53720995 Soil Dynamics and Machine Foundations Swami Saran 2

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SOIL DYNAMICS. AND" '..

MACHINE FOUNDATIONS

By

Dr. SWAMI SARANDepartmentof Civil Enginemng

University of RoorkeeRoorkee-247 667

(INDIA)

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Though much care has been taken by the author and the publishers to make the book error (factual orprinting) free. But neither the author nor the publisher takes any legal responsibility for any mistakethat might have crept in at any stage.

Published by .-Suneel Galgotia for Galgotia Publications (P) Ltd.5, Ansari Road, Darya Ganj, New Delhi-ll0 002.

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, PREFACE

During the last 25 years, considerable work in the area of soil dynamicsand machine foundationshas beenreported.Courseson soil dynamicsandmachinefoundationsalreadyexistat graduatelevelinmanyinstitutions,and its inclusion at undergraduate level is progressingfast.

The author is engaged in teaching the courseon soil dynamicsandmachine foundationsatgr'duate levelfrom last fLfteenyears. The text of thisbookhas been developed mainlyout of my notes preparedfor teachingthe students.The consideration in developingthe text is its lucide presentationfor clear understandingof thesubject.The material has been arrangedlogicallyso that the readercan follow the developmentalsequenceofthe subject with relative ease. A number of solved examples have been included in each chapter. All theformulae,charts and examples are given in SI units.

Someof the material included in this textbook has been drawnfrom the works of other autors. Inspiteofsincereefforts,somecontributionsmaynothavebeen acknowledged.The authorapologisesforsuchomissions.

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

The author would be failing in his duty it he does not aclaiowledgethe support he received from hisfamilymembers who encouraged him through the various stagesof study and writing.

. .

The book is dedicated to author's Sonin law, (Late) Shri AkhilGupta asa token of his love,affectionandregards to him.

(Dr. Swami Saran)

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CONTENTS.

1.PREFACE

INTRODUCTION1.1 General1.2 Earthquake Loading1.3 Equivalent Dynamic Load to an Actual Earthquake Load1.4 Seismic Force for Pseudo-staticAnalysis

Illustrative ExamplesReferencesPractice Problems

1-12I369

121212

13-6613141518323639485364

2. THEORY OF VIBRATIONS2.1 General2.2 Defmitions2.3 Harmonic Motion2.4 Vibrations of a SingleDegreeFreedom System2.5 Vibration Isolation2.6 Theory of Vibration MeasuringInstruments2.7 Vibration of Multiple DegreeFreedom Systems2.8 Undamped Dynamic VibrationAbsorbers

Illustrative ExamplesPractice Problems

3. WAVE PROP AGATION IN AN ELASTIC, HOMOGENEOUS..AND ISOTROPIC MEDIUM3.1 General3.2 Stress, Strain and Elastic Constants3.3 Longitudinal Elastic Waves in a Rod oflnfmite Length3.4 Torsional Vibration ora Rod of Infmite Length3.5 End Conditions3.6 Longitudinal Vibrations of Rods of Finite Length3.7 Torsional Vibrations of Rods of Finite Length3.8 Wave Propagation in an lnfmite, HomogeneousIsotropic, Elastic Medium3.9 Wave Propagation in Elastic, Half Space3.10 Geophysical Prospecting3.11 Typical Values of CompressionWave and Shear Wave Velocities

Illustrative ExamplesReferences..

'. . ;

Practice Problems

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108108116117

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viii Soil Dynamics & Machine Foundations

DYNAMIC SOIL PRO~ER~5. '-. ."' .4.1 General4.2 LaboratoryTechinques4.3 FieldTests4.4 FactorsAffecting Shear Modulus, ElasticModulus and Elastic Constants

IllustrativeExamplesReferencesPracticeProblems

ëò DYNANnCEARTHPRESSUREëò ï Generalëòî Pseudo-static Methods

5.3 Displacement AnalysisIllustrative ExamplesReferencesPracticeProblems

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187187201221236237

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301306309314319323

6. DYNAMIC BEARING CAPACITY OF SHALLOW FOUNDATIONS6.1 General

6.2 Pseudo-static Analysis6.3 Bearing Capacity of Footings6.4 Dynamics Analysis


7. LIQUEFACTION OF SOILS7. 1 General7.2 Definitions

7.3 Mechanism of Liquefaction7.4 Laboratory Studieséòë DynamicTriaxial Test7.6 Cyclic Simple ShearTest7.7 Comparisonof Cyclic Stress Causing Liquefactionunder Triaxial and

SimpleShear Conditions7.8 StandardCurves and Correlations for Liquefaction7.9 Evaluationof Zone of Liquefactionin Field7.10 VibrationTable Studies7.11 Field Blast Studies7.12 Evaluationof LiquefactionPotentialusing Standard Penetration Resistance7.13 Factors Affecting Liquefaction -

Contents '

8.

9.

ix

7.14 AntiliquefactionMeasures7.15 Studieson Use of Gravel Drains

IllustrativeExamplesReferencesPracticeProblems

324326332336339

340-351340340347348349

. .GENERAL PRINCIPLES OF MACIDNE FOUNDATION DESIGN8.1 General

8.2 Types of Machines and Foundations8.3 General Requirements of Machine Foundation8.4 Perimissible Amplitude8.5 Allowable Soil Pressure8.6 Permissible Stresses of Concrete of Steel8.7 Permissible Stresses of Timber

References

349350351

FOUNDATIONS OF RECIPROCATING MACHINES9.1 General

9.2 Modes of Vibrationof a Rigid FoundationBlock9.3 Methodsof Analysis9.4 Linear Elastic WeightlessSpring Method9.5 Elastic Half-space Method9.6 Effect of Footing Shape on VibratoryResponse9.7 Dynamic Response of Embedded BlockFoundation9.8 Soil Mass Participating in Vibrations9.9 Design Procedure for a Block Foundation


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10. FOUNDATIONS OF IMPACT TYPE MACIDNES10.1 General10.2 DynamicAnalysis .

10.3 Design Procedure for a Hammer FoundationIllustrative ExamplesReferencesPractice Problems

11. FOUNDATIONS OF ROTARY MACHINES11.1 General11.2 Special Considerations11.3 Design Criteria

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, INTRODUCTION

1.1 GENERAL

Geotechnical engineers frequently come across two types of problem in relation to the analysis and de-sign of foundations namely (i) foundations subjected to static loads and (ii) foundations subjected todynamic loads. The characteristic feature of a static load is that for a given structure the load carried bythe foundation at any given time is constant in magnitude and direction ~.g. dead weight of the structure.Live loads such as weight of train on a bridge and assembly of peopl{in a building are also classified asstatic load, The characteristic feature of a dynamic load is that it varies with time. Dynamic loads onfoundations and engineering structures may act due to earthquakes, bomb blasts, operation of machines,pile driving, quarrying, fast moving'traffic, wind or sea waves action. The nature of each dynamic load isdifferent from another. Figure 1.1 shows the variation of dynamic load with time in some typical cases,Purely dynamic loads do not occur in nature. Loads ar~ always combinations of static and dynamic loads.Static loads are caused by the dead weight of the structure, while dynamic loads may be caused throughthe sources mentioned above. ' .

0-3

01 0.1..c:0-0...CII- 0.1CIoJ .vv

et 0.2

0-3 .

I I I I I I I I I I I I I I I

0 ~ 10'L...t-I.

15 ,"Timt. . 5

(a) A.f~el~r9gramof F;.LCentro earthquake of May 18,1940NScomponent

Fig. 1,.1:.y,,-:i~J.iono(dyn_m'c load with time in some typical cases (...Contd.)

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Soil Dynamics & Machine Foundations

Time

(b) Dynamic load due to steady state vibration

Ud0 +vEdC>-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 with time in IOmetypical cases

Jntroduttion ¶

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Vibrations of earth's surface caused by waves coming from a source of disturbance inside the earth aredescribed 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 thusreleased is propogated in the form of waves which impart energy to the media through which they passand vibrate the structures standing on the earth's..surface. The point inside the earth mass where slippingor fracture begins is termed as focus and the point just above the focus on the earth's surface is termed asepicentre. The position of the focus is determined,withthe help of seismographrecords (Fig: 1.2]'u:ti't'isingthe 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

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Epic.entric. distance ~

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1.2.1 Intensity. The severity of shaking of an earthquake as felt or ob!jervedthroughdamage is'describedas intensity ata certain place on an arbitrary scale. For this purpose modified Mercalli scale is morecommon 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 notrecognize it as an earthquake; standing motor cars may rock slightly, and vibration may be feltlike 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 ofcracked plasters; unstable objects overturned; disturbance of trees, poles and other tall objectsnoticed sometimes and pendulumclocks may stop. .

Felt by all; many frightened and run outdoors; some heavy furnituremoved; a few instances offallen plaster or damaged chimneys;damage slight.

Everybodyruns outdoors, damagenegligible in buildings of good design and construction; slightto moderate in well built ordinary structures; considerable in poorly built or badly designedstructures; some chimneys broken; noticed by persons driving motor cars.

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 framedstructure; 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 thrownout of plumb; very heavy in substantial buildings with parti~1collapse; buildings shifted offfoundations;ground cracked conspicuously; 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 andsteep 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",

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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 earthquakeA = 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 ErgsA 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, VpVII,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 ojEarthquake(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 "

20,00060,000:,..., \

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15

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For teleseism$surface waveswith time:period 20 s '

c0.+:v 2~I-0U

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

1 10 100Distance in km

1000 10000 \

. .

Fig. 1.4: Distaoce corredio.o for magnitude determination

1.3 EQUIVALENT DYNAMIC LOAD TO AN ACTUAL EARTHQUAKE LOADFigure 1.1 (a) shows the variation of dynamic load wi+htime observed during El Centro earthquake. Theloading is not periodic and the.peaks in anY two cycles are different. For the analysis and design offoundations such a random variation is converted into equivalent number of cycles of uniformly varyingload [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 analysesand laboratory teStingare 'carried out using this concept.' .

According to Seed and Idriss (1911), the average equivalent uniform acceleration is about 65 percentof 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 earthquakes ofmagnitudes 1, 1.5 and 8respectively.

Lee and Chan (1912) suggested the following procedure for ..convertingthe irregular stress-timehistory to the equivalent number of cycles of cyclic shear stresses of maximum magnitude equal toK '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 accelerationrange (0 to amax)into convenient number ofleveIs and note the mean stress or mean accelerationwithin' each level as mentioned in column no. 2 of Table 1.4. Then the number of cycles withpeaks 'Yhichfall within each of these levels is counted and recorded. Note that because the actualtime history is not symmetric about the zero stress axis, the number of peaks on both sides arecounted and two peaks are equivalent to one cycle. For example, an earthquake record shown inFig. 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:.Ie:.IUu

<X

9mox,= -0.12. 0 2 4 6Time ( s )

8 10 12 14

Fig. 1.5 :, A typical earthquake record. '

(ii) Seed et al. (1975) gave aplot between stress ratio and conversion factor as sh!)wnin Fig. 1.6.Conversion factor is defined as'the ratio of equivalentnumber of cycles for 0.65 'tmaxto equivalentnumber of cycles for K . 'tmax'Referring to this curve (Fig. 1.6) determine the conversion factorto each average stress level (Col. 4 of Table 1.4).

(iii) Determine the equivalen~number of unifofm cycles,at a maximum stress level of 0.65 'tma.xbymultiplying the values listed in Cols. 3 and 4. 'These are listed in Col. 5.

, , . . . . t , ' ,', , ,

(iv) Determine the total.,number of equivalent stress ,cycles at .0..65'tmaxby adding the values listed inCol. 5. ,," , - ". " .< ,.', >

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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 ofcycles obtained for 0.65 'tmaxas illustrated in Table 1.4 is divided by this conversion factor to obtainequivalent 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 specifically for lique-facti~mstudies. More details of these procedures have been.discussed in Chapter 7.

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..

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 '6580 - 60 70 3/2 = 1.5 1.2 1.860 - 40 50 7/2 = 3.5 0.20 0.7040 - 20 30 5/2 = 2.5 negligible 0.020 - 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

Introduction j' ..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 inFig. 1.7. Two methods namely (i) seismic coefficient method and (ii) response spectrum method are usedfor 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 aboveVl1IVIIVILess than VI

Bombay

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0 0

~ Vo0 po rt

Blair0

.:0

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Fig. 1.7 : Seismiczonesof 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

.-

:/ <",

10

~- t ~:,- -. -_. ----


The vertical seismic coefficient, <Xv shall be considered in the case of structures in 'which stability isa criterion of design or for overall stability of structures. It may be taken as half of the horizontal seismiccoefficient. Therefore, . .

a<x=---1L

v 2In response spectrum method, the response acceleration coefficient is 'first obtained for the natural

period and damping of the structure and the design value of horizontal seismic coefficient is computedusing the following expression:

...(l.4)

Tab{e 1.5 : Values of Basic Seismic Coefficien~ <X/1

Zone No. ao

VIVIII11

I

0.080.050.040.020.01

,-Table 1.6 : Values of Importance Factor, I

S No. Va/lIeoJIType oJ Structure

2.3.4.

I. 3.02.02.0

Containment structure of seismic power reactor for preliminary designDam~ (all types)Containers of inflammable or poisonous gases or liquidsImportant service and community structures, such as hospitals, water towersand tanks, schools, important bridges, emergency buildings liketelephone exchange and fire brigades, large assembly structures likecinemas, assembly halls and subway stationsAll others

\.51.05.

.Table 1.7 : Values of Pfor Different Soil-Foundation System

Soil

Rock orHard soil

Mediumsoils.Softsoils

Values off3Jor-----------------------------Raftfoundation

Pilefoundation

Wellfoundation

Isolated Combinedorfootings isolatedwithout footingstie beams with tie beams

-\.0 1.0

1.2 1.0

1.5 1.2

\.0 1.0 1.0

1.0. 1.0 1.2

1.0 1.2 1.5

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Introduction

where

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, S,ab ,= ~ . I . F .;.,JL, 0 g

...( L 1)

F0 = Seismic zoni1)gfactor for average acceleration spectra (Table 1.8)S .

~ = average acceleration coefficient as read from Fig. 1.8 for appropriate natural period andgdamping of the structure.

-c~u--~0uc0

0.6

0.5....0L..~-~uu0~010L..~><!

0'4

0 .3___-

0.2----

0.111

0101 0

If) 0

. ,

0.4Natural

1-6 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

V

IV

III

U

I .

0.40

0.25

0.20

0.10

0.05,.

12 Soil Dynamics & Machine Foundations

tILL USTRATIVE EXAMPLES'

Example 1.1The srandard torsion seismograph recorded an average trace amplitude of 8.0 mm. The distance to theepicentre 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 = 3.9ÎÛÚÛÎÛÒÝÛÍ

HousnCf, G. W. (1965), "Intensity of earthquake ground shaking near the causative fault", Proceedings 3rd WorldConference on Earthquake Engineering, New Zealand, Vo\. 1.

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

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

DC

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THEORY OF VIBRATIONS

2.1 GENERAL

In order to understand the behaviour of a structure subjected to dynamic load lucidly, one must study themechanics of vibrations 'caused by the dynamic load. The pattern of variation of a dynamic load withrespect to time may be either periodic or transient. The periodical motions can be resolved into sinusoi-dally varying components e.g. vibrations in the case of reciprocating machine foundations. Transientvibrations may have very complicated non-periodic time history e.g. vibrations due to earthquakes andquarry blasts.A structure subjected to a dynamic load (periodic or transient) may vibrate in one of the following fourways of deformation or a combination there-of:

(i) Extensional (Fig. 2.1 a)(iii) bending (Fig. 2.1 c)

(ii) Shearing (Fig. 2.1 b)(iv) torsional (Fig. 2.1 d)

t ~ . c:

-~J(a) Extenslonal,- .: , (b) Shearing, : ,." (c) Bending (d) Torsional

',:' "Fig. 2.1 : Different types oCvlbratlons. '-1\";01, ' ," ,.!

, . ,,{-';',"

" .H ..

14 s.u /JyruuIfics & Machine Foundations

The forms of vibration mainly depend on the mass, stiffness distribution and end conditions of thesystem.

To study the response of a vibratory system, in many cases it is satisfactory to reduce it to an idealizedsystem of lumped parameters. In this regard, the simplest model consists of mass, spring and dashpotThis 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 of the 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 theposition of a system during vibration, is called degrees of freedom (Fig. 2.2).

~D:

m

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

Z2

. .~

KI

Z,

.- -,

~ ¢óãÝí-óååó¢¢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

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 freevibrations. 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 calledforced 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 systemduring 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 resonanceoccurs. 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 radiationdamping.

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)

NL T:2!!-r- Go)

Timq.t

'. .,

c

Fig. 2.3 : Quantities describing harmonic motion

;. :'~f,t;,\r.j'~~!.


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 represents thepeak-to-peak displacementamplitude,sometimesreferred to as doubleamplitude, 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 by f

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

ZI = AI sin rotMotion at any other point in the system might be expressed as

Z, = A, sin (rot-'e, )I I I

1t ~ 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 peakmotion occurs at point 1. The angle 8 is then called phase lag. For negative values of 8 the peak motionat 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 21tT=-=-f ro

The velocity and acceleration of motion are obtained from the derivatives of Eq. (2.1.).dZ .

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

= roA sin (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 ol A; 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.. .,.,~;<r'l!\"k. .',", . ,~Ii<i"

Theory of Vibrations

N

z,z,z..

+'C~E~v0a.UI

0

oN...

>-+'v0~>

c0-0....c:,I

c:,Iv0.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 harmonicmotion as shown in Fig. 2.5 occurs. It appears to be harmonic except for a gradual increase and decreasein 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:,IE

,~v0a.III

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

.'J' ,.,~T, b ~

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

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

...(2.7)

TimtZ (t)

;;" 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<01-<021fb = Tb = 21t ...(2.8)

The frequency of the combined oscillations is the average of the frequencies of the two componentsand is given by

f = i = (2~)(0) 1;1t0) 2 ) ...(2.9)

The maximum and minimum amplitudes of motion are the sum and difference of the amplitudes ofthe two sources respectively.

Zmax = AI + A2 ...(2.10a)

"Zmin = IAI - A21 ,...(2.10b)If the drive systems of two machines designed to operate at the same speed are not synchronized, they

may result vibrations having the beat frequency.

2.4 VIBRATIONS OF A SINGLE DEGREE FREEDOM SYSTEM1

The simplest model to repre~ent a single degree of freedom system consisting of a rigid mass m supportedby a spring and dashpot is shown in Fig. 2. 6a. The motion of the mass m is specified by one co-ordinateZ. Damping in this system is represented by the dashpot, and the resulting damping force is proportionalto the velocity. The system is sabject to an external time dependent force F (t).

----

Z - Dj splac(ZmentZ - V(ZlocityZ - Ac c(z l(Zration

c

KZ+ Cl +1

..mz L - - -'-Z-, -

mm

f F(t)

(I) Spring-mlss-dashpot system (b) Frcc-body diagram

~c ~ Pl8- 2.6, SI..,. ..,...' .,....... .-.:.., . . . .

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

>-":, ,;;[; /, '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, where Kis 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 andgiven 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 stillused 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 acts through the centre of gravity of the body in the direction opposite to that of accelera-tion. " '

The equilibrium of mass m givesmZ + 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 excitingforce are equal to zero. Therefore the'"equation of motion of the system becomes .."

m Z + 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 contwhere AI and Az are both constant~ and conis undamped natural frequency.

Substituting Eq. (2.13) in Eq. (2.12), we get? ,

-(j)~ (AI cos (j)i + Az sin (j)nt j+(~) (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 = V0

Substituting 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")Jiti..j}iJ'iI.J'!,';~"; >is:.:,,,, '

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

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

, ...(2.16)

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 = Azcos 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)

( )

22 Vo

Az = ,/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>

'reory ilf Jl"l6iatiOns 21

At time t equal to Displacement Z is ..

08

Az cos 8

Az(J)n

1t +8L- 00) n

1I+8

0)3-1t+82 .

-AZ

0(J)n

21t + 8O)n AZ

It is evident from Fig. 2.7 that nature of foundation displacement is sinusoidal. The magnitude ofmaximum 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- =-=0K K st .

Where g = Acceleration due to gravity, 9.81 mIs2W = Weight of mass m°st = staticdeflectionof the spring

Therefore

...(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,

- .

,.,....~ ~.~ .nI

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 excitationforce 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 = 02 (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 + 2eThe 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 m

The roots AI and A2are real and negative. The motion of the system is not oscillatory but is anexponential subsiden~~(Fig. 2. 9). Because.of the relatively large damping, so much energy is dissipated

'-..----

,....N 20:I:-c

.....

10

Theory of Vi!'rations ,

23

by the damping force that there is sufficient kinetic energy left t~ carry the mass and pass the equilibriumposition. 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 isgiven 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. " '

zc2=l"km

Time,t

.,

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

(~ ) = K. C = C2m m' c

Then 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 asdamping ratio:

When ...(2.34)

'Now

C~=-

CcC - C Cc - C 2JK"m_c:'fK2m - Cc . 2m - Cc' 2m - Cc'Vm

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

..,..(2.36)

...(2.37)

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

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


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

Case III : (~ )2 < K2m m

The roots Al and Al are complex and are given by .

AI,2 = [-;:!:i~I-;2 ]COnThe 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>"t A 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)

or Z = 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<m

Fig. 2.11 : Free vibrations of a viscously underdamped system

As a convenient measure of damping, we may compute the ratio of amplitudes of the successivecycles of vibration

or

-0> f,1Z e "---L = -0> f,(t+Zn/o>"cI)Z2 e n

ZI 0> f,.21t/o)ncl- = en

Zz r:2ZI Znl;!"l-f,-- =eZzZI - 21t;

loge 22.- ~

...(2.44 a)

or ...(2.44 b)

or ...(2.44 c)

...(2.44 d)

~ }, ",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:-:2or ~= 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,hcycle, thenZo Zl Z2 Zn-I 0- = - = - = . . . = - = e where~=2x ~Z\ Z2 ZJ Zn

Zo, =[

Zo] [

~] [

Z2]

..[

Z"-I]

= enoZn Z, Z2 Z) Z"

Therefore,

or

Z1 I 0 ..~ = - oge Zn n

Z1 I 0..}: = - oge.z~ 2xn n

Hence ...(2.441)

...(2.44 g)

Therefore, a system isover damped if ~ > 1;

critically damped if ~ = 1 andunder damped if ~ < 1.

2.4.3 Forced Vibrations Of Single Degree Freedom Syst~m. In many cases of vibrations caused byrotating parts of machines, th~ systems are subjected to periodic exciting forces. Let us consider the caseof a single degree freedom sys~.:mwhich is acted upon by a steady state sinusoidal exciting force havingmagnitude 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 thisequation consists of two parts namely (i) complementary function, and (ii) particular integral. Thecomplementary 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 forcing

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

m 2:2+ C 22 + K Z2 = Fo sin rotwhere Z2 = displacement of mass m a~~nYinstant t when vibrating with forcing function.

~.;Y,

...(2.48)

8111;1'

"

26"--H,' " "

Soil Dynamics & Mac/line Foundations

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

, 22 = AI sin 00t + A2cos 00'twhere 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 + A2 cos 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) =0Solving Egs, (2,52 a) and (2.52 b), we get

(K-moo2) FoAt- - 2(K - mm2) + C2m2

and A2 = - CmFo

" ". '<c, "',, >, (K-~~2)2+c"2m2,"."By substitutingJhe values 'of Al and A~"inEq::2'.49;:'" , ' """ , : ' ," ',' , "',.."

...(2.49)"...~ ,'" . . .:'

, "

",'

. ...(2,51 a),..(2.? 1 b)

.)F. 0,

m = Q. 0: ',n ",(2,52 a)

...{2.52 b)

0.. '.)2.53 a)

"

...(2,53 b).t '

~, ':22':°2" ~ 2 {(K'-mm2)sinmi-cwcosmt}(K - m m ) + Cm"

'...(2.54)

let, ' tan e = 'C 0)K-l~cd2

-, ' .,.(2.55)

By substituting Eg. (2.55) in Eg. (2.54), one can obtain

22 = Fo . sin(mt - e)

~(K~';'m2t+'c2 cii1' :" :,., "Eg. (2.56) may be written as t . . " , ,

22 = . '.ccFo!K , . 'sin{~t -"'er~(I:!12)+(2Tl~)~.. "I , -"~; . '"

. W ,',

11 = Frequency ratio =;- "',,;, ".,:.' ,n-,' ':Ci"","'G'}""::"',',_:;,~'-',~ = Dam ping ratio = -c .:;=

,

' ';JKi{i . '.., 0"'2' 'Km 'c

...(2.56),,' ,

, ,'! " ..~. " ...(2.57)

Where,,'

""""'~!i1I"1!",?' ~"" "".",..1d."" "~~",...""."..",""."""",."""""""",-,,i""'-~-' .

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~venby

F. /K '.Az = 0 " oi

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

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

"I

+'C~

E~u0a.UI ....

....,.,..

Transi~nt , ,

0

~211"(;) ~

NN

..+'C~E~u0a.UI

0

Time.t. "

-,'h ..':'1 i." ..;.

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

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

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

""

N\

'I\ " .+'C~Ec:.Iu0a.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)

~-

" .". ,( '. ;: , "" ~>;;;<~i:«.~..,'

28 SoU Dyrul",ks & Maehi"e Foundlltions

The quantity FelK is equal to the static deflection of the mass under force Fo' Dynamic magnificationfactor, f.1is derIDedas the ratio of the dynamic amplitude Az to the static deflection. and is given by

~ = ~ 1 . -- ...(2.59)-(1-112)2 +(211~)2

The variation of f.1versus 11is shown in Fig. 2.13 for different values of damping ratio ~.It would beseen that near 11= 1, the value of f.1is maximum. This is called resonance and the forcing frequencyJ atwhich it occurs is called the resonant frequency.

5

~=o

t.

00 1.5

Frc&quc&ncy ratio. "\.

Fie. 1.13 : Macnlficatlon factor (J&)vcnus freqllcacy ratio (11)

"0 1

I 1 "0.1-...30

u

I 11 0)c

I....

c0.-..cu 2.-.....-cClc

1

Differentiating Eq. (2.59) with respect to 11and equating to zero, it can be shown that resonance willoccur at a frequency ratio given by

11= ~1-2~2which 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°

<D120°

c:.J

C'IC0

c:.J 90°' r =0.707\11

0.J:a..

600

00 '1.0 2.0 3.0

FrczquQncy ratio# -rz.

Fig. 2.14: Phase Jagversus frequency ratio for different amounts of damping, . ' . , : " ..,' ~ .;':J '

<'., ~ ,";~,'~',{f.(;"..


'By substituting Eq. (2.60) in Eq. (2.59), the maximum value of magnification factor is obtained. Itis given by ,

I

J.lmax = 2E,~I-E/I

- 2E,Assuming a damping of 5% in a structure, its amplitude at resonance will be 10 times the static

deflection. This indicates that systems will be subjected to very large amplitudes at resonance whichshould be avoided.

...(2.61 )

(For small values of;) ...(2.62)

. '. --- - -The phase angle e given by Eq. (2.55) indicates the phase difference between the motion and the

exciting force: It can}e.-writt~.n~s .

-I(

211E,

)e = Tan _1- 2'" . T\

Variation of e with respect to 11is shown in Fig. 2.14

...(2.63 )

2.4.3.1 Rotating mass type excitation. Machines with unbalanced rotating masses develop alternatingforce as shown in Fig. 2.15 a. Since horizontal forces on the foundation at any instant cancel, the netvibrating force on the foundation is vertical and equal to 2 me ero2sin rot, where me is the mass of eachrotating element, placed at eccentricity e from the centre <,>frotating shaft and ro is the angular frequencyof masses. Fig. 2.15b shows such a system mounted on elastic supports with dashpot representing viscousdamping. .

12m~. ~~Forc~ gczn~rat~d

(a) Rotating mass type excitation (b) Mass-spring-das~ot system

Fig. 2.15 : Single degree freedom system with rotating mass type excitation

The equation of motion can be written as-. 2m Z + C Z + K Z = 2 111ero sin rot .. . - e. .

where m is the mass 'of foundation including 2 me' Equations (2.64) imd (2.48) are similar, except that2 Ill" ero2 appears in Eq. (2.64) in place of Fo' The solution of Eq. (2.64) may therefore be written as,

...(2.64)

'.'; :~,(i:', '.'),' '<'F '" ". ,." ',', , ". .(\~.tJi;"":


where

Since

or

.---.

';Iile

31

Z = A; sin (0)t + 0') .

(2mee/ m)'T}2Az = I 2 2

(1-T}2) +(2~T})2

F=2m .eO)0 e

F (J) 2 0)2

K = 2 me . e K = 2 me . e (mro~)

...(2.65)

..(2.66)

= (2 me :}T}2

e = Tan-I ( 2T}\)1-11 ...(2.67)

3.00.10

2.0

5.0

1.0

0-0 4.01..0 ' ,2.0, '. 3.0

Frequency ratio. 1) .(a) Az 1(2m~elm) versus Irequency rauo 11

0° -0 4.0 5.01.0 2.0 3.0

Frqquqncy ratio., - 't(b) Phase angle versus frequencYT&tio11

Fig. 2.16 : Response,oh system with rotating unbalance

180°.0.05

'(DI 0.25'... 0.50aI

C7Ic:0 90°aIU\0.s::.a..


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 similarin shape to those in Fig. 2.13 except that these starts from origin. The variation of phase angle e with 11is shown in Fig. 2.16 b. Differentiating Eq. (2.68) with respect to 11and equating to zero. it can be shownthat resonance will occur at a frequency ratio given by

11"\=-

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 thefoundation 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 anyother rotating machinery should be isolated from the foundation so that the adjoining structure is not setinto heavy vibrations. Another example may be the isolation of delicate instruments from their supportswhich 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 forceisolation 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 withunbalanced 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 as22m 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)

':;0:"<\',:,-;' '-' ,'t "t" "',"',' '1

Theory of Jlibrlltions;;::

