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1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Mechanical VibrationsChapter 3
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
2 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
SDOF Definitions
• lumped mass• stiffness proportionalto displacement
• damping proportional tovelocity
• linear time invariant• 2nd order differentialequations
Assumptions
m
k c
x(t)
3 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
Consider the SDOF system witha sinusoidally varying forcingfunction applied to the mass asshown F=F0sinωt
From the Newton’s Second Law,
∑ ω=++⇒= tsinFkxxcxmmaf 0&&& (3.1.1)
4 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
The solution consists of the complementarysolution (homogeneous solution) and the particularsolution. The complementary part of the solutionhas already been discussed in Chapter 2.The particular solution in the one of interest here.Since the oscillation of the response is at thesame frequency as the excitation, the particularsolution will be of the form
( )φ−ω= tsinXx (3.1.2)
5 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
Substituting this into the differential equation,the solution is of the form
Note that this is also seen graphically as
(recall that the velocity and acceleration are 90 and 180 degrees ahead of the displacement)
(3.1.4)
( ) ( )2220
cmk
FXω+ω−
=
ω−ω
=φ −2
1
mkctan(3.1.3)
6 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
This is expressed in nondimensional form as
and can be further reduced recalling the followingexpressions for a SDOF
(3.1.6)(3.1.5)
( )222
0
kc
km1
kF
Xω+
ω−
=
ω−
ω=φ −
km1k
ctan 2
1
mk
n =ω nc m2c ω=ccc
=ζ
7 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
The nondimensional expression is
(3.1.7)
(3.1.8)
2
n
22
n
021
1FXk
ωω
ζ+
ωω
−
=
2
n
n1
1
2tan
ωω
−
ωω
ζ=φ −
8 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
This yields the popular plot of forced response
9 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
The complex force vector also yields usefulinformation for interpretation of the results
10 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Harmonic Vibration
The differential equation describing the system
and the complete solution of this problem is givenas
(3.1.10)
(3.1.11)
tsinmFxx2x 02
nn ω=ω+ζω+ &&&
( )1n2t
1
2
n
22
n
0
t1sineX
21
)tsin(kF)t(x
n φ+ωζ−+
ωω
ζ+
ωω
−
φ−ω=
ζω−
11 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Complex Frequency Response Function
The Complex FRF - real and imaginary parts
(3.1.17)
2
n
22
n
n
2
n
22
n
2
n
21
2j
21
1)j(h
ωω
ζ+
ωω
−
ωω
ζ−
ωω
ζ+
ωω
−
ωω
−=ω
12 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
The effects of unbalance is a common problem invibrating systems.
Consider a onedimensional systemwith an unbalancerepresented by aneccentric mass, m,with offset, e,rotating at somespeed, ω, as shown
13 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
Let x be the displacement of the non-rotatingmass (M-m) about the equilibrium point, then thedisplacement of the eccentric mass is
and the equation of motion becomes
tsinex ω+
( ) xckxtsinexdtdmx)mM( 2
2&&& −−=ω++−
14 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
This can easily be cast as
which is essentially identical to (3.1.1) with thesubstitution of F0=meω2
The steady-state solution just developed isapplicable for this solution
(3.2.1)
(3.2.3)
( ) tsinmekxxcxM 2 ωω=++ &&&
( ) ( )222
2
cMk
meXω+ω−
ω= (3.2.2)
ω−ω
=φ −2
1
Mkctan
15 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
The differential equation describing the system
and the complete solution of this problem is givenas
(3.1.10)
(3.1.11)
tsinmFxx2x 02
nn ω=ω+ζω+ &&&
( )1n2t
1
2
n
22
n
0
t1sineX
21
)tsin(kF)t(x
n φ+ωζ−+
ωω
ζ+
ωω
−
φ−ω=
ζω−
16 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
Manipulating into nondimensional form
(3.2.4)
(3.2.5)
2
n
22
n
2
n
21eX
mM
ωω
ζ+
ωω
−
ωω
=
2
n
n1
1
2tan
ωω
−
ωω
ζ=φ −
17 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
This yields the popular plot of forced response
18 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
The differential equation describing the system
and the complete solution of this problem is givenas
(3.1.10)
(3.1.11)
tsinmFxx2x 02
nn ω=ω+ζω+ &&&
( )1n2t
1
2
n
22
n
0
t1sineX
21
)tsin(kF)t(x
n φ+ωζ−+
ωω
ζ+
ωω
−
φ−ω=
ζω−
19 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Rotating Unbalance
The complete solution of this problem is given as
(3.2.6)
( ) ( )( )1n
2t1
222
2
t1sineX
cMk
)tsin(me)t(x
n φ+ωζ−+
ω+ω−
φ−ωω=
ζω−
20 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Support Motion
Many times a system is excited at the location ofsupport commonly called ‘base excitation’
21 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Support Motion
With the motion of the base denoted as ‘y’ andthe motion of the mass relative to the intertialreference frame as ‘x’, the differential equationof motion becomes
Substitute
into the equations to give
)yx(c)yx(kxm &&&& −−−−= (3.5.1)
yxz −= (3.5.2)
(3.5.3)tsinYmymkzzczm 2 ωω=−=++ &&&&&
22 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Support Motion
This is identical in form to equation 3.2.1 where zreplaces x and mω2Y replaces meω2
Thus the solution can be written by inspection as
(3.5.4)
(3.5.5)
( ) ( )222
2
cmk
YmZω+ω−
ω=
ω−ω
=φ −2
1
mkctan
23 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Support Motion
The steady state amplitude and phase from thisequation can be written as
(3.5.8)
(3.5.9)
( )( ) ( )222
22
cmk
ckYX
ω+ω−
ω+=
( ) ( )
ω−ω−ω
=φ 22
3
cmkkmctan
24 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Forced Response - Support Motion
25 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Vibration Isolation
Dynamical response can be minimized through theuse of a proper isolation design.
