Fig.(a) below.
Since friction force varies with the direction of velocity, we need to
consider two cases as indicated in Fig.(b) and (c).
114 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Equation of Motion:
Case 1.
When x is positive and dx/dt is positive or when x is negative and
dx/dt is positive (i.e., for the half cycle during which the mass
moves from left to right) the equation of motion can be obtained
using Newton’s second law (Fig.b):
mx  kx  N
Hence
or
mx  kx   N
N
x(t )  A1 cos nt  A2 sin nt 
k
where ωn = √k/m is the frequency of vibration
A1 & A2 are constants
115 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.126)
(2.127)
2.9 Free Vibration with Coulomb Damping
•
Equation of Motion:
Case 2.
When x is positive and dx/dt is negative or when x is negative and
dx/dt is negative (i.e., for the half cycle during which the mass
moves from right to left) the equation of motion can be derived
from Fig. (c):
 kx  N  mx
or mx  kx  N
(2.128)
The solution of the equation is given by:
N
x(t )  A3 cos nt  A4 sin nt 
k
where A3 & A4 are constants
116 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.129)
2.9 Free Vibration with Coulomb Damping
•
Equation of Motion:
Motion of the mass with Coulomb damping
117 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Solution:
Eqs.(2.107) & (2.109) can be expressed as a single equation using
N = mg:
mx  mg sgn( x )  kx  0
(2.130)
where sgn(y) is called the sigum function, whose value is defined
as 1 for y > 0, -1 for y< 0, and 0 for y = 0.
Assuming initial conditions as
x(t  0)  x0
x (t  0)  0
118 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.131)
2.9 Free Vibration with Coulomb Damping
•
Solution:
The solution is valid for half the cycle only, i.e., for 0 ≤ t ≤ π/ω n.
Hence, the solution becomes the initial conditions for the next half
cycle. The procedure continued until the motion stops, i.e., when x n
≤ μN/k. Thus the number of half cycles (r) that elapse before the
motion ceases is:
2 N N

k
k
N 
x0 
k  (2.134)
2 N 
k

x0  r


r


119 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Solution:
Note the following characteristics of a system with Coulomb
damping:
1.
The equation of motion is nonlinear with Coulomb damping, while it
is linear with viscous damping
2.
The natural frequency of the system is unaltered with the addition
of Coulomb damping, while it is reduced with the addition of
viscous damping.
120 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Solution:
Note the following characteristics of a system with Coulomb
damping:
3.
The motion is periodic with Coulomb damping, while it can be
nonperiodic in a viscously damped (overdamped) system.
4.
The system comes to rest after some time with Coulomb damping,
whereas the motion theoretically continues forever (perhaps with
an infinitesimally small amplitude) with viscous damping.
121 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Coulomb Damping
•
Solution:
Note the following characteristics of a system with Coulomb
damping:
5.
The amplitude reduces linearly with Coulomb damping, whereas it
reduces exponentially with viscous damping.
6.
In each successive cycle, the amplitude of motion is reduced by the
amount 4μN/k, so the amplitudes at the end of any two
consecutive cycles are related:
X m  X m 1 
122 © 2011 Mechanical Vibrations Fifth Edition in SI Units
4N
k
(2.135)
2.9 Free Vibration with Coulomb Damping
•
Torsional Systems with Coulomb Damping:
The equation governing the angular oscillations of the system is
J 0  kt  T
J 0  kt  T
(2.136)
(2.137)
The frequency of vibration is given by
n 
123 © 2011 Mechanical Vibrations Fifth Edition in SI Units
kt
J0
(2.138)
2.9 Free Vibration with Coulomb Damping
•
Torsional Systems with Coulomb Damping:
The amplitude of motion at the end of the rth half cycle (θr) is
given by:
2T
r  0  r
kt
(2.139)
The motion ceases when

T


 0 k
t
r
 2T

kt
124 © 2011 Mechanical Vibrations Fifth Edition in SI Units


(2.140)



