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C. R. Yang, NTNU MT
Chapter 4 Vibration Under General Forcing Condition
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Learning Objectives

Find the response of single-DOF systems subjected to general
periodic forces using the Fourier series

Use the method of convolution or Duhamel (杜哈明) integral to solve
vibration problems of systems subjected to arbitrary forces

Find the response of systems subjected to earthquakes using
response spectra

Solve undamped and damped systems subjected to arbitrary forces,
including impulse, step, and ramp forces, using Laplace transform

Understand the characteristics of transient response, such as peak
time, overshoot, settling time, rise time, and decay time, and
procedures for their estimation

Apply numerical methods to solve vibration problems of systems
subjected to forces that are described numerically

Solve forced-vibration problems using MATLAB
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Chapter Outline
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Introduction
Response Under a General Periodic Force
Response Under a Periodic Force of Irregular Form
Response Under a Nonperiodic Force
Convolution Integral
Response Spectrum
Laplace Transforms
Numerical Methods
Response to Irregular Forcing Conditions Using Numerical
Methods
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4.1
Introduction
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4.1 Introduction
•
Many practical systems are subjected to several types of forcing
functions that are not harmonic.
•
The general forcing functions may be periodic (nonharmonic) or
nonperiodic. The nonperiodic forces include forces such as :
A step force: a suddenly applied constant force
A ramp force: a linearly increasing force
An exponentially varying force



•
A nonperiodic forcing function may be acting for short, long, or
infinite duration.
•
Shock: a forcing function or excitation of short duration compared
to the natural (固有的) time period of the system.
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4.1 Introduction
•
Some examples of general forcing functions include the motion
imparted (傳遞) by a cam to the follower; the vibration felt by an
instrument when its package is dropped from a height; the force
applied to the foundation of a forging press; the motion of an
automobile when it hits a pothole; and the ground vibration of a
building frame during an earthquake, etc.
•
If the forcing function is periodic but not harmonic, it can be replace
by a sum of harmonic functions using the harmonic analysis
procedures (Section 1.11). Using the principle of superposition, the
response of the system can then be determined by superposing the
responses due to the individual harmonic forcing functions.
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The vibration in a cam to the follower
The typical force fluctuation in a valve mechanism during the cam
operating cycle. The dynamic force affecting valvecam follower of a large
diesel engine during the cam operating cycle at engine running speed 500
r/min is shown.
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•
The response of a system subjected to any type of nonperiodic force
is commonly found using the following methods:
(1) Convolution integral; (2) Laplace transform; (3) Numerical
methods
•
The first two methods are analytical ones, in which the response is
expressed in a way that helps in studying the behavior of the
system under the applied force with respect to various parameters
and in designing the system.
•
The third method can be used to find the response of a system
under any arbitrary force for which an analytical solution is difficult
or impossible to find. However, the solution found is applicable only
for particular set of parameter values, which makes it difficult to
study the behavior of the system when the parameters are varied.
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4.2
Response Under a General Periodic Force
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4.2 Response Under a General Periodic Force
•
Fourier Series Expansion:
When the external force F(t) is periodic with period =2/, it can
be expanded in a Fourier Series (Section 1.11)
aj 
a0 
F (t )    (a j cos jt  b j sin jt ) (4.1)
2 j 1
2


0
F ( t )cos j tdt
j  0, 1, 2, ...
2 
b j   F ( t )sin j tdt

0
j  1, 2, ...

The response of systems under general periodic forces is considered
herein for both first- and second-order system
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Displacement
excitation
Fig. 4.1 Examples of first-order systems
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Fig. 4.2 Examples of second-order systems
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4.2 Response Under a General Periodic Force
• Consider a spring-damper system subjected to a periodic excitation as shown
in Fig. 4.1(a). The equation of motion of the system is given by
k
k
cx  k ( x  y )  0 (4.4)  x  x  y
c
c
where y(t) is the periodic motion (or excitation), which is expressed as
Fourier series


