Vibration and Shock Handbook 02

Vibration and Shock Handbook 02 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

2 Frequency-Domain

Analysis

Clarence W de Silva

The University of British Columbia

2.1 Introduction 2-12.2 Response to Harmonic Excitations 2-2Response Characteristics † Measurement of Damping Ratio (Q-Factor Method)

2.3 Transform Techniques 2-14Transfer Function † Frequency-Response Function

(Frequency Transfer Function)2.4 Mechanical Impedance Approach 2-25Interconnection Laws

2.5 Transmissibility Functions 2-31Force Transmissibility † Motion Transmissibility †

General Case † Peak Values of Frequency-Response Functions2.6 Receptance Method 2-37Application of Receptance

Appendix 2A Transform Techniques 2-40

Summary

This chapter presents the frequency-domain analysis of mechanical vibrating systems In the frequency domain,the independent variable is frequency The response of a vibrating system to harmonic excitations under variouslevels of damping (overdamped, underdamped, and critically damped) is analyzed Frequency transfer functiontechniques including impedance, mobility, force transmissibility, motion transmissibility, and receptance arestudied Transform techniques (Fourier and Laplace) are applicable in the frequency-domain analysis TheQ-factor method of measuring damping is derived Component interconnection laws are established for frequency-domain analysis The emphasis is on single-degree-of-freedom (single-DoF) systems

2.1 Introduction

In many vibration problems, the primary excitation force typically has a repetitive periodic nature, and

in some cases this periodic forcing function may be even purely sinusoidal Examples are excitations due

to mass eccentricity and misalignments in rotational components, tooth meshing in gears, andelectromagnetic devices excited by AC or periodic electrical signals In basic terms, the frequency-response of a dynamic system is the response to a pure sinusoidal excitation As the amplitude andthe frequency of the excitation are changed, the response also changes In this manner, the response

of the system over a range of excitation frequencies can be determined This represents the frequencyresponse In this case, frequency ðvÞ is the independent variable and hence we are dealing with thefrequency domain

2-1

Frequency-domain considerations are applicable even when the signals are not periodic In fact, a timesignal can be transformed into its frequency spectrum through the Fourier transform For a giventime signal, an equivalent Fourier spectrum, which contains all the frequency (sinusoidal) components ofthe signal, can be determined either analytically or computationally Hence, a time-domainrepresentation and analysis has an equivalent frequency-domain representation and analysis, at leastfor linear dynamic systems For this reason, and also because of the periodic nature of typical vibrationsignals, frequency-response analysis is extremely useful in the subject of mechanical vibrations Theresponse to a particular form of “excitation” is what is considered in the frequency-domain analysis.Hence, we are specifically dealing with the subject of “forced response” analysis, albeit in the frequencydomain.

2.2 Response to Harmonic Excitations

Consider a simple oscillator with an excitation

force f ðtÞ; as shown in Figure 2.1

The equation of motion is given by

m€x þ b_x þ kx ¼ f ðtÞ ð2:1ÞSuppose that f ðtÞ is sinusoidal (i.e., harmonic)

Pick the time reference such that

f ðtÞ ¼ f0cosvt ð2:2Þwhere

v ¼ excitation frequency

f0¼ forcing excitation amplitude

For a system subjected to a forcing excitation, we have

Total Response

T ¼ Homogeneous Response xh

H ðNatural ResponseÞ

þ Particular Response xp

P ðEnforced ResponseÞ

X ðDepends only on initial conditionsÞ does not contain enforced response and depends entirely on the natural=homogeneous response

F ðDepends only on f 0 Þ but contains a natural=homogeneous component

Using these concepts, we analyze the forced problem, which may be written as

vn¼ undamped natural frequency

We will consider several important cases.