33

--'-~, '

~~~ Ma chine

K2'

c K2

Iso lata r

Foundation Ground surfac~

Fig. 2.17 : Machine-isolator-roundation systcm

The only force which can be applied to the foundation is the spring force KZ and the damping force. '

C Z; hence the total force tqmsmitted to the foundation during steady state forced vibration is

Ft = KZ + CZSubstituting Eq. (2.73) in Eq. (2.75), we get

22m em .F = e .sm(mt- e)+t~ 2 2(1-1l) +(211~) ,

..,(2.75)

C.2me em2 /K '00 cos(mt-e)2 2(1-112) +(211~)

...(2.76)Equation (2.76) can be written as:

2 Jl+(211~)2F = 2me em , . sin(m t - P)t 2 2

. (1-112) +(211~)

where p is the phase difference between the exciting force and the force transmitted to the foundation andis given by , ' i

- P ~-e --=~-~~-l[

coo

]'" r K,, ", J -

...(2.77)

...(2.78)~

;"": ','"o/,,'n1.."n" ',' i J " ".

34 Soil 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~)2IlT is called the transmissibility of the system. A plot of IlTversus 11for different values of ~is shown

in 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 ofdamping 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 stabilityunder 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 ~=0

1.0 2.0 3.0

Frczquczncyr(:itio I 1'\.

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

1- 3.0=<.

»....-.-oDU\U\.- 2-0EIIIC0...I-

,f "c"'" f ','" ,,': ,r:,,<"f:' ';:£'4/ l',,' tt~~7T{~:,:~";';:s:",.:';


, . . : ';' , ' ' '

2.5.2. Motion Isolation. In many situations, it would be necessary to isolate structure or mechanicalsystems from vibrations transmitted from the neighboring machines. Again we require a suitable mount-ing system so that least vibrations are transmitted to the system due to the vibrating base. We consider asystem mounted through a spring and dashpot and attached to the surface which vibrates harmonicallywith frequency (I)and amplitude Y0 as shown in Fig. 2.19.

Machina

z" .

Foundation. , ,

Iso lator

v = Yo Sin GJt

Vi brating ground.d u (l to n (l i9 h bo u r in 9machines

Fig. 2.19: Motion isolation system

Let Z be the absolute displacement of mass; the equation of motion of the system can be written as:

m Z + C (2 - Y) + K (Z -:- Y) = 0 ...(2.80)

or m Z + K 2 .+ K Z = C Y .+ K Y = C (I) Y0 cos (I) 1 .+ K Y0 sin (I) 1

m Z'+ C 2.+ KZ = Yo ~K2.+(Cro)2 sin (rol'+ ex) ...(2.81)or

h T -I CO)ere ex=, an K

l:he solution of Eq. (2.81) will give the maximum amplitude as:

Z - . ~1.+(21l~)2 ,

max - Vo'~,. , " (l'-1l2)2+(21l~l", . ) ..,' -'. ,~

...(2.8?)

...(2.83)

" " . ,,' ",

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 systemcan also be studied from the response curves shown in fig. 2.18. It is again noted that for the vibrationisolation 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 dampingand stiffness properties vary widely with applied load, temperature, shape factor, excitation frequencyand the amplitude of vibration. The maximum temperature upto which rubber can be used satisfactorilyis about 65°c. It must not be used in presence of oil which attacks rubber. It is found very s' ".,ble for highfrequency vibrations.

2.5.3.2.Felt. Felt is used in compressfun only and is capable of taking extremely high loads. It has veryhigh damping and so is suitable in the range of low frequency ratio. It is mainly used in conjunction withmetallic 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 notaffected 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 desiredconditions. 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 INSTRUMENTSThe purpose of a vibration measuring instrument is to give an output signal which represents, as closelyas 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 thedirectly 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 any of the vibrationphenomena. 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.

. .(~

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 themass and the structure, since it is this relative motion which is detected and amplified. let 2 be theabsolute 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- TJ2Equation(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

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

2.6.1. Displacement Pickup. The instrument will read the displacement of the structure directly if1121.1= I and 8 = O.The variation o{Tl~ with-~'aiid'~-is shown 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 isapproximately equal to 180°. Therefore to design a displacement pickup, 1'\should be large which meansthat 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 issensitive, flimsy and can be used in a weak vibration environment. The instrument can not be used formeasurement of strong vibrations.

3.0

,- -tI \I \0II

. .

-- . -

- -"

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 2X = 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 forsmall 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

........--.

<~


natural frequency of the instrument itself should be high compared to the frequency to be measured. Inother words, the instrument should have a stiff spring and small mass. The instrument is less sensitiveand suitable for the measurement of strong motion. The instrument size is small.

2.6.3 Velocity Pickup. Equation (2.88) can be rewritten as1

X = - TIJ!Y0(J)sin «(J)1- 0)COn

The output of the instrument will be proportional to velocity of the structure if ~ 111l is a constant.O)n

At 11= 1, Eq. (2.90) can be written as1 1 . .. .1

X = - 2 ~ Yo(J) sm (0)1- 0) .: atTl = 1, f.1= - ...(2.91)con ':1 .. . . 2~Since O)nand ~are constant, the instrument will measure the velocity at 11= 1.It may be noted that the same instrument can be used to measure displacement, acceleration and

velocity in different frequency ranges. . ,

X a Y if TI» 1 Displacement pickup (Vibrometer~)

...(2.90)

..X a Y if 11 « lAcceleration pickup (Accelerometers)

X a Y if 11= 1 Velocity pickup (Velometers) .

Displacement and velocity pickups have the disadvantage of having rather a large size if motionshaving small frequency of vibration are to be measured. Calibration of these pickups is not simple. Fur-ther. corrections have to be made in the observations as the response is not flat in the starting regions.From the point of view of small size, flat frequency response, sturdiness and ease of calibration, accelera-tion pickups are to be favoured. They are relatively less sensitive and this disadvantage can easily beovercome by high gain electronic instrumentation..

. .

2.6.4. Design of Acceleration Pickup. The relative displacem~nt between, the mass an~ the supportwould be a measure of the support acceleration ifTl is less than 0.75 an4 ~is of the order of 0.6 to 0.7.Of the various methods of measurement of relative displacement, .electrical gauging,:in 'whIch {he me-chanical quantity is converted into an equivalent electrical quantity is best suited for a~.<;elerationpick-ups. Electrical gauging offers the possibility of high magnification of ~e signals which are usually weakbecause the spring is stiff and the displacements are small. The mechanica,l quantity alters either theresistance, or capacitance or the inductance of the circuit which consequently alters the current in thecircuit. '

2.7 VIBRATION OF MULTIPLE DEGREE FREEDOM SYSTEMS

In the preceding sections, vibrations of systems having single degree of freedom have been discussed. Inmany engineering problems, one may come across the systems which may have more than one degree offreedom. Two degrees freedom cases arise when the foundation of the system is yielding thus addinganother degree of freedom or a spring'mass system is attached to the main system to reduce itsvibrations.In systems when there are a number of masses con~ected with each other, even if each mass is con-strained to have one degree of freedom, the system as a whole h~s as many degrees of freedom as thereare masses. Such an idealization is done for carrying out dyn,!mic analysis of multistoreyed buildings.ô ø

40

(a) Four storeyedframe

(b) Idealisation (c) First mode

'--'i.


(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 thefloor levels and the lumped mass has a value corresponding to weight of the floor, part of the supportingsystem (columns) above and below the floor and effective live load. The restoring forces are provided bythe supporting systems. Figure 2.22b shows such an idealization and it gives a four degrees of freedomsystem. In free vibration a system having four degrees of freedom has four natural frequencies and thevibration of the any point in the system, in general, is a combination of four harmonics of these fournatural 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 ofvibration. Figure 2.22<.to 2.221'show the four modes of vibration. If all the masses vibrate in phase (Fig.2.22c), the mode is termed the first or lowest or fundamental mode of vibration and the frequency asso-ciated with this mode would be the lowest in magnitude compared to other modes. If all adjacent massesvibrate out of phase with each other (Fig. 2.22£), the mode is termed the highest mode of vibration andthe frequency associated with this mode would be highest in magnitude compared to other modes.

2.7.1. Two Degrees of Freedom Systems.

2.7.1.1. Undampedfree vibration: Figure 2.23 shows a mass-spring system with two degrees of free-dom. Let Z\ be the displaceuent of mass ml and Z2 the displacement of mass m2' The equations of motionof the system can be written:

In t Z\ + Kt ZI + K2 (Z\ - Z2) = 0

1nl Z2 + K3 Zl + K2 (Z2 - Z,) = 0The solution of Eqs. (2.92) and (2.93) will be of the following form:

ZI=A\sinro,,( .

Z2 = A2 sin ro" tSubstitution of Eqs. (2.94) and (2.95), into Eqs. (2.92) and (2.93), yields:

(KI + K2 - ml C1)~)Al - K2A2 =0

- K2 Al + (~ + KJ - m2 C1);)A2 = 0

...(2.92)

...(2.93 )

...(2.94 )

...(2.9S)

...(2.96)

...(2.97)

, ,

I .~heory of Vibrations

~!'\Wm

ìï

Æï

×¦ô

ÕîøÆÆóÆ´÷

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) -= 0m) ~ 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

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) sin (0 t + A(2) sin (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 2

Al - K2 - K2 +KJ -"'2 ffinJ(i)- 2- KA2 KI+K2-mlffinl 2

(2)' 2A I - K2 - K2 + KJ - "'2 ffin2(2) - 2 - KA2 KJ + K2 - m) ffin2 2

2.7.1.2. Undamped forced vibrations. Consider the system shown in Fig. 2.24 with excitation force Ft

sin (0 t acting on mass ml. In this case, equations of motion will be:

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

1n2 Z2 + KJ Z2 + K2 (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

leory of Vibrations 43

For steady state, the solutions will be as

21 =Al sin 00t

2z = Az sin 00tSubstituting Eqs. (2.107) and (2.108) in Eqs. (2.105) and (2.106), we get

Z(KI + Kz - ml 00) AI - Kz Az = Fo

z- Kz AI + (Kz + K3 - mz 00) Az = 0Solving for AI and Az from the above two equations, we get

z(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

Ã

mlrnz co - + co +ml rnz m\rnz

The above t\VOequations give steady state amplitude of vibration of the ~womasses respectively, asa function of 00. 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 thesame 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 for0 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

mz 2z - 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

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

"'.'~..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 cont

Z) = AJ sin cont. . . . . . . . . . . . . .

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

Zn = An sin contSubstitution 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) - mJ ID~] A) - K4 A4 = 0

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

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

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

[ (KI + Kz) - ml ID;]

-K20

-Kz

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

-K)

0

00

0

00 =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 of con correspondingto n naturalfrequencies. The mode shapes can be obtained from Eq. (2.121) to (2.124) by using, at one time, one ofthe 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<1pesbecomes tedius. Numerical tech-

. J'iG.~esare invariably resorted to in such cases.. ,

Holzer's numerical technique is a convenient method of solving the problem for the system idealizedas sho~ in Fig. 2.26. By sUI11II1iPgJfotcesat free end, .

'-, . --.'O".::.,:-=C°'-:"~",_,- ;: " ~-::--===.::.~-~:---==E'~--,'::.._~

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 mi (- 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 variousval~~s.pfQ);.~ode $ape.can be obtained.. .. -. .

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

...(2.129)or

iI

m1.JK1

Iml

K2


1.0

~

t' ..-J +J c:~«~1:

0(..)2n.

-1.02

wn12

""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 subjected to forced vibration,and Fj (t) represents the for~e on mass mr The equation of motion for the mass mj will be

nm. 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 variousmodes. The individual modal response would be some fraction of the total response with the sum offractions 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 )

nZ. = 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 A j - oonr mj 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~'

~,


Since the left hand side is a summation involving different modes of vibration, the right hand sideshould 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 = - J fr 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 ofsingle degree of freedom. The d's are termed as normal co-ordinates and this approach is known asnormal 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 toresonance. 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 theexcitation 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) =0The forced vibration solution will be of the form

ZI = Al sin rot~ = Azsin rot

Substitution of Eqs. (2.142) and (2.143) in Eqs. (2.140) and (2.141) yields

Al (-M0}2 õ Õ õ Ka)-KaAZ = Fa

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

...(2.142)

...(2.143)

...(2.144)

(2.145)

óóþó

þ þþóþùòóþóò

Subtituting:

Z2

maAbsor bersyst <zm

Ka

Z1

MMainsyst<zm

Fig. 2.28 : Vibration absorber

FZSl = -2... = Staticdeflectionof main systemK

2 Karo = - = Natural freqeuncy of the absorberna ma

K .ro~ = M = Natural frequency of mam system

mIlm = -.Jl.. = Mass ratio = Absorber massIMain massM

The Eqs. (2.144) and (2.145) can be written as

(1

K{/ <i

)K{/A I +--- -A 2 - =Z

K <.02 K SIn

A2 =

(1:~: )äò±²¿

and

Solving Eqs. (2.146) and (2.147) for At and A2' we getCil1-- 2

~t. =

(

L 2

)

'

(

; -roK' no - 2

)

'KSi ro ro'

, <, n' " "--, ~-- -~, " .,".. ld"2. 1+ K" 2 K'- rona ron

, J

.

.49

...(2.146)

...(2.147)

...(2.148)I ".

i , "..",.-, "-~t'u'",.'."',,

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

or oona = ,oon ...(2.151 a)

Ko = Kma M

...(2.151 b)

K mor -!L=-!!..=Ii ...(2.151 e)K M t"'mWhen the condition enumerated in Eqs. (2.151) is fulfilled, the absorber is known as a tuned

absorber. .

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

(002

J1- -

~,: ~ ( 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 denominatorsbecomes zero. For this condition the expression for the denominators can be written as

..(2.153)

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

(

OOnt

)4-<2+llm)

(OOnt

)2 +1 =0

OOna <Ona...{2.154)

The Eq. (2.154) is quadratic in <0;1'and therefore there are two values of oonlfor which the denomi-nators ofEqs. (2.152) and (2.153) become zero. These two frequencies are the natural frequencies of the3ystem. Solution of Eq. (2.154) gives:

(:J ~ (1+~2m)~J~m+~4~1.6

...(2.155)

1.2

_1.4

_

1

°c c33

1.0

0.8

0.60 0.2 0.4 0.6 0.8

Mass ratio }.ImFig. 2.29: Natural frequency ratio versus mass ratio

The relationships of Eq. (2.155) is plotted in Fig. 2.29. From this plot, it is evident that greater themass ratio, greater is the spread between the two resonant frequencies. The frequency response curve forthemainsystemisshownin Fig.2.30fora valueof ~'"=0.2. The dotted curves shown actually mean thatthe amplitude is negative or its phase difference with respect to the exciting force is 1800. It can benoticed from this figure that by attaching a vibration absorber {oona= (On)to the main system vibratingat resonance reduces its vibration to zero. Now if the exciting frequency is absolutely constant, thesystem will work efficiently. Any change in the exciting frequency will shift the operating point from theoptimum point and the main system response will no longer be zero. It may be noted that by adding thevibration absorber, we have introduced two resonant points instead of one in the original system. Now

I --:::w;-' -Hr.

52

!IIZ~


if the variation of the exciting frequency is such that the operating point shifts near one of the newresonant points, then amplitudes will be excessive. Thus depending upon the variation of the excitingfrequencies the spread between the two resonant frequencies has to be decided to remain reasonably awayfrom 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 forvarying forcing frequency excitation. To make the vibration absorber effective over an extended range offrequencies of the disturbing force, it is advantageous to introduce a damping device in the absorbersystem. Such an absorber system is called a damped dynamic vibration absorber.

8

6

...<I~ 4,

z

1.0

00

}Jm=0.20

\

'---2.0 2.51.50.5 1.0

G.)n løòË²¿

The Eq. (2.154) can also be written as

Fig. 2.30 : Response versus frequency of a vibration absorber

~m J~)-lr( )

2(J) nl

(J) //(/

...(2.156)

t ILLUSTRATIVE EXAMPLEs!

Example 2.1The motionof a particle is representedby the equationz = 20 sin rot. Show the relative positions andmagnitudes of the 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 theacceleration vector leads the velocity vector by another 1t/2. Figures 2.31 a and 2.31b show the threevectors 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 se cFig. 2.31 : Vector diagram (Example 2.1)

21t 21tTime period = - = - = 1t sill 2

for ro = 2.0 rad/s

21t 21tTime perIod = - = - = 41tsill (0.5)

for ro = 0.5 rad/s

Example 2.2A bodyperforms,simultaneously,the motions

Zl (mm) = 20 sin 8.0 tZ2 (mm) = 21 sin 8.5 t

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

~~tm~f~dJ "T'f:ifrir" """"-";"~'.'.<f:;-:F;;':.;;,:Jijj~ .,.-,'!.,!,,~,~~'111""

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 0.51= 21t = 21t = 0.0795 Hz, and

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 naturalfrequency 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 2.4For the system shown in Fig. 2.32, determine the naturalfrequency of the system if

K1

Kt = 1000 N/mKz = 500 N/mKJ = 2000 N/mK4 = Ks = 750 N/m

Mass of the body = 5 kgSolution:

Let Ket and Ke2represent respectively the effective stiffnessesof 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

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

.- ..... ... -Theory of Vibratimrs 55

Kel = 285.7 N/mKe2 = 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 = 3.0Hzn 21tV;; 21t 5.0Example 2.5A vibrating system consists of a mass of 5 kg, a spring stiffness of 5 N/mm and a dashpot with a dampingcoefficient 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 =2.07eZ 2

~ = 7.92Z2

Therefore the free amplitude in the next cycle decreases by 7.92 times.Example 2.6A 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 consecutiveamplitudes was 10/3. Determine the amplitude and phase angle when a force F = 3 sin 4 t acts on thesystem. 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:<,'..'""

(i)

..

'.

----., -=., -,.;".,'~ -""'~-.~ ~~,,~,..-'>,:;<..."".","..;g""~,,;,;~,,,,:.u ,.~~

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 = 141.4°1-11 1-1.273Example 2.7Show that, in frequency - dependent excitation the damping factor ~is given by the followingexpres-S1On:

): -.: ..!.(

12~ 11J~ - 2 2/n

Where 11 and12 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

EE 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

-.------

I I II II I I

I II I

f1 I fn Ifl14 18 22 24

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

"

Th~o,.,ofv~,!s 57

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 . =2

In In

Now [for small values of~]

Also

~ = ! (Iz - 11

)2 InThis methodfordeterminingviscousdamping 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 producesan 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.---

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 2.9The rotor of a motor having mass 2 kg was running at a constant speed of 30 c/s with an eccentricity of160 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 (21t x 30/ = 22716 N= 22.72 kN

:ii) Force transmittedJl = = 0 15T unbalancedforce .

. ~1 +4112~2 .

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

or 1 + 4112 x (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 2.10A 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 massas 0.05 mm. Determine the amplitudes of displacement: velo'city and acceleration of the vibrating ma-chine. Neglect damping.

~;,:,~ .;,~

'/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 = 0.33337.71

~ = - for ~=01- ,,2 ,

1= = 1.1251- (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 2.11Determine the natural frequencies and mode shapes of the system represented by a mathematical mod~1shown in Fig. 2.34 a.

,/

. , co:" '1,,';

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

, 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 systemhas 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

)= K

III 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 of amplitudes Al and ~ 'for the two modescan be obtainedusing Eqs.(2.103) and (2.104). . . '

,}~~r .

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

.~."

eory of Vtbtalions_/ 6i

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

A~2) 2KA (2) = K + 2 K - m x 5K / m =- 12

. The mode shapes are shown in Fig. 2.34 band 2.34 c.

B:xample2.12

Determine the 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 solutionswit be as

2t = 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

'11,"

62Soil Dynamics & Machine Follndations

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

(3 K - 111oo~)Al - 2 K A2 = 0

óîÕßø +(3K-moo~)A2-KA3'=O

- K A2 + (K - 111 co;) A3 = 0For nontrivial solutions of 00/1in Eqs. (2.159)

...(2.159a),

...(2.159 b)

...(2.159 c)

...(2.160)

.," . IIIW-

Pllttll1~ A -= !l Eg . (2.160) become as~ K

or

...(2.161 a)

...(2.161 b)Eg. (2.161 b) is cubic in A. The values of A are worked out as

A( = 0.238; A2= 1.637; and A) = 5.129

00/1( = .J0.238Kl m; CO/l2= .J1.637 Kl m; and 00/13= ~5.129 Kl mTherefore,

(ii) Egs. (2.159) in terms of A can be written as

(3 - A) Al - 2 A2 = 0

- 2 AI + (3 - A)A2 - A) = 0- A2 + (1 - A) A3 = 0

A = 0.238

...(2.162a)...(2.162 b)...(2.162 c)

For r mode:

Eg. (2.162 a) gives

" AI(3 - 0.238) AI - 2 A2 = 0 or A = 0.7242

Eg. (2,162 h) gives

- 2 x 0.724 A2 -t-(3 - 0.238) Az - AJ = 0

A2 '

A = 0.7613

, ,.'

or

-

",'.1 "iI; -' ", (..'

.,03K-11l00 -2K/I

-2K "'K 2-K 1=0-' -moo/l

0 -K 2K-mcon

3-A -2 0-2 3-A

-1 I =00 -1 I-A

3 2f. - 7 A + lOA - 2 = 0

:," 'h, , .\


Assuming Al = a, A2 = 1.381 a and A) = 1.815 a:. Al : A2 : A) = a : 1.381 a : 1.815 a

= 1: 1.381: 1.815

;::;0.551 : 0.761 : 1

Similarly,

For II mode; A. = 1.637

Al : A2 : A) = - 0.933: - 0.635 : 1, andFor III mode: A.= 5.129

Al : A2 : A) = 3.891 : - 4.14 : 1The mode shapes are plotted in Figs. 2.35 b, 2.35 c and 2.35 d,

Example 2.13A small reciprocating machine weighs 50 kg and runs at a constant sp~~d of 6000 rpm. After it wasinstalled, it was found that the forcing frequency is very close to th~}latural frequency of the system.What dynamic absorber should be added if the nearest natural frequency of the system should be at least20 percent from the forcing frequency.Solution:

(i) 21t N 2 Tt x 6000 = 628 rad/sill = W "" 60

At the time of installation of machine,Forcing frequency ==Natural frequency of system

~ = 6282K = m x 628

= 50 x 6282 = 201 x 105 N/m

Therefore,

or

(ii) Aner adding the vibration absorber to the system, the natural frequency becomes (1 :i:0.2) 628i.e. 753.6 rad/s or 502.4 rad/s

For tuned absorber:

ma KaM = K = JIM

Now from Eq. (2.156)

J(:,:J-IfJInl- ~

)l;;:)illnl = 0.80)//(/

.


2}2

{(0.8) -1 = 0.2025- 2J-lm- 0.8

and when (Onl = 1.2(Ona

{(1.2)2 -1}2J-l = 2 = 0.134

m 1.2Adopting the higher value of J-lm

Ka = 0.2025 x 201 x 105 = 40.7 x 105 N/m

ma = 0.2025 x 50 = 10.12 kg

PRACTICE PROBLEMS

2.1 A single degree (mass-spring-dashpot) system is subjected to a frequency dependentoscillatoryforce (m eo (02sin (0f). Proceeding from fundamentals, derive the expression of the amplitudeof the system.

2.2 'Presence of damping reduces the effectiveness of the isolation system'. Is this statement true?If yes, explain with neat sketches. .

2.3 Give two methods of determining 'damping factor' of,a single degree freedom system.2.4 Starting from fundamentals~explain the principles involved in the design of (i) Displacement

pickup, (if) Velocitypickup,and (iif) Accelerationpickup. Illustrateyour answerwithneatsketches.2.5 Describe the principles involved in a 'tuned dynamic vibration absorber'. Illustrate your answer

with neat sketches. Discuss clearly its limitations.2.6 A mass of 25 kg when suspended from a spring, causes a static deflection of 25 mm. Find the

natural frequency of the system. Ans. (20 rad/s)

2.7 A spring mas system (K\, m) has a natural frequency of f\. ~fa second spring of stiffness K2 isattached in series with the first spring, the natural frequency becomes f\/2. Determine K2 interms ofK\. Ans. (K/3)

2.8 A mass of 5 kg is attached to the lower end of a spring whose upper end is fixed. The nawralperiod of this system is 0.40s. Determine the natural period when a mass of 2.5 kg is attachedto the mid point of this spring with the upper and lower ends fixed. Ans. (0.14 sec)

2.9 Determine the differential equation of motion of the system shown in Fig. 2.36. The moment ofinertia of weight W about the point 0 is Jo' Show that th~ system becomes unstable when:

K.ab >-- W

...'1

-..-

Theory of Vlbrations 65

KWt") T

b

0 c-1

A

~a -1

l .~

Fig. 2.36 : Mass-spring system

2.10 A body vibrating in a viscous medium has a period of 0.30 s and an inertial amplitude of30 mm.Determine the logarithmic decrement if the amplitude after 10 cycles is 0.3 mm. Ans. (0.46)

2.11 A vibration system consists of mass of 6 kg, a spring stiffness of 0.7 N/m and a dashpotwith a. dampingcoefficientof 2 N-s/m.Determine(a) Damping ratio

(b) Logarithmic decrement Ans. (0.488, 3.55)2.12 Write a differential equation of motion for the .system shown in Fig. 2.37 and determine the

.natural frequencyof dampedoscillationsand the criticaldampingcoefficient.

., .

w .,b

a r K

Fig. 2.37 : Mass-spring dashpot system

2.13 A mass is attached to a spring of stiffness 6 N/mm has a viscous damping device. When the masswas displaced and released, the period (jfvibration was found to be 1.8s and the ratio of consecu-tive amplitude was 4.2 to 1. Determine the amplitude and phase angle when a force F = 2 sin3tN actson the system. Ans. (0.708mm, 56.4°)

2.14 A sp~ingmass system is excited by a force Fa sin (J)t. At resonance the amplitudewas measured. td be.100mill..At 80%resonantfrequencythe amplitudewasmeasured80 mm.Determinethe

. d~mpingfactorof the system. Ans (0.1874)2~r5:AssUnlfngsm~ll ~plitudes, set up differential equation of motion for double pendulum using

the coordinates shown in Fig. 2.38. Show that the natural frequencies of the system as given bythe equation. .

co'I 2 = ~, g.(2 :1:/2)'\11,. vi

Determine. the ratio of the amplitudes xI/x2';.:':!Z~,"'~.'T"""7~.. ,.. .. I . I

~~...~..,.:",,!..,:.;..~ rt .."'. ..~ ':.."h.e. ! ,,- "- - ..J

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

'1'1' '\""I\~ .,t;\.~;...t-, )J .J ~ ~"V :; ~,. :.>'1;'"; 1./uJ j

..

...

:i:

66 Soil Dynamics & Machine "Foundations

m

Fig.2.38: Doublependulumsystem

2.16 A motor weighs 220 kg and has rotating unbalance of 3000 N-mm. The motor ~s running atconstant sped of 2000 rpm. For vibration isolation, springs with damping factor of 0.25 is used.Specify the springs for mounting such that only 20 percent of the unbalanced force is transmittedto the foundation. -Also determine the magnitude of the transmitted force. .

Ans. (Ka = 931.22 kN/m, 26.3 kN)2.17 A small reciprocating machine weighs 60 kg and runs at a constant speed of 5000 rpm. After it

was installed, it was found that the forcing frequency is very close to the natural frequency of thesystem, What dynamic vibration absorber should be added if the nearest natural frequency of thesystem should be at least 25 percent from the forcing frequency? 6

Ans. (15.3 kg, 4.2 x 10 N/m)2.18 A mass of 1 kg is to be supported on a spring having a stiffness of 980 N/m. The damping

coefficient is 6.26 N-s/m. Detelmine the natural frequency of the system. Find also the logarith-mic decrement and the amplitude after three cycles 'if the initial displacement is 0.3mm.

Ans. (31.14 rad/s, 0.628, 0.0456 mm)2.19 A machine having a mass of 100 kg and supported on springs of total stiffness 7.84 x 105 N/m

has an unbalanced rotating element which results in a disturbing force of 392 N at a speed of3000 rpm. Assuming a damping factor of 0.20 determine(a) the amplitude of motion due to the unbalance,(b) the transmissibility, and(c) the transmitted force. Ans. (0.043 mm, 0.148, 58.2 N)

2.20 The static deflection of the vibrometer mass is 20 mm. The instrument .when attached to a ma-chine vibrating with a frequency of 125 cpm records a relative amplitude of 0.03 mm. Find outfor the machine,(a) the amplitude of vibration,(b) the maximum velocity of vibration, and ,

(c) the maximum acceleration of vibration. Ans. (0.0576 mm, 0.754 mmlsec, 9.86 mmlsec2),

DD. "

,::~,"'.,':"'.":"'V ",.. ,,;,,)' ~"<i:;','<' ",}~",,"'~., ':~,{""J""'""",,,--- --- ..---

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

WAVE PROP AGA TION IN AN ELASTIC, HOMO-GENEOUS AND ISOTROPIC MEDIUM

3.1 GENERAL

A sudden load applied to ~ body does not disturb the entire body at the instant of loading. The partsclosest to the source 'of disturbances are affected first, and the deformatio~s produced by the disturbancesubsequently spread through out the body in the form of stress waves. 'The propagation speed of seismicwaves through the earth depends on the elastic properties and density of materials. The phenomenon ofwave propagation in an elastic medium is of great importance in the study of foundations subjected todynamic loads.

In this chapter wave propagation in (i) an elastic bar, (ii) an elastic infinite medium and (iii) anelastic half space have been discussed.

3.2 STRESS, STRAIN AND ELASTIC CONSTANTS

3.2.1 Stress. The external forces acting on a body constitute what is called the "load", No material isperfectly rigid, therefore the application of a load on a body causes deforn1ation. In all cases internalforces are called into play in the material to resist the load and are referred to as "stresses". The intensityof the stress is estimated as the force acting on unit area of cross-section, and is expressed in such unitsas N/mm2, KN/m2. etc. At right angles to the direction of the stress, the body dilates or contracts,depending on whether the stress is compressional or tensile. The stresses preserve the shape of the bodybut change the volume. Shear stress is said to exist on a section of body if on opposite faces of the sectionequal and opposite forces exist.

3.2.2. Strain. Strain is a measure of the deformation produced by the application of the external forces.In Fig, 3.1a, the deformation is an elongation of a bar by the amount t11, and if 1is the initiai length ofthe bar, then

, Tensile strain = £,= ~1 , ...(3.1a)Similarly in Fig. 3.tb the deformation is a shortening of the bar by the amount t11, therefore

. . t11 3 1 bCompresslve steam = £c =T ...(. )

------

- R:.'

68 Soil Dynamics & ~~e FlHUldations

F

..i6l

... f

l

1III ; 1 ;TL ~

tF

l

(a) Tensile strain (b) Compressive strain

Fig.3.1 :Axialstrain

3.2.2.1. A transverse strain: A transverse strain tw is defined as the ratio of the expansion-or contrac-tion 6 w perpendicular to the direction of the stress to the original width w of the body (Fig. 3.2). Thus

6wEw = -;- ...(3.2)

w

l- llW--I.-- --7I,

II

II

II

II

I

F

Fig. 3.2 : Transverse strain

3.2.3. Elastic Constants. An elastic material is one which obeys Hook's ~aw ofpruportionally betweenstress and strain. For an isotropic elastic material subjected to normal stress Oxin the x-direction, thestrains in x, y, z directions are

C1t =--!.x E

t =e =-~Y Z E

~

Wave'P,;op'agat;On"l" lin Ettistii!/llo",:rigeneous and Isotropic Medium , 69

If the element cif material i~:Subjected to normal stress O'x'O'y"O'z'then by superposition we obtain1

Ex = E [O'x - Il (O'y + O'z)]1 ; ,

Ey = E [O'y - Il (O'x + 0')]I".

Ez = E [O'z - Il (O'x + O'y)] ...(3.4 c)

In the above expressions, E is the modulus of elasticity and 11is Poisson's ratio. It may be noted thathere E is dynamic modulus of elasticity and ,not the static modulus.

'Equations (3.4 a), 3.4 b), an:d (3.4 c) can be rearranged so, that the stresses are expressed in terms ofthe strains as follows: (Timoshenko and Goodier, 1951; Kolsly, 1963).

...(3.4 a), "

...(3.4 b)

,'., ' IlEO'x = (1+'Jl)(I-21.1)

. ~..

JlEay = (1 + Jl)(I-2Jl)

E(Ex + Ey+ Ez) + 1 + Il Ex ...(3.5 a)

E(E + E + E) + 1+ 11 Ey.t y r ...(3.5 b)

.' JlE,.. E

az = (1 + Jl)(I-21.1) (Ex+ Ey+ Ez)-t: 1+ fl Ez

For simplicity the equations may be writtena =A. E+2GEx xa =A. e+2Gey ya =A. E+2Gez z

...(3.5 c)

...(3.6 a)

...(3.6 b)

...(3.6 c)in which -

E = Ex+ Ey+ EzI.1E

')..-. - (1+fl)(I-2Jl).. .,,'. .

. E

G.= 2(1 + 11) ...(3.9)Similarly in an isotropic elastic material, there exists linear relation between shear stress and shear

strain. Thus

...(3.7)

...(3.8)

'txyYxy= G

'tyz. I'yz =. G

...(3.10 a)

...(3.10 b)

CrY:x=6

G is the shear modulus or rigidity modulus and is the same as given by Eq. (3.9).Equations (3.6) and (3.10) comprise six equations that define the stress-strain relationship.

,',

...(3.10 c)

,

8B:,1

70 Soil DyIUUlfics 4 Machint! Foundations

3.3 LONGITUDINAL ELASTIC WAYES IN A ROD OF INFINITE LENGTH

ax~ ~ OX + oOX b.xox

~ 6x ~

A = Area of crosseciion

Q

x

a b

x~~x ~~ u (Displacement)

Fig,3.3 : Longitudinal vibration of a rod

Consider the free vibration of a rod with crossectional area A, Young's modulus E and unit weight r

(Fig. 3.3). Now let the stressalong sectionaa is crx and the stress on section bb is (crx+ °ocr;.6.x).Assuming that the stress is uniform over the entire crossectional area and the crossection remain planeduring the vibration. the summation of force in x-direction is given by:

(ocr

)ocr'

IF =-cr A+ cr + :L6,x A=---L.6,x.Ax x x ox ox

If the displacement of the element in x-direction is u, the equation of motion for element can beNritten by applying Newton's second law of motion as given below:

ocrx.6,x.A =(

6,x.A. 'Y)

.cluox g at2

...(3.11)

or2

ocrx - 'Y aufu - g'dt2

...(3,12)

Form stress-strain relationship E = stre~s~x -d~ection = ~stram m x -direction oulaxoucr = E.-

x oxDifferentiatingEq, (3.13)withrespectto x

ocr.t - E a2uox - ox2

...(3.13)

...(3.14)

, --- - ".,.::',

',',-Y-;'P{,,',-,-*"r'1,h~"";;:':'..~,""-'""".*: ,,'*'" ""':;""':-' -,/ .,' <r'::" ,- .;o.,ft~:t-\>:~,:'

Wave Prop~gationin an Elastic, H.omogeneous and Isotropic Medium, " -- ' ." . .'71

'Y a2u = E clug at2 ax2

Using the mass density p = 'Y, Eq. 3.15 can be writteng ,

, a2u ,E clu- .at2 = p ax2

a2u 2 cluor - = v -at2 c ax2

Ewhere v; = P ...(3,18)

Vc is defined as the 10ngitudinal-wave-propagation-velocityin the rod. Equation (3,17) has the exactform of the wave equation, and it indicates that during 10ngitudiIia~vibrations, displacement patterns arepropagated in the axial direction at the velocity vc~' : :; ;' '

The solution ofEq. (3.17) may be written'in the form' .

u=/l(vct+x)+/2(vct-x) .r,. ...(3.19)

where I1 and 12 are arbitrary functions. In this equation, the first term ?iepresents the wave travelling inthe positive x direction, and the second term represents the wave travelling in the negative x direction.

If the wave propagation in a rod is considered at SOIpt::interrpediate point in the bar, it can be notedthat at the instant a wave is generated, there is compressive stress of the fac~ in the positive direction ofx and tensile stress in the negative direction of x. Hence, when the compressive wave travels in onedirection, the tensile waves travel in the opposite direction.

To understand the difference between the wave propagation velocity Vcand the velocityof particlesin the stressed zone it, consider the stressed zone at the end of the bar as shown in Fig. 3.4 a. When auniformly distributed compressive stress pulse of intensity crxand duration: t".(Fig, 3.4 b) is applied tothe end of the bar, initially only a small zone of the rod will experience the compression. This compres-sion will be transmitted to the successive zones of the bar as time increases. The transmission of the

, '

compressive stress from one zone to another occurs at the velocity of the wav~ propagated in the medium,i.e. vc' During a time interval .1.t, the compressive stre~s will trave,l along the bar a distance (.1.x = vc'~ f).At any time after In' a segment of the bar of length, xn = vctn, constitutes the compressed zone. Theamount of the elastic shortening?f this, zone is given by

Therefore, ...(3.15)

, ,

, ,-;- i ...(3.16)

...(3,17)

cr cru=-Lx =-Lv ,tE n E c n ...(3.20)

u cror t;: = E Vc ..' .., < , , ...(3.21)The displacement u divided by In represents the velocity of.the ~nd of the bar or particle velocity.Hence ,

. crxu =-vE c

It is evident from Eq. (3.22), .that the particle waVevelo~ity it .dependson ,theintensity of stress but. .' .. , , .the wave propagation velocity V (Eq. 3.18), is only a function of mate~ial properties. Further both wavec '" ',' . '. ,~,,' 'propagation velocity and particle velocity are in the same direction when a compressive stress'is appliedbut the wave propagation velocity is opposite to the particle ve19cj~.when,a tensile,stress is applied.

. ,,', !', ," ( " . t.. ; " .'

...(3.22)

~

72 's~u Dynamics & Machine Foundtiiions

~x

-.J I+-x

x

(a) Stressed zone at the end of the rod

er

erX

, "

- ttn

(b) Uniformly distributed compressive stress

Fig. 3.4: Wave propagation velocityand particle velocity in a rod. "

3.4.<TORSIONAL VIBRATION OF A ROD OF INFINITE LENGTH

In!if:.3.S G, a rod subjected to a torque T which produces angular rotation e is shown. The expressionfo~'t.: :toraue car. be written as

1. -T = G I Be

. " P Bxwhere G = shear modulus or"thc'material of rod -,'

. I '= Polar moment o:f inertia of the crossection Of rod,p ." , ! .~ ,ae'. _..,.'.;;A'""~,,,. jf,.'..«ft'll!"-ex =:Angle of twist per unit length of rod.

...(3.23)

It- .-,.--'-'

OX....(

Xn = Vc tli,"

)(

.-

u

la B - _3i.;'1£h

,!,-~e;Pr~g~!p,.;II#E.l!.t~~J.J:lfI'RlJIIeneous an41sotrQpic Medium. . ,.,., ';' , .; '. ;.

73

L

' '-'-'---

.\ .\\,I x

x~X r-

. (a) A rod siabjectedto torque T , ,

;';.'

T + aT t::.xoX

r fix I';,," '... ,

, .. , ": ': '

.: (b) Motion of the element of rod of length Ax

Fig. 3.5 : Torsional vibration of a rod

The torque due to rotational inertia of an element of rod of length 6.x can be written as#eT = pI~x~

P :ot2By applying Newton's,second law of.mo~onto.arielement oflength 6.x shown in Fig. 3.5b,

.. ,,

(, aT-- ,) . #e" ".

- T + T+ Ox 6.,x = p!p*x-y',' .. at

.,,(3.24)

or aT =,oI,Jre;" ox P p:!> 2

~ ,1'01 l\'1'.I ; "vt

" ,.,,(3.25) .

,o.

" ,'.', ,,": '."': "'. " """. ' ""," "'" .,

Soil Dynamics & Machine' Foillldations74

or

, oT '.. a2a- =GI-OX p ax2

or a2a ,= pI ciaP ax2 P 8t2

ia - 0 ciaa? - P ox2

, a2e a2e- 2'~ - Vs --z-at", . ax

t. '

From Eq. (3.23), ...(3.26)

Therefore,

,J3.27)'~':' ..-.

"

,J3.28)

--

"" is the shear wave velocity of the material of the rod.

3.5 END CONDITIONS

Free End Conditions, Consider an elastic rod in which a compression wave is travelling in the positive,r-direction and an identical tension wave is travelling in the negative x-direction (Fig. 3.6 a). Wh;n thet\\/O\vaves pass by each other in the crossover zone, the portion of the rod in which the two waves aresuperposed has zero stress wit:.1:ltwice the particle veioc.~tyof either wave (Fig. 3.6 b). After the two waveshave passed the crossover zone the stress and velocity return to zero at the crossover point and both the-:ompressive and tensile waves return to their initial shape and magnitUde (Fig. 3.6 c). It will thus be seenthat on the centre line cross-section, the stress is zero at all time. This stre~s condition is the same as that\\hi-:h exists at the free end of the rod. By removing one-half of the rod, the centre line cross-section canbe considered a free end (fig. 3.6 d). Hence 'it can be seen that a compre,ssion wave is reflected from a freel'l1d as a tension wave of the same m,agnitude and shape. Similarly, it can be observed that a tension waveIS reIkcled from a free end as a compression wave of the same magnitude and shape.

(a) Compression and tension wave~ travelling in opposite directions

~

~~ 0-'=0

~,. .00 ,,' U = 2uO.--q-nrn00'---

, Vc

y ~Xt1

(b) Waves at the crosso~'er zone

Fig. : 3.6 : Elastic waves in a rod with free end conditions (...Contd.)

'"..,'..., ,..' --,

Vc.tId-: 0----

oo

r=oXt

ien sion "00

-v c

l4~' \~~)'" ,'" "'.~ :,. , y.~:,~. "~ ;i0: <~~- "';':;:~~!-""i"" ',' 'c: :,'"

f'ave.Propagation in ,an El~$tit;, Homogeneous and Isotropic Medium 75

.h ~ . er

. "0' er = a' 0- Vc I u = 0

~ .. Xt2

(c) Waves after passing the crossover zone

Vc~

.~~

Free enq .. Xt3

-4Vc

(d) Waves considering one half of the rod

Fig. : 3.6 :'Elastic waves in a rod with free end condWons

Fixed End Conditions. N°'Y,~onsider an,e.lasticrod in which a compression wave is travelling in positivex-direction and an identical compression wave is travelling in the negative x-direction (Fig. 3.7 a),Whe'nthe two waves pass by each other in the crossover zone, the centreline cross-section has stress equalto twice the stress in each wave and zero particle velocity (Fig. 3.7 b). After the waves pass each other,they return to their original shape and magnitude. The centre line cross-section r:~mainsstationary dur-,ing the entire process and hence, behaves like a fixed end of the rod. Considering left half of the rod (Fig.3.7 d), it can be observed that a compression wave is reflected from a fixed a fixed end of a rod as acompression wave of the same magnitude and shape, and that at'the fixed end the stress.is doubled.

'v-4

~~o.

I. er=0

~u=oI

Vc-~~ .Xto

, (a) Two Identical waves travelling in opposite directions

er = 20-0

, j' U = 0

Xt1

(b) Waves at crossover zone

Fig. 3.7: Elastic waves in a rod with fixed end conditions (...Contd.)

- -,- -- - - - -----

III'JIE,J.

76 $Dil Dynamics & Machine Foundations.

G

Vc : Vc~, I '--

"0 ~.. ..~'. 0-::0 '

"/ u :: 6".~ '-'- Xtz

(c) Waves after passing"the crossC)verzone

Vc , Vc 'l~" " ~

Fixed endxt3

(d) Waves considering one halfofthe rod. " "

Fig. 3.7: Elastic waves in arod ~ith fixed end conditions'

3.6 LONGITUDINAL VIBRA TIONSOF RODS OF FINITE LENGTH "

Consider a rod of length L vibrating' in, one of its normal mode~ The solution of t~e wave equation(Eq. 3.17) can be written as '

; u = U (AI cos (J)nt + Az ~ii1con1)where U = Displacement amplitude along the length of rod

AI' A2 = Arbitrary Constants

(I) 1/ = Natural frequency of therodSubstituting Eg. (3,29) in Eg. (3.17), we get

...(3.29)

.,

d2 U + oo~U = 0~ 2dx Vc

...(3.30)

OOnX OOIlX 'The solution of Eq. (3.30) is U = A) cos - + A4 Sin -'"--, ~ ~.'\, and A4 are arbitrary constants which are detenrtined by satisfying the boundary conditions at the

ends of the rod.Three possible end conditions are:

1. Both ends free (free-free)

2. One end fixed and on~,end fr~e,(fixed-:fre~~

3. Both en~s fixed (fixed-fixed)', .' :" ,", .

...(3.31)

Free-Free Condition... " '"..

The stress and ,strainat both ends of ~ rod of finite lertgthin ~ree-freecondition (Fig. 3.8 a) will be;/l:ro, Th'is means that dUMx= 0 at x ~'O x :i,;L-. ,- . '

, ,; ,.:' '" :: ' " . ' \ .

. ~',' ,. ,',' , ':' ,,'~-j ,.0

,,-

I.' ;; ",_:::_, :;; ":_'~,;

,:<., .,

- " ",,'!.. "

Wave ,PropagaC;{}1Jin, an Elastic, H~m,9.geneous and Isotropic Medium

~T

u ..-x

(a) Rod of finite length with free-free end conditions

. ' ,'. 0'

U1= A~ cos ~x (n = 1 )

U2 =A3 cos 2Tfx (n = 2)L

U3=A3 cos 3Tfx (n = 3 )L

(d) Third harmonic

Fig. 3.8 : Normal modes of vibration of a rod of ~nite lengt~. ~Ith free-free ,end conditions

"

Differentiating Eq. (3.31) w:!-t"x,we getdU ron

(A . 'ro"x <o"X

)- = - - 3sm-+A4cos-dx \! ' V Vcc' c

The condition ~~= 0 at .~ .=;? g.ives.A~=,O.,,' "'.J! -l1:>,..!fi 't"L}",t...

0-1 El

l-x -I~dX

Ô

ßÖ

Ìß́í

ø¾÷ First harmonic

,'~ (c) Second harmonic

, ;'.

~l~k'

77

0

AJ

I

78 ", - Soil Dynamics & Machine -Foundations

PuttingdU- = 0 at x = L we getdx '

. ro LA sIn -!! = 0J vc

roLFor a nontrivialsolution, -!! = n1t

Vc

...(3.33)

- n1tvcor ron - -r.-. n = 1, 2, 3 ... ...(3.34)Equation (3.34) is the frequency equation'for the rod in free-free case. By substituting Eq. (3.34) in

Eq, (3,31). we getn1tx

Un = AJ cos L ...(3.35)l::c,:;,.' L::q,(3.35), the distribution of displacement along the rod can be found for any harmonic. The

rirsl three harmonics are shown in Figs. 3.8b, c and d.

Fixed-Free Condition.

[11Fixed-free case (Fig. 3.9 a), the end conditions of the ro<1are:(i) At x = 0, Displacement i.e. U = 0, and

(ii) At x = 1, Strain i.e. d U/dx = 0

L -tx

(a) Rod of finite length with fixed-free end conditions

~I~A4

TU)

. TTx )= A4 sin =2L(n = 1

(b) First harmonic

U2=A s' 31Tx4 In- 2L (n=2)

(c) Second harmonicFig.3.9: Normal modes of vibrations of a rod offinite length with fixed-free end conditions (...Contd.)

"<' >,"'"

Wave Propagation in tin Elastic, Homogeneous and Isotropic Medium 79

, ; 'SlTXU3=A 4 slO (n=3)

, ,,2L

(d) Third harmonic

Fig. 3.9 : Normal modes ofvibrations ora rod of finite length with fixed-free end conditions

Putting the first end condition in Eq. 3.31, yields AJ = O. By substituting the second end conditionin Eq. 3.32, , " .

" W LA cos -2L = 0

4 vc...(3.36)

wnL 1t- = (2 n - 1) -v 2c '

1tVcor wn = (2 n - I) 2 L 'The displacement amplitude can be written as

U - A . (2n-I)1tXn - 4 sm 2L

The first three harmonics described by Eq. (3.38) are shown in Fig. 3.9 b, c and d.

Fixed - Fixed Condition.In this case (Fig. 3.10 a) the end condition are:(i) At x = O. U = 0 , and '

(ii) At x = L , U = 0

or" .

,. '....(3.37)

...(3.38)

~

~

:L ..,

I ..x

(a) Rod ornnlte length with fixed-fixedend conditions

U -A . TTx( 1 )

, 1- 4..SIn L n =( .

(b) First harmonic

Fig. 3. to: Normal modes ofvlbration oh rod of finite length with fixed-fixed end conditions (...Contd.)

80

f

(c) Second harmonic

(d) Third harmonic

~, ".~~£~

Soil, D.ynamics ~ Machine f~u~cJptions

. 21TxU2=A4Sln-C-(n =2)

U _ A . ,3TIx ( 33 - 4sIn L n = )

Fig. 3.10: Normal modes of vibration ora rod of finite length with fixed-fixed end conditions

Putting these end conditions in Eq. (3.31), we get

A3 = 0, andoonL

A4 sin - = 0Vc

orn1tx

oo=-n-1 23n L' - , , , ...

Th d 'I

.. b U A. n1tx

e ISp ac,ement IS gIven y, n = 4 sm LThe first three harmonics described by Eq. (3.41) are shown in Fig. 3.10 b, c and d.

...(3.39)

...(3.40)

...(3.41)

3.7 TORSIONAL VIBRATIONS .oF JlODSOF FINITE LENGTH

As the wave Eq. (3.28) is identic<\~t~ wave Eq. (3.17) , the proble'm' of torsional vibr~tions of rods of finitelength can be solved in the saITi~manner as foJ. tJ:le-case <?flo~gi,tudinal vibrati.ons discussed in theprevious section, Th,e solution .of Eq. (3.28) can be written as :

. e = eA(A I cos OOnt + A2 sin oont)

where eA =Rotational amplitude of angular vibrationI "

A I' A2 = Arbitrary constants(j) n =Natural frequency of the rod

Substituting Eq. (3.42) in Eq. (3~28) , we get,2 2~ ~ e '

2 + 2...=0. dx' v's

...(3.42)

...(3.43)

Wave-Propagation'in'an"Elilstic/ Homoggneous and Isotropic Medium 81-

The solution of Eq. (3.43) is

e - A ,conx . COnXA - 3cos -+A4 sm-

Vs: Vs

The values of arbitrary constants A3 andA4 can~eobtained by puttingoappropriate end conditions.The solutionfor the three types.of end c.onditions.are givenbelow: .

...(3.44)

Free-Free Condition.

n 1tVscon = L ...(3.45)

e - An7tx

An - 3cos- L ...(3.46)

Fixed-Free condition.

(2n-I)7tvscon= 2 L ...(3.47)

" ,

eAn =A4sin (2n-l)7tx2L ...(3.48)

Fixed - Fi,,<:edcondition.

n 1tVs(On = L ...(3.49)

eAn = A4 ~1tXL ...(3.50)

3.8 WAVE PROPAGATION IN AN INFINITE, H~MOGENEOUS~ISOTROPIC,ELASJICME-DIUM '

In:this section,;theprop~gation of stress waves,in:anj,nfinite"homogeneoou$~~~Qpic. elaiUc ,~di.um, ~presented.'Eig 3.ll 'showsthe,stresse5Jicting:on a;S0il,eletneni.witb.sides~aucingdx, dy.'il~d<ftZ.Thesolid vectors are acting on the visible faces ofthe element and the dotted vectors areacting,on:t.h~hidQ.~nfaces. For obtaining the differential equations of motion; the.sum of the forces acting parallel to each axisis considered~In the x.,direction the ~quilibrium equation is

[", -(", + aa"; d X) ]<dYodz)+[V'( 'n + a;;)dz ]<dxody)

( (U't

)]ilu

+ 'ty:c- 'tyx+ a;'dY. (dx.dz)+p(dx.dy.dz)at2:::10...(J..5.0

or,,cf." aox' a~.xy':"~.tt'p- =- +- +............

. ':0l2 ox ay' j)z..~{3.52 a)

,', .-

" ".-..-....

8~~ ':' ,"', ~" .,' " ";":- , ~ . SIJU Dln9nUc;s. & Machine F o.ullllations

z ...I ,~.

'

(1':+ ~'tn dZ}ZI az

-." ~.."dy, itr..

, , crz+ ~ ,d-z ), ,

t" I er.yl I -r,. ('tu+ ~ IY ('

("", It

ax' dl) I

cry II

I , tt I ~IZ

1:yz

"" ..

~-ryz't'yz+ ~,!lY),

(GY + ~cry dy)C)y

('t'yx + ~f)~YX .dy)

I

~ax(cr. + ~.d.)

Fig. 3.11 : Stress on an element of an infinite elastic medium

Equations similar to Eq. (3.51) can be written for the y -and z -directions. These will give

;iv (h yx ocry o't yzp- = -+-+-ot2 ox ay oz2 .'

a w o't o'tzy' ocrp- - --M...+-+---L (3 52 c)ot2 - ox ay 07 ' ... .In the above expressions,p is the'imiss density of the soil, u, y'and ware :displacementsin the x, y,

and z directions respe'ctively.To express the right hand sides of Eqs. (3.52) , the relationship for anelastic medium given by,Eqs.,(3.6) to' Eqs. (3.10) are used. The equations for strains and rotations ofelastic and isotropic:materials in terms of displacements are' as follows,:' 'Axial strains, " :;'

...(3.52 b)

, . ouEx = ox .

avEy ='ay

awE;: = ay

. ;,

,...(3.53 a)

...(3.53 b)

...(3.53 c)

Shear strains,

ov ouYxy'" 0;C+ ay

ow' av','--+-Yyz- ay oz

ou owyzx= OZ+ ox

...(3.54 a).

...(3.54 b)

~, ...(3.54c)

,-) .., ,

~< 2.,:,'

"',,',"'l,:,". :,.",~,~: ":",),,'0' , :' ..Wave Propagation in an ElastIc, Homogeneous and IsotropIc MedIum

J.....8'3

Rotations.ow 8v'

2w = ---Xo y oz.. ';,

iJu' .;iw2w - ~--y - oz oxov ou2w =---z ox oy

In E:q,'(3.:~5);"wx' w/ ,wz 'repie~~Ilt'the i~tati~rt~ 'aborttx;y~rid'~iaxes'respettivetY..': ,,' """"""""":'-"""':""'/:~";",""i' "",1"", ",' ',',"

...(3.55 a)"

.:.(3.55 b)

,..(3,55 c)

3.8.1.CornpressiQD Waves. Subst.itut~onoff:qs.p,. 6 a), (3.10 a), ,an~;(~,lO b) into Eq. (3. 52 a) gives',' """j;U"d" (~'-"' 2G ' )

'd(O

"'

)d

(G

'

)', '" '

, p-,= - ",E+ E.. +~ '1,' +-- '1, "",ot2 "d.,J." ,X,dy,xY"dZ ,xy

, ," .., " .. ' ", ' 0,' ",. " ,

Now on substitutionof Eqs. (3.54 a) and (3.54,b) in Eq. (3.56), ~e get,

, ' ,', '~~i' ~'~(A<'~';~E,)+GM~: ~~;FG:z(~>~;r, ,', "", ' co" , -/,'

"'

[

'2 2 '2' 2 2 2

], 02u - A.d E + G a u + ~ + d w' + a u + a u + a up 0(2 ,-: ax . ax2 dX,ay' dX'az ax2 al di. . ,,'.. '

02u 02v azw" OE-+-+- --ox2 OX'ay ox.fJz - ox

The Eq. (3. 57) can be written as , ..

azu ' o'E 2

p ot2 = (A. + G) ox +? ~ u

or ...(3.57)

As ...(3.58)

...(3.59)where

" ,

2 az az azV u = -+-+---'-ox2 ay2 fJz2,

Similarly Eqs. '(3.51 b) and (3.52 c) can be expressed asazv ... aE 'p- = (A. + G) - + G V20(2 ay

...(3.60)

.'.~(3.61)and " "

02W aE 2P-y = (A. + G) ~ + G V w

, &~, '

Equations (3.59) , (3.61) and (3, 62) are the equations of motion of an infinite homogeneous, iso-tropic, and elastic medium. On differentiating these equations with respect to x, y and z, respectively, and

, adding

...(3.62)

;'"

fil

(au Ov Ow

) (a2E 02£ file

)2

(au Ov Ow

), P 0(2 Ox+ ay + Oz' ',= (A + G)'fix2 + aji2 t"fJz2' + G V at'+ ay + fJz: ",."'::"}J, :";"; !"..i-~),':J.' ','.' 'c,.'~';y:!..;, ';". '.:.',:~(TE

(k G) 2 ' 2 ',. ."",..",,-,,'-=',' ""', """"""""q,,, , '.."'. v ;" . ,~,p "2'" "" +-, (V E) ,+ ,(GV E), '..d" ',J. ,".. '. ",C "I, ~ ':'J",)",;,q~, \,-O~i'~',_",\,,;v~' .".'" ""J r"!'" "" ,,' .~f..'

,P"",)..".I,;t"",'jl',ttf.~<l;.U;4;"d~,'H. '1-"'>""1';'_',,1,,1'1.,;,,,., ,'." ';. n,...O~;'J1

"

or.,. ;,

'., < "," ,,', L«'\' ;,

84 "s.iJ ,J)yIUVflil:s..& MadMe' F~IIRd4tU1ns

or

where

p ilE = (A.+ 2G) V2E&2

<YE - (A.+2G) (V2£)= V;V2f,. at2 - . p

...(3.63 a)Hem:e

..~(3.63 b)

2 A.+2GVp = P . ...(3~64)

vp is the ve1ocityofcompre~on waves which are: a1soreferredJls primary wave or:P-wave.It is important to note the.difference in the wave velocities for an intmite.elasticmedium with those

obtainedfor an elasticrod. In the rod Vc= ~E/p: but in the infinite medium vp-~ ~(A.+2G)/p. Thismeans that Vp >Vcthat is 'compressionwave travels faster in infmite medium. It is due to the fact that ininfinite medium, there are no lateral displacements, while.in the rod lateral displacements are possible.

. '" ,

3.8.2 Shear-Waves; DiffererniatingEq.'(3:61) with:resJrect to z and Eq. (3. 62) with respect to y, we get2

( ) a-a av E 2avp at2 az: = (A.+ G) (ay)(az) + G V az

a2p

(aw

)ae 2 aw

P;z -;- = (A.+ G) -;-;-+GV -ut ay ayuZ aySubtracting Eq. (3.65) from Eq. (3.66), we get

p~ (aw- av

)= G V2

(aw- av

.

.

)at2 ay az ay azaw av -

From Eq. (3.55 a), ay - az = 2 wx' Therefore

rJw 2,p--f = GV"wxat

...(3.65) .

and ...(3.66)

...(3.67)

or rJwx G 2- 2 2-- =.-"11 w = V' V wat2 p x s x

Similar expressions can be obtained for Wy and Wz as below:

...(3.68)

rJWy . G "11 2- 2 2--=- Wy =vVw.8t2 P s y

'rJwz' = G V2w = v2V2wat2 p, z s z

The above expressioMindicate"thafithe'Totation .is.propagated with velocity Vs which is equal to

~G I p. Shear wave is also referred as distortion wave or s-wavoe.}tmay be noted that shear wave propa-gates at the same velocity in both the-rod!andthe..infinite-niedium/Fig. 3.12 shows plots of shear wavevelocity and void ratio at several confuii,ng pressures for sands (Hardin and Richart, 1963).

...(3.69)

...(3.70)

W~ve :ProplIgiitionin 4" E/~'stic;"H~mogeneollS alid Isotropic Medium

- -(

,1.'85

- -J ;.

--

,-'N

,E'"

"~ "26.'0- '.lA::I-::I ,- ,

"tJ, 19:50E ,,-

390 Con'1ini'ng(pres'sore

300,+./11

/~<

0.5

360

--\11-

330

Round grains - Ottawa sand--- Angular grains -.Crushed quartz

E>....u~ 270~>~

,> 2400~~. .

.'0 210'~ 's:.V1

180

,150.'I '

120 L..:0.3 1.0 1. 1 1.2 '1.3

Void ,ratio(a)

I'.g. 3.12: Variation of shear wavev~locity IIndshear modulus with void ratio," and coririnlng'press,jre for dry sands (Hardin ani:!Richart. 1963)

. ,I,.. .~ -.' ,. i . -'" . ; .,.

, -, .,', ,'. ' .., "'".

, .,..~---

.' :-86

Et.

. .i SoU :!)yntUlli~ &I Machine Foundtllions

3.9WAVEPROP AGATION INELASTIC IL\LF,SP ACE .' . -

In an elaspc~lly homogeneous gr~und, stressed shddenly'at a point 'S' near its surface (Fig. 3.13), threeelastic w~ves travel outwards at different speedS.Two are body waves; that is, they are propagated asspherical,fronts affected only a minor extent by the free sUrfaceof the ground, and the'third is a surfacewave which'is confined to the region. near ~ fiee surface. ;

~G

x,

-I'~,vst

vp t

Vrt t"

5

Fig.3.13: Pulsefrontsofthe P,Sand R waves

The two b.ody waves as already d,escribed in the previous section differ in that !he ground motionwithin t~e pulse is in the direction of propagation (i.e. radial) is the faster 'f' (primary) wave, but normalto it (i.e. tangential,to the pulse front) is the slower 'S' (secondary) wave (Fig. 3.13). The stresses in the

I ,"

P wave~ which is ~fl°!lgitudinal wave like a sound wave in a~~,are thus due to uniaxial compression,while during VIepassage' of an S wave the medium is su~jected to shear stres~. The surface wave travelsmore slowly.than either body wave, and is generally complex.,This wave was' first studied by Rayleigh(1885)and laterwas,describedin detailby Lamb(1904), It is refe,rred'asRayleighwaveor R-wave.Theinfluence of Raleigh'wave decreases rapidly with depth.

Equations (3.59) ,(3.61) and'(3.62) may be used to study the characteristics of Rayleigh wave, Thehalf space i,~defined as the x-y plane with z assumed to be positive toward the interior of the half-space(Fig. 3.14) .Let u and w represent the displacements in the directions x and z, respectively and areindependent of y, then

~ " a",u=~+-ax ,az_a~ O\V,w----" az. ax

',,' ",', ' av' ",.,". .

where ~ and 'Ifare two pot~nti~l,functions. As ay = 0, th~'dh~ti~n.e"~f the wave can be written as

...(3.71)

...(3.72)

atH'~'j;

" ,""

Wav'i"Propagaitoidli an ElaStic;'Ho,,;(/geneous and Isotropic Medium 87

'0

-,

.. '- au,.:.,aw,

' :c.:..

(d2

,

4> d,

2'41

J

"

(#cpa2'V

Ö

,1:;'--+-- -+- + ,---dX dZ dx2 dXdZ ai, oxOz

- a24>a2cp' 2' ' "I:; ãóõóãªùþ

~ ~1 '

,,0

2 '. ' 'I':-: 'U.A.' Z '. ,

...(3.73)'or

,,'

Plane wave frontIr----L..-

,)/

/I

Ix I

I ',.I

/

//.//

./, /'/

//

//'

.//'

Z .///'

//..

, Fig. '3.14: Wave prop~gat!on in ~Iastic half space

Similarly the rotation in x-z plane is given by. Z 2 .

'" " ""' 2 """-"""& aw d'V + d'V oz'" " "",'7'",W~;'=az-ox =~ ~=,v'V

,. ?..:'~,.;.j .",..",d?",dz".., "

Substituting u and w from Eqs. ~3.71) and (3.72) i~ e~s:,(~' 59) and (3.62), we get

p l(

,~2~)

+ p ~(

az~l:'~ki\:,~:Cj)' ~;(V.Z~y~.p!G ,-2- (V2'V)',OX -,' ,':.",az...2".i;J"",\",;,,.~-,ox, :"""":'" :;Oz.,'

, D.).~tj;.i:>~f};~:#~,alk-. '6'!-'f,~'~"""J'" L rJ'1~rl"~ ll.":i!i'."~'," ,'" : ',' 't' i"f:"

, ,<"::"',>:":'{,~F,,>':r:'{..'~,:J:;~~: ~;" '" <-:' ",~/ ::":""'" ", , '" 0,' ""

;..(3.74)

...(3.75 )

,.. ~~,..~,.,...~~~=~..~ ,.,.,.., ""~""'j-'.,"~..-,. '-="..,

""-...

88 Soil DylUlmics & Machine Foundations

and p ~(

&$)

- p ~(

&'V

J=(A+ 2 G) ~ (V2$) - G ! (V2'V)

Oz 8t2 ax 8t2 v£. UA.

Equations(3.75) and (3. 76) are satisfiedif

&$ = A+2G V2$ = lV2$8t2 P P

&\jI G - 2 2 - 2- = -'V 'V = v 'V 'Vat2 p s

Now, consider a sinusoidal wave travelling in positive x direction. Let the solutions of q>and 'V beexpressed as

...(3.76)

...(3.77)

and ...(3.78)

~ = F (z) ei (wr- nx)'V = G (z) ei (Wt - nx)- --

F (z) and G (z) are the functions which describe the variation in amplitude of the wave with depth,and 12is the wave number given by

...(3.79)

...(3.80)

21tn =T ...(3.81)

Where L is the wave length.

Substituting Eq. (3.79) in Eq. (3.77), and Eq. (3.80) in Eq. (3.78), we get

- 0./ F (z) = v~ [Fit (l) -' n2 F (z)]2 2 2

- Cl) G (z) = v [G" (z) - n G (z)]s

Equations (3.82) and (3.83) can be rearranged as2

F" (z) - q F (z) = 02

G" (z) - 5 G (z) = 0

...(3.82)

...(3.83)

...(3.84)

...(3.8S)

where2

2 2 (J)q =n - 2

vp. 2

2 2 (J)5=n-2

VsThe solution of Eqs. (3.84) and (3.85) can be expressed in the form

F (z) = AI e-qz + A2 eqz

( )-5Z sz

G z = B I e + B2 ewhere A I' A2' B I and B2 are Constants.

A solution that allows the amplitude of the wave to become infinity is not possible; thereforeA2 = B2 = O.HenceEqs.(3. 79) and (3. 80) can be writtenas

'" - A [-qz + i(CDI-nx»)'t' - le'" - B [-J'z + j (wr - nx)]

- le. - ," .Now at the surface of the half space i.e. at z = 0, °z, 1::xahd~1:~.are equal to zero.

...(3.86)

...(3.87)

...(3.88 )

...(3.89)

...(3.90)

...(3.9] )

IIi IIIi iI;{~d'~

"

Wave Propagation in an Elastic, Homogeneous and Isotropic Medium 89

crz = A E + 2 G Ez==1 E + 2 G ;= 0,

(aw 00)'t =G'" =G -+- =0, , zx IZX , ' Ox az

Combining Eqs. (3.71) and (3.72) and the solutions of cl>and 'I' from Eqs. (3.90) and (3.91). Eqs.(3.92) and (3.93) can be written as' '

AI - 2iGns- - 2 2BI ' (A + 2G)q - An

, '2 2AI - -(Tt + S )BI - 2 i n q

Equating the right hand sides of Eqs (3. 94) and (3.95)2422 2 22 2 "I 22

16G n s q =(s +n) [(l,+2G)q -lI.n] ...(3.96)Substituting q and s from Eqs. (3.86) and (3. 87) in Eq. (3. 96), and dividing both sides by G2 n8,

we get

Therefore, (3.92)

and ...(3.93)

...(3.94)

and ...(3.95)

, .

16(

1- ~\)(

1- ~\)

=(

2- (A+2G). ~\)

2

(2- ~\

J

2

vpn vsn G vpn vsn...(3.97)

From Eq. (3.81)

Wave length = 21t = velocityof Wave - vrn (ID/2n) - (ID/21t) ...(3.98)

orID

n =-vr

...(3.99)

where vr is the Rayleigh wave velocity.Using the following relationships:

2 2 2ID ro vr 2 2

2'2=2 2 2=2=a~vpn vs(ro /vr) ,Vp

...(3.100) ,

and2 2 2

ID - ro - vr - 22'2- 2 2 2 -2-~Vs n vs(ro Ivr) Vs

22 Vs¿ãó®

Yp

...(3.10t)

where ...(3.102 a)

or

GIp= (A,+2G)/p

2.. G'~a"='(A,+2G)

. ,: ";} ,:J.~:~r2t.;~(:

t '

(3.102 b)"

, .t' ". " ',. ,. ", ':'i' ",+ . '/,: ::~:;,,; i;~'liJi,~n lJ'rU

..: i . :; ,',;.0 J-. ,. .'",. . , '", I ~..

.c..=;--'-' ,- -- .,,' '; ,.-

=- --~ === It.. c"

" , .. "" '.,. --'-' .,. '.-' .',- ,},

90 Soil Dynamics & Machine Foundations, , ,, , ,, ,-- ..'

Using the relations (3.8) and (3.9) , Eq. (3.10.2) can be written as2 1-2)l

øÈ =2-2)lSustituting Eqs. (3. 10.0.), (3. 10-1) and (3. 10.3) in Eq. (3.97),

16 (l-a2~2)(1-~2) ~ (2-~2)2 (2-~2)2 ...(3.10.4)

or ~6 -,8 ~4 - (16 a2 - 24) ~2 - 16 (1 - (X2):::;0. ...(3.10.5)

Equation (3. 10.5) is cubic in ~2. For a given value of ')l' Poisson's ratio, the value of ~2 can be

determined. Using Eqs. (3.10.0.)and (3.10.1), we may then obtain the values ofv,Jvp and v,Jvs' It may benoted that the value of ~2 is independent to the frequency of the wave. Therefore the Rayleigh wavevelocity is also independent to the frequency and dependent only on the elastic properties of the medium.

S

...(3.10.3)

vr/vs

0 -0 0.1 0.2 0.3 0.4 0.5

. J .Polssons ratlo)v

Fig. 3.15: Variations ofv r / v. and vp / v. with Poisson's ratio, Jl

Figure 3.15 shows the variation of v,Jvs an~ v/vs withPoisson ratio Jl.The three types of wave appear in order on idp..alizedseismogram (Fig. 3.16), which is a graph of

ground motion against time at a particular geopho~e ~ at a distance x from the source '0'. The time zero'is, of course, the time of the shot, and it is clear that ~he thr~e ve!ocities vp' Vsand vr could be found fromthis record. In practice, this determinatiords made'byco'ITl.,j.ningihe}nformationJromseveral geophonesat various distances from the source on time-distance graph as shown in Fig. 3.17." ' ,0

4

>I'0c0 3

"-I> >

....0 2tII:J-0>

a..,

rave Propagation in an Elastic, Homogeneous and Isotropic Medium91

Groundmotion

'"

X/Vs ~

Time

x/vp --x / vr ~

.'Fig. 3.16 : I~alised seismogram ofthe ground motion at a distance x from the source

, ,

..

Stope 1 /vr

u(\)\J\

Slope 1/vs

-

stope 1/vp

X,) m

Fig. 3.17 : Travel time graph constructed from a set of seismograms

3.9.1. Displacement of Rayleigh Waves. Substituting the relations developed for 4>and 'V {Eqs. (3.90)and (3. 91)} in Eqs. (3.71) and (3.72), we get .'

(. A - q=+

.B - s=

) i(rot - not)11 =- In le . (se e .

(A- q= B . :.s:

) i{oot - not)

w=- Iqe - "ne,. e

...(3.106)

...(3.107)

- -~ '. ~ .-

"...'. ., .11... "'dO".... . ' "0 . ," , i

. Now, substituting the value of BI in terms of Al from eq. (3.95) the above expressions can bewntten as

- A .(

-qz+ 2qs -sz)

i(oot-nx)U - 1m -e . 2 2 e es +n

(

. . 2

)- A -qz+ 2n . -sz i(OOt-nx)

W - I q -e2. 2 e es +nFrom Eqs. (3.108) and (3. 109), the variation of u and w with depth can be expressed as

U( } - <-qln)(l/z)+(

2(qln)(Sln»)

-(sIn) (nz)Z -e . 2 2 es In +1

W ( ) - -(qln) (nz) .+ 2 (-sin) (nz)Z --e 2 2 es In +1

Uc;:Tlg Eqs. (3.99), (3.100) and (3. 101), eqs. (3.86) and (3. 87) can be written as :2 . 2 2

q ro Vr 2 22" =1 -22=l-2"=l-a ~n n vp Vp

2 2 2s ~ 3:.. 22" =1- 22 =1- 2 =1-~" n Vs Vs

...(3.108)

...(3.109)

...(3.110)

...(3.111 a)

...(3.111 b)

and ...(3.112)

0 :H. rizontalcc mponent

Amplitude at depth zAmplitude at surface

0.6 .0.8- 6

O.41-~(Z)J

~

.I:.N..

0 g' 0.6= ~ V: 0.25..

)) - 0.33.. > -0 C 1.J- O.40

'" o..t ~~0.50

vertical

component

[W(Z>]

1.0

1.2

1.4

Fig. 3.18 : Amplitude ratio versus dimensionless depth for Rayieigh wave


~or a given value of Poisson's ratio, the values of v/vr and v/vs can be obtained from Fig. 3.15(Richart, 1962). Hence values of qln and sin are determinable for a given value of Po isson's ratio. As n= (21t/wave length), the variation of U (z) and W (z) can then be studied with respect to a non-dimen-sional term (z/wavelength). ,Such variation is shown in Fig. 3.18 for Poisson's ratio of 0.25, 0.33, 0.40and 0.50.

The amplitude of body waves, which spread out along a hemispherical wave front, are proportionalto lIx,x being the distance from the s()urce. The amplitude of the rayleigh waves, which spread out in a

cylindrical wave front, are proportional to l/.r;. Thus the attenuation of the amplitude of the Rayleighwaves is slower then that of the body waves. '

3.10 GEOPHYSICAL PROSPECTING

3.10.1. General. Geophysical exploration is relatively new area of technology. It involves measurementof some physical field such as electrical, magnetic etc. on the earth's surface and interpreting the data soobtained in terms of properties of subsurface layers of soil. The geophysical techniques most widelyemployed for exploration work are the Seismic, gravity, magnetic and electrical methods. Less commonmethods involve the measurement of radioactivity and temperature at or near the earth's surface and in. '.an.

In this section, the seismic method of geophysical exploration had been discussed. This methodutilises the propagation of elastic waves through the earth and is based on three fundamental principlesnamely (a) the waves are propagated with different velocities in different geological strata, (b) the con-trast between the velocities is large, and (c) the strata velocities increase with depth. It consists of gen-erating an elastic pulse or a more extended elastic vibration at shallow depth, and the resulting motionof the ground at nearby points on the surface is detected by seismic instruments known as geophones.Measurements of the travel-time of the pulse to geophones at variQus distances give the velocity of prop a-gation of the pulse in the ground. The ground is generally not homogeneous in its elastic properties andthis velocity therefore vary both with depth and laterally. ,

The real stratum, which in fact often consists of stratified material, is usually best approximated bya layered medium, each layer having a constant velocity or one changing in a simple and regular way.with depth. The interfaces between layers may be inclined at an angle to the horizontal and to each other.In this section few simple cases have been discussed.

Let us considerthe case of one horizontalinterfaceat a depthh1between media in which the com-pression wave (P-wave) velocities are vplare vp2'vp2being greater. Figure 3.19 shows the.possible pathsof the body waves generated from the source S. " - '

The first path as indicated by ray 1 is the same as the path of surface wave (Le. Rayleigh wave). Acompression or shear wave (ray2) striking an interface will generate two reflected (P and S) and tworefracted (P and S) waves (Fig. 3.19). According to the laws of reflection and refraction:, ,

,~in ip = sin rp = sin.rs,:= sin Rp = sin Rsvpl vpl vsl Yp2 . vs2

The equality of the angles of incidence and reflection (e.g. ip = rp) holds only if incident a'rtdreflected waves are of the same type.

...(3.113)

c;.on'"~ ...:.J.: ~'.

lE.

94 .. " "'Soil Dynamics & 'Machine~ Foundations' ',:

source <1)

p-wave

" "

"

v P1

I ' ,~' b

' a.

J

....~

-a..-. -..,,;" , .,.,.

'a.'

I

Q:-,

ir p

h1

"

..

',>' ,

vp2 '~"f"0 ~(-

"Cl ~... ~Ov"

Fig.3.19: Possiblepathsofbodywaves

In seismic refraction only compressional waves (P~waves) are considered and the interpretation isbased mainly on the first arrival times derived from the se'ismograms, This is due to the fact that P-wavetravels much faster than any other wave, Therefore for this case, Eq, (3.i 13)' can be written as, ,-, '

sinip =vplsin Rp vp2

The above equation is Snell's law or the law of re~action. Since vp2> vpJ' angle of refraction Rp isgreater than angle of incidence ip: Whenip increases; there is a 'unique' case where Rp = 90° andsin Rp = 1. Then -,' '

, v, . - pi - . .sm 'p' - ~ -.5m 'pc

vp2, ,, ,

, Angle ipc is called critical angle of incidence. For ip >ipc' the energy is totally reflected in ~he upperlayer. If vp2is less than vpl so that the ray path is refracted away from the normal this critical refractioncannot occur. . ~, ',. -- ~" " , :' : d' :

..,,(3.114)

...(3.115)

"" : -; -

. "

jt}cj~~ óôñ¢ñóñ¢Ö. ,,", ,- .

",- , .f

", ',,'

~;,'.f?'

Wa!'.e,Propagatil!n i~ an Elastic" Homogeneolls and Isotropic Medillm '

9S

It can be shown that the trajectory based on critical angles give the shortest time. Let a geophone isplaced at a distance x from the source (Pig. 3.2'0). ' "

.~(Sourccz)

"S .- .. ~ .. .,' ,

-1(Gczophone)

G .

x

.. '

hlVp1

Intczrtaccz

A BFig. 3.20 : Typical trajectory of a compression wave

h.SA = BG = --.!

. cos ip

. AB = x-2 hI tan ipTotal time, taken by the wave ~eaching from S to G will be:

. '

, 2h l . x - 2nl tan ipT= +Ypl cos ip vp2

...(3.116)

j3.117)

...(3.118), .

The time T will be minimum when

d T - 2hI sin ip , 2 hI --- - -0di 2. . 2 .P Vpl cos 'p Yp2 cos 'p

...(3.119)

. '. Vpl '.sIn, ' = - = SIn,P v . pc

, - p2

Therefore the travel tiITlefrom ~ toG via the second layer is minimum when slant ray paths throughvpl layer make angle ipc with the normal to the surface. Hence a geophone on the surface at any distance

, greater than the critical range 2 h I tan ipc from S will lie on one of these rays and will record the arrivalof the wave at the a,ppropri~tetime (Fig. ,321} Th~ ,refractedwav.esshown by d~~ed lines are known ashead waves. " ' . .

or ...(3.120)

,1.-- --.

",,; .. " '

~,;..'",' """,._"Hi;j; .,,'- -, " "t-~;." -,.. ,'" ..

" '

, ','.~,: " " _.'


.. 2h1tan 1pc ..

h1

.5

vp1

vp2 - - - Head waves

Fig. 3.21 : Critically refracted ray path and head waves

3.10.2. Depth Formulae.

3.10.2.1. Two layer soil medium. If the first arrivals ofthe elastic waves are recorded by detectors plantedin the ground, the times from the impact instant to the detectors can be plotted on a time distance graphas shown in the (Fig. 3.22). The slope of the lines yield the reciprocal of velocities, namely lIvp' There-fore, the lower the velocity, the steeper the slope of time-distance line. ..

The intersection, break point between the two velocity lines is obviously the point where the timesare equal. The distance between the impact point S and the break point is called the critical distance.The break point corresponds to the emergence of the wave front contact at the ground surface. Intercepttime is the total arrival time of the refracted waves minus the time'x / vp2, x being the distance betweenthe impact point and the receiving station. It is the intersection between the prolongation of the time-distance segment corresponding to the second medium and the time axis through the impact point.

For calculating the depth at an impact point two different approaches are available either using theintercept time or critical distance (Milton, 1960; Parasnis, 1962).

~Sou r S(ZS

x 1

ipc ipc h, vp1

(a) Ray paths VP2Fig. 3.22: Fundamentatprinc:ip'e ofrefrac:tlon shooting (...Contd.)

G1 G2 G3

.9 -'/ /

/ //

1 //

I //

/ //

I/

/ /

. I. // I I

'pc I'pc // /

/1hI

/ // /

11/

/ /I /

/


0>

"'"

----------

....

01

E+"

......T2 '0 I+"cIi...LL.

, .Xc Dj'stance) X

(b) Time-distance plot

Fig, 3,22: Fundamental principle of refraction shooting

Intercept Time. Refer Fig. (3.22a)

Arrival time T I of direct surface waves = ~vpl

2 h x - 2 hI tan i cArrival time T2 of refracted waves = I. + P

vpl cos 'pc vp2

...(3.121)

...(3.122)

, . Vpl

sin 'pc - vpZ

2hl

Tz = vpl cos ipc

...(3..123)

2hl sin ipc sin ipc +~. vvpl cos 'pc p2

. z2 hl(1- cos ipc) +~

. vvpl cos, pc pz= 2hl

. .

Vp l cos ipc

T - x 2hi cos iz - -+ pcVpz vpIT - x 2hl ,jZ

Z - -+ 2v -vvp2 . Vpl Vp2 p2 pI

vZcos " = .

,1 pI

pc 1-2.vp2

...(3.124 a)or

or ...(3.124 b)

. 98 Soil Dynamics & Machine Foundations

.This is the equation of a straight .line with slope l/vp2 and an intercept on the time axis through

impact point is given by putting X = O. '

2hl cos ipcT2i =

vpl

, T2ivpi

hi = 2cos'i~c

h = Tzi vpI vp2

I ~ 2 2Vp2 -V pI

Critical Distance. Let the critical distance is Xc at which TI = Tz (Fig. 3.22b)2 hi + .-'c - 2 hi tan ipc = Xc

V ') Vp lp-

...(3.125)

or ...(3.126)

or ...(3.127)

...(3.128)\',,;COSlpCI'

or 2h,

vpl cosipc- 2hl(l-cos2 ipc) = x (~-~ I

vplcosipc cl Vpl Vp2)

. -.

("'pI

J2hl cos lpc - Xc 1--

vp2or

or X (l-sinipc)c

hi = 2cosipc...(3.129)

orX

JVp2 -V pI

hI =; vp2 +vpl...(3.130)

3.10.2.2. Three layer soil medium. Consider a three-layered soil medium with compression wave veloci-ties of layers 1, 2 and 3 as v I, v ') and v 3 with the condition that v I < v , < v 3 (Fig. 3.23a). If S is ap p~ p "p p- psource of disturbance, the direct wave travelling through layer 1 will arrive first at A, which is locateda small distance away from S. The travel time for this can be given by Eq. (3.121) as T I""x/vpl' At agreater distance x, the first arrival will corresponds to the wave taking the path SDEB. The travel timefor this can be obtained from Eq. (3.123b) as

I 2 2X 2 hlVvp2 -V pI

T2 = -+ '

,vp2 vpl vp2

At a still larger distance, the rust arrival corresponds to the path SFGHIC. According to Snell's law

sini = vplsin ic2 ' vp2

vp2sin ic2 = -;-p3

...(J.131 )

...( 1.132)

Therefore, v I vp2 - vpIp .--sin; = vp2 vp3 vp3

'~-"':'~",F"'" '~')""~ ',",;" ,,' '

Wave Propagation in an Elastic, Hom~geneous and Isotropic Medium 99

The equation of the time of arrival, TJ for the wave from S to C,through thirdlayerjs

= 2 hI 2 I? x -2 ~ tani - 2 I? tan ie2TJ .+ . + " .

vpl cos I vp COSle2 vpJ .

Substituting the values of cos i and cos le2from Eqs. (3.132) & (3.133) in eq, (3. 134). We get

, x 2hl~v;J-v;1 2~~v;J-V;2 .TJ = - +, , +v v v v v '

, pJ pJ pI .pJ p2A B

...LaY<lr 1

vp1

ic1 = sin-1 (vp1 /vp2)

,..( 1,35)

s

si n-1 (v P1 / v P3 )(}

h1

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

, . ~ ,...' .E

, .-1 ( I )IC2=stn, vp2 vp31I 1

.' ',.. -. ' '" . "..- " ,

LaY<lrvp2 hl

,f' ,.. .' ,.. -' . .,, "',' , ,.,.; ',':,:,.":",, .,.~,

,~.'~.., .', -,:' ~-.', Or< ,:' : ,.. ,"

(a)Ray °¿¬¸

'# " " . ,

: 'Layer: 3 ",Vp3

0>

t. TZi0 '

" - - ~,-- -,-

,--.

----- 'I" ' I. , , IIII

.1IIII,I::

. C" .

...

tII

,~ T3i...

u.

I

. IIIIII

: ,'1. ,.' ,I "

...\J\....

(b) ",'P;' "',~, ôòØ¢Ý´ùòÄ ,'. ,xC2 'Oistance,'x,.;.,j,.'..\;,

"!;"Time-dis'tan~e ,if' 1~i~.:t:ttf ".t., gi;!:o, ,~. tf."'.l ,:1 ,'. ,.t'.:,)~-1" .FJ~v~~\ '-:',:.,: ,; c;"""','- """"",y-.,: I"",~"P.<:~ ~"",L :.4;,;"..,.,..,., ;.,.,.",',' ',..""";-..,, ,,"')J';"""';"-'-""~"" (er ",.1.,,"" ",."., , .;:,<;( 1:,..",,'d~.J ";""<::~"-'~'>;J!Lf""}Wd,Ll;),I;>",.q"i'1> ~.,..if"""":.;'i:' ,..,l1.".~.1 !U --<;,.'" ,rf::;.:.tj.",.ff' ,I,-,JJ..J. ~.J\J,', J,c.di;'

, ," ,Flg..3~~3..:"'~efraction shooting an three layered sod medium, "..' ,,~' .

.' :,.\:.';~t<:,".. . ,,'," ' . '",.

".._~ "" , , ,.." " ", "..

100 Soil Dynamic..& Machine Foundations

The records of geophones placed at various distances from the source may be plotted on time versusdistancegraph as shown in Fig. 3.23 b. The line OA corresponds to Eq. (3.121), OBto Eq. (3.123 b), andBC to Eq. (3.135): The thickness of the fIrst layer can be obtained either by Eq. (3.127) us-ingintercepttimeT2i or by Eq. (3.130) using critical distance xci' The thickness of the second layer h2can be obtainedby the two approaches as given below:

Intercept time: (Fig. 3.23)

or

2 hI cosi 2 ~ cosic2TJ i = +

vpl vp2

h2 = .!.(

T3' 2 hl~V~3 - V~I}

vpJ vp22 I V

p J Vp l . ~2V vpJ - vp2

Critical distance: (Fig. 3.23)At distance xc2' T2 = T3 . Therefore .

~ 2 2 ~ 2 2 ~ 2x 2 hI vp2 - vpI x 2 hI vpJ - vpI 2 ~ vpJ - vp2-£f.. + = -£f.. + +vp2 vpl vp2 vp3" vp3 vpl vpJ vp2

. v -v. hl (Vp2~V~3-V~I-Vp3~V~2-V~I )h = xc2 p3 p2-

2 2' v +v ~ 2 2p3 p2 Vpl vp3 -vp2

...(3.136 a)

...(1.37)

...(1.38)

or ...(1.139)

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

Layer 11 vp1..

toIE....

La yer 2 I vp20>~~0 Scigment n

Slope=1/vpn....IJI~

u.

Layer n, vpnSczgment 1Slope = 1/ vp1

Di stance J x

(a) Ray paths (b) Time-distance plot

Fig. 3,24 : Refraction shooting In multilayer soil

3.10.2.3. Multi/ayer soil system. If there are n number oflayers, the fIrst arrival time at various distancesfrom the sources of disturbance will pl~t as shown in Fig. 3.24. There ~iII be n segments on the t versusx plo,t.Using either intercept time or critical distance ~pp~oac~,the thickness of various layers can be

",!",,;,1"1 ¢òþå þþæôòþþþþôôùæôô¶¢ùòòº¢. J

u

Wave,Propttgatioll in all Elastic,-HolIIfIgelleous alld Isotropic Medium 101

obtained by detennining hI, h2"""'" hn - 1 in sequence. The general equation can be written as : -

Tni vpn Vp(n-l)

h =~ 2n - 1 2 v~n- vp(n-l)

Vpn vp(n-l) "j=n-2h_

Ö

´ÁóÔ~ 2 2 ~j=1 iv- 2Vpn - Vp(n-l) Pi Vpn

...(3.140)

The eq. (3.140) is based on intercept time approa,ch.

3.10.2.4. Sloping layer system. Fig. 3.25 a shows two soil layers. The interface of soil layers 1 and 2 isinclinedat an angle 0.with respect to horizontal.Therays such as ABCDmakingthe criticalangle ipc(sin ipc = Vp/Vp2)wIth the normal to the refractor, take the shortest time from A to D and are thereforefirst arrivals.

Referring Fig. 3.25 a,Zd

- AB = Cl = cos ipcDK = x sin et

- DK x sin aID = . = .

COSlpc COSlpc

...(3.141)

...(3.142)

(3.143)

(Zd + x sin a)CD = .cos Ipc -

AK = EG = x cos a ...(3.145)

EB = Zd tan ipc ...(3.146)- CG = (Zd + x sin a) tan ipc (3.147)

If we assume that point A to be the energy source and D the detector station, the time from A to Dfor the ray ABCD, i.e. the downdip time T2d is

AB BC -CDT = -+-+-2dvpl vp2 vpl

Zd x cos a - Zd tanipc';"(Zd + x sin a) tan ipc Zd + x sin a+ +vplcosipc vp2 vplcosipc

...(3.144)

...(3.148)

= ...(3.149)

2 Zd cos ipc x . .T = +-sm(lpc+a)2d v I v Ip p-

The Eq. (3.150) represents a straight line with slope sin (ipc + a)/vp\, and whenx = 0, intercept is T2 d;'

. 2 zd cos ipcTu; =--- -. vpl

The apparent velocitYvpll s~ (ipc +a) is equal to vp2sin i pcI sin (ipc+ a) and is smaller than thetrue velocity vp2since ~in ipcI sin (i pc+ a ) < 1.0

or ...(3.150)

where ...(3.151)

- '0-' ,.~.. ~; "".:u ;. 1H.:;;..r..,JII.

R

102 SoU Jiynamics '& Machine FtlflllilawnS':

:/: ;;";:;';:;'~"

I-A

It --i

hu

'..,

(a) Ray paths

.' ':,

""-,:,,,-

0

T2ui

..e-0>

-:;;...

T2di I"

IIIItcu.

11incrtosing tor TdDistance.- .

x increasing tor Tu

. ;,

!~ " ,""', ;, ,--:--'

Fig. 3.25 : RefractloD'Surveyjn soiis \vit~~~~n~~dJ.a!erin1:;...'

.. ,

I ~

/tJ1Iehopagatioll in an Elastic, Homogeneous and Isotropic Medium 103

For up-dip recording the equation of-time can be obtained by replacing Zd by Zu' and a by - a ,then'e get, .

2zucos ipc -x -T = + -'- sin (i - a )2u pc ' -

vpl vpl

In this caSe, apparent velocity is vI/sin (ipe - et.) which is equal to v2 sin ipjsin (ipe - a) . Thiselocity will be greater than the true velocity vp2

...(3.152 a)

, Therefore,

For x = 0, intercept is T 2ui' where

2Zu cos ipeT2ui =

vpl

T2divpIZd =

2 cos ipe

T2divpih =d 2 cosipc cosa

.t3.152 b)

...(3.153 a)

, ,

...(3.153 b)

Tu; vpIZ =u 2 cos ipe

Tu; vpih =u 2 cos ipccos et.

...(3.154 a)

...(3.154 b)

The vertical depths hd and hu can be obtained by dividing Zd and Zu by cos a.For critical distances: (Fig. 3.25 b)

xed = 2 Zd cos ipe + xed sin (ipe + a)vpl vpl vpl

...(3.155)

similary

xcd {I-sin (ipc +a)}Zd = 2'

COSlpe

c : ,. , Xcd{l-sin(ipc+a}h =d 2 cosipccosa -

Xcu {1- sin(ipc- a}h =,iU . 2cosipc COSa. .

It'is evident from Eq. (3.150), that apparent down dip velocity in the second layer is given by

...(3.156)

or ...(3.157)

...(3.158)

\ ,,-

',}'\" '

",' . Y" J',,'" p

- Yu ':=::":sin(iF'~ a.,>~. r """"', -'

...(3.159)

',' p, "~;I?-.J'}'I'~ I

if .." (

!;;~9~.titIi)2~JV;;{!f~.),~ -!", ~'>"Hi! co..;', nb:; ,':'-'.

.--

104 So,il l)ynamics. & .M~dJine ,Foundatio/l

Similarly the apparent up dip velocity will be

vpl

v2u = sin Upc- a)The true velocity vp2 in the refractor can be derived as follows:

vsin Upc+ a) = 1 = sin ipccos a + cos ipc sin aVu

v 1sin (ipc - a) = ~= sin ipccos a - cos ipcsin av2u

v I v I2 sin i cos a. = ~ + J!......pc Vu v2u

= vpl(vU +v2u)Vu Vu

...(3.16f

...(3.16'

...(3.16.

orV I

2 --L- cos etvp2

...(3.16

,:( sin ipc = VPI

]...(3.16

vp2

or

V VV = 2 cos a. 2u 2dp'

- v2u+ Vu

3.10.2.5. Refraction in a medium having continuous change of speed with depth. The problem maysolved by dividing the strata in infinitesimally thin members i.e. layers and each of higher speed than tabove (Fig. 3.26).

or ...(3.16

VK-2

~Z K-l VK-1

~ZK vK

~ ZK+1 VK+1

vK+2

Vmax

Fig. 3.26: Refraction s~rvey in soil having continuous change of velocitywith depth

Wave Propagation in an Elastic, Homogeneous and Isotropic Medium105

The solution may be obtained for a given velocity-depth functions such asv = v + k'z0 '

where v = speed at depth zv0 = speed at zero depthk' = constant

For any particular ray there will be a layer having a speed v max in which the path becomes horizontal.

Therefore vmax is actually a parameter that itself identifies the particular ray under consideration.Consider kth layer:

Travel Time for passing the ray through kth layer~Z

AT - k°k--'

vk cos ik

...(3.166)

As

= ~Zkr-.1 . 2.vkV1-sm ik

. . -1

(vk

)ik = SIn -----vrnax

~k

~Tk =Vk1/1- (l

)

2

vrnax

Horizontal distance traversed by this ray in kth layer

t:!.xk = /).Zk tanik

...(3.167)

...(3.168)

, , ,

...(3.169)..

...(3.170 a)

~vrnax

or ~xk =~Zk Ft' 2vk1-- 2

vrnaxTotal time for a wave to travel through N-such layers:

N ~t=L, k 2k=\ ~Vk

)v 1------k 1

Vrnax

j3.170 b)

...(3.171)'

and net horizontal distance

N ~Zkl

x = t;-I(

vrnaX

)

2

1- lv 'rnax

...(3.1n)

According to Snell's law:

sinik = ~sin 90 vrna.x ,

""l. = sinik = sinio =p (constant) ray parameter.'-. ,. "v ' -vk --"'~"'-~'O N'" ~rnax

...(3.173)

'o!.;I"',.. ;, . it, ,:> ,il M 4,::.~(; 1,~ ','"kP 'i",1iII '~f'A.!iftoj ,~qn '..' ., .; 115!.; k :;"", ) t

--"


Considering v~locity-depth function i.e. v = Vo+ KZ, and replacing vk by V and lIvrnaxby p.Zmu .

t = !.~1 ~VZ

r pv dZx= 0 v~l-ii

Zmax

J pv dZt = 0 (vo+k'z)~I-p2(vo+k'z)2

Zmax

x = J p(vo+k'z}dz0 ~1-p2(vo+k'z)2

Integration of the above two expressions are as under:- 1

[(I2 2

)\12

{I2

( + k ')2

}1/2

]x - - - p Vo - - p Vo zk'p

1 (vo+k'z) (I+~I- i v~)t = -Ink' .

( ~2 2)Vo 1+ I-p (vo+k'z)

...(3.174)

and ...(3.175)

then ...(3.176)

and ...(3.177)

...(3.178)

and ...(3.179)

Equation of x can be written as.:

(

~ 1- iv~)

2

(Vo

)2 1

x- k'p + z+/1 = k,2iThis is a equation of a circle of radius 1/ k'p and centre at - Vo/ k' and ~1- iv~ / k' p from x and

z axis respectively. Figure 3.27 a shows a family of such circles for a number of rays having differentangles of immergence into the earth.

C, C2 C3

...(3.180)

Cl. Line of centers-- ----"'%'~

)(

e>+

'\-~\~

Vok'

Surface of ~a rth x

z

(a) Ray pathsFig. 3.27 : Ray paths and time distance curve for linear Increase of speedwith depth (...CoDtd.)

--

'~~'~~

Wave Propagation in an Elastic, Homogeneous and Isotropic Medium

. (a) Ray paths107

to!E 2 . -1T = - Sin hK'

Kx'2vo

r-

r-

Distance.. x(b) Time-distance plot

Fig. 3.27 : Ray paths and time distance curve for linear increase of speed with depth

1 V

Radius of circle = k' p = k~+Zmax ...(3.181)

or r vozmax= k'p -F ...(3.182 a)

vo vo= k'sini -F0 '

,

{

sinio

}p=-

, Vo

or Z =. '\lo (co'

t)max ,1C' sec 107" ...(J.182 b)

Putting the value of Z as zmaxin Eqs. (3.178) imd (3.179) and eliminating p using Eq. (3.181), we get

t ~ 3. In v. -, (1 + ~p' v~ )k Vo

= 2 vl114Jt ll.+ ~( VV~ J

2

)-In ' v1-l~)k, \In

...(3.183 Cl)

where

2 v

~ (vo'

)2 ~v/\ .

X =.:..:..nwl 1- ~ = ~ cot 10.ok. . vmax, k.'\ k 'ZVmax = Vo+ . ~'- \; "... .

...(3.183b)

...(3.184)");' '! .1' "~L

," .

-

#

108h

Soil Dynamics & Mllchine Foundatitlas.

The time-distance relation for a such a circular ray between entry into and emergence from the earthcan be obtained-by eliminatingp and z from Eqs. (3.178) and (3.179). This can be expressed as

- 1- Sinh-Ik' xT - kt. 2v0...(3.185)

The time-distance curve applying for a linear increase of speed with depth is shown in Fig. 3.27b.The mverse slope of the time-distance curve at any point is equal to the velocity at the depth of maximumpenetration for the ray reaching the surface at that point.

3.11 TYPICAL VALUES OF COMPRESSION WAVE AND SHEAR WAVE VELOCITIES

Some typical values of compression wave and shear wave velocities (vp and vs)are given in Table 3.1Table 3.1 : Compression and Shear Wave Velocities

i.iaterial vp,m/s Vs' m/s

1. Moist clay and Wet Soils2. Sand3. Sand stone4 Lime stone5. Granite

600-1750300-800

1500-45003500-65004600- 7000

80-160100-500 .

750-22001800-38002500-4000

I ILLUSTRATIVE EXAMPLES t"~'Xamp'" 3.1

Wave pro{Ju".,'Jtiontests were conducted near Mathura Refinery, Mathura, for determining the insitu velocities of wave propagation and dynamic elastic moduli. Seismic waves were generated by the impact (

a h:':PlHlcrfalling through a height of 2.0 m. Two geophones were placed in the ground respectively.I.Om and 5.0 m from the source. The analysis of data gave the velocity of compression wave, V as 3CpmIs. The soil at the site was cohesionless and the position of water table was at 1.0 m depth below tJground surface. Determine E, G, VSand vr'Solution:

1. Assume suitable values of Poisson ratio, ).1.and submerged density of soil, Pb say

).1.= 0.25 and Pb = 1000 Kg/m32. From Eq. (3.64)

î A+2GV =-

p PPutting the values of Aand G from Eqs. (3.8) and (3.9) in the above equation, we get

p(I+).1.)(1-2).1.) 2E = (1-).1.). vp

- 1000(1+0.25)(1-2 x 0.25) x (300)'2 - 2- (1-0.25) . - 76453000 N/m

2= 76453 kN/m

- -~. LiIIiIft 11 -

ve Propagation in an Elastic, Homogeneous and Isotropic Medium 109

From Eq. (3.9)

G = E - 764532(1+ J.!)- 5(1+ 0.25) = 30581kN/m2

3. From Eq. (3.103),2 1- 2 J.! 1- 2 x 0.25 1

Cl = 2 - 2J.1= 2 - 2 x 0.25 = 3"

From eq. (3. 105),6 4 2 2 1.~ - 8 ~ - (16 Cl - 24 ) ~ - 16 (1 - Cl ) = 0"

For J.1= 0.253 ~6- 24 ~4 + 56 ~2- 32 ='0

(~2 - 4 ) (3 ~4 - 12 ~2+ 8) = 02 2 2~ =2 2+- 2--

. ' J3' J32

s 22" =1-~n

For ~2 = 2 and (2 + 21 J3), value of sIn will work out imaginary. Therefore

~2 = (2-2/J3) = 2 - 1.155 = 0.8452 2 1

Cl ~ =0.845 x 3" = 0.282

2L =~a2~2 =0.531vpv =300 x 0.531 = 160rnlsI"

vI" = JP2 = .J0.854 = 0.9192Vs

r 160Vs = 0.9192 = 0.9192 = 174 rn/s

or

Therefore,

From Eq. (3.112), , ,

From Eq. (3.100),

From Eq. (3.1~1),

Example 3.2The data of a refractor test is given below:

Distance of Geophonesfrom Source (m)

048122025 . :J

Travel Time(milli s)

024689

'. ..

Determine the depth of the refractor using time"intercept approach and critical distance approach.

,- dO"dd~ 0'_-"""",'<~iL:

äþ

þ ù ô ù½ ùåùôù¢ù

ò òòòù

ô òòòù ù

óó þó òô óòòòòò ó óóòóó

110 Soil DyntUllics & Machine FoundlZlUJr.

Solution:

1. The data given in the example has been plotted on time-distance graph as shown iFig. 3.28. From this figure,.

10

- 6

8

-u1/,

E ""./,

""."'" ,

//

41- //

I

""."'"

. ""./r ."

.......~

E1--

00 4 8

I

IIIIIIIIIII,xc

12 16 20 24

'2i

2

Distance (m)Fig. 3.28: Time-distance plot for the data given Inexample 3.2

Also,

TZi = 0.003 and Xc = 12 m.12

vpl = 0.006 = 2000 mls8 '

vp2 = 0.002 = 4000 mls2. From time-intercept approach (Eq. 3.127)

h = Tu vpl vp2I 42 22 vp2-vpl

.

= 0.003x 2000 x 4000 = 3.46 ID.' 2~4oo02_-20002.{

';....


3. From critical distance approach (Eq. 3. 130)

xJ

Vp2-VpI

hi =.{ vp2 -v pI

= Q /4000-2000 = 3 462 ~4000+2000 '. mExample 3.3

In Fig. 3.29, determine the time of the flIst arrival wave from source Sand geophone G.

t5.0 m

-t10.0 m

J

Is..30m ~

~= 500 m I sec.Vp1

, .vp2 ~=~ 2000 mlsec

vp3 = 3000 mlsec

Fig. 3.29 : Data for example 3.3Solution:

(i) Arrivaltime T1 of the directsurfacewave= ~I = :o~ =0.06S(ii) Arrival timeT2 of refracted wave fro~ flIst layer (Eq. 3.124 b)

x 2hl ~ 2 2=-+ v -vVp2 Vpl Vp2 p2 pi30 2 x 5 ~ 2 2

= 2000 + 500 x 2000 "2000 -500 = 0.015 + 0.01936= 0.03445 .'

(fU) Arrival time T) of refracted wave from second layer (Eq. 3.135)

x 2hl~v~) -V~l 2~ ~v~) - V~2=-+ +~) ~)~l ~)~2

30 2 x 5~30002 -5002 2 x 10~30002-20002= 3000 + 3000 x 500 + 3000x 2000= 0.01 + 0.01972 + 0.00745= 0.03715 s

M_inimum ~avel ~e ~, O:O~'f4, .s:' ; -. -," \" ; ;" i: '-.' .

: ..

,'.'-

112 Soil Dynamics & Machine Foundations "

Example 3.4Between two points A and B (Fig. 3.30) two seismic refraction tests were performed and the data ob-tained is given below: .

Source at A Source at B

Distance ofGeophonesfrom A (m)

Travel time(milli sec)

Distance ofGeophnes

form B (m)

Travel time(milli sec)

05

101520253035

03.1.6.39.4

12.6. 15.0

15.916.5

05

101520253035

03.16.19.3

11.112.914.816.5

Determine the slope of the refractor, depth of refractor at points A and B, and the two velocities usingcritical distance approach and time intercept approach.

A

r

~B

Vp1

hu

Vp2

Fig. 3.30: Sloping layer (example 3.4)

Solution:1. Plot the data on time-distance graph as shown in Fig. 3.31. In this figure

. 15v 1 = Inverse slope of AC == 3 = 1600 mIs, orp 9.4 x 10-

10= Inverse slope of BE = 6.1 x 10-3 = 1640 m/s

Adopt ,;' "1600+ 1640 = 1620 ~vpl 2 ."" s

. ,35 ',. .~.

v2u = Inverse slope of CD = (16.5-11.2) x 10-3 ~ 6603 m/s

~~.

Wave Propagation in an Elastic, Homogeneous and Isotropic Medium 113 J'~I

35Vu = Inverse slope of EF = 3 = 2756 m/s(16.5-3.8) x 10-

T2u; = 11.2 x 1O-3s,Xcu= 23.5 mTu; = 3.8 x 1O-3s,xcd = 14.75 m

18

2

D

16

~

".-'-"-

"...-

4

I

III

I

I

,Ii. "-

"-"-

"-"- , ,

"-"-"

, ,

T2di

3.8 sec

Distance (m)

Fig. 3.31 : Time-distance plot for the data given in example 3.4

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

14 I

'"..........

..........12 l-- ..........

..-..........

T2ui,.....

10uCII'"I--.-E 8-CiIJE.-t-

6

114Soil Dynamics & Machine Foundation$.

. v I 16002. FromEq. (3.160),sm(i - a) = ...L = - = 0.2423pc vzu 6603 .

oripc- a = 14°

Sin (i + a) = vpl = 1600 = 0.5805pc vZd 2756

or i + a = 35.5°pc

Therefore, ipc = 24.7° and a = 10.75°3. Time-intercept approach:

From Eq. (3.159),

From Eqs. (3.153b) and (3.154b)

h - TZdivpld- 2 cosi cosapc

3.8 x 10-3 x 1620= 2cos24.7°coslO.75° = 3.4 m

h = TZujVplu 2 cosi cos apc

- 11.2 x 10-3 x 1620- 2 cos 24.7° cos 10.75° = 10.03 m

4. Critical distance ApproachFrom Eqs. (3. 157) and (3. 158)

xcd {1-sin(ipc +a)}hd = 2 cos ipccos a

14.75(1-0.5805)= 2 cos 24.7° cos 10.75° = 3.4 m

Xcu {1- sin (ipc - a)}h =u 2 cos ipc cos a

- 23.5(1- 0.2423)- 2 cos24.7° cos 10.75° =9.97 m

. vzu vZd 6603 x 27565: True velocIty (Eq. 3.165), vp2 = vzu +v2d 2 cos a = 6603+2756 x2 x cos 10.75°

= 3820 m/se£~

" .. .' ~I ...,.. ;;" ::

,"".~,

Wave Propagation i" a" Elastic, Homogeneous and Isotropic Medium 115

Example 3.5Following is the data obtained from a seismic refraction test

Geophone No.

...

Distance from Shot Point(m)

O'1002003004005006007008009001000

Travel Time(rriilli sec)

GOGlG2G3'G4G5G6G7G8G9

GI0

0200399592780.96311381306146516181763

Assuming that the velocity is varying linearly with depth, determine the values of depth of deepestpoint of the wave paths from ground surface corresponding to the waves reachtng to geophones G7 andG9. Also determine the coefficient 'K' in both cases.

Solution:

(i) Initial velocity, vo = 100200 x 10-3 = 500 m/sVelocity of the wave reaching to geophone G7 i.e.

200v = = 611 m/s

7 (1465 -1138) x IO-~

. - . -I 500 - 54 90

'0 - Sill . 611 - .2vo .

K' = -cot '0x

(ii) Similarly

..

From Eq. (3.183), I

(iii) Similarly

2 x 500 °= cot 54.9 = 1 0700 .

~ 500Zmax= k' (cosec io- 1) = 1.0 (cosec 54.9° - 1) = I11m

From Eq. (3.182 b)

200v = = 671m/ s9 (1763-1465) x 10-3. . -I 500 0'0 ~ Sill . 671=48.16

2 Vo cot io 2 x 500 cot 48.16°.k' = ~ = 900 = 0.995

Zmtu:=.f,(cosec io-1) = 0~::5 (cosec48~16°-1)= 172m. -

. . ..; ..' "." ."-. .:,," .'~~-= ~~..

116 Soil sy,..",ics & MlIchilre Fo""dations

zo

5 10 15 20

(m)

25

-uQ,/\1'1

-'E 10,~E~

DistanceFig. 3.32 : Time distance plots

ÎÛÚÛÎÛÒÝÛÍ

Hardin, B.O. and Richart,F. E.,Jr. (1963), "Elastic wave velocities in granular soils," J. Soil, Mech. Found. Engg.Div., Am. Soc. Civ. Engg. , vo\. 89, SM-I, pp. 33.65.

Kolsky, H. (1963). "Stress waves in solids." Dover, New York.

Lamb, H. (190.:l), "On the propOlgationof tremors over the surface of an elastic solid," Philos. Trans. Royal Soc.London, Ser.A. 203, pp. 1-42.

Milton, O.B. (1960). "Introduction to geophysical prospecting", McGraw-Hill,New York.

Parasnis,D.S. (1962). "Principles of appliedgeophysics",Methuen.Rayleigh,L. ( 1885), "On wavespropagatedalong the plane surfaceof an elastic solid." Proc. London Math. Soc..

vol. 17, pp. 4-1 \.

Richart F.E., Jr. (1962), "Foundation vibrations". Trans. Am. Soc. Civ Engg., vo!. 127 , Part I. pp. 863-8lJ8.

Timoshenko, S. and Goodier, J.N.(1951), "Theory of elasticity", McGraw Hill Book Co. Inc., Tokyo.

---

Wave Propagation in an Elastrc, Homogeneous and Isotropic Medium 117

I "

PRACTICE PROBLEMS

. ?.1 Explain the generation of (a) compression wave, (b) shear wave and (c) Rayleigh wave. Describetheir relative magnitudes.

--. " . . " 0 , , ,0'

. .}.2; ~scijb~ t,he)V.aveprppag~tion in an infinite, homogeneous, isotropic and elastic medium.

.' 7.",3.3""AssumingPoisson'S"rntioif(J.t-jas-tt35'and"denslt)r ofs'oifils1800'kgliri3 ,dete~i~~ E:G::~~ ~~dvr if compression wave velocity is 450 mls.

3.4 What are the informations obtained by a seismic survey? Give its economic aspects.3.5 Give precisely the three principles which makes the basis of Seismic wave propagation theory,3.6 A shot is fired at the ground surface on a particular"location and 'observations from-geophories

gave the fol~owing,data.

Distance of Gephonefrom shot point (m)

5101520253040

Travel Time"'(MiIli sec)

.. ~ 42.5085.00127.50170.00187.50197.50215.00

c ~~;

Plot the time-distance graph and determine

(i) Velocities of two layers

(ii) Values of E and G of the two layers assuming ~ = 0.25. ,

(iii) Time intercept and critical distance,, '

(iv) Depth of layer from.time intercept and criticcit"distance,,' .' .

(v) Verification of travel time for 30'm geophone.

3.7 Prove that in a dipping layer interface trajectory based on tritical angle~ giv~ sho'rt~st' time. "

3.8 In a dipping interface travel time-distance plot is' obtained as shO\vn in FIg. 3::n. Deteimine thefollowing: .; , '

(a) Apparent velocities

(b) Slope of interface

(c) Depth of layer at points A and B using critical distance concept

(d) Depth of layer at points A and B using time intercept concept

(e) True velocity V2

00

"-' - u -- ._. ,-- - ~--~.,.........~. .' -:-?

.:

DYNAMIC SOIL PROPERTIES

4.1 GENERAL

The problems involving the dynamic loading of soils can be divided into two categories:(i) Having large strain amplitude response-Strong motion earthquakes, blasts and nuclear explosions

can develop large strain amplitudes of the order of 0.01% to 0.1%, and(ii) Having small strain amplitude respon~e-Foundationsof the machineshave usually low strain ampli-

tude ofthe order of 0.0001%to 0.001%.

The soilproperties which areneeded in analysis and design of a structuresubjected to dynamic loadingare:

(a) Dynamic moduli, such as Young's modulus E, shear modulus G, and bulk modulus K

(b) Poisson's ratio J.1

(c) Dynamic elastic constants, such as coefficient of elastic uniform compression CII' coefficient ofelastic uniform shear Ct, coefficient of elastic non-uniform Compression C$ and coefficient of elas-tic non-uniform shear C'p'

(d) Damping ratio~.

(e) Liquefaction parameters, such as cyclic stress ratio, cyclic deformation and pore pressure response.

if) Strength-deformation characteristics in terms of strain rate effects.

Since the dynamic properties of soils are strain dependent various laboratory and field techniques havebeen developed to measure these properties over a wide range of strain amplitudes.

4.2 LABORATORY TECHNIQUES .

The laboratory methods used for determining the dynamic properties of soils are:

(I) Resonant column test,

(ii) Ultrasonic pulse test,

. (iii) Cyclic simple shear test,

(iv) Cyclic torsional simple shear test, and

(v) Cyclic triaxial compression test.

"/-"

'm,-, ..,

Jynamic Soil Properties' 119

t.2.1. Resonant Column Test. The resonant columntest isused to obtain the elasticmodulusE, shearmodu..;IusG and damping characteristics of soils at Iow strain amplitudes.This test is based on the theory of wavepropagation in prismatic rods (Richart, Hall and Woods, 1970).Either a cyclically varying axial load or tor-sionalload is applied to one end of the prismatic or cylindrical ~pecimenof soil. This in turn willpropagateeither a compression wave or a shear wave in the specimen. In this technique the excitation frequency gen-erating the wave is adjusted until the specimen experiencesresonance. The value of the resonant frequencyis used in getting the value ofE and G depending on the type of the excitation (axial or torsional).

The resonant column technique was used for testing of soils by many investigators (Ishimotoand Iida,1937;lida, 1938, 1940;Wilsonand Dietrich, 1960;Hardinand Richart, 1963;Hall andRichart, 1963;Hardinand Music, 1965;Drnevich, 1967;Anderson, 1974;Lord et ai, 1976;Woods, 1978).Severalversionsof tor-sional resonant column device using different end conditions to'conc;traintthe tt'st specimenare available.Some common end conditions used in developing the equipment are discussed below:

(I) Fixed-free: Hall and Richart (1963) described the apparatus with fixed-free end c~ndition. In thisarrangement one end of the specimen is fixed against rotation and the other end is free to rotate under theapplied torsion (Fig. 4.1a). A node occurs at the fixed endand the distnbution of angular rotatione along thespecimenis a 1/4sinewave. '

x ., ,

e ~l, t) Y" :'" X

80,t) e(l,t)

//,

//

/ '/.I. ~

/A ~--~/ (0) J/Jo =00

J JiJ

/

//«Y4SintZ wavtZ~

8 e(b) J/Jo=O.S

Fig. 4.1 : Schematic of resonant column with fixed-free end conditions(After Hardin, t 970 and Drnevich, t 967)

As shown in Fig. 4.1b, by adding a mass at the free end, the variation of e along the specimenbecomesnearly linear. J and Jo are respectively'the polar moment of inertias of the specimen and the added massrespectively.Dmevich(19~7)usedthe~oncePtof addedmasstoobtaina uniformstraindistribution'throughoutthe length of the specimen. .

, (il) Spring-base model: Figure 4.2 shows the configu~ationof an apparatus which canbe described asthe spring.:.basemodel. ,Depending on the stiffness of the spring compared to specimen's stiffness, it canrepresent either fIXed-free airangem~nt,~rfree-freeconfiguration.In the case when thespringis stiff in com-

, parison ~ospecimen, the configuI'~ti,~~may ~~~onsideredas fix~d-free. In such a case, a nodewill occur atmid height of the specimen, and the distribution of angular rota~on would then be a 1/2cosine wave.

, " ,"".', " . ,

. -."";t""' "".,', ".'. ,..,..,0.;", t<.'.I ,~"'-:;""'J1J.)'.IJ'..~",. ',';.;'Ii..:o.",:,:L.."f,i;.'i{}I.,,'il., :.'1'\II ,:-,\

"-;::";"":"~'~;""

~ 'r,- '-:--:.

~'I.''. -; ----- ,-, ",- ~=. ::'::::,=:T-

if~

'~


½±²²»½¬»¼ to platen

passiv<z <znd platczn

Soil sp<zcimen

Activll Ilnd platen

/ W<zightl<z"sc;torsional springf.- ~~~

WQ:ightlQ:ss torsional da shpot

"

{

portion of vibration Q:xcitation dllvicllrigidly connQ:ctlld to platlln

~L :';9hil...

Wllightlllc;c;longitudinal dashpot

longitudinal spring

Fig. 4.2 : Schematic of resonant column with spring-base model (Woods, 1978)

(iil) Fixed-partially restrained: Figure 4.3 showsthe configurationof anapparatusin whichthe top capis partially restrained by springs acting against an inertial mass. The apparatus used by Hardin and Music(1965) is of this type. " "

Dmevich (1967, 1972)developed a hollow cylinder apparatus as shown in Fig. 4.4 to study the effectof shear strain amplitude on shear modulus and damping. It may be noted that in the usual torsional reso-nant column test, the shear strain is not constant but varies from zero at the centre of the sample to amaximum at the outer surface. In hollow specimen, the variation in shearing strain acrossthe thickness ofthe cylinder wall becomes relatively very small. The configuration of this apparatus is similar to Fig. 4.1band therefore the shearing strain is almost uniform along the length of specimen. Using Drnevich's appa-ratus, Anderson(1974) tes.tedclaysupto 1%shearingstrainand Woods(1978)testeddense sandsupto shear-ingstrainof 0.5%.4.2.1.1. Calibration and determination olG and~. Hardin (1970) suggested the followingprocedure ofcalibration of the apparatusdescribed in Fig. 4.1b:

(I) For this model the vibration excitation device itself,without a specimenattachedis a single degreeof freedom system.Firstly remove the specimencap and the additional rigidmass,connect the sinewave generator to the vibration excitation device and vary the excitation frequency to determinethe resonant frequencyf"I of the device. .

(il) Attach theadditionalrigid massof polar momentof inertia Jo' anddeterminetheresonantfrequency!"A of the new system.

';C~

ynamic Soil Properties 121

Driving \, ~for C(l --~

. Sp(lcimqn. non- rigiddis'turbqd mass

" .

'"

~Fixqd'.

Fig. 4.3 : Schematic of resonant column with fixed-partially restrained end conditions (Woods, 1978)

The rotational spring constant (torque per unit rotation), Ko,of the spring about the axis of specimencan be obtained using Eq. 4.1.

2 1"2

Ko = ;1t -(

JAi n

)

:1- InA

1nl

...(4.1)

(iii) With the added mass removed and with the specimen cap, specimen and all apparatus, determinethe resonant frequency/"0' The value of mass polar moment of inertia of the rigid mass, Jocan becomputed using Eq. 4.2.

"Ko '

. Jo = 41tln20

;,/,-.0 ".

... -

"'-_'__.0"'_"-

122

Taring springSoil Dynamics & Machil,e Foundations

Long, LVDT/

Rot .accelerometerRot.LVDT

Top 'cap lid

,""

.:,

~,.:

!:~j" 'I, ..

'" M em branes.-:,."',,

.~ '

,':I,'~,:

!:'.:-.. ;

,'.

,

, '

;:-'

,,-.~'. :

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

;' :--,'

'J pore-pressure transdueczrI

",~ ...:I" ''I.,'.,'..-

....-To bae kpressure

BoHom cap

/

Fig, 4,4 : Hollow specimen resonant column apparatus (Ornevich, 1972)

Now at resonance cut off the power and record the decay curve for the vibration, From the decay curvecompute the logarithmic decrement for the apparatus, °A' as follows:

1 AI0 = - log -A n eA "

where AI =Amplitudeof vibrationfor thefirstcycleAn = Amplitude of vibration for the nthcycle.

Under steady state vibrations, the apparatus damping constant, DAis given by

...(4,3)

- °A WTDA - -"Ko 101t ..,(4.4

.~

Dynamic Soil Properties 123

(iv) To measure the torque current constant, Kt' excite the apparatus successively at frequencies (J2 / 2)/"0' .fi Ino and 2Ino' During the steady state vibration at each of these frequencies measure thecurrent flowing through the coils, C in amperes, and the displacement amplitude of vibration, e inradians. For each frequency compute the torque-current constant Kt' as follows:

eKo~ = CMf

...(4.5)

where Mfis given in Table 4.1

Table 4.1 : Value ofMf

The value of Kt shall be taken as the average of three measured values. The three measured valuesshould not differ by more than 10%.

After calibration, the specimen shall be placed in the apparatus with minimum disturbance. A knownvalue of ambient pressure is applied as done in triaxial compression test. With the power as Iow as is prac-tical, the resonant frequency of the system'/"R' is obtained by varying the frequency of excitation. At reso-nant frequency, the amplitude of vibration, eR, in radians and the current flowing through the coils of thevibration excitation device, CR are measured. Now the power is cut off and the record of decay curve for thefree vibration of the system with specimen is obtained. Using this decay curve, the value of the logarithmic'decrement,Oscan be obtained employing Eq. (4.3). '

The procedure of obtaining G and ~has been explained in the following steps:(i) Calculate the mass density of the specimen, p, from Eq. (4.6),

4W

P = rtd2f gwhere W = Total weightof specimen

I = Length of specimend = Diameter of specimen

-, . ~ ,- ,g =Accelerationdue to gravity,

(it) Calculate the inertia of the spe~imen_ab<iutits_~xis,J, as follows:- , .,. 4 "

1tpd 1J=- 32

f)

...(4.6)

...(4.7)

,. ;-,

,'.

Frequency Mf

. ()f.. 2-

. r.,

(.fi)1110I ,.y

I2/110

-3

~.

124 SoU Dynamics cl Machine Foundations

(iii) Calculate the system factor, T ~s follows:

10 KoT = ---I 41t21fni

where 10 = Mass polar moment of inertia of the apparatus, derIDedby Eq. (4.2).Ko = Rotational spring constant, derIDed by Eq. (4.1)

1 = Inertia of the specimen, defined by Eq. (4.7)

fnR = Resonant frequency of the complete system.(iv) Using Fig. 4.5, determine the dimensionless frequency F for the value ofT computed in step (iii).

...(4.8)

100

t>I::I0>

543

SO4030

20

~ 10'I-0

2

0.60.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

value of F

Fig. 4.5 : System factor T versus F

G =4.2p (I,; ./)'

(v) Forsteadystatevibrations,the dampingfactorof thesystem,Ds is givenby

"- 1(

KICR'

)Ds - 2 2 2 --21tDA FnR41t J ftiR ~~

...(4.9

...(4.Hi '

.,.0 ,

. .:~,0...

\\

f- \i\

""

" '"- '-- "

"""""

..........'"-

- ...........-- I I I I I I I I

'liSDYllamic Soil Properties

.8)

(vi) Compute O/T . Using Fig. 4.6, determine the va.lue ofR corresponding to the value of T computedin step (iii). Then damping ratio is given by

Os~=0.5 TR ...(4.11)

(vii) For the free vibration method, using Fig. 4.7, determine the value of mode shape factor Cmfor thevalue of T computed in step (iii)

100

ii).

~01-0

~::ICJ>

.(4.9)

4.10)

~

50

20

10

5

2

0.7 -

1.0 1.1 1.2 1.3Value of R

1.4

Fig. 4.6 : System factor T versus R

--

1.5

----

125

11

~,I:

'11!I!I

126 Soil DYllamics & Mac/tille Fou1Idatio/ls

I' 100

soI'I1I

1'1

20

lil

.....10

!I

I;I:11

'+-0~::J0>

5

jj

2

I,I I'

! :I Ii

I'

0.61.00 1.01 1.02 1.03

value of CM

f

II

Fig. 4.7 : System factor T versus mode shape factor Cm(viii) Compute the system energy ratio, S as follows:

<> 32KolS = 4

1tC",GdThe value of damping ratio ~is then given by

~= ~[Os(l+S)-OA S]21t

4.2.2. Ultra Sonic Pulse Test. The theory of ultrasound is similar to that of audible sound. Sound is theresult of mechanical disturbance ora material; that is. a vibration. Ultrasonic pulses of either compression orshear waves can be generated and received by suitable' piezoelectric crystals. Using elastic theory, a rela-tionship between the speed of propogatiQn and wave amplitude of these waves and certain properties of themedia through which they are travelling can be determined as follows:

2(1+11)(1-211)E=pvc (1-jl)G - 2

- pvs

...(4.12)

...(4.13)

1I

III

II1

r

l

...(4.14)

...(4.15)

- - .~

---

~ 2.302 I Aou = - ogl0-

n AnVc = Velocity of compression wavep. = Poisson's ratioG = Shear Modulus() = Logarithmicdecrement

An = Amplitude after n oscillations.

Lawrence (1963) described the basic apparatus required to measure the propagation velocities (i.e. Vcand vs) through sand. Stephenson (1978) described an equipment for conducting the ultrasonic tests. Hisequipment includes a pulse generator, an oscilloscope, and two ultrasonic probes (transmitter and receiver).The pulse generator delivers-a variable-voltage direct CUITentpulse to the transmitting probe simultaneouslywith a 7 volt trigger pulse to th~ time base of the oscilloscope. The gen_e.ratorwas designed such that thepulse interval and pulse width can be varied. Stephenson caITied out the tests on silty clay samples.

One of the main advantage of ultrasonic test is that it can be perfoITned on very soft sea floor sedimentswhile still retained in the core liner.

The drawback of this approach is that it is very difficult to identify the exact wave aITival times. Sec-ondly the strain amplitudes which can be achieved by this technique, are only in the very low region.

G.L.

Dynamic Soil Properties

( )2

I-! VC

J..L-= 2 Vs

1-(::]'

where

¨

Í±· × ³¿

""""-

~~-

127

...(4.16)

...(4.17)

Vs = Velocity of shear waveE = Young'sModulusp = Mass density = y /g

Ao = Initial value of amplitude

", ".Typical ±·´ IZlem~nt

Rocl<Bascz met ion

(a) Shear wave induced in soil by horizontal earthquake vibrations

CJo er

! + t-x---x.---1:.B

er

A

--+"tC

A- Soil element at restand in mean position

B- Maximum d<zformationio the. le ft

C- Maxi m u m dlZtormationto tht right

(b) Shear deformations resulting from propagated wave (for a single cycle)

Fig. 4.8 : Stress condition on an element of soli below ground surface


4.2.3. Cyclic Simple Shear Test. During an earthquake or other source of ground vibrations, a soil elementbelow a foundation or in an embankment is subjected to an initial sustained stress together with a superim-posed series of repeated and reversals of shear stresses (Fig. 4.8). The magnitude of induced shear stressesdepend on the magnitude of acceleration of the dynamic force. In a direct shear box test, uniform state ofshear strain occurs only on either side of failure plane. The simple shear device was designed to overcomethis limitation of direct shear box by enabling a uniform state of shear strain throughout the specimen. Thissimulates the field conditions in a much better way.

Kjellrnan (1951), Hvorslev and Kaufman (1952), Roscoe (1953), and Bjerrum and Landra (1966) havedescribed simple shear apparatus. The Roscoe simple shear device has a box for a square shaped samplewith side length of 60 mm and a thickness of 20 mm. The box is provided with two side walls and two hingedwalls (Fig. 4.9). Peacock and Seed (1968) have modified Roscoe's simple shear apparatus for dynamic test-ing. The dynamic shear forces were applied by a double acting piston with controlled compressed air pres-sure using solenoid valves. Figure (4.10) illustrates how the end walls rotate simultaneously at the ends ofthe shearing chamber to deform the soil uniformly. Prakash, Nandkumaran and Joshi (1973) have also devel-oped similar type of simple shear apparatus which has the facility of applying sustained normal stress, sus-tained shear stress and oscillatory shear stress.