x  ax  ay  A0   Aj sin  j t   B j cos  j t (4.5)
j 1
(與Eq. (4.1)做比較)
j 1
where
aa0
k
a  , A0 
, Aj  aa j , B j  ab j ,  j  j , j  1, 2, 3,...(4.6)
c
2
a0 
F (t )  y    (a j cos jt  b j sin jt ) (4.1)
2 j 1
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Example 4.1
Response of a First-Order System under Periodic Force
Find the response of the spring-damper system, shown in Fig. 4.1(a)
subjected to a periodic force with the equation of motion given by Eq.
(4.5)


j 1
j 1
x  ax  ay  A0   Aj sin  j t   B j cos  j t (4.5)
Solution
Using the principle of superposition, the steady-state solution of Eq.
(4.5) can be found by summing the steady-state solutions
corresponding to the individual forcing terms on the right-hand of Eq.
(4.5)
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a
aa
k
, A0  0 , Aj  aa j , B j  ab j ,  j  j , j  1, 2, 3,...(4.6)
c
2
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用以求出C
j
j
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4.2 Response Under a General Periodic Force
Example 4.3
Response of a Second-Order system Under Periodic Force
• The equation of motion of Fig. 4.2(a) can be expressed as

a0 
mx  cx  kx  F (t )    a j cos jt   b j sin jt
j 1
2 j 1
(4.8)
Where the forcing function F(t) is periodic.
• To determine the response of Eq. (4.8) using the principle of
superposition.
a0
a
( E .1)  the solution is x p ( t )  0
2
2k
mx  cx  kx  a j cos j t ( E .2)
mx  cx  kx 
mx  cx  kx  b j sin j t
( E .3)
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Using the results of Section 3.4, we can express the
solutions of Eqs.(E.2) and (E.3), respectively, as
where  j  tan1 (
2 jr

)
and
r

1  j 2r 2
n
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•Thus the complete steady-state solution of Eq.(4.8) is given by

(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 ) ( E .9)
Operation frequency
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4.2 Response Under a General Periodic Force
Example 4.5 Total Response Under Harmonic Base Excitation
Find the total response of a viscously damped single-DOF system subjected to
a harmonic base excitation (3.6 Section) for the following data:
m  10kg, c  20 N - m/s , k  4000 N/m,
y (t )  0.05 sin 5t m, x0  0.02 m, x0  10 m/s.
Initial conditions
Solution
The equation of motion of the system is given by (see Eq. (3.65)):
mx  cx  kx  ky  cy  kY sin t  cY cos t
(E.1)
Eq. (E.1) is similar to Eq.(4.8) with a0  0, a1  cY , b1  kY , and ai  bi  0; i  2, 3...
The steady-state response of the system can be expressed as
b1
 a1

x p (t ) 
cos(

t


)

sin(

t


)
1
1 

k

(1  r 2 ) 2  (2r ) 2  k
1
(using Eq.(E.9) of
(E.2)
example 4.3)
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4.2 Response Under a General Periodic Force
Example 4.5
Total Response Under Harmonic Base Excitation
Solution
For the given data, We find
4000
 20 rad/s,
10

5
c
20
r

 0.25,   

 0.05,
 n 20
cc 2 km 2 (4000)(10)
Y  0.05 m,   5 rad/s,  n 
k

m
c
d  1   2  n  19.975 rad/s
a1  cY  (20)(5)(0.05)  5, b1  kY  (4000)(0.05)  200,
 2 r 
1  2(0.05)(0.25) 

tan

  0.02666 rad
2 
2
1

r
1

(0.25)




1  tan 1 
(1  r 2 )2  (2 r )2  (1  0.252 )2  (2(0.05)(0.25))2  0.937833
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4.2 Response Under a General Periodic Force
Example 4.5
Total Response Under Harmonic Base Excitation
Solution
  d  1   2  n
The solution of the homogeneous equation is given by (see Eq. (2.70)):
xh (t )  X 0e nt cos(d t  0 )  X 0e  t cos(19.975t  0 )
(E.3)
where X0 and Φ0 are unknown constants
The total solution can be expressed as the superposition of xh(t) and xp(t) as:
x(t )  X 0 e t cos(19.975t  0 ) 
1
200
 5

cos(
5
t


)