It can be easily verified that xpgiven by Equation 2.10 satisfies the forced system Equation 2.4, with

z ¼ 0: Hence, it is a particular solution

Complete solution:

x ¼ A1cosvnt þ A2sinvnt

|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

H Satisfies the homogeneous equation

ðv22v2Þcosvt

|fflfflfflfflfflffl{zfflfflfflfflfflffl}

P Satisfies the equation with input

Hence, the complete response is

þ v2 a

n2v2cosvt

|fflfflfflfflfflffl{zfflfflfflfflfflffl}

P Particular solution

pWill exhibit a beat phenomenon for v n 2 v ; i:e:;

ð v n þ v Þ=2 wave modulated by ð v n 2 v Þ=2 wave

in general, as is clear from Equation 2.14b

Case 2: Undamped oscillator withv5vn(resonant condition)

In this case, the xpthat was used before is no longer valid This is the degenerate case, because otherwisethe particular solution cannot be distinguished from the homogeneous solution and the former will becompletely absorbed into the latter Instead, in view of the “double-integration” nature of the forcedsystem equation whenv ¼ vn; we use the particular solution ðPÞ:

pSinusodal at v

þ |fflfflffl{zfflfflffl}2v sin vtat

F Forced response ðdepends on inputÞ pAmplitude increases linearly

ð2:18Þ

Since the forced response increases steadily, this is an unstable response in the BIBO sense.Furthermore, the homogeneous solution H and the free response X are identical, and the particularsolution P is identical to the forced response F in this case

Note that the same system (undamped oscillator) gives a bounded response for some excitations, whileproducing an unstable (steady linear increase) response when the excitation frequency is equal to itsnatural frequency Hence, the system is not quite unstable, but is not quite stable either In fact, theundamped oscillator is said to be marginally stable When the excitation frequency is equal to the naturalfrequency it is reasonable for the system to respond in a complementary and steadily increasing mannerbecause this corresponds to the most “receptive” excitation Specifically, in this case, the excitationcomplements the natural response of the system In other words, the system is “in resonance” with theexcitation, and the condition is called a resonance We will address this aspect for the more general case of

a damped oscillator, in the sequel

Figure 2.2 shows typical forced responses of an undamped oscillator when there is a large differencebetween the excitation and the natural frequencies (Case 1); a small difference between the excitation andthe natural frequencies when a beat phenomenon is clearly manifested (also Case 1); and for the resonantcase (Case 2)

Case 3: Damped oscillator

The equation of forced motion is

Particular Solution (Method 1):

Since derivatives of both odd order and even order are present in this equation, theparticular solution should have terms corresponding to odd and even derivatives of the

FIGURE 2.2 Forced response of a harmonically excited undamped simple oscillator: (a) for a large frequency difference; (b) for a small frequency difference (beat phenomenon); (c) response at resonance.

forcing function (i.e., sinvt and cos vt) Hence, the appropriate particular solution will be of theform:

Substitute Equation 2.20 into Equation 2.19 We obtain

2v2a1cosvt 2 v2a2sinvt þ 2zvn½2va1sinvt þ va2cosvt þ v2½a1cosvt þ a2sinvt ¼ a cos vtEquate like coefficients:

2v2a1þ 2zvnva2þv2a1¼ a2v2a22 2zvnva1þv2a2¼ 0Hence, we have

" #

ð2:22Þwith the determinant

This is the method of undetermined coefficients

Particular Solution (Method 2): Complex Function Method

Consider

where the excitation is complex (Note: ej v t ¼ cosvt þ j sin vt.)

The resulting complex particular solution is

Note that we should take the real part of this solution as the true particular solution

First substitute Equation 2.26 into Equation 2.25:

X 2v2þ 2zvnjv þ v2 ej v t¼ aej v t

Hence, since ej v t– 0 in general,

xp¼ a

It can easily be verified that this result is identical to what was obtained previously by Method 1, asgiven by Equation 2.20 together with Equations 2.23 and 2.24

In passing, we will note here that the frequency-domain transfer function (i.e., response and excitation

in the frequency domain) of the system Equation 2.19 is:

The particular section ðPÞ is equal to the steady-state solution, because the homogeneous solution diesout due to damping

The particular solution (Equation 2.33) has the following characteristics:

1 The frequency is the same as the excitation frequencyv:

2 The amplitude is amplified by the magnitude 1 lDl ¼ lGðjvÞl:

3 The response is “lagged” by the phase anglef of D (or “led” by the phase angle of GðjvÞ; denoted

Equation 2.28c is D ¼v22v2þ 2zvnv j

Hence,

lDl2¼ ðv22v2Þ2þ ð2zvnvÞ2¼ D ð2:35ÞThe resonance corresponds to a minimum value of D; or

dD

dv ¼ 2ðv22v2Þð22vÞ þ 2ð2zvnÞ2v ¼ 0 For a minimum: ð2:36ÞHence, with straightforward algebra, the required condition for resonance is

2v2þv2þ 2z2v2¼ 0or

The magnitude and the phase angle plots of GðjvÞ are shown inFigure 2.3.These curves correspond tothe amplification and the phase change of the particular response (the steady-state response) with respect

to the excitation input This pair, the magnitude and phase angle plots of a transfer function with respect

to frequency, is termed a Bode plot Usually, logarithmic scales are used for both magnitude (e.g.,decibels) and frequency (e.g., decades) In summary, the steady-state response of a linear system to asinusoidal excitation is completely determined by the frequency transfer function of the system The totalresponse is determined by adding H to P and substituting initial conditions, as usual

For an undamped oscillator ðz ¼ 0Þ; we notice from Equation 2.34 that the magnitude of GðjvÞbecomes infinity when the excitation frequency is equal to the natural frequency ðvnÞ of the oscillator.This frequency ðvnÞ is clearly the resonant frequency (as well as natural frequency) of the oscillator Thisfact has been further supported by the nature of the corresponding time response (see Equation 2.18 and

Figure 2.2(c)),which grows (linearly) with time

2.2.2 Measurement of Damping Ratio (Q-Factor Method)

The frequency transfer function of a simple oscillator (Equation 2.19) may be used to determine thedamping ratio This frequency-domain method is also termed the half-power point method, for reasonsthat should become clear from the following development

First we assume thatz , 1=pffiffi2: Strictly speaking, we should assume thatz , 1=2pffiffi2:

Without loss of generality, consider the normalized (or, non-dimensionalized) transfer function

It is clear from Equation 2.39a that the phase angle of GðjvÞ at v ¼ vnis 2p=2:

When power is half of the peak power value (e.g., because the displacement squared is proportional topotential energy), then the velocity squared is proportional to kinetic energy, and power is the rate of

G D =

FIGURE 2.3 Magnitude and phase angle curves of a simple oscillator (A Bode plot).

change of energy Therefore, when the amplification is 1=pffiffi2of the peak value we have half-power points,given by:

12

The Q-factor, which measures the sharpness of resonant peak, is defined by

Example 2.1

A dynamic model of a fluid coupling system is shown inFigure 2.5.The fluid coupler is represented by arotatory viscous damper with damping constant b: It is connected to a rotatory load of moment ofinertia J; restrained by a torsional spring of stiffness k; as shown We now obtain the frequency transferfunction of the system that relates the restraining torque t of the spring to the angular displacementexcitation aðtÞ that is applied at the free end of the fluid coupler If aðtÞ ¼ a0sinvt; what is themagnitude (i.e., amplitude) oft at steady state?

Solution

Newton’s Second Law gives

J €u ¼ bð _a 2 _uÞ 2 kuHence,

Note that the frequency transfer function is obtained simply by setting s ¼ jv:

The restraining torque of the spring ist ¼ ku: Hence,

Then, the corresponding frequency-response function (frequency transfer function) is

tanf ¼ v2zv22nvv2; f ¼ phase lag:

Particular solution P is also the steady-state response

with k=J ¼v2; b=J ¼ 2zvn; andvnequal to the undamped natural frequency of the load.

