linear vibrations of systems with one degree of … · 2014-05-23 · self-excitedvibrations....

4L INEAR VIBRATIONS OF SYSTEMS WITH ONEDEGREE OF FREEDOM

4.1 general classification of vibrations

Different possibilities for a classification exist1:

• Classification with respect to the number of degrees of freedom: f

– 1 DOF or known as Single Degree Of Freedom (SDOF)

– n DOFs or known as Multiple Degrees Of Freedom (MDOF)

– infinite numbder of DOFs (continuous system: beams, plates, . . .)

Spatially discrete vibration systems with f = n DOFs can result fromdiscrete multi-body systems but from finite-element systems, where thecontinuum was discretized to some representative nodal points leadingto a finite number of DOFs.

• Classification with respect to the character of the describing equation ofmotion:

– linear vibrations

– non-linear vibrations (e.g. oscillators with non-linear spring or dampercharacteristics, contact or play, etc.)

Even for an apparently simple non-linear system with one DOF the ana-lytical solution can be very complicated, if not impossible. In the lattercase, a solution is only possible by numerical methods.

1 S. z.B. Magnus, K., Popp, K. : Schwingungen, Teubner, 1997.

79

80 linear vibrations of systems with one degree of freedom

As shown in the last chapter non-linear systems can be linearized abouta reference trajectory or a reference point in many cases. The descriptionby linear differential equations has many methodological advantages.

In the context of non-linear differential equations we can get new, verycomplex dynamic behaviour, which does not occur with linear systems.An interesting phenomenon is deterministic chaos. Although the motionof the system is described by deterministic equations and the input exci-tation is deterministic (e.g. a sine-excitation) the response of the systemseems to be random (which we call chaotic in this context). Starting withtwo adjacent initial conditions which differ only slightly we may end upin totally different system states. This phenomenon is called the "but-terfly effect". However, order can be found when we observe the systemover a very long time span. The system converges to a so-called "strangeattractor".

• Classification with respect to the origin of the vibration

– free vibrations

– self-excited vibrations

– parameter-excited vibrations

– forced vibrations

– coupled vibrations

If the system performs free vibrations (linear or non-linear) it starts from aninitial state and is left alone, there is no other influence from outside thesystem.

self-excited vibrations can occur if the system has access to an exter-nal reservoir of energy other than forced vibrations where the rhythm ofthe excitation is prescribed, here the system itself determines the rhythmof the energy transfer into the system. The system excites itself. It takesas much energy as it needs to maintain the vibration (e. g.as in the caseof a pendulum clock where the energy storage is an elastic spring). Thesesystems are called autonomous systems. If more energy per cycle flowsinto the system than energy is dissipated due to friction etc. then insta-ble vibration can occur leading to strongly increasing amplitudes as inthe case of unstable feedback control loops. This must be avoided underany conditions.

4.1 general classification of vibrations 81

This may also occur if there is an interaction of a structure with a fluidas in the case of an airplane wing which tends to unstable vibrationwhen reaching a critical speed. This phenomenon is called "wing flutter".Usually, this speed is beyond the travelling speed of the airplane, but hasto be checked for every new prototype.

Also the destruction of the Tacoma-Bridge in the 40th in the US is a"good" example how dangerous self-excited vibrations by fluid-structureinteraction can be. Other examples are unstable vibrations of turbo ma-chinery coming from or friction induced vibrations in brakes (squealnoise) or slip-stick phenomena with tool machines leading to marks ofthe cutting tool at the surface of the workpiece. Also unstable motionof the bogie of rail vehicles at very high speeds belong to the group ofself-excited vibrations.

parameter-excited vibrations occur if one or more coefficients ofthe differential equations are not constant but periodically time-varying.The frequency of the parameter change is prescribed explicitely as a func-tion of time, e.g. by the rotational speed of a shaft. Examples are: pendu-lum with periodically varying length, rotating shaft with unsymmetriccross-section, periodically varying stiffness of gear-wheels.

forced vibrations emerge from external disturbances (e.g. periodicaldisturbances from unbalance). The rhythm of the vibration here is notprescribed by varying parameters but by a time-dependent disturbanceterm on the right hand side of the equation of motion.

coupled oscillations occur by the fact that two oscillators can influ-ence and excite each other.

All these origins for vibrations can occur also in combined form.

Another possibility of classification is due to the character of the excitation:

• deterministic

– harmonic / periodic

– transient (e.g. vibrations after an impact)

• stochastic (or random) vibrations

– stationary


– instationary

4.2 free undamped vibrations of the linear oscillator

4.2.1 Equation of Motion

Newton’s resp. Euler ’s equations or the Lagrange’s formalism can be used herein order to derive the equation of motion. The example shows a mass m withelastic foundation (total stiffness k)

Newton’s law:

mx =∑i

Fi = −2k2x (4.1)

mx+ kx = 0 (4.2)

Division by the mass yields:

x+ ω20x = 0 (4.3)

withNatural circularfrequency[ω0] = s−1 ω2

0 =k

m(4.4)

ω0 is the natural circular frequency of the free undamped vibration.

xm

k

2k

2

xm

k

2x

k

2x

k

2x

k

2x

Figure 4.1: Vibration system with one DOF

4.2 free undamped vibrations of the linear oscillator 83

Tx

A

t

Figure 4.2: Example for a solution x (t) of the undamped free oscillator

4.2.2 Solution of the Equation of Motion

This differential equation has the solution: Basic approach tosolve a vibrationequationx (t) = Ac cosω0t+As sinω0t (4.5)

The constants Ac und As follow from the initial conditions:

x0 = x (t = 0) (4.6)v0 = x (t = 0) (4.7)

For t = 0 this leads directly to Ac = x0 and after differentiation of eq. (4.5) toget the velocities, the constant As = v0

ω0so that we can express the constants

in terms of the initial displacement and the initial velocity:

x (t) = x0 cosω0t+v0ω0

sinω0t (4.8)

Further important quantities are the time of a cycle or period: Period[T ] = s

T0 =2πω0

(4.9)

And the natural frequency : Natural frequency[f0] = s−1 = Hz

f0 =1T0

=ω02π (4.10)

Another kind of representation of the solution is:

x (t) = A sin (ω0t+ ϕ) (4.11)


Using trigonometric theorems:

sin (α+ β) = sinα cos β + sin β cosα . (4.12)

Here:

x (t) = A (sinω0t cosϕ+ sinϕ cosω0t) (4.13)= A cosϕ︸︷︷︸

As

sinω0t+A sinϕ︸︷︷︸Ac

cosω0t (4.14)

with

Ac = A sinϕ (4.15)As = A cosϕ (4.16)

Furthermore:A2c +A2

s = A2(sin2 ϕ+ cos2 ϕ

)(4.17)

andAcAs

=A sinϕA cosϕ (4.18)

so thatA =

√A2c +A2

s (4.19)and

tanϕ =AcAs

. (4.20)

x

j

A

2p

w t

Figure 4.3: Phase Angle ϕ

4.2.3 Complex Notation

Starting with the equation of motion (eq. (4.3))

x+ ω20x = 0 (4.21)


and solve it using an exponential approach: Exponentialapproach tosolve a vibrationequation

x (t) = eλt (4.22)x (t) = λeλt (4.23)x (t) = λ2eλt (4.24)

we get

λ2eλt + ω20eλt = 0 (4.25)

λ2 + ω20 = 0 (4.26)λ2 = −ω2

0 . (4.27)

which has the two conjugate complex solutions λ:

λ1,2 = ±iω0 (4.28)

The general solution for x (t) is:

x (t) = A+eiω0t +A−e−iω0t (4.29)

where the constants A+, A− are conjugate complex, too, so that the solu-tion x (t) becomes real again. The constants can be derived from the initialdisplacement and velocity, respectively.

4.2.4 Relation Between Complex and Real Notation

We start with the well-known general formula:

e±iλ = cosλ± i sinλ (4.30)

which we can use to express sin and cos by:

cosλ =eiλ + e−iλ

2 (4.31)

sin λ =eiλ − e−iλ

2i (4.32)

Putting this into eq. (4.29) yields:

x =Ac2(eiω0t + e−iω0t

)+As2i(eiω0t − e−iω0t

)(4.33)

=Ac2(eiω0t + e−iω0t

)− iAs2

(eiω0t − e−iω0t

)(4.34)

=Ac − iAs

2︸︷︷︸A+(=A−

∗)

eiω0t +Ac + iAs

2︸︷︷︸A−

e−iω0t (4.35)


where (...)∗ denotes the conjugate complex number of (...).complex conjugateof A is A∗

This delivers the relation between the real constants Ac, As and the complexconstants A+ and A−:

A+ =Ac − iAs

2 = A∗− (4.36)

A− =Ac + iAs

2 (4.37)

Furthermore, it follows that :

ReA+eiω0t

=

12 (Ac cosω0t+As sinω0t) (4.38)

ReA−e−iω0t

=

12 (Ac cosω0t+As sinω0t) (4.39)

so that

ReA+eiω0t +A−e−iω0t

= (Ac cosω0t+As sinω0t) (4.40)

ImA+eiω0t +A−e−iω0t

= 0 (4.41)

4.2.5 Further Examples of Single Degree of Freedom Systems

There are many applications, where the elastic element of a vibration systemis not a simple spring but e.g. an elastic beam. The lowest natural frequency ofsuch a system can be calculated by a relatively simple approximation, wherewe assume that the deflection shape of the vibrating beam or shaft etc. isapproximately equal to the static deflection shape (which is approximatelytrue for the lowest vibration mode).

example Flexural Vibration of a Beam:

The mass of the beam is assumed to be small: m >> mB

Figure 4.4: Flexural vibration of a beam


The stiffness constant is k = 48EIl3 where E is the Young’s modulus, I is the

area moment of inertia. The constant follows from the elementary theory ofbeam bending, where the deflection x of the beam due to a static force F in themiddle of the beam is x = Fl3

48EI . The stiffness coefficient k follows immediatelyfrom F = kx. The natural circular frequency is:

ω0 =

√k

m=

√48EIml3

(4.42)

example Torsion Vibration of a Disc:

The equation of motion is:Jϕ+ k

_

ϕ = 0 (4.43)Here the stiffness coefficient of the shaft is k

_

= GIp

l , where G is the shearmodulus and Ip is the polar moment of inertia. A static torque MT at theposition of the disc will cause a twist angle ϕ of the shaft:

ω0 =

√k_

J=

√GIpJl

f0 =ω02π (4.44)

Here, again the mass moment of inertia of the bar has been neglected.

