MECHATRONICS

Lecture 07

Slovak University of TechnologyFaculty of Material Science and Technology in Trnava

MECHANICAL VIBRATION

Linear model systems

0 ukubum

02 20 uuu

m

b

2

m

k0

0

pb

Free damped oscillation of the system with single DOF

Free oscillation of a damped system is described by the solution of the homogenous equation of motion

where m, b, k are parameters of the mass, damping and stiffness,

u(t) is the solution (response of the model).

The equation can be modified into the form

is the decay rate,

is the own (natural) angular frequency of the undamped model,

is the damping ratio.

tt eCeCtu 2121)(

02 20

2

The solution of the is well known equation

where both constants of integration can be solved from the initial conditions for

u(0) and ů(0)

or from another conditions for

u(t) and ů(t) for known time.

Values λ1 and λ2 are roots of so called characteristic equation:

Model and time courses of the linear system with single DOF

It is known from the mathematics course that the form of solution for ODR is the sum of homogenous solution and the particular solution.

The homogenous solution is valid for F(t) = 0. The form of the particular solution is given by the function F(t). It is obvious from the previous paragraph and Tab. that the homogenous part of the solution disappears after a time.

For us the settled damped oscillation (particular part of the solution) is more important and will be analyzed in the following text.

Forced damped oscillation of the system with single DOFExcited oscillation of damped system with single DOF is described by the equation

)(tFukubum

)(1

2 20 tF

muuu

or

]Re[cos)( 00tieFtFtF

titi eim

FeutF

20

20

0 2

1)(

Let us discuss now the case of excitation by a harmonic force

with the solution (response, time response) has the form

where

u0 is the amplitude of the response,

F0 is the amplitude of the exciting force,

is the oscillating frequence.

ti

p

st eb

utu

222 )2()1(

)(

222 )2()1(

)cos()(

p

st

b

tutu

The absolute value of the expression within the brackets is the transfer function

(see also the frequency characteristics) H(ω).

Substituting u0 = F0/k - static sag and η = ω/ω0 - frequency tuning we have

The real part is the responce to the exciting force F0cos ωt and is equal to

where the phase angle φ means the delay of the response to the excitation due to damping of the model system.

222 )2()1(

1)(

pst bu

tu

;0,1

2)(

2

pbarctgt

The amplitude frequency characteristics

The phase frequency characteristics

The frequency characteristics (i.e. amplitude vs. frequency and phase vs.frequency) for two types of excitation force:

Characteristics for harmonic force

The excitation by rotating mass

trm

muuu n

n sin2 220

222

2

0)2()1(

p

nn

b

rmu

Very often the excitation is caused by unbalanced rotating mass. Such case is shown on Fig. The unbalanced rotor is represented by the mass mn placed on the eccentricity rn from the axes of rotation. m is the total mass of the equipment. The resultant stiffness is k and the damping is b. The vertical inertia force is

F = mnrnω2sinωt.

The equation of motion is

When we compare this equation with basic equation for forced vibration, we see that both equation

F(t) = mnrnω2sinωt

The solution is identical to that excited by harmonic force. The amplitude of sustained vibrations is given by following formula

This expression is possible to plot in the amplitude and phase diagram

The frequency transfer and the phase frequency characteristics is defined as

222

20

)2()1(

pnn b

m

rmu

;0,1

2)(

2pb

arctgt

Exciting force is a periodic function of time

1

21 )sin()cos()(i

ii tiFtiFtF

dttFT

FFT

F0

10 )(1

dttitFT

FFT

Fi

0

1 )cos()(2 dttitF

TF

FT

Fi

0

2 )sin()(2

n

iii tiFtiF

muuu

121

20 )sin()cos(

12

In many cases the exiting force is a periodic function of time.It means that its value repeat after the period TF

F(t) = F(t+TF) = F(t+n TF) for n = 1, 2, ... , n

In such case it is possible expand the force into Fourier series

The equation

where = 2/TF .

The determination of Fourier coefficients is well known from mathematics

In practical applications we do´nt take infinity number of Fourier coefficients, but only n.

i

iFiiii F

FarctgFFF

2

122

21 ,

n

iFii tiF

mm

Fuuu

1

1020 )sin(

12

The right hand side of we arange when used

F1i = Fi sin Fi, F2i = Fi cos Fi, for i = 1, 2, ...

So it is

Now we can re-write the equation in the form

If holds the law of superposition we can determine the response for each component of the force separately and then the resultant response is given by adding all particular calculated responses due to separate harmonic terms.

n

ipih uuuu

10

20

10100

m

F

k

Fu

)sin( 0 th eCu

)sin(0 iFiipi tiuu

22220)2()1(

ibik

Fu

p

ii

2)(1

2

i

ibarctg p

Fi

The general solution is obtained again from the homogeneous and particular solutions

The amplitude of particular solution is done by

In this equation is

N

j

N

Tj

Tt F

Fj

22

N

jjY

NF

010

1)

2cos(

2

01 N

jiY

NF

N

jji

)2

sin(2

02 N

jiY

NF

N

jji

From we see that, that particular harmonic components magnified the

response according the value of Fi and the order i.

Very often the course of forces is known from measurements. In such case the

components of Fourier series is also possible to get from measured values.

We consider the period of the force is TF and the number of of measurements

is N+1. The time interval will be Δt = TF/N, and the time from the beginning of

force action is tj = jΔt.

We introduce the value

The measured function will be denoted by Y(tj) = Yj .

The coefficients of Fourier series are determined by

for i = 1, 2, ...

The kinematical excitation

umtuuktuub zz )()(

)()()( tftuktubukubum zz

The exciting, considered so far has been done by the force acting on the moving mass. Now we shall consider that the frame move harmonically according the formula

uz(t) = h sinωt.

Such case is sometimes called seismic excitation.

The differential equation of motion of the moving mass will be

It is seen that the motion is harmonic.

If we consider that the base move according the function f(t) is

f(t) = bhω cosωt + kh sinωt

After arrangement of we get

m

k0 m

b

2

ththuuu sincos22 20

20

zph sin2 0

zph cos020

220

40

22220 )2(14 pbhhhp

)sin(2 20 zo tpuuu

Using the notation and the equation obtains the form

From here we get

By using of these expressions the equation obtains the form

The right hand side of is possible to simplify by notation

)sin(0 zp tuu

222

2

0)2()1(

)2(1

p

p

b

bhu

222

2

0

)2()1(

)2(1

p

p

b

b

h

u

20for

)2()1(

222

3

p

p

b

barctg

The particular solution of this equation will be

.

with the amplitude of harmonic motion of the mass

The course of frequency transfer λ is shown in Fig.

The phase is given by the formulas

.

Kinematical excitation

The force is general function of time

Very often the excitation force is a general function of time.

The particular solution is given by Duhamel integral:

The analytical solution of this integral is possible for simple functions of general force.

dteFm

u d

tt

dp )(sin)(

1

0

)(