#### Transcript response of a damped system under the harmonic

```RESPONSE OF A
DAMPED SYSTEM
UNDER THE
HARMONIC MOTION
OF THE BASE
y(t) = Displacement of the base
 From
the figure, the equation of motion is
mx  c( x  y )  k ( x  y )  0
(3.64)
If y (t )  Y sin t ,
mx  cx  kx  ky  cy  kY sin t  cY cos t
 A sin( t   )
where
 c 
A  Y k  (c ) and   tan 
 k 
2
2
1
(3.65)

The steady-state response of the mass can be
expressed as
Y k 2  (c ) 2
x p (t ) 
sin( t  1   )
2 2
2 1/ 2
(k  m )  (c )

where

(3.66)
 c 
2 
 k  m 
1  tan 1 
OR using trigonometric identities,
x p (t )  X sin( t   )
(3.67)
where
Displacement
transmissibility

X 
k  (c )


Y  (k  m 2 )  (c ) 2 
2
2
1/ 2
 1  (2r )


2 2
2
 (1  r )  (2r ) 
2



mc 3
2r 3
1 
  tan 

tan

2
2
2
2
k
(
k

m

)

(

c
)
1

(
4


1
)
r




1
1/ 2
(3.68)
(3.69)
 If
expressed in complex form, response of the
system can be expressed as
 1  i 2r  it 
Ye 
x p (t )  Re 
2
 1 - r  i 2r 

 and
(3.70)
the displacement transmissibility as


X
2 1/ 2
 Td  1  (2r )
H (i )
Y
(3.71)
Displacement transmissibility :
The ratio of the amplitude of the response x p (t ) to that of the motion y(t).
 The
variations of displacement
transmissibility is shown in the figure below.
EXAMPLE
VEHICLE MOVING ON A ROUGH ROAD

The figure below shows a simple model of a motor
vehicle that can vibrate in the vertical direction while
traveling over a rough road. The vehicle has a mass of
1200kg. The suspension system has a spring constant
of 400 kN/m and a damping ratio of ζ = 0.5. If the
vehicle speed is 20 km/hr, determine the displacement
amplitude of the vehicle. The road surface varies
sinusoidally with an amplitude of Y = 0.05m and a
wavelength of 6m.
SOLUTION
The frequency can be found by
 v 1000  1
  2f  2 
 3600  6
For v = 20 km/hr, ω = 5.81778 rad/s. The natural
frequency is given by,
k  400  10
n 
 
m  1200
3
1/ 2



Hence, the frequency ratio is
 5.81778
r

 0.318653
n 18.2574
7
SOLUTION CON’T
The amplitude ratio can be found from
Eq.(3.68):
8
1/ 2

X  1  (2r )


Y  (1  r 2 ) 2  (2r ) 2 
 1.469237
2
1/ 2


1  (2  0.5  0.318653)

2
2
(
1

0
.
318653
)

(
2

0
.
5

0
.
318653
)


2
Thus, the displacement amplitude of the vehicle
is given by
X  1.469237Y  1.469237(0.05)  0.073462 m
This indicates that a 5cm bump in the road is
transmitted as a 7.3cm bump to the chassis and
the passengers of the car.
VIBRATION UNDER GENERAL
FORCING CONDITIONS
INTRODUCTION
A general forcing function may be periodic
(nonharmonic) or nonperiodic. A nonperiodic forcing
function may be acting for a short, long, or infinite
duration.
 Shock is defined as the small forcing function or
excitation as compared to the natural time period of the
system.
 Some examples of general forcing functions include the
motion imparted by a cam to the follower; the vibration
felt by an instrument when its package is dropped from
a height; etc.
 The transient response of a system can be found by
using what is known as the convolution integral.

RESPONSE UNDER A GENERAL PERIODIC
FORCE
 The
equation of motion can be expressed as

a0 
mx  cx  kx  F (t )    a j cos jt   b j sin jt
2 j 1
j 1
(4.4)
 The
steady-state solution of the equation is
derived as:
(a j / k )
a0 
x p (t ) 

cos( jt   j )
2
2
2
2
2k j 1 (1  j r )  (2jr)


j 1
(b j / k )
(1  j r )  (2jr )
2 2 2
2
sin( jt   j )
(4.13)
RESPONSE UNDER A NONPERIODIC FORCE
 When
the exciting force F(t) is nonperiodic, such
as that due to the blast from an explosion, a
different method of calculating the response is
required.
 Various
methods can be used to find the
response of the system to an arbitrary excitation.
Some of these methods are as follows:
1.
2.
3.
4.
5.
Representing the excitation by a Fourier integral
Using the method of convolution integral
Using the method of Laplace transforms
First approximating F(t) by a suitable interpolation model
and then using a numerical procedure
Numerically integrating the equation of motion.