An isolation system attempts either to protectdelicate equipment from vibration transmitted to itfrom its supporting structure or to preventvibratory forces generated by machines frombeing transmitted to its surroundings.
The basic problem is the same for these twoobjectives - reducing transmitted force.
26 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Vibration Isolation - Force Transmitted
Notice that motion transmitted from thesupporting structure to the mass m is less thanone when the frequency ratio is greater thatsquare root 2.
This implies that thenatural frequency of thesupported system must bevery small compared to thedisturbing frequency.A soft spring can be usedto satisfy this requirement.
27 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Vibration Isolation - Force Transmitted
Another problem is to reduce the force transmittedby the machine to the supporting structure whichessentially has the same requirement.
The force to be isolated is transitted through thespring and damper as shown
28 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Vibration Isolation - Force Transmitted
The force to be isolated is transitted through thespring and damper is
With the disturbing force equal to F0sinwt thisequation becomes
(3.6.1)( ) ( )2
n
22T
21kXXckXF
ωζω
+=ω+=
(3.6.1a)2
n
22
n
0
21
kF
X
ωω
ζ+
ωω
−
=
29 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Vibration Isolation - Force Transmitted
The transmissibility TR, defined as the ratio ofthe transmitted force to the disturbing force, is
and when damping is small becomes
(3.6.2)
2
n
22
n
2
n
0
T
21
21
FFTR
ωω
ζ+
ωω
−
ωω
ζ+==
1
1FFTR 2
n
0
T
−
ωω
== (3.6.3)
30 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Sharpness of Resonance
The peak amplitude of response occurs atresonance. In order to find the sharpness ofresonance, the two side bands at the half powerpoints are required.
At the half power points,
(3.10.1)2
n
22
n
2
21
121
21
ωω
ζ+
ωω
−
=
ζ
31 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Sharpness of Resonance
Solving yields
and if the damping is assumed to be small
Letting the two frequencies corresponding to theroots of 3.10.3 gives
(3.10.2)
(3.10.3)
( ) 222
n1221 ζ−ζ±ζ−=
ωω
ζ±=
ωω 21
2
n
n
122n
21
22 24
ω
ω−ω≅
ωω−ω
=ζ
32 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
Sharpness of Resonance
The Q factor is defined as
(3.10.4)ζ
=ω−ω
ω=
21Q
12
n
33 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
MATLAB Examples - VTB2_3VIBRATION TOOLBOX EXAMPLE 2_3
function VTB2_3(z,rmin,rmax,opt)% VTB2_3 Steady state magnitude and phase of a% single degree of freedom damped system.
>> vtb2_3([0.02:.02:.1],0.5,1.5,1)>> vtb2_3([0.02:.02:.1],0.5,1.5,3)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.510
-1
100
101
102
Frequency Ratio
Nor
mal
ized
Am
plitu
de
Normalized Amplitude vers us Frequency Ratio
ζ = 0.02ζ = 0.04ζ = 0.06ζ = 0.08ζ = 0.1
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.50
22.545
67.590
112.5135
157.5180
Frequency Ratio
Pha
se la
g ( °
)
P has e vers us Frequency Ratio
ζ = 0.02ζ = 0.04ζ = 0.06ζ = 0.08ζ = 0.1
34 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3
MATLAB Examples - VTB1_4VIBRATION TOOLBOX EXAMPLE 1_4
>> clear>> x0=0; v0=0; m=1; d=.1; k=2; dt=.01; n=10000;>> t=0:dt:n*dt; u=[sin(t)];>> [x,xd]=VTB1_4(n,dt,x0,v0,m,d,k,u);>>>> plot(t,u); % Plots force versus time.>> plot(t,x); % Plots displacement versus time.
0 10 20 30 40 50 60 70 80 90 100-1.5
-1
-0.5
0
0.5
1
1.5
2