Wire wound membraneor stacked disks

~

t:-- '- Hinge

(a) (b)

Fig. 4.9 : Schematic arrangement of Roscoe simple shear apparatusPrakash, Nandkumaran and Bansal (1974) have conducted tests on three artificial soils (SM, CL and

CH) using oscillatory shear apparatus developed by Prakash et al (1973). A more systematic study has beendone by Dass (1977) on clay of high compressibility (CH, LL = 65.5%, PL = 28.0%, qu=78 kN/m2) using thesame dynamic simple shear apparatus. The salient features of her work are reported here to understand thebehavior of soil under dynamic load. The static strength of soil was 36 kN/m2. She performed the testskeepingthe'variationof variousparametersasgivenbelow:.

Sustained normal stress, on (kN/m2) = 15,21,26,29Sustained shear stress, tst (~of static strength) = 10,25,50,60.


Oscillatory shear stress, 'Cdy n ( kN/m2) = 13,19,22,25Number of cycles of oscillatory shear stress = 1to 1100The oscillatory shear stress was applied at the rate of one cycle per second. The f::lilurecriteria was

chosen corresponding to 12mm displacement.

Sh ea ring chamber Soil samplePian , ,

D~==4JùÄ÷óããð÷ \ ~-

ÅÃóã ==(]

(0==0)

L 7....

End plate rotation Soi I deformation

Elevation

Fig. 4.10 : Schematic diagram illustrating rotation o(hinged end plates andsoil deformation in oscillatory simple shear (After Peacock and Seed. 1968)

""""


A typical plot oftest data in terms of number of cycles versus shear displacement is shown in Fig, 4.11

for "Cdy n equal to 13 kN/m2. Similar plots were obtained for other values of 'Cdy n . From these plots, numberof cycles and oscillatory shear stress corresponding to 12 mm displacement have been obtained and plottedas shown in Fig. 4.12. In Fig. 4.13, plots between dynamic shear stress and sustained stress are shown fordifferent number of cycles. It can be seen from Figs. 4.12 and 4.13 that for a fixed sustained shear stress, theamount of dynamic shear stress decreases as number of cycles increases for causing 12 mm displacement.As sustained shear stress increases, less number of cycles and dynamic shear stress is needed to causefailure.

Number of cyo;l(S

101

10 100 1000

Undisturb(don: 0.21 kg/cm2, 2

tdyn~ 0.131 kglcm

...cOIl

s:.1/1

8

'[sf in '1. of normal strength2

~ L.EE

cOIl

~ '6vca..1/1

"

10

12

Fig. 4.11 : Shear displacement versus number of cycles

"

";1";",

Dynamic Soil Properties ~ 131

120

Sustained shearstress in percent of normal strength

.s::. 100

....Cl>C...........III

2CIn= 0.21 kg' cm

cE 80....0c....0

...CtJ~ 60...a.cIIIIIItJ....t; !oO....ClQI

.s::.III

>-....0

E 20'-uIII0

01 10 100 1000

No. of cycle at failure

Fig. 4.12 : Oscillatory shear stress versus number of cycles

4.2.4. Cyclic Torsional Simple Shear Test. In cyclic simple shear apparatus, it is not possible to measure theconfining pressure and the test is performed under Ko consolidation conditions. Torsional simple shear de-vices have been developed to overcome these difficulties. Ishihara and Li (1972) modified the triaxial appa-ratus to provide torsional strain capabilities. This has the disadvantage that the shear strain in the samplevaries with maximum at the outer circumference and zero at the centre. This problem has been minimised byusing hollow cylinder torsional shear apparatus (Hardin, 1971; Dmevich, 1972; Yoshimi and Oh-aka, 1973;Ishibashi and Sherif, 1974, Ishihara and Yasuda, 1975; Cho et aI, 1976; Iwasaki, Tatsuoka and Takagi, 1978).The apparatus used by Drnevich (1972) has already been described earlier and shown in Fig. 4.4. This hasthe advantage that both resonant column and cyclic torsional shear tests can be performed in the samedevice and on the same sample.

, ' ,{

-. ,._.,----


. . Undistu rbed.s::..-01C It\I....

V1 100

0---0 RemouIded

0E....0C

D Number of atcycle at failure

-0....

20

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

0-0 20 40 60 80 100

Su stained sh ear s tress in percent no rmal str 12ng t h

Fig. 4.13 : Dynamic shear stress versus ~ustained shear stress

Some of the above mentioned investigators have used long hollow cylinder to obtain uniform condi-tions atthe test section (Hardin, 1971;Drnevich, 1972,Ishihara and Yasuda, 1975;Iwasakiet aI, 1977).It isdifficult to prepare samplesfor longhollowcylinder devices.Keeping this fact in view,Yoshini and Oh-Oka(1973); Ishibashi and Sherif( 1974),andCho et al (1976) useda short cylinderin whichthe taper waspropor-tional to the insideand outsideradii (Fig.4.14). The internalandexternalradii rl andrz are selectedsuch thatthe difference in average shear stressescomputed considering two extremeconditions isminimum.The twoextremeconditions are:

(i) Shear stress varies linearlywith radius as for an elastic material (Eq. 4. 18)(ii) Shear stress is constant as for pure plastic deformation (Eq. 4. 19)

C.-11\11\t\I.... 40....11\

....0t\I

.s::.11\

u'-E0C>-

Q

"~1~?~~~~~\;;~~:i~;~~:..'~~'.:,'~::ii~;.~;i:~~S~;,;"'~i~i :~:{{;~w'..,. ",,)!~:~'~.-i~:'x'i'~'~i:1ff~f:t;:~~i6!;r{"~f\ti~,.Dynamic Soil Properties 133

[

3 3

]

4T r2 -Tt'tmoe= - Z 2 4 4

31t (rZ -rt )(rZ -rt)...(4.18)

3T

(I

J- - 3 3

tavp - 21t r2 -rt...(4.19)

where T is the applied torque.

x-- x

~/~~"~Boundaries indicated by heavy lines

Fig. 4. t 4 : Cross-section for short specimen

4.2.5. Cyclic Triaxial Compression Test. In general the stress-deformationand strength characteristicsofa soil depend on the following factors:

1. Type of soil

2. Relativedensityin caseofcohesionlesssoils;consistencylimits, watercontentandstateofdisturbancein cohesive soils

3. Initial static stress level i.e. sustained stress4. Magnitude of dynamic stress

5. Number of pulses of dynamic load

6. Frequency of loading

7. Shape of wave form ofloading

8. One directional or two directional loading

In one directional loading only compression of the sample is done while in two directional loading bothcompression and extension is done. All the factors listed above can be studied lucidly on a triaxial set up.; ,


Prof. Arthur Casagrande of Harvard university was refered a problem of studying the effect of vibra-tions created by bomb explosion on the stability of the Panama canal projects. For this Casagrande andShannon (1948, 1949) developed the follo~ing three types of apparatus for studying the strength of soilsunder transient loading (Table 4.2).

Table 4.2 : Type of Apparatus

Type of apparatus Time ofloading (seconds) Remarks

(i) Pendulum loading 0.05 to 0.01 Suitable for performing fasttransient tests

(ii) Falling beam

(iii) Hydraulic loading

0.5 to 300

0.05 to any desired largervalue

Time ofloading was defined as the time between the beginning oftest and the point at which the maxi-mum compressive stress is reached (Fig. 4. 15). The pendulum loading apparatus (Fig. 4. 16) utilizes theenergy of a pendulum which, when released from a selected height, strikes a spring connected to the pistonrod of a hydraulic lower cylinder. This lower cylinder is connected hydraulically to an upper cylinder, whichis mounted on a loading frame.

"000J

TimeI" ~Timeof loading

FiOg.4.15 : Time of loading in transient tests

The falling beam apparatus consists essentially of a beam with a weight and rider, a dashpot to contr-the velocity of the fall of the beam, and a yoke for transmitting the load from the beam to the specimen (Fi4. 17). A small beam mounted above the yoke counter-balances the weight of the beam.

The hydraulic loading apparatus (Fig. 4.18) consists of a constant volume vane-type hydraulic purconnected to a hydrauliC cylinder through valves by which either the pressure in the cylinder or the volUlof the liquid delivered to the cylinder can be controlled. The peak load that can be produced by this .apTratus is much greater than can be obtained by either the pendulum type or falling beam apparatus.

.' d~,",_,_,,' d',~'- " - ~~\~~i~:~;, -'If,~~t{t'~::~J~'~i~1if~f,t;;~~~;r{\;f'~*~~,- t'


[

3 3

]

4T r2 -r)'tave = - 2 2 4 4

31t (r2 -Ij )(r2 -r\)...(4.18)

3T

(1

)'tavp = 21t ri - r\3 ...(4.19)

where T is the applied torque.

x-- x

~/~~"~Boundaries indic a t ed by heavy Iin es

Fig. 4.) 4 : Cross-section for short specimen

4.2.5. Cyclic Triaxial Compression Test. In general the stress-deformationand strength characteristicsofa soil depend on the following factors:

1. Type of soil

2. Relativedensityin caseof cohesionlesssoils;consistencylimits, watercontentandstateofdisturbancein cohesive soils

3. Initial static stress level i.e. sustained stress4. Magnitude of dynamic stress5. Number of pulses of dynamic load6. Frequency of loading7. Shape of wave form ofloading8. One directional or twodirectional loading

In one directional loading only compressionof the sample is done while in two directional loadingbothcompression and extension is done. All the factors listed above can be studied lucidly on a triaxial set up.


Prof. Arthur Casagrande of Harvard university was refered a problem of studying the effect of vibra-tions created by bomb explosion on the stability of the Panama canal projects. For this Casagrande andShannon (1948, 1949) developed the follo",:,ingthree types of apparatus for studying the strength of soilsunder transient loading (Table 4.2).

Table 4.2 : Type of Apparatus

Type of apparatus Time of loading (seconds) Remarks

(i) Pendulum loading 0.05 to 0.01 Suitable for performing fasttransient tests

(ii) Falling beam

(iii) Hydraulic loading

0.5 to 300

0.05 to any desired largervalue

Time ofloading was defined asthe time between the beginning of test and the point at which the maxi-mum compressive stress is reached (Fig. 4. 15). The pendulum loading apparatus (Fig. 4. 16)utilizes theenergy of a pendulum which, when released from a selectedheight, strikes a spring connectedto thepistonrod of a hydraulic lowercylinder.This lower cylinder is connectedhydraulicallyto anupper cylinder,whichis mounted on a loading frame.

"0

00-I

TimeJ.. ~Time of loading

Fig. 4.15 : Time of loading in transient tests

The falling beam apparatus consists essentially of a beam with a weight and rider, a dashpot to contr-the velocity of the fall of the beam, and a yoke for transmitting the load from the beam to the specimen (Fi4. 17). A small beam mounted above the yoke counter-balances the weight of the beam.

The hydraulic loading apparatus (Fig. 4. 18) consists of a constant volume vane-type hydraulic purconnected to a hydrauliC cylinder through valves by which either the pressure in the cylinder or the volurof the liquid delivered to the cylinder can be controlled. The peak load that can be produced by this apfratus is much greater than can be obtained by either the pendulum type or falling beam apparatus.

'$ffj.."f;;:!dfj:,'~~W~(,%"~~.Q$:,~",<,'.r:(\i.i"'.~ ;F?';.,tl~i">"


HydrauliccylindtZr

LowtZrcylindtZr

135

~

UpptZr cylinderAdjustablereaction

DeformationgageTe stsptZcimen

Fig. 4.16 : Pendulum loading apparatus (Casagrande and Shannon. 1948)

Countczrwcz ight

Fixed fulcrum

.Load gagczD<zformation gogczTtZst sptZcimtZn

Doshpot

RidtZr

Spring

Fig. 4.17 : Falling beam apparatus (Casagrande & Shannon, 1948)


pr«ssure r«gulator(pressure of dischargemay b. quickly variedby manual control.b«twun 10 and 1000lb. p«r sq. inch)

5 HP motor220V,AC1200 RPM

Three position nlec torvalve (with Pto A. Bis Conn«cted to T andvice verso; with Pto T.

and 8 ore blocked)

F Iow control valve(valves will maintain Aconstant flow betwEenSand 1250 CU. in. permin. rega rdless ofpussure)

van«-type hydraulicpump 100 lb. persq. inch 2.5 GPM

Filhr

Hyraulic cylindu3 inch bore, 3 inch stroke(used to apply load topiston of triaxialcompression apparatus)

Twenty gallonoil resuvoir

Fig. 4.18 : Hydraulic loading apparatus (Casagrande & Shannon, 1948)

For measuring load, a load gage of rectangular or cylindrical shape is used, with four strain gages mountedon the inside face. For measuring deflection, a thin flexible steel spring cantilever is used with strain gagesmounted on the cantilever, the base of which is clamped to the loading piston.

Casagrande and Shannon (1948) performed both static and transient compression tests on six differentmaterials. Out of these, two typical materials named as Cambridge clay and Manchester sand having prop-erties as given in Table 4.3 are selected for illustration. The transient compression tests were performed withdifferent time of loading both in confined and unconfined states.

Table 4.3 : Properties of Soils used in Tests

Cambridge clay Manchester sand

Natural water

Content

Grain Size 0.21 mm to 0.42 mm

Liquid limit

Plastic limit

30-50%

37-59% el113X

20-27% el11in

0.88

0.61

In Fig. 4. 19, a simultaneous plot of stress and strain versus time from an unconfined compression testwith a time of loading of 0.02 s on cambridge clay is shown. Similar plots were prepared for other times of10adll1gand on Manchester sand. Using this data, stress-strain plots were obtained as shown in Figs. 4. 20aand 4.20b. In these figures, stress-strain curves for corresponding static tests are also shown. Typical plotsof maximum compressive stress versus time of loading (or unconfined and confined transient tests on Cam-bridge clay are shown in Fig. 4. 21 a and b respectively. A typical plot in terms of principal stress ratio afailure and time ofloading for Manchester sand is shown in Fig. 4.22. .

~&.

)ynamic Soil Properties137

4 8

NEu

3 I-Sh<zar tailurlZ_-at 0.02 s6

0~

Cl.:ill: 2

~

IIIIIIt:i........V)

4. .::0.......tJ)

2

00 0.04 0.08 0.12

TimlZ~ s0.16 0.20

00.24

Fig. 4.19 : Time vs stress and strain in an unconfined transient test on Cambridge clay(Casagrande & Shannon, 1948)

From the typical test data presented above, it may be concluded that the strength of clay decreases withthe increases in time of loading and for time ofloadingequal to 0.02 s, the strength of clay is approximately1. 75 to 2.0 times greater than the static strength. The strength of sand is almost independent to the time ofloading. Transient strength of sand increased only about 10 percent.

0

a..2

static t<zst,timlZof 10ading,46S S

0-0

3 II

I

I

dTransi<znt test,time of load ing0.02 s

2.0 2.40.4 0.8 1.2 1.6stress, kg/cm2

(a) Cambridge clay

Fig. 4. 20 : Stress vs strain curves (...Contd.)

c 4-0... S.....(j')

6

7

80


0

static te'st. timeof fa i lure 2100 s

Shear fa i lu reI

Transiqnt test tim~to failurq 0.03 s

1.750.50 0.75 1.00stress. kg/cm2

(b) Manchester sand

Fig. 4. 20 : Stress vs strain curves

0

).4

1.0

0.81000 100 10 1

Ti m q of 10a din g. s0.1 0.01

(a) Unconfined compression test

Fig. 4.21 : Maximum compresslve stress versus time 01 loading for Cambridge clay (...Contd.)

0 2;;-

..c 3.-0...+' 4(/)

5

6

70

N 2.6Eu- 2.4C7'oX

,2.2IJ'I

IJ'ItII... 2.0-1/1tII

1.8>-1/11/1tII 1.6...a.E0u

E 1.2::JE.-x0:E

)ynamic Soil Properties 139

tr 5.2Ib 5.0

I I I I I I 1111 I I I[TlTIIII :1\11111 illll

//

//

//V 0

/ " -

//

/V

0 .//1

II I IIIII1I I": I 1 hili I I I IIIII I I I I100 10 1

Time of loading I S

0.1 0.01

-11'11' 4..8~...~ 4.6...E 4,40>~ 4.2~

E 4.0::JE.- 3.8x0~ 3.6

1000

(b) Confined compressin test

Fig. 4.21 : Maximum compressive stress versus time of loading for Cambridge clay

7

~

0.0+-

E 0::J ....

E lA.- IIIX ~0 ...~~

5

100 0.1 0.01

0.~ bMv-CO 6

10Time of loading, S

Fig. 4.22 : Maximum principal stress ratio versus time of loading for Manchestor sand

41000

Modulus of deformation is defined as the slope of a line drawn from the origin through the point on thestress-deformation ~urve and correspon~ing to stress of one-half the strength. It is found that in case ofclays, modulus of deformation in fast transient tests was about two times that obtained in static tests. Incase of sands, modulu.sof deformation was found independent ~othe time of loading.

6.0

5.8N

Ev 5.6-etoX

5.4..-

,..- 0 (1

00 0.... v

-0- 0 0 0

11111I I I 111\1 I I ,Ill' I I 11111 'I' 1111\,,°,

140

""'}f><;C


As evident from above discussions, in transient loading test, an initially unstressed sample of soil isloaded to failure in a short period of time. Under earthquake loading conditions, an initially stressed soilelement is subjected to a series of stress pulses, none of which would necessarily cause failure by itself, butthe cumulative effect of which is to induce failure or significant deformations. Seed and Fead (1959) devel-oped the oscillatory triaxial apparatus as shown in Fig. 4. 23 to study the effect of number of stress reversalsand other factors on the deformation in soils.

Counter- ba Ion c efor loading yoke

Deformationgauge

Dynomometer

Triaxialcompression ceH

Soil specimen inrubb(zr membrane

Countczr to recordnumber of loadapplications

pressure cylindarBellofrom seat

pistonLoading yoke

Air pressure equal to desiredconfining pressur on specimen

r- -EI-;ct;icalconnections: to timing unitII ,I .I . Compressed airAir

Ipre ssu re fr .

Air pre c;sure:re9ula,tor ' ~gauge

r.J1I

J

Air pressure reservoir

Fig. 4.23 : Apparatus for oscillatory triaxial test (Seed and Fead, 1959)

'-"'-"'.""-' '--.--.-.------.-

Seed (1960) performed tests on Vicksburg silty clay(wn = 22%, S =93%) for studying the effect ofvari-ous factors on its dynamic strength characteristics. A typical stress strain- relationship is shown in Fig.4.24. It pertains tQthe sustained static stress of2 kg/cm2. The magnitude of dynamic stress was 35% of thesustained static stress. It gives the magnitude of dynamic stress as :t 0.70 kg/cm2. The static stress-straincurve indicating the static strength as 3.0 kg/cm2 is also ~hown in the figure. Therefore the initial factor ofsafety for this typical case will be 3.0/2.0 = 1.5. Transient strength of the soil corresponding to 0.02 s t~meofloading was found as 4.0 kg/cm2. In this figure, B represents the point which is obtained after 100 cycles ofloading. It may be noted that at point 'B', the strains are increased but factor of safety still remains more thanunity (3.0/2.7). It means that failure will not occur but the strains may become excessive. Hence in design,one should examine whether these strains are within permissible limits.

Deviator st res s) kg/cm 20

01 2 43 5

Fig. 4.24 : Stress versus strain for Vicksburg silty clay (Seed, 1960)

The deviator stress versus strain curves similar to as shown in Fig. 4.24 were drawn for different valuesof sustained static stress and dynamic oscillatory stress. Fig. 4. 25 shows the effects of single transientstress applications of the various intensities for initial factors of safety ranging between 1.0 to 2.0. Theshaded portion of the figure shows the deformation of the specimens induced at stress levels correspondingto the different factors of safety in normal strength test. The upper curves show the increase in deformationcaused by single transient pulses corresponding to 20, 40 and 60 percent of the initial sustained stress. Figs

4

8

12a-a,

c:16.-

0..........\JI

- 200.-X<X: 24

28

32

36

str<z.ssinduced du ring.ormQI strength te st

- - -fat factor safety, 1.5"i Strain induced

\ 10 during earthquakeload ing

- \SO 1 I70

100 - - - - - - IB \ I

stress curve \for sample in normalstr<zngth test Stress strain curve

\ for sample after

Confining pressure,'l kg/crl \earthquake loading

Transient strll"gthfor time of loading0.02 s

4.26 and 4.27 show similar data for a series of 30 and 100pulses. It is interesting to note that for this soil asingle transient pulse equal to 20 percent of initial sustained stress causesno appreciable deformation eventhough the factor of safety maybe as lowas 1.1.However it maybe seen that a series of 100such pulses forthe same initial factor of safety will cause an increase in axial strain of 10percent. The significance of in-crease numbers of stress pulses in producing increased deformation of soil is readily apparent from thesefigures.

,......;;. 20.......

.:: 180

~ 16III

0 14x<t 12

10

30

28

26

2l..

22

Sing!(z transient pulse

IncrllaSIl in straindull to transillnt stress

Transient stress'in c rlla se = 60°/.

= 40%

6

4

20-1.0

Fig. 4.25 : Soil deformations induced byvarious combinations of sustained and

transient stresses (Seed. 1960)

Simulahd earthquake:30 pulses at 2 cyclesp /lr se c .In c r/lase In stre ss

during simu latedearthquake = 60 °/0

= 40 °/0= 20 °/0

strain inducedduring simu lated.czarthqu a keload ing

2.0

Fig. 4.26 : Soil deformation induced by variouscombination of susbined and vibratory

stresses for 30 pulses (Seed. 1960)

It may be seen from Fig. 4.24 that in a normal strength test the maximum resistance of the soil is reachedat an axial strain of about 25 percent. If this strain is adopted as a criterion of failure, then a variety'of com-binations of initially sustained stress and earthquake stress intensities which will cause failure of the soilmay be obtained from Figs. 4.25,4.26 and 4.27. Figure 4.28 shows the combination of initial stress and earth-quake stress intensity causing 25% axial strain (i.e. failure). For single transient stress pulse the combina-tions of sustained and transient stresses would have to approach about 140 percent of the normal strengthto induce failure. For earthquake inducing 100 majQJ.pulses failure would occur when the combined stressestotalled only 100 percent of the normal strength. -, ,

20

18c

16.-0...+-

14III-0 12.-)(et 10

8

6

4

20-1.0

;;~f&~;i.:~t~,"%?r~~~t,+;i0't~~'~£~~~:!,,~~~},~~{~;.c'j} ,,:~;':i''~:'i, ~


.\ Simula-ted earthquak e:100 pulses at 2cyclesper sec

.Inerescz in stressduring simula-tczd

. earthquake =60ol:: 40O}

/ / :: 20°/01

Fig. 4.27 : Soil deformation induced by variouscombination of sustained and vibratory stresses

for 100 pulses (Seed. 1960)

143

160 Number of transient stress, pulses1/

~:x

g 400-.c:........

::1, 20

Fig. 4.28 : Combination of sustained and vibratorystress intensities causing failure in compacted

silty clay (Seed. 1960)

Seed (1960) reported that the conditions producing failure shown in Fig. 4. 28 for compacted silty claydo not in any way represent the characteristics of conditions producing failure in other types of soils. Forexample in sensitive undisturbed clays, repeated deformations will lead to increase in pore pressure and aresulting reduction in strength. Consequently a series of transient pulses is likely to induce failure at lowerstress levels than for the silty clay.

On the same oscillatory triaxial test set up Seed and Chan (1966) carried out more elaborate study ondifferent types of soils. Here, few typical results are presented. Figure 4. 29a compares combinations ofsustained and pulsating stresses that induce failure in soft and compacted clays in one transient pulse. Thestrength exhibited by undisturbed Silty clay was greater than that displayed by compacted soils. As thenumber of pulses increases to 30 (Fig. 4.29b), failure occurs in sensitive soils at considerably lower stresslevel than that in compacted soils.

3028

26

24

22

20

18,c:.-

160........11\ 14-0 12x

<{ 10I

8

6

4

20 -

1.0

0--0c 80---11\11\.... 60...-er.

144

160

c 14-0...CJ\Cc:I"-I-11 120-0E"-g 100+-00-

0 80III'"c:I...t-'"C'IC

..60

-....0'"::Ja.

40

'"20

StrClss pulse torm ---0-

0 20 40 60 80 100Su~tain<Zdstress ,010ot normal st rength

(a) For single pulse


160

L:. 140.....0'1Ct.I...~ 120

Compactedsilty clay

....0

0

~ 1000c Compacted

sandy clay800-

0 "'"'"t.I...+-'"C'1C

60

.....0'"::Ja.

40undisturbedsilty clay

20

stress pulse to rm ,0

0 20 40 60 80 1Sustained str~ss/O/o of normal strengt

(b) For 30 pulses

Fig. 4.29 : Comparison of stres£ conditions causing failure fo. different soils (Seed and Chan. 1960)

Symmetrical stress pulses of two directionalloadings resulted in a reduction in strength of all the soilstested. Typical results with San Francisco Bay mud are shown in Fig. 4.30. Below the dotted line drawn at 45°from origin the stress conditions in one-or two-d:rectionalloadings are same since the pulsating stress iseither smaller than or equal to sustained stress.

Figure 4.31 shows the results of pulsating load tests with one-directional loading on duplicate speci-mens of San Francisco Bay mud using the two forms of stress pulse. The longer dwell period under maximumstress for the flat peaked pulse causes larger deformations and induces failure in a smaller number of stressapplications than in comparable tests using the triangular pulse form.

A typical total stress versus total strain curve under pulsating loads is shown in fig. 4.32. For compari-son, the stress strain relationship obtained from normal strength test is also shown. It may be noted that insituations involving 10 stress pulses, total stress versus total strain is somewhat higher than the stressstrain relationship of a normal test; if there are 100 stress pulses, this is slightly below the normal plot. Fur-ther it was noted that with an initial factor of safety between 1.5 and 2, and 30 pulses of dynamic stress, therelationship between total stress and total strain will coincide almost exactly with the normal stress versusstrain relationship.

180San Francisco Bay mudWatflT cont~nt ~ 91 °/0Str<zss puls~ form -'1r- Symm~trical

stress pulses. _. - Non-symm~tricali)--r stress puls~s

(no axial exhnsion)

, ,

20

0 -0 20 40 60 80 100Sustained strQ55,o/. of static strQn9th

Fig. 4.30 : Combination of sustained and pulsating stresses causing failurein one- and twodirectional loadings (Seed and Chan, 1966)

0E'-g 160-0...C

~ 120'-;;::,c.I...a.C'ICC.- t:.I~';;801ft 1ftc.I.......1ft

C'I 40C...01ft:J£l.

200San froncisco boy mud

-Wote co nten t ~ 91 0/0-Unconsolidoted undroinedte sts

'-Contin ing pressu re = 1 kgl cm2-No sustained strczss

For m' of.M..stress pulse

IFo rm ofstress pulse

01 10 100 1000 10,000 100,000

No. of pulses to couse foi lureFig. 4.31 : Effect of pulse form on number of pulses causing failure (Seed and Chan. 1966)

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

- 1200E....0c 100....00.. 80.

IIIIII

....60III

Cl>C....

400III:;Cl.

146

140

20

Et:...


Total stress (initial+pulsating) for initialfactor of safety between 1.5 and 2.0 vs. totalstrain after 10 pulsczs

0'0.-'- ." ~"'--;-'~

..' ." 0 ". stress vs. strain relationshipin normal compression tczst

~::'"J:.o'\' 1" ,.'t." ",

o

, '£"-"':":::.:";:,\;: (::~::..Y.,r.:'.:~:;.:.::- ',..-:::;;:.' ',::',::

Total stress {initial + pulsating)for initialfactor of safety betweczn l.S and 2,0 vs,totalstrain after 100 pulses

San Francisco Bay mudWote r co ntent z 91 %

Un conso Iidatczd -und ra ined te stf ' . 2

Con Inln9 pressure" 1 kg/cm

Stress pulse form J1.r0-

0 16 186 8 10 12 142 4strain,c/.

Fig. 4.32 : Relationship between total stress and total strain under pulsating load conditions(Seed and Chan. 1966)

4.2.6. Final Comments on Laboratory Testing, The laboratory techniques discussed above and more com-mon in use are listed in Table 4.4. The specific soil properties measured by each test are also indicated, Therange of strain amplitude over which each technique is applicable is shown in Fig, 4. 33.

Table 4.4 : Laboratory Techniques for Measuring Dynamic Soil Properties (Woods, 1978)

I" ~ ,'- ~",......

.J:120

....C\C

100....III

-0E... 800C

'+-

601

I

0 I0

I;;--

I:IJ\IJ\t.I....... 40U1

Techniques Shear Young's Material Cyclic stress Attenuationmodulus modulus damping behaviour

Resonant column x x x

With adaption

Ultrasonic pulse )( x - - x

Cyclic triaxial - x x x

Cyclic simple shear x - x x

Cyclic torsional Shear x - x xDynamic1 D compression x

- DZI!BIll !!IB\Im .. .. :am111:


104Shearing

103strain amplitude (°/.)

102 10'

Fig. 4.33 : Shearing strain amplitude capabilities or laboratory apparatus (Woods, 1978)

Each laboratory test has some merit when compared with the other. The resonant column device is bet-ter suited for determining shear modulus at Iow strains and the hollow cylinder device at higher strain levels.The cyclic simple shear device is suitable for determining shear modulus and damping characteristics ofsoils. The cyclic triaxial test is more suited to obtain the Young's modulus of the material.

4.3 FIELD TESTS

Field methods generally depend on the measurement of velocity of waves propagating through the soil oron the response of soil structure systems to dynamic excitation. The following methods are in use for deter-mining dynamic properties of soil:

1. Seismic cross-bore hole survey2. Seismic up-hole survey'3. Seismic down-hole survey4. Seismic refraction survey5. Vertical block resonance test6. Horizontal block resonance test

7. Cyclic plate load test8. Standard penetration test

In this section, the above listed field methods have been described briefly along with the typical testsetups and methods of interpretation.

Q%!III!'It;j'~"-- III.. .

I I I

Resonant column (solid samples)- - --Resonant column (hollow samples)

Torsional shear (hollow samples)

Pulse methods Cyclic triaxial--Cy cl ic simple sh ea r

Shaking tabler .

Ty pie a I motion cha racteristicsstrong ground Close-in

Properly designed shaking from nuclearmachine foundation earthquake explosion

I I I

148 Soil Dynamics & Machine Foundatio!q l

4.3.1. SeismicCross-borehole Survey. Thismethodis based on themeasurementof velocityof wave propa-gation from one borehole to another. Figure 4.34 shows the essentials of seismic cross-holesurvey as out-lined by Stoke and Woods (1972). A source of seismic energy is generated at th~ bottom of one borehole.'"and the time of travel of the shear wave from this borehole to another at known distance is measured. Shearwave velocity is then computedby dividing the distance between the boreholes by the travel ~e.

Reciver boreholes Source borehole

o~ ~

0 ~o 0(a) Plan view

Oscilloscope

@ TriggerInput

Ver tical impulse

~Verticalvelocitytransducer

Assum edpath ofbo dy waves

-- /~ ~

Generation ofbody wave s

3D velocity transducerwedged in place

Casing andgrout

(b) Cross-sectional view

Fig. 4.34 : Multiple hole seismic cross-hole survey (Stoke and Woods, 1972)

As discussed above, seismic cross-borehole survey can be done using two boreholes one having thesource for causing wave generation and another having geophone for recording travel time. However, forextensive investigations andbetter accuracy, three or more boreholes arranged in a straight line should beused.

.. !Ri!.II ..


In this case the wave velocities can be calculated from the time intervals between succeeding pairs ofholes, eliminating most of the concern over triggering the timing instruments and the effects of bore holecasing and backfilling (Stokoe and Hour, 1978). Also this arrangement of bore holes in a straight line over-comes problems of site anisotropy by examining one qirection only at a time.

For better results, the following points should be kept in view.(i) The diameter of the boreholes should be small to cause least disturbance in the soil. Casing in the

boreholes will provide good coupling with the soil and transmission of waves. Void spaces aroundthe casing should be filled with weak cement slurry grout or dry sand.

(ii) Boreholes should be vertical for the travel distance to be measured properly. In general any bore-hole 10 m or more in depth should be surveyed using an inclinometer or other logging device fordetermining verticality (Woods, 1978).

(iii) Boreholes should be spaced as close as possible within the time resolution characteristics of therecording equipment. Large spacings can lead to difficulties with refracted waves arriving beforethe direct transmission through the intervening soil. Spacings as close as 2-3 m can be used satis-factorily (Woods, 1978).

(iv) The seismic source must be capable of generating predominantly one kind of wave. Further it mustalso be capable of repeating desired characteristics at a predetermined energy level. Miller, Troncoscoand Brown (1975) have described a source which is capable of .developing high amplitude shearwaves. It consists of a falling weight which impacts on an hydraulically expanded borehole anchor.

(v) The receivers must be oriented in the shearing mode and should be securely coupled to the sidesof the borehole.

Wfl ightRod

Shot detector

Recorder

l/

// S wave

//

//

SPT sampler

Fig. 4.35 : Seismic up-hole survey with SPT (Goto et al., 1973)

III _1...°'" "..,. . p;;;aB/\

4.3.2. Seismic Up-Hole Survey. Seismicup-hole survey is done by usingonly oneborehole. In this method.the receiver is placed at the surface, and shear waves are generated at different depths within the borehole.Figure 4.35 shows the schematic presentation of the arrangement used in seismic up-hole survey (Goteet al., 1977). This method gives the average value of wave velocity for the soil between the excitation andthe receivers if one receiver is used, or between the receivers.

The major disadvantage in seismic up-hole survey is that it is more difficult to generate waves of thedesired type. .

4.3.4. Seismic Down-hole Survey. In this method, seismic waves are generated at the surface of the groundnear the top of the borehole, and travel times of the body waves between the source and the receivers whichhave been clamped to the borehole wall at predetermined depths are obtained. The arrangement used inseismic down-hole survey is shown schematic ally in Fig. 4.36. This also requires only one borehole.

The main advantage of this method is that low velocity layers can be detected even if trapped betweerlayers of greater velocity provided the geophone spacings are close enough.

R<zcord<2r

W<zight

fExpand<zr

pump

Rubb<zr<zxpand<2r

,Backplate

woodenplo te

3-componqntgQophonq

Fig. 4.36 : Seismic: down-hole survey (Woods, 1978)

(.t ,------------

Dynq.mic Soil Properties 151

4.3.4. Seismic Refraction Survey. The seismic refraction survey is frequently used for site investigations.It enables the detennination of elastic- wave velocity in each layer, the thickness of each layer, and the dipangle of each layer as long as the wave velocities increase in each suceedingly deep layer.

The details of this method has already been described in Art. 3.10 of chapter 3.

4.3.5. Vertical Block Resonance Test. The vertical block resonance test is used for determining the valuesof coefficient of elastic uniform compression (C), Young's modulus (E) and damping ratio (~) of the soil.According to IS 5249: 1984, a test block of size 1.5 m x 0.75 m x 0.70 m high is casted in M 15 concrete in a pitof pIan dimensions 4.5 m x 2.75 m and depth equal to the proposed depth of foundation. Foundation boltsshould be embedded into the concrete block at the time of casting for fixing the oscillator assembly. Theoscillator assembly is mounted on the block so that it generates purely vertical sinusoidal vibrations. Theline of action of vibrating force should pass through the centre of gravity of the block. Two acceleration ordisplacement pickups are mounted on the top of the block as shown in Fig. 4.37 such that they sense thevertical motion of the block. A schematic diagram of the set up is shown in Fig. 4.37.

Motor osci \latorassembly

" ,

Concr(te(M 150)

For V te stAdc e le ra t iontransduc Clrs

77777// /,(a) sectional view

4.5Qm

1

to- --i

1 m min

~ 1.Sm, 1&:::11

I C:Jj ~2.75 m

T0.75 m

,11m min

(b)' plan view",

Fig. 4.37 : Set up for block ~esonance test

- ..- - - - -. --"'-.-"-------..-


The mechanical oscillator works on the principle of eccentric masses mounted on two shafts rotating inopposite directions (Fig. 2.15). The force generated by the oscillator is given by

2Fd = 2mee CJ.) ...(4.23)

The oscillator is fIrstset at aparticular eccentricity (e). As evident from Eq. (4.23)higher the eccentricitymore will be the force level. It is then operated at constant frequency, and the acceleration of the oscillatorymotion of the block is monitored. The oscillator frequency is increased in steps, and the signals of monitor-ing pickups are recorded. At any eccentricity and frequency the dynamic forceshould not exceed 20 percentof the total mass of the block and oscillator assembly. The amplitude of vibration (A) at a given frequencylis given by

a-A- (mm) = ---:;=- 2- 41t-f

az = Vertical acceleration of theblock, mm/if = frequency, Hz.

Amplitude versus frequency curves are plotted for each eccentricity to determine the natural frequencyof the foundation-soil system (Fig. 4.38). The natural frequency,f. ~, at different eccentricity (i.e. force level)n~

is different because different forces cause different strain levels of the block which may be accounted forwhen appropriate design parameters are being chosen.

The coefficient of elastic uniform compression (Cu) of the soil is then determinedusing Eq. (4.25)

4 2 1. 2 m1t n-C =-~u A

...(4.24)

where

...(4.25)

where fnz = Natural frequency of foundation-soil system, Hzm = Mass of the block oscillator and motor, Kg -sec2/m

2A = Base contact area of the block, m

From the value ofC obtained from Eq. (4.25) for the test block of contact area A the value ofC I for theU u

actual foundation having contact area A I may be obtained from Eq. (4.26)

- rECuI = Cu V A; ...(4.26)

The Eq. (4.26) is valid for base areas of foundations up to 10 m2. For areas larger than 10 m2, the value 01Cu obtained for 10 m2 is used.

The coefficient of elastic uniform compression (Cu) is related to the elastic Young's modulus (E) by Eq(4.27) which is in the form of Boussinesq relationship for the elastic settlement of a surface footing

E Cs

Cu = (1- ~2) . JBL...(4.27

where ~ = Poisson's ratioB = Width of base of the blockL = Length of base of the block

Cs = CoefficientdependingonLIBratio

,amic. So.il Properties153

200

so

Vertical vibration.0/8 ~O'

0 e

-A e~ 8

.. '8

00

: 35

: 70°

: 1050

: 1400

150

IIIc:0~v

tt

It-

E I:rQI" 100::J.....a.E~ il

r-8 ---4

15 . 45 50Frequency, cps

Fig~4~38 : Amplitude versus frequency plots from vertical resonance test"

--


Barkan (1962) recommended the values of Csfor various LIB ratios as listed in Table 4.5

Table 4.5: Values orcs(After Barkan, 1962)

The value of damping ratio ~is determined using Eq. (4.28)

~=f2-f12fnz

...(4.28)

Awhere f2,Jj = Two frequencies at which amplitudes is equal to If

Amax = Maximumamplitudehzz = Resonant frequency

This is also illustrated in example 2.6.

4.3.6. Horizontal Block Resonance Test. Horizontal block resonance test is also performed on the block setup as shown in Fig. 4.37. In this test, the mechanical oscillator is mounted on the block so that horizontalsinusoidal vibrations are generated in the direction of the longitudinal axis of the block. Three acce~erationor displacement pickups are mounted along the vertical centreline of the transverse face of the block tosense horizontal vibrations (Fig. 4.37a). The oscillator is excited in steps starting from rest condition. Thesignal of each acceleration pick up is amplified and recorded. Rest of the procedure is same as desqribed forvertical block resonance test. Similar testscan be performed by exciting the block in the direction oftrans-verse axIs.

The amplitude of Horizontal vibrations (Ax) is obtained using Eq. (4.29).

axA = ~

f2x 4n ...(4.29)

whereax(mm) = Horizontal acceleration in the direction under consideration in mmli

f = Frequency in Hz

Amplitude versus frequency curves are plotted for each force level to obtain the natural frequency,flLlof the block soil system as done in vertical resonance test. A typical frrquency versus amplitude plot i~;shown in Fig. 4.39. It may be noted that the case of horizontal vibrations.is a problem of two degrees o(

" "

I.,

f

,

'

,

'

,

.

",,

'

. . .

lJB Cs

1.0 1.06-

1.5 1.07

2.0 1.09-3.0 1.135.0 12210.0 1.41

2:,;:r',£{';:;~~{;I;!ffl~'$(>".

[,00

lIB i2;.;:~ ',.

lie Soil Properties 155

)m. The mode of vibration is obtained by plotting amplitude versu: -:y of the system from the analysis of data from the pickups mounttblock. Typical plots are shown in Fig. 4.40. If the plot correspOl

~ncycorresponds to first mode or lower natural frequency. On the (g. 4o40b, then natural frequency corresponds to second moc

~8-t

. , ,

IIIC0'-uE 200Q,I

~::J...Q.Ecd:

1-~ 1.

100

0-10 15 20 25 30 35 40

1[,5

Frequency I CpS

Fig. 4.39 : Amplitude versus frequency plots rr~m honzon~al resonance test

'amic Soil Properties 157

The coefficient of elastic uniform shear (Ct) of soil is given by the following equation:î î

èï¬®·³æ

Ct= ~ 2(Ao+Io):t (Ao+Io) -4rAolo...(4.30)

where Mmr = M

1110

hx = Horizontal resonant frequency of block soil systemAo = A/MA = Contact area of block with soil

M = Mass of block, oscillator and soil

10 = 3.46 I/Mmo

Mm = Mass moment of inertia of block, oscillator, motor, etc. about the horizontal axis pass-ing through the centre of gravity of block and perpendicular to the direction of vibration

Mmo = Mass moment of inertia of the block, oscillator; motor etc. about L1ehorizontal axis passingthrough centre of contact area of block and soil and perpendicular to the direction ofvibration.

I = Moment of inertia of the foundation contact area about the horizontal axis passing throughthe centre of gravity of area and perpendicular to the direction of vibration.

In Eq. (4.30), negative sign is taken when the system vibrates in first mode and positive sign when theystem vibrates in second mode.

For the size of the block recommended in IS 5249-1977 and for first natural frequency, the Eg. (4.30)'educes to

2Ct = 92.3 fnxl ...(4.31)

In Eq. (4.31), Ct is in kN/m3.

The coefficient of elastic uniform shear (Ctl) for actual area of foundation (At) is given by

C" ~C,~AAlIS 5249 : 1977reconunends the followingrelations between Cu'and Ct, Ccpand CII':

Cu = 1.5to 2.0 Ct ...(4.33)

Ccp= 3.46Ct ...(4.34)CIjI= 0.75Cu ...(4.35)

4.3.7. Cyclic Plate Load Test. The cyclicplate load test i,sperformed in a testpit dug upto the proposedbaselevel of foundation. The equipme.nt is same as used in static plate load test. Circular or square bearingplates of mild steel not less than 25 mm,thickness and varying in size from 300 to 750 mm with chequeredorgrooved bottom are used. The test pit should be at least five times the width of the plate. The equipment isassembled according to details given in IS 1988-1982.A typical set up is shown in Fig. 4.41.

...(4.32)

158 Soil Dynamics & Machine Foundations'

Sand bags to give areaction of SOOkN ò

ðòêð³õó 1.0m 1 0 m -+ 0.60m-4

Wooden sleeper2m xO.30mxO.15m2n05. on either side

Prov jng rl n,9

Foundation level

f--- 1.Sm Plate 300 mm x 300mm

Sectional elevation

-,100mm

-L

J,Sm

3nos R.S.J. 300mm x lS0mm ot LoOO mm clctop row of joists not shown in pion

Plan

Fig. 4.4t : Set up for cyclic plate load test

To commence the test, a seating pressure of about 7 kPa is first applied to the plate. It is then removedand dial gauges are set to read zero. Load is then applied in equal cumulative incrementsof not more than 100kPa or of not more than one fifth of the estimated allowable bearing pressure. In cyclic plate load test, eachincremental load is maintained constant till the settlement of the plate is complete. The load is then releasedto ~ero and the plate is allowed to rebound. The reading of fmal settlement is taken. The load is then in-

1" I

f-Tr~ ,.,.".:.,..,.,.""II\~- !IIiI

mic Soil Properties

- III

159

;ed to next higher magnitude ofloading and maintained constant till the settlement is complete. which.l is recorded. The load is then reduced to zero and the settlement reading taken. The next increment ofis then applied. The cycles of unloading andreloading arecontinuedtill the required final load is reached.The data obtained from a cyclic plate load test is shown in Fig. 4.42. From this data, the load intensityus elastic rebound is plotted as shown in Fig. 4.43, and the slope of the line is coefficient of elasticJITn compressiOn.

where

Ð íÝ ã -(kN/m)ïï Í»

° ã Load intensity in kN/m2

Se = Elastic rebound corresponding to p in m.

Load intflns ity, p

p, pz .p 3

-f-

l_--r-z2 ---1-j__-Sfl3 - ---

l~i-Se4-r-'- -

_t_-SflS

-------

Fig. 4.42 : Load intensity versus settlementin a cyclic plate load test

...(4.36)

s.~

a.

p

...>-~

\1\CC:II

~C

"0"0J

Elastic rflbound, Sfl

Fig. 4.43 : Load intensity versus elastic reboundfrom cyclic plate load test

3.8. Standard Penetration test (SPT). The standard penetration test (SPT) is the most extensively usedsitu test in India and many other countries. This test is carried in a bore hole using a split spoon sampler.s per IS: 2131-1981, steps involved in carring out this test are as follows:

(i) The borehole is advanced to the depth at which the SPT has to be performed. The bottom of theborehole is cleaned.

ºæ®¬¢òò¢ô¢óó -""" ..mic Soil Properties

91: III

159

;ed to next higher magnitude ofloading and maintained constant rill the settlement is complete, which1is recorded. The load is then reduced to zero and the settlement reading taken. The next increment ofis then applied. The cycles of unloading andreloadingare continuedtill the required final loadis reached.The data obtained from a cyclic plate load test is shown in Fig. 4.42. From this data, the load intensityus elastic rebound is plotted as shown in Fig. 4.43, and the slope of the line is coefficient of elasticJrm compreSSiOn.

where

Ð íÝ ã -(kN/m)ïï Í»

° ã Load intensity in kN/m2

Se = Elastic rebound corresponding to p in m.

Load int<zns ity, p

-f-

p, pz 'p 3

l_--r-z2 ---1-j__-S<Z3 ---

1--_t-Se4-r-'- -

_t_-S<zs -------

Fig. 4.42 : Load intensity versus settlementin a cyclic plate load test

...(436)

s.~a.

p

...>--4J\11CC:II

-4JC

"0

00..J

Elastic r<zbound, S<z

Fig. 4.43 : Load intensity versus elastic reboundfrom cyclic plate load test

3.8. Standard Penetration test (SPT). The standard penetration test (SPT) is the most extensively usedsitu test in India and many other countries. This test is carried in a bore hole using a split spoon sampler.s per IS: 2131-1981, steps involved in earring out this test are as follows:

(I) The borehole is advanced to the depth at which the SPT has to be performed. The bottom of theborehole is cleaned.

--~ mE -i

160 Soil Dynamics & Machille Foulldations

(ii) The split-spoon, attached to standard drill rods of required length is lowered into the borehole andrested at the bottom.

(iii) The split -spoon sampler is seated 150 mm by blows of a drop hammerof 65 kg fallingverticallyandfreely from a height of750 mm.Thereafter, the split spoonsamplershallbe further driven300 mm intwo steps each of 150m. The number of blows required to effect each 150 mm of penetration shallbe recorded. The first 150 mm of drive may be considered to be seating drive. The total blowsrequired for the second and third 150 mm of penetration is termed the penetration resistance N.If the split spoon sampler is driven less than 450 mm (total), then N-value shall be for the last300 mm penetration. In case, the total penetration is less than 300 mm for 50 blows, it is entered asrefusal in the borelog.

(iv) The split spoon sampler is then withdrawn and is detached from the drill rods. The split barrel isdisconnected from the cutting shoe and the coupling. The soil sample collected inside the barrel iscollected carefully and preserved for transporting the same to the laboratory for further tests.

(v) Standard penetration tests shall be conducted at every change in stratum or intervals of not morethan 1.5 m whichever is less. Tests may be done at lesser intervals (usually Q.75m) if specified orconsidered necessary.

The penetration test in gravelly soils requires careful interpretation since pushing a piece of gravel cansreatly change the blowcount.

4.3.8.1. Corrections to obsetYed SPTvalues (N) ill cohesionless soils. Following two types of correctionsare normally applied to the observed SPT values (N) in coh~sionless soils:

Corrections due to dilatancy:Invery fine, or silty, saturatedsand, Terzaghi and Peck (19~7) recommendthat the observed N-valllesbe

corrected to N' if N was greater than 15as1

N' = 15+ - (N-15)2Bazaraa (1967) recommendedthe correction as

N' = 0.6 N (forN > 15) ...(4.38)This correction is introduced with the view that in saturated dense sand (N > 15); the fast rate of

application of shear through the blows of drop hammer, is likely to induce negative pore pressures and thustemporary increase in shear strength will occur. This will lead to a N-value higher than the actual one. Sincesufficient experimental evidence isnot available to confirm this correction, many engineers are not applyingthis correction. However this correction has also been recommended in IS: 213 1-1981.

...(4.37)

Correction due to overburden pressure:

On the basis of field tests, corrections to the N-value for overburden effects were proposed by manyinvestigators (Gibbs and Holtz 1957; Teng 1965; Bazaraa 1967; Peck, Hanson and Thornburn 1974). Themethods noffi1ally normaly used are:

Bazaraa (1967)

For 0"0< 75k Pa4N

1--N = 1+0.040"

I

D11IIII --


For aa-> 75kPa

-~.miiYL~.M£!,,:

161

4N

N' = 3.25+0.01ao

Where a 0 = Effective over burden pressure, kPa

...(4.40)

Peck, Hanson and Thornbum (1974)

N' = 0.77 N loglO 2000ao

Figure 4.44 gives the correction factor based on EqJ..4Al). Use of this figure has been recommended in~2 131-1981. In this figure,

Ct>1,;:)'-:J.D'-e:,I>

.0t>1>

t 400t>1

'4-'t-W

...(4.41)

. 2000~ = Correction factor = 0.77log10 ~ao

...(4.42)

correction foetor CN" .

1.2 1.6 2.0

300

500

Fig. 4.44 : Over burden correction

There is a controversy whether the correction due to dilatancy should be applied first and then thecorrection due to over burden pressure or vice-versa. However in IS 2131-1981, it is recommended that thecorrection due to overburden should be applied first.

A typical set of o~~eryed N-value~ are shown in Fig. 4.45. C_orre~t~d~-values as per IS Code recommen-dations are also shown in the figure. .

. H

0.4 0.80

0CL.:.:.

... 100t>1'-:JIIIIIIt>1'-Co 200

c,,: ..;..i.:.,,:;~:t,'4':~',:V'c,:e;\.;~';:'.:,,:;:f.l'!.:;.:;X!!::*,:~,/'::":t::\i"l//</";'::\i',!"".b~..';;":~;!';"'":."~";';'.of"t.:'.' ~',,;;"..{"':"i:."<'" ':.,'.L" . ,""'c',','?,,,,:..."" ;'" ,"~j,htc\.U~~.:.

162

E.s::Q. 6Cl0

. '

00

4

2It = 20 KN1m2

--Yl-~sub = ~OKN 1m2

3

4

5

7

8

9

10

11

vf


Standard' penetration resistane. N8 12 16 20 24 28 32

",, ,"\\.."-"-"-"-

'\\\\\\\\t"-"-"-""-"-"-"-

')I

II

III

Obse rvedCorrected

Fig. 4.45 : Typical SPT data

~-_.- III.~

y"amic Soil Properties 163

The SPT is es.sentiallyundrained test for the duration of each blow and the energy generated by thePT hammer isprincipally shearing energy.Therefore the test maybe useful to predict the dynamicbehaviourfsoils. Seed el et. (1983) presented.correlations between SPT and observed liquefaction. Ohasaki (1970)escribes a useful Japanese rule of thumb that says liquefaction is not a problem if the blow count from a;PTexceeds twicethe depthof sampleinmeters~ .

Imai (1977) reported the following correlation between N (observed) and shear wave velocity, Vsmls):

Vs = 91~.337 ...(4.43)Bowles (1982) has given a number of equations to obtain stress-strain modulus Eson the basis ofSPT

md cone-penetration test (CPT). These equations are given in Table 4.6.

Table 4.6 : Equations for Stress-Strain Modulus Esby SPT and CPT(After Bowles, 1982)

SPT CPT

Sand

Clayey sandSilly sandGravelly sandSoft clay

Es = 500(N+ 15)Es = 180oo+750NEs = (15000to22000)inN

Es =Es =Es =

.,~ = 2 t04Qc2Es = 2(1+ Or )Qc

320(N + 15)300 (N + 6)1200(N+6)

Es = 3 to 6QcEs = 1 to 2Qc

Es = 6 to 8 Qc

Using the undrained shearstrength: CIt

I > 30, or organicpI < 30, or stiffpI < OCR < 2

OCR> 2

Es = 100 to 500 Cu

Es = 500 to 1500 Cu

Es = 800 to 1200 CII

Es = 1500 to 2000 CII

Notes: 1.Unit ofEs is KPa in correlations with N.

2 Es will have the same unit as of Qc3. Es will have the same unit as of Cu

4.4 FACTORS AFFECTING SHEARMODULUS,ELASTICMODULUSANDELASTICCONSTANTSHardin and Black (1968) have given the following factors which influence the shear modulus,elastic modu-

.lus and elastic constants:

(l) Type of soil including grain characteristics, grain shape, grain size, grading and mineralogy;(ii) Void ratio;

(iii) Initial average effective confining pressure;

'}


(iv) Degree of saturation;(v) Frequency of vibration and number of cycles ofload;. .

(VI) Ambient stress history and vibration history;(vii) Magnitude of dynamic stress; and

(viii) Time effectS;

The shear modulus G, elastic modulus E, and dynamic elastic constants (ClI' C'Ct and C\jI)are relatedwith each other directly as evident from Eqs. 3.9,4.27,4.34 and 4.35. Therefore the factors listed from (i) to(viii) will effect G, E, ClI'C'Ct' C\jI in similar way. Keeping this in view, the effect of the factors have beendiscussed only on dynamic shear modulus G and damping ratio,

Soil behavior over a wide range of strain amplitudes is nonlinear and on unloading follows a differentstress-strain path forming a hysteresis loop as shown in Fig. 4. 46. The area inside this loop represents theenergy absorbed by the soil during its deformation and is a measure of the internal damping within the soil.At very low strain amplitudes « 0.0001 %) the soil acts essentially as a linear elastic material with little orno loss 9f energy. The s~ear modulus under these c.onditions is maximuI? but as the strain amplitudes isincreased, the shear modulus decreases and the damping within the soil increases.

óïð

îð

ñ

¶Ù³¿¨/ 1

//

/

IIIIII,IIIL

10

Yx 104 mm/mm

lOO cyc les

la cycles1 cycle0

CL.:tr.

~ 10

Dry cl~(jn sand

<z = G 57I

ac = 25 kPa- 20Ù³¿¨= f)') M po

Fil!. 4.46 : Stn'ss-strain 111°1)and points at 1st, 10th and 100th cycles of lo:ldin~ (After Hoadll'Y, 19115) w'~'

J

"'11!''.,.i

"

>i>t---'.-tt!,;

namic Soil Properties 165

Ishihara (1971) presented Fig, 4.47, which shows strain levels associated with different phenomenon in~field and in corresponding field and laboratory tests. Prakash and Puri (1980) presented the data ofG

)m different insitu tests as shown in F~. 4.48. It is evident from this fi~e that G decreases significantly:tenstrain amplitude is higher than 10. . For lower strain amplitude«10.5), G may be consideredconstant.

Fig. 4.47 : Strain levels associated with different insitu laboratory tests (After Ishihara, 1971)

The machine foundations are usually designed for very low strain amplitudes so that the behaviour ofsoil is elastic under vibrations, It is to avoid the building up of any residual strains in the soil due to theoperation of the machine, Large strain amplitudes may be developed by commercial blasting, earthquakes,nuclear blasts, pile driving operations, compaction devices or excessive vibrations of the machinery,

The subsequent discussions have been made under two heads (i) Shear modulus for low strain ampli-tudes, and (ii) Shear modulus for large strain amplitudes.4.4.1. Shear Modulus for low Strain Amplitudes in Cohesionless Soil. In the case of cohesionless soils,shear modulus G is dependent on effective confining pressurecro and void ratio e. The effect of other fac-tors on G is negligible (Hardin and Richart, 1963). They have reported the results of several resonant col-umn tests in dry Ottawa sands as shown in Fig. 4.49. Straight lines have been fitted through the test pointscorresponding to different confining pressures.,Similar results are also shown inFig. 3.12 (of chapter 3) assolid lines. In order to 'extend the lines for wider range of void ratios, 'dotted liiiesnave been drawn in Fig.3.12 to represent the results from tests using clean angular grained materials. The peak to peak shear strainampli~de for the~e

.tests was 10.3rad, It can be seen from these figr. . es~~t

.

v,~

..

and~.~r:.

i~,d.~~en

.

~en,

.

t Of

.

thegradatIon and gram slze'dlstnbutlOn. The effect of confinIng press fi m<rramr-on"\:i/antfG lSXSIcant ', ' .ff",'. I?"!"""'j'.!' S

"

. . ,.11. I, ..' ", . ",Ic."""""V,.." 1_"i:.'>J.)."$O""",,,~ '-' . J ',..".."t, ',,'""" """" ~ co': .

.t1y ' .~:~,.,., ,','.:; f/i,~ \ I -:"C>-"':i.j>;""Aj " ,:.;~;"",'""",..; ,,-7. -'..'.;/ \:00.>" '-.,.7 ",,;;I "J

Magnitude of train-6 -5 -4 -3 -2 -1

'9 1 10 10 10 10I I

Cra cks,ditferllntia I-S[ld1

Phenomena wav propagation. vibration compac tion,se'ttlczmcznt liquifacation

Mechanical Elastic Elastic-plastic Failurecharacter itics

Constants Shear modulus, poisson's ratioJdamping ratioAngle OT

inNmal friction,cohcsion

seismi c waye I I+"' methodc::::J W

.,.E In situW vibration test I ,..,., '...

c::::J- III RcpIlotedg IE load ing test J

Iwavtl propa\)ation I I>- te st... c0 W+' E R sonant0 W I I""" column testo:J.a'"0 0 Repeated...JCII r IE loading test

- -~~---~"UIIII ~~ &

)ynll1ltic SoU Properties 167

,Sy'm.

" ,::-.

(J' 'H' ,0 .. 3.. OOk",/ 2A 'h

et

e, So kN/tr] 2-----0

.

0.65 ,0.75Void ratio~ e

Fig. 4.49 : Variation of shcar-wa\'c ve!ocity wit,h void ratio for various confining pressures. grain SiLl', . ' . " J,',' .

and gradation in dry' OUa\V8 sand (Hardin and Richart, 1963)"" :,.i"':' .~,"';';")' ô÷òò¢ôøþò¢þòò·ååÖùæùå C,"',' '..

C'""',' .', '"," ,'", "',',,;",:';, ,', ".

0',", ,,'

,>

Void. ratio

Max. Minl

0.71 ' , 0.495

0.89 0.54

0.66 0.32

0.76 0.42

.0A

11I

390

360u11\ 330-E

..11\>

300..>--.-u0 270->

240>0

14... 2100

.I:.If)

180 "150

1200.35

168

~.

Soil Dynamics & MachiIJe ,Foundations

Following empirical expressions have bp.endeveloped for Vsand G for round-grained sands and angu-lar-grained crushed Quartz (Hardin and Black, 1968). '

For round grained sands (e < 0.8)

Vs = (11.36-5.35 e) (ao}O.32

G= 6908(2.l7-e) (a )°-',l+e 0

For angular grained sands (e~0.8)

Vs = (18.43-6.2e)(ao)0.252

G = 3230(2.97 - e) Ca )°.5l+e 0

where

...(4.45)

...(4.46)

...(4.47)

...(4.48)

Vs = shear wave velocity in m/sG = Shear modulus in kN/m2

a 0 = Mean effective confining pressure in N/m2for Eqs.(4.45) and (4.47) ; and in kN/m2forEqs. (4.46) and (4.48)

e = Void ratio

. Hardin and Richart (1963) have shown that the effect of degree of saturation on Vs is insignificant(Fig. 4.50).

420

360

~ 300~ 270~ 240>-.~ 210u0-; 180>

ISO. Dry. Drained0 Saturated

120ZO 70 100 150

Pressure, cr, kN/mZ0

ZOOSO 500

Fig. 4.50 : Variation of shear wave velocity with confining pressure for a specimen of Ottawa sand in thedry, saturated and drained conditions (Hardin and Richart, 1963)

Seed and Idriss (1970) have suggested the following equation

G = lOOOK (ao)05 ...(4.49)

where G is in kN/m2units, K is an empirical factor which variesaccording to relative density of sand. and cr0is the mean effective confini~g stress in kN/m2 units. Table 4.7 gives some values ofK obtained from fieldmeasured values of shear modulus.

vnamic Soil Properties 169

Table4.7: FieldMeasured.ValuesofK .

(Seed and Idriss, 1970)

Soil K

7.510.013.014.0

16.019.0

Loose moist sand

Dense dry sandDense saturated sand

Dense saturated silty sandDense saturated sand

Extremely dense silty sandDense dry sand (slightly cemented)Moist clayey sand

.36.0

26.0

~.4.2. Cohesive Soils. Few investigators (Lawrence, 1965; Hardin and Black, 1968; Humphries and Wahls.1968) have performed tests on cohesive soils using resonant column devices to get the shear modulus. Onhe basis of analysis of experimental data, Hardin and Black (1968) have ootained the variation of G with e

md cr0 as shown in Fig. 4.51 for normally loaded clays.They have also reported the shear modulus tor somelndisturbed clay specimens collected from the field. The following relation has been suggested:

.,(2.97- er (

-)0.5

G=C cr .1 l+e 0

Nhere Cl is constant; and G and <10are both in kN/m2 units.

Equation 4.51 has been suggested by Hardin and Drnevich (1972b) for both clays and sand.

...(4.50)

- - 2= 3230(2.97-e) . OCR)

k« 1 )

0.5

G (l+e) ( 0

where K is function of plasticity index (Table 4.8), and OCR is over consolidation ratio.

...(4.5 -

Table4.8: ValuesofK-. .

(Hardin and Drnevich, 1972b)

00

20

40

(j)

00

> 100

k

0.00

0.18 .

0300.410.480.50

Plasticity index PI -

-170 Soil Dynamics & Machille Foundations

552,000

. Tap wahr kaolinite

A Salt flocculated kaolinite

0 Di spczrsed clay

420,000v Flocculated clay

o~:

(i\~ .

(6QO) \" ...: ,'-...'W:' . '"~:. \

00.5 0.7 0.9 1.1 1.3 1.5

Void ratio e

Fig. 4,5 t : Experimental values of G for some normally consolidated clays (Hardin and Black, t 968)

Hardin (1978) has suggested the following expression:

625 (OCRl

(ao

)

0.5

G= 2 -O.3+0.7e Pa

where Po is the atmospheric prt:ssure in the same unit as of ao' G is in kN/m2units.

I;j\

,-..0Cl..oX'-'V\ 280,000::J::J"'00E....0

.J:.If)

'amic Soil Properties 171

For clays, Seed and Idriss (1970) suggested an equation of the form:

Glcll = K ...(4.)3)~reCllis the undrained shearing strength of soil. K is a constant whose value lies between 1000 and 3000.

Ohasaki and Iwasaki (1973) developed a relationship by correlating the shear modulus obtained in asshole survey (SCS) with SPT 'N' values.

G = 12000No.8 ...(4.54). ere G is in kN/m2units, and N is the N-value recorded in standard penetration test. This equation applies

both sands and clays.

t.3. Shear Modulus for Large Strain Amplitudes. Figure 4.52 shows a plot between shear stress (t) and~arstrain (y) . The stress-strain curve is approximated by a hyperbolic function defined in terms of initial:ar modulus Glllaxand a reference shear strain Yrwhich is defined by Eq. (4.55)

Ì

1max

Yr = Gmax...(4.55)

'. .

'"Cma X - -,- -11 ---

J/JI -------

, --I 16 san d, I max --------I .--// --

/1 I // ---------I y/ (.\0'/

I .If.. //1/ // .'/ V

'; /1~ / It / I/ I

r; :I

T= ¥'1 }>- +-6maxI'max

'¥r yFig. 4.52 : Hypcrbolic strcss-strain relationship (Hardinand Drncvich. 1972 b)

The reference strain is equal to the strain at which a line drawn through the origin with a slope equal toimaxintersects the horizontal line at t =tmax:tmax is the shear stress at failure in the soil. It can be obtained1the following manner. Figure 4.53a shows a soil element at a given depth being subjected to vertical andorizontal effective stresses of ery andKo cryrespectively.Koisthecoefficientofearthpressureat rest.Thelohr circle corresponding to stresses cryand Ko cryis shown as circle 1 in Fig. 4.53c.

From the geometry of the circles 1and 2, we get

'm" = [{~(l+Ko)cr.Sin++ccos+r-H(l-Ko)cr..n...(4.56)

erV

172

""( maxcv

KoO=-v'tmax

Ko erv.~.'t,

I..

( a) (b) .-' ~

,U\III<:)I'-+'III

Koo-v ay E f f <zc t i VIZnor mal¬®» er(c)

Fig. 4.53 : Determination of 'tnl2x

Gmaxis the value of G applicable for very low strain amplitude, and therefore can be obtained using theappropriate equation from Eqs. (4.46), (4.48) to (4.54). Thus the value of reference shear strain Yrcan beevaluated.

The hyperbolic stress-strain curve 'which defines the initial loading curve and also the end point of thecomplete stress reversal loop is given by Eq. (4.57).

't = Y1 1-+-Gmax 'Ymax

~;

...(4.51),.

As't

G=- rCombining Eqs. (4.55), (4.57)and (4.58), we get

GG = ---1!!illL..

l+lYr

...(4.58)ft,

)ynamic SoU Properties 173

Using Eq. (4.59), one c~nobtainthe value of G at any strain amplitude, y. For every small strain ampli-udes, Y/Yr= 0; and the Eq. (4.59) reduces toG=Gmax'

"4.4. Estimation of Dampling Ratio. Hardin and Drnevich (1972) presented a relation betWeendamping'atio~andthemaximumvalueof dampingratio~maxasbelow:

~= ~max( 1- GG )max

Combining Eqs. (4.59) and (4.60), we get,

...(4.60)

l

I

Yr.!: =~max ---;;;-., 1+-

Yr

...(4.61)

Typical values of ~max are given in Table 4.9.

Table 4.9: Typical values of~max(Hardin & Drnevich), 1972)

, ,

Soil type Value of~max

Clean dry sandsClean saturated sands

33-1.5 (log n)

28-1.5 (logn)

Saturated Silt 26- 4 a~2 + 0.7/12 -1.5(logn)

Saturated cohesive soil -1/2 11231-(3+0.03/) ao + 1.51 -1.5(logn)

n - Number of cycles

a 0 - Mean effective principle stress, kg/cm2

1 -Frequency of loading, Hz

Forcohesionlesssoils, thevalueof ~maxis dependentonlyon thenumberof cyclesof loadingn whilefor silts and clays;the frequency of loading/(Hertz) and the mean eff~ctive principal stress (ao in kgf/cm2)influencethe maximum damping ratio.

ô þ¶¢":'~;~2;;:\æå¢¢áå¢æ·¢¥·º¬ùï·ô¶æ¢¢¢îæ¢¢åô¢æ¢¢¢ô¢¢¢¢¢¢·¢¢Û·Ð¢Öëæù'.':2:~~:'; ··¢ù æùæ ååùôæ£å·ø~~eIit


I~USTRATIVEEXAMPLES -Example4.1 ,

A soil specimen was tested in a resonantcolumn device (torsional vibration, Fixed -free condition) for deter-mination of shear modulus. Given a specimen length of90 mm, diameter 35 mm, mass of 160g, and a fre-quency at a normal mode of vibration (n = 1)of800 cps, determine the shear modulus of the specimen.Solution:

I. From Eq. (3.47)

1 (2n -1)1tvsCl) =--n 2 /

, .+t;, " n = 1.:>

2/ro 2.(0.090) .(27t.800) = 288m/sn -V =-- 1tS 7t

2. Mass density of Soil in the specimen

p = ðòïêð

ø

òðíë

÷

î ãïèìèòékg/m3

7t 2 .0.090

3. G = pv; = 1848.7.2882

= ïòëíí x 1O8N/m2

Example4.2A vertical vibration test was conducted on a 1.5 m x 0.75 m x 0.70 m high concrete block in an open pitha\'ing depth 2.0 which is equal to the anticipated depth of actual foundation. The test was repeated atdifferent settings (8) of eccentic masses.

The data obtained from the tests are given below:

The soil is sandy in nature having angle of internal friction, = 35° and saturated density Ysa/= 20 kN/m3.The water table lies at a depth of3.0 m below the ground surface. Probable size of the actual foundation4.0 x 3.0 x 3.5 m high. Determinethe valuesofC", E and G to be'adopted for the design of actualfoundation.Limiting vertical amplitude of the machine is 150microns.

,~b:.,'."

ÍòÒ± èº´¬Æ Amplitude at

(Deg) Resonance (Microns)

ïò íê ìïòð ïíòð

îò éî ìðòð îìòðò

íò ïðè íìòð íîòð

ìò ïìì íïòð ìðòð


× ¢ËÍÌÎßÌ×ÊÛ EXAMPLEsjExample 4.1A soil specimen was tested in a resonant columndevice (torsional vibration, Fixed -free condition) for deter-mination of shear modulus. Given a specimen length of90 mm, diameter 35 mm, mass of 160 g, and a fre-quency at a normal mode of vibration (n = 1) of800 cps, determine the shear modulus of the specimen.Solution:

I. From Eq. (3.47)

1 (2n -1)1tvs(J) =--n 2 /

" dt; '. n = 1.::>

2/ffin - 2.(0.O90).(21t.800) =288m1sv =-- 1ts 1t

2. Mass density of Soil in the specimen

p = 0.160

(.035

)

2 = 1848.7kg/m)

1t 2 .0.090

3. G = pv; = 1848.7.2882

= 1.533x 108N/m2

Example 4.2A vertical vibration test was conducted on a 1.5 m x 0.75 m x 0.70 m high concrete block in an open pithaving depth 2.0 which is equal to the anticipated depth of actual foundation. The test was repeated atdifferent settings (e) of eccentic masses.

The data obtained from the tests are given below:

The soil is sandy in nature having angle of internal friction ~=35° and saturated density Ysa/=20 kN/m). The water table lies at a depth of3.0 m below the ground surface. Probable size of the actual foundation4.0 x 3.0 x 3.5 m high. Determinethe valuesofCII' E and G to be'adopted for the design of actualfoundation.Limiting vertical amplitude of the machine is 150 microns.

'"

,\,}":~~>'it.iji-,~~.;,'f,Ij.

SoNo e /nz Amplitude at(Deg) Resonance (Microns)

1. 36 41.0 13.02. 72 40.0 24.0.3. 108 34.0 32.04. 144 31.0 40.0

"m:t.~

mic SoU Properties 175

tion:

Mass of oscilator and motor

Mass of block, oscillator and motor

2= 1.5 x 0.75 = 1.125 m

= (1.125x 0.70) x 2400= 1890 kg

= 100 kg (assumed)= 1890+ 100

= 1990 Kg.2 2

, 41t fnzmC =-U A

1. Area of block

Mass of block

2.

4 2 2

C = 1t fnz' 1990U 1.125.1000 =71.1 fn~ kN/m3

The calculated values of Cu for different observed resonant frequencies are listed in column No. 5 of',e4.10.

l.13E 1

Cu = (I-Jl2)'.fAAssuming Jl =0.35

F = .JiJ2s(1-0.352) Cu=0.8236 Cu kN/m21.13E 0.8236 C 2G = , = u =0.3050C kN/m

2 (1 + Jl) 2(1+0.35) ,u

For different values of Cu (col. No. 5), calculated values ofE and G are listed in Cols. 6 and 7 of Table 4.10)ectively. '

3. Correction for confining pressure and area

The mean effective confining pressure aO!at depth of' one -half the width below the centre of block isen by

crO! = cry (1 + 2 Ko)3where cry = cr~l+ crv2

,cry! =, Effective overburden pressure at the depth under consideration

crv2 = Increase in vertical pressure"~u~t() the weight of block ",

Assuming that the top 2.0 m soil has a moist unit weight of 18 kN/m3, and the nex~1.0 m soil i.e. uptoter table is satUratedthen ."c -'

.,. - i . ' ' 0 70 . . '

;;; .", " , :~vJ; :::;:..J.~~~8iZ+aO.x ~~ =~3 ~~m2\,) ~, ,: : '::", ,,;

"-;,' ,T~"'I'J,r;~'. I " ;: ) : ',f'

;~j, ') " ':.

" ~,..'

.'

176 Soil Dynamics & Machine Foundati'ms'

From Taylor (1948):

- 4q

[

2mn~m2+n2+1 m2+n2+2 . -12mn~m2+n2+1

]a - - . +sm

v2 - 41t m2 +n2 + l+m2n2 m2 +n2 + 1 m2 +n2 + l+m2n2

L12 1.5/2³ =2=0.70/2=2.14

B/2 0.75/2n = 2 = 0.70/2 =1.07

2q = 24 x 0.70= 16.8 kN/m

[Assuming unit weight of concrete =24 kN/m3]

Substituting the above values of rn, nand q in the expression of av2' we get- 2crY2= 13.44kN/m

cry = 43 + 13.44 = 56.44 kN/m2

Ko = 1- sin <jI= 1 - sin 35° = 0.426

- (1+2 x 0.426

)2

0'01 = 56.44 , 3 =34.84kN/m

For the actual foundation

aYI = 18x2.0+20x 1.0+(20-10)xO.5=61kN/m24.0/2

m = 3.012 = 1.3343.Q/2

n=3.0/2=1.02q = 24x3.5=84kN/m

Substituting the above values of m, nand q in the expression of crY2'we get

crY2 = 63.76 kN/m2-

kN 2av = 61+63.76= 124.76 Im

cr02 = 124.76 (1 + 2'x] 0.426) =77.01 kN/m2 .

Area of actual foundation = 4.0 x 3.0= 12.0m2 (> 10m2).

(

-)

05

( )

05' 05 05'Cu2 = E2 = G2 = 0'02 . ~ =

(77.0

) (1.125

)' = 0.4986

Cui El Gl 0'01 A2 34.84 10

The values ofCu' E and G of the actual foundation at different strain levels (= amplitude at resonancelwidth of test block) are given in cols. 8,9 and 10of Table 4.10 respectively. The corresponding values ofsn:ainlevels are listed in col. 11.

- ~'..""

.

'~.~iitfIIJ

~ynamic SoU Properties 177

4. Strain level Correction

. . 150 x 10-6 -4Strain in Actual foundation = 3.0 = 0.5 x 10

Thevaluesorcu' E and G correspondingtostrainlevelof 0.5 x 10-4canbeobtainedby interpolation.

=(4.10-(4.10-3.40) 0.50-0.427 )x 4

Cu 0.533 - 0.427, 10= 4.10-0.7 x 0.6886=3.62 x 104kN/m3

E = [3.37 - (3.37 - 2.80) x 0.6886) x 104 = 2.98 x 104 kN/m2

G = [1.25-(1.25-1.04) x 0.6886] x 104= 1.10 x 104kN/nl

Hence the response of the proposed foundation block should be checked using

Cu = 3.62 x 104 k1~/m3E = 2.98 x 104kN/m2

G = 1.10 x 104kN/m2

I.

2.

3.

4.

S.No.

( I.)

continued

Table 4.10 : Analysis of Data for Cu' E and G

For test blockAmplitude

e In;; at resonance Cu E G

(Deg.) (microns) x 104 kN/m2 x 104 kN/m2 x 104 kN/m2

(2) (3) (4) (5) (6) (7)

36 41 13 11.95 9.84 3.64

72 40 24 11.38 9.84 3.47

108 34 32 8.22 6.77 2.51

144 31 40 6.83 5.62- 2.08

For Actual foundation

Cu E G Strain level

x 104 kN/m2 x 104 kN/m2 x 104 kN/m2 x 10-4

(8) (9) (10) (11)

5.96 4.90 1.81 0.173

5.67 4.67 1.73 0.320

4.10 3.37 1.25 0.427

3.40 2.80 1.04 0.533

Iynamic SoU Properties 177

4. Strain level Correction

. . 150 x 10-6 -4Strain in Actual foundation = 3.0 = 0.5 x 10

ThevaluesofCu' E and G correspondingtostrainlevelof 0.5 x 10-4canbe obtainedby interpolation.

=(4.10- (4.10-3.40 )

0.50-0.427 )x 4Cu 0.533 - 0.427. 10

= 4.10-0.7 x 0.6886=3.62 x 104kN/m3

E = [3.37-(3.37 -2.80) x 0.6886) x 104=2.98 x 1041cN/m2

G = [1.25-(1.25-1.04) x 0.6886] x 104= 1.10 x 104kN/m2

Hence the response of the proposed foundation block should be checked usingC = 3.62x 104kN/m3u

E = 2.98 x 104kN/m2

G = 1.10 x 104 kN/m2

\.2.3.4.

S.No.

( \.)

continued

!'

Table 4.10: Analysis of Data for Cu' E and G

For test blockAmplitude

e In;; at resonance Cu E G

(Deg.) (microns) x 104 kN/m2 x 104 kN/m2 x 104 kN/m2

(2) (3) (4) (5) (6) (7)

36 41 13 1\.95 9.84 3.64

72 40 24 11.38 9.84 3.47

108 34 32 8.22 6.77 2.51

144 31 40 6.83 5.62 - 2.08

For Actual foundation

Cu E G Strain level

x 104 kN/m2 x 104 kN/m2 x 104 kN/m2 x 10-4

(8) (9) (10) (11)

5.96 4.90 1.81 0.173

5.67 4.67 1.73 0.320

4.10 3.37 1.25 0.427

3.40 2.80 1.04 0.533

:~~tt~r~~tii\Wi': ~~;t~~;,i:j~;~~'f~;';~;1c~h~ "~'K(",oiJ(ii~~7"I,.!;,i;',(..t,j; IIf;..~~'

"iizinic Soil Properties'. ,

179

, , 8~2xO.6194f. nx2

c =t (0.5546+0.9926) x IQ-3 I~(0.5546+0.9926)2 x 10-6 -4 x 0.6194 x 0.5546 x 0.9926 x 10-6

248.856 ht'( 3 3

= (1.5283):t(0.9989) x 10 N/mTherefore, , <

Ct = 92.3/;'\ kN/m3(First mode)

Ct = 19.3 1;'2 kN/m3(Second mode)

~

(b) Ct=92.3 In~1

The Calculated values of Ct for different observed resonant frequencies are listed in col. 5 ofTable4.11.As in example4.2.

C'--1f. = 0.4986Ct!

The values ofCt for the actual foundation are given incol. 6. The corresponding strain levels a re listedin coL7.

'r.

(I)

1

2

3

4

S.No.

-6 '

S.. '

I ~ d . 100 x 100 333 10-4tramm actua loun atlOn= . " =. x

, 3.0Therefore, the value of Ct for actual foundation

[0.373-0.333

]3

= 2.03-(2.03- L66) x 0.373-0.280 x 104= 1.87 x 104kN/m

Example 4.4The soil profile at a site is shown in Fig. 4.54 . Two cross borehole tests were conducted at the site to deter-mine the values of shear wave velocities in the small areas around points A and B. The average values ofshear wave velocities were obs'ervedas 11Omlsand130 m/srespecnvely. Determine the values of dynamicshear modulus G for points A, B, C and D. '

Table 4.11 : Analysis of Data for C't

"e fn.d Amplitude 'Ct Ct Strain level

(Deg.) (Hz). " (microns), x 104kN/m2 x 104 kN/m2 x 10-4

(2) (3) (4) (5) (6) (7) "

36 25 10 5.77 2.87 0.13372 23 16 4.88 2.43 0.213108 21 21 '4.07 2.03 0.280144 19 28 333 1.66 0373

180

0.0

- 4.0 m

- 10.0m

Ae

c .


f2.0m

+1.Qm--L

if = 18 kN 1m3

Fine grained soil

,1.0m .

~Lf-- --------1.0m ~ = 21 kN/m3t 862.0m Satu re ted san d

-1 .0

Fig.4.54: Soil profiles (example 4.4)

Solution:

1. Fine grained soil stratum (0.0 to - 4.0 m)

Observed shear wave velocity at point_Av = 1l0rn/s

S .

G = P v;

18G= 9.81 X(110)2=2.2xl04kN/m2

G1'A= 18x 2 = 36kN/m2

ave = 18x 3 = 54kN/m2

(

-

)

05(G) G 54 05~ = g:~ = (36) . = 1.~247

(G)c ~ i.i~ 104x1.2247=2.7X 1.04kN/m2"if.

'~:;~ViY;i'"t\tr"~;\:cf~~7W;j;:'~}j;",' '9/~f.{kt;~:~-it\i~:~\~h;: .)l~t]~,;:;:..;)~ft~i~~\;f~"";- r(~1-+1~it~i;.~t~f.'~~f~~'f"!!t&[~\~; "" ,j:Jr

vnamic Soil Properties 181

.2. Saturated sand (..,,4.0~to ~ to.OOm) . " . ,.

(vS>B= 130mls

(G~ = 9~~lx (130)2 =.3.6 x 104kN/m2

crvB= 18x4:0~21 x 1.0+(21 -10) x 1.0 ~104kN/m2- kN 2crvD = 104+ (21-10) x 2.0= 126 /m

, ,

(26

)0,5 .

(G>O = J.6x 104 x 104 =2.96 x 104kN/m2

'xample 4.5 ,

.t a particular site, the top 10.0m soil ismedium grained sandhaving dryunit weight as 17kN/m3.The waterlble is 6.0 m below the ground surface. The value of specific gravity of soil grains is 2.67. The direct shear:st gave the value of as 36°. Determine the value of shear modulus of the soil at depth of 7.0 m belowround surface.,olution:

1. Yd= 17= 2.67 x 10l+ee = 0.57

(2.67 + 0.57) x 10 = 20.6 kN/m3Ysat= 1+0.57 '

(O'vhOm = 17x 6.0+(20.6-10) x 1.0. .

2= 112.6kN/m '

Ko = I-sin<l>=I-sin36°=0.412

- -(

1+2Ko)

-' _(

1+2XO.412

)x '

0'0 - 3 ,crv ~ 3 112.62= 68.46 kN/m

2. From Eq. (4.34)

26908 (2.17 - e) (

-)0.5

G = . 0'0l+e

2 '

G~' 6908(2.17-0.57) .(68.46)°.51+ 0.57

.' 4 2= 9.3 x10 teN/m

, >':: .. ,:' ,t.

;182 SoU Dynamics & Machine Foundations


Anderson, D. G. (1974), "Dynamic modulus of cohesive soils", Ph. D Dissertation, University of Michigan, Ann Ar-bor.

Barkan, D. D. (1962), "Dynamics of bases and foundations." McGraw-HilI, New York.

Bjerrum, L, and Landra, A.'(1966), '-'Direct simple shear tests on a Norwegian quick clay", Geotechnique 26(1), pp. "-1-20.

Casagrande, A. and shannan, W.L. (1948a), "Stress deformation and strength characteristics of soils under dynamicloads", Proc. Second Int Conf. Soil Mech. Foundation Engg., Vol. 5, pp. 29-34.

Casagrande, A. and Shanan, W.L (1948b), "Research on stress deformation and strength characteristics of soils and soft.rocks under transient loading," Harvard Soil Mechanics Series No. 31.

Cho, Y.,Rizzo, P. C, and Humphries, W. K. (1976), "Saturated sand and cyclic dynamic tests", Am. soc. Civ. Eng.,. Ann. Cony. Expo., Philadelphia, rA, Meet. Prepr. 2752, pp. 285-312.

Dass, B.( 1977), "Behavior of Clayunder oscillatory loading", M.E. Thesis, U. O. R, Roorkee.Drnevich, V. P. (1967), "Effect of strain history on the dynamic properties of sand", Ph.D. Dissertation, university of

Michigan, Ann. Arbor.

Drnevich, V. P. (1972), "Undrained cyclic shear of saturated sand", J. Soil. mech. Found. Div., Am., Soc. Civ. Eng., 98(SM-8), pp. 807-825.

Goto, N., Kagami, H., Shiono, K. & Ohta, Y. (1977), "An easy-capable and high precise shear wave measurement bymeans of the standard penetration test". Proceedings sixth world conference on earthquake engineering,pp. 171-176.

Hall, 1. R., Jr., and Richart, F. E., Jr. (1963), "Dissipation of elastic wave energy in granular soils", J. Soil Mech. Found.Div., Am. Soc. Civ. Engg., 89 (SM-6), pp. 27-56.

Hardin, B.C (1971), "Program of simple shear testing of soils", Univ. Ky., Soil Mech., Ser. No. 8, pp. 1-14.

Hardin, B. O. & Black, W.L., (1968), "Vibration modulus of normally consolidated clay", Journal of the soil Mechanicsand Foundations Division, ASCE, 94 (SM2).

Hardin, B. O. & Richart, F. E. Jr., (1963), "Elastic wave velocities in granular soils", Journal of the Soil Mechanics andFoundations division, ASCE 89 (SM 1), pp. 33-65.

Hardin, B. O. & Music, 1. (1965), "Apparatus for vibration of soil specimens during the triaxial test." Instruments andapparatus for soil and rock mechanics, ASTM, STP 392, pp. 55-74.

Hoadley, P. J. (1985), "Measurement of dynamic soil properties", Chapter of a book on Analysis and Design offoun-dations for vibration, pp. 349-420.

Hvorslev, M. 1., and Kaufman, R. 1(1952), "Torsion shear apparatus and testing procedure", USAE Waterways expoStn., Bull 38, pp. 1-76.

Iida, K. (1938), "The velocity of elastic waves in sand," Bull. Earthquake Res. Inst., Tokyo Imp. Univ., 16, pp. 131-144.

lida, K. (1940), "On the elastic properties of soil particularly in relation to its water content", Bull. Earthquake Res.Inst., Tokyo imp. Univ., pp. 18,675-690.

Imai, T. (1977), "Velocity of P-and S-waves in subsurface layers of ground in Japan", Proc. 9th Int. Conf. Sol Mech...Found., Tokyo, Vol. 2, pp. 257-260. . . .

IS : 5249 (1978), "Deterrrtination of dynamic propertIes' of soil".

~ ,f"

~"a",ic Soil Properties 183

: 1888 (1982), "Method of load test".

: 213I (1981), "Method for standard penetration test for soils"hibashi, I., and Sherif, M.A. (1974), "Soil liquefactionby torsional simple shear device", J. Geotech. Eng. Div., Am,

soc. civ. Eng., 100(GT-8), pp. 871-888. ,

hihara, K. (1971), "Factors affecting dynamic properties of soil", Proc. Asian Reg. Conf. Soil Mech. Found. Eng.. 4th,Bangkok, Vo!. 2.

hihara, K., and Li, S. (1972), "Liquefaction of saturated sand in triaxial torsion shear test": Soils Found (Jpn.) 12 (2),pp. 19-39.

hihara, K., and Yasuda, S. (1975), "Sand liquefaction in hollow cylinder torsion under irregular excitation", SoilsFound. (Jpn.), 15(1), pp. 45-59.

himoto, M., and lida, K. (1937), "Determination of elastic constants of soils by means of vibration methods", Bull.Earthquake Res. Inst., Tokyo Imp. Univ., 15, p. 67.

vasaki, T., Tatsuoka, F., and Takagi, Y. (1977), "Shear moduli of sands under cyclic torsional shear loading," Tech.Memo. No. 1264, Public Works Res. Inst., Ministry of Construction, Chiba-Shi, Japan.

jellman, W. (1951), Testing of shear strength in Sweden, "Geotechnique, 2, pp. 225-232.- , .awrence, F. V., Jr., (1963), "Propagation velocity of ultrasonic waves through sa.n.~",MIT Research Report, R63-8.ord, A. F., Jr., Curran, 1.W., and Koemer, R.M. (1976), "New transducer system for determining dynamic mechanical

properties and attenuation in soil", 1. Acoust. Soc. Am. 60 (2), pp. 517-520.

filler, R. P., Troncosco J. H. & Brown, F. R. (1975), "In situ impulse test for dynamic shear modulus of soils", Pro-ceedings of the conference on in situ measurement of soil properties, Geotechnical Engineering Division.ASCE. Specialty conference, Raleigh, North Carolina, Vo!. 1, pp. 319-335. .

~acock, W. H., and Seed, H. B. (1968), "Sand liquefaction under cyclic loading simple shear conditions", 1. Soil Mech.Found. Div., Am. Soc. Civ. Eng. 94 (SM-3), pp. 689-708.

rakash, S., and Puri, V. K. (1981), "Dynamic properties of soils from in situ tests", 1. Geotech. Eng. Div.. Am. Soc.Civ. Eng. 107 (GT-7), pp. 943-963. .

rakash, S., Nandkumaran, P., and Joshi, V. H. (1973), "Design and performance of an oscillatory shear box", 1. IndianGeotech. Soc., 3 (2), pp. 101-112.

rakash, S., Nandkumaran, P., and Joshi, V. H. (1974). "Behaviour of soils under oscillatory shear stress", Proc. ythSymposium on Earthquake Eng., Roorkee, V01. 1, pp. 101-12.

ichart, F. E., Jr., Hall, 1. R., and Woods, R. D. (1970), "Vibrations of soils and foundations." Prentice-Hall, EnglewoodCliffs, New Jersey.

oscoe, K. H. (1953), "An apparatus for the application of simple shear to soil samples," Proc. Int. Conf. Soil Mech.Found. Eng., 3rd, Zurich, vot. I, pp. 186-191.

eed, H. B., Idriss, I. M., and Arango, L. (1983), "Evaluation of liquefaction potential using field performance data",J. Geotech. Eng. Div., Am. Soc. Civ. Eng., 109 (GT-3), pp. 458-482.

eed, (1960), "Soil strength during earthquakes", Proc. Second World conf. Earthquake Engg., vo!. I, pp. 183-194.

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~ed, H. B. and Fead, J. W. N. (1959): "Apparatus forrepeated load tests on soils," SpecialTechanical Publication No.20~, ASTM, Philadelphia,

5ilver, M. L., Chan, C. K. et a!. (1976), "C~c1ic triaxiatstrengtllofstandard testsand", 1. Geot. Engg., Div. ASCE,vol., 102, no. aT 5, pp. 511-523. - : ",..\ " ." /'. " '. ",' .. /

~

184 Soil Dynamics & Machine Foundations .

Stephenson, R. W., (1978), "Ultrasonic testing for determining dynamic soH modulii", Denver: Dynamic. Geotechnical '

testing, ASTM, STP 654., pp. 179-195.

Stokoe, K. H., 1I, and Hoar, R. J. (1978), "Variables affecting in situ seismic measurements", Proc. Am. Soc. Civ. Eng.Spec. Conf. Earthquake Eng. Soil Dyn., Pasadena, CA, vot. 2.

Stokoe, K. H., and Woods, R. D. (1972), "In situ shear wave velocity by cross-hole method", J. Soil Mech. Found. Diy.,Am. Soc. Civ. Eng., 98 (SM-5), pp. 443-460..

Taylor, D. W. (1948): "Fundamentals of soil mechanics:' John Willey sons, Inc., New York.

Terzaghi. K. and Peck, R. B. (1967), "Soil mechanics in engineering practice", John Willey and Sons, New York./<

Wilson, S.D., and Dietrich, R. J. (1960), "Effect of consolidation pressure on elastic and strength properties of clay",Proc. am. Soc. Civ. Eng. Res. Conf. Shear Strength Cohesive Soils, Boulder,CO, pp. 419-435.

Woods, R.D. (1978), "Measurement of dynamic soil properties: State-cf-the-Art", Proc. Am. Soc. Civ. Eng. Spec.Conf. Earthquake Eng. Soil Dyn., Pasadena, CA, Vot. I, pp. 91-180.

Yoshimi, Y., and Oh-Oka, H. (1973), "A ring torsion apparatus for simple shear tests", Proc. Int. Conf. Soil Mech.Found. Eng., Moscow, Vol. I, Pt. 2, pp. 501-506.

r:>

"

-".,-

>ynamic Soil Properties 185

PRACTICE PROBLEMS- I

4.1 Describe the salient features. of a resonant column apparatus. 'How is calibration done and theI ., value of shear modulus determined?

4.2 A clayey soil specimen was tested in a resonant column device (torsional vibration, free-free endcondition) for determination of shear modulus. Given a specimen length of 100 mm, diameter 36mm, mass of 180 g, and a frequency at normal mode of vibration of 900 cps, determine the shearmodulus.

4.3 Explain the difference between simple shear and direct shear tests. What is the principle involvedin oscillatory shear box test? Give the salient features of a study made on clay under dynamic loadsusing oscillatory shear box.

4.4 List the factors affecting shear strength of cohesive soils under static and dynamic loads. Explainwith neat sketches the effect of dynamic stress level, initial factor of safety and number of pulses.

4.5 Draw typical transient strength (single impulse) characteristics of sandandclaytestedfor the fol-lowing time of loading: ' ,0.' . .

(i) 0.02 s (ii) 0008sand (iii) Static

4.6 Describe briefly the following:(a) Seismic cross-borehole survey(b) Seismic up-hole survey(c) Seismic down-hole survey

4.7 A vertical vibration test' was conducted on a MI5 concrete block 1.5 x 0.75 x 0.70m high usingdifferent eccentricities (8) of the rotating mass of the oscillator. The data obtained are given asfollows:

Determine the value of co~fficient of elastic uniform compression, Cu for confining pressure of100 kN/m2 and base contact area ~f 10 nl.. Also give the values of strain levels at different eccen-tricities. '. .'

, "

4.8 A horizontal vibration test was conducted on a M 15 concrete block 1.5 x 0.75 x 0.7 m high. Thedata obtaint:;d is shown in Fig. 4.5.5, Determine the value of coefficient of elastic uniform shear, C

"'"'~"" ' C'., r.and the"cbrres "ortding$ -, 'n leyL~.:" ' . .

';K~~il;C " ,. ,., ye'" ,

",. " '

SoNo. 8 /nz - Amplitude at resonance(Deg) (Hz) (microns)

. ",. ,

1. 36 41 13

2, 72 40 24

30 108 34 32

4, 144 31 " 40c

'--':-<"""il""s;";,;;~';<'N.';:3I'.I"'.:'.J,i".':;,,; '!'{,'"",;,:';';.:t"""':;""f,-,. ';,-,,'c:i,'c,"',,"""",' C"',>,',, '" ,.,>"z,>;,h,,';~~

186 SoU Dynamics & Machine FoundlJtions

0.32

0.24

e = 1050

0.28

,..."E 0.20E.........~

"'C

-=0.16

a.E«

0 .12

2

0.08

0.04

010 15 20 25 30 35

Frequency (cps)

Fig. 4.55 : Amplitude versus frequency plot obtained from a horizontal vibration test

4.9 A cyclic plate load test was performed on a plate of 600 mm x 600 mm size. The elastic settlementcorresponding to a loading intensity of80 kN/m2was 2 mm. Using this data, determine the coeffi-cient of elastic uniform compression, Cufor a foundation block of base area 15m2.

4.10 Discuss the factors affecting shear modulus and damping. Illustrate the procedure of obtainingthe shear modulus and damping at given strain amplitude from Gmaxand ~ma)('

,: DD

"

';1;\< H

,"-' ..

78 Soil Dynamics & Mat;hineFoundations

-:xample 4.3

(a) Determine the expressions of coefficient of elastic uniform shear, C'!:in temis of resonant frequencyfor the block of size 1.5 m x 0.75 x,0.70 m high, tested wider horizontal vibrations.

(b) Determine the value ofCt for the foundation mentioned in example 4.2, if the block tested in hori-zontal vibrations give the following results: .. " ,

Pelmissible value of horizontal amplitude is 100 microns.Solution:

(a) Value ofCt is given by Eq. 4.32 2

8n r f roe

Ct = ~ 2(Ao+lo):t (Ao+Io) -4rAoIoMoment of inertia of the base contact area of block about axis of rotation

I = 0.75 x 1.53 =0.2109m412

Height of combined centre of gravity of block, oscillator and motor from base.

1890 x 0.35 + 100 x 0.85L = 19.9 =0.375m

The e.g. of oscillator and motor is assumed at a height of 0.15 m from the top of the block. Mass momentof mertia Mm about an axis passing through the combined centre of gravity and perpendicular to the planeof vibrations is given by .

(1.52 + 0.702

JMm = 1890 . + 1890(0.375-0.35)2+ 100(0.85-0.375)212 " - , ' ,

= 431.55 + 1.184-22.56=455.29 Kg-m2M = M +mL2= '455.29+ 1990x (0.375)2 = 735.13 Kg-m2.

/nO In. '

~ Mill ==455.29 =0 6194r 735.13'MI1/0,

A 1.125 , . -3 2A = -=-=0 5546x 10 m !kg0 m 1990 .

(

I.

) (0.2109

)-3 2

10 = 3.46 M11/0 = 3.46 735.13 = 0.9926x 10 m /kg

S.No. e (Deg.) !,,\(Hz) Amplitude atResonance (Microns)

1. 36 35 102, 72 34 16. .

3. 108 30 214. 144 T7 28

-,,- ':';:"-.;",.!"".;;,i,~;,'.~J",1:}",.~.%/I';<,""<7'%"4"..:"";,,,",-.-;.'-' ~::;,:F',""" ", ",..':"h"';;;""+~;" 'i-';', ':',/;::;;i,'L -- , -- - ..,,"'l--. ~ ':"~

+ (f»)

DYNAMIC EARTH PRES-5.1 GENERALIn the seismic zones, the retaining walls are subjected to dynamic earth pressure, the magnitlis more than the static earth pressure due to ground motion. Since a dynamic load is repetitithere is a need to determine the displacement of the wall due to earthquakes and their dama,:This becomes more essential if the frequency of the dynamic load is likely to be close tefrequency of the wall-backfill-foundation-base soil system. This essentially,consists in writiequation of motion of the system under free and forced vibrations. This requires the informdistribution of backfill soil mass and base soil mass participating in vibrations. It is ofterassess these. Therefore, more often, pseudo-static analysis is caf!ied out for getting dynamisure. In this method, the dynamic force is replaced by an equivalent static forcl

In this chapter, firstly various methods of computing the magnitude and point of at:dynamic earth pressure based on pseudo-static analysis have been discussed. It is followmethods of predicting the displacement of the retaining wan:

5.2 PSEUDO-STATIC METHODS

5.2.1. Mononobe-Okabe Theory. Mononobe-Okabe (1929) modified classical Coulomb'evaluating dynamic earth pressure by incorporating the effect of inertia force.

Ct

Wl

::jj'o1h ~.9'Wl '

Wt(t :totv1

ol.hton~=1:toLv

B

(0)

Fig. 5.1 : (a) Forces acting on raU!-'rewedge In active state (b) Force Polygon(c) Dynamic earth pressure versus wedge angle 9 plot

188 SoU Dynamics & Machine Foundations 1\.:

Figure 5.1 shows a wall of height H and inclined vertically at an angle a retaining cohesionless soilwith unit weight yand angle of shearing resistance cjI.BCI is the trial failure plane which is inclined tovertical by an angle 8. The backfill is inclined and making an angle i with horizontal.

During an earthquake the inertia force may act on the assumed failure wedge ABCI both horizontally'and vertically. If ah and av are the horizontal and vertical accelerations caused by the earthquake on thewedge ABCp the corresponding inertial forces are WI . Gig horizontally and WI . ajg vertically, WIbeing the weight of the wedge ABCl' During the wor~t condition, WI . ah/g acts towards the fill andW, . aig may act vertically either in the' downward or upward direction. Therefore the direction thatgives the maximum increase in earth pressure is adopted in practice.

If a"and ai, are respectively the horizontal and vertical seismic coefficients, then, .<~

ahah = g ...(5.i)'

ai,a=-v g

For the failure condition the soil wedge ABC, is in equilibrium under the following forces:

(i) W" weight of the wedge a~ting at centre of gravity of ABCI.

(ii) Earth pressure P, inclined at an angle 0 to normal to the wall in the anticlockwise direction.

(iii) Soil reaction R, inclined at an angle cl»to the normal 011the face BC"

(iv) Horizontal inertia force (W, a/,) acting at the centre of gravity of the wedge ABC,.

(v) Vertical inertia force :f: WI av acting at the centre of gravIty of the wedge ABC,.

Weight WI and the inertia forces W I all and :f:WI av can be combined to give a resultant WI' where

...(5.2) :.

-[

2 2]

1/2W, = W, (l:t:av) +a" ...(5.3)

The resultant W, is inclined with vertical at angle '1', such that .;;'~

',' --

-,(

ah)'" = tan l:t:av

The directions of all the three forces WI' P I and R, are known but the magnitude of only one force

W, is known. The magnitude of the other forces can be obtained by considering the force polygon as

shown in Fig. 5,1 b. P, is the value of dynamic earth pressure corresponding to the trial wedge ABCl'~More trials are made and the values ofP2,PJ etc, are obtained. Variation ofP and 8 is shown in Fig, 5.1c..The maximum value of P is the dynamic active earth pressure (PA)dyn' :,

Mononobe and Okabe (1929) gave the following relation for the comp~tation of dynamic active earthpressure [(PA)dyn] : . -.. '

0' ..-"- ;""""-1

, '" ::~5.~r--

1 2(PA)dyn,="2 yH (KA)dyn

where (KA)dynis coefficient of dynamic active earth pre~~ure and given by :

...(5.5)

, ' -,; "1:;,.. "tf i;"U "

: ~.., ;:./., ';;!C"i":~;~li,;..,:~//-:.-. - ,...~, ;!'> ~,

.or,

' "' "I,~,5>,' ".,.",:r..~,,$

,.,' ,

",

'"

",

J,.,F'" ,-.;~"" ,~{.~",..~:~..<;~'..".c',":,,<'.' 1""',, .. '.., .:'!'-" ",'",. ,"""""" , ",""" ,

'"~

" ... " ...' ,," """"'-", .--

.,.

;, "'. '.J

IIIiI:S

Dynamic. Earth .Pressure 189

,2

(l:tay) cos2(~-'V -al . 1(K) = 2- x

{

.( A. s::)

."(A.' )

}

1/2A dYII COS'V COS a COS(cS+ a + \V) 1+ srn 'f + ~ srn 'f -l - \V

cos (a - I) cos (cS+ a + \V)

The expression of (KA)dyngives two values ~dependingon the sign of ay. For design purposes thehigher of the two values shall be taken.

Mononobe and Okabe also gave the expression for the computation of dynamic passive earth pres-sure (PP)dynwhich is

...(5.6)

1 2(PP)dyn = -2 rH (KP)dyn

where (KP)dyn is coefficient of dynamic passive earth pressure and given by :

...(5.?)

2

2(1:t ay) cos (~+ a - \V)

I

1(Kp)d n = 2 x 112

y cos \jI cos a cos (cS- a + \V) 1-{

srn ( ~ + cS)sm ( ~ + l - \V)}cos (a - i) cos (cS- a + \jI

For design t'urposes, the lesser value of (Kp)dYIIwill be taken out of its two values corresponding to:!:ay.

...(5.8)

5.2.2. Effect of Uniform Surcharge. The additional active and passive dynamic earth pressures [(PAq)dynand (PPq)dyn]against the wall due to uniform surcharge of intensity q per unit area on the inclined earthfill surface shall be :

q H cos a(P Aq)dyn = cos (a - i) (KA)dyn ...(5.9)

q H cos a(PPq)dyn= cos(a - i) (KP)dyn ...(5.lO)

5.2.3. Effect of Saturation on Lateral Dynamic Earth Pressure. For saturated earth fill, th~ saturatedunit weight of soil shall be adopted..

For submerged earth fill, the dynamic active and passive earth pressures during earthquakes shall befound with the following modifications (IS:1893-19.84).

(i) The value of 8 shall be taken as 1/2 the value of the 8 for dry backfill.(ii) The value of 'V shall be taken as

. ' -Ir

- 'Ys

)ah

'V = tan 'Ys - 1 1:t ay- -

. where 'Ys= Saturated unit weight of the soil.(Ui) Submerged unit weight shalJ be adopted.

Hydrodynamic"pressure on account of water contained in earth fill shall not be considered separatelyas the effect of acceleration on water has been taken indirectly. .

...(S.II)

. ~,~~


5.2.4. Partially Submerged backfill. The ratio of lateral dynamic increment in active pressures to thevertical pressures at various depths along the height of wall may be taken as shown in Fig. 5.2 Thepressure distribution of dynamic increment in active pressures may be obtained by multiplying the ver-tical effective pressures by the coefficients in Fig. 5.2 at corresp.ondingdepths'

H

hw

r- 3 [(KA)dyn - KA]--\ 7

-- I/

//

//

//

//

/ V

z

1

IZ

1-/

//

//

// '

/C

I I

]hw

3 [(KA)dyn - KA H

I IKA and (KA)dyn a re th Clvalues of KA and (XA)dynin submerged condition

lateral dynamic IncrementFig. 5.2: Distribution of the ratio' ve!1ical effective ~ressure with height o(wall

, ,

The value of lateral dynamic increment in active ca~e'can be obtaineq by.integrating it in the portiopabove water level and below water level separately. By doing this we get (Refer Fig. 5.2) ':

(P) = r(H-hw)3[(K ) -K' ] .H-Z .z dz +I'h~.3[(KA)Dyn-KA]hw.hw-z'.AIDyn Ö± Adyn;, A ' H !, , 0)(: 'H' ..' h'

, , W

[1 (H - hW> + 1h . z1 dz' " .

., ,-'",: ,

" , '., ",:,'

.. ~. :.' -.~ f:f'::>:/:"'~';

Dynllmic Ellrth Pressure 191

1 (H-h )2 1 h2= -[(KA)d yn-KAT W.y (H +2hw) + -[(K~Jd yn-KA1'~ [3 Y (H - hw)+ Ybhw]2 H 2 H

...(5.12)where

(PAI)Dyn= Lateral dynamic incrementin active casehw = Height of water level above base of wall

KA = Coefficientof static activeearthpressure in dry/moist/saturatedcondition(KA)dyn= Coefficientof dynamicactiveearth pressure in dry/moist/saturatedCondition.

KA = Coefficient of static active earth pressure in submerged condition

(KA )dyn = Coefficient of dynamic active earth pressure in submerged condition

Y = Density of soil above water level

Yb = Submerged density of soilValues of active earth pressure coefficients shall be obtained using Eq. (5.6) as described below.

KA - By putting aft = av = 'If= 0 in Eq. (5.6) . -

(KA)dyn- Eq. (5.6)

KA - By putting ah = ,fJ.1'~ ~ = 0 and 8 as ~ in Eq. (5.6)

(KA)dyn - By putting 8 and ~, and 'Ifgiven by.Eq. 5.11 in Eq. (5.6)

The additional dynamic increment due to the uniform surcharge of intensity q per unit area on theinclined earth fill shall be:

J(H"':hw) . cosa H-z d= 3 K -K]. '-q z

(PAqI)dyn 0 [( A)dyn A cos (a-i) H

Jftw3[(KA)dyn-KA]hw. cos a . hw-z'. qdz'+ 0 H cos(a:-i) cos (a-I)

[

2 2 2

]

3q cos a H - hw , , hw= cos (a-i) {(KA)dyn-KA}' 2H +{(KA)dyn-KA}'2H ...(5.B)

A similar procedure as described above may be utilized for determining the dynamic decrement inpaSSIvepressures.

5.2.5. Modified Culmann Construction. Kapila (1962) modified the Culmann's graphical for obtainingdynamic active and passive earth pressures.

. ,.

A Cl Cz C3

t.i.

I:'~,'it~i"1I~

1!~1!

192, '

Soil Dynamics & Machine FollndtitiollS

1

ModifiedCulmann's line

sH

,~.

~

Fig. 5.3 : Modified Culmann's construction for dyna~ic active earth pressure

Different steps in modified construction'for determining dynamic active earth pressure are as follows(Fig. 5.3)

(i) Draw the wall section along with backfill surface on a suitable scale,

(ii) Draw BS at an angle «1>- W) with the horizontal.(iii) Draw BL at an angle of (90 - <X- 8 - W)below BS,

(iv) Intercept BDI equal to the resultant of the weight WI of first trial we,dgeABCl and inertialforces (:I:WI <Xvand WI <Xh)'The magnitude of this resultant is WI'

w =I f - 2 2WlV(l+CXv) +CXh

(v) Through Dl draw DI El parallel to BL intersecting BCl at El'

(vi) Measure Dl El to the same force scale as BDl' The Dl El is the dynamic earth pressure for trial~~. . .

(vii) Repeat steps (iv) to (vi) with BC2, BC3 etc. as trial wedges,

(viii) Draw a smooth curve through BEl E2E3' This is the modified Culmann's line.(ix) Draw a line parallel to BS and tangential to this curve. The maximum coordinate 'm the direction

, of.BL isq1?tained from the !Jointof tan~ency ~~4 i~.:th:Hyn~c,activeearth pr~ssure, {PA)dyn'For determining the passive earth pressure draw BS at «1>-~'V) below horizontal. Next Draw BL at

(90 - <X- 8 - W)below BS, The other steps for constructionremain unaltered(Fig, 5.4)0

--'--d--' -!

>ynamic ~~rlh Pressure 193, .cu Imann s line

plane of rupture

(;

sMinimum pas~ivepressure vector

Fig. 5.4: Modified Culmann's construction for.dynamic passive earth pressure

Effect of ~niformly distributed load and line load on the back fill surface may be handled in thesimilar way as for the static case.5.2.6. Dynamic Active Earth Pressure for c - Cl»Soils. The solutions so far discussed consider the soilto be cohesionless. A general solution for the determination of total (static plus"dynamic) earth pressuresf~r a c - cl>soil has been developed by Prakash and Saran, 1966 and Saran and Prakash, 1968.

q/Unit .area

Hl

ho

:.

H

B

Fig. 5.5 : Forces acting on failure. wedge in active state for seismic condition ~nc-+ soil

C?

Figure 5.5 shows a section of wall whose face AB is in contact with soil. The soil retained is hori-zontal and carries a uniform surcharge. The inclination of the wall AB with vertical is <X and inclinationof the trial failure surface is 9 I' AEC 1D1is the cracked zone in clayey soils, EC I being at depth h0 belowAD I .ho is given by expression

h = n (H - h ) = nH0 I 0 ...(5.14)where

HI = Total height of retaining wallH = Height of retaining wall in which backfill is free from cracks

In this analysis only horizontal inertia force is considered. All the forces acting on the assumed.failure wedge AEBCIDI are listed in Table 5.1 along with their horizontal and vertical components.

Table 5.1: Computation of Forces Acting on Wedge AEB Ct Dt (Fig. 5.5)

Designation Horizontal ComponentVertical Component

1. Weight of wedge

ABCI°1

2. Cohesion, C = c H sec 813. Adhesion, C = c' H seca

4. Surcharge

5. Soil Reaction RI6. Inertia force

7. Earth pressure PI

r 2- 'Y H (tan IX+ tan 81)21 ~2

+ 'Yn H2 (tan IX+ tan 81) + - 'Y n2cH

2H (tan IX)

cHtan81c' H tan IXc'H

q H [tan IX+ tan 81) + n H tan IX]RI sin (81 + q,) RI cos (81 + q,)

(W + Q) IXhP I cos (IX+ 0)P I sin (IX+ 0)

A summation of all the vertical components gives12 2 1222" 'YH (tan <X+ tan 9I) + 'Yn H (tan <X+ tan 9I) + 2" 'Yn H tan <X- cH - c' H + qH (tan <X+

tan 91 + n tan a) = PI sin (a + 8) + RI sin (91 + $) ...(5.15)A summation of all the horizontal components gives

cH tan 91 + c'H tan a + (W + Q) ah = PI cos (a + 8)'+ RI cos (91 + $) ...(5.16)

Multiply Eq. 5.15 by cos (91 + $), Eq. 5.16 by sin (91 + $), substitute for Wand Q from Table 5.1,assuming c = c', and adding, we get

PI sin (~ + 8) = 'YH2[(n + 1/2) (tan <X+ tan 91) + n2 tan a] [cos (91 + $) ,. ah sin (91 + $)] + qH[(n + 1) tan <X+ tan 9d [cos (91 + $) + <Xhsin (91 + $)] - cH [cos ~sec <X+ cos $ sec 9d ...(5.1?)where ~ =91 + $ + <X

Introducing the following dimensionless parameters:

- cos J3sec IX+ cos ~sec 91(Nac)dynsin(J3+o) .

[(n+l) tan a + tan9t1 [cos (91+~)+ah sin (91+~)](Naq)dyn = , sin (p + 0)

.'., ., i'; ; ,

<,14

...(5.18)

...(5.19)'~

;.,1

~

vnamic Earth Pressure 195

. 2

-'"- (N) = [(n+ 1/ 2)(tan a+tan el) +ntan a][cos (e1+~) +ah sin (e1+<1»]--ay dyn sin(~+O)

We get (Pl)dyn = yH2 (Nay)dyn+ qH (Naq)dyn- cH (Nac)dyn ...(5.21)'here $Nac)dyn' (Naq)dyn and (Nay)dyn are earth pressure coefficients which depend on a, n, <1>,() and e1.or given parameters of wall and soil, the values of (Nac)dyn' (Naq)dyn and (Ntrf)dyn are computed forifferent wedge angles e2, e3 etc. The variation of these coefficient with respect to wedge angle e arehown in Fig. 5.6, and the maximum values of (Naq)dyn and (Nay)dynand minimum value of (Nac)dyn arebtained.

...(5.20)

...01;\"u0Z

...-0~eT0Z

(Naqm )5~t

...0+-loO"

)""0z......

(Narm~t

(Nacm)o;tat

ecm ac aqm( b)

aq arm

(c)

, er(0)

Fig. 5.6: (a) (Nac)dyn versus eplot; (b) (N.q)dynversus e plot; (c) (N.-r>dynversus e plot

For such condition, the earth pressure corresponds to dynamic active earth pressure. Eq. 5.21 can b~.vritten as

2. '(P A)dyn = yH (Naym)dyn+ qH (Naqm)dyn- cH (Nacm)dyn

For static case, et.h= 0; earth pressure coefficients then become

cos ~ sec et. + cos ~ sec el(Nac)stat = , sin (~+8)

...(5.22)

...(5.23)

(N) = [(n+1)(tana+umel)]cos(el+<\)aqstat sin(~+()

...(5.24)

2[(n+1/2)(tana+tane)+n tana]cos(el +$)

(Nay)stat = sin(~+8), , (5.25)

Minimum value of (NaC>stat'and maximum value of (Naq)statand (Nay)statcan be obtained in similarmanner as illustrated above for getting the ~tatic earth :pressure coefficients. It is found convenient toobtain the dyna~~c ea~th pressure coeffi,cten~~JrQmP1efollowing constaq.ts : '-,

',d,' , ', '..'., ",

'

(N ,il'\0""1, c'"

. ,. ,', ," ,::...1', aqiit'ldyn' ' , '

AI = (N{I~",)stat ',:' { , .~;, ..

('f. i'~CJi"

i!. ~,

"~.$.'~~~

Machi"e .F oundations!1~t~ !,,'~ I

, (Naym)dyn, ' .

I~' 1

, " ,"-2 = (N ,)-, ...(5.27)

.

'

.

.'.

...

'.

.

.

.,

.

"

,arm. stat " ',!

Nacmboth for the static and dynamic case is same and bas' been plotted in Fig. 5.7 for different .~iinclination of the wall varying + 20° to - 20° with the vertical. As evident from the Eq. 5.23, Nacmfactor ,~1-1

is also independent of n., " , " , , '~II

~i

196 Soil Dynamics &

, ~.S

~.O+'0+-III

" 3.5Eu0Z 3.0-..........t)I0 2.5u

t)I....::J~ 2.0t)I....a.

.c 1.5+'....0w

1.0

0.50 s 10 20 25

r/J (deg)

lS 30 35 40 45

Fig. 5.7: (Nacm)statversus ellfor all n (Prakash and Saran, 1966 and Saran and Prakash, 1968)

""

..

, '

~

,'f

~

".

"

t~~

,}

Figures 5.8, 5.9 and 5.10 show plots of (Naqm)statversus.<1>for n of 0,0.2 and 0.4 respectively. These"plots consider the inclinationof the wall from + 20° to - 2.0°.Plots of (NaY11l)statfor the same range of n, ~<1>and Clhave been drawn in Figs. 5~11,,~.12 an~ 5.~3.

1

...

, It is found that the values of Al and A2 alter. slightly with incr~a~e in n. It is therefore recommended.

...

that the effec~of n on Al and "-2may,I'.otbe considered. Secondly, It ISobserved that Al and A2are almost.same (Prakash and Saran, 1966; andSanln and Prakash, 1968). Hence o~ly one value of-A(= Ali=~) is Jrecomm~nd,ed(Fig. 5:14). Since.~ is the .ratio o~:e~rth.pre~s~re~oeffic~ent~'in.(i) dynamic and (ii) static . J~case, and both the ~oefficients decrease with <1>.the ~s?ape of-the curves ~or.different f!-h values indicate .~;the rate of decrease of one in relation to the other.:: '~~:-::"'\::',;' ",!' "', " '":.,,.,. '" ,:.' 'c " " .,', - 'J~

rJynamic Eart" !ressure

:J11\11\

~ 0.4c..c....I-0 0.1w

'

0 .~ 1.0,...Eer0z 0.8...........~ 0.6u~I-:J::: 0.4~L-c.

~ 0.2L-0W

" ,

n =0

'r.

00 30 4015 20 15

(/J{dq'g)

35 455 10

Fig, 5.8: (N.qm)stat versus ellfor n = 0 (Pr&kash and Saran. 1966 and Saran and Prakash, 1968)

n =0.21

"

"

0-0 4030, , ..35 , _,45.' '25 ,

.' 'C/J {d q g) .

Fig. 5.9 : (N8qm).t8t versus' for n = 0.2 (pta,iuisb':and Saran, 1966 and Saran" and Prakash, 1968)

:fJ ',' ',:- ..'",.v'"',,,;-. "..ii";;'~';;~'~,~,#:!~,t:'i"i'~i:;"!o{,h,.J'Y,:;r~;'J'":>i':~'" ,,:. :.:,..:

10 15 20 "5, "r' --' . '

197

....1.2

0....III,...E 1.0er0z-.... 0.8....ti0U

0,6I-

198

Eer0Z 0.8........1>1

~ 0.6Col...:JIIIIII~ 0.4a..s::.......0LIJ 0.2

SoU Dynamics & Machine Foundations

1.4

n =0.4

00 3S'5 10 15 20 2S

</J (deg)

30 40 4S

Fi~. 5.to: (N.<]m),'.'versus ~ for n = 0.4 (Prakash and Saran. t 966 and Saran and Prakash, t 968)1.0

~0~11'

E 0.8)0

0Z

-: 0.6..........~0v~L- 0.4:J11'III~L-a. 0.2

.r::......L-0UJ 0

0

n=O

s 10

0-10--200

1S 20 2S 30 3S 40

,. ~ (d~Q).Fig. 5.11 : (~II'JII1)Jt.tversus' for It = 0.0 (Prakash and Saran, t966 and Saran and Prakash, 1968)

f~I, ,

4S

~

J,t

ii,

1.0mamic Earth Pressure 199

0 -

0 ,\, 5' .,- 10 15 20 25 '30" .; -;. ,'.. C: :1;);1;. ': ,.}

", .-', .. "'" . 'r'.""';> J"~; tI. (dczg) ;",;',,"" , , ':;.',"':"'~ ':\J .., ,,'fJ .. .

Fig:5.13'~ {N'~)...;vers~;'. for n'" 0.4 (Prak~sb and Saran, 1966 and Saran and Prakash, 1968)

+-0-

~ 0.8E

>00:z:; 0.6....~0u

t:.I

~ 0-41'11\11\~....a..r. 0.2+'....0lLI

-0 1.0+'III

,.....

E~0 0.8z........~ 0.6u~....:J11\ 0.411\~....a..r.t: 0.20UJ

n :: 0.2

00

-~-L~ '30 35 40 4S20 2S

~ (deg)

10 1S5

Fig. 5.12 : (Naym)statversus 4jIfor n =0.2 (Prakash and Saran, 1966 and Suan and Praliash. 1968)

1.2

n:: 0-4

3S 40 4S

200 Soil Dynamics & Machine Foundtltio.

2.0

A

so

~.~J

5.2.7. Point of Application. According to Indian standard (IS : ~893-1984) specifications, the pressuresare located as follows:

From the total pressures computed from Eqs. 5.? and 5.9 or from graphical construction, subtract thestatic pressure obtained by putting ab = ay = O. The remainder is the dynamic increment in active caseand dynamic decrement in passive case. The Static component of the total pressure shall be applied at an

1.9

1 .8

1.6

1.5

1.4

1.3

40

Fig. 5.14: A.versus, (prakash and Saran, 1966)

"

/,.,

1.2 r

0.10 {I

1.1.0.05 {

I1.0

0 10 20 30

rjJ (deg)

i"i'

'mic Earth Pressure 201

ltion Hl3 above the base of wall. The point of application of the dynamic increment and dynamic;::mentshall be assumed to be at an elevation R/2 and 2R/3 respectively above the base of the wall.

The static component of total active and passive earth pressure due to uniformly distributed sur-ge on the backfill surface obtained by putting ah = av = 0 in Eqs. 5.9 and 5.10 shall be applied atabove the base of the wall. The point of application of both the dynamic increment and dynamicement in this case shall be assumed to be at an elevation 2R/3 above the base of the wall.

The static and dynamic active earth pressures due to cohesion only (q = y = 0) are same. The pointpplication of this pressure shall be assumed to be at an elevation of R/2 above the base of the wall.

I',rj'H

DISPLACEMENT ANALYSIS

re are very few methods available to compute displacements of rigid retaining walls during earth-<es. They are;

(i) Richard-Elms Model based on Newmark's Approach(ii) Solution in pure Translation(iii) Solution in pure rotation(iv) Nadim-whitman modd(v) Saran-Reddy-viladkar Model,..,

1. Richard-Elms Model. Newmark (1965) proposed a basic procedu~e fer evaluating the potentialJrmation that would be experienced by an embankment dam shaken by an earthquake by consideringsliding block-on-a-plane mode as shown in Fig. (5.15 a). In this important development, it wasisaged that slope failure would be initiated and movements would begin to occur if the inertial forcesthe potential sliding mass were reversed. Thus by computing an acceleration at which the inertial:es become sufficiently high to caus~ yielding to begin, and integrating the effective acceleration onsliding mass in eXcess of this yield acceleration as a function of time (Fig. 5.15 c), velocities andmately the displacements of the sliding mass could be evaluated.This analysis is based essentially upon the rigid plastic behaviour of materials Fig. (5.15 b). Though

; method was developed for a sliding analysis of an earth dam, it has been used by Richard and Elms79) to compute the displacements of retaining walls. They have proposed a method for design ofvity retaining walls based on limiting displacement considering the wall inertia effect. The proce--edeveloped by them is described below. .

" .

FaHu r~ sf re ss

k(t)wIIIIII~......-

(/'I/

w

strain

. . '(a), ~>, " ,- .:/- (b)--Fig. S.1S: (a) Forces on sliding block (b) Rigid plastic stress strahi behaviour ora materia",""'; t ,. 'c :, " . , ,"" .'<l,

202

-iT:'i,7.-'~<)~:;jji

~

Soil Dynamics & Machine FolUUlati4- D

>-...

Time

>-coX

nuu«

Ti mq

u0ti>

+'CtiEtiU0

a.VI Timq0

(c)

Fig. 5.15: (c) Integration of effective acceleration time history to determine velocities and displacements

H

F' h Ww ,

\dWw< 1 :t d...v) \ I,

TFig. 5.16: Forces ~n a gravity 'wail

~t)t-:~

. .-:~.:h q.

t'\' "

rJk;, ~if

,;.

'4,f<

~:'

,)

I,,','!,"J'.,~:i"

--

Earth Pressure 203

ravity retaining wall is shown in Fig. 5.16, along with the forces acting on it during an earth-t1this figure various terms used are :

Ww = Weight of the retaining wall

lJ.h'av = Horizontaland vertical seismiccoefficients

PA)dyn= Dynamic active earth pressure, Eq. 5.5a = Inclination of wall face with vertical

8 = Angle of wall friction

<Pb= Soil-wallfriction angle at the base of the wallN = Vertical component Qfthe reaction at the base of the wallT = Horizontal component of the reaction at the base of the wall

,mming the forces in the vertical and horizontal directions, we get

N = Ww:t:.ay Ww + (PA)dynsin (a + 8)T = ah Ww+ (PA)dyncos (a + 8)

t slidding ~ = N tan <PbDIvingEqs. (5.28), (5.29) and (5.30), we get <".

(PA)dyn[cos(a +8) - sin (a + 8). tan ~b]W -w- (l:tay)tan~b-ah

...,5.28)

...(5.29)

...(5.30)

...(5.31 )

'utting (PA)dyn= ~ yH2 (KA)dyn and ah = (1 :t:.a) tan 'V, the eq. (5.31) can be written as

1 2Ww = 2" yH (KA)dyn' CIE

cos (a + 8) - sin (a + 8). tan ~b"here CIE = (l:tay)(tan~b-tan \jI)