4.2.6 Approximate Consideration of the Spring Mass

If the spring mass (which can be the mass of a beam, bar or shaft dependingon the actual case) cannot be neglected compared to the main mass m, we canconsider the effect of the spring mass on the natural frequency.

Figure 4.5: Torsion vibration of a disc


example Mass m connected to a bar:

Figure 4.6: Mass m connected to a bar

The bar has the mass mS , length l, a cross section area A and Young’s modu-lus E. The stiffnes coefficient is k = EA

l .

The real deflection shape is approximated by the static deflection shape. At acertain position z of the bar we have a static deflection:

u (z,x) = z

lx (4.45)

where x is the deflection of the mass m. At the top z = 0 and hence u = 0, atz = l we get u = x. Now we consider the kinetic energy of both masses:

Ekin =12mx

2 +12

∫ l

0u2dm =

12mx

2 +12

∫ l

0

(z

lx)2ρAdz (4.46)

=12mx

2 +12ρA

l2x2∫ l

0z2dz = 1

2mx2 +

12ρA

l2x2 1

3 l3 (4.47)

The mass of the bar is mS = ρAl with ρ as the mass density so that

Ekin =12mx

2 +12

(13mS

)x2 (4.48)

Finally, we get:

Ekin =12

(m+

mS

3

)x2 =

12(meff )x

2 (4.49)

As we can see (for this example) that the mass of the bar is considered by onethird of its total mass.

The natural frequency is:

f0 =1

2π

√√√√ k(m+ mS

3

) =1

2π

√√√√ EA/l(m+ mS

3

) (4.50)

4.3 free vibrations of a viscously damped oscillator 89

xm

k

2k

2

xm

k

2x

k

2x

k

2x

k

2x

ccx

cx

Figure 4.7: SDOF oscillator with viscous damping

which is smaller as in the case when we neglect the mass of the bar.

The factor 13 applies also in the case of mass connected with a helical spring, or

in the case of a disc mounted on a torsion bar. In the case of flexural bendingof beams the static deflection shape is more complicated as in this examplebut can be calculated by means of the beam theory or taken from tables. Thegeneral procedure to calculate the kinetic energy is identical to the example.

4.3 free vibrations of a viscously damped oscillator

4.3.1 Equation of Motion

We consider viscous damping which means that the damper force is propor-tional to the relative velocity of the two connection pins of the damper. Theforce represents fairly well the conditions of damping due to the oil in a dash-pot. Other, more complicated damping laws can be considered which lead tonon-linear equations of motion. Non-linear Equation Of Motion (EOM) (EOM)require a much more complicated mathematical treatment. In most cases onlya numerical solution is possible.

The free body diagram and application of Newton’s law yields:

mx =∑i

Fi = −2k2x− cx (4.51)


ormx+ cx+ kx = 0 (4.52)

The mass m, the viscous damping coefficient c and the stiffness k are constants(as in the previous case) andm, c, k > 0. The viscous damping reduces the totalenergy of the vibrating system. Division by m yields:

x+ 2δx+ ω20x = 0 (4.53)

whereδ =

c

2m (4.54)

and

ω0 =

√k

m

Transition to the dimensionless time (or an angle, depending on the interpre-tation)

τ = ω0t (4.55)

leads to the following derivatives (using the chain rule):

d (. . .)

dt =d (. . .)

dτdτdt =

d (. . .)

dτ ω0 . (4.56)

If we distinguish the derivatives with respect to time t by the dot, and thex = dxdt

x′= dx

dτ2nd derivativesanalogically!

derivatives with respect to τ by the prime, we find that:

ω20x′′+ 2δω0x

′+ ω2

0x = 0 . (4.57)

Division by the square of the natural circular frequency:

x′′+

2δω0x′+ x = 0 (4.58)

orx′′+ 2Dx

′+ x = 0 (4.59)

The parameter D is the only quantity which describes the behavior of the freeDimensionlessdamping damped oscillator. It is called the dimensionless damping (in Germany also

called Lehr’s damping):

D =δ

ω0=

c

2mω0=

c

2√km

(4.60)

As we can see, D contains all three parameters m, c and k.


Rel

Rel

Rel

Rel

Iml

Iml

Iml

Iml

D = 1

D = 0 0 < D < 1

D > 1

No Damping Weak Damping

Critical Damping Strong Damping

Figure 4.8: Solutions of the characteristic equation depending on D

4.3.2 Solution of the Equation of Motion

The solution to eq. (4.59) can be found by the assumption that the solutionhas the form:

x (τ ) = Aeλτ (4.61)where A and λ are in general complex. Putting this into eq. (4.59) we get the Characteristic

equationcharacteristic equationλ2 + 2Dλ+ 1 = 0 (4.62)

which has two solutions: i =√−1

λ1,2 = −D±√D2 − 1 = −D± i

√1−D2 (4.63)

Eq. (4.63) shows that the oscillator changes its behaviour depending whetherDleads to real or complex solutions of λ (see Fig. 4.8). The general mathematicalsolution for x (τ ) resp. x (t) for λ1 6= λ2 is:

x (τ ) = A1eλ1τ +A2eλ2τ (4.64)

and if λ1 = λ2 = λ:

x (τ ) = A1eλτ + τA2eλτ = (A1 + τA2) eλτ (4.65)


The constants A1,2 are determined from the initial conditions (initial displace-ment and velocity, respectively). They can be real or complex.

For different D we get different types of solutions, which we will discuss in thefollowing sections.

1. Overdamped Case: Strong Damping:D > 1D > 1 represents strong damping, we do not get a vibration, but the masscreeps back to the equilibrium position. The mathematical solution forthis case is:

λ1,2 = −D±√D2 − 1

we obtain two real (and different) values. The solution is:

x (τ ) = A1e(−D+√D2−1)τ +A2e(−D−

√D2−1)τ (4.66)

or, if we go back to the t-time scale:

x (t) = A1e(−D+√D2−1)ω0t +A2e(−D−

√D2−1)ω0t (4.67)

Because the real values of λ are always negative we find the creep motionback to the static equilibrium.

2. Critical Damping:D = 1For D = 1 we get the transition from vibration to creep motion (hencewe call this state critical). The two solutions of λ are identical and realwhich follows immediately from eq. (4.63):

λ1,2 = −D±√D2 − 1 = −1 = λ

The resulting motion is

x (τ ) = A1e−τ + τA2e−τ = (A1 + τA2) e−τ (4.68)

and on the t-time scale:

x (t) = A1e−ω0t + ω0tA2e−ω0t = (A1 + ω0A2t) e−ω0t (4.69)

3. Weak Damping, Damped Vibrations:0 < D < 1For 0 < D < 1 we get:

λ1,2 = −D± i√

1−D2

which are conjugate complex. The solution is:

x (τ ) = A1e(−D+i√

1−D2)τ +A2e(−D−i√

1−D2)τ (4.70)

On the t-time scale we obtain:

x (t) = A1e(−D+i√

1−D2)ω0t +A2e(−D−i√

1−D2)ω0t (4.71)


0 10 20 30 40 50 60 700

0.05

0.1

0.15

0.2

0.25

Time t

Am

plit

ude

x

(a) D = 2

0

0.1

0.2

0.3

0.4

Am

plit

ude

x

0 10 20 30 40 50 60 70

Time t

(b) D = 1

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(c) D = 0.5

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(d) D = 0.1

0 10 20 30 40 50 60 70-1.5

-1

-0.5

0

0.5

1

1.5

Time t

Am

plit

ude

x

(e) D = 0.01

0 10 20 30 40 50 60 70-1

-0.5

0

0.5

1

Time t

Am

plit

ude

x

(f) D = 0

Figure 4.9: Free vibrations of a viscously damped oscillator for different D


If we separate the real- and imaginary parts in the exponential function (eq. (4.71))we get:

x (t) = e−Dω0t(A1e+i

√1−D2ω0t +A2e−i

√1−D2ω0t

)(4.72)

The exponential term standing left of the bracket describes the decaying be-havior of the vibration (for D = 0 we obtain the undamped case withoutreduction of the amplitudes).

The expression with the imaginary exponent show us the oscillation because

e±ωDt = cosωDt± sinωDt

and using eq. (4.72) the circular frequency of the damped oscillation is:

ωD =√

1−D2ω0 (4.73)

The frequency is reduced by the influence of the damping. However, if thevalues for D are in the range 0.01 - 0.1, the change of the frequency comparedto the undamped case is negligible.