sin(
5
t


)
1
1 
0.937833  4000
4000

 X 0 e t cos(19.975t  0 )  0.0013333 cos(5t  0.026666)
 0.053314 sin( 5t  0.026666)
(E.4)
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4.2 Response Under a General Periodic Force
Example 4.5
Total Response Under Harmonic Base Excitation
Solution
The velocity of the mass can be expressed from Eq.(E.4)
dx
(t )   X 0e  t cos(19.975t  0 )  19.975 X 0e  t sin(19.975t  0 )
dt
0.006665sin(5t  0.02666)  0.266572cos(5t  0.02666)
( E .5)
x( t ) 
(使用初始條件，求出X0與0)
Using Eqs.(E.4) and (E.5), we find
x0 (t )  x(t  0)  0.02  X 0 cos 0  0.001333 cos(0.02666)  0.053314 sin( 0.02666)
 X 0 cos 0  0.020088
(E.6)
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4.2 Response Under a General Periodic Force
Example 4.5
Total Response Under Harmonic Base Excitation
Solution
x0  x (t  0)  10   X 0 cos 0  19.975 X 0 sin 0
0.006665 sin( 0.02666)  0.266572 cos(0.02666)
  X 0 cos 0  19.975sin 0  9.733345
(E.7)
The solution of (E.6) and (E.7) yields X0=0.488695 and Φ0=1.529683 rad.
Thus the total response of the mass under base excitation, in meters, is given
by
x(t )  0.488695e t cos(19.975t  1.529683)
 0.001333 cos(5t  0.02666)  0.053314 sin( 5t  0.02666)
(E.8)
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4.3
Response Under a Periodic Force of Irregular Form
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4.3 Response Under a Periodic Force of Irregular
Form
• In
some cases, the force acting on a system may be quite irregular
and may be determined only experimentally. Examples of such
forces include wind and earthquake-induced forces.
• In
such cases, the forces will be available in graphical form and no
analytical expression can be found to describe F(t).
• Sometimes,
the value of F(t) may be available only at a number of
discrete points t1, t2, t3…tN. In all these cases, it is possible to find
the Fourier coefficients by using a numerical integration procedure,
as described in Section 1.11.
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4.3 Response Under a Periodic Force of Irregular
Form
• If
F1, F2, F3…FN denote the values of F(t) at t1, t2, t3…tN, respectively,
where N denotes an even number of equidistant (等距離) points in
one time period  (=N△t), as shown in Fig. 4.5. The application of
trapezoidal (梯形的) rule gives:
x p (t ) 
2 N
a0   Fi
N i 1
2 jti
2 N
a j   Fi cos
, j  1,2,...
N i 1

2 jti
2 N
b j   Fi sin
, j  1,2,...
N i 1

(4.9)
(4.10)


j 1

(4.11)

j 1
a0
2k
(a j / k )
(1  j r )  (2 jr )
2 2 2
2
cos( j t   j )
2
sin( j t   j ) ( E .9)
(b j / k )
(1  j r )  (2 jr )
2 2 2
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4.3 Response Under a Periodic Force of Irregular
Form
• An
irregular forcing function:
• Once
the Fourier coefficients a0, aj, and bj are known, the steadystate response of the system can be found using Eq. (E.9) of
Example 4.3 with
2
 2 


r



n
n  n 
( 4.12)
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4.3 Response Under a Periodic Force of Irregular
Form
Example 4.6
Steady-State Vibration of a Hydraulic Valve
Find the steady-state response of the valve in the figure (see example
4.4) if the pressure fluctuations in the chamber are found to be
periodic. The valves of pressure measured at 0.01 second intervals in
one cycle are given below.
period
pressure
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4.3 Response Under a Periodic Force of Irregular
Form
Example 4.6
Steady-State Vibration of a Hydraulic Valve
Solution (see Example 1.20)
Since the pressure fluctuations on the valve are periodic, the Fourier
analysis of the given data of pressures in a cycle gives:
p(t )  34083.3  26996.0 cos 52.36t  8307.7 sin 52.36t
1416.7 cos104.72t  3608.3 sin 104.72t
乘上A變成力
5833.3 cos157.08t  2333.3 sin 157.08t  ... N/m 2
(E.1)
Other quantities needed for the computation are
2
2