~orstatic condition, the weight of wall W is given by :1 2W=- y H K .C2 A I

cos (a + 8) - sin (a + 8) . tan ~bwhere Cl = t J..an 'l'b

W (KA)dyn tan ~b~= .W '. KA (l:tay)(tan ~b -tan \jI)

...(5.32)

...(5.33)

...(5.34)

...(5.35)

rherefore,

>ubstituting...(5.36)

...

n Eq. (5.36),

FT = Ratio of earth pressure coefficients in dynamic and static cases(KA)dyn

FT = KA

. . tan <Pb.FI =:= Wall InertIa factor = (l:tav) (tan ~b-tan \jI)

W~=F F=FW T I w

...(5.37).e.

md ...(5.38)

...(5.39)

204

y,~. i

Soil Dynamics & Machine Foundations~ ;, 1

Fw is factor of safety applied to the weight of the wall to take into account the effect of soil pressure -~

and wall inertia. Figure 5.17 shows a plot of Fp F: and Fw for v~ous values of ah. From this figure for t.' ,F( = 1.0 and Fw =1.5, the value ah works out to be 0.18. However, if the wall inertial factor is considered, ;~$i

the critical horizontal acceleration corresponding to Fw = 1.5 is equal to 0.105. Thert:fore, if a wall is :' I

designed such that Ww = 1.5 W, the wall will start to move laterally at a ~alue of ah = 0.105. Hence forno lateral movement, the weight of the wall has to be increased by a considerable amount over the staticcondition, which may prove to be uneconomical. Keeping this in view, the actual design is carried forsome lateral displacement of wall.

i. j

II,. I... I

14

'"

fw

12

10

2 FT = 1.5

y/

./ññóä

L"/ T

':.--- ==- - -=- -..::;:::- ~. ~. .

0.105 I II 10.18.

4

o-D 0.1 0.2 0.6 "

~

Fig. 5.17: Variation of Fr F. and Fw with ab (Richards and Elms, 197?).,,;1i

~,. ,~~.~

3'8

u.;..

u:-.. 6

u:

zm;c Eartll Pressure 205

Richards and Elms (1969) have given a design procedure based on a limited allowable wall move-t, rather than on the assumption that the wall will not move at alL Such procedure is as follows:

(i) Decide upon' an acceptable maxilllum displacement, d. .. .

(ii) Determine the design value of ahd from Eq. (5.40) [Franklind and chang, 19771

-(

5ah)*

ahd - ah d ...(5.40)

where

ah = Acceleration coefficient from earthquake recordd = Maximum displacement in mm

(Hi) Using ahd' determine the required wall weight, Ww by substituting it in Eq. (5.31). The value

ahdof avd may be taken as 2'

/(iv) Apply a suitable safe~yfactor, say 1.5, to Ww'

There are three limitations to Richard-Elms analysis (Prakash, 1981): ~hese are:1~The soil is assumed to be a rigid plastic material. The walls do undergo reasonable displace-

ments before the limiting equilibrium conditions (active) develop and experience very largedisplacements before the passive conditions develop.

2. The physical properties of the system and its geometry (particularly its natural period) are notconsidered.

3. Walls may undergo displacements by either sliding or tilting or both. This method does notapparently consider this difference in their physical behaviour, although it is logical to concludethat displacements computed by this method are in sliding only.

.2. Solution in Pure Translation. A method for computation of displacement in translation only, ofd retaining wall under dynamic loads had been developed by Nandakumaran (1973).

Earth tpressure

(p)

ActiveC

--+BF

".Disptaceme nt

(a)

Fig.S.tS: (a) Earth pressure (P) versus displacement of wall

206

GtJ

(Rep + P r~p!

RSA(R8A- PA) J

Ix .I

I

,I. :

Basefriction {B.FJ

RBP--,';

H oXI

Displacement Displace m ent

(b)/

(c)

x ,°2 z

TB

"2 tan ~-11..

X,Ot

.~B

(d) (e)Fig. 5.18: (b) Base-Friction (B.F.) versus displacement

(c) Resultant of'P' and B.F. versus displacement(d) Simplified bilinear forces-displacement diagr.am(e) Computation of base resistance

The force-displacement relationships considered in this analysis are shown in Fig. 5.18. Fig. 5.18ashows the variation of earth pressure with displacement. In Fig 5.18b, variation of base resistance withdisplacementis given. The net force away from the fill is the differenceof active earth pressurePA andthe base resistance, RBA(Fig. 5.18c). The net force towards the wall is the sum of the passive earthpressure, Pp and the base resistance, RBP(Fig. 5.18c). The resulting bi1ine~rforce-displacement relation-ship is shown in Fig. 5.18d and is characterized by the following parameters:

tI

.t

I

I

namic Earth Pressure 207

. (i) Slope of force displacement relationship on the active and passive sides as Kl and K2 respec-tively, where K2 = n . Kt.

(ii) Yield displacement, ZyFor the resistance of the base, it is assumed that a column of soil of height (B/2) tan $ provides all

the resistance in a passive case (Fig. 5. 18e), B being the width of the wall at its base.The mathematical model is shown in Fig. 5.19. The parameters that are needed to define the system

r displacement analysis are: 1) the mass of system, rn, 2) period of the wall-soils system, 3) yieldsplacement, 4) damping in the system, and 5) parameters of ground motion.

ºÖ´ñ×ùºþþùºþòòô

K

m

., .c

xy = Y sin GJt z = x-yFig. 5.19: Mathematical model considered for the analysis (Nandkumaran, 1973)

The vibrating mass of the system consists of the mass of the wall and that of the soil vibrating witLe wall. Nandakumaran (1973) conducted vibratory tests on .translating walls and found that for thelIposes of matching the computed frequency of the wall with the measured natural frequency, the soilass participating in the vibrations is 0.8 times the mass of soil on the Ranking failure wedge.

Yield displacement for a given wall can be determined by considering the force-displacement rela-onships.

The ground motion is considered to be a sinusoidal motion of definite magnitude and period.The equation of motion can be written in the following form (Fig. 5.19) :

rnx + C (x - y) + K (x - y) = 0 .

. rnl + Ci + Kz = - myZ + 2 11~i 4-112z = - y

where z = (x - y),112= Kfm where K has been defined as the stiffnesson the tension side and

. C~ = Damping ratio = 2~Km

For ease in computations, all the three equations obtained by linear acceleration method (Biggs,~63) to be satisfied at each instant of time or at the end of each time interval selected, c~n be divided byy, the relative displacement on the tension side at which the resistance becomes constant (yield displace-lent) to obtain the following relations:

2

'l'n+_1= 'l'n+ 'i'n: t + t6 (\jJn-1+2\j1n)

~n+1 = ~~ ~.~-«(Pn+1+2\Vn)

...(5.41 a)

...(5.41 b)

...(5.41 c)

..:(5.42)

...(5.43).: "

JI.~. j

:~.;t j, I

\J. ;j\ !

, !

where\Jin+\ =-YIl-21l~ \Jin+\-1l2.K.Zy('I'n+ I)

Z'1'=-

Zy. Z'1'=-

ZyZ

...(5.44)

...(5.45)

...(5.46)

'1'=-Zy

With these relationships, the analysis is performed for the range of variables listed in Table 5.2Table 5. 2 : Range of Variables Considered in DisplacementAnalysis in Translation

...(5.47)

Variable Range of values

Ground acceleration amplitude (a) galsPeriod of ground motion (T)sDamping (~) %Natural period (TII)see.

Yield displacement (Zy) mm

100, 200 and 3000.5, .3, 0.2, and 0.15, 10, 151.0,0.5,0.3 and 0.21.0, 2.0, 3.0, 5.9, aJ1d10.0

To study the response characteristics of the system, two cases were considered; one in which plasticdeformation does not take place and the other in which it does. Figure 5.20 shows the response of theelastic system. It is evident from this figure that steady state conditions are attained in about 6 cycles andalso that displacements on the tension side are larger than those on compression side. The response of thesystem wherein slips take place has been plotted in Fig. 5.21. This shows that even when plastic defor-mations occur, a sort of steady state is achieved in the sense that slip per cycle becomes a constant afterabout 6 cycles.

12

8

EE 4-

+'C~E~u.E -4-a.III

0Tim<z ~

(5)

0-8

tI

-12

A =300 ga[5T = 0.3 5Tn =0.3 s

Zy = 200 mmTt. = 2.0

Damping = 10°/0

0.3 sL.; ;

Fig. 5.20 : Response of an elastic system with different stiffnesses of tension and compression sides(Nandkumaran,1973)

f

I4:

'namic Earth Pressure

A = 300 gals.T = 0 3 5Tn = 0.3 5

Zy= 10 mmn = 2.0

0

Damping =10%

Tima (5) --4 0.3si, ,

Fi~. 5.21: Displacement versus time (Nankumaran, 1973)-.-.

80

70

Zy =Smm; =10°'0n =2.0

Zy =10 mm~ =10°,.n =2.0

60EE

~ SO

A 1 T0.39.0.55

u>u

0.6

A IT0.39 . 0.5 r

~ 4.0..c:toIE~ 300a-lii

0 20

10I

0'29,0.3)-0.1 , Q.3 -

0.20.4. 0'.6 0.8 0.4.

Natural ptlriod.S

Fig. 5.22: Natural period versus slip per cycle (Nandkumaran, 1973)

209

..J

,-... 12-0c0'"c 8E'-

"0

IC0

4-c---+-C

E

u0-a.\11.-Q

!10 Soil Dynamics & Machine Foundations

Fig. 5.22 shows typical set of results in the form of slips per cycle versus the natural period of theNaIl in seconds for the yield displacement Z= 5.0 mm. and 10:0 .mm,~= 10% and n = 2 for differenty~round motions. The ground motion is considered to be an equivalent motion of uniform peak accelera-:ion of well derIDedcycles. .

Any problem can be solved with the following stepes :1. Determine the natural period of the wall using the following equation:

T=21t~...(5.48)

where K = stiffness on the tension side and In = mass of soil and the wall.

2. Determine the yield displacement.3. Determine the slip per cycle from Fig. 5.22 or similar other plots corresponding to the yield

displacement, the natural period of the wall and the ground motion considered.4. Compute the total slip during the ground motion.

This method of analysis is better than the one proposed by Richards and Elms (1979) in that (i)definite procedure for determining the natural period of the soil-wall system in translation has beenformulated, and (ii) physical behaviour of the retaining wall is considered in developing the force-dis-placement relationships. The method, however, suffers from the-fact that the tilting of the wall has notbeen considered.

5.3.3. Solution in Pure Rotation. A method of analysis for computing the, rotational displacement ofrigid retaining walls under dynamic loads has been presented by Prakash etal (1981) and it is based onthe following assumptions:

(i) Rocking vibrations are independent of sliding vibrations and the rocking stiffness is not affectedby sliding of the wall.

(ii) The earthquake motion may be considered as an equivalent sinusoidal motion having constantpeak acceleration.

(Hi) Wall may be assumed to rotate about the heel.(iv) Soil stiffness for rotational displacement of wall away from the backfill may be computed corre~

sponding to average displacement for development of fully active conditions.(v) Soil stiffness for rotational displacement of the wall towards the backfill may be computed cor-

responding to average displacement for development of fully passive conditions.(vi) The stiffness values computed in (iv) and (v) remain unchanged during phases of wall rotation

towards and away from backfill respectively.(vii) Soil participating in vibrations may be neglected.The mathematical model base upon these simplifying assumptions is shown in Fig. 5.23a. Figure.

5.23 b shows the scheme for calculation of side resistance corresponding to active and passive conditions.If fully active conditions are assumed to develop at a displacement of 0.25% of height of wall, then soilstiffness Kt in active state is giv~n by ,

K = Po-PA =I average displacement

KpyH2 - KoyH22 2

(2.5 H

). 100

...(5.49)

.~1

t'fIi1,,11».'1"""'

'namic Earth Pressure 211

t-=a-b t --t B. - ~ .r-,,

II 'IIIJ

//R<ztaining \wall

~

kA

ti'.xL..:::J +'IJ' CUt C:>I~ .-L.. Ua.:;:~

.t:1

C:>I

+' 0... u

kp

(kA or kP)H

Backfill

Lf-"[rotation

. b -1

Q.2SH F, Z.5H~ lOO."t- 100 ~

Di splacement

(a) (b)Fig. 5.23: (a) Mathematical model for rotation of rigid walls,

(b) Scheme for computation of spring stiffnesses (After Prakash et al., 1981)

Similar if fully passive conditions are assumed to develop at 2.5 % of wall height, soil stiffness Kz1passive state may be computed as :

Pp - PoK - -, z - averagedisplacement-

Z zKp rH - Ko rH

2 2

(2.SH

)100...(5.50)

where::

PA = Active earth pressurePp = Passive earth pressurePo = Earth pressure at rest

,KA = Coefficient of active earth pressureKp = Coefficient of passive earth pressureKo = Coefficient of earth pressure at rest

The rotation resistances of the base, in active and passive states (MRAand MRP)may be given by

M~A=.C,.I'CPA ,', ...(5.51 a)

- . MRP = c,. I . CP~ ...(5.51 b)n which C, is coefficient of elastic.non-uniform compression, I is moI1!e~t'ofin~rtia ~fthe base about anlxis through. the heel of.the!walland.perpendicularto the plane ofvibra.tiQns;,an~~A and CPBitre angles)frotation'away and towards!he~backfill. '1'-;," U' ,):~ "!'; 'i'n!>d':~1; ..' '-;' 0:

212.,.

SoU Dynamics & Machine Fou1tdtztio1$!$ ,..,

The equations of motion for rotation of wall away and towards the backfill are respectively:

(

. K H2

)Mmo ~A + C.I- 13 4>A= M (t)-"

...(5.52 a) -i.

(

Z

J. .. KzH

and Mmo Ij>p+ C. I - 3 4>p= M (t) ...(5.52 b)

Since the stiffnesses Kl and Kz are different, the period of the wall for the two conditions i.e. towardsthe backfill and away from backfill would be different. This would result in different values of 4>Aand 4>pfor each half cycle of motion and net rotational displacement of (4>A - 4>p)for one cycle of ground motion.The maximum displacement of wall for any number of cycles may be computed as :

4>T = n (4>A - 4>p) . H ...(5.53)where

n = Number of equivalent uniform cycles of ground motionH = Height of Wall

Based on the above, a parametric study was made considering the range of variables listed in Table 5.3

It was observed that the contribution of rotational displacement may be significant. The contributionof rotational displacement using the above approach was compared with the sliding displacement for a 3m high wall with backfill having angle of internal friction, 4>,equal to 36°, period of ground notion of 0.3 s,

Table: 5.3: Range of Variables considered in Displacement Analysis in Rotation

Variable Range a/values

Height of wall (m)Angle of internal friction for backfill (degrees)Period of ground motion (5)Damping (~)

C$ base kN/m3Base width/Height of wall

3.0, 5.0, 7.5 and 10.030, 33, 360.30, 5, 10, 15

43, 4, 5, 6 and 8 (x 10 )1/3

horizontal seismic coefficient CJ.hequal to 0.25 and C$ equal to 3 x 104kN/m3. The total slip in 15 cyclesdue to sliding was 213 mm. Displacement of top of wall due to rotation found by this analysis was147 mm.

This illiustrates that the rotational displacement may not be negligible and an attempt should bemade to account for it. The displacement analysis for rotational displacement is highly simplified. Nev-ertheless it shows explicitly that in some cases neglecting rotational displacement may seriously under-estimate the total displacement. In actual practice it may be essential to account for combined effects ofrocking and sliding that will affect the overall response of the system.

1j

5.3.4. Nadim- whitman Analysis. The Richard -Elms model assumes a constant value of wall accelera- ~~

tion (C1.h. g) when slippage is taking place. But once the backfil1 beings to slip, compatability of move- ?ment requires the backfill to have a vertical acceleration, thus causing change in wall acceleration. ~

. ~

I

ynamic Earl/. Pres~ure 213

Zarrab~(1979) considered the equilibrium of the.wall and the backftll wedge separately and.satisfiedle continuity requirements at failure surfaces as shown in Fig. 5.24a. An iterative procedure was devel-?ed for computing the instantaneous values of the inclination of failure plane, the dynamic active earth:essure and the acceleration of the wall, given the input of horizontal and vertical ground accelerations.he horizontal acceleration of the wall and the inclination of failure plane in the backfill are not constantI Zarrabi's model.

w.k~

~WJ

Ww

IW.kn

w.~-w

Rw tan ~bRw

(a)

3f 1f1

(c)

Contact element with Cn=lE8kN/mlm C5=0 ,

Slip element with Cn=lEBkN/mlm,C5=12SkN/m/m

Slip element with,Cn = 1E 12 kN I m/mCs = 1E S kN/mIm

Cn = Normal stiffness of slip elementsC~ = Shear stiffness of slip elements

(b)

Rigid boundary'4m

sea le ~

Fig. 5.24: (a) Force resolution of wall and soil wedge in Zarrabi's model(b) Retaining wall and its finite clement idealization(c) Effect of ground motion amplification on permanent wall displacement (Zarrabi, 1979)

. ,

Generally, displace:nents computed with Zarrabi's model. are slightly lower than those computedlith the Richard- Elms model. Dynamic tests on model retaining walls performed by Lai (1979) sh~wlat Zarrabi's model predicts the movement of the wall more accurately than Richard- Elms model. Lai,Iso, obserVed, a single rupture plane in the backfill in contrast to Zarrabi prediction. Later pn;-Zarrabi110delhas been modified to have a coiistanHncIlnationof failure plane 'in the b~ckfilt> ',.. ~ ,

. '1: <~ ':i...;:'!'>' ":", '

7

6

SI :1 1

:r I \A= 0.2

R N= 0.112Uniform G

.2

1

""

'i


The Richard, - Elms,model and Zarrabi's model assume a rigid - plastic behaviour of the backfillmaterial. Hence the input ground acceleration is constant throughout the backfill. But due to more-or-less elastic behaviour of soil at stress level below failure, the input acceleration is not constant. Henceamplification' of motion cannot be taken into account in these models.

Nadim and Whitrnan (1983) used a two -dimensional plane - strain finite element model for comput-ing permanent displacements taking into account the ground motion amplification. The slip element atthe ,base of the wall has been assigned a very large v<llueof normal stiffness, thus restraining the wallfrom vertical and rotational movements relative to its base. Thus, the wall undergoes only translationalmovements. In their paper, the results of finite element mesh us,edby them is shown is Fig. 5.24 b. Tounderstand the effect of ground motion amplification, typical results are shown in Fig. 5.24 c. In thisfigure, R is the ratio of permanent displacement from the FE model to the permanent displacement fromrigid - plastic (Richard - Elms or Zarrabi's) model. I, is the fundamental frequency of wall and I is thefrequency of 'ground motion. It can be seen that effect of amplification of motion on displacement isgreater when I II, is greater than 0.3 The FE model predicts zero permanent displacement ,at high fre-quency, because in the analysis only three cycles of base motion are considered during which steady -stateconditions can not be achieved. However, it can be said that large values off If, are not of great p'racticalinterest because displacements are very small.

Nadim and Whitman (1983) suggested the following simple procedure for taking into account the '

effects of ground motion amplification in the seismic design:

(i) Evaluate the fundamental frequency I, of the backfill fQf the d~sign earthquake using one-di-mensional amplification theory by using the following equation and estimate the ground motionfrequency, f

I, = VI/4 H ...(5.54)where

H = Height at retaining wall in m

VI = Peak velocityof earthquakein m/s(ii) If I Ifl is less than 0.25, neglect the amplification of ground motion. IfI II, is in the vicinity of

0.5, increase the peak acceleration, A and the peak velocity, V of the desjgn earthquake by 25-30%. Ifll I, is between 0.7 and 1, increase A and V by 50%. Obtain ah as A/g.

(iii) Use the value of ah from the previous step in the Eq. (5.40) given by Richard - Elms model forgetting ahd for known value of the displacement.

(iv) The value of ahd estimatedin step (iii) is used as the value of horizontalseismiccoefficientinthe Mononobe - Okabe analysis to calculate the lateral thrust for which the wall is designed. The

value of vertical seismic coefficient may be taken as a~d ..

5.3.5.Saran, Reddy and viladkar Model. Saran et al (1985) have chosen the mathematical model insuch way that it results translation and rotation simultaneously and therefore it has two degrees of free-dom. " ' , ,f

, '11 'I,

In practice, cro,

~~ ':' section pf rigid ret~iiningwall vari,

~s,

'to cl great exte!l.~' Areas<?n~,

1?leap,

pro~j,

.tQ,

,.:~,;rk.tion is, theref6~e, made by lumping the. mass of. the rigi~ retaining wall at its centre :<?fgravity. '\:IJ.t~J

,b~ckfin soil is replaced by closely spaced independent elastic springs shown in Fig. 5.25. " .",.;

.nic, Earth PresslIre

¸

ï¢ ¾ ¢ -Dynamic"'", ---

.QC tlve "

215

Displacedposition

III . . IInltla~ positionIIII

X JIIIII

.Dynamic. ,

pas slve

Fig. 5.25: Mathematical model for displacement analysis under dynamic condition

To determine the spring constants soil modulus values have oeen used" The s?~lmodplus depends ontype of soil. It varies linearly with depth in sands and n?rmally consolidated clays, but remains

lstantwith depth in case of over consolidatedclays.For linearform of variation- k =:,11It;'h, where 11fthe constant of horizontal subgrade reaction and h is the depth below ground surface. Value of 11ftalso)ends on the type of movement namely (i) wall moving away from backfill (active) an (ii)'wall movinglards backfill (passive.). P,rpbable range of 11ft.in cohesionles .soils is given in Table, 504 '

Table 5.4: Rangle of,values of Modulus of Subgrade Reactions 1h' . , ,.-

TIlt KNlm3 -

Soil'Active passive

Loose 'sand

Medium dense sand

'200-300 ;

400-600

800-1200Dense sand

'400-60d'

800-1200

1600~2400, -, .

In case of soil modulus linearly-varying with depth, the soil-reaction -is assumed'to 'act as a loadingtensity. Treating this load to be acting on a-beam of length, equal to the 'height 'of retaining w~ll,-theactions-at different points are evaluated treatIng this'beam'to be siniply' suppo'ited:at tIi{spring points,Jr the retaining wall of height H~d'ivided into ~ t6nvenie-nf~~mber ot ~qual ~egrI{erits'of,height -~Ji,the

'actions hence'the.' spring constants vallies' at"~a:fiou~'ciivi'si~h p'oTnts\vould'oe a~'~n~~r..." " .' ,

t--°'--t K,- I

:IThl K..,

hl K...

h,Retaining

I illwall

-- - - - - - - - jKn-"

- I --.F=F sin wt

0

16 Soil Dynamic.s & Machine Foundations

k - 11 - 6 l1h(1:1hi

k2 = l1h (1:1h)2

k3 = 2 l1h (1:1h)2

...(5.55 a)

...(5.55 b)

...(5.55 c)

ki = (i - 1) l1h (1:1h)2 ...(5.55 d)

1 2kn = "'6 (3 n - 4) T\h(1:1h) ...(5.55e)

where kl and kn are the spring constants at the top most an bottom most points, ki, the spring constant atmy division point 'f.

In case of soil modulus constant with depth, the soil reaction is assumed to act as uniformly distrib-uted loading intensity. Treating this uniformly distributed load to be acting on a beam of length, equal tothe height of retaining wall, the reactions at different points are evaluated treating th,isbeam to be simplysupported at these points. The spring cons'tants would be as under, .

For the top most spring,1k = - k (1:1h)1 2

k; = k (1:1h)

1k = - k(l:1h)n . 2 ...(5.56 c)

For any intermediate spring,

...(5.56 a)

..(5.56 b)

For the bottom most spring,

The method is based on the following assumptions:1. The earthquake motion may be considered as an equivalent sinusoidal motion with uniform peak

acceleration and the total displacement is equal to residual displacement per cycle multiplied bynumber of cycles.

2. Soil stiffnesses (or spring constants) for displacement of wall towards the backfill and away fromthe backfill are different.

3. Soil participating in vibration, damping of soil and base friction are neglected.Assumpations 1 and 2 are usually made in such as analysis while assumption 3 needs justification.It is difficult to determine analytically the soil mass that would participate in vibrations along with

wall when it undergoes translational and rotational motions simultaneously. Neglecting this mass, themethod gives higher displacements and the solution is conservative. However, the mass of vibrating soilcan be found out by carefully conducted experim ~nts. For the case of pure translation, Nandakumaran(1973) has conducted experiments to determine the vibrating soil-mass and concluded that itcan be takenequal to 0.8 times the mass of Rankine's wedge. By adopting similar technique, the soil mass vibratingalong with rigid retaining wall under combined rotational and translational motion~cim be found out. ,

Then it is added to the mass of the wall to lump at centre of gravity'and the analysis can be carried out'without any changes. .'. '.-

-

,~#'

Dj'

laren,.da:W2

dehasedr

IS

deS)

5.u

., iI

~;)"~J I

11& III it

0'

Earth Pressure 217

oils, it is customary to consider valUes of damping such' as 15%'or 20% of critiCal in view ofergy absorption compared to other engineering structural materials. In the present analysis however,bsorption in the form of plastic displacement of the wall has been considered. Therefore smaller; values would be appropriate. Neglecting even this smaller damping, the displacement of thethis method will be more than the actual displacement.

~displacementof retainingwall is greatly influencedby base friction. In case of walls in alluvial; and' at the waterfront, translation'al'motion,is predominant.In some other cases, the wallsmayedominant rotational motion. But in general for any type of foundation soil, retaining wall pos-'ansl..tional and rotational motions simultaneously. For'rigid retaining walls, the stability ismainlyts gravity, hence base friction, the analysis will lead to an overestimation of the displacement.wever, refinment of the model by including vibrating soil,mass, damping of soil and base friction~d so that the analysis can predict displacement close to the actual displacements,study the response characteristics of the 'system, two casoesare considered, one in which plastic

ations do not occur (elasti~ system) and the other in which plastic deformations do occur (plastic).

, Analysis of an elastic system active condition. The equations of motion of the retaining wall)' Alemberts principle can be written in general terms as .follows;

;=\

:'(' + Lk;[x+ {H -h)-(i-I)Md 9] = Fa sin (J) t11

;=\

+ Lk;[x+{H-h)-(i-I)LlhJ 9) (H-h)-(i-l) LllzJ]=0

...(5.57)

...(5.58)11

hereM = Mass of reitaining wall

J = Polar mass moment of inertia of the wall about the axis of rotation(J) = Frequency of the excitation forceH = Height of retaining wallh = Height of centre of gravity of wall from its basex = Translatory displ~cemente = Rotational displacement

,etting :;=\"f.k;n ,= a

"1T '..

F...JL = aM a

;=\ ,:;

I,kj {(H-h)-(i-l)L\h}

...(5.59)4

...(5.60)

n =b ...(5.61)M;=\ - '2 t ' , ~ \

"f. kj {(H - h ) - (i - 1)6.h}, - - '. - ,n '! ,) '\

~ J =c ...(5.62)

-::rI

218 Soil Dynamics & Machine Foun~

The equations of motion of the rigid retaining wall cai1thu~ be written,as~, ,

, ", , '

: x' + ax ~ be+ 'A" sih ro l' " ,

, ". 0., , ,',' , ..:.(S.6?).

( )""

, , ' " b' '"

e+ ca = "7 x'; ", , ...{S.64), ,

, , ,

where J = Mr2, r being the radius gyration and 'b',can be called as coupling coeffici~ntbecause if b.=,o,the two equationsbe~ome,independentof each other. ,. "

The solutions of Eqs. (5.63) and' (S.64) can be written as,,' X = X 'sinrot

, e = ~ sin (0 1. " "',,

where X and ~are arbitrary constants.,SubstitutingEqs.'(5.65):and(S.66) in Eqs. (S.63) and (S.64) , we get

(- (02 + a) X = b~ + ao'

(ch c) ~ ~ (:,)x

.':.(5.6S)

...(S~66)

"',(5:67)

" ., ...(S'.6'8)

Solving Eqs. (S.67) and (5.68), we get"

ao2X = b2

) 2( a - 00 -,.2 (c - 00 )

...(S.69)

ao~ = '

2 22,

, .' ,., (a-m Hc-oo )---b

u b '

...(S.70)

Hence the solution becomes -"

X = ao

(a-002)- b2 ,sinro(2,. (c-002)

...(S.7!)

e = a0

(a-002) (c- 2 1'2 sin rot00 )-T-b

..,(S.72)

Therefore, the displacement of the top of rigid retaining wall is given by -Xtop = X + (H - h ) e

{

2 2 -

}

r (c-ro) +btH-h) . .x = a srn ro(top(

2) (

2)

2 b2 0a-ro c-ro r-

...(S.73 a)

or

.<0 I"" !

(S,73b) I

t,..ii

!icEarth Pressure

? Natural frequencies. Under free vibration condition, the equations of motion are:

. 'x' +ax :dba .. .-- . -;~;~,-..

(b

J

-

9 + ca = ; xubstituting the solution:

x = A sin ro tn

a = B sin ron t~ A and B are arbitrary constants~quations 5.74 and 5.75 become:

(-00; +a) A = b . B

(-<o;+c)B ~ (:,)AFrom these we get,

bA =- 2B a -OOn

and'2

~ ~ (::)Equating,

2b c-oo

a-oo; - ( ~ )\r

(

2

)4 2 b

OOn-(a+c)OOn+QC-; =0

and solving we get,00;\ = ;(a+c)+J( c;a r +(~ r

( )2

()2

2 I c-a boon2 = 2(a+c)-J -y- + r

219

...(5.74)

...(5.75)

...(5.76)

...(5.77)

...(5.78)

...(5.79)

...(5.80)

...(5.81)

...(5.82)

...(5.83)

...(5.84)

.5.3. Passive condition. The ratio of stiffnesses on the compression and tension sides is denoted by n.nce in the passive condition, the values of a, band c change and these can be given by :

a = n (a)ab = n (~)aC = n (c)a-

The solution for this condition is similar to active condition described above.

...(5.85)

...(5.86)

...(5.87)

220 Soil Dynamic'S &

~

Mad<ineF...d_~ I

5.3.5.4. Analysis of a plastic system-active conditio,n. Assume that Zyand 9yare the yield displaceme~~\:occurring simultaneously in all springs: the equations of motion can be written as:

x. + a Zy = b 9y + ao sin Cl)t..

(b

)e + ce = - Zy r2 y

Integrating the above equations twice, we get

t2 ao sin Cl)tx = (b ey - a Z) T - 002 + Cl t + C2

"

...(5.88)

...(5.89)

...(5.90)

(b

)t2

e = ? Zy -coy "2 + C3 t + C4 ...(5.91)

Let 'fe' be the time after which displacement of top of wall (yto~ becomes greater than yield displace-ment (Yd) and plastic system starts. Let xe' xc' ee' 8e be the values corresponding to time te and can becalculated by using the equations developed for elastic system. The following boundary conditions can beapplied to evaluate the constants of integration:

(i) t = re' X = xe

(ii) t = re'X = xe

...(5.92 a)

...(5.92 b)

...(5.92 c)

...(5.92 d)

(iii) t = re' 8 = 8e

(iv) t = re' e = ecTherefore, we have

Zy = xeey = ee

a cos Cl)tC = x - (be - a Z ) t + 0 e

I e y y e Cl)2 .

. te ao sm 00teC2 = xe - xete + (bey - a Z) T+ 002

C =8 _ ( b2Zy-Cey ) te3 e r

(b

)t;C4 = 9 - ZZy-Cey_2 - 8 t

ere e

Displacementof the top of rigid retaining wall is given ~yx =x - (H - h) etop

...(5.93)

...(5.94)

...(5.95)

leao cosoote ...(5.96)00

...(5.97)

...(5.98)

...(5.99)

5.3.5.5. Passive condition. The ratio of stiffnesses on the compression and tension sides is denoted by nHence in the passive condition, the values of a, band c, change and these can be given by:

a = 11(a)ab = 11(b)ac = 11(c)a

...(5.100 a)

...(5.100 b)

...(5.100 c)

l'I.

nic Earth Pressure 221

'he solutiOltissimilar to the above procedure for active condition except the values of Zy and ay"Inression side (passive condition), the displacements for achieving yield condition are very large,~in most of the cases plastic system for l"assive case is not considered.

lILL USTRA TIVE EXAMPLE~

mple 5.1 .

0 m high -retainin~ wall with back face inclined 20° with vertical retains cohesionless backfill33°, 1/ = 1~_KN/m and 8 = 20°), The backfill surface is sloping at an angle 10° to the horizontal.(a) Determine the total active earthpressure using Coulomb's theory and Culmann's graphical con-

struction.

(b) If the retaining wall is located in a seismic region (ah = 0.1), determine total active earth pres-sure using Mononobe's equation and modified Culmann's graphical construction.

. -ltion :

-,. .

Static active earth pressure ,.-.,,0

2) I

- 1 2 cos «I>- a . 1/2

}

2PA = 2"1 H cos2 a cos (8 + a)

{ [sin ( + 8). sin ( - i)

]- I + cos (a - i) cos (8 + a)

2 cos2 (33-20) x= ~ x 18 x 6.0 x cos220 cos (20+20)

1

{I +

[sin (33 + 20) sin (33 -10)

]

1/2

}

2

cos (20 -10) cos (20 + 20)= 168.42 kN/m

Refer Fig. ".26 for Culmann's graphic,alconstructionfor getting static activepressure.Ds Es gives~ total active earth pressure. ' . .

P = 17 x 10 = 170 KN/mA

(b) Dynamic active earth pressure

- 1 2 col «I>- 'V - a)( I:!: ay) x ,1 2(P)d - - 1 H 2)

'

{ ]

1I2

}

A yn 2 cos'I' cos a cos(8+ a + 'I'

[sin ( 4>+ 0) sin ( 4>- i - \jI )

1+ cos (a - i) cos (0 + a + \jI)

ah 0.1Assuming a = - = - = 0.05v 2 2 .

-I ah -I 0.1'V = tan - = tan -

l~av' 1:!:0.05= 5.44° with ;:. ay and .

= 6.0° with'\- ay .

222 Soil Dynamics & Machine Foundatioi$' 'w,. . . . : i i,

6.0m

Linear scale - 1 : 100F'orctZ scaltZ, -1mm = 10KN

'0

Pr~ ssurtZ I in~

Fig. 4.26: Culmann's graphical construction

Value of (f A)dynwith (+) ay2

= .!." 18 x 60 2 cos (33-5.44-20)(1+0.05) 1A. 2 X ' 22 cos 5.44cos 20 cos (20+ 20+ 5.44)

{I +

[sin (33+ 20) sin (33-10-5.44)

]

1I2

}cos (20-10) cos(20+20+5.44

= 214.26 kN/m

Value of (PA)dynwith (-) ay2

- 1 18 6 02 cos (33-6-20)(1-0.05) 1--x x x2 . cos6cos220cos(20+20+6)

{

1+[

~in(33+20)Sin(33-10-6)]

1/2

}

2

cos (20-10) cos (20+ 20+ 6)= 198.05 kN/m .

Therefore (+) ay case governs the value of dynamic active earth pressure.

Hence, (PA)dyn= 214.26 kN/m .:.Refer Fig. 5.27 for modified Culmann's graphical constructi9n for getting dynamic active earth pres-, " . '. .

sure

",ic Earth Pressure 2'3"

m

Lin~ar, scalCl -1 :100Forc~ scalCl -1mm:10KN

Fig. 5.27 : Culmann's graphical construction

DsEs gives the total dynamic activ~,earth pressure.(PA)dyn= 21 x 10 = 210 kN/m

,ample 5.2retaining wall 8.0 m high is inclined 200 to the vertical and retains horizontal backfill with followingaperties :

y/= 18 kN/m3, ~=300 and c = 6.0 kN/m2There is a superimposed load of intensity 15 kN/m2 on the backfill. The wall is located in seismic

gion having horizontal seismic coefficient of 0.1. Compute the dynamic active earth pressure and de-cmine the percentage increase in pressure over the static earth pressure.

Solution:

(i).~ 2c 1

ho = Y ~KA2x 6.0 x..[j = 1.15 m= 18 .

1- sin 30 1where KA = l+sin30 = 3'

224 Soil Dynamics & Machine FoundatiiiJl;

(ii)

Figs 5.7 to 5.23 give:

H = 8.0 - 1.15 = 6.85 m

ho 1.16n=1[=-=0.167.. 6.85.

For = 30°, a. = - 20°and n = 0.167

(i ii)

(Naym)stat = 0.33, (Naqm)stat= 0.512 and (Nacm)stat= 1.22

(PA)stat.= 18 x 6.85 x 0.33 + 15 x 6.85 x 0.512 - 6 x 6.85 x 1.2= 282.0 kN/m

For = 30°, a.,= :- 20°, dh = 0.1, Fig. 5.14 givesA. = 1.18

Therefore, (Naqm)dyn= 1.18 x 0.512 = 0.609 and(Naym)dyn= 1.18 x 0.33 = 0.393

(PA)dyn= 18 x 8.852 x 0.393 - 15 x 6.85 x 0.609 - 6 x 6.85 x 1.2= 345.18 kN/m

(iv) Percentage increase over static pressure

345.10- 282.00 x 100 = 22.4 %= 282.01

Example 5.3A 5.0 m high retaining wall with backface inclined 20° with vertical retains cohesionless backfill«I>= 30°,Yr= 18 kN/m3and 0 = 20°). The backfill surface is sloping at an angle 15°to the horizontal.Determine the weight of the retaining wall.

«(1)For static condition,

(b) For zero displacement condition under earthquake loading(c) For a displacement of 50 mm under earthquake loading.

Solution:

(a) ,For static condition. .

. cos2 (cp- a) 1KA = cos2acos(o+a)'

{ [Sin(cp+O)Sin(cp-O

]

l/2

}

2

1+ cos(a -0 cos (8 +a.) . ,

cos(30-20) . '. 1= cos2 20 cos (20+ 20) .

{ [sin (30+ 20) sin (3P-15)

]

I/2

}

2

. 1+ cos(30-20)cos(20+20). .'" .

= 0.6225. ','

.

!amic Earth Pressure

From Eq. (5.35)

Cos (a +0) -Sin (a +0) tan ~bCl = tan ~b

Cos (20 + 20) -Sin (20 + 20) tan 30tan 30

=

= 0.6840Thus from Eq. (5.34)

1 2W = "2y H KACl

= .!. x 1800 x 52 x 0.6225 x 0.6840 = 9765 kg/m2(b) For zero displacement condition

cos2(~- \If - a) 1(KA)dyn = 2 ~ )

x

{

1/2

}

2cos \If cos a cos ( u + a + \If

[sin (~ + 0) sin (~ - i - \If)

]- 1+ cos (a-i) cos(o+a+\jI)

Assuming ay = a2h = 021 = 0.05

-I ah 0.1\jI = tan l:tay = 1:t0.05

= 5.440 with + ay and= 6.00with - ay

With + ay

cos2 (30- 5.44 - 20)(1 + 0.05) 1(K) = x 1/2

}

2Adyn cos5.44cos220cos(20+20+5.44)

{ [sin(30+20)Sin(30"""15-5.44)

]1+ cos(20-15)cos(20+20+5-44) -= 0.8311

--

With - ay:

cos2 (30-6-20)(1-0.05) 1 .(KA)dyn = cos6cos2 20cos(20+20+6) x

{ [~in(30+20)Sin(30-15-6)

]

1/2

}

2

. 1- cos(20~15) cos(20+20+6)= 0.7727

From Eq. (5.36)With + ay :

W 0.8311 tan30 -

W = 0.6225 x (1+ 0.05)( tan 30- tanSAM= 1.524

225-

~.

226. Soil Dynamics & Machine Foundations

With - av :W 0.7727 tan 30---!!. = xW 0.6225 (1- 0.05)( tan 30- tan 6)

= 1.596Therefore

Ww = 1.596 x 9765 = 15585 kg/m(c) For displacement condition, d = 50 mmFrom Eq. (5.40)

[5ah

]a"d = ah -cl

-[

5 x 0.1

]

1/4

- 0.1 50= 0.03162

Assuming

= ahd = 0.3162 = 0.1581avd 2 2With (+) avd

-I[

ahd

]-I

[0.03162

]'" = Tan l+ahd = Tan 1'+0.01581= 0.780

cos2 (30-1.78 - 20)(1 + 0.01581) 1(K) - x 1/2

}

2A dyn - cos 1.78 cos2 20 cos (20+ 20 + 1.78)

{ [sin (30+ 20) sin (30 -15-1. 78)

J1+ cos (20-15) cos (20+ 20+ 1.78)

= 0.684

With (-) a,'d-I

[0.03162

] 840

'" = Tan 1-0.01581 = 1.It gives

= cos2(30-1.84-20)(1-0.01581) x - 1 2(KA)t(,'n cos1.84cos220cos"(20+20+1.84)

{ [sin-(30+20)sin(30-15-1.84

J

1/2

}1+ cos (20-15) cos (20+ 20+ 1.84)

= 0.672

Therefore, for (+) ve a,'dWw - 0.684 Tan30W - 0.6225' (1+0.01581) (tan 30-tan 1.78)

= 1.1434

,

:am;c Earth-Pressure"".'

For (- )ve avd

Ww - 0.672 . Tan 30W - 0.6225 (1-0.01581) (tan30-tan 1.84)

= 1.162- "

Hence

227

w w = 1.162 x 9765 = 11347 kg/m "'

If a factor of safety of 1.5 is used, then

" W = 9765 x 1.5 = 14647 kg (Static condition) ~Ww = 15585 x 1.5 = 23377 kg (Earthquake condition - zero displacement)

Ww = 11347x 1.5 = 17020kg/m (Earthquakecondition-50 mm displacement) "

It may be noted that the weight of wall gets re~uced significantly if the wall is designed for somesplacement. "

-.

xample 5.4ompute the displacement of a vertical retaining wall baving section and.~ackfill properties as shown inig. 5.28. The characteristics of the ground motion are:

Period = 0.50 s

"," 1'Ranki'ne's,,,

"

II,,,

Average ground accelerationNumber of significant cyclesYield displacement

= 0.2 g= 10 "r

= 5 mm; n= 2 and ~ =10%

-tl.0m ~

6.0 m

"K.4s- 0/2f 3.0.m ..-~-t "

",I, "

I,

'tt ="18 kN I in3

~ = 30°

Fig.5.28 :'Section or retaining wall"

wedge"

. .


Solution:

(i) Refer Fig. 5.261 .

Weight of wall = 2" (1.0 + 3.0) x 6.0 x 24 = 288 kNWeight of soil vibrating along with wall

1 ( 30)= 0.8 x 2" x 6.0 Tan 45-T x 6.0 x 18 = 149.6 kN

Total weight = 288 + 149.6 = 437.6 KN .

3m = 437.6xl0 = 44.61x 103kg9.81

Let the coefficient of base friction = 0.6. 1 . 1

288xO.6--xI8x-x62K = 2 3

1 0.005= 12960 kN/m = 1290 x 103 N/m

T=21t~n VK;.

(Fig. 5.18d)

32

V

44.61 x 10T = 1t

n 12960 x 103= 0.368 s

(ii) From Fig. 5.22 (a), for Zy = 5 mm, TII= 0.368 s; n = 2, T = 0.5s, A = 0.2 gSlip per cycle = 24 mm

Total slip = 24 x 10 = 240 mm

Example 5.5Determine the displacement of a model wall shown in Fig. 5.29 retaining medium dense sand ($ = 36°,Y= 18 kN/m2, and T)h= 520 kN/m\ The wall is subjected to following dynamic conditions:

Yield displacement: = 6mmGround acceleration: =.0.25 g

Time period: = 0.3 sSolution:

1. The wall is divided into foul equal number of segments with 11ftequal to 0.75 m and the backfillsoil is idealized by using springs as shown in Fig. 5.30. The mass of retaining wall is assumed to belumped at its e.g. which is at a distance of 1.23 m above the base (Fig. 5.30).

2. Consider the backfill characteristics, the spring constants in active and passive cases determinedusing Eqs. 5.55 and are given below in Table 5.4. The ratio of stiffnesses in passive and active states istaken as 2.0.

. ~

..-~..' -," --~~if'.t?~f~

nic, Earth Pressure

3.0m

H= 3.0 m

yII .

:-i°.3m~ .

A, : reIIIIII,,I,IIIII

'" ,Retaining I

I- - - ~ .?ll- -;- - +- - - X

I,IIII,0

f4 1.0 m

229

00 = 36'( = 18 kN 1m3

7th: 520 I<N/m3M~dium dense sand

c--i

Fig. 5.29 : Section of retaining wall

v

~a = a sin wt0

io......

Ib = 1.0 m

M

c

Fig. 5.30 : Mathematical model adopted for solution

- '0 0---

.," .'":': ::~;::~'~Yf~'~t*~11~{!',~~:r~i;:';'c::?0t~~,1;\q~i01;-J;{:;:':;~ "ii14:Wf'.o~';.:'.~':,,'~.-'>c',"',",,',,';'.:'-',,.','!,m,'.-'t:~'.:';,-~:",'-::'1'r'.:~\r-{<;'i-l;;:.:,';-Ji-;'-::,;~'rt','!;,<'-"':';:i',~"-;:~';',.'i-H~';''-,i~!t=~"/i.'."-~t,~-,-'.,~'..~:,':t'.~..,-'::;~.-,(;',,'.-."i"""'~i,.:!tP1(F't-..':~'r;-;:;.;'.--',-, ~".~-.'i~'~-,,'~-',~"";¢

230"

Soil Dynamics &' Machine Founilations

.~.".

Table 5.4 : Values of Spring Constants

Spring - Spring locationw,r, to c,g"hj

(m)

, Spring constant

f active case, kj(KN/m)

,Spring constantpassive case, 'kj

(KN/m)

KJ

K2

K3

K4

Ks

1.77

1.02

0.27

- 0.48

-1.23

48.8292.5585.0877.5

536.3

77.6585.0

1170,01755,0

1072.6

L kj = 2340.1 L k, = 4680.2I

3. Equation of Motion

The quantities required in the analysis are calculated as shown below (Fig. 5.29), 2 x 0.3 + 1.0 3 1.6 2DIstance of e.g. from CD = 0.3+ 1.0 x'3 = 1.3 = 1. 3 m

.

( 0,7), 3xO.3xO.15+0.5x3xO,7 0.3+3Distance of e.g. from BC = 1+0.3 .

2 x 3.0

= 0.13;;5°.56 = 0.3563 3

0.3 x 3 0.7 x 3 I 2Ivv = .- .+0.3x3.0(1.5-1.23)+ +-xO.7x3x(1.23-1),~ 12' . 36 2

= 0.675 + 0.06561 + 0.525 + 0.05554= 1.321155 m

3 . ' . 3 2

I = 0.3 x3+0.3X3.0(0.356-0.15)2+0.7 X3+.!.XO.7X3X (0.7 -0.056

)YJ 12 36 2 3

= 0.00675 + 0.0381924 + 0.285833 + 0.0330194= 0.01065451 m4

Ixy = 1.321155 + 0.1065451 = 1.4277002 m4

A = 1+2°.3x 3.0 = 1.95 m2

r = ~ 1.~~~7 = 0.8556 m

M = (°:3; 1.°) x 3.6 x 23.~~~03.~ 4.58, x 103'k~ - . .

J = Mr2= 3.35 x 103 kg-m2 ,'.1 ,

'"

l,

f

"'r:i;'j.:'::,'C":':,i',iL;;;";;[!';;;;:',',,;;,..,.j'..,,", .'<' I11III ..

lmic Earth Pressure 231

The values of a, band c in the active and passive cases can be detennined as below:Active case Passive caseI.k.a = ---L = 51094M .

I. k. h.b = I I = 117.51M

I.k. h~ 1513.393c = I I = = 451.76J 3.35

The natural frequencies of the wall by considering the tension side (i.e. active case) are given by :

a = - 2.0 x 510.94 = - 1021.88

b =-2.0 x 117.51 =-235.02

c = - 2.0 x 451.76 = - 903.52

'

( )2

()2

2 1 c-a boonl,2=2(a+C)I~ 2 +-;:

Puttmg the values of a = 510.14, b = 115.51 and c = 451.76, we get00/11= 24.94 rad/soon2 = 18.46 rand/s

The natural time periods are therefore,Tnl = 0.25 sTn2 = 0.34 s

Earthquake motions are erratic and no two accelerograms are similar. The two main parameters oflY ground motion are the amplitude of acceleration and the number of zero crossings in unit time. A~rysimple and convenient form of ground motion including the above two parameters, is a sinusoidalotion. Moreover, while proposing a method for analysing the liquefaction potential of sand deposits,eed and ldriss (1967) contended that any given accelerogram can be considered equivalent to someefinite number of cycles of loading of equal magnitude. Such idealization have the advantage that aftertudying the effect of two parameters, the effect of a probable earthquake motion at any site, can benalysed. Because of the above advantages, sinusoidal ground motions are utilised in the present study.

Given ao = 0.25 g = 2.45 mIs221t

T p = 0.3 s ; ro= 0.3 = 20.94 rad/sa = 2.45 Sin (20.94 t)

Predictioll of Displacements ill Elastic SystemThe displacements in passive case (t = 0 to tpl2) can be cal~ulated as shown below:

x = X SinWte = ~ Sin Wt

where X = ao 2(a-ro2)- b

,2 (c - ro2 )

= 2.45.(-1021.88- 20.94')- (-235.02)' =- 1.7448 x 10-3 m. 2

j. 0.8556 (-903.52-20.942)

H' """"'i'-;:,:;"",:~',J,.""~',~,,J.':';',, '""~';;,~,::,,,)':',.,:,;,:";'< ':":""":,/","",\i,, """ .,C-.;--,",:-.c ,', '" ',' '.' """..'..n'-"':{!\;;{k,',',;.",'A" i

232 Soil Dynamics & Machille Foundations

Î ó ù ¿±¬óùó î

î î ®ø¿ó³ ÷¨ø½ó³ ÷óó¾¾

óìî ãóìòïéìï ¨ ïð ®¿¼

øóïðîïòèèóîðòçìî÷̈ øóçðíòëîóîðòçìî÷ðòèëëêòóøóîíëòðî÷óîíëòðî

Hence, we havex = - 1.745 x to-3 Sin (20.94 ¬÷

» ãó ìòïéì ¨ ïðóì Sin (20.94 t)

The computed values of displacements from time 0 to 0.15 s are given in Table 5.5

Table 5.5 : Values of Displacement in Passive State (Elastic-Condition)

îòìëã

Ì¸» displacement in active case (t = t; to tp) can be calculated as shown below -x = X Sin rot

e = ~Sin rot

where x = ao/ b2

(a-m2)-\ ,.2c-m2

= 2.45 ,

(510,94-20,94')- 117,25' ~-1.817 x \0-3 m0.85562 (451.76-20.942)

~ = ao

=

2., ., r(a -ro-).(c-ro-) --bb

2.452

. (~10.94 - 20.942).( 451.76- 20.942) °ig~561

x =-1.817 x Sin (20.94 t)e = -2.197 x Sin (20.94 ¬÷

óî ¼ãóîòïçé¨´ð ra

117.51Hence, we have

. :.:

\i

Time Translational Rotation IJisp. of wall at top Total disp. at topt(s} displacement x (mm) e (rad) due to rotation xe (mm) xro/mm)

0 0 0 0 0

0.0375 - 1.23 - 2.9510 x 10-4 - 0.52 - 1.75

0.0750 - \.74 - 4.1740 x 10-4 - 0.74 -2.48

0.1125 - 1.23 - 2.9510 x 10-4 - 0.52 - 1.75 .

0.15 0 f) 0 0

- -- -'namic Earth Pressure 233

The values of displacements in active state considering elastic condition are given in Table 5.6.Table 5.6: Value of Displacement in ActiveCondition (Elastic-State)

The dynamic response of the retaining wall under elastic system is shown in Fig. 5.31 which indi-:ates that the slip (permanent displacement) after one cycle of ground motion is zero. It can be concluded:hat in elastic system, after any number of cycles, the residual displacement would be zero.

70~

- 5-

0

."

60

50EE-0~ 40....0-c:tiE 30to>u0a.III

0

20

10

~-=Ji.!!'..!!'-- - - - -- - -- - --

0

0.0375 0.0750 0.11251

0.15001

0.18751

0.22501

0.26251

0.3000

Time. s

Fig. 5.31 : Dynamic response of retaining wall

Time Translational Rotation Disp. of wall at top Total disp. at topt(s) displacement x (mm) e (rad) due to rotation xB (mm) xla/mm)

0.15 0 0 0 00.1875 1.28 0.02197 27.48 mm 28.760.2250 1.82 0.01552 38.89 mm 40.710.2625 1.29 0.02197 27.48 mm 28.760.3 0 0 0 0

234 - Soil Dynamics & Machitie Foundtitions

Check for Plastic Conditions -

When the displacement ()fwall (Xtop)is greater than yield displacement (Yd)' the system would beplastic. Therefore the equations of plastic system should be used. In passive case, yield displacement (Yd)-is so large that plastic conditions do not arise and elastic system is considered. In' active case, to identifythe time 'le' afterwl}ich plastic conditions- exist, a line has been drawn cof!"espondingto Yd = 6mm(Fig. 5.31).

It gives,te.=0.1631 sec '

xe = - 1.817 x 10-3 sin (20.94 le) = 4.9116 x 10-4 m. '

ae = - 2.197 x 10-2 sin (20.94 le) = 5.9388 x 10-3 rad.

Based on the analysis, - -

Z =x =4.9116 x 10--4 m\. e - -, - - ,

ay "- ae= 5.9388x 1O-3rad.

xe = 0.3663 m/s

ae = 0.4429 rad/sec

Cl = 0.3663-(117.51 x 5.9388x 10-3-510.94 x 4.9116 x 10-4) 0.1631+ 2.45 cos (20.94x 0.1631)= - 0.1489' . 20.94

C2 = 4.9116 x 10-4 -0.3663 x 0.1631+(117.51 x 5.9388x 10-3

-510.94 x 4.9116 x 10-4) x 0.16312+245 sin (20.94 x 0.1631) cos (20.94 x 0.1631)?' 2 0.1631 x 2.45- 20.94 20.94= - 0.0173

C3 = 0.4429-[ 117.512 x 4.9116 x 10-4 -451.76 x 5.9388 x 10-3]

x 0.1631= 0.86760.8556

-3 0.1631C4 = 5.9388 x 10 + (0.4247) x - 0.4429 x 0.16312

= - O.1009

Hence the governing equations for displacement become,x = 0.2234 [2 - 5.587 x 10-3sin(20.94 t) - 0.1489 t - 0.0173a = - 1.302 t2 + 0.8676 [- 0.1009

From time 0.1631 s to 0.2963.5, the computations of displacement are given in the following Table.

Table 5.7: Values of Displacements in Active Condition (Plastic-State)

Time Translational Rotatiol! Disp. of wall at top Total disp. at topt(s) displacement x (mm) e (rad) due to rotation Xo (mm) x/ (mm)(JP

0.1531 0.47 5.97026 x 10-3 10.57 11.040.1875 1.18 16.0015 x 10-3 28.32 29.500.2250 0.69 8.396 10-3 50.26 51.01

f0.2625 - 2.44 37.129 x 10-3 65.72 63.28 ,;,). " " -,t,0.2963 - 6.77 41.8624 x 10-- 74.10 67.33

)ynamic Eartl, Pressure 235

,.From t =e;2963 s to t = 03 s;the- displacements 'are computed using the expressions obtained byolving the equations of motion under elastic condition taking boundary conditions satisfying the previ-msly computed values t = 0.2963 s.

The complete solution can be expressed as :

AI. A l A t. A lXe = I sm rot + 2 cos rot + 3sm rot + 4 cos rot

I '11 Iee = BI sin ro t + B2 cos ro t + B3 sin ro t + B4 cos ro t

Superscripts of A and B indicate the mode of vibration. Therefore Constants AI' A2' BI and B2::orrespond to the mode when system is vibrating with wnl' and A3, A4' B3 and B4 for the second mode.

xe = AI sinrot+A2 cosrot+A3 sinrot+A4 cosrotA A A A

e = -1.. sin ro t + --1.. cos ro t + J sin rot + -1. cos ro te m m m m

Putting boundary conditions, we get

- 0.0067669 = - 0.0786 AI - 0.9969 A2 + 0.0786 A3 + 0.9969 A40.0418624 = 0.0763 AI - 0.9679 A2 - 0.1139A3 + 1.445 A4

- 0.133143 = 20.875 AI + 1.6459 A2 + 20.075 A3 + 1.6459 A40.09603 = - 20.267 AI + 1.5979 A2 + 29.094 A3 + 2.385 A4

AI = - 0.00376A2 = - 0.02171A3 = - 0.00187

. A4 = 0.0144~ ,'"

Therefore for range of tp from 0.2963 to 0.3 s the equations of displacements will be :xe = - 0.00376 sin ro t - 0.02171 cos rot - 0.00187 sin ro t + 0.01448 cos ro tee = 0.003'65 sin rot + 0.02107 cos ro t - 0.002710 sinro t + 0.02099 cos ro t

, Values of displacements in ~lastic condition from time ,0.2963 s to 0.3 s are given in Table 5.8

Table 5.8 : Values of Displacements in Elastic State

Total disp. at topXI (mm)op

67.33

67.34

67.33

67.29

67.21, .

Hence it is fo~rid that after 'one cycle, the disphi~ement ~sequal to 67.21 ,mmand it can be called asslip. The total displacement after n cycles is e'qual'to n times the slip. The 'final translational displace-ment and r~tation of the retaining wall are therefore known. The displacement curve in palstic state isalso shown in Fig. 5.31.

Time TraJ;slational Rotation Disp. of wall at topt(s) displacement x (mm) e (rad) due to rotation xe (mm)

0.2963 - 6.77 41. 8624 x 10-3 74.10

0.2975 - 6.85 41.916 x 10-3 74.191. ,0.298 - 6.98 41.98359 x 10-3 74.3109

0.299 - 7.10 42.0320x 10-3 74.39670

42.062 x 10-3,,'

0.3 - 7.23 74.4498

.:.'..' ;'r:;,;"i:;":.~'d~~.'; .:;..,.-/:{;i;' ..,.,..,rtif~;~;8t."'



Biggs, J. M. (1963), "Introduction to structural dynamics", McGraw Hill Book Co., New York.Coulomb, C. A. (1776), "Essai sur une application des regles des maximis et minimis a quelque problems de

statique relalifs a I'architecture", Mem. acad. roy. press. disversavants, vol 7, Paris.Culmann, K. , (1866), "Die graphische statik", Zurich.

IS 1893 : I. P. (1962), "Earthquake resistant design of retaining walls", Proc. Symposium in Earthquake Engineer-ing, Universityof Roorkee, Roorkee.

Lai, C. S. (1979), "Behaviour of retaining walls under seismic loading", M. E. Report, University of Canterbury,New Zealand.

Mononobe, N. (1929), "Earthquake proof construction of masonry dam", Proceedings. World EngineeringCongress,vol. 9, p. 275.

Nadim, F., and Whitman R. V. (1983), "Seismically induced movement of retaining walls", Jour. of Geot. Engg.DiYn.,ASCE, Vol. 109, No. 7, pp. 915-931.

Nandakumaran, P. (1973), "Behaviour of retaining walls under dynamic loads", Ph.D. Thesis, Universityof Roorkee,Roorkee.

Newmark, N.M. (1965), "Effect of earthquakes on dams and embankments", Geotechnique, Vol. 15, No. 2, pp. 129-160.

Ohde, S. 91926), "General theory of earth pressures", Journal, Japanese Society of Civil Engineers, Tokyo, Japan,Vol. 12, No. I.

Prakash, S., and Saran, S. (1966), "Static and dynamic earth pressures behind retaining walls," Proc. 3rd Sympo-sium 011Earthquake Engineering, University of Roorkee, Roorkee, Vol. 1, pp. 277-288.

Prakash, S., Puri, V. K. and Khandoker J. U. (1981), "Rockingdisplacements of retainingwalls during earthquakes",Int. Conf. on Recent Advances in Geotcchanical EarthquilkeEngineering and Soil Dynamics, Vol. 3,St. Louis, U.S.A.

Prakash S. (1981), "Analysis of rigid retaining wallsduring earthquakes", Int. Conf. on RecentAdvances inGeotech.Earthquake Engg. and Soil Dynamics, Vol. 3, St. Louis U.S.A.

Reddy, R. K., Saran.,S., and Viladkar, M.N. (1985), "Prediction ofdisplacements of retaining walls under dynamicconditiOns", Bull. of Indian Soc. Earth. Tech., Paper No.-239, vol. 22, No. 3.

Richard, R. k, and Elms, D. G. (1979), "Seismic behaviour of gravity retaining walls", Journ. Geotech. Engg.Divn., ASCE, Vol. 105, No. GT4, pp. 449-464.

Saran, S., and Prakash, S. (1968), "Dimensionless parameters for static and dynamic earth pressures behind retain-ing walls", Jour. Indian National Society of Soil Mech. and Found. Engg., July pp. 295-310.

Seed, H. B., and Whitman, R. V. (1970), "Design of earth retaining structures tor dynamic loads", ASCE Specialityconference on Lateral Stresses in Ground and Design of Earth Retaining Structures, pp. 103-147,Ithaca, New York.

Zarrabi, k. (1979), "Sliding of gravity retaining wall during earthquakes considering vertical accelera-tion and changing inclination of failure furface", M. S. Thesis, MIT, USA. <.,"

:.~.jJ

4

;;:\r£;;:;i,~;~~.;[;:J~1~;t!fi ,;~:~~-~ <:v~?';;:;~\.;':', i~'w'~~f~il!¥~~ j' .: . . . ,~>.-"

Dynamic Eartlt Pressure 237

PRACTICE PROBLEMS

5.1 Explain with neat sketches the following:(a) Mononobe- Okabe's approach, and(b) Modified Culmann's graphical construction for getting dynamic active earth pressure.

5.2 How is the effect of partly submerged backfill considered in computing dynamic earth pres-sure?

.5.3 Explain the salient features of the following:

(a) Richard-Elms model(b) Nadim-Whitman model

(c) Reddy-Saran-viladkar modelfor getting displacement of rigid retaining wall.

5.4 A vertical retaining wall is 8m high and retains noncohesive backfill with y = 18 kN/m3, 4> = 30°8 = 20° .The backfill is inclined to the horizontal by 15°.The wall is located in a seismic areawhere the design seismic coefficients are

ah = 0.10 ; al, = 0.05Compute the static and dynamic earth pressure on the wall using both modified Coulomb's ap-proach, and Culmann's graphical construction.

5.5 If the retaining wall (Problem 504)is to incline at lO°with the vertical, would you recommendits inclination towards or away from the fill. Justify your answer fully.

5.6 The backfill of retaining wall (Problem 5.4) is carrying a surcharge of 50 kN/rn2. Estimate theincrease in static and dyn~mic earth pressures.

5.7 The backfill of retaining wall (Problem 5.4) is submerged upto 4.0 m from the base of wall.Estimate the total pressure on wall both in static and dynamic cases.

5.8 Compute the displacement of the wall (problem SA) f~r the following condition:Period of wall = 0.25 s

Zy = 5.0 mmPeriod of ground motion = 0040s

Equivalent number of cycles in an earthquake o~magnitude 7.0 will not exceed 15.

DD

t

DYNAMIC BEARING CAPACITY OFSHALLOW FOUNDATIONS.

6.1 GENERAL

Foundations may be subjected to dynamic loads due to earthquakes, bomb blasts and operations ofmachines. The dynamic loads due to nuclear blasts are mainly vertical. Horizontal dynamic loads onfoundations are mostly due to earthquakes. Basically there are two types of approaches namely (i)pseudo-static analysis and (ii) dynamic analysis for getting the solution. In this chapter, pseudo-staticanalysis is first presented and it is followed by dynamic analysis. Design of foundations of differenttypes of machines have been given in detail in chapters 8 to 10.

6.2 PSEUDO-STATIC ANALYSIS

Pseudo-static analysis is more commonly used for designing foundations subjected to earthquake forces.Adopting appropriate values of horizontal and vertical seismic coefficients, equivalent seismic forcescan be conveniently evaluated. These forces in combination of static forces make the foundationsubjected to eccentric inclined load. In Secs. 6.3 and 6.4, the procedure of detennining bearing capacity.s.:ttlement, tilt and horizontal displacement of shallow foundations subjected to eccentric - inclined loadsha\'e been prL'scnted. It is preceded by brief description on fundamental concepts involved in bearingcapacity analysis.

6.3 BEARING CAPACITY OF FOOTINGS

6.3.1. Modes of Shear Failure. The maximum load per unit area that can be imposed on a footingwithout causing rupture of soil is its bearing capacity (some times termed critical or ultimate bearingcapacity). It is usually denoted by quoThis load may be obtained by carrying out a load test on thefooting which will give a curve between average load per unit area and settlement of the footing. Basedon pressure-settlement characteristics of a footing and pattern of shearing zones, three modes of shearfailure have been identified as (i) general shear failure, (ii) punching shear failure and (iii) local shearfailure (Caquot, 1934; Terzaghi, 1943; DeBeer and Vesic, 1958; Vesic, 1973).

In general shear failure, well defined slip lines extend from the edge of the footing to the adjacentground. Abrupt failure is indicated by the pressure-settlement curve (Fig. 6.1a). Usually in this type,failure is sudden and catastrophic and bulging of adjacent ground occurs. This type of failure occursin soils having brittle type stress-strain behaviour (e.g. dense sand and stiff clays).

,..,:",,;,,"d:"~'tX,~,'.","",r,d.:"Y""::'-" "J:'.~'f.,,'"~~:"~:" w,::":'~.' ",r;1',;;~"f"':"", "",'" "'" "",.0, ':"':'<"",," ,>,'[:";':""'0, ..,.""...;~.;t~A",. ,,';"~',:':},,~~, -Dynamic Bearillg Capacity of Shallow Foulldatiolls 239

In punching shear failure, there is vertical shear around the footing perimeter and compressionof soil immediately under the footing, with soil on the sides of the footing remaining practicallyuninvolved, The pressure-settlement curve indicates a continuous increase in settlement with increasingload (Fig, 6,1 b). .

Fig. 6.1 : Typical modes of failure (a) General shear, (b) Punching shear and (c) Local shear

The local shear failure is an intermediate failure mode and has some of the characteristics of boththe general shear and punching shear failure modes. Well defined slip lines immediately below thefooting extend only a short distance into the soil mass. The pressure-settlement curve does not indicatethe bearing capacity clearly (Fig. 6.1 c). This type of failure occurs in soils having plastic stress-stramcharacteristics (e.g. loose sand and soft clay).

In Fig. 6.2, types of' failure modes that can be expected for a footing in ~ particular type of sandis illustrated (Vesic, 1973). This figure indicates that the type of failure depends on the relative densityand depth-width ratio (D/B) of the footing. There is a critical value of (D/B) ratio below which onlypunching shear failure occurs.

+- - Loadc

(a)

<::.I

. EI

<::.I-+-+-<::.I.-: Load

(b) c<::.I. "E

,"

'<::.I

+-+-<::.I

lI'I

I

.... .,' Load+-

/ C

(c)

", <::.I

--- ,,-...." E<::.I+--<::.IIf)

, 240 Soil Dynamics & Machine Foundations

00

Relative density, Or0.2 0-4 0.6 0.8 1.0

Pun chingshearfai lu rezone

Gen eralsh earfailur(zont

10

Fig. 6.2 : Region for three different modes of failure

The criteria given in Table 6.1 may also be followed for identification of type of failure:

Table 6.1 : Identification of Type of Failure

6.3.2. Generalized Bearing Capacity Equation. In the design of foundation usually net bearingcapacity is computed' and used. It is defined as the maximum net intensity of loading at the base ofthe foundation that the soil can support before failing in shear. It is denoted by,"qnu'Therefore

qnu=qu-YIDf ...(6.1)where, qu = Ultimate bearing capacity

The equation of net bearing capacity developed for strip footing considering general shear failure(Terzaghi, 1943; Meyerhof, 1951) is extended to consider variations from the basic assumptions byapplying modification factors that account for the effect of each variation (Hansen, 1970). It may bewritten as :

1"q =cN.S.d'i.b + y .D . (N -1 ) óÍò¼ù·ò¾ò® +-'Y 2 'B.N y .S y .d .i y .b y .r'

nu c c c eel f q q q q q w 2 y w

...(6.2)

elcc..

.J::...a. 5'"0

>.--0

!'t:

Type of failure Relative density Dr (%) (Deg) Void ratio e

1. General shear failure 70 36° S 0.55'2. Local shear failure or punching shear failure S 20 S 29° 0.75

Dynamic Bearing Capacity of Shallow FoundatioJIs 241

where

qllu = Net ultimate bearing capacityc = Undrained cohension of soil

B = Width of footing

Df = Depth of foundation below ground surfaceNc- Nq, Ny = Bearing capacity factors

Sc' Sq, Sy = Shape factors for square, rectangular and circular foundationsdc' dq, dy = Depth factors

ie' iq, iy = Inclination factorsbc' bq' by = Ground inclination factors

rw' r:v = Ground water table factors

6.3.2.1. Bearing capacity factors. Nc' Nq and Ny are non-dimensional factors which depend on angleof shearing resistance of soil (Terzaghi, 1943; Terzaghi and Peck, 1967). Their values may be obtainedfrom Table 6.2. "

6.3.2.2. Shape factors. Approximate values of shape factors which are sufficiently accurate for mostpractical purposes are given in Table 6.3.

Table 6.3 : Shape Factors

Table 6.2 : Bearing Capacity Factors

cp Ne Hq HyDeg

0 5.14 \.00 0.005 6.49 .; 1.57 0.45

10 8.35.'

2.47 1.22

15 10.98 3.94 2.6520 14.83 6.40 5.3925 20.72 10.66 10.88

30 30.14 18.40 22.4035 46.12 33.30 48.03

40 75.31 64.20 109.4145 138.88 134.88 27 I. 76

50 266.89 319.07 762.89

S.No. Shape of footing Se Sq Sy

(i) Continuous strip 1.00 1.00 1.00

(ii) Rectangle 1+ 0.2 B/L 1+ 0.2 B/L 1-0.4 B/L

(Hi) Square 1.3 1.2 0.8

(iv) Circle 1.3 1.2 0.6

(B = diameter)

242 Soil Dynamics- & Machine Foundations

6.3.2.3. Depth factors. The bearing capacity factors given in Table 6.2 does not consider the shearingresistance of the failure plane passing through the soil zone above the level of the foundation base.If this upper 'soil zone possess significant shearing strength, the ultimate value of bearing capacity wouldbe increased (Meyerhof, 1951). For this case, depth factors are applied, whereby

Dd = 1 + 0.4 --L

C B ...(6.3)

2 DJd = 1 + 2 tan <1>(1 - sin <1» . (6.4)q , B

dy = 1 ...(6.5)The use of depth factors is conditional upon the soil above foundation level being not significantly

inferior in shear strength characteristics to that below this level.

6.3.2.4 Factors for eccentric-inclined loads. The effect of eccentricity can be conveniently and con-servatively considered as follows:

One way eccentricity (Fig. 6.3 a) - If the load has an eccentricity e, with respect to the centroidof the foundation in only one direction, then the dimension of the footing in the direction of eccentricityshall be reduced by a length equal to 2e. The modified dimension shall be used in the bearing capacityequation and in determining the effective area of the footing in resisting the load.

Two way ecc?ntricity (Fig. 6.3 b) - If the load has double eccentricity (eL and eB) with respectto the centroid of the footing then the effective dimensions of the footing to be used in determiningthe bearing capacity as~ell as in computing the effective area of the footing in resisting the load shallbe determined as given below:

L' = L - 2 eL

B' = B-2 eB

...(6.6)

...( 6.7)

...(6.8)A' = Lt X B'

~I

I

, I

Q h,t"'.

,\7 '«'\~1\Z2?r" .,,:~..,.,. t ." ""':"""-"""';"""

," :":';:'::"""':""~." " .., , .W'«)".'. : ::.::~

J.. B or L ~

QvG.

e

Fig. 6.3 : Eccentrically-Obliquely loaded footings (a) One way eccentricity

a- :;!:f

Dynamic Bearing Capacity of Shallow Foundations 243

~ for B- axis........

Fig. 6.3 : Eccentrically-Obliquely loaded footings (b) Two way eccentricity

In computing the shape- and depth factors for eccentrically-obliquely loaded footings, effective width

(B') and e_ffe~tive length__(~? will be used in place of total width (B) and total length (L).For a design,;'eccOentricity' should be limited to one-sixth of the foundation dimension to prevent

the condition of uplift occurring under part of the foundation,

Inclination factors - Following inclination factors may be adopted in design.

[ ]

m+l

, - 1 Qh"ly - - Q + BLc cot '

...(6.9)

.

[Q

]

m

Iq = 1- hQ + BLc cot 

V iq - [N:~~$] (when p O.

= 1 mQh- cNcBL (when = 0°)

...(6.10)

...(6.11)

...(6.12)

where Qh is the horizontal component of the load Q acting on the foundation at inclination i withvertical. Values of m are taken as given below:

(i) If the angle of inclination i is in the plane of L-axis

111=(2+~) (1+~) ..(6.13)

(ii) If the angle of inclination is in the plane of the B-axis

~, y m'= (2+oB

) (1+B

) ,. ~',./. L L- -0" ,'" .' " ,.'

As P~!)~,':640].:)981,:tlfernclination factors are give,n"by-~~:-~:-,:::--".-.- - ------ -,-- - ,--,~.- -,-' -- - ,.--. -' -

...(6,14)

-""...'

.) 'q~cq: ~' . '; ..'

2440ID-

..0'-"0Lt)

00

0~

11

0~ .-

ColV

Z

0N

0

0~

000

~0N

0.-(6~p)q,

00.--

d'-"

0co

°0 0ID11

ColV

Z0~

0N

0~

00

0'0~

0N

( 6~p) <t>

'..~


00M

11

0t

0~ 0

N 0

(6~p) cp

00N

11

0t

0N

00~

(6~p) et>

-1J 0

~'-"

0 .-N tII

VZ

0.-

00

-&-."."...'""'E

7

""':-.::;;1

.....

- 0~u

0M

0 t>IN V

Z

0

00

.. I ~. "}:Jl

Dynamic Bearing Capacity of Shallow Foundations01.0

'".0.........

0~

- o.et) ~~ 0

N -0,'.

(6<>p) p) et>

00

1-tI I '" i"0

.........0 tIJC') eT

Z 100C')

1 °gtI0 1\ ,-N

Z

0 to... '\.u. 18r.

10"0 II0 0 0 0 0 0 0 0 0 0 er..t C') N .... M N "" '"::'"(60p) p) cp

'-'"',.7...

'.Ir:-.::>O£

'.

'" 1l

0

I '"u

.........

00

1 i11

00N.-

\0 tIJ 11

10 tIJeT ,- N eTZ

Z

246

"........c ,

°0JI

0t

I.t'IM

0M

00N

I.t'IN

(6ap)cp

".......0'-"

°0JI

::>.t

0M

0N

0

øê¿°÷ et>

0N

000

0-.t

00

0U)

00


_°"tJ ,

°.0

M

"C:.I')0

Z

0...t

I.t'IM

( 6ap ) et>

0N

C:.I)0z

0co

0...t

°0NJI

b-..t

0M

'1f)M

1f)N

(6ap) et>

..."".

0-..t

0M

0 ,-"N C:.I

)0Z

0

00M

~'":::I~...;>.~

7.

'c:~~

'.

".......0U)

u ,

0...:t

0N

00N

ID ,< .<' -,' ,,' ,,(i' 'f" ';- ,.;.r"','.,,171.

J)ymimii: 'Bearing 'CapJCity"of'Sh~llow Foundations "247---

,

(

"

)2

ic = iq = 1 ~;0

iy= (1- ~J

, ::': '( ,t,

, ":..(6.15)

...(6.16)

For more accurate estimation of bearing capacity of a eccentrically-obliquely loaded footing, bealingcapacity fa~tor:; as shown ,in Figs. 6.4 to 6.6 developed by Saran and Agarwal (1991) may ,b~,u$ed.These factors have been obtflined by carrying out a theoretical analysis based on limit equilibrium andlimit analysis approaches'. ~":":

As evident from these figures, bearing capacity factors (Nyei'Nqei and Ncei) depend on <j),i ande/B. Values of these bearing capacity factors are substituted in (Eq. 6.2) in place of Nc' Nq and NI forgetting the bearing capacity of eccentrically-obliquely loaded footing. If use of these bearing capacity,factors is made, then inclination factors, and reduced dimensions of B' and L' for accounting the effectof eccentricity and inclination are not us,ed.The!efore the mod~fiedbe~ring capacity equation will beas given below: '

~, 1, -.qnu = cNeei . Se . de + YiD! (Nqei -1) Sq . dq . 'rw + 2"'Y2 . B '.~'Yei. S1' dy' r'w

Use of Eqs. 6.6 to 6.16 make more'cons~rv~tiveqestimate of~he bearing capacity.,

...(6.17)

6.3.2.5. Base inclination factors. If the base of foundation is inclined from the horizontal and an appliedload acts nonnal to the base (Fig. 6.7), the pattern of rupture surface beneath the foundation will bedifferent from the pattern that develop beneath the level footing carrying a vertical load (Meyerhof,1953). For this condition base inclination factors as given below may be used:. , '"

, (l-b) ,

b =' b - ,q fer tP > 0°,c q N tantPe '

= 1- 0.0067 a for 4>= 0°,

)

2b = b = (1 - ~ tan'" " "

.q 1 , 57.3 If , .'['

where a represents the angle of the base inclination in degree with horiz~ilta1.. '

...(6.18)

...(6,19)

:;' ...(6.20)

~

"

_'0 "

rJ ,.,c Fi~.6.7 : Inclined, footing ',," ",~s

248

,:

Soil Dynamics & Machine Foundatitnu,:..". "~

6.3.2.6. Water table factors. Correction factors rw and r'w may be computed using following Eqs.(Fig. 6.~). '

darw = 1.0 - 0.5 D (For da ~ DJ ...(6.21):'f Y "

d' .r~ = 0.5 + 0.5 ~ (For, db ,~ B) . ...(6.22},

where do and db represent the position of water table with respect to the base of the footing as shown:in fig. 6.8. For the position of water level 'below the base of footing, do = 0 i.e. rw = 1; and for theposition of water level at depth more than B, db = B i.e. r'll' = 1. '

"'"B

Footi ng

~ 1.0L.L." 0.90..u.B 0.8c0 0.7..u::J'0 0.6tICl:

0.50 o. Z O. 4 0.6 0.8

do/Of

(b)

,~L.

..~," 'f

1.0~

.:: 0.90..u0 0.8Of-

?

c:.~ 0.7..u::J

'0 0.6tIa:

j-.""..f

0.4 0.6 0.8¼¾ñÞ

1.0

(c)

..

,J..I,. ,

.-,,"~

Fig. 6.8 : Correction factors 'cor position of water table

;'~,,;'~,0";;.~',:'

vwot.rzr IrzveltOf '

do

-t- I I water 'dbI I (rzv<z1

LB I I \!L L______J(0)

~ ~-_.- J,~II1I~-., -- ID ~,,' <;~"",.;.

Iynamic Bearing Capacity of Shallow Foundations 249

).3.3. Local and Punching Shear Failure. The assumption that the soil behaves as a rigid materials satisfied for the case of general shear but is not appropriate for punching and local shear. Comparison)f the relative pressure-settlement curves (Fig. 6.1) indicates that, for punching and local shear failure;ases, the ultimate pressure is less and the settlement is greater than for the condition of general shear:'ailure.For design purposes, the general shear, local shear and punching shear failures can be identified1Sper the criterion given in Table 6.1.

Terzaghi (1943) proposed empirical adjustments to shear strength parameters c and <{Ito cover thecase of local and punching shear failure. Shear strength parameters cm and <{Imshould be used in thebearing capacity equation and the bearing capacity factors are obtained on'the basis of <Pminstead of:j>, where'

2c = - C

III 3

<{Im = tan -1 (~ tan ~)

...( 6.23 a)

...(6.23 h)

If the failure lies between general shear and local shear failure, then linear interpolation is done

to evaluate the value of bearing capacity factors. For example, value of Ny for ~ = 34° will be

(N ) (N ) " ,

'Y 4>=36° ym CP=29°

(NY)4>= 34°= (Nym )$ = 29° + 36° - 29° x (36° - 34° ) ...(6.24 )

where (Nyrn)$ = 29° value of Ny factor for <{I::;:29° considering local shear" failure condition. Thereforeits value will be obtained using Table 6.2 for <1>m= tan -1 [(2/3) tan 29°] = 20.29°.

, .

6.3.4. Factor of Safety. The net bearing capacity of the soil is divided by a safety factor to obtain thenet safe bearing capacity. It is denoted by qnF' ' .

A factor of safety is used as a safeguard against

(i) natural variations in shear strength of soil,

(ii) assumptions made in theoretical methods,

(hi) inaccuracies of empirical methods and

(iv) excessive settlement of footings near shear failure.

A factor of safety of 2.5 to 3.0 is generally used to cover the variation or uncertainties listed above.Therefore

qnF = qnuF

6.4 SETTLEMENT, TILT AND HORIZONTAL DISPLACEMENT

...(6.25)

An eccentrically-obliquely loaded rigid footing settles as shown in Fig. 6.9 in which Se and Sm representrespectively the settlement of the point under load and edge of the footing. If' t' is the tilt of the footing,then Srn is given by : '

Srn = Se + (BI2 - e ) sin tIn Fig. 6.9, Ho represents the horizontal displacement of ' the footing.

...(6.26)

---....----

"

,,250 Soil Dynamics &, Machine Foundations

{1:~,

e!---.' , " " ,~ I

" -.","'IQd(Zi,

I., . (e/2-~) i I ~ e/2 -1l_.l.

I I ' ' -E

I Se 5,0 "S ..--rnl

, l-:HD~ ' " - --- EFinal position' I ~) ~

~~el1.-Fig,6.9: Settlement, tilt and horizontal displacement of eccentrically-obliquely loaded footing

Agarwal (1986) carried out modeltests on eccentrically-obliquely loaded footings resting on sandFootings of different widths and shapes were used. In each test, for a pressure increment observationswere taken to record Se' t and Ho' Effect of relative density of sand was also studied. In addition tothese tests, pressure.settlement and pressure-tilt characteristics of eccentrically-obliquely loaded foot-ings resting on clay and Sand beds were also obtained using non-linear constitutive laws of soils (Saran& Agarwal, 1989). From the model test data and results of analysis based on constitutive laws, plotsof SeI So versus e I Band S ISo versus elB were prepared for different load inclinations (Figs. 6.10. m& 6.11). So represents the settlement of the footing subjected to central vertical load (i.e. elB = 0 =i) and obtained corresponding to the pressure intensity giving the same factor of safely at which Seand Sill values are taken.

These plots were found independent to the type of soil, factor of safety, size and shape of footingThe average relationships can be represented by the following expressions:

~e = Ao + AI (~)+A2(~r0

(,

- ~:=Bo + BI ~)where.A" ~ 1 - 0.56 (~ )-0,82 (~)'

A, ~.. 3.51 + 147 (~ )+5,67( ~)'

A, ~ 4,74 - \.38 (~ )-lZ,4S( ~r(

,

) (,

)

2I ',-

Bo = I - 0.48 ~,...~.~;. q, ",...', '

...(6.27

...(6.28

,..(6,29

...(6.3(

..:(6.31

-" .

,/""

,..(63

i_h,"""'"

ilntlmic Bearing Capacity of Shallow Foundations 251

B\ ~ ~ 1.8~ + 0,94 W- L63( ~)' ...(6.33)

0.8

0-4

i ::150

0.6 ..SeSo

.

0.2

0 ~-

0 0.1 0.2 '.,

ù´×Þ

0.3

Fig. 6.10 : 8/80 versus e/B for j = 15°

1.2

, 0I = 15

0.8

0.4

Srn-So

00 0.1 0.2

e/B0.3

, . ',; " < :., ,{,;;-~Fig.6.11 : Sm/SOversus elB for i-15° 'Co):, ..


Values of So can be obtained using the data of plate load test or standard penetration test incohesionless soils, and consolidation test data in clays in conventional manner.

Similarly a unique correlation was obtained between Ho / B and i/~.Ho i i 2 i 3 i 4B =0.121 (~)-0.682(~) +1.99(~) +2.01(~) ...(6.34)

The above correlation is also found independent to the type of soil, factor of safety, size and shapeof the footing. The effect of eIB was found small and the displacement value decreased little with theincrease in eccentricity: This effect is neglected considering the results slightly on the safe side.

6.5 DYNAMIC ANALYSIS

. The dynamic bearing capacity problem attracted attention of the investigators in 1960 when theperformance of foundations under transient loads became of concern to the engineering profession(Wallace, 1961; Cunny and Sloan, 1961; Fisher, 1962; Johnson and Ireland, 1963; Mckee and shenkman.1962: White, 1964; Chummar, 1965; Triandafilidis, 1965). All analytical approaches are based on theassumption that soil rupture under transient loads occurs along a static rupture surface. In this sectionthe sailent features of th~ analysis developed by Triandafilidis (1965) and Wallace (!961) for transientvertical load; and by Chummar (1965) for transient horizontal load have been presented.

~ ~ (enter of rotationI (Fczllllnius)

8 -Tqu I 0.43 8 .

: -ZPrandtl's rupture surfacll

~-r, \ C Ilntllr of rotation

~ (Fllllllnius)B ---rq I', r =2.20SB/rr -u I , - --7, - -L- ---

w WCO5~

Fig. 6.12 : Illustrations of mode of failure, and dynamic equilibrium of moving soil mass

-",.-""

')ynamic Bearing Capacity of Shallow Foundations 253

5.5.1. TriandafiIidis's Solution. Triand~rllidis (1965) has presented a solution for dynamic responseof continuous surface footing supporting by saturated cohesive soil (I\>= 0 condition) and subjec,ted tovertical transient load. The analysis is based on the following assumptions:

(i) The failure surface of soil is cylinderical for evaluation of bearing capacity under staticcondition (Fig. 6.12) .

(ii) The saturated cohesive soil (I\>= 0) behaves as a rtid plastic material (Fig. -6.13).

(iii) The forcing function is assumed to be an exponentially dec~ying pulse (Fig. 6. 14)(iv) The influence of strain rate on the shear strength is neglected.

(v) The dead weight of the foundation is neglected.

U\U\C:II~..U\

"0er

CJ\CJ\ I.C:II.....t/) v

E0c>-0

, ,'.

strain rimaFig. 6.13 : Assumed stress-strain relationship Fig. 6.14: Transient vertical load

Analysis

Let the transient stress pulse be expressed in the formq = q e-~ I = A. q e-~ Idoli ...(6.35)

where, qd = Stress at time t~ = Decaying function

qu = Static bearing capacity of continuous footingqo = Instantaneous peak intensity of the stress pulse

A. = Over load factor = qo. . qu ".

. The rupture surface is shown in Fig. 6.12 with centre of rotation at point 0 loc~tedat a heightof 0.43 B above the ground surface. The equation of motion is written by equating the moment of thedisturbing and restoring forces taken about the point O. The only disturbing and restoring force is anexternally applied dynamic pulse. The restoring forces consist of shearing resistance along the rupturesurface, the inertia of the soil :rp~sspa~icipating in motion and the resistance caused by the ~isplacement

111&1::,

254r. , Soil Dynamics ~ Machine Foundations

of c~ntre of gravity of soil mass.

Driving moment Md due to applied dynamic pulse is. P- I . 2

Mdp = "2qd B ...(6.36)where, B= Width of footing

The static bearing capacity of a continuous footing along the failure surface (Fellenius, 1948) is

qu = 5.54 Cu ...(6.37 a)where, Cu= Undrained shear strength

Resisting moment Mrs due to shear strength is

Mrs = ; qu B2 ...(6.37 b)An applied pulse imparts an acceleration to the soil mass. The resisting moment Mri due to the

rigid body motion of the failed soil mass is

Mri = JoeJo = .Polarmass moment of inertia

WB2--1.36 g

W = Weight of the cylinderical soil mass= 0.31 Y1tB2

Y = Unit weight of soil2WB ..

Mri = 1.36,geThe displaced position of the soil mass generates a restoring moment Mrw ' which may be

expressed as

...(6.38)where,

...(6.39)

...(6.40)

Therefore, ...(6.41)

M = W r sin erwM =Wrerw

2.205 Br =- 1tBy equating the moments of driving forces to those of the restoring forces, the following equation

of motion is obtained.

where,

...(6.42 a)

...(6.42 b)

...(6.43)

For small rotations,

Mdp = Mrs + Mri + MrwSub'5tituting for moments and rearranging, we get

...(6.44)

.. 3g[

O.68g] [, -~t,

1]e +-e= q ""e -1tB W u ...(6.45)

Equation (6.45) is a second order, nonhomogeneous, linear differential equation with constantcoefficients. The natural frequency and the time period of the system are given by

co. ~ ~;~

T ~ 2~ ~ ~:

...(6.46)

.,.(6.47)

.. ~.JI,,~'{!:

DynU'tJic Bear;n~ C9pacity of S~lllloHJ.Fpundatiolls 255

Solution of Eq. (6.45) gives the following relation

W 'T2

[{~2T2

} (21tt) ~).T . (2~t\ -f}t ,~2T2

](8) == 1-).+- cos - +-sm - J +).e óóóï ...(6.48)

0.68gqu' 41t2+~2T2 - .41t2 T 21t' T 41i2 -,

The ab<?vere1ation can be used to trace the history of m~tion of the foundation. For determination ofthe maximum angular deflection 8, Eq. (6.48) can be differentiated with respect to time. Thus

~(e) ~ 21tT[{

A-I- ~2T2}

Sin{

21tt}

+ ~ATcos{

21tt}

- ~)'T e-J3t]0.68gqu 4~2+~2r 41t2, T 21t T 21t

For obtaining the critical time t = te which corresponds to 8 = 8max' the right-hand side ofEq. (6.49)is equated to zero. Since 2 1tT/ (41t2+ ~2 T2) :t=0

['\ 1 ~2T2] . [21tt] ~1tT .[21t

.

t] ~).T -J3t ~ 0/\,- -- SIn - +-cos - --e -41t2 T T T 21t ,'h. ,

By using small increments of time t in Eq. (6.50), the value of le can be obtained. This value oft = le can 'then be substituted in to Eq. (6.48) with known values of ~, ). and B' to obtain(W/0.68 g qll )8max = K, dynamic load factor. Figures 6.15, 6.16 a~~ 6.i7 give the values of K (s2)for B = 0.6, 1.5 and 3.0m, respectively, with), = 1-5 and ~= 0-50 s-1 '

10

...(6.49)

...(6.50)

~

-I10

~-------- - - - - - -::::.--- ; --- -- ... -------'" ---...-

",""""'" .- --,,/ ~~ . .-'-'/ ,, /- .. ---,-- "

, ~ '. /,/

' /' -- " -"f I /' -''-I, / /' "/"'"

f,' , /"

1/", /

/ .'.// . /<'

, 1

I .' /,f

..

.f

, ---,.,"

/",

-N

V\

..I-0 -2+0 10u0....

'000

-310

u -4- 10E0c

,>. -50 '10

-610

......0 0.0+0 "

"u- 0.5 --------oT 1.0 -----en 2.0 ---->.-OQ1. 5,0 -.-.-u 10.0 ~., ;...g 50.0 ~

B = O.Gm

-7' .10

1.0 2.0 3.0 4,0 5.0Ov(Z,rload ratio»" F'

.) "0'- " ---,-"" .. _. "

Fig. 6.15 : Relationship between overload ratio and dynamic load factor for continuous footings 0.6 m wide,. , " 1..1 .,,;,i

--- '..u. ~~,~-~~ -

256

10

-I10

NIII...,

----;::.::::'::===:.-,' -" - -------","'" " -------'" ..," ' .

-' -' ,",.' -,,'"/ ' /,' ,---' -"

I ;' , """ ..-"I ' /' /"

I 'I /' """,

/{/ /../" ./'/ --"

I,' ,/."'/ "

f: ,//./

/

-2~ 10.....0

t -12 10-000 -10- 10u - S'=1.Sm

0 0.0t 0.5--- ----2:- 1.0----->0.1111 2.0----0"" 5.0-,-,-u CQ, 10,0 - ., -050.0-"'-

e0

~ 1050

,

/

-,10-710

1.0 2.0Qverlood

Fig. 6.16 : Relationship between overload ratio anddynamic load factor for continuous

footings 1.5 m wide

Soil Dynamics & Machine Foundations'

10

,...N -I

III 10~

i 102u000-

-- - - ;:;.::;-:..-;::.::;=~-/-: ------.. .. ------,'."'" -' - ,-""

"/- /- ..-',,'/ .,' /,/' " ---",

;"/ /./. ../.....-'/' ,,/

I // "'-".' .......

;/ ..- /r .'

f' ./ i B = 3.0m" +- 0.0/ ~ 0.5---00-1J.O- ---->o..!!!.2,0 --=---

0 CQ,5.0 - ---

/ I ~ 10.0 ~_.-0 50.0 - ..-,..-

-3'0 1000

-/0..~ 10E0cl; 105

-c10

1071.0

Fig. 6,17 : Relationship between overload ratio anddynamic load factor for continuous

footings 3.0 m wide

6,5,2, Wallace's Solution, Analysis presented by Trianadafilidis (1965) is based on rotational modeof failure. However, it is possible that a foundation may fail by vertically punching into the soil massdue to the application of vertical transient load. Wallace (1961) presented a procedure for the estimationof the vertical displacement of continuous footing considering punching mode of failure. The analysisis based on the following assumptions:

(i) The failure surface in the soil mass is assumed to be of similar type as suggested by Terzaghi(1943) for the evaluation of static bearing capacity of strip footings. This is shown in Fig. 6.18.

~B

. 9 ton ~r: ro t <'

Fig, 6,18 : Failure surface

(ii) The soil behaves as a rigid plastic material (Fig. 6. 13)(iii) The ultimate shear strength is given by

s=c+crtan<l>where, s = Ultimate shear strength

c = Cohesion

...(6.51)

(j = Normal stress

<p= Angle of internal friction

(iv) The dynamic load applied to the footing is initially peak triangular force pulse (Fig. 6.19).

(v) The footing is assumed to be weightless and to impart uniform load to the soil surface.

-Peak intensity, q "t:!.....

CT\c:,-

"U(10..J

c:0

>-.-'"c:(y..-c:

..-(1....:J

-0-0(10..J

'.,

Tim e

Fig. 6.19 : Loading function

Analysis

The applied load is assumed to be an initial-peak triangular force which decays to zero at timetd (Fig. 6.19). The peak load q is expressed in pressure units. Since the function is discontinuousat time td' two equations are necessary

For 0 ,; t ,; td' Loading function ~ qB(1- ~ )For t ~ td' Loading function = 0In Fig. 6.18, BD is an arc of a logarithmic spiral with its centre at O. It is defined by the

Eq. (6.54).

...(6.52)

...(6.53)

r = ro eatan' ...(6.54)

where, "0 = Distance OB (Fig. 6.18)<p = Angle of internal friction

.. .!

-.-..-....... ....

--_Uu_--~--- --- _';::'77~-- --- I IT'

<

258 - Soil Dy~ic$. & ¥lIchine Foundations -

The static bearing capacity -qu for such a' failure surface is -given by

1...(6.55)¯« ã ½Ò½õ¯Ò¯õóþøÞÒ§2 -

where, c = Cohesion

q = "(DJ ...(6.56)"( = Density of soil

DJ = Depth of footingNe, Nq, and Ny = Bearing capacity factors

The bearing capacity factors depend on and K, K being 2 (Distance OA)/H, Fig. 6.18. The valueof K locates the centre of the spiral which is the centre of rotation. Obviously the correct value of Kis that which yields the minimum value of the bearing capacity. It is obtained by trial and error foreach set of problem parameters. The values of N"{'Ne and Nq for various values of -and K are givenin columns 3, 4 and 5 of Table 6.4.

Table 6.4 : Bearing Capacity Factors (N'Y'Ne, Nq' NI' NR)

(deg) K Ny Ne Nq NI' NR NI

(1) (2) (3) - (4) (5) (6) (7) (8)

0 - 0.05 0.0000 5.7277 1.0 0.0633 2.0125 5.6366

0.00 0.0000 5,7124 1.0 0.0631 1.9723 5,5887+ 0.05 0,0000 5.7258 1.0 0.0633 1.9433 5.5394

5 - 0.65 0.,1454 79.6255 7.9664 0.3755 8.9076 4.8709

- 0.60 0,1445 29.8163 3.6086 0.2280 6.4362 5.3126- 0.55 0,1481 18.9958 2,6619 0.1579 5.0332 5.6460

- 0.50 0.1553 14.3469 2.2552 0.12\3 4.1699 5.8636- 0.45 0.1655 11.8179 2.0339 0.1011 3.6088 5.9750- 0.40 0,1786 10.2699 1.8985 0.0897 3.2299 6.0020- 0.35 0.1945 9.2580 1.8100 0.0833 2.9674 5.9698

- 0.30 0.2\31 8.5723 1.7500 0.0799 ,- 2.7828 5.9005- 0.25 0.2344 8.1007 1.7087 0.0786 2.6523 5.8108- 0.20 0.2585 7.7778 1.6805 0.0785 2.5604 5.7116

- 0.15 0.2855 7.5629 1.6617 0.0793 2.4969 5.6099

- O.IQ 0.3154 7.4291 1.6500 0.0809 2.4547 5.5096

- 0.05 0.3483 7.3580 1.6437 0.0829 2.4288 5.4128

0.00 0.3843 7.3366 1.6419 0.0853 2.4155 5.3205

+ 0.05 0.4233 7.3553 1.6435 '0.0881 2.4122 5.233010 - 0.60 0.5700 53.9491 10.5127 0.1120 5.7922 7.1922

Dynamic Bearing Capacity -OfShallow Foundations : 259

- 0.55 0.5588 28.9945 ' 6.1125 ' 0:0935 4.8411 7.1948

- 0.50 0.5645 20.5266 4.6194 '0.0833 4.2238 7.1228- 0.45. 0.5832 16.3539 3.8837 ',0.0779 3.8095 6.9932~ 0.40 0.6127 13.9337 3.4569 0.0757 3.5264 6.8273-:-0.35 0.6521 12.4031 3.1870 0.0755 3.3323 6.6445

;,

- 0.30 0.7008 11.3881 3.0080 0.0767 3.2008 6.4587- 0.25 0.7586 10.7004 2.8868 0.0790 3.1147 6.2781- 0.20 ,0.8253 10.2345 2.8046 0.0821 3.0625 6.1071.:...0.15' 0.9012 ' 9.9267 2.7503 0.0858 3.0360 5.9474

- 0.10 0.9863 '9.7361 ,2.7167 0.0901 3.0294 5.7994

- 0.05 1.0807 9.6352 ' 2.6990,0.0948 3.0386 5.6676

0.00 1.1848 9.6049 2.6936 0.0999 3.0604 5.5360

+" 0.05 1.2986 9.6313 2.6983 0.1053 3.0923 3.418715 - 0.55 1.5462 46.5473 13.4724 0.0707 5.2677 8.6324

- 0.50 1.519830.2759 9.1124 0.0696 4.7177 8.2310- 0.4'5 1.5342 23.2038 7.2175 0.0707 4.3564 7.8481- 0.40 1.5806 19.3483 6.1844 0.0734 '4.1189 7.4903- 0.35 1.6540 16.9964 5.5542 '0.0773 3.9669 7.1622- 0.30 1.7520 15.4722 5.1458 0.0823 3.8766 6.8645- 0.25 ,1.8730 14.4550 4.8732 0.0881 3.8322 6.5961- 0.20 2.0166 13.7730 4.6905 '0.0947 3.8232 6.3542- 0.15 2.1825 13.3257 4.5706 0.1020 3.8418 6.1361-0.10 2.3710 13.0501 4.4968 0.1101 3.8825 5.9388"-0.05 2.5823 12.9048 4.4579 0.1183 3.9413 5.7596

0.00 2.8168 12.8613 '4.4462 0.1282 4.0149 5.5961

+ 0.053.0750 '12.8991 :'4.4563 0.1383 4.1008 5.4463

20 - 0.50 3.6745' 46.2884 17.8477 ' ,,0.0673 5.6658 9.1768

- 0.45 3.6419 33.8986 , 13.3381 0.0728 5.3067 8.5380- 0.40 3.6943 27.6099 ,. 11.0492 0.0796 ' 5.0886 7.9941' , " ,

, -.0.35 ' 3.8151 23.9713 . 9.7067 0.0~77 .4.9684 7.5267

- 0.30 3.9952" 2'i.5'S75 . 8.&572 ., '0.0970 4.9199 7.1214. , . """,,', i " , '

',- 0.25 4.2298' 20.0542 '8.2992' 0.1076 4.9258 6.7672

- 0.20 4.5161 ' 19.0369\"7~9289 ' ' 0.1194 :4.9746 6.4552- 0.15 5.8533 18.J742: '. ' '7 :6f,77 ,< 0.1325 5.0582 6.1783

~ 0.10 . 5.2413; 17.?678 ;' '1.5-398 ,', ,,0.1470 5.1704 5.9309

-0.05 5.6804 "'.17:~542 ".7.4620 ,-..:0.,1629 5.3068 5.7084:. O~OO, "",:6.1717 :~~'.J7.6903 7.4368 "",-.0.1802 5.4638 5.5072'",~,',.,?'r"".,..,. ., '..",,", , ,- ,

"',," :. . .,'tS"t,~p{~2. .~~tt~:l...§r,' l:~...!l.<lt5.7, ; \r/j~89 '.\,.;,~~~l,989 ,,~5,~486 ,~:324325 '. '::-" 0.50 ' ';; ': "8.5665", 73.8778 "35.4499 ',0.0732, ' , 7.23469.9384'

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

,/~;:~{~~r~,'1frY~/*~;;:~:''/;2d'i~:!, ; ,";"J'\ ',,' ";;"':~~!.;'" ;'.l';'t,',,~:~ - ~'i""\ .,;c,-'", , ...'~.,


- 0.45 8.3599 51.2706 24.9079 0.0835 6.8363 9.0503- 0.40 8.3728 40.7056 19.9814 0.0954 6.6214 8.3291- 0.35 8.5541 34.7663 17.2119 0.1094 6.5339 7.7297- 0.30 8.8760 31.1015 15.5029 0.1254 6.5404 7.2223- 0.25 9.3230 28.7315 14.3977 0.1437 6.6199 6.7864- 0.20 9.8871 27.1750 13.6720 0.1646 6.7584 6.4075- 0.15 IQ.5646 26.1681 13.2024 0.1882' 6.9462 6.0748- 0.10 11.3542 25.5533 12.9157 0.2148 7.1761 5.7803- 0.05 12.2569 25.2309 12.7654 0.2445 7.4429 5.5178

0.00 13.2745 25.1345 12.7205 0.2775 7.7423 5.2825+ 0.05 14.4095 25.2180 12.7594 0.3139 8.0710 5.0704

30 - 0.45 19.3095 80.8644 47.6872 0.1064 9.3123 9.3540,- 0.40 19.1315 62.4470 37.0539 0.1267 9.0899 8.4705- 0.35 19.3718 52.5548 31.3426 0.1506 9.0494 7.7518- 0.30 19.940 46.6067 27.9084 0.1787 9.1446 7.1533- 0.25 20.1887 42.8208 25.7226 0.2116 9.3473 6.6458- 0.20 21.9566 40.3597 24.3017 0.2500 9.6392 6.2095- 0.15 23.3512 38.7778 23.3884 0.2944 10.0081 5.8303- 0.10 24.9984 37.8159 22.8330 0.3456 10.4452 5.4979- 0.05 26.8993 37.3127 22.5425 0.4041 10.9441 5.2044

0.00 29.0580 37. 1624 22.4558 0.470<: 11.4998 4.9436+ 0.05 31.4810 37.2926 22.5309 0.5457 12.1084 4.7107

35 - 0.45 46.2942 134.3023 95.0397 0.1527 13.4981 9.4021- 0.40 45.4427 100.6609 71.4837 0.1887 13.2639 8.3844 ,- 0.35 45.6687 83.4477 59.4308 0.2323 13.3114 7.5703- 0.30 4.6.7356 73.3676 52.3727 0.2849 13.5708 6.9017- 0.25 48.5145 67.0529 47.9511 0.3481 14.0015 6.3419- 0.20 . 50.9356 62.9887 45.1052 0.4237 14.5786 5.8661- 0.15 53.9640 60.3926 43.2874 0.5133 15.2895 5.4569- 0.10 57.8568 58.8199 42.1862 0.6191 16.1127 5.1018- 0.05 61.805 I 57.9989 41.6113 0.7428 17.0515 4.7911

0.00 66.6296 57.7539 41.4398 0.8868 18.0970 4.5175+ 0.05 72.0773 57,9662 41.5884 1.0529 19.2451 4.2753. -

'40 - 0.40 115.7097 112.8231. 146.0161 0.3229 20.8738 8.0404. , ,

where, ~ == Displacement at any time tNI == Coefficient of dynamic inertial shear resistance

The coefficient NI. depends on cl>and K, and its values are listed in column no. 6 of Table 6.4.Displacement of the soil mass within the failure surface due to downward movement of the footing

will increase the restoring moment about the point 0, and the increase in moment will be proportionalto the displacement provided the rotation is not excessive. It is expressed as

RF = NRB Y~ ...(6.58)The coefficeint NR also depends on qi and K. Its values are listed in column no. 7 of Table 6.4.'. . " ,.

The differenti~J" equati~~s are setup",~Y,_~~uatingtheJour 'vetiCalfo~ces to z~ro. There m~st be'; "'.." J ' . " ""," .'" , .". . " - ~, \",.. . ." ."

separ~te equation,s for be(qre!lP,~i~fteF >titpe id' sin~e the loading ~uncn~m is def11}edin !ha,t man!1er.~ , . ">,,,) ,..,.",.1"""," U,. ._, ,.'~ , . 0" ~~"" ".h' 'J,,~".' )ForO < t < t " "', "" - - d' ,,'

Dynamic"Bearing ,Capacity of Shallow Foundations 261

- 0.35 115.5504 141.1002 119.3973 0.4107 21.1138 7.1701

- 0:30 117.6386 123.0124 104.2199 0.5195 21.7125 6.4650

- 0.25 121.5875 111.8576 94.8599 0.6536 22.6077 5.8817

- 0.20 127.1879 104.7472 88.8935 " 0.8175 23.7619 5.3914

-0.15 134.3346 100.2323 85.1051 1.0168 25.1570 4.9741

'- 0.10 142.9868 97.5069 82.8181 1.2572 26.7775 4.6 I52

- 0.05 153.1451 96.0866 81.6263 1.5450 28.6173 4.3038

0.00 164.839 95.6630 81.2709 1.8870 30.6724 4.0317

+ 0.05 178.1176 96.0303 81.5791 2.2904 32.9409 3.7924

45 - 0.40 327.6781 322.2748 323.2752 0.6576 36.2961 7.4295

- 0.35 325.4943 259.1345 260.1349 0.8611 37.0113 6.5559

- 0.30 329.9752 224.0769 225.0772 1.1194 38.3965 5.8568

- 0.25 339.8627 202.7837 203.7840 1.4447 40.3468 5.2846

- 0.20 354.4804 189.3358 190.3361 1.8515 42.8070 4.8083, .

- 0.15 373.4971 180.8450 181.8452 2.3565 45.7496 4.4062

- 0.10 393.7473 175.7358 176.7361 2.9784 49.1634 4.0628

- 0.05 424.2605 173.0775 174.0778 3.7386 53.0475 3.7669

0.00 456.1177 172.2851 173.2853 4.6607" 57.4067 3.5096

+ 0.05 492.4763 172.9729 173.9732 5.7709 62.2499 3.2843, '

Any acceleration of the soil mass ACDBA due to the downward movement of the footing will causeinerti::1 forces which will resist jmch movement. The inertial forces are directly proportional to theacceleration of each individual soil mass and thereby dependent on displacements. The effective totalinertial force is obtained by combining the inertial forces on ech separate mass using energyconsiderations.

2

The inertial force is given by, If ==NI Y B d ...(6.57)" dt

óóÁò ô ùóóóùó þþôù

óîêî Soil Dymimics &: MaclrineFouniJiitions

ùî

ø

ò

÷

î¼¬ÿòô ôó ¬Ò×§Þ óóó®õÒÎ§Þ¬ÿòõ¯«Þó¯Þ£óó ãðô ¼Ö ó ùò óô¬¿

¼î¬ÿòõ¢¬ÿò ãó̄ ó¯« ¯ ô ¬¼¬î Ò×Þ Ò×§Þ Ò×§Þ¬¼

...(6.59 a)

or, ...(6.59 b)

For t ~ td

2 d2A "

NI yB ---r+NR yBt!.+qu B = 0dt "

d2t!. NR qu-+- t!. = --d t2 NIB NI YB

The solution of the differential equations will yield equations of footing displacement versus time.The forms of the particular solutions ofEq. 6.59 (b) and Eq. 6.59 (d) are found to be

t!. = Cl cos(K't)+C2 Sin(K't)+ (¯ó¯«

]-

(q

)t ...(6.60 a)

NR"t, NRytd -

A = C3cos (K't) + C4 sin (K't) !k-, NRy

respective! y, in which K' ~ ~~~~ ;and ~.. C2C3and C4are coefficients of integration, The coefficients

Cl and C2 are evaluated by the initial conditions. The coeffIcients C3 and C4 are evaluated by theconditions of displacement and velocity at td as defined by Eq. 6.60 (a). Solution and substitution ofthe coefficients yield nondimensional Eqs. 6.60. .

For 0 ~ t ~ td

...(6.59 c)

or, ...(6.59 d)

and, ...(6.60 b)

q

(NR "t

.)t!. =

(L-I

) [I-COS(Klt)]+ t q~/[Sin(K1t)-:-(K1t)]qu qu d...(6.60 c)

For t ~ td

(N:, Y) ~ ~ \(1- æ÷ ÖÕù Í·ÑøÕþ¼ÝÑÍøÕùId) + [JK' {1-COS(K"d)+in(K")-1...(6.60 d)

The coefficientsNY'Nc' N , NI and NR are dependent only on values of and K. Using magnitudesq .

of from 0° to 45° and ofK for the region where the ultimate static shear resistance could be a minimum,

these coefficients were evaluat.ed. The' values obtai~ed are given ,inTable 6,.4for every fifth degree.The maximum displacement from Eq. 6.60 (a) and Eq. 6.60 (b) is the predicated permanent footing

displacement,sitice-doWnward niotion ceases at the.time'ofma?ci~Um displacemetit'imd rebdund is not

- ~onside~ed. The'se data aI6ng' ;ith th'etim~s'~;of~~rifu'fdl~pIac:ement are giv'e~ ID'Fi~'-'6;20.-..

"amic Bearing Capacity of Shallow Foundations 263

(p\) JlZ =i~ Pl- UO!~OJ np pOO1 IOUO!SU2>W!PUON0 .-

00..-

U")0

xN C1ci E

<I..---.

--; :,

:J0 ~ er

='"

........ ..c,;'"C-.:!!

0'<'. '~

\\'),d-

d

"0

E::IE...'"::'"

.~...~

...

07.

c.......iOD~

0c0

.~Cto'EN .-

0 "'00 C. 0

Z.-00.

6-U")0

-:000 .\l' - U")

.'0

N

( P~ ) lLZ = >I P ~- uoHoJnp POOl )OUOISU2>WIPUON

If' .~~

+-U") c0 to'-I. E

to'U0a.N U\

0 .-"0

E:JE.-x:0E

,)~f


6.5.3. Chummar's Solution. Chummar (1965) presented a solution for dynamic response of a stripfooting supported by c - 4>soil and subjected to horizontal transient load. The analysis is based onthe following assumptions: .

(i) The failure of the footing occurs with the application of ~horizontal dynamic load acting ata certain height above the base of the footing. .

(ii) The resulting motion in the footing is of a rotatory nature. The failure surface is a logarithmicspiral with its centre on the base corner of the footing, whic.h is also the centre of rotation(Fig. 6.21).

(iii) The rotating soil mass is considered to be a rigid body rotating about a fixed axis.

(iv) The soil exhibits. rigid plastic, stress - strain characteristics.

Q

~ ... AQr

C2

/c

Log spiral r = ro eetan ~ - Re~u1tant friction

Fig. 6.21 : Transient horizontal load on a continuous footing resting on ground surface.

Analysis

The static bearing capacity of the footing is calculated by assuming that the footing fails whe'acted upon by a vertical static load, which causes rotation of the logarithmic spiral failure. The ultimatstatic bearing capacity qu is given by

1 .

q =cN + - yBNU c 2 y ...(6.61

where, c = Cohesion

B = Footing width and equal to the initial radious of spiraly = Unit weight of the soil

Ne and Ny = Bearing capacity factors fOl:the assumed type of failure

I .. (~:' ",':m

Clmic Bearing Capacity of Shallow Foundatiolls 265

Considering moment of the forces about 0, the centre of rotation:

2 '

d . c B 27ttan~ B2Moment ue to cohesIOn c, MRC = 2 tan cl>(e -1) = 'I' C

...(6.62a)

( 27ttan~ 1)where \jI= e -

, 2 tan cl>

Moment due to weight W of soil wedge,

M = 'YB3 tan 4>(e31ttan<l>+ 1)RW 9tan24>+1

tan 4>(e37t tan tjI+ 1)where, E = 2

9 tan 4>+ 1

...(6.62 b)

= E 'YB3 ...(6.63 a)

...(6.63 b)

4> = Angle of in~ernal friction

B2Moment of qu about 0 = qu 2 = MRC + MRW

...(6.64)

It gives,37t tan 

- c ( 21t tan 1) 2'YB tan 4>(e + 1)q -- e - + '

u tan 4> 9 tan2 4>+1

Combining Eqs. 6.61 and 6.65, we get

...(6.65)

'.,

N = 4 tan 4>(eh tan~ + 1)y 9 tan2 4>+ 1

...(6.66)

27ttan~ 1e -N =c tan cl>

With a suitable factor of safety F, the static vertical force on the foundation per unit length can~ gIven as

and, ...(6.67)

Q = ~(cNc+i'YBNy)...(6.68)

The variation of dynamic force considered in the analysis is shown in Fig. 6.22. In tnIS

Qd (max) = AQ ...(6.69)where

Qd (max)= Maximumvalue of horizontaltransientload per unit length actingat height H abovebase of the footing

A = Over load factor

"-="~'.'~..~ ~.."'~..'.~~~ ,.. ~

166 . Soil Dynamics & Machine FolJndations

"U00

.~ PE0c>-Cl

Pmax

~ tdTimq

~Fig. 6.22 : Loading function

For considering of the dynamic equilibrium of the foundation with the horizontal transient load,the moment of each of the forces (per unit length) about the centre of the log spiral needs to beconsidered:

1. Moment due to the vertical force Q

1M = - Q BI 2 ...(6.70

2. Moment due to the horizontal' force Qd at any time t

Qd(max)HtM2 = QdH = td

tMd(max)td

...(6.71

where

Md (max)= Qd (max)H3. Moment due to the cohesive force acting along the failure surface is given by 'Eq. 6.620

4. Moment due to weight of soil mass' in the failure wedge is' given by Eq. 6.630.

5. Moment of the force due to displacement of the centre of gravity of th~ failure wedge (C'Figure 6.21) from its initial position:

M) = W d..Xwhere W is the weight of the failure wedg~, and given by

'YB2 (i It tan $ - 1)W=(4 tan cp)

/1 X = R cos (11- ex) - R cos 11

...(6.,

...(6.-

...(6.

and R = QCI (Fig. 6.21). When ex is small, Eq. (6.74) can be written as

/1 X = (R sin 11)ex ...(6

(e41ttancj)-I)where ~ =c 16 tan ~

Moment due to the frictional resistance along the failure surface will be zero as its resultant will,pass through the ce'ntre' of log spiral.

Now for the equation of motion,'M\ + M2 = MRC+ MRW+ M3 + M4

.Substitution of the 'proper terms for the moments in Eq. (6.85) gives

(~'t~)+ K' a ~ A [( Md;;")} ; QB-E]-

,gp sin 11

k - - B~c

Oynamic Beating Capacity of Shallow Foundations

R = ~(x)2 + ("2)2

- -4 B tan 2 <1>( e31t tan cj)+ 1)X =

(9 tan2 <1>+l)(i1t tancj)-1)

- 4Btan(e31ttan~+1)z =

3( ~9tan2 <1>+l)(i1ttan~-l)Combining Eqs. (6.72) - (6.78) . .

. 3M3 = PB (sin 11)a

p =(e31ttan~+1)

3 ( ~9 tan2 <1>+ 1)6. Moment due to inertia force of soil wedge:

(d2a

)M = -Z J

4 d [

where J is the mass moment of inertia of the soil wedge about the axis of rotation

J =[

1 B4]

(i 1t tan $ - 1)16gtan<1> .

and g is the acceleration due to gravity. Substitution of Eq. (6.62) into Eq. (6.81) yields

~c 1 B4 d2aM = .-4 g d [2

However,

where,

where,

where

A=~(1 B4 ~c)

E = \jIC132+ E 1 B3

Ilj,~"r:~!)"

267

...(6.76)

...(6.77)

...(6.78)

...(6.79)

...(6.80)

...(6.81 )

...(6.82)

...(6.83 )

...(6.84)

...(6.85)

...(6.86)

...(6.87)

...(6.88)

...(6.89)

,.-fj:

III'II:~

268 Soil Dy"amics & Machine Fou"datio"s

Solution of the differential equation of motion [Eq. (6,86)] with proper boundary conditions yieldsthe following results:

For I ~ Id

A ( 1 ) A Md(max) , A

(Md(max).t 1

)a = - E--QB cos(kl)--' srn(kt)+- +-QB-E ...(6.90)

k2 2 k3 td - kZ Id 2

For I > Id

a = ( ~ ) [G 1 k cos (k Id) - G 2 sin {k td) ] cos ( k t) + (~ )[G 1 k sin ( kId)

-Gz COS(kld)]Sin(kt)+(~ )(~QB-E)

h G A(

. 1)

A Md (max) . k A Md (max)were, I = 2" E - - Q B cos (k Id) - -3" srn ( Id) + zk 2 k Id k

A(

1 ) . A Md(max) A Md(max)and, G2 = -k' E--QB sm(ktd)-Z' cos(ktd)+ 2

2 k ~ k ~The procedure of computations have been discuss~d in example 6.4.

...(6.91

...(6.92

...(6.93

! ILL USTRA TIVE EXAMPLES I

Example 6.1Proportion an isolated footing for a column of 500 mm x 500 mm size subjected to a vertical 10of 2400 kN. The structure is located in seismic region. The earthquake force results a moment of 4kN-m and shear load of 360 kN at the base of the footing.

The soil properties are as follows:. z 3

C = 6 kN/m , = 39° and y = 18 kN/m

A plate load test was performed at the anticipated depth of foundation on a plate of size 600 rx 600 mm and a pressure settlement record as given below was obtained. The permissible valuessettlement, tilt and lateral displacement are 50 mm, 1 degree and 25 mm respectively.

2Pressure (kN/m )

Settlement (mm)Solution:

(i) Safe bearing capacity

0.0

0.0

240

2.0

480

5.0

720

7.5

960

12.0

1200

16.0

1440

23.0

l(

2

Eccentricity of load, e = ~= ~OOOO = 0.1667 m

Inclination of load ~ i ~ tan-I ( ~:) ~ tan- t U4~O ) = 8.5 3'

~n


Let the size of footing is 2.0 m x 2.0 m and is located at 1.0 m depth below ground surface.Hence, B' = B-2 e = 2 - 2 (0.1667) = 1.667 mand L' = L = 2 m

. qnu = [e Ne Se de ie + YIDlNq - 1) Sq dq iq rw + 1. Y2B NySydy iy r'J2

Since <1>= 39° it is the case of general shear failure. For <1>= 39°, Ne = 70.79, Nq = 59.62 andNy = 100.71

Depth factors are given as-

and

- DJ - ~ - "de - 1+ 0.4 I3'"" - 1+ 0.4 1.667 - 1.24

. 2 DJdq = 1+2 tan (I-sm°<l» I3'""

= 1+ 2 tan 39 (J - sin 39)2 1.;67 = 1.13d = 1y

Shape factors are calculated as :

and

Br 1.667Se = 1+0.2T = 1+0.2~ = 1.167

0" B' 1.667Sq = 1+0.2T = 1+0}-2 = 1.167, ,

B' 1.667Sy = I-OAT = 1-0.4~ = 0.667

Q = ~Q~ + Q~ = J3602 + 24002 = 2426.85 kN

m =(2+~)(1+~) = (2+1.~67)(1+1.~67)= 5.195

[

.

]

mi-I Qq Q + B' Le cot 

[360

]

50195

= 1- 2426.85+ 1.667(2)( 6) cot 39 = 0.438

i =i - (l-iq) = 0.438- (1-0.438) = 0.428e q Ne tan 70.79 tan 39

Inclination factors:

and

Therefore,

[Q

]

m+1

. - 1- h'y - Q + B Lc cot 

[

"

]

6.195

= 1 360 = 0.3742426.85 + 1.667 (2) (6) cot 39

".'

270 Soil Dynamics & Machine Follndation~

Assuming water table to be below the ground surface at a depth greater than (Dj + B), hencer = r' = 1 and also say Y = Y = Y'" '" , 1 2

hence,qllll = CNe Se de ie +YI Dj (Nq -1) Sq dq iq rw+~ Y2 B Ny Sy dy iy r~

qllll= 6 (70.79) (1.167) (1.24) (0.428) + 18 (1) (5~.62 - 1) (1.167) (1.13) (0.438) (1)+ (0.5) (18) (1.667) (100.7) (0.6677) (1) (0.374) (1)

= 263,06 + 609.45 + 376.92 = 1249.43 kN/m2

According to Meyerhof-

(.

)2

( )2

" I 853le =Iq = 1-90 = 1- 90 = 0.82

(

.

)

2. - I 8.53 2Iy - 1-~ = (1-39) = 0.61

Therefore,

ql1l1= 6 (70,79) (1.167) (1.24) (0.82) + 18 (1) (59.62 - 1) (1.167) (1.13) (0.84)

+ I (18)(1.667)(100.71)(0.667)(1)(0.61)(1)

= 503.99 + 1140.98 + 614.76 = 2259.74 kN/m2

From charts - (According to Saran & Agarwa1, 1991)

 = 39° , ~ = 0.1g67 = 0.083, and i = 8.53°

Hence for ~ = 0.083 and i = 8.53,

Ne = 50.89, Nq = 34.62 and Ny = 42.43In this case - [As effect of eccentricity and inclination has considered already]

;=;=;=1e q yB =2m

For =0.0 (Fig. 6.4 to 6.6) For =0.10 (Fig. 6.4 to 6.6)

i = 0° i = 10° i = 8.53° ; = 0° ; = 10° i = 8.53"

Ny 144 58 70.64 75 42 46.85

N 68 41 44.97 47 30 32.50qN 88 48 53.88 58 37 40.09. c

-~'~<I!!.......- .. .---


Depth factors will be :D . 1

de = 1+ 0.4 £ = 1+ 0.4"2 = 1.2

dq = 1+ 2tan 39 (1- sin 39)2 ~ = 1.11d = Iy

and shape factor will be-

S = 1.3, S = 1.2 and Sy = 0.8 (As footing is square)< e q

Thus qnu = 6 (50.89) (1.3) (1.2) (1) + 18 (1) (34.62 - 1) (1.2) (1.11) (1) (1)

+ ~ (18) (2) (42.43) (0.8) (1) (1) (1)= 476.33 + 806.07 + 610.99

= 1893.39 kN/m2

Therefore, value of qnufrom charts lies between values of qnuobtained by Eq. 6.17 and Meyerhofsmethod.

Hence Qnu = qnu x Area of footing

Qnu = 1893.39x 2 x 2 = 7573.56 kN

Factor of safety = 7;:t~6 = 3.16 > 3, Therefore foundation is safe against shear.(2) Settlement computation-

When footing is subjecting to a central vertical loa~.only, in that case e / B = 0 and i = 0For '" = 390 N = 88 N = 144 and N = 68't' 'e 'y q

qnu =(6)(88)(1.33)(1.2)(1) + 18(1)(68-1)(1.2)(1.11)(1)(1)

+ ~ (18) (2) (144) (0.8) (1) (~) (I)= 823.68 + 1606.39 + 2703.6 = 4503.67 kN/m2

Pressure on footing corresponding to a F.O.S. = 3.16,-will be = 453~~667= 1425.22 kN/m2From plate load test data corresponding to pressure 1425.22 kN/m2,

thus

Sp = 22.2 mm

[ ]

2. Sf . Bf (Bp+30)

Now smce S = B (B + 30)p f f

Thus settlement of footing when it is subjecting to central vertical load will be

Now,

[200(60+30)

]2 x 22.2 = 37.77 mm

So = Sf = 60 (20 + 30) .

-!... - 8.53,,;,0.2187~ - 39

II

- I.;'!

~

272 Soil Dynamics & Machine Foundations. .

and

AD ~ 1-0.56W-O.82(~J= 1 - 0.56 (0.2187) - 0.82 (0.2187)2 = 0.838

AI ~ -3.51+ 1.47 (~ )-5.67 (~)'=-3.51 + 1.47(0.2187) + 5.67(0.2187)2 = -2.917

A2 ~ 4.74-1.38(~)-12.45(~)'= 4.47 -1.38 (0.2187) - 12.45 (0.2187)2 = 3.843

Se (e ) (e)

2

So = Ao+AI B +A2 B.. 2

= 0.838-2.917(°.1;67)+3.843(°.1;67) = 0.6216

Se = 0.6216 So ~ 0.6216 x 37.77 = 23.48 ³³

ÞÜ ¢ ïóðòìèÉóÑèîø¢®

= 1 - 0.48 (0.2187) - 0.82 (0.2187)2 = 0.856

B, ~ -1.80+0.94(~)+1.63(~)'

= - 1.80 + 0.94 (0.2187) + 1.63 (0.2187)2 = - 1.516

i: ~ Bo+BI(~)= 0.856+(-1.516)(°.1;67) = 0.7296

Srn = 0.7296 So = 0.7296 (37.77) = 27.56 mm < 50 mm (safe)S -S ". = m e = 27.56- 23.48 = o0049sm t B 2000 .T-e ~-166.7

t = 0.28° < 1° (safe)

Hence

andH-1l = 0.12 (0.2187) - 0.682 (0.2187)2 + 1.99 (0.2187)3- 2.01 (0.2187)4B .

= 0.0101

Ho = 0.0101 B = 0.0101(2000)= 21.21 mm"< 25 mm (safe)Example 6.2A 1.5 m wide strip foundation is subjected to a'vertical transient stress pulse which can be given asqd = 650 e-IOt kN/m2. The soil supporting the foundation is saturated clay with Cu= 60 kN/m2. Theunit weight of soil is 19 kN/m3. Determine the maximum angular rotation of the footing might undergo.

~~".<

Dynamic Bearing Capacity of ShaUow Foundations 273

Solution:

1. 2qu =.5.54 Cu= 5.54 x 60 = 332.4 kN/m

650A.= 332.4 = 1.955

2. Reffering Fig. 6.16, For A.= 1.955 and ~ = 10s-1

(W

) 8max = 0.002980.68 g qll

2W=0.311tyB

=0.311t19 x (1.5)2= 41.61 kN

Therefore., 8max = (0.00298) (0.68 x 9.81 x 332.4)41.61

= 0.158 rad = 9.1°

Example 6.3A 2.5 m wide continuous footing located at 1.5 m below the ground surface is subjected to a verticaltransient load (qd (max) = 3000 kN/m2, td = 0.3s). The properties of the soil are y = 18 kN/mJ,~ = 30° and c = 50 kN/m2. Calculate the maximum vertical movement of the foundation.Solution:

11. qu = CNe + y DJ Nq + 2" y B Ny '.

1

= 50 Ne + 18 x 1.5 Nq + 2" x 18 x 2.5 Ny

= 50 Nc + 27 Nq + 22.5 Ny

The computations of qu are done for 'I>= 30° and different values of k by taking Nc' Nq and Nyfactors from Table 6.4. These are given below in Table 6.5. .

Table 6.5: Computations of qu for Different Values of k

k

- 0.45- 0.40- 0.35- 0.30- 0.25- 0.20- 0.15

. - 0.10- 0.05- 0.00+0.05

2qu (kN/m )

57654553391035333304251222043070308031183181

274 '. . Soil Dynamici & Machine' Foundations

The minimum value of qu is obtained at k = - 0.15 as 2204 kN/m2.2. For k = - 0.15 and = 30°, from Table 6.4.

{N;:NR = 10.0081 and V~ = 5.8303

. .

NR Y = 10.0081 x 18 = 0 08174qu 2204 .

k' ~ ~~~~ - ~~~ .~ .

= 5.8303 x ~ 2~5 = 5.22

qd (max) 3000- = 2204= 1.36qutd K' = 0.3 x 5.22 = 1.566

3.

For qd (max) = 1.36 and td k' = 1.566, Fig. 6.20 givesqu

Y NR A = 0.034-'Umaxqu

Therefore, t:. = 0.034max 0.08174 =0.416m.

Example' 6.4A 2.5 m wide continuous surface footing is subjected to a horizontal transient load of duration 0.4 sapplied at a height of 4.0 m from the base of footing. The properties of the soil are y = 17 kN/m3,c = 30 kN/m2 and = 32°. Determine the value of the maximum horizontal load that can be appliedon the footing. Also compute ,the rotation at time equal to 0.6 s.

Solution:

(i) Determine Q using a suitable factor of safety (= 2.0). c = 30 kN/m2, = 32°, Y = 17 kN/m:md B = 2.5 m .

21t tan, 1 21t tan 32° 1.

e - e -. N = = . = 79.4c tan ~ tan 32°

4.32° ( 31t tan 32°

1)N = 4 tan~ (e31ttanljl+ 1) = tan e + = 200y 1 + 9 tan2 ~ 1 + 9 tan2 32°

Q = ! B (c N +! YB N)2 c 2 y

= ; x 2.5(30 x 79.4; ; x 17 x 2.5 x 200)= 8290 kN

Dynamic Bearing Capacity of Shallow' Foundations 275

(ii) Determine '11, E, /le' ~ and sin 11H = 4.0 m

td=O.4si1t tan' - 1 79.4 .

V = 2 tan ~ = 2 = 39.7

- tan ~(e31ttan, + 1)E=

(1 + 9 tan2 ~)

= 200 = 50441t tanl\» 1e -

~c = 16 'tan~ = 256

e31ttan, + 1~ = =' 56.6

3[ ~9 tan2~+I] '....

- -4Bta~2~ (e31ttan'+I)X =(1+9 tan2~) (i1ttan, -1)

50= (- 2 B) 39.7 ~,- 2.52 B

" ,

z = .' 4B tan (e31t tan cl»+ 1)

3[ ~9 tan2 + 1] (i1t ta~ cl»-1)

= (2B) ~~:~ = 2.85 B. Z, 2.85 B

SIll 11 = = . " = 0.75~x-2+z-2 ~(-2.52B)2+(2.85B)2

(iii) Determine k, A and E,

k ~ ~g Psin 1] =lleB9.81 x 56.6 x 0.75 = 807

(256x 2.5), "

A = g = 9.81 = 0.0000577'YB4 Il 17 x 2.~ x 256 .

3 3E = 'IIcB + E 'YB

= 39.7 x 30 x 2.52 + 50 x 17 x 2.53 = 20700 kN


(iv) Determine Md (max)in terms of A.

Md (max) = H . Qd (max)= H . A. . Q= 4 x 8290 A.= 33160 A.

(v) Determine A.crwhich corresponds to a. = 0

A(E

1)

A Md(max). A[

Md(maX)1 1

]a. = - 2 óóÏÞ cos(kt)--. sm(kt)+- õóÏÞóÛk 2 k3 td k2 td 2

For t = td

a. = 0.00005277(20700-.!. x 8290 x 2.5)cos (0.8070 x 0.4) 0.00005377.33~~~A sin (0.8070 x 0.4)

0.807 2 0.807 .

+ 0.0000577 .(33160 A x 0.4 + 1. x 8290 x 2.5 - 20700 ]0.8072 0.40 2

= 0.9159 cos (0.3228) - 9.10 A.sin (0.3228) + 2.94 A.+ 0.9181 - 1.834= 0.05 A. - 0.0474

For a. = 0, A = 0.948 = Aa

(vi) Determine Md (max)for A. = A.cr

Md(max)= 33160Acr = 33160 x 0.948 = 31436kN-m(vii) Determine Gland G2

A ( 1 ) A Md(max) . (k )A M

G1 = 2 E-- 2 Q.B cos(ktd)--r sm td +2 d(max)k k td k

= 0..8685 - 2.89 x 0.948 + 2.94 x 0.948= 0.9159

G - A ( 1 ) . A Md (max) A Md (max)2 - -- E--Q.B sm(ktd)--' cos(ktd)+-.k 2 k2 td k2 td

= - 0.9159x 0.807 sin (0.3228) - 9.1 x 0.948 x 0.807 cos (0.3228) + 2.94 x 0.948

= - 0.2346 - 6.60 + 2.79= - 4.05

(viii) Determine a. for t = 0.6 s

CL ~ G }GI k cos (kid) -G, sin (kid)] cos(k 'd )+G}GI k sin (kid )-G, cos(k 'd)] sin (k I)

+ ~ (.!.QB- E)K2 2

Oynamic Bearing Capacity of Shallow Foundations 277

1 "= 0.807 [0.9159 x 0.807 cos (0.807 x 0.4) + 4:05 sin (0.807 x 0.4)] cos (0.807 x 0.6)

+ 0.;07 [0.9159 x 0.807 sin (0.807 x 0.4) - 4.05 cos (0.807 x 0.4)] sin (0.807 x 0.6)

+ 0.0000577 [1.x 8290 x 2.5-20700

]0.8072 2 '

1 ' ' 1 ' ,

= 0.807 [0.701 + 1.285] x 884 + 0.807 [0.2346 - 3.841] x 0.466 - 0.9159= 2.175 - 2.082 - 0.9159 = - 8229 º¿¼


Agarwal, R, K. (1986), "Behaviour of shallow foundations subjected to eccentric - inclined loads". Ph,D. Thesis,University of Roorkee, India. " '

Caquot, A. (1934), "Equilibre des massifs a frottement interene", Ganthiev-Villars, Paris.

Chummar, A.V. (1965), "Dynamic bearing capacity of footings", Master of E'ngineering Dissertation, Universityof Roorkee, India. ' ,

Cunny, R.W. and'Sloan R. C. (1961). "Dynamic loading machine and results of preliminary small-scale footingtests", A.S.T.M. Symposium on Soil Dynamics, Special Technical Publications. No. 3.5, pp. 65-77.

De Beer, E. and Vesic, A. (1958), "Etude experimental de la capaciteportante du sable sons des fondations directedetablies en surface", Annales des Travaux Public ,~e Delgigue, .59 (3), pp 5-58.

Fellenius, W. (1948) "Erdstatische bcrchnungen", 4th ed., W. Ernst Und Sohn, Berlin.

Fisher, W. E. (1962). "Experimental studies of dynamically loaded footings on sand", Report to U. S. Army EngineerWaterways Experiment Station, University of Illinois, Soil Mechanics Series No.. 6.

Hansen, J. B. (1970), "A revised and extended formula for bearing capacity", Bull. No. 28, Danish GeotechnicalInstitute, Copenhegen.

IS : 6403 (1981), "Code of practice for determination of bearing capacity of shallow foundations".,Johnson, T. D, and Ireland H. O. (1963), "Tests on clay subsoils beneath statically and dynamically loaded spread

footings", Report to U. S, Army Engineer Waterways Experiment Station, University of Illinois, SoilMechanics Series No. 7.

M<:.Kee,K, E., and Shenkman S. (1962). "Design and analysis of foundations for protective structures", Final Reportto Armour Research Foundation, Illinois Institute of Technology.

Meyerhof, G. G. (1951), "The ultimate bearing capacity of foundations", Geotechnique, Vol. 2, No. 4, pp. 301-331.

Meyerhof, G. G. (1953), "The bearing capacity of footings under eccentric and inclined loads", Proc. Third Int.Conf. Soil Mech. Foun. Engg. , Zurich, vol. I, pp. 440-445.

Saran, S. and Agarwal R. K. (1989), "Eccentrically -obliquely loaded footings", AS~E, Journal of Geot. Engs., Vol. 115, No. 11, pp. 1673-1680. .

Saran, S. and Agarwal R. K. (1991), "Bearing ~apacity of eccentrically -obliquely loaded footings", ASCE, Journalof Geot. Engs., Vol. 117, No. 11, pp. 1669-1690. '

;!&~t'

0

