Soil Dynamics and Machine Foundations Swami Saran

7/28/2019 Soil Dynamics and Machine Foundations Swami Saran

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SOIL DYNAMICS

. AND" '..

MACHINE FOUNDATIONS

By

Dr. SWAMI SARAN

Departmentof Civil Enginemng

University of RoorkeeRoorkee-247 667

(INDIA)

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1999

iF GalgotiaPublicationspvt.ltd.5,AnsarIRoad, Daryaganj,New Delhl-110 002

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No matter infull or part may be reproduced or transmitted in any form or by any means (except for review

or criticism) without the written permission of the author and publishers.

Though much care has been taken by the author and the publishers to make the book error (factual or

printing) free. But neither the author nor the publisher takes any legal responsibility for any mistake

that might have crept in at any stage.

Published by .-

Suneel Galgotia for Galgotia Publications (P) Ltd.

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, PREFACE

During the last 25 years, considerable work in the area of soil dynamicsand machine foundationshas been

reported.Courseson soil dynamicsandmachinefoundationsalreadyexistat graduatelevelin manyinstitutions,

and its inclusion at undergraduate level is progressing fast.The author is engaged in teaching thecourseon soil dynamicsand machinefoundationsat gr'duate level

from last fLfteenyears. The text of thisbookhas been developed mainlyout of my notes preparedfor teachingthe students.The considerationin developingthe text is its lucide presentationfor clearunderstandingof thesubject.The material has been arrangedlogicallyso that the readercan followthe developmentalsequenceof

the subject with relative ease. A number of solved examples have been included in each chapter. All the

formulae,charts and examples are given in SI units.

Someof the material included in thistextbook has been drawnfrom the works of other autors.Inspiteofsincereefforts,some contributionsmay nothavebeen acknowledged.The authorapologisesforsuchomissions.

The author wishes to expresshis appreciationto Km. LataJuneja,Sri RaJeevGrover and Sri S. S. Guptafor typingand drawing work. Thanks arealso due to the many collegues,friends and studentswho assistedin

wittingof thisbook. . .

The author would be failing in his duty it he does not aclaiowledge the support he received from his

familymembers who encouraged him through the various stagesof study and writing.. .

Thebook is dedicated to author's Sonin law, (Late) ShriAkhilGupta as a token of his love,affectionandregards to him.

(Dr. Swami Saran)

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CONTENTS.

1.

PREFACE

INTRODUCTION

1.1 General

1.2 Earthquake Loading

1.3 Equivalent Dynamic Load to an Actual Earthquake Load

1.4 Seismic Force for Pseudo-staticAnalysis

Illustrative ExamplesReferences

Practice Problems

1-12

I

3

6

9

12

12

12

13-66

13

14

15

18

32

36

39

48

53

64

2. THEORY OF VIBRATIONS

2.1 General

2.2 Defmitions

2.3 Harmonic Motion

2.4 Vibrations of a SingleDegree Freedom System2.5 Vibration Isolation

2.6 Theory of Vibration MeasuringInstruments

2.7 Vibration of Multiple Degree Freedom Systems

2.8 Undamped Dynamic VibrationAbsorbers

Illustrative ExamplesPractice Problems

3. WAVE PROP AGATION IN AN ELASTIC, HOMOGENEOUS.. ANDISOTROPIC MEDIUM

3.1 General

3.2 Stress, Strain and Elastic Constants

3.3 Longitudinal Elastic Wavesin a Rod oflnfmite Length

3.4 Torsional Vibration ora Rod of Infmite Length

3.5 End Conditions

3.6 Longitudinal Vibrations of Rods of Finite Length

3.7 Torsional Vibrations of Rods of FiniteLength

3.8 Wave Propagation in an lnfmite, HomogeneousIsotropic, Elastic Medium

3.9 Wave Propagation in Elastic, Half Space

3.10 Geophysical Prospecting3.11 Typical Values of CompressionWave and Shear WaveVelocities

Illustrative Examples

References..'. . ;

Practice Problems

67

67

70

72

74

76

80

81

86

93108

108

116

117

i,.,~


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4.

viii Soil Dynamics & Machine Foundations

DYNAMIC SOIL PRO~ER~5. '-. ."' .

4.1 General

4.2 LaboratoryTechinques

4.3 FieldTests

4.4 FactorsAffecting Shear Modulus, ElasticModulus and Elastic Constants

IllustrativeExamples

References

PracticeProblems

DYNANnCEARTHPRESSURE

General

Pseudo-static Methods

5.3 Displacement Analysis


References

PracticeProblems

--.118-186

118

118

147

163174

182

184

187

187

201

221

236

237

238

238

238 .

249

268

277

278

279

2.79

281

283

288

296

300

301

306309

314

319

323

6. DYNAMIC BEARING CAPACITY OF SHALLOW FOUNDATIONS

6.1 General

6.2 Pseudo-static Analysis

6.3 Bearing Capacity of Footings

6.4 Dynamics Analysis


References

Practice Problems

7. LIQUEFACTION OF SOILS7. 1 General

7.2 Definitions

7.3 Mechanism of Liquefaction

7.4 Laboratory Studies

DynamicTriaxial Test

7.6 CyclicSimple Shear Test

7.7 Comparisonof Cyclic Stress CausingLiquefactionunder Triaxial and

SimpleShear Conditions

7.8 StandardCurves and Correlations for Liquefaction

7.9 Evaluationof Zone of Liquefactionin Field

7.10 VibrationTable Studies

7.11 Field Blast Studies

7.12 Evaluationof LiquefactionPotentialusing StandardPenetrationResistance

7.13 Factors Affecting Liquefaction -


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Contents '

8.

9.

ix

7.14 AntiliquefactionMeasures

7.15 Studies on Use ofGravel Drains

IllustrativeExamples

ReferencesPracticeProblems

324

326

332

336339

340-351

340

340

347

348

349

. .GENERAL PRINCIPLES OF MACIDNE FOUNDATION DESIGN

8.1 General

8.2 Types ofMachines and Foundations

8.3 General Requirements ofMachine Foundation

8.4 Perimissible Amplitude

8.5 Allowable Soil Pressure

8.6 Permissible Stresses of Concrete of Steel

8.7 Permissible Stresses of Timber

References

349

350

351

FOUNDATIONS OF RECIPROCATING MACHINES

9.1 General

9.2 Modes of Vibrationof a Rigid FoundationBlock

9.3 Methods of Analysis

9.4 Linear Elastic WeightlessSpring Method

9.5 Elastic Half-spaceMethod

9.6 Effect of Footing Shape on Vibratory Response

9.7 Dynamic Responseof Embedded BlockFoundation

9.8 Soil Mass Participatingin Vibrations

9.9 Design Procedure for a Block FoundationIllustrative ExamplesReferences

Practice Problems

352

~ 352

353

354

370

392

394

400

402408

419 .

420

423

426

432

436

442

442

10. FOUNDATIONS OF IMPACT TYPE MACIDNES

10.1 General

10.2 DynamicAnalysis .

10.3 Design Procedure for a Hammer Foundation


References

Practice Problems

11. FOUNDATIONS OF ROTARY MACHINES

11.1 General

11.2 Special Considerations

11.3 Design Criteria

443

444

445

,{"


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2

1:)

00

u

Edc>-a

+

Period of loading

T usually large

_.

Soil Dynamics & Machine Foundations

Time

(b) Dynamic load due to steady state vibration

Ud0 +

v

Ed

C>-a

~.T.I

(c) Multiple impulse loading

Vertical

High frequencypredominates

Time

.'

(d) Trice ofvertical acceleration of ground due to pile driving

. Fig. 1.1: Variation of dynamic load withtime in IOmetypical cases


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Jntroduttion

-, ".-

Vibrations of earth's surface caused by waves coming from a source of disturbance inside the earth are

described as Earthquakes and are one of the ri1ostdestructive forces that nature unleashes on earth.

When, at any depth below tile gro~d surfa~e,the strain ene~gy'ac~~ulated due to deformations in earthmass exceeds the resilience of the storing material, it gets release through rupture. The energy thus

released is propogated in the form of waves which impart energy to the media through which they pass

and vibrate the structures standing on the earth's..surface. The point inside the earth mass where slipping

or fracture begins is termed as focus and the point just above the focus on the earth's surface is termed as

epicentre. The position of the focus is determined,with the help of seismographrecords (Fig: 1.2]'u:ti't'ising

the average velocities of different waves and time difference in reaching the waves at the ground surface.

Figure 1.3 explains the various terms in simple manner.

ITrace

1amplitude)

Fig. 1.2. :A typical earthquake record

~

E

Epic.entric. distance ~

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7. 'y.. .."" .... "

Epic.entre // .

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Soil Dynamics cl Machine Fo"nd4tio.ns

1.2.1 Intensity. The severity of shaking of an earthquake as felt or ob!jervedthroughdamage is'described

as intensity ata certain place on an arbitrary scale. For this purpose modified Mercalli scale is more

common in use. It is divided into 12 degrees of intensity as presented in Table!.L

Table 1.1 : Modified MereaIli Intensity Scale (Abridged}

ClassofEarthquakes Description

"

fII

IV

V

VI

VII

VIII

IX

X

XI

XII

Not felt except by a very few under specially favourable circumstances.

Felt only by a few persons at rest, specially on upper floors of buildings; and delicately sus-

pended objects may swing.

Felt quite noticeably indoors, specially on upper floors of buildings but many people do not

recognize it as an earthquake; standing motor cars may rock slightly, and vibration may be felt

like the passing of a truck.During the day felt indoors by many, outdoors by a few; at night some awakened,dishes, win-

dows, doors disturbed, walls make cracking sound, sensation like heavytruck striking the build-

ing; and standing motor car rocked noticeably.

Felt by nearly everyone; many awakened; some dishes, windows, etc. broken; a few instances of

cracked plasters; unstable objects overturned; disturbance of trees, poles and other tall objects

noticed sometimes and pendulum clocks may stop. .

Felt by all; many frightened and run outdoors; some heavy furnituremoved; a few instances of

fallen plaster or damaged chimneys; damage slight.

Everybodyruns outdoors, damagenegligible in buildings of gooddesign and construction; slight

to moderate in well built ordinary structures; considerable in poorly built or badly designed

structures; some chimneys broken; noticed by persons driving motorcars.Damage slight j!, spe~ially designed structures; considerable in' ordinary substantial buildingswith partial collapse; very heavy it) poorly built structures; panel walls thrown out of framed

structure; heavy furniture overturned; sand and mud ejected in small amounts; changes in wellwater; and disturbs persons driving motor cars. .

