1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Mechanical VibrationsChapter 4
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
2 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Impulse Excitation
Impulsive excitations are generallyconsidered to be a large magnitudeforce that acts over a very shortduration timeThe time integral of the force is
When the force is equal to unity and the timeapproaches zero then the unit impulse exists andthe delta function has the property of
(4.1.1)∫= dt)t(FF̂
( ) ξ≠=ξ−δ tfor0t
3 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Impulse Excitation
Integrated over all time, the delta function is
If this function is multiplied times any forcingfunction then the product will result in only onevalue at t=ξ and zero elsewhere
(4.1.2)∫∞
∞<ξ<=ξ−δ0
01dt)t(
∫∞
∞<ξ<ξ=ξ−δ0
0)(fdt)t()t(f (4.1.3)
4 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Impulse Excitation
Considering impact-momentum on the system, asudden change in velocity is equal to the actualapplied input divided by the force.Recall that the free response due to initialconditions is given by
tcos)0(xtsin)0(xx nnn
ω+ωω
= &
5 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Impulse Excitation
Then the velocity initial condition yields
and it can be seen that the solution includes h(t)
(4.1.4)
(4.1.5)
)t(hF̂tsinm
F̂x nn
=ωω
=
tsinm
1)t(h nn
ωω
=
6 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Impulse Excitation
When damping is considered in the solution, thefree response is given as (x(0)=0)
which can be written as
or as
(4.16)
tsine)0(xt1sin1e)0(xx d
d
t2
n2n
tn
ωω
=ζ−ωζ−ω
=σ−ζω− &&
t1sine1m
F̂x 2n
t2
n
n ζ−ωζ−ω
= ζω−
)t(hF̂tsinem
F̂x dt
d=ω
ω= σ−
7 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Arbitrary Excitation
Using the unitimpulse responsefunction, theresponse due toarbitrary loadingscan be determined.The arbitrary forceis considered to bea series of impulses
8 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Arbitrary Excitation
Since the system is considered linear, then thesuperposition of the responses of each individualimpulse can be obtained through numericalintegration
This is called the superposition integral. But it isalso referred to as the
Convolution Integralor
Duhammel’s Integral
(4.2.1)∫ ξξ−ξ=t
0
d)t(h)(f)t(x
9 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Step Excitation
Determine the indamped response due to a step.For the undamped system,
which is substituted into (4.2.1) to give
(4.2.2)
tsinm
1)t(h nn
ωω
=
( ) ξξ−ωω
= ∫ dtsinm
F)t(xt
0n
n
0
( )tcos1kF)t(x n
0 ω−=
10 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Step Excitation
This implies that the peak response is twice thestatical displacement
(4.2.2)( )tcos1kF)t(x n
0 ω−=
0 5 10 15 20 25 300
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
Dis
plac
emen
t
Dis placement vers us Time
Note: Force selected such that F/k ratio is 1.0
11 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Step Excitation
When damping is included in the equation, then
and
(4.2.2)t1sin
1me)t(h 2
n2n
tn
ζ−ωζ−ω
=ζω−
ψ−ζ−ω
ζ−ω−=
ζω−t1cos
1me1
kF)t(x 2
n2n
t0
n
(4.2.3)
12 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Step Excitation
This can be simplified as
tsinme)t(h d
d
tω
ω=
σ−
ψ−ω
ω−=
σ−tcos
me1
kF)t(x d
d
t0
M=1 ; K=2
C=0
C=0.1
C=0.5C=1.0
13 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Base Excitation
For base excitation,
the equation of motion is expressed as z=x-y andwill result in
Notice that the F/m term is replaced by thenegative of the base acceleration (ie, F=ma)
(3.5.1)
(4.2.4)yzz2z 2nn &&&&& −=ω+ζω+
)yx(c)yx(kxm &&&& −−−−=
yxz −= (3.5.2)
14 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Base Excitation
For and undamped systeminitially at rest, thesolution for the relativedisplacement is
(4.2.5)( ) ξξ−ωξω
−= ∫ dtsin)(y1)t(zt
0n
n&&
15 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Ramp Excitation and Rise Time
This solution must always be considered in twoparts - the time less than and greater than t1
The ramp of the force is
and h(t) for the convolution integral is
(4.4.1)
=
10 t
tF)t(f
(4.4.1)tsink
tsinm
1)t(h nn
nn
ωω
=ωω
=
16 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Ramp Excitation and Rise Time
The response for the first part of the ramp is
and the response of the step portion after t1 is
(4.4.2)
11n
n
1
0
n
t
0 10
n
ttt
tsintt
kF
d)t(sint
Fk
)t(x
<
ωω
−=
ξξ−ωξω
= ∫
( )1
1n
1n
1
10 ttt
ttsint
ttkF)t(x >
ω
−ω−
−−=
17 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Ramp Excitation and Rise Time
The superposition of the two pieces of the solutiongives the total response due to the force as
(4.4.3)( )
11n
1n
1n
n0 ttt
ttsint
tsin1kF)t(x >
ω
−ω+
ωω
−=
18 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
Rectangular Pulse
The rectangular pulse is the sum of two differentstep functions - one positive and one negativeshifted in time
Step Up
Step down
Combined
(4.4.4)( ) 1n0
tttcos1F
)t(kx<ω−=
(4.4.5)
(4.4.6)
( )( ) 11n0
ttttcos1F
)t(kx<−ω−−=
( )( ) ( )( ) 11nn0
ttttcos1tcos1F
)t(kx<−ω−−ω−=
( )( ) ( )( ) 11nn0
ttttcostcosF
)t(kx<−ω+ω−=
19 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
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-0.4
-0.2
0
0.2
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Dis
plac
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Dis placement vers us Time
MATLAB Examples - VTB3_1VIBRATION TOOLBOX EXAMPLE 3_1
>> m=1; c=.1; k=2; tf=100; F0=1;>> vtb3_1(m,c,k,F0,tf)>>
20 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
0 10 20 30 40 50 60 70 80 90 1000
0.5
1
Time
Dis
plac
emen
t
Dis placement vers us Time
MATLAB Examples - VTB3_2VIBRATION TOOLBOX EXAMPLE 3_2
>> m=1; c=.1; k=2; tf=100; F0=1;>> VTB3_2(m,c,k,F0,tf)>>>>
21 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
MATLAB Examples - VTB1_4VIBRATION 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);>>
0 5 10 15 20 25 30 35 40-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
22 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 4
0 5 10 15 20 25 300
0.05
0.1
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0.3
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MATLAB Examples - VTB1_4VIBRATION 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);>>