~~~';';'-;";;;>;".~~;:."-' ' --


Terzaghi, K. (1943), "Theoretical soil mechanics", John Wiley and Sons, New York.

Terzaghi, K. and Peck, R. B. (1967), "Soil mechanics in engineering practice", 1st Ed. , John Wiley and Sons,New York.

Triandafilidis, 'G. E.' (1961), "Analytic~l study of dynamic bearing capacity of foundations", Ph. D. Thesis,University of Illinois, Urbana, Illinois.

Triandafilidis, G. E. (1965), "Dynamic response of continuous footings supported on cohesive soils", Proc. sixthInt. Conf. Soil Mech. Found. Engin., Montreal, Vol. 2, pp. 205 - 208.

Vesic, A. S. (1973), "Analysis of ultimate loads of shallow foundation", J SMFD, ASCE, Vol, 99, SMI,pp. 45-73. .

Wallace, W.L. (1961), "Displacement of long footings by dynamic loads", ASCE, Journal of the Soil Mechanicsand Foundation Division, 87, SM5, pp. 45-68.

White, C. R. (1964), "Static and dynamic plate bearing tests' on dry sand without overburden", Report R 277,U. S. Naval Civil Engineering Laboratory.

PRACTICE PROBLEMS

1. Describe stepwise pseudo-static analysis of designing footing subjected to earthquake loading.

2. Differentiate between Triandafilidis and Wallace analyses of dynamic bearing capacity offooting subjected to transient vertical load. Give the salient features of anyone.

3. Describe the method of obtaining the maximum horizontal dynamic load that can be appliedon the footing. Give the expression of determining the rotation of the footing.

4. A 2.0 m wide strip footing is subjected to a vertical transient pulse (qu = 600 e-st). The soilsupporting the foundation is Clay with Cu= 50 kN/m2. The unit weight of soil is 18 kN/m3.Determine the maximum angular rotation of footing that it might undergo.

5. A 2.0 m wide footing located at 1.0 m below ground surface is subjected to a vertical transientload (qd(max) = 2000 kN/m2, td = 2 s). The properties of the soil are y = 17 kN/m3, = 32°and c = 30 kN/m2. Using Wallace's approach, determine maximum vertical movement of thefoundation.

6. A 3.0 m wide surface footing is subjected to a horizontal dynamic load having duration 0.3 s.The properties of soil are y= 18 kN/m3, = 35 and c = 20 kN/m2. Using Chummar's approach,determine the value of maximum horizontal load that can be applied on the footing. Alsodetermine the rotation of footing after 0.2 sand 0.4 s'

DD

LIQUEFACTION OF SOILS.7.1 GENERAL

\1any failures of earth structures, slopes and foundations on saturated sands have been attributed in theliterature to liquefaction of the sands. The best known cases of foundation failures due to liquefaction are[hose that occurred during the 1964 earthquake in Niigata, Japan (Kishida, 1966). Classical examples of'iquefaction are the flow slides that have occurred in the province of Zealand in Holland (Geuze, 1948;Koppejan, et al. 1948) and in the point bar deposits along the Mississippi river (Waterways experiment'itation, 1967). The failures of Fort Peck Dam in Montana in 1938 (Casagrande, 1965;Corps of Engineers,1939; Middlebrooks, 1942), the Cal~veras Qam in California in 1920(Hazen, 1920) and the Lower LaB\Jorman Dam during the 1971 San Fernando Earthquake (Seed et al., 1975) in California provide typical~xamples of liquefaction failures of hydraulic-fill dams,

Liquefaction often appears in the form of s~nd fountains, and a large number of such fountains haveJeen observed during Dhubri Earthquake in Assam in 1930 and Bihar Earthquake in -1934(Housner,,958; Dunn et al., 1939). When soil fails in this manner, a structure resting on it simply sinks into it. Thenost recent Koyna earthquake of 1995 is an illustration of liquefaction phenomenon causingcatastrophicJamages to structUres and resulting in loss of life a,nd property.

7.2 DEFINITIONS

7.2.1.Liquefaction. It denotesa conditionwherea soilwillunde~gocontinueddeformationat clconstantow residual stress or with no resIdual resistance, due to the build-up and maintenance of high pore waterJressure which reduces the effective confining pressure to a very low value; pore pressure build-up lead-ng to true liquefaction of this type may be due either to static or cyclic stress applications.

7.2.2. Initial Liquefaction. It denotes a coI1ditionwhere, during the course of cyclic stress applications,he residual pore water pressure on completion of any full stress -cycle becomes equal to the applied'onfining pressure, the development of initialliquafaction has no implications concerning the magnitude)f the deformations which the soil might subsequently undergo; however, it defines a condition which is: usefulbasis for assessingvariousposs,ibleformsof subsequentsoil behaviour.

'.2.3. Initial Liquefaction with Limited Strain Potential, Cyclic Mobility or Cyclic Liquefaction. Itlenotes a condition _inwhich cyclic stress applications develop a condition of initial liquefaction andubsequent cyclic ~tress applicati?ns c~u~e limi~edstrains to develop either because of the remainingesistance ~f the ,soil to-deformatio~ .or because the soil dilates, the pore pressure drops. and the soiltabilizes under the applied loads. !J, ('i, j'. c' c, ;: :.' -,' ,.. ,,'.- I. . ~

-i'

;'

--- - -~-="-~~11

..-..::...-

~.""~<.. <~--, ..~~~""'~ --"7 Ih,


In laboratory undrained cyclic tests (triaxial, direct simple shear and gyratory shear) on saturatedsands, cyclic mobility has been ob8erved to develop and to result in large strains (Lee and Seed, 1967;Seed and Lee, 1966). It is controversial whether cyclic mobility occurs in dilative sands in situ duringearthquakes to the same extreme degree as has been observed in the laboratory. A simple means forunderstanding the difference between liquefaction and cyclic mobility as observed in the laboratory isthrough the use of the state diagram, which is shown in Fig. 7.1 (Castro and Poulos, 1976). The axes arevoid ratio and effective minor principal stress" The steady state line shown represen~sth~ 10c1,1sq[ statesin which a soil can f1owOa(c~onstanteffectiv.erilinorprincipal stre"ssa)'and constant shear stress. The voidratio at the steady state is the same a:>the critical void ratio.

Ftow at con stant volume

L ique fa c t ion.. ,CII Contractive: soils (loose)IJI sta te , .- 'neJICyclic mobility

( Larg e strains andsoftening causedby cyclic loading,state may reach B)

I .I II I.L I . IDI at Ive I

soiLs I(dense) :I

I I

D - ~--/.

MonotonlcLoad ing

{Cyclic or monotonic Loadingof dilative soil s1arts here

. I

0 - - -C3t <Tic 03c

( During) Effective minor principal stress, OJflow

Fig. 7.1: Undrained tests ~n fully saturated sands depicted ODstate diagram (Castro and Poulos, 1976)

LIquefaction is the result of undrained failure of a fully saturated, highly contractive (loose) sand,'1\"1

example starting at Point C and ending with steady state flow at constant volume and constant <1)a'Point A. During undrained flow, the soil remains at Point A in the state diagram.,

The quicksand condition that is so familiar through the use of quicksand devices for instruction ilsoil mechanics is depicted by points on the zero effective stress axis at void ratios above Q. In this statesand has zero strength and is also neither diiative nor contractive. At void ratios above Q the sand grainare not in close contact at all times.

'TICd\J).x.u.-:J

Q_0

0 ()I- "-d cI- ()I

'TI E.- '-0 u> ()I()I Cl.0111\0I- (\)(\).1:.> -<C 00 B

IDi1!1iII

Liquefaction ,of SoUs

~'

281

.The mechanics of cyclic mobility may also be illustrated with the aid of Fig. 7.1. Consider first thebehaviour in Fig. 7.r when a fully saturated dilative sand starting, for example, at Point D is loadedmonotonically (statically) in the undrained condition. In that case the point on the state diagram maymove slightly to the left of Point D but then it will move horizontally towards the steady state line as loadis applied. If one now starts a new test at Point D, but this time applies cyclic loading, one can follow thebehaviour by plotting the average void ratio and the effective stress each time the applied cyclic loadpasses through zero. In this case the state point moves horizontally to the left, because the average voidratio is held constant and the pore pressure rises due to cyclic loading.

The magnitude of pore pressure build up in the cyclic test will depend on the magnitude of the cyclicload, the number of cycles, the type of test, and the soil type, to name a few variables. In particular, it hasbeen observed in the laboratory that in-triaxial tests for which the hydrostatic stress condition is passedduvng' cycling, and if a large enough number of cycles of sufficient size are applied, the state point fort1w d'\f,erageconditions in the specimen eventually reaches zero effective stress at Point B each time thehydrostatic stress state is reached. Subsequent application of undrained monotonic loading moves the

.state point to the right toward the steady state line, and the resistanceof the specimenincreases. :During cycling in the test described above, strains develop and the specimen becomes softer. If these

strains are large enough, one can say that the specimen has developed .~yclicmobility. Adequate evidencehas been presented to show that most of the strains measured in cyclic load tests in the laboratory are dueto internal redistribution of void ratio in the laboratory specimens. For example, at the completion ofsuch tests the void ratio at the top of the specimen is much higher than at the bottom (Castro, 1969). Thusthe horizontal line D-B in Fig. 7.1 is fictitious in the sense that it represents average conditions. Near thetop of the specimen, the void ratio increases, and near the bottom the void ratio decreases. The porepressures that build up and the strains measured in the lab~ratory are due to the formation of such loosezones (Castro and Poulos; 1976). '

In summary then, specimens that lie above the steady state line on Fig. 7.1 can liquefy if the loadapplied is large enough. Such liquefaction can be triggered by monotonic or cyclic undrained loading.The further to the right of the steady state line that the starting point is, the greater will be the deforma-tion associated with the liquefaction. If the initial point is above Q, the strength after liquefaction will be.zero. If the starting point is below Q, the strength after liquefaction will be small but finite. Saturatedsands starting at points on or below the steady state line, will be dilative during undrained monotonicloading in the triaxial cell and the state point will move to the right. If cyclically loaded the state pointswill shift to the left as strains occur and the specimen softens. If enough cycles are applied, if they arelarge enough, and if the hydrostatic stress condition is passed during each cycle, then the zero effectivestress condition (i.e. initial liquefaction) can ultimately be reached in the laboratory.

7.3 MECHANISM OF LIQUEFACTIONThe strength of sand is primarily due to internal friction. In saturated state it may be expressed as (Fig.7.2).

S = on tan cpwhere, S = Shear strength of sand

on = Effective normal pressure on any plane xx at depth Z = Yhw+ Ysub(Z - kw)cp = Angle of internal friction . .

Y = Unit weight ~f soil above water tableYsub=Submergedunit weightof soil

...(7.1)

~.. ". ."0_"""",'.",, 0... - _. , ,., ..., .

282 Soil Dynamics & Machine Fo!,ndations

Ù®±«²¼ Surface

t---------

. hw~-[- - ~ -1l.. -

z'(sub

1 xx

#

Fig. 7.2: Section of ground showing the position of water table

If a saturated sand is subjected to ground vibrations, it tends to compact and decrease in volume, ifdrainage is restrained the tendency to decrease in volume results in an increase in pore pressure. Thestrength may now be expressed as,

Sdyn. = (an - Udyn)tan $dyn ...(7.2)where,

Sdyn= Shear strength of soil.under vibrations

U~yn= Excesspore water pressuredue to groundvibrations$dyn= Angleof internal frictionunder dynamicconditions

It is seen that with development of additional positive pore pressure, the strength of sand is reduced.In sands, $dynis almost equal to $, i.e. angle of internal friction in static conditions.

For complete loss of strength i.e. Sdynis zero.

Thus, a" - Udyn= 0or

(Jn = Udyn

or Udyn = 1an ...(7.3)

Expressing Udynin terms of rise in water head, hw and Y5ubas (G - 1/1 + e) Yw' the Eq. (7.3) can bewritten as :

Yw.hw = 1G-l.'Yw'Zl+e

or h G-l-1f. =-=iZ l+e er

G = Sp~ific gravity of soil particlese = Void ratio

'-;0 ". .

ier = err Healhydraulic gradient.

...(7.4)

where,

iquefltct;oll of Soils 283

It is seen that, because of increase in pore water pressure the effective.stress reduces, resulting in lossf strength. Transfer of intergranular stress takes place from soil grains to pore water. Thus if this'ansfer is complete, there is complete loss of strength, resulting in what is known as complete liquefac-.on. However, if only partial transfer of stress from the grains to the pore water occurs, there is partialJSS of strength resulting in partial liquefaction.

In case of complete liquefaction, the effective stress is lost and the sand-water mixture behaves as a;iscous material and process of consolidation starts, followed by surface settlement, resulting in closerJacking of sand grains. Thus the structures resting on such a material start sinking. The rate of sinking)f structures depends upon the time for which the sand remains in liquefied state.

Liquefaction of sand may develop at any zone of a deposit, where the necessary combination of in-,itu density, surcharge conditions and vibration characteristics occur. Such a zone may be at the surface,w at some depth below the ground surface, depending only on the state of sand and the induced motion,

However, liquefaction of the upper layers of a deposit may also occur, not as a direct result of the~roltnd motion to which they are subjected, but because of the development of liquefaction in an under-lying zone of the deposit. Once liquefaction develops at some depth in a mass of sand, the excess porewater pressure in the liquefied zone will dissipate by flow of water in an upward direction. If the hydrau-lic gradient becomes sufficiently large, the upward flow of water will induce a quick or liquefied condi-tion 111the surface layers of the deposit.

Thus, an important feature of the phenomenon of liquefaction is the fact that, its onset in one zoneof deposit may lead to liquefaction of other z~nes, which would have remained stable otherwise.

7.4 LABORATORY STUDIES , .

7.4.1. Field Conditions for Soil Liquefaction. An element of soil located at depth Z below the horizon-tal ground surface will be subjected to vertical effective stress er, which is equal to <Jvi'and horizontal

effective stress Ko er, where Ko is the coefficient of earth pressure at rest (Fig. 7.3a). There is no initialshear stress acting on the element. Due to ground shaking during an earthquake, a cyclic shear stress "("will be Imposed on the soil element (Fig. 7.3b). In the case of sloping ground, an element of soil will also'have initial shear stress, "(hi(Fig. 7Aa). During earthquake, the stresses on the element will be as shownin Fig 7Ab. The presence of the initial shear stresses can have major effect on the response of the soil toa superimposed cyclic stress condition and in general, the presence of initial stresses tends to reduce therate of pore pressure generation due to cyclic stress applications. Since the most critical conditions arelikely to be those associated with no initial shear stresses on horizontal planes, a condition analogus i:1earthquake problems to soil response under essentially level ground is considered. Hence, soil elementscan be considered to undergo a series ofcyc\ic stress conditions as illustrated in Fig. 7.5. The actual stressseries are somewhat random in pattern but nevertheless cyclic in nature as shown in Fig. 7.6

..,

. .

-

i}:;"",':>','" :.,\'~ ,//~,' ,;.,.. ""',',.,;;",",' ',,'-"'.'" ',',"' -,"'-,..~- ,,", --". _.. -".- ,.- '

28~

=+ ~ /~ H' '/"/' f ~-- --- --.2.-- ---. ..

z

'1 DA

'//1'/1'1' 1////1' "'1' Ill' " "

(a) No gr,ound shaking

=+ '" 'YI "" ~ Jh;"".2- -- -óóóóó--z

'L OB

"" , /" '" "'./ , / "" "" """"" " '/' ,. ""./+(b) Earthquake loading

Soil Dynamics & Marhine FOUlJdlltioll...

0-= OVi

OVi= ohw + lsub (z-hw)

Ko ()

Stresses on element A

0-= OVi

Ko 0-

~h

Stresses on element 8

Fig. 7.3 : Stress conditions for soil element below horizontal ground in cyclic loading conditions

rh

~~.if.~i~.~,.~;:,...~<w<'""!J

Liquefactiollof Soils / 285

0'" " "" / / r " / / / / .I'" ".I"/

(0) No ground shaking Stresses on element A

cv= av i

--L -r;h=r= Chi

1HDHPo-~

I'

DB

'" >"'+"O' uu"

(b) Earthquake loading Stresses on element B

Fig. 7.4: Stress conditi ons for soil ele me nt below sloping ground in cycliiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiii

() = Uvi

! ..:hi

DA .jor- KO

I

-- k J.

-.., ..1

r


"""V/I\I' "A' 11/'\\

00t -C---

--11 1!-KoCTO---t

00-C_t

-+- \\ ~\-- Ko0-0---t

Cyclic toad sequence

Fig. 7.5: Idealised field loading conditions

II.

I

,

11

1:max.

Dept h:; 14.0m

Gav

I.

JI illili li'

aiffi

12

Ti.me (5)

18 24 30

IIII

rII

r

l101,.

Fig. 7.6 : Sh~ar stress variation determined by response analysis

7.4.2. Different Laboratory Tests. Simulating cyclic shear stress conditions, following types of testprocedures have been adopted for liquefaction studies:

(1) Dynamic trhixial test (Seed and Lee, 1966; Lee and Seed, 1967).(2) Cyclic simple shear test (Peacock and Seed, 1968, Finn et aI., 1970, Seed and Peacock, 1971).

(3) Cyclic torsid~al shear teSt (Yoshimi and Oh-Oka, Ishibashi and Sherif, 1974).(4) Shaking. table test (frakash and Mathur, 1965; Yoshimi, 1967~Finn et aI., 1970).Typical studies 011above mentioned laboratory tests are described herein.

11 IIIm

.. I!Io- ----

III

I

. 286I

/I'/l1I/'" \I

I 00J tI 1I ---c=J--K0 0;,

I

tI .I t,I

Initial stresses

40

30

N 20E--z 10-

0

.....1I1

- 30

óìðÔ0 6

288 , Soil Dynamics & Machine 'Foulldations

7.5 DYNAMIC TRIAXIAL TEST

Seed and lee (1966) reported the first set of comprehensive data on liquefaction characteristics of sandstudied by dynamic triaxial test. They used the concept of developing cyclic shear stress on section xx ofthe soil sample as shown by the stress conditions I and 11on the sample given in Col. 1 QfFig. 7.7. Thestress condition I is achieved by increasing axial stress on the specimen by an amount ad' keeping lateralstress a3 constant, and simultaneously reducing the all round pressure on the specimen by an amountadl2 (Cols. 3 and 4 of Condition I). Similarly the desired stress condition 11can be induced by reducingthe vertical stress by ad' and simultaneously applying an increase in all round pressure equal to ad 12(Cols. 3 and 4 of Condition II). It may be noted that during testing the pore pressure should be correctedby reducing it by ad 12 in condition I, and increasing by ad 12 in condition H.

Seed and Lee (1966) performed several undrained triaxial tests on Sacramento river sand (emax =1.03,emin = 0.61). The grain size ranged between 0.149 mm and 0.297 mm. The results of a typical testin loose sand (e = 0.87, DR = 38%) are shown in Fig. 7.8. In this testthe initial all-around pressure andthe initial pore water pressure were 196 kN/m2 and 98 kN/m2 respectively, thus giving the value ofeffective confining pressure as 98 kN/m2. The cyclic deviator stress ad of magnitude 38.2 kN/m2 wasapplied with a frequency of 2 Cpg.The test data in Fig. 7.8 shows the variation of load, deformation andpore-water pressure with time.

"0

8~ DR~38°/o 1 eo=0'87, OJ=98,OkN/m201-.S z 100-0""" 0I/)

"3 ~ 100Go .

Time'"

h 38.2 kNjm£

~::J - 100I/)N

~ EL.. '-o..z ' 0~x..L..-0 Ia.

Time

. '-Time

- - .,' -, f sec.t--Fig. 7.8: Typical pulsating load test on loose saturated Sacramento river sand

(Seed and Lee, 1966)

62St- Compression'15

EL.. -.2E

0"0 E.0-)( 'H::L Extension<C

Juefactioll of Soils

:ompression 30- OR = 38 °/0

. . - 2~=. 98.0kNjm

2at Od =(+)38-2 kNj m

at Od = 0

C:Xten sion - 301 2 4

Number

. 2at Od =(-)38'2 kN/mI r

10 20 40of cycles

L..-~Nd E~ -~zL..~0-a.::J.S ~

L..~ ::Jen\/)C IIId ~

£. L..ua.

(a) Axial strain versus number of cycles- .

150

100 InitiaL e tfe ctive - - -confining pressure

,-atCJd,=O-. at-Od =+38,2 kN/m2

50 2- at Od :-38'2 kNjm

0 1 2. 4 1°--' 20Number of cycLes

40

(b) Observed change in pore water pressure and number of cycles

10 4020

Number of cycles. (c)Changein porepressure and numberof cycles

Fig. 7.9 : Typical pulsating load test on loose Sacramento river sand (Seed and Lee, 1966)

289

100

lOO

'"

100

-0

201-0-c

..

'- 10o'L........ 04\/)

.i..#,0 -10)(« -20

150

L..-10°r-Initial effective

"""Nd E- confining pre ssure

zL..0- 50a. ::Jc.-

L..::J

en\/)C \/) ,0.I:. L..ua.

- SO, . 2 4


From this data, variation of axial strain amplitude, the observed changes in pore-water pressures,and the pore water pressure changes correct~d to mean extreme principal stress conditions with numberof stress cycles have been plotted in Fig. 7..9.It was observed that, during the first eight cycles of stressapplication, th~ sample showed no noticeab!7.d~formati()n although pore-pressure increased gradually.The pore pressure became equal to O"}during the ninth cycle, indicating zero effective confining pressure.During the tenth cycle, the axial strain exceeded 20% and the soil liquefied.

Similar tests as described above were performed by Seed and Lee (1966) for different values of 0"d'The relationship between O"dagainst the number of cycles of pulsating load applications is shown inFig. 7.10. It is evident from this figure that number of cycles of pulsating load application increases withthe decrease of the value of 0"d'-

NE 60

III

.

OR ::. 38 0/0

eo ::. 0-87

03= 98,OkN/m2

-z~-I;' 50

..

~ 40~~-0\ 30c:-dIII 20-:)Cl.

~ 10d~Q.

0, 3 10 30 100 300 1000

:~ .

Number .of cycles to cause faiLureFig. 7.10: Relationship between pu~s~tingdeviator stress and number of cycles required

to cause ~ailure in Sacramel1to river sand (Seed and Lee. 1966)

Test data obtained on dense Sacramento river sand is shown in Fig. 7.11. It may be noted that thechange in pore water pressure become equal to o"}after about 13 cycles, however the axial strain ampli-tude did not exceed 10% even after 30 cycles. This is due to the fact that in dense condition soil dilates,and the pore water pressure reduces which in turn stabilises the soil under load. As discussed earlier thiscorresponds to cyclic mobility (Seed, 1976; Castro, 1976).

.iquefaction of Soils 291

11P re sSlon 15

~:::I111111~L-0..L-~.....d-~N~ E~ ........0 Zo...xc-<:I)CJ)cd£.U

OR = ' 78 0/0 '

.,0)= _9~'.0~N/m~r-

",,'" at Od =" +68'7 kN/nf

at Od = 0

"

'.. "at Od = 2" -68'7kN/m' ~,

2 4 10 20Number of cycles

40 100

(a) Axial strain versus number of cycles

150"

100 AA-- -a-- -A~at Od=0Initial effectiveconfining pressure

0

50 u at Od =

- -\--:r..s-..t::~ 68.7 kNjm2uat Od-- 68'7 kNim2

- 501 20 40 10010

Number of cy cl e s

2 4

(b) Corrected change in pore water pressure

Fig. 7.11 : Typical pulsating load test on dense Sacramento river sand (Seedand Lee, 1966)

Lee and Seed (1967) have extended this work for studying the various factors affecting liquefactionand identified the followings :

Relative density, Figure 7.12 shows the plots 'of peak pulsating stress ( Le.. the stress causing liq-'uefaction) against number of cycles of stress in loose and dense sands. The initial liquefaction corre-spondsto the conditionwhen the pore waterpressurebecomesequal to the confIDingpressure0')" Cri-terion for complete liquefaction is taken corresponding to 20% double amplitude $train. The figure indi-cates that in loose sand, initial liquefaction and failure occur simultaneously (Fig. 7. 12a). With the

- " .1'"",,'

-0-0,-c:0 0L--111.... - 5d-x« -:-10

tension -15 1

.iquefaction of Soils

TtpreSSlon 15

~::J111111c:J1...a.

150

1001...c:J.....0-~Nc:J E1..."-0 Za..x

50

c-c:J0'1C0

L:.U

- 501

291

OR = .78 0/0 '

.~,:03'= _98.0~~/m~r... ."," at Od =

.-K"" +68'7 kN/m2

at Od = 0

" .

'.. at Od = 2" -68.7kN/m. ~,

2 4 10 20Number of cycles

40 100

(a) Axial strain versus number of cycles"

Initial effectiveconfining pressure

AA- - -a-- -6~at Od=0

0

u at Od =

-..:\--:r:s--t.:~68.7 kN/m2u .at Od -- 68.7 kN/-m2

2 10

Number of cy cle S

20 1004 40

(b) Corrected change in pore water pressure

Fig. 7.11 : Typical pulsating load test on dense Sacramento river sand (Seed and Lee. 1966)

Lee and Seed (1967) have extended this work for studying the various factors affecting liquefactionand identified the followings :

Relative density. Figure 7.12 shows the plots .of peak pulsating stress ( Le.. the stress causing liq-uefaction) against number of cycles of stress in loose and dense sands. The initial liquefaction corre-spondsto the conditionwhenthe pore waterpressurebecomesequal to the confming pressure0"3'Cri-terion for complete liquefaction is taken corresponding to 20% double amplitude strain. The figure indi-cates that in loose sand, initial liquefaction and failure occur simultaneously (Fig. 7. 12a). With the. ..

- " .#",'..'

- 100-0- 5c:.-0 O\-..... I111.... - 50-)(« -:-10

tension -15 1

0292 Soil Dynamics ~ Machine Foundations

"O""'-'O~~"-'-""-'o"""o~"' "o , """_0"-' . "':'~'"

incre~se in relative density, the difference betWeen the number of cycles to cause initialliquefacticin and'~f?ilure increases.-

.NE

. -Z

'.:;100

-0VI-::Ja.x0

..~ .

11'" 80

~tIJL.-111

0'\C

00

'~.r

..

10

'..';

strain

100 1000Number of cycles

(a) Loose sand

eo = 0 -87OR = 38 0/0

OJ = 9 8.0 k N/ m~

010,000

eo = 0.61OR= 1000/0

OJ = 98'OkN/m2\\ Initial

\~ Liquefaction, , " ..........

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

...... -- ----

01

..~...

10 100 1000 10,000 100..000Number of cycles

(b) Dense sand

Fig. 7.12: Peak pulsating stress venus number of cycles(Lee and Seed, 1967)

0 ._--..-.......-.....

L \,

- _.0--

100,000

..

-N 200E

..........zx 160-'"

VI 120VIL.-\I)0'\ 80c ...-0\I)- 40::Ja.x0

a.

[que/action of Soils 293

Confining pressure. Figure 7.13 shows the influence of confIning pressure on initial liquefactionmd failure conditions. At all relative densities for a given peak pulsating stress, the number of cycles to:ause initial-liquefaction tFigs:-7:B"u and b}-or-fai1ure-(i-.e:-26%strain, Figs:-7:i3 c and d) increasedvith the increase in confIning pressure-

,N

E-~ 180

'OR = 7 8 0/0

eo = 0.71

DJ (kNfm2)1500

~~ .-It - . 500A '. 100I I I

100 1,000 10~000 100,000Number -of cycles'

(a) For initial tiquefaction in medium dense sand

10

'A

, -

OR = 100 0/0

eo = 0.61

OJ (kNfm2)1500

(b) For initial liquefaction in dense sandFig. 7.13: Peak pulsating stress versus number of cycles (Lee and Seed. 1967) (...Contd.)

...:t-.;",,!1;~,,!"J)'.. ,f; <.,'~"";..!",,..~ -1.> 'b\':;:"""';?" i';~k1t~;!;-';:".';';t;r;ft"';+,;:-;;~("i,,(,t~~l't.~;;, , '.:..Y""f.'i.':"~,;;;". .' '~ii-" ;;":8"';f ",;;. ~f~:ii:',~

-...

120L.

....

CTIc:.- 60-0-':J' -Co

d 0a. 1

2000-N

E............z-- '500l5'

"

1000....

CTIc:.-....0

500-':JCo

0 1- &a.

0-,

- 5004 . OO

10 - 100 1,060 ' 10.000 100,000.,Number of cycles

180

120

60

01

2000N

E......z-"

~ 1500~IIIIII1:>1

;: 1000III

C7Ic:

0III

.....:JQ

500

-

SoU Dynamics & Machine Foundations, . ", ,.,. ,..', i~

,',-

OR : 78°/0- eo :,0,71 ". .. ,-

10 100 1,000Number of cycles

~ (kN/m2)1500

50010°110~000 100,000

-"C1:>1

'Q.0

1

(c:)20% strain (i.e. failure) in medium-dense sand

OR = 100 °'0eo = 0.61

10

500100L-

10.000 100,000

-,'

Fig. 7.13: Peak pulsating stress versus number ofc:yc:les(Lee and Seed,1967)- ,

,"" :. .";.

100 1,000Number of cycles

(d) 20% strain (i.e. failure) in dense sand

Peak pulsating stress. Figures 7.14 a and b show respectiyely the vari~t!onof peak pulsating stres~Odwith confining pressure for initial liquefaction and 20% axial strain in 100cycles. It may be noted thafor a giverirelative density and number of cycles ofload applicafion;the"odmcreases linearly with 03 foinitial liquefaction, while for 20% axial strain condition, similar linear trend exists only in loose sand5

;- . " ". . I . . ~;'~;'~

294

-N.

E--.,Z..¥-H'

"'VI

t.I..VI

0'\c:-CSVI-:Ja.CSt.Ia.

Liquefaction of Soils

2000........

NEz 1600oX

-\11

~ 1200L...-\11

01C 800

Ini.tiol voidratio.)e6

0 '61.-0\11-:Ja. 400.xd(:,)Cl.

0.71

-00 400 800 1200

03 (kNfm2)

3600 2000...

(a) Initial liquefaction in 100 cycles

2000.........

NE---Z 1600oX

, .'

.......

\11\11~ 1200.....-\11

01C

";:: 800d\11-':Ja. 400

::s:d~Cl.

Initial voidratio.) eo

0.61

0 -71

ðóéè0'81

00 400 800 1200

03 (kNjm2)

1600 2000

(b) 20% strain in 100 c)'desFig. 7.14 : Influence of pulsating stress on the liquefaction of Sacramento river sand (Lee anti Seed. 1967)

-- -..'." '--".-'

---~ _u_-,,--"-- -

296 Soil Dynamics & Machine FOlllldations

Number of cycles of pulsating stress. From the Figs. 7.12 and 7~13,one can conclude that for agiven p\llsating stress, number of cycles needed for causing initial liquefaction and failure increases withthe increase in relative density"and confining pressure. -

7.6 CYCLIC SIMPLE SHEAR TEST

The cyclic simple shear test device is already described in C;:haptet 4:with a mention that it simulatesearthquake. condition in a better way. Peacock and Seed (1968) reported the first set of comprehensivedata on liquefaction studies by using this test. They performed tests on Monterey sand (SP, emax= 0.83,emin = 0.53, DIO = 0.54 mm). The sample size was 60mm square and 20mm thick. The samples weretested at relative densities (DR) of 50%, 80% and 90% giving the sand in loose, medium dense and densestates respectively. The oscillatory shear stress was applied at a frequency of 1 cps or 2 cps keeping thenormal stress constant.

Initial relative density °R=50oloInitial void ratio, eo = 0-68Initial confining pressure,OV = 500 kN/m2Frequency ã ï Ø¦

¢Äïï î\11-C>lN;: EÄïïóòòòòòòòòò

1- Zd ~ 2C>I ---

.I:.if)

4

4

(a) Applied cyclic shear stress

20

0

2

(b) Shear strain response

Fig. 7.15 : Record of typical pulsating load test on loose sandin simple shear conditions (Peacock and Seed. t 968)..

0 I'

0 I I1111

M0

mumf---' cum-

0 -

0

-0-0

1----;-c.01--\11 11-dC>I

.J::.If)

I

j1ftttt

0

124,eyc: le su

iquefaction of Soils 297

~1-:J\1111'1~- 'L-N 60Q.E1-~-oz4~~~ -21-:::::>0

0.. 0

(c) Pore water pressure response

Fig. 7.15 : Record of typical pulsating load test on loose sand in simple shear conditions (Peacock and Seed, 1968)

The typical test data in Fig. 7.15 show the variation of shear stress, shear strain and pore waterIressure with time. As evident from Fig. 7.I5b, there was no significant shear strain of the sample duringlle application of the first 24 cycles of stress. During the twenty-fifth stress cycle, the shear strain sud-.enIy increased to a value of about 15% and become 23% in the next cycle. Pore pressure increased;radually until the effective confining pressure is reduced to zerc. (Fig. 7.15e). At this point the resultingleformations became extremely large, and the soil had essentially liquefied. Similar trend was also ob-erved in triaxial test.

>eackock and Seed (1968) have also studied the effects of followll1g factors on liquefaction:

Relative density. Figure 7.16 shows a plot of peak pulsatingyress (1:h)dJUsmg initial liquefaction\ith number of cycles of application for different relative densities and confll1ing pressures. From this19ure it can be concluded that for a given value of confining pressure and number of cycles of stresslpplication, 1h increases with the increase of relative density'. A mo;"e clear presentation is shown in:ig. 7.17.

'\.

","'"

2-""""'",-

Qy (kNjm ) ------800500

~11'1

I/)

~1-

80

-11'1

1-a 60~N.c E111"-

cnZC x 40'- --a .c.III~

:JQ.::ca -~ --

Q

01

90°/0

10 100 1,000 10-,000Number of cycles

Fig. 7.16 : Initialliqu'efaction of cyclic simple shear test on Monterery sand (Peacock and Seed, 1968)

... ' ,

0 f"" '"

-_ ..rW lr".-..- ..-Cc

...

V

100.........

NE

~ 80~

For 100 cycles of pulsating stress

±ª (kN/m2)

800

-

100

Fig. 7.17 : Effe~t of relative density on cyclic stress causing initial liquefaction (Peacock and Seed, 1968)

Confining pressure. From the data presented in Figs. 7.16 and 7.17, plot of'th versus aI, was prepared as shown in Fig. 7.18. For a given value of DR and number of cycles of stress application, t,increases linearly with the increase in av'

\/IIII~L.............

V)NL... E0"-..

~ Z 50\II~-enNc.- -...........~~

75 \\

Initial relative. density, DR= 50°/0Initial void ratio eo = 0,68

:J I-0.0

~ .!:o~~Cl.

-.. 2DV (kN/m )

800500300

<....}

'1~

01 10 100 1000

Num b e r of c.y c l e sFig. 1.18: (a) Cyclic stresses required to cause initi~lliqueraction at different confining pre~sure

to

I{t,t

'j"If

..\/I 60\/I

L......III

en 40c"-...0III

:J 200. I 200oX0

Cl. 00 20 40

Relative density (°/0)

uefacti~n of Soils

11\11\~'-- .......

,II\N'- E0..........~z1I\.x0'1-c: N.;: --0 '~11\ 1:)-::J '-0.0

75

50

25

.x0 .r=~I'-JQ. 0

0 100 200 300 400 500 600 700 800

Initial effective confining pressure (kNfm2)

Fig. 7.18: (b) Effect of confining pressure on cyclic stress to cause failure in 10 cycles and 100 cycles(Peacock and Seed, 1968)

Peak pulsating stres!> and number of cycles of stress application. From Fig. 7.16, it can be seen

at for a given value of o"v and relative density DR' a decrease ~f "Chrequires an increase of number ofdes ~o cause liquefaction. Further for a given value of "Ch:"number of cycles of stress applicationquired to cause liquefaction increases with the increase in relative density DR and confining pres-re 0"v .

Frequency ofload application. Tests wererformed at frequencies of 1Hz, 2 Hz, and 4~, and the effect of frequency on the stressusing liquefaction was found negligible.

Seed and Peacock (1971) have studied thefeet of coefficient of earth pressure (Ko) one peak pulsating shear stress "Ch causing liq-:faction in cyclic simple shear test.The value.Ko depends on the overconsolidation ratio)CR).

Figure 7.19 shows a plot of stress ratio,la,,) with number of cycles of stress appli-.tion for different values of Ko' For a givenlative density and number of cycles of stress'plication, the value of ("Cia,.) decreasesIth the decrease of K ..0

299

OR =50%eo = 0.68

10 c.ycles

100 cycle

0.4

It) 0.)---s::.

~0

0~

0.2 OCR: 8ko : ,1/1

1/1~~

~ 0.1OCR:"Ko : 0.75OCR: 1

Ko : 0.4

J"

01 10 100 '.000 10.000Number ot cycles causing initial liquefaction

Fig. 7.19 : Influence of overconsolidation ratio (OCR) on stresscausing liquefaction in smiple shear tests

. (Seed and Peacock, 1971)

.iJ;

300

.1;'\


7.7 CO~IP ARlSON OF CYCLE STRESSES CAUSING LIQUEF~.\CTION UNDER TRlAXIAL ANDSIMPLE SHEAR CONDITIONS ,PeaCock and Seed (1968) performed both cyclic triaxial and cyclic simple shear tests for liquefactionstudieson Montereysandwith a relativedensityof 50%and confiningpressures(cr3or crv)of 300, 500and 800 KN/m2. Rbults are plotted in Figs. 7.20 and 7.21. It may be seen fro~ these figures that thecyclic stress required to cause liquefaction of loose sands under simple shear condition ("Ch) is about 35percent of the cyclic stress required to cause liquefaction in triaxial condition (crd/2).

150 "".......

~ Triaxial test results

~ OJ{kN/m2).- 800

-N

E.........

z~

N--l5'

.'~--~

~0 100

(-.JL:

, 500

lI\lI\~~

IfI

""""--~L-a~

.J::.IfI

CJ) 50c

'\ ". e- 300

..... Simple shear test resultsaVI--:Ja.

, "-...... DV (kN/m2)800500300

oXa~0.. OR= 500/0

eo = 0-68

01 10 lOO 1000"

Number of cycles 'iFig. 7.20: Cyclic stress required to cause liquefaction of Monterery sand at different confining pressures

in triaxial and simpl~ shear tests (Peacock and Seed, 1968)

.;quefact;on of Soils 301

2: 150~

~0~Q.

Relati ve density} OR = 500/0Initial void ratio,eo =0.68

L-e.£:.

~U\U\

~ 100-U\-N

L- E0--~z.c.~U\-

CJ'I

.S 50-0U\::)a.

10 cycles

100 cycles

~

100, 200 300 400 500 600 700 800

Initial effective confining pressure 03 or OV (kNfm2)

Fig. 7.21 : Comparison of pulsating shear strength of I~ose Monterey sand under cyclicloading-simple shear and -triaxial conditions (Peacock and Seed. 1968)

7.8 STANDARD CURVES AND CORRELATIONS FOR LIQUEFACTION

For evaluation of liquefaction potential, Seed and Idriss (1971) developed standard curves between cyc licstress ratio (crd/2/a)) versus mean grain size (Dso) for 10 and 30 number of cycles of stress applicationfor an initial relative density of compaction of 50% (Figs. 7.22a and b). These curves were prepared bycompiling the results of various tests conducted by several investigators on various types of sand.

The values of stress ratio ("th/a~.)causing liquefaction, estimated from the result of simple shear

tests, have ~hown that the value of"th/av is less than the corresponding value ofad/2 a) (Fig. 7.20,7.21and 7.22). The two stress ratios may be expressed by the relation.

("th

) (ad

)=- = - ,CI ...(7.5)\, av simple~hear 2 cr) triax.

where, Cl = Correction factor to be applied to laboratory triaxial test data t? obtain stress conditionscausing'liquefaction in the field "

". ~,",.' ';

"",, "'. --.:"..,~..~:

Soil Dynamics & Machine Foundatiolls-----

. rTriaxi cl. compression testdata for Ode./203 at

~ 9 liquefaction~

.

0-- vField value at -Ch/ay causinglique taction esti mated tramre su Us ot sim pie sh ear tests

Relative density = 50%No. of stress cycles = 10

0.3 0.1 0-03 0.01

Mean grain size 050' mm

(a) In 10 cycles

'VTriaxial compression testdata for 0Ct /2 CJj atliqu e ta e.tion

0 ,- Field value at7:h/r:sv causingliquefaction estimated fromresul ts at simple shear tests

Re lative densi ty = 50 °/.No. of stress cycles =30

0-3 0.1 0.010-03

Me an grain size 050, mm

(b) In 30 cyclesFig. 7.22 : Stress conditions causing liquefaction of sands (Seed and Idriss, 197\)

-----

3021:5'N

'-......

1:5' 0.30...lfo

U>- 0.25u

c:.-c:

0.200.--uC-C>i

0 -15:Jv.--(J)C1I1

0 -10:JduG-

0 -05d

1I11I1C>i

0L..-1.0If)

-N------

tj' 0.30...

1I1C>i .-u>- 0-25u0MC

c: 0.200-ud.....C>i:J 0 - 15CT

(J)c:.-1I1 0 - 10:Jd\J

0.....

0.05d...\fI\fIC>iL.. 0.....

(f) 2.0

Liquefactioll of Soils 303

Seed and Peacock (1971)have proposedthe followingth,reealternativecriteriaof obtaininge 1 :

(i) Maximum ratio of shear stress developed during cyclic loading to the r.ormal stress. Theinitial stress conditions of a specimen in simple shear device are shown in Fig. 7.2"?a; the correspondingMohr's circle is shown in Fig. 7.23 b. Figure 7.23 c and d show respectively the stress conditions on thesoil specimen during cyclic simple shear test and corresponding Mohr's circle. It can be seen fromFig. 7.23 d, that the maximum ratio of shear stress to normal stress in cyclic simple shear stress is't/Ko ay. This ratio in triaxial test is ad/2a3 (Fig. 7.7). Therefore,

~ = adKoal, 2a3

...(7.6)

now/- /-

e = 'th ai, = 'th' av -1 /2 I K - - Koad . a3 'th I oal,

...(7.7)

GV

KoOV

(a.)

ay-rh

(c)Ko CYy

Th

V1tfI~L-

.....V1

L-a~

s:::.if)

QV(b)-

KoOV

NormaLs t re.ss

.. ", -

V1V1~L-

Normalstress

(d)

.....V1L-a~

s:::.VI

Fig. 7.23 : Maximum shear stress for cyclic simple shear tests

(ii) Ratio of maximum shear stress to the mean principal stress. In simple shear test (Fig. 7.23 d)

Maximum shearslress. ~max= ,/~; + [~a, (1- Ko)J

-... -----_.

~1

\


Mean principal stress during consolidation (F'ig. 7.23 a)

.1 - - -="3 [O"v + Ko o"v + Ko O"v]

1=.- 0" (1 + 2 K )3 v 0

...(7.9)

In Triaxial test

aMaximum shear stress = J

2

Minor principal stress = 0"3

...(7.10)

...(7.11)

Therefore,

2 l{- 2"r

h + - 0" (1- K )}. 2 v 0

[~crv(1 + 2 Ko)]

= .:!.L20'3

...(7.12)

It gives(-.

'th

J=

(.:!.L

)~.!.(1+2K )2_.!.(1-Ko)2 /(ad/2a3)2- . 2 9 0 4O"v 0'3...(7.13)

Hence 1 2 1 2 2Cl = ,/-(I+2Ko) --(I-Ko) /(ad/la3)9 4

...(7.14)

(Hi)Ratio of maximum change in shear stress to the mean principal stress during consolidation.

"rh

[crv(1 +32Ko)] a3. (1+2 Ko)

It gives Cl = 3 ...(7.16)Finn et al. (1970) have shown that for initialliquefactiQn of normally consolidated sands

(1+ Ko)Cl = 2 ...(7.17)

Castro (1975 has prop<?sed that the initial liquefaction is better repr~sented by the criteria of the ratioof the octahedral shear stress during cyclic loading to the effective octahedral normal stress during con-solidation. It gives the value of Cl as . . . , . .

- 2 (1+2 Ko)

Cl -. (3.[3) ...(7.18)Values of Cl computed from the above equations are given in Table 7.1 Weighted average values of

CI are given in the lastcolumnof the table.In normallyconsolidatedsands,valueof Korangesfrom 0.3to 0.5 which in turn gives the value of Cl varying from 0.45 to 0.55.

Ja;) ,...(7.l5)

'.

. . , ..

II1II -,-- ,.;"~'".;:,:'<,.""'".",,,;, . .,.,'i., ..'..', 0'".1)<: ",ih""".<.(,.'~:<.A.,I'A\:;t",-.,.j,~,"',',''"i1.

Liquefaction of Soils 305

Table 7.1 : Values of Cl

Value orc! using----------------------------

ad* For - = 0.42 0'3

In simple shear test equipment, there is always some nonuniformity of stress conditions. This causesspecimens to develop liquefaction under lower horizontal cyclic stresses as compared to that in the field.Seed and Peacock (1971) demonstrated this fact for a uniform medium sand (DR = 50%) in which thefield values were about 1.2 times the laboratory values. It can be expressed by the following relation:

("Ch

)-

("Ch

)"

- -C-cr ~. cr

v field v simple shear

where Cz = Constant to account the nonuniformity of stress conditions in simple shear testCombining Eqs. (7.5) and (7.19), we get

(

"C

) (a

) (a

)-1:L =C C -IL =C-Lcr I z 20' r 20'v field 3 triax. 3 triax.

...(7.19)

...(7.20)

where

Cr = Cl Cz

Seed and Idriss (1971) suggested the values of Cl' as given in Table 7.2.Table 7.2 : Values of Cr

...(7.21)

Relative densityOR (%)

Cr

0-50

6080

0.570.60 .

0.68

As evident .from Fig. 7.17, upto a relative densitY of 80%, the peak pulsating shear stress causingliquefaction increases almost linearly with the increase in relative density. Keeping this fact in view, thefollowing general relation is suggested:

("Ch

)=

(.5!L

).C. DR

cry . field 2 0'3 triax. r 50DR . 50 .

'. ...(7.22)

Ko Equation Equation Equation Equation Equation Average7.7 7.14* 7.16 7.17 7.18 value

0.3 0.3 Negative 0.53 0.65 0.61 0.45value

0.4 0.4 Negative 0.6 0.7 0.69 0.53value

0.5 0.5 0.25 0.67 0.75 0.77 0.550.6 0.6 0.54 0.73 0.8 0.85 0.680.7 0.7 0.71 0.80 0.85 0.92 0.780.8 0.8 0.83 0.87 0.90 1.00 0.87

306 Soil Dynamics & Machine Follndatiolls

where

(~h

)= Cyclic shear stress ratio in the field at relative density of DR percent

crv fieldOR

(5!iL

J= Stress ratio obtained from triaxial test at relative density of 50%. It can be determined

2 cr3 triax.50using Fig. 7.22.

7.9 EVALUATION OF ZONE OF LIQUEFACTION IN FIELDAt a depth below the ground surface, liquefaction will occur if shear stress induced by earthquake is morethan the shear stress predicted by Eq. 7.22. By comparing the induced and predicted shear stresses atvarious depths, iiquefaction zone can be obtained.

In a sand deposit consider a column of soil of height h and unit area of cross section subjected tomaximum ground acceleration Qmax(Fig. 7.24). Assuming the soil column to behave as a rigid body, themaximum shear stress 't at a depth h is given bymax

-

(rh

)'tmax - g .Qmax

where g = Accelerationdue to gravityr = Unit weight of soil

...(7.23)

¿³¿¨ --. ..'. . ::.~~::~.:':I

Unit crosssectional¿®»¿h

~

Tmax=("(hf9) amax.

Fig. 7.24 : Maximum shear stress at a depth for a rigid soil column

Since the soil column behaves as a deformable body, the actual shear stress at depth h, ('tmax)actistaken as

( ) - -(rh

).

'tmaxact - rd. 'tmax- rd g .omax

where, rd = Stressreductionfactor . .'

...(7.24)

;qllefact;on of Soils 307

Seed and Idriss (1971.)recommended the use of charts shown in Fig. 7.25 for obtaining the values)f rd at various depths. In this figure the range of rd for different soil profiles alongwith the average valuelpto depth of 12 ni is shown. The critical depth for development of liquefaction is usually less than12 m.

rd0

00'2 0-4 0.6 1'0

£.-Q.~

0 18

Range ofdifferent soilprofiles

Average value6

- 12E......

24

30

Fig. 7.25 : Reduction factor rd versus depths (Seed and Idriss, 1971)

The actual time history of shear stress at any point in a soil deposit during an earthquake will be asshown in Fig. 7.6. According to Seed and Idriss (1971), The average equivalent uniform shear stress 10\'is about 65 percent of the maximum shear stress 1max'Therefore

rh'tav = 0.65 -.amax .rd

'. gThe corresponding number of significant cycles Ns for 'tavis given in Table 7.3.

Tabl~ 7.3: Significant Number of CyclesNs Corresponding to 'ta\'-

...(7.25)

Earthquake magnitude, Mon Richter's scale

.- Ns

77.5

t... .~ '.~ 8 . -. ..

102030

-~-

~

308 Soil Dy/tamics & Machi/te Follllda/i~ns

The procedure of locating liquefaction zone can be summarised in following steps:(i) Establish the design earthquake, and obtain peak ground acceleration amax'Also obtain number

of significant cycles Ns corresponding to earthquake magnitude using Table 7.3.(ii) Using Eq. 7.25, determine 'tav at depth h below ground surface.

(iii) Using Fig. 7.22, determine the value of «Jd/2 <J3)for given value ofDso of soil and number ofequivalent cycles Ns for the relative density of 50%.

(iv) Using Eq. 7.22, determine the value of (th

) for the relative density DR of the soil at site.crv field. OR

Multiplying (t h

) with effectives stress at depth h, one can obtain the value of shear stresscrv field

OR'th required for causing liquefaction.

(v) At depth h, liquefaction will occur if

'tav > 'th

(vi) Repeat steps (ii) to (iv) for other values of h to locate the zone of liquefaction. 'tavand 'th can beplotted as shown in Fig. 7,26,

.

Equivalent peak shear stress¬æòææòþæóææþóþôô¢ôô ,,;,-,,-,,",.,.-,- ..'.-' '- - '- ." . ,- -, - - .' ,.,..

~. . - Critical depth for.~,~.Oepthof water- liquefactionJd --"',.tab~~ -'-. GWT -.::---

--c::.I

U

0-~

:J\11

"0C:J0~0'1

c::.I£.-

Zone of initialLique faction

~0....~I

\£. Equivalent cyclica. shear stressc::.I developed due~0 earthquake in-N°s \

cycLes

L Peak cyclic shear stressneeded to causeliquefaction in Nscycles Oaboratory)

, '

Fig. 7.26 : Zone of initiai liquefaction in field

fuefaCtion of Soils 309

1.0VIB~TI9,N TABLE STUDIESl vibrationta~le studies,a large specimenof saturatedsand is preparedin a tank which is placedon abration' table~fig. 7.27 sho:ws atypical setup of a horizontal shake table available at University ofoorkee.Ji conSlslsofa rigid platform on which the test tank (LOSm x 0.6 m x 0.6 m high) is mounted.he platform with wheels' rest~ on four knife edges being rigidly fixecl.on two pairs of rails anchored tole foundation. The platforin is connected with motor and brake assembly for imparting vibrations. Somernportantcharacteristicsof the table are: .

.. . .

.Amplitude ~ 0-10 mm .

.Frequency - 0-20 HzAcceleration - 0-20 g

"

Facilities are available for measuring pore/pressures at different depths in the sample placed in tank.[he procedure of carrying out test is simple. Firstly the sand is placed in the tank under saturated con-jition. The table is then excited with the desired amplitude and acceleration. Variation of pore pressurewith time and number of cycles are then noted. .

The main advantages of vibration table studies are:

(i) It simulates field conditions in better way as the size of sample is'large, prepared and consoli-dated under anisotropic conditions.

(ii) It is possible to trace the actual pore-water pressure distribution during liquefaction.

(iii) Deformation occurs under plane strain conditions.

(iv) Visual examination of sample during vibration is possible.

..(a) General view

'Fig. 7.27 :.Horizontal shaking table (..~Contd.)

1. Vibratory system

2. Horizontal shakingtable with a tankmounted on it.

3. Settlement measur-ing Device. .

4. Pore-Pressure mea-suring system

3tO

1. Motor

2.Pulley3. Eccentric wheels4. Crank-shaft

5. Connecting rod6. Fly wheel7. Hand break8. Revoluation counter

,~,

Soil Dynamics & Machine Foundations~"..,

(b) Vibratory system

Fig. 7.27 : Horizontal shaking table

Since 1957, many investigators have studied lIquefaction characteristics of sand using vibrationtable on different sizes of soil samples and dynamit: characteristics of load. The effect of the followingaspects have been studied:

(i) Grain size characteJ:istics of soil.

(ii) Relative density.

(iii) Initial stress condition i.e. overburden pressure.

(iv) Intensity and character of excitation force.

(v) Entrapped air.

The important conclusions drawn from vibration table studies are:

1. For a given sand placed at a particular density, there is sudden increase in pore pressure at adefinite acceleration. This is termed as 'critical acceleration'. Critical acceleration is not uniqueproperty of sand. It depends on the type of sand, its density, the amplitude and frequency ofoscillation and the overburden pressure (Maslov, 1957; Matsuo and Ohara. 1960: Florin andIvanov,1961).

2. If sand is subjected to shock loading, the whole stratum liquefied at the same time, while understeady-state vibrations, the liquefaction starts from the top and proceeds downward (Florin anpIvanov,1961). .',

3. As the surcharge pressure incre~sed, the number of cycles required to cause liquefaction 'in.creased (Fig. 7.28; Finn, 1970). Tests have shown that, even small drainage surcharge wilreduce the time of the liquefied state tenfold (Fig. 7.29; Florin and Ivanov. 1961).

le/action of Soils

; 70E

..........zoX. 5.6-~I-:J 42VIVI~I-a. 28~0'1I-a.c\JI-:JI.f'I

~

6

x

First liquefactionthin membraneFirst liquefactionthi c.k mem braneFirst liquefaction resultsfor old container A

100

.'

Ac c el era tron 0.25 9Frequency 2 Hz

eo 0.67

A

1000 10,000to first liquefaction "

Fig. 7.28 : Effect of surface pressure on resistance to Initial liquefaction In vibration table studies (Finn, 1972)

22.5

20

/-:~.~r. , ..'

~C}-()

..-.tr 6)f--j(

00 50

- ---

, .'

q

q (k NI m2)O.2

2050100

75 10025

Fig. 7.29 : Influenceoftbe intensity-or dl'alnlng~w:cbarge on tbeperiod-of1lme within whichthe sand remain liquid (Florin"and Ivanov, i961i

-

Number of impacts

311

~--

15-1/1-C0

v10c-

C>I::ItT

-0

'" 5E

'""~-, --

312 -,'-' , ,.-, , - '. -' -' - So.il Dynamics & Ma~hine Foundaiiolls'l-'.-.~-

4, The time during which the ~iquefiedstate lasts is much less for coarser grained soils than for finegrained' sbiiS (Fig: 7,'30.,Gupta,'l979). He carried out liquefaction studies on four sands namely(i) Ukai sanq (Dso=,l.& mDJ.}i(it) Obra sand (Dso = 1.0 mm), (iii) Tenughat sand (Dso = 0.47mm) and Solani sand (Dso = 0.15 mm). The maximum pore water pressure developed in about6 to 10 cycles. It started dissipating immediately after attaining maximum valu~. The total timerequired for dissipation was about 6s for Ukai sand, and 20s for 01;>raand Tenughat sands. Thecorresponding value for Sohmi sand was 12Us,it"re~ained c9nstant for ~bout 35s: thu~ ,the timerequired for,dissipation decreases with the increa~e in coarseness.

Time,s

Eu

0,80'220

0.4 1.2 '--,., ?- ", '

"12 200-04

<:I}L-::I

':: 10<:I}L-a.

Accl. 10010 9,5Hz '. ,

155mm depth

-/",Sotani.sand,

';~'~ ',:- " ,",:-::-'!, ~"

ñÌ»²«¹¸¿¬ sandObra sandUkai sand<:I}

L-ell. 01 40

Number of cvcles200 400 600 1OC

Fig. 7.30: Pore pressure versus number of cycles for different sands (Gupta, 1979)

Since the liquid state lasts for only a short time, the liquefied masses of soil have no time fordisplacements, so that there is practically no indication that the phenomenon of liquefactionoccurs in coarse-grained soils.

5. The excess pore-water pressuresdec!e,ases with the increase in initial relative density (Maslov,1957; Gupta, 1979). Figure 7.31'shows a typical test data indicating the effect of relative densityon an increase in pore pressure at 10 percent g for Solani sand (Gupta, 1979). In this case, nopore-water pressure increase was observed when initial density became 62 percent. Tests per-formed on other types of sand with different accelerations gave the values of relative densities aslisted in Table 7.4 beyond which no pore-water pressure was observed.

Table 7.4 : Initial Relative Density Beyond which no Excess Pore Pressure Develops (Gupta, 1979), -.

Intia/ Re/ative DensityAcceleratioll - - - - - - - - "7",...- - - - - ~ -:-- - - - - - - - - - -

(g) So/ani,sa.~ld TefJughatsand Obra sand Ukai sandPercent 65Q--(O.15m'!IL ,<:~~-{O.4!),. .,' p.Omm) (1,8 mm)

10 62.5 52.0 51.5 50.520 62,S 61.5 60.0 59.540 66.0 64.0 62.5 62.050 ":','. ..0', "'66.5' .) 65.0 ,;".,~...;';'; ',..- 64'.0 '63.0

. .. " -;', " . ." '

t

efaction of Soils

360

32

- 240EE-~ 200\-::JIfjIfj~ 160L.-

a.

~L.-aCl.

280

Oepth(mm). 600 1550 250

u/CTv /> 1 completeliquefaction

1-5

1.0

0-5

0

-90

is;;

313

u

ay

Fig.7.31: Pore pressurevs.initialrelativedensityin Solanisand(Gupta,1979)

DeAlba, Seed and Chan . (1976) presented the results of shake table tests in the form of stress ratiola) and number of cycles required for causing liquefaction as shown in Fig. 7.32. It indicates that foriven value of'th' more number of cycles are required for liquefying a sand having more relative density.is is a similar conclusion as obtaine~ by ~eed and Peacock (1971) by cyclic simple tests. In shake tableIS, the value of'th is given by : . -

. ..- . W 'o'tho=-.amg

where W = Tot~l p~essure exert~d OJ?-the bas~ of t~nk p~ac_edon t!1eshake table 00-am = P~ak ac~eleration of die uniform cyclic mQtion

" ---~- -.......~,--""" "~-- '~""'0..., "

"CA... "" '."C\

---Excess porepressure

u/ov

120

80

40

020 30 40 50 - 60 70 80

Initial relative density (°/0)

...(7.26)

314 Soil Dynamics & Machine Foundationsa

0L.

0.4+-

11'1111~ 0.3~ ~ '

,

'U .r::.

~ P 0.2u~L.L.au

OR (0/0)

¢ êèe--e. 54

0.11 10 100 1000

Num be r 0 fey cl e S J Ns

Fig. 7.32: Corrected 'Ch/cry versus Ns for initial liquefaction from shake table studies (DeAlba, Seed and Chad, 1976)

7.11 FIELD BLAST STUDIESIn blast tests, predetermined charge (like ammonia, gelatin etc.) with electric detonators is installed atpredetermined depth in a cased bore hole. The hole is later filled with sand and the casing ~swithdrawn.Lead wires from detonators are connected with blaster so that the charges may be fired at any desiredmoment. Acceleration pickups are placed at regular intervals from the blast point to record horizontaland vertical acceleration at the time of blasting. Similarly porewater pickups and settlement gauges areplaced at certain distances from the blast point to record the increase in pore water pressure and groundsettlements, Accelerations, porewater pressure and ground settlement at the blast point are then obtainedby extrapolation. The data is then interpreted to obtain the liquefaction potential.

One of the main purpose of carrying out blast tests is to ascertain whether the soil at the site willliquefy under simulated earthquake loading. Field data using small explosives at some depth at the sitealongwith pore pressure and settlement observations for predicting liquefaction potential are available irliterature from few investigations (Florin and Ivanov, 1961; Kummeneja and Eide, 1961; Krishna amPrakash, 1968; Prakash and Gupta, 1970; Arya et aI., 1978; Gupta and Mukerjee, 1979).

For examining the chances of liquefaction at barrage site, Gupta and Mukerjee (1979) performe<blast tests in a river bed having the soil profile as given below:

Depth . Description of Soil Average N-va/ue Remarks

0-4 m4m-7m7 m - 20 m

Fine sandClaySilty sand mixed,with kankars

'5 .", -., ... ' . .

- Cu ='2.47, °50 = 0.12 mmPosition of water table near the surfal

21-60(increases.with depth) Cu = Uniformity coefficient

The critical hydraulic gradient of top loose sandy deposit works out,to be 0.8. Special gelatin ((percent, 2 kg) was installed at 4m depth in 150 mm diameter cased bore holes. Blasting was done withe help of an electric exploder. Horizontal surface accelerations were measured using acceleration pieups placed at various distances fro,mthe source point. The depth of each acceleration pickup was 200 ITbelow the ground surface. The porewater pressures at vari~us .dls~ancesfrom sourc~ point were measurat 2.5m depth from the ground surface. ù

iIIIb.,


812

B:flOm , .

Srn~ B130 Srn

¢º»®»²½»point

814. .

B15 -. 816

Fig. 7.33: Site layout for field blasting tests (Gupta and Mukerjee. 1979)

,Figure 7.33 shows the sketch of layout of the tests at the site. A typical acceleration record obtained

lt 35 m distance from the blast point is .shown in Fig. 7.34. Variation of maximum horizontal accelera-jon and porewater pressure' with dis;tan~~'from blast point are shown respectively in Figs. 7.35 and 7.36.

Fig. 7.34: Surface ac:c:eleration.(Guptaand Mukerjee. 1979)

B7

I10m

I 86

r

1am I

8S

I

10m I

84I

10m

, 316 Soil Dynamics & Machine Foundations

On the basis of past earthquakes, the maximum possible acceleration record at the site is assumed asshown in Fig. 7.37. Using the method of Lee and Chan (1972) this earthquake record is worked out tobe equivalent to 19.6 cycles of 0.075 g acceleration (Table 7.5)

00 10 20 30 40 50

Distance tor blast (m)8060 70

Fig. 7.35 : Acceleration versus distance (Gupta and Mukerjee, 1979)

II

112'9 I 1 I I2'5 5,0 7'5 10.0 12.5

'Distance trom blast (m). Fig. 7.36: Port: jJressure vs. distance (Gupta and Mukerjee, 1979)

).0

2.5E

C>I~ 2.011\ 11-740/1C>I

:i 1.5C>I...0a. 1.0uE0 0-5c>-

0

00

Depth of blast = 4 mPore pressure measuredat 2.5 m

15'0 17-5 20-0

" ,

24

20

-0'10 16

;---c0 12;::.0...C>I- 8C>Iuu0

.3C 40C>Ia.

\

\\

0 (

"((

"

i'-. "'.

""'("-- .-.....-'--

iquefactionof ~oils

-0

,."

:,;.

. ..

/'

-,"",~~",

317

','""

N<-

0.--

., .

.;

0

0 c-,"

0 0 (3 0 00

'C~('6 '~hruo q DJ ~l:~"J"\1'- -'"'.y"

., "'J lP.!)J:;1 .;:r>"'r ,:.; . Yh"."h..V ;~(!.,.J;"k\/t,:"J .JVi..,.at~fl:t j. 11.'-,

, "

-" . ,

co0

.0

0

.,,-"

"'

t.' "

co 0--r--0---.;.......

.::0:=

:::.--- "Cc

If) <":<":-- C.=

E- .:.

.2c

...:r -E......'"'"I- 0"

..r--'-:r--oD

N

318 Soil Dynamics & Machine FoundatWd

Table 7.5 ~EquivaleDt Cy~les for Anticipated Earthquake

Acc!' level Average acc!. Number Conversion Equivalentin percent in percent of cycles factor number ofof peak acc!. (Fig. 1.6) cycles at 0.651: .- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _"J.iL

(1) (2) (3) (4) (5)100-80 90 17/2 = 8.5 2.6 22.180-60 70 8/2 = 4.0 1.2 4.860-40 50 25/2 = 12.5 0.20 2.5

40-100 20 > 1000 negligible 0.0Total 29.4

Total number of cycles for 0.75 1:max= 29.4/1.5 = 19.6

22

20 -~19'6i\1\I .,.......

0- 16~z-

0"1

~r--00

12

111

~-u>-u

\-0

0

0

8...c~0>:JerUJ

4IIIIJ

00 2.9 . 10 20 30. Distance from bLast (m)Fig. 7.38: Equivalent cycles versus distance (Gupta and Mukerjee, 1979)

40

. Liquefaction of Soils '319

Similarly the blast records at different distances are also converted into equivalent number of cyclesof 0.075 g acceleration (Fig. 7.38). From this figure it can be observed that vibrations generated due toblast at a distance of2.9 m are equivalent to 19.6 cycles of 0.075 g, the expected earthquake i.e. the blasthas the same severity as the design earthquake at a distance of 2.9 m from the blast hole.

From Fig. 7)6, the pore pressure developed at a distance of 2.9 m, and at a depth of 2.5 m is1.74 m of water column. The critical hydraulic gradient for this site is 0.8, therefore at a depth of 2.5 mthe critical porewater pressure or hydraulic head is 0.8 x 2.5 = 2.0 m. The pore pressure developed is1.74 m. The actual porewater pressures developed will be larger than the measured value of 1.74 m,

'~ecause there will'be a time lag in rise of water level in piezometer pipe. Hence under the above condi-tions, a larger pore pressure is expected to be developed and complete liquefaction of site is expectedduring the earthquake.

7.12 EVALUATION OF LIQUEFACTION POTENTIAL USING STANDARD PENETRATIONRESIST ANCE

The standard penetra~ion test is mostcommonly used insitu test in a bore hole tohave fairly ~ood estimation of relative den-sity of cohesionless soil. Since liquefactionprimarily depends on the initial relativedensity of saturated sand, many research-ers have made attempt to develop correla-tions in liquefaction potential and standard...penetration resistance. IS: 2131-1981 givesthe standard procedure for carrying outstandard penetration test. SPT values (N)obtained in the field for sand have to becorrected for accounting the effect ofovberburden pressure as below:

NI = CN . N ...(7.27)NI = Corrected value of stan-

dard penetration resis-tance

CN = Correction factor(Fig. 7.39)

The correlation between ~1 values andrelative density of granular soils suggestedby Terzaghi and Peck (1967) is given inTable 7.6.

~. .",

...

0.40

Corr~ction factor, CN0.8 la 1.2 1-4 1-6 2-00.6 1.8

'"'N

E-2oX

-- 100t>I~:JIIIIIIt>I~

c. 200ct>I"~:J

.DLt>I

~ 300-0v-~t>I>t>I> 400Ut>I--tU

500

Fig. 7.39: Chart for correction ofN-values in sand for inOuenceofoverburden pressure(pecketal., 1974)

-

J .