 52.36 rad/s ,

0.12
n  100 rad/s ,

r
 0.5236
n
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4.3 Response Under a Periodic Force of Irregular
Form
Example 4.6
Steady-State Vibration of a Hydraulic Valve
Solution
We have also
  0.2
A  0.000625 m 2
 2r 
1  2  0.2  0.5236 

tan


  16.1
2
2
1 r 
 1  0.5236 
 4r 
1  4  0.2  0.5236 
2  tan 1 

tan
 77.01

2 
2 
 1  4r 
 1  4  0.5236 
 6r 
1  6  0.2  0.5236 
3  tan 1 

tan
 23.18


2
2 
 1  9r 
 1  9  0.5236 
1  tan 1 
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4.3 Response Under a Periodic Force of Irregular
Form
Example 4.6
Steady-State Vibration of a Hydraulic Valve
Solution
(Using Eq. (E.9) of Example 4.3)
The steady-state response of the valve can be expressed as
x p (t ) 
34083.3 A
(26996.0 A / k )
(8309.7 A / k )

cos(52.36t  1 ) 
sin( 52.36t  1 )
2
2
2
2
2
2
k
(1  r )  (2r )
(1  r )  (2r )


(1416.7 A / k )
2
cos(104.72t  2 ) 
2
cos(157.08t  3 ) 
(1  4r )  (4r )
2 2
(5833.3 A / k )
(1  9r )  (6r )
2 2
  0.2, A  0.000625 m 2 ,  
 n  100 rad/s, r 
2


(3608.3 A / k )
2
sin( 104.72t  2 )
2
sin( 157.08t  3 )
(1  4r )  (4r )
2 2
(2333.3 A / k )
(1  9r )  (6r )
2 2
2
 52.36 rad/s ,
0.12

 0.5236, K  ??? (see example 4.4)
n
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4.4
Response Under a Nonperiodic Force
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4.4 Response Under a Nonperiodic Force
•
We have seen that periodic forces of any general waveform can be
represented by a Fourier series as a superposition of harmonic
components of various frequencies. The response of a linear
system is then found by superposing the harmonic response to
each of the exciting forces.
•
When the exciting force F(t) is nonperiodic, such as that due to the
blast (爆破) 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:
Representing the excitation by a Fourier integral
Using the method of convolution integral
Using the method of Laplace transforms
4. Numerically integrating the equation of motion
1.
2.
3.
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5
Convolution Integral
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral

A nonperiodic exciting force usually has a magnitude that varies
with time; it acts for a specified period and then stops. The
simplest form is the impulsive force – a force that has a large
magnitude F and acts for a very short time  t. From dynamics,
we know that impulse can be measured by finding the change it
causes in momentum (動量=質量×速度) of the system.

If x1 and x2 denote the velocities of the mass m before and after
the application of the impulse, we have
Impulse  F t  mx2  mx1 (4.12)
台灣師範大學機電科技學系
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4.5 Convolution Integral
• By
designating (指定) the magnitude of the impulse
can write, in general,
t  t
F  t
•A
Fdt
Ft
by F, we
(4.13)
unit impulse acting at t=0 (f) is defined as
t  t
f  lim t
t 0
Fdt  Fdt  1
(4.14)
It can be seen that in order for Fdt to have a infinite value, F tends
to infinity (Since dt tends to zero)
台灣師範大學機電科技學系
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• The unit impulse, f=1, acting at t=0, is also denoted by the
Dirac delta function as
f  f  (t )   (t )
( 4.15)
  if t=a
 0 otherwise
 (t  a )  
and




0
The impulse of magnitude F , acting at t=0
 F  F  (t )
0
( 4.16)
 ( t  a )dt  1
g( t ) ( t  a )dt  g(a )
L{ ( t  a )}  e  as
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
• Response
to an impulse
Consider a viscously damped spring-mass system subjected a unit
impulse at t=0, as shown in Figs. 4.6(a) and (b). For an
underdamped system, the solution of the equation of motion
mx  cx  kx  0
(4.17)
is given by Eq. (2.72)(free vibration) as
x(t )  e
where
 n t
 


x0  n x0
sin d t 
 x0 cos d t 
d


c
2mn
k
n 
m
( 4.19),
d  n 1   2 
(4.18)
k c
 
m m
2
( 4.20)
( 4.21)
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4.5 Convolution Integral
Fig. 4.6 A single-degree-of-freedom system subjected to an impulse
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
If the mass is at rest before the unit impulse is applied (x= x =0 for
t<0 or at t=0-), we obtain, from the impulse-momentum relation,