The period (from one maximum to the next maximum) then is:

T =2πωD

(4.74)

Using trigonometric functions instead of the exponential function eq. (4.72)becomes:

x (t) = e−Dω0t[A cos

(√1−D2ω0t

)+B sin

(√1−D2ω0t

)](4.75)

Introducing the initial displacement x0 and initial velocity v0 at time instantt = 0 yields the constants:

A = x0 (4.76)

andB =

v0 +Dω0x0

ω0√

1−D2 . (4.77)

or alternativelyx (t) = Ce−Dω0t sin (ωDt+ ϕ) (4.78)

The maximum amplitude C and the phase angle ϕ can be determined fromthe following equations:

C =√A2 +B2 =

√(x0ωD)

2 + (v0 +Dω0x0)2

ωD(4.79)

4.4 forced vibrations from harmonic excitation 95

andtanϕ =

A

B(4.80)

An important quantity to describe the damping behavior is the logarithmic Logarithmicdecrementdecrement:

ϑ =1n

ln x1xn+1

= 2π D√1−D2 (4.81)

The xi are maxima of the damped vibration and can be determined frommeasured curves of the free damped oscillations of a vibrating system. Becausethe use of two subsequent maxima can lead to inaccurate results we observe thedecaying process over a longer time (e.g. n = 5− 10 periods, if the damping issmall enough). From the logarithmic decrement we can immediately determineD. From the nT on the time axis from maximum xi to the maximum xn+i wecan calculate the damped natural circular frequency ωD.

4.4 forced vibrations from harmonic excitation

As discussed earlier, forced vibrations are one very important practical mech-anism for the occurrence of vibrations. The equation of motion of the dampedlinear SDOF oscillator with an external force (see fig. 4.10) is:

mx+ cx+ kx = F (t) (4.82)

The general solution of this differential equation is: Approach to theequation of motionin case of forced vi-bration

x (t) = xhom (t)︸︷︷︸free vibrations

+ xpart (t)︸︷︷︸results from external force

(4.83)

x

m

F(t)

ck

Figure 4.10: SDOF oscillator with viscous damping and external force


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

which consists of the homogeneous part resulting from the free vibration andthe particular part resulting from the external disturbance F (t) (see fig. 4.11).The homogeneous solution has already been treated in the last chapter.

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

4.4.1 Excitation with Constant Force Amplitude

The excitation function is harmonic, Ω is the frequency of excitation:In the script onlythe real approachis presented,however, thereis an equivalentcomplex approachthat could befollowed. Complexapproach isillustrated inappendix D.1.

F (t) = F cos Ωt (4.84)

Eq. 4.82 becomes:mx+ cx+ kx = F cos Ωt (4.85)

Dividing by the mass m:

x+c

mx+

k

mx =

F

mcos Ωt . (4.86)

Introducing again the dimension less damping (eq. (4.60)) and the naturalcircular frequency (eq. (4.4)):

2D =c

mω0, (4.87)

andω2

0 =k

m(4.88)

and the amplitude

f =F

m(4.89)


This yields:x+ 2Dω0x+ ω2

0x = f cos Ωt . (4.90)

To solve this differential equation, we make an approach with harmonic func-tions

x (t) = A cos Ωt+B sin Ωt (4.91)

This covers also a possible phase lag due to the damping in the system.Differentiating eq. (4.91) to get the velocity and the acceleration and puttingthis into eq. (4.90) leads to:

−Ω2A cos Ωt−Ω2B sin Ωt+ . . .

. . .+ 2Dω0 (−ΩA sin Ωt+ ΩB cos Ωt) + . . .

. . .+ ω20 (A cos Ωt+B sin Ωt) = f cos Ωt (4.92)

After separating the coefficients of the sin- and cos-functions and comparingthe coefficients we get:

−Ω2A+ 2Dω0ΩB + ω20A = f (4.93)

−Ω2B − 2Dω0ΩA+ ω20B = 0 (4.94)

From the second equation we see that

−Ω2B + ω20B = 2Dω0ΩA , (4.95)

which leads toB =

2Dω0Ωω2

0 −Ω2A (4.96)

and we put this result into eq. (4.93):

−Ω2A+ 2Dω0Ω2Dω0Ωω2

0 −Ω2A+ ω20A = f (4.97)[(

ω20 −Ω2

)+

4D2ω20Ω2

ω20 −Ω2

]A = f (4.98)[(

ω20 −Ω2

)2+ 4D2ω2

0Ω2]A = f

(ω2

0 −Ω2)

(4.99)

This yields the solution for A and B:

A =f(ω2

0 −Ω2)

(ω20 −Ω2)

2+ 4D2ω2

0Ω2(4.100)

andB =

f (2Dω0Ω)

(ω20 −Ω2)

2+ 4D2ω2

0Ω2(4.101)


Introducing the dimensionless ratio of frequencies:

η =Excitation frequenzNatural frequenz =

Ωω0

(4.102)

leads to:

A =

(fω2

0

) (1− η2

)(1− η2)2 + 4D2η2

(4.103)

and

B =

(fω2

0

)(2Dη)

(1− η2)2 + 4D2η2. (4.104)

with A and B we have found the solution for x (t) = A cos Ωt+B sin Ωt.

Another possibility is to present the solution with amplitude and phase angle:

x (t) = C cos (Ωt− ϕ) (4.105)

The amplitude is:

C =√A2 +B2 =

1√(1− η2)2 + 4D2η2

f

ω20

(4.106)

Considering that f = Fm und ω2

0 = km we get:

C =1√

(1− η2)2 + 4D2η2

F

k(4.107)

Introducing the dimensionless magnification factor V 1 which only depends onthe frequency ratio and the damping D:

V1 (η,D) =1√

(1− η2)2 + 4D2η2(4.108)

we get the amplitude as

C = V1F

k(4.109)

and the phase angle (using trigonometric functions similar as in 4.2.2 on p. 84):

tanϕ =B

A=

2Dη1− η2 (4.110)


In fig. 4.12(a) 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 damp-ing D. The maximum of the magnification curve for a given D can be foundat

ηres =√

1− 2D2 =Ωresω0

(4.111)

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

Cmax = V1 (ηres,D)F

k=

12D√

1−D2F

k(4.112)

η → 0 V1 ≈ 1 The system behaves quasi-statically.η →∞ V1 → 0 The vibrations are very small.


0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

V1

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2D

2 2η η

1V1=

(a) Amplitude

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η ω= /Ω0

D=0,05D=0,1

D=0,2D=0,3

D=0,5D=0,7071

1- 2η

φ =2Dη

arctan

D=0

(b) Phase

Figure 4.12: Magnification factor V1 and phase angle to describe the vibration behav-ior of the damped oscillator under constant force amplitude excitation


c

Ω Ω

mu

2

k

2k

2

ε ε

mM

xmu

2

Figure 4.13: SDOF oscillator with unbalance excitation

4.4.2 Harmonic Force from Imbalance Excitation

The total mass of the system in fig. 4.13 consists of the mass mM and the tworotating unbalance masses mu:

m = mM + 2mu

2 (4.113)

The disturbance force from the unbalance is depending on the angular speedΩ, ε the excentricity:

FUnbalance (t) = Ω2εmu cos Ωt (4.114)

Now, following the same way as before (real or complex) leads to the solution:

x (t) = C cos (Ωt− ϕ) (4.115)

with Amplitude

x = C =η2√

(1− η2)2 + 4D2η2εmu

m= V3 (η,D) ε

mu

m, (4.116)

Phasetanϕ =

2Dη1− η2 (4.117)

and Magnification factor

V3 (η,D) =η2√

(1− η2)2 + 4D2η2. (4.118)

The phase has the same expression as in the previous case (eq. (4.110)), how-ever, the magnification factor is different (see fig. 4.14), because the forceamplitude is increasing with increasing angular speed.


V3

0 0,5 1 1,5 2 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

η ω= /Ω0

2,5

√(1- ) +42 2D

2 2η η

η2

V3=

Figure 4.14: Magnification factor V3 for the case of imbalance excitation

As can be seen, for:η → 0 V3 ≈ 0 There is no force if the system is not rotating or

rotates only slowlyη →∞ V3 → 1 The Mass mM is vibrating with an amplitude

(εmum ), but the common center of gravity of total

system mM and mu does not move.

4.4.3 Support Motion / Ground Motion

4.4.3.1 Case 1

The equation of motion for this system depicted in fig. 4.15 is:

mx+ cx+ kx = ku (t) (4.119)

Under harmonic excitation:

u (t) = u cos Ωt (4.120)


u(t)

c

k

m

x

Figure 4.15: Excitation of the SDOF oscillator by harmonic motion of one springend

The mathematical treatment is nearly identical to the first case, only the exci-tation function is different: the excitation F

k is replaced by u here. This leadsto the result for the amplitude of vibration:

x =1√

(1− η2)2 + 4D2η2︸︷︷︸mag. function V1

u = V1 (η,D) u (4.121)

The magnification factor again is V1 (eq. (4.108)). Also, the phase relation isidentical as before (eq. (4.110)):

tanϕ =2Dη

1− η2 (4.122)

4.4.3.2 Case 2

The equation of motion now (see fig. 4.16) also contains the velocity u:

mx+ cx+ kx = cu+ ku (4.123)

Amplitude of vibration and phase shift become:

x =

√1 + 4D2η2√

(1− η2)2 + 4D2η2︸︷︷︸mag. function V2

u = V2 (η,D) u (4.124)

and

tanϕ =2Dη3

1− η2 + 4D2η2 . (4.125)


ck

mx

u(t)

Figure 4.16: Excitation of the SDOF oscillator by harmonic motion of the spring/-damper combination

As can be seen the phase now is different due to the fact that the damper forcedepending on the relative velocity between ground motion and motion of themass plays a role. The amplitude behaviour is described by the magnificationfactor:

V2 =

√1 + 4D2η2√

(1− η2)2 + 4D2η2(4.126)

Notice that all curves have an intersection point at η =√

2 (see fig. 4.17 whichmeans 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 dampingforces.Further cases of ground motion excitation are possible.