--/.

I /'. /

/

II

1/

//I .

320 Soil Dynamics & Machine Foundatio"

Table 7.6 : NI and, related to Relative Density

NI

0-44-1010-3030-50> 50

- After the occurence of Niigata earthquake, Kishida (1966), kuizumi (1966), and Ohasaki (1966studied the areas in Niigata where liquefaction had not occured and developed criteria for differentiatintbetween liquefaction and nonliquefaction conditions in that city, based on N-values of the sand deposit(Seed, 1979). The results of these studies for Niigata areshown in Fig. 7AO.Ohasaki (1970) gave a usefurule of thumb that says liquefaction is not a problem if the blow count from a standard penetration tesexceeds twice the depth in meters.

0

-N

E.........

z~ 50-~L-~VI\11~L-0.

100

III Light damage &no liquefactionI

Heavy damage.and liquefaction

u~--

UJ

--- Boundar y determined by damage survey (Kishida)---Boundary determined by field observation(Kuizumi)- Ohasaki

10 30 40

c(:,)1JL-:J

.DL-(:,)>0

(:,)>

150

20S tanda rd penet ra t ion resi stance (N blows)

Fig. 7.40 : Analysis of liquefaction potential at Niigata for earthquake of June t 6, t 994 (Seed, 1979)

...

200

CompaCtness Relative Density q,DR (%) (Deg.)

Very loose 0-15 < 28Loose 15-35 28-30Medium 35-65 30-36Dense 65-85 36-41

Very dense > 85 > 41

On the basis o(more comprehensive study on the subject and data presented by other investigators(Seed and Peacock, 1971; Christian and Swiger, 1976; Seed, Mori et aI., 1977), Seed (1979) proposed thefollowing procedure for liquefaction analysis:

(i) Establish the design earthquake, and obtain the peak ground acceleration Qmax'Also obtain numberof significant cycles corresponding to the magnitude of earthquake using Table 7.3.

(ii) Using Eq. 7.25, determine "Cavat depth h below ground surface.(iii) Determine the value of standard penetration resistance value (N) at depth h below ground sur-

face. Obtain corrected NI value after applying overburden correction to N using Fig. 7.39.

(iv) Using Fig. 7.41, determine ('Ch/crv)for the given magnitude of earthquake and NI value obtained

in step (iii). Multiplying ('Ch/crv) with effective stress at depth h below ground surface, one canobtain the value of shear stress "Ch required for causing liquefaction.

NE 0-6.........z..x

~ 0:J 0 .1.1\-1.1\ 11 0 5KI~ -

~ L..L.. 0

8.- ~

\J .9'- - 0 - 4- C

\J c::v>--\J 0

..:.:: Q.Cl cc::v .-Cl. 2 0.30"1 -C 1.1\

'- L..1.1\ Cl:J c::vCl ..c.\J III

113 ~ O.2--- .-r== .~

0 ..c..- -

Liquefaction of Soils

- .-ClL..

0--III 01.1\ 0c::v 0L.. ....-III 0.~ 0- .-v ->-ClU L..

-'.~:

321

Solid poi nts indicate sites andtest conditions showing liquefaction

Open points indicate siteswhere no liquefaction0 cc urred.

86'5

.6.5

07.8

07.8

. Based on field data

;}

Extrapolated from results11 of.large scale laboratorytests

00 10 20 30 1,0 50Modified1.i:)~netration resistance> Nl-blows/ft.

Fig. 7.41 : Correlation between field liquefaction behaviour or'sands for level ground conditionsand penetration resistance (Seed, 1979)

iI... ,I;

322 Soil Dynamics & Machine Foundation..

(v) At depth h, liquefaction will occur if

'tav > 'th

(vi) Repeat steps (ii) to (v) for other values of h to locate the zone ofliquefaction.

Iwasaki (1986) introduced the concept of liquefaction resistance factor FLwhich is defmed asR

F =-L L

R is the ratio of insitu cyclic strength of soil and effective overburden pressure. It depends on relativedensity, effective overburden pressure and mean particle size. It is given by

For 0.02 < Dso < 0.6 mm

...(7.28)

~(

0.35

)R = 0.882 V~~-+70 + 0.225 10glO DSQ...(7.29 a)

For 0.6 < Dso < 2.0 mm

R ~ 0.882~a, ~70 - 0.05 ...(7.29b)

where N = Observed value of standard penetration resistance

crv = Effectiveoverburdenpressureat the depthunderconsiderationfor liquefactionexami-nation in kN/m2

D50 = Mean grain size in mm

L = is the ratio of dynamic load induced by seismic motion and effective overburden pres-sure. It is given by

where

a crL - max v '"- -.=:-. d

g crvamax = Peak ground acceleration due to earthquake

= 0.184 x 10°.320 M (Dfo.s. g

M = Magnitude of earthquake on Richter's scale

D = Maximumepicentraldistance in km. (Fig 7.42)°v = Total overburden pressure

rd = Reductionfactor to accountthe flexibilityof the groundg = Acceleration due to gravity, mIs2rd = 1 - 0.015 h

. h = depth of plane below ground surface in m

For the soil not to liquefy FL should be greater than unity.

...(7.30.

...(7.31

Liquefaction of Soils 323,

9

5

Nu mbers represen t theearthquake numbers8

~ 30..g 7:J-C

0'1 60~

Mean line

L09100 = 0-87 M-4-S

Lower bound

l0910 0 =0.77M - 3'6(M>6)

41

'r 0

2 5 10 20 50 lOO" 200 500 1000

Maximum epicentral distance ot liquefied sites,D (km)Fig. 7.42: Relationship between the maximum epicentral distanceofliquefied sites (D)

and earthquake magnitude (M) (Kuribayashi, Tatsuoka and Yoshida, t 977)

7.13 FACTORS AFFECTING LIQUEFACTION ~ -Although the factors affecting liquefaction have been discussed during the laboratory and field studies onliquiefaction, they are summarised below: '

7.13.1. Soil Type. Liquefaction .occursin cahesionless soils as they lose their strength completely undervibration due ta the development of pore pressures which in turn reduce the effective stress to zera. ,Liquefactian daes not occur in case of cohesive soils. Only highly sensitive clays may laase their strengthsubstantially under vibration.

7.13.2. Grain Size and Its Distribution. Fine and uniform sands are more prone to liquefaction thancoarser ones. Since the permeability of coarse sand is greater than fine sand, the pore pressure develapedduring vibrations can dissipate faster.

7.13.3. Initial Relative Density. It is .one .ofthe mast important factars contralling liquefactian. Bathpare pressures and settlement are canstderably reduced during vibratians with increase in initial relativedensity and hence chances of liquefaction and excessive settlement reduce with increased relative den-sity.

'7.13.4. Vibration Characteristics. Out .ofthe four parameters .ofdynamiuc load namely (i) frequency;(ii) amplitude; (iii) acceleratian; and (iv) velacity; frequency and acceleratian are mare impartant. Fre-quency .of the dynamic laad plays vital rale ,.only if it is clase ta the natural frequency .of the system.Further the liquefactian depends .onthe type of the dynamic laad i.e. whether'it isa transient laad .orthe

, . ,,', . . ..11 I" . ""', " .laad causing steady vibratians.'.'" , ' '0.' . . r, ' :, , . , . '" ~L'>+::' ' . .:

324 ' Soil Dy"amics & Macll ille Folllldatiolls

Whole stratum gets liquefied at the same t~eunder tra~sient loading, while it inayproceed fromtop to lower layers u?der steaqy state vibrations (Florin and Ivanov, 1961). ,For a given acceleration,liquefaction occurs only after ~ certain number of cycles imparted to the deposit. Further, horizontalvibratiorls have more severe em~-c.tthan vertical vibrations. Multi directional shaking is more severe thanone directional loading (Seed ~t al.~ 1977), as the pore water pressure build up is much faster and thestress ratio required is about 10 percent less than that required for unidirectional shaking.

7.13.5. Location of Drainage and Dimension of Deposit. Sands are more pervious than fine grainedsoil. However, if a pervious deposit has large dimensions, the drainage path increases and the depositmay behave as undrained, thereby, increasing the chances of liquefaction of such a d~posit.The drainagepath is reduced by the introduction of drains made out of highly pervious materiaL"

7.13.6. Surcharge Load. If the surcharge load, i.e. the initial effective stress is large, then transfer ofstress from soil grains to pore water will require higher intensity vibrations or vibrations for a longerduration. If the initial stress condition is not isotropic as in field, then stress condition causing liquefac-tion depends upon Ko (coefficient of earth pressure at rest) and for Ko > 5, the stress condition requiredto cause liquefaction increases by at least 50%.

7.13.7. Method of Soil Formation. Sands tmlike clays do not exhibit a characteristics structure. Butrecent investigations show that liquefaction characteristics of saturated sands under cyclic loading aresignificantly influenced by method of sample preparation and by soil structure.

7.13.8. Period Under Ssustained Load. Age of sand deposit may influen<:eits liquefaction characteris-tics. A 75% increase in liquefaction resistance has been reported on liquefaction of an undisturbed sandcompared to its freshly prepared sample which may be due to some form of cementation or welding atcontact points of sand particles and associated with secondary compression of soil.

7.13.9. Previous Strain History. Studies on liquefaction characteristics of freshly deposited sand and ofsimilar deposit previously subjected to some strain history reveal, that although the prior strain historycaused no significant change in the density of the sand, it increased the stress that causes liquefaction bya factor of 1.5.

7.13.10. Trapped Air. If air is trapped in saturated soil and pore pressure develop, a part of it is dissi-pated due to the compression of air. Hence, trapped air helps to reduce the possibility to liquefaction,

7.14 ANTILIQUEFACTION MEASURES

A comprehensive study is required to find out various possible measure to prevent liquefaction. Tho.ughit depends on -a number of factors, however, few can be controlled in field. Based on these, certainmethqds have been suggested (Lew, 1984). Liquefaction resistance to some extent can be improved by :

7.14.1. Compaction of Loose Sands. As has been indicated earlier, loose saturated sands are more proneto liquefaction than dense saturated s~nds. Therefo~e!the liquefaction potential can be,redu~ed by coI?-patting the loose sand deposit before any structure is constructed. The various '-methodssuggested forcompaction of loose' sands in'situ are: ' . .' .h ,. .- . ' . - '

7.14.1.1. Rolling with rubber tyrerollers: It may be accompli~hedby excavati~g some depth, then care-fully backfilling in controlled lift thickness and compacting the soil. When ruJ>bertyres are used, lifts arecommonly 150 mm to 200 mm. This method, however cannot be used for compacting deep sand deposits.

;qllefactioll of Soils 325

,14.1.2. Compaction with vibratoryplates and vibratory rollers: Compacti'onof cohesionless soils canc achieved using s~ooth wheel rollers commonly with a vibratory devIce inside. Lift depths upto ahout.5m to 2m can be compacted with this equipment (Bowles, 1982). Also plates mounted with vibratory~semblycan be used; however, small thifkness of soils ca~ be compacted by these methods and they can'ot be used for large deposits.' . ,.

.14.1.3. Driving of piles: Piles when driven in loose deposits, compacts the sand within an areaovered by eight times around it. This concept may be utilized in compacting the site having loose sand\eposits. As pile remains in the sand, the overall stiffness of the soil stratum increases substantially.

'.14.1.4. Vibrofloatation : The method is most commonly used to densify- cohesionless deposits of sandsL11dgravel with having not more than 20% silt or 10% clay. Vibrofloatation utilizes a cylindrical penetrator.t is an equipment of about 4m long and 400mm in diameter. The lower half is vibrator and upper halfs stationary part. The device has water jets at top and bottom. Vibrofloat is lowered under its own weightvith bottom jet on which induces the quick sand condition, when it reaches the desired depth, the flow)f water is diverted to upperjet and vibrofloat is pulled out slowly. Top jet aids the compaction process.\s the vibrofloat is pulled out a crater is formed. Sand or gravel is added to the crater formed.

7.14.1.5. Blasting: The explosion of buried charges induces liquefaction of the soil mass followed by~scape of excess pore water pressure which acts as a lubricant to facilitate re-arrangement and thusleading the sand to a more compacted state.

The earliest use of detonating buried charges of explosive for compacting loose cohesionless soils in[heir natural state has been reported by Lymari. (1942). He concluded that

(i) Lateral distribution of charges should be based on results obtained from a series of single shots.,~

(ii) Where loose sands greater than 1Omthick are to be compacted, two or more tiers of small chargesare preferred.

(iii) For deposits less than 10m thick, charges placed at 2/3rd depth from surface will generallysuffice.

(iv) There is no apparent limit of depth that can be compacted by means of explosive. .

Later Hall (1962) reported that

(i) Repeated blasts are more effective than a single blast of several small charges detonated simul-taneously.

(ii) Very little compaction can be achieved in top 1m.

(iii) Small charges are more effective than large charges for compacting upper 1.5m of sand. .'

(iy) The compaction gained by repeating the blasts more than 3 times is small.(v) The relative densities can be increased to 80%.

. -. . ..

7.14.2 Grouting and Chemical Stabilization. Grouting is a technique of inserting some kind of stabi- '-lizing agent iI:1tothe soil mass under pressure. The pressure forces the agent into the soil voids in a limitspace around the injection tube. The agent reacts with the soil and/or itself to form a stable mass. Themost common grout is a mixture of cement and water, with or without sand. Generally grout can be usedif the permeability of the deposit is greater than 10-5 m/s. Chemical stabilization is in the form of lime,cement, flyash or combination of these.

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

326 Soil Dynamics & MJlcmne Ftnmdations

7.14.3. Application of Surcharge. Application of surcharge over the deposit liable to liquefy can also beused as an effective measure against liquefaction. Figure 7.43 shows a plot between rise in pore pressureand effective over burden pressure at an acceleration of ten percent of g. It indicates that pore pressureincreases with increase in overburden pressure till a maximum value of pore pressure is reached, afterwhich it starts decreasing with further increase in surcharge. Thus an overbuden prssure above this value,depending upon the situation, makes the deposit safe against liquefaction.

24Dead weight surcharge250 mm depth

_Acct. 10% g-No. of cyc les 10

Solani sard201- DR=20 %E 15Hz"z

-"- 16Cl!L-:J~ IZone ot liquefaction~ 12 .L-a.

- -8 12 16 20- 24 28 32 36

Effective overburden pressure (kN/m2 )

40 44 48

Fig. 7.43: Excess pore water pressure versus initial pressure on Solani sand (Gupta, 1979)

7.14.4. Drainage Using Coarse Material Blanket and Drains. Blankets and drains of material withhigher permeability reduces the length of drainage path and also due to higher coefficient of permeabil-ity, speed up the drainage process (Katsumi et a1. 1988 and Susumu et a1. 1988).

7.15 STUDIES ON USE OF GRAVEL DRAINSYoshimi and Kuwabara (1973) were first to introduce gravel drains to stabilize a potentially liquefiablesand deposit. Seed and Brooker (1976) have proposed an analytical procedure for designing such drains(Fig. 7.44). These drains are considered fully effective if the permeability of material of drains, ktJ' isabout 200 times the permeability, ks' of the soil in which they are installed i.e. (kd/ks) > 200. The effectiY,edrainage path is reduced by ,the introduction of number of artificial drams. Seed and Brooker (i97~)developed nondimensional charts as shown in Fig. 7.45 for determining the spacing of drains. Vario~5. .' '1terms shown on this figure are as below: ., . .'

." '. .< . ill(

. . , , .. ~

.~

Cl!L-80

a.IIIIII

-.CI!4v

)(

UJ

00 4

'--~q~~e!aftioll; °t'~~Lf'.,C,', \, "

327f:~;,

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

c,:

, "

e'

,,-'.

.e' . '-"",'..'""

(0 ) Plan

."

'..

. ':,,', ":-',~-,','~":",.',,-.

~ "::~~LG_WT ~':::'{. ',.-',-

",' I:',,~"'-::-

, :. ','...";- :;""",-,-, ',.',-,-.-..- ': : . .' ~ "'.: "':, ~~~I'..','- - ,~ ,-,..' ,..-.

t ' 'r"1 /'~:'"

" "',, ,',- ,.

,',, '

" .i, , c '

, ..

J -',

Re' '", ''..' ,

"r, :,',

t"" ""'N

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

, Re, .

... ; "~

.:2"::::., .'~';;':!"":~',~~ J-".': ,:;.,,-y,~~:,-J'.:-) Rd~ $"

""', ~ ,". .,., ,~'

:: 'G :,/ " .; ;')

(b) Sec tior.' ot, S-S.. .'

¥' \ ,,-, \~\

: i."".', . , ,;"::,. ;'"" {',,"-," t' '"It> -".' ,fi.g,.7.4,4 :,.:Gravel drains,","'- ,-.".",..:',.,'"""c.<,"..,-,,, -";;'"""",,"'-!"~'~;,,,,~;'\"""

328

0-6rg

0.4

SoU Dynamics & Madine Founda~s

<-

0 .21-1f:J°Tad-:''. '

0.2 0-3 0.4R.d/ Re - -

(b) NiNl =2Fig. 7.45: Relation between gr~3test perewater pressure ratio and drain system parameters (...Contd.)

1.0

rg0-4

0-2

-:.

00

00 0.1 0-5 0-6

0.1 0'2 0'3 0-4 O'S 0-6

Rd / Re

(a) iNl = 1

I! I

lefaaion- of Soils

0-8

.

,.0

0 -6 t- 110

T - 200ad-

rg0.4.

0.2

1.0

0.8

0,6

rg0.4

0.2

00 2.0 3-0 ~.O

Rd / Re5.0 6.0

0.6

Fig. 7.45: Relation between greatest porewater pressure ratio and drain system parameters

1.0

(c) N/Nt = 3

00 0'2.:.0-1 0.30,4 .' . 0,5

Rd / Re

(d) .NslNl = 4

, . .'

o~

~19

"

'J:

'" I

" .' c' .:.

330 Soil Dynamics & Machine FiJII;idat;o;u

Limiting value of Ugchosen for designr = ,- , ..g .' cr -y

-~ -<'~-'~'-

Uu = Excess porewater pressure build up in ~.cyclic 'simple sh~ar test (Fig. 7.46)0 ,

cry = Initial consolidation pressure .

Ns = Number of cyclic stress applicationsNI = Number of stres~cycles needed for liquefaqiion

Rd = Radius of rock or gravel drains, : '

Re = ~fec

(

tiV; rad

J

iUSof the rock or gravel drains

T - h ,d '

ad - '¥ro 'my R~ -

Kh = Coefficient of permeability of sand in horizontal direction''¥ro= Unit ~eight of water' ,'-'

mv = Coefficient,of volume compre~~ibili~_.~fsand,1d = Duration of earthquake" ,

-'

0.2 , 0.4 0~6N/N, ,

0.8 . 1.0

Fi,g. 7.46: Rate of pore water pressure buildup in cyclic simple s,hear test

Yosufumi et al. (1~84) had,developed a method to evaluate liquefacti~~resistance under partiadrained condition, They assumed that the dissipation of excess pore water pressure induced by an ea!quake will occur according to Darcy's-'law.Dynamic triaxial apparatus was used to conduct tests unperfectly undrained and perfectly drained conditions. In case of perfectly uildrained condition, liquettion resistance was neither influenced by permeability of sample nor by the frequency of loading. Thireasonable, because, no drainage of water is involved and hence permeability of soil does not have a I

to play, It was also concluded that in case of p'artial~ydrained condition, the effect of drainagefrequency is remarkable for soils with relatively largef1relative density, OR' and not so significantsoils with smaller relative density: But for loosesoils'rat~ of generation' of pore pressure depend'number of cycles of stresses which in turn depends on frequency. Rate of pore pressure built up deplon the rate of dissipation of pore pressure, which is based on draInage.

1.0

0.8

0.6Ug-ay 0.4

0.2

00

'luefactidn of Soils ",' . 331

Yasushi and Taniguchi (1982) carried out large scale model tests to confirm the effectiveness of'avel drains for preventing liquefaction of sand deposits. The purpose of the tests, as stated by them,as : .

(i) to know the generation and dissipation characteristics of pore water pressure,,. " , '"',-' '".- ',-, " , ,0" ", "

(ii) to clarify the effective are~foithe g~aveldra~fromthe view-p~intof preventing liquefaction and(iii) to know whether the grayel drain.is effective in preventing the liquefaction of subsoils under a

road that is partially buried. ' :' - ,,"", .They performed tests on shaking table of 12m x 12 m x':3ffi(deep) size, filled with cohesionless soiL

'he acceleration of loading was 200 gals, the duration of shaking w,as,one minute and the frequency was, cps. They concluded that pore water pressure within 500 mm from the edge of a gravel drain is muchmaller than that, for away from the gravel drain. '

Wang (1984) made experimental study on liquefaction inhibiting effect of gravel drains. A shaking)oXof size 1.5 m x 0.28 m x 0.5 m was used. He used gravel drains walls under the foundation and it waslssumed that under plain strain condition the walls ar~ referred as drains. It was noted by him that the,ection of non liquefied zone of deposit was basically a trapezoid in which pore pressure ratio (i.e. the"atio of excess pore w<l.terpressure to the effective overburden pressure) was generally below 0.6. Basi-~ally this zone did not reduce<with ~ncreasingvibration time. ' ' , .

As the number of d!'ainsin~talled is increased, the non liquefied zone increases. As the accelerationincreases, the zone reduces gradually but the increase in time does not reduce the non liquefied zone. Theangle of trapezoid was found to be 15° to 17°in the 'direction of depth. The zone is about 40 mm outsid,ethe drains. It was also observed by him that the surface drains may effectively prevent foundation settle-ment. In order to obtain good effect in reducing foundation settlement it must be ensured that adequatedept.I1,~ndwidthQf drains qe inst~lled when ~n~tallingshallow dr~ins and outside drains.

, ,

a-hara and Tamamoto (1987) pi"ese'nteda fundamental 'study on gravel pile for preventing liquefac-tion. They used a shaking box of size 1.0 m x 0.35 x 0.65 m (deep). Radii of gravel piles were 0.75 m,0.15 ~ etc. The flow of pore water was assumed to be horizontaL They measured the pore water pressureat points near the dr~irisa~d away from the 'drai~s.They concluded that liquefaction occured at points toofar from the drain and that at points close to the gravel drain, liquefaction did not occur. Results obtainedby them presented -in form of -optimum radius of pile and optimum spacing between gravel piles.Figure 7.47 shows the effective circle whiChis defined as the circle with area equal to area of square withsides equal to the line joining mid p~ints of the ~pacing between adjacent gravel piles. The sides of thesquares are tak~n as optimum spacing between gravel piles. They found that the effective area of gravelpile increases in proportion to the diameter of the gravel pile and the permeability. For a fixed diameterof pile and permeability of'soil, as the optimum distance decreases, the pore pressure ratio decreases. Asthe permeability increases, pore pressure ratio decrea~es very sharply. Highly permeable gravel are muchmore effective even at higher optimum distance and smaller diameter of drains.

, AJlexible , ve~tical drain formed by us'ing organi,c fibres like jute or coir has been used in severalprojects. The 'most important properties of.such drains are permeability and tensile strength, The jutefilter cover h.~spermeability better than 10-5m/sec. This facilitates the flow of water from pervious lensespresent in the seams and layer of sand and speeds up the pore water pressure dissipation. They have theadvantage of decaying and getting mixed with the soil without harming the environment. When the filterpermeability 'is large, the clogging of the drain has to'be considered. - - - - - -


C ire le of et f e cti ve a re a..

-'-EL:~ l'77'r~'-i" . ~I

Fig. 7.47: Radius and distance (or spacing) of gravel pi/es (O-hara, 1987)

Geotextiles are used fairly widely in surface and subsurface installations, (Krishnaswamy and Issac,1995), Crushed stone wrapped in geotextiles have often been used as surface and subsurface drains.Perforated plastic pipes too may be used for this purpose. They may be filled with crushed stones, ifnecessary.

I ILLUSTRATIVE EXAMPLES I

Example 7.1At a given site, a boring supplemented with standard penetration tests was done upto 15.0 m depth. Thtresults of the boring are as given below:

Remarks

(i) Position of watttable lies 1.5 mbelow the groursurface

(ii) 'YmOiSI= 19 kN/r

'Ysub = 10 kN/m

Depth Classification D50 N-Value DR(m) of soils (mm) (%)1.5 SP 0.18 3 193.0 SP 0.20 5 304,5 SM 0.12 6 356.0 SM 0.14 9 40

7.5 SM 0.13 '. 12 459.0 SP 0.16 17 52

10.5 SW 0.20 20 5212.0 ' SW 0.22 18 46

- 13.0 SW 0.22 24 6015.0 SW 0.24 30 65

.'

iquefaction' of Soils 333..

The site is located in seismic ally ,active region, and is likely to be subjected by 'an earthquake ,of1agnitude 7.5. Determine the zone of liquefaction using ,

(a) Seed and ldriss (1971) method(b) Seed (1979) method(c) lwasaki (1986) method

>olution :

(a) Seed and Idriss (1971) method(i) From Fig. 7.42, For M = 7.5, D = 106 Km

From Eq. 7.31, amax = 0.984 x 10(0.302x 1.5) x 10~.8 = 0.083g 9.81

Number of significant cycles (Table 7.3), Ns = 20 (For M = 7.5)amax .

'tav =0.65yh. - . rdg= 0.65 x Yh x (0.083) rd= 0.054 Yh rd

It may be noted that y h represents the total stress at depth h below ground surface.Value of rd are readfrom Fig. 7.25. Values of total stress, ad and 'tavat different depths are given in Cols, 3 and 5 of Table 7.7.

(iii) For 50% relative density, the stress ratio ad/2a3 is read from Figs. 7.22a and'b for given valuesof,Dso' Average of the two values is the stress ratio for number of significant cycles equal to 20.The stress causing liquefactionat any depth is then c9mputedusing Eq. 7.22. '., .

(ad

)DR -'

'th= 2a "50.Cr.av3 50%

Values of Cr are .obtained using Table 7.2. The details of computations of 'rh are summarised inTable 7.8. .

(ii) From Eq. (7.25)~

Table 7.7 : Computation of 'ray

S.No.(m)

Depth(kN/m2)

Total stress rd tal'(kN/m2)---------------------------------

(I) (2) (3) (4) (5)

1. 1.5 28.50 0.99 1.522. 3.0 58.50 0.98 3.103. 4.5 88.50 0.97 4.644. 6.0 118.50 0.97 6.215. 7.5 148.50 0.96 7.696. 9.0 178.50 0.95 9,147. 10.5 208.50 0.94 10.548. 12.0 238.50 0.92' 11.909. 13.5 r 268.50 0.91 13.23

1p. .15.0 298.50 0.90 14.51

3:J~;,

I.

2.

3,

4.

5.

6.

Id .),' :"

S. No

7.

8.

9,

1O,

(b) S,eed (1979) Method.

" . '(i) In. this method, the value of shear' stress at any depth induced 'by the earthq",lake is obtained, ,.;'. exactly in th,e same manner as illustrated in Seed and ldriss (1971) method (Table 7.7)

(ii) To de,termine 1:hfirstly N-values are corrected for effective overburden pressure using Fig. 7.39.

The stress rati<?1:hla v is then obtained using the relevant curve of Fig. 7.41 for the given value

of corrected N. The details of computations are ~iven in Table 7.9.

Table 7.9 : Detail of Computations of'th by Seed (1979) Method

Soil Dynamics & Machine Foundatiolls

';1'able 7.8 Computations' of 'thfrom Seed and Idriss (1971) Metbod"

Depth Effective stress. cr DR , Cr , .rh/a!, .rh

(kN/m2);,

(m) (kN/nO

1.5 28.50 19 0.55 0.2198 1.313.0 43.50 30 0.55 0.2260 3.24. .'

4.5 58.50 35 ,0.55, 0.2012 . ' 4-.53,6.0 73.50 40 0.55 0.2074 6..717.5 88.50 45 0.555 0.2043 9.039.0 103.50 52 0.573 0.2136 13.17

10.5 118.50 52 0.573 0.2260 15.9612.0 133.50 46 0.556 0.2298 15.6913.5. 148.50 60 0.60 0.2298 24.57

15.0 163.50 65 0.61 0.3336 30.29

S.No. Depth N- Va/ue Corrected Effective .rh/av .rh,(m) ", N stressóóóóóóóóóóóóóóóóùóóóóóóóóóóóóóóóóóóóó

(1) (2) (3) (4) (5) - (6) (7)

1. > 1.5 3 4 28.50:, 0.045 ' '1.28. ." " - -2. 3.0 5 6 43.50 0.067 2.91

3. 4.5 6 7 58.50 0.079 4.624. 6.0 9 10 73.50 0.113 8.305. 7.5 12 13 88.50 0.146 12.926:- 9.0 17 17 103.50 0.191 19.777.. 10.5 20 ' 18 118.50 0.194 22.998. 12,0 18 17 133.50 0.183 24.439. 13.5 24 " 20 148.50 0.225 31.93

I

10. 15.0 30 25 . 163.50 0.269 43.98

l'UeTattwn ofSolls: ,'" ,<""

I Iw-asaki's Method

.--'( i) Fir'stIy the'value of factor R is obtained using the following Eel.7.29a ;. , . , ,,' .. '

R .= 0.882~ +0.225 loglO (0.35

J.. V~:+ 70, °50

The details of computations of factor R are given in Table 7.10.

.,

" Table 7.10 :' DetaH~ of Computations for obtaining Factor R in Iwasaki's Method

a a, 'L = max '~'r- d

. gall

The details of computations are given in Table 7.11. The ratio of factor of safety RlL is listed in theastcolumnofTable7.11. ",' '.,

, .," ' ,

Table 7.11 : 'Details :of Computations of Obtaining Liquefaction Potential by Iwasaki's J\1ethod

S.No., - ., ,Depth (m). , . crJcrv'" rei L FL---------------------------------

. (I).. (2) '. (3),

L.2.3.4. .5,

. 6:7,

, 8. .,9.

10,,- , ::" i'C,:

,).5., 1.0003.0 1.3454.5 1.5136.0 '!' .' .1.6127.5 . ~, '1.678

. ; 9.0 ' '1.72510.5 i,75812.0., :,,-,'~."'L78713.5 1.80815.0. ,'L ; ...;,;',1.8,26 .

;:..\:; ."

, - ...

S,No, Depth ay 'D5o N R

(m)2

(mm)(kN/m )- - - - - -- - '- - - - - - - - - - - - - - - - - - - - - - - - -(I) ,(2) (3) (4) (5) (6)

1. 1.5 28.50 0.18 3 0.21892. 3.0 43.50 0.20 5 0.23983. ' 4.5 58.50 ' 0.12 6 0.29504, 6.0 - 73.50 ,0.14 ,9 0.31045. 7.5 88.50 0.13 12 0.3394"6. 9.0 103.50 .0.16 17 0.35257.',' 10.5 118.50 0,20 20 0.34198. 12.0 133.50 0.22 18 0.30779. 13.5 148.50 0.22 24 0.3377

"1'0. -,':,. ". 15:0 163.50,' .' 0.24 " 30 0:3464

(ii) The factor L is then obtained using Eq. 7.30 :/.

(4) (5) (6), ,

0,9,9 0.0820 2.6700.98 0.1092 2.1960.97 0.1215 2.4270.97 ',0.1295 2.3970.96 0.1334 2.5440.95 0.1357 2.5970.94 . . , 0.1369 2.4980.92 "

0.1362 2.2600.91 0.1363 : 2.4780.90 , 0.136,1 2.545

-,

Soil Dynamics & Ml1ChineFoullda .

In Table 7.12, summary of different methods are given. It is evident from this table that liquefaction ,tloccurs only uptoL5 m depth according to Seed and Idriss (1971) method, 3.0 m depth according to Seed $'(1979) method; and no liquefaction according to Iwasaki's method.

336

Tabl-e 7.12 : Summary of Different Methods

S.No, t'avDepth

(m) (kN/m2)

1.2.3.4.5.6.7.8.9.

10.

1.523.104.646.217.699.14

10.5411.9013.2314.51

1.53.04.56.07.59.0

10.512.013.515.0

/wasakimethod

FL (R/L)

2.6702.1962.4272.3972.5442.5972.4982.2602.4 782.545


Arya, A. S., ~andakumaran, p" Puri, V. K. and S. Mukerjee, (1978), "Verification of liquefaction potential by fieldblast tests", Proc. 2nd. International Conference on Microzonation, Seattle-, USA, Vol. II, p. 865.

Bowles, 1. E. (1982), "Foundation analysis and design", McGraw Hill Book Co. SIngapore.Casagrande, A. (1965), "The role of the calculated risk in earthwork and foundation engineering", 1. Geotech, Engg.

Div., ASCE, Vo\. 91, No. SM4, pp: 1-40, Proc. Paper 4390.

Castro, G. (1969), "Liquefaction of sands", Harvard Soil Mechanics Series No. 81, Harvard University, Cambridge,Massachusetts: . . .

Castro, G. (1975), "Liquefaction and cyclic 'mobility of saturated sands", Journal of the Geotechnical EngineeringDivision, ASCE, 10L(GTS), pp. 551-569.

Castro, G. and S. 1. Poulos (1976), "Factors affecting liquefaction and cyclic mobility ", Symposiumon Soil lique-faction, ASCE, National Connvention, Philadelphia, pp. 105-138.

Christian, J. T. and W. 'F. Swiger (1976), "Statistics of liquefaction and S.P.T. results", 1. Geotech. Engg. Div.,ASCE, Vol. 101,No. G T 11, pp. 1135-1150. . .

Corps of Engineers, U. S. Dept. of the Army (1939), "Report of the slide of a portion of the upstream face at fortpeck dam", U. S. Government Printing Office, Washington, D. C.

DeAlba, P., H. B., Seed, and C. K. Chan (1976), "Sand liquefaction in large scale simple shear tests", J. Geotech.Engin. Div., ASCE, Vol. 102, No. GT9, pp. 909-927. .

Dunn, J. A., 1. B. Auden and A. M. N. Ghosh (1939), "The Bihar Nepal earthquake of 1934", Mem. Geol. Surv.,-' India, Vo\. 73, p. 32. .

Finn, W. D. L., P. L. Bransby and D. 1. Pickering (1970), "EffeCtof strain history on liquefactionof sands", 1. SoilMech. Found. Div., ASCE, Vat. 96, No: SM 6, pp. 1971':J934.

t'h thSeed and Idriss Seed (1979)method (1971)

(kNlm2) (kN/m2)

1.31 1.283.24 2.914.53 4.626.71 8.309.03 12.92

13.17 19.7715.96 22.9915.69 24.4324.57 31.9330.29 43.98

'I


Finn, W. D. L, J. J. Emery and Y. P. Gupta (1970), "A shaking table study of the liquefaction of saturated sandsduring earthquakes", Proc. Third Europ. Symp., Earthquake Engin., pp. 253-262.

Florin, V.A., and P. L Ivanov .(1961), '~Liquefactionof saturated sandy soils", Proc. Fifth Into Conf. Soil mech.Found. Engin., Paris, V01. 1, pp. 107-111.

Gupta, M. K. and S. Mukherjee (1979), "Blast tests for liquefaction studies", Proc. International Symposium onInsitu Testing of Soil and Rock and Performance of Structures, Roorkee, India; vel. I, p. 253.

Geuze, E. (1948), "Critical density of some dutch sands", Proc. 2nd ICSMFE, vol. 1Il, pp. 125-13{). ...Gupta, M. K. (1979), "Liquefaction of sands during earthquakes", Ph. D. Thesis, University of Roorkee, India.Hall, E. C. (1962), "Compacting of a dam foundation by blasting", ASCE Journal, Vol. 80.Hazen, A. (1920), "Hydraulic fill dams", ASCE Transactions, Vol. 83, pp. 1713-1745.Housner, G. W. (1958), "Mechanics of sand blows", Bull. Seismol. Soc. Am., Vo1.48, No. 2, pp. 155-168.lshibashi, 1.and sherif, M.A. (1974), "Soil liquefaction by torsional simple shear device", Journal of the Geotechnical

Engineering Division, ASCE, 100, G T 8, pp. 871-888.Katsumi, M. M. Maraya and T. Miteuru (1988), "Analysis of gravel drain.against liquefaction and its application to

design", IXth WCEE, Tokyo, vol. Ill, pp. 249-254.Kishida, H. (1966), "Damage' of reinforced concrete bu,ildings in Niigata city with special reference to foundation

engineering", Soil Found. Engin. (Tokyo), Vo\. 9, No. \, pp. 75-92. ..~, .

Koppejan, A. W. ,Wamelen, B. M. and L J. Weinberg (1948), "Coastal flow slides in the dutch province of seeland",Proc. 2nd lCSMFE, Vo\. 5, pp. 89-96, Rotterdam.

Krishnamaswamy,N. R. and N. T. lssac (1995), "Liquefaction analysis of saturated reinforced granular soil", ASCE,Vol. 121, No. 9, pp. 645-652.

Krishna, J. and S. Prakash (1968), "Blast tests at obra dam site", J. Inst. Engin. (India), Vol. 47, No. 9, pt. CI5, pp.1273-1284. ..,'

Kuizumi, Y.{1966), "Change in density of Sand subsoil caused by the Niigata earthquake", Soil Found. Engin.(Tokyo), Vol. 8, No. 2, pp. 38-44.

Kuribayshi, E., Tatsuoka, F.and Yoshida, S. (1977), "History of earthquake induced soilliquefactiori in Japan",Bulletion of PWRI, 31.

Kuwabara, F. and Yoshumi, Y. (1973), "Effect of sub surface liquefaction on strength of surface soil", ASCE, JGE,.VoI.19,No.2. .

Lee, K. L , and C. K. Chan (1972), "Number of equivalent significant cycles in strong motion earthquakes", Proc., First lnt. Conf., Microzonation, Seattle, Vol. 2, pp. 609-627.

Lee, K. L. and H. B. Seed (1967), "Cyclic stress conditions causing liquefaction of sands", J. Soil Mech. Found,Div., ASCE, Yo\. 93, No. SMI, pp. 47-70.

Lew, M. (1984), "Risk and mitigation of liquefaction hazard", Proc. YIIlth WCEE, Yol. I, pp. 183-190.

Lyman, A: R. N. (1942), "Compaction of cohesionless foundation soil by explosive", ASCE Trans., Yo\. 107.Maslov, N. N. (1957), "Questions of seismic stability of submerged sandy foundations and structures", Proc. Forth

Int. Conf. Soil Mech. Found. Engin., London, Vol. 1, pp. 368-372.Matsuo, H. and S. Ohara (1960), "Lateral earth pressure and stability of quay walls", Proc. Second World Confer-

ence on Earthquake Engineering, Tokyo, Vol. 1, pp. 165-182. ,

Middlebrooks, T. A. (1942), "Fort peck slide", ASCE Transactions, Vol. 107, pp. 723-764.Ohasaki, Y. (1966), "Niigata earthquake 19'64,.buildingdamag~ and soil conditions", Soil ~ound.(Tokyo), Vol. 6,

No. 2, pp. 14-37. ,', ' , :Ohasaki, Y. (1970), ;"Effectsof sand.c~mpacti~n on liquefaction during the Tokachioki earthquak~", ,S()il.fQ.und.

. (Tokyo),Yol.;.lO,No.-2,'pp.";t112-J28... ';, '.I' ; ! i. ',! "',

338 Soil Dynamics & Machine .Foundation

O-hara, S. and T. Tamamoto (1987)""Fundamental study on gravel pile.method for preventing liquefaction", ECEE87, pp. 5.3/41-48.

Peacock, W. H. and H. B. Seed (1968), "Sand liquefaction under cyclic loading simple shear conditions", 1. SoilMech. Found., Div., ASCE, Vol. 94, No. SM 3, pp. 689-708.

Prakash, S. (1981), "Soil dynamics", McGraw HillBook Co.Prakash, S. and M. K.Gupta (1970a), "Final report on liquefaction and settlement characteristics of loose sand

'" under vibrations", Proc. International Conference on Dynamic Waves in Civil Engineering, Swansea,. pp. 323-328~' . .

Prakash, S, and M. K. Gupta(1970b), "Blast tests at Tenughat dam site", 1. Southeast Asian Soc. Soil mech. Found.Engin (Bangkok), Vol. I, No. 1, pp. 41-50. .

Prakash, S. and M. K. Gupta (1970c), "Liquefaction and settlement characteristics of Ukai dam sand", Bull IndianSoc. Earthquake Tech. (Roorkee), Vol. 7, No. 3, pp. 123-132,

Prakasn, S. and mathur, 1.N. (1965), "Liquefaction of fine sand under dynamic loads", Proc. 5th Symposium ofthlCivil and Hydraulic Engineering Department, Indian Institute of Science, Bangalore., India

Seed, H. B. and K. L. Lee (1966), "Liquefaction of saturated sands during cyclic loading", ASCE, JGE, VoL 92, NoSM 6, pp, 105-34,

. .

Seed, H. B. and Idriss, L M, (1971), "Simplified procedure for evaluating soil liquefaction potential", Journal of So;mechanics and Foundations Division, ASCE, 97, SM9, pp, 1249-1273. .

Seed, H. B" Lee, K, L. , Idriss, L M. and F. L Makdisi (1971), "The slides in the San Fernando dams during tl-earthquake of Feb, 7, 1971", Journal of the Geotechnical Engineering Division, Proceedings, ASCIVoL 101,No.GT7. .

Seed; H. B. and Booker, 1.R. (1976), "Stabilisation of potentially liquefiable sand deposits using Gravel drasystem", Report No. EERC 76-10, Earthquake Engineering Research Centre, University of Califcnia, Berkeley.

Seed, H, B. (1976a), "Some aspects of sand liquefaction under cyclic loading", Conference on Behaviour of 0'shore Structures, The Norwegian Institute of Technology, Norway.

Seed, H. B. (1976b), "Evaluation of soil liquefaction effects on level ground during earthquakes", State of the fPaper,Symposiumon Soil Liquefaction,ASCENationalConvention,Philadelphia,pp. 1-104.

Seed, H.B. (1979), "Soil liquefaction and cyclic mobility evaluation for level ground duringearthquakes", 1.GeoteEngin. Div., ASCE, Vol. 105, No. GT2, pp. 201-255.

Seed, H. B., I.Arango and C. K. Chan (1975), "Evaluation of soil liquefaction potential during earthquakes", ReJ:on EERC, 75-28, Earthquake Engineering Research Center, University of California, Berkeley.

Seed, H. B. and L M, Idriss (1967), "Analysis of soil liquefaction Niigata earthquake", J. Soil mech. Found. DASCE, yoL 93, No. SM 3, pp. 83-108. '

Seed, H. B. , K. Mori and C. K. Chan (1977), "Influence of seismic history on liquefaction of sands", J.GeoltEngin. Div., ASCE, VoL 103, No. G T 4, pp. 246-270.

Seed, H. B., and W. H. Peacock (1971), "Test procedures for measuring soil liquefaction characteristics", 1.Mech. Found. Div., ASCE, Vol. 97, No. SM 8, pp. 1099-1199. .

Susumu, I. A. I. , Koizimi, K., Node, S. and H. Ysuchia (1988), "Large scale model tests and analysis of Grdrains", IXth WCEE, Tokyo, Vol. III,pp. 261-266.

Terzaghi, K. and R. B. Peck (1967), "Soil mechanics in engineering practice", John Wiley and Sons, Inc., New 'rWang, S. (1984), "Experimental study on liquefaction inhibiting effect of gravel drains", Proc. VIII WCEE, Cal

. nia, Vol. 1, pp. 207-214. .

.WaterwaysExperimentStationU. S. Corpsof Engineers(1967),"Potamologyinvestigations,report 12-18,\cation of empirical methods for determining river bank stability", 1965Data, Vicksburg, Missis'

×¬åå¢³


Yasushi, S. and Taniquchi, E. (1982), "Large scale shaking table tests on the effectiveness of Gravel drains", Earth-" quake Engg,Conference,Southamopton,pp. 843-847.

.Yosufurni, T., Kokusho, G. and Matsui (1984), "On preventing liquefaction of level ground using Gravel pnes",<' ',. ' Proc. JASCE, No. 352, pp. 89-98.

,Yoshimi, Y. 'and H. Oh-aka (1973), "A ring torsion apparatus for simple shear tests", Proc. 8th International Con-'- . -' ference on Soil mechanics and Foundation Engineering, Vol. 12, Moscow, USSR.. .

Yoshimi, Y. (1967), "Experimental study of liquefaction of saturated sands""Soil Found. (Tokyo), Vol. 7, No. 2 pp.'.", ,20-32. ,,' " " :

, Y~~h{~i,:' Y. ;a~d H. Ohaka (1975),- "I~fl~ericeoi dJgT~e'of sh~~t str~ss revers~l on the liquefaction potential of" .. saturated sands"" Soil FolJnd.(Tokyo), Vol. 15, No. 3, pp. 27-40.

, , ' ' ", '- . ' ' '

!'..'"

"'.. ,."'" ,,', ,'C' ,', ..' ',' :-, ,..',,',; ,---: .." , ,--,- L, '

PRACTICE PROBLEMS, I

:- ,'..

7.1 Explain the terms 'initial liquefaction', 'liquefaction' and 'cyclic mobility'. Illustrate your an-,s~e-r 'with neat ~ketches. ' ,. " ' "...

7.2 listand discuss the factors on which liquefaction of,saturated sand depends., ~, ,

7.3 Give the salient features of the liquefaction studies made by (a) triaxial tests, (b) shake tabletests, and (c) blast tests. '

7.4 Describe briefly the following methods of predicting liquefaction potential(a) Seed and ldriss (1971) -method ",' ,'"

(b) Seed (1979) method(c) lwasaki (1986) method ' '/>'

7.5 At a given site boring supplement with SPT was done' upto 20.0 m depth. The results of theboring are as given below': , ". , : ' . '. ~ '

Depth Classification D50 N-Value 'DR" , Remarks"

(m) of soils (mm) , ,(%) , ,"I, ;, .f

;'.2,0 SM , 0.20 . ,,4 25 (i) Position of water4.0 SM 0.18 4 30 tabie lies 2.0 m~.O SP 0.16 5 33, below the ground8.0 SP 0.18 7 40 surface"

10.0 SP 0.19 9 43

12.0 SP 0.19 I0 ~' 4514.0, , SP' 0.19 ',' 12 " 52 (ii) 'YmOiSI=20kN/m)16,0 SW 0.30 14 56

, ::' 318.0 SW 0.32 16 60 'Ysub= 10 k~/rn20.0 SM 0.20 18' .' ,~3 , . ,'. .~

This ~jte is 10catedTD.s~{smically,actIve region, ,and the .likely to be subject~4 by an ~aitliquak~,ofmagnitude 7.5. Determine',:ih~"zoneof liquefaction ,by the three methods mentioned in Prob. 7.4. ;-, ' , ,"',. " . '

" ,.. <':1 ':'.. t' , ; , ,.,..'" DLJ

, , . '; " u-.; , '. :.

GENERAL PRINCIPLES OF MACHINEFOUNDATION DESIGN

8.1 GENERAL

For machine foundations which are subjected to dynamic loads in addition to static loads, the conven.tional considerations of bearing capacity and allowable settlement are insufficient to ensure a good design. In general, a foundation weighs several times as much as machine (Cozens, 1938; Rausch, 1959)Also the dynamic loads produced by the moving parts of the machine are small in comparison to thstatic weight of the machine and foundation. But the dynamic load acts repetitively on the foundatiorsoil system over long periods of time. Therefore, it is necessary that the soil behaviour be"elastic undtthe vibration levels producedby the machine, otherwisedeformationwill increasewith each cycle (

loading and excessive settlement may occur. The most important parameters for the design of a machirfoundation are: (i) natural frequency of the machine-foundation-soil system; and (ii) amplitude of ID!tion of the machine at its operating frequency.

8.2 TYPES OF MACHINES AND FOUNDATIONS

There are various types of machines that generate different periodic forces. The main categories are:

8.2.1. Reciprocating Machines. These include steam, diesel and gas engines, compressors and pum;The basic mechanism of a reciprocating machine consists of a piston that moves within a cylinderconnecting rod, a piston rod and a crank. The crank rotates with a constant angular velocity. Figureshows the outline of a typical Gangsaw in which the out of balances forces may lead to vibration pr'lems.

The operating speeds of reciprocating machines are usually smaller than 1000 rpm. Large feci!cating engines, compressors and blowers generally operate at frequencies ranging with in 50-250 rReciprocating engines such as diesel and gas engines usually operate within 300-1000 rpm. ~

The magnitude of the unbalanced forces and moments depend upon the number of cylinders iDmachine, their size, piston displacement and the direction of mounting. The mechanism developiDfof balance inertia forces for a single crank is ~hownin Fig. 8.2. It consists of a piston of mass mp~'within a cylinder,"a connecting rod AB of ma~smr and crank AO of mass mewhich rotates abou~JI

. at ~freq~ency 00.The centre of g~avity of the connecting"rod is lo~ated at a distance L) from PO1~'"the rotatmg masses are to be partially or fully balanced, counterweights of mass mwmay be locat~their centre of gravity at point c"

c~

Genel-lll P,inciples. 0/ Machine Foundation Design 341

,,' ,

" . , '..

- Uppu slide b ao-ck,EInC0litC.,E"

log teed

l Saw 'Modes

lower said~ block

- ... ' ,'r',0u c.onn~ctin9 roda.>-~ c.ounter

weight Fly wheel

. .'" '.,. .. .Foundation

block ~

. "'.. - " . . . '. .. .

Fig. 8.1 : Outline of a typical Gang-saw machiDe

""" "

Piston

mp,""'., '"

,-,, 0, ..-

'..

'-vCOUnt,,-...~ttt

->

z

Fig. 8.2 : Crank mechanism

L

1RUP...&4i.

GeneJ!.al,P,inciples- of Machine Foundation Design 341

,,'

" '

,EIn

c:0

, litc:.,E'0

- UppuSl~,de., blO'ckLogteed

l. Saw 'blad~s

Lower slide block

- -'-0'1'uQ.>-~

c.onnecting rod

c.ounterw~ight Fly whul-.

. .'" .':: :

foundationblock . . - ~

. "'.- - '" ,..',".

Fig, 8.1 : Outline of a typical Gang-saw machiDe

'-'J' ..

Piston

mp,"

"'., ".

,'," 0, ..

"

"-vCount~.. '7~t

,~'

z

Fig, 8,2 : Crank mechanism

~,"C.".'>i.,

L

342,-" .~oil Dy"ami~. & ~a.c~i"e F,°'!"d.f!..tiO/,'s- "~

In order to simply the analysis of the motion of the connecting rod, the mass mr is replaced by twoequivalent masses; one rotating with the crank pin A, the other translating with the wrist pin B. Theinertia forces can then b~ e~pre~sed in.t(:OI1sof thtHotal rotating 'mass (mrot)and the total reciprocatingmass (mrec)'The total rotating'mass is a'ssumedto be con~entrated at the crank pin A.

r2 L2' r;'m =,--'-me+- mr~- mw. rot. - "i ' -L ,ri'

. '. ' L I'

== +~mrec mp L mr

...(8.1)

...(8.2)

I '

F: (m rot:+: mrec ),,~1~2c:,os "'t, C .'

Jt r. 2CJ:)2F : mre c: 1 cos 2 u.J tL

Fy : mro t r1",2sin u:>t",

'.~

';~.-0./

0 1r 2'1r ' .. 00' 31T- "

L.1T.'

Fig. 8.3: Variation ofinertiaJorces with time

,:,,;.' . ' ...\ ,.' ,. ;; ~:!'~

'"-

r!ral Principles of Machine Foundation Design 343

The inertia force (Fz) in the z direction may be shown to be2 ,

2 '~ ' 2 '

F~ = (mrot + mrec) r} 0> cos (0 t + mrec L (0 cos 2 0> t, " ..,.(S.3)lchhas a primary component (F1 acting at the frequency of rotation, and a secondary component (F")[ngat twice the rotation frequency.

F = Ft + F"z ...(S.4)And in the y direction

2 . . , ,,':- -,,' , ""'-' I

Fy = mrot . rl 0> sm 0>t ' .,.(S.5)The time variations of these inertia forces are illustrated in Fig. S.3. ,,' ,',

If the rotating mass is balanced, the inertia force in the y direction disappears and that. in the z. ' " -, '1

ectlon becomes ' -, ' " ',: -.". " ", '

F = m r ,.,2(

COS Cl)t + rl cos 20) t)

,- ..' - 0' '-," "'

(-S"

6); ree 1 VJ L' .., ,

The amplitude of the primary (F:nax)and secondary (F;ax) inertia forces are then relat~da~J.ollQWs'r,

F" = ~ Ftmax L max . '; ..:(8,7f

The preceding development relates to a single cylinder machine, which possesses unba~ancedpri-ary and secondary forces. As more cylinders are added the unbalanced forces and couples are modifiedshown in Table 8.1 (Newcomb, 1951). With a six cylinders machine complete balance is achieved.

Different crank arrangements pertaining tot,able S.l are shown in Fig. S.4.

x ~ ~ lM(a)

In-tincz cytindczr Qpposczd cytindczr

(b) (c)

y ~ h1u-(d) (e) (I)

1~ +~ A J:m,1;t0 0Cranks at 180 C ran ks at 90

(g) (h)

Fig. 8.4 Different crank arrangements: (a) Single cr;ank(b) Two cranks at 180°(c) Two cranks at 90°(d) Two cylinders at 90° on one crank (e)T,io'op~s;d ~yllnders ~~one crank (g) Four cylinders (h) Sixcylinders

-', .".. ...:\, .! 1 .'~..:. :_--~,-~ ,-".", ,,-..

-",!}",'1-..t"'.'"."'l't';.

-

"""""~>""">"';";""';""'. "i" ":C

:If'j

344,

Soil Dynamics & Mac/,ine Foundations,~, "

,.

Reciprocating machines are very frequently encountered in practice. Usually the following two typesof foundations are used for such machines:

," ; (a) Block type foundation consisting of a pedestal of concrete on which the machine rests (Fig. 8.5).; ; "Cb) Box or Caisson type foundation consisting a hollow concrete block supporting the machinery on

, its top (Fig. 8.6). ': "',..., .

Fig. 8.S : Block type foundation Fig. 8.6 : Box type foundation

8.2.2. Impact Machines. These include machines like forging hammers, punch presses, and stampingmachines which produce impact loads. Forge hamIners are divided into two groups: drop hammers fordie stamping and forge hammers proper. These machines consist of falling ram, an anvil, and a frame(Fig. 8.7). The speeds of operation usually range from 50 to 150 blows per minute. The dynamic load~attain a peak in a very short interval and then practically die out. '

Anvil

Fig. 8.7 :Dro'p biu~~er wit" frame mounted-on anvil

---

:etll!rilf'Principles 'ofMddtitte'Ft1tIiidation Design '345

:able' 8.1 : Unbalanced Forc:esalid Couples for Different CFank Arrangements (Newcomb, 1951)

Crack arrangements Forces ' --- Couples-------------------------, .

(Fig. 8.4) Primary Secondary ,Primary , :-Secondary

l. --Single 'crank ..F' without counter wts.(0.5) F' with counter wts.

7. Two cranks at 1800

In-line cylinders 0Opposed cylinders 0

Two cranks at 900 (1.41) F' without counter wts.(0.707) F' with counter wts.

Fit 0 :0

2F" {

F' 0 without cou~terwts.' "

(0.5) F' 0 with counterwts.0

0

:00

...0

0 (1.41) F' 0 without counter wts. E" 0(0.707) F' 0 with counter wts.

1. Two cylinders on, one crank, cylin-

ders at 90°

F' without counter wts.

0 with counter wts.

1.41 F" 0 0

-, .e. Two cylinders on

one crank,

2 F' without counter wts.F' with counter wts.

0 0 0

,'opposed cylinders

r Three cranks at 1200 0 0 (3.46) F' 0 without counterwts. (3.46) F 11 0(1.73) F' 0 without counterwts.

g. Four CylindersCranks at 1800Cranks at 90°

,-00

00

0 0

h. Six cylinders 0

(1.41) F'O without counterwts. '(4.0) F" 0(0.707) F' 0 with counterwts.

0 00

F' = primary force; F" = secondary force; 0 = cylinder-centre distance

Impact machines may also be mounted on block foundations, but their details would be quite differ-ent from those of reciprocating machines.

8.2.3. Rotary -Machines. These include high speed machines such as turbogenerators, turbines, androtary compressors which operate at frequencies of the order of 3000 rpm to 10000 rpm. Associated withthese machines there maybe a consider~ble amount of auxiliary equipment such as condensers, coolersand pumps witl1connectingpipework and ducting. To accomodate theseauxiljary 'equipments a commonfoundation arrangement is a two storey frame structure with the turbine located on the upper slab and 'theauxiliary equipment placed beneath, the upper slab being flush with the floot .level of machine ?all(Fig. 8.8).

~

,0, .' ~--"" ,.__'0' '0'-. eo "-' ._- -,", -"~, '... 0" '0'" .. 0',.

i +.tt~JJ\; 0.) , ,. ,> r+,.I,~I':"'," 0./ '

+ '

'1I)'J£:nh;£!"!l1l{~~,,!. n ;!' ,uP '> \' ,it

-

- Co< .-. .~.. ~

Turbin~ G~ n~ ra tor

SoU Dynamics & Machine Foundatiolls346

==:J

Floor

L1m~Upperstab

':-:';.'~'::..u ': -: ::.-, ,- . . .:'

Base stab

. "....

c::.

- '0"' ",. :.':"'"'-,.

Fig. 8.8 : Concrete frame turbogenerator foundationRotating machinery is balanced before erection. However, in actual operation some inbalance al-

ways exists. It means that the axis of rotation lies at certain eccentricity with respect to principal axis ofinertia of the whole unit. Although the amount of eccentricity is small in rotary machines the unbalancedforce may be large due to their high speed. Figure 8.9 a shows a typical rotating mass type oscillator inwhich a single mass me is placed on a rotating shaft at an eccentricity e from axis of rotation. Theunbalanced forces produced by such a system in vertical and horizontal directions are given by

Fy = me e ol sin rot ...(8.8 a)FH = me e cos rot ... ...(8.8 b)

3cInU-N F :: me ~(A)l

F::me~w2/

f

'F( .)F, /

(a) Single shaft (b) Double shaft

Fig. 8.9 : Rotating mass type oscillator

"" ,.c.

meral Principles of Machine Follndation Design 347

Figure 8.9b shows two equal masses mounted on two parallel shafts at the same eccentricity, the1aft rotating in opposite directions with the same angular velocity. Such an arrangement produces an;;cillating force with a controlled direction. For the arrangement shown in Fig. 8.9b, horizontal force)mponents cancel and the vertical components are added to give

F = 2 me e 002 sin 00t ...(8.9)

.3 GENERAL REQUIREMENTS OF MACHINE FOUNDATIONS

or the satisfactory design of a machine foundation, the following requirements are met:1. The combined centre of gravity of the machine and foundation should as far as possible be in the

same vertical line as the centre of gravity of the base plane.2. The foundation should be safe against shear failure.

. 3. The settlement and tilt of the foundation should be within permissible limits.4. No resonance should occur; that is the natural frequency of the machine-foundation-soil system

should not coincide with the operating frequency of the machine. Generally, a zone of resonanceis defined and the natural frequency of the system should lie outside this zone.

If 00represents the operating frequency of the machine and oonas the natural frequency of thesystem, then

(a) In reciprocating machines (IS: 2974 pt 1-1982)

For important machines: 0.5 > ~ > 2.0(On

For ordinary machines:(0

0.6 > - > 1.5(On

"

(b) In impact machines (IS: 2974 Pt 11-1980)(0

0.4 > - > 1.5(On

(c) In rotary machines (IS: 2974 Pt 111-1992)(0

0.8 > ;- > 1.25n

It may be noted that where natural frequency of system oonis below the operating frequency ofmachine 00,the amplitudes during the transient resonance should be considered. For low speedmachines, the natural frequency should be high, and vice versa. When natural frequency islower than the operating speed, the foundation is said to be low tuned or under tuned, when thenatural frequency is higher than the operating speed, it is high tuned or over tuned.

5. The amplitude of motion at operating frequencies should not exceed the permissible amplitude.In no case the permissible amplitude should exceed the limiting ampli~~e of the machine whichis prescribed by the manufacturer., . . . .

6. 'The vibra'tiOli~must not be amloying to the persons working in the fact~iy or be damaging toother precis;.on machines. Th~ '~ahiie 'of vibrations that are perceptible, -annoying, or harmfuldepends on the frequency of the vibrations and the amplitudes of motion. Ri~rt (1962) devel-

ùæ{æ

~-~~":"-.:,.':;;;:';';;"';":,:";::,L_:~,_-':"'".:1:L'-'~U:t "'" " '";,,, ."", ; ,""" "

348

tJ1

"'rig""I~

.DC .....~ 0::J 0n 0

'<'"n N" 03 00

Soil Dy",a,mics &. Machine. FollndationS"

oped a plot for vibrations (Fig, 8,10) that gives various limits of frequency and amplitude fordifferent purposes, In this figure, the envelop described by the shaded line indicates only a limitfor safety and not a limit for satisfactory operation of machines,

V'1000.....0..000

Fig, 8,10 : Limiting amplitudes ofvibratlon for a particular frequency (Richan, 1962)

8.4 PERMISSIBLE AMPLITUDE

For the design of machine foundation, the values of permi~sible amplitudes suggested by BureauIndian St~ndards for the foundations 'of different typ~s of~chines.are given in Table 8,_2,,,t, -'i' "," -,. "',;', ",. ',"",','" :"", '

,x/A (~,N~ ;: :~ , . .."- , , .'0 . - . .',.,; ',""

, .

Disp la c.emen t ampl i tude AI mm0 0 0 '-/ A. 0 ,0 00 0 0 . 0 0 00 0 0 0 0 0 . 0

0 0 0 0, 0.

.....g 0 0N U1 ..... N U1 ..... N '" -

0o.

o"r ,o -oCO0 ') , '1»

0.<> \--':''b" .0

0" ,,0 ,'"0 ')', \--,'

)"D -0'1>0; 0.;>0 "D""- ",q 1")')( 0" "D

0 &>)"D "D"""'"

,>0 0"" 0)"D

0'"",Y>

eral Principles of Machine Foundation Design 349

Table 8.2: Values of Permissible Amplitudes for Foundations of Different Machines

, No. . Type of machine ReferencePermissible amplitudemm

I.2,

Reciprocating machinesHammer.(a) For foundation block(b) For anvilRotary machines(a) Speed < 1500 rpm(b) Speed 1500 to 3000 rpm

3.

(c) speed> 3000 rpm

. 0.2 IS: 2974 (Pt - I)

1.0 to 2.01.0 to 3,0

IS: 2974 (Pt - 11)

0.20.4 to 0.6Vertical vibration0.7 to 0.9Horizontal vibration0.2 to 0.3Vertical vibration0.4 to 0.5Horizontal vibration

IS: 2974 (Pt - IV)

IS: 2974(Pt -III): 1992

Permissible amplitude dependents on the weight of tup, lower value for 10kN tup and higher value for the tupweight equal to' 30 kN or higher. ..

,5 ALLOW ABLE SOIL PRESSURE

he allowable soil pressure should be evaluated by adequate sub-soil exploration and testing hi accor-mce with IS: 1904-1978, The soil stress below the foundation,s,hallnot exceed 80 percent of the allow-)le soil pressure. When seismic forces are considered, the a1f6wablesoil pressure may be increased as~ecified in IS: 1893-1978.

.6 PERMISSIBLE STRESSES OF CONCRETE AND STEELor the construction ofthe foundation of a machine MI5 or M20 or M25 concrete in accordance with IS:.56-1978 shall be used. The allowable stresses of concrete and steel shall be reduced to 40 percent foroncrete and 55% for steel, if the detailed design of foundation and components is limited to static load)f foundati'Jn and machine. Considering temperature and all other loadings together, these assumed,tresses may be exceeded by 33.5 percent. Alternatively, full value of stresses for concrete and steel as;pecified in IS: 456-1978 may be used if dynamic loads are separately considered in detailed design bylpplying suitable creep and fatigue factors.

The following dynamic moduli of concrete may be used in the design:

Grade of Concrete

MI5M20M25M30

Dynamic elastic modulus (kN/m2)

2.5 x 107

3.0 x 107

3.4 x 107

3.7 x 107

....."',

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

(~~.

350- -~

SoU Dynamics & Machine Foundatioll$'"-.

8.7 PERMISSffiLE STRESSES OF TIMBER

The timber is generally used under the anvil of hammer foundations. Grade of timber is specified accord--\'>ing to the size of defects like knots, checks etc. in the timbe~ Timber is thus classified into three grades."Select, Grade I and Grade H. The best quality timber having minimum or no defects at all is of the select

~rade. ?rade I timber is one hav~g. defects not larger than the ~peci~ed ones. Grade H t~mberi.spoore~In quahty than grade I. The permissible values of stresses are given IDTable 8.3 for species of timber ofgrade I. In machIne'foundations timber of select grade is used. The permissible stresses of timber givenin Table 8.3 may be multiplied by 1.16 to get the permissible stress of timber of select grade.

Table 8.3 - MinimumPermissibleStress Limits (N/mm2)in Three Groups of Struc,turalTimbers(For Grade I Material)

t

S.No.

(i) Bending and tension along grain

(ii) Shear(I)Horizontal

Inside(2)

Strength Character Location of Use

Along grain

(iii) Compression parallel to grain

(iv) Compression perpendicular to grain

(v) Modulus of elasticity (x 103 N/mm2)

All locationsAll locationsInside(2)Inside(2)

All locations and grade

(I) The values of horizontal shear to be used only for beams. In all other cases shear along grain to be used":

(2) For working stresses for other locations of use, that is out side and wet, generally factors of 5/6 and2/3 are applied ~

The permissible bearing pressures on other elastic materials such as felt, cork and rubber are gener-.Jally given by the manufactureres of these materials. No specific values are recommended here since theY4vary in wide limits. .

, 1...

.. .

", "'~}

- ..I;..

~~J

""'1<--J

i

Group Group GroupA B C

18.0 12.0 8.51.05 0.64 0.49

11.1 1.8 4.94.0 2.5 1.112.6 9.8 5.6

-,

-

'

ó ùþôòþôæôùôùþù¬ùô¢åþùæùâþôæùåôôôôùôòþùô òþôþùþ ò îÿåô

Õù

FOUNDATIONS OF RECIPROCATING MACHINES..

9.1 GENERAL

Reciprocating machines are common in use. Steam engines, internal combustion engines (e.g. diesel,and gas engines), pumps and compressors fall in this category of machines. Block type or box typefoundations are used for reciprocating machines.

For the satisfactory performance of the machine-foundation system, the requirements given in sec.8.3 should be fulfilled. For this, one has to obtain (i) the natural frequency of the system, and (ii) theamplitude of foundation during machine operation. In this chapter, methods have been presented toobtain these two parameters in different modes of vibration. The basic assumptions made in the analysesare: (i) the foundation block is considered to have only interial properties and to lack elastic properties,and (ii) the soil is considered to have only elastic properties and to lack properties of interia. Design stepsand illustrative examples are given at the end of the chapter.

9.2 MODES OF VIBRATION OF A RIGID FOUNDATION BLOCK

A rigid block has, in general, six degrees of freedom. Three of them are translations along the threeprincipal axes and the other three are rotations about the three axes. Thus, under the action of unbalancedforces, the rigid block may undergo .vibrations as follows (fig. 9.1) :

1. Translation along Z axis - Vertical vibration2. Translation along Y axis - Longitudinal or sliding vibration3. Translation along X axis - Lateral or sliding vibration4. Rotation about Z axis - Yawing motion5. Rotation about Y axis - Rocking vibration6. Rotation about X axis - Pitching or rocking vibration

The vibratory modes may be 'decoupled' or 'coupled'. Ofthe six modes, translation along Z axis androtation around the Z axis can occur independently ot any other motion and are called decoupled modes.However, translation along the X or Y axis and the corresponding rotation about the Y or X axis, respec-tively, always occur together and are called coJlpled modes. Therefore the dynamic analysis of a blockfoundation should be carried out for the following cases:

(i) Uncoupled translatory motion along Z axis i.e. vertical vibration.(ii) Coupled sliding and rocking motion of the foundation in X- Z and Y -Z planes passing through

the common centre of gravity of machine and foundation.(iii) Uncoupled twisting motion about Z axis.

oundatiQns of .Recip~oCQtingM achines 353

A rigid block being ~ problem of six degrees of freedom has six natural frequencies. The naturalrequency is determined in a particular mode (decoupled or coupled) and compared with the operatingrequency. Similarly, amplitude is worked out in a particular mode and compared with the permissible~. '

vczrtical

Rocking

~--', ~ +X

Latczral

/ "

Fig. 9.1 : Modes ofvibration oh rigid block foundation

9.3 METHODS OF ANALYSIS

The following two methods are commonly used for analysing a machine foundation:(i) Linear .elastic weightless spring method (Barkan, 1962)

(ii) Elastic half - space method (Richart, 1962)In the first method which is proposed by Barkan (1962), soil is replaced by elastic springs. In devel-

oping the analysis the ,effects of damping andpartidpating soil 'mass are neglected. Damping does notaffect the natural.frequency of the-system appreciably, but. it affects resonant amplitudes considerably:Since the zone of" resonance is avoided in designing!I1achinefoundations,the effect of dampingon '

amplitudes computed at operating frequency is also $mall. Hence neglecting, damping may not affect thedesign apPI:eciably,and if any that on the conservative side. Empirical methods have been suggested toobtain the soil ma~s ,participating'in vibration. ' . ,

In the elastic half-space method,;:,!he"machine:foundation is idealised as a vibrating mechanicaloscillator with.a circular' base resting ~Il,the~surface;ofgroumLThe .ground is assumed to bean elastic,homogeneous, isotropicf;semi~infiriit~ody~~whichis referredto.asanelastic half-space. This approachis apparently more rational, but relatively more complicated.

'-"','

.,"'.

. >i


In the above two methods, the effect of side soil resistance is not considered that is the foundation isassumed to rest on the ground surface.

9.4 LINEAR ELASTIC WEIGHTLESS SPRING METHOD

Barkan (1962) has given the analysis of block foundation in following modes of vibration:(i) Vertical vibration(ii) Pure sliding vibration

(iii) Pure rocking vibration(iv) Coupled sliding and rocking vibration(v) Yawing motion

Machinq;

Fou ndation

Ground L<zv<z1'"

Of

-LSoil

Fig. 9.2: Block foundation

Let us consider a block foundation of base contact area A placed at a depth Dfbelow the ground level(Fig. 9.2). Neglectingthe effect of side soil resistance and considering soil as weightless elastic material,the machine - foundation soil system can be idealised to mass-spring system shown in Fig. 9.3 a to 9.3 efor different modes of vibration. Barkan (1962) had introduced the following soil parameters which yieldthe spring stiffnesses of soil in various modes:

(a) Coefficient of elastic uniform compression (C,) : It is defined as the ratio of compressive stressapplied to a rigid foundation bl~ck to the 'elastic' part of the settlement induced consequently. Thus

Cu =!!.L (Fig. 9.3 a)Sez...(9.2)

It is used in vertical vibration mode. From defmition, spring constant Kz

Load qz AK= . =-=C.A

z elastic deformation Sez u...(9.3 )

.-

""datioilS of ReCiprocating Machines

.1 Fz<t) t Fz(t)

-- ---------

m

L-- ~-- ~.-- J Sf ez

(a) Vertical vibration

..,IIIIII .

III

-+I~'SexII

m

Fx(t),.. . ..,.'

Kx-"'7jN...-

(b) Pure sliding vibration---My(t) ---""""

f:\

II

II

II

II

I'--....

-""-....

My(t)

F:)-'"7

II

II

II

I

'U:{f

rp'

(c) Pure rocking vibration

Fig. 9.3 : Types of motion of a rigid foundation (...Contd.)

355

Fz

K.Z = Se2

Fx

KX = Sex

My

K,= T

K~

IIII Fx (t)II . . .III,I

I ,.--.

....I ----

356 , ';Soil"'vDymnnics-'>& ~Mfld,ine '-Fl1wuJtltions

, nMZ{t), 'i = Angle ot

. torston ottoun'da tier.(not shown)

~MZ{t)

/ /

Fd (t)r-, a.......

(d) Yawing vibrationFd (t)

. ..,

I/

I,I,,

i.,.... .... ... ....'...

.........., , , , "

I,,,,I

I-

'Kx Kx'.'-.~"'-.-i t'r' ~~ ",

Sex "...,

>."'"

K~

(e) Coupled slidding and rocking vibration

'.Fig~'9.3': Types of motion of a rigid foundation

(b) Coefficient of elastic uniform shear (Cr): It is defined as the ratio of the average shear stress athe foundation contact area to the elastic part of the sliding movement of the foundation.

qCt = Sx (Fig. 9.3 b)ex

...{9.4

It is used in analysing sliding vibration mode. The spring constant Kx is given by

. K = Shear load = qx .A = C . A (9 "x t ... "-

Sex Sex .

(c) Coefficient of elastic non-uniform Compression (C~ : It is used in rocking vibratic(Fig. 9.3 c). In this case the elastic settlement of the block is not uniform over the base. Itdefined as the ratio of intensity of pressure at certain location from the centre of the base of blocto the corresponding elastic settlement. If cpis the angle of rotation of block, then at a distaD-I from the centre of the base of block, the elastic-deformation will be I cp.Taking the intensity ,pressure at this location as q, C. is given by

qC~ =[cp

...(9.

FoLlm/lIlilJl;,. (~rJ:ecfproclItillg Machines J57

where,

The sllj},:f.::';sKIj>is'defil1E".1as the motnent per unit rotation,a~d is given byM

K =-=C .1<jI cp

J = Moment of irf'rtia of the base of block about the axis of rotationM = Moment caused due to soil reaction

...(9.7)

(d) Coefficient of elastic non-uniform shear (CljIJ: It is used in yawing motion. If a foundation isacted upon by a moment with respect to vertical axis, it will rotate about this axis (Fig. 9.3 d).Tests have shown that the angle of rotation \j/ of the block is proportional to the external mo-ment.

Therefore, M =K \j/z 'I' ...(9.8)

...(9.9)where K = C . J'I' 'I' zJ- = Polar moment of the intertia of contact base area of foundation.

In the rotation of a' foundation around a vertical axis, the base of the foundation undergoes nonllni-form sliding, hence the term "Coefficient of elastic nonuniform shear". is applied to the Coefficient C .IjI

Barkan (1962) derived the £q. (9.10) for determining the value of C . It is based on theory of elas--. U

ticity. .

where,

1.13£ 1

Cu=I-~2'JA£ = Young's modulus of soil

...(9.10)

~ = Poisson's ratioA = Area of base of the foundation

, .'

He also developed the relationships between Cu' Ccp'Ct and C'I" For analysis and design of machinefoundation, he recommended that

Cu=2Ct ...(9.11)

Ccp=.2Cu. ...(9.12)Ct = 1.5 C'I' ...(9.13)

For preliminary design, Barkan (1962) recommended the values of ClIas listed in Table 9.1.

The procedure of determining th.evalues of cu' Ct £ and G have been given in detail in sec 4.3 ofchapter 4. As discussed in that section; dynamic elastic' constants depend on (i) base area of foundation,(ii) confining pressure and (iii) strain level. The method of converting the value of adynamic elasticconstant obtained from a field test for using in the de~ign of actual foundation has been illustrated inexamples 4. 2 and 4.3.

j . .,'... ,..:-: " , U, !' " ',. ~ c :".; r . ~.

. .~,~. ,t. "': . n.!,i;.'. -I


Table 9.1 : Recommended Design Values for the Coeficient of Elastic Uniform Compression Cu2*for A = 10 m

* After Barkan (1962)

9.4.1. Vertical Vibrations. For the purpose of analysis, the machine -foundaton-soil system shown inFig. 9.3a is represented by the idealised mass-spring system shown,in fig 9.4. Let the unabalanced forceis representd by

Fz.(t) = Fz sin rot (91.4)

³³ ã Ó¿ of machine

plus foundation

Kz = Cu.A

Fig: 9.4 : Equivalent model for vertical vibration

If the centre of gravity of the foundation and machine and the centroid of the base area of the foundation in contact with the soil lie on a vertical line that coincides with the line of action of the excitintforce Fz, then foundation will vibrate vertically only.

The equation of motion of the system is

mz + Kz' Z = Fzsin IDt' (9.15

where, M = Mass of machine and foundation

Kz = Equivalent spring constant of the soil in vertical direction for base area A of the foUl

dation = CuA

Soil Soil group Permissible static load,3

Cu' kN/mkN/m2

2 3 4

Weak soils (clays and silty clays with sand in plastic up to 1504

up to 3 x 10

state; clayey and silty sands; also soils of categories II

and III with laminae of oraganic silt and of peat)11 Soil of medium strength (clays and silty clays with sand 150 to 350 (3 - 5) x 104

close to the plastic limit; sand)III Strong soils (clays and silty clays with sand of hard con- 350 to 500 (5 - 10)x 104

sistency; gravels and gravelly sands; loess and loessial soils)IV Rocks > 500 > 10 x 104

., ,-,

,tidaiiiJns of Reciprocating' MaCih;n~ 359

Cu = Coefficient of elas~c uniform compression., .

Therefore, the natural frequency IDnz of the system is

ID.z = ~~z =r~ A.

The amplitude .of motion Az is given by

. Fz sin IDt Fz sin 0)tA = = 2

. Z . K -m0)2 C A-mO)z Il

Fz sin 0)tA = 2 2

Z m (0) - 0) )nz '

Maximum amplitude of motion Az is given by- Fz

Az - 2 2

. .' m(Ct>nz.-ro ) . .3.2. Sliding Vibrations of a Block. In practice, rocking and sliding occur simultneously. But if thebration in rocking can be neglected ,then only horizontal displcement of the foundation would occurtderan excitingforce Fx(t) on the block of area A (Fig.9.3b). This system can be indealised as shownFig.9.5.

;.

(9.16)

(9.17a)

or (9.17b)

(9.18)

-,. I

III,IIIII

~x ~/ ,\..

m

vhere,

Fig. 9.5: (a) Block foundation in pure sliding vibration(b) Equivalent model

The equation of motion of the system is'mx + K.~ = Fx sin rot

x = Sliding displacement of,the foundationKx = Equivalent spring c~nstap.t of the soil in sliding for base area A of the

foundation= Ct. A, '",

Ct = Coefficient of elastic uniform shear. :

Therefore, the natural frequency Ct>nxof the system is

: ' ~;',~ fKX; =' rc;Aq

nx" f;; v-;;-Maximum amplitude of motion A is given by. x

. '~"-'..'>;:).,:' """:~:':,'-:I:A l_~I"'-<'Fx":. .,.; ,:~f':, ,.

", ,.1, .. " x'~'~~;x'_~2)"""'J"_: '."

(9.19)

: (9.20)

", ...(9)J). . . .. ",

,

360 SoU Dynamii;s ~,Mac;4,ine, Found~tiof}s.

9.4.3. Pure Rockmg Vibrations of a Block. Consider only the rock~ngy~bri,J,tionsin~uced in a founda-tion block by an externally exciting moment Mv(t) (FIg. 9.3c). this "isalso a hypothetical case asrocking vibrations are coupled with sliding vibrations. Let the unbalanced moment be given by -

My (t) = Mysin IDt --where, My = Moment acting in the X - Z place'

At any time t, considering that the applied moment is actIng in clockwise -direction the"displacedposition of the block will be as shown in Fig. 9.6. In ~achine foundations, as the rotation <pis small,tan <p= <pin radians. The equation of motion can be obtained by applying Newton's second law of motion.

(9.22)

~

My (t) = My Sin GJt

~'

('-,- "

...

x-----

( a)

qst

IElem<znt dA

(b)

Fig. 9.6 : Block foundation under pure rocking vibrations

The various moments acting on the foundation about the centre of rotation are obtained as describelow:

(i) Moment MRdue to soil reaction: Consider an element dA of the foundation area in contact'the soil and located ar ..: :..tance / from the axis of rotati~n (Fig. 9.6 b). At any time, the soil wil

ndations of Reciprocating Machines. . ' '" 1>." " -", .361

npressed nonuniformly. From t~e de~n~t~on of ~oef5cient of dastic un}form compression, ,C~~sd R I dA

C~ = 14> ...(9.23 a)

d R = Soil reaction force acting on dement dAcl> = Angle of rotation

I~~hefoundation does not lose contact with soil, then the soil reaction will be as shown in Fig. 9.6 b.le total reactive moment MR against the foundation area in contact with soil is given by ,

A '

MR = J C~.l~dA.l ==C~.cI>JPdA = C~<\> I

.ere,

...(9.24 )

here, I = Moment of inertia of tpe foundation area in contact with the soil with respect to the axisof rotation. .

This moment acts is the anticlockwise direction. '

(ii) Moment Mw due to .the displaced position of centre of gravity of the block: As shown inig. 9.6 a, the centre of gravity of the block is shifted from point 0 to 0' . As angle of rotation <\>is small,:le moment Mw of Weight W will be ' .

M =WLti\w 't'

vhere, L = Distance between the centre of gravity of block and axis of rotation.This moment acts in the clockwise direction.

(iii) Moment Mj caused by intertia of foundation:

...(9.25 )

, "

It is given by Mj = Mmo ~ ...(9.26). .

.vhere, Mmo =; Moment of inertia of the'mass of the foundation and machine \"":"ithrespect of axis ofrotation. '

This moment acts in the anticlockwise direction. .The equation of motion can be written by equating clockwise moments to anticlockwise moments

Therefore, My sin' (tJt + WL cl>= c cl>14> + Mmo ~

or , Mmo 4> + (C~ 1- WL) <\>= My sin Cl)t

The natural frequency wn«/>of this system is given by

_,t.l- WLwn4>- Mmo

...(9.27)

...(9.28)

and maximum displacement A ~ i~ given by. . . ' , . '.

. MyA = 2 2

~. Mmo(ron,~ro)

. ,

...(9.29)".

. '. "", . ' .' . ,--" .'

l

-- .-'-

362 Soil Dynamics. &Mac1iin~ Foundatiolt/;

In practice, C. I is many times WL ; hence Eq. (9.28) may be written:'

~.Icon.=I Mmo

If the dimensions of the footing at the base are a and b in the X andY direction~, respectively,ba31=-12

'.0;, 'l

...(9.30),;~

, ...(9.31), -.

Cell ba3conI/! = "MU ...(9.32)

Y /nO

It is seen from Eq. (9.32) that the linear dimension of the contact area perpendicular to the axis ofrotation exercises a considerably greater effect on the natural frequency of rocking vibrations than theother dimension. This principle is sometimes used in proportioning the sides of the machine foundationundergoing predominantly rocking yibrations.

, '

The amplitude of the vertical motipn of the edge of the footing is

Azr = ~xA,Mya/2

= Mmo(O)~'- 0)2)Similarly, the contribution of rocking, towards the horizontal amplitude is

Axr = h. A,where, h = Height of the point above the base where amplitude is to be determined.

Azr and Axr are added to Az (Eq. 9.18) and Ax (Eq.9.21) respectively to obtain total vertical ansliding amplitudes when rocking is combined with vertical and sliding vibrations.

9.4.4. Yawing Vibrations of a Block. A foundation is subjected to yawing motion if it is subjected totorsional moment Mz (t) about Z-axis (Fig. 9.7a).The positionof the foundationat any time t maytdefined in terms of angle of rotation "'.

Let the unbalanced moment is given by

Mz(t) = Mzsin (J)t

As explained in Sec. 9.4, the resistive moment due to soil is C'II'Jz "'.The equation of motion is written by taking moment about Z- axis. It gives

...(9.33'

...(9.34

...(9.3

Mmz iV + C'IIJz '" = Mz sin rot ...(9.:

where,, ~~

Mmz = Mass moment of inertia of the machine and foundation about the axis of rotat(Z-axis)

Jz = Polar moment of inertia of foundation contact area

C1jf = Coefficientof elasticnonu~iformsheart

, .'4

,~1'.."i.-

óóóòóóó

Foundations of Reciprocating Machines 363

'. z

x

v (a) Isom~teric vi<zw

" .'

, ", .'

( b) pia n

Fig. 9.7: (a) Yawing motion ora rigid block .

(b) Development of non-uniform shear below the base

The' expressions' for natural, frequency an{ma~imux:n angular displacements are as follows:

- tw J,(J)nl/f- Mmz...{9,37)

Mz,A = M

.

(2 . 2

)'I' mz IDnljl- CD

The horizontal displacement AhlJlc~used, by .t,orsion is

Ah'l' = r A'I' ...(9.39)r = Horizontal distance of the point on the foundation from the axis of motion (Z-axis)where,

'9.4.5. Simultaneous'Vertical, Sliding and Rocking Vibrations. In general, a machine -foundation issubjected to time dependent vertical force, horizontal force and moment, and therefore it simultaneouslyslide, rock and vibrate vertically. In Fig. ~.8, a foundationblock subjected to a vertical force (Fzsin 00I),

....

,~i:~\

364 SoU-Dynamics & Machine Foundations

a horizontal force (Fx sin (J) t) and an oscillatory moment (My sin (J)t) is shown. These forces and momentare considered to act at the combined centre of gravity 0 of the machine and the foundation, which is also-taken as the origin of coordinates. At any time t, considering the vertical force acting in downwarddirection, horizontal force in right-hand side direction, and moment in the clockwise direction, the foun-dation block will be displaced as shown in Fig.'9.8. It is therefore subjected to (i) displacement z, in thevertical direction (ii) displacement Xo in the horizontal direction at the base and (Ui) rotation <1>of thebase.

~z

zI '

F Sin C.Jtz

Initial position

x

Fig. 9.8: Block foundation subjected to simultaneous vertical, sliding and rocking vibrations

The equations of motion can be written by evaluating the resisting and actuating forces and.moment- - ,acting on the foundation in the displaced position. These forces and moments are obtained as give:below: .

(i) Upward soil reaction Rv due to vertical displacement z :Rv = Cu A z

(ii) Horizontal soil reaction Rx due to horizontal displacement xo:Rx = C't A Xo

As the origin is at 0, Xocan be expressed in terms of x and 4>as below:xo=x-Lcp

where L = Height of Centre of gravity 0 from base of the block(tU) Moment MRdue to resistance of soil induced by rotation:of the foundation by $ :The MR about point 0 is given by

M -=C "'IR .'"

...(9.4(

...(9.4

..,(9.4

jJ

+ .-

.:.(90'

--"

, ' , '<, ',' , "," . ,

uni/ations of Reciprocating Machines 365

(iv) Moment Mwdue to displaced position 'of the centre' of gravity of block :The moment Mw about point 0 is given by

Mw = W L 

(v) Moment MxRdue to horizontal resisting force Rx :Moment MxRabout point 0 is'give~ ~y .

M = R . L = C A (X - L"') . LxR x ~ ~

(vi) Interial forces and moment:

..(9.44)

...(9.45)

(a) In the Z - direction(b) In the X - direction

Fiz = mi

Fix = mx

...(9.46)

...(9.47)

(c) In the rotational mode Mi$ = Mm ~ ...(9.48)"here' M = Mass moment Of inertia of the machine and foundation about an axis passing throughm '

combined centre of gravity 0 and in the direction of Y-axis

Cheequations of the motion 'can now be written as below: . ,

In the Z - direction:. mi + Cu Az = Fz sin cot ...(9.49)

...(9.50)In the X - Direction: mx + CtA Xo= Fxsin cotSubstituting the yalue of Xo from Eq. (9.42) in Eq. (9.50), we get

mx + C A (x --L<I»= F sin COtt X

In the rotational mode.....(9.51)

, "

or

Mm + C$ I - WL - Ct A (x - L<I» = My sin cot.. 2

Mm - Ct A Lx + (C$ I - WL + C~ AL ) = My sin COt ...(9.52)Equation (9.49) contains only the terms of z, therefore the motion in Z - direction is independent of

any other motion. The solution of this equation is already given in Eqs. 9.16 to 9.18. Equations (9.51) ,

and (9.52) contain both x and and are interdependent. Therefore, sliding and rocking are coupledmodes. A solution for simultaneous rocking and sliding vibrations is presented below.

9.4.5.1.Naturalfrequencies of coupledrocking and sliding.The systemrepresentedby Eqs. (9.51) and(9.52) is a two-degree-of-freedom system. The solutions for natural frequencies are obtained by consid-ering the free vibrations of the system.

Hence, ...(9.53)

...(9.54)

mx + C~ Ax - c.~ AL= 0, '

and Mm~ - Ct A Lx + (C$ I - W L + C~A L2 ) ==0Particular solutions of these equations may be assumed as

x = xI sin (con t + ex) ...(9.55)and

= <1>1sin (cO n t + ex). ...(9.56)in which x I' <1>1and exare arbitrary constants whose values depend upon the initial conditions of motion.

i::~*I.',

366 Soil Dynamics &I,M~chine Ft!""~

By substituting Eqs. (9.55) and (9:56) into Eqs. (9.53) and (9.54) and dividing by sin (con t + a), we get,2

-mOOn Xl + Ct A xl - Ct A LI= 0 ".,

2or Xl (CtA -moon)-CtAL1 =0

and -Mmoo~1+I(CtAL2+CcpI -WL) -C'tALxI =0

CtAL1xl = 2

CtA -mOOn

By substituting the value of xI from Eq. (9.59) into Eq. (9.58), we get

:", ...(9.57)

~

...(9.58)

From Eq. (9.57), ...(9.59)

<1>1[-C~ A 2 L2+ (Cc!>I - WL + Ct AL2 - Mm oo~)( CtA -m oo~)] = 0 ...(9.60)

For a nontrivial solution, <1>1can not be zero. Hence the expression within the parentheses mu~tJ)(zero, This gives' , I ~

2 2 2 2 2 2-CtA L + (C. I óÉÔõÝ¬ßÔ óÓ³×Ü²÷øÝ¬ß-mOOn) =0 ...(9.6~

The term (On'which represents the natural frequency in combined sliding and rocking, is the onI,unknown in Eq. (9.61), which can now be solved. Equation (9.61) may be rewritten as follows:

-C~A2L2+C~A2L2+CtA(C.I -WL) -CtAMmoo~-CtAL2moo~-(C.I-WL)moo~+Mm1nro: =...(9.6

By dividing by mMm and rearranging, one obtains

4 2

[(

CA.I-WL

]C A(M +mL2)

]

C A

(CA.I-WL

)00 -00 'f + t m +-L- 'f =0 ...(9,(

n n Mm mMm m Mmb

By definition, the quantity (Mm+ mL2) is the mass moment of inertia of the foundation and machabout an axis that passes through the centroidof the base contact area and is perpendicular to the'~1:of vibrations. This is denoted by M . Thus, , ij~mo

2 ,M IM = M + mL .:;t9.mo m '

M ~Mm = r where 1> r > 0 , , (1~mo . -.J

,

'~.~.Equation (9.63) may be rewritten as .',.:h ,c

4 oo~

(C.I-WL CtA

)CtA C.I- WL

,

',,

"

OOn-- +- +- =0 .t't.;.(Sr Mmo m m tMmo 'i

'i 'f~iCtA 2 ,.~- = 00nx

"

,\,

~,"'".

'4m ~,~f:' ..:0' ," ,

Further, by denoting

Now,

C.I-:-WL: 2M' '; ~ OOn,mo

~, ""

~ --- .. --.

Foundations of !leciproC/lting Machi,!es 367

'2y writting the Eq. 9.66 in terms of OOn,and CJ)n~we ?et

(

2 2

)

2 22 0>nx + 0>ncp 2 0>nx Cl)n+

, 0> - 0> + =0n r n r .

The Eq. (9.69) has two positive roots, CJ)nland CJ)n2'which correspond to two natural frequencies ofthe system: The ro~ts of Eq. (9.69) are:

.

[(

2 '2

J (

2 2

J

2 2 2

]

0>2 =1. O>nx + O>nq,:t (Onx+ O>nq, - 4(J)n."(J)ncilnl,2 2 r r r

, Equation (9.70) may be rewritten as .