Impulse  f  1  mx (t  0)  mx (t  0 )  mx0
1
(4.22)  x0 
m
Thus the initial conditions are given by
x(t  0)  x0  0
x (t  0)  x0 
1
m
(4.23)
(4.24)
In view of Eqs. (4.23) and (4.24), Eq.(4.18) reduces to
e  nt
x(t )  g (t ) 
sin d t
md
(4.25)
Eq. (4.25) gives the response of a single-DOF system to a unit
impulse, which is also known as the impulse response function,
denoted by g(t), as shown in Fig. 4.6(c).
台灣師範大學機電科技學系
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4.5 Convolution Integral
• Response
to an impulse
If the magnitude of the impulse is F instead of unity, the initial
velocity x0 is F / m and the response of the system becomes
Fe n t
x( t ) 
sin d t  Fg( t )
md
(4.26)
The displacement due to a change in the velocity at time 0
If the impulse F is applied at an arbitrary time t = , it will change
the velocity at t = by an amount F/m (下頁). Assuming that x=0
until the impulse is applied, the displacement x at any subsequent
time t, cause by a change in the velocity at time , is given by Eq.
(4.26) with t replaced by the time elapsed after the application of
the impulse  that is , t-. Thus we obtain
x(t )  Fg (t   )
(4.27)
The displacement due to a change in the velocity at time 
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
 mx2  mx1  m( x2  x1 )  mV
V 
F
m
Impulse Response
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C. R. Yang, NTNU MT
4.5 Convolution Integral
Example 4.7
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.4.8(a). Assuming m = 5kg, k = 2000 N/m, c = 10 N-s/m and
F = 20 N-s, find the response of the system.
Example 4.8
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
Example 4.7
Response of a Structure Under Impact
Solution
From the known data,
n 
k
2000
c
c
10

 20 rad/s,   

 0.05,
m
5
cc 2 km 2 2000(5)
d  1   2 n  19.975 rad/s
Assuming that the impact is given at t = 0, the response of the system
e  nt
20
x1 (t )  F
sin d t 
e 0.05( 20)t sin 19.975t  0.20025e t sin 19.975t m
md
(5)(19.975)
(From Eq. (4.26))
(E.1)
(Continued to Example 4.8)
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
• Response
to General Forcing Condition
Consider the response of the system under an arbitrary external
force F(t), shown in Fig. 4.9, the impulse response, for the impulse
force F() at time , is given by
x( t )  F ( )  g( t   )
(4.28)
This force may be assumed to be
made up of a series of impulses of
varying magnitude.
An arbitrary (nonperiodic) forcing function
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
• Response
to General Forcing Condition
The total response at time t can be found by summing all the responses
due to the elementary impulses acting at all times  :
x(t )   F ( ) g (t  )
Letting
  0
(4.29)
and replacing the summation by integration, we obtain
x(t )   F ( ) g(t   )d
t
0
(4.30)
By substituting Eq. (4.25), Eq. (4.27) into Eq. (4.30), we obtain
1 t
 n ( t  )
x(t ) 
F
(

)
e
sin d (t   )d
0
md
(4.31)
which represents the response of an underdamped single-DOF system to the
arbitrary excitation F(t) and is also called the convolution or Duhamel integral
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
• Response
to Base Excitation
If a spring-mass-damper system is subjected to an arbitrary base
excitation described by its displacement, velocity, or acceleration, the
equation of motion can be expressed in terms of the relative displacement
of the mass z=x-y as follows (see Section 3.6.2)
mz  cz  kz   my
(4.32)
This is similar to the equation
mx  cx  kx  F
(4.33)
All of the results derived for the
force-excited system are
applicable to the base-excited
system only for x replaced by z
and the term F is replaced by  my
For an undamped system subjected to base excitation, the relative
displacement can be found from Eq. (4.31)
z (t )  
1
d
 n ( t  )


y
(

)
e
sin d (t   )d

t
0
(4.34)
where the variable z replacing x
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.5 Convolution Integral
Example 4.12 Compacting Machine Under Linear Force
Determine the response of the compacting machine shown in Figure
(a) when a linearly varying force (shown in Figure (b)) is applied due
to the motion of the cam.
X(t)
台灣師範大學機電科技學系
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4.5 Convolution Integral
Example 4.12
Compacting Machine Under Linear Force
Solution
Figure (b) is known as the ramp function
F ( )   F  .
Where F denotes the rate of increase of the force F per unit time. By
substituting this into Eq. (4.31), we obtain
F t  (t  )
x(t ) 
sin d (t   )d
0 e
md
F t
 ( t  )

