99

0 5 10 15 20 25 30 35 40-3

-2

-1

0

1

2

3

4.4.1 Forced Vibrations From Harmonic Excitation As discussed earlier, forced vibrations are one very important practical mechanism for the occurrence of vibrations.

x

m

F(t)

ck

Fig. 4.10: Sdof Oscillator with Viscous Damping and External Force The equation of motion of the damped linear sdof oscillator with an external force is: ( )tFkxxcxm =++ &&& (4.4.1) The general solution of this differential equation is: ( ) ( ) ( )

321321forceexternalfromresults

partvibrationsfree

txtxtx += hom (4.4.2)

which consists of the homogeneous part resulting from the free vibration and the particular part resulting from the external disturbance F(t). The homogeneous solution has already been treated in the last chapter.

Fig 4.11: Homogeneous and particular part of the solution and superposition

While the homogeneous part of the solution will decay to zero with time we are especially interested in the stationary solution.

4.4.2 Excitation with Constant Force Amplitude 4.4.1.1 Real Approach The excitation function is harmonic, Ω is the frequency of excitation tFtF Ω= cosˆ)( (4.4.3) Eqn. 4.4.1 becomes

tFkxxcxm Ω=++ cosˆ&&& (4.4.4) Dividing by the mass m

tmFx

mkx

mcx Ω=++ cos

ˆ&&& (4.4.5)

Introducing again the dimension less damping and the natural circular frequency

0

2ωmcD = and

mk

=20ω

and the amplitude

mFfˆˆ = (4.4.6)

This yields: (4.4.7) tfxxDx Ω=++ cosˆ2 2

00 ωω &&&

To solve this differential equation, we make an approach with harmonic functions tBtAtx Ω+Ω= sincos)( (4.4.8) This covers also a possible phase lag due to the damping in the system. Differentiating (4.4.8) to get the velocity and the acceleration and putting this into eqn. 4.4.7 leads to

tf

tBtAtBtADtBtA

Ω=

Ω+Ω+ΩΩ+ΩΩ−+ΩΩ−ΩΩ−

cosˆ)sincos()cossin(2sincos 2

0022 ωω

(4.4.9) After separating the coefficients of the sin- and cos-functions and comparing the coefficients we get: (4.4.10a) fABDA ˆ2 2

002 =+Ω+Ω− ωω

100

(4.4.10b) 02 200

2 =+Ω−Ω− BADB ωω From the second equation we see that

ADBB Ω=+Ω− 020

2 2 ωω which leads to

( )AD

B 220

02Ω−

Ω=ω

ω

and we put this result into eqn.(4.4.10a):

fAADDA ˆ)(

22 2

0220

00

2 =+Ω−

ΩΩ+Ω− ω

ωω

ω

fAD ˆ)(

4)( 22

0

220

222

0 =

Ω−

Ω+Ω−

ωω

ω

[ ] )(ˆ4)( 22

022

02222

0 Ω−=Ω+Ω− ωωω fAD This yields the solution for A and B:

[ ]220

22220

220

4)()(ˆ

Ω+Ω−

Ω−=

ωωω

Df

A (4.4.11a)

and

[ ]220

22220

0

4)()2(ˆ

Ω+Ω−

Ω=

ωωω

DDf

B (4.4.11b)

Introducing the dimensionless ratio of frequencies

0ωη Ω

==frequencyNaturalfrequencyExcitation

(4.4.12)

2222

220

4)1()1)(/ˆ(ηη

ηωD

fA

+−

−= (4.4.13a)

and

2222

20

4)1()2)(/ˆ(ηη

ηωDDf

B+−

= (4.4.13b)

With A and B we have found the solution for tBtAtx Ω+Ω= sincos)( .