2 I[(

2 2) + ~ 2 2 ,2

4 2 2

1(J)nl.2= 2r (J)nx+(J)ncp 7" (O>nx+(J)ncpo - r(J)nx (J)nq,)

From the property of a quardratic equation:, 2 2

. 2 2 - 0>nx + 0>ncpO>nl+O>n2 -

...(9.69)

...(9.70)

...(9.71)

...(9.72)r

2 2 0>2 0>2(J)nl x (J)n2 = ~ ncp.r

...(9.73)

22-1. 22_"221/2and (J)nl-(J)n2 - r [«(J)nx+ (J)ncp) 4r(J)ncp(J)ncp] ...(9.74)

It can be proved that OOnxand 00ncpwill always lie bet~~en limiting natural frequencies 0001and 0002'

9.4.5.2. Amplitudes of coupled rocking a,nd sliding. The amplitudes of vibration are determined in thefollowing three cases:

Case I. If only the horizontal force Fx sin ootis acting: Eqs. (9.51) and (9.52) may be rewritten asfollows: ' .

mx+ Ct Ax - Ct AL~ = Fx sin (J)t

- 2"

Mm~+ ~(Ct AI.: + Ccp1- WL)-Ct ALx = 0Assume that the particular solution to these equations are

, x ~ A 'sin (J) tx

...(9.75)

...(9.76)

~ ::;: Acp sin (J) tin which Ax and Acpare the maximum sliding and rocking amplitudes respectively. By substituting thesesolutions into the above equations, we get ,

. 2Ax (Ct A - moo) - Ct ALA. = Fx ...(9.77)

-Ct A LAx + A. (Ct A L2+ C. I - WL - ~m 002),= 0 ...(9.78)

- (C~L2 +C.I-WL-Mmoo2)Ax - .A. ...(9.79)CtA~

or

~

:'~~'(~

,.t~SDi/Dynamics '& Machine FtiuiidQiio~368

By substituting for Ax in eq. (9.77), we get2 2 2

(Ct A L + C, 1- WL-Mm 00 )(CtA-moo ) - .A~ -Ct ALA, - 0

CtAL

."«i

.~~

~~~"

. "

. , ):'j. . " C AL' . '.

A~ =..

[

CtA+(C",I-WL) 2

[C tA(mL2+M ) (C",I-wt)

]

4

]

x.~t'f";.;".

.

t.mM . 't' -00 t m + 't' . +00

. m mMm mMm Mm .('

=

[

2 2 Ct~L

]

FX

M OOIlX 00 Il~ - ~ ( 2 + 2 ) + 4m m 00Il~ 00n.l. 00r ' r . 't'

-:-':--~-By using the relations gives in (9.72) and (9.73) into'Eq. (9.80), we get

C AL ,..A - t xF~ -

[2 2 2 2 2 4

]

xmMm oon1 oon2 -00 (oon1+oon2)+00 "

...(9.80)

.

CtAL F2 2 2 2 xmMm(OOn1-OO )(OOn2 -00 )

2 2 2 2 2Let, mMm(OOn1 -00 )(OOn2 -00 ) = ~(oo )

...(9.81)

...(9.82)

C ALA = t F

~ ~ (00)2 x

By substituting for A~ in Eq. (9.79), we get

...(9.83)

A = Ct AL2 +Cq. I -WL-Mm 002Fx ~ (002) x

Case 11. If only moment My sin rot is acting: Eqs. (9.51) and,(9.52) may be rewritten as. ,

mx+ Ct Ax- Ct ALcp = 0.. ..,

and Mmlj>-Ij>(CtAe+C4»I-WL)-CtALx =Mysinrot ...(9.86By assuming solutions as for Eqs. (9.75) as (9.76), it can be shown that the following expression

hold:'. ,'. . . '. 11.

...(9.84)

,.v

...(9.85'

,

CALM

t yAx = ~ (002) - -2

C A-moo M- t 2 yA. - ~ (00.)

..:(9.B';

and ...(9.B', ,

'i.

.. .. ..- --oii

Poundations of Reciprocating Machines 369

Case Ill.: If both the unbalanced force Fx aJ:?dmo~ent MJLare acting, the ~plitudes of m~tion aredetermined as follows:

0 - 02 0 0 0 0 . 2 . 0 -

A= (C'tAL + CtPI-:WL-Mm(O )Fx + (c.~) Myx. 0 A (

2)

0 0

0 L.l (0 0 0 0 0

. 2(CorAL)Fx+ (CorA -mm) MyA -

4>- oL\(ro2)

...(9.89)

and ...(9.90)"

The total amplitude of the vertical and horizontal vibration are given bya

. Ay = Az+2A~

and Ah = Ax + h ArpH__,__-where, h = Height of the top of the foundation above the combined center of gravity.

In foundations with two degrees of freedom, specific forms of vibrations correspond to the frequen-cies (0/11and (0112'These vibrat~ons are characterised by a certain interrelationship between the ampli-tudes Ax and A.pwhich depends on the foundation size and the soil propet:t~es,but does not depend on theinitial conditions of foundation motion. ..- ..

Let us examine the case when the foundation is subjected to exciting moment My only. The ratio ofamplitudes Ax and A4>obtained using Eqs. (9.87) and (9.88) is given by

. 2Ax CtA L ronx Lp = -= 2 = 2 2AA C A - m ro «) - «)'I' or 0 nx

...(9.91)

...(9.92)

...(9.93), o'

z z ..."\",/ \

,. \:\

\ ..."\v \

)-- '....

---II

\\

\~( /\ ",/

V

,,,0 ,

..J

--L

--- -- /

(a) (b)Fig. 9.9: (a) Rocking and sliding in phase with each other

(b) Rocking and sliding In opposite phl1se

The following cases are important for consideration of form of vibration:

(i) If (0 « (Onx'then p ==L . It means that the.-.foundationrotates about an axis that passes throughthe centroid of the base contact area and slidi~g is absent!

~

370 v Soil Dynamics & Machine Fou"datiol

'(U) If 00'= (On2' (O~2 being the lower limiting natural frequency, then oo~ - 00;2> 0 . It m~ans that

during vibration at frequency (0n2' when the centre of gravity deviates from the equilibriumposition, for example, the positive dir~ction of the X -axis, the rotation of the foundation willalso be positive, and changes of amplitudes Ax and A, will be in phase. The form of vibrationwill be as shown in Fig, 9.9 a, i.e. the foundation will undergo rocking vibrations with respectto a point situated at a distance PI from,the centre of gravity of foundation, The value of PI isdetermined by the absolute value of expression (Eq. 9.93) if 0>n2 is substituted for O>n'

(Ui) If 0>= O>nl then oo~ - 00;1< 0, P will be negative, and Ax and A~ will be out of phase, Figun9.9 b illustrates the form of vibrations around,a point which lies higher than the centre of gravit;and at a distance P2 determined from expression (Eq. 9.93) if 0>nl is substituted for O>n'

9.5 ELASTIC HALF-SPACE METHOD

. . ,r<tJ .

~':~~':.:::::,',::~,..':Gp)J

F (t)

'., ; ". ' " .-4, pe .':::.:'~":','. :':'::':~"""':;"~::',<;,

GPJJ

(a) (b)

Fig. 9.10: Oscillating force on the surface of elastic half-space

9.5.1. Vertical Vibrations. Lamb (1904) studied the problem of vibration of a single oscillating fon(Vertical or horizontal, Fig, 9,10) acting at a point on the surface of an elastic half space. Reissner (193.developed the analysis for the problem of vibration of a uniformly loaded flexible circular area (Fig. 9,1by integration of Lamb's solut~on for a point load, Based on his work; the vertical displ<icementan]centre of the circular area is given by

where,

F ;rotoe I' ' fZ = (J 1+ I 2)

0 Gr 0

F0 = Magnitude of oscillatory force0> = Forcing frequency, rad/sG = Dynamic shear modulus of the medium

r0 = Radius of the footing/1' /2 = Displacement functions

i=H

,..(9.9

,..,"

III .Fountiatitins'0/ Rlc;procatin'g"Mai:h in iii 371' '-

'",.., , , .j:" , . ,-:,-;,-; ~ -: .', ::-,: -' ,- -

fF.- i

__

(..;t 'r ,: .-""

- 0 (Z - ' .>-.

óÐ¢ýññ¢ '-

: ",-,-.':, .: -, :.i??//(/4 -"-. .' - . ~,. ~ '. ,",

"

,

~

Go '00'0" '0 l°":::"-:':

, o':! i: ;

.

p)J

load p(Zr F. (ZiG.)tunit or(Zo = ~

11"r20

z

Fig.9.11:Vibrationora uniformlyloadedcircular flexiblearea

Reissner introduced following two non-dimensional tenus :(a) Mass ratio, b : It is given by

m Wb=~=~

pro yroIt describes the relation between the mass of vibrating footing and a certain mass of the elastic half-space.

(b) Dimensionless frequency, ao : It is given by

a =00,

r Jp =,

Cllr Iv° ,oVa.. O,S_,

m = Mass of the machine and foundation

Vs = Shear wave velocityy = Unit weight of soilp = mass density of soil

Using the displacement Eq. (9.94), and solving the equation ofeguilibrium forces, Reissner obtainedthe following expression for the amplitude of foundation having circular base:

, ...,(9,95)

...(9,96)

where,

Az =Azn ~. ,(FofGro)

fl2 + f222 2 2 2

(~-:-~ao 11),.+, (bao f2)...(9.97)

where; ,Az = Amplitude of foundatiQn,.> ' " ' :,

,'A = Dimensionlessamplitude; i ; ,::; ,', ;,zn . ,"'" , """:"',""""")"':""'j"'.'", ,", "'"",,,"

. V~lue,s,of disp~a~einentlu~ctio~fi'ai!2f2 ~~ ~o.~d dep~~dien(op. P~is~o~~sr,a~o ~..an~ ~~~~.~n~i?~~less frequency factor a '.Their~values fo~"f1exible'"'circular'foUndation are' giveIi'iii Table 9.2:;:'.' :-'-'.,, J!

0 . '.-it';~. ,:.[

. , ,,' ,: .. '.1 : i

: ,':' i .. {J':.."

372SoU, DJ!n.am.ics.&, M.~~hine Fou~d~tion~;~

Table 9.2 :Values of Displacement Function 'of Flexible Foundations (Bowles, 1977}

Poisson's Ratio J.1

0

0.25

0.5

Valus of(fl)

0.318310 - 0.092841 a/ 1-0.07405 a04

0.238733: - 0.{)59683'a02 + 0.004163 a04

0.159155", 0.039789 a/ + 0.002432 a04

Values of (- 12)

0.214474'ao'~'. 0.029561 'a03+3

0.148594 Go- 0.017757 Go +3

0.104547 Go- 0.011038 Go +

0.001528 a 505

0.000808 Go50.000444 a0

0

0.25

0.5

The classical work of Reissner (1936) for circular loaded area was extended by Quinlan (1953) andSung (1953) for the following three contact pressure distributions:

F ;w/e :(i) Rigid base (Fig. 9.12 a), fz = O~ 2 for r::;; ro2rrr r - r0 0

F ;w/.. e

I~ = 0 2 for r ::;;ro~ rrro'

...(9.98)

(ii) Uniform (Fig. 9.12 b), ...(9.99)

It is the same as considered by Reissner i.e. for flexible foundations with circular l>ase

2 (r2 - r2) F iwt(iii) Parabolic (9.12 c), f = .0 4 0 for r ::;;ro

Z : 1t ro...(9.100)

, I

Cl ~ ~/'

. ~-d'

(0) (b) (c) ~.j

Fig. 9.12: Contact pressure distribution under a circular foundation

In the above equations, fz is the contact pressure at a distance r measured from the centre of foun,tfdation. Equation (9.97) holds good for all the three types oJ conta9t pressure distributions with change4:val~es of/l a~d/2. The v~lues off} an4/2f~~ rig~dbas~ foundatio;nswere comp~ted by Sung (19~~)?Jt1l/the assUinpt~o!l~at the.press.ur~.di~~b~~~~ri:r~ma,inunchang'e~ with,frequency. Their values ~re ~~v. '~A

.,

in Table 9.3. . . .,' '" . . ~~~......'I

'

.

'~

..r,;

datio/fs o/,Reciprocating M/lcMnes

{-; ,": ',: ',' '.. .

0.7

0.6

O.S

c:N

<t

c:,I'U;:)

+'

,0.4

0.1

. .' ,

373,

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

poisson's ratio) j.J = 0.25,', ,

,,-.

0 ~

0, , 0; S '. a ,-' 1.0,'. 0 .. "f.. ',', : ,- , "

1.S" . >\'

Fig. 9.13: Plot of AZAversus ao for a rigid circular foundationsubj~te~ to:constantexcitati.on,r~rce (Ric~art,I,962)i..r' ',', ;, , ,

.. ,

-- ,--

a.E0

\11 0.3VIc:,I-c01/1CtIE 0.2b

,~

374 Soil Dy,u,inics-&' Machilre -Fourrdations \,

Table 9.3 : Values of Displacement Functions for Rigid Foundations~'.'<' ..' '-~'-"'-(Bowres~"f977r-'--'- 0'0 .",- .

0

0.25

0.5

V alues' of 112 4

0.250000 - 0.109375 ao + 0.010905 ao0.187500 - 0.070313 a02 + 0~006131a040.125000 - 0.046875 a/ + 0.003581 a04

Values of -/23 5

0.21447 ao - 0.039416 ao + 0.002444 ao0.14894 ao - 0.023677 a03+ 0.001294 a05

3 50.104547 ao- 0.014717 ao + 0.00717 ao

Poisson's ratio J.1.

00.250.5

Figure 9.13 shows a typical plot of A versus a for various values of mass ratio b for a rigid basezn 0 .

circular footing subjected to a constant force excitation F0 e/(JH. A high mass ratio (greater height offooting and smaller contact radius) implies a large amplitude of vibration for a given set of conditions.

Manytimes foundations are subjected to a frequency dependent excitation (Fig.2.1~). The amplitudeof the external oscillating force is given by

2F=2m em0 e

where, 2 me= Total rotating mass

For this condition, the amplitude of vibration Aze is given by2 2 2

2 me e CO I1 + 12

Aze= Ora (l-ba~/I)2+(ba;fz)2

...(9.101)

...(9.102)

or~

2 2Az I1 + 12

Azen = (2meero2)/Oro (l-ba; 11)2+ (ba; 12)2:..(9.103)

where, Azen = Non-dimensional amplitude in frequency dependent excitation. .

Figure 9.14 shows a typical plot of Azenversus ao for various values of mass ratio b for rigid basecircular footing subjected to frequency dependent excitation.

It may be noted that the curves shown in Figs.9.13 and 9 .14 are similar to the frequency-amplitudecurves shown in Figs.2.13 and 2.16 resp~<::tively.

Richart and Whitman ( 1967) have studied the effect of the shape of contact pressure distribution andPoisson's ratio on amplit~d'~'~freq~en~y;esponseof rigid circular footing ~~bjectedto frequency depen-dentexcitation.Figure9.15demonstratesthenatureof variationof Azenwith aofor three types of contactpressure distribution; i.e. uniform, rigid and parabolic. Parabolic and uniform pressure distributionsproduced higher displacement than a rigid base. The effect of Poisson's ratio on the variation of Azencapbe seen in Fig. 9.16. The peak value of Azendecreases with the increases in the value of J.1.;but thecorresponding value of ao increases with increase in J.1.. . i'

l;.4",

"daiions ut Reciprocating' MilChin~s 375

0.28"--"--""'-"""""""'-'-""'-""""-'- "-'.-- '~.

0.24

0.20

c~N

«... 0.16

~"0

::::J....

0-E0

0.12IIIIII"C0

IIIC~

E 0.08(:)

0.04

poisson'5 ratio, ).J = 0.25

.'....

0 -0

--

0.5°0

1.0 1.5

Fig. 9.14: Plot of AzcDversus.o for. ~igid ~in:ular foundationsubJ,ectedto frequencydependentexcitation. " (Rlchart,1962) "

-..

"" . ", ...

- .. -, -_~_::c,i:- -'-<-':":' ~'-- -'- ;-- -- .- - .---------- -

376

0,6

c:~ O.S

<t~

-g 0.41,--a.E0 0.3U1U1b.I

C 0.20U1C\:11

E 0.''0

Soil Dynamiq_&Machine,Founda.(~(Jf3

b :'5

J.l : 0.25

0-0

parabolicpnz ssu rq ,

distribution

1.0 1.50.5°0

Fig. 9.t 5 : Effect of c~ntact pressure distribution on the variation of Aze,nwith ao(Richart and Whitman. 1967)

c~ 0.4-

<t...

~"U

~ 0.3a.E0 0.2U1III~

c:0

'jA 0.1c:~E0

Rigid basqb : 5

0...0

0.5

0.5 1.0 1.5°0

Fig. 9.16 : Effect of Poisson's ratio on the variation of Ann with ao (Richart and Whitman, t 967)

F~omthe amplitude~frequencycurves.in,Figs.9.J3 ,and9,.14,on~can obtain:-(i)~axffI1umamItude, and (ii) value of ao for maximum amplitude (i:,e;.resonance condition) for different values 01Richart (1962) have plotted this data in graphical form as shown in Figs. 9.17 and 9.18 whichconvenient for design use. -

',dations of Reciprocating Machines100

Rotating maS5Constant forc(Z

Rigid bas(Z

:--.......'.......

"""'.......

"""'0

= 0.5

< ,

1-0 .2 0.4- 0.6 0.8 1.0 1.2 1.4 1.6

ya lu (Z of ao at r (Z50no nc (ZFig. 9.17 : Plot of mass ratio b versus ao for reso!,ance condition for vertical vibrations (Richart, 1962)

,Rigi d ba C;(Z

Rota'ting ma5S. ", . .

..~ Con s'tant" torc(Z

. .

,,; 1--.0

,-

1.0 1.2 '1.4

. ~. Dim(lnsio'rde-ss',,:omplitude ot re5.0nanC(l ..~Fig. 9.18: Plot of mass ratio bversus, dim~nsio.nless~mplitude at resonance (Ricbart, 1962)

--;)""\"";*~;,'J..~.~, ~'",

, ""'.. '.~\!;,A ,~,{~,:'H"...;,,~) ':!l~t:...n~iv~ir~l1'i' ~(1'i 'r,;',,~~ ~, 1r~ Q...(m'\\i<l~}{ : 1l1'" ,,~4 {

.£).

0.-...0 10....

IIIIII0

100 r\ '\\ '\\ \,, \\\ ", \\\ \\

.£) I \ \\\ \\

0 \ \\.;: ,"

10 '\ \\

Ji<0 fJ =0.5 \"0.25

0

Soil Dynamics & Machine Found4ti.o.,

óó¢þò¢ù ùòù æþþùóù¢ùå¢öôºLysmer and Richart (1966) propose4 a .siplplifi~dmass-~pring-dashpot analog foL-calculating,the

response of a rigid circular footing subjected to vertical oscillations. The values of spril1gconstant Itand damping constant Czwere taken as giv~n ~elow :" :

. , 4GroK=-

z I-J.t2

3.4 ro er;GC = -vpvz I-J.I.The equation of motion may thus pe written as

2 -

.. 3.4 ro er;. 4 G ro F iwtmz+-vpG.z+_.z = eI-J.i. I-J.i. Z

Lysmer and Richart (1986) also suggested the modified mass ratio as1 I-J.I. m

B -J.I.b ~.--- - 3z - 4 - 4 pro

.The damping ratio; z is obtained asC C 3.4 r; JPG

~, ~ C: ~ 2J~, m ~ îøïóñ´÷ùtoTomI-J.I.Putting the value of m in terms ofBz from Eq. (9.107), Eq. (9.108) becomes

; - 0.425z - ùÖÐææ

The response of the system can be studied using Eq. (2.58).The dashed curves in Fig. 9.19 illustrate how well the response curves for the analog agree with t!

response curves for half space method. The derivation of magnification factor, Mz is given below.'3

378

and

H -:E

~ 2L-e....u0-c0....0u-c(JI0~

'.f-,,<'"

...(9.104)

...(9.105)

....(9.106)

"

...(9.101'

,...(9.108

...(9.t'0'

}J = 1/3

Halt-space theory

--- - Si mptitied ono log

Bz :: 5.1

Constant torce excitation '"

',1 ~~J-)Z

-..0...:t!?..",."I!.

00.5 . L 0 1.S ~,

, Dimen'sioritesS' freq"uency, 'cro" "l , "- , ~Fig. 9.19: Response of a rigid circular footID~' for vertical vibrations (Lysmer and Rlchart, 1966) ",

"i"

1

"-'~~:

.I

lions of ReciproCllting Machines 379

e natural frequency of the system described by Eq. (9.106), i.e.

ri "" '."

. ',".K' 2"

. <.0 = -L. ~. m m ~l-~ .

tting the valu'e~'of Kz, ;n a~d~~zfrom-Eqs:' (9.104), (9.107) and (9.109) into Eq.(9.110), we get

<.0 ~ f(Bz - 0.36)0 .~nz' 0 B r'. z 0

...(9.110)

...(9.111)

Id at resonance the amplitude is given by Eq. (2.61). i.e.2

Fo / Kz

(Az)max = ~ 22 ~ 1-~zutting the values of ~z from Eq. (9.109) into Eq. (9.112), we get

Fo Bz(Az)max = Kz 0.85 ~Bz - 0.18

" Fo.(I-J.1) Bz

r (Az)max= 40'0 . 0.85~Bz - 0.18. (A) .4G, B

vlagnificationfactor, M- = z max . 0 = z"- Fo(1-J.1) 0.85.JBz-0.18

~or a frequency - dependent excitation ,the resonant frequency and the maximum amplitude are1 by : H

...(9.112)

...(9.113a)

...(9.113b)

Magnification factor,

0.90 1

<.onz = ,/(Bz - 0.45)P . '02 mee Bz

(Az)max = -;;;- .0.85~Bz- 0.18B'(Az)max - z

Mze = 3mee - 0.85~Bz-0.18)

-,...(9.114)

and ...(9.115)

...(9.115 a)m

2 me = Unbalanced rotating masse = Eccentricity of mass from the axis of rotation

.2. Pure Sliding Vibrations. Amold et at. (1955) have obtained theoretical solutions for slidingrationsof rigiq circular foundations(Fig. 9.20) subjectedto an oscillatoryhorizontalforce F0 ei(J)(.~yhave presented the solutions for two cases namely (i) constant force excitation, and (ii) frequency)t:nderit excitation.

:re

In constant force excitation (F0 = constant), the amplitude Ax is gi~en by. . - . 'F. .. ... -"' .

Ax = O~ Axn0lere, Axn = No!1-dimensional amplitude factor

...(9.116)" ..

.-~- "~ --. ., ."" ... """'-' -=.""---==~ ;;:;;~

380 Soil Dynamics & Machine ,Eo. .

... .. F""(Z iQt0otT f

Foundation

.;"-,~': -'". -. . "-'f

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

x f<i

,

s.t '

I:~'"", I1}\ I,G

P}.J

r0 1 1.Sf

-h'J'J.'I..~;

z

Fig. 9.20: Rigid circular foundation subjected to sliding oscillations ";J

The variation of amplitude versus frequency is shown in Fig. 9.21 by dotted lines. The envelop <Jrav.Tto these curves is used to define the frequency at maximum amplitude, The definition of mass ratio.b 1same as given in Eq. (9.95). The plot of b versus Go for resonant amplitude is given in Fig. 9.22~

:xi10 f-

t,ti"0::J....

}J = 0~

III C)(

××Ää¥IIIti

C0IIICti

E,-c 0.1-

0.2

--- - Axnß¨»²

..~,a.E0 cti0'1 x

.= <{ 11:) .....- 0

~"~

'":I-" Iit,~I

'"T'

1.0°0

Fig. 9.21 : Plots of AsDand AscIIversus aofor sliding vibrations (RIchart, 1962)

tionsof ReCiprocating Machines íèï

ïðð

ïð

£òÖãÑ

Eccentric ³¿oscillator

×ô

Öó

Ý±² ¢¬¿² ¬

ï

ðòî ðòìò ðòê ðòè ïòð ïòî ïòìò ïòê

pð

Ú·¹òçòîîæ Plot of mass ratio b versus a for resonance condition for sliding vibrations (Richart. 1962). 0 .

For the case Fo = 2 m e ci ,the value of A is given by: .ex. /,

- 2 meeAx - -rAxen

pro

The firm line in Fig. 9.21. shows the envelop of Axenversus ao for resonant condition. The variationhe mass ration b versus ao for resonant condition is given in Fig. 9.22.Hall (1967) propos~d a simplified mass-spring dashpot analog for calculating the response of a rigid

:ular footing subjected to sliding vibrations. The values of the spring constant Kx and damping con-1t Cx were taken as given below. ;

...(9.117)

32(1-J.1) GK = ro. x . 7-8J.1

18.4 (1 ~ J.1) 2 c:;:;oC = rovPux 7-8J.1

The equation of motion thus can be written as

.. 18.4 (1-J.1f 2 JPG. ' 32(l-J.1) G - F ;CJ)lmx+ r P .x+ r .x - e

7 - 8J.1- 0 . . 7 - 8J.1 0 0

He also suggested the modified mass ratio B -for sliding vibrations as. \. " X '..

. 7 -8J.1 mB = -x 32(1- Ji) pr3 '.' .".',.,..".."",J,,'I, "O,,}. "".

...(9.118)

and ...(9.119)

..,(9.120)

...(9.120 a)

382Soil Dynamics &: ,Mllchine Foundlltitt~ 1

. I

...(9.121)

The damping ratio ~x is given by

~ - Cx - Cxx - Cc - 2 ~kx m

Putting the values of Cx and kx from Eqs. (9.118) and (9.119) in Eq. (9.121), we get

~ - 0.2875x- Fx

Figure 9.23 illustrates how well the response C\lrvesfor the analog agree with the response curves forthe half space model.

4Bx = 5

Exact solution- - -Analog solution

00.5 1.0

,°0Fig. 9.23 : Response of rigid circular footing for pure sliding (Hall, 1967)

I

...(9.122) I

It

i1~.j

.' T.:!i;

..

tii..,'

",'

1.5,:;,,

~.~.~~

(1/1.1'

~.

,J''1i~;rjI.,.~

3

x::E I I \ 2... -L-0.....u0.....

c 20.-.....0u.-......-C0'10

::E

"

ùïïä¥ù

,dotions -of RecipI'ocatingMachines 383

The natural frequency <.olltand maximum value of amplitude (Ax}m~ can be' computed using Eqs.23) and (9.124) respectively. . . .

f-: g 2 .

<.0 = -2.. 1-~nx m x

. Ft / kx(AJ X

=.R

...(9,124)" ma 2 ~ 1- ~x x

,2. Pure RockingVibrations. Amold et al. (1955) and Bycroft (1956) have obtained theoretical so-ons for rigid circular foundations subjected to pure rocking vibrations (Fig. 9.24). The contact pres-~ below the foundation is varied according to

. . 3 My r cosa. jro(q - e- 3~

27tro 'Vo -rMy = Exciting moment about Y-Y axis

a = Angle of rotation

...(9,123)

and

(for r ~ ro) ...(9.125)

ere

\I

/\rjJ I fJ /

\~I- I

i\T7\

\ ~ / My Cli~,t

Footing,

- /"",, " L' ',,' .,...

~~:, ':"',:" ,'", " " r'. ., ,..' ., -: 0.". ."" , . '. .' . '"', . . - I-' - ,G

i.~1I

p}J

x

. ~,;" " y t',. "'"~~~-~ptali~;bt tootin9'~ .~.., : ....~ i .1..1 ; ';} ,

Fig. 9.24 : Rigid clrcuJar foundation subjected to rocking oscillations"

'.;." ;,I':(V;.t.~'H;tL""'i 4;':' ,.)"~:;> Ö¼

384 Soil,Dynamics &:.Machine Fou"dtltioM .Borowicka (1943) gave the following equation for computing static rotation of the foundation under

the static application of moment My'3(1-J.l) Mmo

A~ = 8 ~proUnder dynamic conditions the amplitude of rocking is a function of the inertia ratio B. which is

given by

...(9.126)

3(1-J.l) MmoB. =8 5pro

where, Mmo = Mass moment of interia of machine and foundation about the axis of rotationFor the dynamic moment My, the amplitude of angular rotation A. can be expressed as

A. =(

MY3)

'A,nGro

where A.n = Non-dimensionalrotational amplitudeFig. 9.25a shows the variation of A.n with dimensionless frequency ao for various values of inertia

ratio B.. The envelop curve shown by the flfm line can be used to define the relation between ao atmaximum amplitude (resonant condition). The plot of inertia ratio B. versus ao for resonant amplitudeis shown in Fig. 9.25 b. .

Hall (1967) proposed an equivalent mass-spring-dashpot analog for calculating the response of arigid circular footing subjected to rocking vibrations. The plot of spring constant k. and damping con-stant C. were taken as given below:

'"'t):I..

40

a.E0-0c0

10

.. c0"6-+-40~

0c0'"CtIIE0

...(9.127)

...(9.128)

38G1~

K. = 3(1-J.l)

0.8r: ,fGPCcp= (l-f.l)(I+B,)

...(9.129)

...(9.130)

10.2

'f ':'11,I -

I "1 , ,,I , ,'I , \ ,'I ,\ '\

" " / \, \ I \ I \'\ I \ 1\, " \

I 20 \ I 10 \ I S \I \1 \ I\I 1\ (, I

JJ = 0

0.4 0.6 0.8 1.4

°0Fig. 9.25 : (a) Plot of A." versus 10 (Richlrt. 1962) (...Contd.)

'j.j...

H'I\!~

Foundations of Reciprocatipg Machines 385

60

10.2

)J = 0

0.4 0.6 1.2 1.4

°0(b)

Fig. 9.25: (b) Plot of B. versus. ao for resonance condition for rocking oscillations of rigid circular foundation(Richart. 1962),

The equation of motion can be written as4 ~ 3

M ;j, 0.8ra "Gp .h 8Gra '" _ M i(f)/",+ '",+ '", - e

m$ (l-fl)(l+B~) 3(1-fl) YFor critical damping,

/-

...(9,131)

c~c = 2~K~MmaTherefore damping ratio ~~ will be

C~ 0.15~~ = -c

=) In ..,(9.133)

~c (1+ B$ "B~Figure 9.26 illustrates how well the response curves for the analog agree with the response curves for

the half space model. The undamped natural frequency ron~and amplitude A$ in rocking vibrations aregivenby .

...(9.13'2)

~~ro =n~ Mina

...(9.134)

My . ..

A. ~ k'[{l-:~r+~'ro:ni

...(9,135)

1

-e.m

...0 10.-....0....

0.-.....L..C

386

50

0.50

Soil Dynamics & Machine Foundations ".

B~=S Exact solu tio n

- - - Analog solution

0.5 t.'1.0°0 "

l<

Fig. 9.26: Response of rigid circular foundations subjected to pure rocking vibrations (Hall, 1967)

9.5.3. Torsional Vibrations. Reissner and Sagoci (1944) have obtained theoretical solutions for rigidcircular foundations subjected to torsional vibrations (Fig. 9.27). The variation of tangential shear stressis given by

where

3 Mr't 9

= -:-.~ for 0 < r < rZ 41t 3 2 2 0. ro ro - r.

'tz9 = Tangential shear stressM = Mz ei(J)( = Moment at any time t

Mz = Maximummomentabout Z-axisFor a static moment Mz.the angle of rotation A'IIsis given by

A -[

3

]M

",s~ 16G r; z

...(9.136)"'.

~I>

Wo

'if

~

."!

~.u(9.I31~i

".,.~.~,.c"'":

20

9--.. 10-)-J:)-- 5'--:>0u-+--c:0\ 20z

---------..-.'j'

Foulidiltions '6/ Ri!~iprdcating 'Mai:htnes 387

The amplitude of the angle of rotation Av can be expressed as.-"-"" '-',"'. "-"'--'."-"-~"'""

'. MAv = ~Awn

Gro

where Awn = Non-dimensional amplitude factorUnder dynamic condition the am~litude'of torsion is a function of inertia ratio Bw which is given by

MmzBw = -S ,..(9.139)pro

where M = Polar mass moment of inertia of the machine and foundation about the' axis of rotationm= .

...(9.138)

.(

:', :,,~(,::::~(~<'<~ ,

: ..."",

."r,-. ,"'00"'- ',_..

G " - ro ~ .,z' . "

FootingTZ9

Flaxibl<zfoundation

, , ro(b)

Ize

Rigidto undat ion

ro(c)--'.. - -.. , ,.,..,-,,-'"

(a ).-~'

Fig. 9.27: Rigid circular foundation subjected to torsional vibrations

Figure 9:~8 _ashows the ~ariation of Awn with~ime,n~ionless frequency aofor ~~rious values ofinertia ~atioB",:'~e' ~nveIop'cur~f's,ho~'~y !~~,:~~..lirie'c~n'~.~.~s~ato,defme 't~e relation between aoat maXimum amplItude (resonant condItion) ana the'values of Intena ratio B", (FIg. 9.28 b)

-,

388 Soil Dynamics & Machine F:oulldatioRS

-~.~,,--

10tII

'0::J....

0c.~ c'" r... <t0.... 1

II1\I1I Ir, I ,

I I ,\ I'

B4' = 10! \ 5, \I \,' \ .-

/ \, \ /'. /;\ )// / \ / \

/ / \ /' \/ // ~;.\./ \

/./ "'" \ \ññ¢óóþùþ I \ I \

,"-"-" Flczxiblcz, .

"V'....... "

Rigid" ".......

Co

E0

'"'"to'

C0'"CtIIEQ 0.2

0(a)

0..4 0.800

1.2 1.6 2.0

60

10

(b)0.8 1.2 1.6 2.0

7m

.~ 10....0L.'"'"0~

0.400

Fig. 9.28 (a) Plot of ~ versus ao; (b) Plot of B. at resonance versus ao for torsional oscillations ofa rigid circular foundation (Rlchart,1962)

'..,'f;''w

."iO'~

",. .......

FiJundlltions of ReCiprocating Miichines 389

Richart and Whitman (1967) propose~ an equivalent mass-spring-dashpot analog for calculating theresponse of a rigid circular footing subjecred to torsio~al vibrations. The values of spring constant kw anddamping constant CIjIwere taken as below;

, = 16G r3 ~ ,~ J 0 ~::

":;;-~~_'d

...(9.140)

',.'"

...(9.141)and

The equation of motion can be written as

.. 1.6ro4.Jp G." 16 G 3 - M ;m(M III + . \V+ - r. \V - e,

mz 'Y I+B "", 3-: 0 ZIjI .

1

...(9.142)

The damping r~tio ~1j1~ili be , , "

C,o.s

~1j1 = 2 ~K: . m = (1 + 2 BIj1),..(9.143)

The undamped natural frequency wnlj1and amphtude AIj1of the torsional vibration are given by

'Kwnlj1= vM:;

" ,...(9.144)

A = Mz

W " KW[{I- ~:r +~Wro:Jr

...(9.145)

9.5.4. Coupled'Rocking and Sliding'Vibrations. Figure 9.29 a shows a rigid circular foundation rest-ing on the surface of elastic half-space and subjected to an oscillatory moment Mv /(JJlan a oscillatoryhorizontalforce Fx e,m{.The sign convention chosen is illustrated in Fig. 9.29 b, which indicates that + xand + F act to the right and that + 4>and + Mare clockwis'e.Figure9.29c showsthe motionof the footingwhen both translation of the e.g. and rotation about c.g. are positive (i.e. in phase). In this case. the centreof rotation lies below the base of the footing, and the motion ,is termed as first mode of vibration. If thetranslation is positive while the rotation is negative, then the centre of rotation lies above the e.g. and themotion is designated as the second mode' of vibration.

': "',1/ ,~}') ,/1' "

Nl J > 1 t

."..

390 Soil, Dynamks " ,MilehineFOUIIdtltions IJ

...~

Er MR +x~+Fx

( b)

~

~~leA]

Lc,&

Fig. 9.29 : Coupled rocking and sliding vibrations of a rigid circular block on an elastic half space(After Richart and Whitman, 1967)

From Fig. 9.29a

wherex 0 = x - L<\>

XO = Displacement of basex = Displacement of c.g,

<\>= Angle of rotation in radians

..,(9,146)

Therefore Xo = x- L ...(9.14'n

The equations of motion are written in terms of x and <\>.The equation of motion for sliding is ~ ,

or

. .. C . K F irotm x + x Xo + x Xo = x e

m x+ Cx (x- L~) + Kx(x- Lcp)= F.tirot

.1 ~I.

~4

r...(9:148

, .[)"\Ormx+C x +K -x-LC d.-LK '" =F irotx x x'f x'f .t

The equation of motion for rocking'is, '

.. . irotMill cp+C~ <\>+K~ <\>+C:c,xo'L+Kx,xo.L = M.ve

Sustituting the value of Xo from Eq, (9.147), the above equation simplifies to

Mm ~+(C~ +L2tx) ci>+(K~+ L2K:c}cp- L Cx x- L Kx'x = My irol

or

.,.1~aiJ

., ~:y;t ,;,.,,<:;

...(9,Jf~'::

'.'~:r:".,..t~

'~:',;~!

M (l iwty II I

I !. I Fx(li6.>t

I ILm

, .. FxfMR

(a)

I Ir , I I

:[1I 1/ / = II I +1-.- I

Xo1r- '--r I-- x

I :

(c)

'Idations of Reciprocating~Machines 391

The natural frequencies of coupled rocking and sliding are obtained by putting forcing functions in, (9.148) and (9. 149)equal to zero. Thus

³¢õÝ ¨õÕ ¨óÔÝ n.-LK .n. = 0x x x 'I' x 'I' ...(9.150)"

( 2 ).

( 2)Mmcp+ C.+L Cx cp+ ÕòõÔÕ¨ cp-LCxx-LKx'x =0 ...(9.151)

The solutions of above equations can be obtained by substituting

x = A eiCiJndt ...(9.152)

...(9.153)cp = B eiCiJndt

which A and B are arbitrary constants. By doing this the Eqs (9.150) and (9.151) become

A - LKx +i LCx,oondB-2 K . c-m OOnd+ x + I x OOnd

...(9.154)

d A -Mm OO;d+(KcjI+L2 Kx)+i(CcjI+L2 Cx)OOnd

B L Kt -i Ct OOnd, - #.'

Equating Eqs. (9.154) and (9.155), and substituting

...(9,155)

if x - ff.x = ~ ~ = Cx

oonx = -, oonq» - M' r M ' x 2 ~m mo mo vL'\o.x rH

C .

~4I = ~ x , on simplification we get2 K. Mmo

[ (

2 2

)

22

1

2

4 roncP+ronx 4~x~~ronxron~ 2, roncpron..trond- - . rond+ +r ~ r

J

4[

~xronxrond ( 2 - 2.)+~~ron~rond ( 2- 2

)]2-ro~ ro~ ronx ro~ -0r r

It may be noted that Eq. (9.156) reduces to Eq. (9.69) when ~x= ~cjI= O.As the effect of damping onlaturalfrequency is small, the undamped natural frequencies for coupled sliding and rocking vibrationsanbe computed using Eq. (9.70). .

The damped amplitudes of rocking and sliding of a foundation subjected to a horizontal force For/°)1aregiven by

,d

...(9.156)

]

lfJ

A ~ Fr ÅøóÓ³®±îõ¢õÕ®Ôî÷î +4ro2(~,,~KrMmo+L2~x~Kxm)2 -xl -m Mm, ~ (002)

( 2' 2)112

F L ronx ronx+4~x roand A =--L-.1 Mm ~ (ro2)

...(9.157)

...(9.158)

...

392 Soi/Dynamics.& Machine Foun

where

[

,

{

2 2

}

2 2

]

2

lI. (CO2)~ ",4_",2 '" ,+ :"'M , 4 ~x~+;M '" ,+ + "',. r"'M

;~'." '

[

, 2

]

ln

+ +x m"; m (m~ -m2)+ ~.m~.. m(m;. -m2)} , "

~~I.," 't

...(9-.IS~~"':' .

, ,The dampedamplitudesof rockingand slidingof the foundationsubjectedto an excitingmomentiw . ,c

M, e are gIven by '$I

[2 2 2

]

1/2

Mv (OOnx) +(2;x OOn.t) ,A-'

x2 -. M " (2

)III D. 00

:4

."(9.Jf~1",iI

!I1'r

and [2 2 2 2

]

112 t- My (OO"x-OO) +(2;xoo/lxoo)

~,,

",

"

,

'

A<j>2 - - 2 ...(9.1~!Mm f.. (00 )'.

When a footing is subjected to an oscillatory moment Myirot and a horizontal force Fx im/ simulb-neously, then the resultingamplitudesof slidingand rocking are . "j

A = A 1 + A 2 ...(9.162x x x

A<1>= Acjll + A<1>2 ...(9.163

9.6 EFFECT OF FOOTING SHAPE ON VIBRATORY RESPONSE fyElastic half-space theory was developed for a footing with circular contact area. Response of a footing~1influenced by the shape of contact area. The usual practice is to transform area of any shape to an equjy:lent circle of same area (for translational modes) or equivalent moment of interia (for rocking and)bsional modes) (Richart and whitman, 1967 ; Whitman and Richart, 1967). Thus

r = i!b0 -

7t

(

3

J

1/4

For rocking vibrations: ro = b3:

[ ]

114ab (a2 +b2)

For torsional Vibraf .ns : ro = 61t

where ro = Equivalentradiusa = Width of foundation (parallel to the axis._ofrotation for rocking)b = Length of foundation (perpendicular to the axis of rotation for 'rocking)

~'

For translatory vibrations: ..(9.U,;;:.1

~~~1..:~(~~t

~aj,yt~ ..",./I

, " ~.j91,'. ~'.~if

tt'

t ,1"

,,

"

,,

-,

'

,

'

"d1I .' "

["'~,,

' ,~,'

i<) "4!t"l:'1~~.h..

~,5

,'

,

"

"""

'

",

'

~

'

,

.'

,

'

,

'

't\', 'ft,

'"r

394 Soil Dynamics & Machine Foun~ ,.9.7 DYNAMIC RESPONSE OF EMBEDDED BLOCK FOUNDATIONS - - ~iFor an embedded foundation, the soil resistances are mobilised both below the base and on the sides. Th~additional soil reaction that comes into play on the sides may have significant influence on the dynamitresponse of embedded foundations. Typical response curves showing the effect of embedment are pre~sented in Fig. 9.30. It gives that as a result of embedment, the natural frequency of the foundation- soil,system increasesand the amplitude of vibration decreases (Novak, 1970, 1985;Beredugo(1971) ; Beredugoand Novak, 1972; Fry, 1963 ; Stokoe, 1972 ; Stokoe and Richart, 1974 ; Chae, 1971 ; Gupta, 1972 iVijayvergiya,1981).

. The problem of embedded foundations has been analysed by both linear elastic weightless springapproach (Prakash and Puri, 1971, 1972 ; Vijayvergiya, 1981) and elastic half-space theory (Anandkrishanand Krishnaswamy, 1973; Baranov, 1967; Berdugo and Novak, 1972; Novak and Beredugo, 1971).Th~analysis developed by Vigayvergiya Cl981) is simple and logical, and therefore selected for presentati9Phere. On the basis of theoretical analysis, he had recommended the.equivalent spring stiffnesses in dif-ferent types of motion as given below: .

9.7.1. Vertical Vibrations (Fig. 9.31)

Fig. 9.31 : Embedded block foundation subjected to pure vertical vibration

Õ¦» ã Ý«Ü ßõîÝ¬¿ªø¾Üõ¿Ü÷

Õ¦»ãÛ¯«·§¿´»²¬ spring stiffness of the embedded foundationÝ«Ü ãCoefficient of elastic uniform compression obtained at the base of foundation

Ý¬õÝ¬ÜÝ¬¿ª= Averagevalue of coefficientof elasticuniformshear=- 2

! Fz5in 6>1

Machin<z

-~,~

hFoundationø±¨ bxh)

0

1f..

1~a

where

Ct = Coefficient of elastic uniform shear at the ground surfaceÝ¬ Ü ã Ceofficient of elastic uniform shear at the base of foundatiqn

D = Embedment depthb = Width of base of foundation

a = Length of base of foundation

¢

¬ Ú¦ Í·² êâù

³

Õ¦ä¦

...(9.!

~

d

.

Foundations of Reciprocating Machines

The equation of motion will be

mz+Kze'z = Fzsin rot

The natural frequency oonzeand maximum amplitude Aze of motion are given by

fK:oonze = v--;;;

FzAze = ( 2 2). m ronze - ro

9.7.2. Pure Sliding vibrations (Fig. 9.32)

Machin<z

. , ,

Fx Sin wt -FxSin wt

~ ..h...-. ~ m

1~

Foundationø¿¨¾¨ h)

0

1 '.

a -iFig. 9.32: Embedded block foundation subjected to pure sliding vibration

K = Ct D A + 2 Cllavb D + 2 Ctav aDxe

where Kxe = Equivalent spring stiffness of the emebedded foundationC +C

Cl/a v = Average value of coefficient of elastic uniform compression = !!.- uD2

CII = Coefficient of elastic uniform compression at the ground surface.Similarly, the equation of motion will be

mx + Kxe ..i = ~t sin rot

The natural frequency and maximum amplitude of vibration are given by

fK::oonxe = v--;-

FxA = 2 2)xe m (ronxe - ro

~.

..

~

~""...

.. .... ....""..

~.."L.

395

...(9.168)

...(9.169a)

...(9.169b)

Õ¨ä¦

...(9.170) .

...(9.171)

...(9.172)

...(9.173)

396 Soil Dynamics & Machine Follndatioi;.\- - -,

9.7.3~ Pure Rocking Vibrations (Fig. 9.33) ,.

Machinq

m-

~

rFig. 9.33 : Embedded block foundation subjected to rocking vibration

fC~av b 3 2 Db a2

Kljle =C~D.I-W.L+ 24 (16D -12hD )+2C~avlo+Ctav 2

where Kljle = Equivalent spring stiffness of the embedded foudationCIjID= Coefficientof elasticnon-uniformcompressionat the base levelof foundation

Cljlav= Averagevalue of coefficientof elasticnon-uniform compression= C~+C~D2

CIjI = Coefficient of elastic non-uniform compression at the ground surface

L = Height of the combined e.g. of machine and foundation from the centre of the base J

W = Weight of foundation ,b 3,

I = ~ '~J12 ,

-t...(9.£7<

:1, l"1

3I = aD

0 3The equation of motion will be

MI1IO«\>+Kljle.$ = My sin rotThe natural frequency and maximum amplitude of vibration are given by

~~eCt)lIljIe= M mo

MyA =

(2 2

):ce M",O ro119- ro

.,.

...(9.IJ';J

...(9.,.}

-- i

..~t}~,,I

.i;J

j.

.-;i',.f

My Sin GJt-::.'\\,: ×

×¸

ï

Foundation 0

(axbxh) 1I.. a ./

Foundations

]>

~K~t2~~

,..(9,174)

on

D

the base

,..(9)75)

...(9.176\ '"

...(9.

...

~ IJI~

,.Fo/llldatiolls of Reciprocating Machines 397

". 9.7.4. Coupled sliding and rocking Vibrations. The equations of motion in coupled vibrations aregiven as

and

mx+K.u,x+Kxtp' = Fxsinrot

Mm~+KcI>'+Ktpx'x = My sin rot

}\There Kxx = C'tD A + 2 Cl/aVbD + 2 C'tavaD

Kxtp = Ctpavb (D2 - 2 DL) - C'tD . AL

2 . a2 2 3 3Kq,tp = CtpD1+ C'tD AL - WL + 2 CljlavIy + C'tavbD 2 + 3' Ctpav[L + (D - L )]

Ktpx = - [ C'tD AL + 2 Cl/a v bD ( L- ~ ) + 2 C'ta v ( L - ~ ) aD]

...(9.178)

...(9.179)

...(9,180)

...(9.181)

...(9.182)

...(9.183)

where CljIlIV = Average value of coefficient of ~Iasticnonunifonn shear

ly = Moment of inertia of area a x Dlying in the plane of vibration about axis of rotation

Da3 aDb2=-+-12 4

Mm ~i + Ktptp. + ~tpx X = 0

The solution of the above equations can be represented by

x=Adsinoot ...(9.187)

= Atpfsin 00t ...(9.188)By substituting x and from Eqs, (9,187) and (9. 188) in Eqs, (9.185) and (9.186), we get

(Kxx - m 002) Axl + Kxtp Atpl = Fx sin 00 t ...(9.189)2

KtpxA.d + (Ktptp- Mm 00 ) Atpl = 0 ...(9.190)By solving Eqs, (9, 189) and (9,190), we get

.

1'he natural frequencies of the system can be obtained by solving Eq. (9,184)

m Mm co~e- (r:zKtptp+ Mm Kx.~)co~e+ (K~~K~ - K~x x Kxtp)= 0The amplitudes of vibration ofthesystem can be obtained as below:

(a) Only the horizontal force Fx sin ootis acting: The equations of motion will be :

m x + Kxx . x + Kfttp . = Fx sin 00t

l

tI

2. - .(K~~- Mm 00 )Axl - 2 2 Fx

(Kxx-moo )(Ktp~-Mmoo )-KxtpK<I1~

- -K~ '.,Atpl - 2 2 / r.

(K.n-moo ) (Ktptp-Mm 00 )-KxtpK<I1'

~O//II~ ~""i"'""

----u

cl!~

...(9.184)

...(9.185)

,..(9.186)

...(9.191)

...(9.192)

~-

~

f''"I'

,

-::.11.411

II,'1I

III, ,

~I1'1.I ~\j

'11I.. .

JI

d .

ItlI

F t j

Q,

III~.I

ó¢ùó±

íçè Soil DYllamics & Machifle Foufldatiofls

11

(b) Only the moment My sin cot is acting: The equations of motion will be :m x + Kxx . x + Kcpx= 0

Mm ~+ K.p.p . + K.pxx = My sin COtThe solutions of the above equations can be represented as:

X =Ax2 sin COt ...(9.195)

 = A.p2 sin co t ...(9.196)Substituting the values of x and from Eqs. (9. 195) and (9.196) in Eqs. (9. 193) and (9.194), we get

2(Kxx - m CO) Ax2 + Kx.pA.p2 = 0 ...(9.197)

2K.px' Ax2 + (K.p.p - Mm CO) A.p2 = My ...(9.198)

By solving Eqs. (9.197) and (9.198), we get

-- -Kx.pAd - 2 2 My

(Ku-mco )(K.p.p-Mmco )-KcpxKx.p2

(Kxx -m CO)A.p2= 2 2 My

(K,X,X- mCO ) (K.p.p- Mm ro )- KcpxKx.pIf both Fx sin cot and My sin cot are acting simultaneously, then

Ax = Axt + Ax2 ...(9.201)A.p =A.pl + A.p2 ...(9.202)

Sometimes to screen the vibrations, some .air gap is left between the pit and the foundation block(Fig. 9.34). Figure 9.35 shows the comparison between the response of embedded foundation with air gapand without air gap. From this it can be concluded that if air gap is provided around the foundation theamplitude of vibration increases whereas the natural frequency decreases when compared with corre-sponding foundation with no air gap around it. The response of embedded foundation with air gap canbe obtained by analysis given in sec. 9.4 by using CuD'CtD,C.pDand C",Din place of CII'Ct' CeI>and CIjIrespectively.

...(9.193)

...(9.194)I.i

1I

f.i

;.

4:

If', ...(9.199)

~" I

I~

and ...(9.200)

H:

Machinq

'" ..,\ ..//..

hFoundation(axbxh)

li Ai r gap

t- a ~Fig. 9.34 : Embedded block foundation with air gap

III

ïï

åô®

ô

ô

×ô

òº

×æ×ù

¬×

º

ïïæ

¬

¶ò

'1111 da tio 11s

.(9.193)

.(9.194)

..(9.195)

..(9.196)), we get..(9.197)

..(9.198)

...(9.199)

...(9.200)

...(9.201)

...(9.202)lion blockith air gap.dation thevith corre-ir gap canCQ>and C'v

(-'i,..'

~!Ill,

,..

Foundations of ReciprocatillgMacltines

I

~

t,I

f

I

,~

1I

iI

j

,I

1

I

u

w

~

Ii

c Cl0 ...-'- Cl

L..0 ...-~ Ln> "'Cl- CCl Clv tf)

.....L.. >-01 ...-> :::

(/)

00

0r.f)

SUOJ:>!W c apn~!ldwV'.. -- --

~

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

0L.()

L.()--.j'

0--.j'

L.()M

L.()

0

L.()

00

399

-QC)'"

..,.,...........>:;C-c.'".....'"0::'i'0c'"c.'"OJ)...'"::'i"'-fc,~';....Q'>'",;:t:...>

~~:=u"uC...:::!0'...

.::,/,

'CBc.a«!'.or,~0-:..~

Ii:

I

~

r"1:1

~~!

IIll'

11

1,\

"

\I

11

I'~

,

1111

III

Ill!

I!!

I c-L.() L.() ClN N C7). c- .0 Cl 0 l-II C1 " .-m CD Cl

L........ ...... 00« 0' Z

0M tf)

c-V

...>-v

L.() c('4

::Jer01l-LL

0N

ij

rJ~,.i I

.

I

'I..,

"'I

400 Soil DYllamics & Mac/lille Foulldatiolls

.,

9.8 SOIL MASS PARTICIPATING IN VIBRATIONS

In both the methods of analysis and design of foundations of reciprocating machines described in sec. 9.4and 9.5, the effect of soil mass participating in vibrations has not been considered.

J!~I

I:

,;1

.

'f

11i 11

Fig. 9.36 : Stress distribution in soil mass

Pauw (1953) developed equations for the apparent soil mass by equating the kinetic energy of theaffected zone to theCkineticenergy of a mass assumed to be concentrated at the base of the foundation. Hegave the following expression for apparent soil mass ms for translatory modes of vibration:

b3m =Lcs ga. m

where a. = Factor which defines the slope oftnincated pyramid (Fig. 9.36). It is generally taken'unity.Cm = Functionwhich dependson sand r

a.he~=-. b

.

t If

P,...(9.203)

j ar="b

he = Equivalent surcharge defined by the ratio of foundation pressure to unit weight of soil.For non-cohesive soils, Cmis obtained from Fig. 9.37 (a). No graphical data is suggested by Pauw for

cohesive soils.The expression for mass moment of inertia of soil in rotational vibrations is given by

'Yb5C.M =---'-ms 12g a.

whereC. canbe obtainedfromFig. 9.37 b to e. In these figures, C~, C~ and c~ denote the factors of massI I r I

moments of inertia about X, Y and Z axes respectively. These factors can be obtained from Figs. 9.37bto d for cohesionless soils and from Fig. 9.37e for cohesive soils.

..(9.204)

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

'Ill

IJI

-

Foundations of Reciprocating Machinesìðï

Ý±¸¬·±²´¬ soil0.2

"' - 'tb3óóÝgJ- m 0.5M )f'b5mz = CX5 12g-" i

1

1.0- - - - - -- -- - -.\"'&.a

-11Cl) 2.0

- - -- -- ---

S~O

JCm

r

0Cb)

r=12JLr0:6 . 0~8

0.25 Y;rh C.--IMm ys - 12'3 aC.. 0.5

5 z- rJL Cj,Mmzs- 129cl

~'1.a1.0

V) 2.0

5.0

0Cc)

0.5CjY--;:-3

r::1--'--0.5 cf

rJ+r

1.5

M.mx$-= ¥b5 x12 gel Cj

Yb5 Y1290£ Ý¶Mmys=

Cj

(t)

Fig. 9.37: Apparent mass factors for horizontal contact surfac£..>-' :.,' '

I';. ' ;

L

0.2

0.5

I 1 0.a

'""6.11 2.0Cl)

5.0

0

(Ca)

0.2

0.5

I.a 1.011 2.0Cl)

5.0

2

CJI.a,"...

5


Balkrishna (1961) has developed the following expression for the apparent soil mass in verticalvibrations:

( )

3/2

m = ~ . 0.4775 Q 1t P.. s 3 4 P

where Q = Sum of the static and dynamic load

Barkan (1962) has suggested that the apparent soil mass may be taken as 23% of the mass of machineplus foundation.

Hsieh (1962) gave the expressions for getting apparent soil mass as given in Table 9.9.

...(9.205)

Table 9.9 : Effective Mass and Mass Moment of Inertia for Soil below a Vibrating Footing(Hsieh, 1962)

Mode .of vibrationEffective mass or mass moment of inertia of soil

ïïãð 11= 0.25 11= 0.5

Vertical translation 30.5 pro

30.2 pro

3\.0 pro

30.2 pro

, 32.0 -p ro

31.0 proHorizontal translation

Rocking 50.4 pro

Torsion (about vertical axis) 0.3 P r5 0.3 P r5 0.3 pr50 0 0

The apparent soil mass/mass moment of inertia is added to the mass m/mass moment of inertia Mmor J0 to get the natural frequency and amplitude of vibration.

.9.9 DESIGN PROCEDURE FOR A BLOCK.FOUNDATIONThe design of a block foundation provided for a reciprocating machine may be carried out in followingsteps:

9.9.1. Machine Data. The following information shall be obtained from the-manufactures of tl:1ema-chine for guidance in designing: -

(a) A detailed loading diagram comprising the plan, elevation and section showing details of con-nectionsand point of applicationof all-loads on foundation; -

(b) Distance between axis of the main shaft of the machine and the top face of foundation;(c) Capacity or rated.output of machine; -(d) Operating speed of machine; and(e) Exciting forces of the machine and short circuit moment of motor, if any.

9.9.2. Soil Data. The following information about the subsurface soil should be ~own :(a) Soil profile and data (including soil properties generally for depth equal to twice the width of ;;

the proposed foundation or up to hard stratum). ~(b) Soil investigation to ascertain allowable soil pressures and to determine the dynamicproperties 4. 'it

of the sOIL.. ,>1

(c) The relative position of the w~ter tablebelo~'-g~o~nd ~t differe~t times of the year. 'JThe minimum distance to any important foundation in the vicinity of the machine foundation should,

alsobe accertained. .. }- I

-, ~,~-

Foundations of Reciprocating .Machines 403'

9.9.3. Trial size of the Foundation. '.

Area of Block-The size of the foundation block (in plan) should be larger than the bed plate of themachine it supports, with a minimum all-round clearance of 150 mm.Depth- In all cases, the depth of foundation should be such as to rest the foundation on good bearingstrata and to ensure stability against rotations in vertical plane.

Centre of Gravity"::The combined center of gravity of the machine and the block shall be as much belowthe top of foundation as.possible, but in no case it shall be above the top of foundation.

Eccentricity-The eccentricity shall not exceed 5 percent of the least width in any horizontal section.Sharp corners shall be avoided, whenever possible, praticularly in the openings.

9.9.4. Selecting Soil Constants. The values of dynamic elastic constants (Cu' Ccfj1Ct' C'V'G, E and J.l)are obtained from relevant tests and corresponding strain levels are noted. ,These values are reduced to10m2 contact area and 10 kN/m2 confining pressure. A plot is then prepared between dynamic elasticconstants and strain level. The value of dynamic elastic constants are picked up corresponding to thestrain level expected in the actual foundation. These values of dynamic elastic constants are then cor-rected for the actual area of the foundation (if < 10m2), and confining pressure. The details of this hasalreadybeen discussedin illustrativeexample4.2 ,.-' .

9.9.5. Centering the Foundation area in Contact with Soil. Determine the combined center of gravity(Table 9.4) for the machine and the foundationin X, Y and Z planes and check to see that the eccentricityalong X or Y axis is not over 5 percent. This. is the upper limit for this type of analysis. If eccentricityexceeds 5 percent, the additional rocking due to vertical eccentric loading must be considered in theanalysis (Barkan, 1962) "

The static pressure should be checked; it should be less than 80 percent of the allowable soil pressureunder static conditions. This condition is met in most practical foundations. .

Table 9.4 : Determination of CG of the System

Elementof

system

DimensionsCoordinate e.g. Static moments ofof the element mass of elements

a.t azWeight ofelement

Mass ofelementay Xj . Yj Zj mjXj mjYj mj

2

3

9.9.6. Design Values of Exciting Loads and Moments. The fmal values of force and resulting momentsare now obtained with respect to the combined center of gravity of the system. The relative magnitudesof the unbalanced forces and moments will decide the nature of vibrations in the block foundation.

9.9.7. Determination of Moments of Inertia and mass Moment~' of Jnertia. The moments of inertiaand mass moments of in~rtia may be obtained using the formulae given in Tables 9.5 and 9.6.

.

404 Soil Dynamics & Machine Foundllliolls

Table 9.5 : Moments of Inertia

Shape of the area FigureFormula for

Ix Iz~

Rectangle

,

ffi" I Y T

--f::.G. bX I X Jl

y1---0 ~

ab312

ba312

ab(a2 +b2)12

y

Circle X

T....N

X 11'0

J..

.!!:...ct64

1td4

64

1td4

32

)

Table 9.6 : Mass Moments of Inertia'1;1

Shape of Elements FigureMm:

,MmT

Formula for

Mmy

z 1'J'

Rectangular block

""-.

m 2-- 12 (b + h2)

m12 (a2 + h2)

m12 (i + b2) ~

"t

!j".""1

!!!-

(3d2 + h2

J!!!-

(3d2 + h2

)12 4 12 4

,.~f'

z'"

TX hCircularblock 1/ 1- 1y,u

t-d-j

Foundations of Reciprocating Machines- 405

whereM - M + L2-: mo- m m

Mmo = Mass moment of inertiaof machineand foundationabout the axis of rotation passingthrough base.

L = Distance of combined centre of gravity above base.

Mmr=-Mmo .

9.9.8. Determination of Natural Frequencies and AmplitudesLinear weightless spring approach

(i) Vertical Vibration

(0 =W.Anz ~ m .-,.

(ii) Torsional Vibration

FzAz = 2 2

m (OOnz-00 )

. -~cw J,(On",- MmzMz

A = 2 2'" Mmz,,(OOnw-00 )

(iii) Combined Rocking and SlidingSliding and rocking are coupled modes of vibration.

follows:

and

The natural frequencies are determined as

'lihere,

co ~ ~c, Anx m

- ~C.I-WL<.oncp - Mmo

2 1

[(

2 2 ) ( 2 2 )2 2 2

]and OOnl,2 = 2; OOnx+OOn+:t OOnx+OOn, -4r OOnxOOn,

The amplitudes of vibration can be computed with the following equations:2 2

(Ct AI.;+C.I- WL- Mm00 ) FJ:+(Ct AL) MA - v Y

x - ~(oo2)2

(CtAL)Fx+(CtA-moo )MyA -~ - ~ (002)

Ax = Linear 4orizontal amplitu_de of the combined center of gravityAcp = Rotational amplitude in radians around the combined center of gravity. -

2 ( 2 - 2)(, 2 ,. 2) '-L\«o)=mMm (J)nl-(J) (J)n2-~

.