4.5 excitation by impacts

4.5.1 Impact of Finite Duration

We consider an impact of finite length Ti and constant force level during theimpact (see fig. 4.18). The impact duration Ti is much smaller than the periodof vibration T .

Ti << T =2πωD

(4.127)

4.5 excitation by impacts 105

η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure 4.17: Magnification factor V2 for the case of ground excitation via spring and

damper

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

p = mv0 =∫ Ti

0Fdt = F Ti (4.128)

This leads to the initial velocity:

v0 =F Tim

(4.129)

Ti

F(t)

t

F(t)

kc

x

m

F(t)^

Figure 4.18: SDOF oscillator under impact loading


Using the results of the viscously damped free oscillator for D < 1 (eq. (4.75)),

x (t) = e−Dω0t

A cos

√1−D2ω0︸︷︷︸ωD

t

+B sin

√1−D2ω0︸︷︷︸ωD

t

(4.130)

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

x0 = 0 ⇒ A = 0 (4.131)

and

v0 =F Tim

⇒ B =v0

ω0√

1−D2 =v0ωD

(4.132)

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

x (t) =v0ωD

e−Dω0t sin (ωDt) (4.133)

4.5.2 DIRAC-Impact

The DIRAC-Impact is defined by

F (t) = F δ (t)→ δ (t) =

0 t 6= 0

∞ t = 0but

∫ ∞−∞

δ (t) dt = 1 (4.134)

δ is the Kronecker symbol. The duration of this impact is infinitely shortbut the impact is infinitely large. However, the integral is equal to 1 or F ,

F

t

Figure 4.19: DIRAC-Impact

4.6 excitation by forces with arbitrary time functions 107

respectively. For the initial displacement x0 = 0 and calculation of the initialvelocity following the previous chapter, we get:

x (t) =F

mωDe−Dω0t sin (ωDt) (4.135)

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

h (t) =1

mωDe−Dω0t sin (ωDt) (4.136)

The IRF is an important characteristic of a dynamic system in control theory.

4.6 excitation by forces with arbitrary time functions

F( )τ

τ τ τ+Δ

F

t

t

x

τ

Figure 4.20: Interpretation of an arbitrary time function as series of DIRAC-impacts

Using the results of the previous chapters we can solve the problem of an arbi-trary time function F (t) (see fig. 4.20) 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:

x (t) =∫ t

0

1mωD

e−Dω0(t−τ ) sin (ωD (t− τ ))F (τ ) dτ (4.137)

=∫ t

0h (t− τ )F (τ ) dτ (4.138)

As can be seen, the integral contains the response of the SDOF oscillator withrespect to a DIRAC-impact multiplied with the actual force F (τ ), which isintegrated from time 0 bis t.

4.7 periodic excitations

4.7.1 Fourier Series Representation of Signals

Periodic signals can be decomposed into an infinite series of trigonometricfunctions, called Fourier series (see fig. 4.21). The period of the signal is T andthe corresponding fundamental frequency is:

ω =2πT

(4.139)

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

x (t) =a02 +

∞∑k=1

ak cos (kωt) + bk sin (kωt) (4.140)

Figure 4.21: Scheme of signal decomposition by trigonometric functions

4.7 periodic excitations 109

The Fourier-coefficients a0, ak and bk must be determined. They describe howstrong 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

∫ T

0x (t) dt (4.141)

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

ak =2T

∫ T

0x (t) cos (kωt) dt (4.142)

andbk =

2T

∫ T

0x (t) sin (kωt) dt (4.143)

The individual frequencies of this terms are

ωk = kω =2πkT

(4.144)

For k = 1 we call the frequency ω1 fundamental frequency or basic harmonicand the frequencies for k = 2, 3, . . . the second, third, . . . harmonic (or generallyhigher harmonics).

The real trigonometric functions can also be transformed into complex expo-nential expression:

x (t) =∞∑

k=−∞Xke

ikωt (4.145)

TheXk are the complex Fourier coefficients which can be determined by solvingthe integral:

Xk =1T

∫ T

0x (t) e−ikωtdt (4.146)

or

Xk =1T

∫ T

0x (t) [cos kωt− i sin kωt] dt (4.147)

which clearly shows the relation to the real Fourier coefficients series given byeq. (4.142) and eq. (4.143):

Re Xk =ak2 ; Im Xk = −

bk2 (4.148)

The connection to the other real representation (chap. 4.7.1) is:

|Xk| = ck tanϕk =(

Im XkRe Xk

)(4.149)

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

X−k = X∗k (4.150)


4.7.2 Forced Vibration Under General Periodic Excitation

Let us use once more the SDOF oscillator (see fig. 4.22) but now the force isa periodic function which can be represented by a Fourier series

F (t) =F02 +

∞∑k=1

Fck cos (kΩt) + Fsk sin (kΩt) (4.151)

The Fck and Fsk are the Fourier coefficients which can be determined ac-

x

m

F(t)

ck

Figure 4.22: SDOF oscillator with viscous damping and external force

cording to the last chapter (eqns. 4.141 - 4.143). The response due to such anexcitation is:Here, the stiffness

of the system iscalled k∗. x (t) =

F02k∗ +

∞∑k=1

V1 (ηk,D)Fckk∗

cos (kΩt− ϕk) + . . .

. . .+ V1 (ηk,D)Fskk∗

sin (kΩt− ϕk)(4.152)

with the frequency ratio:

ηk =kΩω0

, k = 1, 2, . . .∞ (4.153)

Each individual frequency is considered with its special amplification factor Vand individual phase shift ϕ, which in the present case can be calculated from(see eq. (4.108)):

V1 (ηk,D) =1√

(1− η2k)

2+ 4D2η2

k

tanϕk =2Dηk1− η2

k

For the other cases of mass unbalance excitation or ground excitation theprocedure works analogously. The appropriate V -functions have to be usedand the correct pre-factors (which is in the present case 1

k ) have to be used.


Figure 4.23: Perturbation function

example: Graphical determination of the complete solution for a givenperturbation function F (t) (see fig. 4.23)

F (t) = F1 (t) + F3 (t) = F1 sin (Ω1t) + F3 sin (Ω3t)

Fig. 4.24 shows the graphical representation of both partial functions of theperturbation function (only the first two not vanishing parts of the Fourierseries) where:

F1 (t) = F1 sin (Ω1t) F3 (t) = F3 sin (Ω3t)

F1 = 1 F3 =13 F1

Ω1 = Ω Ω3 = 3Ω

The frequency-dependent amplitude (magnification factor) including the val-ues V1 and V3 which are determined for η1 and η3 is depicted in a resonancediagram shown in fig. 4.25(a) (dimensionless damping amounts D = 0.2). Fig.4.25(b) shows the according phase angle diagram. The graphical representa-tions of the partial solutions

x1 (t) = A1 sin (Ω1t− ϕ1)

x3 (t) = A3 sin (Ω3t− ϕ3)

are illustrated in fig. 4.25(c). In fig. 4.25(d) one can study the complete solution

x (t) = x1 (t) + x3 (t)

with A1 > A3.


The difference in case of resonance is shown by the graphical representation ofthe frequency-dependent amplitude (magnification factor) including the values

Figure 4.24: Graphical representation of both sub-functions

(a) Magnification factor

0 1

h

φ3

φ1

h1

h3

p/2

Pha

se a

ngle

(b) Phase angle

Time t

Dis

plac

emen

t x

(c) Partial solution

Time t

Dis

plac

emen

t x

(d) Complete solution

Figure 4.25: Graphical determination of the complete solution for a given perturba-tion function F (t) without resonance


Amplitude

(a) Magnification factor

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η = Ω / ω0

D=0,05

D=0,1

D=0,2

D=0,3D=0,5

D=0,7071

21-ηφ =

2Dηarctan

D=0

(b) Phase angle

Time t

Dis

plac

emen

t x

(c) Partial solution

Time t

Dis

plac

emen

t x

(d) Complete solution

Figure 4.26: Graphical determination of the complete solution for a given perturba-tion function F (t) in case of resonance

V1 and V3 which are determined for η1 and η3. This is depicted in a resonance di-agram in fig. 4.26(a). Here we have resonance, dimensionless damping amountsD = 0.2. Fig. 4.26(b) shows the according phase angle diagram. The graphicalrepresentations of the partial solutions

x1 (t) = A1 sin (Ω1t− ϕ1)

x3 (t) = A3 sin (Ω3t− ϕ3)

are illustrated in fig. 4.26(c). In fig. 4.26(d) one can study the complete solution

x (t) = x1 (t) + x3 (t)

with A1 < A3.


4.8 vibration isolation of machines

In dynamics of machines and systems the linear vibrations of SDOF systemsintroduced at the beginning of chapter 4 are often applied to isolate machineryfrom vibrations.

• Requirements:

– machine can be idealized as rigid

– vibration process can be considered to be a SDOF system

• Vibration ssolation:

– important assignment of dynamics of machines and systems

– concerning machine manufactures, project planers, operators

– any installation site whether ground, floor or another carrying struc-ture is elastic. That is why excitation by imbalanced inertia forcescan occur

• Purpose:

– installing the machine in order to minimize forces (inertia forces, ...)diverted into the foundation / structure

– mounting sensitive devices in a way so that they are well shieldedfrom their vibrating environment

We differ between:

1. Low tuning:The highest natural frequency of the ground vibration is less than thelowest excitation frequency.

2. High tuning:The natural frequencies of the ground vibration are above the range ofthe excitation frequencies.

3. Mixed tuning:The range of natural frequencies and the range of excitation frequenciespartially overlap each other, but there is no resonance.

4.8 vibration isolation of machines 115

c

Ω

k

2k

2

ε

x

mu

Figure 4.27: Imbalanced machine - substituted mechanical system

These terms are generally valid for MDOF systems too.