(
t


)
e
sin d (t   )( d )
0
md
F  t t  (t  )

sin d (t   )( d ) (See problem 4.28)
0 e
md
n
n
n
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
Problem 4.28
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
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C. R. Yang, NTNU MT
4.5 Convolution Integral
Example 4.12
Compacting Machine Under Linear Force
Solution
These integrals can be evaluated and the response expressed as
follows:
x(t ) 
F 
2
e
t 
k  n
 n t
 2

d2   2n2 

 sin d t 
  cos d t    2
n d


 n

(E.1)
For an undamped system (d=n), Eq.(E.1) reduces to
F
nt  sin nt 
x(t ) 
n k
(E.2)
Fig. 4.13(c) shows the response given by Eq.(E.2)
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C. R. Yang, NTNU MT
4.6
Response Spectrum
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• The graph showing the variation of the maximum response
(maximum displacement, velocity, acceleration, or any other
quantity) with the natural frequency (or natural period) of a single
degree of freedom system to a specified forcing function is known
as the response spectrum.
• Since the max. response is plotted against the natural frequency
(or natural period), the response spectrum gives the max. response
of all possible single-DOF systems.
• Once the response spectrum corresponding to a specified forcing
function is available, we need to know just the natural frequency of
the system to find its max. response.
• Example 4.14 illustrates the construction of a response spectrum.
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.11
Response Spectrum of a Sinusoidal Pulse
Find the undamped response spectrum for the sinusoidal pulse force
shown in the figure using the initial conditions x(0)  x(0)  0
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.14
Response Spectrum of a Sinusoidal Pulse
Solution
The equation of motion of an undamped system can be expressed as
F0 sin t ,
mx  kx  F (t )  
0,
where  

t0
0  t  t0
t  t0
(E.1)
Free vibration
(E.2)
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.14
Response Spectrum of a Sinusoidal Pulse
Solution
Superimposing the homogeneous solution xc(t) and the particular
solution xp(t), as 0  t  t0
x(t )  xc (t )  x p (t )
(E.3)
 F0 
x(t)  Acosn t  B sin nt  
sin t
2 
 k  m 
(E.4)
where n 
2
n

k
m
(E.5)
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.14
Response Spectrum of a Sinusoidal Pulse
Solution
Using the initial conditions, the constants can be found:
A  0, B  
Thus,
x(t ) 
x( t )
 st
F0 / k
1  ( / n ) 2
F0
n (k  m 2 )



sin

t

sin

t

n ,
n


(E.6)
0  t  t0
(E.7)
  t n
2 t 

sin 
sin
(E.8)
 , 0  t  t0
2 
t 0 2t 0
n 
 n  
F0
1 

where


(E.9)
st
 2t 0 
k
1
台灣師範大學機電科技學系
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4.6 Response Spectrum
Example 4.14
Response Spectrum of a Sinusoidal Pulse
Solution
Since there is no force applied for t>t0, the solution can be expressed
as a free vibration solution
x(t )  A cos n t  B sin n t , t  t0 (E.10)
where the constants A’ and B’ can be found by using the values of x (t  t0 )
and x (t  t0 ) , given by Eq. (E.8), as initial conditions for t>t0:
 
2t0 
x(t  t0 )    n sin
  A cos nt0  B sin nt0
n 
 2t0
(E.11)
 
