Soil Dynamics and Machine Foundations Swami Saran_2

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SOIL DYNAMICS . AND"

MACHINE

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FOUNDATIONS

By Dr. SWAMI SARAN Department of Civil Enginemng University of Roorkee Roorkee-247 667

(INDIA)

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Galgotia Publications pvt.ltd.

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First Edition 1999

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No matter in full or part may be reproduced or transmitted in any form or by any means (exceptfor review or criticism) without the written permission of the author and publishers.

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

Published by .Suneel Galgotia for Galgotia Publications (P) Ltd. 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 dynamics and machine foundationshas been reported.Courseson soil dynamicsandmachinefoundationsalreadyexistat graduatelevelin many institutions, and its inclusion at undergraduate level is progressing fast. The author is engaged in teaching the course on soil dynamicsand machine foundationsat gr'duate level from last fLfteenyears. The text of this book has been developed mainlyout of my notes preparedfor teaching the students.The consideration in developingthe text is its lucide presentationfor clear understandingof the subject.The material has been arrangedlogicallyso that the reader can follow the developmentalsequenceof the subject with relative ease. A number of solved examples have been included in each chapter. All the formulae,charts and examples are given in SI units. Some of the material included in this text book has been drawn from the works of other autors. Inspiteof sincereefforts,somecontributionsmay nothavebeen acknowledged.The authorapologisesfor suchomissions. The author wishes to express his appreciationto Km. Lata Juneja, Sri RaJeevGrover and Sri S. S. Gupta for typing and drawing work. Thanks arealso due to the many collegues,friends and studentswho assistedin

wittingof thisbook.

.

.

The author would be failing in his duty it he does not aclaiowledge the support he received from his family members who. encouraged him through the various stages. of study and writing. The book is dedicated to author's Sonin law, (Late) Shri Akhil Gupta as a token of his love, affectionand regards to him. (Dr. Swami Saran)

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CONTENTS

. PREFACE

1.

INTRODUCTION 1.1 General 1.2 1.3 1.4

2.

THEORY OF VIBRATIONS 2.1 General 2.2 Defmitions 2.3 Harmonic Motion 2.4 Vibrations of a Single Degree Freedom System 2.5 Vibration Isolation 2.6 2.7 2.8

3.

Earthquake Loading Equivalent Dynamic Load to an Actual Earthquake Load Seismic Force for Pseudo-staticAnalysis Illustrative Examples References Practice Problems

Theory of Vibration Measuring Instruments Vibration of Multiple Degree Freedom Systems Undamped Dynamic VibrationAbsorbers Illustrative Examples Practice Problems

WAVE PROP AGATION IN AN ELASTIC, HOMOGENEOUS. . AND ISOTROPIC MEDIUM 3.1 General 3.2 Stress, Strain and Elastic Constants 3.3 Longitudinal Elastic Waves in a Rod oflnfmite Length 3.4 Torsional Vibration ora Rod of Infmite Length 3.5 End Conditions 3.6 3.7 3.8 3.9 3.10 3.11

Longitudinal Vibrations of Rods of Finite Length Torsional Vibrations of Rods of Finite Length Wave Propagation in an lnfmite, HomogeneousIsotropic, Elastic Medium Wave Propagation in Elastic, Half Space Geophysical Prospecting Typical Values of CompressionWave and Shear Wave Velocities Illustrative Examples References..'. . ; Practice Problems

1-12 I 3 6 9 12 12 12 13-66 13 14 15 18 32 36 39 48 53 64 êéóïïé 67 67 70 72 74 76 80 81 86 93 108 108 116 117

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Soil Dynamics & Machine Foundations

DYNAMIC SOIL PRO~ER~5. 4.1 General

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4.2 4.3

LaboratoryTechinques Field Tests

4.4

FactorsAffecting Shear Modulus, ElasticModulus and Elastic Constants IllustrativeExamples References PracticeProblems

ïèéóîíé 187 187 201 221 236

DYNANnCEARTHPRESSURE ëò ï

General

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Pseudo-static Methods

5.3

Displacement Analysis Illustrative Examples References

237

PracticeProblems

6.

7.

DYNAMIC BEARING CAPACITY OF SHALLOW FOUNDATIONS 6.1

General

6.2

Pseudo-static Analysis

6.3 6.4

Bearing Capacity of Footings Dynamics Analysis Illustrative Examples References Practice Problems

LIQUEFACTION OF SOILS 7. 1 General 7.2 Definitions 7.3 7.4

--.118-186 118 118 147 163 174 182 184

Mechanism of Liquefaction Laboratory Studies

DynamicTriaxial Test 7.6 Cyclic Simple Shear Test 7.7 Comparisonof Cyclic Stress Causing Liquefactionunder Triaxial and Simple Shear Conditions 7.8 StandardCurves and Correlations for Liquefaction 7.9 Evaluationof Zone of Liquefactionin Field 7.10 VibrationTable Studies 7.11 Field Blast Studies

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7.12

Evaluationof LiquefactionPotentialusing Standard Penetration Resistance

7.13

Factors Affecting Liquefaction

-

îíèóîéè 238 238 238 . 249 268 277 278 îéçóííç 279 2.79 281 283 288 296 300 301 306 309 314 319 323

Contents

ix

'

324 326 332 336 339

7.14 AntiliquefactionMeasures 7.15 Studies on Use of Gravel Drains IllustrativeExamples References Practice Problems

8.

9.

.

GENERAL PRINCIPLES 8.1 General 8.2 8.3

Types of Machines and Foundations General Requirements of Machine Foundation

8.4 8.5

Perimissible Amplitude Allowable Soil Pressure

8.6 8.7

Permissible Stresses of Concrete of Steel Permissible Stresses of Timber References

FOUNDATIONS OF RECIPROCATING MACHINES 9.1 General 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

10.

Modes of Vibrationof a Rigid FoundationBlock Methods of Analysis Linear Elastic Weightless Spring Method Elastic Half-space Method Effect of Footing Shape on Vibratory Response Dynamic Response of Embedded Block Foundation Soil Mass Participating in Vibrations Design Procedure for a Block Foundation Illustrative Examples References Practice Problems

FOUNDATIONS OF IMPACT TYPE MACIDNES 10.1 General

11.

340-351 340 340 347 348 349 349 350 351 íëîóìîî 352 ~

352 353 354 370 392 394 400 402 408 419 . 420

ìîíóììî 423

Design Procedure for a Hammer Foundation

426 432

Illustrative Examples References Practice Problems

436 442 442

10.2 DynamicAnalysis 10.3

.

OF MACIDNE FOUNDATION DESIGN

.

FOUNDATIONS OF ROTARY MACHINES 11.1 General 11.2 Special Considerations 11.3 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 design of foundations namely (i) foundations subjected to static loads and (ii) foundations subjected to dynamic loads. The characteristic feature of a static load is that for a given structure the load carried by the 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 as static load, The characteristic feature of a dynamic load is that it varies with time. Dynamic loads on foundations 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 is different 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 through the sources mentioned above. . '

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High frequency predominates (d) Trice ofvertical acceleration of ground due to pile driving .

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Vibrations of earth's surface caused by waves coming from a source of disturbance inside the earth are described as Earthquakes and are one of the ri1ostdestructive forces that nature unleashes on earth. When, at any depth below tile gro~d surfa~e,the strain ene~gy'ac~~ulated due to deformations in earth mass exceeds the resilience of the storing material, it gets release through rupture. The energy thus released is propogated in the form of waves which impart energy to the media through which they pass and vibrate the structures standing on the earth's..surface. The point inside the earth mass where slipping or fracture begins is termed as focus and the point just above the focus on the earth's surface is termed as epicentre. The position of the focus is determined,with the help of seismograph records (Fig: 1.2]'u:ti't'ising the average velocities of different waves and time difference in reaching the waves at the ground surface. Figure 1.3 explains the various terms in simple manner.

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Fig. 1.2. : A typical earthquake

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1.2.1 Intensity. The severity of shaking of an earthquake as felt or ob!jervedthrough damage is'described as intensity ata certain place on an arbitrary scale. For this purpose modified Mercalli scale is more common in use. It is divided into 12 degrees of intensity as presented in Table!.L Table 1.1 : Modified MereaIli Intensity Scale (Abridged} Classof Earthquakes

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

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Felt only by a few persons at rest, specially on upper floors of buildings; and delicately suspended objects may swing.

fII

Felt quite noticeably indoors, specially on upper floors of buildings but many people do not recognize it as an earthquake; standing motor cars may rock slightly, and vibration may be felt like the passing of a truck.

IV

During the day felt indoors by many, outdoors by a few; at night some awakened,dishes, windows, doors disturbed, walls make cracking sound, sensation like heavytruck striking the building; and standing motor car rocked noticeably.

V

Felt by nearly everyone; many awakened; some dishes, windows, etc. broken; a few instances of cracked plasters; unstable objects overturned; disturbance of trees, poles and other tall objects

noticed sometimes and pendulum clocks may stop.

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VI

Felt by all; many frightened and run outdoors; some heavy furniture moved; a few instances of fallen plaster or damaged chimneys; damage slight.

VII

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

VIII

Damage slight j!, spe~ially designed structures; considerable in' ordinary substantial buildings with partial collapse; very heavy it) poorly built structures; panel walls thrown out of framed structure; heavy furniture overturned; sand and mud ejected in small amounts; changes in well water; and disturbs persons driving motor cars. .

IX

Damage considerable in specially designed structures; well designed framed structures thrown out of plumb; very heavy in substantial buildings with parti~1collapse; buildings shifted off foundations; ground cracked conspicuously; and underground pipes broken.

X

Some well built wooden structure~ destroyed; most masonry and framed structures with foundations destroyed; ground badly cracked; rails bent; land-slides considerable from river banks and steep slopes; shifted sand and mud; and water splashed over banks.

XI

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

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Total damage; waves seen on ground surface; lines of sight and '" , lever distorted; and objects thrown

upward into the air.

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1.2.2 Magnitude. Magnitude of an earthquake ,.._" is a measure of the size of an earthquake, based on the -.-". . ~"."""""", ,.,..,.._~,-,, ".. -',", amplitude of elastic waves it generates. Richter (1958) suggested the following relation. M = loglOA -loglO Aa . ...(1.1) where M = Magnitude of earthquake A = Trace amplitude in mm (Fig. 1.2) Aa = Distance correction (F:ig.1.4) ,. A relationship between strain energy released py an earthquake and its magnitude is given by Richter (1958) as follows ~

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= 11.4 +

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1.5 M

where

E = Energy released in earthquake in Ergs A comparison of the magnitude M of an earthquake with maximum i tensity of the Modified Mercalli 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, Modified Mercalli Scale

2

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3

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XI

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

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Table 1.3 : Fault Length, Affected Area,~nd Duration of.Eart~quake Magnitude oj Earthquake (Richter scale)

Fault Length"

5

1-2

6

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,(km)

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25-50 .

8

>250

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20,000

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1.3 EQUIVALENT DYNAMIC LOAD TO AN ACTUAL EARTHQUAKE LOAD Figure 1.1 (a) shows the variation of dynamic load wi+htime observed during El Centro earthquake. The loading is not periodic and the.peaks in anY two cycles are different. For the analysis and design of foundations such a random variation is converted into equivalent number of cycles of uniformly varying load [Fig. 1.1 (b)]. It means that the structure-foundation-soil system subjected to Ns cycles of uniformly varying load will suffer same deformations and stresses as by the actual earthquakes. Most of the analyses and laboratory teSting are 'carried out using this concept.' . According to Seed and Idriss (1911), the average equivalent uniform acceleration is about 65 percent of the maximum acceleration. The number of significant cycles, Ns depends on the magnitude of earthquake. They recommended the values ofNs as 10, 20 and 30 for earthquakes ofmagnitudes 1, 1.5 and 8 respectively. Lee and Chan (1912) suggested the following procedure for ..converting the irregular stress-time history to the equivalent number of cycles of cyclic shear stresses of maximum magnitude equal to K 'tmax'

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a constant less th~.1JP.ity :

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(i) Let Fig. 1.5 shows ~}Yl'icalea~CJ.uakereco.rd.Divide the s;t;essrange (0 to 'tmaX> or acceleration range (0 to amax)into convenient number ofleveIs and note the mean stress or mean acceleration within' each level as mentioned in column no. 2 of Table 1.4. Then the number of cycles with peaks 'Yhichfall within each of these levels is counted and recorded. Note that because the actual time history is not symmetric about the zero stress axis, the number of peaks on both sides are counted and two peaks are equivalent to one cycle. For example, an earthquake record shown in Fig. 1.5 has number of cycles in various ranges of acceleration levels as listed in Col. 3 of Table 1.4. Om ox :; + 0 . 12

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a plot 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 equivalent number of cycles for K . 'tmax'Referring to this curve (Fig. 1.6) determine the conversion factor to each average stress level (Col. 4 of Table 1.4).

(ii) Seed et al. (1975) gave

(iii) Determine the equivalen~number of unifofm cycles,at a maximum stress level of 0.65 'tma.xby multiplying the values listed in Cols. 3 and 4. 'These are listed in Col. 5. , , . . . t ,

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Table 1.4 : Equivalent Cycles for Anticipated Eartbquake Acceleration level il/

Average and level in

percent of

percent of

(I)

(2)

100- 80

Number

Conversion

of. cycles

factor

(3)

Equivalent number of cycles at 0.65 1"ma:c

(4)

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(5)

90 70

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2.6 1.2

1.8

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0.20

0.70

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50 30

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negligible

0.0

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10

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For getting the equivalent number of cycles for 0.75 'tmax'read the yalue of conversion factor (Fig. 1.6) corresponding to an ordinate value of 0.75. It comes out as 1.5. The value of equivalent number of cycles obtained for 0.65 'tmaxas illustrated in Table 1.4 is divided by this conversion factor to obtain equivalent number of cycles corresponding to 0.75 'tmaxi.e. 9.0/1.5= 6.0 cycles. Seed and Idriss (1971) and Lee and Chan (1972) developed the above concepts specifically for liquefacti~mstudies. More details of these procedures have been.discussed in Chapter 7.

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1.4 SEISMIC FORCE FOR PSEUDO-STATIC ANALYSIS For the purpose of determining seismic force, the country is classified into five zones as shown in Fig. 1.7. Two methods namely (i) seismic coefficient method and (ii) response spectrum method are used for computing the seismic force. For pseudo-static design of foundations of buildings, bridges and similar structures, seismic coefficient method is used. For the analysis of earth dams and dynamic designs, response spectrum method is used (IS- 1893 : 1975).

Equivalent modi fide

Q) Iv.

mercalli intensity IX and above Vl1I VII VI Less than VI

Bombay

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method, the design value of horizontal

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= ~I ao

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= Basicseismiccoefficient,Table 1.5

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

~ = Coefficient

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

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The vertical seismic coefficient,
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In 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 computed using the following expression: Tab{e 1.5 : Values of Basic Seismic Coefficien~ Zone No.

ao

V IV III

0.08

11

0.02 0.01

0.05 0.04

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Table 1.6 : Values of Importance Factor, I S No.

Va/lIeoJI

Type oJ Structure

I. 2.

Containment structure of seismic power reactor for preliminary design

3.0

Dam~ (all types)

2.0

3.

Containers of inflammable or poisonous gases or liquids

2.0

4.

Important service and community structures, such as hospitals, water towers and tanks, schools, important bridges, emergency buildings like telephone exchange and fire brigades, large assembly structures like

5.

\.5 1.0

cinemas, assembly halls and subway stations All others

.Table 1.7 : Values of P for Different Soil-Foundation System Values off3Jor -----------------------------

Isolated footings without tie beams

Combinedor isolated footings with tie beams

Raft foundation

Rock or Hard soil

\.0

1.0

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1.0

1.0

Medium soils .Soft soils

1.2

1.0

1.0.

1.0

1.2

1.5

1.2

1.0

1.2

1.5

Soil

Pile foundation

Well foundation

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

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Table 1.8 : Values of Seismic Zoning Factor. Fo Zone No.

Fo

V

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IV

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III

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U

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Foundations

tILL USTRATIVE EXAMPLES' Example 1.1 The srandard torsion seismograph recorded an average trace amplitude of 8.0 mm. The distance to the epicentre is. estimated about 100 km. Determine the magnitude of earthquake. Solution: From Fig. 1.4, the distance correction for 100 km is 3.0. Hence,

M = 10glO8.0 + 3 = 3.9

ÎÛÚÛÎÛÒÝÛÍ HousnCf, G. W. (1965), "Intensity of earthquake ground shaking near the causative fault", Proceedings 3rd World Conference on Earthquake Engineering, New Zealand, Vo\. 1. Housner. G. W. (1970), "Design spectrum", in EarthquukeEngineering Cliffs, New Jersey, pp. 97-106.

(R. W. Wiegel, Ed.), Prentice-HalI, Englewood

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", Proceedings, 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 Engineering Research Center, University of California, Berkeley. Seed, H. B., and Idriss, 1. M. (1971), "Simplified procedure for evaluating soil liquefaction J. Soil Mech. Found. Engg., ASCE, Vo\. 97, No. SM9, pp. 1249-1273.

F

potential"

PRACTICE PROBLEMS

1.1 Explain the terms 'Intensity' and 'Magnitude' irt relation to earthquake. How are fault length an, duration of earthquake depend on magnitude? 1.2 Describe a method of getting equivalent number of cycles of uniformly varying load for an actur earthquake record, 1.3 Determine the equivalent number ef cycles for 0.75 Tmaxfor El Centro earthquake. DC

"".'\'-<-,'~.",:~,,",:;""-',/'.'~',e:

,",-

:",:;""

:..'

,'",""

!!JI!!'!!'

'..".'

:R'

THEORY OF VIBRATIONS

2.1 GENERAL In order to understand the behaviour of a structure subjected to dynamic load lucidly, one must study the mechanics of vibrations 'caused by the dynamic load. The pattern of variation of a dynamic load with respect to time may be either periodic or transient. The periodical motions can be resolved into sinusoidally varying components e.g. vibrations in the case of reciprocating machine foundations. Transient vibrations may have very complicated non-periodic time history e.g. vibrations due to earthquakes and quarry blasts. A structure subjected to a dynamic load (periodic or transient) may vibrate in one of the following four ways of deformation or a combination there-of: (i) Extensional (Fig. 2.1 a) (ii) Shearing (Fig. 2.1 b) (iii) bending (Fig. 2.1 c) (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 the system. To study the response of a vibratory system, in many cases it is satisfactory to reduce it to an idealized system of lumped parameters. In this regard, the simplest model consists of mass, spring and dashpot This chapter is framed to provide the basic concepts and dynamic analysis of such systems. Actual field problems 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 the position of a system during vibration, is called degrees of freedom (Fig. 2.2).

D:

~

m

(a) One degree of freedom

(b) Two degrees offreedom

KI

Z,

.

Z2

- -, Z)

. .~

~ óååó¢¢ ¢óãÝí -

- 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

15

Theory of Vibrations

2.2.3 Periodic Motion: If motion repeats itself at regular intervals of time, it is called periodic motion. 2.2.4 Free Vibration: If a system vibrates without an external force, then it is said to undergo free vibrations. Such vibrations can be caused by setting the system in motion initially and allowing it to move ~~~~~. . 2.2.5 Natural Frequency: This is the property of the system and corresponds to the number of free oscillations made by the system in unit time. 2.2.6 Forced Vibrations: Vibrations that are developed by externally applied exciting forces are called forced vibrations. These vibrations occur at the frequency of the externally applied exciting force. 2.2.7 Forcing Frequency: This refers to the periodicity of the external forces which acts on the system during forced vibrations. This is also termed as operating frequency. 2.2.8 Frequency Ratio: The ratio of the forcing frequency and natural frequency of the system is referred as frequency ratio. 2.2.9 Amplitude of Motion: The maximum displacement of a vibrating body from the mean position is

amplitudeof 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}cyof excitation coincides with anyone of the natural frequencies of the system, the condition of resonance occurs. The amplitudes of motion are very excessive at resonance. 2.2.12 Damping: All vibration systems offer resistance to motion due to their own inherent properties. This resistance is called damping force and it depends on the condition of vibration, material and type of the system..If the force of damping is constant, it is t&med Coulomb damping. If the damping force is proportional to the velocity, it is termed viscous damping. If the damping in a system is free from its material property and is contributed by the geometry of the system, it is called geometrical or radiation damping. 2.3 HARMONIC MOTION Harmonic motion is the simplest form of vibratory motion. It may be described mathematically by the following equation: ...(2.1) Z = A sin (rot - 0)

N

L r-

T:2!!Go) Timq.t '.

.,

c Fig. 2.3 : Quantities describing harmonic motion

;.

16

:'~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 as follows: Z = Displacement of the rotating mass at any time t A = Displacement amplitude from the mean position, sometimes referred as single amplitude. The distance 2 A represents the peak-to-peak displacementamplitude,sometimesreferred to as double amplitude, and is the quantity most often measured from vibration records. ro = Circular frequency in radians per unit time. Because the motion repeats itself after 21tradians, the' frequency of oscillation in terms of cycles per unit time will be ro/21t.It is denoted by f 8 = Phase angle. It is required to specify the time relationship between two quantities having the same frequency when their peak values ha'ving like sign do not occur simultaneously. In Eq. (2.1) the phase angle is a reference to the time origin. More commonly, the phase angle is used as a reference to another quantity having the same frequency. For example, at some reference point in a harmonically vibrating system, the motion may be expressed by ...(2.2) ZI = AI sin rot Motion at any other point in the system might be expressed as Z,I

with

= A,I sin ( rot-'e,

1t ~ 8 ~

...(2.3 )

)

I

- 1t.

For positive values of 8 the motion at point i reaches its peak within one half cycle after the peak motion occurs at point 1. The angle 8 is then called phase lag. For negative values of 8 the peak motion at i occurs within one half cycle ahead of motion at 1, and 8 is called as phase lead. The time period, T is given by 1 21t T=-=f ro The velocity and acceleration of motion are obtained from the derivatives of Eq. (2.1.). dZ . Velocity = -dt = Z = roA cos (rot- 8)

Acceleration

= roA sin (rot- 8 + ~) 2 d Z .. 2 = -rdt = Z = ro A sin (rot- 8)

...(2.4)

...(2.5)

...(2.6)

= 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 vector representation of harmonic displacement, velocity and acceleration is presented considering the displacementas the referencequantity(8 = 0).

, .J

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

",C.. ., .,~;< r'l!\"k. . ',", . ,~Ii

17 N .. +' C ~

z,z,z

E ~

TimtZ,t

v 0

a. UI 0

oN

... >+'

Ti mtZ,t

v 0 ~

>

Timcz,t

c 0

0

.... c:,I c:,I

v 0

.et Fig. 2.4: Vector representation

of harmonic displacement. velocity and acceleration

When two harmonic motions having little different frequencies are superimposed. a non harmonic motion as shown in Fig. 2.5 occurs. It appears to be harmonic except for a gradual increase and decrease in amplitude. The displacement of such a vibration is given by: Z = AI sin (0011- 91) + A2 sin (0021- 92)

..

-

D,

N .,/

...(2.7)

2A max

2Am\n

./

.,/

+' C c:,I

E

TimtZ (t)

,~

v 0

a. III

c

,--.'J'

---

'-'" ,.,

~T , b

"""'-

~ "

:'

3!

j,;I',:

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

---

','"

;;"

C,"

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

18


The dashed curve (Fig. 2.5), representing the envelop of the vibration amplitudes oscillates at a frequency, called the beat frequency, which corresponds to the difference in the two source frequencies: I 1<01 -<021 fb = Tb = 21t ...(2.8) The frequency of the combined oscillations is the average of the frequencies of the two components and 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 of the 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 SYSTEM 1

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

Z - Dj splac(Zment

Z-

V(Zlocity

Z-

Ac c(z l(Zration

c KZ+ Cl

.. mz -,

----

LZ

-

+1

- -

m m f F(t)

(I) Spring-mlss-dashpot

system

(b) Frcc-body diagram

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

. .-.:..,

~

.

..,...'

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

...

.,......

-'-

., >-":,

,;;[; /,

'1\", ",

;,.,

"',c,...'"

"-":

,,'

,.'r:,/';

~:.:

"'1~F"",';

19

Theory of Vibrations . ,

,

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 the

spring constant and indicates the stiffness. This force always acts towards the equilibrium position of the system. (iii) Damping force, Fi The damping force is considered directly proportional to the velocity and given by C . Z where C is called the coefficient of viscous damping; this force always opposes the motion. In some problems in which the damping is not viscous, the concept of viscous damping is still used by defining an equivalent viscous damping which is obtained so that the total the energy dissipated per cycle is same as for the actual damping during a steady state of motion. (iv) Inertia force, F.: l It is due to the acceleration of the mass and is given by m Z. According to De-Alemberfs principle, a body which is not in static equilibrium by virtue of some acceleration which it possess, can be brought to static equilibrium by' introduculg on it an inertia force. This force acts through the centre of gravity of the body in the direction opposite to that of accelera,

tion.

"

'

The equilibrium of mass m gives mZ + CZ + KZ = F (t) 2.4.1 Undamped

Free Vibrations.

,

For undamped free vibrations, the damping force and the exciting

force are equal to zero. Therefore the'"equation of motion of the system becomes m Z + KZ = 0:

.."

...(2.12a)

'

.. K Z+mZ=O (

)

or

.::(2.11)

,

which is the equation of motion of the system.

...(2.12b)

The solution of this equation can be obtained by substituting"

Z = A I cos

con

...(2.B)

t + Az sin cont

where 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

~- .

"

or

'co ,,'

,

"

,-,'

n

=:1:

,

...(2.14)

m

The values of constants A I and A2 are obtained by supstituting proper boundary conditions. We may

nave the following two boundary conditions: '" . (i) At time t

' "' ' ~

= 0, displacement Z = Zo' and

Z = V0

(ii) At time 1 = '0, velocity

Substituting the first boundary condition in Eq. (2.13) . "/, ',,'

Now,

;

"'"..',',

"'

Z :.

"""':.!'",I;'j,d",.

,',:'}..,

:';"h'

,,',",

,-"""".."Ar-r;"'O:iI'i),+.'nji;~:J}'i"..ql.d")Jiti..j}iJ'iI.J'!,';~";

':,; 'z ,=: -:' AI"

00,;

si~ cont + A2 C1)n'~os cont

,

"

:!':"'"

>is:.:,,,,

'"

'

C.

'

:; ,

"

... ( 2.15 ) j

...(2.16)

20

Soil Dymunics & Machine Fo"ndations

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

A

Hence

2

=--2..

...(2.17)

~n

Vo

.

-con

2 = 20 cos oont +.

Now let.

...(2.18)

sin oont

20 = Az cos 9 V --2.. = AZ sin 9 co

and

...(2.19) ...(2.20)

n

Substitution of Eqs. (2.19) and (2.20) into Eq. (2.18) yields 2 = Az cos (oont - 9) where

9

...(2.21 )

~

= tan-I

...(2.22)

( con20) 2 2

Az = ,/20 + .

Vo

-

...(2.23)

( con )

The displacement of mass given by Eq. (2.21) can be represented Fig. 2.7. It may be noted that

graphically

as shown in

c+)

~

One cycle

Acceleration

\

+Az

y,%

""

:N

0'

1',"

..

/.-0,

/.

. '

3

oN .. N

'

\

e\ --

"~/. /

, "2~\

/ 2

/

/

0

,/ TI.r -A

Z

"IT.~,

9

'\'.

2lT +9

,

'-

/ 0

-'"

V"

/ 0

0 isplacement

Time,t

/

"

1'/

velocity

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

21

'reory ilf Jl"l6iatiOns

At time t equal to

Displacement Z is

0 8

Az cos 8

(J)n

Az

..

1t +8

L-

0

0) n 1I+8

-AZ

0)

3 -1t+8 2

.

0

(J)n

21t

+8 AZ

O)n

It is evident from Fig. 2.7 that nature of foundation displacement is sinusoidal. The magnitude of maximum displacement is Az. The time required for the motion to repeat itself is the period of vibration, T and is therefore given by. . T

=

21t O)n

...(2.24)

The natural frequency of oscillation, J. =1- =~ T

n

W -mg =-=0 K K

Now Where

g

1"

21t

is given by

=...!..

(K

...(2.25)

21t v-;;

st

...(2.26)

.

= Acceleration due to gravity, 9.81 mIs2

W = Weight of mass m °st

= staticdeflectionof the spring

Therefore

In

-

I

-

21t

rg

...(2.27)

Vfut

Eq. (2.27) shows that the natural frequency is a function of static deflection. The relation ofIn and Os!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.

I, -.

...~

,.,. ~.~

.n I

22

Soil Dynamics & Machine Foundiuions .

40

30

,....

-

N :I: 20

c

.....

'-..

10

00

---4

2 .

6 6stat

8

10

(mm)

Fig. 2.8 : Relationship between natural frequency and static deflection

2.4.2 Free Vibrations

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

force Fo sin (J)t on the system is zero), the differential equation of motion can be written as mZ + Cl + KZ = 0 ...(2.28) where C is the damping constant or force per unit velocity. The solution of Eq. (2.28) may be written as '),.t . .

Z

=A e

...(2.29)

where A and A are arbitrary constants. By substituting the value of Z given by Eq. (2.29) in Eq. (2.28), we get m A A2it + C A AIt + K A it = 0 C K 2 or A +

By solving Eq. (2.30)

()

...(2.30)

ni A + m = 0

C

F) C

2

K

= - 2m :i: V~~) -;; The completesolutionof Eq.(2.28) is givenby .

A,1,2 .

Z -- A le Alt + A 2e A2t

...(2.31 )

...(2.32) 2 The physical significance of this solution depends upon the relative magnitudes 'of (C/2m) and (K/m), which determines whether the exponents are real or complex quantities. '

Case I :

'

(~2m) > Km 2

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

:~

Theory of Vi!'rations

23 ,

by the damping force that there is sufficient kinetic energy left t~ carry the mass and pass the equilibrium position. Physically this means a relatively large damping and the system is said to be over damped,

z

2 C > 4 km

...

-"

Tim(l,t

Fig. 2.9 : Free vibrations ofviscously overdamped system

Case 11 :

(~ ) = K 2

2m

m

,-

The roots Al and Az are equal and negative. Since the equality must be fulfilled, the solution is given by

Z = (AI.+ Az t) le = (AI + Az t) e-Ct/Zm ...(2,33) In this case also, there is no vibratory motion. It is similar to oyer damped case except that it is possible for the sign to change once as shown in Fig. 2010.This,case is of little importance in itself; it assumes greater significance as a measure of the damping capacity of the system.

"

'

z c2=l"km

Time,t

.,

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

When

damped system

(~2m) = K.m' C = C

...(2.34)

c

Then Cc "=,2 ~Km". ...(2.35) The system in this conditioon is known as ~ritically damped system anaC ~ is known as critical damping constant.' The ratio of the actual damping constant to the critical damping constant. is. defined as damping ratio: .. C

~=-

'Now

,..(2.36)

Cc

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

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

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

...(2.37)

-

,

.. ".'

r

, .

[.',

24


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

Case III :

(~ ) 2m

Foundations

...(2.38)

COn

2 < K

m

The roots Al and Al are complex and are given by

.

...(2.39)

= [-;:!:i~I-;2 ]COn The complete solution of Eq. (2.28) is given by AI,2

(-~+j~I-~2 )O>i (-~-j~l-e )(J)"I Z - AI e + Az e r:-:2 r , Z = e-~o>"t A j"I-~2 0>,,1+ A e-;~I-~- O),i'

or

the

Eq. (2.41) can be written as

I

e

z

Z = e-~O)", [Cl sin( (J)n~ or

Z = e-~O)II' [Cl sin(J)ndt+CZ

where

wild

t)+Cz cos( (J)1I~t)]

...(2.40) ...(2.41 )

...(2.42) ...(2.43 )

COS(J)ndt]

= (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.

z

2

C < 41
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 successive cycles of vibration

e-0> "f,1

Z ---L Z2

or or or

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

-Z I

= en0> f,.21t/o)ncl Zz r:2 -ZI =e Znl;!"l-f,Zz ZI loge 22.-

21t;

~

...(2.44 a)

...(2.44 b) ...(2.44 c) ...(2.44 d)

~ },

",inF1':

j'/,"

" 't~..

' .~t

.o,~;

',.

,-:)

Tlreory of Vtb",tiolU

. . Natural logarithm mk. ,decrement.

of ratio of two successive

peak

amplitudes

{i,e,

log,

(~)} is called as logarith.

1

Z\

r:-:2

~= 2x loge ~ ' As for small valuesof~, V1- ~-

or

::

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 for a number of cycles, say n. In such a case, if Z" is the peak amplitudes of the n,h cycle, then Zo Zl Z2 Zn-I 0 - = - =- =...== e where~ = 2x ~ .'

Z\

Z2

Zo, Zn

Therefore,

~ = -n

Z 0 oge Zn

I

1

or ~}:

Zn

Z2 = Zo ~ .. Z"-I = eno [ Z, ] [ Z2 ] [ Z) ] [ Z" ] 1

Hence

ZJ

= -2xn

I

..

...(2.441)

Z0..

...(2.44 g)

oge.z n

Therefore, a system is over damped if ~ > 1; critically damped if ~ = 1 and under damped if ~ < 1. 2.4.3 Forced Vibrations Of Single Degree Freedom Syst~m. In many cases of vibrations caused by rotating parts of machines, th~ systems are subjected to periodic exciting forces. Let us consider the case of a single degree freedom sys~.:mwhich is acted upon by a steady state sinusoidal exciting force having magnitude F and frequency 0>(i.e. F(t) = Fosin rot). For this case the equation of motion (Eq. 2.11) can be written as : .. . 111Z + C Z + K Z = Fo Sin ro t ...(2.45) Eq-;(2.45) is a linear, non-homogeneous, second order differential equation. The solution of this equation consists of two parts namely (i) complementary function, and (ii) particular integral. The complementary function is obtained by considering no forcing function. Therefore the equation of motion in 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 rot where Z2 = displacement of mass m a~~nYinstant t when vibrating with forcing function.

...(2.48)

~

.;Y,

8111;1' "

"--H,'

26

"

"

Soil Dynamics & Mac/line Foundations

;

The,solution

of Eq. (2A8).,\~

,,',

"

.

gi'{en by'.

~

',",

,,'

" ,

'"

t,;

"

22 = AI sin 00t + A2cos 00't where AI and A2 are two, arbitrary constants.

...(2.49)

,

"...~ ,'"

.

. .:' , "

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':.

:

,

Considering .sine and Cosine functions in Eq. (2.50) separately, 2

(-

"

,;

"

, ,

'

(,.

'

','

",

,"~

,

'

'

,..'(2.50: ,

,'.,

.

. ...(2,51 a)

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),

,

",'

,..(2.? 1 b)

.,

.

Al(~ - o}) - A2(~ and from Eg. (2,51 b) A{~-W

)

F.

m =

)+A2(~-w2)

0,

Q.

:

",(2,52

0

',n

a)

...{2.52 b)

=0

Solving Egs, (2,52 a) and (2.52 b), we get (K-moo2) Fo

At- -

'.)2.53

0..

2

a)

(K - mm2) + C2m2 "

and

A2 =

- CmFo

...(2,53 b) t

"

".

'
"',, >,

. '

(K-~~2)2+c"2m2,"."

By substitutingJhe values 'of Al and A~"inEq::2'.49;:' " , ,: , "',.." '

'

"""

~,

,"

','

':22 ':°2"

~

(K - m m ) + Cm" let,

'

'...(2.54)

2 {(K'-mm2)sinmi-cwcosmt} -

tan e =

'C 0)

,

'

.,.(2.55)

K-l~cd2

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

22 =

Fo

...(2.56)

. sin(mt - e) ,,'

~(K~';'m2t+'c2

Eg. (2.56) may be written as

t

. '.ccFo!K

22 =

..

cii1' "

,

,

.

11

-,'

..,

""""'~!i1I"1!",?'

~""

"".",..1d.""

"~~",...""."..",""."""",."""""""",-,,i""'-~-'

,

"I

W

= Frequency ratio =;-

~ = Dam p ing

, ,'!

'sin{~t -"'er

.

,

:,., "

,

~(I:!12)+(2Tl~)~.. Where

:"

n

-"~;

"

..~

. " ...(2.57)

. '"

,,'

,',

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

':Ci"","'G'}""::"',',_:;,~'-',

c c .:;= 0"'2' ';JKi{i 'Km

ratio = -

,

'

.'

'

.

~J'(,,~\'k

27

Theory of Vibrations ,'..,

. .

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 the motion 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~ven by F. /K 0

Az =

",

. .

'.

"

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

...,

N

..,

...(2,58)

oi

;;"

'~','--'

,~

'

"I +' C ~

E ~ u 0

a. UI ,.,..

0

....

,

Transi~nt

....

211"

~

(;)

~

,

N N .. +' C ~

Time.t

E

.

~ u 0

a.

"

-,

'h

UI

..'

:'1 ." ..i

0

;. . . /' .

'.'. -' ,.;"

"

'---'-

N +' C ~

'I\

"

,-

#"

"

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

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

"

\

-; ..i ,

~,'

,

:~~.::

. Time,t

E

c:.I u 0

a. III

'..,

q

0

. "! ".'

" .

, ','" .), < ,"

'.', ,',

"

.

Co mpl~t~

solu tion

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

i,

-

~ rr"'-

"'-T--'1iiiiii[-'

~-

"

.".

,(

'.

~>;;;<~i:«.~..,'

, ""

;:

SoU Dyrul",ks

28

& Maehi"e

Foundlltions

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

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

.

--

...(2.59) -

The variation of f.1versus 11is shown in Fig. 2.13 for different values of damping ratio ~.It would be seen that near 11= 1, the value of f.1is maximum. This is called resonance and the forcing frequency J at which it occurs is called the resonant frequency. 5

~=o t.

... 0

I

1

I

I

"0.1

3

u

c

....

c

...c

11

0)

0

u .....

.-

2

c Cl

c

1

0 0

1.5 Frc&quc&ncy

ratio.

"\.

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

"0

1

29


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

...(2.60 a)

11 = ~1-2~2

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

where

...(2.60 b)

ffind

= ffin ~1-2~2

ffind

= Damped resonant frequency

180°

150°

120°

c:.J C'I C 0

c:.J

r =0.707

90°'

\11

~ = 0-5

0 .J: a..

600

°

30

0 0

2.0

'1.0 FrczquQncy

ratio#

-rz.

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

3.0

<'.,

30

~

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


Foundations

'By substituting Eq. (2.60) in Eq. (2.59), the maximum value of magnification factor is obtained. It is given by , I ...(2.61 ) J.lmax =

2E,~I-E/ I - 2E,

(For small values of;)

...(2.62)

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 which should be avoided. .

'.

---

-

-

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

.

_

211E,

e = Tan-I 1- 2 ( T\ ) '"

...(2.63 )

.

Variation of e with respect to 11is shown in Fig. 2.14 2.4.3.1 Rotating mass type excitation. Machines with unbalanced rotating masses develop alternating force as shown in Fig. 2.15 a. Since horizontal forces on the foundation at any instant cancel, the net vibrating force on the foundation is vertical and equal to 2 me ero2sin rot, where me is the mass of each rotating element, placed at eccentricity e from the centre <,>f rotating shaft and ro is the angular frequency of masses. Fig. 2.15b shows such a system mounted on elastic supports with dashpot representing viscous damping. .

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 -. 2 m Z +. C. Z- + K Z = 2 111ero sin rot e.

..

...(2.64)

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

'.';

:~,(i:',

'.'),'

'<'F

'" ".

,."

',', ,

".

.(\~.tJi;"":

31


Z = A; sin (0)t + 0') (2mee/

where Az

=

m)'T}2

I

F=2m 0

e

.eO)

F

or

..(2.66)

2

(1-T}2)

Since

...(2.65)

.

2

+(2~T}) 2

(J)

2

K = 2 me . e K

0)2

= 2 me . e

e = Tan-I 2T}\ 1-11

(

= (2 me :}T}2

(mro~)

...(2.67)

)

3.0 0.10

2.0 .---.

';Iile 1.0

00

1..0

,2.0,

'

Frequency

'. 3.0 ratio.

5.0

4.0 1)

.

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

180°. '(D ... aI C7I c: 0 90° aI

0.05 I

0.50 0.25'

U\

0

.s::.

a..

0° 0

1.0

2.0

3.0

4.0

Frqquqncy ratio., - 't (b) Phase angle versus frequencYT&tio11 Fig. 2.16 : Response,oh system with rotating unbalance

5.0

32

Soil Dynamics

& Machine Foundations

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

1"\2

/

(2mee m)

= ~ (1-1"\2)2+(21"\1;)2

...(2.68)

The value of A=/(2me elm) is plotted against frequency ratio 1"\in Fig. 2.16 a. The curves are similar in shape to those in Fig. 2.13 except that these starts from origin. The variation of phase angle e with 11 is shown in Fig. 2.16 b. Differentiating Eq. (2.68) with respect to 11and equating to zero. it can be shown that resonance will occur at a frequency ratio given by 1 ...(2.69 a) 1"\=-

F-2e 0011

ro =-

or

...(2.69 b)

nd .JI=21;2 By substituting Eq. (2.69 a) in Eq. (2.68), we get .

Az

-

( 2meelm ) max -

l' 21;~1-1;2

~

2.5 VIBRATION

...(2.70)

2\

for small damping

...(2.71 )

ISOLATION

In case a machine is rigidly fastened to the foundation, the force will be transmitted directly to the foundation and may cause objectionable vibrations. It is desirable to isolate the machine from the foun'dation through a suitably designed mounting system in such a way that the transmitted force is reduced. For example, the inertial force developed in a reciprocating engine or unbalanced forces produced in any other rotating machinery should be isolated from the foundation so that the adjoining structure is not set into heavy vibrations. Another example may be the isolation of delicate instruments from their supports which may be subjected to certain vibrations. In either case the effectiveness of isolation may be measured in terms of the force or motion transmitted to the foundation. The first type is known as force isolation and the second type as motion isolation.

2.5.1 Force Isolation. Figure 2.17 s~ows a machine of mass m supported on the foundation by means of an isolator having an equivalent stiffness K and damping coefficient C. The machine is excited with unbalanced vertical force of magnitude 2 me eci sin 00t . The equation .ofmotion of the ~achine can be written as: ... 2 m Z + C Z + KZ = 2 me eoo sin 00t

...(2.72)

The steady state motion of the mass of machine can be worked out as 2m

Z =

w~ere

eoo

2

/K

.

e2 .sm(oot-8) (1-'12). +(21"\1;)2

r

8 = Tan-I

2'11;2

[ 1-.1"\ ]

...(2.73)

...(2.74)

':;0:"<\',:,-;'

'-'

,'t

"t"

"',"','

'1

33

Theory of Jlibrlltions ;;::

--'-~,

' ~~~

K

Ma chine

2'

Iso lata r

K 2

c

Ground surfac~

Foundation

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 + CZ

..,(2.75)

Substituting Eq. (2.73) in Eq. (2.75), we get 2

.

2m em

F =

. sm(mt- e)+

e

t~

(1-1l) 2 +(211~)2

C.2me em2 /K '00 cos(mt-e) 2 2

(1-112)

,

+(211~)

...(2.76) Equation (2.76) can be written as: F t = 2me em

2

Jl+(211~)2 ,

2

. (1-112) where

2

. sin(mt - P)

...(2.77)

+(211~)

p is the phase difference between the exciting force and the force transmitted to the foundation and

is given by

,

i

'

-

P ~-e --=~-~~-l '" , r",[ ;"":

','"o/,,'n1.."n"

coo

-]

K, J

...(2.78) ~

','

i

J

" ".

34

Soil Dynamics

& Machine. Foundations

Since the force 2 m e e ol is the force which would be transmitted if springs were infinitely rigid, a measure of the effectiveness of the i~olation mounting system is given by .

' Ft

.

~1+(211~)2

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

transmitted to the foundation will be less than the exciting force. However in this case, the presence of damping reduces the effectiveness of the isolation system as the curves for damped case are above the undamped ones for 11>12. A certain amount of damping, however, is essential to maintain stability under transient conditions and to prevent excessive amplitudes when the vibrations pass through resonance 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 .

f

~ =0

~ =0

=0.125

~ =0.125

1- 3.0 =<. » ....

.-

oD U\

.-U\ E

2-0

III C 0 ... I-

~ =0.5 ~ =1.0 ~ =2.0 1.0

~ 0

: 0.125'

I

~ =0

-

0

1.0

Frczquczncyr(:itio Fig. 2.18: Transmissibility

3.0

2.0 I

1'\.

(J.1-r)versus freqeuncy ratio (Tt>

,f

"c"'"

f

','"

,,': ,r:,,<"f:'

';:£'4/

:~";

tt~~7T{~:,';:s:",.:';

l',,'

35

Theory of Vibrations , . .

:

';'

,

'

'

'

2.5.2. Motion Isolation. In many situations, it would be necessary to isolate structure or mechanical systems from vibrations transmitted from the neighboring machines. Again we require a suitable mounting system so that least vibrations are transmitted to the system due to the vibrating base. We consider a system mounted through a spring and dashpot and attached to the surface which vibrates harmonically with 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 9 machines 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

h

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

...(2.80)

or

m

or

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

Z + K 2 .+

K Z = C Y .+ K Y = C (I) Y0 cos

(I)

1 .+ K Y0 sin (I) 1

...(2.81)

T -I CO) ex =, an K

ere

...(2.8?)

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

-

Z ,.

max

.

~1.+(21l~)2

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

,~

...(2.83)

,

"

"

. ,,' ",

18E:.

36

SoU Dynamics & Machine Foundations

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

~T - -y;

~

~1 + (2T\~)2

...(2.84)

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

Equation (2.84) is the same expression as Eq. (2.79) obtained earlier. Transmissibilityof such system can also be studied from the response curves shown in fig. 2.18. It is again noted that for the vibration isolation to be effective, it must be designed in such a way that T\>

.fi.

2.5.3. Materials Used In Vibration Isolation. Materials used for vibration isolation are rubber, felt. cork and metallic springs. The effectiveness of each depends on the operating conditions. 1.5.3.1. Rubber. Rubber is loaded in compression or in shear, the latter mode gives higher flexibility. With loading greater than about 0.6 N per sq mm, it undergoes much faster deterioration. Its damping and stiffness properties vary widely with applied load, temperature, shape factor, excitation frequency and the amplitude of vibration. The maximum temperature upto which rubber can be used satisfactorily is about 65°c. It must not be used in presence of oil which attacks rubber. It is found very s' ".,ble for high frequency vibrations. 2.5.3.2.Felt. Felt is used in compressfun only and is capable of taking extremely high loads. It has very high damping and so is suitable in the range of low frequency ratio. It is mainly used in conjunction with metallic springs to reduce noise transmission. 2.5.3.3. Cork. Cork is very useful for accoustic isolation and is also used in small pads placed underneath a large concrete block. For satisfactory working it must be loaded from 10 to 25 N/sq mm. It is not affected by oil products or moderate temperature changes. However, its properties change with the frequency of excitation. 1.5.3.4. Metallic springs. Metallic springs are not affected by the operating conditions or the environments. They are quite consistent in their behaviour and can be accurately designed for any desired conditions. They have high sound transmissibility which can be reduced by loading felt in conjunction with it. It has negligible damping and so is suitable for working in the range of high frequency ratio.

2.6 THEORY OF VIBRATION MEASURING INSTRUMENTS The purpose of a vibration measuring instrument is to give an output signal which represents, as closely as possible, the vibration phenomenon. This phenomenon may be displaceme~t, velocity or acceleration of the vibrating system and accordingly the instrument which reproduces signals proportional to these are called vibrometers. velometers or accelerometers. There are essentially two basic systems of vibration measurement. One method is known as the directly connected system in which motions can be measured from a reference surface which is fixed. More often such a reference surface is not available. The second system, known as "Seismic system" does not 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 vibration phenomena. It consists of a frame in which the mass ~ is supported by means of a spring K and dashpot C. The frame is mounted on a vibrating body and vibrates al~ng with it. The system reduces to a spring mass dashpot system having base on support excitation as discussed in Art. 2.5.2 illustrating motion isolation.

.

.(~

37

Thtory of VibratiOns

z m

c

K

y

=

Yo Sin '->t

Fig. 2.20 : Vibration measuring instrument

Let the surface S of the structure be vibrating harmonically with an unknown amplitude Y 0 and an unknown frequency (0. The output of the instrument will depend upon the relative motion between the mass and the structure, since it is this relative motion which is detected and amplified. let 2 be the absolute displacement of the mass, then the output of the instrument will be proportional to X = 2 - Y. The equation of motion of the system can be written as

m

Z + C (Z - Y) + K (2 - Y) = 0

Subtracting m Y from both sides,

...

...(2.8S)

..

,

2

m X + C X + K X = - m Y = m Y0(0 sin (0 t The solution can be written as

..:.(2.86)

2

...(2.S7) where

X = ~ (1- TJ2)2+ TJ (2TJ~)2 Yo sin «(0 t(0. 11 = = frequency ratio (On

-

~ = damping

and e = tanEquation(2.S7)ca~ be rewrittenas: . X

,ratio

2 TJ~

1

where

e)

( 1- TJ2) ,

= 1)2.J! Y0 sin

;°1 ">;(1..;>'

«(0 t - 8)

(2.,SS)

,

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

ill

38.

Soil Dynamics

2.6.1. Displacement 1121.1=

Pickup. The instrument will read the displacement

& Machine - -." '-" Foundlltions .' . -. .

of the structure directly

if

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 is approximately equal to 180°. Therefore to design a displacement pickup, 1'\should be large which means that the natural frequency of the instrument itself'shou~d be low compared to the frequency to be measured. Or in other words, the instrument should have a soft spring and heavy mass. The instrument is sensitive, flimsy and can be used in a weak vibration environment. The instrument can not be used for measurement of strong vibrations. ,-

-t

\

I

3.0

\0

I

--

I I

. -

-

-"

2 0 ,

.

1 0

.

0 0

3.0

2.0

1.0

.

FrequClncy

ratio,

4.0

5.0

'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 X =2 (J,)n

2

1.1

Yoro sin (rot - 8) .

...(2.89)

'

The output of the instrument will be proportional to the acceleration of the structure if J.1is constant. Figure 2.13 shows the variation of J.1with 1'\and;. It is seen that J.1is approximately equal to unity for small values of 1'\.Therefore to design an acceleration pickt!p, 11should be small which means that th~' " ",.~'n'7""~:"'"""'"

.'1::'

'lr~-r\f

........--.

<~

39


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

2.6.3 Velocity Pickup. Equation (2.88) can be rewritten as 1 X = - TIJ!Y0 (J)sin «(J)1 - 0)

...(2.90)

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 as

1 . .. .1 ~ Yo(J) sm (0)10) .: atTl = 1, f.1= ...(2.91) con 2 ':1 .. . . 2~ Since O)nand ~are constant, the instrument will measure the velocity at 11= 1. X

=-

1

It may be noted that the same instrument can be used to measure displacement, acceleration and velocity in different frequency ranges. ., Xa Y if TI» 1 Displacement pickup (Vibrometer~) .. 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 motions having small frequency of vibration are to be measured. Calibration of these pickups is not simple. Further. 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, acceleration pickups are to be favoured. They are relatively less sensitive and this disadvantage can easily be overcome by high gain electronic instrumentation.. .

.

2.6.4. Design of Acceleration Pickup. The relative displacem~nt between, the mass an~ the support would 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 mechanical quantity is converted into an equivalent electrical quantity is best suited for a~.<;elerationpickups. Electrical gauging offers the possibility of high magnification of ~e signals which are usually weak because the spring is stiff and the displacements are small. The mechanica,l quantity alters either the resistance, or capacitance or the inductance of the circuit which consequently alters the current in the circuit. '

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

'--'i.


40

, (a) Four storeyed frame

,

(c) First mode

(b) Idealisation

(d) Second mode

(e) Third mode

(t) Fourth mode

Fig. 2.22 : A four storeyed frame with mode shapes

Figure 2.22 a shows the frame work of a four storeyed. building. It is usual to lump the masses at the floor levels and the lumped mass has a value corresponding to weight of the floor, part of the supporting system (columns) above and below the floor and effective live load. The restoring forces are provided by the supporting systems. Figure 2.22b shows such an idealization and it gives a four degrees of freedom system. In free vibration a system having four degrees of freedom has four natural frequencies and the vibration of the any point in the system, in general, is a combination of four harmonics of these four natural frequencies respectively. Under certain conditions, any point in the system may execute harmonic vibrations at any of the four natural frequencies, and these are known as the principal modes of vibration. 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 associated with this mode would be the lowest in magnitude compared to other modes. If all adjacent masses vibrate out of phase with each other (Fig. 2.22£), the mode is termed the highest mode of vibration and the 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 freedom. Let Z\ be the displaceuent of mass ml and Z2 the displacement of mass m2' The equations of motion of the system can be written: In t Z\ + Kt ZI + K2 (Z\

-

Z2)

...(2.92)

=0

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

ZI=A\sinro,,( Z2

.

= A2 sin ro" t

...(2.93 ) ...(2.94 ) ...(2.9S)

Substitution of Eqs. (2.94) and (2.95), into Eqs. (2.92) and (2.93), yields:

=0

...(2.96)

- m2 C1);)A2 = 0

...(2.97)

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

-

K2 Al +

(~

+ KJ

.

I

~!'\Wm

ìï

~heory of Vibrations

×¦ô

Æï ÕîøÆÆóÆ´÷

Jz~Z2

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

For nontrivial solutions of

oon

in Eqs. (2.96) and (2.97), 2

or

Kt + K2 -mt ron

-K2

- K2

K2 + K3 - ~ ron

=0

.- f2 9;!)

00: _ Kt + K2 + K2 + K3 O)~+ K) K2 + K2 K3 + K3 K) - = 0

...(2.99)

[

m)

~

21

ml~

]

Equation (2.99) is quadratic in ro2, n and the roots of this equation are: 1/2

ro~= .!. K) +K2:t- K2 ~K3 + 2

[(

m)

"'2

) {(

KI +K2 m)

K2 +K3 2 + 4 K~

~

)

mj

~

}

] ...(2.100)

From Eq. (2.100), two valuesofnatura!~!!e9~~ncies oon)and oon2can be obtained. con)is corresponding to the fIrst mode and COn2 is of the second mode. .,

-

11iiii

..,

42

Soil Dynamics & Machine Foundatiofls The general equation of motion of the two masses can now be written as Z I = A (I) I sin (0nl t + A (2) I sin (0n2 t

...(2.101)

Z 2 = A(I) ...(2.102) 2 sin (0n I t + A(2) 2 sin (0n2 t 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 and

(0 2 Al K2 - K2 +KJ -"'2 ffinJ (i)2K A2 KI+K2-mlffinl 2 (2)' 2 AI K2 - K2 + KJ - "'2 ffin2 (2) 2 K A2 KJ + K2 - m) ffin2 2

...(2.103j

...(2.10{

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: (2.105)

ml Zt + Kt Zt + K2 (Zl - Z2) = Fo sin (0 t 1n2 Z2 + KJ Z2 + K2 (Z2 - Z\)

...(2.106

= 0

F0 sin G.)tl 21

22

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

.. ',",y

43

leory of Vibrations

For steady state, the solutions will be as 21

= Al

...(2.107)

sin 00t

...(2.108)

2z = Az sin 00t Substituting 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 = 0 Solving for AI and Az from the above two equations, we get z AI

...(2.110)

(Kz +K3 -rnz co ) Fo

= 4

ml rnz co -

[

KI+Kz

(

+

ml

rnz

...(2.111 a)

z KIKz+KzK3+K3KI

Kz+K3

)

co +

mlrnz

]

K3Fo

Az =

and

...(2.109)

4 mlrnz

co

[

-

KI+Kz

(

ml

+

Kz+K3 rnz

-

Z'

)

co +

...(2.11Ib)

KIKz+KzK3+K3KI

m\rnz

Ã

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

a 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 the

same type as the expression of natural frequency given by Eq. (2,99). Therefore at 00 = oolll and Cl)

= 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

Cl) =

0

/K2 rnz +KJ)

...(2.112)

Thus it makes the mass ml motionless at this frequency. No such stationary condition exists for 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 vibration absorbers which will be discussed in Art. 2.8. '

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 motion are:

=0

...(2.113)

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

...(2.114)

m3 23 - K3 (2z - 23) + K4 (23 - 24)

...(2.115)

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

;:

0

'. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... .. ... .... ........... mn 2n

",'"

- Kn (2n -

"' . ,c-""

I -'- 2n)

'; ';

. "',

=0

..; . :--';co,,---,

...(2.116)

".,'~'.,-..~,,... n

181:.J 44 Soil Dynamics & Machine

Z1

Z2

Z3

Kn -1

Zn-1

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

Foundations

"'.'~..2,,~,.."t;,~.>,

:"""',;,n

rA';".!n~";'F."'"

-

"'.,.,

~

45


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

...(2.117)

Z2 = A2 sin cont

...(2.118)

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

...(2.119)

......... ..... Zn

= An

...(2.120)

sin cont

Substitution of Eqs. (2.117) to (2.120) into Eqs. (2.113) to (2.116), yields: [(KI+K2)-mIID~] AI-K2~ -K2A1 + [(K2+K))-~ID~]

A2-KJAJ

=0

...(2.121)

=0

...(2.122) ...(2.123)

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

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

........ ........... .. ............. -

Kn An

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

...(2.124)

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

[ (KI

-Kz

+ Kz) - ml ID;] -K2 0

[(Kz+KJ)-~ID;] -K)

0

0

'"

0

0

0 0

0 =0

0

(2.125) .

2 '"

-Kn

(Kn-mnIDn)

Equations (2.125) is of nthdegree in CI);and therefore gives n values of con correspondingto n natural frequencies. The mode shapes can be obtained from Eq. (2.121) to (2.124) by using, at one time, one of the various values of conas obt1incd from Eq. (2.125).

.

When the numht.'Tvi degreeS of freedom exceeds three, the problem of forming the frequency equation and s01";~jgit for determ41ation of frequencies and mode sh<1pesbecomes tedius. Numerical techJ'iG.~esare invariably resorted to in such cases. . ,

Holzer's numerical technique is a convenient method of solving the problem for the system idealized as sho~ in Fig. 2.26. By sUI11II1iPgJfotces at free end, .

'-,

. --.

'O".::.,:-=C°'-:"~",_,- ;: "

~-::--===.::.~ -~:---==E'~--,'::.._~

If

46 Soil Dynamics & Machine Foundations

.. m1

.J K1

m1 Z1

I ml

m2 Zz

K2

m3 Z3

. m.1-

1

~_1

K.

1-1 m.

Kj-1(Zj-Zi-1)

.

I

K.J m i+1

Fig. 2.26 : An idealised multi-degree freedom system Inertia force at a level below mass mi - I .. = ".;-1 m. Z. L...J=I }

}

...(2.126)

Spring force at that level corresponding to the difference of adjoining masses = K.1- I ( Z.1 - z.11)

Equating Eqs. (2.126) and (2.127) i-I

...(2.127)

.

..

Lj=lmjZi

= Ki-I (Zi- Zi-I)

...(2.128)

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

(

I'i~'t mi - Ai

U)~ sin U)n

t) = Ki -

I (Ai sin (Unt - Ai - I sin (Unt) '2

or .'

Ai

= Ai-I -

U)n "i-I A i-I £...j=lmj j

K

Equation (2.129) gives a relationship b~tween any two succ~sive

...(2.129) amplitudes. Starting with any

arbitraryvalueof AI' amplitudeof all othermassescanbe deterinined.A plot of An + 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 $ape.can be obtained .. various val~~s.pfQ);.~ode .. -. . by substituting the correct value of (O~in Eq. (2.129).

iI

47


1.0

' ~

J

+

J c: ~« ~ 1:

(..)2 n.

..-

t

0

2 ""nz

2 wn1

-1.0

--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 n

I

m. Z. + I

=1

I

= F. ( t)

K.. Z.

IJ )

where i = 1,2,3,

...(2.130)

,

,n

The amplitude of vibration of a mass is the algebraic sum of the amplitudes of vibration in various modes. The individual modal response would be some fraction of the total response with the sum of fractions being equal to unity. If the factors by which the modes of vibration are multiplied are represented by the coordinates d, then for mass mj' Z.'" -- A.(1) d 1 + A.(2) d2

+ ... + A.1(r) dr +... + A.(n) ,n d

...(2.131 )

Equation (2.131) can be written as n

Z. -'

= I

r=l

A~r)

'

d

...(2.132)

r

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

(r) ..

n

Im. I A.,r d + I

r=1

n

(r)-

I K..'I A.1 d r - F., ( t)

r=1 j=l

.

...(2.133)

.

Under free vibrations, it can be shown n

(r)

~ Kij A j 1=1

-

2

(r)

...(2.134)

oonr mj Aj

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

ImjAj

r=1

or

(r)

.. n 2 (r) dr + I oonrmjAj dr r=l

n (r)'. 2 ImjAj (dr+oonr.dr) r=l

-- Fj(t)

..(2.135)

-Fj(t)

...(2.136)

l'

-.-'T .~

""}'-

-")J'!!I~'

~,

48


Since the left hand side is a summation involving different modes of vibration, the right hand side should also be expressed as a summation of equivalent force contribution in corresponding modes. Let F; (t) be expanded as: Fj (t)

i

= r=1mj A~r) fr

...(2.137 a)

(t)

where fr (t) is the modal force and given by n

fr(t) =

Ili

(t) .A~r)

i~1

Z

(2.137 b)

Lm[(A~r»)] ;=1 Substituting Eq. (2.137 a) in Eq. (2.136), we get .. z dr + O}nrd,. =fr (t)

(2.138)

Equation (2.138) is a single degree freedom equation and its solution can be written as 1 dr

=-

I

J fr Ct) sinO}nr(t -1:) dt

where 0 < 1:< I

..(2.139)

0}nr 0

It is observed that the co-ordinate d, uncouples the n degree of freedom system into n systems of single degree of freedom. The d's are termed as normal co-ordinates and this approach is known as normal mode theory. Therefore the total solution is expressed as a sum of contribution of individual modes. 2.8 UNDAMPED DYNAMIC VIBRATION ABSORBER A system on which a steady oscillatory force is acting may vibrate excessively, especially when close to resonance. Such excessive vibrations can be eliminated by coupling a properly designed spring mass sytem to the main system. This forms the principle of undamped dynamic vibration absorber where the excitation is finally transmitted to the auxiliary system, bringing the main system to rest. Let the combination of K and M be the schematic representation of the main system under consideration with the force F0 sin CJ}t acting on it. A spring - mass (auxiliary) absorber system is attached to the main system as shown in Fig. 2.28. The equations of motion of the complete system can be written as: MZ1 + KZI + Ko (ZI-ZZ)

moZZ+Ka(ZI-ZZ)

= Fa sin rot

...(2.140)

=0

...(2.141)

The forced vibration solution will be of the form ZI = Al sin rot

...(2.142)

~

...(2.143)

= Az sin 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

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

óóþ

ó

...(2.144)

= Fa

(2.145)

=0

þ

þþóþùòóþóò

-

"J

.49


Z2

ma Absor ber syst

Ka Z1 M Main syst
Fig. 2.28 : Vibration absorber

Subtituting:

F ZSl 2

ro

na

= -2... K = Staticdeflectionof main system = -Ka = Natural freqeuncy of the absorber ma

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

m Ilm

= -.Jl.. M = Mass ratio = Absorber massIMain mass

The Eqs. (2.144) and (2.145) can be written as K{/ +---

AI (1

K

<.02 n)

-A 2 -K{/

and

...(2.146)

=Z

K

SI

...(2.147)

=

A2

(

1:~: äò±²¿

)

Solving Eqs. (2.146) and (2.147) for At and A2' we get Cil '

,

<,

, "

~t. = Si n'

.,".. -

L

" "--,

(

ro2

ld"2. rona

1-- 2 ;

-ro ' no

K

~--

)(

1+ K"

.

' K -~

ro'2

-

2 ron

)

...(2.148) I

,J

".

K'

i

, "..",.-,

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

50

SoU Dynamics & Machine Foundlltions

K ~~

...(2.149)

i - oo~ ) -~ (1- OO~a ro')(1+ ~..~'

~

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. Further Eq. (2.149) gives

Az - -K Zst - 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 does not vibrate. The addition of a vibration absorber to a main system is not much meaningful unless the main system is operating at resonance or at least near it. Under these conditions, 00= oon'But for the absorber to be effective, 00should be equal to 00na . Therefore, for the effectiveness of the absorber at the operating frequency corresponding to the natural frequency of the main system alone, we have or ...(2.151 a) oona = ,oon

or

Ko

=K

ma

M

K m -!L=-!!..=Ii K M

...(2.151 b)

...(2.151 e)

t"'m

When 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:

1- -002

( ~,:

1- --y~

(

OOna 00')

(

J

...(2.152)

1+ Jlm- --y- - Jlm OOnaJ OO~a00' ..(2.153)

~: = (

1- --y-

00') (

OO"a

1+ Jlm- --y- - Jlm OOna) I 00'

The denominators of Eqs. (2.152) and (2.153) are identical. At a value of 00when these denominators are zero the two masses have infinite amplitudes of vibration. Let when-00= oonl'the denominators becomes zero. For this condition the expression for the denominators can be written as

-

.. .~

...,

~- .-.--

~.-

51


OOnt

( OOna)

OOnt

4-<2+llm)

(

)

2 +1 =0

...{2.154)

The Eq. (2.154) is quadratic in <0;1'and therefore there are two values of oonlfor which the denominators ofEqs. (2.152) and (2.153) become zero. These two frequencies are the natural frequencies of the 3ystem. Solution of Eq. (2.154) gives:

(:J

~

...(2.155)

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

1.6

_1.4

1.2

°

_

c c

33 1

1.0

0.8

0.6 0

0.2

0.4 Mass

ratio

0.6

0.8

}.Im

Fig. 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 the mass ratio, greater is the spread between the two resonant frequencies. The frequency response curve for the mainsystemis shownin Fig.2.30 fora valueof ~'" = 0.2. The dotted curves shown actually mean that the amplitude is negative or its phase difference with respect to the exciting force is 1800. It can be noticed from this figure that by attaching a vibration absorber {oona= (On)to the main system vibrating

at resonance reduces its vibration to zero. Now if the exciting frequency is absolutely constant, the system will work efficiently. Any change in the exciting frequency will shift the operating point from the optimum point and the main system response will no longer be zero. It may be noted that by adding the vibration absorber, we have introduced two resonant points instead of one in the original system. Now

I

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

!IIZ~

52

Soil Dynamics

& Machine Foundations

if the variation of the exciting frequency is such that the operating point shifts near one of the new resonant points, then amplitudes will be excessive. Thus depending upon the variation of the exciting frequencies the spread between the two resonant frequencies has to be decided to remain reasonably away from the resonant points. After deciding the spread between the resonant frequencies, a proper value of !J.mcan be chosen from the curve of Fig. 2.29. Undamped dynamic vibration absorbers are not suitable for varying forcing frequency excitation. To make the vibration absorber effective over an extended range of frequencies of the disturbing force, it is advantageous to introduce a damping device in the absorber system. Such an absorber system is called a damped dynamic vibration absorber.

8

6

...

4, }Jm=0.20

z

\

1.0 0

1.5

1.0

0.5

0

'--2.0

2.5

G.)n l øòË²¿ Fig. 2.30 : Response versus frequency of a vibration absorber

The Eq. (2.154) can also be written as

~m

J~)-lr 2

(J)

nl

( (J) //(/ )

...(2.156)

53


t ILLUSTRATIVE EXAMPLEs! Example 2.1

The motion of a particle is representedby the equationz = 20 sin rot. Show the relative positions and magnitudes of the displacement, velocity and acceleration vectors at time t = 0, and ro = 2.0 rad/s and 0.5 rad/s. Solution: Z = 20 sin rot

l

Z = 20

= 20 ro sin rot + ~ ) -Z =- 20ro rocos2 sinrot rot 2 = 20 ro sin (w t + 1t)

The magnitudes of displacement, velocity and acceleration vectors are 10, 10 ro and 10 ro2 respectively. The phase difference is such that the velocity vector leads the displacement vector by 1t/2 and the acceleration vector leads the velocity vector by another 1t/2. Figures 2.31 a and 2.31 b show the three

vectors for ro = 2.0 and O.?Orad/s respectively.

I

20 (V el.)

40 (Accln.) 10(oispl.) (a) G..)= 2.0

rod/sec

s(Vel.)

. ~/z 2.S(Acc!n.)

". 10(OlspL)

( b) CV= 0.5 rod I se c Fig. 2.31 : Vector diagram (Example 2.1) 21t

21t

Time period

= -ill = -2 = 1t s

Time perIod

= -ill = -(0.5) = 41ts

21t

21t

for ro = 2.0 rad/s for ro

= 0.5 rad/s

Example 2.2 A body performs,simultaneously,the motions Zl (mm)

= 20 sin 8.0 t

Z2 (mm) = 21 sin 8.5 t

Determine the maximwn an~ minimum )~1itude periodic motion.

of the combuled motion, and the time period of the

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

,.-,'!.,!,,~,~~ '111""

54

Soil Dynamics & Machine 'Foundations

Solution: Z max = 21 + 20 = 41 mm Z mm . = 21 - 20 = 1 mm

The beat frequency is given by 8.5-8.0 1=

21t

T = 2.. -

I

Example

0.5 = 21t = 0.0795 Hz, and

21t

- 0.5 = 41t = 12.57 s

2.3

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

Solution: Stiffness of the spring, K = W ~\t 20 x 9.81 ::::104 N/m K = 20xl0-J

1[K Natural frequency,In

= 21t

V;

21t 1 ~1O420

~

= 3.6 Hz.

K1

Example 2.4 For the system shown in Fig. 2.32, determine the natural frequency of the system if

KZ

Kt = 1000 N/m Kz = 500 N/m KJ = 2000 N/m K4 = Ks = 750 N/m Mass of the body = 5 kg

K3

Solution: Let Ket and Ke2represent respectively the effective stiffnesses of the top three springs and the lower two springs, then

-1 Kel

=

1

1

1

Kt

Kz

KJ

-+-+-

=~+~+~. 1000 500 2000 -0.0035

~ .....-

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

K4,

KS

". 'Fig. 2.32:"MuHpriags system

...

.- .....