CONVOLUTION INTEGRAL
Impulse  Ft  mx2  mx1
(4.17)
By designating the magnitude of the impulse
by F, we can write, in general,
F 
t  t
t
~
Fdt
(4.18)
A unit impulse ( f ) is defined as
~
f  lim
~

t  t
t 0 t
Fdt  Fdt  1
(4.19)
CONVOLUTION INTEGRAL
• Response to an impulse
For an undamped system, the solution of the
(4.20)
equation of motion mx  cx  kx  0
is given by Eq.(2.27) as
x(t )  e
where
 nt


x0  n x0
sin d t 
 x0 cos d t 
d


(4.21)
c
 
2mn
d  n
n 
k
m
k c
2
1  
 
m m
(4.22)
2
(4.23)
(4.24)
14
CONVOLUTION INTEGRAL
Figure 4.3: A single degree of freedom system subjected to an impulse
15
CONVOLUTION INTEGRAL
If the mass is at rest before the unit impulse is
applied, we obtain, from the impulsemomentum relation,
Impulse  f  1  mx (t  0)  mx (t  0  )  mx0
(4.25)
~
Thus the initial conditions are given by
x(t  0)  x0  0
1
x (t  0)  x0 
m
(4.26)
Hence, Eq.(4.21) reduces to
e  nt
x(t )  g (t ) 
sin d t
md
(4.27)
16
CONVOLUTION INTEGRAL
which is also known as the impulse response
function. The Eq.(4.27) is shown in Fig.4.3(c). If
the magnitude of the impulse is F instead of
~

x
unity, the initial velocity 0 is F/ m and the
~
response of the system becomes
x(t ) 
Fe
~
 n t
md
sin d t  F g (t )
~
(4.28)
If the impulse is applied at an arbitrary time t =
τ, it will change the velocity at t = τ, shown in
Fig.4.4(a). Thus,
x(t )  F g (t   )
~
(4.29)
17
CONVOLUTION INTEGRAL
Figure 4.4: Impulse Response
18
EXAMPLE 4.4
RESPONSE OF A STRUCTURE UNDER IMPACT
In the vibration testing of a structure, an impact
hammer with a load cell to measure the impact
force is used to cause excitation, as shown in
Fig.4.5(a). Assuming m = 5kg, k = 2000 N/m, c =
10 N-s/m and F~ = 20 N-s, find the response of the
system.
19
SOLUTION
From the known data, we can compute,
k
2000
c
c
10

 20 rad/s,   

 0.05,
m
5
cc 2 km 2 2000(5)
20
n 
d  1   2 n  19.975 rad/s
Assuming that the impact is given at t = 0, we
find the response of the system as
e  nt
x1 (t )  F
sin d t
~ m
d
20

e 0.05( 20)t sin 19.975t
(5)(19.975)
 0.20025e t sin 19.975t m
(E.1)
CONVOLUTION INTEGRAL
• Response to General Forcing Condition
Consider the response of the system under an
arbitrary external force, shown in Fig.4.6.
Hence, the response is given by
x(t )  F ( )g (t   )
Fig.4.6: An arbitrary (nonperiodic) forcing
function
(4.30)
21
CONVOLUTION INTEGRAL
The total response at time t can be found by
summing all the responses due to the
elementary impulses acting at all times τ:
x(t )   F ( ) g (t  )
(4.31)
Letting  0 and replacing the summation by
integration, we obtain
1 t
 n ( t  )
x(t ) 
F
(

)
e
sin d (t   )d (4.33)

md 0
which is called the convolution or Duhamel integral.
22
TWO DEGREE OF FREEDOM
SYSTEMS
INTRODUCTION
 As
is evident from the systems shown in
Figs.5.1 and 5.2, the configuration of a system
can be specified by a set of independent
coordinates termed as generalized coordinates,
such as length, angle, or some other physical
parameters.
 Principle coordinates is defined as any set of
coordinates that leads a coupled equation of
motion to an uncoupled system of equations.
24
EQUATIONS OF MOTION FOR FORCED VIBRATION
Consider a viscously damped two degree of
freedom spring-mass system, shown in Fig.5.3.
Figure 5.3: A two degree of freedom spring-mass-damper system
EQUATIONS OF MOTION FOR FORCED VIBRATION
m1x1  (c1  c2 ) x1  c2 x2  (k1  k2 ) x1  k 2 x2  F1
m2 x2  c2 x1  (c2  c3 ) x2  k2 x1  (k2  k3 ) x2  F2
26
The application of Newton’s second law of motion
to each of the masses gives the equations of
motion:
(5.1)
(5.2)
Both equations can be written in matrix form




as
[m]x (t )  [c]x (t )  [k ]x (t )  F (t )
(5.3)
where [m], [c], and [k] are called the mass,
damping, and stiffness matrices, respectively,
and are given by
EQUATIONS OF MOTION FOR FORCED VIBRATION
c1  c2  c2 
[c ]  


c
c

c
2
2
3

k1  k 2  k 2 
[k ]  


k
k

k
2
2
3

And the displacement and force vectors are
given respectively:

 x1 (t ) 
F1 (t ) 

x (t )  
F (t )  


 x2 (t )
F2 (t )
It can be seen that the matrices [m], [c], and [k]
are symmetric:
27
m1 0 
[ m]  

0
m
2

EQUATIONS OF MOTION FOR FORCED VIBRATION
[m]T  [m],
[c]T  [c],
[k ]T  [k ]
where the superscript T denotes the transpose
of the matrix.
The solution of Eqs.(5.1) and (5.2) involves four
constants of integration (two for each equation).
We shall first consider the free vibration
solution of Eqs.(5.1) and (5.2).
28
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