4.4.1 Forced Vibrations From Harmonic Excitation
As discussed earlier, forced vibrations are one very important practical mechanism for the
occurrence of vibrations.
F(t)
m
x
k
c
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:
m&x& + cx& + kx = F (t )
(4.4.1)
The general solution of this differential equation is:
x(t ) =
(t ) +
x
1hom
23
free vibrations
(t )
x
1part
23
(4.4.2)
results from external force
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.
3
2
1
0
-1
-2
-3
0
5
10
15
20
25
30
35
40
Fig 4.11: Homogeneous and particular part of the solution and superposition
99
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
F (t ) = Fˆ cos Ωt
(4.4.3)
Eqn. 4.4.1 becomes
m&x& + cx& + kx = Fˆ cos Ωt
(4.4.4)
Dividing by the mass m
&x& +
c
k
Fˆ
x& + x = cos Ωt
m
m
m
(4.4.5)
Introducing again the dimension less damping and the natural circular frequency
2D =
c
mω 0
and
ω 02 =
k
m
and the amplitude
Fˆ
fˆ =
m
This yields:
(4.4.6)
&x& + 2 Dω 0 x& + ω 02 x = fˆ cos Ωt
(4.4.7)
To solve this differential equation, we make an approach with harmonic functions
x(t ) = A cos Ωt + B sin Ωt
(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
− Ω 2 A cos Ωt − Ω 2 B sin Ωt + 2 Dω 0 (−ΩA sin Ωt + ΩB cos Ωt ) + ω 02 ( A cos Ωt + B sin Ωt )
= fˆ cos Ωt
(4.4.9)
After separating the coefficients of the sin- and cos-functions and comparing the coefficients
we get:
− Ω 2 A + 2 Dω 0 ΩB + ω 02 A = fˆ
(4.4.10a)
100
− Ω 2 B − 2 Dω 0 ΩA + ω 02 B = 0
(4.4.10b)
From the second equation we see that
− Ω 2 B + ω 02 B = 2 Dω 0 ΩA
which leads to
B=
2 Dω 0 Ω
(ω
)
A
− Ω2
and we put this result into eqn.(4.4.10a):
2
0
− Ω 2 A + 2 Dω 0 Ω
2 Dω 0 Ω
(ω 02
2
−Ω )
A + ω 02 A = fˆ
 2
4 D 2ω 02 Ω 2 
2
A = fˆ
(ω 0 − Ω ) + 2
2 
(ω 0 − Ω ) 

