Transcript lecture 2 (cont)

ERT 452
VIBRATION
Lecture 2
Free Vibration of Single Degree
of
Freedom Systems
MUNIRA MOHAMED NAZARI
SCHOOL OF BIOPROCESS
UNIVERSITI MALAYSIA
PERLIS1
Page
COURSE OUTLINE
Free Vibration with Viscous Damping
Free Vibration with Coulomb Damping
Page 2
Free Vibration with Viscous Damping
Page 3
Free Vibration with Viscous Damping
• Equation of Motion:
F  cx
(2.58)
where c = damping constant
From the figure, Newton’s law yields the equation
of motion:
mx  cx  kx
or
mx  cx  kx  0
(2.59)
Page 4
Free Vibration with Viscous Damping
Assume a solution in the form:
x(t )  Cest
(2.60)
Hence, the characteristic equation is
ms  cs  k  0
2
(2.61)
the roots of which are
 c  c  4mk
c
k
 c 


 
 
2m
2m
 2m  m
2
s1, 2
2
(2.62)
These roots give two solutions to Eq.(2.59)
x1 (t )  C1e s1t and x2 (t )  C2e s2t
(2.63)
Page 5
Free Vibration with Viscous Damping
Thus the general solution is:
x(t )  C1e s1t  C2e s2t
 C1e
2
 c


 c  k 
 
  t

2
m
2
m

 m



 C2 e
2
 c


 c  k 
 
  t

2
m
2
m

 m



(2.64)
where C1 and C2 are arbitrary constants to be
determined from the initial conditions of the system.
Page 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
or
k
 cc 

  0
 2m  m
k
cc  2m
 2 km  2mn
m
(2.65)
The damping ratio ζ is defined as:
  c / cc
(2.66)
Page 7
Free Vibration with Viscous Damping
Thus the general solution for Eq.(2.64) is:
x(t )  C1e
    2 1  t

 n
 C2e
    2 1  t

 n
(2.69)
Assuming that ζ ≠ 0, consider the following 3 cases:
Case1. Underdamped system
(  1 or c  cc or c/ 2m  k / m )
For this condition, (ζ2-1) is negative and the roots are:


    i 1   
s1     i 1   n
2
s2
2
n
Page 8
Free Vibration with Viscous Damping
and the solution can be written in different forms:
x(t )  C1e
e
   i 1 2  t

 n
 n t
C e

 C2 e
i 1 2  n t
1
  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


(2.70)
where (C’1,C’2), (X,Φ), and (X0, Φ0) are arbitrary
constants to be determined from initial conditions.
Page 9
Free Vibration with Viscous Damping
For the initial conditions at t = 0,
C1  x0 and C2 
x0   n x0
1   2 n
(2.71)
and hence the solution becomes
x(t )  e
 n t


x0  n x0


2
2
sin 1   nt  (2.72)
 x0 cos 1   nt 
2

1   n



Eq.(2.72) describes a damped harmonic motion.
Its amplitude decreases exponentially with time,
as shown in the figure below.
Page 10
Free Vibration with Viscous Damping
The frequency of damped vibration is:
d  1   2 n
(2.76)
Underdamped Solution
Page 11
Free Vibration with Viscous Damping
Case2. Critically damped system (  1 or c  ccor c/ 2m  k / m )
In this case, the two roots are:
s1  s2  
cc
 n
2m
(2.77)
Due to repeated roots, the solution of Eq.(2.59) is given by
x(t )  (C1  C2t )e nt
(2.78)
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
(2.80)
Page 12
Free Vibration with Viscous Damping
It can be seen that the motion represented by Eq.(2.80) is
aperiodic (i.e., nonperiodic). 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
Page 13
Free Vibration with Viscous Damping
Case3. Overdamped system (  1 or c  ccor 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:
    2 1  t

 n
    2 1  t

 n
x(t )  C1e
 C2e
For the initial conditions at t = 0,
C1 
C1 
(2.81)


x0n    2  1  x0
2n  2  1


 x0n    2  1  x0
2n   1
2
(2.82)
Page 14
Free Vibration with Viscous Damping
• Logarithmic Decrement:
Using Eq.(2.70),
 n t1
cos(d t1  0 )
x1 X 0 e

 n t 2
x2 X 0 e
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
x2
 d 2m
1 
(2.85)
Page 15
Free Vibration with Viscous Damping
For small damping,
  2
Hence,
 
or
if

2    2
2

 
2
  1
(2.86)
(2.87)
(2.88)
Thus,
1  x1 

  ln 
m  xm1 
where m is an integer.
(2.92)
Page 16
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
2
 dx 
W  
c  dt   cX 2d cos 2 d t  d (d t )
t 0
0
 dt 
 cd X 2
(2.94)
( 2 /  d )
Page 17
Free Vibration with Viscous Damping
Consider the system shown in the figure below. The total force
resisting the motion is:
F  kx  cv  kx  cx
(2.95)
If we assume simple harmonic motion:
x(t )  X sin d t
(2.96)
Thus, Eq.(2.95) becomes
F  kX sin d t  cd X cos d t
(2.97)
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
(2.98)
Page 18
Free Vibration with Viscous Damping
Computing the fraction of the total energy of the vibrating
system that is dissipated in each cycle of motion,
 2  c 
cd X 2
W


 2
  2  4  constant
1
W
 d  2m 
m d2 X 2
2
(2.99)
where W is either the max potential energy or the max
kinetic energy.
The loss coefficient, defined as the ratio of the energy
dissipated per radian and the total strain energy:
loss coefficien t 
(W / 2 ) W

