Mechanical Vibrations
Chapter 4
Peter Avitabile
Mechanical Engineering Department
University of Massachusetts Lowell
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Impulse Excitation
Impulsive excitations are generally
considered to be a large magnitude
force that acts over a very short
duration time
The time integral of the force is
F̂ = ∫ F( t )dt
(4.1.1)
When the force is equal to unity and the time
approaches zero then the unit impulse exists and
the delta function has the property of
δ(t − ξ ) = 0 for t ≠ ξ
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Impulse Excitation
Integrated over all time, the delta function is
∞
∫ δ(t − ξ)dt = 1
0<ξ<∞
(4.1.2)
0
If this function is multiplied times any forcing
function then the product will result in only one
value at t=ξ and zero elsewhere
∞
∫ f (t )δ(t − ξ)dt = f (ξ)
0<ξ<∞
0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.1.3)
Impulse Excitation
Considering impact-momentum on the system, a
sudden change in velocity is equal to the actual
applied input divided by the force.
Recall that the free response due to initial
conditions is given by
x& (0)
x=
sin ωn t + x (0) cos ωn t
ωn
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Impulse Excitation
Then the velocity initial condition yields
F̂
x=
sin ωn t = F̂h ( t )
mωn
(4.1.4)
and it can be seen that the solution includes h(t)
1
h(t) =
sin ωn t
mωn
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.1.5)
Impulse Excitation
When damping is considered in the solution, the
free response is given as (x(0)=0)
x=
x& (0)e −ζω t
n
ωn 1 − ζ
2
sin ωn
x& (0)e −σt
1− ζ t =
sin ωd t
ωd
2
which can be written as
F̂
x=
e −ζω t sin ωn 1 − ζ 2 t
mωn 1 − ζ 2
n
or as
x=
F̂ −σt
e sin ωd t = F̂h ( t )
mωd
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.16)
Arbitrary Excitation
Using the unit
impulse response
function, the
response due to
arbitrary loadings
can be determined.
The arbitrary force
is considered to be
a series of impulses
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Arbitrary Excitation
Since the system is considered linear, then the
superposition of the responses of each individual
impulse can be obtained through numerical
integration
t
x ( t ) = ∫ f (ξ)h ( t − ξ)dξ
(4.2.1)
0
This is called the superposition integral. But it is
also referred to as the
Convolution Integral
or
Duhammel’s Integral
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Step Excitation
Determine the indamped response due to a step.
For the undamped system,
1
h(t) =
sin ωn t
mωn
which is substituted into (4.2.1) to give
t
x(t) =
F0
sin ωn (t − ξ )dξ
∫
mωn 0
F0
x ( t ) = (1 − cos ωn t )
k
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.2.2)
Step Excitation
This implies that the peak response is twice the
statical displacement
F0
x ( t ) = (1 − cos ωn t )
k
(4.2.2)
Dis placement vers us Time
2
1.8
1.6
1.4
Dis placement
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
Time
20
25
30
Note: Force selected such that F/k ratio is 1.0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Step Excitation
When damping is included in the equation, then
h(t) =
e −ζω t
n
mωn 1 − ζ 2
sin ωn 1 − ζ 2 t
(4.2.2)
and

F0 
e −ζω t
2
cos ωn 1 − ζ t − ψ 
x ( t ) = 1 −
k  mωn 1 − ζ 2

n
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.2.3)
Step Excitation
This can be simplified as
h(t) =
−σt
e
sin ωd t
mωd

F0  e −σt
x ( t ) = 1 −
cos ωd t − ψ 
k  mωd

C=0
C=0.1
C=0.5
M=1 ; K=2
22.457 Mechanical Vibrations - Chapter 4
C=1.0
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Base Excitation
For base excitation,
m&x& = − k ( x − y) − c( x& − y& )
z=x−y
(3.5.1)
(3.5.2)
the equation of motion is expressed as z=x-y and
will result in
(4.2.4)
&z& + 2ζωn z& + ωn 2 z = −&y&
Notice that the F/m term is replaced by the
negative of the base acceleration (ie, F=ma)
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Base Excitation
For and undamped system
initially at rest, the
solution for the relative
displacement is
t
1
z( t ) = −
&y&(ξ) sin ωn (t − ξ )dξ
∫
ωn 0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.2.5)
Ramp Excitation and Rise Time
This solution must always be considered in two
parts - the time less than and greater than t1
The ramp of the force is
t
f ( t ) = F0  
 t1 
(4.4.1)
and h(t) for the convolution integral is
ωn
1
sin ωn t =
sin ωn t
h(t) =
mωn
k
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.4.1)
Ramp Excitation and Rise Time
The response for the first part of the ramp is
t
ωn
ξ
x(t) =
F0 sin ωn ( t − ξ)dξ
∫
k 0 t1
F0  t sin ωn t 
=  −
 t < t1
k  t1 ω n t1 
(4.4.2)
and the response of the step portion after t1 is
F0  t − t1 sin ωn (t − t1 ) 
−
x(t) = − 
 t > t1
ω n t1
k  t1

22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
Ramp Excitation and Rise Time
The superposition of the two pieces of the solution
gives the total response due to the force as
x(t) =
F0  sin ωn t sin ωn (t − t1 ) 
+
1 −
 t > t1
ω n t1
ω n t1
k

22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.4.3)
Rectangular Pulse
The rectangular pulse is the sum of two different
step functions - one positive and one negative
shifted in time
Step Up
Step down
kx ( t )
= (1 − cos ωn t ) t < t1
F0
kx ( t )
= −(1 − cos ωn (t − t1 )) t < t1
F0
(4.4.4)
(4.4.5)
Combined
kx ( t )
= (1 − cos ωn (t )) − (1 − cos ωn (t − t1 )) t < t1
F0
kx ( t )
= (− cos ωn (t )) + (cos ωn (t − t1 )) t < t1
F0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
(4.4.6)
MATLAB Examples - VTB3_1
VIBRATION TOOLBOX EXAMPLE 3_1
>> m=1; c=.1; k=2; tf=100; F0=1;
>> vtb3_1(m,c,k,F0,tf)
>>
Dis placement vers us Time
0.8
0.6
Dis placement
0.4
0.2
0
-0.2
-0.4
-0.6
0
22.457 Mechanical Vibrations - Chapter 4
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Time
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB3_2
VIBRATION TOOLBOX EXAMPLE 3_2
>> m=1; c=.1; k=2; tf=100; F0=1;
>> VTB3_2(m,c,k,F0,tf)
>>
>>
Dis placement vers us Time
Dis placement
1
0.5
0
0
22.457 Mechanical Vibrations - Chapter 4
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Time
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB1_4
VIBRATION TOOLBOX EXAMPLE 1_4 – pulse
>> clear; p=1:1:1000; pp=p./p; ppp=[(p./p-p./p) pp (p./p-p./p) (p./p-p./p)];
>> x0=0; v0=0; m=1; d=.5; k=2; dt=.01; n=4000;
>> u=ppp; [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);
>> t=0:dt:n*dt; plot(t,x);plot(t,x);
>>
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory
MATLAB Examples - VTB1_4
VIBRATION TOOLBOX EXAMPLE 1_4 – ramp up (basically a static problem)
>> clear; x0=0; v0=0; m=1; d=.5; k=2; dt=.01; n=3000;
>> t=0:dt*100:n; u=t./3000; [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);
>> t=0:dt:n*dt; plot(t,x);
>>
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
22.457 Mechanical Vibrations - Chapter 4
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Dr. Peter Avitabile
Modal Analysis & Controls Laboratory