[(ω
2
0
]
− Ω 2 ) 2 + 4 D 2ω 02 Ω 2 A = fˆ (ω 02 − Ω 2 )
This yields the solution for A and B:
A =
[(ω
fˆ (ω 02 − Ω 2 )
2
0
− Ω 2 ) 2 + 4 D 2ω 02 Ω 2
]
(4.4.11a)
]
(4.4.11b)
and
B =
[(ω
fˆ (2 Dω 0 Ω)
2
0
− Ω 2 ) 2 + 4 D 2ω 02 Ω 2
Introducing the dimensionless ratio of frequencies
η=
Excitation frequency
Ω
=
ω0
Natural frequency
A =
(4.4.12)
( fˆ / ω 02 )(1 − η 2 )
(4.4.13a)
(1 − η 2 ) 2 + 4 D 2η 2
and
B =
( fˆ / ω 02 )( 2 Dη )
(4.4.13b)
(1 − η 2 ) 2 + 4 D 2η 2
With A and B we have found the solution for x(t ) = A cos Ωt + B sin Ωt .
101
Another possibility is to present the solution with amplitude and phase angle:
x(t ) = C cos(Ωt − ϕ )
(4.4.14)
The amplitude is
C=
A2 + B 2 =
fˆ
1
2 2
2 2
(1 − η ) + 4 D η
ω 02
(4.4.15)
Considering that fˆ = Fˆ / m and ω 02 = k / m we get
C=
1
(1 − η 2 ) 2 + 4 D 2η 2
Fˆ
k
(4.4.16)
Introducing the dimensionless magnification factor V1 which only depends on the frequency
ratio and the damping D :
V1 (η , D) =
1
2 2
(4.4.17)
2 2
(1 − η ) + 4 D η
we get the amplitude as
Fˆ
C = V1
k
(4.4.18)
and the phase angle (using trigonometric functions similar as in chap.4.2.2) :
B
2 Dη
tan ϕ = ( ) =
A 1−η 2
(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
η res = 1 − 2 D 2 =
Ω res
(4.4.20)
ω0
If D is very small then η res ≈ 1 . The maximum amplitude for this D then is
C max =
Fˆ
Fˆ
1
V1 (η res , D) =
k
k 2D 1 − D 2
(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
10
9
D=0,05
8
V1 =
7
1
(1− η2 ) 2 + 4 D 2 η 2
6
V1
5
D=0,1
D=0,2
4
3
D=0,3
D=0,5
D=0,7071
2
1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
η= Ω
ω0
ϕ
D=0,05
D=0,1
D=0,2
D=0,5
D=0,7071
180
D=0
150
120
90
ϕ = arctan
60
2Dη
1 − η2
30
0
0
0.
1
1.
2
2.
3
3.
4
4.
5
η= Ω
ω0
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
e iΩt = cos Ωt + i sin Ωt
From this we can derive that
cos Ωt =
e iΩt + e −iΩt
2
(4.4.22)
sin Ωt =
e iΩt − e −iΩt
2i
(4.4.23)
and
The harmonic force is
Fˆ
Fˆ
Fˆ
F (t ) = Fˆ cos Ωt = (e iΩt + e −iΩt ) = e iΩt + e −iΩt
2
2
2
(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
x1 (t ) = xˆ1e + iΩt
x 2 (t ) = xˆ 2 e
(4.4.25a)
−iΩt
(4.4.25b)
Putting both approaches into the equation of motion yields
(−Ω 2 m + iΩc + k ) xˆ1e iΩt =
(−Ω 2 m − iΩc + k ) xˆ 2 e −iΩt
Fˆ iΩt
e
2
Fˆ
= e −iΩt
2
(4.4.26a)
(4.4.26b)
Dividing by k and introducing the frequency ratio η (eqn.(4.4.12))
Fˆ
Fˆ
and
(1 − η 2 ) + 2 Dη i xˆ1 =
(1 − η 2 ) − 2 Dη i xˆ 2 =
2k
2k
[
]
[
]
(4.4.27)
The solution for x1 and x2 are
xˆ1
=
Fˆ
(1 − η 2 ) 2 + 4 D 2η 2 2k
(4.4.28a)
xˆ 2
=
Fˆ
(1 − η 2 ) 2 + 4 D 2η 2 2k
(4.4.28b)
(1 − η 2 ) − 2 Dη i
(1 − η 2 ) + 2 Dη i
104
As we can see, the solution of one part is the conjugate complex of the other:
xˆ1 = xˆ 2
(4.4.29)
The solution for x(t) is combined from the two partial solutions, which we just have found:
x(t ) = xˆ1e iΩt + xˆ 2 e −iΩt
(4.4.30)
This can be resolved:
x(t ) = xˆ1 cos Ωt + ixˆ1 sin Ωt + xˆ 2 cos Ωt + ixˆ 2 sin Ωt
and using the fact that xˆ1 = xˆ 2 , we finally get
x(t ) = 2 Re{xˆ1 }cos Ωt − 2 Im{xˆ1 }sin Ωt
(4.4.31)
The factor of 2 compensates the factor ½ associated with the force amplitude. All the
information can be extracted from x̂1 only so that only this part of the solution has to be
solved.
x(t ) =
2 Dη
Fˆ
Fˆ
sin Ωt
cos
Ω
t
+
(1 − η 2 ) 2 + 4 D 2η 2 k
(1 − η 2 ) 2 + 4 D 2η 2 k
(1 − η 2 )
(4.4.32)
which is the same result as eqn. (4.4.8) with (4.4.13).
Also the magnitude ( x̂1 ) and phase can be obtained in the same way and yield the previous
results:
xˆ =
Magnitude:
1
1 ˆ
1
F = V1 (η , D ) Fˆ
k
(1 − η ² )² + 4 D ²η3² k
14442444
(4.4.33)
magnificat ion factor V1
tan ϕ = −
Phase:
Im{xˆ1 } 2 Dη
=
Re{xˆ1 } 1 − η ²
(4.4.34)
4.4.1.3 Complex Approach, Alternative
Instead of eqn. (4.4.24), we can write
 Fˆ