55

Theory of Vibratimrs

Kel = 285.7 N/m Ke2 = K4 + Ks = 750 + 750 = 1500 N/m Now Kel and Ke2are two springs in parallel, therefore effective stiffness,

= Kel

Ke

f. n ~

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

~. /K = ~~1785.7 21tV;; 21t 5.0 = 3.0Hz

Example 2.5 A vibrating system consists of a mass of 5 kg, a spring stiffness of 5 N/mm and a dashpot with a damping coefficient of 0.1 N-s/m. Determine (i) damping ratio and (ii) logarithmic decrement. Solution: (i)

Cc

= 2 ~km =

2J5x 10-3 x 5 = 0.319

N-s/m

J: C 0.1 ~=C=0.319=0.313 c ,,

27t~

(in Lograthimic

decrement

= ~ 1-

~2

- 27t.x0.313 = 2.07 - ~ 1- 0.3132

Z log ::,..l = 2.07 eZ 2

~

= 7.92 Z2 Therefore the free amplitude in the next cycle decreases by 7.92 times. Example 2.6 A mass attached to a spring of stiffness of 5 N/mm has a viscous damping device. When the mass was displaced and released, the period of vibration was found to be 2.0 s, and the ratio 01 the consecutive amplitudes was 10/3. Determine the amplitude and phase angle when a force F = 3 sin 4 t acts on the system. The unit of the force is Newton. . Solution: i.e.

. 27t~

(i)

ZI

10

.

= loge Z2 = loge 3 = 1.2 ~ = 0.195 TII = 2.0 S

1- ~2

or, (ii)

21t (lJ n (lJ

21t

= T = '2 = 3.14 = 4.0 rad/s

T1 ='~'= .

.' ..

,

.

'.'

'.

4.0 3.14

IDn i.:'

Fo

.,',

radls

= 1.273 F

3.0

=.;..",fi..,;'! 3.0N;AsI= -9..= -5.0 = 0.6mm. "."K. t:<,'..'""

----.

, -=., -,.;".,'~ -""'~-.~

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

56

Soil Dynamics & Machine' Foundations

From Eq. (2.58), A

Az =; ~ (1-11)2

=

t+(2~11)

. 2;.

'.

Asl

= Static Deflection "

0.6

.

,

.

,

'

~Q.755llll11

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

-1'

211~

e = T an ( --r = Tan ( 1-11 )

2 x 1.273 x 0.195

) = 141.4°

2

1-1.273

Example 2.7 Show that, in frequency

S1On:

- dependent excitation the damping

factor ~ is given by the followingexpres-

): -.: ..!. 12 ~ 11 ~ 2 ( 2/n J

-

Where 11 and 12 frequencies at which the amplitUdeis 1/.J2 times 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

Amox

=

0.084

0.08

E E" 0.07

C:I "'0

-::J c. E 0.06

et

0.05

I I

I I

I

I I

I

I

I

I f1 I 14

0.04 10

I

Ifl

fn 18

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

24

22 . n,

,

Hz

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

-.------

"

57

Th~o,.,ofv~,!s At resonance, 11 = 1 ana A~I Zst

= 1/2 ~ (for small values of ~). If the frequency ratio is T\when

amplitude 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

Now

11~-11i

Also

1- 2 112 111

= 4~~1-~2 = 4~

[for small values of~]

= Il- 112 = 12- 11

(

In2

= 2 I 2 - J;.

12+ I.

)(

In

)

In

12 + f . = 2

since

( In )

In

Iz - 11 2 ( In )

~=!

Therefore

This methodfor determiningviscous damping is knownas the band width method. Example 2.8

.

A machine of mass 100 kg is supported on springs of total stiffness of 784 N/mm. The machine produces an unbalanced disturbing force of 392 N at a speed 50 c/s. Assuming a damping factor of 0.20, determine (i) the amplitude of motion due to unbalance, (ii) the transmissibility, and (iii) the transmitted force. 'Solution: , 184

( i)

(J)II

= ~KI m = ..'/ ,

, 00=

,'n

V

21t

~st ,

Now

Fi) '-

100

,

392

',

'= 0.5 'innl

'

2

-

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

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

..-

==87.7 rad/s

= 3.58

~ ~ - ,;784 , ;. : :'?it . '.

~'\,

"

= 314 rad/s

00 - ,314 =;., UJln; ~ 87.7"

, "

x 50

x 103

-

=

,0.5 '~(1-'3582)2 +(2 x 3.58 x

.' ':;' cc:

, ---

0.2)2

= 0.042

mm

. - -

.~

"'-;

~~~,~"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 Force transmitted = 392 x 0.1467 = 57.5 N.

(hi)

Example 2.9 The rotor of a motor having mass 2 kg was running at a constant speed of 30 c/s with an eccentricity of 160 mm. The motor was mounted on an isolatorwith damping factor of 0.25. Determine the stiffness of the isolator spring such that 15% of the unbalanced force is transmitted to the foundation. Also ~etermine the 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/ = 22.72 kN

:ii)

Jl = T

.

I.e.

~1 +4112~2

~ (1-

or or 114 -

= 22716 N

Force transmitted

unbalancedforce = 0.15

.

112)2+ (211~)2 .

= 0.15

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

12.84112 - 43.44 = 0 (0

It gives Therefore

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

= 0.15

x

22.72 = 3.4 kN.

Example 2.10 A seismic instrument with a natural frequency of 6Hz is used tomeasure the vibration of a machine running at 12.9rpm. The in~trumentgives the reading for th~r~lative displacement of the seismic mass as 0.05 mm. Determine the amplitudes of displacement: velo'city and acceleration of the vibrating machine. Neglect damping.

~;,:,~ .;,~

59

'/reory 01 v.ibrations

';olution : CJJ

(i)

n

= 6 Hz = 37.7

rad/s

120 x 21t =' 12.57 rad/s 60

(J)

= 120 rpm =

'1

= 12.57 37.7 = 0.333 1

~ =-

for ~= 0

1- ,,2 1

=

,

= 1.125

1- (0.333)2 (ii) For displacement pickup, Eq. (2.88) gives 2

X=,,~yo

0.05 = (0.333)2 x 1.125 x Yo or or

Yo

(iii) For velocity

= 0.40

mm

pickup, Eqo (2.91) gives' 1 X = -00 11~ (Y 0 00) n 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 ,

or

0.05

=

(YOOO2)

OOn

'

1.125 (Y 002) (37.7)2

I.e,

(Y() 002)

,

0

= Acceleration = (37.7)2 x 0.05 1.125

= 63.17

mm/S2

Example 2.11 Determine the natural frequencies and mode shapes of the system represented by a mathematical mod~1 shown in Fig. 2.34 a. ,/

. ,

co:"

'1,,';

,-

~:::."'O'~,""""",

,~>=j"~

, i

J

60

Soil Dynamics & Machi'Je FOUlrilatiolls j

+1

+ 1

(c) Second mode

(b) First mode

(a) Two degrees freedom system

Fig. 2.34 : Two degrees freedom system with mode shapes

Solution: (i) The system shown in Fig. 2.34a is a two degree freedom

has already been described in Art. 2.7.

system. The solution of such a system

'

(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 \l2

(02 = .!. III 2

(02 112

Hence,

[(

3K + 3K _ 4 ~ (2 K)2 m m) { ,m2 }

= .!.r-6K + 4K 2 l m, m ]

=K )

m

= 5.K ,

m,

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

(iii) The relative values of amplitudes Al and

,

"

.,'

'

~ 'for the two modes can be obtainedusing Eqs.

(2.103) and (2.104). .

. '

,}~~r . . ~."

,,"

-;, '

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

eory of Vtbtalions_/

6i A (1)

K

-1-= A(l) 2

.

A~2)

2K

A (2) 2 = K + 2 K - m x 5K / m .

2K

2 = =+1 K1+ K2 -m 1 o:lnl K+2K-m x K/m

=-

1

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 moqel shown 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 shapes

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

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

=0

...(2.157 a)

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

.~:(2.157 b)

- 22) = 0

...(2.157 c)

m 23'+ K (23

For steady state, the solutionswit be as

= At

sin ront

~ =~

...(2.158 a)

sin ron t

...(2.158 b)

2t 23

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

= A3 sin

.

ront

...(2.1"58 c)

;-:-

.1'

~;-.

'-~""~"~-='

~

:::-."'::::""'c.::--

.:"

_._~

.~-

~

--

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

'11

,"

62

Soil Dynamics & Machine Follndations

substituting Eqs. (2.158) in Eqs. (2.157), we get ., . (3 K - 111oo~) Al - 2 K A2 = 0

óîÕßø

...(2.159a),

+(3K-moo~)A2-KA3'=O

- K A2 + (K -

...(2.159 b)

co;) A3

111

=0

...(2.159 c)

For nontrivial solutions of 00/1in Eqs. (2.159) 3K-11l00

.,

-2K

/I

-2K 0

0

-'"'K -moo/l2 -K

-K 2 1=0 K-mcon

...(2.160)

.,

" . IIIWPllttll1~ A -= !l ~

Eg .

K

3-A -2 0

f.3 -

or

(2.160) become as

-2 3-A -1

0 -1

I-A

I =0

...(2.161 a)

2 7 A + lOA - 2 = 0

...(2.161 b)

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

Therefore,

00/1(

= 0.238; A2= 1.637; and A) = 5.129 = .J0.238Kl m; CO/l2= .J1.637 Kl m; and 00/13 = ~5.129 Kl m

(ii) Egs. (2.159) in terms of A can be written as (3 - A) Al - 2 A2 = 0

...(2.162a)

- 2 A I + (3 - A) A2 - A) = 0

...(2.162 b)

=0

...(2.162 c)

- A2 + (1 - A) A3 For r mode:

A = 0.238

Eg. (2.162 a) gives AI

"

(3 - 0.238) AI - 2 A2

= 0 or A2 = 0.724

Eg. (2,162 h) gives

- 2 .'

or

x

0.724 A2 -t-(3 - 0.238) Az - AJ

=0 ,,

A2

A3

'

= 0.761 ,'.1 "iI;

-'

" ", (..'

:,"

,

.\

63


Assuming Al

'h,

= 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, and For III mode: A.= 5.129 Al : A2 : A) = 3.891 : - 4.14 : 1 The mode shapes are plotted in Figs. 2.35 b, 2.35 c and 2.35 d, Example 2.13 A small reciprocating machine weighs 50 kg and runs at a constant sp~~d of 6000 rpm. After it was installed, 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 least 20 percent from the forcing frequency. Solution: (i)

ill

21t N

=W

""

2 Tt x 6000 = 628 rad/s

60

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

~

= 628 K = m x 628 2 = 50 x 6282 = 201 x 105 N/m

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

Ka

= K = JIM

Now from Eq. (2.156)

J(:,:J-If JInl-

illnl 0)//(/

~

= 0.8

l;;:))

. 64

Soil Dynamics & Machine Foundations 2

2

- {(0.8) 2-1

J-lmand when

}

0.8

= 0.2025

= 1.2

(Onl (Ona

J-l

m

=

{(1.2)2 -1}2 2

1.2

= 0.134

Adopting 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 dependent oscillatory force (m eo (02sin (0f). Proceeding from fundamentals, derive the expression of the amplitude of 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 is attached in series with the first spring, the natural frequency becomes f\/2. Determine K2 in terms 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 nawral period of this system is 0.40s. Determine the natural period when a mass of 2.5 kg is attached to 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 of inertia of weight W about the point 0 is Jo' Show that th~ system becomes unstable when: K.a b >- W .. .'1

-..Theory of Vlbrations

65

Wt")

K

0

b

c a

-1

~

T

A l

-1

.~

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.

b

.,

w

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 mass was displaced and released, the period (jf vibration was found to be 1.8 s and the ratio of consecutive amplitude was 4.2 to 1. Determine the amplitude and phase angle when a force F = 2 sin3t N acts on 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.100 mill..At 80%resonantfrequencythe amplitudewas measured80 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 by the equation. .

co' I

\11,.2

= ~, g.(2 vi

:1:/2)'

Determine. the ratio of the amplitudes xI/x2' ..

.. I.

;.:':!Z~,"'~.'T"""7~..

.I

t .."'. ..~ ~~...~..,.:",,!..,:.;..~

'1'1'

, )J .J ~ .:.

,:.:;.

..-'.

",.

'\""I\~

':.."h. r

~"V :; ~,.

,.

,- "-

e.

- ! ..J

,

.,t;\.~;...t-

:.>'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 at constant 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 transmitted

to 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 the system, What dynamic vibration absorber should be added if the nearest natural frequency of the system 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 logarithmic 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 has an unbalanced rotating element which results in a disturbing force of 392 N at a 3000 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, 2.20 The static deflection of the vibrometer mass is 20 mm. The instrument .when attached chine vibrating with a frequency of 125 cpm records a relative amplitude of 0.03 mm. for the machine, (a) the amplitude of vibration,

105 N/m speed of

58.2 N) to a maFind out

(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

,.,'", .."

.

",..

,,;,,)'

~"
",}~",,"'~.,

---

--..':~,{""J""'""",,,

..

WAVE PROP AGA TION IN AN ELASTIC, HOMOGENEOUS AND ISOTROPIC MEDIUM

3.1 GENERAL A sudden load applied to ~ body does not disturb the entire body at the instant of loading. The parts closest to the source 'of disturbances are affected first, and the deformatio~s produced by the disturbance subsequently spread through out the body in the form of stress waves. 'The propagation speed of seismic waves through the earth depends on the elastic properties and density of materials. The phenomenon of wave propagation in an elastic medium is of great importance in the study of foundations subjected to dynamic loads. In this chapter wave propagation in (i) an elastic bar, (ii) an elastic infinite medium and (iii) an elastic 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 is perfectly rigid, therefore the application of a load on a body causes deforn1ation. In all cases internal forces are called into play in the material to resist the load and are referred to as "stresses". The intensity of the stress is estimated as the force acting on unit area of cross-section, and is expressed in such units as 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 body but change the volume. Shear stress is said to exist on a section of body if on opposite faces of the section equal 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 1 is the initiai length of the 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 1b Compresslve steam = £c = T ...(. )

------

--

-

R:.'

68

Soil Dynamics & ~~e

FlHUldations

F

..i 6l ... f l l

1

III ; 1 ;

TL

t

~

F

(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 contraction 6 w perpendicular to the direction of the stress to the original width w of the body (Fig. 3.2). Thus 6w Ew

...(3.2)

= -;-

l- llW--I

.-- --7I

F

I

,

I I I

w I

I

I

I I

Fig. 3.2 : Transverse strain

3.2.3. Elastic Constants. An elastic material is one which obeys Hook's ~aw ofpruportionally between stress and strain. For an isotropic elastic material subjected to normal stress Oxin the x-direction, the strains in x, y, z directions are t t

C1

x Y

=--!. E

=e =-~ 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 obtain 1

Ex

E

=

, "

[O'x

-

1 ; Ey

=E

Ez

= E

...(3.4 a)

Il (O'y + O'z)] ,

[O'y - Il (O'x + 0')]

...(3.4 b)

[O'z - Il (O'x + O'y)]

...(3.4 c)

I".

In the above expressions, E is the modulus of elasticity and 11is Poisson's ratio. It may be noted that here 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 of the strains as follows: (Timoshenko and Goodier, 1951; Kolsly, 1963). ,'.,

E

IlE

'

O'x =

ay

=

(1 +'Jl)(I-21.1) (Ex + Ey+ Ez) + 1 + Il Ex JlE (1 + Jl)(I-2Jl)

.'

az

...(3.5 a)

. ~..

E (E.t + Ey + E) + 1 + r11 Ey

JlE,..

...(3.5 b)

E

= (1 + Jl)(I-21.1)

(Ex+ Ey+ Ez) -t: 1 + fl Ez

...(3.5 c)

For simplicity the equations may be written

a x =A. E+2GE a

y

=A. e+2Ge

a z =A. E+2Ge

in which

x

...(3.6 a)

y

...(3.6 b)

z

...(3.6 c)

-

= Ex+ Ey+ Ez

...(3.7)

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

...(3.8)

E .. ..,,'. .

E

G.= 2(1 + 11)

...(3.9)

Similarly in an isotropic elastic material, there exists linear relation between shear stress and shear strain. Thus 'txy G

...(3.10 a)

'tyz . I'yz =. G

...(3.10 b)

Yxy=

Cr

,',

Y:x=6

...(3.10

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.

,

c)

8B:,1

70

Soil DyIUUlfics 4 Machint! Foundations

3.3 LONGITUDINAL ELASTIC WAYES IN A ROD OF INFINITE LENGTH

~ OX + oOX

ax~

ox

b.x

~ 6x ~ A = Area of crosseciion Q

x a

x

b

~

~~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 plane during the vibration. the summation of force in x-direction is given by: IF

x

=-cr

x

A+

(

cr + x

ocr

:L6,x

ox

)

ocr' A=---L.6,x.A

ox

...(3.11)

If the displacement of the element in x-direction is u, the equation of motion for element can be Nritten by applying Newton's second law of motion as given below: ocrx .6,x.A

ox

=

6,x.A. 'Y

ocrx - 'Y

or

fu

.clu

g ) at2

(

a2u

...(3,12)

- g'dt2

= stre~s stram ~ m xx --d~ection direction= ~ oula x cr = E.-ou

Form stress-strain relationship E

x

ox

...(3.13)

DifferentiatingEq, (3.13)with respectto x ocr.t - E a2u ox - ox2

...(3.14)

.,.::', ,

--- - "

',',-Y-;'P{,,',-,-*"r'1,h~"";;:':'..~,"" -'""".*:

Wave , " Prop~gationin -' ."

,,'*'"

""':;""':-'

-,/

.,'

,-

71

an. Elastic, H.omogeneous and Isotropic Medium .'

'Y a2u

Therefore,

g at2

.;o.,ft~:t-\>:~,:'

= E clu

...(3.15)

ax2 , ,

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

,

, ,-;-

a2u

,E clu

i

= p ax2

- .at2

...(3.16)

or -a2u = v2 -clu ...(3,17) at2 c ax2 E where v; = ...(3,18) Vc is defined as the 10ngitudinal-wave-propagation-velocityin the rod. Equation (3,17) has the exact form of the wave equation, and it indicates that during 10ngitudiIia~vibrations, displacement patterns are propagated in the axial direction at the velocity vc~' : :; ;' '

P

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 in the 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 noted that at the instant a wave is generated, there is compressive stress of the fac~ in the positive direction of x and tensile stress in the negative direction of x. Hence, when the compressive wave travels in one direction, the tensile waves travel in the opposite direction. To understand the difference between the wave propagation velocity Vcand the velocityof particles in the stressed zone it, consider the stressed zone at the end of the bar as shown in Fig. 3.4 a. When a uniformly distributed compressive stress pulse of intensity crxand duration: t".(Fig, 3.4 b) is applied to the end of the bar, initially only a small zone of the rod will experience the compression. This compression 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. The amount of the elastic shortening?f

or

this, zone is given by

cr cr u=-Lx =-Lv ,t E n E c n u cr t;: = E Vc

...(3.20)

..'

..,

<

, ,

...(3.21)

The displacement u divided by In represents the velocity of.the ~nd of the bar or particle velocity.

Hence

,

.

crx

u =-v ...(3.22) E c It is evident from Eq. (3.22), .that the particle waVevelo~ity of stress but . .'it .dependson .. , , ,theintensity . the wave propagation velocity Vc (Eq. 3.18), is only a function of ','mate~ial wave '" . '. ,~,,' properties. Further both '

propagation velocity and particle velocity are in the same direction when a compressive stress'is applied but the wave propagation velocity is opposite to the particle ve19cj~. when,a tensile,stress is applied. .

,,',

!',

,"

( "

. t.. ; "

.'

~

72

's~u Dynamics & Machine Foundtiiions

~x

-.J I+x

OX.... (

Xn

= Vc tli ,"

)

.-

(

x

u (a) Stressed zone at the end of the rod

er

er X ,

tn

"

t

(b) Uniformly distributed compressive stress Fig. 3.4: Wave propagation velocityand particle velocity in a rod .

"

3.4.
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 expression fo~'t.: :toraue 1. - car. be written as

. where

"

T = G I Be P Bx

...(3.23)

G = shear modulus or"thc'material of rod -,' . I,p '= Polar moment o:f inertia of the crossection Of rod ." , ! .~ , ae'.

_..,.'.;;A'""~,,,.

jf,.'..«ft'll!"-

ex =:Angle of twist per unit length of rod.

- .-,.--'-'

It

la

-

B

_3i.;'1£h

73

.

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

;.

an41sotrQpic Medium

L

'

'-'-'---

.\

.

\

\

,I

x

x

r-

~X

. (a) A rod siabjectedto torque T

,, ;';.'

T + aT oX t::.x

r

,," , ..

'...

,

,

I'; '

": ':

fix

.: (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

T = pI~x~

#e

.,,(3.24)

P :ot2 By applying Newton's,second law of.mo~onto.arielement oflength 6.x shown in Fig. 3.5b, ..

,,

-

, aT--

.

,

#e"

) = p!p *x-yat

T + ( T+ Ox 6.,x

or ;" ~

','

..

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

,1'01 l\'1'.I

o.

".

" ,

,.,,(3.25)

.

,

"

'.',

74

,,":

'."':

"'.

"

""".

'

"","

"'"

.,

Soil Dynamics & Machine' Foillldations

,

From Eq. (3.23),

or

Therefore,

oT

t. '

- =GI- a2a OX p ax2 '..

...(3.26)

a2a ,= pI cia P ax2

P 8t2

ia - 0 cia

or

a? - P a2e -

,

ox2

'~':'

at", -

Vs

,J3.27)

"

2 a2e

'~

..-.

. --zax

,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 the t\\/O\vaves pass by each other in the crossover zone, the portion of the rod in which the two waves are superposed has zero stress wit:.1:l twice the particle veioc.~tyof either wave (Fig. 3.6 b). After the two waves have 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 seen that 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 can be considered a free end (fig. 3.6 d). Hence 'it can be seen that a compre,ssion wave is reflected from a free l'l1d as a tension wave of the same m,agnitude and shape. Similarly, it can be observed that a tension wave IS re Ikcled from a free end as a compression wave of the same magnitude and shape.

t

----

Id-:

Vc.

0

oo ien r=o sion

-v

Xt0

"0

c

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

~~ 00 ~

0-'=0

~,.,,' .

U = 2uO .

y

~Xt1

--q-nrn '---00 ,

Vc

(b) Waves at the crosso~'er zone Fig. : 3.6 : Elastic waves in a rod with free end conditions

..,'...,

'"

,..' --,

(...Contd.)

l4~'

\~~)'"

f'ave.Propagation

,'" "'.~

:,.

, y.~:,~.

"~ ;i0: <~~- "';':;:~~!-""i"" ','

'c:

:,'"

75

in ,an El~$tit;, Homogeneous and Isotropic Medium

~

.h

.

-

.

er

.. Xt2

~

"0'

er

Vc

u

I

=

a'

0

0

=

(c) Waves after passing the crossover zone

Vc

~

Free enq .~~

.. Xt3

-4 Vc (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 positive x-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 equal to 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 a compression wave of the same magnitude and shape, and that at'the fixed end the stress.is doubled. 'v

-4

I

. er=0

~

~u=o I

~o.

,

-

Vc

~ ~

.Xto

(a) Two Identical waves travelling in opposite directions

Xt1 er = 20-0 ,

j' U = 0

(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.

Vc

:

I '--

~,

G "0

'.

~..

"/

Vc

'-'-

".~

..~

0-::0 u ::

Xtz

'

6

(c) Waves after passing"the crossC)ver zone

Vc

Vc

,

l

~

'

Fixed end "

xt 3

~

"

(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) U = Displacement amplitude along the length of rod ;

where

...(3.29)

AI' A2 = Arbitrary Constants (I) 1/

= Natural frequency of therod

Substituting Eg. (3,29) in Eg. (3.17), we get .,

~d2 U + oo~ 2 U =0 dx

...(3.30)

Vc OOnX

The solution

of Eq. (3.30) is ,

U

=

A)

cos

-~

OOIlX

+ A4 Sin

-'"-~

'

...(3.31)

.'\, 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)',

.' :" ,",

.. "

. '"..

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.'

,_:::_,

;;

"

:;;

":_'~,;

., ,:<

.,

-

"

~l~k'

77

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

u 0-1

..

-x

El

l-x

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

Ô

~

´

ßÖ

T

ßí

ø¾÷

Ì

U1= A~ cos ~x (n

=

1)

U2 =A3 cos 2Tfx L (n

=

2)

U3=A3 cos 3Tfx L (n

=

3)

First harmonic

(c) Second harmonic

,'~

0

AJ

I (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

-dU dx The condition

get ron

= -\!

cc'(

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

'.

0'

"

. 'ro"x

- A 3sm-+A4cos' V , ;'.

)

", -

78

-dU dx

Putting

. ro L A sIn -!!

Soil Dynamics & Machine -Foundations

= 0 at x = L we get '

=0

...(3.33)

vc roL For a nontrivialsolution, -!! = n1t J

Vc

- n1tvc or 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 get n1tx 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

-t x

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

~I~

.

A4

TTx

= A4 sin =2L(n

=1

)

TU) (b) First harmonic

U2=A 4 s' In- 31Tx 2L

(n=2)

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

"<'

Wave Propagation in tin Elastic, Homogeneous

79


'SlTX

;

,

U3=A, 4 slO ,,2L

(n=3)

(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 condition in Eq. 3.32,

, "

W L

A cos -2L 4 vc

wnL

or

"

.

= 0

...(3.36) 1t

-v

= (2 n - 1) -2

c

"

'

1tVc

or

wn

= (2 n -

.

,. '.

...(3.37)

I) 2 L '

The displacement amplitude can be written as U - A . (2n-I)1tX n

-

4 sm

...(3.38)

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 '

~

:L

.., ..x

I

~

(a) Rod ornnlte length with fixed-fixed end conditions

U ,

1

-A

-

.

4..SIn . (

TTx

L

(

n

=

1)

(b) First harmonic Fig. 3. to: Normal modes ofvlbration

oh rod of finite length with fixed-fixed end conditions (...Contd.)

> ,"'"

~, ".~~£~

80

Soil, D.ynamics ~ Machine

f~u~cJptions

f

. 21Tx U2=A4Sln-C-(n

=2)

(c) Second harmonic

_ . ,3TIx 3 - A4sIn L ( n

U

=

3

)

(d) Third harmonic 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, and oonL

A4 sin

...(3.39)

- Vc = 0 n1tx oo=-n-1 n L'

or

Th d e

.. I b U ISp ac,ement IS gIven y, n '

=

A

23

-

, , , ...

. n1tx 4 sm

...(3.40) ...(3.41)

L

The first three harmonics described by Eq. (3.41) are shown in Fig. 3.10 b, c and d. 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 finite length can be solved in the saITi~manner as foJ. tJ:le-case
. e = eA(A I cos where eA

OOn t + A2 sin oont)

...(3.42)

= Rotational amplitude of angular vibration I

"

A I' A2 = Arbitrary constants (j) n

= Natural frequency of the rod

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

2

~ 2 + ~2...=0 e . dx'

v's

'

...(3.43)

Wave-Propagation'in'an"Elilstic/

81-

Homoggneous and Isotropic Medium

The solution of Eq. (3.43) is ,conx eA -- A3cos -+A4

. COnX sm-

Vs:

...(3.44)

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:

.

Free-Free Condition. n 1tVs

L

...(3.45)

eAn -- A3cos-n7tx L

...(3.46)

con

=

Fixed-Free condition. con =

(2n-I)7tvs 2L

eAn =A4sin

...(3.47) " ,

...(3.48)

(2n-l)7tx 2L

Fixed - Fi,,<:edcondition. n 1tVs (On

=

...(3.49)

L

...(3.50)

eAn = A4 ~1tXL

3.8 WAVE PROPAGATION IN AN INFINITE, H~MOGENEOUS~ISOTROPIC,ELASJICMEDIUM '

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
[",

-(",

+ aa"; d X) ]
'n + a;;)dz

U't

(

+ 'ty:c-

( 'tyx+

]
ilu

a;'dY.

...(J..5.0

)] (dx.dz)+p(dx.dy.dz)at2:::10 ,cf." p. ':0l2

or,

aox' ox

=-

a~.xy':"~.tt' +............ ay' j)z

+-

,', .

-

"

".-..-....

..~{3.52 a)

8~~

':'

,"',

~" .,'

"

";":-

,~ .

SIJU Dln9nUc;s. & Machine F o.ullllations

z

... I

.."

-."

dy '

( 1':+ ZI

~'tn az

,

dZ} ,

,

itr..

~ ""

crz+

~

,d-z ),

..

,

't'yz+

I ,. I('tu + ~ax'

t" yl

cry

("",

I

t

,

II

,~.

-r

~-ryz ~,!lY),

er.

IY dl)

('I I

t

t

I

(GY + ~cry C)y

~IZ

dy)

1:yz

('t'yx

+

~f)~YX . dy)

~ax (cr. + ~.d.)

I

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

p-;iv

ot2

(h yx ocry = -+-+ox ay

2

o't yz

...(3.52 b)

oz

.'

o't o'tzy' ocr p-a w - --M...+-+---L

ot2 - ox

ay

(3 52 c)

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 an elastic medium given by,Eqs.,(3.6) to' Eqs. (3.10) are used. The equations for strains and rotations of elastic and isotropic :materials in terms of displacements are' as follows,:' '

Axial strains,

:;'

"

, .

.

ou

;,

,...(3.53 a)

= ox . av Ey ='ay aw

Ex

E;:

...(3.53 b) ...(3.53 c)

= ay

Shear strains,

ov

ou

Yxy'" 0;C+ ay ow' av','

...(3.54 a) .

--+Yyz- ay oz ou ow yzx = OZ + ox

...(3.54 b)

~, ,-

)

..,

,

...(3.54 c)

~< 2.,:,'

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

:,.",~,~:

":",),,'0'

,

:'

J..... 8'3

..

Wave Propagation in an ElastIc, Homogeneous

and IsotropIc MedIum

Rotations.

ow 8v' 2w = --Xo y oz.. iJu' .;iw

';,

...(3.55 a)

"

2wy -- ~-oz ox ov ou 2w =--z ox oy

.:.(3.55

b)

,..(3,55 c)

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

,,'

""""""""":'-"""':""'/:~";",""i'

"",1"",

",'

',',"

3.8.1.CornpressiQD Waves. Subst.itut~onoff:qs.p,. 6 a), (3.10 a), ,an~;(~,lO b) into Eq. (3. 52 a) gives 'd d ~'-"' ( ",E+ 2G E..)+~ (O '1,' ) +-- (G '1 ) , p-,= "'

','

,

"",ot2

,

'

',

"""j;U"d"

'"

'

'

,"

..,

"

"d.,J."'

..

,X,dy,xY"dZ

",

'

0,'

",.

,xy ,

"

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

'

,', '~~i'

,

,',

,

'2

02u - A.d E + G a u +

p 0(2

co"

"", "'

or

~~;FG:z(~> ~;r

~'~(A<'~';~E,)+GM~:'

,-: ax . .

,,'..

~

2

,

-/,'

'2'

+ d

2

w'

+

2

2

au+au+au

al [ ax2 dX,ay' dX'az ax2 '

...(3.57)

di ]

.

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

As

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

...(3.58) , ..

azu

'

o'E

2

...(3.59)

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

where

" ,

2 az az az V u = -+-+---'ox2 ay2 fJz2, Similarly Eqs. '(3.51 b) and (3.52 c) can be expressed as azv ... aE

p-

0(2

= ( A.

+ G)

-ay

...(3.60)

'

+ G V2

.'.~(3.61)

and

" "

02W

= (A. +

P-y ,

G)

aE

2

+ GV w

~

&~,

Equations

(3.59)

...(3.62)

'

, (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

fil "

;'

,

" ;,

Ov

:";";

"",..",,-,,'-=',' , , '.."'. v ;" ",;,q~,

a2E

Ow

P 0(2 ( Ox+ ay + Oz') "'::"}J,

or.,.

au

. ,~,p

"2'"

','.'

""(k

+-, G)

\,-O~i'~',_",\,,;v~'

'.,

< "," ,,', L«'\'

'

""', E) ,+ ,(GV .".'"

,P"",)..".I,;t"",'jl',ttf.~
2

'c,.'~';y:!..;,

2 (V

file

""J

',.

2

E),

r

"!'"

Ov

Ow

ay + fJz:) ",.

';".

'.:.',:~

.

""""""""q,, '..d" ',J. ,"..

'1-"'>""1';'_',,1,,1'1.,;,,,.,

;,

au

+ aji2 t"fJz2' ) + G V ( at'+

(A + G)'fix2 (

',=

!"..i-~),':J.' (TE

02£

""

,'."

'. ",C ,,'

';.

"I,

~ ':'J",) .~f..'

n,...O~;'J1

84

"s.iJ ,J)yIUVflil:s..& MadMe'

Hem:e

...(3.63 a)

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

or .

where

F~IIRd4tU1ns

..~(3.63 b)

2

A.+2G 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 Vp

=

.

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

3.8.2 Shear-Waves;

DiffererniatingEq.'(3:61) 2

a-E

a av

()

2av

p at2 az: = (A.+ G) (ay)(az) a2p

and

'"

,

with:resJrect to z and Eq. (3. 62) with respect to y, we get

+ G V az

ae

aw

2

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

aw

-

ay

...(3.65) . ...(3.66)

.

aw- av = G V2 aw- av at2( ay az) ( ay az) aw av -

p~

...(3.67)

.

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

rJw

p--f at or

2,

= GV"wx

G 2-w = V'2 V2-w -rJwx =.-"11 at2 p x s x

...(3.68)

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

G

2 2y =vVw s y

...(3.69)

'rJwz' = G V2w = v2V2w

...(3.70)

rJWy -=.8t2

at2

.

P

p,

"11

2-W

z

The above expressioMindicate"thafithe'Totation

s

z

.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 propagates at the same velocity in both the-rod!andthe..infinite-niedium/Fig. 3.12 shows plots of shear wave velocity and void ratio at several confuii,ng pressures for sands (Hardin and Richart, 1963).

W~ve :ProplIgiitionin -

,1.'85

4" E/~'stic;"H~mogeneollS alid Isotropic Medium -

-(

-J

;.

,-

'N ,E '"

"~ - '."26.'0

-

lA ::I

--

::I

,-

0

E

,,-

Con'1ini'ng (pres'sore

390 30

360

--

,

19:5

"tJ,

0.5

0,+

./11

/~<

---

330

-

Round grains Ottawa Angular grains -.Crushed

sand quartz

-E

\11

>

.... u

~

270

~

> ~ ,>

0

240

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

.,.

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

, ., ..~---

Et.

.'

.

:-86

SoU :!)yntUlli~

.i

3.9WAVEPROP AGATION INELASTIC IL\LF,SP ACE

&I Machine Foundtllions

.'

.-

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

~

x

G

~,vst

,

-I'

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 motion within t~e pulse is in the direction of propagation (i.e. radial) is the faster 'f' (primary) wave, but normal to it (i.e.I tangential,to the pulse front) is the slower 'S' (secondary) wave (Fig. 3.13). The stresses in the ," P wave~ which is ~fl°!lgitudinal wave like a sound wave in a~~,are thus due to uniaxial compression, while during VIe passage' of an S wave the medium is su~jected to shear stres~. The surface wave travels more slowly.than either body wave, and is generally complex.,This wave was' first studied by Rayleigh

(1885)and later was,describedin detailby Lamb(1904), It is refe,rred'asRayleighwaveor R-wave.The influence 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, The half 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 are independent of y, then ~

u=~+-

" ',,'

where

",',

"

a",

...(3.71)

ax ,az _a~ O\V ,w---az.

'

...(3.72)

ax

av'

",.,".

.

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

atH'~'j;

"

,""

Wav'i"Propagaitoidli

an ElaStic;'Ho,,;(/geneous -

.. 'au,.:.,aw ,1:;'--+-, '

dX dZ

"

( dx2

dXdZJ

I:;

~

'.

'

:-: 'U.A.'

,,0

( ai,

,

,

a24>a2cp' 2' ãóõóãªùþ 2 'I' ~1

-

'or

#cpa2'V d2'41 + ,---

d24>

-+-

:c.:..

,

'0

87


oxOzÖ

' "

...(3.73)

'

Z

'.

Plane wave

,

front

I r---L..-

,,'

,)

/

x

I

I 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 .

'"

"

"

"

,.

2

""'

"""-"""&

"",'7'",W~;'=az-ox

?..:'~,.;.j

Z

aw

2

d'V + d'V =~

.

oz'

;..(3.74)

~=,v'V

.",..",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~

l

az~ :'~ki\:,~:Cj)' ~;(V.Z~y~.p!G ,-2- (V2'V) (

',OX -,' ,':.",az...2".i;J"",\",;,,.~-,ox, , D.).~tj;.i:>~f};~:#~,alk-. ,

'6'!-'f,~'~"""J'"

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

L ~;"

'"

:"""":'" rJ'1~rl"~ <-:'

",~/

:;Oz.,' ll.":i!i'."~'," ::":""'"

...(3.75 ,'"

:

",

, '"

',' 0,'

't'

i"f:"

""

, .. ~~,..~,.,...~~~=~..~

,.,.,..,

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

'-="..,

)

""-

...

88

Soil DylUlmics & Machine Foundations

and

p

~

&$

Oz ( 8t2 )

- p

~ax

&'V ( 8t2 J

= (A + 2 G) ~ (V2$) v£.

G

! (V2'V)

...(3.76)

UA.

Equations(3.75) and (3. 76) are satisfiedif &$ = A+2G V2$ = lV2$ 8t2

P

&\jI

and

G

-

= -'V

a t2

2

-

p

...(3.77)

P

2

-

2

'V = v 'V 'V

...(3.78)

s

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

-

= F (z) ei (wr- nx) =--G (z) ei (Wt - nx)

...(3.79) ...(3.80)

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 21t

n

=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 ...(3.82)

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

- Cl) G (z)

2

= vs

2

[G" (z) - n G (z)]

...(3.83)

Equations (3.82) and (3.83) can be rearranged as 2 F" (z) - q F (z) = 0 2

G" (z) - 5 G (z)

where

...(3.84)

=0

2

...(3.8S) 2

2 (J)

q =n - 2 vp . 2

...(3.86)

5=n-2

...(3.87)

2

2

(J)

Vs The solution of Eqs. (3.84) and (3.85) can be expressed in the form F (z)

=

...(3.88 )

AI e-qz + A2 eqz

-5Z sz G (z ) = B I e + B2 e

...(3.89)

where A I' A2' B I and B2 are Constants. A solution that allows the amplitude of the wave to become infinity is not possible; therefore A2 = B2 = O.HenceEqs.(3. 79) and (3. 80) can be writtenas

'" - A 't' - le - B '" -

le.

[-qz + i(CDI-nx»)

...(3.90)

[-J'z + j (wr - nx)]

-

,"

.

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

...(3.9] )

IIi

IIIi

iI;{~d'~

"

Wave Propagation in an Elastic, Homogeneous

89


Therefore,

crz = A E + 2 G Ez==1 E + 2 G

and

't

aw -+-

,

,

=G'"

zx

,

IZX

=G ,

00

( '

;=

(3.92)

0

)=0

...(3.93)

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

BI

' (A + 2G)q

AI

- -(Tt + S )

,

and

'2

- An

...(3.94)

2

2

...(3.95)

BI 2i n q Equating the right hand sides of Eqs (3. 94) and (3.95) 16G

2422

2

n s q =(s

+n)

22

2

[(l,+2G)q

"I

-lI.n]

22

...(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 ,. 2

16 1-

(

~\

vpn

)(

=

~\ )

1-

(

vsn

2- (A+2G). G

~\

vpn

)(

2

2-

~\

...(3.97)

vsn J

From Eq. (3.81)

Wave length = 21t = velocity of Wave vr n (ID/2n) - (ID/21t) n =-

or

...(3.98)

ID

...(3.99)

vr

where vr is the Rayleigh wave velocity. Using the following relationships: 2 ID 2'2=2 vpn

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

2 ID 2'2Vs n

and

-

2

2 ro 2 2 2 vs(ro Ivr) 2 Vs

2

2 ...(3.100)

2 - vr -2-~ Vs

2

...(3.10t)

¿ãó®

where

,

...(3.102 a)

Yp

GIp = (A,+2G)/p t

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

or . ,: ";}

t'

".

"

',.

,.

'

(3.102 b) "

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

",

,

':'i' ",+ ..: i

-

. :; ,',;.0

. '/,:

::~:;,,;

J-. ,.

.

i;~'liJi,~n lJ'rU .'",. . , '", I ~..

.c..=;--'-'

,- -- .,,'

'; ,.

=-

--~ ===

It

.. c"

"

,

'.,.

.. ""

90

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

.',-

,},

Soil Dynamics & Machine Foundations ,

--

,

,

,

,

,

,

,

..'

Using the relations (3.8) and (3.9) , Eq. (3.10.2) can be written as 2 1-2)l øÈ =2-2)l Sustituting Eqs. (3. 10.0.), (3. 10-1) and (3. 10.3) in Eq. (3.97), 16 (l-a2~2)(1-~2)

~ (2-~2)2

...(3.10.3)

(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 be noted that the value of ~2 is independent to the frequency of the wave. Therefore the Rayleigh wave velocity is also independent to the frequency and dependent only on the elastic properties of the medium. S

4 >I '0

c 0

3

>"-I >

.... 0

2

tII

-:J 0

>

vr/vs 0 0

0.1

0.2

.

Polssons

0.3 J

.

0.4

0.5

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..alized seismogram (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 from this record. In practice, this determinatiords made'byco'ITl.,j.ningihe}nformationJromseveral geophones at various distances from the source on time-distance graph as shown in Fig. 3.17. " ' ,0

a..,

91 rave Propagation in an Elastic, Homogeneous


Ground motion '" ~

X/Vs

x/vp

Time

-x / vr

~

.' Fig. 3.16 : I~alised

seismogram ofthe ground motion at a distance x from the source , ,

u (\) \J\

Stope

1 /vr

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 .' . - s= i(rot - not) . - q= + B ...(3.106) A ) 11 =- ( In le (se e . . :. s: i{oot not) - q= B ...(3.107) ) (A .

w=-

Iqe

-

"ne,.

e

- -~ '. ~ ."...'.

.,

.11... "'dO"....

.'

"0

. ,"

,i

92

Soil Dynamics & Machine Foundations .

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

wntten

as

- A . -qz+ 2qs -sz e i(oot-nx) 1 m -e 2 2 e s +n ) (

U -

...(3.108)

.

.

-

2

.

2n -sz i(OOt-nx) -qz ...(3.109) W - A I q -e2. + 2 e e s +n ( ) From Eqs. (3.108) and (3. 109), the variation of u and w with depth can be expressed as -

U( Z} -e

.

<-qln)(l/z)

W (Z) ---e

2(qln)(Sln» 2 2 +( s In +1

-(sIn) (nz)

)

.

-(qln) (nz) .+

2 2

...(3.110)

e

(-sin) (nz) 2

e

s 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

2"

=1

n

s

and

2

2" "

=1-

ro

~

22 =1n Vs

Amplitude Amplitude

-

0 :

6

Vr

-22=l-2"=l-a n vp 2

2

Vp 2

3:..

2

...(3.111 b)

~

2

...(3.112)

2 =1-~ Vs

at depth z at surface 0.6

.0.8

H. rizontal cc mponent

O.41-~(Z)J vertical

.I:.

N.. 0

component

g' 0.6

= ~ ..

V: 0.25

.. >

))

- 0.33

[W(Z>]

1.J - O. 40 '" o..t ~~ 0.50

~C

0

...(3.111 a)

1.0

1.2

1.4

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


93

~or a given value of Poisson's ratio, the values of v/vr and (Richart, 1962). Hence values of qln and sin are determinable for = (21t/wave length), the variation of U (z) and W (z) can then be sional term (z/wavelength). ,Such variation is shown in Fig. 3.18 and 0.50.

v/vs can be obtained from Fig. 3.15 a given value of Po isson's ratio. As n studied with respect to a non-dimenfor Poisson's ratio of 0.25, 0.33, 0.40


The amplitude of body waves, which spread out along a hemispherical wave front, are proportional to 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;. waves is slower then that of the body waves.

Thus the attenuation of the amplitude of the Rayleigh

'

3.10

GEOPHYSICAL PROSPECTING

3.10.1. General. Geophysical exploration is relatively new area of technology. It involves measurement of some physical field such as electrical, magnetic etc. on the earth's surface and interpreting the data so obtained in terms of properties of subsurface layers of soil. The geophysical techniques most widely employed for exploration work are the Seismic, gravity, magnetic and electrical methods. Less common methods 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 method utilises the propagation of elastic waves through the earth and is based on three fundamental principles namely (a) the waves are propagated with different velocities in different geological strata, (b) the contrast between the velocities is large, and (c) the strata velocities increase with depth. It consists of generating an elastic pulse or a more extended elastic vibration at shallow depth, and the resulting motion of 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 agation of the pulse in the ground. The ground is generally not homogeneous in its elastic properties and this velocity therefore vary both with depth and laterally.

,

The real stratum, which in fact often consists of stratified material, is usually best approximated by a 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 depth h 1between media in which the compression wave (P-wave) velocities are vplare vp2'vp2being greater. Figure 3.19 shows the.possible paths of 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). A compression or shear wave (ray2) striking an interface will generate two reflected (P and S) and two refracted (P and S) waves (Fig. 3.19). According to the laws of reflection, and refraction: , ,~in ip = sin rp = sin.rs,:= sin Rp = sin Rs vpl

vpl

vsl

Yp2

...(3.113)

. vs2

The equality of the angles of incidence and reflection (e.g. ip = rp) holds only if incident a'rtd

reflected waves are of the same type.

c;.on'"~

...:.J.: ~ '.

lE.

94

"'Soil

.. "

Dynamics

& 'Machine~

Foundations'

',:

<1)

source

"

"

"

I

'

,~

b a.

'

'

p-wave "

J -.

'

v P1

,

.... ~ -a.. -..,,; 'a.

.,.,.

h1

Q:-, I

ir

p

"

..

',>'

,

'~"f

vp2

"0 ~("Cl ~ ...

~Ov "

Fig.3.19: Possiblepathsofbodywaves In seismic refraction only compressional waves (P~waves) are considered and the interpretation is based mainly on the first arrival times derived from the se'ismograms, This is due to the fact that P-wave travels much faster than any other wave, Therefore , for this case, Eq, (3.i 13)' ',can be written as -, sinip

=vpl

sin Rp The above equation

..,,(3.114)

vp2

is Snell's law or the law of re~action.

greater than angle of incidence ip: Whenip

sin Rp = 1. Then

Since vp2> vpJ' angle of refraction

increases; there is a 'unique' case where Rp

-,'

,

Rp is

= 90°

and

'

,

v

. -

pi

sm 'p' - ~ vp2 ,

- . .

...(3.115)

-.5m 'pc

,

,

,

, Angle ipc is called critical angle of incidence. For ip >ipc' the energy is totally reflected in ~he upper layer. If vp2 is less than vpl so that the ray path is refracted away from the normal this critical refraction cannot

occur.

.

~,

',. --

~"

"

" . "

"

. ,,", ,- . jt}cj~~

óôñ¢ñóñ¢Ö "

,-, ", ',,'

, :'

.f

:

d'

: -;

:

-

~;,'.f?'

9S

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

'

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

x

.~ (Sourccz) "S

-1

..

.-

~

.. .,'

(Gczophone) G

,

Vp1

.

hl

..

Intczrtaccz

'

B

A Fig. 3.20 : Typical trajectory

of a compression wave

h. SA = BG = --.! .

...(3.116)

cos ip j3.117)

AB = x-2 hI tan ip Total time, taken by the wave ~eaching from S to G will be: . x - 2nl tan i 2h .

.

'

T=

,

l

,

p

+

Ypl cos ip

.

...(3.118)

vp2

The time T will be minimum when d T - 2 hI sin ip ,-di

.

or

,

sIn,

2 hI --0 2. . 2. Vpl cos 'p Yp2 cos 'p

P '.

Vpl '. P, ' =- -vp2 = SIn,pc

...(3.119)

...(3.120)

.

Therefore the travel tiITlefrom ~ toG via the second layer is minimum when slant ray paths through vpl 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 arrival

of the wave at the a,ppropri~tetime (Fig. ,321} Th~ ,refracted wav.esshown by d~~ed lines are known as head waves. ' . . "

,

1.--

--.

"

,,; .. ~,;..'",'

""",._"Hi;j;

.,,'-

"

'

-,

"

"t-~;." -,.. ,'"

..

" ' , ',' .~,:

"

"

_.'

96


..

2h1tan

5

..

1pc

G1

. I

1

.9/

/

/

/

/ /

.

vp1

/

I.

/

/

'pc I'pc /

/

hI

-- -

vp2

I

/

/ /

/

11

/

/

I

/

/

/

/

/

I

/ /

I

Head

-'

/

/

/

h1

G3

G2

/

/1

/

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 planted in the ground, the times from the impact instant to the detectors can be plotted on a time distance graph as shown in the (Fig. 3.22). The slope of the lines yield the reciprocal of velocities, namely lIvp' Therefore, 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 times are 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. Intercept time is the total arrival time of the refracted waves minus the time'x / vp2, x being the distance between the impact point and the receiving station. It is the intersection between the prolongation of the timedistance 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 the intercept time or critical distance (Milton, 1960; Parasnis, 1962).

~Sou

x

1

r S(Z

S

ipc

h,

ipc

(a) Ray paths

vp1 VP2

Fig. 3.22: Fundamentatprinc:ip'e ofrefrac:tlon shooting (...Contd.)


97


.... 01

E +"

"'" 0

> ...

--- -------

...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)

...(3.121)

Arrival time T I of direct surface waves = ~ vpl

Arrival time T2 of refracted waves =

2h

x - 2 hI tan i c

I. + vpl cos 'pc

P

Vpl

,. sin 'pc

...(3..123)

- vpZ 2hl

Tz = vpl cos ipc

2hl sin ipc sin ipc +~ . vp2 vpl cos 'pc .

2hl

= V

p l cos ipc

x T z -- -+ V z p x T

or or

Z

- -+

vp2

..

cos

=.1 pc 1-2 ,

"

...(3.122)

vp2

2 hl(1- cos ipc) +~ vpz vpl cos, .pc

2 hi cos i pc vpI 2hl vZ -v 2 . Vpl Vp2

vZ

pI

.vp2

z

,j

p2

pI

...(3.124 a) ...(3.124 b)

. 98


. This is the equation of a straight .line with slope l/v p2 and an intercept on the time axis through impact point is given by putting X = O. T2i

'

2hl cos ipc

=

T2i vpi

,

or or

...(3.125)

vpl

...(3.126)

hi

=

h

= Tzi vpI vp2

2cos'i~c ...(3.127)

~ Vp2 2 -V pI 2

I

Critical Distance. Let the critical distance is Xc at which TI = Tz (Fig. 3.22b) 2 hi \',,;COSlpC I' or

2h, vpl cosipc

or

+ .-'c - 2 hi tan ipc

=

...(3.128)

Xc Vp l

V p-')

(~-~ I

- 2hl(l-cos2 ipc) = x vplcosipc cl Vpl .

-.

Vp2)

"'pI

2hl cos lpc - Xc 1--

(

vp2 J

Xc (l-sinipc)

or hi

= X

or hI

...(3.129)

2 cosipc Vp2 -V pI

J vp2

=;

...(3.130)

+vpl

3.10.2.2. Three layer soil medium. Consider a three-layered soil medium with compression wave velocithat v I < v , < v 3 (Fig. 3.23a). If S is a ties of layers 1, 2 and 3 as vp I , vp~ and vp 3 with the condition "p pp source of disturbance, the direct wave travelling through layer 1 will arrive first at A, which is located ')

a small distance away from S. The travel time for this can be given by Eq. (3.121) as T I""x/vpl' At a greater distance x, the first arrival will corresponds to the wave taking the path SDEB. The travel time for this can be obtained from Eq. (3.123b) as

I2 X

T2

= -+

2

2 hlVvp2 -V pI

,vp2

'

vpl

vp2

At a still larger distance, the rust arrival corresponds to the path SFGHIC. According to Snell's law sin ic2

sini

= vpl vp2

sin

= -;p3

...(J.131 )

'

vp2

ic2

...( 1.132)

v pI .--vp2 - vpI

Therefore, sin;

= vp2 vp3

vp3

'~-"':'~",F"'"

'~')""~

Wave Propagation

',",;"

,,'

'

99

in an Elastic, Hom~geneous and Isotropic Medium

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

TJ

2 hI

2 I?

.+

vpl cos I

. +

~ tani - 2 I? tan ie2

x -2

"

vp COSle2

.

vpJ

.

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

= -x

TJ

,

v

pJ

,

s

2hl~v;J-v;1 ,

+,

v

+

v

pJ

2~~v;J-V;2

pI

.pJ

v

.

v

,..( 1,35)

'

p2

A

B ...

LaY
=

h1

sin-1 (vp1 /vp2)

si n-1 (v P1 / v P3 )

., ,

(}

'--",'~"'_:

, . ~ ,...' ,

,f'

,..

.'

,..

-'

,

I

I

.'.

"

.E

.-1 ( vp2 vp31)

IC2=stn,

, . ',..

. "..-'

-

'

,

"

LaY
1

. .,,

,.,.;

"','

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

,~.'~..,

.',

-,:'

~-.',

Or< ,:'

:

,.. ,"

hl .,.~,

"

"

. ,

: 'Layer: 3 ", Vp3

(a)

°¿¬¸-

Ray

.

-- -,---.,

... tII

,~ T3i ...

" -" - ~, -----

.,

0 >

t. TZi 0 '

I I I I I I

...

\J\ ....

u.

: ,'1.

",'P;'

"!;"Time-dis'tan~e

,.' ,I

"

"',~,

ôòØ¢Ý´ùòÄ

,if' 1~i~.:t:ttf

".t.,

'I II

,'.

,xC2

gi;!:o,

'Oistance,'x,.;.,j,.'..\;,

,~. tf."'.l ,:1 ,'.

,.t'.:,)~-1"

""','""""",y-.,: I"",~"P.<:~ ~"",L :.4;,;"..,.,..,., ;.,.,.",',' ',..""";-..,, ,,"')J';"""';"-'-""~"" ,,"" ",."., , .;:,<;( 1:,..",,'d~.J ";""<::~"-'~'>;J!Lf""}Wd,Ll;),I;>",.q"i'1> ~.,..if"""":.;'i:' ,..,l1.".~.1 !U --<;,.'" ,rf::;.:.tj.",.ff' , ," ,Flg..3~~3..:"'~efraction shooting an three layered sod medium, .'

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

.

,,',"

'

""

,

,

,.."

"

",

.FJ~v~~\ '-:',:.,: ,; c;" ,I,-,JJ..J. "..' .

" .._~

.

I I I .1 I I I I ,I::

I .

(b)

'

,

C"

"..

(er ",.1. ~.J\J,', J,c.di;' ,,~' . '",.

100

Soil Dynamic..& Machine Foundations

The records of geophones placed at various distances from the source may be plotted on time versus distancegraph as shown in Fig. 3.23 b. The line OA corresponds to Eq. (3.121), OB to Eq. (3.123 b), and BC to Eq. (3.135): The thickness of the fIrst layer can be obtained either by Eq. (3.127) us-ingintercept timeT2i or by Eq. (3.130) using critical distance xci' The thickness of the second layer h2can be obtained by the two approaches as given below: Intercept time: (Fig. 3.23) TJ i

=

2 hI cosi

+

2 ~ cosic2

vpl

or

...(3.136 a)

vp2

h2 = .!. 2 T3'I

2 hl~V~3 - V~I V p J Vp l

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

.

}

vpl vp2

vp3" .

h =

or

2

:':"'~"'.:

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

Layer

11 vp1

~2

V vpJ -

...(1.37)

vp2

.

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

vpJ vp2

vp3 vpl v

xc2

2'

~

2

vpJ - vp2

...(1.38)

vpJ vp2

(

-v. p3

~

hl Vp2~V~3-V~I-Vp3~V~2-V~I p2-

vp3 +v p2

)

...(1.139)

Vpl ~ vp3 2 -vp22

...... .. toI

E ....

La yer

2 I vp2

0 > ~ ~ 0

Scigment

.... IJI

n

Slope=1/vpn

~

u. Sczgment Layer

n, vpn

1

Slope = 1/ vp1 Di stance

Jx

(b) Time-distance plot

(a) Ray paths

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 distances from the sources of disturbance will pl~t as shown in Fig. 3.24. There ~iII be n segments on the t versus x 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

obtained by detennining hI, h2"""'"

101

alld Isotropic Medium

hn - 1 in sequence. The general equation can be written as : Tni

hn -

1

vpn

Vp(n-l)

Vpn vp(n-l)

2 v~n - v =~

~ Vpn 2 - Vp(n-l) 2

p(n-l) 2

´ÁóÔ

"j=n-2h_ ~j=1

Ö iv-

Pi

Vpn 2

...(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 is

inclinedat an angle 0.with respect to horizontal.The rays such as ABCDmakingthe criticalangle ipc (sin ipc = Vp/Vp2)wIth the

first arrivals.

normal to the refractor, take the shortest time from A to D and are therefore Zd

Referring Fig. 3.25 a,

-

AB = Cl = cos ipc DK = x sin et DK

-

ID =

=

.

COSlpc

...(3.141) ...(3.142)

x sin a .

(3.143)

COSlpc

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

...(3.144) -

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 D for the ray ABCD, i.e. the downdip time T2d is AB BC -CD T 2d = -+-+...(3.148) vpl

= or

T 2d =

vp2

vpl

Zd

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

...(3.149)

2 Zd cos ipc x . . +-sm(lpc+a) vp I v I p-

...(3.150)

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

where

Tu; =-- -.

2 zd cos ipc

...(3.151)

vpl

The apparent velocitYvpll s~ (ipc +a) is equal to vp2sin i pc I sin (ipc+ a) and is smaller than the true velocity vp2since ~in ipcI sin (i pc+ a ) < 1.0 - '0-'

.~..

, ~; "".:u

;. 1H.:;;..r..,JII.

R

102

SoU Jiynamics '& Machine FtlflllilawnS': :/:

It

I-

--i

A

hu

'..,

(a) Ray paths

.' ""-,:,,,-

..

e

0 > 0

T2ui

-:;; ...

T2di

I "

I

. 11incrtosing tor Td .

I I

x increasing

!~ " ;,

.

Itcu

Distance.-

Fig. 3.25 : RefractloD'Surveyjn

soiis

,""',

;,

tor Tu

,--:--'

\vit~~~~n~~dJ.a!erin1:;

...' .. ,

':,

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

I

/tJ1Iehopagatioll

~

in an Elastic, Homogeneous


103

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

.

'e get, T

2u

-x = 2zu cos ipc + -'- sin (ipc - a ) -

...(3.152 a)

'

vpl

vpl

(ipe - et.) which is equal to v2 sin ipjsin (ipe - a) . This

In this caSe, apparent velocity is vI/sin

elocity will be greater than the true velocity vp2 For x = 0, intercept is T 2ui' where 2Zu cos ipe

T2ui =

.t3.152 b)

vpl T2divpI

, Therefore,

Zd =

...(3.153 a)

2 cos ipe , ,

T2divpi h = d 2 cosipc cos a

=

Z

...(3.153 b)

Tu; vpI

...(3.154 a)

2 cos ipe

u

Tu; vpi h = u 2 cos ipccos et.

...(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)

xcd {I-sin (ipc +a)} Zd c

or

=

:

h ,. =,

2'

Xcd{l-sin(ipc+a}

...(3.157)

2 cosipccosa

d

h =

similary

...(3.156)

COSlpe

Xcu

,iU

-

{1- sin(ipc- a} ...(3.158)

. 2cosipc COSa. .

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

",'

',,'". Y" p J

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

',}'\" '

~. r """"', -'

if

'.

','

p,

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

.."

(

!;;~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

...(3.16f

= sin Upc- a)

The true velocity vp2 in the refractor can be derived as follows: v ...(3.16'

sin Upc+ a) = 1 = sin ipccos a + cos ipc sin a Vu v 1 sin (ipc or

or or

- a) = ~= v2u

...(3.16.

sin ipccos a - cos ipcsin a

v I v I 2 sin i cos a. = ~ + J!...... pc Vu v2u

...(3.16

V I 2 --L- cos et = vpl(vU +v2u) Vu Vu vp2 V

p' -

= 2 cos a.

V

,:

(

sin ipc

= VPI vp2

...(3.16 ]

V

2u 2d

...(3.16

v2u+ Vu

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

VK-2

~Z K-l

~ZK

~ Z K+1

VK-1

v

K

VK+1

v

K+2

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


105


The solution may be obtained for a given velocity-depth functions such as v = v0 + k'z

...(3.166)

'

where

v

= speed at depth z

v0 = speed at zero depth k' = 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 k = ~Zk A T °k--' . 2. r-. 1 vk cos ik vkV1-sm ik .

As

ik

. -1

= SIn

...(3.167)

vk

...(3.168)

vrnax) ( ----,, ,

~k 2

..

l

...(3.169)

= Vk1/1- ( vrnax)

~Tk

Horizontal distance traversed by this ray in kth layer t:!.xk =

...(3.170 a)

/).Zk tanik

~ vrnax

or

~xk =~Zk

'

Ft 1--

j3.170

2 vk

2 vrnax

Total time for a wave to travel through N-such layers: N ~ t=L, k 2 k=\

vk

and net horizontal distance N

x

=

~

1

b)

Vk 1------

...(3.171)'

)

Vrnax

~Zkl

vrnaX

t;-I1- l

...(3.1n)

2

( vrnax) '

According to Snell's law:

sinik =

~

sin 90

vrna.x

""l.

v rnax

...(3.173) ,

='-.sinik = sinio ,. = p"(constant) ray parameter. ' -vk --"'~"'-~'O N'" ~ 'o!.;I"',

.. ;,

. it,

,:> ,il M 4,::.~(; 1,~ ','"kP 'i",1iII '~f'A.!iftoj

,~qn

'..'

., .;

115!.; k :;"",

) t

--"

106

SoU Dynamics & Machine

Foundations

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

t

.

...(3.174)

= !.~1 ~V Z

and

x=

r v~l-ii

...(3.175)

pv dZ

0

Zmax

then t and

=

x =

J

...(3.176)

pv dZ

0 (vo+k'z)~I-p2(vo+k'z)2 Zmax

J

p(vo+k'z}dz 0 ~1-p2(vo+k'z)2 Integration of the above two expressions are as under: 2 2 \12 2 - 1 ' 2 1/2 x - - [(I - p Vo) - {I- p ( Vo+ k z) } ] k'p

i

1 (vo+k'z) (I+~It = -In

and

k'

Equation of

v~)

.

Vo 1+ ~2 I-p

(

x can be written as.:

(vo+k'z)

...(3.177)

...(3.178)

...(3.179)

)

2

2

~ 1- iv~

Vo 2

k'p

) + ( z+/1 ) = k,2 i

( x-

1 ...(3.180)

This 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 different angles of immergence into the earth.

--

C,

C2

C3

Line of ----

Cl.

centers

Vo

"'%' ~

k'

)(

Surface

e>+

of ~a rth

'\-~ \~

z (a) Ray paths Fig. 3.27 : Ray paths and time distance curve for linear Increase of speed with depth (...CoDtd.)

-

-

x

'~~' ~~


. (a)

107


Ray

paths

rto!

E

r-

Kx'

. h-1

-K'2

T=

Sin

2vo

Distance.. x (b) Time-distance plot Fig. 3.27 : Ray paths and time distance curve for linear increase of speed with depth

1

V

...(3.181)

Radius of circle = k' p = k~ +Zmax

r vo zmax= k'p -F

or

...(3.182 a) ,

vo vo = k'sini 0 -F ' or

Z

=. '\lo

max

,1C'

p=,{

sinio Vo

}

'

(co

sec 107"

t)

...(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 +

k

~ p'

~(

l.+

= -In 2

l

vl114Jt '

k

where

Vmax

=

);'

vo' 2 ) . ( vmax,

.ok. ~ '\ k ' Z Vo+ . ~ .

'! .1' "~L ," .

-

2 VV~

J

v1-l~) \In

=.:..:..nwl 1-

'- \; "...

) )

...(3.183

Cl)

,

2v X

v~

Vo

~

"

~

=~

v/\

k.

.

cot 10

...(3.183b)

...(3.184)

# h

108

Soil Dynamics & Mllchine

Foundatitlas.

The time-distance relation for a such a circular ray between entry into and emergence from the earth can be obtained-by eliminating p and z from Eqs. (3.178) and (3.179). This can be expressed as - 1- Sinh-I k' x T - 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 maximum penetration 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 3.1 Table 3.1 : Compression and Shear Wave Velocities i.iaterial 1. 2. 3. 4 5.

vp,m/s 600-1750 300-800 1500-4500 3500-6500 4600- 7000

Moist clay and Wet Soils Sand Sand stone Lime stone Granite

I

Vs' m/s

80-160 100-500 . 750-2200 1800-3800 2500-4000

ILLUSTRATIVE EXAMPLES t

"~'Xamp'" 3.1 Wave pro{Ju".,'Jtiontests were conducted near Mathura Refinery, Mathura, for determining the insitu ve locities 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, Vp as 3C mIs. The soil at the site was cohesionless and the position of water table was at 1.0 m depth below tJ ground 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/m3 2. From Eq. (3.64) V

î

=-A+2G

P Putting the values of A and G from Eqs. (3.8) and (3.9) in the above equation, we get p(I+).1.)(1-2).1.) 2 E = (1-).1.). vp 2 - 1000(1+0.25)(1-2 x 0.25) x (300)'2 (1-0.25) . - 76453000 N/m p

= 76453 kN/m 2

-

-~. LiIIiIft

11

ve Propagation in an Elastic, Homogeneous

109


From Eq. (3.9) G =

E

-

76453

2(1 + J.!)- 5(1+ 0.25)

= 30581kN/m2

2 1- 2 J.! 1- 2 x 0.25 1 Cl = 2 - 2 J.1= 2 - 2 x 0.25 = 3"

3. From Eq. (3.103),

From eq. (3. 105), 6 4 2 2 ~ - 8 ~ - (16 Cl - 24 ) ~ - 16 (1 " -

1. Cl

For

) =0

J.1= 0.25

or Therefore,

3 ~6- 24 ~4 + 56 ~2- 32 =' 0 (~2 - 4 ) (3 ~4 - 12 ~2+ 8) = 0 2

~ =2 2+-

'

.

2

s 2" =1-~ n

From Eq. (3.112),

2

J3'

2--

2

J3

2 ,,

For ~2 = 2 and (2 + 21 J3), value of sIn will work out imaginary. Therefore

~2 = (2-2/J3) 2 Cl

From Eq. (3.100),

2

~ = 0.845

= 2 - 1.155 = 0.845

1 x 3" = 0.282

2L =~a2~2 =0.531

vp vI" =300 x 0.531 = 160rnls

From Eq. (3.1~1),

vI" = Vs

JP2 = .J0.854 = 0.9192

r 160 = 0.9192 = 0.9192 = 174 rn/s

Vs

Example 3.2 The data of a refractor test is given below: Distance of Geophones from Source (m)

Travel Time (milli s)

0

0

4

2

8

4

12

6

20

8

25

'.

. :J

9

..

Determine the depth of the refractor using time"intercept approach and critical distance approach.

,-

dO"dd~

0'_-"""",

'<~iL:

ò

äþ þ

ù ô

ù½

óó

þó

ùåùôù¢ù

òòòù

ôòòùò

ó óóòóó

òô óòòòòò

110

ù

Soil DyntUllics & Machine FoundlZlUJr.

Solution: 1. The data given in the example has been plotted on time-distance Fig. 3.28. From this figure,.

graph as shown i

10

8

u

1/,

-

6

E

""./,

.......

""."'"

,

~

//

E 1--

41.r ." I

//

""."'"

I

I I I I

""./

'2i

I

2

I I I I

I I

0 0

4

,xc 12

8

16

Distance

(m)

Fig. 3.28: Time-distance plot for the data given In example 3.2

. Also,

TZi = 0.003 and Xc = 12 m. 12 vpl = 0.006 = 2000 mls

8

=

vp2

0.002

'

= 4000

mls

2. From time-intercept approach (Eq. 3.127) h = Tu vpl vp2 2 vp2-vpl2 42

I

= 0.003 x 2000 x 4000 = 3.46 ID .'

2~4oo02_-20002.{

20

24

';....



111

3. From critical distance approach (Eq. 3. 130) x hi

Vp2-VpI

=.{ J vp2 -v

pI

= Q2 ~/4000-2000 4000+2000 = 3'.46 m Example 3.3 In Fig. 3.29, determine the time of the flIst arrival wave from source Sand geophone G.

t 5.0

~

30m

..Is

~

m

Vp1

10.0 m

vp2

=

500

m I sec.

-t ,. ~=~

2000 mlsec

J vp3

=

3000 mlsec

Fig. 3.29 : Data for example 3.3

Solution:

(i) Arrivaltime T 1 of the direct surfacewave = ~I = :o~ = 0.06S (ii) Arrival timeT2 of refracted wave fro~ flIst layer (Eq. 3.124 b) =-+ x Vp2

v -v 2hl ~ p2 2 2 pi

Vpl Vp2

30

2x5

~

= 2000 + 500 x 2000 "2000 = 0.03445

2

2

-500

= 0.015

+ 0.01936

.'

(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 x 500 + 3000 x 2000 = 0.01 + 0.01972 + 0.00745

= 3000 +

= 0.03715 s M_inimum ~avel ~e " ; ;" i: '-.'

~, O:O~'f4, .s:' .

: ..

; -. -," \

,'.' -

112


Example 3.4 Between two points A and B (Fig. 3.30) two seismic refraction tests were performed and the data obtained is given below: . Source at B

Source at A Distance of Geophones from A (m) 0 5 10 15 20 25 30 35

Travel time (milli sec)

Distance of Geophnes form B (m)

Travel time (milli sec)

0 3.1 6.1 9.3 11.1 12.9 14.8 16.5

0 5 10 15 20 25 30 35

0 3.1. 6.3 9.4 12.6 . 15.0 15.9 16.5

Determine the slope of the refractor, depth of refractor at points A and B, and the two velocities using critical distance approach and time intercept approach.

A

~B

r

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

15

.

vp 1 = Inverse slope of AC ==9.4 x 10- 3 = 1600 mIs, or 10

= Inverse slope of BE = Adopt

vpl

,;' "1600+ 1640 = 1620

2

.""

.

v2u

6.1 x 10-3

= 1640 m/s

~s ,35

= Inverse slope of CD = (16.5-11.2)

',.

.~.

x 10-3 ~ 6603 m/s

"

~~.

Wave Propagation in an Elastic, Homogeneous and Isotropic Medium

Vu

= Inverse

slope of EF

=

113

35 3

(16.5-3.8) x 10-

= 2756 m/s

T2u; = 11.2 x 1O-3s,Xcu= 23.5 m Tu; = 3.8 x 1O-3s,xcd = 14.75 m 18

D 16 ~

14

'" I

"-

".-'-

"...-

I

..........

12 l--

I I

..........

..-

..........

..........

T2ui

, ,

I

,.....

u CII '"

10

I

-.--

I

I

E

,I

8

i.

CiIJ

E .t-

"-

"""-

6

"-

,,

"-

"-

4

" T2di 3.8 sec

2

Distance

(m)

Fig. 3.31 : Time-distance plot for the data given in example 3.4

-

'~--. - ~--~..

~,-.---

J '~ I

114

Soil Dynamics & Machine Foundation$.

.

v I

1600

vzu

6603

2. FromEq. (3.160),sm(i - a) = ...L = pc

or

From Eq. (3.159),

.

ipc- a = 14° Sin (i

+ a) = vpl = 1600 = 0.5805

pc

or

= 0.2423

vZd

2756

ipc + a = 35.5°

Therefore, 3. Time-intercept approach:

ipc = 24.7° and a = 10.75°

From Eqs. (3.153b) and (3.154b) hd- -

TZdivpl 2 cosi pc cosa 3.8 x 10-3 x 1620

= 2cos24.7°coslO.75°

= 3.4 m

h = TZujVpl u 2 cosi pc cos a 11.2 x 10-3 x 1620 - 2 cos 24.7° cos 10.75° = 10.03 m

4. Critical distance Approach From Eqs. (3. 157) and (3. 158) xcd {1-sin(ipc +a)} hd = 2 cos ipccos a 14.75(1-0.5805)

= h = u

2 cos 24.7° cos 10.75°

= 3.4 m

Xcu {1- sin (i pc - a)}

2 cos ipc cos a

23.5(1- 0.2423) - 2 cos 24.7° cos 10.75° =9.97 m . 5: True velocIty (Eq. 3.165),

vzu vZd 6603 x 2756 vp2 = vzu +v2d 2 cos a = 6603+2756 x2 x cos 10.75°

= 3820 " ..

m/se£~ .' ~I ...,.. ;;" ::

,"".~,

115

Wave Propagation i" a" Elastic, Homogeneous and Isotropic Medium

Example 3.5 Following is the data obtained from a seismic refraction test Geophone No. GO Gl G2 G3' G4 G5 G6 G7 G8 G9 GI0

Distance from Shot Point

Travel Time

(m)

(rriilli sec)

O'

0 200 399 592 780 .963 1138 1306 1465 1618 1763

...

100 200 300 400 500 600 700 800 900 1000

Assuming that the velocity is varying linearly with depth, determine the values of depth of deepest point of the wave paths from ground surface corresponding to the waves reachtng to geophones G7 and G9. Also determine the coefficient 'K' in both cases. Solution: (i) Initial velocity,

vo = 200100 x 10-3 = 500 m/s

Velocity of the wave reaching to geophone G7 i.e. 200 v = = 611 m/s 7 (1465 -1138) x IO-~ (ii) Similarly

. - . -I 500 - 54 90

.. From Eq. (3.183),

'0 - Sill . 611 -

2vo K' = -cot x

I

.

. '0

= 2 x 500 cot 54.9 ° = 1 0

700

From Eq. (3.182 b)

.

~ 500 Zmax= k' (cosec io - 1) = 1.0 (cosec 54.9° - 1) = I11m (iii) Similarly 200

=

v 9

= 671m/ s

(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

u

Q,/ \1'1

'E , 10 ~

E ~

5

10

20

15 Distance

25

(m)

Fig. 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

I

PRACTICE PROBLEMS

"

.

117


?.1 Explain the generation of (a) compression wave, (b) shear wave and (c) Rayleigh wave. Describe their relative magnitudes. --. " . . " , 0

,

,0'

. .}.2; ~scijb~ t,he)V.aveprppag~tion in an infinite, homogeneous, isotropic and elastic medium. .' 7.",3.3""Assuming Poisson'S"rntioif(J.tjas-tt35'and"denslt)r ofs'oifils1800'kgliri3 ,dete~i~~ vr if compression wave velocity is 450 mls.

E:G::~ ~ ~~d

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. Travel Time

Distance of Gephone from shot point (m)

"'(MiIli sec) .. ~

5 10 15 20 25 30 40

42.50 85.00 127.50 170.00 187.50 197.50 215.00

c

~~;

Plot the time-distance graph and determine (i) Velocities of two layers ~ = 0.25

(ii) Values of E and G of. the two layers assuming ,

(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 the

following:

.;

,

'

(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 machines have usually low strain amplitude ofthe order of 0.0001%to 0.001%. The soilproperties which are needed in analysis and design of a structuresubjected to dynamic loading are: (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 of elastic uniform shear Ct, coefficient of elastic non-uniform Compression C$ and coefficient of elastic 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 have been 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 ,-,

119

Jynamic Soil Properties'

t.2.1. Resonant Column Test. The resonant columntest isused to obtain the elastic modulusE, shearmodu..; IusG and damping characteristics of soils at Iow strain amplitudes.This test is based on the theory of wave propagation in prismatic rods (Richart, Hall and Woods, 1970).Either a cyclically varying axial load or torsionalload is applied to one end of the prismatic or cylindrical ~pecimenof soil. This in turn willpropagate either a compression wave or a shear wave in the specimen. In this technique the excitation frequency generating the wave is adjusted until the specimen experiencesresonance. The value of the resonant frequency is 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 (Ishimoto and Iida, 1937;lida, 1938, 1940;Wilson and Dietrich, 1960;Hardin and Richart, 1963;Hall and Richart, 1963;Hardin and Music, 1965;Drnevich, 1967;Anderson, 1974;Lord et ai, 1976;Woods, 1978).Severalversionsof torsional resonant column device using different end conditions to'conc;traintthe tt'st specimen are 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 this arrangement one end of the specimen is fixed against rotation and the other end is free to rotate under the applied torsion (Fig. 4.1a). A node occurs at the fixed end and the distnbution of angular rotatione along the

specimenis a 1/4sinewave.

'

., ,

x

e ~l,

t) Y"

:'" X

80,t)

e(l,t)

/

J I.

~-~/

/

J

/ /

/A

/,

/ '

/.

~

~

iJ

//«Y4SintZ

wavtZ

8 (0)

J/Jo

e

=00

(b)

Fig. 4.1 : Schematic (After

of resonant Hardin,

column

with fixed-free

t 970 and Drnevich,

J/Jo=O.S end conditions

t 967)

As shown in Fig. 4.1b, by adding a mass at the free end, the variation of e along the specimenbecomes nearly linear. J and Jo are respectively'the polar moment of inertias of the specimen and the added mass respectively.Dmevich(19~7)usedthe ~oncePtof addedmasstoobtaina uniformstraindistribution'throughout the length of the specimen. . (il) Spring-base model: Figure 4.2 shows the configu~ationof an apparatus which canbe described as the spring.:.basemodel. ,Depending on the stiffness of the spring compared to specimen's stiffness, it can represent either fIXed-free airangem~nt,~rfree-freeconfiguration.In the case when the springis stiff in comparison ~ospecimen, the configuI'~ti,~~may ~~~onsideredas fix~d-free. In such a case, a nodewill occur at mid height of the specimen, and,"".',the distribution of angular rota~on would then be a 1/2cosine wave. , " " . , ,

,

. -. "";t""' '-:;""'J1J.)'.IJ'..~",.

"".,',

".'. ,..,..,0.;", ',';.;'Ii..:o.",:,:L.."f,i;.'i{}I.,,'il.,

~ 'r,-

t<.'.I

,~" :.'1'\II

"-;::";"":"~'~;"" ,:-,\

'-:--:.

~'I.''. -;

-

----

,-,

",-

~=.

::'::::,=:T

-

..,

if~

'~ 120


½±²²»½¬»¼ to platen

passiv

Soil sp
Activll

/

Ilnd

platen

W
f.~~

~ WQ:ightlQ:ss

"

portion { rigidly

~

of

vibration

connQ:ctlld

torsional

Q:xcitation to

da shpot

dllvicll

platlln

Wllightlllc;c;longitudinal dashpot

L:';9hil...longitudinal Fig. 4.2 : Schematic

spring

of resonant

column

spring with spring-base

model (Woods, 1978)

(iil) Fixed-partially restrained: Figure 4.3 showsthe configurationof an apparatusin which the top cap is 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 effect of shear strain amplitude on shear modulus and damping. It may be noted that in the usual torsional resonant column test, the shear strain is not constant but varies from zero at the centre of the sample to a maximum at the outer surface. In hollow specimen, the variation in shearing strain acrossthe thickness of the cylinder wall becomes relatively very small. The configuration of this apparatus is similar to Fig. 4.1b and therefore the shearing strain is almost uniform along the length of specimen. Using Drnevich's apparatus, Anderson(1974) tes.tedclaysupto 1% shearingstrainand Woods(1978)testeddense sandsupto shearing strainof 0.5%. "

"

4.2.1.1. Calibration and determination olG and~. Hardin (1970) suggested the following procedure of calibration of the apparatus described in Fig. 4.1b: (I) For this model the vibration excitation device itself, without a specimen attachedis a single degree of freedom system.Firstly remove the specimen cap and the additional rigidmass, connect the sine wave generator to the vibration excitation device and vary the excitation frequency to determine the resonant frequencyf"I of the device. . (il) Attach the additionalrigid mass of polar momentof inertia Jo' and determinetheresonantfrequency !"A of the new system.

';C~

121

ynamic Soil Properties

Driving

~

\,

for C(l --~

.

Sp(lcimqn.

non- rigid

dis'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 specimen can be obtained using Eq. 4.1. 2

Ko

-

= ;1t 1-

J Ai

:

1"2 n

...(4.1)

InA

( 1nl )

(iii) With the added mass removed and with the specimen cap, specimen and all apparatus, determine the resonant frequency/"0' The value of mass polar moment of inertia of the rigid mass, Jocan be computed using Eq. 4.2. "Ko . Jo

'

= 41tln20

;,/,.0

".

... "'-_'__.0"'_"-

122

Soil Dynamics & Machil,e Foundations

Taring

spring

Long, LVDT

/

Rot .accelerometer Rot.LVDT

Top 'cap

.:,

lid

~,.:

,""

" 'I

!:~j

, ..

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

M

em branes

!:'.: -.. ;

,': I,' ~,:

, , '

,,-. ~'. :

,'. .

;:-' ".

,-.. ~ .

;' :-

~

, ..

Sand

",-

",

-,'

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

'J

pore-pressure

....-

BoHom

To bae k pressure

transdueczr

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 curve compute the logarithmic decrement for the apparatus, °A' as follows: 1

AI

0A = -n logeA" where

...(4,3)

AI = Amplitudeof vibrationfor thefirstcycle An

= Amplitude

of vibration for the nth cycle.

Under steady state vibrations, the apparatus damping constant, DAis given by

- °A WT DA - -"Ko 10 1t

..,(4.4

.~ 123

Dynamic Soil Properties

(iv) To measure the torque current constant, Kt' excite the apparatus successively at frequencies

/"0'

(J2 / 2)

.fi Ino and 2 Ino' During

the steady state vibration at each of these frequencies measure the current flowing through the coils, C in amperes, and the displacement amplitude of vibration, e in radians. For each frequency compute the torque-current constant Kt' as follows:

eKo

~

...(4.5)

= CMf

where Mfis given in Table 4.1

Table 4.1 : Value ofMf

Frequency

.

Mf 2

-

()f..

. r.,

I ,.

(.fi)

y

1110

-I

2/110

3

The value of Kt shall be taken as the average of three measured values. The three measured values should not differ by more than 10%. After calibration, the specimen shall be placed in the apparatus with minimum disturbance. A known value of ambient pressure is applied as done in triaxial compression test. With the power as Iow as is practical, the resonant frequency of the system'/"R' is obtained by varying the frequency of excitation. At resonant frequency, the amplitude of vibration, eR, in radians and the current flowing through the coils of the vibration excitation device, CR are measured. Now the power is cut off and the record of decay curve for the free 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= where

f)

...(4.6)

rtd2f g

W = Total weightof specimen I = Length of specimen d = Diameter of specimen -, . ~ ,- ,

g = Accelerationdue to gravity, (it) Calculate the inertia of the spe~imen_ab
,

4

.,.

J=- 1t pd 1 32 ,. ;-,

,'.

"

...(4.7)

~.

124

SoU Dynamics cl Machine Foundations

(iii) Calculate the system factor, T ~s follows: 10 Ko T = --I 41t21fni

...(4.8)

10 = Mass polar moment of inertia of the apparatus, derIDedby Eq. (4.2).

where

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). 100

\

SO 40 30

\ f-

20

~

\ i\

""

10

'I0 t>I

::I 0

>

5 4 3

" '"

-

'-

-

"

"""""

2

..........

--

0.6 - I 0.1 0.2

I

I

I

0.3

I

I

0.5

0.4

value

'"

0.6 of

........... I

I

0.7

0.8

0.9

1.0

F

Fig. 4.5 : System factor T versus F

G = 4.2p

...(4.9

(I,; ./)'

(v) For steadystatevibrations,the dampingfactorof the system,Ds is givenby Ds

"-

-

1 KICR' 2 2 2 --21tDA

FnR

)

41t J ftiR ( ~~

.0

,0...

...(4.H i '

., , . .:~

----

125

'liS

DYllamic Soil Properties

.8)

(vi) Compute O/T . Using Fig. 4.6, determine the va.lue ofR corresponding to the value of T computed in step (iii). Then damping ratio is given by

~ = 0.5

Os TR

...(4.11)

(vii) For the free vibration method, using Fig. 4.7, determine the value of mode shape factor Cmfor the value of T computed in step (iii) 100 ii).

50 11

20

~ 010

10

~, I: '11

~

::I CJ >

!I !I

5

2

.(4.9)

0.7 1.0 -

4.10)

1.1

1.2 Value

1.4

1.3 of R

Fig. 4.6 : System factor T versus R

~

--

1.5

126

Soil DYllamics & Mac/tille Fou1Idatio/ls

100

I'

so I' I1I 1'1

20 lil

.....

'+0 ~ ::J

!

10

5

0 >

I I;

I: 11

2 jj

I, I !

'

I:

0.6 1.00

I Ii I'

1.02

1.01 value

1.03

of CM

Fig. 4.7 : System factor T versus mode shape factor Cm (viii) Compute the system energy ratio, S as follows: <>

32Kol

S =

...(4.12)

4

1tC",Gd f

The value of damping ratio

I

I I I

given by

~= ~[Os(l+S)-OA 21t

I

1 I

~ is then

S]

4.2.2. Ultra Sonic Pulse Test. The theory of ultrasound is similar to that of audible sound. Sound is the result of mechanical disturbance ora material; that is. a vibration. Ultrasonic pulses of either compression or shear waves can be generated and received by suitable' piezoelectric crystals. Using elastic theory, a relationship between the speed of propogatiQn and wave amplitude of these waves and certain properties of the media through which they are travelling can be determined as follows: 2(1+11)(1-211)

I

E=pvc

G -- pvs2

I

1

...(4.14)

(1-jl)

...(4.15)

r

-l-

...(4.13)

.~

---

~~-


127

2

I-!

VC

J..L-=

2

(

Vs

)

...(4.16)

1-(::]' ~

u

where

Ao 2.302 I ogl0n An

=-

...(4.17)

= Velocity of compression wave p. = Poisson's ratio G = Shear Modulus

Vc

Vs =

Velocity of shear wave

E = Young'sModulus p = Mass density = y /g Ao = Initial value of amplitude

() = Logarithmicdecrement An = Amplitude after n oscillations.

Lawrence (1963) described the basic apparatus required to measure the propagation velocities (i.e. Vc and vs) through sand. Stephenson (1978) described an equipment for conducting the ultrasonic tests. His equipment 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 simultaneously with a 7 volt trigger pulse to th~ time base of the oscilloscope. The gen_e.ratorwas designed such that the pulse 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 sediments while still retained in the core liner. The drawback of this approach is that it is very difficult to identify the exact wave aITival times. Secondly the strain amplitudes which can be achieved by this technique, are only in the very low region. G.L. ", ".

Typical

-±·´

IZlem~nt

¨ Í±· × ³¿--

Rocl<

""""-

Bascz met ion (a) Shear

er

wave induced

in soil by horizontal

CJo

!

er

-t

+

--1: .

B

x---x. --

earthquake

vibrations

A- Soil element at rest and in mean position B- Maximum d
+"t

A

(b) Shear deformations

C resulting from propagated wave (for a single cycle)

Fig. 4.8 : Stress condition on an element of soli below ground surface

128


4.2.3. Cyclic Simple Shear Test. During an earthquake or other source of ground vibrations, a soil element below a foundation or in an embankment is subjected to an initial sustained stress together with a superimposed series of repeated and reversals of shear stresses (Fig. 4.8). The magnitude of induced shear stresses depend on the magnitude of acceleration of the dynamic force. In a direct shear box test, uniform state of shear strain occurs only on either side of failure plane. The simple shear device was designed to overcome this limitation of direct shear box by enabling a uniform state of shear strain throughout the specimen. This simulates the field conditions in a much better way. Kjellrnan (1951), Hvorslev and Kaufman (1952), Roscoe (1953), and Bjerrum and Landra (1966) have described simple shear apparatus. The Roscoe simple shear device has a box for a square shaped sample with side length of 60 mm and a thickness of 20 mm. The box is provided with two side walls and two hinged walls (Fig. 4.9). Peacock and Seed (1968) have modified Roscoe's simple shear apparatus for dynamic testing. The dynamic shear forces were applied by a double acting piston with controlled compressed air pressure using solenoid valves. Figure (4.10) illustrates how the end walls rotate simultaneously at the ends of the shearing chamber to deform the soil uniformly. Prakash, Nandkumaran and Joshi (1973) have also developed similar type of simple shear apparatus which has the facility of applying sustained normal stress, sustained shear stress and oscillatory shear stress.

Wire wound or stacked

membrane disks

~

'-

t:--

Hinge (b)

(a) Fig. 4.9 : Schematic

arrangement

of Roscoe simple shear

apparatus

Prakash, 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 been done by Dass (1977) on clay of high compressibility (CH, LL = 65.5%, PL = 28.0%, qu=78 kN/m2) using the same dynamic simple shear apparatus. The salient features of her work are reported here to understand the behavior of soil under dynamic load. The static strength of soil was 36 kN/m2. She performed the tests

keepingthe'variationof variousparametersas givenbelow: . Sustained normal stress, on (kN/m2) Sustained shear stress, tst (~of static strength)

= 15,21,26,29 = 10,25,50,60.

129


shear stress, 'Cdy n ( kN/m2) = 13,19,22,25 Number of cycles of oscillatory shear stress = 1to 1100 The oscillatory shear stress was applied at the rate of one cycle per second. The f::lilurecriteria was chosen corresponding to 12mm displacement. Oscillatory

Sh ea ring chamber

Pian

Soil , , sample

D~==4J ùÄ÷óããð÷

\

-~

ÅÃóã ==(]

(0==0

L ....

7

)

End plate

rotation

Soi I deformation

Elevation Fig. 4.10 : Schematic diagram illustrating rotation o(hinged end plates and soil deformation in oscillatory simple shear (After Peacock and Seed. 1968)

""""

130


A typical plot oftest data in terms of number of cycles versus shear displacement is shown in Fig, 4.11 for "Cd y n equal to 13 kN/m2. Similar plots were obtained for other values of 'Cdy n . From these plots, number of cycles and oscillatory shear stress corresponding to 12 mm displacement have been obtained and plotted as shown in Fig. 4.12. In Fig. 4.13, plots between dynamic shear stress and sustained stress are shown for different number of cycles. It can be seen from Figs. 4.12 and 4.13 that for a fixed sustained shear stress, the amount 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 cause failure.

Number 1 01

of

cyo;l(S

10

1000

100 Undisturb(d

on: , tdyn~

0.21 kg/cm2

2

0.131 kglcm

2 '[sf

in '1. of normal strength

L.

~

E E

c ~ v OIl

'6

c a..

1/1

"

... c s:. 8 OIl

1/1

10

12

Fig. 4.11 : Shear

displacement

versus

number

of cycles "

";1";",


~

131

120 Sustained stress

shear

in percent

.s::. 100 .... Cl> C ... .... .... III

of normal

strength

CIn= 0.21 kg' cm

2

c E .... 80 0 c

.... 0 ... C tJ

~ 60 ... a. c III III tJ

....

t;

!oO

.... Cl QI .s::. III >-

.... 0

E '-

20

u III 0

0 1

10 No. of Fig. 4.12 : Oscillatory

100 cycle shear

at

1000

failure

stress versus

number

of cycles

4.2.4. Cyclic Torsional Simple Shear Test. In cyclic simple shear apparatus, it is not possible to measure the confining pressure and the test is performed under Ko consolidation conditions. Torsional simple shear devices have been developed to overcome these difficulties. Ishihara and Li (1972) modified the triaxial apparatus to provide torsional strain capabilities. This has the disadvantage that the shear strain in the sample varies with maximum at the outer circumference and zero at the centre. This problem has been minimised by using 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 has the advantage that both resonant column and cyclic torsional shear tests can be performed in the same device and on the same sample. , '

-.

,{

,._.,---

132


.

.s::.

.01 C t\I ....

V1

. Undistu r bed

0---0

I

100

RemouIded Number of at cycle at failure

D

0 E .... 0 C

0

....

.-C 11\ 11\ t\I ....

.... 40 11\

.... 0 t\I .s::. 11\

'-u E 0 C >Q

"-

"-"-

20

-.;;;:

"

0-

0

20

Su stained

40

sh ear s tress

60 in percent

80

100

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 conditions atthe test section (Hardin, 1971;Drnevich, 1972,Ishihara and Yasuda, 1975;Iwasaki et aI, 1977).It is difficultto prepare samplesfor longhollowcylinder devices.Keeping this fact in view,Yoshini and Oh-Oka (1973); Ishibashi and Sherif( 1974),and Cho et al (1976) used a short cylinderin whichthe taper wasproportionalto the insideand outside radii (Fig. 4.14). The internaland externalradii rl and rz are selected such that the difference in average shear stressescomputed considering two extremeconditions is minimum. The two extremeconditions 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)

"~1~?~~~~~\;;~~:i~;~~:. .'~~'.:,'~::ii~;.~;i:~~S~;,;"'~i~i

Dynamic

.

:~:{{;~w'. .,. ",,)!~:~'~.-i~:'x'i'~'~i:1ff~f:t;:~~i6!;r{"~f\ti~,

Soil Properties

133 3

3

4T r2 -Tt 'tmoe= Z 2 4 4 31t [ (rZ -rt )(rZ -rt) ]

3T

- -

...(4.18)

I 3

tavp - 21t ( r2 -rt

3

...(4.19)

J

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-deformation and strength characteristics of a soil depend on the following factors: 1. Type of soil 2. Relativedensityin caseof cohesionlesssoils;consistencylimits, watercontentand stateof disturbance in cohesive soils 3. Initial static stress level i.e. sustained stress 4. 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 both ; compression and extension is done., All the factors listed above can be studied lucidly on a triaxial set up.

134


Prof. Arthur Casagrande of Harvard university was refered a problem of studying the effect of vibrations created by bomb explosion on the stability of the Panama canal projects. For this Casagrande and Shannon (1948, 1949) developed the follo~ing three types of apparatus for studying the strength of soils under transient loading (Table 4.2). Table 4.2 : Type of Apparatus Type of apparatus

Remarks

Time ofloading (seconds)

(i) Pendulum loading

0.05 to 0.01

(ii) Falling beam

0.5 to 300

(iii) Hydraulic loading

0.05 to any desired larger value

Suitable for performing fast transient tests

Time ofloading was defined as the time between the beginning oftest and the point at which the maximum compressive stress is reached (Fig. 4. 15). The pendulum loading apparatus (Fig. 4. 16) utilizes the energy of a pendulum which, when released from a selected height, strikes a spring connected to the piston rod of a hydraulic lower cylinder. This lower cylinder is connected hydraulically to an upper cylinder, which is mounted on a loading frame.

"0 0 0 J

I"

~

Time

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 contrthe velocity of the fall of the beam, and a yoke for transmitting the load from the beam to the specimen (Fi 4. 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 pur connected to a hydrauliC cylinder through valves by which either the pressure in the cylinder or the volUl of the liquid delivered to the cylinder can be controlled. The peak load that can be produced by this .apT ratus 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'~*~~,


133

4T

'tave

3

3

2

4

r2 -r)

=-

2

'tavp

=

3T

1

21t

( ri - r\3)

...(4.18)

4

31t[ (r2 -Ij )(r2 -r\)

] ...(4.19)

where T is the applied torque.

x--

x

~/~~"~ Boundaries

indic a t ed

Fig. 4.) 4 : Cross-section

by heavy for short

Iin es

specimen

4.2.5. Cyclic Triaxial Compression Test. In general the stress-deformation and strength characteristics of a soil depend on the following factors: 1. Type of soil 2. Relativedensityin caseof cohesionlesssoils;consistencylimits, watercontentand stateofdisturbance in cohesive soils 3. Initial static stress level i.e. sustained stress 4. 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 compressionof the sample is done while in two directional loading both compression and extension is done. All the factors listed above can be studied lucidly on a triaxial set up.

t'

134


Prof. Arthur Casagrande of Harvard university was refered a problem of studying the effect of vibrations created by bomb explosion on the stability of the Panama canal projects. For this Casagrande and Shannon (1948, 1949) developed the follo",:,ingthree types of apparatus for studying the strength of soils under transient loading (Table 4.2). Table 4.2 : Type of Apparatus Type of apparatus

Time of loading (seconds)

(i) Pendulum loading

0.05 to 0.01

(ii) Falling beam

0.5 to 300

(iii) Hydraulic loading

0.05 to any desired larger value

Remarks Suitable for performing fast transient tests

Time ofloading was defined asthe time between the beginning of test and the point at which the maximum compressive stress is reached (Fig. 4. 15). The pendulum loading apparatus (Fig. 4. 16)utilizes the energy of a pendulum which, when released from a selected height, strikes a spring connected to the piston rod of a hydraulic lower cylinder.This lower cylinder is connectedhydraulicallyto an upper cylinder, which is mounted on a loading frame.

"0

0 0

-I

J..

~

Time

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 contrthe velocity of the fall of the beam, and a yoke for transmitting the load from the beam to the specimen (Fi 4. 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 pur connected to a hydrauliC cylinder through valves by which either the pressure in the cylinder or the volur of the liquid delivered to the cylinder can be controlled. The peak load that can be produced by this apf ratus 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">"

135


~

UpptZr cylinder Adjustable reaction Deformation gage

Hydraulic cylindtZr

Te st sptZcimen

LowtZr cylindtZr Fig. 4.16 : Pendulum

loading

apparatus

(Casagrande

and Shannon.

1948)

Countczrwcz ight Fixed

fulcrum

. Load gagcz D
gogcz

TtZst sptZcimtZn RidtZr

Spring Doshpot Fig. 4.17 : Falling

beam apparatus

(Casagrande

& Shannon,

1948)

136


pr«ssure r«gulator (pressure of discharge may b. quickly varied by manual control. b«twun 10 and 1000 lb. p«r sq. inch)

Three position nlec tor valve (with Pto A. B is Conn«cted to T and vice verso; with Pto T. and 8 ore blocked)

F Iow control valve (valves will maintain A constant flow betwEen Sand 1250 CU. in. per min. rega rdless of pussure)

van«-type hydraulic pump 100 lb. per sq. inch 2.5 GPM 5 HP motor 220V,AC 1200 RPM Hyraulic cylindu 3 inch bore, 3 inch stroke (used to apply load to piston of triaxial compression apparatus)

Filhr Twenty gallon oil 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 mounted on the inside face. For measuring deflection, a thin flexible steel spring cantilever is used with strain gages mounted 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 different materials. Out of these, two typical materials named as Cambridge clay and Manchester sand having properties as given in Table 4.3 are selected for illustration. The transient compression tests were performed with different time of loading both in confined and unconfined states.

Table 4.3 : Properties of Soils used in Tests Manchester sand

Cambridge clay Natural water

Grain Size

0.21 mm to 0.42 mm

Content

30-50%

Liquid limit

37-59%

el113X

0.88

Plastic limit

20-27%

el11in

0.61

In Fig. 4. 19, a simultaneous plot of stress and strain versus time from an unconfined compression test with a time of loading of 0.02 s on cambridge clay is shown. Similar plots were prepared for other times of 10adll1gand on Manchester sand. Using this data, stress-strain plots were obtained as shown in Figs. 4. 20a and 4.20b. In these figures, stress-strain curves for corresponding static tests are also shown. Typical plots of maximum compressive stress versus time of loading (or unconfined and confined transient tests on Cambridge clay are shown in Fig. 4. 21 a and b respectively. A typical plot in terms of principal stress ratio a failure and time ofloading for Manchester sand is shown in Fig. 4.22. .

~&.

137

)ynamic Soil Properties

8

4

6

3 I-Sh
at 0.02 s

N

E

0

~

u

Cl .:ill:

2

4. .:: 0... ....

~

III III t:i

tJ)

...

..... V)

2

0 0.24

0 0

0.04

0.08

0.12 TimlZ~ s

0.16

Fig. 4.19 : Time vs stress and strain in an unconfined (Casagrande & Shannon,

0.20

transient 1948)

test on Cambridge

clay

From the typical test data presented above, it may be concluded that the strength of clay decreases with the increases in time of loading and for time ofloadingequal to 0.02 s, the strength of clay is approximately 1. 75 to 2.0 times greater than the static strength. The strength of sand is almost independent to the time of loading. Transient strength of sand increased only about 10 percent. 0

static t
2

0

3

-0

4

0

c

a..

I

I

I I

d

... ..... S

Transi
(j')

6 7

8 0

0.4

0.8 stress,

1.2 1.6 kg/cm2

(a) Cambridge clay Fig. 4. 20 : Stress vs strain curves (...Contd.)

2.0

2.4

138


0

0

static te'st. time of fa i lure 2100 s

2

;;..

Shear fa i lu re

.-c 3 0 ... +' (/)

I Transiqnt test tim~ to failurq 0.03 s

4

5 6 7 0

0.50

1.75

0.75 1.00 stress. kg/cm2 (b) Manchester sand

Fig. 4. 20 : Stress vs strain curves N

-

E

u

C7' oX

, IJ'I IJ'I tII

-... 1/1 tII

2.6 2.4 2.2 2.0

> -

1.8

tII

... a. E 0

1.6

E ::J .-E x 0

1.2

1/1 1/1

u

).4 0

1.0

:E 0.8

1000

100

10 1 T i m q of 10a din g. s

0.1

0.01

(a) Unconfined compression test Fig. 4.21 : Maximum compresslve stress versus time 01 loading for Cambridge clay (...Contd.)

139

)ynamic Soil Properties

6.0 [TlTIIII N

:1\11111

I

illll

I

I I

I

I

1111

I I

5.8

-

/

E

v

5.6

/

et

-tr oX

..

I b

11' 11' ~

5.4

/

5.2 5.0

...

/V

~ 4.6

~

E 4.0 ::J .-E 3.8

/

V ./

0

x 0

-

"

//

4.2

~

~

0

/

...

E 4,4 0 >

/

/

4..8

I

/1

3.6 ": 1000

I 1

hili

I I

I

IIIII

Time

I

I II

compressive

loading compressin

stress versus

I

time of loading

00

0.0

E

::J

+0

.-E X

0

~~

....

for Cambridge

clay

,..

-

6

~

0.01

test

bM

CO

I

S

7

v-

IIIII1I

0.1

0

.~

I

1

of

(b) Confined Fig. 4.21 : Maximum

I I

10

100

0

(1

v

0....

-0

5

-

0

0

0

lA III ~

...

4 11111I 1000

I I

111\1 I I

100

,Ill'

11111 'I'

I I

10 Time of loading,

Fig. 4.22 : Maximum

principal

stress ratio versus

1111\,,°,

0.1

0.01

S

time of loading

for Manchestor

sand

Modulus of deformation is defined as the slope of a line drawn from the origin through the point on the stress-deformation ~urve and correspon~ing to stress of one-half the strength. It is found that in case of clays, modulus of deformation in fast transient tests was about two times that obtained in static tests. In case of sands, modulu.sof deformation was found independent ~othe time of loading.

""'}f><;C

140


As evident from above discussions, in transient loading test, an initially unstressed sample of soil is loaded to failure in a short period of time. Under earthquake loading conditions, an initially stressed soil element is subjected to a series of stress pulses, none of which would necessarily cause failure by itself, but the cumulative effect of which is to induce failure or significant deformations. Seed and Fead (1959) developed the oscillatory triaxial apparatus as shown in Fig. 4. 23 to study the effect of number of stress reversals and other factors on the deformation in soils.

Counter- ba Ion c e for loading yoke Deformation gauge

Air pressure equal to desired confining pressur on specimen

Dynomometer Triaxial compression

r- -EI-;ct;icalconnections ceH

to timing

:

I

Soil specimen in rubb(zr membrane

unit

I

I I Air .

Compressed

.

,

air Air pre c;sure

Ipre ssu re fr ~gauge :re9ula,tor '

.

r.J1

I J

Countczr to record number of load applications pressure cylindar Bellofrom seat Loading

piston yoke

Air pressure Fig. 4.23 : Apparatus

for oscillatory

triaxial

reservoir

test (Seed and Fead, 1959)

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

'--.--.-.------.-

141


Seed (1960) performed tests on Vicksburg silty clay(wn = 22%, S =93%) for studying the effect ofvarious factors on its dynamic strength characteristics. A typical stress strain- relationship is shown in Fig. 4.24. It pertains tQ the sustained static stress of2 kg/cm2. The magnitude of dynamic stress was 35% of the sustained static stress. It gives the magnitude of dynamic stress as :t 0.70 kg/cm2. The static stress-strain curve indicating the static strength as 3.0 kg/cm2 is also ~hown in the figure. Therefore the initial factor of safety for this typical case will be 3.0/2.0 = 1.5. Transient strength of the soil corresponding to 0.02 s t~meof loading was found as 4.0 kg/cm2. In this figure, B represents the point which is obtained after 100 cycles of loading. It may be noted that at point 'B', the strains are increased but factor of safety still remains more than unity (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 0

0

1

st res s) kg/cm 2

2

3

4

5

str
4

-

- -fat factor safety, 1.5

8

"i

-a

a

12

-

,

.-c: 0 ....

-

\

70

SO 100

16

stress curve for sample in normal str
20

24 Confining

pressure,'l

kg/crl

- - 1 - - - - II I

B \

...... \JI

.-0X

Strain induced

10 during earthquake load ing

\

\ \ \

Stress strain curve for sample after earthquake loading

28 32 36

Transient strll"gth for time of loading 0.02 s 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 values of sustained static stress and dynamic oscillatory stress. Fig. 4. 25 shows the effects of single transient stress applications of the various intensities for initial factors of safety ranging between 1.0 to 2.0. The shaded portion of the figure shows the deformation of the specimens induced at stress levels corresponding to the different factors of safety in normal strength test. The upper curves show the increase in deformation caused by single transient pulses corresponding to 20, 40 and 60 percent of the initial sustained stress. Figs

142


Foundations

4.26 and 4.27 show similar data for a series of 30 and 100pulses. It is interesting to note that for this soil a single transient pulse equal to 20 percent of initial sustained stress causes no appreciable deformation even though the factor of safety may be as low as 1.1.However it may be seen that a series of 100such pulses for the same initial factor of safety will cause an increase in axial strain of 10percent. The significance of increase numbers of stress pulses in producing increased deformation of soil is readily apparent from these figures. 30

Sing!(z transient

28

Simulahd earthquake: 30 pulses at 2 cycles p /lr se c . In c r/lase In stre ss during simu lated earthquake = 60 °/0

pulse

26 2l..

IncrllaSIl in strain dull to transillnt stress

22

=

=

,...... ;;. 20 .......

20

.::

18

18

0

~III 16 0

14

x
12

Transient stress 'in c rlla se = 60°/. = 40%

.-c0

... +-

III

.-0 )(

et

10

.

16

40 °/0 20 °/0

strain induced during si mu lated czarthqu a ke load ing

14

12 10 8

6

6

4

4

2

2

01.0

0-

1.0

Fig. 4.25 : Soil deformations induced by various combinations of sustained and transient

stresses

(Seed.

1960)

2.0

Fig. 4.26 : Soil deformation induced by various combination 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 reached at an axial strain of about 25 percent. If this strain is adopted as a criterion of failure, then a variety'of combinations of initially sustained stress and earthquake stress intensities which will cause failure of the soil may be obtained from Figs. 4.25,4.26 and 4.27. Figure 4.28 shows the combination of initial stress and earthquake stress intensity causing 25% axial strain (i.e. failure). For single transient stress pulse the combinations of sustained and transient stresses would have to approach about 140 percent of the normal strength to induce failure. For earthquake inducing 100 majQJ.pulses failure would occur when the combined stresses totalled only 100 percent of the normal strength. -, ,

;;~f&~;i.:~t~,"%?r~~~t,+;i0't~~'~£~~~:!,,~~~},~~{~;.c'j}

,,:~;':i''~:'i,

143


.

30 28

\

26 24

/

22 20

.-0c:

.... ....

-0

11\

x <{

160

Simula-ted earthquak e: 100 pulses at 2cycles per sec .Inerescz in stress during simula-tczd . earthquake =60ol

,

Number of transient pulses

stress

1/

:: 40O}

/

:: 20°/01

18,

--0

0

16

c ---

14

11\ 11\

12

.... ...er. 60

80

~ :x

10

I

g0-

8

40

.c: .... ....

6

::1,

20

4 2 0 1.0 -

Fig. 4.27 : Soil deformation induced by various combination of sustained and vibratory stresses for 100 pulses

(Seed.

1960)

Fig. 4.28 : Combination of sustained and vibratory stress 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 clay do not in any way represent the characteristics of conditions producing failure in other types of soils. For example in sensitive undisturbed clays, repeated deformations will lead to increase in pore pressure and a resulting reduction in strength. Consequently a series of transient pulses is likely to induce failure at lower stress levels than for the silty clay. On the same oscillatory triaxial test set up Seed and Chan (1966) carried out more elaborate study on different types of soils. Here, few typical results are presented. Figure 4. 29a compares combinations of sustained and pulsating stresses that induce failure in soft and compacted clays in one transient pulse. The strength exhibited by undisturbed Silty clay was greater than that displayed by compacted soils. As the number of pulses increases to 30 (Fig. 4.29b), failure occurs in sensitive soils at considerably lower stress level than that in compacted soils.

~

144


160

160

c... 14-0

L:. 140

CJ\ C c:I

"I-

-11

..... 0'1 C t.I

120

~ 120

0 E "-

g

0

~

100

0

+0

-

.... 0

80

'"

80

t.I ... +-

60

0

..

Compacted sandy clay

0

III

'" c:I ...

t-

100

c

0

0

Compacted silty clay

...

"

'" '"

60

'"

C'I C

....

C'1 C

0

'" 40

.....

'" 40 undisturbed ::J 0

::J a.

'"

silty

a.

20

clay

20

StrClss pulse torm ---

stress

00 20 40 60 80 100 Su~tain
of stres£ conditions

,

0 20 40 60 80 1 Sustained str~ss/O/o of normal strengt (b) For 30 pulses

(a) For single pulse Fig. 4.29 : Comparison

pulse to rm

0

causing

failure

fo. different

soils (Seed and Chan.

1960)

Symmetrical stress pulses of two directionalloadings resulted in a reduction in strength of all the soils tested. 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 is either smaller than or equal to sustained stress. Figure 4.31 shows the results of pulsating load tests with one-directional loading on duplicate specimens of San Francisco Bay mud using the two forms of stress pulse. The longer dwell period under maximum stress for the flat peaked pulse causes larger deformations and induces failure in a smaller number of stress applications 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 comparison, the stress strain relationship obtained from normal strength test is also shown. It may be noted that in situations involving 10 stress pulses, total stress versus total strain is somewhat higher than the stress strain relationship of a normal test; if there are 100 stress pulses, this is slightly below the normal plot. Further it was noted that with an initial factor of safety between 1.5 and 2, and 30 pulses of dynamic stress, the relationship between total stress and total strain will coincide almost exactly with the normal stress versus strain relationship.

Dynamic

145

Soil Properties

180 San Francisco WatflT cont~nt

-

Bay mud ~ 91 °/0

Str
stress pulses

. _. - Non-symm~trical i)--r stress puls~s (no axial exhnsion)

-

120

0

E

....

0 c

....

100

0

0

... III III

80

....

III

60

Cl> C

, ,

.... 0

III

40

:;

Cl.

20 00 20 Sustained

40 60 80 100 strQ55,o/. of static strQn9th

Fig. 4.30 : Combination of sustained and pulsating stresses causing failure in one- and two- directional loadings (Seed and Chan, 1966)

200

San froncisco boy mud -Wo te co nten t ~ 91 0/0 -Unconsolidoted undroined te sts '-Contin ing pressu re = 1 kgl cm2 -No sustained strczss

0 E

'-

-g

160

0

... C

~

120

'-;;::, c.I... a.C'I .-CCt:.I

For m' of .M..stress pulse I Fo rm of stress pulse

~';;80

1ft 1ft c.I ....

... 1ft C'I

C

40

...

0 1ft :J £l.

0 1

10 100 1000 10,000 100,000 No. of pulses to couse foi lure

Fig. 4.31 : Effect of pulse form on number

of pulses causing

~-,

failure

(Seed and Chan. 1966)

,------..-.-.....--.

Et:...

146


140 Total stress (initial+pulsating) for initial factor of safety between 1.5 and 2.0 vs. total strain after 10 pulsczs

120 .J:

.... C\ C .... III

0'0.-'- ." ..'

."

~"'--;-'~

0

".

100

stress vs. strain relationship in normal compression tczst o

-

~::'"J:.o'\'

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

0

E ... 0 C

1"

(::~::..Y.,r.:'.:~:;.:.::-

I

0

-

IJ\ IJ\ t.I ... .... U1

,

'£"-"

',::',::

Total stress {initial + pulsating)for initial factor of safety betweczn l.S and 2,0 vs,total strain after 100 pulses

I

0

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

80

'+-

;;-

,.'t."

I

601

40

I: San Francisco Bay mud Wo te r co ntent z 91 % Un conso Iidatczd -und ra ined te st ' . 2 Con f Inln9 pressure" 1 kg/cm

20

Stress

00

2

6

4

Fig. 4.32 : Relationship

between

8 strain,c/.

10

pulse

form

14

12

total stress and total strain (Seed and Chan. 1966)

under

J1.r

pulsating

18

16

load conditions

4.2.6. Final Comments on Laboratory Testing, The laboratory techniques discussed above and more common in use are listed in Table 4.4. The specific soil properties measured by each test are also indicated, The range 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) Techniques

Resonant

column

Shear modulus

Young's modulus

damping

x

x

x

)(

x

Material

Cyclic stress behaviour

Attenuation

-

x

With adaption Ultrasonic

pulse

-

Cyclic triaxial Cyclic simple shear Cyclic torsional Shear Dynamic 1 D compression

x x

-

x

-

x x x

x x x

x

I"

~

,'-

~",......

-

BIll DZI!

.. :am

..

!!IB\Im

111:

147


strain

Shearing 104

103 column

(°/.)

102

10' I

I

I

Resonant

amplitude

(solid samples)- - --

Resonant

column

Torsional

shear

(hollow (hollow

--

samples) samples) Cyclic triaxial

Pulse methods

sh ea r

Cy cl ic simple Shaking r

Ty pie a I motion Properly machine

cha racteristics

designed foundation I

I

table

.

strong ground shaking from earthquake

Fig. 4.33 : Shearing strain amplitude capabilities or laboratory

I

apparatus

Close-in nuclear explosion (Woods, 1978)

Each laboratory test has some merit when compared with the other. The resonant column device is better 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 of soils. 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 or on the response of soil structure systems to dynamic excitation. The following methods are in use for determining dynamic properties of soil: 1. Seismic cross-bore hole survey 2. Seismic up-hole survey' 3. Seismic down-hole survey 4. Seismic refraction survey 5. Vertical block resonance test 6. Horizontal block resonance test 7. Cyclic plate load test 8. Standard penetration test In this section, the above listed field methods have been described briefly along with the typical test setups and methods of interpretation.

--

III..

.

Q%!III!'It;j'~"

148

Soil Dynamics & Machine Foundatio!q

l

4.3.1. SeismicCross-borehole Survey. Thismethod is based on the measurementof velocityof wave propagation from one borehole to another. Figure 4.34 shows the essentials of seismic cross-holesurvey as outlined 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. Shear wave velocity is then computed by dividing the distance between the boreholes by the travel ~e.

Reciver

o~

Source

boreholes ~

0

~o

0 (a)

Plan

view

Oscilloscope

@

Ver tical impulse

Trigger

Input

borehole

~

Vertical velocity transducer

Assum ed path of bo dy waves

--

~

/

~

Generation of body wave s 3D velocity transducer wedged in place (b)

Casing grout Cross-sectional

and

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 the source for causing wave generation and another having geophone for recording travel time. However, for extensive investigations andbetter accuracy, three or more boreholes arranged in a straight line should be used.

..

..

!Ri!.II

149


In this case the wave velocities can be calculated from the time intervals between succeeding pairs of holes, eliminating most of the concern over triggering the timing instruments and the effects of bore hole casing and backfilling (Stokoe and Hour, 1978). Also this arrangement of bore holes in a straight line overcomes 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 around the 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 borehole 10 m or more in depth should be surveyed using an inclinometer or other logging device for determining verticality (Woods, 1978). (iii) Boreholes should be spaced as close as possible within the time resolution characteristics of the recording equipment. Large spacings can lead to difficulties with refracted waves arriving before the direct transmission through the intervening soil. Spacings as close as 2-3 m can be used satisfactorily (Woods, 1978). (iv) The seismic source must be capable of generating predominantly one kind of wave. Further it must also be capable of repeating desired characteristics at a predetermined energy level. Miller, Troncosco and Brown (1975) have described a source which is capable of .developing high amplitude shear waves. 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 sides of the borehole. Wfl ight

Rod

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/

150


Foundations

4.3.2. Seismic Up-Hole Survey. Seismicup-hole survey is done by using only one borehole. 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 (Gote et al., 1977). This method gives the average value of wave velocity for the soil between the excitation and the 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 the desired type. . 4.3.4. Seismic Down-hole Survey. In this method, seismic waves are generated at the surface of the ground near the top of the borehole, and travel times of the body waves between the source and the receivers which have been clamped to the borehole wall at predetermined depths are obtained. The arrangement used in seismic 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 betweer layers of greater velocity provided the geophone spacings are close enough.

R
W
f

Expand
Rubb
,Back plate

wooden plo te

3-componqnt gQophonq

Fig. 4.36 : Seismic: down-hole survey (Woods, 1978)

(.t

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

151

Dynq.mic Soil Properties

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 dip angle 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 values of 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 pit of pIan dimensions 4.5 m x 2.75 m and depth equal to the proposed depth of foundation. Foundation bolts should be embedded into the concrete block at the time of casting for fixing the oscillator assembly. The oscillator assembly is mounted on the block so that it generates purely vertical sinusoidal vibrations. The line of action of vibrating force should pass through the centre of gravity of the block. Two acceleration or displacement pickups are mounted on the top of the block as shown in Fig. 4.37 such that they sense the vertical motion of the block. A schematic diagram of the set up is shown in Fig. 4.37. Motor osci \lator assembly ,

"

For V te st Adc e le ra t ion transduc Clrs

Concr(te (M 150)

77777// /, (a) sectional

1

view

4.5Qm

to-

--i

1 m min

T

1.Sm

~

1&:::11

,

0.75 m

I

,1

(b)'

C:Jj ~

plan

2.75 m

1m min

view",

Fig. 4.37 : Set up for block ~esonance test

-

..- - - -

-.

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

152


The mechanical oscillator works on the principle of eccentric masses mounted on two shafts rotating in opposite directions (Fig. 2.15). The force generated by the oscillator is given by 2

Fd = 2 mee CJ.)

...(4.23)

The oscillator is fIrstset at a particular eccentricity (e). As evident from Eq. (4.23)higher the eccentricity more will be the force level. It is then operated at constant frequency, and the acceleration of the oscillatory motion of the block is monitored. The oscillator frequency is increased in steps, and the signals of monitoring pickups are recorded. At any eccentricity and frequency the dynamic forceshould not exceed 20 percent of the total mass of the block and oscillator assembly. The amplitude of vibration (A) at a given frequency lis given by

a-

...(4.24)

A- (mm) = ---:;=- 2

-

where

az

41t-f

= Vertical acceleration of the block, mm/i

f = frequency, Hz. Amplitude versus frequency curves are plotted for each eccentricity to determine the natural frequency of the foundation-soil system (Fig. 4.38). The natural frequency,f. n~~, at different eccentricity (i.e. force level)

is different because different forces cause different strain levels of the block which may be accounted for when appropriate design parameters are being chosen. The coefficient of elastic uniform compression (Cu) of the soil is then determined using Eq. (4.25)

4 1t2 1.n2m

C =-~ u

where

fnz

...(4.25)

A

= Natural frequency of foundation-soil system, Hz

m = Mass of the block oscillator and motor, Kg -sec2/m 2 A = Base contact area of the block, m From the value ofC U obtained from Eq . (4.25) for the test block of contact area A the value ofC u I for the actual foundation having contact area A I may be obtained from Eq. (4.26) CuI

rE

= 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 01 Cu 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 Cu

where

Cs

= (1- ~ 2) . JBL

~ = Poisson's ratio B = Width of base of the block

L = Length of base of the block Cs = Coefficientdependingon LIBratio

...(4.27

153

,amic. So.il Properties

200

150

Vertical vibration .0/8

~O' 0

0

0 e

c: 0

tt

-A

e

v

It-

~

8 : 1050

E

I:r

..

'8 : 1400

III ~

:

35

: 70°

QI

"::J 100 ..... a. E ~

il

r-

8 ---4

so

15

. 45 Frequency,

50

cps

Fig~ 4~38 : Amplitude versus frequency plots from vertical resonance test "

-

154


Barkan (1962) recommended the values of Csfor various LIB ratios as listed in Table 4.5 Table 4.5: Values orcs (After Barkan, 1962) lJB

Cs

1.0

1.06

-

1.5

1.07

-

2.0

The value of damping ratio

1.09

3.0

1.13

5.0

122

10.0

1.41

~is determined using Eq. (4.28) ~=f2-f1 2fnz

...(4.28)

A where f2,Jj = Two frequencies at which amplitudes is equal to Amax

= Maximumamplitude

hzz

= Resonant frequency

If

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 set up as shown in Fig. 4.37. In this test, the mechanical oscillator is mounted on the block so that horizontal sinusoidal vibrations are generated in the direction of the longitudinal axis of the block. Three acce~eration or displacement pickups are mounted along the vertical centreline of the transverse face of the block to sense horizontal vibrations (Fig. 4.37a). The oscillator is excited in steps starting from rest condition. The signal of each acceleration pick up is amplified and recorded. Rest of the procedure is same as desqribed for vertical block resonance test. Similar testscan be performed by exciting the block in the direction oftransverse axIs.

The amplitude of Horizontal vibrations (Ax) is obtained using Eq. (4.29). A = ~ x

ax

4n f

2

...(4.29)

where ax(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,flLl of 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( ' '' " " f . ,

.. ..,", , I, ,

lIB

2:,;:r',£{';:;~~{;I;!ffl~'$(>" .

i2;.;:~

155

lie Soil Properties

)m. The mode of vibration is obtained by plotting amplitude versu: :y of the system from the analysis of data from the pickups mountt block. 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

-

[,00

., ,

III C 0

'u

E 200 Q,I ~

::J

... Q. E cd:

1.

1-~ ~8-t

100

010

1 15

20

25

30

Frequency

I CpS

35

40

[,5

Fig. 4.39 : Amplitude versus frequency plots rr~m honzon~al resonance test

',.

157

'amic Soil Properties

The coefficient of elastic uniform shear (Ct) of soil is given by the following equation: î î èï¬®·³æ

Ct= (Ao+Io):t where

~ (Ao+Io)

...(4.30) 2

-4rAolo

Mm r = M 1110

hx = Horizontal resonant frequency of block soil system Ao = A/M A = 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 passing 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 passing through centre of contact area of block and soil and perpendicular to the direction of vibration. I = Moment of inertia of the foundation contact area about the horizontal axis passing through the 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 the ystem 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 2 Ct

= 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,~AAl IS 5249 : 1977reconunends the followingrelations between Cu' and Ct, Ccpand CII': Cu = 1.5to 2.0 Ct Ccp =

...(4.32)

...(4.33)

3.46Ct

...(4.34)

CIjI= 0.75Cu

...(4.35)

4.3.7. Cyclic Plate Load Test. The cyclic plate load test i,sperformed in a test pit dug upto the proposed base level of foundation. The equipme.nt is same as used in static plate load test. Circular or square bearing plates of mild steel not less than 25 mm,thickness and varying in size from 300 to 750 mm with chequered or grooved bottom are used. The test pit should be at least five times the width of the plate. The equipment is assembled according to details given in IS 1988-1982. A typical set up is shown in Fig. 4.41.

1" I 158

Soil Dynamics & Machine Foundations'

Sand

bags to give a of SOOkN

reaction

ðòêð³õó

ò

1.0m

10 m

-+

0.60m-4

Wooden sleeper 2m xO.30mxO.15m 2n05. on either

side

Prov jng rl n,9

Foundation

f---

1.Sm

level

Plate 300 mm x 300 mm

elevation

Sectional

-,

100mm -L J,Sm

3nos R.S.J. 300mm x lS0mm ot LoOO mm clc top 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 removed and dial gauges are set to read zero. Load is then applied in equal cumulative incrementsof not more than 100 kPa or of not more than one fifth of the estimated allowable bearing pressure. In cyclic plate load test, each incremental load is maintained constant till the settlement of the plate is complete. The load is then released to ~ero and the plate is allowed to rebound. The reading of fmal settlement is taken. The load is then in-

f-Tr~

,.,.".:.,..,.,.""II\~-

-

!IIiI

III

159

mic Soil Properties

;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 of is then applied. The cycles of unloading andreloading are continued till 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 intensity us elastic rebound is plotted as shown in Fig. 4.43, and the slope of the line is coefficient of elastic JITn compressiOn.

Ð

Ý ã -(kN/m) Í»

í

...(4.36)

ïï

° ã Load intensity in kN/m2

where

Se

= Elastic rebound corresponding to p in m.

s.~ Load intflns ity, p pz .p 3 p,

>-

z2l_--r---

~

\1\ C C:II ~ C

1-j__Sfl3 - ---

p

"

"0

l~i-

0 J

Se4 -r-'-

a. ...

-

_t_SflS

-------

-fElastic Fig. 4.42 : Load intensity versus settlement in a cyclic plate load test

rflbound,

Sfl

Fig. 4.43 : Load intensity versus elastic rebound from cyclic plate load test

3.8. Standard Penetration test (SPT). The standard penetration test (SPT) is the most extensively used situ 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 the borehole is cleaned.

ºæ®¬¢òò¢ô¢óó

..

-"""

91:

III

159

mic Soil Properties

;ed to next higher magnitude ofloading and maintained constant rill the settlement is complete, which 1is recorded. The load is then reduced to zero and the settlement reading taken. The next increment of isthen applied. The cycles of unloading andreloading are continued till 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 intensity us elastic rebound is plotted as shown in Fig. 4.43, and the slope of the line is coefficient of elastic Jrm compreSSiOn.

Ð

Ý ã -(kN/m) Í»

í

...(436)

ïï

° ã Load intensity in kN/m2

where

Se

= Elastic rebound corresponding to p in m.

s.~ Load int
p,

pz

'p 3

>-

z2l_--r---

-4J

\11 C C:II -4J C

1-j__S
p

"0

_t-

0 0

..J

Se4 -r-'-

a....

-

_t_S
-------

-fElastic Fig. 4.42 : Load intensity versus settlement in a cyclic plate load test

r
S
Fig. 4.43 : Load intensity versus elastic rebound from cyclic plate load test

3.8. Standard Penetration test (SPT). The standard penetration test (SPT) is the most extensively used situ 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 the borehole is cleaned.

--~

-

mE

160

Soil Dynamics & Machille Foulldations

(ii) The split-spoon, attached to standard drill rods of required length is lowered into the borehole and rested at the bottom. (iii) The split -spoon sampler is seated 150 mm by blows of a drop hammerof 65 kg fallingvertically and freely from a height of750 mm. Thereafter, the split spoon samplershallbe further driven 300 mm in two steps each of 150 m. The number of blows required to effect each 150 mm of penetration shall be recorded. The first 150 mm of drive may be considered to be seating drive. The total blows required 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 last 300 mm penetration. In case, the total penetration is less than 300 mm for 50 blows, it is entered as refusal in the borelog. (iv) The split spoon sampler is then withdrawn and is detached from the drill rods. The split barrel is disconnected from the cutting shoe and the coupling. The soil sample collected inside the barrel is collected 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 more than 1.5 m whichever is less. Tests may be done at lesser intervals (usually Q.75m) if specified or considered necessary. The penetration test in gravelly soils requires careful interpretation since pushing a piece of gravel can sreatly change the blowcount. 4.3.8.1. Corrections to obsetYed SPTvalues (N) ill cohesionless soils. Following two types of corrections are normally applied to the observed SPT values (N) in coh~sionless soils: Corrections due to dilatancy: In very fine, or silty, saturatedsand, Terzaghi and Peck (19~7) recommendthat the observed N-valllesbe corrected to N' if N was greater than 15 as 1 ...(4.37) N' = 15 + -2 (N-15) Bazaraa (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 thus temporary increase in shear strength will occur. This will lead to a N-value higher than the actual one. Since sufficient experimental evidence is not available to confirm this correction, many engineers are not applying this correction. However this correction has also been recommended in IS: 213 1-1981. Correction due to overburden pressure: On the basis of field tests, corrections to the N-value for overburden effects were proposed by many investigators (Gibbs and Holtz 1957; Teng 1965; Bazaraa 1967; Peck, Hanson and Thornburn 1974). The methods noffi1ally normaly used are: Bazaraa (1967) For 0"0< 75k Pa 4N 1-N = 1+0.040"

I

i

D11IIII

--

-~.

miiYL~.M£!,,:

161


For aa-> 75kPa

4N ...(4.40)

N' = 3.25+0.01ao Where

a 0 = Effective over burden pressure, kPa

Peck, Hanson and Thornbum (1974) ...(4.41) N' = 0.77 N loglO 2000 ao 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,

~

. 2000 = Correction factor = 0.77log10 ~ ao

correction

foetor

CN "

0.8

0.4 0

1.2

1.6

...(4.42)

.

2.0

0 CL .:.:.

... t>1

'-

100

:J III

III t>1 'Co

200

C t>1 ,;:)

'-

:J .D 'e:,I >

300

.0 t>1

>

t

t>1 '4'tW

400

500 Fig. 4.44 : Over burden correction

There is a controversy whether the correction due to dilatancy should be applied first and then the correction due to over burden pressure or vice-versa. However in IS 2131-1981, it is recommended that the correction 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

c,,: ..;..i.:.,,:;~:t,'4':~',:V'c,:e;\.;~';:'.:,,:;:f.l'!.:;.:;X!!::*,:~,/'::":t ::\i"l//
~',,;;"..{"':"i:."<'" ':.,'.L"

. ,""'c',','?,,,,:..."" ;'"

,"~j,htc\.U~~.:.v

f

162


0

0

4

Standard' penetration 8 12 16

", It = 20 KN1m2 2

--Yl~sub = ~OKN 1m2

3

,,

"\

\

resistane. 20

.."-

"-""'\ \

\

\

\

\ \t "-

"-

"-"

5 E .s:: Q.

28

Obse rved Corrected

\

4

N 24

"-

"-""')

6

Cl

0

I I I I

7

8

9

10

11 Fig. 4.45 : Typical SPT data

.

'

I

I

32

~-_.-

III .~

163

y"amic Soil Properties

The SPT is es.sentiallyundrained test for the duration of each blow and the energy generated by the PT hammer isprincipally shearing energy.Therefore the test maybe useful to predict the dynamicbehaviour fsoils. 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, Vs mls): Vs = 91 ~.337 ...(4.43) Bowles (1982) has given a number of equations to obtain stress-strain modulus Es on the basis ofSPT md cone-penetration test (CPT). These equations are given in Table 4.6. Table 4.6 : Equations for Stress-Strain Modulus Es by SPT and CPT (After Bowles, 1982) SPT Sand

Es = 500(N+ 15) Es = 180oo+750N Es = (15000to22000)inN

Clayey sand Silly sand Gravelly sand Soft clay

Es = 320(N + 15) Es = 300 (N + 6) Es = 1200(N+6)

CPT .,~ = 2 t04Qc 2 Es = 2(1 + Or )Qc

Es = 3 to 6Qc Es = 1 to 2Qc Es = 6 to 8 Qc

Using the undrained shear strength: CIt

Notes:

Ip > 30, or organic

Es = 100 to 500 Cu

Ip < 30, or stiff

Es = 500 to 1500 Cu

I < OCR < 2 OCR> 2

Es = 800 to 1200 CII Es = 1500 to 2000 CII

1.Unit ofEs is KPa in correlations with N. 2 Es will have the same unit as of Qc 3. Es will have the same unit as of Cu

4.4 FACTORS AFFECTING SHEARMODULUS,ELASTICMODULUSANDELASTIC CONSTANTS Hardin 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;

'}

164


(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 related with 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 been discussed 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 different stress-strain path forming a hysteresis loop as shown in Fig. 4. 46. The area inside this loop represents the energy 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 or no loss 9f energy. The s~ear modulus under these c.onditions is maximuI? but as the strain amplitudes is increased, the shear modulus decreases and the damping within the soil increases.

