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
)(