Fˆ
F (t ) = Fˆ cos Ωt = (e iΩt + e −iΩt ) = 2 Re e iΩt 
2
 2

or
{
F (t ) = Fˆ cos Ωt = Re Fˆe iΩt
}
(4.4.35)
105
According to this approach, we formulate the steady state response as
{
x(t ) = Re Xˆe iΩt
}
(4.4.36)
The complex amplitude X̂ is determined from the equation of motion, solving
{(
)
} {
Re − Ω 2 m + iΩc + k Xˆe iΩt = Re Fˆe iΩt
}
(4.4.37)
The real parts are equal if the complex expression is equal:
(− Ω 2 m + iΩc + k )Xˆe iΩt = Fˆe iΩt
(4.4.38)
Elimination of the time function yields:
(− Ω 2 m + iΩc + k )Xˆ = Fˆ
(4.4.39)
The expression in brackets is also called the dynamic stiffness
(
k dyn (Ω) = k − Ω 2 m + iΩc
)
(4.4.40)
Now we solve (4.4.39) to get the complex amplitude:
Xˆ =
Fˆ
(k − Ω 2 m + iΩc)
(4.4.41)
The expression
H (Ω) =
1
(k − Ω 2 m + iΩc)
=
Xˆ Output
=
Input
Fˆ
(4.4.42)
is the complex Frequency Response Function (FRF). Introducing the dimensionless frequency
η as before yields:
Xˆ =
Fˆ
1 − η 2 + i 2 Dη k
(
1
Because
)
{
(4.4.43)
} {
} {
xˆ cos(Ωt − ϕ ) = xˆ Re e i (Ωt −ϕ ) = Re xˆe −iϕ e iΩt = Re Xˆe iΩt
}
(4.4.44)
we take the magnitude x̂ and phase lag ϕ of this complex result
Xˆ (Ω) = xˆe −iϕ
(4.4.45)
which leads to the same result as before, see (4.4.33) and (4.4.34):
106
xˆ =
1
1 ˆ
1
F = V1 (η , D ) Fˆ
k
(1 − η ² )² + 4 D ²η3² k
14442444
(4.4.33)
magnificat ion factor V1
tan ϕ = −
4.4.2
{}
{}
2 Dη
Im Xˆ
=
ˆ
1
−η²
Re X
(4.4.34)
Harmonic Force from Imbalance Excitation
Ω
ε
Ω
ε
mk
2
mk
2
x
mM
c
k
2
k
2
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 :
m
(4.4.46)
m = mM + 2 U
2
The disturbance force from the unbalance is depending on the angular speed Ω , ε is the
excentricity:
FUnbalance (t ) = Ω ² ε mU cos Ωt
(4.4.47)
Now, following the same way as before (real or complex) leads to the solution:
x(t ) = C cos(Ωt − ϕ )
where
Amplitude:
xˆ = C =
η²
(1 − η ² )² + 4 D ²η ²
ε
mU
m
= V3 (η , D ) ε U
m
m
(4.4.48)
107
tan ϕ =
Phase:
2 Dη
1−η²
Magnification factor V3 (η , D ) =
(4.4.49)
η²
(4.4.50)
(1 − η ² )² + 4 D ²η ²
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.
10
9
D=0,05
8
η2
V3 =
2 2
(1−η ) + 4 D 2η2
7
6
V3
D=0,1
D=0,2
D=0,3
D=0,5
D=0,7071
5
4
3
2
1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
η= Ω
ω0
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)
k
m
x
c
Fig. 4.15: Excitation of the sdof oscillator by harmonic motion of one spring end
The equation of motion for this system is
m&x& + cx& + kx = ku (t )
(4.4.51)
Under harmonic excitation:
u (t ) = uˆ cos Ωt
(4.4.52)
The mathematical treatment is nearly identical to the first case, only the excitation function is
different: the excitation Fˆ / k is replaced by uˆ here. This leads to the result for the amplitude
of vibration
Amplitude:
xˆ =
1
(1 − η ² )² + 4 D ²η3²
14442444
⋅ uˆ = V1 (η , D ) uˆ
(4.4.53)
mag . functionV1
The magnification factor again is V1. Also, the phase relation is identical as before:
Phase:
tan ϕ =
2 Dη
1−η²
(4.4.54)
109
4.4.3.2 Case 2
m
x
c
k
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& :
m&x& + cx& + kx = cu& + ku
(4.4.55)
Amplitude of vibration and phase shift becomes
Amplitude:
xˆ =
1 + 4 D ²η ²
(1 − η ² )² + 4 D ²η3²
14442444
⋅ uˆ = V2 (η , D ) uˆ
(4.4.56)
Magn. function V2
Phase:
tan ϕ =
2 Dη ³
(1 − η ² ) + 4 D ²η ²
(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
D=0,05
1
V2 =
8
V2
6
(1−η2 ) 2 + 4 D 2η2
D=0,1
D=0,2
4
D=0,3
D=0,5
D=0,7071
2
0
1+ 4 D 2η2
0
1
2
2
3
4
5
η= Ω
ω0
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 η = 2 which means that for η > 2
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
F(t)
F(t)
m
x
F̂
c
k
t
Ti
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:
Ti << T =
2π
ωD
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
Ti
p = mv0 = ∫ Fˆ dt =Fˆ Ti
0
This leads to the initial velocity :
v0 =
FˆTi
m
(4.5.1)
Using the results of the viscously damped free oscillator for D < 1,
x(t ) = e