2t 
x (t  t0 )     cos 0   n A sin nt  n B cos nt
n 
 t0 t0
(E.12)
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.14
Response Spectrum of a Sinusoidal Pulse
Solution
Where
Hence,

 st

1 n
 2t0



(E.13)
2

A 
sin nt0 ,
nt0
Therefore,
x( t )
 st



1  cos nt0 
B  
n t0
( n / t0 )
2 1  ( n / 2t0 )2

(代回(E.10)整理)
(E.14)

 t0
t 
t 
sin 2     sin 2  ,
n 
n n 

t  t0
(E.15)
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• Response Spectrum for Base Excitation
For a harmonic oscillator (an undamped system under free
vibration), we notice that
x max  n2 x max
and
x max  n x max
Thus, the acceleration and displacement spectra can be obtained:
Sd 
Sv
n
,
S a  n S v
(4.38)
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• Response Spectrum for Base Excitation
The velocity response spectrum can be obtained:
Sv  z(t ) max 
e  nt
1 
2
P2  Q2
(4.44)
max
Thus the pseudo response spectra are given by:
S d  z max 
Sv
n
;
S v  z max ;
S a  z max  n S v
(4.45)
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• Earthquake Response Spectra
The most direct description of an earthquake motion in time
domain is provided by accelerograms that are recorded by
instruments called strong motion accelerographs.
A typical accelerogram is shown in the figure below.
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• Earthquake Response Spectra
A response spectrum is used to provide the most descriptive
representation of the influence of a given earthquake on a structure
of machine. It is possible to plot the maximum response of a single
degree freedom system using logarithmic scales.
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4.6 Response Spectrum
Example 4.17
Derailment of Trolley of a Crane During Earthquake
The trolley of an electric overhead traveling (EOT) crane travels
horizontally on the girder as indicated in the figure. Assuming the
trolley as a point mass, the crane can be modeled as a single degree
of freedom system with a period 2 s and a damping ratio 2%.
Determine whether the trolley derails under a vertical earthquake
excitation.
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.17
Derailment of Trolley of a Crane During Earthquake
台灣師範大學機電科技學系
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.17
Derailment of Trolley of a Crane During Earthquake
Solution
For  n = 2 s and ζ = 0.02, Fig.4.16 gives the spectral acceleration as
Sa = 0.25 g and hence the trolley will not derail.
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4.6 Response Spectrum
• Design Under a Shock Environment
When a force is applied for short duration, usually for a period of
less than one natural time period, it is called a shock load.
A shock may be described by a pulse shock, velocity shock, or a
shock response spectrum. The pulse shocks are introduced by
applied forces or displacements in different forms.
A velocity shock is caused by sudden changes in the velocity. The
shock response spectrum describes the way in which a machine or
structure responds to a specific shock.
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C. R. Yang, NTNU MT
4.6 Response Spectrum
• Design Under a Shock Environment
Typical shock pulses
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
A printed circuit board (PCB) is mounted on a cantilevered aluminum
bracket, as shown in Figure (a). The bracket is placed in a container
that is expected to be dropped from a low-flying helicopter. The
resulting shock can be approximated as a half-sine wave pulse, as
shown in figure (b). Design the bracket to withstand an acceleration
level of 100g under the half-sine wave pulse shown in figure (b).
Assume a specific weight of 30 kN/m3, a Young’s modulus of 70 GPa,
and a permissible stress of 180 MPa for aluminum.
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4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
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4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
The self-weight of the beam is given by
w  (0.3)(0.015  d )(30 103 )  135d
The total weight is
W  Weight of beam  Weight of PCB  135d  2
The area moment of inertia of the cross section of the beam is
I
1
 0.015  d 3  0.00125d 3
12
台灣師範大學機電科技學系
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4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
The static deflection of the beam can be obtained
Wl 3
(135d  2)(0.3)3
10 135d  2
S st 

 (1.0286 10 )
9
3
3EI 3  (70 10 )(0.00125d )
d3
We adopt a trial and error procedure to determine the values of
unknown.
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4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
Assuming d = 10 mm,
We have
Hence,
 st  (1.0286 10 10 )
 n  2
135  0.01  2
4

3
.
446

10
m
3
0.01
 st
3.466 104
 2
 0.03726 s
g
9.8
t0
0.1

 2.6841
 n 0.03726
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
The dynamic load acting on the cantilever is given by:
 3.35 
(100g)  368.5 N
Pd  Aa Mas  (1.1)
 g 
The maximum bending stress at the root of the cantilever bracket can
be computed as:
 max
 0.01 
(368.5  0.3)