101

Another possibility is to present the solution with amplitude and phase angle: )cos()( ϕ−Ω= tCtx (4.4.14) The amplitude is

202222

22 ˆ

4)1(

1ωηη

f

DBAC

+−=+= (4.4.15)

Considering that and we get mFf /ˆˆ = mk /20 =ω

kF

DC

ˆ

4)1(

12222 ηη +−

= (4.4.16)

Introducing the dimensionless magnification factor V1 which only depends on the frequency ratio and the damping D :

222214)1(

1),(ηη

ηD

DV+−

= (4.4.17)

we get the amplitude as

kFVCˆ

1= (4.4.18)

and the phase angle (using trigonometric functions similar as in chap.4.2.2) :

212)(tanηηϕ

−==

DAB

(4.4.19)

We can see that as η approaches 1 the amplitude grows rapidly, and its value near or at the resonance is very sensitive to changes of the damping D. The maximum of the magnification curve for a given D can be found at

0

221ω

η resres D

Ω=−= (4.4.20)

If D is very small then 1≈resη . The maximum amplitude for this D then is

21max

12

1ˆ),(

ˆ

DDkFDV

kFC res

−== η (4.4.21)

For η→ 0: V1≈1: the system behaves quasi-statically, for very large values of η: V1→ 0: the vibrations are very small.

102

0ωΩ=η

22221 D4)1(1V

η+η−=

D=0,7071D=0,5D=0,3

D=0,2D=0,1

D=0,05

54.543.532.52 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10

V1

0ωΩ=η

21D2arctanη−η

=

D=0,7071D=0,5D=0

D=0,05 D=0,2D=0,1

ϕ

ϕ

180

150

120

90

60

30

054.43.32.21.10.0

Fig.4.12: Magnification factor V1 and phase angle to describe the vibration behavior of the damped oscillator under constant force amplitude excitation

103

4.4.1.2 Complex Approach Let us first recall that we can represent a real harmonic functions by a complex exponential function using tite ti Ω+Ω=Ω sincos From this we can derive that

2

costiti eet

Ω−Ω +=Ω (4.4.22)

and

ieet

titi

2sin

Ω−Ω −=Ω (4.4.23)

The harmonic force is

titititi eFeFeeFtFtF Ω−ΩΩ−Ω +=+=Ω=2

ˆ

2

ˆ)(

2

ˆcosˆ)( (4.4.24)

This means that we have to solve the equation of motion twice, for the exp(iΩt) and the exp(-iΩt) term. For the first step we make the approach (4.4.25a) tiextx Ω+= 11 ˆ)(

(4.4.25b) tiextx Ω−= 22 ˆ)( Putting both approaches into the equation of motion yields

titi eFexkcim ΩΩ =+Ω+Ω−2

ˆˆ)( 1

2 (4.4.26a)

titi eFexkcim Ω−Ω− =+Ω−Ω−2

ˆˆ)( 2

2 (4.4.26b)

Dividing by k and introducing the frequency ratio η (eqn.(4.4.12))

[ ]k

FxiD2

ˆˆ2)1( 1

2 =+− ηη and [ ]k

FxiD2

ˆˆ2)1( 2

2 =−− ηη (4.4.27)

The solution for x1 and x2 are

kF

DiDx

2

ˆ

4)1(2)1(ˆ

2222

2

1ηηηη

+−

−−= (4.4.28a)

kF

DiDx

2

ˆ

4)1(2)1(ˆ

2222

2

2ηηηη

+−

+−= (4.4.28b)

104

As we can see, the solution of one part is the conjugate complex of the other: 21 ˆˆ xx = (4.4.29) The solution for x(t) is combined from the two partial solutions, which we just have found:

titi exextx Ω−Ω += 21 ˆˆ)( (4.4.30) This can be resolved: txitxtxitxtx Ω+Ω+Ω+Ω= sinˆcosˆsinˆcosˆ)( 2211 and using the fact that 21 ˆˆ xx = , we finally get

txtxtx Ω−Ω= sinˆIm2cosˆRe2)( 11 (4.4.31)

The factor of 2 compensates the factor ½ associated with the force amplitude. All the information can be extracted from only so that only this part of the solution has to be solved.

1x

tkF

DDt

kF

Dtx Ω

+−+Ω

+−

−= sin

ˆ

4)1(2cos

ˆ

4)1()1()( 22222222

2

ηηη

ηηη (4.4.32)

which is the same result as eqn. (4.4.8) with (4.4.13). Also the magnitude ( 1x ) and phase can be obtained in the same way and yield the previous results:

Magnitude: ( )

( ) Fk

DVFkD

x

Vfactorionmagnificat

ˆ1,ˆ1²²4²²1

1ˆ 1

1

ηηη

=+−

=444 3444 21

(4.4.33)

Phase: ²1

2ˆReˆImtan1

1ηηϕ

−=−=

Dxx

(4.4.34)