Now, define the normalized frequency

r ¼ v

Then, from (ix) we have

t0¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ka0zrð1 2 r2Þ2þ ð2zrÞ2

k

FluidCoupling

b

LoadInertia

2.3 Transform Techniques

Concepts of frequency-response analysis originate from the nature of the response of a dynamic system to

a sinusoidal (i.e., harmonic) excitation These concepts can be generalized because the time-domainanalysis, where the independent variable is time ðtÞ; and the frequency-domain analysis, where theindependent variable is frequency ðvÞ; are linked through the Fourier transformation Analytically, it ismore general and versatile to use the Laplace transformation, where the independent variable is theLaplace variable ðsÞ which is complex (i.e., non-real) This is true because analytical Laplace transformsmay exist even for time functions that do not have “analytical” Fourier transforms But with compatibledefinitions, the Fourier transform results can be obtained from the Laplace transform results simply bysetting s ¼ jv: In the present section, we will formally introduce the Laplace transformation and theFourier transformation, and will illustrate how these techniques are useful in the response analysis ofvibrating systems The preference of one domain over another will depend on such factors as the nature

of the excitation input, the type of the analytical model available, the time duration of interest, and thequantities that need to be determined

ðs þj1

The integration is performed along a vertical line parallel to the imaginary (vertical) axis, located at

s from the origin in the complex Laplace plane (s-plane) For a given piecewise-continuous function

1

Magnitude ofRestraining

Torque t0

ka0

FIGURE 2.6 Variation of the steady-state transmitted torque with frequency.

f ðtÞ; the Laplace transform exists if the integral in Equation 2.46 converges A sufficient condition forthis is

ð1

Convergence is guaranteed by choosing a sufficiently large and positive s: This property is anadvantage of the Laplace transformation over the Fourier transformation (For a more completediscussion of the Fourier transformation, see later in this chapter and in Chapter 4.)

By the use of Laplace transformation, the convolution integral equation can be converted into analgebraic relationship To illustrate this, consider the convolution integral which gives the response yðtÞ of

a dynamic system to an excitation input uðtÞ; with zero initial conditions By definition (Equation 2.46),its Laplace transform is written as

t , 0: Again using the definition of Laplace transformation, the foregoing relation can be expressed as

of the system The transfer function of a linear and constant-parameter system is a unique function thatcompletely represents the system A physically realizable, linear, constant-parameter system possesses aunique transfer function, even if the Laplace transforms of a particular input and the correspondingoutput do not exist This is clear from the fact that the transfer function is a system model and does notdepend on the system input itself

Note: The transfer function is also commonly denoted by GðsÞ: However, in the present context we useHðsÞ in view of its relation to hðtÞ: Some useful Laplace transform relations are given inTable 2.1.Also,note that the Fourier transform (set s ¼ jvÞ is given by

Forward:

FðjvÞ ¼ð1

0 f ðtÞ expð2jvtÞdtInverse:

f ðtÞ ¼ðj1

2j1FðjvÞ expðjvtÞdvConsider the nth-order linear, constant-parameter dynamic system given by

anddtnyn þ an21ddtn21n21y þ · · · þ a0y ¼ b0u þ b1duðtÞdt þ · · · þ bmdmdtuðtÞm ð2:53Þ

for physically realizable systems, m # n: By applying a Laplace transformation and then integrating byparts, it may be verified that

of the response under zero initial conditions This is quite logical because the analytical model of a system

is independent of the system’s initial conditions

The denominator polynomial of a transfer function is the system’s characteristic polynomial Its roots arethe poles or the eigenvalues of the system If all the eigenvalues have negative real parts, the system is stable.The response of a stable system is bounded (that is, remains finite) when the input is bounded (which is theBIBO stability) The zero-input response of an asymptotically stable system approaches zero with time.2.3.2 Frequency-Response Function (Frequency Transfer Function)

The Fourier integral transform of the impulse-response function is given by

2 p j

Ðsþj1FðsÞ expðstÞds Ð10 f ðtÞ expð2stÞdt

Step function, UðtÞ 1s

In view of the fact that hðtÞ ¼ 0 for t , 0; the lower limit of integration in Equation 2.56 could be madezero Then, from Equation 2.52, it is clear that Hð f Þ is obtained simply by setting s ¼ j2pf in HðsÞ: Hence,strictly speaking, we should use the notation Hð j2pf Þ and not Hð f Þ However, for notational simplicity,

we denote Hðj2pf Þ by Hð f Þ: Furthermore, since the angular frequency v ¼ 2pf ; we can express thefrequency-response function by Hð jvÞ; or simply by HðvÞ for notational convenience It should be notedthat the frequency-response function, like the Laplace transfer function, is a complete representation of alinear, constant-parameter system In view of the fact that both uðtÞ ¼ 0 and yðtÞ ¼ 0 for t , 0; we canwrite the Fourier transforms of the input and the output of a system directly by setting s ¼ j2pf ¼ jv inthe corresponding Laplace transforms Specifically, we have, according to the notation used here