ñ îð

¶ Ù³¿¨

0 CL

/

.:tr.

~

10

lOO cyc les

/

la

/ 1

I I I I I I

/

cycles

1 cycle

, I I I L

óïð

10

Y x 104

Dry

cl~(jn

mm/mm

sand

G 57

I

Fil!. 4.46 : Stn'ss-strain

ac

20

111°1)and points

= 25 kPa

Ù³¿¨ =

f)') M po

at 1st, 10th and 100th cycles of lo:ldin~ (After

Hoadll'Y, 19115)

w' ~' "'11!' '., .i " J

>i>t---

'.-tt!,;

165

namic Soil Properties

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 considered constant.

Magnitude of

-6 '9

train

Phenomena

wav

Constants

Shear modulus,

seismi c waye method ::::J W In situ E W ... test vibration

I

+"'

c

I

I

ratioJdamping

I

-S[ld1

I compac tion, liquifacation Failure

Elastic-plastic poisson's

10

10

10

Cra cks,ditferllntia se'ttlczmcznt

Elastic

-1

-2

-3

-4

10

propagation. vibration

Mechanical character itics

-

-5

1

Angle OT ratio inNmal friction, cohcsion

.,.

I

,..,.,

'

c::::J

III

RcpIloted

E

load ing test Iwavtl propa\)ation

...

c

te st

+'

E

>-

g

0 W

0"""W

o:J .a'"

0 0 ...JCII

E

R sonant

column

I I

I

I

I

test

Repeated

loading

I

r

test

Fig. 4.47 : Strain

J

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 of soil is elastic under vibrations, It is to avoid the building up of any residual strains in the soil due to the operation 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 amplitudes, 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 factors on G is negligible (Hardin and Richart, 1963). They have reported the results of several resonant column tests in dry Ottawa sands as shown in Fig. 4.49. Straight lines have been fitted through the test points corresponding to different confining pressures.,Similar results are also shown inFig. 3.12 (of chapter 3) as solid 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 strain : i~,d.~~en ~en, t Of the ampli~de for the~e tests was 10.3rad, It can be seen from these figr. es t v,~ and~.~r gradatIon and gram slze'dlstnbutlOn. The effect of confinIng press fi m
.

cant .

',

'

~~

S

.ff",'. I?"!"""'j'.!' . ..' . . . . ,.11. I, ", .J .',..".."t, ", ',,' Ic."""""V,.." 1_"i:.'>J """ """" .)."$O""",,,~ '-' ~ .co': . ..

.t1

/i,~

y

' .~:~,.,., ,','.:; f -:"C>-" ':i.j>;""Aj ,:.;~;"", '""",..; ,,-7."J -'..' .;/\ I\:00.>" '-.,.7 " ",,;;I

"

-

~~

-~~---~"UIIII

&

167

)ynll1ltic SoU Properties

Void. ratio

,Sy'm.

Max.

I

Minl

.

0.71

' , 0.495

0

0.89

0.54

A

0.66

0.32

11

0.76

0.42

390 360 u

-

" ,

::-.

11\

330

E

(J'

'H'

0 .. 3

,

.. AOOk",/ 'h 2

..

11\

>

300

..

->.u

-0

270

>

et

240

> 0

14

...

210

0

e,

.I:. If)

So kN/tr] -----

2

0

180

150

.

"

120 0.35

0.65 Void

.

Fig. 4.49 : Variation ,

""

of shcar-wa\'c

'

.

and gradation :,.i"':'

C'""','

ve!ocity

"

ratio~

wit,h void ratio for various

J,','

'","

and Richart, C,"','

,'",

"',',,;",:';,

0' ,", ,>

confining

pressures.

grain

SiLl'

.

ô÷òò¢ôøþò¢þòò·ååÖùæùå .',

e

in dry' OUa\V8 sand (Hardin

.~,"';';")'

,0.75

,,'

1963) '.. ,',

".

~.

168

Soil Dynamics & MachiIJe ,Foundations

Following empirical expressions have bp.endeveloped for Vsand G for round-grained sands and angular-grained crushed Quartz (Hardin and Black, 1968). '

For round grained sands (e < 0.8) ...(4.45)

= (11.36-5.35 e) (ao}O.3

Vs

2 G= 6908(2.l7-e) ,l+e For angular grained sands (e~0.8) Vs

(a )°-'

...(4.46)

0

...(4.47)

= (18.43-6.2e)(ao)0.25 2

G = 3230(2.97

l+e

where

Vs

- e)

...(4.48)

Ca )°.5 0

= shear wave velocity in m/s

G = Shear modulus in kN/m2

a 0 = Mean effective confining pressure in N/m2 for Eqs.(4.45) and (4.47) ; and in kN/m2for Eqs. (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 >.~ u 210 0

-; 180 >

. Dry

. Drained

ISO

0 Saturated 120 ZO

70

SO

100

Pressure,

150

ZOO

500

cr, kN/mZ 0

Fig. 4.50 : Variation of shear wave velocity with confining pressure for a specimen of Ottawa sand in the dry, saturated and drained conditions (Hardin and Richart, 1963) Seed and Idriss (1970) have suggested G

=

the following lOOOK (ao)05

equation ...(4.49)

where G is in kN/m2units, K is an empirical factor which varies according to relative density of sand. and cr0 is the mean effective confini~g stress in kN/m2 units. Table 4.7 gives some values ofK obtained from field measured values of shear modulus.

169

vnamic Soil Properties

Table4.7: FieldMeasured.ValuesofK

.

(Seed and Idriss, 1970) K

Soil

7.5

Loose moist sand Dense dry sand Dense saturated sand

10.0

Dense saturated silty sand Dense saturated sand

14.0

Extremely dense silty sand

19.0

Dense dry sand (slightly cemented)

.36.0

Moist clayey sand

26.0

13.0 16.0

~.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. On he 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 some lndisturbed clay specimens collected from the field. The following relation has been suggested: G=C Nhere Cl is constant; Equation

(2.971

l+e

.,

er (-cr ).0.5

...(4.50)

0

and G and <10 are both in kN/m2 units.

4.51 has been

suggested

by Hardin and Drnevich --

(1972b)

for both clays and sand.

2

= 3230(2.97-e) G (l+e)

.

(

OCR)

k «

1 )0.5 0

where K is function of plasticity index (Table 4.8), and OCR is over consolidation ratio. Table4.8:

ValuesofK -. .

(Hardin and Drnevich, 1972b)

Plasticity index PI -

k

00

0.00

20

0.18

40

030

(j)

0.41

00

0.48

> 100

0.50

.

...(4.5 -

170

Soil Dynamics & Machille Foundations

552,000

.

Tap wahr

A Salt 0

kaolinite

flocculated

Di spczrsed

clay

v Flocculated

420,000

kaolinite

clay

o~:

,-.. 0 Cl.. oX

'-' V\

::J ::J

280,000 (i\ ~ .

"'0 0

(6QO)

E

\

.\

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

.... 0

"~:.

.J:.

If)

0 0.5

0.7

1.1 0.9 Void ratio e

1.3

1.5

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: 0.5

G=

625 (OCRl

O.3+0.7e

ao

2

( Pa)

where Po is the atmospheric prt:ssure in the same unit as of ao' G is in kN/m2units.

;I j \

171

'amic Soil Properties

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 a sshole survey (SCS) with SPT 'N' values. G = 12000No.8

...(4.54)

. ere G is in kN/m2 units, 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) 1max

Yr

Ì '"Cma

=

...(4.55)

Gmax

'. .

- -,- - ---

X

11

-------

/I

,J I 16

-------san d

, I max I J /1 I

---------

.--//

//

I I

y/

.If..

1/

'/

V

~ /

I

/

I

---

(.\0'/

//

//

.

'; /1

t

/

r;

T=

I

-1

¥'

+- }>

6maxI'max

:

I

'¥r Fig. 4.52 : Hypcrbolic

y

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 to imaxintersects the horizontal line at t

= tmax:tmax

is the shear stress at failure in the soil. It can be obtained

1the following manner. Figure 4.53a shows a soil element at a given depth being subjected to vertical and orizontal effective stresses of ery andKo cryrespectively.Koisthecoefficientofearthpressureat rest.The lohr circle corresponding to stresses cryand Ko cryis shown as circle 1 in Fig. 4.53c. From the geometry of the circles 1 and 2, we get ...(4.56)

'm" = [{~(l+Ko)cr.Sin++ccos+r-H(l-Ko)cr..n

172

cv

erV

max

""(

'tmax KoO=-v

Ko erv .~.'t,

I..

( a)

(b)

.-'

~

, U\ III <:)I

'-

+' III

ay

Koo-v

E f f
nor 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 the appropriate equation from Eqs. (4.46), (4.48) to (4.54). Thus the value of reference shear strain Yrcan be evaluated. The hyperbolic stress-strain curve 'which defines the initial loading curve and also the end point of the

complete stress reversal loop is given by Eq. (4.57). 't =

1 -+-

Gmax

As

G=-

Y

~;

1 'Ymax

...(4.51)

,.

't

...(4.58)

r

ft,

Combining Eqs. (4.55), (4.57)and (4.58), we get G G = ---1!!illL ..

l+l Yr

173

)ynamic SoU Properties

Using Eq. (4.59), one c~n obtainthe value of G at any strain amplitude, y. For every small strain ampliudes, 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

...(4.60)

max ) ( 1- GG

Combining Eqs. (4.59) and (4.60), we get,

l Yr

.!: .,

...(4.61)

=~max ---;;;1+I

Yr

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 sands

33-1.5 (log n)

Clean saturated sands

28-1.5 (logn)

Saturated Silt

26- 4 a~2 + 0.7/12 -1.5(logn)

Saturated cohesive soil

31-(3+0.03/)

-1/2 112 ao + 1.51 -1.5(logn)

n - Number of cycles a 0 - Mean effective principle stress, kg/cm2

1 -Frequency

of loading, Hz

Forcohesionlesssoils, the valueof ~max is dependentonlyon thenumberof cyclesof loadingn while for silts and clays;the frequency of loading/(Hertz) and the mean eff~ctive principal stress (ao in kgf/cm2) influencethe maximum damping ratio.

ô

~~eIit

":'~;~2;;:\ æå¢¢áå¢æ·¢¥·º¬ùï·ô¶æ¢¢¢îæ¢¢åô¢æ¢¢¢ô¢¢¢¢¢¢·¢¢Û·Ð¢Öëæù '.':2:~~:'; ··¢ù æùæ

ååùôæ£å·ø

þ¶¢

174


I~USTRATIVEEXAMPLES Example4.1

-

,

A soil specimen was tested in a resonant column device (torsional vibration, Fixed -free condition) for determination of shear modulus. Given a specimen length of90 mm, diameter 35 mm, mass of 160 g, and a frequency at a normal mode of vibration (n = 1)of800 cps, determine the shear modulus of the specimen. Solution: I. From Eq. (3.47) Cl) =--1 (2n -1)1tvs , .+t;,

" .:>

n

2

VS

=--

/

n=1 2/ro n - 2.(0.090) .(27t.800) = 288 m/s 7t

1t

2. Mass density of Soil in the specimen

p=

î ðòïêð

òðíë

7tø

3.

2

÷

ã ïèìèòékg/m3

.0.090

G = pv; = 1848.7.2882

=

ïòëíí

x 1O8N/m2

Example 4.2 A vertical vibration test was conducted on a 1.5 m x 0.75 m x 0.70 m high concrete block in an open pit ha\'ing depth 2.0 which is equal to the anticipated depth of actual foundation. The test was repeated at different settings (8) of eccentic masses. The data obtained from the tests are given below: è

ÍòÒ±

º´¬Æ

(Deg)

Amplitude at Resonance (Microns)

íê

ìïòð

ïíòð

éî

ìðòð

îìòð

íò

ïðè

íìòð

íîòð

ìò

ïìì

íïòð

ìðòð

ïò

îò ò

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 foundation 4.0 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: .,'."

174


×

¢ËÍÌÎßÌ×ÊÛ EXAMPLEsj

Example 4.1 A soil specimen was tested in a resonant column device (torsional vibration, Fixed -free condition) for determination of shear modulus. Given a specimen length of90 mm, diameter 35 mm, mass of 160 g, and a frequency at a normal mode of vibration (n = 1) of800 cps, determine the shear modulus of the specimen. Solution: I. From Eq. (3.47) (J) =--1 (2n -1)1tvs n 2 / "

dt;

'. n .::>

=1 2/ffin - 2.(0.O90).(21t.800) 1t

vs =-- 1t

=288m1s

2. Mass density of Soil in the specimen

p =

3.

0.160 2 = 1848.7kg/m) .035 1t( 2 ) .0.090

G = pv; = 1848.7.2882

= 1.533x 108N/m2 Example 4.2 A vertical vibration test was conducted on a 1.5 m x 0.75 m x 0.70 m high concrete block in an open pit having depth 2.0 which is equal to the anticipated depth of actual foundation. The test was repeated at different settings (e) of eccentic masses. The data obtained from the tests are given below:

e

SoNo

/nz


36

41.0

13.0

72

40.0

24.0

108

34.0

32.0

144

31.0

40.0

(Deg) 1. 2. 3. 4.

.

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 foundation 4.0 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.

"m:t.~

175

mic SoU Properties

tion:

= 1.5 x

1. Area of block

= 1.125 m

0.75

2

Mass of block

= (1.125 x 0.70) x 2400= 1890 kg

Mass of oscilator and motor

= 100 kg (assumed)

Mass of block, oscillator and motor

= 1890+ 100 = 1990 Kg. 2

2.

=-41t

,

C U

C =

4

2

fnzm

A

2

2

1t fnz'

1990

1.125.1000 = 71.1 fn~ kN/m3

U

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)'.fA Assuming

Jl

=0.35 F = .JiJ2s(1-0.352) 1.13 G=

E ,

Cu=0.8236 Cu kN/m2

= 0.8236 Cu =0.3050C kN/m2

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 is en by crO!

where

= cry (1 + 23Ko)

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. upto ter table is satUratedthen ."c -' .,.

-i .

''

;;; .", " , :~vJ; :::;:..J.~~~8iZ+aO.x "-;,'

,T~"'I'J,r;~'

0 70 . .

~~

=~3 ~~m2\,)

' ,: : '::",

~, . I

" ;~j,

,,;

;: ) : ') " ':.

',f'

~ ,..' "

176

.'

Soil Dynamics & Machine Foundati'ms' From Taylor (1948):

-a - -4q

2mn~m2+n2+1

v2 - 41t m2 +n2 + l+m2n2

. -12mn~m2+n2+1 .m2+n2+2 +sm m2 +n2 + 1

[

m2 +n2 + l+m2n2

]

L 12 1.5/2 ³ =2=0.70/2=2.14 B/2 0.75/2 n = 2 = 0.70/2 =1.07 q = 24 x 0.70= 16.8 kN/m

2

[Assuming unit weight of concrete

= 24 kN/m3]

Substituting the above values of rn, nand q in the expression of av2' we get

-

2

crY2= 13.44kN/m cry = 43 + 13.44 = 56.44 kN/m2 Ko = 1 - sin
-

1+2 x 0.426

0'01

(

= 56.44 ,

2

) =34.84kN/m

3

For the actual foundation aYI = 18x2.0+20x 1.0+(20-10)xO.5=61kN/m2 4.0/2 m = 3.012 = 1.334 3.Q/2 n=3.0/2=1.0

q = 24x3.5=84kN/m

2

Substituting the above values of m, nand q in the expression of crY2'we get crY2

av

= 63.76 kN/m2 2

= 61+63.76= 124.76kNIm

cr02 = 124.76 (1 + 2'x] 0.426) =77.01 kN/m2

.

Area of actual foundation = 4.0 x 3.0= 12.0m2 (> 10m2) .

Cu2

Cui

= E2 = G2 = El

Gl

-

0'02

( 0'01)

05

.

~ ( A2)

05'

= 77.0

05

1.125

( 34.84) ( 10 )

05' '

= 0.4986

The values ofCu' E and G of the actual foundation at different strain levels (= amplitude at resonancel width of test block) are given in cols. 8,9 and 10 of Table 4.10 respectively. The corresponding values of sn:ainlevels are listed in col. 11.

-

"" ~' .. ~ '~

.

.

iitfIIJ

177

~ynamic SoU Properties

4. Strain level Correction .

150 x 10-6

.

Strain in Actual foundation =

3.0

-4

= 0.5 x 10

Thevaluesorcu' E and G correspondingto strainlevelof 0.5 x 10-4canbe obtainedby interpolation. Cu

= 4.10-(4.10-3.40) (

0.50-0.427

)

x

0.533 - 0.427,

= 4.10-0.7 x 0.6886=3.62

4

10

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~/m3

E = 2.98 x 104kN/m2 G

=

1.10 x 104 kN/m2

Table 4.10 : Analysis of Data for Cu' E and G For test block Amplitude S.No.

e

In;;

at resonance

(Deg.)

E

G

(microns)

Cu x 104 kN/m2

x 104 kN/m2

x 104 kN/m2

( I.)

(2)

(3)

(4)

(5)

(6)

(7)

I.

36

41

13

11.95

9.84

3.64

2.

72

40

24

11.38

9.84

3.47

3.

108

34

32

8.22

6.77

2.51

4.

144

31

40

6.83

5.62-

2.08

continued For Actual foundation Cu x 104 kN/m2

E

G

Strain level

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

177

Iynamic SoU Properties

4. Strain level Correction .

150 x 10-6

.

Strain in Actual foundation =

-4

= 0.5 x 10

3.0

ThevaluesofCu' E and G correspondingto strainlevelof 0.5 x 10-4canbe obtainedby interpolation. Cu

= 4.10- ( 4.10-3.40 )

(

= 4.10-0.7

0.50-0.427 0.533 - 0.427.

)

x

4

10

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 using Cu = 3.62 x 104kN/m3 E

= 2.98 x 104kN/m2

G = 1.10 x 104 kN/m2 Table 4.10: Analysis of Data for Cu' E and G For test block Amplitude S.No.

e

E

G

(microns)

Cu x 104 kN/m2

x 104 kN/m2

x 104 kN/m2

at resonance

In;;

(Deg.) ( \.)

(2)

(3)

(4)

(5)

(6)

(7)

\.

36

41

13

1\.95

9.84

3.64

2.

72

40

24

11.38

9.84

3.47

3.

108

34

32

8.22

6.77

2.51

4.

144

31

40

6.83

5.62 -

2.08

continued For Actual foundation Cu x 104 kN/m2

E

G

Strain level

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;

I

If;..~~'

. ,

179

"iizinic Soil Properties'

c

=

,

(0.5546+0.9926)

t

,

8~2xO.6194f.

x IQ-3 I~(0.5546+0.9926)2

2 48.856 ht'(

3

2 nx

x 10-6 -4 x 0.6194 x 0.5546 x 0.9926 x 10-6

3

= (1.5283):t(0.9989) x 10 N/m Therefore, ,

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 of Table4.11. As in example 4.2. 'r.

C'

= 0.4986

--1f.

Ct!

The values ofCt for the actual foundation are given in col. 6. The corresponding strain levels a re listed in coL7. Table 4.11 : Analysis of Data for C't

S.No.

"e

Amplitude

fn.d

Strain level

x 104kN/m2

Ct x 104 kN/m2

(4)

(5)

(6)

25

10

5.77

2.87

0.133

72

23

16

4.88

2.43

0.213

3

108

21

21

'4.07

2.03

0.280

4

144

19

28

333

1.66

0373

.

100 x 10

(Deg.)

(Hz).

(I)

(2)

(3)

1

36

2

..

'

(microns),

"

-6

S tram m actuaI ~lound atlOn = ,

'Ct

.

"

3.0

x 10-4 (7) "

'

=.0 333 x 10-4

Therefore, the value of Ct for actual foundation = [ 2.03-(2.03-

0.373-0.333 3 L66) x 0.373-0.280 ] x 104= 1.87 x 104kN/m

Example 4.4 The soil profile at a site is shown in Fig. 4.54 . Two cross borehole tests were conducted at the site to determine the values of shear wave velocities in the small areas around points A and B. The average values of shear wave velocities were obs'ervedas 11Omlsand130 m/srespecnvely. Determine the values of dynamic shear modulus G for points A, B, C and D. '

180


0.0

f

2.0m

if = 18 kN 1m3

+

Ae

1.Qm

c.

Fine grained

--L

- 4.0 m

soil

,

1.0m

f--

--------

1.0m

t

.

~L ~

=

21 kN/m3

86

2.0m

Satu re ted san d

-1 .0 - 10.0m 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_A vS = 1l0rn/s .

G = P v; 18 G= 9.81 X(110)2=2.2xl04kN/m2 G1'A= 18x 2 = 36kN/m2 ave = 18x 3 = 54 kN/m2 (G)

~

G

= ( g:~ )

05

54 05

= (36) . = 1.~247

(G)c ~ i.i~ 104x1.2247=2.7X 1.04kN/m2

" if.

.)l~t]~,;:;:..;) ~ft~i~~\;f~"";- r(~1-+1~it~i;.~t~f.' ~~f~~'f"!!t&[~\~;

'~:;~ViY;i'"t\tr"~;\:cf~~7W;j;:'~} j;",' '9/~f.{kt;~:~-it\i~:~\~h;:

""

181

vnamic Soil Properties

.2. Saturated sand (..,,4.0~to

~

to.OOm) (vS>B =

.

"

130mls

= 9~~lx

(G~

. ,.

(130)2 =.3.6 x 104kN/m2

crvB= 18 x 4:0 ~21 x 1.0+(21 -10) x 1.0 ~104kN/m2

-

2

= 104+ (21-10) x 2.0= 126kN/m

crvD ,

,

26 = J.6x 104 x ( 104)

(G>O

0,5

.

=2.96 x 104kN/m2

'xample 4.5 .t a particular site, the top 10.0m soil is medium grained sand having dry unit weight as 17 kN/m3.The water lble 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 below round surface. ,

,olution: 1.

Yd= 17= 2.67 l+ex 10 e = 0.57 Ysat=

(2.67 + 0.57) x 10 = 20.6 kN/m3

1+0.57

'

= 17 x 6.0+(20.6-10)

(O'vhOm .

x 1.0

.

2

= 112.6kN/m Ko

'

= I-sin=I-sin36°=0.412

- - 1+2Ko -' _ 1+2XO.412 x 0'0 - ( 3 ),crv ~ ( 3 ) 112.6 '

= 68.46 kN/m 2 2. From Eq. (4.34) 2

- 0.5 6908 (2.17 - e) G = . ( 0'0 ) l+e 2

G~' 6908(2.17-0.57) 1+ 0.57 .' 4 2 = 9.3 x 10 teN/m , >'::

..

'

.(68.46)°.5

,:'

,t.

,j:Jr

;182


Foundations

ÎÛÚÛÎÛÒÝÛÍ Anderson, D. G. (1974), "Dynamic modulus of cohesive soils", Ph. D Dissertation, University of Michigan, Ann Arbor. 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 1-20.

26(1), pp.

"-

Casagrande, A. and shannan, W.L. (1948a), "Stress deformation and strength characteristics of soils under dynamic loads", 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 by means 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 Mechanics and Foundations Division, ASCE, 94 (SM2). Hardin, B. O. & Richart, F. E. Jr., (1963), "Elastic wave velocities in granular soils", Journal of the Soil Mechanics and Foundations 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 and apparatus 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 offoundations for vibration, pp. 349-420. Hvorslev, M. 1., and Kaufman, R. 1(1952), "Torsion shear apparatus and testing procedure", USAE Waterways expo Stn., Bull 38, pp. 1-76. Iida, K. (1938), "The velocity of elastic waves in sand," Bull. Earthquake Res. Inst., Tokyo Imp. Univ., 16, pp. 131144.

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".

~

183

~"a",ic Soil Properties

: 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", Soils Found. (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", Proceedings 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. Indian Geotech. Soc., 3 (2), pp. 101-112. rakash, S., Nandkumaran, P., and Joshi, V. H. (1974). "Behaviour of soils under oscillatory shear stress", Proc. yth Symposium 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, Englewood Cliffs, 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. eed, H. B. and Chan, C. K. (1966), "Clay strength under earthquake loading conditions", 1. Soil Mech. Found. Div.. ASCE, vo\. 92, SM 2, pp. 53-78. ~ed, H. B. and Fead, J. W. N. (1959): "Apparatus forrepeated load tests on soils," SpecialTechanical 20~, ASTM, Philadelphia, 5ilver, M. L., Chan, C. K. et a!. (1976), "C~c1ic triaxiatstrengtllofstandard ,

102, no.. aT 5, pp. 511-523. "

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

Publication No.

testsand", 1. Geot. Engg., Div. ASCE,vol. /

/'

,f"

~

184


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:>

"

'

-".,-

185

>ynamic Soil Properties

PRACTICE PROBLEMS

-

I

4.1 Describe the salient features. of a resonant. column apparatus. 'How is calibration done and the I value of shear modulus determined? ,

4.2 A clayey soil specimen was tested in a resonant column device (torsional vibration, free-free end condition) for determination of shear modulus. Given a specimen length of 100 mm, diameter 36 mm, mass of 180 g, and a frequency at normal mode of vibration of 900 cps, determine the shear modulus. 4.3 Explain the difference between simple shear and direct shear tests. What is the principle involved in oscillatory shear box test? Give the salient features of a study made on clay under dynamic loads using oscillatory shear box. 4.4 List the factors affecting shear strength of cohesive soils under static and dynamic loads. Explain with 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 sandand claytestedfor the following 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 using different eccentricities (8) of the rotating mass of the oscillator. The data obtained are given as follows: SoNo.

8

- Amplitude at resonance

/nz (Hz)

(Deg) ",.

.

(microns)

,

1.

36

41

13

2,

72

40

24

30

108

34

32

144

31

4,

c

"

40

Determine the value of co~fficient of elastic uniform compression, Cu for confining pressure of 100 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. The data obtaint:;d is shown in Fig. 4.5.5, Determine the value of coefficient of elastic uniform shear, C r "'"'~""

.and

the"cbrres

'

"ortding$

C'.,

-, 'n leyL~.:"

'

.

. ",. " '

';K~~il;C

"

,.

,.,

ye'"

,

'--':-<"""il""s;";,;;~';<'N.';:3I'.I"'.:'.J,i".':;,,;

'!'{,'"",;,:';';.:t"""':;""f,-,.

';,-,,'c:i,'c,"',,"""",'

C"',>,',,

186

'"

,.,>"z,>;,h,,';~~

SoU Dynamics & Machine FoundlJtions

0.32

e = 1050 0.28

0.24

,..."

E 0.20 E

......... ~ "'C

-=0.16

a. E

2

«

0 .12

0.08

0.04

0 10

15

20 Frequency

25

30

35

(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 settlement corresponding to a loading intensity of80 kN/m2 was 2 mm. Using this data, determine the coefficient 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 obtaining the 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 frequency for 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:

..

"

,

S.No.

e (Deg.)

!,,\(Hz)


1.

36

35

10

72

34

16

108

30

21

144

T7

28

2, . 3.

.

4.

Pelmissible value of horizontal amplitude is 100 microns. Solution: (a) Value ofCt is given by Eq. 4.3 2

2

8n r f roe Ct

= (Ao+lo):t ~ (Ao+Io)

2 -4rAoIo

Moment of inertia of the base contact area of block about axis of rotation I

= 0.75 x

12

1.53 =0.2109m4

Height of combined centre of gravity of block, oscillator and motor from base. 1890 x 0.35 + 100 x 0.85 L = 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 moment of mertia Mm about an axis passing through the combined centre of gravity and perpendicular to the plane of vibrations is given by .

1.52 + 0.702

Mm = 1890(

12

. "

100(0.85-0.375)2 ' , J + 1890(0.375-0.35)2+

- ,

= 431.55 + 1.184-22.56=455.29 Kg-m2 M/nO = MIn. + mL2= '455.29+ 1990x (0.375)2 = 735.13 K g-m2 '

.

r A

~

MI1/0, A

0

==455.29

Mill

= 0 6194

735.13'

1.125

= -=-=0 m 1990 I.

, . -3

2

.5546x 10 m !kg

0.2109

-3

2

10 = 3.46 ( M ) = 3.46 ( 735.13) = 0.9926 x 10 m /kg 11/0

-,,-

':';:"-.;",.!"".;;,i,~;,'.~J",1:}",.~.%/I';<,""<7'%"4"..:"";,,,",-.-;.'-' ~::;,:F',""" ", ",..':"h"';;;""+~;"

'i-';', ':',/;::;;i,'L

--

,

--

DYNAMIC EARTH PRES

-

5.1 GENERAL In the seismic zones, the retaining walls are subjected to dynamic earth pressure, the magnitl is more than the static earth pressure due to ground motion. Since a dynamic load is repetiti there 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 te frequency of the wall-backfill-foundation-base soil system. This essentially,consists in writi equation of motion of the system under free and forced vibrations. This requires the inform distribution of backfill soil mass and base soil mass participating in vibrations. It is ofter assess these. Therefore, more often, pseudo-static analysis is caf!ied out for getting dynami sure. 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 follow methods 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 Wl

~.9' '

Wt(t :totv1

ol.h ton~=1:toLv

+ (f»)

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

-

--.

..,,"'l

~

':"~

188


Foundations 1\ .:

Figure 5.1 shows a wall of height H and inclined vertically at an angle a retaining cohesionless soil with unit weight yand angle of shearing resistance cjI.BCI is the trial failure plane which is inclined to vertical 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 the wedge ABCp the corresponding inertial forces are WI . Gig horizontally and WI . ajg vertically, WI being the weight of the wedge ABCl' During the wor~t condition, WI . ah/g acts towards the fill and W, . aig may act vertically either in the' downward or upward direction. Therefore the direction that gives the maximum increase in earth pressure is adopted in practice. If

<~

ah

ah = g

...(5.i)'

a=- ai,

...(5.2)

v

.

,

a" and ai, are respectively the horizontal and vertical seismic coefficients, then

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: W I av acting at the centre of gravIty of the wedge ABC ,. Weight W I and the inertia forces W I all and :f: WI av can be combined to give a resultant WI' where

-

2 1/2

2

W,

= W, [ (l:t:av) +a"

...(5.3)

]

.;;

The resultant W, is inclined with vertical at angle '1', such that

'~

0' ..

-, '"

= tan

(

ah l:t:av

',' --

)

-"-

;""""-1

, '"

::~5.~r--

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 earth pressure [(PA)dyn] :

. -..

1

'

2

(PA)dyn,="2 yH

...(5.5)

(KA)dyn

where (KA)dyn is coefficient of dynamic active earth pre~~ure and given by : ,

'

-,;

"1:;,..

"tf i;"U

"

.,.

: ~..,

...~, ;!'> ,F'" '.'

.

;:./.,

~ .or ' , , ,-.;~"" 1""',,

"

';;!C"i":~;~li,;..,:~//-:.-. "I,~,5> ' ".,.",:r..~,,$ .,' " , , ", """, , , . ,~{.~",..~:~..<;~'..".c',":,,< , .:'!'" ",'",. ,," ,"""""" """"'-", , ...' "'

. '..

'"~

' "

J,.-

" ,

"

..

.--

,

;, "

'.

'.J IIIiI:S

189

Dynamic. Earth .Pressure ,2

(K)

=

A dYII

(l:tay) cos2(~-'V -al

.

1."

.

( A. s::) ( A.' ) 1+ srn 'f + ~ srn 'f -l - \V { cos (a - I) cos ( cS+ a + \V) }

2x COS'V COS a COS(cS+ a + \V)

1/2

...(5.6)

The expression of (KA)dyngives two values ~dependingon the sign of ay. For design purposes the higher of the two values shall be taken. Mononobe and Okabe also gave the expression for the computation of dynamic passive earth pressure (PP)dynwhich is 1 (PP)dyn

2

= -2 rH

...(5.?)

(KP)dyn

where (KP)dyn is coefficient of dynamic passive earth pressure and given by : 2 2

1

(1:t ay) cos (~+ a - \V)

(Kp)d n = 2 x y cos \jI cos a cos (cS- a + \V) 1-

112

...(5.8)

srn ( ~ + cS)sm ( ~ + l - \V)

I

{ 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. 5.2.2. Effect of Uniform Surcharge. The additional active and passive dynamic earth pressures [(P Aq)dyn and (PPq)dyn]against the wall due to uniform surcharge of intensity q per unit area on the inclined earth fill surface shall be :

q H cos a (P Aq)dyn = cos (a - i) (KA)dyn

q H cos a (PPq)dyn= cos(a - i) (KP)dyn

...(5.9)

...(5.lO)

5.2.3. Effect of Saturation on Lateral Dynamic Earth Pressure. For saturated earth fill, th~ saturated unit weight of soil shall be adopted.. For submerged earth fill, the dynamic active and passive earth pressures during earthquakes shall be found 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

.

' 'V

.

-I - 'Ys

= tan -

ah

r -'Ys - 1) 1:t ay

...(S.II)

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 separately as the effect of acceleration on water has been taken indirectly. .

.

~,~~

190


5.2.4. Partially Submerged backfill. The ratio of lateral dynamic increment in active pressures to the vertical pressures at various depths along the height of wall may be taken as shown in Fig. 5.2 The pressure distribution of dynamic increment in active pressures may be obtained by multiplying the vertical effective pressures by the coefficients in Fig. 5.2 at corresp.ondingdepths'

r- 3 [(KA)dyn

-

KA

]--\

--

z

/

I

7

/

1

/ / / / / /

V

/ H Z

/

I

1/ / hw

I

/

I

3 [(KA)dyn

-

KA]

hw

H

/ / /

/

I

I

KA and (KA)dyn a re th Cl values of KA and (XA)dyn in submerged condition

'

/ C

lateral dynamic Increment Fig. 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 portiop above water level and below water level separately. By doing this we get (Refer Fig. 5.2) ': (P)

AIDyn

= r(H-hw)3[( K ) -K' ] .H-Z Ö± Adyn;, A '

[1 (H -

hW>

H

I

.z dz + 'h~ 3[(KA)Dyn-KA]hw.hw-z'. !,

+ 1h . z1 dz'

,

.

0)(:

'H'

.

"

'",: , "

.,

,-

..

~. :.' -.~

,

'.,

f:f'::>:/:"'~';

",:,'

..',

,

h'W

191

Dynllmic Ellrth Pressure

1

= -[(KA)d yn -KAT 2

(H-h

)2

1

H W.y

h2

(H +2hw) + -[(K~Jd yn-KA1'~ H [3 Y (H - hw)+ Ybhw] 2 ...(5.12)

where (PAI)Dyn= Lateral dynamic incrementin active case hw = Height of water level above base of wall KA = Coefficientof static active earthpressure in dry/moist/saturatedcondition (KA)dyn= Coefficientof dynamicactive earth pressure in dry/moist/saturatedCondition. KA (KA )dyn

= Coefficient of static active earth pressure in submerged condition = Coefficient of dynamic active earth pressure in submerged condition

Y = Density of soil above water level Yb = Submerged density of soil Values of active earth pressure coefficients shall be obtained using Eq. (5.6) as described below. KA (KA)dynKA -

(KA)dyn -

By putting aft = av = 'If= 0 in Eq. (5.6)

. -

Eq. (5.6)

~ = 0 and 8 as

By putting ah = ,fJ.1'~

~ in Eq. (5.6)

By putting 8 and ~, and 'If given by.Eq. 5.11 in Eq. (5.6)

The additional dynamic increment due to the uniform surcharge of intensity q per unit area on the inclined earth fill shall be:

(PAqI)dyn

=J

(H"':hw)

0

.

3 K [(

A)dyn

-K].

A

cosa cos (a-i)

H-z

'-q

H

dz

ftw3[(KA)dyn-KA]hw. cos a . hw-z'. + J0 H cos(a:-i) cos (a-I) 3q cos a = cos (a-i)

2

[

2

qdz' 2

H - hw , , hw {(KA)dyn-KA}' 2H +{(KA)dyn-KA}'2H

...(5.B) ]

A similar procedure as described above may be utilized for determining the dynamic decrement in paSSIvepressures. 5.2.5. Modified Culmann Construction. Kapila (1962) modified the Culmann's graphical for obtaining dynamic active and passive earth pressures. . ,.

t

.i.

I:

'~,'

,

192

'

Soil Dynamics & Machine FollndtitiollS

it ~i "1

I~

Cz

Cl

A

1 !~

C3

1!

Modified Culmann's line

s

H

1

,~.

~

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 -
(iv) Intercept BDI equal to the resultant of the weight WI of first trial we,dgeABCl and inertial forces (:I:W I
wI =

f 2 2 WlV(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 E2 E3' 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 !Joint of 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 -
193

>ynamic ~~rlh Pressure

,

.

cu Imann s line

plane of rupture

(;

s Minimum pas~ive pressure 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 the similar way as for the static case.

5.2.6. Dynamic Active Earth Pressure for c - Cl» Soils. The solutions so far discussed consider the soil to be cohesionless. A general solution for the determination of total (static plus"dynamic) earth pressures f~r a c - cl>soil has been developed by Prakash and Saran, 1966 and Saran and Prakash, 1968. q/Unit

.area

:.

ho

H

Hl

B Fig. 5.5 : Forces acting on failure. wedge in active state for seismic condition ~nc-+ soil C?

194


Figure 5.5 shows a section of wall whose face AB is in contact with soil. The soil retained is horizontal and carries a uniform surcharge. The inclination of the wall AB with vertical is
h0 = n (H I - h 0 ) = nH

where HI

= Total

...(5.14)

height of retaining wall

H = 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

Vertical Component

r

-2

1. Weight of wedge

Horizontal Component

2

'Y H

(tan IX+ tan 81)

1 ~2 2 + 'Yn H2 (tan IX+ tan 81) + - 'Y n H (tan IX)

ABCI°1

2

2. Cohesion, C = c H sec 81 3. Adhesion, Ca = c' H sec

cH

4. Surcharge

q H [tan IX+ tan 81) + n H tan IX]

5. Soil Reaction RI 6. Inertia force

RI sin (81 + q,)

7. Earth pressure PI

P I sin (IX+ 0)

cHtan81 c' H tan IX

c'H

RI cos (81 + q,) (W + Q) IXh P I cos (IX+ 0)

A summation of all the vertical components gives 12 2 122 2" 'YH (tan
...(5.15)

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
Introducing the following dimensionless parameters: - cos J3sec IX+ cos

(Nac)dyn (Naq)dyn

.'.,

=

~ sec 91

sin(J3+o)

.

[(n+l) tan a + tan9t1 [cos (91 +~)+ah sin (91+~)] , sin (p + 0)

., i'; ; ,

;.,1

4

...(5.18) ...(5.19) '~

~

vnamic Earth Pressure

195 2

.

-'"-

(N)

ay dyn

= [(n + 1/ 2)(tan a+tan el) +ntan a][cos (e1+~) +ah sin (e1+<1»]--

...(5.20)

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 for

ifferent wedge angles e2, e3 etc. The variation of these coefficient with respect to wedge angle e are hown in Fig. 5.6, and the maximum values of (Naq)dyn and (Nay)dynand minimum value of (Nac)dyn are btained.

-0

... ...

0 +loO

~eT 0 Z

" )"" 0

)5~t

......

... 0

"1;\ u

(Naqm

z

0 Z

(Nacm)

(Narm~t

o;tat

ecm

ac

aqm

(0) Fig. 5.6:

arm

aq

( b)

(a) (Nac)dyn versus

e plot;

,

er

(c)

(b) (N.q)dynversus e plot; (c) (N.-r>dyn versus 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

'

...(5.22)

(Naym)dyn+ qH (Naqm)dyn- cH (Nacm)dyn

For static case, et.h= 0; earth pressure coefficients then become cos

(Nac)stat

=

~ sec et. + cos ~ sec el sin (~+8)

...(5.23)

,

(N) = [(n+1)(tana+umel)]cos(el+<\) aqstat sin(~+() 2

[(n+1/2)(tana+tane)+n (Nay)stat

=

...(5.24) tana]cos(el

+$) ,

sin(~+8),

(5.25)

Minimum value of (NaC>stat'and maximum value of (Naq)statand (Nay)statcan be obtained in similar manner as illustrated above for getting the ~tatic earth :pressure coefficients. It is found convenient to obtain the dyna~~c ea~th pressure coeffi,cten~~JrQmP1efollowing constaq.ts : '-, ',d,' , ', '..'., " ,il'\0""1 c'" (N aqiit'ldyn' ' . ,. ,', ," ,::...1', , ' '

,

AI

= (N{I~",)stat

,

',:'

{ , .~;,

..

('f. i '~CJi "

.

196

Soil Dynamics &

i! ~,

"~.$. '~~~

Machi"e .F oundations!1

~t

~ ! ,,'~ I (Naym)dyn,

,

,

"

,"-2

=

'

(N arm. ) stat

...(5.27)

-,

,

independent ,

of

n.,

~' '

I.~i

,

"

,

,

.

"

-

~i

~.S

" " ..

3.5

,

'

~ ,'f

~

3.0

..... ..... t)I

u0

,

'~II

E

u 0 Z

. ',! .

,~1-1

~.O +' 0 +III

"

.,

.

.

" ..

"

1

.

.'.

.

Nacmboth for the static and dynamic case is same and bas' been plotted in Fig. 5.7 for different inclination of the wall varying + 20° to - 20° with the vertical. As evident from the Eq. 5.23, Nacmfactor is also

'.

"

.

2.5

t)I ....

::J

~ t)I

2.0

....

a.

.c +'

.... 0

1.5 "

w

1.0

0.5 0

s

10

t~

lS

20

25

30

35

40

45

~

r/J (deg) Fig. 5.7: (Nacm)statversus ellfor all n (Prakash and Saran, 1966 and Saran and Prakash, 1968)

,}

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)stat for the same range of n,

~

<1> and Clhave been drawn in Figs. 5~11,,~.12 an~ 5.~3.

...

, 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. 1 same (Prakash and Saran, 1966; andSanln and Prakash, 1968). Hence o~ly one value of-A(= Ali=~) is J recomm~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

"

197

,

1.2

....

0 .... III

,...

E 1.0 er 0 z

-

n

=0

0.8 .... .... ti 0 U

0,6

I-

:J

11\ 11\

~ 0.4 c. .c .... I-

0.1

0

w

'r.

0 0

5

15

10

20

15

30

35

40

45

(/J{dq'g) Fig, 5.8: (N.qm)stat versus ellfor n = 0 (Pr&kash and Saran. 1966 and Saran and Prakash, 1968)

'

. ~ ,... 1.0 0

E er 0 z 0.8

n =0.21

.

..... .....

~ u

0.6

~

I-

"

:J ::: ~

0.4

L-

c.

~L-

0.2

0 W

" 0-

0

5, "

15

10

20 "

.' '25

,

30,

, ..35

.' 'C/J {d q g) Fig. 5.9 : (N8qm).t8t versus' :fJ ','

for n = 0.2 (pta,iuisb':and Saran,

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

40

, _,45

. '

r' --' .

1966 and Saran" and Prakash, 1968)

"..ii";;'~';;~'~,~,#:!~,t:'i"i'~i:;"!o{,h,.J'Y,:;r~;'J'":>i':~'"

,,:.

:.:,..:

198


f ~I,,

1.4

n =0.4

E er 0 Z

0.8

.... ....

1>1

~

0.6

Col

... :J III III

~ 0.4 a. .s::.

... ... 0 LIJ 0.2

0 0

5

10

15

20

2S

30

3S'

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

0.8 n=O

0 Z

-:

..... ..... ~ 0

0.6

v

~L-

0.4

:J

11' III ~

L-

a.

0.2

.r::. .....

L0 UJ

~

0

-10-200

0 0

s

10 ,.

Fig. 5.11 : (~II'JII1)Jt.tversus'

1S

20 ~

2S

30

3S

40

4S

J ,t

(d~Q).

for It = 0.0 (Prakash and Saran, t 966 and Saran and Prakash, 1968)

i

i ,

199

mamic Earth Pressure

1 .0 +-

0

~ 0.8 E

>0 0

:z

:; ....

n ::

0.2

0.6

~ 0 u

t:.I

~11\

0-41'

11\ ~ ....

a.

.r. 0.2 +' .... 0 lLI

0

-~-L~

0

5

1S

10

20

'

35

30

2S

40

4S

~ (deg) Fig. 5.12 : (Naym)statversus 4jIfor n = 0.2 (Prakash and Saran, 1966 and Suan

and Praliash. 1968)

1.2

-0 1.0 +' III

,.....

n:: 0-4

E ~ 0 0.8

z

.... ....

~ 0.6 u ~

....

:J 11\ 11\

0.4

~ ....

a.

.r.

t: 0.2 0 UJ

0

-

0 "

.;

",

.-', ,""

,\,

-;.

5'

.,-

,'.. .. "'",

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

10 C:

15 :1;);1;. J"~; ':\J

Fig:5.13'~ {N'~)...;vers~;'.

20 ': tI. (dczg) .., ,,'fJ

25

3S

'30

40

,.} ;",;', ..

.

for n'" 0.4 (Prak~sb and Saran, 1966 and Saran and Prakash, 1968)

4S

200

Soil Dynamics & Machine Foundtltio.

2.0

1.9

1 .8

1.6 A

1.5

1.4

1.3

1.2 r 0.10

{

I

1 .1

.

1.0

I 0

0.05 { 10

20 rjJ

Fig. 5.14: A.versus,

30

40

so

(deg) (prakash

and Saran, 1966)

5.2.7. Point of Application. According to Indian standard (IS : ~893-1984) specifications, the pressures are located as follows: From the total pressures computed from Eqs. 5.? and 5.9 or from graphical construction, subtract the static pressure obtained by putting ab

= ay = O. The

remainder is the dynamic increment in active case

and dynamic decrement in passive case. The Static component of the total pressure shall be applied at an "

/,.,

~. ~J

i" i' 201

'mic Earth Pressure

I'

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 surge on the backfill surface obtained by putting ah = av = 0 in Eqs. 5.9 and 5.10 shall be applied at above the base of the wall. The point of application of both the dynamic increment and dynamic ement 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 point pplication of this pressure shall be assumed to be at an elevation of R/2 above the base of the wall. DISPLACEMENT

ANALYSIS

re are very few methods available to compute displacements of rigid retaining walls during earth
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 Elms 79) to compute the displacements of retaining walls. They have proposed a method for design of vity retaining walls based on limiting displacement considering the wall inertia effect. The proce-edeveloped by them is described below. ;

.

FaHu r~ sf re ss III III ~

k(t)w

... ...(/'I

/

w

strain .

.

'(a),

~>,

"

,-

.:/-

(b)-

-Fig. S.1S: (a) Forces on sliding block (b) Rigid plastic stress strahi behaviour ora materia" ,""'; t ,. 'c :, " ., ,"" .'
,rj 'H

-iT:'i,7.-'

~

<)~: ;jji ~

Soil Dynamics & Machine FolUUlati4-

202

D

~t )t-:~. .

q.

-:~.:h

>coX

t'\' "

n

rJk; , ~if

u u

Ti mq

«

,;.

>... u 0 ti >

Time +' C ti

E ti U 0

a. VI

Timq

0 (c)

Fig. 5.15: (c) Integration of effective acceleration time history to determine velocities and displacements

H

F

' h Ww

,

\d

Ww< 1 :t d...v) \ I

,

'4, f<

T

~:' ,)

Fig. 5.16: Forces ~n a gravity 'wail ,,', '!, I "J' .,~ :i " --

Earth Pressure

203

ravity retaining wall is shown in Fig. 5.16, along with the forces acting on it during an eartht1this figure various terms used are : Ww

= Weight of the retaining wall

= Horizontal and vertical seismic coefficients PA)dyn= Dynamic active earth pressure, Eq. 5.5

lJ.h'av

a = Inclination of wall face with vertical

8 = Angle of wall friction
Soil-wallfriction angle at the base of the wall

N = Vertical component Qf the reaction at the base of the wall T = Horizontal component of the reaction at the base of the wall ,mming the forces in the vertical and horizontal directions, we get

t slidding

N = Ww :t:.ay Ww + (PA)dynsin (a + 8)

...,5.28)

T = ah Ww+ (PA)dyncos (a + 8) ~ = N tan

...(5.29) ...(5.30)

DIving Eqs. (5.28), (5.29) and (5.30), we get W w-

<".

(PA)dyn[cos(a +8) - sin (a + 8). tan ~b] (l:tay)tan~b-ah

'utting (P A)dyn = ~ yH2 (KA)dyn and ah

1

= (1 :t:.a) tan 'V, the eq. (5.31) can be written as 2

Ww

= 2" yH

CIE

=

...(5.32)

(KA)dyn' CIE

cos (a + 8) - sin (a + 8). tan ~b "here

(l:tay)(tan~b-tan

\jI)

~orstatic condition, the weight of wall W is given by : W=- 1 y H 2 K .C 2 A I where Cl

rherefore, >ubstituting

.e.

- sin (a

(KA)dyn .

...W

'.

KA

...(5.33)

...(5.34) + 8) . tan

tan J..'l'b

W ~=

~b

tan ~b (l:tay)(tan

~b -tan \jI)

...(5.35) ...(5.36)

FT = Ratio of earth pressure coefficients in dynamic and static cases (KA )dyn FT

=

md FI n Eq. (5.36),

cos (a + 8)

=

...(5.31 )

W ~=F W

=:=

KA

. . Wall InertIa factor T

F=F I w

...(5.37)

=

tan
...(5.38) ...(5.39)

y,

~. i

204

Soil Dynamics & Machine Foundations~

;

,

1

F w 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 F w 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 F w = 1.5 is equal to 0.105. Thert:fore, if a wall is :' I designed such that W w = 1.5 W, the wall will start to move laterally at a ~alue of ah

= 0.105.

Hence for

no lateral movement, the weight of the wall has to be increased by a considerable amount over the static condition, which may prove to be uneconomical. Keeping this in view, the actual design is carried for some lateral displacement of wall.

i . j

I II

,. ...

I

14 fw 12 '" 10

8 3'

u.;

..

u:..

u:

6

/

4

L"/ 2

FT = 1.5

':.--- ==.

oD

0.105 I I

0.1

~.

- -

I .10.18

0.2

-=-

..::;:::-

y

./ ññóä T

~ .

0.6

"

~

Fig. 5.17: Variation of Fr F. and Fw with ab (Richards and Elms, 197?)

., , ;1 i ~ . ,, ~ ~.~

205

zm;c Eartll Pressure

Richards and Elms (1969) have given a design procedure based on a limited allowable wall movet, 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 *

-

...(5.40)

- ah ( d )

ahd

where ah = Acceleration coefficient from earthquake record d = Maximum displacement in mm (Hi) Using ahd' determine the required wall weight, Ww by substituting it in Eq. (5.31). The value ahd of avd may be taken as 2'

/ (iv) Apply a suitable safe~y factor, 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 displacements before the limiting equilibrium conditions (active) develop and experience very large displacements before the passive conditions develop. 2. The physical properties of the system and its geometry (particularly its natural period) are not considered. 3. Walls may undergo displacements by either sliding or tilting or both. This method does not apparently consider this difference in their physical behaviour, although it is logical to conclude that displacements computed by this method are in sliding only. .2. Solution in Pure Translation. A method for computation of displacement in translation only, of d retaining wall under dynamic loads had been developed by Nandakumaran (1973).

t

Earth

pressure (p)

--+ BF

Active C

Di sptaceme ".

Fig.S.tS:

nt

(a) (a) Earth pressure (P) versus displacement of wall

206 tJ

G

(Rep + Pp!r~

I

x. I

,I . :

I

Base friction

{B.FJ RBP--,';

H

(R8A- PA) J

RSA

oX

I

Displace m ent

Displacement / (b)

x ,°2

(c)

z

T

X,Ot

B "2 tan

-11..

.~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. 5.18a shows the variation of earth pressure with displacement. In Fig 5.18b, variation of base resistance with

displacementis given. The net force away from the fill is the differenceof active earth pressurePA and the base resistance, RBA(Fig. 5.18c). The net force towards the wall is the sum of the passive earth pressure, Pp and the base resistance, RBP(Fig. 5.18c). The resulting bi1ine~rforce-displacement relationship is shown in Fig. 5.18d and is characterized by the following parameters:

t I .t I

I

namic Earth Pressure

.

207

(i) Slope of force displacement relationship on the active and passive sides as Kl and K2 respectively, where K2 = n . Kt.

(ii) Yield displacement, Zy For 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) yield splacement, 4) damping in the system, and 5) parameters of ground motion. ºÖ´ñ×ùºþþùºþòòô

K

m .,

c y = Y sin GJt

z

=

.

x-y

x

Fig. 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 witL e wall. Nandakumaran (1973) conducted vibratory tests on .translating walls and found that for the lIposes of matching the computed frequency of the wall with the measured natural frequency, the soil ass 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 relaonships. 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) : ...(5.41 a) rnx + C (x - y) + K (x - y) = 0 rnl + Ci + Kz = - my ...(5.41 b) Z + 2 11~i 4-112z = - y ...(5.41 c) .

.

where z = (x - y), 112= Kfm where K has been defined as the stiffnesson the tension side and .

~ = Damping

C 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 by y, the relative displacement on the tension side at which the resistance becomes constant (yield displacelent) to obtain the following relations: 2

'l'n+_1= 'l'n+ 'i'n: t + t6 (\jJn-1+2\j1n)

..:(5.42)

~~ ~. ~-«(Pn+1+2\Vn)

...(5.43)

~n+1 =

.:

"

JI.~ . j j

:~.;t

I

,

\J. j\

208


\Jin+\ =-YIl-21l~ Z

where

\Jin+\-1l2.K.Zy('I'n+

.

'1'=-

!

,

...(5.44)

I)

'1'=-

'1'=-

; !

...(5.45)

Zy Z

...(5.46)

Zy Z

...(5.47) Zy With these relationships, the analysis is performed for the range of variables listed in Table 5.2 Table 5. 2 : Range of Variables Considered in DisplacementAnalysis in Translation Variable

Range of values

Ground acceleration amplitude (a) gals Period of ground motion (T)s Damping (~) %

100, 200 and 300 0.5, .3, 0.2, and 0.1 5, 10, 15 1.0,0.5,0.3 and 0.2

Natural period (TII)see.

1.0, 2.0, 3.0, 5.9, aJ1d10.0

Yield displacement (Zy) mm

To study the response characteristics of the system, two cases were considered; one in which plastic deformation does not take place and the other in which it does. Figure 5.20 shows the response of the elastic system. It is evident from this figure that steady state conditions are attained in about 6 cycles and also that displacements on the tension side are larger than those on compression side. The response of the system wherein slips take place has been plotted in Fig. 5.21. This shows that even when plastic deformations occur, a sort of steady state is achieved in the sense that slip per cycle becomes a constant after about 6 cycles. 12

8 E E

4-

+' C ~

0 Tim
E

~ u

(5)

.E -4a. III 0

-8

-12

t

I

A =300 ga[5 T = 0.3 5 Tn = 0.3 s

Zy = 200 mm Tt. = 2.0 Damping = 10°/0

0.3 s

L.;

; f

Fig. 5.20 : Response of an elastic system with different stiffnesses of tension and compression sides (Nandkumaran,1973)

I 4:

'namic Earth Pressure

-0

,-...

c 0 '" c

12

A = 300 gals .T = 0 3 5 Tn = 0.3 5

8

Zy= 10 mm n = 2.0

'-E "0

I C

0 c

---

209

Damping

=10%

4-

+C E u

-0 a.

.-\11 Q

-

0 Tima (5)

-4

i 0.3s

,, Fi~. 5.21: Displacement versus time (Nankumaran, 1973) -.-.

80 Zy

= Smm

Zy =10 mm

~ =10°,. n =2.0

; =10°'0 n

70

=2.0

60 A 1 T 0.39.0.55

E E ~

u > u

SO

A 0.39

~ 4.0

..c:

toI

E

~

30

0

alii

0

20

I

10

0'29,0.3)-

0.1 , Q.3 -

0.4.

0'.6

0.2

0.8 Natural

0.4.

0.6

ptlriod.S

Fig. 5.22: Natural period versus slip per cycle (Nandkumaran, 1973)

IT . 0.5 r

..J

!10


Fig. 5.22 shows typical set of results in the form of slips per cycle versus the natural period of the NaIl in seconds for the yield displacement Z= y 5.0 mm. and 10:0 .mm, ~= 10% and n = 2 for different ~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~ where K = stiffness on the tension side and In = mass of soil and the wall.

...(5.48)

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 been formulated, and (ii) physical behaviour of the retaining wall is considered in developing the force-displacement relationships. The method, however, suffers from the-fact that the tilting of the wall has not been considered. 5.3.3. Solution in Pure Rotation. A method of analysis for computing the, rotational displacement of rigid retaining walls under dynamic loads has been presented by Prakash etal (1981) and it is based on the following assumptions: (i) Rocking vibrations are independent of sliding vibrations and the rocking stiffness is not affected by sliding of the wall. (ii) The earthquake motion may be considered as an equivalent sinusoidal motion having constant peak 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 corresponding 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 soil stiffness Kt in active state is giv~n by ,

=

K I

KpyH2 - KoyH2 2 2 Po-PA = 2.5 H average displacement ( 100 ) .

...(5.49)

.~ 1

t 'f Ii 1, ,1 1 » .'1"""'


r-,

b t --t .t-=a~

.

211

B

ti'.x

~

, I I

'

L.. :::J IJ' Ut

+' C C:>I

~

.-

L..

I

I

U

a.:;:

~

(kA or kP) H

I J

kp

.t:

C:>I

... u0

+'

1

\

//R
Backfill

kA

rotation

Lf-"[ . b

~

-1

Q.2SH

Z.5H

F,

lOO."t-

100

Di splacement (a)

Fig. 5.23:

~

(b)

(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 Kz 1passive state may be computed as : Z z Kp rH - Ko rH 2 2 Pp - Po K ...(5.50) 2.SH , z - averagedisplacement ( 100 ) where:: PA Pp

= Active earth pressure = Passive earth pressure

= Earth pressure at rest ,KA = Coefficient of active earth pressure Kp = Coefficient of passive earth pressure Ko = Coefficient of earth pressure at rest Po

The rotation resistances of the base, in active and passive states (MRAand MRP)may be given by M~A=.C,.I'CPA

=

c,. I .

,',

...(5.51

a)

...(5.51 b) n which C, is coefficient of elastic.non-uniform compression, I is moI1!e~t'ofin~rtia ~fthe base about an lxis through. the heel of.the!walland.perpendicularto the plane ofvibra.tiQns;,an~~A and CPB itre angles - . MRP

)frotation'away

and

towards!he~backfill.

'1'-;,"

CP~

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: .

Mmo ~A +

.

(

C.I-

13

)

4>A= M (t)

...(5.52 a)

4>p= M (t)

...(5.52 b)

-

i.

Z

..

and Mmo Ij>p+

-"

K H2

KzH

( C.

I-

3

J

Since the stiffnesses Kl and Kz are different, the period of the wall for the two conditions i.e. towards the backfill and away from backfill would be different. This would result in different values of 4> A and 4>p for 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)

...(5.53)

.H

where n = Number of equivalent uniform cycles of ground motion H = 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 contribution of rotational displacement using the above approach was compared with the sliding displacement for a 3 m 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/m3 Base width/Height of wall

3.0, 5.0, 7.5 and 10.0 30, 33, 36 0.3 0, 5, 10, 15 4 3, 4, 5, 6 and 8 (x 10 ) 1/3

horizontal seismic coefficient CJ.h equal to 0.25 and C$ equal to 3 x 104kN/m3. The total slip in 15 cycles due to sliding was 213 mm. Displacement of top of wall due to rotation found by this analysis was 147 mm. This illiustrates that the rotational displacement may not be negligible and an attempt should be made to account for it. The displacement analysis for rotational displacement is highly simplified. Nevertheless it shows explicitly that in some cases neglecting rotational displacement may seriously underestimate the total displacement. In actual practice it may be essential to account for combined effects of rocking 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 acceleration (C1.h. g) when slippage is taking place. But once the backfil1 beings to slip, compatability of movement requires the backfill to have a vertical acceleration, thus causing change in wall acceleration. .

~~

? ~ ~ I

213

ynamic Earl/. Pres~ure

Zarrab~(1979) considered the equilibrium of the.wall and the backftll wedge separately and.satisfied le 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 constant I

Zarrabi's model. 7 6 S

I W.kn

\

:r I

R

w.~ -w

WJ

1

A= 0.2

w.k~

~

:1

I

N = 0.112 Uniform

G

.2 1

Ww Rw tan ~b Rw

3

f 1f1 (a)

(c)

Contact element Slip element

Slip

element

with Cn=lE8kN/mlm C5=0 with Cn=lEBkN/mlm,C5=12SkN/m/m

with,

Rigid

boundary

Cn = 1E 12 kN I m/m

'4m sea le ~

Cs = 1E S kN/m Im Cn = Normal stiffness C~ = Shear stiffness

,

of slip elements

of slip

elements (b)

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 computed lith the Richard- Elms model. Dynamic tests on model retaining walls performed by Lai (1979) sh~w lat 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;-Zarrabi

110delhas been modified to have a coiistanHncIlnationof failure plane 'in the b~ckfilt> .

'1:

<~

':i...;:'!'>'

":",

',..

~

'

,

""

'i

214

Soil Dynamics & Machine Foundations The Richard,

- Elms,model

and Zarrabi's model assume a rigid

- plastic

behaviour of the backfill

material. Hence the input ground acceleration is constant throughout the backfill. But due to more-orless elastic behaviour of soil at stress level below failure, the input acceleration is not constant. Hence amplification' of motion cannot be taken into account in these models. Nadim and Whitrnan (1983) used a two -dimensional plane - strain finite element model for computing permanent displacements taking into account the ground motion amplification. The slip element at the ,base of the wall has been assigned a very large v
- plastic

(Richard

- Elms

or Zarrabi's) model.

I, is the fundamental

frequency of wall and

I is

the

frequency of 'ground motion. It can be seen that effect of amplification of motion on displacement is greater when I I is greater than 0.3 The FE model predicts zero permanent displacement ,at high frequency, because in the analysis only three cycles of base motion are considered during which steady -state conditions can not be achieved. However, it can be said that large values off If, are not of great p'ractical interest because displacements are very small.

I,

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-dimensional amplification theory by using the following equation and estimate the ground motion frequency, f

...(5.54)

I, = VI/4 H 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. If I II, is in the vicinity of 0.5, increase the peak acceleration, A and the peak velocity, V of the desjgn earthquake by 2530%. 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 for

getting ahd for known value of the displacement. (iv) The value of ahd estimatedin step (iii) is used as the value of horizontalseismiccoefficientin the 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 in such way that it results translation and rotation simultaneously and therefore it has two degrees of freedom. '11 'I, "

'

,

,f

,

In practice, cro ~~ ':' section pf rigid ret~iiningwall vari ~s'to cl great exte!l.~' Areas
,

,

,

,

,b~ckfin soil is replaced by closely spaced independent elastic springs shown in Fig. 5.25. .;

,

,

". ",

. 215

nic, Earth PresslIre

t--°'--t

-

K, hl

Displaced position I

I I

K..,

hl

K...

I

:IT

~ I

h,

Retaining

wall

I

-- - - - - - - F=F

¸

ï

0

j -

sin

I

wt

Kn-"

--.

. . I

Inltla

position

I I I

ill

X

J I I I

.

I

I

¢

¾

¢

.

-Dynamic"'" , --.QC tlve "

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?~l modplus depends on type 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 11ft he 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 moving lards 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 Loose 'sand

'200-300

passive ;

'400-60d'

Medium dense sand

400-600

800-1200

Dense sand

800-1200

1600~2400, ,

.

In case of soil modulus linearly-varying with depth, the soil-reaction -is assumed'to 'act as a loading tensity. Treating this load to be acting on a-beam of length, equal to the 'height 'of retaining w~ll,-the actions-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..." " .'

,

,~ #' 16

Soil Dynamic.s & Machine Foundations Dj'

k - 1 1 - 6 l1h(1:1hi

...(5.55 a)

k2

= l1h

k3

= 2 l1h (1:1h)2

...(5.55 c)

ki

= (i - 1) l1h (1:1h)2

...(5.55 d)

kn

=

(1:1h)2

1 "'6

lar en,. da: W2

...(5.55 b)

2

de ha se dr

...(5.55e)

(3 n - 4) T\h(1:1h)

IS

where kl and kn are the spring constants at the top most an bottom most points, ki, the spring constant at my division point 'f.

de S)

In case of soil modulus constant with depth, the soil reaction is assumed to act as uniformly distributed loading intensity. Treating this uniformly distributed load to be acting on a beam of length, equal to the height of retaining wall, the reactions at different points are evaluated treating th,isbeam to be simply supported at these points. The spring cons'tants would be as under,

5. u

.

-21 k (1:1h)

For the top most spring,

k =

For any intermediate spring,

k; = k (1:1h)

..(5.56 b)

For the bottom most spring,

1 kn =. -2 k(l:1h)

...(5.56 c)

1

...(5.56 a)

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 by number of cycles. 2. Soil stiffnesses (or spring constants) for displacement of wall towards the backfill and away from the 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, the method gives higher displacements and the solution is conservative. However, the mass of vibrating soil can 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 taken equal to 0.8 times the mass of Rankine's wedge. By adopting similar technique, the soil mass vibrating along 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. .'. '.-

,

.,

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 of ergy 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 the this 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 walls may

edominant 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 is mainly ts 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

...(5.57)

(J)t

11 ;=\

+ Lk;[x+{H-h)-(i-I)LlhJ

9) (H-h)-(i-l)

LllzJ]=0

...(5.58)

11

here M = Mass of reitaining wall J = Polar mass moment of inertia of the wall about the axis of rotation (J) = Frequency of the excitation force H = Height of retaining wall h = Height of centre of gravity of wall from its base x = Translatory displ~cement e = Rotational displacement ,etting : ;=\ "f.k; n

"1T F ...JL

M

;=\

,=

a

...(5.59) 4

'..

=a

...(5.60)

=b

...(5.61)

a

,:;

I,kj {(H-h)-(i-l)L\h} n

M ;=\ "f.

n

~

-

kj {(H - h ) - (i

J

'2

- 1)6.h}, '!

t ' , ~\ - - '. - , ,)

=c

'\

...(5.62)

-::r I

218

Soil Dynamics & Machine Foun~

The equations of motion of the rigid retaining wall cai1thu~ be written,as~ , , ,

:

",

be+

x' + ax ~ ,

,

'A" sih ro l' ".

,

0.,

'

"

,','

,

..:.(S.6?).

,

"" ,

,

'

"

b'

e + ca where

J

'"

( "7 )

=,

x';

, ...{S.64)

", ,

,

,

,

= 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

,,'

, where X and

= X 'sinrot

~ sin

e = ~ are arbitrary constants.

.':.(5.6S)

(0 1.

"

...(S~66)

"',,

,Substituting Eqs.'(5.65):and(S.66) in Eqs. (S.63) and (S.64) , we get (- (02 + a) X

(ch

~

c)

= b~ + ao'

"',(5:67) ".

(:,)x

~

, ...(S'.6'8)

Solving Eqs. (S.67) and (5.68), we get "

ao

X =

2

(a

~= , ,

- 00

...(S.69)

b2

)

-,.2 ( c -

2 00

)

ao '

2

.'

2

...(S.70)

,.2,

(a-m Hc-oo )---b b u

Hence the solution becomes

'

"

X

=

ao (a-002)-

...(S.7!) 2 b2

,sinro(

,. (c-002)

a0

e=

..,(S.72)

1'2 (a-002) ( c- 002 )-T-b

sin rot

Therefore, the displacement of the top of rigid retaining wall is given by Xtop

or

...(S.73 a)

= X + (H - h ) e 2

x top =

2

.<0 I

-

r (c-ro) +btH-h) . a srn . ro( 2 2 2 2 0 { ( a-ro ) (c-ro ) r- b }

""

(S,73b)

! I

t,.

.ii

!icEarth Pressure

219

? Natural frequencies. Under free vibration condition, the equations of motion are: .

'x'

+ ax

:dba b

..

9 +

ca = ( ;

..

-

.-- . -;~;~,-

...(5.74) ...(5.75)

J x

ubstituting the solution: x = A sin ron t

...(5.76)

a = B sin ron t

...(5.77)

A and B are arbitrary constants ~quations 5.74 and 5.75 become:

~

(-00; + a) A (-
...(5.78)

= b .B

~

(:,)A

...(5.79)

b

From these we get,

A =B

...(5.80)

2

a -OOn 2

and'

...(5.81)

~

~

b

Equating,

(::) c-oo

(\r ~ )

a-oo; 4

2

b

OOn-(a+c)OOn+QC-; ( and solving we get,

2

2

...(5.82)

) =0

00;\ = 2

; (a+c)+J( c;a r +(~ r I

oon2 = 2(a+c)-J

c-a

2

b

...(5.83)

2

(-y- ) +()r

...(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)a

...(5.85)

b = n (~)a

...(5.86)

C = n (c)aThe solution for this condition is similar to active condition described above.

...(5.87)

~ 220

Soil Dynamic'S & Mad
I

5.3.5.4. Analysis of a plastic system-active conditio,n. Assume that Zy and 9yare the yield displaceme~~\: " occurring simultaneously in all springs: the equations of motion can be written as:

...(5.88)

x. + a Zy = b 9y + ao sin Cl)t .. b e + ce = - Z

( r2 )

y

...(5.89)

y

Integrating the above equations twice, we get t2

x = (b ey - a Z) b

e = (?

T -

ao sin Cl)t 002 + Cl t + C2

...(5.90)

t2

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 displacement (Yd) and plastic system starts. Let xe' xc' ee' 8e be the values corresponding to time te and can be calculated by using the equations developed for elastic system. The following boundary conditions can be applied to evaluate the constants of integration: (i) t = re' X = xe

...(5.92 a)

(ii) t = re' X = xe

...(5.92 b)

(iii) t = re' 8 = 8e

...(5.92 c)

(iv) t = re' e = ec Therefore, we have

...(5.92 d) ...(5.93) ...(5.94)

Zy = xe ey = ee C = x - (be - a Z ) t + I e y y e

a cos Cl)t

. C2

Cl) 2 te

e

_

C4 = 9 ere

e

...(5.95)

. ao sm 00te

= xe - xete + (bey - a Z) T+

C =8 3

0

002

leao cosoote ...(5.96) 00

te ( rb2Zy-Cey )

...(5.97)

b t; - 8 t ZZy-Cey _ e ( ) 2

...(5.98)

Displacementof the top of rigid retaining wall is given ~y x top

= x - (H - h) e

...(5.99)

5.3.5.5. Passive condition. The ratio of stiffnesses on the compression and tension sides is denoted by n Hence in the passive condition, the values of a, band c, change and these can be given by: a = 11(a)a

...(5.100 a)

b = 11(b)a

...(5.100 b)

c = 11(c)a

...(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"In ression 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 backfill 33°, 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 construction. (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

PA-

1 H 2 cos2cosa = 2"1

I

)

«I> -(8a+ a) cos

.

-

{ I + [ cos (a

2

=

1/2 2

sin ( + 8). sin ( - i)

cos2 (33-20) ~ x 18 x 6.0 x cos220 cos (20+20)

- i) cos (8 + a) ]

}

1

x

I + sin (33 + 20) sin (33 -10) [ cos (20 -10) cos (20 + 20) ] {

1/2 2 }

= 168.42 kN/m Refer Fig. ".26 for Culmann's graphic,alconstructionfor getting static activepressure. Ds Es gives ~ total active earth pressure.

'

.

.

PA = 17 x 10 = 170 KN/m (b) Dynamic active earth pressure (P)d A yn

--

-1 1 H 2 col «I>-2 'V - a)( I:!: ay)

2

cos'I' cos a cos(8 + a + 'I')

'

x

,1 sin ( 4>+ 0) sin ( 4>- i - \jI )

{ 1+ [ cos (a - i) cos ( 0 + a + \jI) ]

Assuming a

v

'V

ah 0.1 = - = - = 0.05 2 2 .

= tan

-I

- ah

l~av'

= tan

-I

1:!:0.05

= 5.44° with ;:. ay and = 6.0° with'\- ay .

0.1

-

.

2

1I2

}

222

Soil Dynamics & . Machine. .Foundatioi$' . : i i, '0

6.0m

Linear scale - 1 : 100 F'orctZ Pr~ ssurtZ

scaltZ, -1mm

= 10KN

I i n~

Fig. 4.26: Culmann's graphical construction

Value of (f A)dyn with (+) ay 2 = .!." 18 x 60 2 cos (33-5.44-20)(1+0.05)

2

A.

1

2

cos 5.44cos 20 cos (20+ 20+ 5.44)

X

'

1I2 2

I + sin (33+ 20) sin (33-10-5.44) {

[ cos (20-10)

cos(20+20+5.44

]

}

= 214.26 kN/m Value of (P A)dynwith (-) ay 2 - 1 18 x 6 02 cos (33-6-20)(1-0.05) --x

2

.

= 198.05 kN/m

cos6cos220cos(20+20+6)

x

1 1+ ~in(33+20)Sin(33-10-6) [ cos (20-10) cos (20+ 20+ 6)] {

1/2 2 }

.

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 dynamic active earth pres, " for getting . '. . sure

'w,

2'3"

",ic Earth Pressure

m

Lin~ar, scalCl -1 :100 Forc~

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 5.2 retaining wall 8.0 m high is inclined 200 to the vertical and retains horizontal backfill with following aperties : y/= 18 kN/m3, ~ =300 and c = 6.0 kN/m2 There 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 decmine the percentage increase in pressure over the static earth pressure. Solution: .~ 2c

(i)

ho =

=

1- sin 30

1

where KA = l+sin30

Y ~KA 2 x186.0 x..[j = 1.15 m .

1 = 3'

224

Soil Dynamics & Machine FoundatiiiJl;

H = 8.0 - 1.15 = 6.85 m ho 1.16 n=1[=-=0.167 .. 6.85. (ii)

For

= 30°, a. = - 20°and n = 0.167

Figs 5.7 to 5.23 give: = 0.33, (Naqm)stat= 0.512 and (Nacm)stat= 1.2

(Naym)stat

2

(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 (i ii)

For

= A.

Therefore,

30°, a.,= :- 20°, dh = 0.1, Fig. 5.14 gives

= 1.18

(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 282.00 x 100 = 22.4 % = 345.10282.01 Example 5.3 A 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-

KA

a)

1

= cos2acos(o+a)'

{ 1+ [ cos(a cos(30-20)

.'"

.

= 0.6225

.

-0

cos (8 +a.)

. '.

= cos2 20 cos (20+ 20) .

.

l/2 2

Sin(cp+O)Sin(cp-O

]

.

,}

1

sin (30+ 20) sin (3P-15) {

1+ [ cos(30-20)cos(20+20)

.

',' .

I/2 2

] }

!amic Earth Pressure

225-

From Eq. (5.35) Cl

=

Cos (a +0) -Sin (a +0) tan ~b tan ~b

= Cos (20 + 20) -Sin (20 + 20) tan 30 tan 30

= 0.6840 Thus from Eq. (5.34) 1 W

2

= "2y H KACl = .!. 2 x 1800 x 52 x 0.6225 x 0.6840 = 9765 kg/m

(b) For zero displacement condition (KA)dyn =

cos2(~cos \If cos

2

\If -

a)

1

x

1/2 2

~

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 \jI = tan l:tay

0.1 = 1:t0.05

= 5.440 with + ay and

= 6.00with -

--

ay

With + ay cos2 (30- 5.44 - 20)(1 + 0.05)

(K)Adyn =

1 x

cos5.44cos220cos(20+20+5.44)

{ 1+ [ cos(20-15)cos(20+20+5-44)

]

= 0.8311 With - ay: cos2 (30-6-20)(1-0.05) (KA)dyn = cos6cos2 20cos(20+20+6) x .

1

From Eq. (5.36) With + ay :

W

0.8311

tan30

-

W = 0.6225 x (1+ 0.05)( tan 30 - tanSAM

= 1.524

.

~in(30+20)Sin(30-15-6) [ 1cos(20~15) cos(20+20+6) ] {

= 0.7727

1/2

sin(30+20)Sin(30"""15-5.44)

2

1/2

}

2

}-

~.

226.


With - av : W = 0.7727 x tan 30 ---!!. W 0.6225 (1- 0.05)( tan 30- tan 6) = 1.596 Therefore Ww = 1.596 x 9765 = 15585 kg/m (c) For displacement condition, d = 50 mm From Eq. (5.40) 5ah

a"d = ah [ -cl ] 5 x 0.1 - 0.1[ 50 ] = 0.03162

1/4

Assuming avd

= ahd = 0.3162 = 0.1581 2 2

With (+) avd 0.03162 = Tan [ l+ahd ] = Tan [ 1'+0.01581] -I

'"

ahd

-I

= 0.780 cos2 (30-1.78 - 20)(1 + 0.01581) (K)A dyn -- cos 1.78 cos2 20 cos (20+ 20 + 1.78) x

1 sin (30+ 20) sin (30 -15-1. 78)

1/2 2

{

1+ [ cos (20-15)

{

1 1/2 sin-(30+20)sin(30-15-1.84 1+ [ cos (20-15) cos (20+ 20+ 1.84)J

cos (20+ 20+ 1.78) J

}

= 0.684 With (-) a,'d -I

'"

0.03162

= Tan [ 1-0.01581] = 1.84

0

It gives = cos2(30-1.84-20)(1-0.01581) (KA)t(,'n

x

-

cos1.84cos220cos"(20+20+1.84)

2

}

= 0.672 Therefore, for (+) ve a,'d Ww - 0.684 Tan30 W - 0.6225' (1+0.01581) (tan 30-tan 1.78)

= 1.143 4

,

.'

:am;c Earth-Pressure""

227

For (- )ve avd Ww -

0.672 . Tan 30 W - 0.6225 (1-0.01581) (tan30-tan 1.84) = 1.162

"

Hence

ww =

= 11347

1.162 x 9765

kg/m

"'

If a factor of safety of 1.5 is used, then "

-.

W = 9765 x 1.5 = 14647 kg (Static condition) ~ = 15585 x 1.5 = 23377 kg (Earthquake condition - zero displacement)

Ww

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 some splacement. "

xample 5.4 ompute the displacement of a vertical retaining wall baving section and.~ackfill properties as shown in ig. 5.28. The characteristics of the ground motion are: Period = 0.50 s Average ground acceleration

= 0.2 g

Number of significant cycles

= 10

Yield displacement

= 5 mm; n= 2 and

"r

~ =10%

-tl.0m ~ ",

,I

,

I , "

",

" 1'Ranki'ne's

6.0 m

"

I

f

3.0.m

..-~-t

,,

wedge"

'tt ="18 k N I in3

,,,

I

~

"K.4s-

0/2

= 30°

"

Fig. 5.28 :'Section or retaining wall"

. .

228


Solution: (i) Refer Fig. 5.26 1 Weight of wall = 2" (1.0 + 3.0) x 6.0 x 24 = 288 kN Weight of soil vibrating along with wall 30 1 .

(

= 0.8 x 2" x 6.0 Tan 45-T ) Total weight = 288 + 149.6 = 437.6 KN

x 6.0 x 18

= 149.6 kN .

3

m = 437.6xl0 9.81 Let the coefficient of base friction = 0.6 .

= 44.61x 103 kg 1

.

288xO.6--xI8x-x62 K = 2 1 0.005

1 3

(Fig. 5.18d)

= 12960 kN/m = 1290 x 103 N/m T=21t~ n

VK;. 3

44.61 x 10

T = 2 1t n

V

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 g Slip per cycle = 24 mm Total slip = 24 x 10 = 240 mm Example 5.5 Determine 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: = 6mm Ground acceleration: =.0.25 g Time period: = 0.3 s Solution: 1. The wall is divided into foul equal number of segments with 11ftequal to 0.75 m and the backfill soil is idealized by using springs as shown in Fig. 5.30. The mass of retaining wall is assumed to be lumped 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 determined using Eqs. 5.55 and are given below in Table 5.4. The ratio of stiffnesses in passive and active states is taken as 2.0.

.

~

..-~..'

-,"

--~~if'.t?~f~

229

nic, Earth Pressure

y I I

.

:-i°.3m~

A,

.

re

:

I I I I I I

,

,I ,

0

=

36

I

'(

=

18 kN 1m3

I

3.0m

I I I

7th: 520 I
,

'"

Retaining

0

I I

- - - ~ .?ll- -;-I - +- - - X , I I I I

,

0

f4

1.0 m

c --i

v Fig. 5.29 : Section of retaining wall

H = 3.0 m

M ~

a

io

=

a 0 sin wt

c I ......

b

=

1.0 m

Fig. 5.30 : Mathematical model adopted for solution

- '0

.," .'":':

::~;::~'~Yf~'~t*~11~{!',~~:r~i;:';'c::?0t~~,1;\q~i01;-J;{:;:':;~

0---

"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 location

, Spring

w,r, to c,g"hj (m)

f

,Spring

constant

constant

passive case, 'kj (KN/m)

active case, kj (KN/m)

1.77

48.8

77.6

K2

1.02

292.5

585.0

K3

0.27

585.0

1170,0

K4

- 0.48

877.5

1755,0

Ks

-1.23

536.3

1072.6

KJ

L k,I

L kj = 2340.1

= 4680.2

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 2 DIstance of e.g. from CD = 0.3+ 1.0 x'3 = 1.3 = 1. 3 m . ,

3xO.3xO.15+0.5x3xO,7

Distance of e.g. from BC =

= 0.13;;5°.56 Ivv

,~

=

0.3 x 3

.- .+0.3x3.0(1.5-1.23)+

12'

0,7

0.3+3

1+0.3 . 2 x 3.0

3

(

)

= 0.356 0.7 x 3

.

36

3

I

+-xO.7x3x(1.23-1)

2

2

= 0.675 + 0.06561 + 0.525 + 0.0555 = 1.321155 m4 3

I

YJ

=

.

'

.

0.3 x3+0.3X3.0(0.356-0.15)2+0.7

12

3

2

X3+.!.XO.7X3X

36

2

0.7 -0.056 ( 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

l,

,'.1 ,

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 case I.k. a = - 2.0 x 510.94 = - 1021.88 a = ---L = 51094 M . I. k. h. b =-2.0 x 117.51 =-235.02 b = I I = 117.51 M c = - 2.0 x 451.76 = - 903.52 c = I.k. I h~ I = 1513.393 = 451.76 J 3.35 The natural frequencies of the wall by considering the tension side (i.e. active case) are given by : '

1

2

c-a

oonl,2=2(a+C)I~ ( 2

2

b

2

()

) +-;:

Puttmg the values of a = 510.14, b = 115.51 and c = 451.76, we get 00/11= 24.94 rad/s oon2 = 18.46 rand/s

The natural time periods are therefore, Tnl = 0.25 s Tn2 = 0.34 s Earthquake motions are erratic and no two accelerograms are similar. The two main parameters of lY ground motion are the amplitude of acceleration and the number of zero crossings in unit time. A ~ry simple and convenient form of ground motion including the above two parameters, is a sinusoidal otion. 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 some efinite number of cycles of loading of equal magnitude. Such idealization have the advantage that after tudying the effect of two parameters, the effect of a probable earthquake motion at any site, can be nalysed. Because of the above advantages, sinusoidal ground motions are utilised in the present study. Given

ao

= 0.25 g = 2.45 mIs2

21t T p = 0.3 s ; ro= 0.3 = 20.94 rad/s

a = 2.45 Sin (20.94 t) Predictioll of Displacements ill Elastic System The displacements in passive case (t = 0 to tpl2) can be cal~ulated as shown below:

x = X Sin Wt

e = ~ Sin Wt where

X

=

ao (a-ro2)-

2 b ,2 (c - ro2 )

=

2.45

.(-1021.88- 20.94')j.

.

2 (-235.02)'

0.8556 (-903.52-20.942)

=-

1.7448 x 10-3 m

H'

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

'""~';;,~,::,,,)':',.,:,;,:";'<

':":""":,/","",\i,,

"""

232

.,C-.;--,",:-.c

,',

'"

','

'.'

"""..'..n'-"':{!\;;{k,',',;.",'A"

i

Soil Dynamics & Machille Foundations Î ó ¬óù ó

ù

¿±

î

®î

î

ø¿ó³ ÷¨ø½ó³ ÷óó¾ ¾

ã

óì

îòìë

î

ãóìòïéìï ¨ ïð

®¿¼

øóïðîïòèèóîðòçìî÷¨ øóçðíòëîóîðòçìî÷ðòèëëêòóøóîíëòðî÷ óîíëòðî

Hence, we have x = - 1.745 x to-3 Sin (20.94 »

ã ó 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) Time

Translational

Rotation

t(s}

displacement x (mm)

e (rad)

0

0

IJisp. of wall at top due to rotation xe (mm)

0

Total disp. at top xro/mm)

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

Ì¸»

0

0

f)

0

displacement in active case (t = t; to tp) can be calculated as shown below x = X Sin rot

~ Sin rot

e= where

-

x=

ao b2

/

(a-m2)-\

,.2 c-m2

=

2.45 (510,94-20,94')-

~= =

,

117,25' 0.85562 (451.76-20.942)

~-1.817

x \0-3 m

ao 2

., ., r (a -ro-).(c-ro-) --b b 2.45

ãóîòïçé¨´ðóîra ¼

2

. (~10.94 - 20.942).( 451.76- 20.942) °ig~561

117.51

Hence, we have x =-1.817 x Sin (20.94 t)

. :.:

e = -2.197 x Sin (20.94 ¬÷

\i

-- -

-

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) Time t(s)

Translational displacement x (mm)

0.15 0.1875 0.2250 0.2625 0.3

0 1.28 1.82 1.29 0

Rotation

Total disp. at top xla/mm)

Disp. of wall at top due to rotation xB (mm)

e (rad) 0 0.02197 0.01552 0.02197 0

0 27.48 mm 38.89 mm 27.48 mm 0

0 28.76 40.71 28.76 0

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

~ 60

50 E E

."

0

~

40

.... 0

-c: ti

E to> u 0 a.

30

III

0

20

10

~-=Ji.!!'..!!'- - - - -- - --

- --

0

-

5-

0

0.0375

0.0750

0.1125

1 0.1500 Time.

1 0.1875

1

0.2250

s

Fig. 5.31 : Dynamic response of retaining wall

1 0.2625

1 0.3000

234 -

Soil Dynamics & Machitie Foundtitions

Check for Plastic Conditions

-

When the displacement ()fwall (Xtop)is greater than yield displacement (Yd)' the system would be plastic. 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 identify the time 'le' afterwl}ich plastic conditions- exist, a line has been drawn cof!"esponding to 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 e =4.9116 -, x--10--4 , m

ae= 5.9388x

ay

"-

xe

= 0.3663 m/s

1O-3rad.

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 20.1631) 0.1631 x 2.45 cos (20.94 x 0.1631) -?' 20.94 20.94

= - 0.0173 C3 = 0.4429C4

= 5.9388

117.512 x 4.9116 x 10-4 -451.76 x 5.9388 x 10-3 x 0.1631= 0.8676

[ 0.8556

x 10

-3

+ (0.4247)

]

x

0.1631

= - O.1009

2

- 0.4429

x 0.1631

Hence the governing equations for displacement become,

x = 0.2234 [2 - 5.587 x 10-3sin(20.94 t) - 0.1489 t - 0.0173 a = - 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!

t(s)

displacement x (mm)

e (rad)

0.1531

0.47

0.1875

1.18

0.2250

0.69

0.2625

- 2.44

37.129 x 10-3

- 6.77

, 41.8624 x 10--

0.2963

Disp. of wall at top due to rotation Xo (mm)

5.97026 x 10-3 16.0015 x 10-3 8.396

.

10-3 "

Total disp. at top x/(JP(mm)

10.57

11.04

28.32

29.50

50.26

51.01

65.72 "

63.28

74.10

67.33

f

,;,)

-,t

)ynamic Eartl, Pressure

235

,.From t =e;2963 s to t = 03 s;the- displacements 'are computed using the expressions obtained by olving the equations of motion under elastic condition taking boundary conditions satisfying the previmsly computed values t = 0.2963 s. The complete solution can be expressed as : I.

l

t.

l

Xe

= A I sm rot + A 2 cos rot + A 3 sm rot + A 4 cos rot

ee

= BI sin ro t + B2 cos ro t + B3 sin ro t + B4 cos ro t

I

'11

I

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

e

m

sinrot+A2 cosrot+A3 sinrot+A4 cosrot A A A A e = -1.. sin rot + --1.. cos rot + J sin rot + -1. cos rot

m

m

m

Putting boundary conditions, we get - 0.0067669 0.0418624

= - 0.0786 AI - 0.9969 A2 + 0.0786 A3 + 0.9969 A4 = 0.0763 AI - 0.9679 A2 - 0.1139A3

+ 1.445 A4

- 0.133143 = 20.875 AI + 1.6459 A2 + 20.075 A3 + 1.6459 A4

0.09603 = AI

-

20.267 AI + 1.5979 A2 + 29.094 A3 + 2.385 A4

= - 0.00376

A2 = - 0.02171 A3 = - 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 t

ee = 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

Time

TraJ;slational

Rotation

Disp. of wall at top

t(s)

displacement x (mm)

e (rad)

due to rotation xe (mm)

Total disp. at top XI (mm) op

0.2963

- 6.77

41. 8624 x 10-3

74.10

67.33

0.2975

- 6.85

41.916

67.34

0.298

- 6.98

41.98359 x 10-3

74.191. , 74.3109

0.299

- 7.10

42.0320 x

74.39670 ,,'

67.29

0.3

- 7.23

42.062 x 10-3

74.4498

67.21

x 10-3 10-3

67.33

,

.

Hence it is fo~rid that after 'one cycle, the disphi~ement ~sequal to 67.21 ,mm and it can be called as slip. The total displacement after n cycles is e'qual'to n times the slip. The 'final translational displacement and r~tation of the retaining wall are therefore known. The displacement curve in palstic state is also shown in Fig. 5.31.

.:.'..'

;'r:;,;"i:;":.~'d~~.';

.:;..,.-/:{;i;'

..,.,..,rtif~;~;8t. "'

236


Foundations

ÎÛÚÛÎÛÒÝÛÍ

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 Engineering, University of 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 Engineering Congress, 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. 129160. 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 Symposium 011Earthquake Engineering, University of Roorkee, Roorkee, Vol. 1, pp. 277-288. Prakash, S., Puri, V. K. and Khandoker J. U. (1981), "Rocking displacements of retainingwalls during earthquakes", Int. Conf. on Recent Advances in Geotcchanical Earthquilke Engineering and Soil Dynamics, Vol. 3, St. Louis, U.S.A. Prakash S. (1981), "Analysis of rigid retaining walls during earthquakes", Int. Conf. on RecentAdvances in Geotech. 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 dynamic conditiOns", 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 retaining 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 Speciality conference 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 acceleration and changing inclination of failure furface", M. S. Thesis, MIT, USA.

< "., :. ~ .j J 4

;;:\r£;;:;i,~;~~. ;[;:J~1~;t!fi ,;~:~~-~ <:v~?';;:;~\.;':',

i~'w'~~f~il!¥~~

j'

,~

.: . . .

>.-"

237

Dynamic Eartlt Pressure

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 pressure? . 5.3 Explain the salient features of the following: (a) Richard-Elms model (b) Nadim-Whitman model (c) Reddy-Saran-viladkar model for 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 area where the design seismic coefficients are

ah = 0.10 ; al, = 0.05 Compute the static and dynamic earth pressure on the wall using both modified Coulomb's approach, and Culmann's graphical construction. 5.5 If the retaining wall (Problem 504)is to incline at lO° with the vertical, would you recommend its 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 the increase 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

= 5.0 mm Period of ground motion = 0040 s Zy

Equivalent number of cycles in an earthquake o~ magnitude 7.0 will not exceed 15. DD

t

DYNAMIC BEARING CAPACITY OF SHALLOW FOUNDATIONS .

6.1 GENERAL Foundations may be subjected to dynamic loads due to earthquakes, bomb blasts and operations of machines. The dynamic loads due to nuclear blasts are mainly vertical. Horizontal dynamic loads on foundations 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-static analysis is first presented and it is followed by dynamic analysis. Design of foundations of different types of machines have been given in detail in chapters 8 to 10. 6.2 PSEUDO-STATIC

Pseudo-static analysis Adopting appropriate can be conveniently subjected to eccentric

ANALYSIS

is more commonly used for designing foundations subjected to earthquake forces. values of horizontal and vertical seismic coefficients, equivalent seismic forces evaluated. These forces in combination of static forces make the foundation 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

loads

ha\'e been prL'scnted. It is preceded by brief description on fundamental concepts involved in bearing capacity 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 footing without causing rupture of soil is its bearing capacity (some times termed critical or ultimate bearing capacity). It is usually denoted by quoThis load may be obtained by carrying out a load test on the footing which will give a curve between average load per unit area and settlement of the footing. Based on pressure-settlement characteristics of a footing and pattern of shearing zones, three modes of shear failure have been identified as (i) general shear failure, (ii) punching shear failure and (iii) local shear failure (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 adjacent ground. 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 occurs in 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",.

,,';"~',:':},,~~,

239

Dynamic Bearillg Capacity of Shallow Foulldatiolls

In punching shear failure, there is vertical shear around the footing perimeter and compression of soil immediately under the footing, with soil on the sides of the footing remaining practically uninvolved, The pressure-settlement curve indicates a continuous increase in settlement with increasing load (Fig, 6,1 b). .

Load

-

+-

c

<::.I

(a)

(b)

E

.

-+-

.-: I

<::.I

+<::.I Load c

"

<::.I.

E

,"

'<::.I

++<::.I lI'I

I

(c)

---

",

,,-...."

/

.... +C

.,'

Load

<::.I

E

<::.I +<::.I If)

-

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 both the general shear and punching shear failure modes. Well defined slip lines immediately below the footing extend only a short distance into the soil mass. The pressure-settlement curve does not indicate the bearing capacity clearly (Fig. 6.1 c). This type of failure occurs in soils having plastic stress-stram characteristics (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 sand is illustrated (Vesic, 1973). This figure indicates that the type of failure depends on the relative density

and depth-width ratio (D/B) of the footing. There is a critical value of (D/B) ratio below which only punching shear failure occurs.

-

, 240

Soil Dynamics

0

0

Relative density, 0.2 0-4 0.6

& Machine

Foundations

Or 0.8

1.0

Gen eral sh ear failur( zont

elcc ..

..a. .J::.

5

'"0

-

.->

Pun ching shear fai lu re zone

0

!'t:

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 Type of failure

Relative density Dr (%)

(Deg)

Void ratio e

70

36°

S 0.55'

S 20

S 29°

0.75

1.

General shear failure

2.

Local shear failure or punching shear failure

6.3.2. Generalized Bearing Capacity Equation. In the design of foundation usually net bearing capacity is computed' and used. It is defined as the maximum net intensity of loading at the base of the foundation that the soil can support before failing in shear. It is denoted by,"qnu'Therefore qnu=qu-YIDf where, q u = Ultimate bearing capacity

...(6.1)

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 by applying modification factors that account for the effect of each variation (Hansen, 1970). It may be written as : q

nu

=cN.S.d'i.b c

c

1"

c

eel

+ y .D . (N -1 ) óÍò¼ù·ò¾ò® +-'Y 2 'B.N y .S y .d y .i y .b y .r' w f q q q q q w 2 ...(6.2)

241

Dynamic Bearing Capacity of Shallow FoundatioJIs

where qllu

= Net ultimate bearing capacity

c = Undrained cohension of soil B = Width of footing Df = Depth of foundation below ground surface Nc- Nq, Ny = Bearing capacity factors Sc' Sq, Sy = Shape factors for square, rectangular

and circular foundations

dc' dq, dy = Depth factors ie' iq, iy = Inclination

factors

bc' 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 angle of shearing resistance of soil (Terzaghi, 1943; Terzaghi and Peck, 1967). Their values may be obtained

from Table 6.2.

"

Table 6.2 : Bearing cp

Ne

0

5.14

5 10

6.49 8.35

Capacity

Factors Hq

Hy

\.00

0.00

1.57 2.47

0.45 1.22

Deg

.; .'

15

10.98

3.94

2.65

20

14.83

6.40

5.39

25

20.72

10.66

10.88

30

30.14

18.40

22.40

35

46.12

33.30

48.03

40

75.31

64.20

109.41

45

138.88

134.88

27 I. 76

50

266.89

319.07

762.89

6.3.2.2. Shape factors. Approximate values of shape factors which are sufficiently accurate for most practical purposes are given in Table 6.3. Table 6.3 : Shape Factors S.No. (i) (ii)

Shape of footing Continuous strip Rectangle

Se

Sq

1.00

1.00

1+ 0.2 B/L

1+ 0.2 B/L

Sy 1.00 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 shearing resistance 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 would be increased (Meyerhof, 1951). For this case, depth factors are applied, whereby D d = 1 + 0.4 --L

...(6.3)

B

C

2

d = 1 + 2 tan <1>(1 - sin <1» . q

,

DJ

(6.4)

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 conservatively considered as follows: One way eccentricity (Fig. 6.3 a) - If the load has an eccentricity e, with respect to the centroid of the foundation in only one direction, then the dimension of the footing in the direction of eccentricity shall be reduced by a length equal to 2e. The modified dimension shall be used in the bearing capacity equation 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 respect to the centroid of the footing then the effective dimensions of the footing to be used in determining the bearing capacity as~ell as in computing the effective area of the footing in resisting the load shall be determined as given below: -

2 eL

...(6.6)

B' = B-2

eB

...( 6.7)

L' = L A' = Lt

X

B'

...(6.8)

~

Qv G.

e

I

I

,

,t"'.

Qh

"

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

,\7 '«'\~1\Z2?r"

.,,:~..,.,. J..

I

t

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

B or L

.

, ::.::~

~

Fig. 6.3 : Eccentrically-Obliquely loaded footings (a) One way eccentricity

.W'«)

a-

243

Dynamic Bearing Capacity of Shallow Foundations

~

.... ....

Fig. 6.3 : Eccentrically-Obliquely

for B- axis

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 ,

-

ly -

[

1

Qh"

...(6.9)

- Q + BLc cot ]

Q

. Iq

= [ 1-

V

iq

'

m

...(6.10)

Q + BLch cot ]

- [N:~~$] (when p

...(6.11)

O.

= 1 mQh - cNcBL (when = 0°)

...(6.12)

where Qh is the horizontal component of the load Q acting on the foundation at inclination i with vertical. Values of m are taken as given below: (i) If the angle of inclination i is in the plane of L-axis

111=(2+~)

..(6.13)

(1+~)

(ii) If the angle of inclination is in the plane of the B-axis

.

~, y m'= -0" ~',./. ,'"

-

(.'

2+oB 1+B L) ( L)

...(6,14)

,

"

,.'

-" As P~!)~,':640].:)981,:tlfernclination "...' .- - ------ -,-- - ,--,~.- -,-'factors -- are give,n"by-~~:- ,.--.~:-,:::--".-' .)

'q~cq:

~'

. '; ..'

:;!:f

'..~

244

Soil Dynamics & Machine Foundations 0 ID

-

..0 '-"

0 Lt)

0

-

~

00 11

0 ~

1J '-" 0 ~

.-

Col V

0

0 M

Z

0

11

N

0 N

0 ~

.-

0

0

.-

0 N

0

0

tII V

Z

0

0 ~

.-

0

t

0

0 ~

(6~p)q,

0 N

(6~p)

0

0

cp

-&-

." ." ... '" " 'E 7 ""': -.::; ;1 .....

0 0

-

.-

-

0 ~

u

d

'-" 0 co

0 M

0

°0

0 N

0 ID

11

11

Col V Z

0 N

0 ~

0

0 N

0 ~

'0 ~

0 0 N

( 6~p)

0

0

0

t

0

0 N

0 ~

(6~p)

0

et>

0

t>I V Z

.. Dynamic Bearing Capacity of Shallow

I

~.

245

Foundations 0 1.0 '"

.0 .........

0 \J')

-

00

1

tI

I

i

'" "0

.........

0 tIJ C') eT Z

0

1

00 C')

tI

,-

1\

N

0

1

to... '\.u.

18

r. 0

0

..t

0....

N

(60p)

II

"0

0

C')

0

0

0

0

M

10

0

0

""

N

(6<>p)

°g

Z

er '" :: '"

''"' ,.

cp

7.

..

'.

Ir:

-.::> O£

'. '"

1

0

I

l

'"

u

.........

i

00

1

11

0 0 N

.0

11

\

tIJ eT

Z

,-

10N

0

0 ~

- o.

et)

~~

-0

0 N

,'. (6<>p)

-0

0

0 ~,

o.

et)

0 0 N

(6<>p)

0

et>

'0

eT tIJ

Z

"}:Jl

..."".

246


Foundations

0 N

0 -..t

".......

.c

_°

,

"tJ 0 00

°0

, 0 M

° .0

JI

M

"

0

C:.I

,-

"N

')0 Z

C:.I )0

Z

0

-.t 0

I.t'I M

0 t

0 M

I.t'I N

0 N

0

0

0 M

I.t'I M

0

0

...t

(6ap )

(6ap)cp

0

et>

~ '" :::I ~ ... ;> .~ 7.

0

'c: ~ ~ '.

U)

"....... 0

'-"

0 N

°0 JI C:.I )0

"....... u

z

0 U)

,

°0

0 co

N JI

0 ...:t

0

...t

0 ::>

.t

0 M

0 N

øê¿°÷ et>

0

0

0 N

b -..t

0 M

'1f) M

(6ap)

1f) N

et>

00 N

ID

J)ymimii: 'Bearing 'CapJCity"of'Sh~llow

,<

.<'

-,'

,,' ,,(i'

'f"

"247 ---

Foundations ,

ic

"

,

2

= iq = ( 1 ~ ;0

)

::':

'(

,t,

, ":..(6.15) ...(6.16)

~J

iy= (1-

';-

For more accurate estimation of bearing capacity of a eccentrically-obliquely loaded footing, bealing capacity 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 and limit analysis approaches'.

~":":

As evident from these figures, bearing capacity factors (Nyei'Nqei and Ncei) depend on
~,

qnu

= cNeei .

-.

1,

Se . de + YiD! (Nqei -1)

...(6.17)

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. , 6.3.2.5. Base inclination factors. If the base of foundation is inclined from the horizontal and an applied load acts nonnal to the base (Fig. 6.7), the pattern of rupture surface beneath the foundation will be different 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)

,

...(6.18)

b =' b ,q fer tP > 0°, c q N e tantP '

= 1-

0.0067 a

for 4>= 0°

b = b =

...(6,19)

2

,

(1 - ~

tan'"

"

:;'

"

.q 1 57.3 If ) , .'[' the angle of the base inclination in degree with horiz~ilta1.. ' ,

where a

represents

~

" _'0

rJ

,.,c

Fi~.6.7 : Inclined, footing

',," ",~s

"

...(6.20)

,.;.r"','.,, 171.

,: 248

Soil Dynamics & Machine Foundatitnu ,:..".

"

~

6.3.2.6. Water table factors. (Fig.

Correction factors r w and r'w may be computed using following Eqs.

6.~).

' da

rw

Df

= 1.0 - 0.5

(For da ~ DJ Y

...(6.21):'

.

"

d'

~

(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 the r~ = 0.5 + 0.5

position of water level at depth more than B, db = B i.e. r'll' = 1.

'

"'"

B Footi ng

vwot.rzr Of

t

'

Irzvel

do

-t-

I

I water

B

I

I

I (rzv
I

L L______J

'

\!

db

L

(0)

1.0

L.~ 1.0

,~L.

L." 0.9 ..0 u

.:: 0.9 0 .. u

0 Of-

.B 0.8

c 0 0.7

~

?

0.8

c:

.~ ..u 0.7

.. u ::J '0 0.6

::J

'0 tI

tI Cl:

j-. "". .f

0.6

a:

0.5 0 o. Z O. 4 0.6

0.8

do/Of

..

0.4

0.6

0.8

1.0

~, " 'f

¼¾ñÞ ..

(b) Fig. 6.8 : Correction

(c) factors 'cor position of water table

,J..I ,.

,

.-,,"~

;'~,,; '~,0 ";;. ~', :'

~

~-_.-

J,~II1I~-.,

--

ID

~,,'

249

Iynamic Bearing Capacity of Shallow Foundations

).3.3. Local and Punching Shear Failure. The assumption that the soil behaves as a rigid material s 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 identified 1Sper the criterion given in Table 6.1. Terzaghi (1943) proposed empirical adjustments to shear strength parameters c and <{I to cover the case of local and punching shear failure. Shear strength parameters cm and <{Im should be used in the bearing capacity equation and the bearing capacity factors are obtained on'the basis of , where'

2 = - C

c

<{Im

...( 6.23 a)

3

III

= tan

-1

...(6.23

(~ tan ~)

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.

(N )

'Y 4>=36°

For example,

of Ny for "

(N ym ) CP=29°

36° - 29°

(NY)4> = 34°= (Nym )$ = 29° +

value

~=

34° will be

,

...(6.24 )

x (36° - 34° )

where (Nyrn)$ = 29° value of Ny factor for <{I::;:29° considering local shear" failure condition. Therefore its 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 the net safe bearing capacity. It is denoted by qnF' '

.

A factor of safety is used as a safeguard against (i) natural variations (ii) assumptions (hi) inaccuracies

in shear strength of soil,

made in theoretical

methods,

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

...(6.25)

qnF = qnu F 6.4 SETTLEMENT,

TILT AND HORIZONTAL

DISPLACEMENT

An eccentrically-obliquely loaded rigid footing settles as shown in Fig. 6.9 in which Se and Sm represent respectively 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 t

...(6.26)

In Fig. 6.9, Ho represents the horizontal displacement of the footing. '

<;~"",.;.

---....----

"

,,250

Soil Dynamics &, Machine {1:

~ ,

.'

,

" " ,~

I

I

'

I

Se

5,0

..--

I ~ l_.l.

i

(e/2-~)

.

e!--I

-.","'IQd(Zi,

"

I.,

Foundations

'

e/2

-1

-E

"

Srnl , Final position'

l-:HD~

'

"

- --~)

I

E

~

~~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 sand Footings of different widths and shapes were used. In each test, for a pressure increment observations were taken to record Se' t and Ho' Effect of relative density of sand was also studied. In addition to these tests, pressure.settlement and pressure-tilt characteristics of eccentrically-obliquely loaded footings 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, plots of SeI So versus e. I Band Sm IS o versus elB were prepared for different load inclinations (Figs. 6.10 & 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 Se and Sill values are taken. These plots were found independent to the type of soil, factor of safety, size and shape of footing The average relationships can be represented by the following expressions:

~

1 - 0.56

,

~:=Bo + BI (~)

-

where.A"

...(6.27

Ao + AI (~)+A2(~r

~e 0 =

...(6.28

,..(6,29

(~ )-0,82 (~)'

...(6.3(

A, ~.. 3.51 + 147 (~ )+5,67( ~)'

A,

~

(~ )-lZ,4S(

4,74 - \.38

,

,

',-

I

) Bo = I - 0.48 ( ~,...~.~;. ,

,/

""

2

~

( q, )",...' '

r

..:(6.31 -" . ,..(63

i_h,"""'"

ilntlmic Bearing Capacity of Shallow Foundations

B\

~ 1.8~ + 0,94

~

251

W- L63( ~)'

...(6.33)

0.8

i ::150

. .

0.6 Se So

. 0-4

0.2

0

~-

0

0.1

0.2

0.3

'.,

ù´×Þ Fig. 6.10 : 8/80 versus e/B for j

= 15°

1.2 ,

I =

0

15

0.8

-Srn

So 0.4

0

0 ,

.

0.1

0.2

0.3

e/B ',;

" < :., ,{,;;-~Fig.6.11

: Sm/SOversus elB for i-15°

'Co):, ..

252


Values of So can be obtained using the data of plate load test or standard penetration test in cohesionless soils, and consolidation test data in clays in conventional manner. Similarly a unique correlation was obtained between Ho / B and i/~. Ho

B

i i 2 i 3 i 4 =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 shape of the footing. The effect of eIB was found small and the displacement value decreased little with the increase 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 the performance 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 the assumption that soil rupture under transient loads occurs along a static rupture surface. In this section the sailent features of th~ analysis developed by Triandafilidis (1965) and Wallace (!961) for transient vertical load; and by Chummar (1965) for transient horizontal load have been presented.

~

8

qu

(enter

~

I

-T

0.43 8

I

ZPrandtl's

~

\ ~

B

q

u

.

-

:

-r,

of rotation

(Fczllllnius)

---r

rupture

C Ilntllr of rotation (Fllllllnius)

I', r =2.20SB/rr I

, --- - --7 , - -L-

w

surfacll

-

WCO5~

Fig. 6.12 : Illustrations of mode of failure, and dynamic equilibrium of moving soil mass

-",.-""

')ynamic Bearing Capacity of Shallow

253

Foundations

5.5.1. TriandafiIidis's Solution. Triand~rllidis (1965) has presented a solution for dynamic response of continuous surface footing supporting by saturated cohesive soil (I\>= 0 condition) and subjec,ted to vertical transient load. The analysis is based on the following assumptions: (i) The failure surface of soil is cylinderical for evaluation of bearing capacity under static condition (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.

"0

er U\ U\ C:II ~

CJ\ CJ\ I. C:II ...

..U\

..

v

t/)

E 0

c >-

0

'.

,

,

strain

rima Fig. 6.14: Transient

Fig. 6.13 : Assumed stress-strain relationship

vertical load

Analysis

Let the transient stress pulse be expressed in the form qdoli= q e-~ I = A. q e-~ I

...(6.35)

where, qd = Stress at time t ~ = Decaying function q u = Static bearing capacity of continuous footing qo = 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~ted at a height of 0.43 B above the ground surface. The equation of motion is written by equating the moment of the disturbing and restoring forces taken about the point O. The only disturbing and restoring force is an externally applied dynamic pulse. The restoring forces consist of shearing resistance along the rupture surface, the inertia of the soil :rp~sspa~icipating in motion and the resistance caused by the ~isplacement

111&1::,

254 r.

Soil Dynamics

,

~

Machine Foundations

of c~ntre of gravity of soil mass. Driving moment MdP due to applied dynamic pulse is I 2 .

.

= "2qd B

Mdp

...(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 where,

...(6.37 a)

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

= Joe Jo = .Polarmass moment of inertia

Mri

where,

-- WB2 1.36 g W = Weight of the cylinderical soil mass = 0.31 Y1tB2

...(6.38)

...(6.39)

...(6.40)

Y = Unit weight of soil WB2 .. Therefore,

Mri

...(6.41)

= 1.36,ge

The displaced position of the soil mass generates a restoring moment Mrw ' which may be expressed as

Mrw = W r sin e Mrw =Wre

For small rotations,

...(6.42 a) ...(6.42 b)

2.205 B

r =-

where,

...(6.43)

1t

By equating the moments of driving forces to those of the restoring forces, the following equation of motion is obtained. Mdp = Mrs + Mri + Mrw Sub'5tituting

for moments

..

and rearranging,

3g

we get

[,

O.68g

e +-e= 1tB

...(6.44)

-~t,

q ""e - 1] ...(6.45) [ W ] u Equation (6.45) is a second order, nonhomogeneous, linear differential equation with constant coefficients. The natural frequency and the time period of the system are given by co.

T

~

~

...(6.46)

~;~ 2~ ~ ~

:

.,.(6.47)

..

~.JI,,~'{!:

255

DynU'tJic Bear;n~ C9pacity of S~lllloHJ.Fpundatiolls

Solution of Eq. (6.45) gives the following relation W

'T2

(8) ==

0.68gqu'

1-).+-

[{

41t2+~2T2

~2T2

}

- .41t2

cos

21tt

~).T .

2~t\

( )+-sm (-

T

21t'

J

T

+).e

-f}t

,~2T2 óóóï

41i2

-,

]

...(6.48)

The ab
~(e)

~

0.68gqu

4~2+~2r

21tt

21tt

+ ~ATcos ~2T2 Sin - ~)'T e-J3t { T } 21t { T } 21t 41t2,} ]

A-I-

21tT [{

For obtaining the critical time t = te which corresponds Eq. (6.49)is equated to zero. Since 2 1tT/ (41t2 + ~2 T2) :t=0

[

.

]

~1tT -21tt +-cos

/\,- 1-- ~2T2 SIn

'\

[ ]

41t2

T

T

By using small increments t = le can 'then (W/0.68

~).T -J3t ~-

21t t

[ ].--e

.

-

T

in to Eq.

side of

...(6.50)

,'h

,

of time t in Eq. (6.50),

be substituted

to 8 = 8max' the right-hand

0

21t

.

...(6.49)

(6.48)

the value of le can be obtained.

with

known

values

of ~, ). and

This value of B' to obtain

= K, dynamic load factor. Figures 6.15, 6.16 a~~ 6.i7 give the values of K (s2)

g qll )8max

for B = 0.6, 1.5 and 3.0m, respectively, with), = 1-5 and 10

-

N

+0 u

-2

10

.

0 ....

/ f

'0 0 0

-3 10

u

-4

I,

0

/

/

'/

~

/'

"

"

/"

/<'

1 ......

B

0

+0

.'

-" "

, ---,.,

/",

.'

=

O.Gm

0.0

0.5 -------1.0 ---->.-en 2.0 ----

/

"

"u-

oT

OQ1. 5,0 -.-.u 10.0 ~.,

.. .

..

'

"/"'"

,f

-6 10

g

.f

2.0

1.0 .) "0'- "

-

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

/'

/'

I

,

I

-5

10

,/

-

,,/",""""'" ~~ .- -,,

1.// .

'10

-7'

,

/

f,' /" , , /

-E 10

0 c ,>.

'

;

..

I0

s-1

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

10

~

0-50

~-------

-I

V\

~=

Ov(Z,rload

---,-""

..

;...

50.0

~

3.0

4,0

ratio»"

5.0

F' _.

"

Fig. 6.15 : Relationship between overload ratio and dynamic load factor for continuous footings 0.6 m wide ,. , " 1..1 .,,;,i

---

'..u.

-

~~,~-~~

256

Soil Dynamics & Machine Foundations'

10

10

-,'

-I

10

N

","'"

III

..., ~ ...

.. 0

t

-2

10

-1

2

10

0 0

-10

-0 -u e 0

10

/

'

'I

I

{/

/

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

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

;'

/' /'

/" """,

'/

I,' '/

/

,

/

,/

-7

10 1.0

0.0

050.0-"'-

., -

.. ,'."'" -'

~

"/- /-

i

102

'0

10

-3

.~ 10 E 0 c

l; 105

,,'/ .,'

;"/'/'

/./.

I .'// ;/ " ./

-c

10

f/ '

..-'

/,/'

" ---",

../.....-

,,/ ....... ..- /r

i.'

"'-"

B

+~

= 3.0m

0.0 0.5--00-1J.O - ----

>o..!!!. 2,0 --=--0 CQ,5.0 - ---

I 0~

/

-~_.-,....-

10.0

50.0

107 1.0

2.0 Qverlood

Fig. 6.16 : Relationship between overload ratio and dynamic load factor for continuous footings

-- - - ;:;.::;-:..-;::.::;=~-

- ,-"" ../-: ------------

-I

10

-/0.

u CQ, 10,0

/

III

0 0

S'=1.Sm t 0.5--- ---2:- 1.0---->0.1111 2.0---0"" 5.0-,-,0

-, ./

10

-

./

--"

,/." "

,... N

u 0 00-

/../"

f:

~ 105

0

I

I

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

---;::.::::'::===:. -" - -------

Fig. 6,17 : Relationship between overload ratio and dynamic load factor for continuous

1.5 m wide

footings 3.0 m wide

Analysis presented by Trianadafilidis (1965) is based on rotational mode of failure. However, it is possible that a foundation may fail by vertically punching into the soil mass due to the application of vertical transient load. Wallace (1961) presented a procedure for the estimation of the vertical displacement of continuous footing considering punching mode of failure. The analysis is based on the following assumptions:

6,5,2, Wallace's

Solution,

(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

.

Fig, 6,18 : Failure surface

9 ton r: ro t

~ <'

257


(ii) The soil behaves as a rigid plastic material (Fig. 6. 13) (iii) The ultimate shear strength is given by ...(6.51)

s=c+crtan where, s = Ultimate shear c = Cohesion

strength

= Normal stress

(j

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:! ..... c: 0

'" c:

..-

(1 ....

(y

..c:

:J

-0

CT\

-0

,-c:

(1 0 ..J

"U (1 0 ..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 time td (Fig. 6.19). The peak load q is expressed in pressure units. Since the function is discontinuous at time td' two equations are necessary

For 0

,;

t

,;

td' Loading function

~

qB (1-

...(6.52)

~)

...(6.53) For t ~ td' Loading function = 0 In Fig. 6.18, BD is an arc of a logarithmic spiral with its centre at O. It is defined by the Eq. (6.54). r

...(6.54)

= ro ea tan'

= Distance

OB (Fig. 6.18)

where, "0

..

.!

-.-..-.......

....

--_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 ...(6.56)

q = "(DJ "( = Density of soil DJ = Depth of footing

Ne, Nq, and Ny = Bearing capacity factors The bearing capacity factors depend on and K, K being 2 (Distance OA)/H, Fig. 6.18. The value of K locates the centre of the spiral which is the centre of rotation. Obviously the correct value of K is that which yields the minimum value of the bearing capacity. It is obtained by trial and error for each set of problem parameters. The values of N"{'Ne and Nq for various values of - and K are given in columns 3, 4 and 5 of Table 6.4. Table 6.4 : Bearing Capacity Factors (N'Y'Ne, Nq' NI' NR) K

(deg)

(1) 0

(2)

- 0.05

10

(3)

-

Nq

(4)

(5)

NI'

NR

(6)

(7)

NI (8)

0.0000

5.7277

1.0

0.0633

2.0125

5.6366

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

- 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 - 0.25

0.2\31

8.5723

1.7500

0.0799

0.2344

8.1007

1.7087

0.0786

2.7828 2.6523

5.9005 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 - 0.05

0.3154

7.4291

1.6500

0.0809

2.4547

5.5096

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

5.2330

- 0.60

0.5700

53.9491

10.5127

0.1120

5.7922

7.1922

0.00 5

Ne

Ny

,-

Dynamic Bearing Capacity -OfShallow Foundations - 0.55

0.5588

28.9945

- 0.50 - 0.45.

0.5645 0.5832

20.5266 16.3539

0.40

0.6127

-:-0.35

' 0:0935

4.8411

7.1948

4.6194 3.8837

'0.0833 ',0.0779

4.2238 3.8095

7.1228 6.9932

13.9337

3.4569

0.0757

3.5264

6.8273

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

10.2345

2.8046

0.0821

3.0625

6.1071

2.7503

0.0858

3.0360

5.9474

0.0901

3.0294

5.7994

~

- 0.20

,0.8253

.:...0.15'

0.9012

' 9.9267

- 0.10

0.9863

'9.7361

1.0807

9.6352

3.0386

5.6676

9.6049

2.6936

0.0999

3.0604

5.5360

0.05

1.2986

9.6313

2.6983

0.1053

3.0923

3.4187

- 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

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

2.8168

12.8613

3.6745'

46.2884

- 0.45

3.6419

33.8986

- 0.40

3.6943

27.6099

'

.

',-

'

9.1768

13.3381

0.0728

5.3067

8.5380

, ,. 11.0492 "

0.0796

5.0886

7.9941

""",,',

20.0542

4.5161 '

19.0369\"7~9289

5.6804

"'.17:~542

"",:6.1717 .~~tt~:l...§r,'

.,' ,

5.6658

,

4.2298'

18.J742:

-

' ,,0.0673

. 8.&572

17.?678

' ';;

17.8477

. 9.7067

5.8533

0.50

5.4463

23.9713

. 5.2413;

'::-" , ,~"

5.5961

4.1008

2'i.5'S75

0.10

"',," :. . .,'tS"t,~p{~2.

4.0149

0.1383

3.8151

- 0.15

:. O~OO, '",~,',.,?'r"".,..,.

0.1282

,

.

-0.05

'4.4462 :'4.4563

3.9952"

,

0.25

- 0.20 ~

"

'12.8991

- 0.50

- 0.30

. '"

' 2.6990,0.0948

- 0.15

, -.0.35

'. "',

,2.7167

1.1848

0.00

25

;,

0.00

+ 0.053.0750 20

' 6.1125

- 0.05 +"

15

: 259

': "8.5665", "', ':,'~; '. "

, , ." -': , . '

i "

.4.9684

,

7.5267

4.9199

7.1214

4.9258

6.7672

'

0.1076 '

:4.9746

' 0.1194

6.4552

'. ' '7 :6f,77

,< 0.1325

5.0582

6.1783

'1.5-398

,', ,,0.1470

5.1704

5.9309

,-..:0.,1629

5.3068

5.7084

5.4638

5.5072 ,

;'

".7.4620 .,

7.4368

l:~...!l.
73.8778

0.0~77

., '0.0970

'8.2992'

:~~'.J7.6903

'

"35.4499

"",-.0.1802,

'..",,",

,-

'.\,.;,~~~l,989 ,,~5,~486 ',0.0732,

'

,

7.23469.9384'

,~:3243

,/~;:~{~~r~,'1frY~/*~;;:~:' '/;2d'i~:!,

; ,";"J'\

~

',,' ";;"':~~!.;'"

-

260

;'.l';'t,',,~:

~'i""\

.,;c,-'", ,

...'

~.,


30

35

- 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

- 0.45

19.3095

80.8644

47.6872

0.1064

9.3123

- 0.40

19.1315

62.4470

37.0539

0.1267

9.0899

9.3540 , 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

- 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

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

- 0.40

115.7097

112.8231.

.146.0161

0.3229

20.8738

8.0404

- 0.20

'40

.

.

-

,

,

261

Dynamic"Bearing ,Capacity of Shallow Foundations

- 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

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

- 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

0.00 45

,

"

,

.

'

Any acceleration of the soil mass ACDBA due to the downward movement of the footing will cause inerti::1 forces which will resist jmch movement. The inertial forces are directly proportional to the acceleration of each individual soil mass and thereby dependent on displacements. The effective total inertial force is obtained by combining the inertial forces on ech separate mass using energy considerations. 2

The inertial

force

...(6.57)

is given by, If ==NI Y B ddt "

~ == Displacement

where,

NI

==

at any time t

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 proportional to the displacement provided the rotation is not excessive. It is expressed as RF = NRB Y~ '.The

coefficeint NR also depends on .

...(6.58)

and K. Its values are listed in column no. 7 of ,. Table 6.4. The differenti~J" '; "'.." J ' . " equati~~s ""," are .'" setup",~Y,_~~uatingtheJour , .". . " - ~, 'vetiCalfo~ces \",.. to z~ro. .There ." ." m~st be qi

"

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 -

"

"',

,,'

"

ó

óÁò

ô

þþôù

ùóóóùó

Soil Dymimics &: MaclrineFouniJiitions

óîêî

ùî

ò î¼¬ÿòô

¬

ôó

Ò×§Þ óóó®õÒÎ§Þ¬ÿòõ¯«Þó¯Þ£óó ô

¼Ö

ó

ø

ùò

óô¬¿ ÷

¼î¬ÿò õ¢¬ÿò

or,

¼¬î Ò×Þ

...(6.59 a)

ãð

ã ó¯ó¯« Ò×§Þ

¯

ô

Ò×§Þ¬¼

¬

...(6.59 b)

For t ~ td 2 d2A

"

NI yB ---r+NR dt

...(6.59 c)

yBt!.+qu B =0 "

d2t!. NR t!. = -- qu -+...(6.59 d) 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

or,

and,

t!.

= Cl

A

= C3 cos (K't) + C4 sin (K't)

cos(K't)+C2

Sin(K't)+

¯ó¯«

q

( NR"t, ] ( NRytd )

t ...(6.60 a) -

...(6.60 b)

!k-

,

respective! y, in which K' ~

-

NRy

~ ~~~ ;and ~.. C2 C3 and C4 are

coefficients

of integration,

The coefficients

Cl and C2 are evaluated by the initial conditions. The coeffIcients C3 and C4 are evaluated by the conditions of displacement and velocity at td as defined by Eq. 6.60 (a). Solution and substitution of

the coefficients yield nondimensional Eqs. 6.60.

.

For 0 ~ t ~ td

q NR "t t!. ( qu .)

= L-I ( qu ) [I-COS(Klt)]+ tdq~/[Sin(K1t)-:-(K1t)]

...(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 q, NI and NR are dependent only on values of and . K. Using magnitudes 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 ,in Table 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...

If' .~~

"amic Bearing Capacity of Shallow Foundations

(p$

JlZ =i~ Pl-

UO!~OJ np

263

pOO1 IOUO!SU2>W!PUON

.-

0

00

..

U") 0

x C1

N

ci

E

..---. --; 0

:

:J

~ , er

U")

-I.

+E

E ::I E

c to'

to' U 0 N 0

..

c,; '" C.:!!

........

0

= '"

a.

U\ ."0

"0

... '"

:: '" .~ ... ~

E :J .-E x: 0 E

0' <'. ' ~ \\

0 c 0

'),

-d

.~

d

C to'

N 0 0

.

.0 0

-

6

\l'

N

-

U")

0 U") .'

0

( P~ ) lLZ =

>I

P ~- uoHoJnp

.

-:0 00

.

POOl )OUOISU2>WIPUON

E ."'0 C 0

Z

... 0 7.

c .....

..i

OD ~

,)~f

264

Soil Dynamics

& Machine

Foundations

6.5.3. Chummar's Solution. Chummar (1965) presented a solution for dynamic response of a strip footing supported by c - 4>soil and subjected to horizontal transient load. The analysis is based on the following assumptions: . (i) The failure of the footing occurs with the application of

~ horizontal dynamic load acting at

a 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 logarithmic spiral 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 ~

AQ

...

r

C2

Log spiral

r

=

ro eetan

~

/c

-

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 ultimat static

bearing capacity qu is given by

1 + -2 y BN y .

qU =cN where,

c

c = Cohesion B = Footing width and equal to the initial radious of spiral y = Unit weight of the soil

Ne and Ny

= Bearing capacity factors fOl:the assumed type of failure

...(6.61

..

I

(~:'

265

Clmic Bearing Capacity of Shallow Foundatiolls

Considering

moment of the forces about 0, the centre of rotation: 2

d Moment where

'

2 . cB 27ttan~ B ue to cohesIOn c, MRC = 2 tan cl>(e -1) = 'I' C ( 27ttan~ 1) e -

\jI=

,

...(6.62a)

...(6.62 b)

2 tan cl>

Moment due to weight W of soil wedge, M RW

= 'YB3 tan 4>(e31ttan + 1) = E 'YB3 9tan24>+1

...(6.63 a)

tan 4>(e37t tan tjI+ 1)

where,

E = 4>

...(6.63 b)

2 9 tan 4>+ 1

= Angle of in~ernal friction B2

Moment

It gives,

of qu about

0

=

qu

2

=

...(6.64)

MRC + MRW

-

q -u

c

(e

- 1) + 2'YB tan 4>(e

37t tan

+ 1)

21t tan

'

...(6.65)

9 tan2 4>+1

tan 4>

Combining Eqs. 6.61 and 6.65, we get '.,

N = 4 tan 4>(eh tan~ + 1) y 9 tan2 4>+ 1 and,

N = c

...(6.66)

e27ttan~ - 1 tan cl>

...(6.67)

With a suitable factor of safety F, the static vertical force on the foundation per unit length can ~ gIven as

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 above

base of the footing A = Over load factor

",':m

"-="~'.'~..~

~.."'~..'.~~~

166

,..

.

~

Soil Dynamics & Machine FolJndations

"U 0 0

.~ P E 0

Pmax

c >Cl

~

~

td Timq

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 be considered: 1. Moment due to the vertical force Q M = I

-1 Q B

...(6.70

2

2. Moment due to the horizontal' force Qd at any time t Qd(max) Ht M2

= QdH =

td

Md(max) td

t ...(6.71

where Md (max)= Qd (max)H 3. 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 Figure 6.21) from its initial position: M)

of the centre of gravity of th~ failure wedge (C'

= W d..X

...(6.,

where W is the weight of the failure wedg~, and given by

W=

'YB2 (i

It tan $

- 1)

(4 tan cp) /1 X = R cos (11- ex) - R cos 11

and R

= QCI

...(6....( 6.

(Fig. 6.21). When ex is small, Eq. (6.74) can be written as /1 X = (R sin 11)ex

...( 6

Ilj,~"r:~!)"

267

Oynamic Beating Capacity of Shallow Foundations

However, where,

R

=

X

=

...(6.76)

~(x)2 + ("2)2 -4 B tan 2 <1> ( e31t tan cj)+ 1)

(9 tan2 <1>+ l)(i1t

-

4Btan(e31ttan~+1)

z =

...(6.77)

tancj)-1)

...(6.78)

3( ~9tan2 <1>+l)(i1ttan~-l)

Combining Eqs. (6.72) - (6.78)

.

.

= PB

M3

where,

.

3

...(6.79)

(sin 11)a

p =(e31ttan~+1)

...(6.80)

)

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 1 B4

J =

[ 16gtan<1>]

(i

1t tan $

- 1)

...(6.81 )

...(6.82)

.

and g is the acceleration due to gravity. Substitution of Eq. (6.62) into Eq. (6.81) yields M = ~c 1 B4 .-d2a 4 g d [2 where

~c =

...(6.83 )

(e41ttancj)-I)

...(6.84)

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

...(6.85)

...(6.86)

(~'t~)+ K' a where

~

A [( Md;;")}

-

k - -

,

gp sin 11 ~c B

; QB-E] ...(6.87)

...(6.88)

A=~ (1 B4 ~c) E = \jIC132+ E 1 B3

...(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 yields the following results: For I ~ Id A

1 E--QB k2 2

(

a =-

)

cos(kl)--'

A Md(max) k3

td

, A Md(max).t 1 srn(kt)++-QB-E kZ ( Id 2

)

...(6.90)

For I > Id a = (

~

) [G 1 k

cos (k Id) - G 2 sin {k td) ] cos ( k t) +

()

[G 1 k sin ( kId)

~

...(6.91

-Gz COS(kld)]Sin(kt)+(~ )(~QB-E) A

h were, G I = 2" k and,

.

1

A

Md (max) .

( E - -2 Q B) cos (k Id) - -3" k

A 1 G2 = -k' E--QB ( 2

)

.

Id

A

sm(ktd)-Z'

srn ( k Id) +

Md(max)

~

k

A Md (max)

k

...(6.92

z

A Md(max) cos(ktd)+

k

2

...(6.93

~

The procedure of computations have been discuss~d in example 6.4.

!

ILL USTRA TIVE EXAMPLES I

Example 6.1 Proportion an isolated footing for a column of 500 mm x 500 mm size subjected to a vertical 10 of 2400 kN. The structure is located in seismic region. The earthquake force results a moment of 4 kN-m and shear load of 360 kN at the base of the footing. The soil properties are as follows: C

=6

. kN/m

z

, =

39° and y = 18 kN/m

3

A plate load test was performed at the anticipated depth of foundation on a plate of size 600 r x

600 mm and a pressure settlement record as given below was obtained. The permissible values

settlement, tilt and lateral displacement 2 240 Pressure (kN/m ) 0.0 Settlement (mm) Solution:

0.0

are 50 mm, 1 degree and 25 mm respectively. 480 720 1200 1440 960

2.0

5.0

7.5

12.0

16.0

(i) Safe bearing capacity Eccentricity of load,

e =

~

=

~OOOO

Inclination of load ~ i ~ tan-I ( ~:)

= 0.1667 m ~ tan- t U4~O ) = 8.5 3'

23.0

l( 2

~n 269


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, and

B' = B-2 e = 2 - 2 (0.1667) = 1.667 m L' = L = 2 m . qnu = [e Ne Se de ie + YIDlNq - 1) Sq dq iq rw + 1. 2 Y2B NySydy iy r'J

Since <1> = 39° it is the case of general shear Ny = 100.71 Depth factors are given as-

failure. For <1> = 39°, Ne = 70.79, Nq = 59.62 and

DJ ~ " de - 1+ 0.4 I3'"" - 1+ 0.4 1.667 - 1.24

=

dq

. 2 DJ 1+2 tan (I-sm°
= 1+ 2 tan 39 (J - sin 39)2 1.;67 = 1.13 dy = 1

and

Shape factors are calculated as : Br

Se = 1+0.2T 0"

1.667

= 1+0.2~

B'

Sq = 1+0.2T

1.667

= 1+0}-2 , ,

B' Sy = I-OAT Inclination and

factors:

= 1.167

1.667

= 1-0.4~

= 0.667

Q = ~Q~ + Q~ = J3602 + 24002 = 2426.85 kN m

=(2+~)(1+~)

= (2+1.~67)(1+1.~67)= 5.195

.

Therefore,

= 1.167

i-I q

m

Q Q + B' Le cot ]

[

360 = [ 1- 2426.85+ 1.667(2)( 6) cot 39 ] and

i =i - (l-iq) e

q

. 'y -

1-

[

50195

= 0.438- (1-0.438)

Ne tan

Q

[

= 0.428

70.79 tan 39 m+1

h Q + B Lc cot ] "

= 1

= 0.438

360

2426.85 + 1.667 (2) (6) cot 39 ]

6.195

= 0.374

".'

270

Soil Dynamics

& Machine Follndation~

Assuming water table to be below the ground surface at a depth greater than (Dj + B), hence r '" = r' '"= 1, and also sa y Y1 = Y2 = Y 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-

.

I

"

2

( ) (

le =Iq = 1-90

=

853

1- 90

2

)

= 0.82

. 2

. -

I

Iy - ( 1-~

8.53 2 = 0.61 ) = (1-39)

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

From charts -

kN/m2

(According to Saran & Agarwa1, 1991)

=0.0

For

= 2259.74

= 39° , ~ = 0.1g67 = 0.083, and i = 8.53°

(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 N

144

58

70.64

75

42

46.85

68

41

44.97

47

30

32.50

N .c

88

48

53.88

58

37

40.09

q

Hence for

~ = 0.083 and i = 8.53,

Ne = 50.89, Nq = 34.62 and Ny = 42.43 In this case - [As effect of eccentricity and inclination has considered already] ;=;=;=1 e q y B =2m

..

-~'~
.---


271

D

Depth factors will be :

de

= 1+ 0.4

1

.

£ = 1+ 0.4"2 = 1.2

dq = 1+ 2tan 39 (1- sin 39)2 ~ = 1.11 dy = I and shape factor will be<

Thus

S e = 1.3, Sq = 1.2 and Sy qnu

=

0.8

(As footing is square)

= 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 Meyerhofs method. Hence

= qnu x Area of footing Qnu = 1893.39 x 2 x 2 = 7573.56 kN

Qnu

Factor of safety = 7;:t~6 = 3.16 > 3, Therefore foundation is safe against shear. (2) Settlement computationWhen footing is subjecting to a central vertical loa~.only, in that case e / B = 0 and i = 0 For '" 't' = 390'e N = 88'yN = 144 and Nq = 68 thus 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/m2 From plate load test data corresponding to pressure 1425.22 kN/m2, Sp = 22.2 mm 2

. Sf Now smce Sp

.

Bf (Bp+30)

= [ Bf

(B f + 30) ]

Thus settlement of footing when it is subjecting to central vertical load will be

200(60+30)

Now,

So

= Sf = [ 60 (20 + 30) ]

-!...

- 8.53,,;,0.2187

~ - 39

2

x 22.2 = 37.77 mm .

I I

-

I

.;'!

~

272


Hence AD

1-0.56W-O.82(~J

~

= 1 - 0.56 (0.2187) - 0.82 (0.2187)2 = 0.838 AI ~ -3.51+ 1.47

=-3.51 A2

(~ )-5.67 (~)'

+ 1.47(0.2187) + 5.67(0.2187)2 = -2.917

4.74-1.38(~)-12.45(~)'

~

= 4.47 -1.38 (0.2187) - 12.45 (0.2187)2 = 3.843 Se So = Ao+AI

e B +A2

e B)

2

() (

2

..

Se

= 0.838-2.917(°.1;67)+3.843(°.1;67) = 0.6216 = 0.6216 So ~ 0.6216 x 37.77 = 23.48 ³³

and 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) .

sm t

= S m -S B

T-e

e

= "27.56- 23.48 = o 0049 2000 . ~-166.7

t = 0.28° < 1° (safe) and

H -1l

- 0.682 (0.2187)2 + 1.99 (0.2187)3- 2.01 (0.2187)4 B = 0.12 (0.2187) . = 0.0101

Ho = 0.0101 B = 0.0101(2000) = 21.21 mm"< 25 mm (safe) Example 6.2 A 1.5 m wide strip foundation is subjected to a'vertical transient stress pulse which can be given as qd = 650 e-IOt kN/m2. The soil supporting the foundation is saturated clay with Cu= 60 kN/m2. The unit weight of soil is 19 kN/m3. Determine the maximum angular rotation of the footing might undergo.

.

~~".<

273

Dynamic Bearing Capacity of ShaUow Foundations

Solution: 2

1.

qu =.5.54 Cu= 5.54 x 60 = 332.4 kN/m 650 A.= 332.4 = 1.955

2. Reffering Fig. 6.16, For A.= 1.955 and W

( 0.68 g qll )

8max

~ = 10s-1

= 0.00298 2

W=0.311tyB

=0.311t19 x (1.5)2= 41.61 kN Therefore.,

8max = (0.00298) (0.68 x 9.81 41.61x 332.4)

= 0.158 rad

= 9.1°

Example 6.3 A 2.5 m wide continuous footing located at 1.5 m below the ground surface is subjected to a vertical transient 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: 1 1. 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

factors from Table 6.4. These are given below in Table 6.5. Table 6.5: Computations k

- 0.45 - 0.40 - 0.35 - 0.30 - 0.25 - 0.20 - 0.15 . - 0.10 - 0.05 - 0.00 +0.05

of k by taking Nc' Nq and Ny

.

of qu for Different Values of k 2 qu (kN/m ) 5765 4553 3910 3533 3304 2512 2204 3070 3080 3118 3181

274

'. . Soil

The minimum value of qu is obtained at k

2. For

=-

Dynamici & Machine' Foundations

0.15 as 2204 kN/m2.

k = - 0.15 and = 30°, from Table 6.4.

{N;:

NR

= 10.0081 and .

V~ = 5.8303 .

NR Y = 10.0081 x 18 = 0 08174 qu

2204 k'

~

3.