4.8.1 Forces on the Environment Due to Excitation by Inertia Forces

In order to achieve vibration isolation we initially have to calculate the forceson the environment in order to use encountered dependencies for the purposeof minimizing these forces. The determination of these forces is performed bymeans of an example.

example: Imbalanced machine (see fig. 4.27)

Figure 4.28: Imbalanced machine - forces acting on the foundation


Amplitude of the excitation force:

Fu = muεΩ2 = UΩ2

with total mass m and imbalance mass mu.

Forces acting on the foundation like in fig. 4.28:

FFu = kx+ cx

In case of harmonic excitation x (t):

X (t) = XeiΩt

˙X (t) = iΩXeiΩt

¨X (t) = −Ω2XeiΩt

FFu (t) = kXeiΩt + ciΩXeiΩt = (k+ iΩc) XeiΩt

ˆFFu (t) = (k+ iΩc) η2

(1− η2) + 2Dηiεmu

m

=(ω2

0 + 2Dω0iΩ) η2

(1− η2) + 2Dηiεmu

Referring the complex amplitude to the amplitude of the excitation force:

ˆFFuFu

=

(ω2

0 + 2Dω0iΩ)

η2

(1−η2)+2Dηiεmu

Ω2εmu=

1 + 2Dηi1− η2 + 2Dηi

Real amplitude ratio (see. eq. (4.126)):

FFu

Fu=

√√√√ 1 + 4D2η2

(1− η2)2 + 4D2η2= V2 (η,D) (4.154)

Absolute amplitude of the foundation force:

FFu = V2 (η,D) Fu = V2Ω2muε = V2η2muεω

20 = V2η

2mu

mεk (4.155)

FFu = V2η2mu

mεk = V4

mu

mεk (4.156)

V4 = η2V2 (4.157)


η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure 4.29: Magnification factor V2

4.8.2 Tuning of Springs and Dampers

From the real amplitude ration (see eq. (4.154))

FFu

Fu= V2 (η,D) , (4.158)

we can deduce, η should attain higher values so that V2 becomes as small aspossible (see fig. 4.29).

High values of η mean:

Ω = ηω0 ⇒ Ω > ω0

This is called low or "soft" tuning.

Soft springs, heavy foundation block ⇒ small ω0.

Using soft springs, it is necessary to mind the static lowering under dead load,because it can increase to high values (too high!).

Moreover it shows: In order to have V2 as small as possible damping should besmallest possible at fixed Ω. But this brings along disadvantages for passingthe case of resonance (while starting up and shutting down).


Remedy:

• passing resonance frequency very fast (sufficient torque required!)

• connecting a damper while passing resonance frequency

By comparison: rigid installation would lead to FF u

Fu= 1.

numerical example D = 0; Tuning target: we want to tune the foun-dation in a way, so that only 5% of the excitation force is transferred to theenvironment!

FFu

Fu< 5% =

120

V2 (η,D = 0) = 11− η2

η =Ωω0

⇒ 120 = ±

∣∣∣∣∣ 11− η2

∣∣∣∣∣⇒ 1− η2 = −20⇔ −η2 = −21⇔ η =

√21 ≈ 4, 58

that meansΩ ≥ 4, 58ω0 or ω0 ≤

(1

4, 58

)Ω (4.159)

In case of constant force amplitudes we obtain the following isolation effect:Isolation effect incase of excitation

with constantforce amplitude

F (t) = F sin Ωt

or

F (t) = F eiΩt .

Applying the same approach as in section 4.8 we get (see eq. (4.154)):

FFu

Fu= V2 (η,D)

Dealing with machinery installed in way that it is excited by ground motionVibration isolationin case of groundmotion

we have (see eq. (4.126)):x

xa= V2 (η,D)


η ω= /Ω 0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

Hard foundation Soft foundation

Figure 4.30: Magnification factor V2 for tuning the spring

x

m

ck

xa

Figure 4.31: Single-mass oscillator

We notice the most suitable way to manage a vibration isolated installation ofsensitive devices is to use low tuning. The same applies to soundproof roomsetc.

5VIBRATION OF L INEARMULTIPLE -DEGREE -OF -FREEDOM SYSTEMS

5.1 equation of motion

The equation of motion can be derived by using the principles we have learnedsuch as Newton’s/Euler’s laws or Lagrange’s equation of motion. For a generallinear system mDOF system we found that we can write in matrix form:

M x+(C +G

)x+

(K +N

)x = F (5.1)

with the matrices: Note:A general matrix Bcan be decomposedinto the symmetricpart and the skew-symmetric part Band B:

B = B + B

where:

B = 12(B +BT

)B = 1

2(B −BT

)

• M : Mass matrix (symmetric) M = MT

• C: Damping matrix (symmetric) C = CT

• K: Stiffness matrix (symmetric) K = KT

• G: Gyroscopic matrix (skew-symmetric) G = −GT

• N : Matrix of non-conservative forces (skew-symmetric) N = −NT

• F : External forces

In the standard case that we have no gyroscopic forces and no non-conservativedisplacement dependent forces but only inertial forces, damping forces andelastic forces the last equation reduces to:

M x+C x+K x = F (5.2)

121

122 vibration of linear multiple-degree-of-freedom systems

example The system shown in fig. 5.1, where the masses can slide withoutfriction (µ = 0), has them1 0 00 m2 00 0 m3

x1

x2

x3

+c1 0 00 0 00 0 0

x1

x2

x3

+ . . .

. . .+

k1 + k2 + k3 −k2 −k3

−k2 k2 0−k3 0 k3

x1

x2

x3

=

F1 (t)

F2 (t)

F3 (t)

(5.3)

5.2 influence of the weight forces and static equilib-rium

The static equilibrium displacements are calculated by (xstat = xstat = 0):

K xstat = F stat (5.4)

which means for the case of the example shown in fig. 5.2:

K xstat = F stat =

m1g

m2g

(5.5)

F (t)1

F (t)2

F (t)3

x2

k1

c1

x1

x3

k2

k3

m1

m3

m2

m=0

m=0

x1x1x1x1x1

F (t)1

m1

FC1

FK1FK3

FK2

FK3

FK2 m2

F (t)3m3

F (t)2

x1x1x1 x2

x3

Figure 5.1: Multi-Degree-of-Freedom (MDOF) system with free body diagram

5.2 influence of the weight forces and static equilibrium 123

g

k2

k1

m1

m1

m2

m2

k1

k2

x stat1

x stat2

Figure 5.2: Static equilibrium position of a two DOFs system

The dynamic problem for this example is:

M x+K x = F stat︸︷︷︸static force

+F (t) (5.6)

x = xstat + xdyn︸︷︷︸part of the motion

describing the vibrationabout the static equilibrium

(5.7)

From the last equation also follows that

x = xdyn ⇒ x = xdyn (5.8)

so thatM xdyn +K

(xdyn + xstat

)= F stat + F (t) (5.9)

and after rearrangement

M xdyn +K xdyn = F stat −K xstat︸︷︷︸=0

+F (t) (5.10)

M xdyn +K xdyn = F (t) (5.11)

As can be seen the static forces and static displacements can be eliminatedand the equation of motion describes the dynamic process about the staticequilibrium position.

In cases where the weight forces influences the dynamic behavior a simple elim- Noteination of the static forces and displacements is not possible. In the exampleof an inverted pendulum shown in fig. 5.3 the restoring moment is mgl sinφ,where l is the length of the pendulum.


k k

mg

m

Figure 5.3: Case where the static force also influences the dynamics

5.3 ground excitation

Fig. 5.4 shows a MDOF system.

Without ground motion x0 = 0 the equation of motion is:m1 0

0 m2

x1

x2

+c1 0

0 0

x1

x2

+ . . .

. . .+

k1 + k2 −k2

−k2 k2

x1

x2

=

F1

F2

(5.12)

Now, if we include the ground motion, the differences (x1 − x0) and the rel-ative velocity d(x1 − x0)/dt determine the elastic and the damping force, re-

k2

k1

m1

m2

c1

x2

x1

xo

F2

F1

Figure 5.4: 2DOF system with excitation by ground motion x0

5.3 ground excitation 125

spectively at the lower mass. This can be expressed by adding x0 to the lastequation in the following manner:m1 0

0 m2

x1

x2

+c1 0

0 0

x1

x2

+k1 + k2 −k2

−k2 k2

x1

x2

= . . .

. . . =

F1

F2

+k1

0

x0 (t) +

c10

x0 (t) (5.13)

The dynamic force f0 of the vibrating system on the foundation is:

F0 = k1 (x1 − x0) + c1 (x1 − x0) (5.14)

Figure 5.5: Example for ground motion excitation of a building structure (earth-quake excitation)

Figure 5.6: Excitation of a vehicle by rough surface


5.4 free undamped vibrations of the multiple-degree-of-freedom system

5.4.1 Eigensolution, Natural Frequencies and Mode Shapes of the System

The equation of motion of the undamped system is:

M x+K x = 0 (5.15)

To find the solution of the homogeneous differential equation, we make theharmonic solution approach as in the SDOF case. However, now we have toconsider a distribution of the individual amplitudes for each coordinate. Thisis done by introducing (the unknown) vector ϕ:

x = ϕeiωt (5.16)x = −ω2ϕeiωt (5.17)

Putting this into eq. (5.15) yields:(K − ω2M

)ϕ = 0 (5.18)

This is a homogeneous equation with unknown scalar ω and vector ϕ. If we setλ = ω2 we see that this is a general matrix eigenvalue problem1:(

K − λM)ϕ = 0 (5.19)

where λ is the eigenvalue and ϕ is the eigenvector. Because the dimension ofthe matrices is f by f we get f pairs of eigenvalues and eigenvectors:

λi = ω2i for i = 1, 2, . . . f (5.20)

ωi is the i-th natural circular frequency andϕi

the i-th eigenvector which has the physical meaning of a vibrationmode shape.