Damage considerable in specially designed structures; well designed framed structures thrown

out of plumb; very heavy in substantial buildings with parti~1collapse; buildings shifted offfoundations; ground crackedconspicuously; and underground pipes broken.

Some well built wooden structure~ destroyed; most masonry and framed structures with founda-

tions destroyed; ground badly cracked; rails bent; land-slides considerable from river banks and

steep slopes; shifted sand and mud; and water splashed over banks.

Few, if any, masonry structures remain standing; bridge destroyed; broad fissures in ground,underground pipe lines completely out of service; earth slumps and landslips in soft ground; and

rails bent greatly. ~ '-Total damage; waves seen on ground surface; lines of sight and lever distorted; and objects thrown'" ,upward into the air. .1, . 'H",

f'.'.HI",. ...


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1.2.2 Magnitude. Magnitude of an earthquake is a measure of the size of an earthquake, based on the-.-". . ~"."""""", ,.._" ,.,..,.._~,-,- , ".. -',",

amplitude of elastic waves it generates. Richter (1958) suggested the following relation. ~

M = loglOA -loglO Aa . ...(1.1)where

M = Magnitude of earthquake

A = Trace amplitude in mm (Fig. 1.2)

Aa = Distance correction (F:ig.1.4) ,. , ,.A relationship between strain energy released py an earthquake and its magnitude is given by Richter

(1958) as follows

loglo E = 11.4 + 1.5 M ...(1.2)where

E = Energy released in earthquake in Ergs

A comparison of the magnitude M of an earthquake with maximum i

n

tensity of the Modified Mer-

calli Scale is given in Table 1.2., ,

Table 1.2 : Comparison of the Richter Scale Magnitude with the Modified Mercalli Scale

Richter Scale Magnitude (AI) Maximum Intensity, ModifiedMercalli Scale

2

3

4

5

6

7

8

" ;

I, II

m

IV,V

VI, Vp

VII,VIII

, ' . IX, X

XI

The fault length, affected area and duration of earthquake also depend on the magnitude of earth-quake (Housner, 1965; Housner, 1970). Table ,1.3 gives approximate idea about these. '

" '

Table 1.3 : Fault Length, Affected Area,~nd Duration of.Eart~quake

Magnitude oj

Earthquake

(Richter scale)

Fault Length" i '. Affected.Area

(/en?)

Duration of

'Earthquake

(8)

5

6

7

8

,(km)

1-2

, , 2-5 .

25-50 . ; l J "

20,000

60,000:,..., \

: f,2()',000'

2 00 OO',250


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6, SoUDyIUlllfics & Mtrehille Foundations

-c:./":J'c 50\0:

E

.S 4-0

0\

~ 3..!.-

.For nearearthquake

For teleseism$surface waveswith time:period 20 s '

c0

.+:v 2~I-0U

~ 1c0..-11\.-0 0

1 10 100

Distance in km

1000 10000 \

. .

Fig. 1.4: Distaoce corredio.o for magnitude determination

1.3 EQUIVALENT DYNAMIC LOAD TO AN ACTUAL EARTHQUAKE LOAD

Figure 1.1 (a) shows the variation of dynamic load wi+htime observed during El Centro earthquake. The

loading is not periodic and the.peaks in anY two cycles are different. For the analysis and design of

foundations such a random variation is converted into equivalent number of cycles of uniformly varying

load [Fig. 1.1(b)]. It means that the structure-foundation-soil system subjected to Ns cycles of uniformlyvarying load will suffer same deformations and stresses as by the actual earthquakes. Most of the analyses

and laboratory teStingare 'carriedout using this concept.' .

According to Seed and Idriss (1911), the average equivalent uniform acceleration is about 65 percent

of the maximum acceleration. The number of significant cycles, Ns depends on the magnitude of earth-

quake. They recommended the values ofNs as 10, 20 and 30 for earthquakesofmagnitudes 1, 1.5 and 8respectively.

Lee and Chan (1912) suggested the following procedure for..convertingthe irregular stress-time

history to the equivalent number of cycles of cyclic shear stresses of maximum magnitude equal to

K 'tmax' ~ ~eing a constant less th~.1JP.ity :


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Introduction, ',;. ' 7, H,

(i) Let Fig. 1.5 shows ~}Yl'icalea~CJ.uakereco.rd.Divide the s;t;essrange (0 to 'tmaX>or acceleration

range (0 to amax)into convenient number ofleveIs and note the mean stress or meanacceleration

within'each level as mentioned in column no. 2 of Table 1.4. Then the number of cycles withpeaks 'Yhichfall within each ofthese levels is counted and recorded. Note that because the actual

time history is not symmetric about the zero stress axis, the number of peaks on both sides are

counted and two peaks are equivalent to one cycle. For example, an earthquake record shown in

Fig. 1.5 has number of cycles in various ranges of acceleration levels as listed in Col. 3 ofTable 1.4.

Om ox :; + 0 . 12

c:0-0~e:.I

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8 Soil Dynamics & Machine Foundatioiis"

Table 1.4 : Equivalent Cycles for Anticipated Eartbquake

3 1 0-3 0'1

(Ns )0.65 Tmax.

( ~S)k Tmax

0-03 0'01

Conversion factor,

Fig. 1.6 : Conversion factor versus shear stress ratio

For getting the equivalent number of cycles for 0.75 'tmax'read the yalue of conversion factor (Fig.1.6) corresponding to an ordinate value of 0.75. It comes out as 1.5. The value of equivalent number of

cycles obtained for 0.65 'tmaxas illustrated in Table 1.4 is divided by this conversion factor to obtain

equivalent number of cycles corresponding to 0.75 'tmaxi.e. 9.0/1.5= 6.0 cycles.

Seed and Idriss (1971) and Lee and Chan (1972) developed the above concepts specificallyfor lique-

facti~mstudies. More details of these procedures have been.discussed in Chapter 7.

--""" "~.-

..

Acceleration Average and Number Conversion Equivalentlevel il/

level in of. factor number of cyclespercent of percent of cycles at 0.65 1"ma:c

(I) (2) (3) (4) (5)- - ..

100- 80 90 5/2 = 2.5 2.6 '65

80- 60 70 3/2 = 1.5 1.2 1.8

60 - 40 50 7/2 = 3.5 0.20 0.70

40 - 20 30 5/2 =2.5 negligible 0.0

20 - 00 10 >100 negligible 0.00

Total numberof

cycles = 9,0

1.0

r-' 0.8

> -111

0.6:111 .....CI........ ,-

111 x 0.4u I 0

lt5 0.2

010


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Introductionj' ..

9

1.4 SEISMIC FORCE FOR PSEUDO-STATIC ANALYSIS

For the purpose of determining seismic force, the country is classified into five zones as shown in

Fig. 1.7. Two methods namely (i) seismic coefficient method and (ii) response spectrum method are used

for computing the seismic force. For pseudo-static design of foundations of buildings, bridges and similarstructures, seismic coefficient method is used. For the analysis of earth dams and dynamic designs,

response spectrum method is used (IS- 1893 : 1975).

Q) Iv.

Equivalent modi fidemercalli intensity

IX and aboveVl1I

VIIVILess than VI

Bombay

. .. "In :

0 0

~ Vo0 po rt

Blair0

.:0

'I)

Fig. 1.7 : Seismic zonesof India

in seismic coefficient method, the design value of horizontal seismic coefficient CJ.his obtained by the'ollowing expression : ~

ah = ~I ao ...(1.3)vhere

a() =Basicseismiccoefficient,Table 1.5

I. = Coefficient depe.ndingupon the unportance of structure,:Table 1.6

~ = Coefficient depending upon the soil-f~undation system, Table 1.7

.

-


17/500

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Introduction

where

.."".

J~

11

,:

, S,ab ,= ~ . I . F .;.,JL

, 0 g...( L 1)

F 0 = Seismic zoni1)gfactor for average acceleration spectra (Table 1.8)

S .~ = average acceleration coefficient as read from Fig. 1.8 for appropriate natural period andg

damping of the structure.

-c~

u--~0uc0

0.6

0.5....0L..

~-~uu0

~010L..

~>

0'4

0 .3___-

0.2----

0.111

0

1 01 0If) 0

. ,

0.4

Natural1-6 2.0 2.4 2.8 3,0.

vibration in seconds

Fig. 1.8: Average acceleration spectra

Table 1.8 : Values of Seismic Zoning Factor. Fo

Zone No. Fo

VIV

III

U

I .

0.400.25

0.20

0.10

0.05

,.


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12 Soil Dynamics & Machine Foundations

tILL USTRATIVE EXAMPLES'

Example 1.1

The srandard torsion seismograph recorded an average trace amplitude of 8.0 mm. The distance to the

epicentre is.estimated about 100 km. Determine the magnitude of earthquake.

Solution:

From Fig. 1.4, the distance correction for 100 km is 3.0.

Hence,

M = 10glO8.0 + 3 = 3.9

HousnCf, G. W. (1965), "Intensity of earthquake ground shaking near the causative fault", Proceedings 3rd World

Conference on Earthquake Engineering, New Zealand, Vo\. 1.

Housner. G. W. (1970), "Design spectrum", in EarthquukeEngineering (R. W. Wiegel, Ed.), Prentice-HalI, Englewood

Cliffs, New Jersey, pp. 97-106.

IS I:s03-1975. "Criteria for earthquake resistant design of structures", ISI, New Delhi.

Lee, K..l.. and Chan, K. (1972), "Number of equivalent significant cycles in strong motion earthquakes", Proceed-

ings, International Conference on Microzonation, Seattle, Washington, vo\. H, pp. 609-627.

Richter, CF. (1958), "Elementary seismology", W. H. Freeman, San Francisco, California.

Seed. H. B. Idriss, I. M., Makdisi, F. and Banerjee, N. (1975), "Representation of irregular stress - time histories b)equivalent uniform stress series in liquefaction analysis", Report No. EERC 75-29, Earthquake Engi-

neering Research Center, University of California, Berkeley.

Seed, H. B., and Idriss, 1. M. (1971), "Simplified procedure for evaluating soil liquefaction potential"

J. Soil Mech. Found. Engg., ASCE, Vo\. 97, No. SM9, pp. 1249-1273.

F PRACTICE PROBLEMS

1.1 Explain the terms 'Intensity' and 'Magnitude' irt relation to earthquake. How are fault length an,

duration of earthquake depend on magnitude?