A cos 1 − D ²ω 0 t  + B sin  1 − D ²ω 0 t 
1
4
2
4
3
1
4
2
4
3





ω
ω

D
D





− Dω 0t 
we can immediately find the result with the initial conditions x0 and v0:
112
(4.5.2)
x0 = 0 ⇒ A = 0
v0 =
and
Fˆ ⋅ Ti
v0
v
⇒B=
= 0
m
ω 0 1 − D² ω D
(4.5.3)
so that the system response to the impact is a decaying oscillation where we have assumed
that the damping D < 1:
x(t ) =
v0
ωD
e − Dω0t sin (ω D t )
(4.5.4)
4.5.2 DIRAC-Impact
F
t
Fig. 4.19: DIRAC-Impact
The DIRAC-Impact is defined by
0
F (t ) = Fˆ δ (t ) → δ (t ) = 
∞
t≠0
t=0
∞
, but ∫ δ (t )dt = 1
(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 F̂ , respectively. For the initial
displacement x0 = 0 and calculation of the initial velocity following the previous chapter, we
get
x(t ) =
Fˆ
e − Dω0t sin (ω D t )
m ⋅ω D
(4.5.6)
For Fˆ = 1 , the response x(t) is equal to the impulse response function (IRF) h(t)
113
h(t ) =
1
e − Dω0t sin (ω D t )
m ⋅ω D
(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
F(τ)
τ τ+∆τ
t
x
t
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:
t
x(t ) = ∫
0
1
m ωD
e
− Dω 0 (t −τ )
t
sin(ω D (t − τ )) F (τ ) dτ = ∫ h(t − τ ) F (τ ) dτ
(4.6.1)
0
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
ω=
2π
T
(4.7.1)
115
Now, the periodic signal x(t) can be represented as follows
∞
a
x(t ) = 0 + ∑ ak cos(kωt ) + bk sin(kωt )
2 k =1
(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:
a0 =
2T
∫ x(t )dt
T0
(4.7.3)
and represents the off-set of the signal. The other coefficients can be determined from
ak =
2T
∫ x(t ) cos(kωt )dt
T 0
(4.7.4)
bk =
2T
∫ x(t ) sin (kωt )dt
T0
(4.7.5)
The individual frequencies of this terms are
ω k = kω =
2πk
T
(4.7.6)
for k = 1 we call the frequency ω1 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
x(t ) = c0 +
∞
∑ ck cos(kωt + ϕ k )
(4.7.7)
k =1
ck = ak2 + bk2
and
ϕ k = arctan(−
bk
)
ak
(4.7.8)
4.7.1.2 Alternative complex Representation
The real trigonometric functions can also be transformed into complex exponential
expression:
116
x(t ) =
∞
∑ X k eikω t
(4.7.9)
k = −∞
The Xk are the complex Fourier coefficients which can be determined by solving the integral:
Xk =
1T
−ikωt
dt
∫ x(t )e
T0
(4.7.10a)
Xk =
1T
∫ x(t )[cos kωt − i sin kωt ]dt
T 0
(4.7.10b)
or
which clearly shows the relation to the real Fourier coefficients series given by eqns.(4.7.4)
and (4.7.5):
a
b
Re{X k } = k ; Im{X k } = − k
2
2
The connection to the other real representation (chap. 4.7.11) is
X k = ck
tan ϕ k = (
Im{X k }
)
Re{X k }
(4.7.11)
The coefficients with negative index are the conjugate complex values of the corresponding
positive ones:
X − k = X k*
(4.7.12)
4.7.2 Forced Vibration Under General Periodic Excitation
F(t)
m
x
k
c
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
F (t ) =
F0 ∞
+ ∑ Fck cos(kΩt ) + Fsk sin(kΩt )
2 k =1
(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
x(t ) =
F0 ∞
F
F
+ ∑ V1 (η k , D) ck cos(kΩt − ϕ k ) + V1 (η k , D) sk sin( kΩt − ϕ k )
2k k =1
k
k
(4.7.14)
with the frequency ratio
ηk =
kΩ
k = 1,2,...∞
ω0
(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
V1 (η k , D) =
tan ϕ k =
1
(1 − η k2 ) 2
(4.7.16)
+ 4 D 2η k2
2 Dη k
(4.7.17)
1 − η k2
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.
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