-

~~~~

= 5.8303 qd (max)

3000

-qu

= 2204=

.

~~~ .

~

.

~ 2~5 = 5.22

x

1.36

td K' = 0.3 x 5.22 = 1.566

For qd (max) = 1.36 and td k' = 1.566, Fig. 6.20 gives qu Y NR -'Umax qu

= 0.034

A

t:. = 0.034 max 0.08174 =0.416m.

Therefore,

Example' 6.4 A 2.5 m wide continuous surface footing is subjected to a horizontal transient load of duration 0.4 s applied 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 applied

on 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

. .

.N =

e21t

c

-1=

tan,

~

tan

e21t tan 32° - 1 . = 79.4 tan 32° .

4 N = 4 tan~ (e31ttanljl + 1) = y 1 + 9 tan2 ~ Q

= !2 B =

;

x

(

c N

c

tan

31t tan 32°

(e

+

1)

= 200

1 + 9 tan2 32°

+! YB N 2 y

2.5(30 x 79.4;

32 °

)

;

x 17 x 2.5 x 200)= 8290 kN

275

Dynamic Bearing Capacity of Shallow' Foundations (ii) Determine

'11, E, /le'

~ and

sin 11

H = 4.0 m td=O.4s i1t

V= E=

-

- 1

tan'

79.4

2 tan ~

=2

.

= 39.7

tan ~(e31ttan, + 1)

(1 + 9 tan2 ~) = 200 = 50 4 e 41t tanl\» - 1 ~c =

16 'tan~

= 256

e31ttan, + 1

~=

=' 56.6 3[ ~9 tan2~+I]

"

,

'....

- = -4Bta~2~ (e31ttan'+I) (1+9 tan2~) (i1ttan, -1)

X

50 = (- 2 B) 39.7 ~,- 2.52 B z =

.' 4B tan (e31t tan cl»+ 1) 3[ ~9 tan2 + 1] (i1t ta~ cl»-1)

= (2B) . SIll 11

~~:~

Z,

=

= 2.85 B =

~x-2+z-2

2.85 B .

= 0.75

"

~(-2.52B)2+(2.85B)2

(iii) Determine k, A and E,

k ~ ~g

A = E

9.81 x 56.6 x 0.75

(256x 2.5)

P lleB = sin

1]

g = 9.81 'YB4 Il 17 x 2.~ x 256

= 'IIcB = 39.7

3

= 807 ,

"

= 0.0000577 .

3

+ E 'YB

x 30 x 2.52 + 50 x 17 x 2.53

= 20700 kN


276 (iv) Determine Md (max)in terms of Md (max) =

A.

H.

Qd (max) = H .

= 4 x 8290 A. = (v) Determine A.crwhich corresponds to a.

A

a. = - 2

k (E

A.

=0

1

A Md(max).

A

2

k3

k2

óóÏÞ ) cos(kt)--.

td

.Q

A.

33160

sm(kt)+-

Md(maX)1

[

td

1

õóÏÞóÛ] 2

For t = td

)

a. = 0.00005277 20700-.!. x 8290 x 2.5 cos (0.8070 x 0.4)

0.807

(

2

0.00005377.33~~~A sin (0.8070 x 0.4) 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) = 0.05 A. - 0.0474

- 9.10 A.sin (0.3228) + 2.94 A.+ 0.9181 - 1.834

For a. = 0, A = 0.948 = Aa (vi) Determine Md (max)for Md(max)=

33160Acr = 33160 x 0.948 = 31436kN-m

(vii) Determine Gland A

G1 = 2k

(

= A.cr

A.

G2

1

A Md(max) .

)

E-- 2 Q.B cos(ktd)--r

k

td

A

sm (k td ) +2 k

M

d(max)

= 0..8685 - 2.89 x 0.948 + 2.94 x 0.948 = 0.9159 1 G 2 -- --A E--Q.B k 2

(

)

.

A Md (max)

sm(ktd)--' k2

= - 0.9159 x 0.807 sin (0.3228) = - 0.2346 - 6.60 + 2.79

td

A Md (max)

cos(ktd)+-.k2

td

- 9.1 x 0.948 x 0.807 cos (0.3228) + 2.94 x 0.948

= - 4.05 (viii) Determine a. for t = 0.6 s

CL

~

G }GI k cos (kid) -G, sin (kid)] cos(k 'd )+G}GI +

E ~K2(.!.QB) 2

k sin (kid )-G,

cos(k 'd)] sin (k I)

;!&~t'

277

Oynamic Bearing Capacity of Shallow Foundations

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

0.8072

1. x 8290 x 2.5-20700 [2 ] '

0

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 '

'

'

,

ÎÛÚÛÎÛÒÝÛÍ Agarwal, R, K. (1986), "Behaviour of shallow foundations subjected to eccentric

University of Roorkee, India. Caquot, A. (1934), "Equilibre

"

- inclined loads". Ph,D. Thesis,

'

des massifs a frottement interene", Ganthiev-Villars,

Paris.

Chummar, A.V. (1965), "Dynamic bearing capacity of footings", Master of E'ngineering Dissertation, University of Roorkee,

India.

'

,

Cunny, R.W. and'Sloan R. C. (1961). "Dynamic loading machine and results of preliminary small-scale footing tests", 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 capacite portante du sable sons des fondations directed etablies 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 Engineer Waterways 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 Geotechnical Institute, 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, Soil Mechanics Series No. 7. M<:.Kee,K, E., and Shenkman S. (1962). "Design and analysis of foundations for protective structures", Final Report to Armour Research Foundation, Illinois Institute of Technology. Meyerhof, G. G. (1951), "The ultimate bearing capacity of foundations", Geotechnique, Vol. 2, No. 4, pp. 301331. 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

of Geot. Engs., Vol. 117, No. 11, pp. 1669-1690.

-obliquely

loaded footings", ASCE, Journal '

~~~';';'-;";;;>;".~~;:."-'

--

'

278


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. sixth Int. Conf. Soil Mech. Found. Engin., Montreal, Vol. 2, pp. 205 - 208. Vesic, A. S. (1973), "Analysis of ultimate loads of pp. 45-73.

shallow foundation", J SMFD, ASCE, Vol, 99, SMI, .

Wallace, W.L. (1961), "Displacement of long footings by dynamic loads", ASCE, Journal of the Soil Mechanics and 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 of footing 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 applied on 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 soil

supporting 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 transient load (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 the foundation. 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. Also determine 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 the literature 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 have Jeen 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. The nost recent Koyna earthquake of 1995 is an illustration of liquefaction phenomenon causingcatastrophic Jamages to structUres and resulting in loss of life a,nd property. 7.2 DEFINITIONS

7.2.1.Liquefaction. It denotesa conditionwherea soil willunde~gocontinueddeformationat clconstant ow residual stress or with no resIdual resistance, due to the build-up and maintenance of high pore water Jressure which reduces the effective confining pressure to a very low value; pore pressure build-up leadng 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 :

useful basis for assessingvariousposs,ibleforms of subsequentsoil behaviour.

'.2.3. Initial Liquefaction with Limited Strain Potential, Cyclic Mobility or Cyclic Liquefaction. It lenotes a condition _inwhich cyclic stress applications develop a condition of initial liquefaction and ubsequent cyclic ~tress applicati?ns c~u~e limi~ed strains to develop either because of the remaining esistance ~f the ,soil to-deformatio~ .or because the soil dilates, the pore pressure drops. and the soil tabilizes

under. the applied ,,'.- loads. I. ! J,

('i, . ~

j'.

c'

c,

;: :.'

-,' -i'

;'

,.

-

-- - ~-="-~~

11

..-.

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

--"7

Ih,

280


In laboratory undrained cyclic tests (triaxial, direct simple shear and gyratory shear) on saturated sands, 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 during earthquakes to the same extreme degree as has been observed in the laboratory. A simple means for understanding the difference between liquefaction and cyclic mobility as observed in the laboratory is through the use of the state diagram, which is shown in Fig. 7.1 (Castro and Poulos, 1976). The axes are void ratio and effective minor principal stress" The steady state line shown represen~sth~ 10c1,1s q[ states in which a soil can f1owOa(c~onstant effectiv.erilinor principal stre"ssa)'and constant shear stress. The void ratio at the steady state is the same a:>the critical void ratio. 'TI C d

\J)

.x. .-u :J

_0 Q 0-

-

Ftow at con stant volume

()I

"

dI- c()I

L i que fa c t i on

'TI '-E .u

..

>

()I Cl. ()I 0111\ 0 I- (\) (\).1:.

>

-

Contractive

soils (loose)

:

I J

B

- ~--

D I

I

Cyclic mobility

I

(

,C I I

0

Larg e strains and D.I LatI .Ive softening caused soiLs by cyclic loading, (dense) I state may reach B) I

0

-

C3t ( During) flow

. I I :

/.

Monotonlc Load ing

I

I I

-

I sta t e

J I

,

.

'ne

Cyclic or monotonic Loading {of dilative soil s1 arts here

.I

I

-

-

03c minor

principal

stress,

OJ

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 il soil mechanics is depicted by points on the zero effective stress axis at void ratios above Q. In this state sand has zero strength and is also neither diiative nor contractive. At void ratios above Q the sand grain are not in close contact at all times.

~'

IDi1!1iII

Liquefaction

281

,of SoUs

The mechanics of cyclic mobility may also be illustrated with the aid of Fig. 7.1. Consider first the behaviour in Fig. 7.r when a fully saturated dilative sand starting, for example, at Point D is loaded monotonically (statically) in the undrained condition. In that case the point on the state diagram may move slightly to the left of Point D but then it will move horizontally towards the steady state line as load is applied. If one now starts a new test at Point D, but this time applies cyclic loading, one can follow the behaviour by plotting the average void ratio and the effective stress each time the applied cyclic load passes through zero. In this case the state point moves horizontally to the left, because the average void ratio 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 cyclic load, the number of cycles, the type of test, and the soil type, to name a few variables. In particular, it has been observed in the laboratory that in-triaxial tests for which the hydrostatic stress condition is passed duvng' cycling, and if a large enough number of cycles of sufficient size are applied, the state point for t1w d'\f,erageconditions in the specimen eventually reaches zero effective stress at Point B each time the hydrostatic 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 evidence has been presented to show that most of the strains measured in cyclic load tests in the laboratory are due to internal redistribution of void ratio in the laboratory specimens. For example, at the completion of such tests the void ratio at the top of the specimen is much higher than at the bottom (Castro, 1969). Thus the horizontal line D-B in Fig. 7.1 is fictitious in the sense that it represents average conditions. Near the top of the specimen, the void ratio increases, and near the bottom the void ratio decreases. The pore pressures that build up and the strains measured in the lab~ratory are due to the formation of such loose

zones (Castro and Poulos; 1976).

'

In summary then, specimens that lie above the steady state line on Fig. 7.1 can liquefy if the load applied 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 deformation 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. Saturated sands starting at points on or below the steady state line, will be dilative during undrained monotonic loading in the triaxial cell and the state point will move to the right. If cyclically loaded the state points will shift to the left as strains occur and the specimen softens. If enough cycles are applied, if they are large enough, and if the hydrostatic stress condition is passed during each cycle, then the zero effective stress condition (i.e. initial liquefaction) can ultimately be reached in the laboratory. 7.3 MECHANISM OF LIQUEFACTION The strength of sand is primarily due to internal friction. In saturated state it may be expressed as (Fig. 7.2). ...(7.1) S = on tan cp where, 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 table Ysub= Submergedunit weight of soil

.

~..

0...

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

.

- _. , ,., ...,

282

Soil Dynamics & Machine Fo!,ndations

Ù®±«²¼

Surface

t ---------

hw

.

-

~-[-

~

-1l..-

z '(sub

x

1

x

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, if drainage is restrained the tendency to decrease in volume results in an increase in pore pressure. The strength may now be expressed as, ...(7.2)

Sdyn. = (an - Udyn)tan $dyn

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, or

a" - Udyn= 0 (Jn = Udyn

or

Udyn = 1 an

...(7.3)

Expressing Udynin terms of rise in water head, hw and Y5ubas (G

written as :

Yw.hw

-

1/1 + e) Yw' the Eq. (7.3) can be

=1

G-l.'Yw'Z l+e h G-l -1f. =-=i Z l+e where, G = Sp~ific gravity of soil particles e = Void ratio

or

'-;0 ".

ier

.

= err Heal hydraulic gradient.

er

...(7.4)

283

iquefltct;oll of Soils

It is seen that, because of increase in pore water pressure the effective.stress reduces, resulting in loss f 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 partial JSS 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 closer Jacking 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 underlying zone of the deposit. Once liquefaction develops at some depth in a mass of sand, the excess pore water pressure in the liquefied zone will dissipate by flow of water in an upward direction. If the hydraulic gradient becomes sufficiently large, the upward flow of water will induce a quick or liquefied condition 111the surface layers of the deposit. Thus, an important feature of the phenomenon of liquefaction is the fact that, its onset in one zone of 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
..,

. .

i}:;"",':>','" ".-

:.,\'~ ,.-

,//~,'

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

'

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

',', _..