1.2 Describe a method of getting equivalent number of cycles of uniformly varying load for an actur

earthquake record,

1.3 Determine the equivalent number ef cycles for 0.75 Tmaxfor El Centro earthquake.

DC


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"".'\'-


21/500

14s.u /JyruuIfics & Machine Foundations

The forms ofvibration mainly depend on the mass, stiffness distribution and end conditions ofthe

system.

To study the response ofa vibratory system, in many cases it is satisfactory to reduce it to an idealized

system oflumped parameters. In this regard, the simplest model consists of mass, spring and dashpot

This chapter is framed to provide the basic concepts and dynamic analysis of such systems. Actual fieldproblems which can be idealized to mass-spring-dashpot systems, have also been included.

2.2 DEFINITIONS

2.2.1 Vibrations: If the motion ofthe body is oscillatory in character, it is called vibration.. -, -2.2.2 Degrees of Freedom: The number of independent co-ordinates which are required to define the

position of a system during vibration, is called degrees offreedom (Fig. 2.2).

~D:

m

(a) One degree of freedom (b) Two degrees offreedom

Z2

. .~

KI

Z,

.- -,

~ -Z)- J.., .

(c) Three degrees of freedom' . (d) Six degrees 'offreedon~ (e) Infinite degrees offreedom-' , , . .: .',n ,-, t ~ "'_~

Fig. 2.2':'Systems with different degrees of freedom


22/500

Theory of Vibrations 15

2.2.3 Periodic Motion: If motion repeats itself at regular intervals of time, it is called periodic motion.

2.2.4 Free Vibration: If a system vibrates without an external force, then it is said to undergo free

vibrations. Such vibrations can be caused by setting the system in motion initially and allowing it to move~~~~~. .

2.2.5 Natural Frequency: This is the property of the system and corresponds to the number of freeoscillations made by the system in unit time.

2.2.6 Forced Vibrations: Vibrations that are developed by externally applied exciting forces are called

forced vibrations. These vibrations occur at the frequency of the externally applied exciting force.

2.2.7 Forcing Frequency: This refers to the periodicity of the external forces which acts on the system

during forced vibrations. This is also termed as operating frequency.

2.2.8 Frequency Ratio: The ratio of the forcing frequency and natural frequency of the system is re-

ferred as frequency ratio.

2.2.9 Amplitude of Motion: The maximum displacement of a vibrating body from the mean position isamplitudeof motion. . ,

2.2.10 Time Period: Time taken to complete one cycle of vibration is known as time period.

2.2.11 Resonance: A system having n degrees of freedom has n natural frequencies. If the frequel}cyofexcitation coincides with anyone of the natural frequencies of the system, the condition of resonance

occurs. The amplitudes of motion are very excessive at resonance.

2.2.12 Damping: All vibration systems offer resistance to motion due to their own inherent properties.

This resistance is called damping force and it depends on the condition of vibration, material and typeof the system..If the force of damping is constant, it is t&med Coulomb damping. If the damping forceis proportional to the velocity, it is termed viscous damping. If the damping in a system is free from itsmaterial property and is contributed by the geometry of the system, it is called geometrical or radiation

damping.

2.3 HARMONIC MOTION

Harmonic motion is the simplest form of vibratory motion. It may be described mathematically by thefollowing equation:

Z = A sin (rot- 0) ...(2.1)

N

L T:2!!-r- Go)

Timq.t

'. .,

c

Fig. 2.3 : Quantities describing harmonic motion


23/500

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The Eq. (2.1) is plotted as function of time in Fig. 2.3. The various terms of this equation are asfollows:

Z = Displacement of the rotating mass at any time tA = Displacement amplitude from the mean position, sometimes referred as single amplitude. The

distance 2 A representsthe peak-to-peak displacement amplitude,sometimesreferred to as double

amplitude, and is the quantity most often measured from vibration records.

ro = Circular frequency in radians per unit time. Because the motion repeats itself after 21tradians,

the'frequency of oscillation in terms of cycles per unit time will be ro/21t.It is denoted byf

8 = Phase angle. It is required to specify the time relationship between two quantities having the

same frequency when their peak values ha'ving like sign do not occur simultaneously. In Eq.

(2.1) the phase angle is a reference to the time origin.

More commonly, the phase angle is used as a reference to another quantity having the same fre-

quency. For example, at some reference point in a harmonically vibrating system, the motion may be

expressedby

ZI = AI sin rot

Motion at any other point in the system might be expressed as

Z, = A, sin (rot-'e, )I I I1t ~ 8 ~ - 1t.

...(2.2)

...(2.3)

with

For positive values of 8 the motion at point i reaches its peak within one half cycle after the peak

motion occurs at point 1. The angle 8 is then called phase lag. For negative values of 8 the peak motion

at i occurs within one half cycle ahead of motion at 1, and 8 is called as phase lead.

The time period, T is given by1 21t

T=-=-f roThe velocity and acceleration of motion are obtained from the derivatives of Eq. (2.1.).

dZ .Velocity = - = Z = roA cos (rot- 8)dt

= roAsin (rot- 8 + ~ )2

d Z .. 2Acceleration = -r = Z = ro A sin (rot- 8)

dt

= ro2A (sin rot - e + 1t)

Equations (2.5) and (2.6) show that both velocity and acceleration are also harmonic and can be

represented by vectors roA and olA; which rotate at the same speed as A, i.e. ro rad/unit time. These,however, lead the displacement and acceleration vectors by 1tI2and 1trespectively. In Fig. 2.4 vectorrepresentation of harmonic displacement, velocity and acceleration is presented considering the dis-

placementas the referencequantity(8 = 0).

...(2.4)

...(2.5)

...(2.6)

, .J (.,. .~4",t-t

",C.. .,.,~;


24/500

Theory of Vibrations

N

z,z,z

..+'C~

E~v0a.UI

0

oN

...>-+'

v0

~>

c0-0....

c:,I

c:,I

v0

.et

Fig. 2.4: Vector representation of harmonic displacement. velocity and acceleration

17

TimtZ,t

Ti mtZ,t

Timcz,t

When two harmonic motions having little different frequencies are superimposed. a non harmonic

motion as shown in Fig. 2.5 occurs. It appears to be harmonic except for a gradual increase and decrease

in amplitude. The displacement of such a vibration is given by:

Z = AI sin (0011- 91) + A2 sin (0021- 92)

N D,

-

2A max2Am\n

./.,/.. .,/

-+'

Cc:,I

E,~

v0

a.III

c--- """'- ---,--- '-'"

.'J' ,.,

~T, b ~

" :' 3 ! j ,; I' ,: ' ,' "

. ~~ 'i; 'P1>1Flg;'2.5':Motion containi.ng a beat

...(2.7)

TimtZ (t)


25/500

;;" C," 'i'j{':-;,':::;;;~,.


The dashed curve (Fig. 2.5), representing the envelop of the vibration amplitudes oscillates at afrequency, called the beat frequency, which corresponds to the difference in the two source frequencies:

I 1


26/500

>-":, ,;;[; /, '1\", ", ;,., "',c,...'" "-": ,,' ,.'r:,/'; ~:.: "'1~F"",';. ,

Theory of Vibrations . 19

, ,

Figure 2.6 (b) shows the free body diagram offue mass m at allYinstant dunng the course~fvibra-'tions. The forces acting on the mass m are:

(i) Exciting force, F (t): It is the externally applied force that causes the motion of the system.(ii) Restoring force, F,.: It is the force exerted by the spring on the mass emutends to restore the mass

, to its originalposition.For a linear system,restoringforce is equiJ.'to K . Z, whereKis thespring constant and indicates the stiffness. This force always acts towards the equilibrium posi-tion of the system.

(iii) Damping force, Fi The damping force is considered directly proportional to the velocity and

given by C . Z where C is called the coefficient of viscous damping; this force always opposesthe motion.

In some problems in which the damping is not viscous, the concept of viscous damping is still

used by defining an equivalent viscous damping which is obtained so that the total the energydissipated per cycle is same as for the actual damping during a steady state of motion.

(iv) Inertia force, F.: It is due to the acceleration of the mass and is given by mZ.According to De-l,

-Alemberfs principle, a body which is not in static equilibrium by virtue of some accelerationwhich it possess, can be brought to static equilibrium by' introduculg on it an inertia force. Thisforce actsthrough the centre of gravity of the body in the direction opposite to that of accelera-tion. " '

The equilibrium of mass m gives

mZ + CZ + KZ = F (t)

which is the equation of motion of the system. ,

2.4.1 Undamped Free Vibrations. For undamped free vibrations, the damping force and the exciting

force are equal to zero. Therefore the'"equation of motion of the system becomes .."

mZ+ KZ = 0: '

, .::(2.11)

...(2.12a)

or..

( K)Z+mZ=O ...(2.12b)The solution of this equation can be obtained by substituting"

Z = A I cos con t+ Az sin cont

where AI and Az are both constant~ and conis undamped natural frequency.

Substitut ing Eq. (2.13) in Eq. (2.12), we get? ,

-(j)~ (AI cos (j)i + Az sin (j)ntj+(~)(AI ~os oont+ Az sin:oo~t) = 0"

~'co =:1: -, " n m,-,' , .

The values of constants A I and A2 are obtained by supstituting proper boundary conditions. We maynave the following two boundary conditions: ' "' '

'" . ~

(i) At time t = 0, displacement Z = Zo' and

(ii) At time 1 = '0, velocity Z= V0Substituting the first boundary condition in Eq. (2.13)

...(2.B)

"

,,'

...(2.14)or

Now,

. "/, ; "'"..',', "' Z:. """':.!'",I;'j,d",. ,',:'}.., :';"h' ,,',", , " :!':"'" '"',,' ,-"""".."Ar-r;"'O:iI'i),+.'nji;~:J}'i"..ql.d")Jit i..j}iJ'iI.J'!,';~"; >is:.:,,,, '

':,; 'z ,=: -:' AI" 00,; si~ cont + A2 C1)n'~os cont "

C.' ...(2.15 ):; j

, ...(2.16)


27/500

20Soil Dymunics & Machine Fo"ndations

Substituting the second boundary condition.in Eq. (2.16)V.