The solution of the characteristic equation

det(K − ω2M

)= 0 (5.21)

1 The well-known special eigenvalue problem has the form(A− λI

)x = 0, where I is the

identity matrix, x the eigenvector and λ the eigenvalue.

5.4 free undamped vibrations of the multiple-degree-of-freedom system 127

yields the eigenvalues and natural circular frequencies λ = ω2, respectively.The natural frequencies are:

fi =ωi2π (5.22)

The natural frequencies are the resonant frequencies of the structure.

The eigenvectors can be normalized arbitrarily, because they only represent avibration mode shape, no absolute values. Commonly used normalizations are

• normalize ϕiso that

∣∣∣ϕi

∣∣∣ = 1

• normalize ϕiso that the maximum component is 1

• normalize ϕiso that the modal mass (the generalized mass) is 1.

Generalized mass or modal mass:

Mi = ϕTiM ϕ

i(5.23)

Generalized stiffness:Ki = ϕT

iK ϕ

i(5.24)

whereKi = ϕT

iK ϕ

i= ω2

i if Mi = 1 (5.25)

The so-called Rayleigh ratio is: Rayleigh ratio

ω2i =

Ki

Mi=

ϕTiK ϕ

i

ϕTiM ϕ

i

(5.26)

It allows the calculation of the frequency if the vectors are already known.

5.4.2 Modal Matrix, Orthogonality of the Mode Shape Vectors

If we order the natural frequencies so that

ω1 ≤ ω2 ≤ ω3 ≤ . . . ≤ ωf

and put the corresponding eigenvectors columnwise in a matrix, the so-calledmodal matrix, we get: Modal matrix

Φ =[ϕ1,ϕ2, . . . ,ϕ

f

]=

ϕ11 ϕ12 . . . ϕ1f

ϕ21 ϕ22 . . . ϕ2f

. . . . . . . . . . . .

ϕf1 ϕf2 . . . ϕff

(5.27)


The first subscript of the matrix elements denotes the number of the vectorcomponent while the second subscript characterizes the number of the eigen-vector.

The eigenvectors are linearly independent and moreover they are orthogonal.This can be shown by a pair i and j(

K − ωi2M)ϕi= 0 and

(K − ωj2M

)ϕj= 0 (5.28)

Premultplying by the transposed eigenvector with index j and i respectively:

ϕjT(K − ωi2M

)ϕi= 0 and ϕ

iT(K − ωj2M

)ϕj= 0 (5.29)

If we take the transpose of the second equation

ϕjT(KT − ωj2MT

)ϕi= 0 (5.30)

and consider the symmetry of the matrices: M = MT and K = KT andsubtract this equation ϕ

jT(K − ωi2M

)ϕi= 0 from the first eq. (5.28) we get(

ωj2 − ωi2

)ϕjT(M)ϕi= 0 (5.31)

which means that if the eigenvalues are distinct ωi 6= ωj for i 6= j the secondscalar product expression must be equal to zero:

ϕjTMϕ

i= 0 (5.32)

That means that the two distinct eigenvectors i 6= j are orthogonal with respectto the mass matrix. For all combinations we can write:Kronecker

symbol δijϕjTMϕ

i= δijMi

ϕjTKϕ

i= δijωi

2Mi

δij =

0, for i = j

1, for i 6= j(5.33)

or with the modal matrix:

ΦTMΦ = diag Mi =

M1 0 . . . 00 M2

. . . ...... . . . . . . 00 . . . 0 Mf

ΦTKΦ = diagωi

2Mi

=

ω12M1 0 . . . 0

0 ω22M2. . . ...

... . . . . . . 00 . . . 0 ωf

2Mf

(5.34)


Figure 5.7: Mode shapes and natural frequencies of a two storey structure


5.4.3 Free Vibrations, Initial Conditions

The free motion of the undamped system x (t) is a superposition of the modesvibrating with the corresponding natural frequency:

x (t) =f∑i=1

ϕi(Aci cosωit+Asi sinωit) (5.35)

Each mode is weighted by a coefficient Aci and Asi which depend on theinitial displacement shape and the velocities. In order to get these coefficients,we premultiply eq. (5.35) by the transposed j-th eigenvector:

ϕjTMx (t) =

f∑i=1

ϕjTMϕ

i(Aci cosωit+Asi sinωit)︸︷︷︸for i6=j⇒=0

(5.36)

= ϕjTMϕ

j︸︷︷︸Mj

(Acj cosωjt+Asj sinωjt) (5.37)

All but one of the summation terms are equal to zero due to the orthogonalityconditions. With the initial conditions for t = 0 we can derive the coefficients:

t = 0x (t = 0) = x0

ϕjTMx0 = MjAcj

⇒ Acj =ϕjTMx0

Mj(5.38)

x (t) =f∑i=1

ϕiωi (−Aci sinωit+Asi cosωit) (5.39)

t = 0x (t = 0) = v0

ϕjTMx0 = MjωjAsj

⇒ Asj =ϕjTMv0

Mjωj(5.40)

which we have to calculate for modes j.

5.4.4 Rigid Body Modes

As learned earlier the constraints reduce the DOFs of the rigid body motion.If the number of constraints is not sufficient to suppress rigid body motion thesystem has also zero eigenvalues. The number of zero eigenvalues correspondsdirectly to the number of rigid body modes. In the example shown in fig. 5.8 theExample of a rigid

body mode


two masses which are connected with a spring can move with a fixed distanceas a rigid system. This mode is the rigid body mode, while the vibration ofthe two masses is a deformation mode. The equation of motion of this systemis: m1 0

0 m2

x1

x2

+ k −k−k k

x1

x2

=

00

(5.41)

The corresponding eigenvalue problem is:k− λm1 −k−k k− λm2

ϕ1

ϕ2

=

00

(5.42)

The eigenvalues follow from the determinant which is set equal to zero:

det [. . .] =[(k− λm1) (k− λm2)− k2

]= 0 (5.43)

λ2m1m2 − λ (km1 + km2) = 0 (5.44)

Obviously, this quadratic equation has the solution

λ1 = ω21 = 0 (5.45)

andλ2 = ω2

2 = km1 +m2m1m2

(5.46)

The corresponding (unnormalized) eigenvectors are

ϕ1 =

11

(5.47)

which is the rigid body mode: both masses have the same displacement, nopotential energy is stored in the spring and hence no vibration occurs. Thesecond eigenvector, the deformation mode is

ϕ2 =

1−m1m2

(5.48)

which is a vibration of the two masses. Other examples for systems with rigidbody modes are shown in the following figures.

x1x2

m1 m2

k

Figure 5.8: A two-DOFs oscillator which can perform rigid body motion


5.5 forced vibrations of the undamped oscillator underharmonic excitation

The equation of motion for the type of system depicted in fig. 5.12 is:

M x+K x = F (t) (5.49)

Figure 5.9: Examples for systems with torsional and transverse bending motion withrigid body motion

Figure 5.10: Flying airplane (Airbus A318) as a system with 6 rigid body modes anddeformation modes

Figure 5.11: Commercial communication satellite system (EADS) with 6 rigid bodymodes and deformation modes

5.5 forced vibrations of the undamped oscillator under harmonic excitation 133

For a harmonic excitation we can make an exponential approach to solve theproblem as we did with the SDOF system.

F (t) = F︸︷︷︸complexamplitudevector

eiΩt (5.50)

We make a complex harmonic approach for the displacements with Ω as theexcitation frequency:

x = XeiΩt (5.51)The acceleration vector is the second derivative

x = −Ω2XeiΩt (5.52)

Putting both into the equation of motion (eq. (5.49)) and eliminating theexponential function yields (

K −Ω2M)X = F (5.53)

which is a complex linear equation system that can be solved by hand for asmall number of DOFs or numerically. The formal solution is:

X =(K −Ω2M

)−1F (5.54)

which can be solved if determinant of the coefficient matrix:

det(K −Ω2M

)6= 0 (5.55)

If the excitation frequency Ω coincides with one of the natural frequenciesωi we get resonance of the system with infinitely large amplitudes (in theundamped case).

k1

k2

m2

m1

k2

x1

x2

F (t)2

F (t)1

2 2

Figure 5.12: Example for a system under forced excitation


Resonance:det

(K −Ω2M

)= 0⇔ Ω = ωi (5.56)

Part II

APPENDIX

AE INLE ITUNG - ERGÄNZUNG

137

BKINEMATICS - APPENDIX

b.1 general 3-d motion in cartesian coordinates

The position vector in cartesian coordinates with time invariant unit vectorsex, ey und ez:

r (t) = x (t) ex + y (t) ey + z (t) ez (B.1)

The coordinates x (t), y (t) and z (t) are scalar functions of time t. Because theunit vectors are time invariant in the cartesian system, we immediately get thevelocity vector (first derivative) and the acceleration vector (second derivative)of the position vector with respect to time:

v (t) = r (t)

= x (t) ex + y (t) ey + z (t) ez (B.2)a (t) = v (t) = r (t)

= x (t) ex + y (t) ey + z (t) ez (B.3)

with components vx = x, vy = y and vz = z, also ax = x, ay = y and az = z.