-

28~

Soil Dynamics & Marhine

f---

=+ --- --.2.--~

/~

H'

0-= OVi

~

'/"/'

.

OVi= ohw + lsub (z-hw)

..

z

'1

(a)

Ko ()

DA

'//1'/1'1'

1////1'

"'1'

No gr,ound

Ill'

'"

'YI

"

Stresses

on element

A

shaking

óóóóó.2-

--=+

"

""

~

0-= OVi

-- -

Jh;""

rh

z

'L

Ko 0-

OB

~h

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

+

(b)

Earthquake

~ ~. if .~i ~. ~, FOUlJdlltioll... .~ ;:,. .. ~
" '/' ,. ""./

Stresses

on element

8

loading

Fig. 7.3 : Stress conditions for soil element below horizontal ground in cyclic loading conditions

285

Liquefactiollof Soils /

() = Uvi

! ..:hi

.jor-

DA

0'"

" ""

/ /

r

" /

No ground

(0)

/

/

/

.I'"

I

".I"/

Stresses

shaking

KO

on element

A

cv= av i --L

=r=

I'

DB

-r;h Chi

1HDHPo-

~

'"

(b)

>"'+"O'

uu"

Earthquake

Fig.

7.4:

Stresses

loading

Stress

conditi ons

for

soil

ele me nt

below

sloping

ground

in

on element

cycliiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiii iiiiiiiiiiiiiiiiiii iiiiiiiiiiiiii

B

--

k -..,

.

.. 1

I I I

r


286

. I

/I' /l1I/'"

I

\

"""V/I\I'

I

00

00

I J

---c=J--

I

,

K0

t

. t,

I I I

t

t

1I

I

"A' 11/'\\

---

--11 ---

0;

00 -C_t

-C

-+- \\ --- ~\--

1!-KoCTO

t

t

Initial

stresses

Ko 0-0

Cyclic toad

sequence

Fig. 7.5: Idealised field loading conditions

I

40

I

30

I .

1:max. Dept h:; 14.0m

,

20

N

E --

z -

I.

Gav

10 0

..... 1I1 11

JI

ill

ili li' ai ffi

- 30

óìðÔ 0

6

18

12 Ti.me

24

30

(5)

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 test procedures have been adopted for liquefaction studies: (1) Dynamic trhixial test (Seed and Lee, 1966; Lee and Seed, 1967). I

(2) Cyclic simple shear test (Peacock and Seed, 1968, Finn et aI., 1970, Seed and Peacock, 1971).

I

(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.

I I

r

I I

r

l

101,.

11

- ----

..

IIIm

I!Io

J

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 sand studied by dynamic triaxial test. They used the concept of developing cyclic shear stress on section xx of the soil sample as shown by the stress conditions I and 11on the sample given in Col. 1 QfFig. 7.7. The stress condition I is achieved by increasing axial stress on the specimen by an amount ad' keeping lateral stress a3 constant, and simultaneously reducing the all round pressure on the specimen by an amount adl2 (Cols. 3 and 4 of Condition I). Similarly the desired stress condition 11can be induced by reducing the 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 corrected by 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 test

in loose sand (e = 0.87, DR = 38%) are shown in Fig. 7.8. In this testthe initial all-around pressure and the initial pore water pressure were 196 kN/m2 and 98 kN/m2 respectively, thus giving the value of effective confining pressure as 98 kN/m2. The cyclic deviator stress ad of magnitude 38.2 kN/m2 was applied with a frequency of 2 Cpg.The test data in Fig. 7.8 shows the variation of load, deformation and pore-water pressure with time. "0

~

8

DR~38°/o

01.S z

-

0"""

I/)

"3

Go

~

6

-

'15

E L.. .2E

1 eo=0'87,

100 0 100. 2St-

Time '" h 38.2

kNjm£

Compression Time

0

"0 E.

OJ=98,OkN/m2

0-

)(
'H::L

~

Extension

~

::J

I/)N L..

'-E

-

o..z ~x.

.L..

a. 0

100 '

0

. '-Time

I

- - .,' -,

f sec.t--

Fig. 7.8: Typical pulsating load test on loose saturated Sacramento river sand (Seed and Lee, 1966)

289

Juefactioll of Soils

---

:ompression 30 0

0

OR

-

201

.

-

~=. 98.0kNjm

10

o' L.. ...... \/)

38 °/0

.

..

'-c

=

at Od =(+)38-2 kNj m

2

at Od = 0

04

.i..#, 0

2

-10

)(

C:Xten sion

2

.

-20

«

- 30

2

1

4 Number

at Od =(-)38'2 kN/m I r 10 20 40 of cycles

100

(a) Axial strain -versus number. of cycles

L..~N

-

d~ E ~z L..~

0-

150 100

~ L..

~ ::J en\/) C III d ~ £. L..

50 0

ua.

,-atCJd,=O -. at-Od =+38,2 kN/m2

confining pressure

a.::J .S

- - -

InitiaL e tfe ctive

1

2.

4 Number

1°--' of cycLes

at Od :-38'2 20

kNjm

40

2

lOO

(b) Observed change in pore water pressure and number of cycles

150

-

L.. """N

-

d E

z

confiningeffective pre ssure 10°r-Initial '"

L..

0-

a. ::J .-c L.. ::J en\/) C

\/)

0 .I:. L.. ua.

50

,

,

- SO

2

.

4

10

20

40

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)

100

290


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 number of stress cycles have been plotted in Fig. 7..9.It was observed that, during the first eight cycles of stress application, 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 in Fig. 7.10. It is evident from this figure that number of cycles of pulsating load application increases with the decrease of the value of 0"d'

N

-

E 60

z~

OR ::. 38

I;'.. 50 ~ ~

-

eo

.

40

~

::.

0/0

0-87

03= 98,OkN/m2

III 0\

c:

-d

-

III :) Cl.

~

d ~ Q.

30 20 10 0

,

3

10

100

30

300

1000

:~ .

Number

.of cycles

to cause

faiLure

Fig. 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 the change in pore water pressure become equal to o"}after about 13 cycles, however the axial strain amplitude 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 this corresponds to cyclic mobility (Seed, 1976; Castro, 1976).

291

.iquefaction of Soils 11 P re

sSlon 15

--0

OR =

0 ,-

.,0)=

'

78

0/0

",,'"

'

"

_9~'.0~N/m~r-

+68'7 kN/nf

c: 0

-

0

....

- 5

L-

at Od =

at Od

=

0

111

« d x

"

'..

-:-10

"

"at Od =

'

tension

-15

1

2

4 Number

10 of cycles

20

2

~,-68'7kN/m

100

40

(a) Axial strain versus number of cycles "

~ :::I 111 111 ~ L-

0..

150

100

L~ .....

d-

AA-- -a--

Initial effective confining pressure

-A~at Od =0

~N ~ E ~ ........ 50

u at Od =

0 Z o...x

- -\--:r..s-..t::~

c<:I) CJ) c d £. U

68.7 kNjm2 uat Od-

- 68'7 kNim2

0

- 50 1

2

10

4 Number

20

40

100

of cy cl e s

(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 liquefaction and 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 confIDingpressure 0')" Criterion for complete liquefaction is taken corresponding to 20% double amplitude $train. The figure indicates that in loose sand, initial liquefaction and failure occur simultaneously (Fig. 7. 12a). With the

-

" .1'"",,'

291

.iquefaction of Soils

TtpreSSlon 15

-0

10

OR = .78

0/0 '

. .","

0

-

.-c: 0

\..... 111 ....

0 -

.~,:03'= _98.0~~/m~r..

5 O

.-K""

at Od = 0 I

- 5

"

"

'..

)(

« -:-10 tension

.

-15

at Od =

+68'7 kN/m2

1

2

4 Number

10 of cycles

20

.

at Od =

2

~,-68.7kN/m

100

40

(a) Axial strain versus number of cycles "

~

150

::J

111 111 c:J 1... a.

100

Initial effective confining pressure

1...

c:J .....

0-

~N c:J E 1..."0 Z a..x

Od =0

u at Od =

50

-..:\--:r:s--t.:~68.7 kN/m2

c-

c:J 0'1 C 0 L:. U

- -a-- -6~at

AA-

-

0

- 50

1

2

10

4

Number

20

u .at Od 68.7 kN/-m 2

40

100

of cy cle S

(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 liquefaction and identified the followings : Relative density. Figure 7.12 shows the plots .of peak pulsating stress ( Le.. the stress causing liquefaction) against number of cycles of stress in loose and dense sands. The initial liquefaction corre-

sponds to the conditionwhen the pore waterpressurebecomesequal to the confming pressure 0"3'Criterion for complete liquefaction is taken corresponding to 20% double amplitude strain. The figure indicates that in loose sand, initial liquefaction and failure occur. simultaneously .. (Fig. 7. 12a). With the

-

" .#",'..'

0292

Soil Dynamics

"O""'-'O~~"-'-""-'o"""o~"'

"o

,

~ Machine

"""_0"-'

Foundations .

"':'~'"

-

incre~se in relative density, the difference betWeen the number of cycles to cause initialliquefacticin and'~ f?ilure increases. .N .

E

Z

'.:;100

'~.r '..';

11' "

~ tIJ

80

..

strain

L.

111

eo

=

OR

=

OJ =

0'\ C

0 -87 38

0/0

9 8.0 k N/ m~

-

0 VI ::J

a. x 0 ..~ .

0 0

100 1000 Number of cycles

10

010,000

100,000

(a) Loose sand

N

E

-

200

eo = 0.61 OR= 1000/0

..........

z x

'" VI

VI

-c0

160 \

120

-

...

\

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

......

\I) ::J

a. x 0 a.

98'OkN/m2

, , Liquefaction " .... ......

80

=

Initial

\~

L. \I) 0'\

OJ

40

-- - --. .~...

0

1

10 100 1000 Number of cycles

10,000 100..000

(b) Dense sand

Fig. 7.12: Peak pulsating stress venus number of cycles (Lee and Seed, 1967) L \,

0 ._--. .-.......-.....

- _.0--

..

293

[que/action of Soils

Confining pressure. Figure 7.13 shows the influence of confIning pressure on initial liquefaction md 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) increased vith the increase in confIning pressure ,N

E

-~

180 'OR

...

eo

120

=7 8

0/0

= 0.71

L. ....

DJ (kNfm2) 1500

CTI

.-c:

-0

60

-'

:J'

~~

-

Co d

-It 0

a.

. -

'A

10

1

.

A

I

500 '.

I

100 1,000 Number -of cycles'

I

100

10~000

100,000

(a) For initial tiquefaction in medium dense sand

N

-

2000 ,

E

-

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

z

--

'500

OR =

100 0/0

eo

0.61

=

l5' " ....

1000

OJ (kNfm2) 1500

CTI

c: .-....

0

-' 500 :J

Co 0

a.

1-

&

0-

10 -

,

.

4 100 Number

500

-

.,1,060

OO ' 10.000

100,000

of cycles

(b) For initial liquefaction in dense sand Fig. 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:',~

294


---.

,

.

",

,.,.

,..',

i~

,',-

N.

E ,Z..¥ 180

-

H'

"'

OR : 78°/0

VI t.I .. VI

- eo :,0,71

120

0'\

-:J

VI

.. ,-

~ (kN/m2)

c:

-CS

".

1500

60

a.

CS t.I a.

0

1

10

100 Number

500 10°1 10~000

1,000 of cycles

100,000

(c:)20% strain (i.e. failure) in medium-dense sand 2000 N

E

......

z-" ~

~

OR = 100

1500

eo

=

°'0

0.61

III III 1:>1

;:III 1000 C7I c:

0

III ..... :J Q

500 500

-" C

100

1:>1

'Q.

0

1

10

100 1,000 Number of cycles

L10.000

(d) 20% strain (i.e. failure) in dense sand

100,000 -,'

Fig. 7.13: Peak- pulsating stress versus, number ofc:yc:les(Lee and Seed,1967)

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 100 cycles. It may be noted tha for a giverirelative density and number of cycles ofload applicafion;the"odmcreases linearly with 03 fo initial liquefaction, while for 20% axial strain condition, similar linear trend exists only in loose sand5 ;," . "" " .. :. .";. I . . ~;'~;'~

Liquefaction

of Soils

2000 ........ N

E

z 1600 oX

\11

~ 1200 .L..

Ini.tiol void ratio.)e6 0 '61

\11

01 C

.-0

800

0.71

-

\11

:J a.

400

.x d

(:,) Cl.

-0 0

800

400 ...

3600

1200

2000

03 (kNfm2)

(a) Initial liquefaction in 100 cycles

2000 , .'

.........

N

E --Z 1600 oX

....... \11 \11 ~ ....

.-

Initial ratio.)

1200

\11 01 C

";::

d

800

0 -71

-'\11

ðóéè

:J a.

void eo 0.61

400

0'81

::s:

d ~ Cl.

0

0

400

800

1200

1600

2000

03 (kNjm2) (b) 20% strain in 100 c)'des Fig. 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 a given p\llsating stress, number of cycles needed for causing initial liquefaction and failure increases with the 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 simulates earthquake. condition in a better way. Peacock and Seed (1968) reported the first set of comprehensive data 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 were tested at relative densities (DR) of 50%, 80% and 90% giving the sand in loose, medium dense and dense states respectively. The oscillatory shear stress was applied at a frequency of 1 cps or 2 cps keeping the normal stress constant.

Initial

relative

Initial

void

density

ratio,

eo

°R=50olo =

0-68

Initial confining pressure,OV Frequency ã ï Ø¦

¢ Äïï

\11C>lN

;: E Äïïóòòòòòòòòò

1- Z d

C>I

.I:. if)

~ ---

40

I'

î0

I

0

f---' 20 -

=

500 kN/m2

I

1111 cum-

M

mum

40 (a) Applied cyclic shear stress

--

20

0

0 ---;- 1

I

c .0 1-

0

1-

10

-\11

j

1ftttt

d C>I .J::. If)

2u

124,eyc: le s (b) Shear strain response

Fig. 7.15 : Record of typical pulsating load test on loose sandin simple shear conditions (Peacock and Seed. t 968) ..

iquefaction

297

of Soils ~ 1:J \11

11'1

L-N

~Q.E 1-

60' 0

~oz4

W

~~

...

-2 Cc

~ 1-:::::> 0 0..

V

-_ ..r

- ..-

f

""

'"

.-..lr"

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 water Iressure with time. As evident from Fig. 7.I5b, there was no significant shear strain of the sample during lle 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 resulting leformations became extremely large, and the soil had essentially liquefied. Similar trend was also oberved 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 this 19ure it can be concluded that for a given value of confining pressure and number of cycles of stress lpplication, 1h increases with the increase of relative density'. A mo;"e clear presentation is shown in :ig. 7.17. 11'1

I/)

80

~

~

1-

11'1

1-

a

~N

60

90°/0

E 111"-

'\.

.c

-cnZ

'-C x

40

",

a .c.

"'" 2-""""'",-

III~

:J Q

- --

.::c

a

~ Q

0

1

800 500(kNjm

Qy

)

10

Number

-----100

1,000

10-,000

of cycles

Fig. 7.16 : Initialliqu'efaction of cyclic simple shear test on Monterery sand (Peacock and Seed, 1968)

...

'

,

298


100 .........

N

For 100 cycles

E

~

~

.. \/I \/I

of pulsating

stress

80

60 ±ª (kN/m2)

L...

...

800

III

en

c "...

40

0 III

:J

0. oX

20

I

200

0 Cl.

0 0

40

20 Relative

100 density

(°/0)

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 pre

pared 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' \/I III ~ L............. V)N L... E 0"-..

-

75

~Z \II~

\

\

relative. density,

Initial

void ratio

eo

Cl.

0,68

DV (kN/m

I-

2

)

800 500 300

0.0

o~ ~

=

-..

~~

~

DR= 50°/0

50

enN c ......-...... :J

Initial

.!:

0

1

< .... }

'1

~

10 100 1000 Num b e r of c.y c l e s

Fig. 1.18: (a) Cyclic stresses required to cause initi~lliqueraction at different confining pre~sure

to I{t ,t 'j "If

299 uefacti~n of Soils

11\ 11\ ~

-'- ....... 75

,II\N

OR =50% eo = 0.68

'- E

0..........

~z

1I\.x 0'1c: N

10 c.ycles

50 100 cycle

.;: -0

- ~''

11\

1:)

::J

25

0.0

.x

0 .r= ~I'-J Q.

0

0

100

200 300 400

Initial Fig. 7.18:

Peak pulsating

effective

500

600

confining

700

800

pressure

(kNfm2)

(b) Effect of confining pressure on cyclic stress to cause failure in 10 cycles and 100 cycles (Peacock and Seed, 1968)

stres!> and number of cycles of stress application.

at for a given value of

o"v

From Fig. 7.16, it can be seen

and relative density DR' a decrease ~f "Chrequires an increase of number of

des ~o cause liquefaction. Further for a given value of "Ch:"number of cycles of stress application quired to cause liquefaction increases with the increase in relative density DR and confining presre 0"v .

Frequency ofload application. Tests were rformed at frequencies of 1 Hz, 2 Hz, and 4 ~, and the effect of frequency on the stress using liquefaction was found negligible. Seed and Peacock (1971) have studied the feet of coefficient of earth pressure (Ko) on e 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 given lative density and number of cycles of stress 'plication, the value of ("Cia,.) decreases Ith the decrease of K0 .. J"

0.4

It) 0.) --s::.

~ 0

0

0.2

OCR:

~

1/1 1/1 ~ ~

~

8

ko : , OCR:" Ko : 0.75 OCR: 1 Ko : 0.4

0.1

0

1

10

Number

100

ot cycles

causing

'.000

10.000

initial liquefaction

Fig. 7.19 : Influence of overconsolidation ratio (OCR) on stress causing liquefaction in smiple shear tests

. (Seed

and Peacock, 1971)

.iJ;

.1;'\

Soil Dynamics

300

& Machine

Foundations

7.7 CO~IP ARlSON OF CYCLE STRESSES CAUSING LIQUEF~.\CTION UNDER TRlAXIAL AND SIMPLE SHEAR CONDITIONS , PeaCock and Seed (1968) performed both cyclic triaxial and cyclic simple shear tests for liquefaction

studieson Montereysandwith a relativedensityof 50%and confiningpressures (cr3or crv)of 300, 500 and 800 KN/m2. Rbults are plotted in Figs. 7.20 and 7.21. It may be seen fro~ these figures that the cyclic stress required to cause liquefaction of loose sands under simple shear condition ("Ch) is about 35 percent of the cyclic stress required to cause liquefaction in triaxial condition (crd/2).

""

150 N

-

.......

E

Triaxial

~

~

.........

z~ N

'~--~

l5' 0

.-

results OJ{kN/m2)

800

.

-~

test

,

100

500

(-.JL: lI\ lI\ ~ ~ IfI

""""--~

La ~ .J::. IfI

CJ)

c

'\

50

.

e-

" Simple

..... a

, "-

VI -:J a.

......

oX

a ~

OR = 500/0 eo = 0-68

0..

0

1

300

10

shear

test

results

DV (kN/m2) 800 500 300 lOO 1000" Number of cycles

Fig. 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)

'i

.;quefact;on

301

of Soils

2: 150

~Le

Relati ve density} OR

.£:.

Initial

~

=

500/0

void ratio,eo =0.68

U\ U\

-~U\-100 N

E 0-~z L-

.c.~ U\-

10 cycles

CJ'I

-

.S

50

0 U\ ::) a.

100 cycles

~ 0 ~

Q. ~

100,

200

Initial

300

effective

400 confining

500

600

pressure

700

03 or

800

OV (kNfm2)

Fig. 7.21 : Comparison of pulsating shear strength of I~ose Monterey sand under cyclic loading-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 lic stress ratio (crd/2/a)) versus mean grain size (Dso) for 10 and 30 number of cycles of stress application for an initial relative density of compaction of 50% (Figs. 7.22a and b). These curves were prepared by compiling 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. 7.20,7.21 and 7.22). The two stress ratios may be expressed by the relation. ad "th == ,CI ( av ) simple~hear ( 2 cr)) triax.

\,

...(7.5)

where, Cl = Correction factor to be applied to laboratory triaxial test data t? obtain stress conditions causing'liquefaction in the field "

".

~,",.' ';

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

..,~. .~:

302

Soil Dynamics 1:5'

& Machine

Foundatiolls

-----

N '-......

0.30

1:5' ... lfo

U >u

.

0.25

.-c: c: .-0 u C

~

~

r

Triaxi cl. compression test data for Ode./203 at

9

liquefaction

.

-

0.20 0

C>i

:J

0

.--v

--

- 15

v

Field value at -Ch/ay causing lique taction esti mated tram re su Us ot sim pie sh ear tests

(J)

C 1I1

0 -10

:J d

u

G

Relative

d

density

No. of stress

0 -05

=

50%

cycles

10

=

1I1 1I1 C>i

-

L..

0

If)

1.0

0.3 Mean

grain

0.01

0-03

0.1 size

050'

mm

(a) In 10 cycles

-

N

------

tj'

0.30

...

.

1I1 C>i

-u

>u 0

'V

0-25

Triaxial compression data for 0Ct /2 CJj at liqu e ta e.tion

M

C

c: 0

0.20

test

u

d ..... C>i :J CT

15

-

(J)

.-c: 1I1 :J d

-

0

0

0

-

, Field value at7:h/r:sv causing liquefaction estimated from resul ts at simple shear tests

10

\J

Re lative

0 .....

d 0.05 ... \fI \fI C>i L.. .....

(f)

0 2.0

0.1

0-3 Me an

grain

size

=

50 °/. cycles =30

densi ty

No. of stress

0-03 050,

0.01

mm

(b) In 30 cycles Fig. 7.22 : Stress conditions causing liquefaction of sands (Seed and Idriss, 197\)

-----

Liquefactioll

303

of Soils

Seed and Peacock (1971) have proposed the followingth,reealternativecriteria of obtaininge 1 : (i) Maximum ratio of shear stress developed during cyclic loading to the r.ormal stress. The initial stress conditions of a specimen in simple shear device are shown in Fig. 7.2"?a; the corresponding Mohr's circle is shown in Fig. 7.23 b. Figure 7.23 c and d show respectively the stress conditions on the soil specimen during cyclic simple shear test and corresponding Mohr's circle. It can be seen from Fig. 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,

~

=

Koal,

ad

...(7.6)

2a3

//e 1 = 'th ai, = 'th' -- Ko I avK ad /2. a3 'th I oal,

now

...(7.7)

GV V1 tfI ~ L-

..... V1

KoOV

L-

a

(a.)

~ s:::. if)

-

KoOV .. " ,-

ay

(b)

QV NormaL s t re.ss

V1 V1 ~ L-

-rh

..... V1 L-

Ko CYy

a

~

(c)

Normal stress

s:::. VI

(d)

Th

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

~

\

304


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

...(7.9)

0

In Triaxial test a

Maximum shear stress

=J

0"3

...(7.11)

= .:!.L

...(7.12)

Minor principal stress = "r2

h

Therefore,

2 l + - {0" (1- K )} .

2

v

0

...(7.10)

2

20'3 [~crv(1

+ 2 Ko)]

--'th

It gives

=. .:!.L

.!.(1+2K 0 )2_.!.(1-Ko)2 4

( O"vJ

( 2 0'3 )~ 9

Cl

= ,/-(I+2Ko) 9

.

1

Hence

2

1

2

--(I-Ko) 4

...(7.13)

/(ad/2a3)2

2

...(7.14)

/(ad/la3)

(Hi)Ratio of maximum change in shear stress to the mean principal stress during consolidation.

Ja;)

"rh

. It gives

,...(7.l5)

[crv(1 +32Ko)]

a3 (1+2 Ko) 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
solidation. It gives the value of Cl as

.

-

.

.

,

.

.

2 (1+2 Ko)

(3.[3) ...(7.18) Values of Cl computed from the above equations are given in Table 7.1 Weighted average values of Cl

-.

C I are given in the lastcolumnof the table.In normallyconsolidatedsands,valueof Korangesfrom 0.3 to 0.5 which in turn gives the value of Cl varying from 0.45 to 0.55.

'. ..

, ..

1

-,--

II1II

Liquefaction

,.;"~'".;:,:'<,.""'".",,,;, .

'"

.,.,'i., ..'..', 0'".1)< : ",ih""".<.(,.'~:<.A.,I'A\:;t",-.,.j,~,"',','i1.

305

of Soils

Table 7.1 : Values of Cl

----------------------------Value orc!

using

Ko

Equation 7.7

Equation 7.14*

Equation 7.16

Equation 7.17

Equation 7.18

Average value

0.3

0.3

Negative value

0.53

0.65

0.61

0.45

0.4

0.4

0.6

0.7

0.69

0.53

0.5 0.6 0.7 0.8

0.5 0.6 0.7 0.8

Negative value 0.25 0.54 0.71 0.83

0.67 0.73 0.80 0.87

0.75 0.8 0.85 0.90

0.77 0.85 0.92 1.00

0.55 0.68 0.78 0.87

* For -

ad

= 0.4

2 0'3 In simple shear test equipment, there is always some nonuniformity of stress conditions. This causes specimens 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 the

field values were about 1.2 times the laboratory values. It can be expressed by the following relation:

--C-

-"Ch ( cr )

v field

"Ch

"

...(7.19)

~.( cr )

v simple shear

where Cz = Constant to account the nonuniformity of stress conditions in simple shear test Combining Eqs. (7.5) and (7.19), we get "C a a -1:L =C C -IL =C-L I z ( 20' ) r ( 20' ) ( cr ) v field

3 triax.

...(7.20)

3 triax.

where ...(7.21)

Cr = Cl Cz Seed and Idriss (1971) suggested the values of Cl' as given in Table 7.2. Table 7.2 : Values of Cr Relative density OR (%)

Cr

0-50

0.57

60

0.60 .

80

0.68

As evident .from Fig. 7.17, upto a relative densitY of 80%, the peak pulsating shear stress causing liquefaction increases almost linearly with the increase in relative density. Keeping this fact in view, the following general relation is suggested: "Ch

( cry) field DR .

= .5!L .C. DR triax. r 50 ( 2 0'3 ) . 50 .

'.

...(7.22)

306

Soil Dynamics & Machine Follndatiolls where ~h

( crv) field OR 5!iL ( 2 cr3J triax. 50

= Cyclic shear stress ratio in the field at relative density of DR percent = Stress ratio obtained from triaxial test at relative density of 50%. It can be determined using Fig. 7.22.

7.9 EVALUATION OF ZONE OF LIQUEFACTION IN FIELD At a depth below the ground surface, liquefaction will occur if shear stress induced by earthquake is more than the shear stress predicted by Eq. 7.22. By comparing the induced and predicted shear stresses at various 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 to maximum ground acceleration Qmax(Fig. 7.24). Assuming the soil column to behave as a rigid body, the maximum shear stress 'tmax at a de p th h is g iven by

rh - ( g ) . Qmax -

'tmax

where

...(7.23)

g = Accelerationdue to gravity r = Unit weight of soil

¿³¿¨ '.. ...

:

--

:

.~~::~.:':I

Unit cross sectional ¿®»¿

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)actis taken as (

)

-

-

'tmaxact - rd. 'tmax-

where, rd

= Stressreductionfactor

rh rd ( g ) .omax . .'

.

...(7.24)

307

;qllefact;on of Soils

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 value

lpto depth of 12 ni is shown. The critical depth for development of liquefaction is usually less than 12 m.

rd

0 0

0'2

0-4

0.6

Average

value

1'0

6

-

E ......

12

£.

Q. ~ 0

18

Range of different soil profiles

24

30 Fig. 7.25 : Reduction factor r d 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 as shown 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 'tav

rh

= 0.65 -.amax .rd

...(7.25)

g

'.

The corresponding number of significant cycles Ns for 'tavis given in Table 7.3. Tabl~ 7.3: Significant Number of CyclesNs Corresponding to 'ta\'.Earthquake magnitude, M Ns

on Richter's scale 7 7.5 t... .~ '.~

8 . -. ..

10 20 30

-

~-

~

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
th

( crv) field

.

Multiplying 'th

for the relative density DR of the soil at site.

OR

th

( crv ) field OR

with effectives stress at depth h, one can obtain the value of shear stress

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. 'tav and 'th can be plotted as shown in Fig. 7,26,

Equivalent

peak shear

stress

¬æòææòþæóææþóþôô¢ôô -' '- - '- ." . ,- -,,,;,-,,-,,",.,.-,- - .'

.

.

-c::.I

. -

~

--

,.,.. ..'. depth for.~,~.Oepthof water

Critical

-

liquefactionJd --"',.tab~~

.

-'GWT -.::-

U

0 ~

:J \11 "0 C :J 0

Zone of initial Lique faction

~ 0'1 c::.I

L

-

£.

~

0 ....

~I £.

a. c::.I

0 ,

Equivalent

shear

cyclic

stress

developed

earthquake cycLes

\

Peak cyclic shear stress needed to cause liquefaction in Ns cycles Oaboratory)

due~

in-N°s \

'

Fig. 7.26 : Zone of initiai liquefaction in field

309

fuefaCtion of Soils

1.0VIB~TI9,N l

TABLE STUDIES

vibrationta~le studies,a large specimenof saturatedsand is preparedin a tank which is placedon a

bration' table~fig. 7.27 sho:ws atypical setup of a horizontal shake table available at University of oorkee.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 to le foundation. The platforin is connected with motor and brake assembly for imparting vibrations. Some

rnportantcharacteristicsof the table are: .. . .

"

.

.Amplitude

~

0-10 mm

.Frequency

-

0-20 Hz

-

0-20 g

Acceleration

.

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 conjition. The table is then excited with the desired amplitude and acceleration. Variation of pore pressure

with 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 consolidated 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. 1. Vibratory system 2. Horizontal shaking table with a tank mounted on it. 3. Settlement measuring Device. . 4. Pore-Pressure measuring system

(a) General view

..

'Fig. 7.27 :.Horizontal

shaking table (..~Contd.)

,~, 3tO

Soil Dynamics & Machine Foundations ~"..,

1. Motor 2.Pulley 3. Eccentric wheels 4. Crank-shaft 5. Connecting rod 6. Fly wheel 7. Hand break 8. Revoluation counter

(b) Vibratory

system

Fig. 7.27 : Horizontal shaking table

Since 1957, many investigators have studied lIquefaction characteristics of sand using vibration table on different sizes of soil samples and dynamit: characteristics of load. The effect of the following aspects 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 a definite acceleration. This is termed as 'critical acceleration'. Critical acceleration is not unique property of sand. It depends on the type of sand, its density, the amplitude and frequency of oscillation and the overburden pressure (Maslov, 1957; Matsuo and Ohara. 1960: Florin and Ivanov,1961). 2. If sand is subjected to shock loading, the whole stratum liquefied at the same time, while under steady-state vibrations, the liquefaction starts from the top and proceeds downward (Florin anp Ivanov,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 wil reduce the time of the liquefied state tenfold (Fig. 7.29; Florin and Ivanov. 1961).

.'

311

le/action of Soils

;

E

70

..........

z

oX.

5.6

~

I:J VI VI ~

42

~

Frequency eo

2 Hz 0.67

thin membrane 6

I-

x

a .c

0.25 9

First liquefaction

Ia. 28 ~ 0'1

Ac c el era tron

A

First liquefaction thi c.k mem brane First liquefaction for old container

results A

\J

I:J I.f'I

1000 to first liquefaction 100

Fig. 7.28 : Effect of surface

pressure

on resistance

to Initial liquefaction

In vibration

10,000 "

table studies

(Finn, -

1972) ---

22.5

20 , .'

-

15

q

1/1

C 0

-

/-:~.~r

v

. , ..'

c 10 C>I

::I tT

q (k NI m2)

0

~

'"

C}-()

E

5

O. 2

..-. tr )f--j(

0 0

25

Number

50

20 6

50 100

75

100

of impacts

Fig. 7.29 : Influenceoftbe intensity-or dl'alnlng~w:cbarge on tbeperiod-of1lme within which the sand remain liquid (Florin"and Ivanov, i961i

-

--~

"~-,

'"

--

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 fine grained' 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 = 0.47 mm) and Solani sand (Dso = 0.15 mm). The maximum pore water pressure developed in about 6 to 10 cycles. It started dissipating immediately after attaining maximum valu~. The total time required for dissipation was about 6s for Ukai sand, and 20s for 01;>raand Tenughat sands. The corresponding value for Sohmi sand was 12Us,it"re~ained c9nstant for ~bout 35s: thu~ ,the time required for, dissipation decreases with the increa~e in coarseness. Time,s 0'2 20

0.4

0,8

1.2 '--,.,

?

- ", '

"12

-04

200

E u

Accl. 10010 9

<:I}

L::I '::

,5Hz

'.

-/",Sotani.sand,

,

';~'~

155mm depth

10

',:-

"

,",:-::-'!,

~"

ñÌ»²«¹¸¿¬ sand Obra sand

<:I}

L-

a.

Ukai sand

<:I}

L-

e ll.

0

1 Number

40 of cvcles

200

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 for displacements, so that there is practically no indication that the phenomenon of liquefaction occurs 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 density on an increase in pore pressure at 10 percent g for Solani sand (Gupta, 1979). In this case, no pore-water pressure increase was observed when initial density became 62 percent. Tests performed on other types of sand with different accelerations gave the values of relative densities as listed 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 Density

Acceleratioll

- - - - - - - - "7",...- - - - - ~ -:-- - - - - - - - - - -

(g) Percent 10 20 40 50

So/ani,sa.~ld 65Q--(O.15m'!IL 62.5 62,S 66.0 ":','.

..0',

"'66.5'

TefJughat sand ,<:~~-{O.4!),. .,' 52.0 61.5 64.0 .) 65.0 ,;".,~...;';'; . .. " -;', " . ." '

Obra sand p.Omm) 51.5 60.0 62.5 ',..-

64'.0

Ukai sand (1,8 mm) 50.5 59.5 62.0 '63.0

t

is;;

313

efaction of Soils

1-5

360

"

32

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

280

~--

'~

"

"CA...

-

"

" 240

\-

::J Ifj Ifj

~

u

ay 0-5

'. "C\

---

200

1.0

155 250

u/CTv /> 1 complete liquefaction

E E ~

60

0 0

~, "

""'0...,

Oepth(mm)

.

0 Excess pore pressure

u/ov

160

L.-

a. ~ L.-

a Cl.

120 80 40 0 20

30

40 Initial

50 relative

-

60 density

70

80

-90

(°/0)

Fig.7.31: Pore pressure vs.initialrelativedensityin Solanisand (Gupta, 1979)

DeAlba, Seed and Chan (1976) presented the results of shake table tests in the form of stress ratio .

la) and number of cycles required for causing liquefaction as shown in Fig. 7.32. It indicates that for iven 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 table IS, the value of'th is given by : . . ... W '

o'tho=-.am g where W = Tot~l p~essure exert~d OJ?the bas~ of t~nk p~ac_edon t!1eshake table 00am = P~ak ac~eleration of die uniform cyclic mQtion

...(7.26)

314


a

+-

0.4

0 L.

OR (0/0)

11'1 111

~

~

~

¢ e--e.

0.3 '

êè 54

,

'U

.r::.

~u P 0.2 ~ L. L. a

u

0.1

1

10

100

Num be r 0 fey Fig. 7.32: Corrected

'Ch/cry versus Ns for initial

1000

cl e S J Ns from shake table studies (DeAlba, Seed and Chad, 1976)

liquefaction

7.11 FIELD BLAST STUDIES In blast tests, predetermined charge (like ammonia, gelatin etc.) with electric detonators is installed at predetermined 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 desired moment. Acceleration pickups are placed at regular intervals from the blast point to record horizontal and vertical acceleration at the time of blasting. Similarly porewater pickups and settlement gauges are placed at certain distances from the blast point to record the increase in pore water pressure and ground settlements, Accelerations, porewater pressure and ground settlement at the blast point are then obtained by 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 will liquefy under simulated earthquake loading. Field data using small explosives at some depth at the site alongwith pore pressure and settlement observations for predicting liquefaction potential are available ir literature from few investigations (Florin and Ivanov, 1961; Kummeneja and Eide, 1961; Krishna am Prakash, 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 0-4 m 4m-7m

7 m - 20 m

.

Description of Soil

Fine sand Clay Silty sand mixed, with kankars

Average N-va/ue '5

.",

-

21-60 (increases. with depth)

Remarks ., ... ' . . - Cu ='2.47, °50 = 0.12 mm Position of water table near the surfal 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 wi the help of an electric exploder. Horizontal surface accelerations were measured using acceleration pie ups placed at various distances fro,mthe source point. The depth of each acceleration pickup was 200 IT below the ground surface. The porewater pressures at vari~us .dls~ancesfrom sourc~ point were measur at 2.5m depth from the ground surface.

ù

iIIIb.,

Liquefaction

of Soils

315

B7 I

10m

I

86

r

1am

I

8S

812

I

10m 10m

I

I

84

B:f , .

lOm

Srn

~0

Srn

¢º»®»²½» point

B13

.814 .

B15 816

. -

Fig. 7.33: Site layout for field blasting tests (Gupta and Mukerjee. 1979)

,

Fig. 7.34: Surface ac:c:eleration.(Guptaand 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 accelerajon and porewater pressure' with dis;tan~~'from blast point are shown respectively in Figs. 7.35 and 7.36.

, 316


On the basis of past earthquakes, the maximum possible acceleration record at the site is assumed as shown in Fig. 7.37. Using the method of Lee and Chan (1972) this earthquake record is worked out to be equivalent to 19.6 cycles of 0.075 g acceleration (Table 7.5) 24

0'1 0

;-c 0

\

20

\ \

16

0

12

(

"

;::.

0

...

-

C>I

C>I

8

.3C

4

u u 0

0

(

"

( i'-. " '. ""'(

"-- .-.....

C>I

-'--

a.

0 0

10

20 30 40 Distance tor blast

50 (m)

60

70

80

Fig. 7.35 : Acceleration versus distance (Gupta and Mukerjee, 1979)

).0

Depth of blast Pore pressure at 2.5 m

2.5 E

4m measured =

C>I

~

2.0

11\

11-74

0/1 C>I

:i

1.5

C>I

... 0

a. u E 0c

>-

1.0

0-5

0

0 0

I I 112'9 I 2'5 5,0 'Distance

1 7'5 trom

I 10.0 blast

I 12.5

15'0

17-5

(m)

. Fig. 7.36: Port: jJressure vs. distance (Gupta and Mukerjee, 1979)

",

20-0

-,"",~~",

iquefactionof

317

~oils

','""

. .. N

<-

0 .--

co

0-r-0--

.; ....

., . --If)

--

... .::0: = :::. "C c <": <":

C.

=

E -

.:

.

.2 ...:r

c

E ...

-

... '" '"

I

0"

.. r--

'-:

r-oD N

.;

0 /'

-

0

0 ,."

0

0

(3

0

0

c-

co 0 . 0

,"

0

0 .,,-"

:,;.

'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.'-,

t.' "

318

Soil Dynamics & Machine FoundatWd

Table 7.5 ~EquivaleDt Cy~les for Anticipated Earthquake

-

Acc!' level in percent -of-peak - acc!. - -

Average acc!. in percent

Number of cycles

- - - - - - - - - - - - - - -

(1) 100-80 80-60 60-40

(2) 90 70 50

(3) 17/2 = 8.5 8/2 = 4.0 25/2 = 12.5

40-100

20

> 1000

Conversion factor -(Fig. - -1.6)-

- -

Equivalent number of . _"J.iL -cycles - -at -0.651:

(4) 2.6 1.2 0.20

(5) 22.1 4.8 2.5

negligible Total

0.0 29.4

Total number of cycles for 0.75 1:max= 29.4/1.5 = 19.6

22 20

-~19'6

i\ 1\ ,.......

.

I

0- 16

~

z 0"1

~

r-0 0

-u

12

\

0

111

~

>u

0

8

... c

~

0 > :J er UJ

4

I

I I I

0

J

0

0

.

2.9

.

10

Distance

20 from bLast

30 (m)

Fig. 7.38: Equivalent cycles versus distance (Gupta and Mukerjee, 1979)

40

.

'319

Liquefaction of Soils

Similarly the blast records at different distances are also converted into equivalent number of cycles of 0.075 g acceleration (Fig. 7.38). From this figure it can be observed that vibrations generated due to blast at a distance of2.9 m are equivalent to 19.6 cycles of 0.075 g, the expected earthquake i.e. the blast has 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 is 1.74 m of water column. The critical hydraulic gradient for this site is 0.8, therefore at a depth of 2.5 m the critical porewater pressure or hydraulic head is 0.8 x 2.5 = 2.0 m. The pore pressure developed is 1.74 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 conditions, a larger pore pressure is expected to be developed and complete liquefaction of site is expected during the earthquake. 7.12 EVALUATION OF LIQUEFACTION POTENTIAL USING STANDARD PENETRATION RESIST ANCE The standard penetra~ion test is most commonly used insitu test in a bore hole to have fairly ~ood estimation of relative density of cohesionless soil. Since liquefaction primarily depends on the initial relative density of saturated sand, many researchers have made attempt to develop correlations in liquefaction potential and standard... penetration resistance. IS: 2131-1981 gives the standard procedure for carrying out standard penetration test. SPT values (N) obtained in the field for sand have to be corrected for accounting the effect of ovberburden pressure as below: NI = CN . N

0.4 0

t>I ~

:J

'.

III III t>I ~

/ 200

~

:J .D

L t>I

300

>

t>I

-U t>I

(Fig. 7.39)

tU

The correlation between ~ 1 values and relative density of granular soils suggested by Terzaghi and Peck (1967) is given in Table 7.6.

500

",

...

.

1/

/ / I

.

Fig. 7.39: Chart for correction ofN-values in sand for inOuenceof overburden pressure(pecketal., 1974)

~.

/

I I

"

t>I

~ t>I

= Correction factor

2-0

/

oX

-- 100I

cc.

1.8

/.

E

> 400

CN

--

2

0 v

NI = Corrected value of standard penetration resistance

Corr~ction factor, CN 0.8 la 1.2 1-4 1-6

'"' N

-~ -

...(7.27)

0.6

J.

320

Soil Dynamics & Machine Foundatio" Table 7.6 : NI and,

related to Relative Density

CompaCtness

NI

Relative Density

q, (Deg.)

DR (%) 0-4

Very loose Loose

0-15

< 28

4-10

15-35

28-30

10-30

Medium

35-65

30-36

30-50

Dense

65-85

36-41

Very dense

> 85

> 41

> 50

After the occurence of Niigata earthquake, Kishida (1966), kuizumi (1966), and Ohasaki (1966 studied the areas in Niigata where liquefaction had not occured and developed criteria for differentiatint between 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 usefu rule of thumb that says liquefaction is not a problem if the blow count from a standard penetration tes exceeds twice the depth in meters. -

0

N

-z

I I

I

Light damage

& no liquefaction

I

E

.........

~

50

~

L~ VI \11 ~

L0.

c

100

(:,) 1J L:J .D L-

Heavy damage. and liquefaction

(:,)

> 0

(:,)

> ...

150

---

u

-~

Boundar y determined

---Boundary

-

UJ

by damage

survey (Kishida)

determined by field observation(Kuizumi)

Ohasaki

20 0

10

20

30

40

S 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)

-'.~:

Liquefaction

321

of Soils

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 the following procedure for liquefaction analysis: (i) Establish the design earthquake, and obtain the peak ground acceleration Qmax'Also obtain number of 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 surface. 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 can

obtain the value of shear stress N

z 0

KI~ ~

'-

C c::v

\J

0

..:.::

Q.

Cl

.-c

Cl.

0"1

where no liquefaction 0 cc urred.

-

~

.9

\J

c::v

0 5 Open points indicate sites

0

->--\J

.

L..

L..

8.-

Solid poi nts indicate sites and test conditions showing liquefaction

..x

:J 0 1.1\1.1\ 11

0 -4 86'5

-2 0.3

.6.5

C 1.1\ 'L.. 1.1\ Cl :J c::v Cl ..c. \J

for causing liquefaction.

0-6

E

.........

~

"Ch required

07.8

III

-- .-.~ .- .--

113 ~ O. 2

07.8

r== 0 ..c.

.

Cl

Extrapolated from results 11 of.large scale laboratory tests

--0

III

0

1.1\

0

c::v 0 ....

-L.. III

0

->-Cl -

.~ .-0 v U

L..

Based on field data

;}

L..

0 0

10

20

Modified1.i:)~netration

30 resistance>

1,0 Nl-blows/ft.

Fig. 7.41 : Correlation between field liquefaction behaviour or'sands for level ground conditions and penetration resistance (Seed, 1979)

iI...

,I;

50

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 FL which is defmed as R F =...(7.28) L L R is the ratio of insitu cyclic strength of soil and effective overburden pressure. It depends on relative density, effective overburden pressure and mean particle size. It is given by For 0.02 < Dso < 0.6 mm

~

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 = Effectiveoverburdenpressure at the depth underconsiderationfor liquefactionexamination in kN/m2 D50 = Mean grain size in mm L = is the ratio of dynamic load induced by seismic motion and effective overburden pressure. It is given by a max crv " 'd L -- -.=:-. ...(7.30. g

amax

crv

= Peak ground acceleration due to earthquake = 0.184 x 10°.320 M (Dfo.s.

g

where 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 ground g = Acceleration due to gravity, mIs2 rd = 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.

...(7.31

Liquefaction

323,

of Soils

9

Nu mbers represen t the earthquake numbers

8

30

.

~

.g

-

7

:J C

0'1 0

Lower bound

6

~

l0910 0 =0.77M - 3'6 Mean

5

(M>6)

line

L09100

=

0-87 M-4-S 'r

4

1

2

5

Maximum

10

20

epicentral

lOO"

50 distance

0

200

ot liquefied

500

1000

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 on liquiefaction, they are summarised below: '

7.13.1. Soil Type. Liquefaction .occursin cahesionless soils as they lose their strength completely under vibration 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 strength substantially under vibration. 7.13.2. Grain Size and Its Distribution. Fine and uniform sands are more prone to liquefaction than coarser ones. Since the permeability of coarse sand is greater than fine sand, the pore pressure develaped during vibrations can dissipate faster. 7.13.3. Initial Relative Density. It is .one .ofthe mast important factars contralling liquefactian. Bath pare pressures and settlement are canstderably reduced during vibratians with increase in initial relative density and hence chances of liquefaction and excessive settlement reduce with increased relative density. '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. Frequency .of the dynamic laad plays vital rale ,.only if it is clase ta the natural frequency .of the system. Further the liquefactian .,,', depends. .onthe . ..11 typeI" of the .dynamic""',laad i.e. whether'it isa transient laad" .orthe. ,

laad causing steady vibratians.'.'" , ' '0.' , ,. , . '" ~L'>+::' '

..

r,

.

.:

'

:

324

Soil Dy"amics

'

& Macll ille Folllldatiolls

Whole stratum gets liquefied at the same t~eunder tra~sient loading, while it inayproceed from top 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, horizontal vibratiorls have more severe em~-c.tthan vertical vibrations. Multi directional shaking is more severe than one directional loading (Seed ~t al.~ 1977), as the pore water pressure build up is much faster and the stress 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 grained soil. However, if a pervious deposit has large dimensions, the drainage path increases and the deposit may behave as undrained, thereby, increasing the chances of liquefaction of such a d~posit.The drainage path 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 of stress from soil grains to pore water will require higher intensity vibrations or vibrations for a longer duration. If the initial stress condition is not isotropic as in field, then stress condition causing liquefaction depends upon Ko (coefficient of earth pressure at rest) and for Ko > 5, the stress condition required to cause liquefaction increases by at least 50%.

7.13.7. Method of Soil Formation. Sands tmlike clays do not exhibit a characteristics structure. But recent investigations show that liquefaction characteristics of saturated sands under cyclic loading are significantly 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 characteristics. A 75% increase in liquefaction resistance has been reported on liquefaction of an undisturbed sand compared to its freshly prepared sample which may be due to some form of cementation or welding at contact 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 of similar deposit previously subjected to some strain history reveal, that although the prior strain history caused no significant change in the density of the sand, it increased the stress that causes liquefaction by a 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 dissipated 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.ugh it depends on -a number of factors, however, few can be controlled in field. Based on these, certain methqds 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 prone to 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 for compaction

loose' sands . of -

in'situ

are: .

.

' '

. .' .h .

-

,

'

7.14.1.1. Rolling with rubber tyrerollers: It may be accompli~hedby excavati~g some depth, then carefully backfilling in controlled lift thickness and compacting the soil. When ruJ>bertyres are used, lifts are commonly 150 mm to 200 mm. This method, however cannot be used for compacting deep sand deposits.

325

;qllefactioll of Soils

,14.1.2. Compaction with vibratory plates and vibratory rollers: Compacti'on of cohesionless soils can c 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 area overed 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 sands L11d gravel 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 half s stationary part. The device has water jets at top and bottom. Vibrofloat is lowered under its own weight vith 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 thus leading 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 1Om thick are to be compacted, two or more tiers of small charges are preferred. (iii) For deposits less than 10m thick, charges placed at 2/3rd depth from surface will generally suffice. (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 simultaneously. (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 limit space around the injection tube. The agent reacts with the soil and/or itself to form a stable mass. The most common grout is a mixture of cement and water, with or without sand. Generally grout can be used if 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.

......

"","""""""

"'"

Soil Dynamics & MJlcmne Ftnmdations

326

7.14.3. Application of Surcharge. Application of surcharge over the deposit liable to liquefy can also be used as an effective measure against liquefaction. Figure 7.43 shows a plot between rise in pore pressure and effective over burden pressure at an acceleration of ten percent of g. It indicates that pore pressure increases with increase in overburden pressure till a maximum value of pore pressure is reached, after which 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.

24 Dead weight surcharge 250 mm depth _Acct. 10% gNo. of cyc les 10

Solani sard

"E z-"

-

201-

DR=20 15Hz

%

16

Cl! L-

:J

~ I Zone ot liquefaction ~ 12 .

L-

a. Cl! L-

0

a. III III -.CI!

v

8

-

-

4

)(

UJ

0 0

4

8

12 Effective

16

20-

overburden

24

28

pressure

32 (kN/m2

36

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 with higher permeability reduces the length of drainage path and also due to higher coefficient of permeability, speed up the drainage process (Katsumi et a1. 1988 and Susumu et a1. 1988). 7.15 STUDIES ON USE OF GRAVEL DRAINS Yoshimi and Kuwabara (1973) were first to introduce gravel drains to stabilize a potentially liquefiable sand 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' is about 200 times the permeability, ks' of the soil in which they are installed i.e. (kd/ks) > 200. The effectiY,e drainage 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 . '1 terms shown on this figure." are '.as below: ., . .' .< . ill( ..

, ,

.. ~ .~

'--~q~~e!aftioll;

°t'~~Lf

C,',

\, "

'.,

.

327 f:~;,

.

. '-",'"

',~

,".',

c,:

, "

e'

,,-'. .e'

..

.

'"

'-"",'

"

(0 ) Plan , :.

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

- ':

-

: . .' - t ,~

~ "'.:

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

,-,..'

/'~:'"

1

i

,

':::'{. ".

"',

"

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

",'

~ "::~~LG_WT ~

,..-.

'r"

'

. ':,,',

"':, '..',' ~~~I

',.-',-

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

,

c

'

, ..

.

"

'..

J, -' '" , '.' .' , "r

Re' ,

.

,

, :,', t" " "" 'N

. ..

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

; " ~

,..

".':

.:2"::::.,

(b)

.'~';;':!"":~',~~

,:;.,,-y,~~:,-J'.:-)

:: 'G

Rd~

;'"",","'{',,"-,"

$" ""',

" .;

~

,".

.,.,

,~'

;')

Sec tior.' ot, S-S ..

, ,;"::,.

J-

:,/

.'

¥' \ ,,-,

: i."".', .

Re

\~\

t' '"It> -".' ,fi.g,.7.4,4 :,.:Gravel drains ,-.".",..:',.,'"""c.<,"..,-,,,

-";;'"""",,"'-!"~'~;,,,,~;'\"""

I:',

,~"'-::-

328

SoU Dynamics & Madine

Founda~s

<-

0-6 rg

0.4 0 .21-1f:J°

Tad-:' '. '

0 0

0.1

0'2

0'3 Rd

0-4

O'S

0.4

0-5

0-6

/ Re

(a) iNl = 1 1.0

rg

0-4

0-2

0 0

0.1

0.2

0-3 R.d / Re (b) NiNl

-

0-6

-

=2

Fig. 7.45: Relation between gr~3test perewater pressure ratio and drain system parameters (...Contd.)

-:.

I!

.

I

o~

lefaaion- of Soils

~19

,.0 0-8 0 -6 t-

- 200

110

Tad-

rg

0.4. 0.2

0 0

1.0

3-0

2.0

~.O

5.0

6.0

/ Re

Rd

(c) N/Nt = 3

1.0

0.8 0,6 rg

0.4 0.2 "

0

0

.:.0-1

0'2

0.30,4

Rd

.' . 0,5

0.6

/ Re

'J:

(d) .NslNl = 4 Fig. 7.45: Relation between greatest porewater pressure ratio and drain system parameters

,

. .'

'" I " .' c' .:.

330

Soil Dynamics

for design r = Limiting value of,- Ugchosen , .. g

cry

.'

-~

-

& Machine

FiJII;idat;o;u

-<'~

-'~'-

Uu = Excess porewater pressure build up in ~.cyclic 'simple sh~ar test (Fig. 7.46) , 0

= Initial consolidation pressure Ns = Number of cyclic stress applications cry

.

NI = Number of stres~cycles needed for liquefaqiion Rd

= Radius of rock or gravel drains,

:

'

iUSof the rock or gravel drains Re

T

-

h

,d

ad - '¥ro 'my ( Kh

1d

'

'

R~

J-

= Coefficient of permeability of sand in horizontal direction'

'¥ro= Unit ~eight mv

-

= ~fec tiV; rad

of water'

,'-'

= Coefficient,of volume compre~~ibili~_.~fsand,

= Duration

of earthquake"

,

1.0 0.8 0.6

-Ug

ay 0.4 0.2 0 0

0.2

,

0.4

0~6

N/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 partia drained 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 un perfectly undrained and perfectly drained conditions. In case of perfectly uildrained condition, liquet tion resistance was neither influenced by permeability of sample nor by the frequency of loading. Thi

reasonable, 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 drainage frequency is remarkable for soils with relatively largef1relative density, OR' and not so significant soils 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 depl on the rate of dissipation of pore pressure, which is based on draInage.

'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 of ' pore water pressure, ,. " characteristics , "',-'

'".-

',-,

"

,

,0"

",

"

(ii) to clarify the effective are~foithe g~aveldra~fromthe view-p~int of 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 12 m 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 much maller 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 was lssumed 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
'

,

.

As the number of d!'ainsin~talled is increased, the non liquefied zone increases. As the acceleration increases, the zone reduces gradually but the increase in time does not reduce the non liquefied zone. The angle of trapezoid was found to be 15° to 17° in the 'direction of depth. The zone is about 40 mm outsid,e the drains. It was also observed by him that the surface drains may effectively prevent foundation settlement. In order to obtain good effect in reducing foundation settlement it must be ensured that adequate dept.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 liquefaction. 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 pressure at points near the dr~iris a~d away from the 'drai~s.They concluded that liquefaction occured at points too far from the drain and that at points close to the gravel drain, liquefaction did not occur. Results obtained by 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 with sides equal to the line joining mid p~ints of the ~pacing between adjacent gravel piles. The sides of the squares are tak~n as optimum spacing between gravel piles. They found that the effective area of gravel pile increases in proportion to the diameter of the gravel pile and the permeability. For a fixed diameter of pile and permeability of'soil, as the optimum distance decreases, the pore pressure ratio decreases. As the permeability increases, pore pressure ratio decrea~es very sharply. Highly permeable gravel are much more 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 several projects. The 'most important properties of.such drains are permeability and tensile strength, The jute filter cover h.~spermeability better than 10-5m/sec. This facilitates the flow of water from pervious lenses present in the seams and layer of sand and speeds up the pore water pressure dissipation. They have the advantage of decaying and getting mixed with the soil without harming the environment. When the filter permeability 'is large, the clogging of the drain has to' be considered. - - -

,

,

332


C ire le of et f e cti ve a re a

..

-'-EL:~

i"

l'77'r~'.

~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, if necessary.

I

ILLUSTRATIVE EXAMPLES

I

Example 7.1 At a given site, a boring supplemented with standard penetration tests was done upto 15.0 m depth. Tht results of the boring are as given below: Depth

Classification

(m)

of soils

1.5 3.0 4,5 6.0

SP SP SM SM

0.18 0.20 0.12 0.14

7.5 9.0 10.5 12.0 - 13.0 15.0

SM SP SW

0.13 0.16 0.20

' SW

SW SW

N-Value

D50 (mm)

3 5 6 9

19 30 35 40

12 17 20

45 52 52

0.22

18

46

0.22 0.24

24 30

60 65

'.

Remarks

DR (%)

(i) Position of watt table lies 1.5 m below the grour surface

(ii)

'YmOiSI=

19 kN/r

'Ysub = 10 kN/m

.'

333..

iquefaction' of Soils

The site is located in seismic ally ,active region, and is likely to be subjected by 'an earthquake ,of

1agnitude 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 amax = 0.984 x 10(0.302x 1.5) x 10~.8 = 0.083 g 9.81 Number of significant cycles (Table 7.3), Ns = 20 (For M = 7.5)

From Eq. 7.31,

amax

(ii) From Eq. (7.25)~

'tav

.

= 0.65yh. - g . rd

= 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 read from 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 values of,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. . DR

ad 'th=

'.

-'

( 2a 3 ) 50%"50.Cr.av

Values of Cr are .obtained using Table 7.2. The details of computations of 'rh are summarised in

Table 7.8.

.

Table 7.7 : Computation of 'ray S.No. (m)

Depth (kN/m2)

Total stress

(I)

(2)

(3)

1. 2. 3. 4. 5. 6. 7. 8. 9. 1p.

1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 .15.0

28.50 58.50 88.50 118.50 148.50 178.50 208.50 238.50 268.50 298.50

rd

---------------------------------

r

tal' (kN/m2)

(4)

(5)

0.99 0.98 0.97 0.97 0.96 0.95 0.94 0.92' 0.91 0.90

1.52 3.10 4.64 6.21 7.69 9,14 10.54 11.90 13.23 14.51

3:J~;, Id

Soil Dynamics

.),'

& Machine

Foundatiolls

:"

';1'able S.

No

7.8 Computations'

Depth (m)

of 'thfrom

Effective stress. cr

Seed and Idriss

DR

,

(1971) Metbod"

Cr

.rh

, .rh/a!,

(kN/m2)

;,

(kN/nO

I.

1.5

28.50

19

0.55

0.2198

1.31

2.

3.0

43.50

30

0.55

0.2260

3.24.

3,

4.5

58.50

35

,0.55,

0.2012 . '

4-.53,

4.

6.0

73.50

40

0.55

0.2074

6..71

5.

7.5

88.50

45

0.555

0.2043

9.03

6.

9.0

103.50

52

0.573

0.2136

13.17

7.

10.5

118.50

52

0.573

0.2260

15.96

8.

12.0

133.50

46

0.556

0.2298

15.69

9,

13.5.

148.50

60

0.60

0.2298

24.57

1O,

15.0

163.50

65

0.61

0.3336

30.29

.'

(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
Depth

N- Va/ue

of'th by Seed (1979) Method

Corrected

,

Effective

.rh/av

.rh

(6)

(7)

óóóóóóóóóóóóóóóóùóóóóóóóóóóóóóóóóóóóó (m)

(1) 1.

>

",

N

stress (5)

(2)

(3)

(4)

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

4.62

4.

6.0

9

10

73.50

0.113

8.30

5.

7.5

12

13

88.50

0.146

12.92

6:-

9.0

17

17

103.50

0.191

19.77

7..

10.5

20

' 18

118.50

0.194

22.99

8.

12,0

18

17

133.50

0.183

24.43

9.

13.5

24

" 20

148.50

0.225

31.93

163.50

0.269

43.98

. ."

I

10.

15.0

30

25 .

"

l'UeTattwn

ofSolls:

I Iw-asaki's

,'"

,<""

Method

.--'( i) Fir'stIy the'value of factor R is obtained using the following Eel.7.29a ; .

,

.

, ,,'

..

'

0.35

.,

R .= 0.882~

V~:+ 70,

..

+0.225 loglO

(

J

°50

The details of computations of factor R are given in Table 7.10. "

Table 7.10 :' DetaH~ of Computations

S,No,

Depth

ay

(I)

'

(m) ,(2)

(3)

(4)

28.50 43.50 58.50 73.50 88.50 103.50 118.50 133.50 148.50 163.50,'

0.18 0.20 ' 0.12 ,0.14 0.13 .0.16 0,20 0.22 0.22 0.24

4.5 6.0 7.5 9.0 10.5 12.0 13.5 15:0

5. 6. 7.',' 8. 9. "1'0. -,':,. ".

R

N

'D5o

2 (kN/m )

1.5 3.0

4,

Method

'- - - - - - - - - - -(mm) - - - - - - - - - - - - - -

- - - - - -- 1. 2. 3.

for obtaining Factor R in Iwasaki's

(ii) The factor L is then obtained ,

'

.'

using Eq. 7.30 :

L = .

a

(5)

"

(6)

3 5 6 ,9 12 17 20 18 24

0.2189 0.2398 0.2950 0.3104 0.3394 0.3525 0.3419 0.3077 0.3377 0:3464

" 30

/.

a

max '~'r-

d

gall

The details of computations are given in Table 7.11. The ratio of factor of safety RlL is listed in the astcolumnofTable7.11.

",'

,

'

Table 7.11 : 'Details :of Computations S.No.,

-

., ,Depth

(m). ,

--------------------------------.

'.,

.,"

Liquefaction

Potential

rei

L

FL

'. (3),

(4)

(5)

(6)

,

L. 2. 3. 4.

.

5,

. 6: 7,

, 8. ., 9. 10,,- , ::" i'C,:

,).5., 3.0 4.5 6.0 7.5 . ; 9.0

10.5 12.0., 13.5 15.0. ,'L

,

by Iwasaki's J\1ethod

. crJcrv'"

(2)

(I)..

of Obtaining

0.98 0.97 0.97 0.96 0.95 0.94 0.92

i,758 :,,-,'~."'L787 1.808 ; ...;,;',1.8,26 . ;:..\:; ."

,

0.0820 0.1092 0.1215

0,9,9

1.000 1.345 1.513 '!' .' .1.612 . ~, '1.678 ' '1.725

-

...

,

2.670 2.196 2.427 2.397 2.544 2.597 2.498

',0.1295 0.1334 0.1357 "

. . , 0.1369

0.91

0.1362 0.1363

0.90

, 0.136,1

,

2.260 :

2.4 78 2.545

336

Soil Dynamics & Ml1ChineFoullda

.

In Table 7.12, summary of different methods are given. It is evident from this table that liquefaction ,tl occurs 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. Tabl-e 7.12 : Summary of Different Methods S.No,

Depth

(m)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

t'h

th

Seed and Idriss

Seed (1979)

method (1971) (kNlm2)

(kN/m2)

t'av

(kN/m2)

1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 15.0

FL (R/L)

1.28 2.91 4.62 8.30 12.92 19.77 22.99 24.43 31.93 43.98

1.31 3.24 4.53 6.71 9.03 13.17 15.96 15.69 24.57 30.29

1.52 3.10 4.64 6.21 7.69 9.14 10.54 11.90 13.23 14.51

/wasaki method

2.670 2.196 2.427 2.397 2.544 2.597 2.498 2.260 2.4 78 2.545 'I

ÎÛÚÛÎÛÒÝÛÍ Arya, A. S., ~andakumaran,

p" Puri, V. K. and S. Mukerjee, (1978), "Verification of liquefaction potential by field

blast 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 Engineering Division, ASCE, 10L(GTS), pp. 551-569. Castro, G. and S. 1. Poulos (1976), "Factors affecting liquefaction and cyclic mobility ", Symposium on Soil liquefaction, 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 fort peck 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 liquefaction of sands", 1. Soil Mech. Found. Div., ASCE, Vat. 96, No: SM 6, pp. 1971':J934.

.

.

-'

337


Finn, W. D. L, J. J. Emery and Y. P. Gupta (1970), "A shaking table study of the liquefaction of saturated sands during 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 on Insitu 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 Conference 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", ECEE 87, pp. 5.3/41-48. Peacock, W. H. and H. B. Seed (1968), "Sand liquefaction under cyclic loading simple shear conditions", 1. Soil Mech. 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 Indian Soc. 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 ofthl Civil 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, No SM 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 tlearthquake of Feb, 7, 1971", Journal of the Geotechnical Engineering Division, Proceedings, ASCI VoL 101,No.GT7. .

Seed; H. B. and Booker, 1.R. (1976), "Stabilisation of potentially liquefiable sand deposits using Gravel dra system", Report No. EERC 76-10, Earthquake Engineering Research Centre, University of Califc nia, 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 f Paper, Symposiumon Soil Liquefaction,ASCENationalConvention,Philadelphia,pp. 1-104. Seed, H. B. (1979), "Soil liquefaction and cyclic mobility evaluation for level ground during earthquakes", 1.Geote Engin. 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. D ASCE, 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.Geolt Engin. 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 Gr drains", 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 'r Wang, 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

339


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

7.2 listand,

"

neat

~ketches.

'

'

,

"...

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 table tests, 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 the boring are as given below': " . '. ~ . : ,

,

Depth

Classification

(m)

of soils

(mm)

;'.2,0

D50

'

N-Value

'DR"

SM ,

0.20

SM SP SP SP

0.18 0.16 0.18 0.19

4 5 7 9

30 33, 40 43

12.0

SP

0.19

I0

~' 45

SP'

0.19

16,0

,

SW ,

0.30

,

Remarks"

, ,(%)

4.0 ~.O 8.0 10.0

14.0,

'

. ,,4

','

, ,"I,

25

12 "

52

60

14

(i)

SW

0.32

16

20.0

SM '.

0.20

18'

of water

(ii) 'YmOiSI = 20 kN/m)

56 .'

.f

tabie lies 2.0 m below the ground surface"

::'

18.0

Position

;,

'Ysub= 10 k~/rn

,~3

,

.

3

,

.~

This ~jte is 10catedTD.s~{smically,actIve region, ,and the .likely to be subject~4 by an ~aitliquak~,of magnitude liquefaction ,by the three methods mentioned in Prob. 7.4. ;, ' 7.5. Determine',:ih~"zoneof , ,"',. " . ' " ,.. <':1 ':'.. t'

, ;

,, -.; , '.

:.

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

DLJ u

GENERAL PRINCIPLES OF MACHINE FOUNDATION 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 de sign. 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 th static weight of the machine and foundation. But the dynamic load acts repetitively on the foundatior soil system over long periods of time. Therefore, it is necessary that the soil behaviour be"elastic undt

the vibration levels producedby the machine, otherwisedeformationwill increasewith each cycle ( loading and excessive settlement may occur. The most important parameters for the design of a machir foundation 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 cylinder connecting rod, a piston rod and a crank. The crank rotates with a constant angular velocity. Figure shows 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 r Reciprocating 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 iD machine, their size, piston displacement and the direction of mounting. The mechanism developiDf of balance inertia forces for a single crank is ~hown in 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~

341

Genel-lll P,inciples. 0/ Machine Foundation Design

,,'

,

"

.

, '..

Uppu slide

-

b ao-ck

log teed

,E In C 0 lit C

l

Saw 'Modes

., E

lower

said~ block

" -

... '

,'r',

0

c.onn~ctin9

u

rod

a. >- c.ounter ~ weight

. .,. . .. .'"

'

Fly wheel

Foundation block

~

. "'.

.

- " . . . '. .. .

Fig. 8.1 : Outline of a typical Gang-saw machiDe

Piston

mp," "'.,

'"

,-, ,

0,

..-

'..

"""

"

'-v COUnt,,-...~ttt

->

z Fig. 8.2 : Crank mechanism

L

1RUP...&4i.

GeneJ!.al,P,inciples- of Machine

Foundation

341

Design

,,' "

UppuSl~,de., blO'ck

-

Logteed

,E In

c: 0 ,

'

l.

Saw 'blad~s

lit

c: ., E

Lower

slide block

'0

- -'-0'1' u

c.onnecting

Q. >- c.ounter ~ w~ight

-.

. .'" .

':: :

rod

Fly whul

foundation block

..-

~

. "'.

-

- '" ,..',".

Fig, 8.1 : Outline of a typical Gang-saw machiDe

Piston

mp," "'.,

".

,', "

0,

..

"

'-'J' ..

"-v Count~..

,~'

z

Fig, 8,2 : Crank mechanism .".'>i ~,"C L

.,

'7~t

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 two equivalent masses; one rotating with the crank pin A, the other translating with the wrist pin B. The inertia forces can then b~ e~pre~sed in.t(:OI1sof thtHotal rotating 'mass (mrot)and the total reciprocating mass (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

==

mrec

+~

LI

mp

...(8.2)

L mr

I

' C .'

F: (m rot:+: mrec ),,~1~2c:,os "'t,

Jt

...(8.1)

,ri'

'

r. 2CJ:) 2

F : mre c: 1

L

cos 2 u.J t

F y : 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

343

Design

The inertia force (Fz) in the z direction may be shown to be F~

2

2 '~ mrec) r} 0> cos (0 t + mrec

L

= (mrot +

,

'

2 (0 cos 2 0> t,

'

" ..,.(S.3)

lchhas a primary component (F1 acting at the frequency of rotation, and a secondary component (F") [ng at twice the rotation frequency. Fz = Ft + F" ...(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 '

,,',',

'

ectlon becomes

"

'

-, '1

-,

.'

,

,-

F; = mree r 1 ,.,2 VJ

(

COS Cl) t

+ rl cos 20) t L'

..'

-

"

',:

-." '

"

0'

"

'-,"

"'

)

..,

-

"

( S, 6)

The amplitude of the primary (F:nax)and secondary (F;ax) inertia forces are then relat~da~J.ollQWs' F"

max

r, = ~ Ft L max

.

'; ..:(8,7f

The preceding development relates to a single cylinder machine, which possesses unba~ancedpriary and secondary forces. As more cylinders are added the unbalanced forces and couples are modified shown 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

In-tincz

~

cytindczr

Qpposczd

lM

cytindczr

(b)

(a)

y

(c)

~ (d)

1~

h1u(I)

(e)

+~

0

Cranks at 180

C ran ks at 90

A

0

(g)

J:m,1;t (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) Six cylinders -', .".. ...:\ , .! 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 types of 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 stamping machines which produce impact loads. Forge hamIners are divided into two groups: drop hammers for die 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

'345

Design

:able' 8.1 : Unbalanced Forc:esalid Couples for Different CFank Arrangements (Newcomb, 1951) Forces --------------------------,

Crack arrangements

Couples

'

(Fig. 8.4)

Primary

Secondary

l. --Single 'crank

..F' without counter wts. (0.5) F' with counter wts. 7. Two cranks at 1800 0

2F"

Opposed cylinders

0

0

... Two cranks at 900 (1.41) F' without counter wts.