A =--2..2 ~ n ...(2.17)

Hence . Vo2 = 20 cos oont + - sin oont

. con...(2.18)

...(2.19)Now let.

and

20 = Az cos 9V

--2.. = A sin 9co Zn ...(2.20)

where

Substitution of Eqs. (2.19) and (2.20) into Eq. (2.18) yields

2 = Az cos (oont- 9)

9 = tan-I(~

)con20

...(2.21 )

...(2.22)

( )

2

2 VoAz = ,/20 + -

. con

The displacement of mass given by Eq. (2.21) can be represented graphically as shown inFig. 2.7. It may be noted that

...(2.23)

c+)

~ One cycle

Acceleration /.-0,

\ % ." y, '" 0'1'," /. 3

e\ 2~\ TI.r /2 "IT.~, 9 2lT +9 /

\ / / '\ /" / 0 '. 0 /~/ , V' ., / '- -'" "-- , -A

Z 0

0 isplacement "

+Az

:N..

oN

..

N Time,t

"1'/

velocity

(-)

Fig, 2.7 : Plot of displacement. velocity and acceleration for the free vibration of a mass-spring systemI>


28/500

'reory ilf Jl"l6iatiOns 21

At time t equal to Displacement Z is ..

0

8

Az cos 8

Az(J)n

1t +8L- 0

0)n

1I+8

0)

3-1t+82 .

-AZ

0(J)n

21t + 8

O)nAZ

It is evident from Fig. 2.7 that nature of foundation displacement is sinusoidal. The magnitude of

maximum displacement is Az. The time required for the motion to repeat itself is the period of vibration,T and is therefore given by. .

T = 21tO)n

...(2.24)

The natural frequency of oscillation, 1" is given by

J. =1- =~ =...!.. (Kn T 21t 21t v-;;

...(2.25)

Now mg W- =-=0

K K st .Where g = Acceleration due to gravity, 9.81 mIs2

W = Weight of mass m

st = staticdeflectionof the springTherefore

...(2.26)

- I rgIn - 21t Vfut

Eq. (2.27) shows that the natural frequency is a function of static deflection. The relation ofIn andOs!given by Eq. (2.27) gives a curve as shown in Fig. 2.8.

The nature of variation of the velocity and acceleration of the mass is also shown in Fig. 2.7.

...(2.27)

I,

- .

,.,....~ ~.~.n

I


29/500

22 Soil Dynamics & Machine Foundiuions . :~40

30

0-0 2 4 6

. 6stat (mm)

8 10

Fig. 2.8 : Relationship between natural frequency and static deflection

2.4.2 Free Vibrations With Viscous Damping. For damped free vibration system (i.e., the excitation

force Fo sin (J)t on the system is zero), the differential equation of motion can be written as

mZ + Cl + KZ = 0 ...(2.28)

where C is the damping constant or force per unit velocity. The solution of Eq. (2.28) may be written as'),.t . .

Z = A e ...(2.29)

where A and A are arbitrary constants. By substituting the value of Z given by Eq. (2.29) in Eq. (2.28),we get

m A A2it + C A AIt + K A it = 0

2

(C

)K

or A + ni A + m = 0By solving Eq. (2.30)

C FC )2 K. A,1,2 = - 2m :i: V~~) -;;

The completesolutionof Eq.(2.28) is givenby.

Z - A Alt A ' A2t '- le + 2e

The physical significance of this solution depends upon the relative magnitudes 'of(K/m), which determines whether the exponents are real or complex quantities.

...(2.30)

...(2.31 )

...(2.32)2

(C/2m) and

Case I : (~ )2 > K2m mThe roots AI and A2are real and negative. The motion of the system is not oscillatory but is an

exponential subsiden~~(Fig. 2. 9). Because.of the relatively large damping, so much energyis dissipated

'-..----

,....N 20:I:-

c.....

10


30/500

Theory of Vi!'rations ,23

by the damping force that there is sufficient kinetic energy left t~ carry the mass and pass the equilibrium

position. Physically this means a relatively large damping and the system is said to be over damped,

z 2C > 4 km ... -"

Tim(l,t

Fig. 2.9 : Free vibrations ofviscously overdamped system

Case 11 : (~ )2 = K2m m ,-The roots Al and Az are equal and negative. Since the equality must be fulfilled, the solution is

given by

Z = (AI.+ Az t) le = (AI + Az t) e-Ct/Zm ...(2,33)

In this case also, there is no vibratory motion. It is similar to oyer damped case except that it is

possible for the sign to change once as shown in Fig. 2010.This,case is of little importance in itself; itassumes greater significance as a measure of the damping capacity of the system. " '

z

c2=l"kmTime,t

.,

Fig. 2.to': Free vibrations of a vlscouslycritically damped system

(~ ) = K. C = C2m m' cThen Cc "=,2 ~Km". ...(2.35)

The system in this conditioonis known as ~ritically damped system anaC ~ is known as critical damp-

ing constant.' The ratio of the actual damping constant to the critical damping constant. is. defined as

damping ratio:

When ...(2.34)

'Now

C~=-

Cc

C - C Cc - C 2JK"m_c:'fK

2m - Cc . 2m - Cc' 2m - Cc'Vm

By substitutingthis valueof' 2: ' as ~(On in Eq. (2.31), ~~ ~et"

..

,..(2.36)

...(2.37)

.".~.....- ",


31/500

, -.. ".' r , . [.',


AI, 2 =(_;:!:~;2-1) COn...(2.38)

Case III : (~ )2 < K2m mThe roots Al and Al are complex and are given by .

AI,2 = [-;:!:i~I-;2 ]COn

The complete solution of Eq. (2.28) is given by

- (-~+j~I-~2 )O>i (-~-j~l-e )(J)"IZ - A I e + Az e

r:-:2 r ,or Z = e-~o>"tA j"I-~2 0>,,1+ A e-;~I-~- O),i'

the Eq. (2.41) can be written as I e z

...(2.39)

...(2.40)

...(2.41 )

Z = e-~O)",[Cl sin( (J)n~ t)+Cz cos( (J)1I~t)]...(2.42)

orZ = e-~O)II'[Cl sin(J)ndt+CZ COS(J)ndt] ...(2.43 )

where wild = (01/ ~ 1- ;z = Damped natural frequency.The motion of the system is oscillatory (Fig. 2.11) and the amplitude of vibration goes on decreasing

in an exponential fashion.

z2

C < 41 f,1Z e "

---L = -0> f,(t+Zn/o>"cI)Z2 e nZI 0> f,.21t/o)ncl- = enZz r:2ZI Znl;!"l-f,-- =eZz

ZI - 21t;

loge 22.- ~

...(2.44 a)

or ...(2.44 b)

or ...(2.44 c)

...(2.44 d)


32/500

~ }, ",inF1': j' /," " 't~.. ' .~t .o,~; ',.

Tlreory of Vtb",tiolU ,--:)

. . Natural logarithm of ratio of two successive peak amplitudes {i,e, log, (~)} is called as logarith-mk. ,decrement. .

1 Z\ r:-:2

or ~= 2x loge~ ' As for small valuesof~, V1- ~- :: 1 ...(2.44 e)tbus, damping of a system can be obtained from a free vibration record by knowing the successive

amplitudes which are one cycle apart. .'

If the damping is very small, it may be convenient to measure the differences in peak amplitudes fora number of cycles, say n.

In such a case, if Z" is the peak amplitudes of the n,h cycle, then

Zo Zl Z2 Zn-I 0- = - = - = . . . = - = e where~=2x ~Z\ Z2 ZJ Zn

Zo, =[

Zo] [

~] [

Z2

]..

[Z"-I

]= eno

Zn Z, Z2 Z) Z"Therefore,

or

Z1I 0 ..

~ = - oge Zn nZ1

I 0..}: = - oge.z~ 2xn n

Hence ...(2.441)

...(2.44 g)

Therefore, a system is

over damped if~ > 1;critically damped if ~ = 1 and

under damped if ~ < 1.

2.4.3 Forced Vibrations Of Single Degree Freedom Syst~m. In many cases of vibrations caused by

rotating parts of machines, th~ systems are subjected to periodic exciting forces. Let us consider the case

of a single degree freedom sys~.:mwhich is acted upon by a steady state sinusoidal exciting force having

magnitude F and frequency 0>(i.e. F(t) = Fosin rot). For this case the equation of motion (Eq. 2.11) canbe written as :

.. .

111Z + C Z + K Z = Fo Sin ro t ...(2.45)

Eq-;(2.45) is a linear, non-homogeneous, second order differential equation. The solution of this

equation consists of two parts namely (i) complementary function, and (ii) particular integral. The

complementary function is obtained by considering no forcing function. Therefore the equation of motionin this case will be :

.. .m Z, + C Z, + K Z, = 0 ...(2.46)

The solution of Eq. (2.46) has already been obtained in the previous st?ctioIland is given by,

ZI = e-O>/"(C\sinrondt+C2cosrondt) ...(2.47)

Here ZI represents the displacement of mass m at any instant t when vibrating without any forcingfunction. .

The particular integral is obtained by rewriting Eq. (2.45) as

m 2:2+ C 22 + K Z2 = Fo sin rot

where Z2 = displacement of mass m a~~nYinstant t when vibrating with forcing function.

~.;Y,

...(2.48)


33/500

8111;1'

"

26"--H,' " "

Soil Dynamics & Mac/line Foundations

; The,solution of Eq. (2A8).,\~ gi'{en by'. " ,,,', " . ~ ',", ,,' '" t,; "

, 22 = AI sin 00t + A2cos 00't

where AI and A2 are two, arbitrary constants.Substituting Eq. (2.49) in ~q. (2.48) ',' -

m (- At 002sin 00t- A2002cos (0t) + C (AI 00cos 00t- A2 ID;in 00t) + K (AI sin 00t+ A2cos 00t) ,,",,' ""'='Fosin'ro,t':. : , ,..'(2.50:

Considering .sine and Cosine functions in Eq. (2.50) separately, , ' ' , ,'.,2 " " ,; , , ' (,. ' ',' ", ,"~ .

(- m AI 00 + KAt - CA2 00)sin 00t = Fo sin 00t

(- m A2 002+ KA2 + CA! 00).I::osffi t,~ O. 'J"

From Eg. (2.51 a), .,

Al(~ - o}) - A2(~

and from Eg. (2,51 b)

A{~-W )+A2(~-w2) =0

Solving Egs, (2,52 a) and (2.52 b), we get

(K-moo2) FoAt-

- 2

(K - mm2) +C2m2

and A2 = - CmFo

" ". '


34/500

Theory of Vibrations. . ,'..,

~J'(,,~\'k

27

The complete solution is obtained by adding the compJimentary function and the particular integral.

