Vibration and Shock Handbook 01

Vibration and Shock Handbook 01 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.

I Fundamentals and

Analysis

I-1

1 Time-Domain Analysis

Clarence W de Silva

The University of British Columbia

1.1 Introduction 1-11.2 Undamped Oscillator 1-2Energy Storage Elements † The Method of Conservation of Energy † Free Response

1.3 Heavy Springs 1-12Kinetic Energy Equivalence

1.4 Oscillations in Fluid Systems 1-141.5 Damped Simple Oscillator 1-16Case 1: Underdamped Motion ð z , 1 Þ † Logarithmic

Decrement Method † Case 2: Overdamped Motion

ð z 1Þ † Case 3: Critically Damped Motion

ð z ¼ 1 Þ † Justification for the Trial Solution † Stability and Speed of Response

1.6 Forced Response 1-27Impulse-Response Function † Forced Response †

Response to a Support MotionSummary

This chapter concerns the modeling and analysis of mechanical vibrating systems in the time domain Usefulprocedures of time-domain analysis are developed Techniques of analyzing both free response and forcedresponse are given Examples of the application of Newton’s force–motion approach, energy conservationapproach, and Lagrange’s energy approach are given Particular emphasis is placed on the analysis of anundamped oscillator and a damped oscillator Associated terminology, which is useful in vibration analysis, isdefined

1.1 Introduction

Vibrations can be analyzed both in the time domain and in the frequency domain Free vibrations andforced vibrations may have to be analyzed This chapter provides the basics of the time-domainrepresentation and analysis of mechanical vibrations

Free and natural vibrations occur in systems because of the presence of two modes of energy storage.When the stored energy is repeatedly interchanged between these two forms, the resulting time response

of the system is oscillatory In a mechanical system, natural vibrations can occur because kinetic energy(which is manifested as velocities of mass [inertia] elements) may be converted into potential energy(which has two basic types: elastic potential energy due to the deformation in spring-like elements, andgravitation potential energy due to the elevation of mass elements under the Earth’s gravitational pull) andback to kinetic energy, repetitively, during motion An oscillatory excitation (forcing function) is able tomake a dynamic system respond with an oscillatory motion (at the same frequency as the forcingexcitation) even in the absence of two forms of energy storage Such motions are forced responses ratherthan natural or free responses

1-1

An analytical model of a mechanical system is a set of equations These may be developed either bythe Newtonian approach, where Newton’s Second Law is explicitly applied to each inertia element, or

by the Lagrangian or Hamiltonian approach which is based on the concepts of energy (kinetic andpotential energies) A time-domain analytical model is a set of differential equations with respect tothe independent variable time ðtÞ: A frequency-domain model is a set of input–output transferfunctions with respect to the independent variable frequency ðvÞ: The time response will describe howthe system moves (responds) as a function of time The frequency response will describe the way thesystem moves when excited by a harmonic (sinusoidal) forcing input, and is a function of thefrequency of excitation

1.2 Undamped Oscillator

Consider the mechanical system that is

schema-tically shown in Figure 1.1 The inputs (or

excitation) applied to the system are represented

by the force f ðtÞ: The outputs (or response) of

the system are represented by the displacement y:

The system boundary (real or imaginary)

demarcates the region of interest in the analysis

What is outside the system boundary is the

environment in which the system operates An

analytical model of the system may be given by

one or more equations relating the outputs to

the inputs If the rates of changes of the response

(outputs) are not negligible, the system is a dynamic system In this case the analytical model, in thetime domain, becomes one or more differential equations rather than algebraic equations Systemparameters (e.g., mass, stiffness, damping constant) are represented in the model, and their valuesshould be known in order to determine the response of the system to a particular excitation Statevariables are a minimum set of variables, which completely represent the dynamic state of a system atany given time t: These variables are not unique (more than one choice of a valid set of state variables

is possible) For a simple oscillator (a single-degree-of-freedom (DoF) mass–spring–damper system)

an appropriate set of state variables would be the displacement y and the velocity _y: An alternative setwould be _y and the spring force