0

(0.707) F' with counter wts.

, :-Secondary

0

:0

F' 0 without cou~terwts.' "

0

{ (0.5) F' 0 with counterwts.

:0 0

0

(1.41) F' 0 without counter wts. E" 0 (0.707) F' 0 with counter wts.

F' without counter wts.

1. Two cylinders on

,Primary

Fit

In-line cylinders

.

1.41 F"

0

0

0

0

0 with counter wts.

, one crank, cylinders at 90°

e. Two cylinders on

2 F' without counter wts.

-, . 0

F' with counter wts.

one crank, ,'opposed cylinders

r

0

Three cranks at 1200

0

(3.46) F' 0 without counterwts. (3.46) F 0 11

(1.73) F' 0 without counterwts.

g. Four Cylinders Cranks at 1800

,-

Cranks at 90°

0

0

0

0

0

0

(1.41) F'O without counterwts.

'(4.0) F" 0

(0.707) F' 0 with counterwts.

0

h. Six cylinders

0

0

0

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 different from those of reciprocating machines.

8.2.3. Rotary -Machines. These include high speed machines such as turbogenerators, turbines, and rotary compressors which operate at frequencies of the order of 3000 rpm to 10000 rpm. Associated with these machines there maybe a consider~ble amount of auxiliary equipment such as condensers, coolers and pumps witl1connectingpipework and ducting. To accomodate theseauxiljary 'equipments a common foundation arrangement is a two storey frame structure with the turbine located on the upper slab and 'the auxiliary equipment placed beneath, the upper slab being flush with the floot .level of machine ?all (Fig. 8.8). ~

,0,

.'

~--""

,.__'0'

i +.tt~JJ\;

'0'-.

eo

"-'

,

+

._-

-,", -"~, . ,> r+,.I,~I':"',"

0.)

,

'...

0"

0',.

'0'" ..

0./ '

'

'1I)'J£:nh;£!"!l1l{~~,,!. n ;!' ,uP '>

\' ,it

-

-

Co<

.-.

.~.. ~

346

SoU Dynamics & Machine Foundatiolls

Turbin~

G ~ n ~ ra tor

. "....

Floor

L1m~

==:J

c::.

Upper stab

u ':

-: ::.

--, ,.

':-:';.'~':: ..

- '0"' ",

. .:'

. :.':"'"

'-,.

Base stab Fig. 8.8 : Concrete frame turbogenerator foundation

Rotating machinery is balanced before erection. However, in actual operation some inbalance always exists. It means that the axis of rotation lies at certain eccentricity with respect to principal axis of inertia of the whole unit. Although the amount of eccentricity is small in rotary machines the unbalanced force may be large due to their high speed. Figure 8.9 a shows a typical rotating mass type oscillator in which a single mass me is placed on a rotating shaft at an eccentricity e from axis of rotation. The unbalanced forces produced by such a system in vertical and horizontal directions are given by Fy

= me e ol sin rot

FH

= me e cos rot

...(8.8 a) ...(8.8 b)

...

3 c

In UN /

F::me~w2

'

F( , f .)F /

(b) Double shaft

(a) Single shaft

Fig. 8.9 : Rotating mass type oscillator

F ::

me

~(A)l

""

347

meral Principles of Machine Follndation Design

Figure 8.9b shows two equal masses mounted on two parallel shafts at the same eccentricity, the 1aft 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 ...(8.9)

F = 2 me e 002 sin 00t .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 resonance is 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 the system, then

(a) In reciprocating machines (IS: 2974 pt 1-1982) For important machines: 0.5 >

~

> 2.0

(On

For ordinary machines:

0.6 > -

"

(0

> 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 > ;- n > 1.25 It may be noted that where natural frequency of system oonis below the operating frequency of machine 00,the amplitudes during the transient resonance should be considered. For low speed machines, the natural frequency should be high, and vice versa. When natural frequency is lower than the operating speed, the foundation is said to be low tuned or under tuned, when the natural 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 which is, prescribed by the manufacturer. . . . . 6. 'The vibra'tiOli~must not be amloying to the persons working in the fact~iy or be damaging to other precis;.on machines. Th~ '~ahiie 'of vibrations that are perceptible, -annoying, or harmful depends on the frequency of the vibrations and the amplitudes of motion. Ri~rt (1962) devel-

ùæ{æ

,.c.

~-~~":"-.:,.':;;;:';';;"';":,:";::,L_:~,_-':"'".:1:L'-'~U:t

"'"

" '";,,,

."",

;

,"""

"

348

Soil Dy",a,mics &. Machine. FollndationS"

oped a plot for vibrations (Fig, 8,10) that gives various limits of frequency and amplitude for different purposes, In this figure, the envelop described by the shaded line indicates only a limit for safety and not a limit for satisfactory operation of machines,

0

0

Disp la c.emen t ampl i tude AI mm 0 0 ,0 0 . 0 0

.....g 0

0

0

0.

0 0

0

N

0

0

U1

.....

0

0, N

0

.

0 0

0.....

U1

0

.

'-/

0 N

0

A

'"

0

-

o.

"o 0

r

') ,o

'1»-oCO

,

0.<> 'b" 0

tJ1

-0'1> 0; "D ""-

q.;>0 )( ..... 0 0 0

0

.0

,,0 ')',

)"D

"'rig ""I ~

.D C ~ ::J n '< '" n

0"

\--':'

0"

'" )"D

,'" \--,' 0

", 1")' "D &>

"D"""

"

,>0

0

0"

)"D

N

" 0 3 00

0'" ",Y>

V'1 0 0 0 ..... ..0 0 0 0

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 Bureau Indian St~ndards for the foundations 'of different typ~s of~chines.are given:"", in Table 8,_2,' , ,t, -'i' "," -,. "',;', ",. ',"",','" ,x/A

(~,N~

;:

:~

"

,.,; ',"

'0 . - .

,.

.'

-, ., ..", .

349

eral Principles of Machine Foundation Design Table 8.2: Values of Permissible Amplitudes for Foundations of Different Machines ,

No.

I. 2,

3.

.

Type of machine

Reciprocating machines Hammer. (a) For foundation block (b) For anvil Rotary machines (a) Speed < 1500 rpm (b) Speed 1500 to 3000 rpm

(c) speed> 3000 rpm

.

Permissible amplitude mm

Reference

0.2

IS: 2974 (Pt - I)

1.0 to 2.0 1.0 to 3,0

IS: 2974 (Pt - 11)

0.2 0.4 to 0.6 Vertical vibration 0.7 to 0.9 Horizontal vibration 0.2 to 0.3 Vertical vibration 0.4 to 0.5 Horizontal vibration

IS: 2974 (Pt

- IV)

IS: 2974 (Pt

-III): 1992

Permissible amplitude dependents on the weight of tup, lower value for 10 kN tup and higher value for the tup weight 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 accormce 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 STEEL or 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 for oncrete 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 by lpplying suitable creep and fatigue factors. The following dynamic moduli of concrete may be used in the design: Grade of Concrete

Dynamic elastic modulus (kN/m2)

MI5

2.5 x 107

M20

3.0 x 107

M25

3.4 x 107

M30

3.7 x 107

"',

.....

'--~"'--"'-

"'-"""-'-""--'---'-'

'--'

"""'-"

'-"'-

'-,..""

(~~.

-~ SoU Dynamics & Machine Foundatioll$

350

'"-.

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 of grade I. In machIne'foundations timber of select grade is used. The permissible stresses of timber given in Table 8.3 may be multiplied by 1.16 to get the permissible stress of timber of select grade. t Table 8.3 - Minimum PermissibleStress Limits (N/mm2)in Three Groups of Struc,turalTimbers (For Grade I Material) S.No.

(i)

(ii)

Strength Character

Bending and tension along grain

Location of Use

Inside(2)

Shear(I)

Group A

Group B

Group C

18.0

12.0

8.5

1.05

0.64

0.49

Horizontal

All locations

Along grain

All locations

(iii)

Compression parallel to grain

Inside(2)

11.1

1.8

4.9

(iv)

Compression perpendicular to grain

Inside(2)

4.0

2.5

1.1

(v)

Modulus of elasticity (x 103 N/mm2)

All locations and grade

12.6

9.8

5.6

(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 and

2/3 are applied

~

The permissible bearing pressures on other elastic materials such as felt, cork and rubber are gener-.J ally given by the manufactureres of these materials. No specific values are recommended here since theY4 vary in wide limits. . , 1 ...

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

~~J ""'1 <-J

i

-

'

- ,

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 type foundations 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) the amplitude of foundation during machine operation. In this chapter, methods have been presented to obtain these two parameters in different modes of vibration. The basic assumptions made in the analyses are: (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 steps and 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 three principal axes and the other three are rotations about the three axes. Thus, under the action of unbalanced forces, the rigid block may undergo .vibrations as follows (fig. 9.1) : 1. Translation along Z axis

- Vertical

2. Translation along Y axis

-

vibration

Longitudinal or sliding vibration

3. Translation along X axis - Lateral or sliding vibration 4. Rotation about Z axis - Yawing motion 5. Rotation about Y axis - Rocking vibration 6. Rotation about X axis

-

Pitching or rocking vibration

The vibratory modes may be 'decoupled' or 'coupled'. Ofthe six modes, translation along Z axis and rotation 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, respectively, always occur together and are called coJlpled modes. Therefore the dynamic analysis of a block foundation 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.

353

oundatiQns of .Recip~oCQtingM achines

A rigid block being ~ problem of six degrees of freedom has six natural frequencies. The natural requency is determined in a particular mode (decoupled or coupled) and compared with the operating requency. 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 developing the analysis the ,effects of damping andpartidpating soil 'mass are neglected. Damping does not affect 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 the design apPI:eciably,and if any that on the conservative side. Empirical methods have been suggested to obtain the soil ma~s ,participating'in vibration. '

.

,

In the elastic half-space method,;:,!he"machine:foundation is idealised as a vibrating mechanical oscillator 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 approach is apparently more rational, but relatively more complicated.

'-"',' .,"'.

'

.

>i

354


In the above two methods, the effect of side soil resistance is not considered that is the foundation is assumed 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
Of

-L Soil 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 e

for different modes of vibration. Barkan (1962) had introduced the following soil parameters which yield the spring stiffnesses of soil in various modes: (a) Coefficient of elastic uniform compression (C,) : It is defined as the ratio of compressive stress applied to a rigid foundation bl~ck to the 'elastic' part of the settlement induced consequently. Thus Cu =!!.L (Fig. 9.3 a) Sez It is used in vertical vibration mode. From defmition, spring constant Kz K= z

Load

.

elastic deformation

.-

qz A =-=C.A Sez

u

...(9.2)

...(9.3 )

355

""datioilS of ReCiprocating Machines

t Fz(t)

.1 Fz
-- --------Fz

m

K.Z = Se2

L--

--

--

J

~~. f

S

ez

(a) Vertical vibration

.., I I I I I

Fx (t)

. .

I

I I

I I I I I I

.

I I

, ,.

I -"'7jN...-

.... I

I I I I

I I I I I

'--.... 'U:{f

,

-+I~'Sex I

I I

..

.

I

I

--.

m

---- --My(t)

. ..

Fx

Fx (t),

KX = Sex

.'

Kx (b) Pure sliding vibration

---""""

My(t)

-'"7 I I I

f:\

My

F:)

I I

K,=

I I

-""-.... rp'

K~ (c) Pure rocking vibration Fig. 9.3 : Types of motion of a rigid foundation (...Contd.)

T

356

, ';Soil"'vDymnnics-'>& ,

'i = Angle

nMZ{t)

,

.

~Mfld,ine

'-Fl1wuJtltions

ot

torston ot

~MZ{t)

toun'da tier. (not shown)

/

/

(d) Yawing vibration

Fd (t)

I I

,,

I

Fd (t)

.

... ......

....

, , ,

/

,

i., .... '-.~"'-.

,

r, a....

.... ...

....

, "

,,I ,, I '...

tSex

-i 'r' ~~

"...,

-

I

'.

,"

..

'Kx

Kx

>."'"

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 a the foundation contact area to the elastic part of the sliding movement of the foundation. q Ct = Sex x (Fig. 9.3 b)

...{9.4

It is used in analysing sliding vibration mode. The spring constant Kx is given by . K = Shear load = qx .A = C . A x t Sex

Sex

" ... (9 ".

(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. It defined as the ratio of intensity of pressure at certain location from the centre of the base of bloc to the corresponding elastic settlement. If cpisthe angle of rotation of block, then at a distaDI 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 q C~ =[cp

...(9.

J57

F oLlm/lIlilJl;,. (~r J:ecfproclItillg Machines

The sllj},:f.::';sKIj>is'defil1E".1as the motnent per unit rotation,a~d M K =-=C .1
where,

is given by

...(9.7)

cp

J = Moment of irf'rtia of the base of block about the axis of rotation M = Moment caused due to soil reaction

(d) Coefficient of elastic non-uniform shear (CljIJ: It is used in yawing motion. If a foundation is acted 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 moment. Therefore, M z =K 'I' \j/ where K =C .J 'I' 'I' z J- = Polar moment of the intertia of contact base area of foundation.

...(9.8) ...(9.9)

In the rotation of a' foundation around a vertical axis, the base of the foundation undergoes nonllniform 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 CU . It is based on theory of elasticity. . 1.13£

1 ...(9.10)

Cu=I-~2'JA where,

£ = Young's modulus of soil ~ = Poisson's ratio A = Area of base of the foundation

,

.'

He also developed the relationships between Cu' Ccp'Ct and C'I" For analysis and design of machine foundation, 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 of chapter 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 elastic constant obtained from a field test for using in the de~ign of actual foundation has been illustrated in examples 4. 2 and 4.3.

j

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

, U, !' " .

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

',. ~

c :".; r . ,t. "': . n.!,

~.

358


Table 9.1 : Recommended Design Values for the Coeficient of Elastic Uniform Compression Cu 2* for A = 10 m Soil

Permissible static load,

Soil group

Cu' kN/m

3

kN/m2 4

3

2

up to 150

4 up to 3 x 10

150 to 350

(3 - 5) x 104

350 to 500

(5 - 10) x 104

Weak soils (clays and silty clays with sand in plastic 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 close to the plastic limit; sand)

III

Strong soils (clays and silty clays with sand of hard con-

IV

sistency; gravels and gravelly sands; loess and loessial soils) Rocks

> 10 x 104

> 500

* After Barkan (1962) 9.4.1. Vertical Vibrations. For the purpose of analysis, the machine -foundaton-soil system shown in Fig. 9.3a is represented by the idealised mass-spring system shown,in fig 9.4. Let the unabalanced force is representd by (91.4) Fz.(t) = Fz sin rot

³

³

ã

Ó¿-- of machine plus

Kz

=

foundation

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 foun dation in contact with the soil lie on a vertical line that coincides with the line of action of the excitint force Fz, then foundation will vibrate vertically only. The equation of motion of the system is

where,

mz + Kz' Z = Fzsin IDt' M = Mass of machine and foundation

(9.15

Kz = Equivalent spring constant of the soil in vertical direction for base area A of the foUl dation = CuA

.,

359

,tidaiiiJns of Reciprocating' MaCih;n~ ;.

= Coefficient of elas~c uniform compression.

Cu

.

,

Therefore, the natural frequency IDnz of the system is (9.16)

ID.z = ~~z =r~

A.

The amplitude .of motion Az is given by .

or

. Fz sin IDt Fz sin 0)t A = = 2 Z . K -m0)2 C A-mO) z Il A

=

Z

(9.17a)

Fz sin 0)t 2

m ( 0) nz -

0)

2

'

(9.17b)

)

Maximum amplitude of motion Az is given by Az .

Fz

2

m (Ct>nz.-ro

.'

(9.18)

2

)

.

.

3.2. Sliding Vibrations of a Block. In practice, rocking and sliding occur simultneously. But if the bration in rocking can be neglected ,then only horizontal displcement of the foundation would occur tderan excitingforce Fx(t) on the block of area A (Fig.9.3b). This system can be indealised as shown Fig.9.5.

-,. I I I I , I I I I

m

I

~x ~/ 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 vhere,

(9.19)

x = Sliding displacement of,the foundation Kx

= 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.

'",

:

is

Therefore, the natural frequency Ct>nxof the system :

'

~;',~ nx"

fKX; =' rc;A f;; v-;;-

: (9.20)

q

Maximum amplitude of motion . A x is given by .

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

,

,.1,

""":~:':,'-:I:A ..

"

l_~I"'-<'Fx":. x'~'~~;x'_~2)"""'J"_:

.,.;

,:~f':,

,. '."

",

....(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,tions in~uced in a foundation block by an externally exciting moment Mv(t) (FIg. 9.3c). this "isalso a hypothetical case as rocking vibrations are coupled with sliding vibrations. Let the unbalanced moment be given by My (t)

(9.22)

= 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"displaced position of the block will be as shown in Fig. 9.6. In ~achine foundations, as the rotation
radians. The equation of motion can be obtained by applying Newton's second law of motion. My (t) = My Sin GJt

~'

( '-,-

"

...

-----x

~ ( a)

qst I dA

Elem
Fig. 9.6 : Block foundation under pure rocking vibrations

The various moments acting on the foundation about the centre of rotation are obtained as descri below: (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 1>."

npressed nonuniformly.

Machines -",

"

361 .

From t~e de~n~t~on of ~oef5cient of dastic un}form compression, ,C~ ~s d R I dA C~ =

.ere,

14>

...(9.23 a)

d R = Soil reaction force acting on dement dA cl>

= 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

...(9.24 )

= J C~.l~dA.l ==C~.cI>JPdA = C~ I

MR

here,

'

<\>

I = Moment of inertia of tpe foundation area in contact with the soil with respect to the axis of 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 in ig. 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 Mw =WLti\

'

.

...(9.25 )

'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: It is given by .vhere,

Mj

= Mmo ~

,

"

.

...(9.26)

.

Mmo =; Moment of inertia of the'mass of the foundation and machine \"":"ith respect of axis of rotation. '

This moment acts in the anticlockwise direction.

.

The equation of motion can be written by equating clockwise moments to anticlockwise moments Therefore, or

,

My sin' (tJt + WL cl>=

c cl>14> +

Mmo

~

...(9.27)

Mmo 4> + (C~ 1- WL) <\>= My sin Cl)t

The natural frequency wn«/> of this system is given by ...(9.28)

wn4>_,t.lM mo WL and maximum displacement A . . . ' ~ i~ given, by . '. .

A

=

. ,

My 2

2 . '.

l

...(9.29)

".

~. Mmo(ron,~ro) "",

.'

.' .

,--"

.'

-- .-'-

362

Soil Dynamics. &Mac1iin~

Foundatiolt/; '.0;, 'l

In practice, C. I is many times WL ; hence Eq. (9.28) may be written:'

~

.I

con.=I Mmo

...(9.30) ,;~

If the dimensions of the footing at the base are a and b in the X andY direction~, respectively, ba3 1=, ...(9.31) 12

,

-.

Cell ba3

conI/!

= "MU /nO

...(9.32)

Y

It is seen from Eq. (9.32) that the linear dimension of the contact area perpendicular to the axis of rotation exercises a considerably greater effect on the natural frequency of rocking vibrations than the other dimension. This principle is sometimes used in proportioning the sides of the machine foundation undergoing 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.

...(9.33'

...(9.34

Azr and Axr are added to Az (Eq. 9.18) and Ax (Eq.9.21) respectively to obtain total vertical an sliding 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 to torsional moment Mz (t) about Z-axis (Fig. 9.7a). The positionof the foundationat any time t mayt defined in terms of angle of rotation "'. Let the unbalanced moment is given by ...(9.3

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 Mmz iV

...(9.:

+ C'IIJ z '" = Mz sin rot ,

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~iformshear

t

,. '4 ,~1 '.."i .-

óóóòóóó

Foundations

of Reciprocating

363

Machines

'.

z

x

v

(a)

Isom~teric

vi
,

( 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: ...{9,37)

(J)nl/f-

Mmz - tw J,

,A

Mz 2 = M 'I' mz( IDnljl.

.

2 CD

)

The horizontal displacement AhlJlc~used, by .t,orsion is

where,

Ah'l' = r A'I' ...(9.39) r = Horizontal distance of the point on the foundation from the axis of motion (Z-axis)

'9.4.5. Simultaneous'Vertical, Sliding and Rocking Vibrations. In general, a machine -foundation is subjected to time dependent vertical force, horizontal force and moment, and therefore it simultaneously slide, 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 moment are considered to act at the combined centre of gravity 0 of the machine and the foundation, which is alsotaken as the origin of coordinates. At any time t, considering the vertical force acting in downward direction, horizontal force in right-hand side direction, and moment in the clockwise direction, the foundation block will be displaced as shown in Fig.'9.8. It is therefore subjected to (i) displacement z, in the vertical direction (ii) displacement Xo in the horizontal direction at the base and (Ui) rotation <1> of the base.

z I

'

Fz Sin C.Jt Initial

position

x

~ z

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

...(9.4(

(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 MR-=C .'""'I

...(9.4 ..,(9.4 jJ

+ .-

.:.(90'

--" ,

'

,

'<,

','

,

","

uni/ations of Reciprocating (iv)

.

,

365

Machines

Moment Mw due to displaced position 'of the centre' of gravity of block :

The moment Mw about point 0 is given by ..(9.44)

Mw = W L

(v) Moment MxRdue to horizontal resisting force Rx : Moment MxRabout point 0 is'give~ ~y . .L MxR = Rx . L = C~ A (X - L"' ~) (vi)

...(9.45)

Interial forces and moment:

(a) In the Z - direction

Fiz

(b) In the X - direction

Fix

(c) In the rotational mode Mi$ "here'

= mi = mx

...(9.46)

= Mm ~

...(9.48)

...(9.47)

M m = Mass moment Of inertia of the machine and foundation about an axis passing through '

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)

In the X - Direction: mx + CtA Xo= Fx sin cot

...(9.50)

Substituting the yalue of Xo from Eq. (9.42) in Eq. (9.50), we get ...(9.51)

mx + Ct A (x --L
,

Mm + C$ I - WL - Ct A (x - L
..

or

"

= My sin cot

2

Mm - Ct A Lx + (C$ I - WL + C~ AL ) = My sin COt

Equation (9.49) contains only the terms of z, therefore the motion in Z

...(9.52)

- 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 coupled modes. 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,

mx + C~ Ax - c.~ AL =0 ,

...(9.53)

'

and Mm ~ - Ct A Lx + (C$ I - W L + C~A L2 ) ==0 Particular solutions of these equations may be assumed as x = xI sin (con t + ex)

...(9.54) ...(9.55)

and

= <1>1sin

(cO n t

+ ex).

...(9.56)

in which x I' <1>1 and 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

2 -moon)-CtAL1

or Xl (CtA

=0

...(9.57)

,

~

and -Mmoo~1+I(CtAL2+CcpI

-WL)

-C'tALxI

...(9.58)

=0 CtAL1

From Eq. (9.57),

xl =

...(9.59)

2 CtA -mOOn

By substituting the value of xI from Eq. (9.59) into Eq. (9.58), we get

...(9.60)

<1>1 [-C~ A 2 L2+ (Cc!>I - WL + Ct AL2 - Mm oo~)( CtA -m oo~)] = 0

For a nontrivial solution, <1>1 can not be zero. Hence the expression within the parentheses mu~tJ)( zero, This gives' 2

,

I ~

2 2

2

L + (C. I óÉÔõÝ¬ßÔ

-CtA

2

2

...(9.6~

óÓ³×Ü²÷øÝ¬ß-mOOn) =0

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

CA.I-WL

4

2

n

n [(

00 -00

+

'f

Mm

C A(M t

]

m

C A CA.I-WL

+mL2)

mMm

+-L-

m

]

=0

'f

(

Mmb

...(9,(

)

By definition, the quantity (Mm+ mL2) is the mass moment of inertia of the foundation and mach about an axis that passes through the centroidof the base contact area and is perpendicular to the'~1: of vibrations. This is denoted by Mmo. Thus, , ij~ 2 ,M I Mmo = Mm + mL .:;t9. '

M Further, by denoting

~

Mmo m = r where 1 > r > 0

,

,

.(1~ -.J .~. '~

,

Equation (9.63) may be rewritten as 4

r

Now,

C.I-WL

oo~

OOn--

(

Mmo

CtA +m

)

h

CtA C.I+m

WL

'

=0

"

.t't.;.(S ,

tMmo

'i 'i 'f~i ,.~ '4 ,,

CtA

-

.',.: ,c

m

=

2 00nx

"' :0' ~,".. , ,\,

~,~f:' "

",

.

C.I-:-WL: 2 M' mo '; ~ OOn,

~, ""

~

Foundations

---

..

--. 367

of !leciproC/lting Machi,!es

'2y writting the Eq. 9.66 in terms of OOn,and CJ)n~we ?et 2

,

0>

n

2 2 0>nx + 0>ncp

-

r

(

2

2 2 0>nx Cl)n+

n

r

0> +

)

=0

...(9.69)

.

The Eq. (9.69) has two positive roots, CJ)nland CJ)n2' which correspond to two natural frequencies of the system: The ro~ts of Eq. (9.69) are: .

0>2

nl,2

2

=1.2

2

'2

O>nx

+ O>nq,:t

r

2

(Onx+ O>nq,

J

[(

2

(

r

J

2

- 4(J)n."(J)ncil r

...(9.70)

]

, Equation (9.70) may be rewritten as

.

2

2

I

(J)nl.2=

2

2r

2

)+

2 2 ,2 7"~ (O>nx+(J)ncpo

[( (J)nx+(J)ncp

4

2

2

r(J)nx (J)nq,)1

...(9.71)

From the property of a quardratic equation: , .

2

2

O>nl+O>n2 2

2

-

2

...(9.72)

r 0>2

(J)nl x (J)n2

2

0>nx + 0>ncp

=~

0>2

...(9.73)

r ncp.

22-1. 22_"221/2 (J)nl-(J)n2 - r [«(J)nx+ (J)ncp) 4r(J)ncp(J)ncp]

and

It can be proved that

OOnx

...(9.74)

and 00 ncpwill 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 the following three cases: Case I.

If only the horizontal force Fx sin ootis acting: Eqs. (9.51) and (9.52) may be rewritten as follows: ' .

-

mx+ Ct

Ax - Ct AL~ = Fx sin

...(9.75)

(J)t

2"

Mm~+ ~(Ct AI.: + Ccp1- WL)-Ct ALx = 0 Assume that the particular solution to these equations are

x

,

~

Ax'sin

~ ::;: Acp sin

(J)

...(9.76)

t

(J) t

in which Ax and Acpare the maximum sliding and rocking amplitudes respectively. By substituting these solutions into the above equations, we get ,

.

Ax (Ct A

2

- moo) -

Ct ALA.

= Fx

-Ct A LAx + A. (Ct A L2+ C. I - WL - ~m 002),= 0 or

~

...(9.77)

...(9.78)

- (C~L2 +C.I-WL-Mmoo2) Ax .A. ...(9.79) CtA~

~~' (~

:' ,.t~

368

SDi/Dynamics '& Machine FtiuiidQiio~ «i

."

.~ ~

~~~ "

By substituting for Ax in eq. (9.77), we get 2

.,

2

2

(Ct A L + C, 1- WL-Mm 00 )(CtA-moo CtAL .

.

A~

=..

.

.

mM m

OOIlX 00 Il~

r

[ -:-':--~-By

using the relations A

~ -

2

t

[

(C",I-wt) 't'

'.

.

+00

Mm

]

4 x.~t'f";.;". t.

]

- ~ ( 2 + 2 )+ 4 00Il~ 00n.l. 00 r

.

.

't'

]

C AL t 2

mMm [ oon1 oon2 -00

,.. 2

2

4

(oon1+oon2)+00

CtAL

(9.80), we get xF x

] "

F

...(9.81)

2 2 2 2 x mMm(OOn1-OO )(OOn2 -00 ) Let,

.('. ...(9.80)

FX

Ct~L '

2

)

m +

mMm

gives in (9.72) and (9.73) into'Eq. 2

.

2 C tA(mL2+M

-00

= 2

):'j

C AL'

't' mMm

m[

. "

- . A~ -Ct ALA, - 0

"

CtA+(C",I-WL)

mM

)

2 2 2 2 mMm(OOn1 -00 )(OOn2 -00 )

2

= ~(oo )

A = ~

C AL ~

t (00)2

...(9.82)

F

...(9.83)

x

By substituting for A~ in Eq. (9.79), we get A = Ct AL2 +Cq. I -WL-Mm x ~ (002)

Case 11.

...(9.84) ,.v

If only moment My sin rot is acting: Eqs. (9.51) and,(9.52) may be rewritten as . ,

mx + Ct and

002F x

...(9.85'

Ax- Ct ALcp = 0

.. .., Mmlj>-Ij>(CtAe+C4»I-WL)-CtALx

=Mysinrot

...(9.86

By assuming solutions as for Eqs. (9.75) as (9.76), it can be shown that the following expression hold: '.

,'.

.

CAL

My t Ax = ~ (002) -

.

'.

11.

..:(9.B'; -

2

and , ,

,

A.

- Ct A-moo 2 - ~ (00.)

My

...(9.B' 'i.

.. .. .. Poundations

Case Ill.

of Reciprocating

:

- --oii

369

Machines

If both the unbalanced force Fx aJ:?dmo~ent MJLare acting, the ~plitudes

of m~tion are

determined as follows: 0

-

02

A=

0

(C'tAL

x.

0 0

0

.

A 4>-

0 -

+ CtPI-:WL-Mm(O 2 )Fx + (c.~) 0 A L.l( (0 )

My

...(9.89)

0

0

0

0

0

2

.

"

2

0

0

and

.

(CorAL)Fx+ (CorA -mm)

My

...(9.90)

oL\(ro2)

The total amplitude of the vertical and horizontal vibration are given by a .

and

where,

Ay

= Az+2A~

...(9.91)

Ah

= Ax + h ArpH__,__-

...(9.92)

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 frequencies (0/11and (0112'These vibrat~ons are characterised by a certain interrelationship between the amplitudes Ax and A.pwhich depends on the foundation size and the soil propet:t~es,but does not depend on the initial conditions of foundation motion. ..- .. Let us examine the case when the foundation is subjected to exciting moment My only. The ratio of amplitudes Ax and A4>obtained using Eqs. (9.87) and (9.88) is given by . 2 Ax CtA L ronx L ...(9.93) p = -= 2 = 2 2 AA'I' Cor A - m ro «)nx - «) 0

,

o'

z

0

,,, ,

z

I I

---

..."\ ",/

,.

\ \

:\ \

v

..."\ \

-- ' )

..J

....

\

L

--

--- --

\ /

~

\

(

/

\

",/

V

(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 through

the 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 (0 n2' when the centre of gravity deviates from the equilibrium

position, for example, the positive dir~ction of the X - axis, the rotation of the foundation will also be positive, and changes of amplitudes Ax and A, will be in phase. The form of vibration will be as shown in Fig, 9.9 a, i.e. the foundation will undergo rocking vibrations with respect to a point situated at a distance PI from,the centre of gravity of foundation, The value of PI is determined 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, Figun

9.9 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

F (t)

. . ,r
;

"

.

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

~':~~':.:::::,',::~,..': G

pe . :':'::':~"""':;"~::',<;,

G

p

P

)J

JJ

(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. 9,1 by integration of Lamb's solut~on for a point load, Based on his work; the vertical displ
F0 = Magnitude of oscillatory force 0>

= Forcing frequency, rad/s

G = 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

'",.., ,

, .j:" ,

371'

in iii

. ,-:,-;,-;

~ -: .', ::-,: -'

,-

F.- i - 0 (Z

-

__

(..;t

-

,:

'r

,

.-""

'

'-

o':! i: ;

.

.>-.

f

óÐ¢ýññ¢

'-

"

",-,-.':, - .: .-, :.i??//(/4 ~,.

: . .'

Go

'0

'00'0"

p

'. -"," ,

~

~

,

l°":::"-:':

load

p(Zr

unit or(Zo

=

)J

F. (ZiG.)t

~

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 W ..,(9,95) b=~=~ pro yro It 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

,oVa..

Iv

...(9,96)

O,S_,

,

m = Mass of the machine and foundation

where,

= Shear wave velocity y = Unit weight of soil p = mass density of soil

Vs

Using the displacement Eq. (9.94), and solving the equation ofeguilibrium forces, Reissner obtained the following expression for the amplitude of foundation having circular base: fl2 + f22 2 2 2 2 (~-:-~ao 11),.+, (bao f2)

Az = Azn ~. ,(FofG ro) where;

,Az ,'A

"'"

zn ,

= Amplitude

of foundatiQn,.>

= Dimensionless amplitude; . """:"',""""")"':""'j"'.'",

' " i ; ,::;

' ,',

. , ,,'

:, ;,

...(9.97) : .., : ,':' i

, ,",

'.1 : i .. {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';~. ,:.[

372

SoU, 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

Valus of(fl)

a/ 1-0.07405 a04

0

0.318310 - 0.092841

0.25

0.238733: - 0.{)59683'a02 + 0.004163 a04

0.5

0.159155", 0.039789

a/

+ 0.002432 a04

Values of (- 12) 0

0.214474'ao'~'.

0.25

0.148594 Go-

0.5

0.104547 Go-

0.029561 'a03+ 3 0.017757 Go + 3 0.011038 Go +

0.001528 a05 0.000808 0.000444

5 Go 5 a0

The classical work of Reissner (1936) for circular loaded area was extended by Quinlan (1953) and Sung (1953) for the following three contact pressure distributions: F e;w/ (i) Rigid base (Fig. 9.12 a),

fz = ..

(ii) Uniform (Fig. 9.12 b),

I~ = ~

~-

O

2rrr0 r0

r

2

: for r::;; ro

...(9.98)

F e;w/ ...(9.99)

for r ::;;ro

2 rrro'

0

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 = Z

4

.0

:

1t ro

0

...(9.100)

for r ::;;ro

~

~

Cl

I

,

/'

. ~-d'

(b)

(0)

Fig. 9.12: Contact pressure distribution

(c)

~ .j

under a circular foundation

In the above equations, fz is the contact pressure at a distance r measured from the centre of foun,tf dation. 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/ '~A the assUinpt~o!l~at the.press.ur~.di~~b~~~~ri:r~ma,in unchang'e~ with,frequency. Their values ~re ~~v , ' .

in Table 9.3.

.

.

.,'

'"

.

.

.

~~~ ...'r'~

.,;.I. ..

datio/fs o/,Reciprocating {-;

373,

M/lcMnes ,

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

':

, ,'-:

"',.

i ' "

.: -. "."

.

'-',

' ", " ,

.,.

,

~';

-

0.7 poisson's

ratio)

,',

j.J = 0.25

,

0.6

O.S c:

N
,0.4

c:,I 'U ;:) +'

,,-.

a. E 0

0.3

\11 VI

-c

c:,I

0

1/1

C tI

E 0.2 b

0.1

0

~

0,

. .'

,

"

,

.

0; S

'.

a

.

0

,-'

1.S

1.0,' ..

"f..

',',

: ,-

,

"

>\'

Fig. 9.13: Plot of AZAversus ao for a rigid circular foundation subj~te~ to:constantexcitati.on,r~rce (Ric~art,I,962) , , i. .r' ',', ;, .. ,

--

,--

,~

374

Soil Dy,u,inics-&' Machilre -Fourrdations

Table 9.3 : Values of Displacement Functions for Rigid Foundations ~'.'<'

Poisson's

..'

'-~'-"'-(Bowres~"f977r-'--'-

ratio J.1.

0 0.25 0.5

0 0.25 0.5

0'0

.",-

.

V alues' of 11

2 4 0.250000 - 0.109375 ao + 0.010905 ao 0.187500 - 0.070313 a02 + 0~006131a04 0.125000 - 0.046875 a/ + 0.003581 a04 Values of -/2 3 5 0.21447 ao - 0.039416 ao + 0.002444 ao 0.14894 ao - 0.023677 a03 + 0.001294 a05 3 5 0.104547 ao- 0.014717 ao + 0.00717 ao

Figure 9.13 shows a typical plot of Azn versus a0 for various values of mass ratio b for a rigid base . circular footing subjected to a constant force excitation F0 e/(JH. A high mass ratio (greater height of footing 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 amplitude of the external oscillating force is given by 2 F=2m em ...(9.101) 0 e where, 2 me= Total rotating mass For this condition, the amplitude of vibration Aze is given by 2 2 me e CO Aze=

Ora

2 2 I1 + 12

2

Az

or Azen

where,

Azen

...(9.102)

(l-ba~/I)2+(ba;fz)2 2

I1 + 12

= (2meero2)/Oro ~ (l-ba; 11)2+ (ba; 12)2

:..(9.103)

= 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 base circular 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-amplitude curves 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 and Poisson's ratio on amplit~d'~'~freq~en~y;esponseof rigid circular footing ~~bjectedto frequency dependent excitation.Figure9.15 demonstratesthenatureof variationof Azenwith ao for three types of contact pressure distribution; i.e. uniform, rigid and parabolic. Parabolic and uniform pressure distributions produced higher displacement than a rigid base. The effect of Poisson's ratio on the variation of Azencap be seen in Fig. 9.16. The peak value of Azendecreases with the increases in the value of J.1.;but the corresponding value of ao increases with increase in J.1.. . i'

l ;.4 ",

\,

375

"daiions ut Reciprocating' MilChin~s "--"--""'-"""""""'-'-""'-""""-'-

"-'.--

'~.

0.28 poisson'5

ratio,

).J = 0.25

.'....

0.24

0.20 c

~ N

«

... ~

0.16

"0 ::::J

....

0-

E 0 III III

0.12

"

C 0

--

III C ~

E 0.08 (:)

0.04

0 -

0.5

0

°0

1.0

1.5

Fig. 9.14: Plot of AzcDversus.o for. ~igid ~in:ular foundation subJ,ectedto frequencydependentexcitation . " ... (Rlchart,1962) "" . ", "

-..

-- -.. .- -,

-_~_::c,i:- .---'-<-':":' ~'--

-'- ;-

-

--------

376

Soil Dynamiq_&Machine,Founda.(~(Jf3

0,6 b :'5

c:

~ O.S

parabolic

J.l : 0.25

pnz ssu rq distribution

~

-

-g

,

0.41,-

a. E

0.3

0 U1 U1 b.I

C 0.2 0 U1 C \:11

E

'0

0.' 0-

0

0.5

°0

1.0

1.5

Fig. 9.t 5 : Effect of c~ntact pressure distribution on the variation of Aze,nwith ao (Richart and Whitman. 1967)

c

~

0.4-

... ~ "U

~ 0.3

Rigid basq b : 5

a. E 0U1

0.2 0.5

III ~

c:

0 'jA 0.1 c: ~

E 0

0... 0

0.5

°0

Fig. 9.16 : Effect of Poisson's ratio on the variation

1.0

1.5

of Ann with ao (Richart and Whitman, t 967)

F~omthe amplitude~frequencycurves.in,Figs.9.J3 ,and 9,.14,on~can obtain:-(i)~axffI1um amI tude, and (ii) value of ao for maximum amplitude (i:,e;.resonance condition) for different values 01 Richart (1962) have plotted this data in graphical form as shown in Figs. 9.17 and 9.18 which convenient for design use. -

',dations of Reciprocating

Machines

100 Rotating

maS5

Constant

forc(Z

.£)

. .-0

Rigid

bas(Z

... 0 ....

10

III III

0 :--.......

= 0.5

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

<,

10 .2

0.4-

0.6

0.8

ya lu (Z of

1.0 ao

at

1.2

1.4

1.6

r (Z50no nc (Z

Fig. 9.17 : Plot of mass ratio b versus ao for reso!,ance condition for vertical vibrations (Richart, 1962)

100

r

\ '\

\'\ \ \,

, \\

\" .£)

, \\ \ \\ \ \\ \ \\ \ \\

I

0

.;:

,

,"

10 0

'\ fJ

Rigi d

\\

ba C;(Z

Ji< \ =0.5 "

0.25

Rota'ting

ma5S

. .

. ",

0 ..~

Con s'tant"

torc(Z

. . ,,,;

1-

-.0

1.0 . ~. Dim(lnsio'rde-ss',,:omplitude

ot

1.2

re5.0nanC(l

'1.4 ..~

Fig. 9.18: Plot of--;)""\"";*~;,'J..~.~, mass ratio bversus, dim~nsio.nless~mplitude at resonance (Ricbart, 1962) ~'", ,

"'.."

'.~\!;,A

,~,{~,:'H"...;,,~) ':!l~t:...n~iv~ir~l1'i'

~(1'i

'r,;',,~~ ~,

1r~ Q...(m'\\i
,,~4 {

'.f,,<'"

378

Soil Dynamics & Machine Found4ti.o., óó¢þò¢ù

ùòù

æþþùóù¢ùå¢öôº

Lysmer 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 It and damping constant Czwere taken as giv~n ~elow :" : .

K=z

, 4Gro ...(9.104)

I-J.t 2

3.4 ro er;G z I-J.I. The equation of motion may thus pe written as

and

C = -vpv

2

...(9.105)

. ...(9.106)

-

iwt .. 3.4 ro er;. 4 G ro mz+-vpG.z+_.z =F e I-J.i. I-J.i. Z Lysmer and Richart (1986) also suggested the modified mass ratio as 1-J.I. I-J.I. m ~.b 3 B -z

-

4

-

4

"

...(9.101'

pro

.The damping ratio; z is obtained as C

C

3.4 r;

,

JPG

...(9.108

to

~, C: ~ 2J~, m ~ îøïóñ´÷ù I-J.I. To m Putting the value of m in terms ofBz from Eq. (9.107), Eq. (9.108) becomes ~

;

- 0.425 z

-

...(9.t'0'

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 }J = 1/3 Halt-space H

:E ~

L-

theory

--- - Si mptitied

-

ono log .1

Bz :: 5

2

Constant

e

....

torce excitation

u 0

-

'" ',1 ~ ~J

-) Z

c

0

.... 0

-uc

(JI 0 ~ -..0 ... :t!?

."I!. ..",

0

0.5

. L0

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

379

lions of ReciproCllting Machines

e natural frequency of the system described by Eq. (9.106), i.e. .

',".K'

. .

<.0

m

""

= ri-L.

'."

~

m

2"

~l-~

...(9.110)

.

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

nz'

~

f(Bz

...(9.111)

- 0.36) 0 .~B r 0'. z 0

Id at resonance the amplitude is given by Eq. (2.61). i.e. 2 Fo / Kz

...(9.112)

(Az)max = 2 ~~ 1-~z 2 utting the values of ~z from Eq. (9.109) into Eq. (9.112), we get

Fo (Az)max

Fo.(I-J.1)

Bz

40'0

. 0.85~Bz - 0.18

(Az)max= .

vlagnificationfactor,

...(9.113a)

= Kz 0.85 ~Bz - 0.18

"

r

Bz

M- =

(A)

.4G,

z max

"-

.

0

=

Fo(1-J.1)

...(9.113b)

B

z 0.85.JBz-0.18

~or a frequency - dependent excitation ,the resonant frequency and the maximum amplitude are 1 by

:

H

0.90 <.onz

(Az)max

...(9.115)

= -;;;- .0.85~Bz- 0.18

Mze = 3mee

m

:re

Bz B' z

(Az)max -

Magnification factor,

...(9.114)

= ,/(Bz - 0.45)P . '0 2 me e

and

-,

1

...(9.115 a)

- 0.85~Bz-0.18)

2 me = Unbalanced rotating mass e = Eccentricity of mass from the axis of rotation

.2. Pure Sliding Vibrations. Amold et at. (1955) have obtained theoretical solutions for sliding

rationsof rigiq circular foundations(Fig. 9.20) subjectedto an oscillatoryhorizontalforce F0

ei(J)(.

~y have presented the solutions for two cases namely (i) constant force excitation, and (ii) frequency )t:nderit excitation.

In constant force excitation (F0 = constant), the amplitude Ax is gi~en by . .

-

.

'F.

.. ...

-"'

.

...(9.116)

Ax = O~0 Axn

lere,

" ..

Axn = No!1-dimensional amplitude factor

.~-

"~

--. ., .""

...

"""'-'

-=.""---==~

;;:;;~

Soil Dynamics & Machine ,Eo. .

380

.. F""(Z 0 iQt

...

otT f

Foundation f
x

s.t :~'"

'

,

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

P

r0

1

I

I

",

1}\

, I

1.S f

}.J

-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
f-

10

t, ~

ti

}J = 0

"0 ::J ....

a. E 0 0'1

---

III

..~,

ß¨»² c

ti x

.= <{ 1:) .-

- Axn

....

~"~ '":I-

, "I it

1

0

~I '"

C )(

T'

××Ää¥ III ti C 0 III C ti

,-E c 0.10.2

1.0

°0 Fig. 9.21 : Plots of AsDand AscIIversus ao for sliding vibrations (RIchart, 1962)

tionsof

íèï

ReCiprocating Machines

ïðð

£òÖãÑ

³¿--

Eccentric oscillator

ïð × ô

Ö ó

Ý±²

¢¬¿²

¬

ï ðòî

ðòìò

ðòê

ðòè

ïòð

ïòî

ïòìò

ïòê

pð Ú·¹ò çòîîæ

Plot of mass ratio. b versus a 0 for resonance condition for sliding vibrations (Richart. .

For the case Fo = 2 mex.e

ci ,the

value of A is given by:

/,

1962)

.

- 2 me e Ax - -rAxen pro

...(9.117)

The firm line in Fig. 9.21. shows the envelop of Axenversus ao for resonant condition. The variation he 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 con1t Cx were taken as given below.

;

K .

and

x

=

32(1-J.1) G ro

...(9.118)

. 7-8J.1

C = x

18.4 (1 7-8J.1 ~

J.1)

c:;:;

ro vPuo 2

...(9.119)

The equation of motion thus can be written as

.. 18.4 (1-J.1f

mx+

7 - 8J.1-

r

2

0

' P .x+ JPG. . .

32(l-J.1) G r .x -

7 - 8J.1

F e;CJ)l

0

..,(9.120)

0

He also suggested the modified mass ratio B -for sliding vibrations as .

.

\. "

B x

".',.,..".."",J,,'I,

=

X

7 -8J.1

'..

-m

32(1- Ji) pr3 "O,,}.

...(9.120 a) '.'

"".

.

382

Soil Dynamics &: ,Mllchine Foundlltitt~

1

The damping ratio ~x is given by .

~ - Cx Cx x - Cc - 2 ~kx m Putting the values of Cx and kx from Eqs. (9.118) and (9.119) in Eq. (9.121), we get

~

I

...(9.121)

I

- 0.2875

x-

...(9.122)

Fx

I

Figure 9.23 illustrates how well the response C\lrvesfor the analog agree with the response curves for the half space model.

4 Bx = 5 Exact

solution

- - -Analog solution 3 x ::E

I

...

L-

I

\

2

-

0

.....

u 0

..... c

0 ......

2

0

u ......

It

.-

C 0'1

i1~.j

0 ::E

.' T.:!i; "

..

tii.

.,'

",'

0

0.5

1.0 ,°0

Fig. 9.23 : Response of rigid circular footing for pure sliding (Hall, 1967)

.~.~~ ,:;,,

1.5

(1 /1.1' ~.

~

,J' '1i~;r jI.,.~

ùïïä¥ù

383

,dotions -of RecipI'ocatingMachines

The natural frequency <.olltand maximum value of amplitude (Ax}m~ can be' computed using Eqs. 23) and (9.124) respectively. . . . <.0

m

nx

.

and

(AJ

g

= f-: -2.. 1-~

X

.

...(9,123)

x

Ft / kx

=

" ma

2

2

...(9,124)

~x .R1- ~x

,2. Pure RockingVibrations. Amold et al. (1955) and Bycroft (1956) have obtained theoretical soons for rigid circular foundations subjected to pure rocking vibrations (Fig. 9.24). The contact pres~ below the foundation is varied according to .

ere

My

3 My r cosa. ejro( 3~ 27tro 'Vo -r

. q --

...(9.125)

(for r ~ ro)

= Exciting moment about Y -Y axis

a = Angle of rotation \

/

I

\rjJ

fJ /

I

\~I -

Footing,

I

i\T7\ \ , , " '. ., ,..' ., " .". ."" , . G'. .' . '"', . . -

"

-:

~

/

i

My Cli~,t

L

' ',,'

I-'

.,... r

-

-

~ ~:,

0 ,

.~

/"",

1I

':"',:"

,'"

p }J

x

.

~

,;"

"

. "'"~~~-~ptali~;bt 1..1 ; Fig. 9.24 '.;."

y ';} ,

,t'

tootin9'~

.~..,

:

....~

: Rigid clrcuJar foundation subjected to rocking oscillations " ;,I':(V;.t.~'H;tL""'i

4;':'

,.)"~:;>

Ö¼

i

.

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

...(9.126)

~pro

Under dynamic conditions the amplitude of rocking is a function of the inertia ratio B. which is given by 3(1-J.l) Mmo ...(9.127) B. =8 pro5 where, Mmo = Mass moment of interia of machine and foundation about the axis of rotation For the dynamic moment My, the amplitude of angular rotation A. can be expressed as A. = where

...(9.128)

MY3 'A,n

( Gro )

= Non-dimensionalrotational amplitude

A.n

Fig. 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 at maximum amplitude (resonant condition). The plot of inertia ratio B. versus ao for resonant amplitude is shown in Fig. 9.25 b. .

Hall (1967) proposed an equivalent mass-spring-dashpot analog for calculating the response of a rigid circular footing subjected to rocking vibrations. The plot of spring constant k. and damping constant C. were taken as given below: 3

8G 1~

...(9.129)

K. = 3(1-J.l)

0.8r: ,fGP

...(9.130)

Ccp= (l-f.l)(I+B,)

'"

40

't) :I

..

a. E

-

'f :'

0 0

c

" -

, ,, 'I'I ,\, 1I ,I

0

,

\

~

0

c 0

"

"

, \

I

'\

,

'" C tII

0

JJ = 0

11 ,I

10

.. c 0"6+-4 0

E

'

I

I

1 0.2

,I

I

"

20 \ I I

\1 1\

0.4

,

'\

/

\ \

I

1\

10 \ I \I (

0.6

\

\

\

S\ \

1.4

0.8

°0 Fig. 9.25 : (a) Plot of A." versus 10 (Richlrt.

1962) (...Contd.)

'j .j...

.

H'I\!~

Foundations

of Reciprocatipg

385

Machines

60

-e.

m

)J = 0

... 0

10 ..... 0

....

...... 0

L..

C

1 0.2

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 as 4 ~ 3 i(f)/ ;j, 0.8ra "Gp .h 8Gra '" _ M ",+ '",+ '", - M e m$ (l-fl)(l+B~) 3(1-fl) Y For critical damping,

...(9,131)

...(9.13'2)

c~c = 2~K~Mma Therefore damping ratio ~~ will be C~ = 0.15 ~~ = -c ) In

..,(9.133)

(1+ B$ "B~

~c

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 are

givenby

.

ro = n~

~ ~

...(9.134)

Mina My

A.

1

~

.

..

k'[{l-:~r+~'ro:ni

...(9,135)

386

".


50

B~=S

Exact solu tio n

- --

20

Analog

solution

9-.. 10

-)

J -:)

-:>

5'-

0

u -

+c: 0\ 0

2

z

0.5 0

t.' "

1.0

0.5

°0

l<

Fig. 9.26: Response of rigid circular foundations subjected to pure rocking vibrations (Hall, 1967)

Vibrations. Reissner and Sagoci (1944) have obtained theoretical solutions for rigid circular foundations subjected to torsional vibrations (Fig. 9.27). The variation of tangential shear stress is given by 9.5.3. Torsional

3 't 9 Z

Mr

41t

3

. = -:-.~ro

where

'tz9

2

ro

2

- r.

0

for 0 < r < r

...(9.136) "'. ~ I>

= Tangential shear stress

Wo

= Mz ei(J)( = Moment at any time t Mz = Maximummomentabout Z-axis M

'if. "!

For a static moment Mz.the angle of rotation A'IIsis given by A

-

3

",s ~[ 16G r;]

M z

~

~ .u(9.I31~i .,.

~. "

~,. c"' ":

---------..-.

'j'

387

Foulidiltions '6/ Ri!~iprdcating 'Mai:htnes

The amplitude of the angle of rotation Av can be expressed as M Av = ~Awn Gro .-"-""

'-',"'.

"-"'--'."-"-~"'""

'.

where

...(9.138)

Awn = Non-dimensional amplitude factor

Under dynamic condition the am~litude'of torsion is a function of inertia ratio Bw which is given by Mmz Bw = -S ,..(9.139) pro where Mm= = Polar mass moment of inertia of the machine and foundation about the' axis of rotation .

.(

Footing TZ9

-.

,"'00"'-

G " :', :,,~(,::::~(~<'<~

z'

: ...""."r,

',_..

ro ~

,

.

,

., "

Flaxibl
ro

,,

(b)

Ize

Rigid to undat ion

--'..

- -.. (a ).-

,

ro (c)

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

~'

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 of inertia ~atioB",:'~e' ~nveIop'cur~f's,ho~'~y !~~,:~~..lirie'c~n'~.~.~s~ato,defme 't~e relation between ao at maXimum amplItude (resonant condItion) ana the'values of Intena ratio B", (FIg. 9.28 b)

-,

388


F:oulldatioRS

-~.

~,,--

10

tII '0

, "-

::J ....

""

Co

E

II

1\

0

,

I1

rI I

0

c 0 ....

,

I I

\

1

\ \

\

I

'" '" to'

/ \,

C 0 '" C tII

.

/

/ /

E Q

0

/ /

//

/./ ññ¢óóþùþ

0.2

/;\

0..4

"'"

\ \ ~;.\./ I

"

Rigid"

5, \,'

B4' = 10!

Flczxiblcz

"V'.......

, I'

, I

.~ c '" r ...
.

/

\

\ \

\

\

I

0.8

.......

.-

\)/ /'

/'

"

\

\

1.2

1.6

2.0

00

(a)

60

7 m

.~ .... 10 0 L. '" '" 0

~

1

0.4

0 (b)

Fig. 9.28 (a) Plot of

~

0.8

1.2

1.6

2.0

00

versus ao; (b) Plot of B. at resonance versus ao for torsional oscillations of a rigid circular foundation (Rlchart,1962)

'. .,'f;' 'w

." iO' ~

",. .......

389

FiJundlltions of ReCiprocating Miichines ":;;-~~_'d

Richart and Whitman (1967) propose~ an equivalent mass-spring-dashpot analog for calculating the response of a rigid circular footing subjecred to torsio~al vibrations. The values of spring constant kw and damping constant CIjIwere taken as below;

= 16 G r3

~

,

J

~

...(9.140)

,

0

~::

',.'"

and

...(9.141)

The equation of motion can be written as M

.. III

mz 'Y

+

;m( 1.6ro4.Jp G." 16 3 . \V + - G r. \V - M e,

I+B

IjI

.

"",

3-:

0

...(9.142)

Z

1

The damping r~tio ~1j1~ili be

,"

,

C,o.s ,..(9.143) ~1j1

= 2 ~K:

.m

=

(1 + 2 BIj1)

The undamped natural frequency wnlj1and amphtude AIj1of the torsional vibration are given by

'K

"

...(9.144)

vM:;

wnlj1=

A

,

=

Mz

W " KW[{I-

...(9.145)

~:r +~Wro:Jr

9.5.4. Coupled'Rocking and Sliding'Vibrations. Figure 9.29 a shows a rigid circular foundation resting on the surface of elastic half-space and subjected to an oscillatory moment Mv /(JJlan a oscillatory horizontalforce Fx e,m{.The sign convention chosen is illustrated in Fig. 9.29 b, which indicates that + x and + F act to the right and that + 4>and + Mare clockwis'e.Figure9.29c showsthe motionof the footing when both translation of the e.g. and rotation about c.g. are positive (i.e. in phase). In this case. the centre of rotation lies below the base of the footing, and the motion ,is termed as first mode of vibration. If the translation is positive while the rotation is negative, then the centre of rotation lies above the e.g. and the motion is designated as the second mode' of vibration. ': "',1/ ,~}') ,/1' Nl

." ..

J

" >

1t

390

Soil, Dynamks

" ,Milehine FOUIIdtltions

I J

My (l iwt

I I

I

!.

I

... ~ Fx(li6.>t

I

I

IL

m

Er MR ~+Fx

.. Fx

,

fMR

( b)

(a) r

I

I

1

I

/:[1 1-.-

~

I I

,

=

/

II

+

I

I

--r I-- x Xo1r-

+x

'

~~ leA]

I :

(c)

Fig. 9.29 : Coupled rocking

Lc,&

and sliding vibrations of a rigid circular block on an elastic half space (After Richart

and Whitman,

1967)

From Fig. 9.29a x 0 = x - L<\>

where

..,(9,146)

XO = Displacement of base x = Displacement of c.g, <\>

Therefore

= Angle of rotation in radians

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 or

..

m x + Cx

.

Xo

+ Kx Xo = Fx eirot

.1 ~I. ~4

m x+ Cx (x- L~) + Kx(x- Lcp)= F.tirot

mx+C x x +K x -x-LC x'fd.-LK x'f'" =F .t irot The equation of motion for rocking'is .. . ,

irot

= 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

...(9:148 , .[)"\Or .,.1~aiJ

'

Mill cp+C~ <\>+K~ <\>+C:c,xo'L+Kx,xo.L

r

= My

irol

., ~:y;t

,

.,;, ,<:; ...(9,Jf~ ':: ''~:r . :"., ..t~

'~:',;~!

391

'Idations of Reciprocating~Machines

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 x ¨õÕ x ¨óÔÝ x n.-LK 'I' "

.

2

(

2

) (

x .n. 'I'

)

Mmcp+ C.+L Cx cp+ ÕòõÔÕ¨ cp-LCxx-LKx'x

=0

...(9.150)

=0

...(9.151)

The solutions of above equations can be obtained by substituting ...(9.152)

x = A eiCiJndt cp

= B eiCiJndt

...(9.153)

which A and B are arbitrary constants. By doing this the Eqs (9.150) and (9.151) become A LKx +i LCx,oond B-2 -m OOnd+ K x + I. c x OOnd A

d

...(9.154)

-Mm OO;d+(KcjI+L2 Kx)+i(CcjI+L2

Cx)OOnd

...(9,155)

B L Kt, -i Ct- OOnd#.' Equating Eqs. (9.154) and (9.155), and substituting

if

oonx

= -,m oonq» x

ff.

x =~ ~= ~ Cx M'mo r M mo ' x 2 vL'\o.x rH

C

,d ~4I 2 2 roncP+ronx

4

rond-

(

[

4 ~xronxrond r [

-

r

.

= 2 ~ K. xMmo , on simplification we get

4~x~~ronxron~

2,

~.

(ro~2 - ro~2.)+

)

rond+

~~ron~rond

r

2

22 roncpron..t

2(ronx

r

1

+

J

2 2ro~ -0 )]

...(9.156)

It may be noted that Eq. (9.156) reduces to Eq. (9.69) when ~x = ~cjI= O.As the effect of damping on laturalfrequency is small, the undamped natural frequencies for coupled sliding and rocking vibrations an be computed using Eq. (9.70). . The damped amplitudes of rocking and sliding of a foundation subjected to a horizontal force For/°)1 aregiven by lfJ

A xl

~

ÅøóÓ³®±îõ¢õÕ®Ôî÷î +4ro2(~,,~KrMmo+L2~x~Kxm)2 m Mm, ~ (002)

- Fr

(

2'

2

F L ronx ronx+4~x ro

and A =--L.1 Mm

...

~ (ro2)

]

-

...(9.157)

112

)

...(9.158)

392

Soi/Dynamics.& 2

,

2

2

Machine Foun ;~

2

2

'." '

where lI. (CO2)~ [ ",4_",2

{ '" ,+ :"'M

'" ,+} + "',. r"'M

4 ~x~+;M

,

,

]

~

2 ln

+ [ +x m"; m (m~ -m2)+ ~.m~.. m(m;. -m2)} ] ,

~I

., "

't

,.

...(9-.IS~~ "':'

"

,

The dampedamplitudesof rocking and slidingof the foundationsubjectedto an excitingmoment iw . M, e are gIven by '$I ,c

:4

2 Mv

A-'

x2 -.

MIII

[ (OOnx)

2 1/2

2 +(2;x

OOn.t) ]

2 A2

-

."(9.Jf~1 ",

-Mm

iI !I1'r

t

2 112

2 2

- My [ (OO"x-OO)

and

,

" (00 2 ) D. +(2;xoo/lxoo) f..

(00

"' "

]

...(9.1~!

2

,

,

~,,

)'.

When a footing is subjected to an oscillatory moment Myirot and a horizontal force Fx im/ simulb-

neously, then the resultingamplitudesof sliding and rocking are

. "j

Ax = Ax1 + Ax2 A<1>

= Acjll +

...(9.162

A<1>2

...(9.163

9.6 EFFECT OF FOOTING SHAPE ON VIBRATORY RESPONSE

fy Elastic half-space theory was developed for a footing with circular contact area. Response of a footing~1 influenced 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)b sional modes) (Richart and whitman, 1967 ; Whitman and Richart, 1967). Thus ~'

For translatory vibrations:

r0 =

b

i!-

..(9.U

7t

,;;:.1 1/4

3

For rocking vibrations:

ro

~~~1. .:~(~~t

= ( b3: J

~aj,yt ~ .. ",./I

114

ab (a2 +b2) For torsional Vibraf .ns : where

ro

ro

=[

61t

, "

~.j91,

]

'. ~'.~if

tt'

= Equivalentradius

t" ,1 '

a = Width of foundation (parallel to the axis._ofrotation for rocking)

'

b = Length of foundation (perpendicular to the axis of rotation for 'rocking)

-

"

d1,, , ,

.'' " ,,

i<)

I ["'~

"'

,~,

"4!t"l:'1 ,,

' ~~.h.. '' '

,5' " .'' ~ 't\', """ ,

, 'ft, , , '"r",~,

394

Soil Dynamics & Machine Foun~

,.

9.7 DYNAMIC RESPONSE OF EMBEDDED BLOCK FOUNDATIONS - ~i For 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 dynamit response 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) ; Beredugo and Novak,

1972;

Fry, 1963 ; Stokoe,

1972 ; Stokoe and Richart,

1974 ; Chae, 1971 ; Gupta,

i

1972

Vijayvergiya,1981). . The problem of embedded foundations has been analysed by both linear elastic weightless spring approach (Prakash and Puri, 1971, 1972 ; Vijayvergiya, 1981) and elastic half-space theory (Anandkrishan and 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 presentati9P here. On the basis of theoretical analysis, he had recommended the.equivalent spring stiffnesses in different types of motion as given below: .

9.7.1. Vertical Vibrations

(Fig. 9.31) ¢

! Fz5in

6>1

Machin
~,~

Ú¦ Í·²

êâù

³

-

Foundation

h

ø±¨ bxh)

0

1

~

a

f..

Õ¦ä¦

1

Fig. 9.31 : Embedded block foundation subjected to pure vertical vibration

...(9.! ã Ý«Ü ßõîÝ¬¿ªø¾Üõ¿Ü÷ Õ¦» ã Û¯«·§¿´»²¬ spring stiffness of the embedded foundation Ý«Ü ã Coefficient of elastic uniform compression obtained at the base of foundation Õ¦»

where

Ý¬ õÝ¬Ü Ý¬¿ª= Averagevalue of coefficientof elasticuniformshear =-

2

Ct Ý¬ Ü

= Coefficient of elastic uniform shear at the ground surface ã Ceofficient of elastic uniform shear at the base of foundatiqn

D = Embedment depth b = Width of base of foundation a = Length of base of foundation ~

d

.

..

~"" ... ..."".. . ~.."L... . .

395

Foundations of Reciprocating Machines

The equation of motion will be ...(9.168)

= Fzsin rot

mz+Kze'z

The natural frequency oonzeand maximum amplitude Aze of motion are given by

fK: oonze

Aze

...(9.169a)

= v--;;; Fz

=

. m 9.7.2. Pure Sliding vibrations (Fig. 9.32)

2 ronze

(

...(9.169b)

2

- ro )

Machin
.

-

Fx Sin wt

...-.

h

~

m

..

0

ø¿¨¾¨ h)

~

Õ¨ä¦

FxSin wt ~

Foundation

1

, ,

a

-i

1

'.

Fig. 9.32: Embedded block foundation subjected to pure sliding vibration

Kxe = Ct D A + 2 Cllavb D + 2 Ctav aD

...(9.170)

where Kxe = Equivalent spring stiffness of the emebedded foundation

C +C Cl/a v

= Average value of coefficient of elastic uniform compression =

!!.-

2

uD

CII = Coefficient of elastic uniform compression at the ground surface.

Similarly, the equation of motion will be

m x + Kxe

...(9.171)

..i = ~t sin rot

The natural frequency and maximum amplitude of vibration are given by

fK:: oonxe

=

A xe

...(9.172)

= v--;Fx 2 m (ronxe

2

~.

...(9.173)

- ro )

.. ~

.

396

Soil Dynamics & Machine -

Follndatioi;.\ -,

9.7.3~ Pure Rocking Vibrations (Fig. 9.33)

,. Machinq

My Sin

GJt

m

-::.'\\,: ×

¸

-

×

0

Foundation (axbxh)

ï

1 ./

a

I..

~

r

Fig. 9.33 : Embedded block foundation subjected to rocking vibration

C~av b Kljle =C~D.I-W.L+ where

24

3

(16D

f Db a2

2

-12hD

)+2C~avlo+Ctav

2

-t

...(9.£7< :1 ,"1l

Kljle = Equivalent spring stiffness of the embedded foudation

CIjID=

Coefficientof elasticnon-uniformcompressionat the base levelof foundation

Cljlav= Averagevalue of coefficientof elasticnon-uniform compression= CIjI

C~+2C~D

= 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=~ 12

'~J

,

3

I = aD 0 3 The equation of motion will be

.,.

MI1IO«\>+Kljle.$

...(9.I

= My sin rot

The natural frequency and maximum amplitude of vibration are given by Ct)lIljIe =

~

~e M mo

...(9.,. } -- i

My A

= :ce

J';J

2

2

M",O ( ro119- ro )

..~t} ~, ,I .i;J .-;i' ,. f j.

~ IJI ~

Foundations

,.Fo/llldatiolls

.

"

of Reciprocating

397

Machines

9.7.4. Coupled sliding and rocking Vibrations. The equations of motion in coupled vibrations are given as ...(9.178)

mx+K.u,x+Kxtp' = Fxsinrot and

...(9.179)

Mm~+KcI>'+Ktpx'x = My sin rot

}\There

]

= Ctpavb

Kq,tp

= CtpD1+

...(9.181)

(D2 - 2 DL) - C'tD . AL 2

.

a2

C'tD AL - WL + 2 CljlavIy + C'tav bD

2

2

3

3

+ 3' Ctpav[L + (D - L )]

> Ktpx

~K~t2

~ ~

where

= - [ C'tD

CljIlIV =

AL + 2 Cl/a v bD

( L- ~ ) +

2 C'ta v

...(9.182) ...(9.183)

( L - ~ ) aD]

Average value of coefficient of ~Iastic nonunifonn shear

ly = Moment of inertia of area a x Dlying in the plane of vibration about axis of rotation Da3 =-+12

,..(9,174)

(a) Only the horizontal force Fx sin ootis acting: m x + Kxx . x +

the base

Kfttp

.

= Fx sin

...(9.187)

=

...(9.188)

Atpfsin 00t

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 2

KtpxA.d + (Ktptp- Mm 00 ) Atpl

d.

sin 00 t

I tlI F t j

...(9.189)

=0

...(9.190)

By solving Eqs, (9, 189) and (9,190), we get

,..(9)75)

.

Axl ...(9.176\

= Fx

I '11 . . .

,..(9.186)

x=Adsinoot

I

~ I 1'1 . I ~\j

...(9.185)

00t

The solution of the above equations can be represented by

t

I I I , ,

The equations of motion will be :

Mm ~i + Ktptp. + ~tpx X = 0

l

I ,'1 I

JI ...(9.184)

m Mm co~e- (r:zKtptp+ Mm Kx.~)co~e+ (K~~K~ - K~x x Kxtp)= 0 The amplitudes of vibration ofthesystem can be obtained as below:

.

D

aDb2 4

1'he natural frequencies of the system can be obtained by solving Eq. (9,184)

on

-::. 11. , 411 I

...(9,180)

Kxx = C'tD A + 2 Cl/aVbD + 2 C'tav aD Kxtp

f' '" ~I'

2

-

-

2

(Kxx-moo

'" Atpl -

2

)(Ktp~-Mmoo 2

(K.n-moo

...(9.

Q,

.(K~~- Mm 00 ) Fx

...(9.191)

r.

...(9.192)

)-KxtpK
-K~

'., 2 ) (Ktptp-Mm 00 )-KxtpK
I I I~ .I

...

~ O//II~

""i"'""

~

----

u

cl! ~

~-

ó¢ùó±

íçè

Soil DYllamics & Machifle Foufldatiofls

(b) Only the moment My sin cot is acting: The equations of motion will be : 11

I.i

m x + Kxx . x + Kcpx=

...(9.193)

0

1I

...(9.195)

= A.p2 sin co t ...(9.196) 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) 2

;.

Substituting

4:

K.px' Ax2 + (K.p.p - Mm CO) A.p2 = My By solving Eqs. (9.197) and (9.198), we get

,f'

...(9.198)

-Ad and

A.p2=

I~

(K,X,X-

-Kx.p

2

2 My )(K.p.p-Mmco )-KcpxKx.p 2 (Kxx -m CO)

2

2

(Ku-mco ~" I

as:

X =Ax2 sin COt

f.i

I

...(9.194)

~

Mm + K.p.p . + K.pxx = My sin COt The solutions of the above equations can be represented

m CO ) (K.p.p- Mm ro ) - KcpxKx.p

My

If both Fx sin cot and My sin cot are acting simultaneously, then Ax = Axt + Ax2 H:

ïï åô ®

ô

...(9.199)

...(9.200)

...(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 gap and without air gap. From this it can be concluded that if air gap is provided around the foundation the amplitude of vibration increases whereas the natural frequency decreases when compared with corresponding foundation with no air gap around it. The response of embedded foundation with air gap can be obtained by analysis given in sec. 9.4 by using CuD' CtD,C.pDand C",Din place of CII'Ct' CeI> and CIjI respectively.

ô

Machinq × ô

òº

'"

×æ

..,\ ..//..

Foundation (axbxh)

h

×ù

li

Ai r gap

¬

×

t-

º

a

~

ïïæ

Fig. 9.34 : Embedded block foundation with air gap

¬

¶ò

III

u w

~

1111

!Ill,

III Ill!

~

,..

I!!

Ii

'1111

da

tio

Foundations

11 s

399

of ReciprocatillgMacltines 0 L.()

I .(9.193)

I i:

~

.(9.194)

t,

L.()

I

--.j'

I

..(9.195)

-

f

..(9.196) ), we get ..(9.197)

QC)

'"

0 --.j'

..,.,

..... ... ...

r" 1:1

:; Cc. '" .. ... '"

~ ~!

>

..(9.198) I L.() M

...(9.199)

0

::

,~

...(9.200)

1

...(9.201) ...(9.202) lion block ith air gap .dation the vith correir gap can

I Ill' 11

'i '0 c

0 M

I

i

tf) cV ...

>-

I

L.() ('4 L.() N

. 0

j ,I

II

m

1

CQ>and

~

...... 0«

C'v

I

L.() N. 0

cCl C1 L..

v c ::J

c-

er 01 lLL

Cl C7) l-

" .CD Cl ...... 0 0' Z

0 N

'" c. '" OJ)

... '" ::

1,\

'i"' -

f

c

"

,~ ';... .Q '> '" ,;: t:... >

~ ~ : u=

\I

"u

L.()

c '-0 0

~ >

-

Cl v .....

Cl ...Cl

L.. ...Ln "'Cl

0

C

Cl tf)

L.. >01 ...-

> ::: (/)

11

C ... :::! 0' ... .:: ,/, 'C B

c. a «!'. or, ~ 0-: .. ~

L.()

I 0

0 0

0 r.f)

0

SUOJ:>!W c apn~!ldwV'

(

..

-'i ,..'

--

--

~

~,..

~"'--"

~~

-

..- ~'" ,-

,

~

I'

Soil DYllamics & Mac/lille Foulldatiolls

400

9.8 SOIL MASS PARTICIPATING IN VIBRATIONS In both the methods of analysis and design of foundations of reciprocating machines described in sec. 9.4 and 9.5, the effect of soil mass participating in vibrations has not been considered.

.,

J! ~I

I:

;1 'f

.

,

Fig. 9.36 : Stress distribution in soil mass

Pauw (1953) developed equations for the apparent soil mass by equating the kinetic energy of the affected zone to theCkineticenergy of a mass assumed to be concentrated at the base of the foundation. He gave the following expression for apparent soil mass ms for translatory modes of vibration:

11

i

11

.

t

b3

If

m =Lc s ga.

P

,

where

...(9.203) m

a. = Factor which defines the slope oftnincated pyramid (Fig. 9.36). It is generally taken'unity. Cm = Functionwhich dependson sand r ~=- a.he . b

j

a r="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

ij

r

,J~.

'Yb5C. M =---'ms 12g a.

iI

whereC.I canbe obtainedfromFig . 9.37

.

..(9.204)

b to e. In these fi gures, C~, C~ I r and c~I denote the factors of mass

moments of inertia about X, Y and Z axes respectively. These factors can be obtained from Figs. 9.37b to d for cohesionless soils and from Fig. 9.37e for cohesive soils.

I

'I. .,

"'I

~---

-.'-

~'.

-

._-,-,

-

-

--

-, --~

"-'---~

'Ill IJI

~~---'

-

ìðï


Ý±¸¬-·±²´¬-0.2

0.2

"' -óóÝ - 'tb3 gJ-

0.5

I.a

'""6. 11

Cl)

soil Mmz = )f'b5 CX 5 12g-" i

0.5

m

1 0

1.0 - - - - - --

- - -- -- ---

1

-- - -.\"'&.a

2.0

-11

Cl)

2.0

S~O

5.0

r=12JLr (Ca)

0:6

0

J Cm r

0

. 0~8

Cb)

0.2

0.2

;rh 5 CY .

0.5

Mm ys

--I - 12'3 aC..

5 z rJL ,Mmzs- 129cl Cj

0.5 ~'1.a1.0

I.a 1.0

V) 2.0

11 2.0 Cl) 5.0

5.0 0

0.5 Cc)

r::1 --'-0.5 cf

CjY --;:-3

1.5

rJ+r ¥b5

x

Yb5

Y

M.mx$-= 12 gel Cj 2 Mmys= CJI.a ," ...

1290£ Ý¶

5

Cj

(t)

Fig. 9.37: Apparent mass factors for horizontal contact surfac£..> -' :.,' ' ; I';. '

L

402

Soil Dynamics

& Machine

Foundations

Balkrishna (1961) has developed the following expression for the apparent soil mass in vertical vibrations: 3/2

m = ~ 0.4775 Q s 3( 4 P) where Q = Sum of the static and dynamic load .

1t P

...(9.205)

..

Barkan (1962) has suggested that the apparent soil mass may be taken as 23% of the mass of machine plus foundation. Hsieh (1962) gave the expressions for getting apparent soil mass as given in Table 9.9. Table 9.9 : Effective Mass and Mass Moment of Inertia for Soil below a Vibrating Footing (Hsieh, 1962) Effective mass or mass moment of inertia of soil Mode .of vibration

ïïãð Vertical translation Horizontal translation

11 = 0.25

11= 0.5

3

, 3 2.0 -p ro

3 0.5 pro

\.0 pro 3

3

0.2 pro

0.2

3

1.0 pro

pro

5

Rocking

0.4 pro

Torsion (about vertical axis)

0.3 P r5 0

0.3 P r5 0

0.3 pr50

The apparent soil mass/mass moment of inertia is added to the mass m/mass moment of inertia Mm or J0 to get the natural frequency and amplitude of vibration. .

9.9 DESIGN PROCEDURE FOR A BLOCK.FOUNDATION The design of a block foundation provided for a reciprocating machine may be carried out in following steps: 9.9.1. Machine Data. The following information shall be obtained from the-manufactures of tl:1emachine 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.

'J

The minimum distance to any important foundation in the vicinity of the machine foundation should,

alsobe accertained.

..

-

}I

-, ~,~-

Foundations

403'

of Reciprocating .Machines

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 the machine 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 bearing strata 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 below the 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' Ccfj1 Ct' C'V'G, E and J.l) are obtained from relevant tests and corresponding strain levels are noted. ,These values are reduced to 10m2 contact area and 10 kN/m2 confining pressure. A plot is then prepared between dynamic elastic constants and strain level. The value of dynamic elastic constants are picked up corresponding to the strain level expected in the actual foundation. These values of dynamic elastic constants are then corrected for the actual area of the foundation (if < 10m2), and confining pressure. The details of this has

alreadybeen 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 eccentricity along X or Y axis is not over 5 percent. This. is the upper limit for this type of analysis. If eccentricity exceeds 5 percent, the additional rocking due to vertical eccentric loading must be considered in the

analysis (Barkan, 1962)

"

The static pressure should be checked; it should be less than 80 percent of the allowable soil pressure under static conditions. This condition is met in most practical foundations. . Table 9.4 : Determination of CG of the System Element of system

Coordinate e.g. Static moments of of the element mass of elements

Dimensions a.t

ay

az

Weight of element

Mass of element

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 moments are now obtained with respect to the combined center of gravity of the system. The relative magnitudes of 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 inertia and 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 Formula for Figure

Shape of the area

Ix

~

Iz

ab3 12

ba3 12

ab(a2 +b2) 12

1td4

1td4

64

32

,

ffi "

Rectangle

T

Y

I

--f::.G. X

I

1---0

y

b X

Jl

~

y

T.... Circle

X

X

N 11 '0

.!!:...ct 64

J..

)

Table 9.6 : Mass Moments of Inertia

'1;1

Formula for Shape of Elements

,

Figure

MmT

Mm:

Mmy

""-.

z

1'J'

m

Rectangular block

--

2

12 (b + h2)

m

m

12 (a2 + h2)

12 (i

+ b2) ~ "t !j " .""1

z

,.~ f'

'"

X

Circularblock

1/

1-

y,u t-d-j

T h

1

!!!-

3d2 + h2 12 ( 4 J

!!!- 3d2 + h2

12 ( 4

)

Foundations

of Reciprocating

405

Machines-:M mo-

where

-

Mm + m L2

= Mass moment of inertiaof machineand foundationabout the axis of rotation passing

Mmo

through base. L

= Distance of combined

r=-

centre of gravity above base.

Mm Mmo

.

9.9.8. Determination of Natural Frequencies and Amplitudes Linear weightless spring approach (i) Vertical Vibration (0 nz

=

Az

=

.A

W

(ii) Torsional Vibration

~

m

.-,.

Fz 2 2 m (OOnz-00 )

Mmz .(On",-~cw J, and

=

A '"

Mz 2 2 Mmz,,(OOnw-00 )

(iii) Combined Rocking and Sliding Sliding and rocking are coupled modes of vibration. The natural frequencies are determined as follows: conx ~ ~c, mA

<.oncp --

Mmo ~C.IWL 1

2

2

2

2

2

2

) (OOnx+OOn,) -4r OOnl,2 = 2; OOnx +OOn+:t [(

and

2

2

OOnxOOn,

]

The amplitudes of vibration can be computed with the following equations: 2

A x -

2

(Ct AI.;+C.I- WL- Mm00 ) FJ:+(Ct AL) M v

Y

~(oo2) 2 (CtAL)Fx+(CtA-moo

A ~'lihere,

)My

~ (002)

Ax = Linear 4orizontal amplitu_de of the combined center of gravity Acp = Rotational amplitude in radians around the combined center of gravity. 2

L\«o)=mMm

.

ill..

2

-

2

,

2

,.

2

((J)nl-(J) )((J)n2-~ )

-

'-

I j

406


Fou1Jdatio"s

The amplitude of the block should be determined at the bearing level of the foundation as a ... ..'-~."