Since the 'coriipliIne~tarYfti~l(!tioh:lsan'expJnenii~nf'decayin~ function,:iit'will die out'soon and themotion will be des~ribed by only the p~uticula:rmtegral(Fig. 2:i 1)'.:The syStemwill vibrate harmonically

,with the same frequency as the forcing and the pe~ ap11'1!tu4~.,is,g~venbyF. /K '.

Az = 0 " oi

. . ", ..., ~(l~ 1]2)2+(21]~)2 ",,"', ;;"

N.., ,~ '~','--' '

"I

+'

C

~

E~u

0a.UI ........

,.,..

Transi~nt , ,

0

~211"

(;) ~N

N

..+'C

~

E~u0a.UI

0

Time.t

. "

-,

'h ..'

:'1 i." ..;.

. . / ' -;. ..i , ,- #" "

'.'. -' ,.;" ~,: ,~~#"st'~c;!:t>~tatcz

4;i;"':~~.~.",;..' , :~~.::'---'- ~,'

"

"

N\

'I\" .

+'C~

Ec:.Iu0

a.III

Time,t

.,'. " .q , ','" .),

0< , " "

. "! ".' '.', ,', .

Co mpl~t~ solu tion

Fig. 2.12 Superpos,ition of transient and steady state vibrations

i, - ~ rr"'- "'-T--'1iiiiii[-'

...(2,58)

~-


35/500

" .". ,( '. ;:, "" ~>;;;


36/500


Differentiating Eq. (2.59) with respect to 11and equating to zero, it can be shown that resonance will

occur at a frequency ratio given by

11 = ~1-2~2

which is approximately equal to unity for small values of ~.

or ffind = ffin ~1-2~2

...(2.60 a)

...(2.60 b)

where ffind = Damped resonant frequency

30

~ =0-5

180

150


37/500


38/500

'.'; :~,(i:', '.'),' '


39/500


The Eq. (2.66) can be expressed in non-dimensional form as given below:

A~ 1"\2-

/ = ~ ...(2.68)(2mee m) (1-1"\2)2+(21"\1;)2

The value of A=/(2me elm) is plotted against frequency ratio 1"\in Fig. 2.16 a. The curves are similar

in shape to those in Fig. 2.13 except that these starts from origin. The variation of phase angle e with 11

is shown in Fig. 2.16 b. Differentiating Eq. (2.68) with respect to 11and equating to zero. it can be shown

that resonance will occur at a frequency ratio given by1

1"\=-

F-2e

...(2.69 a)

0011or ro =-

nd .JI=21;2 .

By substituting Eq. (2.69 a) in Eq. (2.68), we get

(

Az

)- l'

2meelm max - 21;~1-1;2

~ 2\ for small damping

...(2.69 b)

...(2.70)

...(2.71 )

2.5 VIBRATION ISOLATION

In case a machine is rigidly fastened to the foundation, the force will be transmitted directly to the

foundation and may cause objectionable vibrations. It is desirable to isolate the machine from the foun-

'dation through a suitably designed mounting system in such a way that the transmitted force is reduced.

For example, the inertial force developed in a reciprocating engine or unbalanced forces produced in any

other rotating machinery should be isolated from the foundation so that the adjoining structure is not set

into heavy vibrations. Another example may be the isolation of delicate instruments from their supports

which may be subjected to certain vibrations. In either case the effectiveness of isolation may be mea-

sured in terms of the force or motion transmitted to the foundation. The first type is known as force

isolation and the second type as motion isolation.

2.5.1 Force Isolation. Figure 2.17 s~ows a machine of mass m supported on the foundationby means ofan isolator having an equivalent stiffness K and damping coefficient C. The machine is excited with

unbalanced vertical force of magnitude 2 me eci sin 00t . The equation .ofmotion of the ~achine can bewritten as:

w~ere

... 2m Z + CZ + KZ = 2 me eoo sin 00t

The steady state motion of the mass of machine can be worked out as

22m eoo / K .

Z = r e 2 .sm(oot-8)

(1-'12). +(21"\1;)2

8 = Tan-I[

2'11;2]1-.1"\

...(2.72)

...(2.73)

...(2.74)


40/500

':;0:"


41/500

34Soil Dynamics & Machine. Foundations

Since the force 2 m e ol is the force which would be transmitted if springs were infinitely rigid, ae .

measure of the effectiveness of the i~olation mounting system is given by

. ' Ft ~1+(211~)2

IlT =2 = ~ ...(2.79)2 meem (1-112)2 + (211~)2

IlTis called the transmissibility of the system. A plot of IlTversus 11for different values of~is shownin Fig. 2.18. It will be noted from the figure that for any frequency ratio greater than 12, the forcetransmitted to the foundation will be less than the exciting force. However in this case, the presence of

damping reduces the effectiveness of the isolation system as the curves for damped case are above the

undamped ones for 11>12. A certain amount of damping, however, is essential to maintain stability

under transient conditions and to prevent excessive amplitudes when the vibrations pass through reso-

nance during the starting or stopping of the machine. Therefore, for the vibration isolation system to be

effective 11should be greater than 12.

4.5

4.0

1.0

. ~ =0

f =0.125

~ =0

~ =0.125

0 -0

~ =0.5

~ =1.0

~ =2.0

~ : 0.125'

I ~ =01.0 2.0 3.0

Frczquczncyr(:itio I 1'\.

Fig. 2.18: Transmissibility (J.1-r)versus freqeuncy ratio (Tt>

1- 3.0=


42/500

,f "c"' " f ','" ,,': ,r:,,


43/500

18E:.

36 SoU Dynamics & Machine Foundations

The effectiveness of the mounting system (transmissibility) is given by

- Zmax ~ ~1 + (2T\~)2

~T- -y; - ~(1-T\2)2+(2T\~)2

...(2.84)

Equation (2.84) is the same expression as Eq. (2.79) obtained earlier. Transmissibilityof such system

can also be studied from the response curves shown in fig. 2.18. It is again noted that for the vibration

isolation to be effective, it must be designed in such a way that T\> .fi.

2.5.3. Materials Used In Vibration Isolation. Materials used for vibration isolation are rubber, felt.

cork and metallic springs. The effectiveness of each depends on the operating conditions.

1.5.3.1. Rubber. Rubber is loaded in compression or in shear, the latter mode gives higher flexibility.

With loading greater than about 0.6 N per sq mm, it undergoes much faster deterioration. Its damping

and stiffness properties vary widely with applied load, temperature, shape factor, excitation frequency

and the amplitude of vibration. The maximum temperature upto which rubber can be used satisfactorily

is about 65c. It must not be used in presence of oil which attacks rubber. It is found very s' ".,ble for high

frequency vibrations.

2.5.3.2.Felt. Felt is used in compressfun only and is capable of taking extremely high loads. It has very

high damping and so is suitable in the range of low frequency ratio. It is mainly used in conjunction with

metallic springs to reduce noise transmission.

2.5.3.3. Cork. Cork is very useful for accoustic isolation and is also used in small pads placed under-

neath a large concrete block. For satisfactory working it must be loaded from 10 to 25 N/sq mm. It is not

affected by oil products or moderate temperature changes. However, its properties change with the fre-

quency of excitation.

1.5.3.4. Metallic springs. Metallic springs are not affected by the operating conditions or the environ-

ments. They are quite consistent in their behaviour and can be accurately designed for any desired

conditions. They have high sound transmissibility which can be reduced by loading felt in conjunction -

with it. It has negligible damping and so is suitable for working in the range of high frequency ratio.

2.6 THEORY OF VIBRATION MEASURING INSTRUMENTS

The purpose of a vibrationmeasuring instrument is to give an output signal which represents, as closely

as possible, the vibration phenomenon. This phenomenon may be displaceme~t, velocity or accelerationof the vibrating system and accordingly the instrument which reproduces signals proportional to theseare called vibrometers. velometers or accelerometers.

There are essentially two basic systems of vibration measurement. One method is known as the

directly connected system in which motions can be measured from a reference surface which is fixed.

More often such a referencesurface is not available. The second system, known as "Seismic system" doesnot require a fixed reference surface and therefore is commonly used for vibration measurement.

Figure 2.20 shows a Vibration measuring instrument which is used to measure anyof the vibration

phenomena. It consists of a frame in which the mass ~ is supported by means of a spring K and dashpotC. The frame is mounted on a vibrating body and vibrates al~ng with it. The system reduces to a springmass dashpot system having base on support excitation as discussed in Art. 2.5.2 illustrating motionisolation.


44/500

. .(~

Thtory of VibratiOns 37

zm

cK

y = Yo Sin '->t

Fig. 2.20 : Vibration measuring instrument

Let the surface S of the structure be vibrating harmonically with an unknown amplitude Y0 and anunknown frequency (0. The output of the instrument will depend upon the relative motion between the

mass and the structure, since it is this relative motion which is detected and amplified. let 2 be the

absolute displacement of the mass, then the output of the instrument will be proportional to X = 2 - Y.The equation of motion of the system can be written as

m Z + C (Z - Y) + K (2 - Y) = 0Subtracting m Yfrom both sides,

... .. 2m X + C X + K X = - m Y = m Y0(0 sin (0 t

The solution can be written as

.. .( 2. 8S) ,

..:.(2.86)

where

2

X = ~ TJ Yo sin (0 t- e)(1- TJ2)2+ (2TJ~)2

(0.11 = - = frequency ratio

(On

~ = damping ,ratio

1

(2 TJ~

)and e = tan- 1- TJ2

Equation(2.S7)ca~ be rewrittenas:. ,

X = 1)2.J! Y0 sin (0 t - 8)

...(2.S7)

(2.,SS)

where;1 ">;(1..;>' ,

J! = ~1- TJ2)2+ (2T1~l

ill


45/500

38. Soil Dynamics & Machine Foundlltions- -." '-" .' . -. .

2.6.1. Displacement Pickup. The instrument will read the displacement of the structure directly if

1121.1= I and 8 = O.The variation o{Tl~ with-~'aiid'~-isshown in Fig-.2.21. The variation of8 with 1'\is already given in Fig. 2.14. It is seen"'tnatwneifff is" large, 1'\21.1is approximately equal to 1 and 8 is

approximately equal to 180. Therefore to design a displacement pickup, 1'\should be large which means

that the natural frequency of the instrument itself'shou~d be low compared to the frequency to be mea-

sured. Or in other words, the instrument should have a soft spring and heavy mass. The instrument is

sensitive, flimsy and can be used in a weak vibration environment. The instrument can not be used for

measurement of strong vibrations.