2.9 Free Vibration with Viscous Damping
Example 2.15
Pulley Subjected to Coulomb Damping
A steel shaft of length 1 m and diameter 50 mm is fixed at one end
and carries a pulley of mass moment of inertia 25 kg-m2 at the other
end. A band brake exerts a constant frictional torque of 400 N-m
around the circumference of the pulley. If the pulley is displaced by 6°
and released, determine (1) the number of cycles before the pulley
comes to rest and (2) the final settling position of the pulley.
125 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Viscous Damping
Example 2.15
Pulley Subjected to Coulomb Damping
Solution
(1) The number of half cycles that elapse before the angular motion of
the pullet ceases is:
T



 0 k 
t
r
(E.1)

2
T



kt  where θ0 = 6° = 0.10472 rad,
The torsional spring constant of the shaft given by
kt 
GJ

l
 

(0.05) 4 
 32
  49,087.5 N - m/rad
1
(8 1010 )
126 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Viscous Damping
Example 2.15
Pulley Subjected to Coulomb Damping
Solution
With constant friction torque applied to the pulley = 400 N-m., Eq.
(E.1) gives

400 

49
,
087
.
5

  5.926
800 



 49,087.5 
0.10472  
r
Thus the motion ceases after six half cycles.
127 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.9 Free Vibration with Viscous Damping
Example 2.15
Pulley Subjected to Coulomb Damping
Solution
(2) The angular displacement after six half cycles:
400 

  0.10472  6  2
  0.006935 rad  0.39734
 49,087.5 
from the equilibrium position on the same side of the initial
displacement.
128 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10
Free Vibration with Hysteretic Damping
2.10
129 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10 Free Vibration with Hysteretic Damping
•
Consider the spring-viscous damper arrangement shown in the
figure below. The force needed to cause a displacement:
F  kx  cx
•
(2.141)
For a harmonic motion of frequency ω and amplitude X,

F (t )  kX sin t  cX cos t
 kx  c X 2  ( X sin t ) 2
 kx  c X 2  x 2
130 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.143)
2.10 Free Vibration with Hysteretic Damping
Spring-viscous-damper system
131 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10 Free Vibration with Hysteretic Damping
•
When F versus x is plotted, Eq.(2.143) represents a closed loop, as
shown in Fig(b). The area of the loop denotes the energy dissipated
by the damper in a cycle of motion and is given by:
W   Fdx
2 / 
 0
•
(2.144)
Hence, the damping coefficient:
h
c

•
 kX sin t  cX cos t  X cos t  dt  cX 2
(2.145)
where h = hysteresis damping constant
Eqs.(2.144) and (2.145) gives W  hX
132 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2
(2.146)
2.10 Free Vibration with Hysteretic Damping
Hysteresis loop
133 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10 Free Vibration with Hysteretic Damping
•
Complex Stiffness
For general harmonic motion, x  Xe it , the force is given by
F  kXeit  ciXe it  (k  ic) x
(2.147)
Thus, the force-displacement relation:
F  (k  ih) x
h

where k  ih  k  1  i   k (1  i )
k

134 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.148)
(2.149)
2.10 Free Vibration with Hysteretic Damping
•
Response of the system
The energy loss per cycle can be expressed as W  k X 2
The hysteresis logarithmic decrement can be defined as
 Xj 
  ln
  ln(1   )  
 X 
 j 1 
Corresponding frequency
k

m
135 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.155)
(2.154)
(2.150)
2.10 Free Vibration with Hysteretic Damping
•
Response of the system
Response of a hysteretically damped system
136 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10 Free Vibration with Hysteretic Damping
•
Response of the system
The equivalent viscous damping ratio
  2 eq   
h

h
  eq  
k
2 2k
(2.156)
Thus the equivalent damping constant is

k h
ceq  cc   eq  2 mk    mk 

2
 
137 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(2.157)
2.10 Free Vibration with Viscous Damping
Example 2.17
Response of a Hysteretically Damped Bridge Structure
A bridge structure is modeled as a single degree of freedom system
with an equivalent mass of 5 X 105 kg and an equivalent stiffness of
25 X106 N/m. During a free vibration test, the ratio of successive
amplitudes was found to be 1.04. Estimate the structural damping
constant (β) and the approximate free vibration response of the
bridge.
138 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.10 Free Vibration with Viscous Damping
Example 2.17
Response of a Hysteretically Damped Bridge Structure
Solution
Using the ratio of successive amplitudes,
 Xj 
  ln(1.04)  ln(1   )
  ln
 X 
j 1 