The velocity vector is always tangential to the path of the particle P . In general,this is not true for the acceleration vector.

x

y

z

P

r

e

e

e

z

x y x

y

z

Path

Figure B.1: Cartesian Coordinate System

139

140 kinematics - appendix

The magnitude of the position, velocity and acceleration vector is:

r = |r| =√x2 (t) + y2 (t) + z2 (t) (B.4)

v = |v| =√x2 (t) + y2 (t) + z2 (t) (B.5)

a = |a| =√x2 (t) + y2 (t) + z2 (t) (B.6)

b.2 three-dimensional motion in cylindrical coordinates

The description of motion on a curved path might be easier described in cylin-drical coordinates than in cartesian coordinates.

The position vector is in cylindrical coordinates

r (t) = r (t) er + z (t) ez (B.7)

whereas the unit vector er depends on the angle ϕ? and this on his part againdepends on the time, which can be written as er = er (ϕ (t)) . Due to thistime dependence of the unit vector the derivations with respect to time for thevelocity and the acceleration are not as simple expressions as in cartesian coor-dinates. In fact the following expression can be obtained (without derivation):

v (t) = r (t) = r (t) er + r (t) ϕ (t) eϕ + z (t) ez (B.8)

with the velocity components:

vr = r Radial velocityvϕ = rϕ Circular velocityvz = z Axial velocity

Path

y

x

z

ejez

erj

r

r

z

P

Figure B.2: Cylindrical Coordinates

B.3 natural coordinates, intrinsic coordinates or path variables 141

and the acceleration is:

a (t) = r (t) =[r (t)− r (t) ϕ2 (t)

]er

+ [2r (t) ϕ (t) + r (t) ϕ (t)] eϕ + z (t) ez (B.9)

with the components:

ar = r− rϕ2 Radial acceleration (B.10)aϕ = 2rϕ+ rϕ Circular acceleration (B.11)az = z Axial acceleration (B.12)

Whereas the component −rϕ2 = −rω2 is the centripetal acceleration and ω isthe angular velocity. The component 2rϕ = 2vrω is the Coriolis acceleration.The magnitude of velocity and acceleration can be obtained using the rule ofvector operation:

|v| =√v2r + v2

ϕ + v2z (B.13)

|a| =√a2r + a2

ϕ + a2z (B.14)

The magnitude of the position vector may not be mistaken with the radialcoordinate r

|r| =√r2 + z2 (B.15)

b.3 natural coordinates, intrinsic coordinates or pathvariables

If the path and the velocity with which the particle P moves on the path aregiven, it is of benefit to use path coordinates (so called “natural” coordinates).The coordinate along the path is denoted by s = s (t) and the path velocity by

ej

er

ez

r

j

rj

r

z

ej

er

ez

r

j

rj

r

zrj2

2rj

Figure B.3: Visualisation of the single velocity and acceleration components

142 kinematics - appendix

yx

z

eteb

r

P

Path

s

M

en

0

r

Figure B.4: Natural coordinates / Path variables

v (t) = s (t). The accompanying coordinate system is moving with the particleP on the path. The time variant tangent unit vector et always points in thedirection of the current motion and is always tangential to the curve. Since thevelocity vector is always tangent to the path, the velocity vector can only beexpressed in terms of the tangent vector:

v (t) = v (t) et = set (B.16)

Approximating the path through a circle in the point P (the radius of thecircle ρ is variable, in general), the normal unit vector points to the center ofthis circle. The binormal ~eb is perpendicular to the osculating plane. It can beobtained using the cross product of the tangent unit vector and the normalunit vector:

eb = et × en . (B.17)

The acceleration can be obtained by differentiation of v (t) with respect totime. Note that the tangent unit vector is time variant and has to be taken inconsideration for the derivative of v (t). With the aid of the Frenet’s Formulaof the circle theory, the acceleration is:

a (t) = vet +v2

ρen (B.18)

with the components

at = v = s tangential acceleration, path acceleration (B.19)

an =v2

ρ=s2

ρnormal acceleration (B.20)

B.3 natural coordinates, intrinsic coordinates or path variables 143

You may recognize that for a linear motion with a circle of infinite large radiusthe equations derived earlier can be obtained.

CFUNDAMENTALS OF KINET ICS - APPENDIX

c.1 special cases for the calculation of the angularmomentum

In general, The angular momentum in a ∗-reference frame is given by eq. (3.91)

L0∗ = L0 − r0∗ × p−mr∗S × v0∗

We distinguish the following special cases

1. 0∗ is a fixed point:then the vlocity of this point isv0∗ = 0 which results:

L0∗ = L0 − r0∗ × p

2. 0∗ coincids with the point A which belongs to the moving rigid body, using EularequationvA = vS +ω× rSAand the velocityof the center ofgravity of the rigidbody is includedin the angularmomentumthrough the termvA.

and moves with it:

L0∗ = LA = L0 − rA × p−mrAS × vA

mit r0∗ = rA und p = mvS .

3. 0∗ coincids with the center of gravity of the rigid body A = S:then rAS = 0 and vA = vS which means:

L0∗ = LS = L0 − rS × p

As the angular momentum L0 has two terms, translation and rotation,such as

L0 = rS ×mvS + JSω

this will reduce the angular momentum at the center of gravity of therigid body S to just the rotational term

LS = JSω .

145

146 fundamentals of kinetics - appendix

4. 0∗ is ths instantaneous center of rotation A:at the instantaneous center of rotation, the velocity is given as vA = 0.and this point is not fixed in space, neither fixed to the body, but itsmotion is described by the path of the instantaneous center of rotation.

Using the translational and rotational terms of the angular momentumas in 3, and using vA = 0, we get:

LA = JAω

This final result is also valid if the point A is fixed point.

It is now important to keep in mind what are the terms that change in theEffects on the An-gular momentumtheorem

angular mometum theorem with the change of reference point.

Starting with the angular momentum theorm with reference point 0 (see eq. (3.84)):

L0 = M0

With the change of reference point to an arbitrary moving point 0* the angularmomentom is written according to eq. (3.105) as

L0∗ = L0 − r0∗ × p−mr∗S × v0∗

and the moment with regard to the new reference point is:

M0∗ = r∗ × F = (r− r0∗)× FM0∗ = M0 − r0∗ × F

which leads to the following angular momentum theorm applied at referencepoint 0:

ddt[L0∗ + r0∗ × p+mr∗S × v0∗

]= M0∗ + r0∗ × F (C.1)

Calculating the time derivative and noticing that p = F and

r0∗ × p = r0∗ × F ,

which will result

L0∗ + r0∗ × p+mr∗S × v0∗ +mr∗S × v0∗ = M0∗ . (C.2)

Moreover, we havevS = v0∗ + v∗S

andr∗S = v∗S = vS − v0∗ ,

C.1 special cases for the calculation of the angular momentum 147

so the thrid term on the left side becomes:

mr∗S × v0∗ = m (vS − v0∗)× v0∗

= mvS × v0∗ −mv0∗ × v0∗

= mvS × v0∗

= −v0∗ ×mvS= −v0∗ × p= −r0∗ × p

Finally, we have:L0∗ +mr∗S × v0∗ = M0∗ (C.3)

It is clear from this last relation, that a new term is introduced to the angularmomentum theorm when it is written in term of the moving point ’0∗’, andthe new term, the second term in the left side in eq. (C.3), should always beget in mind. At this point, different special cases could be observered

1. The reference point 0∗ is moving with constant velocity:as v∗0 = konst then v∗0 = 0 and

L0∗ = M0∗

2. The reference point 0∗ is the center of gravity: 0∗ = S:S could also perform accelerated motion. As r∗S = 0, then

LS = MS

3. The point 0∗ is fixed: 0∗ = A or it is the instantanuous center of rotation:for a fixed point or the instantanuous center of rotation we have: vA = 0which results

LA = MA

Normally the center of gravity or the inatatanuous center of rotation is usedin solving problems of classical mechanics, which seems to be a good choice asthe second term in the left side in eq. (C.3) will disappear.

As we have seen, the angular momentum has two terms, the translational andthe rotational. Choosing the center of gravity, or a fixed point ( i.e. instan-tanuous center of rotation) as reference point, this will reduce the angulatmomentum to the rotational part, and the angular momentum theorm basedon such choice takes the following form

LS =ddt(JSω

)= MS (C.4)

148 fundamentals of kinetics - appendix

similarly,LA =

ddt(JAω

)= MA (C.5)

The term ddt

(JSω

)does not simply equal to JSω, as the moment of inertiaNote

of a rotating rigid body in the a fixed reference frame can be time dependent.An exception to this is, for example, a rotating cylinder about its axis that iscoicides with one of the axis of the reference frame, in this situation, we couldwrite that

ddt(JSω

)= JSω

The dependency of the moment of inertia on the actual position (angle) is adeciding factor to the easy and direct application of the angular momentumtheorm. Because of this body-fixed reference frame is used, as the moment ofinertia in the body-fixed frame stays constant and the time derivative of themoment of inertia is cenceled!