In the present section, we will first show that many types of oscillatory systems can be represented bythe equation of an undamped simple oscillator In particular, mechanical, electrical, and fluid systems will

be considered The conservation of energy is a straightforward approach for deriving the equations ofmotion for undamped oscillatory systems (which fall into the class of conservative systems) Theequations of motion for mechanical systems may be derived using the free-body diagram approach, withthe direct application of Newton’s Second Law An alternative and rather convenient approach is the use ofLagrange’s equations The natural (free) response of an undamped simple oscillator is a simple harmonicmotion This is a periodic, sinusoidal motion

1.2.1 Energy Storage Elements

Mass (inertia) and spring are the two basic energy storage elements in mechanical systems The concept

of state variables may be introduced as well through these elements

1.2.1.1 Inertia (m)

Consider an inertia element of lumped mass m; excited by force f ; as shown inFigure 1.2

The resulting velocity is v:

b

k m

y

SystemBoundaryEnvironment

SystemOutputs(Response)System

Inputs(Excitation)

Newton’s Second Law gives

Kinetic energy stored in the mass element is equal

to the work done by the force f on the mass Hence,

we have

1.2.1.2 Spring (k)

Consider a massless spring element of lumped

stiffness k; as shown in Figure 1.3 One end of the

spring is fixed and the other end is free A force f is

applied at the free end, which results in a

displacement (extension) x in the spring

Hooke’s Law gives

ð

2kf2or

Note: f and x are both appropriate state variables for a spring, because both can completely represent theenergy in the spring

From these results, it follows that at finite velocities there cannot be an instantaneous change in the force

of a spring In particular, from Equation 1.9 we see that when the velocities of a spring are finite:

Also, it follows that

1.2.1.3 Gravitation Potential Energy

The work done in raising an object against the

gravitational pull is stored as gravitational

poten-tial energy of the object Consider a lumped mass

m; as shown in Figure 1.4, which is raised to a

height y from some reference level

The work done gives

Hence,

1.2.2 The Method of Conservation of Energy

There is no energy dissipation in undamped systems which contain energy storage elements only Inother words, energy is conserved in these systems, which are known as conservative systems Formechanical systems, conservation of energy gives

These systems tend to be oscillatory in their natural motion, as noted before Also, analogies exist withother types of systems (e.g., fluid and electrical systems) Consider the six systems sketched inFigure 1.5

1.2.2.1 System 1 (Translatory)

Figure 1.5(a) shows a translatory mechanical system (an undamped oscillator) which has just one degree

of freedom x: This may represent a simplified model of a rail car that is impacting against a snubber Theconservation of energy (Equation 1.13) gives

f = mg

mg

FIGURE 1.4 A mass element subjected to gravity.

Here, m is the mass and k is the spring stiffness Differentiate Equation 1.14 with respect to time t: Weobtain

m_x€x þ kx_x ¼ 0Since generally _x – 0 at all t; we can cancel it out Hence, we obtain the equation of motion:

m

FIGURE 1.5 Six examples of single-degree-of-freedom oscillatory systems: (a) translatory; (b) rotatory; (c) flexural; (d) pendulous; (e) liquid slosh; (f) electrical.

Here, m is the lumped mass at the free end of the support and k is the lateral bending stiffness of thesupport structure Then, as before, the equation of motion becomes

1

Differentiate with respect to t after canceling the common ml:

l _u €u þ g sin u _u ¼ 0Since, _u – 0 at all t; we have the equation of motion:

This system is nonlinear, in view of the term sinu:

For smallu; sin u is approximately equal to u: Hence, the linearized equation of motion is

1.2.2.5 System 5 (Liquid Slosh)

Consider a liquid column system shown in Figure 1.5(e) It may represent two liquid tanks linked by apipeline The system parameters are: area of cross section of each column ¼ A; mass density ofliquid ¼r; length of liquid mass ¼ l:

€y þ gðh þ yÞ 2 gðh 2 yÞ ¼ 0

€y þ 2g

1.2.2.6 System 6 (Electrical)

Figure 1.5(f)shows an electrical circuit with a single capacitor and a single inductor Again, conservation

of energy may be used to derive the equation of motion

Voltage balance gives

where vLand vCare voltages across the inductor and the capacitor, respectively

Constitutive equation for the inductor is

Here, v denotes vL: Also,

This equation is identically satisfied for all t:

Hence, the general solution of Equation 1.37 is

indeed Equation 1.39, which is periodic and

sinusoidal

This response is sketched in Figure 1.6 Note

that this sinusoidal oscillatory motion has a

frequency of oscillation ofv (radians/sec) Hence,

a system that provides this type of natural motion

is called a simple oscillator In other words, the

response exactly repeats itself in time periods of T

or a cyclic frequency f ¼ 1=T (Hz) The frequency

v is in fact the angular frequency given by v ¼ 2pf : Also, the response has an amplitude A; which is thepeak value of the sinusoidal response Now, suppose that we shift this response curve to the right throughf=v: Consider the resulting curve to be the reference signal (with signal value ¼ 0 at t ¼ 0; andincreasing) It should be clear that the response shown in Figure 1.6 leads the reference signal by a timeperiod off=v: This may be verified from the fact that the value of the reference signal at time t is the same

as that of the signal in Figure 1.6 at time t 2f=v: Hence, f is termed the phase angle of the response, and

it is a phase lead

The left-hand-side portion of Figure 1.6 is the phasor representation of a sinusoidal response

In this representation, an arm of length A rotates in the counterclockwise direction at angular speedv: This is the phasor The arm starts at an angular position f from the horizontal axis, at time

t ¼ 0: The projection of the arm onto the vertical ðxÞ axis is the time response In this manner,the phasor representation can conveniently indicate the amplitude, frequency, phase angle, and theactual time response (at any time t) of a sinusoidal motion A repetitive (periodic) motion of thetype 1.39 is called a simple harmonic motion, meaning it is a pure sinusoidal oscillation at a singlefrequency

Next, we will show that the amplitude A and the phase anglef both depend on the initial conditions.Substitute the initial conditions (Equation 1.40) into Equation 1.39 and its time derivative We obtain

2

þ Avv0n

2

¼ 1Hence,

Example 1.1

A simple model for a tracked gantry conveyor system in a factory is shown inFigure 1.7

The carriage of mass ðmÞ moves on a frictionless track The pulley is supported on frictionless bearings,and its axis of rotation is fixed Its moment of inertia about this axis is J: The motion of the carriage isrestrained by a spring of stiffness k1; as shown The belt segment that drives the carriage runs over the

FIGURE 1.6 Free response of an undamped simple oscillator.

pulley without slippage, and is attached at the

other end to a fixed spring of stiffness k2: The

displacement of the mass is denoted by x; and the

corresponding rotation of the pulley is denoted by

u: When x ¼ 0 (and u ¼ 0) the springs k1and k2

have an extension of x10and x20, respectively, from

their unstretched (free) configurations Assume

that the springs will remain in tension throughout

the motion of the system

Solution Using Newton’s Second Law

A free-body diagram for the system is shown in

No-Slip Belt And Pulleys

Frictionless Bearings

FIGURE 1.7 A tracked conveyor system.

x

k1

r m

Eliminate the common velocity variable _x (which cannot be zero for all t) We obtain

m þ rJ2 €x þ ðk1þ k2Þx ¼ k2x202 k1x10

which is the same result as before

Solution Using Lagrange’s Equations

ddt

›L

›_qi 2 ›L

›qi ¼ Qi for i ¼ 1; 2; …; nwhere the generalized coordinate qi¼ x and the corresponding generalized force Qi¼ 0 because thereare no nonconservative and external forces We obtain

r2 €x þ ðk1þ k2Þx ¼ k2x202 k1x10which is identical to what we obtained before

ðr2m þ JÞ €u þ r2ðk1þ k2Þu ¼ rk2x202 rk1x10The equivalent moment of inertia:

Jeq¼ r2m þ JThe equivalent torsional stiffness:

Keq¼ r2ðk1þ k2ÞTherefore, the corresponding natural frequency is

Common methods of developing equations of motion for mechanical systems are summarized inBox 1.1.