Uð f Þ ¼ Uðj2p f Þ ¼ UðjvÞand

Yð f Þ ¼ Yð j2p f Þ ¼ Yð jvÞ:

Then, from Equation 2.51, we have

Hð f Þ ¼ b0þ b1j2pf þ · · · þ bmðj2pf Þm

a0þ a1j2pf þ · · · þ anðj2pf Þn ð2:58ÞThis, generally, is a complex function of f that has a magnitude denoted by lHð f Þl and a phase angledenoted by /Hð f Þ:

A further interpretation of the frequency-response function can be given in view of the developmentsgiven in Section 2.2 Consider a harmonic input having cyclic frequency f ; expressed by

In analysis, it is convenient to use the complex input

uðtÞ ¼ u0ðcos 2p ft þ j sin 2p ftÞ ¼ u0expðj2p ftÞ ð2:59bÞand take only the real part of the final result Note that Equation 2.59b does not implicitly satisfy therequirement of uðtÞ ¼ 0for t , 0: Therefore, an appropriate version of the convolution integral, where thelimits of integration automatically account for this requirement, should be used For instance, we can write

for a harmonic excitation, in which the phase lead anglef ¼ /Hð f Þ: It follows from Equation 2.61b that,when a harmonic excitation is applied to a stable, linear, constant-parameter dynamic system havingfrequency-response function Hð f Þ; its steady-state response will also be harmonic at the same frequency,but with an amplification factor of lHð f Þl in its amplitude and a phase lead of /Hð f Þ: This result has beenestablished previously, in Section 2.2 Consequently, the frequency-response function of a stable systemcan be experimentally determined using a sine-sweep test or a sine-dwell test With these methods, aharmonic excitation is applied as the system input, and the amplification factor and the phase-lead angle inthe corresponding response are determined at steady state The frequency of excitation is variedcontinuously for a sine sweep and in steps for a sine dwell The sweep rate should be slow enough, and thedwell times should be long enough, to guarantee steady-state conditions at the output The pair of plots oflHð f Þl and /Hð f Þ against f completely represent the complex frequency-response function, and are Bodeplots or Bode diagrams, as noted earlier In Bode plots, logarithmic scales are normally used for bothfrequency f and magnitude lHð f Þl:

Impulse Response

The impulse-response function of a system can be obtained by taking the inverse Laplace transform of thesystem transfer function For example, consider the damped simple oscillator given by the normalizedtransfer function:

yimpulseðtÞ ¼ hðtÞ ¼ ffiffiffiffiffiffiffiffivn

1 2z2

p expð2zvntÞ sinvdt for z , 1 ð2:70aÞ

yimpulseðtÞ ¼ hðtÞ ¼ ab

ðb 2 aÞ½expð2atÞ 2 expð2btÞ for z 1 ð2:70bÞ

YstepðsÞ ¼ 1

s

v2

which follows from Equation 2.73

To facilitate using Table 2.1, partial fractions of Equation 2.73 are determined in the form

The following results are obtained:

ystepðtÞ ¼ 1 2 ffiffiffiffiffiffiffiffi1

1 2z2

p expð2zvntÞ sinðvdt þfÞ for z , 1 ð2:74aÞ

ystep¼ 1 2 1

ðb 2 aÞ½b expð2atÞ 2 a expð2btÞ forz 1 ð2:74bÞ

ystep¼ 1 2 ðvnt þ 1Þ expð2vntÞ forz ¼ 1 ð2:74cÞ

In Equation 2.74c,

Transfer Function Matrix

Consider the state-space model of a linear dynamic system as given by

where, x ¼ the nth order state vector, u ¼ the rth order input vector, y ¼ the mth order output vector,

A ¼ the system matrix, B ¼ the input gain matrix, C ¼ the output (measurement) gain matrix, and