4.4.1.3 Complex Approach, Alternative Instead of eqn. (4.4.24), we can write

=+=Ω= ΩΩ−Ω tititi eFeeFtFtF2

ˆRe2)(

2

ˆcosˆ)(

or

tieFtFtF Ω=Ω= ˆRecosˆ)( (4.4.35)

105

According to this approach, we formulate the steady state response as

tieXtx Ω= ˆRe)( (4.4.36)

The complex amplitude X is determined from the equation of motion, solving

( ) titi eFeXkcim ΩΩ =+Ω+Ω− ˆReˆRe 2 (4.4.37)

The real parts are equal if the complex expression is equal:

(4.4.38) ( ) titi eFeXkcim ΩΩ =+Ω+Ω− ˆˆ2

Elimination of the time function yields:

(4.4.39) ( FXkcim ˆˆ2 =+Ω+Ω− )

)

)

The expression in brackets is also called the dynamic stiffness

( cimkkdyn Ω+Ω−=Ω 2)( (4.4.40)

Now we solve (4.4.39) to get the complex amplitude:

( cimkFX

Ω+Ω−=

2

ˆˆ (4.4.41)

The expression

( ) InputOutput

FX

cimkH ==

Ω+Ω−=Ω ˆ

ˆ1)(2

(4.4.42)

is the complex Frequency Response Function (FRF). Introducing the dimensionless frequency η as before yields:

( ) kF

DiX

ˆ

211ˆ

2 ηη +−= (4.4.43)

Because ( ) titiiti eXeexextx ΩΩ−−Ω ===−Ω ˆReˆReReˆ)cos(ˆ ϕϕϕ (4.4.44) we take the magnitude and phase lag ϕ of this complex result x (4.4.45) ϕiexX −=Ω ˆ)(ˆ

which leads to the same result as before, see (4.4.33) and (4.4.34):

106

( )( ) F

kDVF

kDx

Vfactorionmagnificat

ˆ1,ˆ1²²4²²1

1ˆ 1

1

ηηη

=+−

=444 3444 21

(4.4.33)

²1

2ˆRe

ˆImtanηηϕ

−=−=

DXX (4.4.34)

4.4.2 Harmonic Force from Imbalance Excitation

c

Ω Ω

mk

2mk

2

k2

k2

ε ε

mM

x

Fig. 4.13: Sdof oscillator with unbalance excitation The total mass of the system consists of the mass mM and the two rotating unbalance masses mu :

2

2 UM

mmm += (4.4.46)

The disturbance force from the unbalance is depending on the angular speed Ω , ε is the excentricity: ( ) tmtF UUnbalance ΩΩ= cos² ε (4.4.47) Now, following the same way as before (real or complex) leads to the solution:

)cos()( ϕ−Ω= tCtx where

Amplitude: ( )( )

mm

DVm

mD

Cx UU εηεηη

η ,²²4²²1

²ˆ 3=+−

==

(4.4.48)

107

Phase: ²1

2tanηηϕ

−=

D (4.4.49)

Magnification factor ( )( ) ²²4²²1

²,3 ηηηη

DD

+−=V (4.4.50)

The phase is the same expression as in the previous case, however, the magnification factor is different, because the force amplitude is increasing with increasing angular speed.

0ωΩ=η

2222

2

3 D4)1(V

η+η−

η=

D=0,7071D=0,5 D=0,3

D=0,2 D=0,1

D=0,05

5 4.5 43.532.521.51 0.5 0 0 1 2 3 4 5V3 6 7 8 9 10

Fig. 4.14: Magnification factor V3 for the case of imbalance excitation

As can be seen: for η→ 0: V1≈0: there is no force if the system is not rotating or rotates only slowly, for very large values of η: V1→ 1: that means that the mass m is vibrating with an amplitude (ε mu/m), but the common center of gravity of total system m and mu does not move.