1.3 Heavy Springs

A heavy spring has its mass and flexibility properties continuously distributed throughout its body

In that sense it has an infinite number of DoF(s), and a single coordinate cannot represent its motion.However, for many practical purposes, a lumped-parameter approximation with just one lumped mass torepresent the inertial characteristics of the spring may be sufficient Such an approximation may beobtained by using the energy approach Here, we represent the spring by a lumped-parameter “model”such that the original spring and the model have the same net kinetic energy and potential energy Thisenergy equivalence is used in deriving a lumped mass parameter for the model Even though damping(energy dissipation) is neglected in the present analysis, it is not difficult to incorporate that as well inthe model

Conservation of energy: T þ V ¼ const:

Differentiate with respect to time t

general incremental motion ðdq1; dq2; …; dqnÞ:

(About centroid or a fixed point)

1.3.1 Kinetic Energy Equivalence

Consider the uniform, heavy spring shown in

Figure 1.9, with one end fixed and the other end

Hence, the equivalent lumped mass concentrated at the free end ¼ ð1=3Þ £ spring mass:

Note: This derivation assumes that one end of the spring is fixed Furthermore, the conditions areassumed to be uniform along the spring

An example of utilizing this result is shown in Figure 1.10 Here, a system with a heavy spring and alumped mass is approximated by a light spring (having the same stiffness) and a lumped mass.Another example is shown in Figure 1.11 In this case, it is not immediately clear which of theapproximations shown on the right-hand side is most appropriate

Example 1.2

A uniform heavy spring of mass msand stiffness k is attached at one end to a mass m that is free to roll on

a frictionless horizontal plane The other end is anchored to a vertical post A schematic diagram of thisarrangement is shown inFigure 1.12

The unstretched length of the spring is l:

Assume that, when the velocity of the connected

mass is v; the velocity distribution along the spring

is given by

vsðxÞ ¼ v sinpx

2lwhere x is the distance of a point along the

spring, as measured from the fixed end We will

determine an equivalent lumped mass located

at the moving end of the spring (i.e., at the

moving mass m) to represent the inertia effects of

the spring

Consider an element of length dx at location x

of the spring Since the spring is uniform, we have

element mass ¼ ðms=lÞdx: Also, according to the

vsinðpx=2lÞ: Hence, kinetic energy of the spring is

2ms

12

l

0¼ 14

ms

l v2l ¼

12

ms

2 v2

It follows that the equivalent lumped mass to be located at the moving end of the spring is ms=2: Thisresult is valid only for the assumed velocity distribution, and corresponds to the first mode of motiononly In fact, a linear velocity distribution would be more realistic in this low frequency (quasi-staticmotion) region, which will give an equivalent lumped mass of ð1=3Þms; as we have seen before Suchapproximations will not be valid for high frequencies say, higher thanpffiffiffiffiffiffik=ms

1.4 Oscillations in Fluid Systems

Fluid systems can undergo oscillations (vibrations) quite analogous to mechanical and electricalsystems Again, the reason for their natural oscillation is the ability to store and repeatedlyinterchange two types of energy — kinetic energy and potential energy The kinetic energy comes fromthe velocity of fluid particles during motion The potential energy arises primarily from the followingthree main sources:

1 Gravitational potential energy

2 Compressibility of the fluid volume

3 Flexibility of the fluid container

A detailed analysis of these three effects is not undertaken here However, we have seen from theexample inFigure 1.5(e)how a liquid column can oscillate due to repeated interchange between kineticenergy and gravitational potential energy Now, let us consider another example

Example 1.3

Consider a cylindrical wooden peg of uniform cross section and height h; floating in a tank of water,

as inFigure 1.13(a).It is pushed by hand until completely immersed in water, in an upright orientation.When released, the object will oscillate up and down while floating in the tank Letrbandrlbe the mass

densities of the body (peg) and the liquid (water), respectively The natural oscillations and the stability

of this system may be studied as below

Suppose that, under equilibrium in the upright position of the body, the submersed length is l: Themass of the body is

where A is the area of cross section (uniform)

By the Archimedes principle, the buoyancy force R is equal to the weight of the liquid displaced by thebody Hence,

l ¼ rb

For a vertical displacement y from the equilibrium position, the equation of motion is (Figure 1.13(b))

m€y ¼ mg 2 Aðl þ yÞrlgSubstitute Equation ii and Equation iii We obtain

m€y ¼ 2ArlgySubstitute Equation i:

Ahrb€y þ Arlgy ¼ 0

y

CM

€y þ rlg

rbhy ¼ 0The natural frequency of oscillations is

vn¼ ffiffiffiffiffiffiffirlg

rbhr

Note that this result is independent of the area of the cross section of the body

Assumptions made:

1 The tank is very large compared to the body The change in liquid level is negligible as the body

is depressed into the water

2 Fluid resistance (viscous effects, drag, etc.) is negligible

3 Dynamics of the liquid itself is negligible Hence, “added inertia” due to liquid motion isneglected

To study stability of the system, note that the buoyancy for R acts through the centroid of the volume ofdisplaced water(Figure 1.13(c)).Its line of action passes through the central axis of the body at point M.The point is known as the metacenter Let C be the centroid of the body

If M is above C; then, when tilted, there will be a restoring couple that will tend to restore the body toits upright position Otherwise the body will be in an unstable situation, and the buoyancy couple willtend to tilt it further towards a horizontal configuration

1.5 Damped Simple Oscillator

Now we will consider free (natural) response of a

simple oscillator in the presence of energy

dissipation (damping)

Assume viscous damping, and consider the

oscillator shown in Figure 1.14 The free-body

diagram of the mass is shown separately

We use the following notation:

vd¼ damped natural frequency

vr¼ resonant frequency

v ¼ frequency of excitation

The concept of resonant frequency will be

addressed inChapter 2

Usually, the viscous damping constant of a single

DoF is denoted by b (but, sometimes c is used

instead of b; particularly for multi-DoF systems)

Apply Newton’s Second Law From the free-body diagram in Figure 1.14, we have the equation ofmotion m€x ¼ 2kx 2 b_x

x b

free-This is a free (or unforced, or homogeneous) equation of motion Its solution is the free (natural) response

of the system and is also known as the homogeneous solution Note thatvn¼pffiffiffiffiffik=m; which is the naturalfrequency when there is no damping, and

Hence,

z ¼ 12

ffiffiffiffimk

rbmor

then Equation 1.50 will represent a solution of Equation 1.47

Equation 1.51 is called the characteristic equation of the system This equation depends on the naturaldynamics of the system, not on forcing excitation or initial conditions

The solution of Equation 1.51 gives the two roots:

These are called eigenvalues or poles of the system

Whenl1–l2; the general solution is

We can identify three categories of damping level, as discussed below, and the nature of the response willdepend on the particular category of damping

1.5.1 Case 1: Underdamped Motion (z < 1)

In this case it follows from Equation 1.52 that the roots of the characteristic equation are

Note:

ejvd t¼ cosvdt þ j sinvdt

e2jvd t¼ cosvdt 2 j sinvdt

So, an alternative form of the general solution would be

Here, A1and A2are the two unknown constants By equating the coefficients it can be shown that