D ¼ the feedforward gain matrix We can express the input–output relation between u and y; in theLaplace domain, by a transfer function matrix of the order m £ r:

To obtain this relation, let us Laplace transform the Equations 2.76a and 2.76b and use zero initialconditions for x; thus,

From Equation 2.77a it follows that,

in which I is the nth order identity matrix By substituting Equation 2.78 into Equation 2.77b, we obtainthe transfer relation

Example 2.2

Consider the simple oscillator equation given by

Note that uðtÞ can be interpreted as a displacement input (e.g., support motion) or kuðtÞ can beinterpreted as the input force applied to the mass Take the Laplace transform of the system equation (i)with zero initial conditions; thus,

The corresponding transfer function is

GðsÞ ¼ YðsÞUðsÞ ¼

k

FðjvÞ ¼ða

21f ðtÞ expð2jvtÞdtNote: May use FðvÞ to denote FðjvÞ

Note: Set s ¼ jv ¼ j2p f to convert Laplace results into Fourier results

v ¼ angular frequency (rad/sec)

f ¼ cyclic frequency (cps or Hz)

Transfer function HðsÞ ¼ output input in Laplace domain; with zero initial conditions:

Frequency transfer function (or frequency-response function) ¼ HðjvÞ

Note: Notation ðGðsÞÞ is also used to denote a system transfer function

UðjvÞ ¼ Fourier spectrum of input uðtÞ

YðjvÞ ¼ Fourier spectrum of output yðtÞ

Note:

lHðjvÞl ¼ response amplification for a harmonic excitation of frequency v

/HðjvÞ ¼ response phase “lead” for a harmonic excitation

Multivariable Systems:

State-Space Model:

_x ¼ Ax þ Bu

y ¼ Cx þ DuTransfer-Matrix Model:

YðsÞ ¼ GðsÞUðsÞwhere

GðsÞ ¼ CðsI 2 AÞ21B þ D

or, in terms of the undamped natural frequencyvn and the damping ratio z; where v2¼ k=m and2zvn¼ b=m; the transfer function is given by

and the output acceleration transfer function is

s2YðsÞUðsÞ ¼ s2GðsÞ ¼

s2v2

In the output acceleration transfer function, we have m ¼ n ¼ 2: This means that if the acceleration ofthe mass that is caused by an applied force is measured, the input (applied force) is instantly felt by theacceleration This corresponds to a feedforward action of the input excitation or a lack of dynamic delay.For example, this is the primary mechanism through which road disturbances are felt inside a vehicle thathas very hard suspension

Example 2.3

Again let us consider the simple oscillator differential equation

By defining the state variables as

in which DðsÞ ¼ s2þ 2zvns þv2 is the characteristic polynomial of the system The first element inthe only column in GðsÞ is the displacement-response transfer function and the second element isthe velocity-response transfer function These results agree with the expressions obtained in theprevious example.

Now, let us consider the acceleration €y as an output and denote it by y3: It is clear from the systemequation (i) that

y3¼ €y ¼ 22zvn_y 2v2y þv2uðtÞ ðviiiÞ

or, in terms of the state variables,

(b) A vibration test setup is schematically shown in Figure 2.7

In this experiment, a mechanical load is excited by a linear motor and its acceleration response ismeasured by an accelerometer and charge amplifier combination The force applied to the load by thelinear motor is also measured, using a force sensor (strain-gauge type) The frequency-response functionacceleration/force is determined from the sensor signals, using a spectrum analyzer

LinearMotor

Instrumentation

ChargeAmplifier

SpectrumAnalyzer

f

ForceSensor

FIGURE 2.7 Measurement of the acceleration spectrum of a mechanical system.