M c
 2   442.2 MPa
 b 
I
1.25 10 9
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
Since this stress exceeds the permissible value, we assume the next
trial value of d as 20 mm. This yields:
 st  (1.0286 10
 n  2
10
135  0.02  2
5
)

6
.
0430

10
m
3
0.02
 st
6.0430 10 5
 2
 0.01560 (s)
g
9.8
t0
0.1

 6.4
 n 0.01560
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C. R. Yang, NTNU MT
4.6 Response Spectrum
Example 4.18
Design of a Bracket for Shock Loads
Solution
The dynamic load can be determined:
 4.7 
(100g)  517 (N)
Pd  Aa Mas  (1.1)
 g 
The maximum bending stress at the root of the bracket will be:
 0.02 
(517  0.3)

M bc
2

  155.1 MPa
 max 

8
I
10
Since this stress is within the permissible limit, the thickness of the bracket
can be taken as 20 mm.
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C. R. Yang, NTNU MT
4.7
Laplace Transforms
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C. R. Yang, NTNU MT
4.7 Laplace Transforms
• The Laplace transform of a function x(t) is defined as:
xt  0  lim sX s 
s 
(4.46)
• The general solution can be expressed as
x(t ) 
x0
(1   2 )1/ 2

e
 n t
sin( d t  1 ) 
x0
d
e  nt sin d t
1 t
 ( t  )
sin d (t   )d
0 F ( )e n
md
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C. R. Yang, NTNU MT
4.8
Numerical Methods
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C. R. Yang, NTNU MT
4.8 Numerical Methods
•
The determination of the response of a system subjected to
arbitrary forcing functions using numerical methods is called
numerical simulation.
•
Numerical simulations can be used to check the accuracy of
analytical solutions, especially if the system is complex.
•
Several methods are available for numerically integrating ordinary
differential equations.
•
The Runge-Kutta methods are quite popular for the numerical
solution of differential equations.
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C. R. Yang, NTNU MT
4.9
Response to Irregular Forcing Conditions Using
Numerical Methods
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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
• Let the function vary with time in an arbitrary manner. The
response of the system can be found:


 n
1 j 1 
 n ( t ti )
x(t )   Fi 1  e
 cos d (t  ti ) 
sin d (t  ti )
k i 1
d



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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
• Thus the response of the system at t = tj becomes



 n
1 j 1
 n ( t j ti )
x j   Fi 1  e
 cos d (t j  ti ) 
sin d (t j  ti )
k i 1
d


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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
Example 4.31
Damped Response Using Numerical Methods
Find the response of a spring-mass-damper system subjected to the
forcing function

t 

F (t )  F0 1  sin
2t0 

(E.1)
in the interval 0  t  t0 , using a numerical procedure.
Assume F0=1, k=1, m=1, ζ=0.1, and t0=τn/2, where τn denotes the
natural period of vibration given by
2
The values of x and
2
n 

 2
1/ 2
 n ( k / m)
x at t=0 are zero
(E.2)
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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
Example 4.31
Damped Response Using Numerical Methods
Solution
For the numerical computations, the time interval 0 to t0 is divided
into 10 equal steps with
ti 
t0 
 ;
10 10
i  2,3,...,11
(E.3)
4 different methods are used to approximate the forcing function F(t).
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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
Example 4.31
Damped Response Using Numerical Methods
Solution
In the figure, F(t) is approximated by a series of rectangular impulses,
each starting at the beginning of the corresponding time step.
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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
Example 4.31
Damped Response Using Numerical Methods
Solution
In Fig. 4.36, piecewise linear (trapezoidal) impulses are used to
approximate the forcing function F(t). The numerical results are given
in Table 4.2. The results can be improved by using a higher-order
polynomial for interpolation instead of the linear function.
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C. R. Yang, NTNU MT
4.9 Response to Irregular Forcing Conditions Using
Numerical Methods
Example 4.31
Damped Response Using Numerical Methods
Solution
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C. R. Yang, NTNU MT
4.10 Examples using MATLAB
• To practice by yourself from Ex. 4.32 to Ex.4.35
• The source codes of all MATLAB programs are given
at the companion website
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