0.04
1    1.04 or

 0.0127

The equivalent viscous damping coefficient is
k k
ceq 

  km

k
m
139 © 2011 Mechanical Vibrations Fifth Edition in SI Units
(E.1)
2.10 Free Vibration with Viscous Damping
Example 2.17
Response of a Hysteretically Damped Bridge Structure
Solution
Using the known values of the equivalent stiffness and equivalent
mass,
ceq  (0.0127) (25 106 )(5 105 )  44.9013 103 N - s/m
Since ceq < cc, the bridge is underdamped. Hence, its free vibration
response is
x(t )  e

 n t
 x0 cos 1   nt 
2

ceq
x 0   n x0
1   n
2
40.9013 103
 

 0.0063
cc 7071.0678 103
140 © 2011 Mechanical Vibrations Fifth Edition in SI Units
sin 1   n t 

2

2.11
Stability of Systems
2.11
141 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
•
•
•
•
Stability is one of the most important characteristics for any
vibrating system
A asymptotically stable (called stable in controls literature) is when
its free-vibration response approaches zero as time approaches
infinity.
A system is considered to be unstable if its free-vibration response
grows without bound (approaches infinity) as time approaches
infinity.
A system is stable (called marginally stable in controls literature) if
its free-vibration response neither decays nor grows, but remains
constant or oscillates as time approaches infinity.
142 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
143 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
Example 2.18
Stability of a System
Consider a uniform rigid bar, of mass m and length l, pivoted at one
end and connected symmetrically by two springs at the other end, as
shown in the figure. Assuming that the springs are unstretched
when the bar is vertical, derive the equation of motion of the system
for small angular displacements of the bar about the pivot point, and
investigate the stability behavior of the system.
144 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
Example 2.18
Stability of a System
145 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
Example 2.18
Stability of a System
The equation of motion of the bar, for rotation about the point O, is
ml 2 
l
   2kl sin   l cos   W sin   0
3
2
 E.1
For small oscillations, Eq. (E.1) reduces to
ml 2 
Wl
2
  2kl     0
3
2
   2  0  E.3
146 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 E.2
2.11 Stability of Systems
Example 2.18
Stability of a System
 12kl 2  3Wl 

Where   
2
2ml


 E.4
2
The characteristic equation is given by s 2   2  0  E.5
The solution of Eq. (E.2) depends on the sign of  2 as indicated
below.
Case 1. When
12kl
2

 3Wl / 2ml 2  0
  t   A1 cos wnt  A2 sin wnt
where
 12kl 2  3Wl 

wn  
2ml


147 © 2011 Mechanical Vibrations Fifth Edition in SI Units
 E.6
1/ 2
 E.7 
2.11 Stability of Systems
Example 2.18
Stability of a System


Case 2. When 12kl  3Wl / 2ml  0
2
2
  t   C1t  C2
 E.8
For the initial conditions   t  0    0 and   t  0   0
  t     t    0
 E.9
Equation (E.9) shows that the system is unstable with the angular
displacement increasing linearly at a constant velocity
148 © 2011 Mechanical Vibrations Fifth Edition in SI Units
2.11 Stability of Systems
Example 2.18
Stability of a System


Case 3. When 12kl  3Wl / 2ml  0
2
2
  t   B1et  B2 e t
For the initial conditions
 t 

 E.10
  t  0    0 and   t  0   0


   E.11
1
 0  0 et   0  0 e t
2
Equation (E.11) shows that increases exponentially with time; hence
the motion is unstable.
149 © 2011 Mechanical Vibrations Fifth Edition in SI Units