Suppose that the mechanical load is approximated by a damped oscillator with mass m; stiffness k; anddamping constant b; as shown inFigure 2.7.If the force applied to this load is f ðtÞ and the displacement

in the same direction is y; show that the equation of motion of the system is given by

m€y þ b_y þ ky ¼ f ðtÞObtain an expression for the acceleration frequency-response function GðjvÞ in the frequency domain,with excitation frequencyv as the independent variable Note that the applied force f is the excitationinput and the acceleration a of the mass is the response, in this case

Express GðjvÞ in terms of (normalized) frequency ratio r ¼ v=vn; wherevnis the undamped naturalfrequency

Giving all the necessary steps, determine an expression for r at which the acceleration response function will exhibit a resonant peak What is the corresponding peak magnitude of lGl?For what range of values of damping ratioz would such a resonant peak be possible?

frequency-Solution

(a) For a single DoF system, apply a sinusoidal forcing excitation at the DoF and measure thedisplacement response at the same location Vary the excitation frequencyv in small steps, andfor each frequency at steady state determine the amplitude ratio of the (displacementresponse/forcing excitation) The peak amplitude ratio will correspond to the resonance For amulti-DoF system, several tests may be needed, with excitations applied at different locations offreedom and the response measured at various locations as well(see Chapters 10and11).In thefrequency domain we have,

lVelocity response spectruml ¼vxlDisplacement response spectrumllAcceleration response spectruml ¼vxlVelocity response spectruml

It follows that the shape of the frequency-response function will depend on whether thedisplacement, velocity, or acceleration is used as the response variable Hence it is likely that thefrequency at which the peak amplification occurs (i.e., resonance) will also depend on the type ofresponse variable that is used

(b) A free-body diagram of the mass element is shown in Figure 2.8

Newton’s Second Law gives

Hence, the equation of motion is

The displacement transfer function is

y

f ¼

1

Note that, for notational convenience, the same

lowercase letters are used to represent the Laplace

transforms as well as the original time-domain

variables (y and f ) The acceleration transfer

function is obtained by multiplying Equation iii

by s2: (From Table 2.1, the Laplace transform of

d=dt is s; with zero initial conditions) Hence

FIGURE 2.8 Free-body diagram.

In the frequency domain, the corresponding frequency-response function is obtained by substituting jvfor s: Hence,

r ¼ 0 or ½ð1 2 r2Þ2þ 4z2r2 þ r2½1 2 r22 2z2 ¼ 0The first result ðr ¼ 0Þ corresponds to static conditions and is ignored Hence, the resonant peakoccurs when

ð1 2 r2Þ2þ 4z2r2þ r22 r42 2z2r2¼ 0which has the valid root

lGlpeak¼

1mð1 2 2z2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2.4 Mechanical Impedance Approach

Any type of force or motion variable may be used as input and output variables in defining a systemtransfer function In vibration studies, three particular choices are widely used The corresponding

frequency transfer functions are named impedance functions, mobility functions, and transmissibilityfunctions These are described in the present section and in the subsequent section, and their use isillustrated.

Through variables (force) and across variables (velocity), when expressed in the frequency domain(as Fourier spectra), are used in defining the two important frequency transfer functions: mechanicalimpedance and mobility In the case of impedance functions, velocity is considered the input variable andthe force is the output variable; whereas in the case of mobility functions, the converse applies Specifically,

dynamic inertia ¼ force/acceleration ¼ impedance/ðjvÞ

acceleration ¼ acceleration/force ¼ mobility £ ðjvÞ

dynamic stiffness ¼ force/displacement ¼ impedance £ jv

receptance ¼ displacement ¼ mobility/ðjvÞ

In these definitions, the variable (force, acceleration, and displacement) should be interpreted as thecorresponding Fourier spectra

The time-domain constitutive relations for the mass, spring, and the damper elements are well known.The corresponding transfer relations are obtained by replacing the derivative operator d=dt by the Laplaceoperator s: The frequency transfer functions are obtained by substituting jv or j2p for s: These results arederived below

and

TABLE 2.2 Definitions of Useful Mechanical Transfer Functions

Transfer Function Definition (in the Frequency Domain) Dynamic stiffness Force/displacement

Receptance, dynamic flexibility, or compliance Displacement/force

Impedance ðZÞ Force/velocity

Mobility ðMÞ Velocity/force

Dynamic inertia Force/acceleration

Accelerance Acceleration/force

Force transmissibility ðT f Þ Transmitted force/applied force

Motion transmissibility ðT m Þ Transmitted velocity/applied velocity