108

4.4.3 Support Motion / Ground Motion 4.4.3.1 Case 1

u(t)

c

k

mx

Fig. 4.15: Excitation of the sdof oscillator by harmonic motion of one spring end

The equation of motion for this system is

( )tkukxxcxm =++ &&& (4.4.51)

Under harmonic excitation: (4.4.52) ( ) tutu Ω= cosˆ The mathematical treatment is nearly identical to the first case, only the excitation function is different: the excitation is replaced by u here. This leads to the result for the amplitude of vibration

kF /ˆ ˆ

Amplitude: ( )

( ) uDVuD

x

functionVmag

ˆ,ˆ²²4²²1

1ˆ 1

. 1

ηηη

=⋅+−

=444 3444 21

(4.4.53)

The magnification factor again is V1. Also, the phase relation is identical as before:

Phase: ²1

2tanηηϕ

−=

D (4.4.54)

109

4.4.3.2 Case 2

ck

mx

u(t)

Fig. 4.16: Excitation of the sdof oscillator by harmonic motion of the spring/damper combination

The equation of motion now also contains the velocity u : & kuuckxxcxm +=++ &&&& (4.4.55)

Amplitude of vibration and phase shift becomes

Amplitude: ( )

( ) uDVuD

Dx

VfunctionMagn

ˆ,ˆ²²4²²1

²²41ˆ 2

. 2

ηηη

η=⋅

+−

+=

444 3444 21

(4.4.56)

Phase: ( ) ²²4²1³2tan

ηηηϕ

DD+−

= (4.4.57)

As can be seen the phase now is different due to the fact that the damper force depending on the relative velocity between ground motion and motion of the mass plays a role. The amplitude behaviour is described by the magnification factor V2.

110

1

0ωΩ=η

2222

22

2 D4)1(

D41Vη+η−

η+=

2

D=0,7071D=0,5D=0,3

D=0,2D=0,1

D=0,05

543210

1 8

V2 6

4 2

0

Fig. 4.17: Magnification factor V2 for the case of ground excitation via spring and damper

Notice that all curves have an intersection point at 2=η which means that for 2>η higher damping does not lead to smaller amplitudes but increases the amplitudes. This is due to the fact that larger relative velocities (due to higher frequencies η) make the damper stiffer and hence the damping forces. Further cases of ground motion excitation are possible.

111

4.4 Excitation by Impacts 4.5.1 Impact of finite duration

Ti

F(t)

t

F

F(t)

kc

x

m

Fig. 4.18: Sdof Oscillator under impact loading We consider an impact of finite length Ti and constant force level during the impact The impact duration Ti is much smaller than the period of vibration T:

D

i Tω

T π2=<<

With the initial condition that there is no initial displacement x0 = 0 we can calculate the velocity by means of the impulse of the force

i

TTFdtFmvp

iˆˆ

00 ∫ ===

This leads to the initial velocity :

mTF

v iˆ

0 = (4.5.1)

Using the results of the viscously damped free oscillator for D < 1,

( )

−+

−= − tDBtDAetx

DD

tD4342143421

ωω

ω ωω 00 ²1sin²1cos0 (4.5.2)

we can immediately find the result with the initial conditions x0 and v0:

112

and 000 =⇒= AxD

i vD

vB

mTF

ωω0

0

00 ²1

ˆ=

−=⇒

⋅=v (4.5.3)

so that the system response to the impact is a decaying oscillation where we have assumed that the damping D < 1:

( ) ( tev

tx DtD

Dω

ωω sin00 −= ) (4.5.4)

4.5.2 DIRAC-Impact

F

t

Fig. 4.19: DIRAC-Impact

The DIRAC-Impact is defined by

( ) ( ) ( ) ( ) 1,000ˆ =

=≠

∞=→= ∫

∞

∞−dttbut

tt

ttFtF δδδ (4.5.5)

δ is the Kronecker symbol. The duration of this impact is infinitely short but the impact is infinitely large. However, the integral is equal to 1 or , respectively. For the initial displacement and calculation of the initial velocity following the previous chapter, we get

F00 =x

( ) ( tem

Ft DtD

Dω

ωω sin

ˆ0−

⋅= )x (4.5.6)

For , the response x(t) is equal to the impulse response function (IRF) h(t) 1ˆ =F

113

( ) ( tem

t DtD

Dω

ωω sin1 0−

⋅= )h (4.5.7)

The IRF is an important characteristic of a dynamic system in control theory. 4.5 Excitation by Forces with Arbitrary Time Functions

F( )τ

τ τ+∆τ

F

t

t

x

τ

Fig. 4.20: Interpretation of an arbitrary time function as series of DIRAC-impulses Using the results of the previous chapters we can solve the problem of an arbitrary time function F(t) as subsequent series of Dirac-impacts, where the initial conditions follow from the time history of the system. The solution is given by the Duhamel-Integral or convolution integral:

∫∫ −=−= −−tt

DtD

DdFthdFte

mt

00

)( )()()())(sin(1)( 0 ττττττωω

τωx (4.6.1)

As can be seen, the integral contains the response of the sdof oscillator with respect to a DIRAC-impact multiplied with the actual force F(τ), which is integrated from time 0 to t.