W
2W
(2.100)
Page 19
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 (a). 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
Page 20
Free Vibration with Viscous Damping
In the underdamped case, the frequency of damped
vibration is given by:
d  1   2 n
where
n 
and
kt
J0
ct
ct
ct
  

ctc 2 J 0n 2 kt J 0
(2.103)
(2.104)
(2.105)
where ctc is the critical torsional damping constant
Page 21
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.
Page 22
Example 2.11
Solution
Approach: We use the equation for the logarithmic
decrement in terms of the damping ratio, equation for the
damped period of vibration, time corresponding to
maximum displacement for an underdamped system, and
envelope passing through the maximum points of an
underdamped system.
Page 23
Example 2.11
Solution
Since
x1.5  x1 / 4,
,
x2  x1.5 / 4  x1 / 16
Hence the logarithmic decrement becomes
 x1 
2
  ln    ln 16  2.7726 
1  2
 x2 
(E.1)
From which ζ can be found as 0.4037. The damped period
of vibration given by 2 s. Hence,
2 d 
n 
2
d

2
n 1   2
2
2 1  (0.4037)
2
 3.4338 rad/s
Page 24
Example 2.11
Solution
The critical damping constant can be obtained:
cc  2mn  2(200)(3.4338)  1.373.54 N - s/m
Thus the damping constant is given by:
c  cc  (0.4037)(1373.54)  554.4981 N - s/m
and the stiffness by:
k  mn2  (200)(3.4338) 2  2358.2652 N/m
The displacement of the mass will attain its max value at
time t1, given by
sin d t1  1   2
This gives:
or
sin d t1  sin t1  1  (0.4037) 2  0.9149
t1 
sin 1 (0.9149)

 0.3678 sec
Page 25
Example 2.11
Solution
The envelope passing through the max points is:
x  1   2 Xe  nt
(E.2)
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 differentiating
the displacement:
x(t )  Xe nt sin d t
x (t )  Xe nt (n sin d t  d cos d t )
as
(E.3)
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
Page 26
Free Vibration with Coulomb
Damping
Page 27
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
(2.106)
where N is normal force,
μ is the coefficient of sliding or kinetic friction
μ is usu 0.1 for lubricated metal, 0.3 for nonlubricated
metal on metal, 1.0 for rubber on metal
 Coulomb damping is sometimes called constant damping
Page 2828
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).
Page 2929
Free Vibration with Coulomb Damping
Case 1. When x is positive and dx/dt is positive or when x is
negative and dx/dt is positive, the equation of motion can be
obtained using Newton’s second law (Fig.b):
mx  kx  N
or
mx  kx   N
(2.107)
Hence,
x(t )  A1 cos nt  A2 sin nt 
N
k
(2.108)
where ωn = √k/m is the frequency of vibration
A1 & A2 are constants
Page 3030
Free Vibration with Coulomb Damping
Case 2. When x is positive and dx/dt is negative or when x is
negative and dx/dt is negative, the equation of motion can be
derived from Fig. (c):
 kx  N  mx
or mx  kx  N
(2.109)
The solution of the equation is given by:
x(t )  A3 cos nt  A4 sin nt 
N
k
(2.110)
where A3 & A4 are constants
Page 3131
Free Vibration with Coulomb Damping
Fig.2.34 Motion of the mass with Coulomb damping
Page 3232
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
(2.111)
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
(2.112)
33
Page 33
Free Vibration with Coulomb Damping
The solution is valid for half the cycle only, i.e., for 0 ≤
t ≤ π/ωn. Hence, the solution becomes the initial
conditions for the next half cycle. The procedure
continued until the motion stops, i.e., when xn ≤ μN/k.
Thus the number of half cycles (r) that elapse before
2N N
r

the motion ceases xis:
k
k
0
N 

 x0  k 
r

 2N 
 k

(2.115)
That is,
Page 3434
Free Vibration with Coulomb Damping
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.
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.
35
Page 35
Free Vibration with Coulomb Damping
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
4N
of any two consecutive
cycles
are(2.related:
X X 
116)
m
m 1
k
As amplitude is reduced by an amount 4μN/k in one
cycle, the slope of the enveloping straight lines
(shown dotted) in Fig 2.34.
Page 3636
Free Vibration with Coulomb Damping
• Torsional Systems with Coulomb Damping:
The equation governing the angular oscillations of the
system is
and
J 0  kt  T
J 0  kt  T
(2.117)
(2.118)
The frequency of vibration is given by
n 
kt
J0
(2.119)
37
Page 37
Free Vibration with Coulomb Damping
and the amplitude of motion at the end of the rth half
cycle (θr) is given by:
2T
r  0  r
kt
T



0

kt
r
2
T

 kt
(2.120)





The motion ceases(2when
.121)
38
Page 38
Example 2.14
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.
Page 3939
Example 2.14
Solution
(1) The number of half cycles that elapse before the angular motion
of the pullet ceases is:
T

 0  k
t
r
 2T
 kt





(E.1)
where θ0 = initial angular displacement = 6° = 0.10472
rad, kt = torsional spring constant of the shaft given by


(8 10 ) (0.05) 4 
GJ
32

  49,087.5 N - m/rad
kt 

l
1
10
40
Page 40
Example 2.14
Solution
and T = constant friction torque applied to the pulley = 400 Nm. Eq.(E.1) gives
 400 
0.10472  

 49,087.5   5.926
r
 800 


 49,087.5 
Thus the motion ceases after six half cycles.
(2) The angular displacement after six half cycles:
 400 
  0.10472  6  2
  0.006935 rad  0.39734
 49,087.5 
Page 4141
Page 42