ill..Ij

406 Soil Dynamics & Machine Fou1Jdatio"s

The amplitude of the block should be determined at the bearing level of the foundation asa

~~~.,,~~:E-C';:,;z::..;~~;_.-,,-~..,_. ... ..'-~." '.' Av = Az+ '2 A.Ah = Ax + h A.

Ah = Horizontal amplitude at bearing levelh = Height of the bearing above the combined center of gravity of the system

Av = Maximumvertical amplitudeElastic half-space approachEquivalent radius, mass ratio, spring constants and damping factors are listed in Table 9.7

Table 9.7 : Values of Equivalent Radius, Mass Ratio, Spring Constants and Damping FactorModeofvibration

where

(1)

Vertica!

Sliding

Rocking

Torsional

Natural frequencies and amplitude of vibrations

(I) Vertical Vibration

OOnz= ~;

(it) Torsfonal VibrationA, ~ K'[{l-(:JrZ+~,:Jr

00 =~'Vn'V Mrn,!,

.'

,,''~,

)~~1

L

and

~ ~ K.[{1_(ro:Jr~(2~.ro:Jr

Equivalent Mass (or inertia) Damping Spring Constantradius ratio factor

(2) (3) (4) (5)

r = f! (l-Il) m 0.425 4GroBz=

--z = Bz

k=-4 3oz 1t pro z 1-1l

f! (7 - 81l) m '0.2875 32(1-J.l)Gror - -Bx= 32(I -Il) pr; x= -a- k = 7-8J.lox 1t x.f

, (ba')'14

3 (I - J.l)Mmo 0.15 38G roB = 8 5

x = (1+ B).JB;k =o 31t $ pro 3 (1- J.l) .!

( rr = ba(a2+b2) Mm- 0.5 16 3 ,tB=---t 'V= 1+2B k'V=3"Gro°'V 6 1t Y pro \jI

--..

='oundations of Recipr-Ocating Machines 407

(iil) Coupled sliding and rocking vibration

0) = fKxnx V-;;

~~O),,~ = Mmo

Damped natural frequencies are obtained as the roots of the following equation:

O

[

4 2

{

(ID~,+ ID;") 4 V.. IDnx"";

}

oo~ 00;,

]

2

00na - 00 nd - +r 0 00 0 To, .0 r 00

4[

~X 00nx 00nd ( 2 2 ) ~'00nd cOnip ( 2 2

)]

2

+ 00n, - 00 lid + 00nx - 00nd = 0r r

Undamped natural frequencies can be obtained by using following Equations:

2 - ~[(

2 + 2 )+ ( 2 + 2 )2 - 2 2

]OOnl,2 - OOnx OOn, - OOn, OOnx

..

4 r oon, oonx2r 0 0

~ ,- .-

Damped amplitudes for motion occasioned by the applied moment, can be obtained as below:

[2

]1/2

A = My (oo~) +(2~x 00n.t)2x Mm" ~(002)

[(2 2)2 -',0, 2

]

112

A =My OOnx-OO +(2~xoonx)..$ M 2111 /),,(00 )

0 '

where /)"(0)2) is given by Eq.

[{

4 2

(

(OO~,+OO~)

~ (ol) = 00 -00 r 0

,

, 0 0... 2

]

1/2

00 00 2 2 ~, 00n, 00, 2 2

+ +x ":-(IDn+-ID)+ r (IDn,-ID)}Damped amplitudes for motion occasioned by an applied force Fx acting at the center of gravity of the

foundation may be obtained as below: 0' ,

[ ]

1/2

- Fx (-Mm 002+k, +L2Kx)2 +40)2(~~~K~Mmo+L2~x.JK7n)2 0

.. ~ -m-~.m .A (0)2) '- . - -.'

( 2 2)1/2 .

F L coIIX coIIX+ 4~xcoA - 0 x '.- ,", "

,.'~:t'~ffl" ( /)".<co2). ..

.. '. 2

4 ;.t ~~ 00nx 00nljl

}

- 00~x 00 ~~

}r. . r

and

6...

408 Soil Dynamies & Machine Foundations

9.9.9. Check for Adequate Foundation. The natural frequencies computed in.step 8 should be awayfrom the resonance zone i.e.

Cl) co- < 0.5 or - < 1.5Cl)n con

The amplitudes computed in step 8 should be less than t~e limiting amplitudes of the machine whichare usually specially by the manufacturer of the machine.

ILL USTRA TIVE EXAMPLE IExample 9.1

A reciprocating machine is symmetrically mounted on a block of size 4.0 m x 3.0 m x 3.5 m high. Thesoil at the site is sandy in nature having cl>= 350and Ysat= kN/m3.The watertable lies at a depth of 3.0m below the ground surface. The block is embedded in the ground by 2.0 m depth. The machine vibratingat a speed of 250 rpm generates

Maximum vertical unbalanced force = 2.5 kN

Torque about Z-axis = 4.0 kN-m

Maximum horizontal unbalanced force = 2:0 kN at a height of 0.2 m above the top of the block.The machine weight is small in comparison to the weight of foundation. Limiting amplitude of the

machine is 150 microns. A block resonance test was conducted at the site to evaluate the dynamic elasticconstants. The data obtained from the test is the same as given in Example 4.2 (Chapter 4).

Determine the natural frequencies and amplitudes by (a) Weightless spring m!=thod,and (b) Elastichalf-space approach.Solution:

1. Machine data (Fig. 9.38)

2.5 Sin G.>t2.0 Sin CiJt

Machincz

4.0 Sin c..>tBlock

:.~... e.g.. .,. Tb =3.0m

1J-

"h = 3.5m

12.om

11-- a =".Om(a)Section

TL = 1.75m

~ a = 4.0 m l-".,.-- - -,..~-, -- -

I"" 0" f. Y"",Fig. 9.3S':, Details oUoundatlon . .

(b) Plan -}..,!'

,,(.~l

.~1

,

,I,---

FoundatitJnsuf R:ecip,octlting Machines 409

Operating speed of machine =500 rpmWeight of machine is small and can be neglected.

F = 2.5 sin Cl)t kNzM = 4.0 sin rot kN mz, ,

Fx = 2.Dsin rot kN acting at a height of 0.2 m above the top of block21t

ro = 250 rpm = 250 x 60 = 26.2 rad/s2. Dynamic elastic constantsRefer example 4.2. The soil data and the size of the actual block are same as in that example. The

procedure of determining the values of dynamic elastic constants for analysing the block foundation isillustrated there. Therefore, following values of dynamic elastic constants may be adopted.

Cu = 3.62 x 104kN/m3G = 1.10 x 104 kN/m2

E = 2.98 x 104kN/m2

" '.

f.1= 0.35

C 3.62C = ---!!.= - x 104 = 1 81 x 104 kN/m3

t 2 2 .C41 = 3.46 Ct = 3.46 x 1.81 x 104 = 6.26 x 104 kN/m3

C", = 0.7S'Cu = 0.75 x 3.62 x 104= 2.71 x 104kN/m33. Foundationdata. "

Let the block is casted in M20 concrete. The unit weight material is taken as 24.0 kN/mJ~Weight of block = 24.0 x 4.0 m x 3.0 x 3.5

= 1008 kN

1008 2m = 9.81 = 102.8x 10 kg

Area of foundationbase = 4.0 x 3.0 = 12.0 m2

In further analysis and design of foundation, the depth of embedment is neglected.

4. Linear weig,htles'sspring app'roach(a) Vertical Vibration

HA-u -(Onz = -;;;-

Fz

(AZ>max = m(oo~ -002), . ., ' ,. \

3.62 x 104 x 12.0 = 65 rad/s102.8

.r ~,-' 2.5 -

.-: 102~~J~?2 -.2~.22J' =:=6.8 x. 10-6 m = 6.8 microns, ',."".:iI...'-':: ".:'r~.~,l'S""" r.

, .'

.

.410 SoU Dynamics & Machinl! Foundatio;

(b) Torsional vibrationab 2 2 .4 x 32 2 4

J = - (a + b ) = - = (4 + 3 ) = 25 mz 12 . 12 .

M =!!!.. (i + b2) = 102..8x 103 (42 + 32) = 214 x 103 Kg-m2mz 12 . . 12

- iC" J, - 2.71 x 104 x 25 -56 2 d/(0 - -- - .ras

nw M 214 .mz(c) Coupled rocking and sliding

(0 = re;;:=nx V-;;-

Therefore

1.81x 104 x 12 = 46 rad/s102.8

- ~C.I-WL(On <p - Mmo

I = ba3 = 3 x 43 = 16 m412 12.

W = 1008 kNL = 1.75 m

Mmo= Mm+ m L2

M ~!!!.. (i + h2) = 102.8 x 102 (32 + 3.52) = 182 x 103 kg-m2m 12 12Mmo = 182 x 103+ 102.8 x 102 x 1.752= 496.8 x 103kg-m2

=V

6.26 x 104 x 16-1008 x 1.75 = 44 8 d/OOn d. . ra s't' 496.8

M 182r = --!!L = - = 0.366

M 496.8mo

2 - 1[(

2 2 )+" ( 2 2 )2 2 2

](J) 11\,2 - - (J)n.\"+ (J)IIcp - (J)ncp+ (J)nx - 4 r (J) ncp(J) n.\"

2r 4~

= x 1 [(462+44.82):t (462+44.82)2 -4 x 0.366 x 462 X44.8~]2 0.366 . . I

= ~ [4123:t 3283]0.732 .

OOnl = 33.8 rad/s and oon2 = 100.6 rad/s2 ( 2 2)( 2 2)!J.(00 ) = m Mm (J)nl - (J) Cl)n2 - Cl)

= 102.8 X 182 (33.82 - 26.22) (100.62 - 26.22)= 102.8 x 182 X 456 x.9434 = 8.05 X 1010 kg2 m2

My = 2 (0.2 + 1.75) =~.~,'kN'ni":,

,cotI

~.

...".''.

, "..; I

------------

lundations of Reciprocating Machines 411

(C411- WL+C-rAL2 -Mmro2) Fx+(C-r AL) MyA =x - ~ (ro2)

(6.26 x 104 x 16-1008 x 1.75+1.81 x 104 x 12x1.752_182 x 26.22) x 2.08.05 x 1010

=

41.81x 10 x 12 x 1.75x (3.9)

+ 8.05 x 1010

= 3080158+ 1482390 = 56.6 x 10-6 m8.05 x 1010

=

2(C-rAL) Fx + (C-rA-mro ) My

~ (ro2)4 4 2

(1.81x10 x12x1.75)x2.0+(1.81x10 x12-102.8x26.2 )3.98.05 x 1010

= 760200+ 571876 = i6.5 x 10-6 rad. 8.05 x 1010

A =$

Hence, a -6 4.0 -6A = A + - A = 6 8 x 10 + - x 165 x 10v z 2 ~. 2'0'

= 39.8 x 10-6 m = 39.8 microns ,1/

Ah = Ax + h' A$ = 56.6 x 10-6 + .1.75 x 16:5 x 10-6= 85.47 x 10-6 m = 85.47 microns

5. Elastic half-space approach(a) Vertical vibration

r = fA = ru. = 1.95m0 V; V-;

Average effective unit weight of soil20+ 10 3= = 15 kN/m2

-' 1008

B = 1-~.m3 = 1-0.35 x W = 1.47Z 4 pro 4

(2-.

)x 1.953

9.81

K = 4Gro = 4 x 1.10x 104 x 1.95 = 13.2 x 104kN/mZ 1-~ 1-0.35

0.425 0.425~ =:JB:' ":=

'

..J.195= 0.304

z. - B °0- .z .,t '

t.

412 Soil Dynamics & Machine Foulldations

.~Kz = ~13.2x 104 = 35.8 rad/sCO - -

8nz - m 101.

Fz

A, ~ k{{l-( :Jf +(:~,~:Jr2.5

~ 13.2 x 1o~[{I - G::~rr + (2 x 0.304 - ~~:~ rr= 29.4 x 10-6 m

(b) Torsional vibration

[ ]

1/4

[ ]

\14ab(a2+b2) 4x3(42+32)r = = = 1.9973 m

0 67t 67t3 2M = 214 x 10 Kg-mmz

B = Mmz = 21~ x 9.81 = 4.40\jI p,; 15x 1.99735

16 3 16 . 4 3 4~ = "3 G,o ="3 x 1.1x 10 x 1.9973 = 46.75 x 10 kNmlrad

0.5 0.5X,"= = 1 2 4 4 = 0.051

T 1+ 2 BIjf + x .

- ~Ijf -~

46.75 x 104 - 46 7 d/co - -- - . ra snljf M 214mz

9.9.10. Coupled sliding and rocking vibration

l ~x3Sliding r = - = - = 1.95m

0 7t 7t4

32(I-Jl)Gr 32(1-0.35)x1.lxl0 x1.95K = 0 =x 7-8Jl 7-8 x0.35

= 10.62 x 104kN/m

7-8Jl m 7-8xO.35 1008B = .- = xx 32(I-Jl) pr; 32(1-0.35) 15x1.953

= 1.83

~ - 0.2875-: 0.2875- ,-~x - 7B: - .J1.83 - 9.212 ,

'1

It';..'~"

r;'oundations of Reciprocating Ma.clJines 413

. - fk:: = /10.62 x 104 = 32.1 rad/smnx - V-;; V 102.8I

Rocking ( 3

)

1/4

(

1/4- ab 4.0x33

ro - \. 3~ ~ ]x) ~ 1.84 ID

K = 8 G r; = 8 x 1.1X 104 x 1.843 - 4cjI 3(1-/l) 3(1-0.35) -28.1x10kNm/rad

BcjI= 3(1-/l) x Mmo = 3(1-0.35)x496.8x9.81-8 Pl~ 8 15x1.845 -3.750.15 0.15

~= (l+B<\»jB; = (1+3.75).J3.75 = 0.016

'" = J K, = 28.1 x 104 = 23.7 radJ,n<l> M 496.8- .1110

9.9.11. Coupled Vibration. Undamped natural frequencies in couples rocking and sliding are given by

2 1

[(2 2 ) ( 2 2 )2 2 2

]m111,2 = h mIIX + mn<l>:t mnx + mn<l> - 4 r mIIX mmp

= 1[(32.12+23.72):t (32.12+23.72)2 -4XO.366X32.12X23.72

]2 x 0.366 "

= ~ [1592:t 1299]0.732(On1 = 20 rad/s anda mn2= 62.8 rad/s .

~ <oh ~ [1co4 -co' ((co;.:co~) 4~x~.:"' co,.}+co;',co;.r2

]

1/2

+4{

~. mnx m (002 -m2 )+ ~<\>mn<\>00(m2 -002)}.\ r II<\> r n.\

~ «(02) =[{

26.24 -26.22(

1592 - 4 x 0.212 x 0.016x 32.1 x 23,7)

+ 32.12 X 23.72}

2

0.366 0.366 0.366

. 2

]

1/2+4

{0.212 x 32.1 x 26.2 (23.72- 26.22)+ 0.016 x 23.7 x 26.2 (32.12 - 26.22) }0.366 0.366:",,'; '. ,;- . .- . .

[ ]In

= (471200-2966467+ 1581341)2+4 (-60772+9337)2 = 919697

"

414l

SoU Dynamics & Machine Foundations ~;;~'

A ~ ~ {(-Mm'" +K. +Kx L,)'12+4"'(~, ~K. Mm.+L' ~x~)'rl2xl mMm ~({i)

= 2.0[(-182 x 26.22 + 28.1 x 104"+10.62 x 104 x 1.752)

2

102.8 x 182 x 919697

2

]

1I2

+ 4 x 26.22 (0.016~28.1 x 104 x 496.8+1.752 x 0.212~1O.62 x 104 x 102.8)

= 2.0 (2.3165x 1011+0.1496 X 1011)1/2102.8 X182 919697

= 58 x 10-6 ID 0

( 2 2)1/2

A = ~'( L COnx COnx+4 ~x CO<1>1 M 2. 111 ~(CO )

r(

") 1/2

)

= 2.0X1.75

l

32.1 321-+4XO.212X26.22)182 919697.

= 26.9 x 10-6 fad '

.\::

[ ]

1/2

My co~'(+(2~xconx)2A - -x2 - Mm ~ (co2)

[2 2

]

1/2

= 3.9 (32.1) +(2xO.212x32.1)182 919697

= 0.81 x 10-6 ID

;'".;,:i,,'" J:;;t11';::;t;~

[

2

]

1/2

- My (co~x-co2) +(2~xconxco)2A(\I2- - 2

Mill ~ (co )

[

2

]

1/2

(32.12_26.22) +(2xO.212x32.1x26.2)23.9182 919697

= 8.3 x 10-6 fad

t~~;J

'(\"

"jft~11i~

,;

,"

'~",.1iij;';i

Ax = A.d + Ax2= 58 x 10-6+ 0:81 x 10-6= 58.81 x 10-6in

A4I= A4II+ A4I2= 26.9 x 10-6+ 8.3 X 10-6= 35.2 X 10-6fad

.. - III . ~..~"

{}undations .of Reciprocating- Machines 415 .

a -6 4.0 -6A = A + - An.= 29.4x 10 + - x 35.2 x 10

4~ Z 2 y 20-6= 99.8 x 1 m

Ah = Ax + h' A, = 58.81 x 10-6+ 1.75 x 35.2 x 10-6= 120.41 x 10-6 m

It may be noted the there are significant differences in the magnitudes of natural frequencies andmplitudes computed by the two approaches. It may be due to the reason that the value of shear modulusJ is computed from the block resonance test data using the relation between Cu' E and G. Actually inlastic half-space theory, it is desirable that the value of G is obtained from wave propagation test. Author'sxperience indicates that the value of G obtained from wave propagation test is much higher than com-'uted from block resonance test data. Use of appropriate value of G may bring the results of the twopproaches closer.

Hence

:xample 9.2

)etermine the natural freque~cies and amplitudes of motion of the foundation (Example 9.1) taking intotccount the embedment effect and apparent soil mass. ~ ,

,olution :

(i) Assume that the values of dynamic elastic constants at a depth of 2.0 m are 20% higher than thevalues at the surface of ground. Therefore

CuD = 1.2 x .3.62 x 104= 4.35 x 104kN/m3C D = 1.2 x 1.81 x 104= 2.18 x 104 kN/m3t ,

CD = 1.2 x 6.26 x 104=7.si x 104 kN/m3

CIjID = 1.2 x 2.71 x 104 = 3.25 x 104 kN/m3The average values of dynamic elastic constants will be

C, = 4.0 x 104 kN/m3./aV- x 4 3Ctav - 2.0 10 kN/m

Ccpav= 6.89 x 104 kN/m3

Cljlav= 3.0 x 104kN/m3D = 2.0 m

(ii) Vertical vibration

Kze = CuD x A + 2 Ctav (bD + aD)= .4.35 x 104 x 12 + 2 x 2.0 x 104 (3 x 2 + 4 x 2)

4= .108.2 x 10 kN/m ..

24 x 3.5he = . 15 = 5.6 m

ah 1.0x5.6 a 4S = -:- = 3.0 = 1.867; r= b = 3" = 1.334

416 , Soil Dynamics & Machine Foundations

, "C '

From Fig. 9.36 a for S = 1.867 and r = 1.334;-!!!.= 1.45r

Therefore

Cm = 1.45 x 1.334 = 1.93

, 'Y b3 15 x 33 3m = - C = x 1.93 = 79.7 x 10 KgS ag m 1.0x9.81

or

4"- -

108.2 x 10- = 75.8 rad/sffinze= VI02.8+79.7

FeA == 2 2)ze (m+ms) (O)nze-O)

= 2.5(102.8+79.7)(75.82_26:22) =2.70 x 10-6m

(iil) Coupled vibration

Kxe = C'tD . A + 2,Cuav bD + 2 Ctav aD= 2.18 x 104 x 12 + 2 x 4 x 104x 3 x 2 + 2 x 2 x 104x 4 x 2

= 106.16 x 104kN/m

From Fig. 9.36 b, for S = 1.867 and r = 1.334, ~ = 0.43r5

b5 15x3M = 'Y CX = 0.43 x 1.334m.'Cs 12ga i 12x9.81xl2= 17.72 kNms

Mmos= Mmxs+ ms x L2 = 17.72 + 79.7 x 1.752= 261.80 kN ms2

Cl) = ~xe = /106.16x 104 = 76.2 rad/snxe m+m ~102.8+ 79.7s "

C .b Db 2K = Cd.D1- WL + ,av (16 D3 - 12 hD2) + 2 C$av x 10+ Ctav x_a~ ~ 24 2

3 3 '

I=ba =3x4 =16m412 11

W = 24 x 3 'x 41<.3.5 = 1008 kN3 4 2} ,

I = aD = ~ ='.10.67 m40 3 3

'-,-~;~

Jj;f

:J-

w,~'!!' i."

Foundations of RecfprocGtingMachilies.. 417

6 89 . 04 .K$e = 7.51 x 104 x 16 - 1008 x 1.75 + . x I. x 3 (16 x 23 - 12 x 3.5 x 22)24 .

2+ 2 x 6.89 x 10.67 + 2.0 x 104 x 2 x 3 x 4

2 -

= 120.16 x 104-0.1764 x 104-34.45 x 104+ 147 x 104+48 x 104

= 280.53 x 104 kN/m

(J)mj>e =K$e

Mmo + Mmos

4280.53 x 10182 + 261.80

= 79.5 rad/s

Kxx = CtD x A + 2 Cuav bD + 2 Ctav aD= 2.18 x 104 x 12 + 2 x 4 x 104 x 3 x 2 + 2 x 2 x 104 x 4 x 2

= 90.16 x -104kN/m2

Kx$ = CqJavb (D - 2DL) - CtD AL= 6.89 x 104 x 3 (22 - 2 x 2 x 1.75) - 2.18 x 104 x 12 x 1.75=-231.81 x 104kN

3 2 3 2I. = D a + a Db = ~ + 4 x 2 x 3 = 28.67 m4y 12 4 12 4

K$$ = C<j>DI + CtD AL 2 - WL + 2 C1jfOl'Iy + Ctav b ~ a2 + ~ C<j>av[L3 + (D - L)3]

= 7.51 x 104 x 16 + 2.18 x 104 x 12 x 1.752 - 1008 x 1.75 + 2 x 3 x 104 x 28.67

+ 2 x 104 x 3 x 22x 42 + ~ x 6.89 x 104 [1.753 + (2.0 - 1.75)3]

= (120.16 + 80.11 - 0.17 + 172.02 + 96 +24.69) x 104= 512.81 x 104kNm

Õ±òô ¢ óÅÝôÜ ßÔõî C"", bD (L- ~ )+2 C"" (L- ~) aD]

~ -[2.18 x 104 x 12 x 1.75+2 x 4 x 104 x 3" 2 (J.75-~ )+2 x 2 x 104 (1.75-~)4 x 2]

= -[45.78+51.84+34.56] 104 = -132.18 x 104 kN."

~

'(.


{KcMI-(Mm +Mmxs) Cil} FxAXl=

{2

}{2

}Kxx -(m+ms) 0) KcMI-(Mm +Mmxs) 0) -Kx4l' K~

= {512.81X 104_(182+17.72) 26.22} 2.0

[{90.16 x 104- (102.8 + 79.7) 26.22} {512.81 ~<104,-(182 + 17.72) 26.22 }

]

---

-(-231.81 X 104) (-132.18 X 104)

4 4998.18 x 10 998.18x10= 77.64 x 104x 498.3 x 104- 30639.32 x 108 = 8048.69 x 108

-6= 12.40 x 10 m

-K .F .A = ~ x

cpl {Kxx -(m+ms).0)2} {Kcpcp-(Mm + Mmxs) 0)2}-Kxcp' K~

= 132.18 X 104 X 2.08048.69 X 108

= 3.2 x 10-6 fad

- Kx . M.vA = ~ .x2

{2

} {2

}Kxx - (m + ms) 0) Kcpcp - (M m + Mmxs ) 0) - Kxcp . K~x

231.81x104x3.9 -6= 8048.69 x 108 = 11.23 x 10 m

{Kxx -(m+ms) 0)2} MyAcp2 =

{2

} {2

}

,

K,n,-(m+ms)O) Kcpcp-(Mm+Mxs)O) -Kxcp K~x

4 2- [90.16 x 10 -(102.8+79.7) x 26.2 ]x3.9- 8048.69 x 108= 3.7 x 10-6 fad

Ax = Axl + Ax2 = (12.40 + 11.23) x 10-6 = 23.63 x 10-6 m

Acp = A4I1+ Acp2= (3.2 + 3.7) x 10-6 = 6.9 x 10-6fadDisplacement of the top of the block

= A + (h - L) Ax qI= 23.63 x 10-6 + (3.5 - 1.75) x 6.9 x 10-6= 35.70 x 10-6 m

.1'

c:

,: ;J:,'""',I

'~;it~.r,,.,"[,

': ~1,'1fJ

'j

",

J;:

Foundations of Reciprocating Machines 419

REFERENCE

Anandakrishnan, M. and Krishnaswamy, N. R. (1973a), "Response of embedded footings to vertical vibrations", J.Soil Mech. Found. Div. , Am. Soc. Civ. Engg; 99, pp. 863-883., '

Arnold R.N. , Bycroft, G.N. and Warburton, G. B. (1955), "Forced vibrations of a body on an infinite elastic solid",Trans. ASME, 77, pp. 391-401.

Balkrishna, R. H. A. (1961). "The design of machine foundations related to the bulb of pressure", Proc. Int. Conf.Soil Mech. Found.' Eng., 5th, Paris, Vo\. 1, pp. 563-568.

Baranov, V.A. (1967), "On the calculation of excited vibrations of an embedded foundation (in Russian)". Vopr.Dyn. Prochn. . 14 pp. 195-209.

darkan. D.D. (1962). "Dynamics of base and foundations", McGraw-Hill, New York.Bereduge, Y.O. , and Novak, M. (1972). "Coupled horizontal and rocking vibration of embedded footings", Can.

Geotech. J. , 9(4), pp. 477-497., ,

Beredugo, Y.O. (197I), "Vibrations of embedded symmetricfootings", Ph. D. Thesis, University of Westem Ontario,London, Canada. ' ,

Bowles. J. E. 91982), "Foundation analysis and design" , McGraw-Hill, New York.Bycroft, G.N. 91956), "Forced vibrations ofa rigid circular plate on a semi-infinite elastic space and on an elastic

stratum", Phi10s.Trans. R. Soc. London, Ser. A, 248, pp. 327-268.

Chae. Y. S. (1971), "Dynamic behaviour of embedded foundation-soil system", Highw. Res. Rec., 323, pp. 49-59.,.Fry, Z. B. 91973), "Development and evaluation of soil bearing capacity, Foundation of Structures", Waterways Exp.

Sta., Tech. Rep. No. 3-632. /, '

Gupta. B. N. (1972), "Effect of foundation embedment on the dynamic behaviour of the foundation-soil system",Geotechnique, 22 (I), pp. 129-137. '

Hall, J. R. (1967), "Coupled rocking and sliding oscillations of rigid circular footing", Proc. Int Symp. Wave Propag.Dyn. Prop. Earth Mater, Albuquerque, NM, pp. 139-148.

Hsieh, T. K. 91962), "Foundation vibrations", Proc. Inst. Civ. Eng., 22, pp. 211-226.Lamb.H. (1904), "On the propagation of tremors over the surface of an elastic solid", Philos.Trans. R. Soc. London,

Ser. A 203, pp. 1-72.Lysmer,J. and Richart, F. E. , Jr. (1966), "Dynamic response of footing to vertical loading", J. Soil Mech. Found.

Div., Am. Soc. Civ. Eng., 92 (SM-I), pp. 65-91.Novak, M. (1970), "Prediction of footing vibrations", J. Soil Mech. Found. Div. , Am. Soc. Civ. Eng., 96 (SM-3),

pp. 836-861. ', ,

Novak, M., (1985), "Experimeflts with shallow and deep foundations". Proc. Symp. Vib. Prob\. Geotech. Eng., Am.Soc. Civ. Eng., Annu. Conv., pp. 1-26.

Novak, M.. and Beredugo, Y.C. (1971), "Effect of embedment on footing vibration", Proc. Can. Conf. EarthquakeEng. ,1st, Vancouver, pp. 1I 1-125. ,

Pauw,A. (1953), "A dynamic analogy for foundation soils system", ASTM Spec. Tech. Pub!., STP, pp. 3-34.Reissner,E.(1936),"StationareAxialymmeterischeduTcheineSchuttelndeMaSseErregteSchwingungeneineHomogenen

,Elastichen Halbraumes", Ing. Arch. 7(6), pp. 38I-39?,"" ',',Rc'ssner, E. (1937), "Freie underzwungene Torsionschwing-ungen des Elastiche~ Halbraumes", Ing. -Arch, 8 (4).

i~,<,..,.!:;.; ...,C,": "",'" ,.(",.L.I>,..',~' ,pp. 229-245. '.':, , , I ". ', \.< ! "','.{' ,..FfAoJ'E:<:l!...\ S } ' . . , ":~

, .

.420 Soil Dynalllics &. Machine Foundations

Reissner~.E.andSagoci, H. E.(1944),~'Forcedtorsional oscillations.ofan.elastichalfspace", 1. Appl. Phys., 15, pp.652- 662.

Richart, F. E. Jr. (1962), "Foundation vibrations", Trans. Am. Soc. Civ. Eng., 127, Part, I, pp. 863-898.Richart, F. E.)r. ,and'Whitman, R. V. ('1967), "Comparison of footing vibrations t~sts with theory", J. Soil Mech.

Found. Div., Am. Soc. Civ. Eng., 93 (SM ~6), pp. 143-168.Stokoe,K.H.II (1972). "Dynamicrespo~seofemb'~ddedfoundations",Ph. D. thesis presented,Universityof Michigan,

Ann Arbor, Michigan.- . .Stokoe, K. H., 11,and Richart, F. E. Jr. (1974), "Dynamic response of embedded machine foundation",1. Geotech.

Eng. Div., Am. Soc. Civ. Eng. , 100 (Gt-4), pp. 427-447 .Sung, T. Y. (1953a), "Vibrations in semi-infinite solids due to periodic sutface loading", Ph. D. Thesis, Harvard

University, Cambridge, Massachusetts.

Vijayvergiya, R. C. (1981). "Response of embedded foundations", Ph. D. Thesis, Univeristy of Roorkee, Roorkee,India.

Whitman; R. C. and Richart, F. E., Jr. (1967) , "Design procedures for dynamically loaded foundations", J. SoilMech. Found. Div., Am. Soc. Civ. Eng., 93 (SM - 6), pp. 169-193. .

PRACTICE PROBLEMS

9.1 (0) List the basic differences in analysing a reciprocating machine foundation by the two clpproaches namely (i) Linear weightless spring-mass system, and (ii) Elastic half-space theot)

(b) Derive the expressions of natural frequency and amplitude of a block foundation subjectelto vertical vibration. ,',)

9.2 Starting from fundamentals, derive the expressions of natural frequencies and amplitud~~l'block foundation subjected to a horizontal force Fx sin rotand a moment My sin rot at the COlTbined e.g. of machine and foundation. , ,~i

9.3 A concrete block shown in Fig. 9.39 is to be used as a foundation for a reciproc~ting e~fioperating at 500 rpm and mounted symmetrically with respect to foundation. The weight O'

.

~.'. .'

.

.

U

.

.

..

engine is 10kN. It is ,likely that the operation of machine exerts the following: . t~Unbalanced vertical force = 1.8 sin rot kN .;~

Unbalanced torsional moment = 6 sin rotkN I "/~1;1't.

The values of the dynamic elastic constants for the design of the foundation may be adop'!given below: . )ni.1.

CII = 4.5 x 104 kN/m3G = 4.8 x 104 kN/m3

J.1.= 0.343

Ysoil~ 17kN/m3

Yconc. = 24 kN/mDeterminethe namralfrequenciesand amplitUdes~f thebiockby (i)Lin~ar elasticispj. -' "~..;:.approach and (il) B~stic half space theory. . -' - . - .-, .:'~

o"ca; c: ,

Fila1IliItfdils "Ol~~i:;ptoCtlii,.g~ac{,,",a 421

, , ,_. - .,", "

, ,

"T'-,2,(0 1ft.

,,1

, " . ,'" -

1..'-Q.5mT-.,'

-0 .0-0 ,'-,

"

~ 8.0 m

Elevation

TIt..Om

1Plan

Fig. 9..~9: Details of foundation

9.4 Design a suitable foundation for a horizontal compressor driven by an electric motor. The fol-lowing data are available: ' , / ,

Weight of compressor,= 160 kNWeightof motor= 60 k;N,Speed of compressor = 250 rpmHorizontal unbalanced' force = 75 kN acting at a height of 1.0 m above top of pedestaL

, , -

The following tests were performed at the site of determine the values of dynamic elastic con-stants :

(a) A verticalvibrationtestwas conductedon anM-I50 concreteblock 1.5x 0.75 x 0.7 m high,using different eccentricities. The data obtained is given in Table 9.8.

Table 9.8

SI.No. Angle, Of setting of. .'.eccentric masses

-,

1,,:Hz

35.0,32.0.31.029.528.0

.,:27.0

Amplitude at resonancemm

-

I23456

15304560

I

120

1~9

0.063750.1500.2100.300.5250.620

-'

...

,.','

422. S,iI DyntUllics &. Machine FollndtUWns. ,., ,

(b) A cyclic-plate load test was.done on plate 300 mm ~.300 mm. The elastic settlement corre-sponding to a load intensity of 250 kN/m2 was 6.00 mm. .

(c) A wave -"propagation test gave an average value of travel time of compression waves as0.02 s, corresponding to a distance between geophones of 6 m.

The water table at the site is 2.0 m below the proposed depth of the foundation 3.0 m. The soilat the site is sandy in nature.

DD

. ~~ft

1::1

'.

~

, " , ' , .FOUNDA TIONSOF IMPACT TYPE MACHINES

10.1 GENERAL

Impact type machines produce transient dynamic loads of short duration. Hammers are most typical ofimpact machines. A hammer-fouridation-soil system consists of a frame, a falling weight known as'tup', the anvil and the foundation block. Figure 8.6 shows a typical foundation for a hammer with itsframe mounted on the anvil. In Fig. 10.1, a foundation for a hammer with its' frame mounted on thefoundation is shown.

~ ,

Fig. 10.1: Typical arrangement ofa hammer foundation with A-frame mounted on foundation< '

,', . . '",'~,'f\" },

.. '";'i~,:;~$~~~4i.~,,'i.t!t:;..!,:~j'.;: ~, ',Lo-'

", ';,'; ~',)' i':(, ~ 0'"

: re .\

"',7; ,c" ,'" ~,..,'

424 Soil Dy,"""ks& 'Mllc"inif' F()Il1I~

',' The foundation of a hammer generally consists of a reinforced concrete block. 'Q1efollowing ar-rangements are used depending upon the size of the hammer:

(i) For small hammers, the anvil may be directly mounted on, the foundation block (Fi~. IO.2a).This system can be modeled as single ,degree of freedom system as sh°W!lin Fig. lO.2b.

',' ,~,',;" ,;:'::~',.:.\,.!','~ /': ~~",.:"::':\:':',:,..",'~"

..- . .. " '" ",'c ;

'9T::":':"', ,- J'~

~'FOUNDATION,BLOCK.RESTINGON SOIL

Tz1ANVIL ANDFOUNDATION'BLOCK""

, , 1

'}

'~

(a) (b)

F;g. 10.2 : Anvil resting directly on the foundation block

(ii) In medium capacity hammer, a vibration isolation layer is placed between the anvil and tlfoundation block (Fig. lO.3a). Usually the isolation Jayer is an elastic pad consisting of rubbefelt, cork, 01 timber adequately protected against water and oil. In case of high capacity hanmers, special elements such as coil springs and dampers may be used in place of elastic pa((Fig. lO.3b). The systems shown in Fig. lO.3a and lO.3b can be modeled as two degrees I

freedom system as shown in Fig. lO.3c.

f.

, ~I

, WI

. .-

.....t

~ -,:1~f&4

,"~

FOUNDA1'1 ON BLOCKRESTING, ON SOIL

(a) (b)

Fig. 10.3 : Anvil resting on elastic pad/spring absorbers (...Contd.)

~,

"oundations of Impact Type Machines

dJ~P

DAMPING1Nl ANvrLABSORBER I

122

SPRINGk2ELASTICPAD

TztFOUNDATIONBLOCK

SOILSPRING

kl< ~ <

DAMPING IN SOIL

(c)

425

Fig. 10.3 : Anvil resting on elastic pad/spring abs'!)rbers

(Ui) For reducing the transmission of vibrations to the adjoining machines or structures, the founda-tion block may also be supported on elastic pads (Fig. lOAa) or on spring absorbers (Fig. lO.4b).In such a case, the foundation is placed in a reinfor<;edconcrete trough. The space between thefoundation and side of trough is filled up with some soft 'materials or an air gap is left. Thesystems shown in Fig. lO.4a and lO.4b can be modeled as three degrees freedom system asshown in Fig. lO.4c. The stiffness of trough (Fig. lO.4a) is very high compared to that of the padbelow the foundation block, the trough may be assumed to be rigidly supported on the soil(Novak, 1983), and therefore a two degree freedom model (Fig. lO.3c) may give sufficiently,accurate results for all practical purposes.

FOUNDATIONBLOCK

FOUNDATIONBLOCK

ELASTIc.PAD BELOWFOUNDATIONBLOCK

TROUGH " .~' i "

~,. (a).' . ,< ,.(b)"

,( ,'r!' ,!1 ' . ,..Fig. 10.4 : Foundation block on elastic pad/spring absorbers (...Contd.)

.......

TROUGH

SPRINGABSORBERBELOWFOUNDATIONBLOCK-::

. 426 Soil Dynamics & M.achine Foun.d,,:tions

~TUP

. TZ)

DAMPINGIN PAD

IANVIL

II

SPRINGkJ OF PAD-. BELOWANVIL.

FOUNDATIONBLOCK

TZ2

SPRINGk2OF PADBELOWTHEFOUNDATOOBLOCK TZ1

SOtLSPRINGk1

DAMPING IN SOIL

(c)

Fig. 10.4 : Foundation block on elastic pad/spring absorbers

In hammer foundations, tup, anvil and foundation are geometrically so aligned that their centres fallon one vertical axis. This will ensure that the loads act on the anvil and foundation without any eccentric-ity.

10.2 DYNAMIC ANALYSIS

10.2.1. Two Degree Freedom System. In general the anvil, pad, foundation, and soil constitute a tWodegree system as shqwn in Fig. 10.5. This model is based on the following assumptions:

(i) The anvil, foundation block, frame, and tup are rigid bodies.(ii) The pad and the soil can be simulated by equivalent weightless, elastic springs.

(iii) The damping of the elastic pad and soil is neglected.

(iv) The time of impact is short compared to the period of natural vibrations of the system.(v) Embedment effects are neglected.' bO.

The notations used in the model have the following meaning: ,~ :r

ml = Mass of foundation and frame if the latter is mounted on the foundation, as in Fig. 'lO~tml = Mass of anvil (with frame if the latter is mounted on the anvil as in Fig. 8.6. ., ":~1$fK) = C'u . AI = Equivalent spring constant of soil under. consideration

.....-. ,;'.

r<'oundations of Impact Type Machines

dJ~P

DAMPING1Nl ANvrLABSORBER I

Tz2

SPRING k2ELASTICPAD

PIFOUNDATIONBLOCK

SOILSPRING

kl'~ .

DAMPING IN SOIL

(c)

425

Fig. 10.3 : Anvil resting on elastic pad/spring abs'!)rbers

(iii) For reducing the transmission of vibratIons to the adjoining machines or structures, the founda-tion block may also be supported on elastic pads (Fig. lO.4a) or on spring absorbers (Fig. 10.4b).In such a case, the foundation is placed in a reinforc;edconcrete trough. The space between thefoundation and side of trough is filled up with some soft materials or an air gap is left. Thesystems shown in Fig. lO.4a and lOAb can be modeled as three degrees freedom system asshown in Fig. 10Ac. The stiffness of trough (Fig. lO.4a) is very high compared to that of the padbelow the foundation block, the trough may be assumed to be rigidly supported on the soil(Novak, 1983), and therefore a two degree freedom model (Fig. 1O.3c)may give sufficiently.accurate results for all practical purposes.

FOUNDA.TlONBLOCK

"" ..,..(a)r', ,. ,"" "- . ,,(b)I.. . , ,. ., .' ., . ., . ,. .

Fig. 10.4: Foundation block on elastic pad/spring absorbers (...Contd.)

........

I'

TROUG H

SPRINGABSORBERBELOWFOUNDATIONBLOCK,:~

FOUNDATION. .,.

BLOCK '..',.

ELASTIc. .:.,-PAD BELOW .'FOUNDATION :BLOCK '8.

ill -~

undations of Impact Type Machines 427

Cu' = A Cu = Modified coefficient of elastic uniform compression, to t,,;keinto account impactcondition which is different from periodic loading

t.. = Multiplying factor that governs the relationship between Cu and Cu' usually 1-2, forimpact depending upon the soil type

Cu = Coefficient of elastic uniform compression -

K2 = (E / b) x Az = Equivalentspringconstantof the pad under the anvilE = Young's modulus of the pad materialb = Thickness of the pad

Al = Area of foundation in contact with soilA2 = Area of the padz\ = Displacement of foundation from the equilibrium positionz2 = Displacement of anvil from the equilibrium position

. TuP--cb

SPRING K2ANVIL

m1

FOUNQATION,," BLOCK

. :~,SPRING~1

I

Fig. to.5 : Two-mass-spring analogy for hammer foundation

IO.l.I.I. Natura/frequencies. The equations of motion in free vibration are

mlz\+Klzl+Kz(Zt-zz) =0

~ 2Z+ Kz (zz --zl) = 0The solution of the above equation can be written as

Let, zl = A sin oont ...(10.3)and,. zz=Bsinco"t. ...(10.4)

where A and.B are arbitrary constants. Substituting the values ofzl and zz fro~ Eqs. (10.3) and (10.4) inEqs. (10.1) and (10.2) respectively, we get

z.!! = .Kz+ Kt -1nl conA KZ

...(10.1)

...(10.2)

...(IO.S)

428 Soil Dynamics & Machine Foundations...

and, B K2- = 2A K2-"'2 con

Equating Eqs. (10.5) and (10.6), on simplification we get

4

{

K2 (ml +"'2) KI

}2 Kl K2 - 0OOn- +-- COn+ -

ml "'2 ml ml "'2

...(10.6)

ffi:-(

l+~)(

KZ+ Kl) 00:+K2x KI (

1+"'2)

=0ml "'2 ml+"'2 "'2 ml +"'2 ml

Let, ffina= Circularnaturalfrequencyof thefoundationof the anvilon the pad

.". "'na ~ ~~ ...(IO.S)(J)nl = Limiting natural frequency of the foundation and anvil on soil

~ IOOnl = ...(10.9)

ml+"'2. .

"'2Ilm = -;;-I

Substituting the values of OOna'oonland Ilmin Eq. (10.7), we get.4 2 2 2 2 2

OOn-(1 +Ilm) (OOna+ffinl) COn+(1+llm) OOnlOOna= 0 ...(10.10 a)

r

]002 = .!.I , 2 2 + . 2 2 2 2 2

nl,2 2l~I+llm)(COna+OOnl)- {(1+llm)(OOna+COnl)} -4(1+llm)OO,IiOOna...(10.10 b)

,,',I

The two natural frequencies of the hammer foundation may be determined by solving the aboveequations,

10.2.1.2.Amplitude of vibration.The generalsolutionof the Eqs. (10.1)and (10.2) is givenby

z, = AI sinronlt +A2 cosronlt +AJ sinron2t +A4 cosron2t ...(10.11)\

Z2 = BI sin ronl t + B2 cos (0nl t + BJ sin (0n2 t + B4 cos (0n2 t

where, AI' A2' AJ' A4' BI' Bz, B3' B4 are arbitrary constants.If system is vibrating at frequency oonl'then from Eq. (10.6)

...(10.12)

2~ - (OnaA - (02 _(O2=a2(Say)na n2

,,1

...( 10.1..

3...

a.rJi

.

")~.

"

~t;;J;"c

."i.",

IJ',J;~4

Similarly when con = con2

~ - K2 (02- =' naA K2 -"'2 CO2 0)2 - 2 = al (Say)nl na 0)nl

rdatioIJsoflmpa.ct Type,Machi~es , 429

It may beno~e~,that values ofal andaz are known from Eqs. (10.13 a) and (10.13 b) respectiv.ely.r.racting az from a I we get . .

. z .

[1

aI - az = Cl)na 2 . 2Cl)na - Cl)nl

( 2 Z ) ZCl)n I - Cl) nZ Cl)na

al - az = ( 2 Z )( 2' 2 )Cl) na - Cl)nl Cl)na - Cl)n2

.Equation (10.12) can therefore be written as :

.~ = al AI sin Cl)nlt +al Az cos Cl)nlt +az AJ smCl)n2 t +az A4 COSCl)nZt

boundary condition:

(i}At t = 0, ZI = Zz = 0

. 1

]2 Z

Cl)na - Cl)nZ

or ...(10.14)

...(10.1S)

and

It gives

(ii) At t = 0, zl = 0; Zz= Va

Az + A4 = 0a I Az + az A4 = 0

A '= A = 0Z 4

(Velocity of anvil)

...(10.16a)

...(10.l6b)...(10.17)

, ,

Zl = AI Cl)nl cos Cl)nl t - Az ronl sin ronl t + AJ ronZ cos ronZt - A4 ronZ sin ronZ t

or AI (Onl + AJ (OnZ= 0Cl)A = - A niJ I-

. Cl)nZ2'z = al Al Conl cos ronl t - al Az Cl)nl sin Cl)nl t + az AJ Cl)nZco£~ nZt - al A4 Cl)nZsin Cl)nZt

or Va = al AI (Onl + az AJ (OnZ

VaA =I (al -az) Cl)nl

-vA = a

3 (al -a2) Cl)nZ

. Va . Va sinCl)nztZ =' SIn ronl ~

I (a\-az) Cl)nl (al -az) Cl)nZPutting the value of(a,- az) from Eq. (10.14) in Eq. (10.21), we get

Z = (CI)~a-CI)~I)(CI)~a-CI)~z)[

sinCl)nlt - sin Cl)n2t]

VI ( ZZ ) Z Cl) Cl) aCl)nl-CI)n2 Cl)na . nl nZ

=. 1 [

(CI)~a-CI)~2)SinCl)nlt (CI)~a-CI)~I)Sinron2t

]v.~ ( 2 Z ) Cl) CI)' a

. Cl)nl-(J)nZ. nl nZ. .

.,' . ...(10.23)

Field observation (Barkan, 1962) of the amplitudes of the anvil and the foundation showed that theibrations occured at the lower frequency only. Therefore, it may be assumed that the amplitude of mo-ion for sin (Onlt= 0 «Onl > (OnZ)'

or ...(10.18)

or ...(10.19)

and ./(10.20)I

I

Therefore, . ...(10.21)

...(10.22)

Similarly,

430 Soil Dynamics & Machine Follndations

Hence approximate expressions for maximum displacement will be as follows (sin Cl)n2t = I): ".. r:

(OO;a-00;1) (OO;a-00;2)ZI = 2 ( 2 2 ) .Va

OOnaOOnl-OOn2 OOn2

(OO;a-00;1) Vaz2 = ( 2 2 )OOnl -OOn2 OOn2

...(10.24}

...(1 0.25} .

10.2.2. Single Degree Freedom System. Sometimes in the case of light hammers, no pad is used be-tween anvil and foundation. rhe system can then be represented as single degree freedom system(Fig. 10.6). In this case the equation of vertical free vibrations of the foundation will be .

m i + Kz = 0 ...(10.26)where,

z = Vertical displacement of centre of mass of foundation and anvil, measured from equilib-rium position

m = Total vibrating mass,K =C' Au 1

TUP--m

m1

SPRING K,

..'

Fig. 10.6: Single-mass-spring analogy for hammer foundation

The natural frequency of the system will be given by Cl) nz =~The Eq. (10.26) is the equation of free vibrationsof the foundationwithoutdamping.ThegeJi~'

solutionof this equationis :(~~

,.

z = A sin COnt + B co~Cl)nt ...(l~t.,

The constants A and B, as usual, are dete~ined from initial conditions of motion. 'JH"irAt t = 0 z = 0 and z = V'., 'a

; ,i2

. .Foundations of rmpllCt Type Machines "

Using these initial conditions., we getVaA = - and E=-OQ)n .

Va .A = - sm 0> tZ Q)n n

The maximum displacement will be

Va(Az)max = IDn

Therefore,

431

...(10.29)

...(10.30)

10.2.3. Determination of Initial Velocity of Anvil, V if

For a single acting drop hammer, the initial velocity of the tup VTi at the time of impact is given by

VTi = 1'\ ~2 g h ...(10.31)

where, h = Drop of tuP in metersg = Acceleration due to gravity, mls21'\= Efficiency of drop (It lies between 0.45 to 0.80. An average value equal to 0.65 may be

adopted) .

For double - acting hammers, operated by pneumatic or steam pressure, VTi is given by

VTi = 1'\.12 gh (Wl+ pAc) .

WlW 1 = Gross weight of the dropping parts, includ~ngupper half of the die in kN.

P = Pneumatic (i.e. steam or air) pressure in kN/m2

Ac = Net area of-.cylinder in m2.The initial velocity of anvil just after the tup' s. impact can be detennined by using the law the of

conservation of momentum. Since the anvil is stationary: .

WMomentum of tup and anvil before impact = ---1 VTi

. g

where

and

. Wl W2Momentum of tup and anvil after impact = - VTa + Vag g

W2= weight of anvil (plus frame if it mounted on the anvil)Va' = Velocity of anvil after impact

VTa ='Velocity oftup after impact

Wl - W, W2v.T ' - VTa + Vag'g. g

where

Therefore,

----

...(10.32)

...(10.33 a)

~ L

432 SoU Dy,,1IIIIics ~ Machine. FoundatiDns

According to Newton's law, the coefficient of elastic restitution, e, is given by

Relative velocity after impacte = Relative-.velocity before impact

Va - VTae=VTi

The value of e depends upon the material of the bodies involved in'iinpact. Theoretically value' of elies between 0 and 1. In forge hammer, usually the value of e does not exceed 0.5 (Barkan, 1962). Sincea larger e gives larger amplitudes of motion, the value of e equal to 0.5 is adopted in designing hammerfoundation. On solving Eqs. (10,34) and (10.35) we get. '

or ...(10.35)

l+eV - 'YT '- W I

a 1+-1.WI

...(10.36)

10.2.4. Stress in the Pad. Maximum compressive stress in the elastic pad below the anvil depends uponthe relative displacements of anvil and the foundation block. The worst case of compression in the paddevelopes when the anvil moves downward, and at same instant of time, the foundation block movesupward. The maximum compressive stress in the pad is thus expressed by

zl + z2 .(Jp = K2 A' , (zl' z2 In absolute values)2

10.2.S. Stresses in the Soil. Stresses transmitted to the soil q through the combined static dynamic loadsare expressed by

...(10.3?)

Wt +W2 +zl Ktq =

Al

10.3 DESIGN PROCEDURE FOR A HAMMER FOUNDATION

...(10.38)

The design of a hammer foundation may be carried out in following.steps:10.3.1. Machine Data. The following information about the hammer is required for the design:

(a) Type and weight of striking part of hammer;(b) Dimensions of base area of anvil and its weight;

(c) Maximum stroke or fall of hammer, mean effective pressure on piston and effective area ofpiston; .

(d) Arrangement and size of anchor bolts; and(e) Permissible amplitudes of the anvil motion and the foundation on block. If this information is

not available, the amplitudes of motion given in Table 8.2 may be considered as limiting values.10.3.2.SoilData. The followinginformationabout the sub-surfacesoil shouldbe known: :

(a) Soil profile: For drop hammers of up to 10 kN Capacity, soil investigations should generally bedone to a depth of 6 m. For heavier impact machines, it is preferable to investigate soil con~~.tions to a depth of 12 m or to a hard stratum. If piles are used, the investigation should ~conducted to a suitable depth, ~

I 1 ~~-

'oundations of Impact Type Machines 433

(b) Soil investigation to ascertain allowable soil pressure and to determine the dynamic propertiesof the soil specifically the value of Cu'

(c) 'The relative position of the water table below ground at different time of the year.

0.3.3. Trial Size of the Foundation.

Nhere,

(a) Weight and area: The weight of the foundation for a hammer and the size of its area in contactwith the soil should be selected in such a way that(i) the static pressure on the soil does not exceed the reduced allowable soil pressure, and

(ii) the foundation does not bounce on the soiL

These conditions may be written as

PsI ~ a qa , andAz < Ap

PSI = Static pressure intensitya = Reduction'factor (= 0.8)

qa = Allowable soil pressure -

Az =Amplitudeof motionAp = Permissible value of amplitude

For Eq. (10.38a)

...(10.38 a)

...(10.38 b)

W /'.

A ~ a qa. '-Considering an average value of A as I mm = (10-3 m), and assuming the system as single degree

f p .reedom system, the Eq. (10.38b) can be written as .

..(10.39)

where,

(I +e) W VI '0 I 3

~C' W A < IO-u .g

Wo = Weight oftup, kNW = Weight of foundation, anvil and frame, kNAl = Base area of foundation in contact with soil, m2VIi = Initial velocity of tup, mls

g = Acceleration due to gravity, mls2C'u = Coefficient of elastic uniform compression for hammer foundation, kN/m3

Substituting the value ofW from Eq. (10.39) into Eq. (10.40), we get

(1+e)WoVTi 32

A. ~ ~ ' x 10 mCuaqa g

...(IOAO)

...(10.41)

."

':

',<,' '""", '"

434 Soil Dynamics &

;i~<,~~df.'1t~."'r;l'

Machine Foundatio~~-" 't~,;Q.'i~

where,

Substituting the value of AI fromEq. (10.39) into (10.40), Weget

(1+e) Wo VIi ~ -3W = ~ x 10 kN

"C~ g

W=W\+W2W\ = Weight of foundationW2 = Weight of anvil

The Eq. (10.42)canbe writtenas

WI (l+e) VIi ~ -3 W2Vi" = ~ x 10 -Vi" ...(10.43)

0 "Cu g 0

From, Eq. (10.42), total weight of the anvil and foundation can be obtained. Knowing the weight otanvil, using Eq. (10.43) weight of the foundation can be worked out. On the basis of experience (Barkan,1962), the weight of the anvil is kept generally 20 times the weight of the tup. Further it is recommended"that the weight of the foundation block should be at least 3 to 5 times that of the anvil.

(b) Depth: The depth of the foundation block shall be so designed that the block is safe both inpunching shear and bending. For the calculations the inertia forces developed shall also beincluded. However, the following minimum thickness of foundation block below the anvil sha,Ifbe provided:

"...(10.42) 'I,:

Let r.;~

l/

~

Mass of Tup kN

Up to 1010to 20

20 to 40

40 to 60

Thickness (Depth) of foundation Block, Min (m)

Over 60

1.0

1.25

1.75

2.25

'2.50..'

10.3.4. Selecting the dynamic elastic Constant C 'u. The procedure of obtaining Cu has already bei4:discussed in Chapter 4 for relevant strain level. The value of C'Umay be taken as A C'u where A vari~between 1 and 2. The value of Cu may also be obtained from the following relation: if'.

4 G r .~.C = -1 0 ...(10.44,)

u -~where, G = Shear modulus

r = Equivalentradius = J AI0 nJ.1 = Poisson' s ratio

~ .... <, I!' ' ,ID ..

, 1&O>na = v-;;;;

10.3.5, Natural frequencies, Compute

(}undations. 0/ Impact Type Machines 435

and " ~oon/ V~

in w~ich,E

K2 = b . A2 '

, ,

E = Yourig's'modulus of pad materialb = ThicIaiess of the pad

A2 = Area of the pad" ,0 ..' ,",", '

K =C' A=A.C A "-.,,,1 u u

Natural frequencies of the combined system are given by:

2 1

[ ( 2 2 ) ['"

( 2 2

)]2 "

( 2 2

)](()nl,2 = '2 (l+~m) OOna+OOnl:t (l+~m) OOna+OOnl -4(1+~m) rona,wnl

10.3.6. Velocity of Dropping Parts to Anvil. Compute the velocity VTiof the tup before impact

~ ~2g (WI+ pA)h .-VTi 11 W

in which, W 1 = Gross weight of dropping parts

p = Steam or air pressure

Ac = Area of the pistonh = Drop of the tup

11 = Efficiency of drop, usually 0.65

Compute the velocity of the anvil Vaafter impact byl+e

V =. , VT 'a W I

1+ ---1-Wl

in which, e = Coefficient of elastic restitution. The value of e may be adopted as 0.6.

.. ", .

10.3.7.Motion Amplitudes of the Foundation and Anvil. Compute the maximum foundation and anvilamplitudes with following equations

(ro~a - (O~I) ( (O~a- 00~2)

Z 1 = 2 ( 2 2 )VI)

OOna (Onl - OOn2 OOn2

(co2,. CO

2)0., "- '

I'

- na, n VZ2 ~ (CO~I: ro~2)cDn2I),;,

where oon2 is the smaller natural frequency.

... ""0"""""" . .. ''~~-~'-_...

436 SoU Dynamics et Machine Foundations

10.3.8. Dynamic Stress in Pad (jp' Compute dynamic stress in ~e pad by

(J = Kz(zJ+zz)P Az

Computed values of natural frequencies should satisfy the criteria for the frequency of operation ofthe hammer. Also, motion amplitudes should be smaller than permissible values, and the stress in the"elastic pad should be smaller than the permissible stress of the pad material.

!ILLUSTRATIVE EXAMPLES' 5

Example 10.1A 15 kN forging hammer is proposed to install in an industrial Complex. The hammer has the followingspecifications:

Weight of tup without die = 11 kNMaximum tup stroke = 800 mmWeight of the upper half of the die = 40 kNArea of piston = 0.12 mzSupply steam pressure = 600.kN/mzWeight of anvil block = 300 kNTotal weight of hammer = 400 kNBearingarea of anvil = 1.8m x 1.8mPermissible vibration amplitude for anvil = 1.5 mmPermissible amplitude for foundation = 1.0 mmIt is proposed to use a pine wood pad of thickness 0.5 m below the anvil. The modulus ,of elastici'

of pad"material is 5 x 105kN/mz, and allowable compressive stress in pad is 3500 kN/mz. 'I'~

A vertical resonance test was conducted on a 1.5 ID x 0.75 m x 0.70 m high concrete block at fproposed depth of fo~ndation. The data obtained are given below: iJ

S. No. 8 (Deg) /"Z (Nz) Amplitude at resonance (mic1'f!'Y

36 41 13 t .,72 40 24

108 34 32'~I

4. 144 31 40 A

.' 'ine soil at the site is sandy in nalnre and water table lies at a depth of 3,0 m below ground ~Allowable soil pressure is 225 kN/mz. Design a suitable foundation for the hammer. Ji;

I.2.3.

.~

Ff!undation';-of Impact'type Machines 437

Solution: , ,

(i) Trial dimensions of foundation.. Let the weight of the block is kept about 5 times the weight ofanvil.' The details of the suggeste~ foundation are shown in Fig. 10.7.

Weight of foundation = 24 x 7 x 5 x 2 = 1680kN.(ii) Evaluation of Cu

Area of test block = 1.5 x 0.75 = 1.125 m2

Weight of test block = (1.125 x 0.75) x 24 = 18.9 kN.Weight of oscillator and motor = 1.0 kN (Assumed),

,-',':'

',.'j ,,-<,' .,,' "; .

L.

438 SoU Dynamics & Machine Foundations'

Total weight of test block, oscillator and motor= 18.9 + 1.0.= 19.9 kN

2 2 ' ,

41t2 fn~ m - 41t x 19.9 x fnz =71.1 fn~ kN / m3CII = A - 1.125 x 9.81

. Amplitudeat resonanceStram level = 'Width of block

Putting the given values of fnz and amplitude at resonance, CIIvalues and corresponding strain levelsare computed and listed in cols. 2 and 3 of Table 10..1respectively.

Table 10.1 : Values of Cu and strain levels-'

S. No. CuI x 104 kN/m3 Strain Level x 10-4 Cu2 x 104 kN/m3

1.

2.

3.

4.

11.95

11.38

8.22

6.83

0.173

0.320

0.427

0.533'

2.96

2.82

2.04

1.69

Correction for confining pressure and area:

The mean effective confining pressure croI at a depthorone-half of the widthof block is givenby

0'01 = cr (1+2K)v- 0

3--where, O'v = crvl + crv2

crvl = Effective overburden pressure at the depth under considerationcc

crv2 = Increase in vertical pressure due to the weight of block "

Assuming that the top 2.0.m soil has a moist unit weight of 18 kN/m3, and the next 1.0.m soil i..eupto water table is saturated, then

- 0.70.crvI = 18 x 2.0.+ 20.2- =43 kN/m2

;-- 4q

[

2mn~m2+n2+1 .m2+n2+2 . -} 2mn~m2+n2+1

]v2 - -4 2 2 2 2 2 2 + srn 2 2 2 21t m +n +1+m n m +n +1 m +n +1+m n

L 1.5- -2 2m = - = - = 2.14z 0..70.

2B 0..75- -

n = .£. = -L = 1.0.7'z ,0..70.""',"2 '

q ='24 x 0..70.= 16.8 teN/ni2, '

"

,'-".

., ,.!

. 'Cl

::..

,,', . . 0' . , ",:, '..~ ,.!,'.., '"'",'"

-,~."~

Ilndations of Impact Type M~~hines 439

Substituting the above values of rn, nand q in the expression of av2, we get

al2 = 13.44 kN/m2

Gv = 43 + 13.44 = 56.44 kN/m2

Assuming <1>= 35°, Ko = 1 - sin 35° = 0.426

-[

1+2XO.426]

2aol = 56.44 3 = 34.84kN/m

For the actual foundation

av! = 18 x 2.0 + 20 x 1.0 + (20 - 10) x 1.5 = 71 kN/m2For the actual foundation

7.02

m = 5.0 = 1.42

5.0- 2 _ 10n - 5.0 - .

2

q = 24 x 2.0 = 48 kN/m3

Substnuting the above values of rn, nand q in the expres~ion of crv2, we get

av2 = 37.2 kN/m2

cry = 71.0 + 37.2 = 108.2 kN/m2

G02 = 108.2[1+2 x30.4266] = 66.8 kN/m2

Cu2 =[

0' 02

]

0.5

[~

]

0.5 .

CuI 0'01 A2

=[

66.80]

°.5

[1.5x 0.75

]°.5 -

34.84 7.0 x 5.0 - 0.248

The values of Cu of the actual foundation for different strain levels are listed in cot 4 of Table 10.1.

1.0 4Stain level in actual foundation = 5.0 x 1000 = 2 x 10-

The strain level in actual foundation is higher than the strain level observed in the tests. Seeing thevariation of Cu with respect to strain level, the value of Cu equal to 1.3 x 104 kN/m3 may be adopted indesign. .

\

440. --, ',.'

"'t~,

Soil DYllamics & Machine Foulldaii~~

(iii) Computations of KI' K2' ml and m2C' = A' C

11 11

= 2.0 x 1.3 x 104 = 2.6 x 104kN/m3(A= 2.0, Assumed)K = C' ,A = 2.6 x 104x 7 x 5 = 91 x 104kN/mI !I I

E 5x 105K, = -b .A2 = 1.8x 1.8= 324 x 104kN/m~ 0.5

Weight of the foundation block

= (7 x 5 x 2 x 1.8 x 1.8 x 0.8) x 24 = 1618 kN1618m = - = 165 x 103 KgI 9.81

Weight of the anvil and frame = 400 kN400 '

m2 = 9.81 =40.8 x 10"'Kg(iv) Natural frequencies of soil-foundation system

(J) = ~ l -~

91 x 1041/1 -

nIl +"'2 '165+40.8 = .J4421 = 66.4 rad/s

r&=(J)/la = v-;;;;

4324 x 10 = J79411 = 281 rad/s40.8

= ~ = 40.8 = 0 24711 1/l\ 165 .

Natural frequencies are given by :

,(J.)~1.2

{

2 .

}

1/2

= ~ [(1 + Ilm) ((J)~a+ (J)~l)J=t~ [(1 + Ilm) (O)~a+ (J)~l)] -4 (1 + 11/11)(J)~a.(J)~l

1 In=: "2[(l + 0.247) (79411 +4421)] ::t ~{[(1+ 0.247)(79411 + 4421)]2 -4 (1 + 0.247)79411 x 4421}

(J) /11.2 = 316 rad/s, 66 rad/s(v) Velocity of dropping parts

VTi ~ ~ ,{W, ~.A, )2gh11 = 0.70 (assumed) ; W\ = 15 kN, p = 600 kN/m2, Ac = 0.12 m2, h = 0.8 " J

..;1

VTi = 0.70 ,iC5+ 60105x 0.12) x 2 x9.81 x0.8 = 6.68 mls-\I~

" .~

'h4.:;i~I

rtions of Impact Type Machines

. l+eVa = w' VTi. 1+-1..

W1

- 1+ 0.5 .6 68- 400'1+-15

= 0.362 m/s

i) Amplitudes of vibration

(ro:a-ro:l) (ro:a-ro:2)ZI = ( 2 )( 2 2 )

.Vurona ronl -ron2 ron2

(79411-100167) (79411-4371)= (79411)(100167-4371)(4371) .(0.362)

= - 1.695x 10-4 m

= - 0.1695 mm «1.0 mm, safe)

(ro:a-ro:l) ,VaZ2 =( 2 -ro\ )ro/12ronl n

(79411-100167) x 0.362= (100167-4371)(4371)= - 1.7944 x 10-5 m

= - 0.0179 mm [< 1.5 mm, safe]

..

vii) Dynamic stress in pad

k2 (zl +z2) .(J2 = A ' (zl' z2 In absolute values)2

= 324 X 104 (0.1695+0.0179) 10-31.8x 1.8

= 187.4 kN/m2 (\

441

(e = 0.5, assumed)

[< 350 kN/m2 , safe]

442Soil Dynamics & Machine Fouml'!!i!:l


Barkan, 0.0. (1962), "Dynamics of bases and foundations", McGraw-HiIl, New York.

Novak, M. (1983), "foundations for shock producing machines", Can, Geotech. J. 20 (1), pp. 141-158.

PRACTICE PROBLEMS-

10.1 Discuss with neat sketches the various possible arrangements ofa hammer foundation to minimistthe vibrations.

10.2 Consideringa two-degree-freedommodel, derivethe expressionof amplitudesofanviland foundatioof a hammer.

10.3 A 20 kN forging hammer is proposed to install in an industrial Complex. The hammer has thfollowing specifications:Weight oftup without die = 12 kNMaximum tup stroke = 900 mmWeight of the upper half of the die = 50 kNArea of piston = 0.15 m2Supply steam pressure = 700 kN/mWeight of anvil block = 400 kNTotal weight of anvil and frame = 500 kNBearingareaof anvil = 2.1 m x 2.1 mPermissible vibration amplitude for anvil = 1.0mmPermissible amplitude for foundation = 0.8 mm

It is proposed to use a pine wood pad of thickness 0.5 m below the anvil. The modulus of elasticitypad material is 6 x 10 kN/m2, and allowable compressive stress in pad is 4000 kN/m2.

A vertical resonance test was conducted on a 1.5 m x 0.75 m x 0.70 m high concrete block at :proposed depth of foundation. The data obtained are given below:

The soil at the site is sandy in nature and water table lies at a depth of 2.0 m below ground surf

Allowable soil pressure = 200 kN/m2. Design a suitable foundation.

C]

.~,i..Ji.

S.No. e (Deg) f nz(Hz) Amplitude at resonance (micron

1. 36 40 142. 72 38 263. 108 35 334. 144 29 41

FOUNDATIONS OF ROTARY MACHINES

.1 GENERAL

le unprecedented burden cast by the importance of oil can be relieved only by exploiting indigenouslergy resources efficiently. Major power energy resources, in long term power plan, incorporate an,timal mix of thermal, hydel and nuclear generation. Power intensity is relatively high in our countryle to various reasons including the substantial substitution among the forms of energy in the variouslportant sectors along with the accelerated programme in rural electrification and assured power sup-y for agricultural sector etc. -Theaim of planning to generate power higher than the demand by at least) percent, calls for a coordinated development of the power supply inqustry. As per the present ratio, theermal sector caters 54 percent; hydro sector 43 percent and 3 percent catered by nuclear sector. The 15:ar National Power Plan from 1980 onwards envisages the installation of additional generating capacityt'almost 100,000 MW in the thermal, hydro and nuclear sectors taken together as against the existing:neration capacity of 31,000 MW. Large capacity thermal power station at coal pit heads called Superhermal Power stations (2000 MW capacity or more) because of their size and sophisticated technology,ill account for as much as 50,000 MW to be commissioned during the next five year plan.

The turbogenerator unit is most expensive, vital and important part in a thermal power plant. Theperating speeds of trubogenerators may range from 3000 rpm to 10000 rpm. Auxiliary equipments suchs condensers, heat exchanger, pipe lines, air vents and ducts for electric wiring are essential features ofturbogenerator installation. Frame foundations are commonly used for turbogenerators with four

:asons :

(i) auxiallary equipment can be arranged more conveniently,(ii) the inspection of and access to all parts of the machine become more convenient,

(ui) less liable to cracking due to settlement and temperature changes, and(iv) more economical due to the saving in material and freedom to add more members to stiffen if

needed. .

The frame foundation is the assemblage of columns, longitudinal and transverse beams. The trans-:erse beams may be often eccentric with respect to the column centre lines and generally have varying:ross-section due to several opening in the top deck and haunches at the junction with columns. Thesometric view of a typical frame foundation is shown in Fig. 11.1. "I' .

. In a power - plant, the long term satisfactory performance of the turbogenerators is affected by theirr'oundations,pence there is vital need to adequately design these foundations for all possible combinations)f static and dynamic loads. Interaction with the mechanical engine~r is also required for any adjustment10the layout of machinery and auxiliary fittings.

... ~.;;;t. n!l""cU~

/

444 Soil Dynamics & Machine Foundatiol'

""'>'

Longitudinalbeam

Topdeck

" ~

Column's, ;.,.

'~

-,-,

..

.1,'.' 1

Fig. 11.1 : Typical frame foundation for a turbo-generator'"

11.2 SPECIAL CONSIDERATIONS

For better perfonnance of aT. G. foundation, following points may be kept in view: . < - ,(i) The entire foundation should be separated from the main building in order to isolate the tfaIf~re

of vibrations from the top deck of the foundation to the building floor of the machine'-roo~~).clear gap should be provided all around. . ,(~

(ii) Other footings placed near to the machine foundation should be checked for non-uniform str~~.

impo')cdby adjacent footing. The Pressure-bulbs under the adjacent footings should not inie1fisignificantlywith each other. ' \wa:<jdi .

4ld

~

lundations of Rotary Machines 445

(iii) All the junctions of beams and columns of the foundation should be provided with adequatehaunches in order to increase the general rigidity of the frame foundation.

(iv) The cross-sectional height of the cantilever elements at the embedment point should not be lessthan 60 to 75 percent of its span, being susceptible to excessive local vibrations.

- .(v) The transverse beams should have their axes vertically below the bearings to avoid torsion. For

the same reason the axes of columns and transverse beams should lie in the same vertical plane.

(vi) The upper platform should be as rigid as possible in its plane.

(vii) Permissible pressure on soil may be reduced by' 20 percent to account for the vibration of thefoundation slab. This slab has much smaller amplitudes of vibration than the upper platfom1.- .- -- -

(viii) The lower foundation shib should be sufficiently rigid to resist non-uniform settlement and heavyenough to lower the common centre of gravity of the machine and foundation. It is therefore:made thicker than required by static computations. For 25 MW machine its thickness is 2m andincreases with the power of the machine to a maximum of 4m. Its weight should not be less thanthe weight of the machine plus the weight of the foundation excluding the base slab and condensers.

(ix) Special reinforcement detailing as laid down in the code IS-2974 Pt III should be followed.

(x) Special care in construction is called for to avoid cracking of concrete. The foundation slabshould be completed in one continuous pouring. In this case the joint between the two concretes,preferably at one-third column height, is specially treated to ensure 100 percent bond.

(xi) Piles may be provided to meet the bearing capacity requirement but then the consideration ofsub grade effect is essential.

(xii) As far as possible the foundation should be dimensioned such that the centre of gravity of thefoundation with the machine should be in vertical alignment with that of the base area in contactwith. the soil. . -- --

(xiii) The ground-water table should be as low as possible and deeper by at least one-fourth of thewidth of foundation below the.base plane. This limits the vibrations propagation, ground-waterbeing a good conductor to wave transmission.

(xiv) Soil-profile and characteristics of soil upto at least thrice the width of the turbine foundation or .till hard stratum is reached or upto pile depth, if piles are provided, should be investigated.

11.3 DESIGN CRITERIA

The design of a T. G. foundation is based on the following design criteria:(i) From the point of view of vibration, the natural frequencies of foundation system should prefer-

ably be at a variance of at least 30 percent from the operating speed of the machine as well ascritical speeds of the rotor. Thus resommce is avoided. An uncertainty of 10 to 20 percent may beassumed in the computed natural frequencies. However, it may not be necessary to avoid reso-nance in higher modes, if the resulting resonant amplitude is relatively insignificant.It is preferable to maintain a frequency separation of 50 percent.

(ii) The amplitudes of vibration should be within permissible limits. Values of permissible ampli-tudes are given in Table 8.2.., -. ( ,

. ~r ':}J! 1.1 , ., > (

-=: --,-,. c . -_.._-

446 Soil Dynamics & Machine Found.

11.4 LOADS ON A T. G. FOUNDATION

The loads acting on a turbogenerator are as given below:

11.4.1. Dead Loads (DL). These include the self weight of the foundation and dead 'Yeight of thechine.

11.4.2. Operation Loads (OL). These loads are supplied by the manufacturer of the machine ancelude frictional forces, power torque, thermal elongation forces, vaccum in the condenser, piping foetc.

The load due to vaccum in condenser, ifnot supplied by the manufacturer, can be obtained fronfollowing equation:

where.Pc = A (pa - Pc)

Pc = Condenser vaccum loadA = Cross-sectional area of the connecting tie between the condenser and turbine

Pa = AtmosphericpressurePc = Vac cum pressure

The value of (p a - Pc) may be taken as 100 kN/m2.

...(1

-8-

TR TA

A

t] Shaf-

H.P.Gcznczrator

Tu r bin czs

Fig. 11.2 : Torque due to normal operation oh multistage turbine-generator unit

The magnitude of the torque depends upon the operational speed and power output capacity (turbines. For a T. G. unit having multistage turbine (Fig. 11.2), the torque may be calculated as be

where,

- 105 PA kNTA - N m

- 105(PB- PA) kNTB - N m- 105(Pc - PB) kNTc - N m

105PT = c kNm

g NTA = Torque due to high-pressure (H. P.) turbine in kNm

TB = Torque due to intermediate-pressure (LP.) turbine in kNm

...( 1

...( 1

...( I

...(1

--

Foundations of Rotary Machines 447

Tc = Torque due to low-pressure (L. P.) turbine in kNm

Tg= Torque due to generator in kNmPA' Pa and Pc = Power transferred by couplings A, B and C respectively in KW

N = Operating speed in rpm

11.4.3. Normal Machine Unbalanced Load (NUL). As mentioned in sec. 8.2, rotary machines arebalanced before erection. However, in actual operation some unbalance always exists. The unbalance isspecified as the distance between the axis of the shaft and mass centre of gravity of rotor, and is known aseffective eccentricity. The magnitude of unbalanced forces can be obtained using Eqs. (8.8) and (8.9).

_l-

(0)

(b)

Fig. 11.3: Unbalanced forces due to rotary machines

For the case of rotors shown in Fig. 11.3a, the resultant unbalanced forces due to the two masses atany time cancel out, but there is a resulting moment M given by

2M=mero./ewhere, / = Distance between the mass centre of gravities of rotors

The components of the moment M in vertical and horizontal directions are given by

Mv=meero2/sinrot ...(11.4 a)2

MH =meero /cosrot ...(11.4 b)When masses have an orientation as shown in Fig. 11.4b, the machine operation will give rise to

both an unbalanced force and a moment. The unbalanced force is given by2

F = 2 me e ro

The unbalanced moment can be computedusing Eq. (11.3),For more than two rotors on a common shaft, combined unbalanced forces and moments can be

computed in similar manner.

...(11.3)

...(11.5)

.

448 Soil Dynamics & Machilre Foundat

11.4.4. Temperature loads in the foundation (TLF). The effect of differential thermal expansion a!shrinkage should be considered in the design of frame foundations. In the absence of the exact data.differential temperature of 200may be assumed between the upper and lower slabs. Besides, a different:temperature of 200 may be: assu\TIedbetween the inner and outer faces of the upper slab. The upper s1;.should be treated as a horizontal closed frame and analysed for the induced moments due to differentitemperature.

To account for the shrinkage of the upper slab relative to the-base slab, a temperature fall of 100C150C may be assumed.. . .

11.4.5. Short circuit forces (SCF). Short circuit condition imposes moment on the turbogeneratfoundation. A fault of this type occur when any two of the three generator phase terminals are shorteThe shock, which is in the form of couple known as "shortcircuit moment", tends to break the stator (the foundation, and this imposes vertical loads on the longitudinal beam supporting the generator state

If accurate information is not available from the manufacturer, the short circuit moment (Msc) m.be taken emipiricallyas four times the rated capacity (in MW) of turbogenerator unit.

Major (1980) has suggested the following fomlula for estimating the short circuit moment:

Msc = 10 r Wr kNmW,. = Capacity ofT. G. Unit in MW

r = Radius of the rotor in m

...( 11.

Where,

11.4.6. Loss of blade unbalance (LBL) or bearing failure load (BFL). One of the buckets or bladesthe turbine rotor may break during the operation of turbo-generator unit. It will increase the unbalancforce. This additional unbalanced forced wi1l depend on the weight of the bucket,. the distance ofcentre of gravity from the axis of rotation and operational speed.. .. .

11.4.7. Seismic load (EQL). The horizontal seismic force is considered both in logitudinal and transve:directions separately. It may be computed from the following equation (IS 1893-1984) :

Fs = ah I ~C S W ...(11.7)

where, Fs =Horizontal seismic forceah = Seismiczone coefficient

I = Importance factor

~ = Soil- foundation factor

C = Numerical base shear coefficient

S = Numerical site structure response coefficient

W = Verticalload due to weight of all permanent components.When earthquake forces are considered in design, the permissible stresses in materials and the

lowable soil pressure may be increased as per IS 1893-1984.

11.4.8. Construction loads (CL). Constructionloads occur only when the machineis being erected.such they are not to be consider~das acting simultaneouslywith dynamicloadswhich occur only durthe operation of the machine. The construction loads are generally taken as uniformly distributed l(varying from 10kN/m2 to 30 kN/m2dependingon the size ofT. G. unit. . .

r.°un~a~io.nso! R~tary M~chines 449

The design of a T. G. unit should be checked for the following load combinations:

(a) Operati?n,: ~~nditionDL+OL+N1J:r..:+JL[ ..:

(b) Short circuit condition

DL + OL + NUL + TLF + SCF} .,'..., .

(c) Loss of blade conditionlbearing failure'doridition .

DL + OL + TLF + LBLIBFL

(d) Seismic condition

DL + OL + NUL + TLF + EQL

11.5 METHODS OF ANALYSIS AND DESIGN

In the case of a frame foundation, it is necessary to check the frequencies and amplitudes of vibration andalso to design the members of frame from structural considerations. The methods for carrying out dynamicanalysis may be divided into two categories:

(a) Two-.dimentional analysis

(b) Three-dimensional analysis

The two-dimensional analysis is based on the following assumptions:(i) The difference between the deformations of individual frame coluinns is insignificalll.

(ii) The deformation of the longitudinal and transverse beams is almost identical.

(iii) The torsional resistance of the longitudinal beams is insignificant in relation to the deformationof the transverse beams. . -.. .. .

(iv) The vertical vibrations of the frames can be determined for each frame individually.

(v) The weight transmitted from the longitudinal beam can be considered as a load supported by thecolumn head, even in case where the transverse beam is eccentrically placed with respect to thecentre line of the column.

(vi) Both the columns and beams can be replaced by weightless elements with the masses lumped ata few points by equating the kinetic energies of the actual and the idealised systems.

(vii) The effect of elasticity of subsoil is neglected, it being relatively much flexible.. .

(viii) When considering horizontal displacement the upper slab is regarded as a rigid plate in its ownplane. .

The two dimensional analysis may be carried out by the following methods:1. Resonance method (Rausch, 1959)

2. Amplitude method (Barkan, 1962) .3. Combined method (Major, 1980)

In subsequent sections, salient features of the above methods are given.

I

i "--'--'.. -" .-

ïïòêÎÛÍÑÒßÒÝÛ METHOD

In this method, the frame foundation is idealized as a single-degree freedom system, and consideration isgiven only to natural frequencies of the system in relation to the operating speed of the machine. Theamplitudes of vibration are not computed in this method.

11.6.1. Vertical Frequency. For obtaining vertical frequency, each transverse frame that consists of twocolumns and a beam perpendicular to main shaft of the machine, is considered separately (Fig. IlAa).

Fz sin cut

Wzt Wz

F sin c..>tz

Wl

ID

I.- -2a ~ ~I lo I lr I

ILl I h, ho1

1I

! 1

ZbGekrmns,- .>71 /" - - - 7 J~ -Y I

Column Base slab I<zndsassumed L - - - - - - - - - - -_J

fixed

m= ql j-W,+2'9

(b)

(a)

11.4: (a) Typical transverse frame;(b) Idealised model

The loads acting on this frame are

(I) Dead load of the machine and bearing, W I

(ii) Load transferred to the columns by longitudinal beams, W2

(iii) Uniformly distributed load due to self weight of cross beam, q per unit length

(Iv) Unbalanced vertical force due to machine operation, Fz sin w t

Foundations of Rotary Machines 451

The frame is modelled as mass-spring system as shown in Fig. II.4b. The stiffness of equivalentspring (K) is computed as the combined stiffness of the beam and columns acting together. It is given by

where,

WK =-

:: 851

W = Total load on the frame

...( 11.8)

or W = W\ + 2W2 + q ./...(11.9)

1 = Effective span

°51 = Total vertical deflection at the centre of the beam due to bending action of beam andaxial compression in columns.

where,o51 = °1 + °2 + °3 +°4

°1 = Vertical deflection of beam due to load WI

°2 = Vertical deflection of beam due to the distributed load q

°3 = Vertical deflection of the beam due to shear84 = Axial compression in column

The magnitudes of 81' 82, °3 and 84 can be obtained using following expressions:

Wj ,3 2K + 181 = 96 E I K + 2b

q 14 5K + 2°2 = 384 Elb' K+2

3 1(

ql)83 ="5 E Ab WI+2

= ~(

W + WI+ql)84 EA 2 2e

Ih hK =-.-le 1

Ab = Cross-sectional area of beamAc = Cross-sectional area of columnIb = Moment of inertia of beam about the axis of bendingle = Moment of inertia of columnE = Young's modulus of concreteK = Relative stiffness factor

1 = Effective span of frameh = Effective height of frame

Values of 1and h are obtained as below:'=1 -2ab0

h =h -2aa0

...(11.10)

...(11.11)

...(11.12)

...(11.13)

...(11.14)

where, ...(11.15)

...(11.16)

...(11.17)

452

where,

Soil Dynamics & Machine Foundatiolls

10 = Centre to centre distance between columns (Fig. 11.4 a)ho = Height of the column from the top of the base slab to the centre of the frame beam

(Fig. 11.4 a)

a = One-half of the depth of the beam for a frame without haunches (Fig. 11.4 a) or thedistance as shown in Fig. 11.5 for a frame with haunches

b = One-half of the column width for a frame without haunches (Fig. 11.4 a) or ~he distanceas shown in Fig. 11.5 for a frame with haunches.

Knowing the values of ho' 10and b, excan be obtained from Fig. 11.6.

0.40

0.30

0<

®óóóIIII

0.20

0

-1b~

0.10

0-0 0.04 0.08 0.12

b/ lo

Fig. 11.5 : Values of a and b for a frame with haunches Fi. 11.6 : CLversus b/lo

The natural frequency of a transverse frame in vertical vibrations is given by

(j) = ~Kz . gnz W

Average vertical natural frequency of the T. G. Foundation is taken as:

...(11.18)

COn:1+cun:2+...+con;n(j) n;a = n ...(11.19)

where,

(J.)I/;!'(J.)1/;2'" = Vertical frequencies of individual transverse frames

....

!.~-

-Foundations of Rotary Machines 453

The average value of vertical amplitude of T. G. foundation may be computed as

IF:A:a

(I Ko),'ll-( oo:Jf+( ~:o:)'

...(11.20)

A:a = Average vertical amplitude of T. G. foundation

L F~ = Total vertical imbalance force

2::Kz = Sum of the stiffness of the individual frames

~ = Damping ratioFor under-tuned foundation, i.e. (() < (() ~, (() = (()n~a should be used in Eq. (11.20). ThenlI~a n -

Aza2::F:

(2::KJ (2~)...(11.20 a)

11.6.2. Horizontal Vibrations. In a T. G. frame formulation, the deck slab undergoes horizontal vibra-tion in the direction perpendicular to the main shaft of the machine. The spring stiffness is provided bythe columns due to their bending action, and for any transverse frame it is given by

K = 12E le(

6K+l

)x 113 3K+2...(11.21)

where,

IfKt = Lateral stiffness of an individual transverse frame

L Kt = Sum of the lateral stiffness of all the transverse frames

WT = Total weight of deck slab and machineThen the natural frequency of the T. G. frame foundation is given by

"'"xa ~ p:~~)g ...(11.22)

The average horizontal amplitude of the foundation may be computed as follows:

A = L~t

'" (LK,),/[I-

(~

)

2

]2 +

(~

J

2

Cl) nxa Cl)nxa

...( 11.23)

For under-tuned foundation, i.e. (() < (()n.m'

(J)11= (J)I/.HIshould be used in Eq. (11.23). Thus

LFtß©ãøÔÕÖøî¢÷

...(11.23 (l)

454 Soil Dynamics & Machille FoUIldatiolls

As mentioned earlier, in this method only the possibility of resonance is checked i.e. the naturalfrequencies computed from Eqs. (11.19) and ( 11.22) should differ by atleast 30 percent from the ope rat-i:1g speed of the machine. The Eqs. (11.20), (11.20a) and (11.23) for determining amplitudes are gi\'ento be used further in combined method.

Resonance method based on idealising each transverse frame to single mass-spring system is anoversimplification of a complex problem. Therefore the values of natural frequencies computed by thismethod are very approximate.

11.7 AMPLITUDE METHOD

In this method also, the vibration analysis is carried out for each transverse frame independently. How-ever, the frame has been idealised as a two-degree-freedom system (Fig. 11.7). The main criterion fordesign is that the amplitudes due to forced vibrations are within permissible limits (Barkan, 1962).

Z2

mz

KZZ1

m,

/

(a) Section of cross frame (b) Mathemetical model

Fig. 11.7: (a) Vertical vibration ora cross frame as a two-degree-of-freec.lom system;

(b) !\Iass-spring model

11.7.1. Vertical Vibration. For the vertical frequency a two-degrce-spring-mass system shown inFig. 11.7 b is adopted. Mass m I lumped over the columns is given by

WI + W2+0.33W3 +0.25W4m =I g

...( 11.24)

,\

,ii

IIIJ

1- t Fzsin wtm,

12c-tzz\\

m, /1 -----"::::::;-mz \ K,1-TII z\

K, \ I :.r:: :"l

IIz \ columns /\

iations of Rotary Machines455

Aass mz acting at the centre of the cross beam is given byWz + 0.45W4m =

Z g

e. W 1 = Dead load of the machine and bearingW2 = Load transferred to be columns by longitudinal beamsW3 = Weight of two columns constituting the transverse frameW4 = Weight of the transverse beam

The stiffness Kl of both the columns of a transverse frame is given by2EAc

KI = hThe stiffness Kz of the frame beam is given by

1K =-

z °SI

...(11.26)

...(11.27)

/ (1+2K) . 3/OS! = 96 E rb (2 + K) + 8G AbG = Shear modulus of beam material

E = Young's modulus of the material of columns

Ac = Cross-sectional area of a columnh = Effective height of the columnI = Effective span of the beam

AiJ = Cross-sectional area of the beamrb = Moment of inertia of the beam

K is defined by Eq. 11.15.The system shown in Fig. 11.7 b is identical to the system shown in Fig 1 18, and therefore can be

alysed by the procedure explained in Art. 2.8. The equations of motion in free vibration will be:

-re, ...(11.28)

ml ZI + KI ZI - Kz (Zz - ZI) = 0

mz Zz + Kz (Zz - ZI) = 0The solution of above equations are:

Z \ = AI sin 0) ...( 11.31 )III

Zz =A z sinO) ...(11.32)III

Substituting Eqs. (11.31) and (11.32) into eqs. (11.29) and (11.30), on simplification. we get

...( 11.29)

...( 11.30)

4 l.z 2 )zz

(I)/I-(I+~l) (1)/111-0)12 +(I+P)(J)IlIIO)IlIZ =0...( 11.3 3)

Kl ...( 11.34)here, 0) III 1 - /Ill + /Il2

~0) Il 1Z = V--;;;;

nlzIlm = ml

...( 11.35)

...( 11.36)

.u-_.'

~56 Soil Dy"amics & Machi"e Fo""dations

The two natural frequencies of the system can be obtained by solving Eq. (11.33). In forced vibra-tion. the equations of motion will be:

/1/] ZI+KI ZI-K2(Z2-ZI)=0 ...(11.37)

1112Z2 + K2 (Z2 - ZI) = F; sin OO{ ...(11.38)

The solution of the above equations can be presented as

ZI = AZI sin OO{ ...( 11.39)

Z2 = An sin OOt .. .(11.40)

Substituting Eqs. (11.39) and (11.40) into Eqs. (11.37) and (11.38), and then solving them we get2

00,,12. F;

1111 [004 - (1 + ~ ,J (00 ~ 1 I + CO~ 12 ) CO2 + (1 + ~ Ill) 00 ~ I 1 CO~ I 2 ][

2 2 2](l+~IIl)OOllll+~IIlOOIl12-CO .F;

[ -+ ( 2 2 ) 2 22 ]1Il2 00 -(1+~m) OOIlII+OOIl12 00 +(I+~/Il)OOIlIIOOIl12

All =

and AZ2 =

...(11.41)

...(11.42)

11.7.2. Horizontal vibration. For analysing the frame foundation in horizontal vibration as t\VO-degree-freedom problem, the upper and lower foundation slabs are assumed to be infinitely rigid. Thecolumns are taken to act as leaf-springs. The stiffness of a leaf spring is considered equal to the lateralstiffness of the individual transverse frame.

A1T m1X

A3~-- - ----L .AZ

~I

r

dl<Z

d k 1 d k 3 j

dm3~IQ

I I ddm1 11 m2

I II t /.I . .

G2 81m3

KX2

01

C<znterline of deckslab (initial)

/

m2

-- \ ... 8z--- 1 "'1'T óóóùèDue to translation 3and rotation

Due to horizontaldisplacement

Fi~. 11.1' : Sprin~-l1Ia" l!1odl'l for combincd horizontal and rotational vibrations of the deck slab

-

Idations of Rotary Machines 457

Figure 11.8 shows a typical mathematical model for a two bays frame foundation. The equivalents ( mi) lumped over the spring i (representing frame i) is given by:

re.

m, = m ,+ mb' + 0.33 m . + m 'I ml I Cl gl

1I111/i = Mass of machine resting on cross-beam of ith frame

IIIbi = Mass of cross-beam of i'h framelIIei = Mass of columns of i'h frame

...(llAJ)

l1lu = Mass transfered from longitudinal girders on either side",I

Using Eq. (11.43), values of ml' 1112and 1".13can be obtained.

K,i represents the lateral stiffness of ith transverse frame. Thus Kxl' Kx2 and Kd can be evaluated.

In Fig. 11.8, points G1 and G2 represent the cent:e of masses (i.e. ml' m2 and 11l3) and centre offness (i.e. Kxl' K,2 and Kd) respectively. Line A1Bl shows the initial position of deck slab. The finalplaced position of the deck slab is represented by the line A3B3 . The deck slab rotates about the massltre G I' dkl' dk2 and dk3 are the distances of different masses flom point G2. The distances of different,sses from point G 1 are shown as d I ' d ~ and d 3. e represents the distance between G I and G ~

. 11/ mL m L

The equations of motion for the system shown in Fig. 11.8 will be:

(Iml).1+ I[Kti (x+dmi 'V)] = F, sin cD( ...(11.44)

(I/1/i dl~lI) \it + I[K'i (x + d,lli'V) dllli] = M. sin cD(...(IIA5)

1ere, F = Horizontal unbalanced forcex

ivL = Unbalanced moment

Denoting

lI/ = 'Ill/, = Total massI ...(II.4())

")

Mm: ='I 11li dl~ll = Polar mass moment of inertia of all the masses about the vertical axis through G I

...{11.47)

IKti(x+dll/i'V) =xIK.ti+(IK.tidmi)'V= x K + K . e 'V = K (x + e 'V)x x x

K, represents the total lateral stiffness.

...( 11.48)

I K, i (x + dm i 'V) dill i = (I Kti dilli) X+ (I Kti dl~1i) \jI

= K, . e. x + [ L Kti( e2 + d; i)] \jI

= K, . e .\'+ K r / \jI + (L K, id; i) \jf

= K, (x+e\jl) e+ KIjI. \jI ...( 11.49)

11


K", represents the equivalent torsional spring stiffness for the frame columns and is given by.,

K", = IK.d dki

It may be noted that IK . dk ' = 0XI I

Making the substitutions from Eqs, (11.46) to (11.50) in Eqs. (11.44) and 911,45), we get

m .~ + K ,\ + K . e \jI = F sin W{.\ X .r

Mmz\jl + K\ex + (K, e2 + KII')\jI = Mz sin W{

...(11.50)

..,(11.51)

...( 11.52)

Equations (11.51) and (11.52) are similar to the Eqs. (9.51) and (9.52). Proceeding exactly in thesame way as discussed earlier, the solutions of Eqs. (11.51) and (11.52) can be obtained from the follow-ing equation:

4( 22 ) 222

CDn - a CDnx + CD/I IjI CDn + CDnx . CD nljl = 0

where fK.Wnx= v-;

~II'W =nII' Mm=

2ea=l+Z r

r = ~Mm=mThe amplitude of vibration in translation and rotation are given by

[

e22 2 2

]

F\ 2 Mc- + - CD ~ - CD . --=---,2 CD/IX CD1lII' m n\ M] m=A =

x L1(CD2)

2 F 2 2)M~e 2 X - - CD ---=-

?CD/I\- (conx M zr- . m III

AIjI = L1(co2)

where (2)

4(

2 2)

2 2 2L\ co = (J) - a (J)n x + con 'v CD + con x con IV

The net amplitude Air is given by

Air = At + y AIjIv '= Distance of the point at which the amplitude

gravity of the system\vhere,

...(11.53)

...(11.54)

...(11.55)

...(11.56)

,..(11.57)

...( 11.58)

...( 11.59)

...( 11.60)

...(11.61)

is being calculated from the centre 01

..,

Juniations of Rotary Machines 459

1.8 COMBINED METHOD

1 fact, the resonance and amplitude methods are complimentary. Since the amplitude method is basedn a system of two degree of freedom, it is obviously an improvement over the resonance method. How-vel', the. fact that in under tuned foundation the increase in amplitude of vibration during accelerationnd deacceleration stages has been ignored in this method. In combined method which is also known asxtended resonance method, the possibilities of resonance and excessive amplitudes both during steadyibration arid acceleration or deacceleration stages are investigated. The analysis may be carried out in

he following steps:

(i) The natural frequencies in vertical and horizontal modes of vibration as computed from Eqs.(11.33) and (11.53) by amplitude method are compared with the operating speed. The naturalfrequency in any mode of vibration should be atleast 30 percent away from the operating speed.

(ii) Amplitude.s of vibration computed from Eqs. (11.41), (11.42) and (11.61) should be within per-missible limits.

(iii) Amplitudes of vibration computed from Eqs. (11.20 a) and (11.23 a) should also be comparedwith permissible amplitudes to take care the possibility of excessive amplitudes during accelera-tion and deacceleration stages. A suitable value of damping ratio may be adopted to use in theseequations.

11.9 THREE DIMENSIONAL ANALYSIS

For turbogenerator foundations of more than 100 MW capacity, a three-dimensional space frame modelis preferred for analysis. The modelling should take into account the basic characteristics of the system,that is, mass, stiffness and damping. Special attention is required while idealising the points of excita-tion.

Nodes are specified to all bearing points, beam-column junctions, mid-points and quarter points ofbeams and columns and where the rhember cross- sections change significantly. Generally "thenumber ofnodes specified on any member should be sufficient to calculate all the modes having frequencies lessthan or equal to the operating speed.

Lumped-mass approach is used having lumped masses at the node points. The machine shall bemodelled to lump its mass together with the mass of the foundation. Equivalent sectional properties ofbeams and columns are used. The computation of equivalent mass moment of inertia of the frame mem-bers pose some difficulty since these depends upon the deflection shape in each mode. These may bediscretised in the first step and considered data in an iterative manner if desired. The columns may beassumed to be fixed at the base, disregarding the base mat.

A typical space frame model is shown in Fig. 11.9.

The dynamic analysis of the frame foundation requires the calculation of Eigen values of the system.The problem can be handled in a systematic manner in the matrix notation. The structure is idealised intoa skeleton system which retains the properties of the original structure. The stiffness matrix of the struc-ture as a whole is assembled from the stiffness matrices of individual members. The resulting equationsare then solved for the time periods and amplitudes.

...

460 Soil Dynamics & Machine Fou/1datio/1s

\

~"/"7

Fig. 1\.9: Space frame model of the found.ation shown in Fig. 1\.1


11.1 BARKAN, D. D. (1962), "Dynamics of bases and foundations," McGraw-Hill Book Co Inc., Ne\\ York.

11.2 IS 2<174(Pt. [[1-1992), "Foundations for rotary-type macines (Medium and high frequency)".

I ,.~ \IAJOR, A. (I %2), Vib:'atlon analysis and design of foundations for machines and turbines, AkademiJI Klado.Budapest, Collet's Hoidlngs Limited. London.

11.4 RAUSCII, E (I 9S(J). ":Y1Jchlnen fundamente und andere dynamisch beanspruchte Baukonstruclionen," VDIVcrlJg. Dusscldorf.

DD

.:,11

VIBRATION ISOLATION AND SCREENING

2.1 GENERAL

n machine foundations, following two types of the problems may arise:

(i) Machines directly mounted on foundation block (Fig. 12.1 a) may cause objectionable vibra-tions.

(ii) Machine foundation suffers excessive amplitudes due to the vibrations transmitted from theneighbouring machines (Fig. 12.1 b).

.~.

Machine!F = Fo s;n "'.

Machine

Z=ZoSin(JtFoundation Foundation

(0) Excessive vibrations ..du<z to machine it se If

(b) Excessive amplitu de due tovibrations transmittedfrom adjacent source

Fig. 12.1 : Machine directly mounted on foundation

The first problem may be tackled by isolating the machine from the foundation through a suitablydesigned mounting system (Fig. 12.2) such that the transmitted force is reduced which in turn willreduce the amplitude. This type of isolation is termed as force isolation. This type of arrangement willalso help in absorbing the vibrations transmitted from adjacent machines. The system used for this pur-:p'ose is termed as motion isolation. For heavier machines, the isolating system may be placed bet\\'~~n'the foundation block and concrete slab as shown in Fig. 12.3. Here the machines are rigidly bolted to the\:Oundation block which is isol1ted from the concrete slab through the mounting system. The mountingsystem is an elastic layer which may be in the form of rubber pad. timber pad, cork pad or metal springs.These have been already discussed in Sec. 2.5.

~62 Soil Dynamics & Machine Foundatioll.

.,'I

Machi ne

Isolator

= Foundation

Fig. 12.2: An isolator placed between machine and foundation

Machine

Founda'tion block

'i/i;Isolator

åóº

¢

ùþ

ßÁòþ ôßù t::>"-.l:i'.~~'~-'O""'I\'"A .' . -, ." . ' .' . -.-

'A. I" 6 Concrete slab'." .'to.:".'

.~ '. A.-., /},': .:..s...".'.;::': ::. .~:~:..~

=

Fig. 12.3 : An isolator placed between foundation block and concrete slab

The systems shown in Figs, 12.2 and 12.3 can be represented by a simple mathematical model sho\'In Fig. 12.4. In this m represents the mass of machine (Fig. 12.2) or mass of machine plus foundati,block (Fig.12.3). The mounting system (i.e. the elastic layer) is characterised by a linear spring witl-spring constant K and dashpot with damping constant C. This mathematical representation involves 0basic assumption that the underlying soil or rock possess infinite rigidity. This system is identical to tane shown in Fig, 2.17 (or Fig. 2,19), and the detailed analysis has already been presented in Sec. =considering both force isolation and motion isolation separately.

.

~ . . ';1',,:,,:;.{

.. .Vibratioll /so/(ltion (llld Screening 463

m

Fig. t 2.4 : Mathematical model

A more realistic model will be that in which the soil or rock is considered as an elastic medium. Thiswill make the system as a two- degree-freedom problem. the solutions of which are presented in the nextsection.

The second problem in which the vibrations are transmitted from the neighbouring machines can besolved by controlling the vibrating energy reaching the desired location. This is referred to as Vibrationscreening. Effective screening of vibration may be achieved by proper interception. scattering. anddiffraction of surface waves by using barriers such as trenches. sheet-pile walls, and piles. If the screen-ing devices are provided near the source of vibration, then it is termed as active screening or activeisolation. In case screening devIces are used by providing barriers at a point remote from the source ofdisturbance but near a site where vibration has to be reduced, it is termed as passive screening or passiveisolation. Both the methods of screening the vibrations have been discussed subsequently.

12.2 FORCE ISOLATION

Since the underlying soil or rock supporting the foundation block (or base slab) does not possess infiniterigidity, the foundation soil should be represented by a spring and not solely by a rigid sllpport as W:iSdone in Sec. 8.2. Then the mathematical model becomes as shown in Fig. 12.5. The various ~termsusedare explall1ed below:

III I = Mass of foundation block or mass of base slab11l1 = Mass of machine if isolator is introduced between machine and foundation block or

mass of machine plus mass of foundation block if isolator is placed between foundationblock and base slab.

K I = Stiffness of the soilK1 = Stiffness of the isolator

111 ~'," .'t_~

..64Soil Dynar.lics & Mac/line Folllltlatiol/s

TZz tF=FOSinGJt

TZ1 rn,

~kl//////// ///////

Fig. 12.5: Two degrees offreedom model

If the machine is subjected to a harmonic force (Fa sin wt), the equation of motion will be:

111121 -rKI2,-K2(22-21) =0 ...( 12.1 )

11/222 + K2 (22 - 21) = Fa sin ((){ ...( 12.2)

The solution ofEqs. (J2.1) and (12.2) gives

(~ )Sinrol21 =

[

nIl n~

]

. . FaW4- K2 + (Kl + K2) w2 + Kl K2

11~ 1nl nIl nIl

[(KI+K~) .,

]

.~ -0)- ~ S111 0) (

1111n~2 - .Fz- 0)4_

[Kz+(Kl+KZ)

]0)2+KlK2 a

~ 1nl nIl 1112

The principal natural frequencies of the system shown in Fig. 12.5 can be obtained by solving thefollowing frequency equation:

...(12.3)

...(12.4)

where

.; I K~ (Kl + K~)]

2 K, Kzc!) I; - L III~ 4- 1111 - W1/ + Ill, 1112 = 0

If (,r) I and CD ~ re p resent the roots of the above equation, Eq . (12.3) can be written as/1 1/-

lK~

).

~ S1l1 0) IIIII III~

2~' - .FI tl (0/ ) 0

2 - 4

[Kz (KI+Kz)

]2 KIKztl(w ) - (J) - -+ (J) +-

~ 1nl 1nl n~

...( 12.5)

...( 12.6)

...(12.7 ll)

ortl(0)2) = (0)2 -0)~1)(0)2 -0)~2) ...(12.7 b)

IiJ

Vibration Isolation and Screening 465

Force transferred to the foundation block or base slab

F{ -= K} Zj

lKj K2

)sin (I)f

11/} / 11-,- - F- ~(c!)2) . a

...( 12.8 j

The transmissibility of the system will be

FtT --F - Fa sin OJf

...(12.9)

Kj K2

1/1}/I~or, T F = ')

Ll((!)~)

The transmissibility TF depends on the system parameters given by Eq.special C3se will be examined in which

KI - K2 - 2- - --p11/, /110

...( 12.9 a)

(12.9). ror il1ustration, a

...(12.10)

Denoting./1/') .-=- = mass ratIO = 1111/, 11/

...( 12.11)

The natural frequency of machine foundation system ((Oil::)in which the isolating spring (k2) ISignored, is given by

KI') -

(J)~:: -= 1/11+ /I~

K)- ')

~=~1+ n~ 1+llm

rn,j

...(12.12)

Using Eqs. (12.10), (12.11) and (12.12), the Eq. (12.9) can be written as1

T, = l ~~- {(I+~ )+(~'" +~f f'r II ~-{(I+~ )-(~'" +~},f't}01'." [ , -= r

.;')

1

1

rI ., ., I.- .,(j)-.

(1 11m. 11~I (0':'

"(1+~;, ó¥ õî Ø¢³õó¼ } l(1+~m)W;

...(1~2.13 a)

{( I+~;' H~",+~I' J"} 1

Figure 12.6 shows the plots ofT I" with l~

) ratio for two values of mass ratio 11 . It is evident from~ " (!) 11::' m

this figure that the frequency ratio ((0 I (Oil::)must exceed a particular value (depending on the magnitudeof 1111/)before the transmissibility fal1s below unity. The particular values of ((0I (OJ,)for which the trans-missibility equals unity are given in Fig. 12.7.

.. . ~

466

108

6

0.10

Soil Dynamics & Machille 'Foulldatiolls

4

\

\\\

0-..5 1-.0 1.5 2.0

Fr~quency ratio, GJ/GJnz

3.0

Fig. t 2.6: TF versus ro/ronzfor different values of mass ratio /JmA desIgner is more concerned in examining whether the isolating system reduces the amplitudes of

the machine and foundation.III case no isolating system is used the maximum amplitude of machine foundation is given by

FaA. =( 2 ) )(/1/ 1 + Ill)) CD. - (I)-- /1-

...(12.14)

)

In most of the machines. the force Fa is frequency dependent (2meeun. The Eq. (12.14) can bewritten as

or

where,

)2 me em-

A; = ( 2 2)(ml + 11/2) mn; - m

- 2 me eA. - 2

11/1(1+~II/) (al -1)CD -a =-2b...I

...(12.15 a)

...(12.15 h)

...(12.16)(J)

l1..

. 2>-+-'--'-.D,-11\11\,- 1.0E11\ 0.8c0....

0.6+-c:.I

).Jm=1.0u....

0-4 ,.urn= 0.5 - --0l1..

}J.m= 0.1 .-------

Z0.2 {,Jnz = kIf (rnl +rnZ)

kllrnl = kZIrn2

VibratiOll lsolatio" a"d Scree"i"g 467

00 3.0 4.0

0.5

1.0GJ

G.)nz

Fig. 12.7 : Mass ratio versus m/mOl

If CDllarepresents the natural frequency of mass m2 resting on isolating spring, then

~2CD - -

IlU IIlz

The values of maximum amplitudes of two degree freedom system obtained from Eqs. (12.3) and(12.4) for frequency dependent force will be

...(12.17)

or

CD~a(2 me e (i)A ==1

[4 ( 2 2 ) 2 2 2 ]ml CD -(I+l1m) CDlla-CDII= CD +(I+l1m)CDlla CDII=

2

A = (2 me e) . (/2=1

[ ( J 2 J J

)]/Ill 1- ( 1+ I1m) (/j + (/2 - (/j (/2

...(12.18 a)

...(12.18b)

where UJ/la(/1 = (I)

...( 12.19)

Similarly. A = [(l+I1I11)(/~+l1m{/i-l](2mee)=2

[ ( 2 2 2 2)]m2 l-(I+l1m) (/( +(/2-(/1 (/2

...(12.20)

~ ~.-- -~. ,-

2.0

I

W =j<2+)Jm )(l+}Jm)GJnzI

1.5

E=<.

1.0

468 Soil DYllamics & Machille Follndations

The Eq. (12.18b) indicates that the amplitude of vibration of foundation will be small if a2 is small.For this oonashould be small. This can be achieved by appropriate selection of the mass above the isolatorspring ("'2) and stiffness of isolator spring (K2)' The efficiency of isolation system is defined as

A:I11 =- A-

Value of A: can be-~omputed from Eq. (12.15a). The value of A:I may be taken equal to the permis-sible amplitude. The value of 11 can also expressed as given below by dividing Eq. (12.18 b) byEq. (12.15 b). Thus

...(12.21)

ai (1 + ~m) (a~ -1)11 =

[ ( 2 2 2")]l-(1+~m) a) +a2 -a] a;

Solving Eqs. (12.21) and (12.22) one can obtain the value of a2' For this a2' appropriate values oflll2and K2 are selected. It will ensnre the amplitude of vibration to be within pf'mlissible limit

Total force on the isolator, Fa = Kz . A:2Fa should be less than the allowable capacity of the Isolator in compression.

...( 12.22)

...( 12.23)

12.3 MOTION ISOLATION

Let us examine a case when a sensitive equipment of mass 11/I ISplaced on a foundation block of mass 1112'The spring K, represents the foundation soil and spring K2 is an isolating spring which is placedbet\\een the masses Ill, and 1112'in order to minimise the transmission of vibrations from the ground to theequipment. If the ground is subjected to a periodic displacement given by Zo sin w l. the equations ofmotIOn \vill be:

nzl ZI + KI Z\ - K2 (Z2 - ZI) = KI Zo sin W f

11/2Z2 +K2(l2-Z,) =0

...( 12.24)

...(12.25)

The vafues of maximum amplitudes of motion are given by2

~-~

A_I = KI lo

[

ml m2 m

]

1

- 4 (KI + K2) K2 2 Kl K200 - +- W +-m, 1112 11I,n/2

~111) Ill.,A = K Z -:2 1 0

[(

K K ) K

]

K K4 I+J J 2)J

W - Ill, - + III~ W + Ill, 111;

The displacement transmissibility of the machine (Tn) is defined as the ratio of displacement ampli-tude of mass 1f/2to the displacement amplitude of the rigid support. Then

KIK2A:2 111,1112T ---

0- Zo - w4_[

(KI+K2)+K2Ã

())2+K)K2

111) 1112 111)11/2

...(12.26)

...( 12.27)

...( 12.28)

~ I

Vibration Isolation alld Screenillg 469

The Eq. (12.28) is identical to Eq. (12.9), and therefore the results shown in Figs. 12.6 and 12.7 holdgood in this case also..

Equation (12.28) can also be written as:

A 2 2- :2 - a( a2 (1 + Ilm)To - l - ( 2 2 2 .,)0 l-(l+llnr) al +°2 -al a2

...( 12.29)

If A-1 is taken as permissible amplitude. and l is the applied dynamic displacement. then the ratio-- 0

A:2i lQ is known. For this value. Cl2can be determined which in turn will give the stiffness of the isolatorspring i.e. K2'

12.4 SCREENING OF VIBRATIONS BY USE OF OPEN TRENCHES

12.4.1. Active Screening. In this case the screening of vibrations is done near the source of vibration.Figure 12.8 shows a circular trench of radius R and depth H which surrounds the machine foundationthat is the source of disturbance. The design of trench barriers is based on some field observations.Barkan ( 1962) mentioned that the reduction in vibration amplitudes occurs only when the trench dimen-sions are sufficiently large compared with the wave length of the surface waves generated by the source ofdIsturbance. Dolling (1966) studied the effect of size and shape of the trench on its ability to screen thevibratIOns.

Amplitude ofsu da c e

displaclZmlZnt

F (t)

L'Footing

Circular opent re n c h otradius R

an d depth H

I

I

R-1

H

1~

Fig. 12.8: Vibration screening using a circular trench surrounding the sourceof vibration-Active screening (Woods. t 968)

The first comprehensive study of screening vibratIOns by use of open trenches was made by Woodsand Richart ( 1967) and \Voods ( 1968). They conducted field tests by creating vertical vibrallons with asmall \"ibrator resting on a small pad at a prepared site. The vibrator could create a maximum force of~O0:. The soil conditions at the site \vere as shown in Fig. 12.9. The water table was below 14.3 m depth.The depth H of trenches was varied from 150 mm to 600 mm, the radius R of annular trench varied from150 mm to 300 mm, and the angular dimension e was varied from 900 to 3600 around the source ofvibration. Frequencies of 200 to 350 Hz were used in the tests. Using velocity transducers. the amplitudes

- . 0;" ," "' i;',C",'. ."0,'. ',c.,

470 Soil Dynamics & Machine FolllldatiOlls

of vertical ground motion were measured at selected points throughout the test site before installation ofthe trench and after installation of the trench. Woods (1968) has introduced a term amplitude reduction.factor which is defined as

ARF = Amplitude reduction factor

- Amplitude of vertical vibration with trench- Amplitude of vertical vibration without trench

T ~ ~

1.2muniform silty fine sand (SM)

Wn = 7 0/0, e = 0.61) td = 16.4.6 k N / m3Vc = 287 rnls

3.1 m

sandy silt (ML)

Wn = 230/0) Cl = 0.68) 'id = 15.77 kN/m3

Vc=534.m/s

Fig. 12.9: Soil stratum at the test site

Some of the resuits of field tests conducted by Woods (1968) are shown in Fig. 12.10 in the form ofARF contour diagrams. The dimensions of the trench are expressed in non-dimensional forms by divid-ing Hand R by the wave length ARof Rayleigh waves. ARis obtained by determining the number of waves(/1L occuring at distancex from the source (AR= x/1I).Wavelengths ARfor different frequenciesare givenin Table 12.1.

Table 12.1: Wavelength and wave velocity for the Rayleigh Wave at the test site (Woods, 1968)

FrequencyHz

AR VRmm m/s

lOO

250

687

513

137

128

300

350

421

336

126

117

'".:,. -~.,,:i,:>},\ . (," .;'if'..":. .~.

Vibratio" Isolatio" a"d Scree"i"g

D 1.25-0.50

f:::~::JO,50-0'25

1::::~::)lo,25-0,125

t~:;:.~ < 0 125:-,:,':,-:- '

H/"'R1.452

RO/AR0.726

(a)

,~ """

471

Boundary of screenedzone

/.. -1,25

~>1.25

D 1.25-0,50F::,::':':lo,50-0,25

1(':,\/:1 0,25 -0.125

p',-::-:,:,)< 0 125,',', '," '

H/AR0,596

Ro/AR0,596

(b)

Fig, 12,10: Amplitude reduction factor contour diagrams for active screening (Woods, 1968)

The field tests of Woods (1968) thus correspond to

A Hr- = 0.222- 0.910 and -;- = 0,222- 1.82R AR'

For satisfactory screening of vibrations, Woods (1968) recommended that ARF shouid be less than orequal to 0,25, The conclusions made on the basis of this study to keep ARF ~ 0,25 are:

(i) For full circle trenches (8 = 360°) , a minimum value of H/AR = 0.6 is required. The zonescreened in this case extended to a distance of atleast 10 wavelengths (I OAR)from the source ofdisturbance

(ii) For partial circle trenches (900 < 8 < 360°), the screened zone was defined as an area outside thetrench extending to at least 10 wave lengths (10 AR)from the source and bounded on the sides by

...

.t72 Soil Dynamics & Machine Fouudations

radial lines from the centre of source through points 45° from ends of trench. In this case also. aminimum value of H/AR = 0.6 is required.

(iii) Partial circle trenches with 8 < 90°, effective screening of vibration is not achieved.

(i,') Trench width is not an important parameter.

12A.2. Passive Screening, Woods (1968) has also performed field tests to study the effectiveness of opentrenches in passive screening (Fig. 12.11), A typical layout of these tests consisting of two vibratione:\citers (used one at a time for the tests), 75 transducer locations. and a trench is shown in Fig, 12 12.The sizes of trenches ranged from 100 mm x 300 mm x 300 mm deep to 2440 mm x 3050 mm x 1220 mmdeep, Frequencies of excitation varied from 200 to 350 Hz.

Source ofdisturbance

Amp! itude0 f s u rfa c evibration

Amplitudeof surfacevibration

Equipmrzntto beprotected

. ' .. ,. . .. ~ . .' , , .., ,

, ~ . '. ."

. - ..' , .t.

:' Hr -.R

,. . ::.1"

. "

Opent rrznch ~ '; .'

Fi~. 1:!.I] : "ihration screening using a straight trench. Passive screening (Woods. I96S)

The values of Hij'I{varied from 0.444 to 3.64 and R/AR from 2.22 to 9.10. It \vas assumed in thesetests that the zones screened by the trench would be symmetrical about the 0° line. Figure 12.13 shows theARf contour diagram for one of these tests.

For satisfactory screening. Woods (1968) recommended that the ARF should be less than or equal to0,25 in a semi-CIrcular zone of radius (1/2) L behind the trench. The conclusions made on the hasis ofthis field study to keep ARF :::0.25 are:

(i) I li'I{ should be atleast 1.33

(id To maintain the same degree of screening, the least area of the trench in the vertical direction((c, 1.11"" AI)' should be a<; follo\\'s:

AI R- -=2 'i at -=7 0, ~ .- ) -.I'R 'R

AI R-, .C6.0 at /' = 7.0,- 'R/'R

.. m

Vibratioll Isolatioll alld Screenillg 473

(hi) Trench width had practically no influence on the effectiveness of screeI'.ing.

Experimentai investigations of Sridharan et. al (1981) indicated that the open unfilled trenches arethe most effective. However the open (unfilled) trenches may present instability problems necessitatingtrenches backfilled with sawdust, sand or bentonite slurry. The performance of open trench \vith sawdustwas found better as compared with sand or bentonite slurry.

OOA ~O 0 ~/". 0

<, "'<:( 0 75 pi c ku P", ',0 0 benches0 ',('", L 0, ' "" , " r>.

0 0 ""' ~",,~ .,..""' ., " Tr en'0 /

0 O

.

0

..

' " -', K ch barriers

0 0 0 ;{~:', ,0 0 ,/ \, v'I 0 0 " \, ,

0 / '."0 0 l.Sm ,," /

/ 0 . // "v :?s'>

0, \, ./ BI 0 0

0

0

0

0

0

/

/l.Sm

vibration exciter footings

Fig. 12.12 : Plan \iew of the field site layout for passive screening (Woods. 1968)

320240/ 0/ 160

/N

[° 800 0 / 006.1 m

0// 0

/'t

000 /

0 00 00

/0 00 00

000 /

474

lliII1J ARF>1.25

0 ARF1.25-0.5

[tJ+ AA Fó¬ó´ó

+ + 0.5-0.25

n ARFQ 0.25-0.125. ARF<0.125

Soil Dynamics & Mac/,ine Forllldatioll.'i

Centqrtin q

4.6m

Trench

~ 1.5 mHI AR = 2.38LIAR = 4.76RI AA = 5.96W/'AR = 0.17

Fig. 12.\3: Amplitude reduction factor contour diagrams for !Jassive screening

12.5 PASSIVE SCREE1\I;\C BY USE OF PILE BARRIERSThere may be situations in which Rayleigh wavelengths may be in the range of 40 to SOm. For such acasc. the open trench will be effective ifits depth range from 53 m to 66 m (i.e. 1.33 AR)'Open tren~hes(filled or unfilled) with such deep depths are not practical. For this reason, possible use of rows of pilesas an energy barrier has been studied by Woods et al. (1974) and Liao and Sangrey ( 1978).

Woods et a1. (1974) used the principle of holography and observed vibrations in a model half-spaceto evaluate the effect of void cylinderical obstacles on reduction of vibration amplitudes. A box of size1000 mm x 1000 mm x 300 mm deep filled with fine sand constituted the model half-space (Fig. 12.14).In this figure. 0 is the diameter of the void cylinderical obstacle and SI! is the net space between twoconsecutive void holes through which energy can pass through the barrier. The numerical evaluation ofbarrier effectiveness was made by obtaining the average of the values of ARF obtained on several linesbeyond the barrier in a section ::I:15° of both sides of an axis through the source of disturbance andperpendicular \0 the barrier. In all tests. H/ARand LIARwere kept as 1.4 and 2.5 respectively. The Isola-tion effectiveness is defined as

Effectiveness = I - ARF ...(12.31)

;"'ii[; ""1iItJ'i

Vibration Isolation and Screeni1lg 475

F(!) t

Fig. 12.14 : Definition of parameters for cylindrical hole barriers (Woods et aI., 1974)

1,0

." " , . l'O" "', , "

0,. , . .: ~', 0: , . '

J"

. '. . . . '0.: . '0:~, :', ': :' '0,. 'J ' ,

"'" 0 "" :'0 :-

" 0,: " ':""

0,15"-

00 0.05 0,10 0.15

Snf"rFig. 12.15: Isolation effectiveness as a function of hole diameter and spacing (Woods et al.. 1974)

"-

"-"

0.20 0.25 0,30

LLa::«I

11 0,5111

I \\\ \ \ I

"'-111t>I "-Ct>I>.;: 0.4ut>I

I ,,\ \\0.125...........W

0.21\ \\

\0.075 \

",","',,' ">,,',,' ,

L Slllg the data of different tests, a non-dimensional plot of the isolation effectiveness versus S/A.Rratio for different vaues ofD/A.R was plotted as shown in Fig. 12.15. Woods et al. (1974) recommendedthat J row of void cylinderical holes may act as an isolation barrier if

D 1- >-6A -RSI! I-<-AR 4

Soil type

.P6

:.1nJ

-


...( 12.32)

...( 12.33)

107Barrier materials

110

Void boreho le

Infinitelyrigid pile

Steel

Concrete

Tim ber

Plasti c foam

010 !

Fi:!. 12.16 : Estimatcd \'aluc\ of Haylcigh wave impedance for ,'arious soils and p:le materials (Liao and Sangrcy. 1978)

I.lao and Sangn:y (1978) used an acoustic modcl employing sound waves in a fluid medium to~'\ ,lIuak thl' possibility of the use of row of piles as passive isolation barriers. They have studied the effectof dlal1l~.t..:r.spacing. and material properties of the soil pile system on the isolation effectiveness They'~\'II1~'iLld~d that:

610

VI-0C0

ZI 5VI 10N

EQJuc

40-0 100.E'-

103>0

..ccrI

2-10>-

0Cl:

f

.

Gravel

0 ens e san dHard clay-Si! tLoose sandVery soft clay

e--

.'---

!Z.'...,'~J

Vibration Isolation and Screelling 477

where.

(i) The Eqs. (12.32) and (12.33) proposed by Woods et al (1974) are generally valid.

(ii) Sn = 0.4 ARmay be the upper limit for a barrier to have some effectiveness.(iii) The effectiveness of the barrier is significantly affected by the material of the pile :md void holes.

Acoustically soft piles (IR < 1) are more efficient than acoustically hard piles (IR > 1). IR isimpedance ratio which is defined as

Pp VRPIR =Ps VRS

Pp = Density of pile materialPs = Density of soil medium

VRP = Rayleigh wave velocity in pile material,\' RS = Rayleigh wave velocity in soil medium

Figure 12.16 gives a general range of the Rayleigh wave impedance (p VR) for various soils and pilematerials. .

(iv) Two rows of barriers are more effective than single row barriers,

...( 12.34)

jILLUSTRATIVE EXAMPLES'

Example 12.1Determine the stiffness of the isolator to be kept between a reciprocating machine and the foundationshown in Fig. 12.17 to bring the vibration amplitude to less than 0.02 mm. The weight of the machine is18 kN, and it produces an unbalanced force of 4.0 kN when operated at a speed of 600 rpm. The dynamicshear modulus and Poisson's ratio of the soil are 2 x 104 kN/m2 and '0.35 respectively,

Solution:

1.24x4.0x3.0x1.5 ')

Mass of the foundation block = = 44 kNs-/m9.81

Stiffness of the soil,

Mass of machine = 9~~1 = 1.83 kNs2/m

Total mass of machine and foundation = 44 + 1.83 = 55.83 kNs2/111

4 G roK=-= 1- ~l

r = fA = ~4 x 3 = 1.95 m0 f; 1t4

K = 4x2xl0 x 1.95 =24 x= 1-0.35

104kN/m

Natural frequency of the whole system without any isolation will be:

478 Soil Dynamics & Machine Foundation

r 4.0 mSection

~

3.0m

~ 4.0 m ~Plan

Fig. 12.17: Machine-foundation system (Example 12.1)

~-CD - --

n; m 1

2 Ttx 550CD= 60 = 57.6 rad!s

(A ) - Fa - 4.0; max - 2 2 - ( 2 .,)m(oo//;-oo) 55.83 65.0 -57.6-

=-07.9̈ 10-501=0.0790101

424 X 10 = 65.0 rad/s

55.83

Machine

Weight = 18kN

T1.5m

///,\,\ Foundation blo c k /\\

1

Vibratioll Isolatioll alld Screellillg 479

The amplitude is greater than the permissible amplitude i.e. 0.02 mm. Hece isolator is requiredbetween machine and foundation.

2. Let the isolator be having stiffness Kz. Adopting the two degrees freedom system as shown inFig. 12.5.

211/2= 1.83kNs Im

211/1 =44 kNs Im

4Kl = K: = 24 x 10 kN/m

(0 - = 65.0 rad/s1/-

AZI - 0.02 = 0.25311= Az - 0.079

Cl)n: = 65.0 = 1.13a I = ---;;- 57.0

'nz - 1.83 = 0.0416/-1./11= 1111- 44

From Eq. (12.22)

a~ (l+/-I.m)(a~-l)11 =

[ ( 2 2 2 2)Jl-(1-llm) al +a2 -al a2

or2 .,

'"' - a2(1+0.0416)(1.13--1)0.k53 -

[ ( 2 2 2 2)J1- (1+ 0.0416) 1.13 + a2 -1.13 a2

or a~ = 0.2310 i.e. a2 = 0.481

(jJ11i!...= 0.481. It gives (0 = 0.481 x 57.6 = 27.7 rad's(j) 11(1

. . 2 1 3Ko = 1112 (jJ = 1.83 x 27.7- = 1.40 x 10 kN/mIl{/

A = [(l+llm)a~+Il",a;-1][2mee(O2]Z2 2

[ ( 2 2 1 1 )1112(0 1-(l+1l",) al +az -aj ai

= [(1+0.0416)x1.l32+0.0416xO.2310-1.-(4.0)1(

~ ~

1.83 x 57.6- 1-(1+0.0416) 1.13- +0.2310-1.13- A 0.2310)-~= - 8.5 x 10 m

Force in the isolator = Ko' An = lAD x 103 x 8.5 x 10-~kN"'1.18kt'

3.

110


A suitable isolation system may be selected which has total stiffness of 1.40 x 103kN/m and allow-able compressive 'load more than 1.18 kN.

Example 12.2Determine the stiffness of the isolation system if it is placed between the foundation block and base slabas shown in Fig. 12.3. Use the data given in example 12.1.Solution:

(i) Let the foundation block of size 4.0 m x 3.0 m x 1.0 m high is rigidly connected with machine.Isolators are placed between this block and base slab as shown in Fig. 12.18. Then

Machine

T1.0 rn Foundation block

iIsolator

TO.Srn

-L

~Base slab

4.0 rn ~Fi~. 12.18: :\-Iachine-foundation isolator system (Example 12.2)

From eq. (12.22),

1//2 == 1.83 + -}x 44 = 31.2 kNs2/mI J

111( == 3" x 44 = 14.7 kNs-/m4 2

Kl = 24 x 10 kNs fm(j) = 65.0 rad/sn::

11 = 0.253

al = 1.1331.2 2 12

1111/= t4.7 = .2 2

a2 (1+2.12)(1.13 -1)0.253 =

[ ( 2 2 22-

)]1-(1+2.12) 1.13 '+a2 -1.13 ([2

. -

WA.--'}..',..".~;:.' ,"'ti'~:,,<. ,." ';",-,:,. "~,<~Atj;!~";,; "'.,:,' ">:!',.,E "1I

bratioll Isolation alld Screellillg 481

or ai = 0.697 i.e. a2 = 0.835

Hence, O)na = 0.835. It gives (J)na= 0.835 x 57.6 = 48 rad/s0)

2 4Ka = 11120)2 =31.2 x 48 = 7.2 x 10 kN/mna

[0+2.12) x 1.132+2.12 x 0.697 -1]4.0- An = .,

r (2 2'

]31.2 x 57.6- Ll-(1+2.12) 1.13 +0.697~1.13 x 0.697)17.85

= 103514[-2.38]--5= - 7.24 x 10 m

Force in the isolator = K A., ==7.2 x 104 x 7.24 x 10- 5 = 5.2 kNa =-

On comparing with the results of example 12.1, it can be concluded that the stiffness of the isolatorand force on it depends significantly on the location of the isolator.

(ii)

Example 12.3It is planned to install a compressor having operating speed of 1000 rpm at a distance of 50m from aprecision machine. Suggest a suitable open trench barrier to provide effective vibration isolation. Thevelocity of shear waves at the site was found as 140 mIs.

Solution:

Active screening. 1000

OperatIng frequency, f = 60 = 16.7 Hz

Rayleigh wave velocity VRmay be taken approximately equal to shear wave velocity i.e. 140 m/so

V 140Therefore, Wave length AI{ == ; == 16.7 ==8.4 mDepth of the trench for active screening is given by

H ==0.6 AI{= 0.6 x 8.4 ==5.04 mA partial ci~cle trench with e = 120° may be located at 4.0m distance from the source (Fig. 12.19 a),

Passive screeningDepth of the trench for passive isolation is given by

H = 1.33 AR= 1.33 x 8.4 = 11.2 m. .Let the trench is provided at a distance of 12 m from the precision machine.

For

R - (50-12) = 4.52 (O.K. as lies between 2 and 7).- - 84AR .R

- = 4.52,AR

AT = 25 4.52-2.0(6.0-2.5)=4.26')..2 . + 7.0-2.0R

Soil Dynamics & Machille Foulldations

24.26 x 8.4Length of trench = 11.2 = 26.8 m say 27 m

The layout of trench with respect to compressor and precision machine is shown in Fig. 12.19b.

12.19.-Layout of trench with respect to compressor and precision machine

~FERENCES

"kan, D. D. (19621 "Dynamics of bases and foundations", McGraw Hill, New York."

Iling, H. J. (1966), "Efficiency of trenches in isolating structures against vibration", Proc. Symp. Vib. Civ. Eng.Butterworth, London.

.0, S, and Sangrey, D. A. (1978), "Use of piles as isolation barriers", J. Geotech. Engg. Div.. Am. Soc. Clv. Eng..104, (GT9), 1139-1152.

dharan. A. , Nagendra, M. V. and Parthasarathy, T. (1981). "Isolation of machine foundations by barriers", 1nLConf. Recent Ad\'. Geotech. Earthquake Eng., St. Louis, Vot. 1,279-282.

lads. R.D. (1968). "Screeningof surfacewaves in soils", J. Soil Mech.Found.Div. , Proc. Am. Soc. Ci\'. Eng., 94.(SM-4),951-979.

Precision machine Precision machine

T12n

Tren ch ) I {27mI

50m

I

38m.0

120

Tre ../ ,/

eññ£³ . -'-Compre ssor Co m pre ssor

(a) Active isolation (b) Passive isolation

'atio" Iso/atio" a"d Scree"i"g 483

)ds, R. D., and Richart, F. E. Jr. (1967), "Screening of elastic wavesby trenches", Proc. Int. Symp. Wave Propag.Dyn. Prop. Earth Mater, Albuquerque., NM, 275-284.

lds,R. D., Barnett, N. E. and Sagessor, R. (1974), "Holography - A new tool for soil dynamics", 1.Geotech. Eng.Div., Am. Soc. Civ., lOO, (GT-I1), 1231-1247.

PRACTICE PROBLEMS

!.1. Explain the difference between 'force isolation' and 'motion isolation'. Sketch a suitable systemfor 'force isolation'. Represent it by a mathematical model and then give the procedure of gettingthe stiffness of the isolator.

2. Starting from fundamentals, derive the expression for the efficiency of isolation system.

3. Explain the difference between "Active screening" and "Passive screening". Give the procedureof designing the open trench barrier in both the cases.

4. Give the salient features of passive screening by use of pile barriers.5. Design a suitable isolation system for keeping the amplitude of the foundation of a reciprocating

machine less than 0.025 mm. The weight of the machine is 25 kN and it produces a sinusoidallyvarying unbalanced force of 4kN in the vertical direction. The operating speed of the machine is800 rpm. The dynamic shear modulus 'of the soil is 2.5 x 104kN/m2. Assume suitably any datanot given.

6. A compressor having an operating speed of 1300 rpm was installed in an industrial unit. Lateron it was planned to place a precision machine at a distance of 60 m from it. It was felt necessaryto protect this precision machine from any damaging vibration caused by the compressor. De-sign open trench barrier to provide effective vibration screening for the cases of (a) active and(b) passive screening. The velocity of shear waves was found as 160 m/s.

DD

-- -----

SUBJECT INDEX

AAmplitude of motion, 15

Amplitude method, 454Antiliquefaction measures, 324

BBandwidth method, 57

Bearing capacityof footings, 238factors, 241

Blasting tests, field, 314Block foundations:

degree of freedom. 352effect of shape on response, 394embedded. 394method of analysis, 353modes of vibration, 352rocking, vibrations, 354, 360, 383, 396rocking and sliding vibrations, 354, 363, 389, 397sliding vibrations, 354, 359, 379, 395vertical vibrations, 354, 358, 370, 394yawing vibrations, 354. 362, 386

Block-resonance test:horizontal, 154vertical,151

cCodTicienr of clastic:

nonllniform col11pn.:ssion,356nonuniform shear, 357uniform compreSSion,354uniform shear. 356

CrItical distance, 96,98, 100

Culmann's construction:modified, 191

Cyclic:mobility, 279simple shear test, 118, 128, 286, 296torsional shear test, 118, 131, 286triaxial compression test, 118, 133

F

f

ff

J

DDamping: '

critical,factor,ratio,

Degrees of freedom, 14

Displacement analysis of retaining wall, 20 Iin pure rotation, 210in pure translation, 205

- using Richard Elms model, 201using Saran et al model, 214

Dynamic bearing capacity, 238dynamic analysis, 252pseudo-static analysis, 238

Dynamic earth pressure, 187effect of saturation, 187effect of submergence, 190effect of uniform surcharge, 189for c-<!>soils, 193Mononobe-Okabe's theory. 200point of application, 200pseudo-static methods, 238

EEarthquake:

epit:enter, 3equivalent dynamic load, 6focus, 3

'ct Index 485

:ensity, 419nitude, 5constants, 68

MMachines

types, 340constants, 68

half-space method, 370ded block foundation, 394

reciprocating, 340impact, 344rotary, 345

Machine foundations

categories of, 340criteria for satisfactory action, 347permissible amplitudes of, 348

Mononobe-Okabe's theory, 187Motion

amplitude of, 15.harmonic, 15periodic, 15isolation, 35

sS ::aring capacity, 238:neralised bearing capacity equation, 240)rizontal displacement, 249ttlement, 249t, 249

ncy:Ircing, 15ltural, 15

N

ysical prospecting, 93 Natural frequency, 15

p

t machine foundations, 421

~sign procedure, 432v'namic analysis, 426

'Pt time, %,97, 100

Periodic motion, 15

Permissible amplitudes, 348

Phase Jag, 16Phase lead, 16

Rspring method, 354

'action, 279Ictors affecting, 323eld tests for, 314.om standard penetration tests, 319litial, 279Iboratory studies, 283lechanism, 281:andard curves and correlatIOns, 30 I.:rr11lnology,279one of, 306

Resonance method, 450

Resonant column test, 118, 119

Rotating mass type excitation, 30

Rotary machinesdesign criteria, 445loads on, 446three dimensional analysis of, 460resonance method of analysis, 450amplitude method of analysis, 454combined method of analysis, 459


sSc\eening of waves, 469Seismic:

coefficients, 9cross-borehole survey, 147, 148down-hole survey, 147, ISOforces, 9zones, 9refraction, 94, 147, 151up-hole survey, 147,150

Shake table tests, 286, 309Shear modulus, 163,165,171Soil mass participation in vibration, 400Standard penetration test, 159Strain, 67

fiee, 19,22,40,43isolation, 32

force, 32motion, 35

measuring instruments, 36acceleratioI" pickup, 38,39displacement pickup, 38velocity pick up, 39

Vibrations of rod

end conditions, 74fixed-fixed, 75. 79fixed-free, 78free-free, 74, 76

of finite length, 76longitudinal, 76torsional, 80

of infinite length, 70longitudinal, 70torsional, 72

TTheory of vibrations, 13

Time period, ISTransient tests

w

u

Wave propagationin elastic half space. 86in elastic rods. 70, 72, 76, 80in infinite medium, 81

WavesUltrasonic pulse test, 118,126

Lndamped free vibration, 19

v

compressIon, 83, 108head, 95Rayleigh, 9\shear. 84, 108

Vibration

absorber, 48forced, 25,42,47

zZone of liquefaction, 306

ODD

if-:" . _'M- _.- ~ '----""~"'-:--~"'--:-'""~'l, ", .\. ,- ',. 1.,

~ .:, \

~ I J

k~, " '- ~L.._":~",-"

~

-,.~.~ :.c ,.:..."""",:"""""""","

53720995 Soil Dynamics and Machine Foundations Swami Saran 2

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Transcript of 53720995 Soil Dynamics and Machine Foundations Swami Saran 2