114

4.7 Periodic Excitations 4.7.1 Fourier Series Representation of Signals Periodic signals can be decomposed into an infinite series of trigonometric functions, called Fourier series.

Fig. 4.21: Scheme of signal decomposition by trigonometric functions

F

t

T

Fig. 4.22: Example of a periodic signal: periodic impacts The period of the signal is T and the corresponding fundamental frequency is

Tπω 2

= (4.7.1)

115

Now, the periodic signal x(t) can be represented as follows

)sin()cos(

2)(

1

0 ∑∞

=++=

kkk tkbtkaatx ωω

(4.7.2) The Fourier-coefficients a0, ak and bk must be determined. They describe how strong the corresponding trigonometric function is present in the signal x(t). The coefficient a0 is the double mean value of the signal in the interval 0…T:

∫=T

dttxT

a0

0 )(2 (4.7.3)

and represents the off-set of the signal. The other coefficients can be determined from

( )∫=T

k dttktxT

a0

cos)(2 ω (4.7.4)

( )∫=T

k dttktxT

b0

sin)(2 ω (4.7.5)

The individual frequencies of this terms are

Tkkk

πωω 2== (4.7.6)

for k = 1 we call the frequency 1ω fundamental frequency or basic harmonic and the frequencies for k = 2,3,… the second, third,… harmonic (or generally higher harmonics). 4.7.1.1 Alternative real Representation We can write the Fourier series as a sum of cosine functions with amplitude ck and a phase shift ϕk

)cos()(1

0 ∑∞

=++=

kkk tkcctx ϕω (4.7.7)

22kkk bac += and )arctan(

k

kk a

b−=ϕ (4.7.8)

4.7.1.2 Alternative complex Representation The real trigonometric functions can also be transformed into complex exponential expression:

116

∑∞

−∞==

k

tikkeXtx ω)( (4.7.9)

The Xk are the complex Fourier coefficients which can be determined by solving the integral:

∫ −=T

tikk dtetx

TX

0)(1 ω (4.7.10a)

or

[∫ −=T

k dttkitktxT

X0

sincos)(1 ωω ] (4.7.10b)

which clearly shows the relation to the real Fourier coefficients series given by eqns.(4.7.4) and (4.7.5):

2

Im;2

Re kk

kk

bXaX −==

The connection to the other real representation (chap. 4.7.11) is

kk cX = )

ReIm

(tank

kk X

X=ϕ (4.7.11)

The coefficients with negative index are the conjugate complex values of the corresponding positive ones:

(4.7.12) *kk XX =−

4.7.2 Forced Vibration Under General Periodic Excitation

x

m

F(t)

ck

Fig. 4.23: Sdof oscillator under periodic excitation

117

Let us use once more the single dof oscillator but now the force is a periodic function which can be represented by a Fourier series

∑∞

=Ω+Ω+=

1

0 )sin()cos(2

)(k

skck tkFtkFF

tF (4.7.13)

The Fck and Fsk are the Fourier coefficients which can be determined according to the last chapter (eqns. 4.7.3.-4.7.5). The response due to such an excitation is

∑∞

=−Ω+−Ω+=

111

0 )sin(),()cos(),(2

)(k

ksk

kkck

k tkk

FDVtk

kF

DVk

Ftx ϕηϕη (4.7.14)

with the frequency ratio

0ωη Ω

=k

k ∞= ,...2,1k (4.7.15)

Each individual frequency is considered with its special amplification factor V and individual phase shift, which in the present case can be calculated from

222214)1(

1),(kk

kD

DVηη

η+−

= (4.7.16)

212

tank

kk

Dηη

ϕ−

= (4.7.17)

For the other cases of mass unbalance excitation or ground excitation the procedure works analogously. The appropriate V-functions have to be used and the correct pre-factors (which is in the present case 1/k) have to be used.

118