DVIBRATIONS - APPENDIX

d.1 excitation with constant amplitude of force - com-plex approach

Let us first recall that we can represent a real harmonic functions by a complexexponential function using

eiΩt = cos Ωt+ i sin Ωt (D.1)

From this we can derive that

cos Ωt =eiΩt + e−iΩt

2 (D.2)

andsin Ωt =

eiΩt − e−iΩt

2i (D.3)

The harmonic force is:

F (t) = F cos Ωt =F

2(eiΩt + e−iΩt

)=F

2 eiΩt +

F

2 e−iΩt (D.4)

This means that we have to solve the equation of motion twice, for the eiΩtand the e−iΩt term. For the first step we make the approach:

x1 (t) = x1eiΩt (D.5)

x2 (t) = x2e−iΩt (D.6)

Putting both approaches into the equation of motion yields:

(−Ω2m+ iΩc+ k)x1eiΩt =

F

2 eiΩt (D.7)

(−Ω2m− iΩc+ k)x2eiΩt =

F

2 e−iΩt (D.8)

Dividing by k and introducing the frequency ratio η (eq. (4.102))[(1− η2) + 2Dηi

]x1 =

F

2k (D.9)[(1− η2)− 2Dηi

]x2 =

F

2k (D.10)

149

150 vibrations - appendix

The solution for x1 and x2 are:

x1 =(1− η2)− 2Dηi

(1− η2)2 + 4D2η2F

2k (D.11)

x2 =(1− η2) + 2Dηi(1− η2)2 + 4D2η2

F

2k (D.12)

As we can see, the solution of one part is the conjugate complex of the other:

x1 = x∗2 (D.13)

The solution for x (t) is combined from the two partial solutions, which we justhave found:

x (t) = x1eiΩt + x2e

−iΩt (D.14)

This can be resolved:

x (t) = x1 cos Ωt+ ix1 sin Ωt+ x2 cos Ωt+ ix2 sin Ωt (D.15)

and using eq. (D.13), we finally get:

x (t) = 2 Re x1 cos Ωt− 2 Im x1 sin Ωt (D.16)

The factor of 2 compensates the factor 12 associated with the force amplitude.

All the information can be extracted from x1 only so that only this part of thesolution has to be solved:

x (t) =(1− η2)

(1− η2)2 + 4D2η2F

kcos Ωt+

2Dη(1− η2)2 + 4D2η2

F

ksin Ωt (D.17)

which is the same result as eq. (4.91) with eq. (4.103) and eq. (4.104).Also the magnitude x = |x1| and phase ϕ can be obtained in the same wayand yield the previous results:

Magnitude: x =1√

(1− η2)2 + 4D2η2︸︷︷︸magnification factor V1

1kF = V1 (η,D)

1kF (D.18)

Phase: tanϕ = −Im x1Re x1

=2Dη

1− η2 (D.19)

D.2 excit. with constant amp. of force - alternative complex app. 151

d.2 excitation with constant amplitude of force - al-ternative complex approach

Instead of eq. (D.4), we can write

F (t) = F cos Ωt =F

2(eiΩt + e−iΩt

)= 2 Re

F

2 eiΩt

(D.20)

orF (t) = F cos Ωt = Re

F eiΩt

(D.21)

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

x (t) = ReXeiΩt

(D.22)

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

Re(−Ω2m+ iΩc+ k

)XeiΩt

= Re

F eiΩt

(D.23)

The real parts are equal if the complex expression is equal:(−Ω2m+ iΩc+ k

)XeiΩt = F eiΩt (D.24)

Elimination of the time function yields:(−Ω2m+ iΩc+ k

)X = F (D.25)

The expression in brackets is also called the dynamic stiffness

kdyn (Ω) =(k−Ω2m+ iΩ

)(D.26)

Now we solve eq. (D.25) to get the complex amplitude:

X =F

(−Ω2m+ iΩc+ k)(D.27)

The expression

H (Ω) =1

(−Ω2m+ iΩc+ k)=X

F=

OutputInput (D.28)

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

X =1

(1− η2 + i2Dη)F

k(D.29)


Because

x cos (Ωt− ϕ) = xReei(Ωt−ϕ)

= Re

xe−iϕeiΩt

= Re

XeiΩt

(D.30)

we take the magnitude x and phase lag ϕ of this complex result

X (Ω) = xe−iϕ (D.31)

which leads to the same result as before, see eq. (D.18) and eq. (D.19):

x =1√

(1− η2)2 + 4D2η2︸︷︷︸magnification factor V1

1kF = V1 (η,D)

1kF (D.32)

tanϕ = −Im x1Re x1

=2Dη

1− η2 (D.33)

d.3 fourier series - alternative real representation

We can write the Fourier series as a sum of cosine functions with amplitude ckand a phase shift ϕk:

x (t) = c0 +∞∑k=1

ck cos (kωt+ ϕk) (D.34)

ck =√ak2 + bk

2 and ϕk = arctan (− bkak

) (D.35)

d.4 fourier series - alternative complex representation

The real trigonometric functions can also be transformed into complex expo-nential expression:

x (t) =∞∑

k=−∞Xke

ikωt (D.36)

TheXk are the complex Fourier coefficients which can be determined by solvingthe integral:

Xk =1T

∫ T

0x (t) e−ikωtdt (D.37)

D.4 fourier series - alternative complex representation 153

or

Xk =1T

∫ T

0x (t) [cos kωt− i sin kωt] dt (D.38)

which clearly shows the relation to the real Fourier coefficients series given byeq. (4.142) and eq. (4.143):

Re Xk =ak2 ; Im Xk = −

bk2 (D.39)

The connection to the other real representation (chap. 4.7.1) is:

|Xk| = ck tanϕk =(

Im XkRe Xk

)(D.40)

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

X−k = X∗k (D.41)


d.5 magnification functions

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

V1

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2D

2 2η η

1V1=

Figure D.1: Magnification function V1

η ω= /Ω0

V2

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

√(1- ) +42 2Dη η

2 2

√1+4D2 2η

V2=

√2Figure D.2: Magnification function V2

D.5 magnification functions 155

V3

0 0,5 1 1,5 2 3 3,5 4 4,5 50

1

2

3

4

5

6

7

8

9

10

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071

η ω= /Ω0

2,5

√(1- ) +42 2D

2 2η η

η2

V3=


V4

0 0,5 1 1,5 2 3 3,5 4 4,5 5

1

2

3

4

5

6

7

8

9

10

η ω= /Ω0

2,5

D=0,05

D=0,1

D=0,2

D=0,3

D=0,5

D=0,7071 D=0,05

D=0,7071

√(1- ) +42 2D

2 2η η

V4=η

2√1+4D2 2η

√2



0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η ω= /Ω0

D=0,05D=0,1

D=0,2D=0,3

D=0,5D=0,7071

1- 2η

φ =2Dη

arctan

D=0

Figure D.5: Phase ϕ - for magnification function V1 and V3

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50°

20°

40°

60°

80°

100°

120°

140°

160°

180°

φ

η = Ω / ω0

D=0,05

D=0,1

D=0,2

D=0,3D=0,5

D=0,7071

21-ηφ =

2Dηarctan

D=0

Figure D.6: Phase ϕ - for magnification function V2 and V4

EL ITERATURE

Die hier vorliegende Zusammenstellung maschinendynamischer Themen um-fasst grundlegende Prinzipien und Arbeitsgebiete. Die dargestellten Inhaltelassen sich in vielen Lehrbüchern wiederfinden. Im besonderen sei jedoch aufdie folgenden Werke hingewiesen:

• Moon, F.C., Applied Dynamics, John Wiley & Sons, 1998.

• Ginsberg, J.H., Advanced Engineering Dynamics, 2nd edition, CambridgeUniv. Press, 1998.

• Weaver, W., Timoshenko, S.P., Young, D.H., Vibration Problems in En-gineering, John Wiley, 1990.

• Inman, D.J., Engineering Vibrations, Prentice Hall, 1994.

• Ginsberg, J.H., Mechanical and Structural Vibrations - Theory and Ap-plications, John Wiley, 2001.

157

FFORMULARY

159

160 formulary

sin(α± β) = sinα cos β ± cosα sin βcos(α± β) = cosα cos β ∓ sinα sin β

sin(2ϕ) = 2 sinϕ cosϕcos(2ϕ) = cos2(ϕ)− sin2(ϕ) = 1− 2 sin2(ϕ)

Table 2: Trigonometric functions

Configuration Equivalent stiffness(bending stiffness EI = const.)

l

l/2

F

wmax

F = keqwmax

kers = 48EIl3

l

F

wmax

F = keqwmax

kers = 12EIl3

l

F

wmax

F = keqwmax

kers = 3EIl3

Table 3: Equivalent stiffnesses for different loaded beams

formulary 161

(1± x)m = 1±mx+ m(m−1)2! x2 ± m(m−1)(m−2)

3! x3 + . . .

+ (±1)n m(m−1)...(m−n+1)n! xn + . . .

m > 0

√1− x = 1− 1

2x−18x

2 − . . . m = 12

Table 4: Reihenentwicklung: Binomische Reihe mit positiven Exponenten (|x| ≤ 1)

Springs connection Equivalent stiffness

k1

k2

Fkn

Parallelschaltungkeq =

n∑i=1

ki

n = 2:

keq = k1 + k2

F

Reihenschaltung

k1

k2

kn

1keq

=n∑i=1

1ki

n = 2:

1keq

= 1k1

+ 1k2

keq =k1k2k1+k2

Table 5: Equivalent stiffness of springs in parallel and serie connection

162 formulary

Rigid body Mass moment of inertia

x y

zl

b

hS

QuaderJxx =

112m

(b2 + h2

)Jyy =

112m

(h2 + l2

)Jzz =

112m

(l2 + b2

)

Stab (b,h<<l)

S

xy

zl Jyy = Jzz =112ml

2

x y

z

l

rZylinder

S

Jxx = Jyy =112m

(l2 + 3r2

)Jzz =

12mr

2

Scheibe ( )h r<<

x y

z

hr

S

Jxx = Jyy =14mr

2

Jzz =12mr

2

x y

z

l

RHohlzylinder r

S

Jxx = Jyy =14m

(R2 + r2 + h2

3

)Jzz =

12m

(R2 + r2

)

KreisringscheibeR r h R≈ , <<

x y

z

R

r

h

S

Jxx = Jyy =12mR

2

Jzz = mR2

Kugel

x y

z

r

SS

Jxx = Jyy = Jzz =25mr

2

Table 6: Mass moment of inertia (given at the center of gravity S)

linear vibrations of systems with one degree of … · 2014-05-23 · self-excitedvibrations....

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