Mechanical Vibrations Chapter 3 - Faculty Server...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.457 Mechanical Vibrations - Chapter 3

Mechanical VibrationsChapter 3

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell


SDOF Definitions

• lumped mass• stiffness proportionalto displacement

• damping proportional tovelocity

• linear time invariant• 2nd order differentialequations

Assumptions

m

k c

x(t)


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)



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)



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)



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



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 −



This yields the popular plot of forced response



The complex force vector also yields usefulinformation for interpretation of the results



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−


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


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



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++−



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





(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−



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 −



This yields the popular plot of forced response





(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−



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−


Forced Response - Support Motion

Many times a system is excited at the location ofsupport commonly called ‘base excitation’



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=−=++ &&&&&



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



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




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.


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.



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



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

−

=



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)


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



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



The Q factor is defined as

(3.10.4)ζ

=ω−ω

ω=

21Q

12

n


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


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

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