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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
Illustrative Examples
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
Illustrative Examples
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
Illustrative Examples
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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'
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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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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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, ' " ' , r.'t~y~~~~
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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; ': ~:"E,j~\ "Or.
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
e:.IUu
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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
.
-
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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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"".'\'-
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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
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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
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;. :'~f,t;,\r.j'~~!.
16 Soil Dynamics & Machine Foundations
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.. .,.,~;
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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)
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;;" C," 'i'j{':-;,':::;;;~,.
18 Soil Dynamics & Machine Foundations
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
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>-":, ,;;[; /, '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)
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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>
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'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
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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
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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)
.".~.....- ",
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, -.. ".' r , . [.',
24 Soil Dynamics & Machine Foundations
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)
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~ }, ",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)
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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
" ". '
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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)
~-
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" .". ,( '. ;:, "" ~>;;;
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Theory of Vibrations 29
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
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'.'; :~,(i:', '.'),' '
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32 Soil Dynamics & Machine Foundations
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)
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':;0:"
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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=
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,f "c"' " f ','" ,,': ,r:,,
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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.
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. .(~
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
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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
........--.
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40
(a) Four storeyedframe
(b) Idealisation (c) First mode
-- .
Soil Dynamics & Machine Foundations
(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
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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
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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
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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
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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
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"'.'~..2,,~,.."t;,~.>, :"""',;,n rA';".!n~";'F."'" "'.,.,
Theory of Vibrations
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
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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
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Theory of Vibrations 47
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~'
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~,
48 Soil Dynamics & Machine Foundations
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)
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Theory of Vibrations
Subtituting:
Z2
ma
Absor ber
syst
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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)
- .. .~ ..., ~- .-.--~.-
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Theory of Vibrations 51
(
OOnt
)4-
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52
!IIZ
Soil Dynamics & Machine Foundations
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
...
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Theory of Vibrations 53
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" """"-";"~'.'.
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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
~.....- -~_'".,-~._........._-_....-
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.- ..... ...
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:,:;
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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
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"
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
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