3.0

,- -t

I \I \0

I

I

. .

-- . -

- -"

2 0,

1 .0

0 -0 1.0 2.0 3.0 4.0 5.0

FrequClncy ratio, '1.

Fig. 2.21 : Response of a vibration measuring instrument to a vibrating base

2.6.2. Acceleration Pickup (Accelerometer). Equation (2.88) can be rewritten as. I 2

X = 2 1.1Yoro sin (rot- 8)(J,)n . '

The output of the instrument will be proportional to the acceleration of the structure if J.1is constant.

Figure 2.13 shows the variation of J.1with 1'\and;. It is seen that J.1is approximately equal to unity for

small values of 1'\.Therefore to design an acceleration pickt!p, 11should be small which means that th~'

...(2.89)

" ",.~'n'7""~:"'"""'"

.'1::' 'lr~-r\f

........--.


46/500


47/500

40

(a) Four storeyedframe

(b) Idealisation (c) First mode

-- .


(d) Second mode (e) Third mode (t) Fourth mode

Fig. 2.22 : A four storeyed frame with mode shapes

Figure 2.22 a shows the frame work of a four storeyed. building. It is usual to lump the masses at the

floor levels and the lumped mass has a value corresponding to weight of the floor, part of the supporting

system (columns) above and below the floor and effective live load. The restoring forces are provided by

the supporting systems. Figure 2.22b shows such an idealization and it gives a four degrees of freedom

system. In free vibration a system having four degrees of freedom has four natural frequencies and the

vibration of the any point in the system, in general, is a combination of four harmonics of these four

natural frequencies respectively. Under certain conditions, any point in the system may execute har-

monic vibrations at any of the four natural frequencies, and these are known as the principal modes of

vibration. Figure 2.22


48/500

I .

~heory of Vibrations

~!'\Wm

Z2

Jz~-

Fig. 2.23 : Free vibration of a two degrees freedom system

For nontrivial solutions of oon in Eqs. (2.96) and (2.97),

2Kt +K2 -mt ron -K2

21 = 0- K2 K2 + K3 - ~ ron

.- f2 9;!)

00: _[

Kt + K2 + K2 + K3

]O)~+ K) K2 + K2 K3 + K3 K) -= 0

m) ~ ml~

Equation (2.99) is quadratic in ro2, and the roots of this equation are:n

ro~= .!.

[(K) +K2:t- K2 ~K3

)+

{(

KI +K2 K2 +K3

)2 + 4 K~

}

1/2

]2 m) "'2 m) ~ mj ~

...(2.100)

From Eq. (2.100), two valuesofnatura!~!!e9~~ncies oon)and oon2can be obtained. con)is correspond-ing to the fIrst mode and COn2is of the second mode.

or

. ,

...(2.99)

- 11iiii


49/500

42

..,

Soil Dynamics & Machine Foundatiofls

The general equation of motion of the two masses can now be written as

Z = A ( I) sin (0 t + A (2) sin ( 0 tI I nl I n2

and Z = A(I) s in (0 t + A(2) s in ( 0 t2 2 n I 2 n2

The superscripts in A represent the mode.

The relative values of amplitudes AI and A2 for the two modes can be obtained using Eqs. (2.96) and(2.97). Thus

(0 2Al - K2 - K2 +KJ -"'2 ffinJ

(i)- 2- K A2 KI+K2-mlffinl 2

(2)' 2A I - K2 - K2 + KJ - "'2 ffin2

(2) - 2 - K A2 KJ + K2 - m) ffin2 2

2.7.1.2. Undampedforced vibrations. Consider the system shown in Fig. 2.24 with excitation force Ftsin (0 t acting on mass ml. In this case, equations of motion will be:

ml Zt + Kt Zt + K2 (Zl - Z2) = Fo sin (0t

1n2 Z2 + KJ Z2 + K 2 ( Z2 - Z\) = 0

F0 sin G.)tl

...(2.101)

...(2.102)

...(2.103j

...(2.10{

(2.105)

...(2.106

21

22

.

Fig. 2.24 : Forced vibration of a two degrees freedom system

.. ',",y


50/500

leory of Vibrations 43

For steady state, the solutions will be as

21 = Al sin 00t

2z = Azsin 00t

Substituting Eqs. (2.107) and (2.108) in Eqs. (2.105) and (2.106), we getZ(KI + Kz - ml 00) AI - KzAz = Fo

z- Kz AI + (Kz + K3- mz00) Az = 0

Solving for AI and Az from the above two equations, we getz

(Kz +K3 -rnz co ) FoA =I

[

4

(

KI+Kz Kz+K3

)

z KIKz+KzK3+K3KI

]ml rnz co - + co +

ml rnz mlrnz

K3Fo -A =z

[

4

(

KI+Kz Kz+K3

)

Z' KIKz+KzK3+K3KI

mlrnz co - + co +ml rnz m\rnz

The above t\VOequations give steady state amplitude of vibration of the ~wo masses respectively, as

a function of00. The denominator of the two equations is same. It may be noted that:

(i) The expression inside the bracket of the denominator of Eqs. (2.1110) and (2.111b) IS of the

same type as the expression of natural frequency given by Eq. (2,99). Therefore at 00= oolll andCl) = Cl)nZ values of A laud Az will be infinite as the denominator will become zero.

(ii) The numerator of the expression for Al becomes zero when

/K2 +KJ)Cl) = rnz ...(2.112)

Thus it makes the mass ml motionless at this frequency. No such stationary condition exists for

0 mass ml' The fact that the mass which is being excited can have zero amplitude of vibration under

certain conditions by coupling it to another spring -mass system forms the principle of dynamic vibrationabsorbers which will be discussed in Art. 2.8. '

...(2.107)

...(2.108)

...(2.109)

...(2.110)

...(2.111 a)

and ...(2.11Ib)

2.7.2. System With n Degrees of Freedom.

2.7.2.1. Undamped free vibrations: Consider a system shown in Fig. 2.25 having n-degree of freedom.

If Z\' 2z, Z3 ... 2n are the displacements of the respective masses at any instant, then equations of motionare:

rn, 2( + K( Z\ + Kz ( ZI - Zz) = 0

mz2z - Kz(Z( - 2z) + K3 (2z - 23) = 0

...(2.113)

...(2.114)

m3 23 - K3(2z - 23) + K4(23 - 24) ;: 0 ...(2.115)

'. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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

mn 2n - Kn (2n - I -'- 2n) = 0 ...(2.116)

".,'~'.,-..~,,-",'" "' . ,c-"" '; '; . "', ..; . :--';co,,---, - ... n


51/500

181:.J

44Soil Dynamics & Machine Foundations

Z1

Z2

Z3

Kn-1

Zn-1

Zn

Fig. 2.25: Undamped free vibrations of a multi-degree freedom system


52/500

"'.'~..2,,~,.."t;,~.>, :"""',;,n rA';".!n~";'F."'" "'.,.,


The solution of Eqs. (2.113) to (2.116) will be of the follow:"'~ IO':n:

ZI = Al sin cont

Z2 = A2 sin contZ) = AJ sin cont

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

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

Zn =An sin cont

Substitution of Eqs. (2.117) to (2.120) into Eqs. (2.113) to (2.116), yields:

[(KI+K2)-mIID~] AI-K2~ =0

-K2A1 + [(K2+K))-~ID~] A2-KJAJ =0

- KJ Az + [(KJ + K4) - mJID~] A) - K4 A4 = 0

, . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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

- Kn An - I + (Kn -mnID~) An = 0

For nontrivial solutions of oonin Eqs. (2.121) to (2.124),

[ (KI + Kz) - mlID;]-K2

0

-Kz

[(Kz+KJ)-~ID;] '"

-K)

0

0

0

0

0

0 =0

0 2'" -Kn (Kn-mnIDn)0

- ~

45

...(2.117)

...(2.118)

...(2.119)

...(2.120)

...(2.121)

...(2.122)

...(2.123)

...(2.124)

(2.125) .

Equations (2.125) is of nthdegree in CI);and therefore gives n values ofcon correspondingto n natural

frequencies. The mode shapes can be obtained from Eq. (2.121) to (2.124) by using, at one time, one of

the various values of conas obt1incd from Eq. (2.125).

When the numht.'Tvi degreeS of freedom exceeds three, the problem of forming the frequency equa-

tion and s01";~jgit for determ41ation of frequencies and mode sh


53/500

If

46Soil Dynamics & Machine Foundations

..m1 Z1

m2 Zz

m3 Z3

.

K.J

m.11-

~_1. Kj-1(Zj-Zi-1)

K.11-

m.I

mi+1

Fig. 2.26 : An idealised multi-degree freedom system

Inertia force at a level below mass mi - I;-1 ..= ".

Im. Z.L...J= } }

Spring force at that level corresponding to the difference of adjoining masses= K. I (Z. - z. 1)1- 1 1-Equating Eqs. (2.126) and (2.127) .

i-I ..

Lj=lmjZi = Ki-I (Zi- Zi-I)

Putting Zi = Ai sin (()t in Eq. (2.128), we get

...(2.126)

...(2.127)

...(2.128)

I'i~'t m i (- Ai U)~ sin U)n t) = Ki - I (Ai sin (Unt - Ai - I sin (Unt)

.'

'2

U)n "i-I AAi = Ai-I - K ...j=lmj ji-I

Equation (2.129) gives a relationship b~tween any two succ~sive amplitudes. Starting with any

arbitraryvalueof AI' amplitudeof all othermassescanbe deterinined.A plot ofAn+ 1 versus (0~ would

have the shape as shown in Fig. 2.27. Finally An + I should worked out to zero' ~ue to fIXityat the base.

The intersection of the curve with (0~ axis would give various val~~s.pfQ);.~ode $ape.can be obtained.. .. -. .

by substituting the correct value of (O~in Eq. (2.129).

...(2.129)or

iI

m1.J

K1

Iml

K2


54/500


1.0

~

t

' ..-J +J c:~~1:

0

(..)2n.

-1.0

2wn1

2""nz

---Fig.-~7: Residual a~a flinction of frequency in Holzer method

2.7.2.2. Forced vibration. Let an undamped n degree of freedom system be subjectedto forced vibration,

and Fj(t) represents the for~e on mass mr The equation of motion for the mass mj will ben

m. Z. + I K.. Z. = F. (t)I I =1 IJ ) ,

where i = 1,2,3, , n

...(2.130)

The amplitude of vibration of a mass is the algebraic sum of the amplitudes of vibration in various

modes. The individual modal response would be some fraction of the total response with the sum of

fractions being equal to unity. If the factors by which the modes of vibration are multiplied are repre-

sented by the coordinates d, then for mass mj'

- (1) (2) (r) (n)Z. - A. d 1 + A. d2 + ... + A. d +... + A. d'" 1 r ,n

Equation (2.131) can be written as

...(2.131 )

n

Z. = I A~r) d-' r=l ' r

Substituting Eq. (2.132) in Eq. (2.130)

...(2.132)

n (r) .. n n (r)-Im. A. d + I I K.. A. d - F. (t)I ,r 'I 1 r ,r=1 r=1 j=l . .

Under free vibrations, it can be shown

n (r) - 2 (r)~ Kij Aj - oonrmj Aj1=1

...(2.133)

...(2.134)

Substituting Eq. (2.134) in Eq. (2.133), we get

n (r) .. n 2 (r) -ImjAj dr + I oonrmjAj dr - Fj(t)

r=1 r=l ..(2.135)

or n (r)'. 2 -ImjAj (dr+oonr.dr) -Fj(t)r=l

...(2.136)

l' -.-'T .~ ""}'- -")J'!!I~'


55/500

~,


Since the left hand side is a summation involving different modes of vibration, the right hand side

should also be expressed as a summation of equivalent force contribution in corresponding modes.

Let F; (t) be expanded as:

Fj (t) = i mj A~r)fr (t)r=1where fr(t) is the modal force and given by

...(2.137 a) -"J

n

Ili (t) .A~r)

fr(t) = i~1 Z

Lm[(A~r)];=1

Substituting Eq. (2.137 a) in Eq. (2.136), we get

.. zdr + O}nrd,.=fr (t)

Equation (2.138) is a single degree freedom equation and its solution can be written as

1 I

dr = - Jfr Ct) sinO}nr(t-1:) dt where 0 < 1:< I0}nr 0

It is observed that the co-ordinate d, uncouples the n degree of freedom system into n systems of

single degree of freedom. The d's are termed as normal co-ordinates and this approach is known as

normal mode theory. Therefore the total solution is expressed as a sum of contribution of individualmodes.

(2.137 b)

(2.138)

..(2.139)

2.8 UNDAMPED DYNAMIC VIBRATION ABSORBER

A system on which a steady oscillatory force is acting may vibrate excessively, especially when close to

resonance. Such excessive vibrations can be eliminated by coupling a properly designed spring masssytem to the main system. This forms the principle of undamped dynamic vibration absorber where the

excitation is finally transmitted to the auxiliary system, bringing the main system to rest.

Let the combination of K and M be the schematic representation of the main system under consid-

eration with the force F0 sin CJ}tacting on it. A spring - mass (auxiliary) absorber system is attached to themain system as shown in Fig. 2.28. The equations of motion of the complete system can be written as:

MZ1 + KZI + Ko (ZI-ZZ) = Fa sin rot ...(2.140)

...(2.141)moZZ+Ka(ZI-ZZ) =0

The forced vibration solution will be of the form

ZI = Al sin rot

~ = Azsin rotSubstitution of Eqs. (2.142) and (2.143) in Eqs. (2.140) and (2.141) yields

Al (-M0}2 Ka)-KaAZ = Fa

-KnAI + Az (-mnO}Z+ Kn) = 0

...(2.142)

...(2.143)

...(2.144)

(2.145)


56/500


Subtituting:

Z2

ma

Absor ber

syst


57/500

50 SoU Dynamics & Machine Foundlltions

~~~

( ro')( ~..~' )

K

1- OO~a 1+ i -oo~

-~lithe natural frequency oonaof the absorber is chosen equal to 00 i.e. frequency of the excitatipn force,it is evident from Eq. (2.148) that Al = 0 indicating that the main mass does not vibrate at all. FurtherEq. (2.149) gives

...(2.149)

Az - -KZst - Ka

or Az Ka = - K Zst ...(2.150)

Thus the absorber system vibrate in such a way that its spring force at all instmts is equal and

opposite to F0 sin 00 t. Hence, there is no net force acting on main mass M and the same therefore doesnot vibrate.

The addition of a vibration absorber to a main system is not much meaningful unless the mainsystem is operating at resonance or at least near it. Under these conditions, 00= oon'But for the absorberto be effective, 00should be equal to 00 .na

Therefore, for the effectiveness of the absorber at the operating frequency corresponding to the natu-

ral frequency of the main system alone, we have

oroona =,oon ...(2.151 a)

Ko = Kma M

...(2.151 b)

K mor -!L=-!!..=Ii ...(2.151 e)

K M t"'m

When the condition enumerated in Eqs. (2.151) is fulfilled, the absorber is known as a tunedabsorber. .

For a tuned absorber, Eqs. (2.148) and (2.149) become:

(

002

J

1- -

~,: ~ ( 00') ( OO~a00' J1- --y- 1+Jlm- --y- - Jlm

OOna OOna

...(2.152)

~: = ( 00') ( I 00' )1- --y- 1+Jlm- --y- - Jlm

OO"a OOna

The denominators of Eqs. (2.152) and (2.153) are identical. At a value of 00when these denomina-

tors are zero the two masses have infinite amplitudes of vibration. Let when-00= oonl'the denominators

becomes zero. For this condition the expression for the denominators can be written as

..(2.153)

- .. .~ ..., ~- .-.--~.-


58/500


(

OOnt

)4-


59/500

52

!IIZ


if the variation of the exciting frequency is such that the operating point shifts near one of the new

resonant points, then amplitudes will be excessive. Thus depending upon the variation of the exciting

frequencies the spread between the two resonant frequencies has to be decided to remain reasonably away

from the resonant points. After deciding the spread between the resonant frequencies, a proper value of

!J.mcan be chosen from the curve of Fig. 2.29. Undamped dynamic vibration absorbers are not suitable for

varying forcing frequency excitation. To make the vibration absorber effective over an extended range of

frequencies of the disturbing force, it is advantageous to introduce a damping device in the absorber

system. Such an absorber system is called a damped dynamic vibration absorber.

8

6

...


60/500


t ILLUSTRATIVE EXAMPLEs!

Example 2.1The motion ofa particle is representedby the equationz = 20 sin rot. Show the relative positions andmagnitudes ofthe displacement, velocity and acceleration vectors at time t = 0, and ro= 2.0 rad/s and0.5 rad/s.

Solution:Z = 20 sin rot

Z = 20 ro cos rot = 20 ro sin lrot+ ~ )- 2 2Z =- 20 ro sin rot = 20 ro sin (w t + 1t)

The magnitudes of displacement, velocity and acceleration vectors are 10, 10 ro and 10 ro2 respec-

tively. The phase difference is such that the velocity vector leads the displacement vector by 1t/2 and the

acceleration vector leads the velocity vector by another 1t/2. Figures 2.31 a and 2.31 b show the three

vectors for ro= 2.0 and O.?Orad/s respectively. I 20 (V el.)

40 (Accln.)

10(oispl.)

(a) G..)= 2.0 rod/sec

s(Vel.)

. ~/z ".2.S(Acc!n.) 10(OlspL)

( b) CV = 0.5 rod I s e c

Fig. 2.31 : Vector diagram (Example 2.1)

21t 21tTime period = - = - = 1ts

ill 2for ro= 2.0 rad/s

21t 21tTime perIod = - = - = 41ts

ill (0.5)for ro = 0.5 rad/s

Example 2.2A bodyperforms,simultaneously,the motions

Zl (mm) = 20 sin 8.0 t

Z2 (mm) = 21 sin 8.5 t

Determine the maximwn an~ minimum )~1itude of the combuled motion, and the time period ofthe

periodic motion.

~~tm~f~dJ "T'f:ifrir" """"-";"~'.'.


61/500

54 Soil Dynamics & Machine 'Foundations

Solution:

Z = 21 + 20 = 41 mmmax

Z . = 21 - 20 = 1 mmmm

The beat frequency is given by

8.5-8.0 0.5

1= 21t = 21t = 0.0795 H z, a nd

T = 2.. - 21tI - 0.5 = 41t = 12.57 s

Example 2.3

A mass of 20 kg when suspended from a spring, causes a static deflection of 20 mm. Find the natural

frequency of the system.

Solution:

Stiffness of the spring, K = W~\t

20 x 9.81 ::::104 N/mK = 20xl0-J

1[KNatural frequency,In = 21t V;

~ 1 ~1O421t 20 = 3.6 Hz.

KZ

Example 2.4

For the system shown in Fig. 2.32, determine the natural

frequency of the system if

K1

Kt = 1000 N/m

Kz = 500 N/m

KJ = 2000 N/m

K4 = Ks = 750 N/m

Mass of the body = 5 kg

Solution:

Let Ket and Ke2represent respectively the effective stiffnesses

of the top three springs and the lower two springs, then

K3

1 1 1 1- = -+-+-Kel Kt Kz KJ

K4, KS

=~+~+~- .1000 500 2000-0.0035 ". 'Fig. 2.32:"MuHpriags system

~.....- -~_'".,-~._........._-_....-


62/500

.- ..... ...

Theory of Vibratimrs 55

Kel = 285.7 N/m

Ke2 = K4 + Ks = 750 + 750 = 1500N/m

Now Kel and Ke2are two springs in parallel, therefore effective stiffness,

Ke = Kel + Ke2:;:: 285.7 + 1500 = 1785.7 N/m

f. ~ ~. /K = ~~1785.7 = 3.0 Hzn 21tV;; 21t 5.0Example 2.5

A vibrating system consists of a mass of 5 kg, a spring stiffnessof 5 N/mm and a dashpot with a damping

coefficient of 0.1 N-s/m. Determine (i) damping ratio and (ii) logarithmic decrement.

Solution:

(i) Cc = 2 ~km = 2J5x 10-3 x 5 = 0.319 N-s/m

J: C 0.1

~=C=0.319=0.313c

(in 27t~ - 27t.x0.313 = 2.07Lograthimic decrement = ~ 1- ~2 - ~ 1- 0.3132

, ,

Zlog ::,..l = 2.07

eZ 2

~ = 7.92Z2

Therefore the free amplitude in the next cycle decreases by 7.92 times.

Example 2.6

A mass attached to a spring of stiffness of 5 N/mm has a viscous damping device. When the mass wasdisplaced and released, the period of vibration was found to be 2.0 s, and the ratio 01 the consecutive

amplitudes was 10/3. Determine the amplitude and phase angle when a force F = 3 sin 4 t acts on the

system. The unit of the force is Newton. .

Solution:

i.e.

or,

(ii)

. 27t~ ZI 10 .

1- ~2 = loge Z2 = loge 3 = 1.2~ = 0.195

TII = 2.0 S21t 21t

(lJ n = T = '2 = 3.14 radls(lJ = 4.0 rad/s

T1 ='~'= 4.0 = 1.273. ID 3.14n

.' i.:' F 3.0, Fo = 3.0N;AsI= -9..= - = 0.6mm.

. '.' .,', .;..",fi..,;'! "."K. 5.0 t:,:;


63/500

56 Soil Dynamics & Machine' Foundations

From Eq. (2.58),A

Az = ~ 2 t. 2; Asl = Static Deflection

;

(1-11) +(2~11). " . . , '

= '. 0.6 , ~Q.755llll11

~(1-1.2732)2 +(2 x.0.195 x 1.273)2

T-I

(

211~

)T

-1'

(

2 x 1.273 x 0.195

)e = an --r = an 2 = 141.41-11 1-1.273Example 2.7

Show that, in frequency - dependent excitation the damping factor~is given by the followingexpres-S1On:

): -.: ..!.

(

12 ~ 11

J~ - 2 2/n

Where 11 and 12 frequencies at which the amplitUdeis 1/.J2times the peak amplitude.Solution:

In a forced vibration test, the system is excited with constant force of excitation and varying frequencies.

A response curve as shown in Fig. 2.33 is obtained.

0.09

0.05

Amox = 0.084

0.08

E

E 0.07"C:I

"'0

::J-

c. 0.06E

et

0.0410

. . n, ,

Fig. 2.33: Determination of viscous damping in forced vibrations by Bandwidth method

-.------

II

II I

I I IIII I

f1 I fn Ifl

14 18 22 24

F r (Zq u (Zn c y .,0f (Zxci to t ion, Hz


64/500

"

Th~o,.,ofv~,!s57

At resonance, 11= 1 ana A~I Zst = 1/2~(for small values of ~). If the frequency ratio is T\whenamplitude of motion is 1/..[i times the peak amplitude, then fr~m Eq. 2.59, we get

I 1 1

.J2' 2~ = 4(1 :"112)2+4~2 ~2

or 114-2112(1 -2~2) + (1- 8~J) = 0

or 11~,2= ~[2(1-2~2):t~4(1-2~2)2_4(1-8~2)]

;, (1 - 2~2):f:2~~1+~2

11~-11i = 4~~1-~2 = 4~

1

-2

= Il- 112= (

12- 11) (

12+I.

)112 111 In2 In In

(

I - J;

)

1 + f= 2 2 . since 2 . =2In In

Now [for small values of~]

Also

~ = !(Iz- 11

)2 In

This methodfor determiningviscousdamping is knownas the band width method.

Example 2.8 .

A machine of mass 100 kg is supported on springs of total stiffness of 784 N/mm. The machine produces

an unbalanced disturbing force of 392 N at a speed 50 c/s. Assuming a damping factor of 0.20, determine(i) the amplitude of motion due to unbalance,

(ii) the transmissibility, and

(iii) the transmitted force.

Therefore

'Solution:

( i), 184 x 103

(J)II = ~KI m = ..'/ " ==87.7 rad/s, V 100

, 00= 21t x 50 = 314 rad/s

Now

00 - ,314 = 3.58,'n =;., ~ 87.7" 'UJln; ,

, Fi) '- 392 '= 0.5 'innl

" ~st ~ ~ - ,;784 ,

, , ' ;. : :'?it .'. - 2

.Az ==4(~.~2l'+(2T\~)

-

= ,0.5

'~(1-'3582)2 +(2 x 3.58 x 0.2)2 = 0.042 mm

" " .'

~'\, ",', ',','. ':;' cc:

, .

..- - - .~ "'-; ~~~,~"F.---


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58 Soil Dynamics & Machine Foulldations

~1+(211~)2

(ii) Transmissibility JlT = ~(1-n2)2 +(211~)2

- ~1+(2X3.58XO.2)2

- ~(1-3.582)2 +(2 x 3.58 x 0.2)2

==0.1467

(hi) Force transmitted = 392 x 0.1467 = 57.5 N.

Example 2.9

The rotor of a motor having mass 2 kg was running at a constant speed of 30 c/s with an eccentricity of

160 mm. The motor was mounted on an isolatorwith damping factor of 0.25. Determine the stiffness ofthe isolator spring such that 15% of the unbalanced force is transmitted to the foundation.Also ~eterminethe magnitude of the transmitted force. '.

Solution:

(i) Maximum force generated by the motOi

= 2 me eo:? = 7. x 2.0 x 0.16 x (21tx 30/ = 22716 N

= 22.72 kN

:ii)Force transmitted

Jl = = 0 15T unbalancedforce .

. ~1 +4112~2 .

I.e. ~ . = 0.15(1- 112)2+ (211~)2

or 1 + 4112x (0.25)2 = (0.15)2 [(1 -r 112)2+ (211 x 0.25)2]

or 114 - 12.84112 - 43.44 = 0

It gives(0

11= 3.95 i.e. ;- = 3.95"

(0 601t

(0" = .JK/ m = 3.95 = 3.95 = 47.7 rad/s

K = m (47.7)2 =:=2.0 x (47.7)2 = 4639 N/m

(iii) Force transmitted to the foundation

Therefore

= 0.15 x 22.72 = 3.4 kN.

Example 2.10

A seismic instrument with a natural frequency of 6Hz is used tomeasure the vibration of a machinerunning at 12.9rpm. The in~trumentgives the reading for th~r~lative displacement of the seismic mass

as 0.05 mm. Determine the amplitudes of displacement: velo'city and acceleration of the vibrating ma-

chine. Neglect damping.


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~;,:,~ .;,~

'/reory 01 v.ibrations59

';olution :

(i) CJJ = 6 Hz = 37.7 rad/sn120 x 21t

(J) = 120 rpm = =' 12 57 rad/s60 .

'1 = 12.57 = 0.33337.7

1~ = - for~= 0

1- ,,2 ,

1= = 1.125

1- (0.333)2

(ii) For displacement pickup, Eq. (2.88) gives2

X=,,~yo0.05 = (0.333)2 x 1.125 x Yo

or Yo = 0.40 mm

(iii) For velocity pickup, Eqo (2.91) gives'

or

1X = - 11~ (Y 00)

00 0n

0.333 x 1.125x (Yo 00)0.05 = (37.7)

or o. 0 (Y 000) = velocity = 5.03 ,mm/s

(iv) For acceleration pickup, Eq. (2.89) gives

X ='4 (YOOO2), OOn ' ,

I.e,

0.05 = 1.125 (Y 002)(37.7)2 0

(Y() 002) = Acceleration = (37.7)2 x 0.05 = 63.17 mm/S21.125

or

Example 2.11

Determine the natural frequencies and mode shapes of the system represented by a mathematical mod~1

shown in Fig. 2.34 a.,/

. , co:" '1,,';

, - ~: :: ." 'O '~," "" "" , , ~>=j "~


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, i

J

60 Soil Dynamics & Machi'Je FOUlrilatiollsj

+1

+ 1

(a) Two degrees freedom system (b) First mode (c) Second mode

Fig. 2.34 :Two degrees freedom system with mode shapes

Solution:

(i) The system shown in Fig. 2.34a is a two degree freedom system. The solution of such a system

has already been described in Art. 2.7. '

(ii) The two natural frequencies of the system can be obtained using Eq. (2.100) by putting KI = K,

Kz = 2 K and KJ = K, and m I = m2 = m. By doing this, we get

(02 = .!.

[(

3K + 3K)

_

{

4 ~ (2 K)2

}

\l2

)

= KIII 2 m m ,m2 m

(02 = .!.r-l6K + 4K

]= 5.K

112 2 m, m , m,

Hence, COni= .JK/ m and CiJ~2?:, [sK/.m , " .,' '

(iii) The relative values ofamplitudes Al and ~ 'for the two modes can be obtainedusing Eqs.(2.103) and (2.104). . . '

,}~~r .

,,"-;, ,:;. ..

' '-'--'

.~."


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eory of Vtbtalions_/ 6i

A (1) K . 2K -1-= 2 = =+1A(l) K + K -m o:l K+2K-m xK/m2 1 2 1 nl

A~2) 2K

A (2) = K + 2 K - m x 5 K / m = - 12. The mode shapes are shown in Fig. 2.34 band 2.34 c.

B:xample2.12

Determinethe natural frequencies and mode shapes of the system represented by the mathematical moqelshown in Fie. 2.35 a.

0.761

1.0

(a) Three degree freedom system (b) First mode (c) Second mode (d) Third mode

Fig. 2.35 : Three degrees freedom system with mod-e shapesSolution:

(i) Equations of motion for the three masses can be written as

m 21 + K 21 + 2 K (21 - 22) = 0

m 7..2 + 2 K (22 - 21) + K (22 - Z3) = 0

m 23'+ K (23 - 22) = 0For steadystate, the solutionswitbe as2t = At sin ront

~ =~ sin ron t23 = A3 sin ront

...(2.157 a)

.~:(2.157 b)

...(2.157 c)

-'-,

.-"

" -;"-. ; -:-

.

...(2.158 a)

...(2.158 b)

...(2.1"58 c)

.1'~;-.

' -~ "" ~" ~- =' ~ : :: -. "' :: :: "" 'c .: :- - . :"_._-

~

.~- ~ --

'W;;i'i~!~:t'jj


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'11,"

62Soil Dynamics & Machine Follndations

substituting Eqs. (2.158) in Eqs. (2.157), we get

., .(3 K - 111oo~)Al - 2 K

Soil Dynamics and Machine Foundations Swami Saran

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