~~~.,,~~:E-C';:,;z::..;~~;_.-,,-~..,_.

where

'.'

Av

= Az+ '2 A.

Ah

= Ax + h A.

Ah = Horizontal amplitude at bearing level h = Height of the bearing above the combined center of gravity of the system Av = Maximumvertical amplitude

Elastic half-space approach Equivalent 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 Factor Modeof vibration (1) Vertica!

Sliding

Rocking

Torsional

Mass (or inertia) ratio (3)

Equivalent radius (2)

roz =

r ox

,o r

°'V

f! f! 1t

(7

-

1t

0.425

m 3

z = Bz '0.2875

pro

- 81l)

m

x= -a- .f

Bx= 32(I -Il) pr; B

31t (ba')'14

=

(l-Il) -4

Bz=

Damping factor (4)

r

ba(a2+b2) 6 1t (

= $

3 (I - J.l)Mmo 5 8 pro

MmB=---t Y pro

0.15 x

Spring Constant (5)

k=-

4Gro 1-1l

kx =

32(1-J.l)Gro 7-8J.l

z

k =

= (1+ B).JB; 0.5 'V= 1+2B

\jI

3 8G ro

3 (1- J.l)

.!

16

,t

3

k'V=3"Gro

Natural frequencies and amplitude of vibrations (I) Vertical Vibration OOnz

=

~; .'

(it) Torsfonal Vibration

A, 00

~

n'V

=

,,' '~,

K'[{l-(:JrZ+~,:Jr

)~

~

~1

'V

M rn,!,

and

~

~

K.[{1_(ro:Jr~(2~.ro:Jr

L

--..

407

='oundations of Recipr-Ocating Machines

(iil) Coupled sliding and rocking vibration 0)

nx

0

'

= fKx V-;;

~

~

Mmo

O),,~ =

Damped natural frequencies are obtained as the roots of the following equation: 2 4

O

(ID~,+ ID;") 4 V..

2

00 na - 00 nd

[

{ ~X 00nx 00nd

4

IDnx"";

-

r

0

00

0

oo~ 00;,

+

To,

r

.0

}

2 00n,

2

(

)

00

] 2

~'

00nd cOnip

(

2

2

- 00 lid + 00nx - 00nd =0 r r [ )] Undamped natural frequencies can be obtained by using following Equations: +

~

2 OOnl,2 -

2 +

)+ ( 2

2

)-

2 2

+

2

2

2r [( OOnx OOn, - OOn, OOnx 4 r oon, oonx ] Damped amplitudes for motion occasioned by the applied moment, can be obtained as below: 0

.

x

.-

1/2

2

[

A = My

~ . ,-

0

]

(oo~) +(2~x 00n.t)2

Mm"

~(002) 112 2 2

2

A =My [( OOnx-OO $

)

M 111

2

-',0,

]

+(2~xoonx).. /),,(00 2 )

where /)"(0)2) is given by Eq. ..

4

~ (ol) =

(OO~,+OO~)

2

00 -00

r

(

[{,

4 ;.t

~~ 00 nx 00 nljl

r.

0

0

+

+x

2

- 00~x 00 ~~

.

r

}

,

00

'.

00

2

2

0...

~, 00 n,

":-(IDn+-ID)+

}

2 1/2

00,

r

2

2

(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

..

and

~

Fx [ (-Mm 002+k, +L2Kx)2 +40)2(~~~K~Mmo+L2~x.JK7n)2

-m-~.m

.A

A

- Fx

(

4 ~xco

0

,.'~:t'~ffl"

6...

(0)2) '- .

2 L coIIX coIIX+ (

/)".
'.-

-

2 1/2 .

)

,", ..

-.' "

]

0

408

Soil Dynamies & Machine Foundations

9.9.9. Check for Adequate Foundation. The natural frequencies computed in.step 8 should be away from the resonance zone i.e. Cl)

-

co

< 0.5 or -

Cl)n

< 1.5

con

The amplitudes computed in step 8 should be less than t~e limiting amplitudes of the machine which are usually specially by the manufacturer of the machine.

ILL USTRA TIVE EXAMPLE

I

Example 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. The soil at the site is sandy in nature having cl>= 350and Ysat= kN/m3.The watertable lies at a depth of 3.0 m below the ground surface. The block is embedded in the ground by 2.0 m depth. The machine vibrating at 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 elastic constants. 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) Elastic half-space approach. Solution: 1. Machine data (Fig. 9.38) 2.5 Sin G.>t 2.0 Sin CiJt

Machincz

4.0 Sin c..>t :.~...

e.g..

h = 3.5 m

2.om

1

11--

Block

T

.,.

T

b =3.0m

L = 1.75m

a =".Om (a)Section

"

1

~

J-

-".,.-- -

-,..~-, -- -

a = 4.0 m (b) Plan

I"" 0" f. Y"", Fig. 9.3S':, Details oUoundatlon

..

l -} ..,!' ,,(. ~l .~1

, ,--- ,I

F oundatitJnsuf

409

R:ecip,octlting Machines

Operating speed of machine =500 rpm

"

'.

Weight of machine is small and can be neglected. Fz = 2.5 sin Cl)t kN Mz, = 4.0 sin rot kN ,m Fx = 2.Dsin rot kN acting at a height of 0.2 m above the top of block 21t ro = 250 rpm = 250 x 60 = 26.2 rad/s 2. Dynamic elastic constants Refer 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 is illustrated there. Therefore, following values of dynamic elastic constants may be adopted. Cu = 3.62 x 104kN/m3 G = 1.10 x 104 kN/m2

E = 2.98 x 104kN/m2 f.1= 0.35 C 3.62 C = ---!!.= x 104 = 1 81 x 104 kN/m3 t

2

.

2

= 3.46 Ct = 3.46 x 1.81 x 104 = 6.26 x 104 kN/m3

C41

C", = 0.7S'Cu = 0.75 x 3.62 x 104= 2.71 x 104kN/m3

3. 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

2

m = 9.81 = 102.8 x 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

-;;;H

A- -

u

(Onz

=

3.62 x 104 x 12.0 = 65 rad/s

102.8

Fz (AZ>max

,

.r

,

,. .

= m(oo~ -002) .

\

'

~,-' .-:

2.5 ,

102~~J~?2

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

-.2~.22J'

=:=6.8 x. 10-6

".:'r~.~,l'S"""

,

.

m r. = 6.8 microns .'

. 410

SoU Dynamics & Machinl! Foundatio;

(b) Torsional vibration ab

2

2

.4 x 32

2

4

= -12 (a + b ). = - 12 = (4 + 3 ) = 25 m

J

z

.

M

=!!!.. (i + b2) = 102..8x 103 (42 + 32) = 214 x 103 Kg-m2 12 . . 12

(0

-

mz

nw -

--

M mz iC" J, -

- 56 .ras 2 d/

-

. 2.71 x214 104 x 25

(c) Coupled rocking and sliding

=

(0 nx

(On

re;;:=

V-;;-

1.81 x 104 x 12 = 46 rad/s

102.8

Mmo -- ~C.I-WL

I = ba3 = 3 x 43 = 16 m4 12 12. W = 1008 kN L = 1.75 m Mmo = Mm+ m L2 M ~!!!.. m 12 Therefore

Mmo OOn d.

(i

+ h2)

= 102.812x 102 (32 + 3.52) = 182 x

103 kg-m2

= 182 x 103+ 102.8 x 102 x 1.752 = 496.8 x 103kg-m2 = 6.26 x 104 x 16-1008 x 1.75 = 44 8 d/ . ra s

496.8 M 182 r = --!!L = = 0.366 Mmo 496.8 't'

2 (J) 11\,2

-

1

- -2r

[(

V

2

2 2 +" (J)n.\"+ (J)IIcp -

2 2

) ((J)ncp+ (J)nx) - 4 r

2

2

(J) ncp (J) n.\"

]

(462+44.82)2 -4 x 0.366 x 462 X44.8~ .I

(462+44.82):t = 2 x 0.366 1 . [

= 33.8 rad/s and oon2 = 100.6 rad/s

!J.(00 ) = m Mm

= = My

102.8

2 (J)nl

2

2

.'.".'

2

..

( - (J) )(Cl)n2 - ) X

Cl)

182 (33.82 - 26.22) (100.62 - 26.22)

102.8 x 182

X

] ,cot I

= ~ 3283] 0.732 [4123:t . OOnl 2

4~

456 x.9434

= 8.05

= 2 (0.2 + 1.75) =~.~,'kN'ni":,

X

1010 kg2 m2

, "..;~.I

------------

lundations

of Reciprocating

=

A

411

Machines

(C411- WL+C-rAL2 -Mmro2) Fx+(C-r AL) My

x -

=

~ (ro2) (6.26 x 104 x 16-1008

x 1.75+1.81 x 104 x 12x1.752_182 8.05 x 1010

x 26.22) x 2.0

4

1.81x 10 x 12 x 1.75x (3.9) + 8.05 x 1010

= 3080158+ 1482390 = 56.6 x 10-6 m 8.05 x 1010

A$ =

=

2 ) My

(C-rAL) Fx + (C-rA-mro ~ (ro2) (1.81x10

4

x12x1.75)x2.0+(1.81x10 8.05 x 1010

4

x12-102.8x26.2

= 760200+ 571876 = i6.5 x 10-6 rad . 8.05 x 1010 Hence,

a -6 4.0 -6 A = A + - A = 6 8 x 10 + x 165 x 10 v z 2 ~. 2' 0'

= 39.8 x 10-6 m = 39.8

microns

1/

,

= 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

Ah

5. Elastic half-space approach (a) Vertical vibration

=

r 0

fA =

V;

ru. = 1.95m

V-;

Average effective unit weight of soil = 20+ 10 = 15 kN/m 3 2 1008

-'

B = 1-~.m3 4

Z

=

K Z

pro

= 1-0.35 4

W

2-.

( 9.81 )

= 1.47

x 1.953

4Gro = 4 x 1.10 x 104 x 1.95 = 13.2 x 104 kN/m 1-~ 1-0.35 0.425 ' 0.425

~z. =-

t.

x

'

":= .

B z °0-..J.195 . :JB: ,t '

= 0.304

2

)3.9

412

Soil Dynamics & Machine Foulldations

CO nz .~Kz

-

m =~

13.2 x 104 = 35.8 rad/s 101.8

Fz

A,

k{{l-( :Jf

~

2.5

13.2

x 1o~[{I -

= 29.4

x 10-6 m

~

r

+(:~,~:J

rr

G::~

+ (2 x 0.304 - ~~:~

r

r

(b) Torsional vibration 1/4

r = 0

M mz

\14

ab(a2+b2) 67t

[

= 214

]

3

4x3(42+32)

=

[

2

x 10 K g -m

67t

= 1.9973 m

]

B = Mmz = 21~ x 9.81 = 4.40

p,;

\jI

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

X,"= T

co -nljf

0.5

0.5 = 1 2 4 4 = 0.051 1+ 2 BIjf + x .

~

-- Ijf M mz

~

46.75 x 104 -- 46 .7 rad/ s 214

9.9.10. Coupled sliding and rocking vibration Sliding

l ~

r = - = 7t

0

K = x

32(I-Jl)Gr

x3

7t 0

7-8Jl

= 1.95m =

4

32(1-0.35)x1.lxl0

x1.95

7-8 x 0.35

= 10.62 x 104kN/m B = 7-8Jl x 32(I-Jl)

.-m

= 7-8xO.35

pr;

32(1-0.35)

x

1008 15x1.953

= 1.83 ~ ~x

- 0.2875-: 0.2875,- 7B: - .J1.83 - 9.212

, '1 t';

..' ~" I

r;'

oundations

of Reciprocating

413

Ma.clJines

.

fk:: = /10.62 x 104 = 32.1 rad/s

-

mnx - I V-;;

(

ab

-

Rocking

V

102.8

3 1/4

ro - \. 3~

)

1/4

4.0x33

(

~

]x)

~

1.84

ID

K = 8 G r; = 8 x 1.1X 104 x 1.843 -

3(1-/l)

cjI

3(1-0.35)

4

-28.1x10kNm/rad

BcjI= 3(1-/l) x Mmo = 3(1-0.35)x496.8x9.818 Pl~ 8 15x1.845 -3.75 0.15 ~=

'" n=

0.15

(l+B<\»jB;

J MK,1110

= (1+3.75).J3.75 = 0.016 496.828.1 x 104 = 23.7 radJ, .

=

9.9.11. Coupled Vibration. Undamped natural frequencies in couples rocking and sliding are given by 1

2

2 IIX

=

2

2

h [(m

m111,2 =

2

2

2

2

IIX

mmp]

(mnx + m ) - 4 r m

) + mn:t

n

1 2 x 0.366 [ (32.12+23.72):t

(32.12+23.72)2 -4XO.366X32.12X23.72 "

]

=~ 0.732 [1592:t 1299] (On1

~

= 20 rad/s anda mn2= 62.8 rad/s

( (co;.:co~)

.

r

4~x~.:"' co,.}+ co;',co;. 2 1/2

+4

{

(

~. mnx m 002 -m2 .\

r

II<\>

00 m2 -002 )+ ~<\>mn<\> ( n.\ r

]

)}

2 ~ «(02)

26.24 -26.22

= [{

1592 - 4 x 0.212 x 0.016x 32.1 x 23,7 + 32.12 X 23.72

( 0.366

)

0.366

0.366

2 1/2

.

+4

x 26.2 (23.72- 26.22)+ 0.016 x0.366 23.7 x 26.2 (32.12 - 26.22) } 0.366 { 0.212 x 32.1 ] :",,';

[

}

'.

,;-

.

.-

= (471200-2966467+ 1581341)2+4 (-60772+9337)2

.

In

] = 919697

.

" l

414


Foundations ~;; ~'

A

~

~

xl

+K. +Kx L,)'12+4"'(~, ~K. Mm.+L' ~x~)'rl2

{(-Mm'"

mMm

=

~({i) 2

2.0 102.8 x 182 x 919697

[ (-182

x 26.22 + 28.1 x 104"+10.62 x 104 x 1.752) 2 1I2

+ 4 x 26.22 (0.016~28.1

=

x 104 x 496.8+1.752

x 0.212~1O.62

0 2 1/2

2

= <1>1

]

2.0 (2.3165x 1011+0.1496 X 1011)1/2 102.8 X 182 919697

= 58 x 10-6 ID

A

x 104 x 102.8)

~'(

.

M

( L COnx COnx+4

~x CO

)

~(CO2 )

111

.\::

r

1/2

")

(

32.1 321-+4XO.212X26.22) 919697.

= 2.0X1.75 182

l = 26.9

)

x 10-6

fad

'

1/2

A

-

My [ co~'(+(2~xconx)2 -

x2 - Mm

]

;'" .;

~ (co2) 2

= 3.9 [ (32.1) +(2xO.212x32.1) 182

= 0.81

,,'",:iJ

2 1/2 ]

:;;

t11

919697

x 10-6

';::;t

;~

ID

t~

1/2

2

~;J

'(\"

"j

- My [ (co~x-co2) +(2~xconxco)2 ]

A(\I2-

-

Mill

2

~ (co ) 1/2

2

3.9 [ (32.12_26.22) 182 = 8.3 x 10-6 fad

+(2xO.212x32.1x26.2)2

]

919697

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

ft~11 ;"

i~

,

,

'~",.1 iij;';i

-

..

III

~..~" 415 .

{}undations .of Reciprocating- Machines

Hence

.

a

= A + -2 An.= 29.4 x

A

4~

Z

y

= 99.8 Ah

-6

x 10

-6 4.0 -6 10 + - x 35.2 x 10

2

m

= 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 and mplitudes computed by the two approaches. It may be due to the reason that the value of shear modulus J is computed from the block resonance test data using the relation between Cu' E and G. Actually in lastic half-space theory, it is desirable that the value of G is obtained from wave propagation test. Author's xperience 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 two pproaches closer. :xample 9.2 )etermine the natural freque~cies and amplitudes of motion of the foundation (Example 9.1) taking into tccount 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 the values at the surface of ground. Therefore CuD = 1.2 x .3.62 x 104 = 4.35 x 104kN/m3 Ct D = 1.2 x 1.81 x 104 = 2.18 x 104 kN/m3 , CD

= 1.2

CIjID

= 1.2 x 2.71 x 104 = 3.25 x 104 kN/m3

x

6.26

x

104=7.si

x

104 kN/m3

The average values of dynamic elastic constants will be C,./aV = 4.0 x 104 kN/m3 Ctav

-- 2.0

x

104 kN/m3

Ccpav= 6.89 x 104 kN/m3 Cljlav = 3.0 x 104 kN/m3 D = 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) = .108.2 x 104 kN/m ..

24 x 3.5 he

=

.

15

= 5.6 m

ah

1.0x5.6

S = -:- =

3.0

a

= 1.867; r=

4

b = 3" = 1.334

416


,

,

From Fig. 9.36 a for S

"C

'

= 1.867 and r = 1.334;-!!!. r = 1.45

or

Cm = 1.45 x 1.334 = 1.93

Therefore

m =-

,

'Y b3

S

ag

C = m

15 x 33 1.0x9.81 4"-

(m+ms)

=

-

= 75.8 rad/s

108.2 x 10ffinze= VI02.8+79.7 Fe A == 2 ze

3

x 1.93 = 79.7 x 10 Kg

2

(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 104 kN/m From Fig. 9.36 b, for S = 1.867 and r = 1.334,

=

M

'Y

b5

CX

m.'Cs 12ga

~r

15x3

=

i

= 17.72 kNms

= 0.43

2

5

0.43 x 1.334

12x9.81xl

Mmos= Mmxs+ ms x L2 = 17.72 + 79.7 x 1.752 = 261.80 kN ms2 Cl)

=

nxe

~

xe

m+m s

C K~ = Cd.D 1- WL + ~

.b

,av 24 3

= /106.16x 104 = 76.2 rad/s ~102.8+ 79.7 "

Db (16 D3 - 12 hD2) + 2 C$av x 10+ Ctav x_a 2 3

2

'

I=ba

=3x4 =16m4 12 11 W = 24 x 3 'x 41<.3.5 = 1008 kN 3 4 2} , I = aD = ~ ='.10.67 m4 0 3 3 '-, -~ ;~ ;f J :J-j

w,~'!!' i."

417

Foundations of RecfprocGtingMachilies.. 6 89 K$e

. 04

= 7.51 x 104 x 16 - 1008 x 1.75 + .

x24 I.

.

x 3 (16 x 23 - 12 x 3.5 x 22) .

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 4

(J)mj> e

280.53 x 10 182 + 261.80

K$e

=

+ Mmos

Mmo

= 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/m 2

Kx$

= CqJavb (D - 2DL) = 6.89 x 104 x 3 (22 =-231.81

x 104kN

3

2

I. = D a + a Db y 12 4

K$$

= CDI + = 7.51

CtD AL 2 x 2 x 1.75) - 2.18 x 104 x 12 x 1.75

=

~ 12

3

2

+ 4 x 2 x 3 = 28.67 m4 4

CtD AL 2 - WL + 2 C1jfOl'Iy + Ctav b

~ a2

+ ~ Cav[L3 + (D - L)3]

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

Õ±òô

x 104kNm

¢

óÅÝôÜ ßÔõî C"", bD

~

-[2.18

x

104

x

12

( L- ~ )+2 C"" ( L- ~) aD]

x 1.75+2 x 4 x 104 x 3" 2 (J.75-~

= -[45.78+51.84+34.56] 104 = -132.18 . "

~

x 104 kN

)+2 x 2 x 104 (1.75-~)4

x 2]

'(.

418


{KcMI-(Mm +Mmxs) Cil} Fx AXl=

2 2 0) }{ KcMI-(Mm +Mmxs) 0) } -Kx4l'

{ Kxx -(m+ms)

=

.1'

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

998.18 x 10

998.18x10

= 77.64 x 104 x 498.3 x 104- 30639.32 x 108 = 8048.69

4

x 108

-6

= 12.40 x 10 m A

-K

= cpl

{Kxx -(m+ms).0)2}

~

.F

. x

{Kcpcp -(Mm + Mmxs) 0)2}-Kxcp' K~

= 132.18 X 104 X 2.0 8048.69 X 108

= 3.2 Ax2

x

10-6 fad - Kx . M.v

=

~

2

{ Kxx

- (m + ms) 0) } { Kcpcp - (M m + Mmxs )

231.81x104x3.9

=

.

8048.69 x 108

11.23 x 10

Kxcp . K~x

c:

m

,: ;J :,' ""',I

0)2} My 2

2 { K,n,-(m+ms)O)

-

-6

=

{Kxx -(m+ms)

Acp2 =

2 0) }

} { Kcpcp-(Mm+Mxs)O)

4

,

} -Kxcp K~x

2

- [90.16 x 10 -(102.8+79.7) x 26.2 ]x3.9 8048.69 x 108

Ax

= 3.7

x 10-6 fad

= 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 10-6fad

Displacement of the top of the block

= Ax + (h - L) AqI = 23.63 x 10-6 + (3.5 = 35.70 x 10-6 m

- 1.75) x 6.9 x 10-6

'~; it ~. r, ,.," [, ': ~ 1,'1 fJ 'j

",

J;:

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. (197 I), "Vibrations of embedded symmetric footings", 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 Sta., Tech. Rep. No. 3-632.

of soil bearing capacity, /, '

Foundation

of Structures",

Waterways

Exp.

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. Earthquake Eng. ,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),"StationareAxialymmeterischeduTcheineSchuttelndeMaSseErregteSchwingungeneine Homogenen ,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. Soil Mech. 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 clp proaches 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 subjectel to 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 rot and a moment My sin rot at the COlT bined 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~fi U operating at 500 rpm and mounted symmetrically with respect to foundation. The weight O' ,

~ '

.

.'.

.

.

.

..

.

.

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/m3

G = 4.8 x 104 kN/m3 J.1.

= 0.34 3

17 kN/m 3 Yconc. = 24 kN/m amplitUdes~f the biock by (i)Lin~ar elasticispj . Determinethe namralfrequenciesand -' "~..; Ysoil~

:.approach and (il) B~stic half space theory.

.

-' - . -

.-,

.:'~

o"ca; c: ,

, , ",,_.

421

"Ol~~i:;ptoCtlii,.g ~ac{,,",a

Fila1IliItfdils "

- .,

.

, "

,

,

"T' -,2,(0 1ft.

-0

.0

,'" -

-0

,'-,

"

,,1

.,'

8.0 m

~

1..'Q.5m T-

Elevation

T

It..Om ,.

1

','

Plan Fig. 9..~9: Details of foundation

9.4 Design a suitable foundation for a horizontal compressor driven by an electric motor. The following data are available: , '

/

,

Weight of compressor, = 160 kN Weight of motor = 60 k;N , Speed of compressor = 250 rpm Horizontal 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 constants :

(a) A verticalvibrationtest was conductedon an M-I50 concreteblock 1.5 x 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

I 2 3 4

35.0 ,32.0 .31.0 29.5

5

15 30 45 60 120

6

1~9

I

28.0 -,

-'

.,:27.0

Amplitude at resonance mm 0.06375 0.150 0.210 0.30 0.525 0.620

422.

S,iI DyntUllics &. Machine .

FollndtUWns ,., ,

(b) A cyclic-plate load test was.done on plate 300 mm ~.300 mm. The elastic settlement corresponding 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 as 0.02 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 soil at 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 of impact 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 its frame mounted on the anvil. In Fig. 10.1, a foundation for a hammer with its' frame mounted on the foundation is shown. ~,

-

Fig. 10.1: Typical arrangement ofa hammer foundation with A frame mounted on foundation <

,',

.. ",'~,'f\"

.. '"

'

' },

;'i~,:;~$~~~4i.~,,'i.t!t:;..!,:~j'.;: ", "',7;

';,';

~',)' ,c"

i':(, ,'" ~,..,'

~

~, 0'"

',Lo-'

:

re

.\

424

','

Dy,"""ks&

Soil

'Mllc"inif'

F()Il1I~

The foundation of a hammer generally consists of a reinforced concrete block. 'Q1efollowing arrangements 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 ;

,-

J'~

'9T::

1

,,

Tz1

~

'FOUNDATION, BLOCK.RESTING

ANVIL AND FOUNDATION

ON SOIL

'BLOCK"" '}

'~

(b)

(a)

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 tl foundation block (Fig. lO.3a). Usually the isolation Jayer is an elastic pad consisting of rubbe felt, cork, 01 timber adequately protected against water and oil. In case of high capacity han mers, 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

. . .-

~

FOUNDA1'1 ON BLOCK RESTING, ON SOIL

-,:1 ~f&4 ,"~

(a)

.... t

(b)

Fig. 10.3 : Anvil resting on elastic pad/spring absorbers (...Contd.)

~,

"oundations

425

of Impact Type Machines

dJ~P DAMPING1N ANvrL

l ABSORBER

122

I

SPRINGk2 ELASTIC PAD

FOUNDATION Tzt BLOCK SOIL SPRING kl < ~

<

DAMPING IN SOIL (c)

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 foundation 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 the foundation and side of trough is filled up with some soft 'materials or an air gap is left. The systems shown in Fig. lO.4a and lO.4b can be modeled as three degrees freedom system as shown in Fig. lO.4c. The stiffness of trough (Fig. lO.4a) is very high compared to that of the pad below 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. FOUNDATION BLOCK

FOUNDATION BLOCK

TROUGH

ELASTIc. PAD BELOW FOUNDATION BLOCK

SPRING ABSORBER BELOW FOUNDATION

BLOCK-:: TROUGH

"

.~'

.'~,. . (a),<

..

i

" (b)

"

,!1 ' . , Fig. 10.4 : Foundation block on elastic pad/spring absorbers (...Contd.)

.......

,.

,(

,'r!'

. 426

Soil Dynamics & M.achine Foun.d,,:tions

~TUP I

. ANVIL I

TZ)

I

SPRINGkJ OF PAD

DAMPING IN PAD

-.

FOUNDATION BLOCK

BELOW ANVIL.

TZ2

SPRING k2OF PAD BELOWTHE FOUNDATOO BLOCK

TZ1 SOtL SPRING k1

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 fall on one vertical axis. This will ensure that the loads act on the anvil and foundation without any eccentricity. 10.2 DYNAMIC ANALYSIS 10.2.1. Two Degree Freedom System. In general the anvil, pad, foundation, and soil constitute a tWo degree 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~t ml

= Mass of anvil (with frame if the latter is mounted on the anvil as in Fig. 8.6. ., ":~1$f

K) = C'u . AI = Equivalent spring constant

of soil under. consideration

425

r<'oundations of Impact Type Machines

dJ~P Tz2

DAMPING1N ANvrL

l ABSORBER

I

SPRING k2 ELASTIC PAD

FOUNDATION PI BLOCK SOIL SPRING kl '~ .

DAMPING IN SOIL (c) 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 foundation 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 the foundation and side of trough is filled up with some soft materials or an air gap is left. The systems shown in Fig. lO.4a and lOAb can be modeled as three degrees freedom system as shown in Fig. 10Ac. The stiffness of trough (Fig. lO.4a) is very high compared to that of the pad below 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. TlON

BLOCK

.,..

FOUNDATION BLOCK

'..'

,.

ELASTIc.

PAD BELOW FOUNDATION BLOCK

.:. ,-.' :

'8.

I'

TROUG H

SPRING ABSORBER BELOW FOUNDATION

B LOCK,:~ ""

.,..(a)r', I.. . ,

,.,""

.

.,"-.' ., . .,. . ,. ,,(b) . Fig. 10.4: Foundation block on elastic pad/spring absorbers (...Contd.)

........

,.

-~

ill

427

undations of Impact Type Machines Cu' =

A Cu = Modified coefficient of elastic uniform compression, to t,,;keinto account impact condition which is different from periodic loading

t.. = Multiplying factor that governs the relationship between Cu and Cu' usually 1-2, for impact depending upon the soil type Cu =

Coefficient of elastic uniform compression

-

K2 = (E / b) x Az = Equivalentspring constantof the pad under the anvil

E = Young's modulus of the pad material b = Thickness of the pad

Al A2 z\ z2

= Area of foundation in contact with soil = Area of the pad = Displacement of foundation from the equilibrium position = Displacement of anvil from the equilibrium position

.

TuP--cb

ANVIL SPRING K2 FOUNQATION ,,"

m1

:~, SPRING~1 I

.

Fig. to.5 : Two-mass-spring

BLOCK

analogy for hammer foundation

IO.l.I.I. Natura/frequencies. The equations of motion in free vibration are mlz\+Klzl+Kz(Zt-zz)

...(10.1)

=0

...(10.2)

~ 2Z+ Kz (zz --zl) = 0 The 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) in Eqs. (10.1) and (10.2) respectively, we get

z

.!! = .Kz+ Kt -1nl con A KZ

...(IO.S)

..

428

Soil Dynamics & Machine Foundations.

and,

B

K2

A

K2 -"'2 con

- =

...(10.6)

2

Equating Eqs. (10.5) and (10.6), on simplification we get 4

OOn-

K2 (ml +"'2)

{

ffi:-

(

l+~ ml

KZ+

)( "'2

KI 2 Kl K2 -- 0 +-COn+ ml } ml "'2

ml "'2

Kl

ml+"'2 )

00:+K2x

KI

"'2

ml +"'2

(

1+"'2 =0 ml)

Let,

ffina= Circularnaturalfrequencyof the foundationof the anvilon the pad

.".

"'na

..

~

~~

(J)nl

...(IO.S) = Limiting natural frequency of the foundation and anvil on soil

OOnl

=

~

I

...(10.9)

ml+"'2

"'2 Ilm

= -;;I

.

Substituting the values of OOna'oonland Ilm in Eq. (10.7), we get 4 OOn -(1

2 2 +Ilm) (OOna+ffinl)

2 COn+(1+llm)

2 2 OOnlOOna =

...(10.10 a)

0

r 002

= .!.I ,

nl,2

2

2

+

2l~I+llm)(COna+OOnl)-

. 2

2

{(1+llm)(OOna+COnl)}

2

2

2

-4(1+llm)OO,IiOOna

...(10.10 b)

]

,,',I

The two natural frequencies of the hammer foundation may be determined by solving the above equations,

10.2.1.2. Amplitude of vibration. The generalsolutionof the Eqs. (10.1)and (10.2) is given by z, = AI sin ronlt +A2 cosronlt +AJ sin ron2t +A4 cosron2t ...(10.11)\ Z2

=

BI sin ronl t + B2 cos (0nl t + BJ sin (0n2 t + B4 cos (0n2 t

...(10.12)

where, AI' A2' AJ' A4' BI' Bz, B3' B4 are arbitrary constants. If system is vibrating at frequency oonl'then from Eq. (10.6)

~

,,1

-

A Similarly

when con = con2

~ -

K2 K2 -"'2 CO2 nl

='

(02 na 0)2 2 = al (Say) na - 0)nl

"

...( 10.13arJi")~ .

.

..

~t;;J ;"c .

...

2 (Ona

A - (02na _(O2=a2(Say) n2 ."i.", IJ',J;~4

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

. 2

Cl) nl

2

Z

(

al

- az =

(

Cl)

2

-

Cl)

Z

- Cl)nZ ]

Cl)na

Z

)

nZ Cl)na

Z

na

1 2

2

.

[ Cl)na Cl)n I

or

.

1

z.

.

a I - az = Cl)na

.

2'

...(10.14)

2

- Cl)nl )(Cl)na - Cl)n2 )

.Equation (10.12) can therefore be written as : .~

= al

...(10.1S)

AI sin Cl)nlt +al Az cos Cl)nlt +az AJ smCl)n2 t +az A4 COSCl)nZ t

boundary condition:

(i}At t = 0, ZI = Zz = 0 and It gives

Az + A4 = 0

...(10.16a)

a I Az + az A4 = 0 A Z '= A 4 = 0

...(10.l6b) ...(10.17) , ,

(ii) At t = 0, zl = 0; Zz= Va (Velocity of anvil) Zl

= AI

Cl)nl cos Cl)nl

or

t - Az ronl sin ronl t + AJ ronZ cos ronZt - A4 ronZ sin ronZ t

AI (Onl + AJ (OnZ =

or

0

Cl) A J = - A I- ni

...(10.18)

. Cl) nZ 2'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

= al

or

Va

or

A =

and

A = a 3 (al -a2) Cl)nZ

I

AI (Onl + az AJ (OnZ

Va

...(10.19)

(al -az) Cl)nl

-v I

Therefore, Putting the value of(a,-

Z

. Va

='

.

SIn ronl

Va sinCl)nzt

I (a\-az) Cl)nl (al -az) Cl)nZ az) from Eq. (10.14) in Eq. (10.21), we get

(

Cl)nl-CI)n2

Similarly,

. ...(10.21)

~

ZI = (CI)~a-CI)~I)(CI)~a-CI)~z) ZZ Z

)

Cl)na

sinCl)nlt - sin Cl)n2t V Cl) Cl) ] a .[ nl nZ

(CI)~a-CI)~2)SinCl)nlt

.~

(

2

Z

=. . .Cl)nl-(J)nZ. 1

)[ .

./(10.20) I

Cl) nl

...(10.22)

v

(CI)~a-CI)~I)Sinron2t CI)' nZ

a

] .,'

. ...(10.23)

Field observation (Barkan, 1962) of the amplitudes of the anvil and the foundation showed that the ibrations occured at the lower frequency only. Therefore, it may be assumed that the amplitude of moion for sin (Onlt= 0 «Onl > (OnZ)'

430

Soil Dynamics & Machine Follndations

Hence approximate expressions for maximum displacement will be as follows (sin Cl)n2t = I):

ZI

=

z2

=

(OO;a-00;1) (OO;a-00;2) . Va 2 2 2 OOnaOOnl-OOn2 OOn2

)

(

" .. r:

...(10.24}

(OO;a-00;1) Va 2

...(1 0.25}

2

(OOnl -OOn2 )OOn2

.

10.2.2. Single Degree Freedom System. Sometimes in the case of light hammers, no pad is used between 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 equilibrium position m = Total vibrating mass, K =C' u A 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 COn t + B co~ Cl)n t The constants A and B, as usual, are dete~ined from initial conditions of motion. z = 0'a and z = V' ., At t = 0

...(l~t.,

,

'JH"ir ; ,i2

. . Foundations of rmpllCt Type Machines

~

431

"

Using these initial conditions., we get Va A = - and E=-O Q)n

Va

A =-

Therefore,

Z

.

.

sm 0> t

Q)n

...(10.29)

n

The maximum displacement will be Va

(Az)max

10.2.3. Determination

...(10.30)

= IDn

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

...(10.31)

= 1'\ ~2 g h

where, h = Drop of tuP in meters g = Acceleration due to gravity, mls2 1'\= 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)

...(10.32)

.

Wl

where

W1

= 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:

.

W

Momentum of tup and anvil before impact = ---1 VTi . g and Wl

...(10.33 a)

W2

Momentum of tup and anvil after impact = - VTa + Va g g where W2= weight of anvil (plus frame if it mounted on the anvil) .

Va' = Velocity of anvil after impact

VTa ='Velocity oftup after impact Therefore,

----

Wl v. - W, W2 VTa + Va T' -

g'g.

g

L

432

SoU Dy,,1IIIIics

~ Machine. FoundatiDns

According to Newton's law, the coefficient of elastic restitution, e, is given by Relative velocity after impact e = Relative-.velocity before impact Va - VTa

e=

or

...(10.35)

VTi

The value of e depends upon the material of the bodies involved in'iinpact. Theoretically value' of e lies between 0 and 1. In forge hammer, usually the value of e does not exceed 0.5 (Barkan, 1962). Since a larger e gives larger amplitudes of motion, the value of e equal to 0.5 is adopted in designing hammer foundation. On solving Eqs. (10,34) and (10.35) we get. ' l+e 'Y ' V ...(10.36) W TI a

1+-1. WI

10.2.4. Stress in the Pad. Maximum compressive stress in the elastic pad below the anvil depends upon the relative displacements of anvil and the foundation block. The worst case of compression in the pad developes when the anvil moves downward, and at same instant of time, the foundation block moves upward. The maximum compressive stress in the pad is thus expressed by zl + z2 (Jp

= K2

A' 2

. , (zl' z2 In absolute values)

...(10.3?)

10.2.S. Stresses in the Soil. Stresses transmitted to the soil q through the combined static dynamic loads are expressed by q = 10.3 DESIGN PROCEDURE

Wt +W2 +zl Kt

...(10.38)

Al

FOR A HAMMER FOUNDATION

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 of piston; . (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.Soil Data. The followinginformationabout the sub-surfacesoil shouldbe known:

:

(a) Soil profile: For drop hammers of up to 10 kN Capacity, soil investigations should generally

be

done 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

~~-

433

'oundations of Impact Type Machines

(b) Soil investigation to ascertain allowable soil pressure and to determine the dynamic properties of 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. (a) Weight and area: The weight of the foundation for a hammer and the size of its area in contact with 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 ...(10.38 a)

PsI ~ a qa , and Nhere,

PSI

a = Reduction'factor

(= 0.8)

q a = Allowable soil pressure Az

...(10.38 b)

Az < Ap = Static pressure intensity

-

= Amplitudeof motion

Ap = Permissible value of amplitude For Eq. (10.38a) W

/'.

..(10.39) 'Considering an average value of Ap as I mm. = (10-3 m), and assuming the system as single degree freedom system, the Eq. (10.38b) can be written as . A . ~ a qa

(I +e) W VI 0

~C' u W A .g

where,

' I

3

...(IOAO)

< IO-

Wo = Weight oftup, kN W = Weight of foundation, anvil and frame, kN

Al = Base area of foundation in contact with soil, m2 VIi = Initial velocity of tup, mls g = Acceleration due to gravity, mls2 C' 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 A. ~

g ~Cuaqa '

32 x 10 m ." ':

...(10.41)

',<,'

'""",

'"

;i~< ,~~df .'1t~. "'r;l'

434


Foundatio~~-"

Substituting the value of AI fromEq. (10.39) into (10.40), Weget (1+e) Wo VIi

W = Let where,

~ "C~ g

~

"

-3

x 10 kN

...(10.42)

'I,:

r.;~

W=W\+W2 W \ = Weight of foundation

l/

= Weight of anvil

W2

't

~,; Q.'i~

The Eq. (10.42) can be written as WI

(l+e) VIi

Vi" =

~

0

"Cu

~ g

-3 W2 x 10 -Vi" 0

...(10.43)

From, Eq. (10.42), total weight of the anvil and foundation can be obtained. Knowing the weight ot anvil, 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 in punching shear and bending. For the calculations the inertia forces developed shall also be included. However, the following minimum thickness of foundation block below the anvil sha,If be provided: ~ Mass of Tup kN

Thickness (Depth) of foundation Block, Min (m)

Up to 10

1.0

10 to 20

1.25

20 to 40

1.75

40 to 60

2.25

Over 60

'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'. 4Gr .~. C = -1 0 ...(10.44,) u -~ G = Shear modulus where, ~.

.<

..

r = Equivalentradius = 0

J.1

J AnI

= Poisson' s ratio

10.3.5, Natural frequencies, Compute ,

O>na

1& = v-;;;;

, I!'

.

,I

'

D

435

(}undations. 0/ Impact Type Machines

and

oon/

~ V~

K2

= b . A2

"

E in w~ich,

,

'

,

E = Yourig's'modulus of pad material b = ThicIaiess of the pad A2 = Area of the pad K 1 =C' u A=A.C u A "

,0

..'

,",",

'

"-.,,,

Natural frequencies of the combined system are given by: 2

'"

1

2 (()nl,2

= '2

2

[ (l+~m)

2

2

[

) (OOna+OOnl:t

2

(OOna+OOnl)]

(l+~m)

"

2

2

(rona,wnl )]

-4(1+~m)

10.3.6. Velocity of Dropping Parts to Anvil. Compute the velocity VTiof the tup before impact VTi

11 ~2g

~

(WIW + pA)h

.-

in which, W 1 = Gross weight of dropping parts p = Steam or air pressure .. "

,.

Ac = Area of the piston h = Drop of the tup 11 = Efficiency of drop, usually 0.65 Compute the velocity of the anvil Vaafter impact by l+e ' Va =. W , VT I

1+ ---1Wl

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 anvil amplitudes with following equations

(ro~a Z1

=

- (O~I) ( (O~a-

2 OOna

(Onl

2,."-

0.,

-

(

(co

2

na,

2

- OOn2 )OOn2 2

CO 'I

n

)

'

V

Z2 ~ (CO~I: ro~2)cDn2I),;, where oon2 is the smaller natural frequency.

...

00~2)

""0"""""" . .. ' '~~-~'-_...

VI)

436

SoU Dynamics et Machine Foundations

10.3.8. Dynamic Stress in Pad (jp' Compute dynamic stress in ~e pad by (J P

= Kz(zJ+zz) Az

Computed values of natural frequencies should satisfy the criteria for the frequency of operation of the 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. 5

!ILLUSTRATIVE EXAMPLES'

Example 10.1 A 15 kN forging hammer is proposed to install in an industrial Complex. The hammer has the following specifications: Weight of tup without die

= 11 kN

Maximum tup stroke

Supply steam pressure

= 800 mm = 40 kN = 0.12 mz = 600.kN/mz

Weight of anvil block

= 300 kN

Total weight of hammer

= 400 kN = 1.8m x 1.8m = 1.5 mm = 1.0 mm

Weight of the upper half of the die Area of piston

Bearingarea of anvil Permissible vibration amplitude for anvil Permissible amplitude for foundation

It 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 105 kN/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 f proposed depth of fo~ndation. The data obtained are given below: S. No.

8 (Deg)

/"Z (Nz)

Amplitude at resonance (mic1'f!'Y

I. 2.

36 72

41 40

13 24

3.

108

34

32'~

4.

144

31

40

.'

.~

iJ

t .,

I

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;

437

Ff!undation';-of Impact'type Machines ,

Solution:

,

(i) Trial dimensions of foundation.. Let the weight of the block is kept about 5 times the weight of anvil.' The details of the suggeste~ foundation are shown in Fig. 10.7.

.

Weight of foundation

= 24

x 7 x 5 x2

= 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

CII =

41t2

2

'

,

x 19.9 x fnz =71.1 Afn~ m -- 41t1.125 x 9.81

fn~ kN / m3

. Amplitude at resonance Stram level = 'Width of block Putting the given values of fnz and amplitude at resonance, CIIvalues and corresponding strain levels are 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.

11.95

0.173

2.96

2.

11.38

0.320

2.82

3.

8.22

0.427

2.04

4.

6.83

0.533'

1.69

Correction for confining pressure and area:

The mean effective confining pressure croI at a depth orone-half of the widthof block is givenby 0'01

= cr v-(1+2K

3

-where,

O'v

crvl

) 0

= crvl + crv2

= Effective overburden pressure at the depth under consideration cc

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..e upto water table is saturated, then "

-

0.70.

crvI = 18 x 2.0.+ 20.2-

;--

4q

"

= 43 kN/m2

2mn~m2+n2+1

.m2+n2+2

v2 - -41t m2 +n 2 +1+m 2 n 2 m2 +n 2 +1 + srn

[

-L m = -2 z

-B n ""',"2

-1.5 =- 2

0..70.

,'-".

. -} 2mn~m2+n2+1

m2 +n 2 +1+m 2 n 2

]

= 2.14 ., ,.!

2 0..75 -

.

= .£.z = ,0..70. -L = 1.0.7'

'Cl ::..

'

q ='24 x 0..70.= 16.8 teN/ni2, '

"

'",'

,,',

. . 0'

. , ",:, '..~

,.!,'..,'"

-,~."~

439

Ilndations of Impact Type M~~hines

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

-

= 56.44 [

aol

3

2

] = 34.84 kN/m

For the actual foundation av!

= 18 x 2.0 + 20 x 1.0 + (20 - 10) x 1.5 = 71 kN/m2

For the actual foundation 7.0 2 m = 5.0 = 1.4 2 5.0 - 2 _ 10 n - 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 cry

= 37.2 kN/m2 = 71.0 + 37.2 = 108.2 kN/m2

G02 = 108.2[1+2 x30.4266] = 66.8 kN/m2 0.5 0' 02

Cu2 = CuI

[

0'01

]

= 66.80 [ 34.84 ]

~

0.5

.

[ A2 ] °.5

1.5x 0.75

[ 7.0 x 5.0 ]

°.5 -

- 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 4 Stain level in actual foundation = 5.0 x 1000 = 2 x 10The strain level in actual foundation is higher than the strain level observed in the tests. Seeing the variation of Cu with respect to strain level, the value of Cu equal to 1.3 x 104 kN/m3 may be adopted in design. .

\ . --,

440

',.' "'t~,

Soil DYllamics & Machine Foulldaii~~

(iii) Computations of KI' K2' ml and m2

C' = A' C 11

11

= 2.0 x 1.3 x 104 = 2.6 x 104kN/m3(A = 2.0, Assumed) K I = C'!I ,A I = 2.6 x 104x 7 x 5 = 91 x 104kN/m E

K, = -b .A 2 = ~

5 x 105 1.8x 1.8 = 324 0.5

x

104kN/m

Weight of the foundation block = (7 x 5 x 2 x 1.8 x 1.8 x 0.8) x 24 = 1618 kN 1618 = 165 x 103 K g 9.81

m =I

Weight of the anvil and frame = 400 kN 400 m2

=

'

9.81 =40.8 x 10"'Kg

(iv) Natural frequencies of soil-foundation system (J)1/1 =

~

l

324 x 10 = J79411 = 281 rad/s 40.8

= v-;;;;

11

=

= .J4421 = 66.4 rad/s

4

r&= (J)/la

91 x 104

-

- ~ nIl +"'2 '165+40.8

~ = 40.8 = 0 247 1/l\

165

.

Natural frequencies are given by : 2

,

.

1/2

(J.)~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)] (J) /11.2

::t

~{[(1+ 0.247)(79411

+ 4421)]2 -4 (1 + 0.247)79411

x 4421}

= 316 rad/s, 66 rad/s

(v) Velocity of dropping parts

VTi

~

~ ,{W, ~.A,

11 = 0.70 (assumed)

)2gh ; W\

=

15 kN, p = 600 kN/m2, Ac = 0.12 m2, h = 0.8 "

J

..;1

-\I~ VTi

"

= 0.70 ,iC5+ 60105 x 0.12) x 2 x9.81 x 0.8 = 6.68 mls .~ 'h4 .:;i ~I

441

rtions of Impact Type Machines

l+e

.

Va = .

-

w' VTi 1+-1.. W1

1+ 0.5 .6 68 400' 1+15

(e = 0.5, assumed)

= 0.362 m/s i) Amplitudes of vibration ZI

=

(ro:a-ro:l) (ro:a-ro:2) 2

2

2

(rona)(ronl -ron2 )ron2

.Vu

(79411-100167) (79411-4371) = (79411)(100167-4371)(4371) .(0.362)

= - 1.695 x 10-4 m = - 0.1695 mm «1.0 mm, safe) Z2 =

(ro:a-ro:l)

2 -ro\n (ronl

,Va

) ro/12

(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

=

A2

. ' (zl' z2 In absolute values)

= 324 X 104 (0.1695+0.0179)

10-3

1.8x 1.8

= 187.4 kN/m2

(\

[< 350 kN/m2 , safe]

442

Soil Dynamics & Machine Fouml'!!i!:l

ÎÛÚÛÎÛÒÝÛÍ 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 minimist the vibrations. 10.2 Consideringa two-degree-freedommodel, derivethe expressionof amplitudesof anviland foundatio of a hammer. 10.3 A 20 kN forging hammer is proposed to install in an industrial Complex. The hammer has th following specifications: Weight oftup without die Maximum tup stroke Weight of the upper half of the die Area of piston Supply steam pressure Weight of anvil block Total weight of anvil and frame

= 12 kN = 900 mm = 50 kN = 0.15 m2 = 700 kN/m = 400 kN = 500 kN = 2.1 m x 2.1 m

Bearingarea of anvil

Permissible vibration amplitude for anvil = 1.0 mm Permissible 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 elasticity pad 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:

S.No.

e (Deg)

f nz(Hz)

1. 2. 3. 4.

36 72 108 144

40 38 35 29

Amplitude at resonance (micron 14 26 33 41

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.

FOUNDATIONS OF ROTARY MACHINES

.1 GENERAL le unprecedented burden cast by the importance of oil can be relieved only by exploiting indigenous lergy 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 country le to various reasons including the substantial substitution among the forms of energy in the various lportant sectors along with the accelerated programme in rural electrification and assured power supy 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, the ermal 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 capacity t'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 Super hermal 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. The perating speeds of trubogenerators may range from 3000 rpm to 10000 rpm. Auxiliary equipments such s condensers, heat exchanger, pipe lines, air vents and ducts for electric wiring are essential features of turbogenerator installation. Frame foundations are commonly used for turbogenerators with four :asons : (i) (ii) (ui) (iv)

auxiallary equipment can be arranged more conveniently, the inspection of and access to all parts of the machine become more convenient, less liable to cracking due to settlement and temperature changes, and 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. The sometric 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 their r'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 adjustment 10the layout of machinery and auxiliary fittings.

...

~.;;;t.

n!l""cU~

/

444

Soil Dynamics & Machine Foundatiol'

""'>'

Longitudinal beam

Top deck ~

"

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 inie1fi

significantlywith each other.

'

\wa:

~

445

lundations of Rotary Machines

(iii) All the junctions of beams and columns of the foundation should be provided with adequate haunches 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 less than 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 the foundation 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 heavy enough 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 and increases with the power of the machine to a maximum of 4m. Its weight should not be less than the 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 slab should 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 of sub grade effect is essential. (xii) As far as possible the foundation should be dimensioned such that the centre of gravity of the foundation with the machine should be in vertical alignment with that of the base area in contact with. the soil. . --(xiii) The ground-water table should be as low as possible and deeper by at least one-fourth of the width of foundation below the.base plane. This limits the vibrations propagation, ground-water being 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 preferably be at a variance of at least 30 percent from the operating speed of the machine as well as critical speeds of the rotor. Thus resommce is avoided. An uncertainty of 10 to 20 percent may be assumed in the computed natural frequencies. However, it may not be necessary to avoid resonance 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 amplitudes 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 the chine. 11.4.2. Operation Loads (OL). These loads are supplied by the manufacturer of the machine anc elude frictional forces, power torque, thermal elongation forces, vaccum in the condenser, piping fo etc. The load due to vaccum in condenser, ifnot supplied by the manufacturer, can be obtained fron following equation: ...(1

Pc = A (pa - Pc) where.

Pc = Condenser vaccum load A = Cross-sectional area of the connecting tie between the condenser and turbine

Pa = Atmosphericpressure Pc = Vac cum pressure The value of (p a - Pc) may be taken as 100 kN/m2.

TA

TR A

-8-

Shaf-

t] 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 - 105 PA kN TA N m - 105(PB - PA) kN

TB -

N

...( 1

m

- 105(Pc - PB) kN Tc N m 105P T = c kNm g

where,

TA

= Torque

N

due to high-pressure (H. P.) turbine in kNm

TB = Torque due to intermediate-pressure

(LP.) turbine in kNm

...( 1 ...( I ...(1

-447

Foundations of Rotary Machines

Tc = Torque due to low-pressure (L. P.) turbine in kNm Tg= Torque due to generator in kNm PA' 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 are balanced before erection. However, in actual operation some unbalance always exists. The unbalance is specified as the distance between the axis of the shaft and mass centre of gravity of rotor, and is known as effective 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 at any time cancel out, but there is a resulting moment M given by 2 M=mero./ ...(11.3) e where,

/ = 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 2 MH =meero /cosrot

...(11.4 a) ...(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 by 2 F = 2 me e ro

...(11.5)

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.

.

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 200 may 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 differenti temperature. To account for the shrinkage of the upper slab relative to the-base slab, a temperature fall of 100C 150C may be assumed.. . . 11.4.5. Short circuit forces (SCF). Short circuit condition imposes moment on the turbogenerat foundation. A fault of this type occur when any two of the three generator phase terminals are shorte The 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 kNm Where,

...( 11.

W,. = Capacity ofT. G. Unit in MW r = Radius of the rotor in m

11.4.6. Loss of blade unbalance (LBL) or bearing failure load (BFL). One of the buckets or blades the turbine rotor may break during the operation of turbo-generator unit. It will increase the unbalanc force. This additional unbalanced forced wi1l depend on the weight of the bucket,. the distance of centre 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

where,

Fs

= ah

I

~C

S W ...(11.7)

= Horizontal seismic force

ah = Seismic zone 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 dynamic loadswhich occur only dur the operation of the machine. The construction loads are generally taken as uniformly distributed l( varying from 10 kN/m2 to 30 kN/m2depending on the size ofT. G. unit. . .

449

r.°un~a~io.nso! R~tary M~chines

The design of a T. G. unit should be checked for the following load combinations: (a) Operati?n,: ~~ndition DL+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 and also to design the members of frame from structural considerations. The methods for carrying out dynamic analysis 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 deformation

of 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 the column head, even in case where the transverse beam is eccentrically placed with respect to the centre line of the column. (vi) Both the columns and beams can be replaced by weightless elements with the masses lumped at a 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 own

plane.

.

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

"--'--'..

-"

.-

450


ïïòêÎÛÍÑÒßÒÝÛ

METHOD

In this method, the frame foundation is idealized as a single-degree freedom system, and consideration is given only to natural frequencies of the system in relation to the operating speed of the machine. The amplitudes of vibration are not computed in this method. 11.6.1. Vertical Frequency. For obtaining vertical frequency, each transverse frame that consists of two columns and a beam perpendicular to main shaft of the machine, is considered separately (Fig. IlAa). Fz sin cut Fz sin c..>t

t

Wz

Wz

Wl ID

I.-

lo

I

~

~

-2a

I

ILl

I

1

I

lr

m= ql j-W,+2' 9

I

h, ho

Zb

,-

Column

assumed fixed

Gekrmns

.>71

1- - -7 /"

J~

!

- 1I

Y- - - - - - - - - - -_J Base slab

(b)

I

L

(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 (iii) Uniformly (Iv) Unbalanced

to the columns by longitudinal

distributed

beams, W 2

load due to self weight of cross beam, q per unit length

vertical force due to machine operation,

Fz sin w t

451

Foundations of Rotary Machines

The frame is modelled as mass-spring system as shown in Fig. II.4b. The stiffness of equivalent spring (K) is computed as the combined stiffness of the beam and columns acting together. It is given by W K =...( 11.8) ::

851

where,

W = Total load on the frame

or

W

= W\

+ 2W2 + q ./...(11.9)

1 = Effective °51

span

= Total vertical deflection at the centre of the beam due to bending action of beam and axial compression

in columns.

o

= °1

51

where,

...(11.10)

+ °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 shear 84 = Axial compression in column

The magnitudes of 81' 82, °3 and 84 can be obtained using following expressions: Wj ,3 2 K + 1 81 = 96 E Ib K + 2

...(11.11)

q 14 5K + 2 °2 = 384 Elb' K+2

...(11.12)

3

1

ql

83 ="5 E Ab ( WI+2 84

=

~

EA e

(

...(11.13)

)

W + WI +ql 2 2

)

Ih h K =-.le 1

where,

...(11.14)

...(11.15)

Ab = Cross-sectional area of beam Ac = Cross-sectional area of column Ib

= Moment of inertia of beam about the axis of bending

le = Moment of inertia of column E = Young's modulus of concrete K = Relative stiffness factor 1 = Effective span of frame h = Effective height of frame Values of 1and h are obtained as below: '=1

0

-2ab

h =h 0 -2aa

...(11.16) ...(11.17)

452

Soil Dynamics & Machine Foundatiolls

where,

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 the distance 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 distance as 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<

®óóó I I

I I

0.20 0

0.10

-1 b~

00

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)

...(11.18)

nz = ~KzW. g

Average vertical natural frequency of the T. G. Foundation is taken as: 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

453

of Rotary Machines

The average value of vertical amplitude of T. G. foundation may be computed as IF:

A:a

(I Ko),'ll-(

...(11.20)

oo:Jf+( ~:o:)'

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 For under-tuned

ratio

foundation,

i.e. (() < (() lI~a ~, (()n = (()n~a should be used in Eq. (11.20). Then

2::F:

A za

...(11.20 a)

(2::KJ (2~)

11.6.2. Horizontal Vibrations. In a T. G. frame formulation, the deck slab undergoes horizontal vibration in the direction perpendicular to the main shaft of the machine. The spring stiffness is provided by the columns due to their bending action, and for any transverse frame it is given by

6K+l

K = 12E le x where,

If

113 ( 3K+2

...(11.21)

)

Kt = 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 machine Then the natural frequency of the T. G. frame foundation is given by

"'"xa

~

...(11.22)

p:~~)g

The average horizontal amplitude of the foundation may be computed as follows: A

L~t

=

...( 11.23) 2

'" For under-tuned

foundation,

(LK,),/ I- ( ~ nxa ) 2 + ~nxa J [ ] ( Cl)

2

Cl)

i.e. (() < (()n.m'

(J)11= (J)I/.HIshould be used in Eq. (11.23).

Thus

LFt

ß©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 natural frequencies computed from Eqs. (11.19) and ( 11.22) should differ by atleast 30 percent from the ope rati:1g speed of the machine. The Eqs. (11.20), (11.20a) and (11.23) for determining amplitudes are gi\'en to be used further in combined method. Resonance method based on idealising each transverse frame to single mass-spring system is an oversimplification of a complex problem. Therefore the values of natural frequencies computed by this method are very approximate. 11.7 AMPLITUDE

METHOD

In this method also, the vibration analysis is carried out for each transverse frame independently. However, the frame has been idealised as a two-degree-freedom system (Fig. 11.7). The main criterion for design is that the amplitudes due to forced vibrations are within permissible limits (Barkan, 1962).

Z2

t Fz sin

1-

wt

m,

\ m, /1

TI K,

z

\

-----"::::::;-mz

KZ

\ K, 1-

\

I

\ \ \

mz

12

c-tzz

I

I I

columns

:.r::

:"l

Z1

z

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 Fig. 11.7 b is adopted. Mass m I lumped over the columns is given by

= WI + W2+0.33W3 +0.25W4

m I

g

system shown in

...( 11.24)

,\ ,ii

IIIJ

455

iations of Rotary Machines

Aass mz acting at the centre of the cross beam is given by m = Z e.

Wz + 0.45W4 g

W 1 = Dead load of the machine and bearing W 2 = Load transferred to be columns by longitudinal beams W 3 = Weight of two columns constituting the transverse frame W4 = Weight of the transverse beam

The stiffness Kl of both the columns of a transverse frame is given by 2EAc KI = h The stiffness Kz of the frame beam is given by 1 K =-

z

-re,

/ (1+2K)

.

...(11.26)

...(11.27)

°SI

3/

...(11.28)

= 96 E rb (2 + K) + 8G Ab G = Shear modulus of beam material

OS!

E = Young's modulus of the material of columns Ac

= Cross-sectional area of a column

h = Effective height of the column I = Effective span of the beam AiJ

= Cross-sectional

area of the beam

rb = 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: ...( 11.29)

ml ZI + KI ZI - Kz (Zz - ZI) = 0

...( 11.30)

mz Zz + Kz (Zz - ZI) = 0 The solution of above equations are: Z \ = A I sin 0)III Zz =A z sinO) III

...( 11.31 )

...(11.32)

Substituting Eqs. (11.31) and (11.32) into eqs. (11.29) and (11.30), on simplification. we get 4 (I)/I-(I+~l)

.z

l

2

(1)/111-0)12

zz

)+(I+P)(J)IlIIO)IlIZ

...( 11.3 3)

=0

Kl

here,

0) III 1 -

~ 0) Il 1Z

...( 11.35)

= V--;;;;

Ilm =

.u-_.'

...( 11.34)

/Ill + /Il2

nlz ml

...( 11.36)

~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 vibration. the equations of motion will be: /1/] ZI+KI

...(11.37)

ZI-K2(Z2-ZI)=0

= F;

1112Z2 + K2 (Z2 - ZI) The solution

Substituting

of the above equations

Eqs. (11.39) All

AZ2

can be presented

...( 11.39)

Z2 = An

.. .(11.40)

and (11.40)

sin OOt

into Eqs. (11.37)

and (11.38), 2 00,,12. F;

=

=

as

ZI = AZI sin OO{

1111

and

...(11.38)

sin OO{

[

(

- (1 + ~ ,J

004

00 ~ 1 I + CO~ 12

) CO2 +

...(11.41)

(1 + ~ Ill) 00 ~ I 1 CO ~ I 2

2 2 [ (l+~IIl)OOllll+~IIlOOIl12-CO

[

and then solving them we get

]

2

] .F; 2 2 2 22 OOIlII+OOIl12 00 +(I+~/Il)OOIlIIOOIl12

-+

)

(

1Il2 00 -(1+~m)

]

...(11.42)

11.7.2. Horizontal vibration. For analysing the frame foundation in horizontal vibration as t\VOdegree-freedom problem, the upper and lower foundation slabs are assumed to be infinitely rigid. The columns are taken to act as leaf-springs. The stiffness of a leaf spring is considered equal to the lateral stiffness of the individual transverse frame. dl
~

IQ

I

r

j

d k3

d k1

dm1

11

I

II I

I

d

It

dm3~

m2 /. .

.

C
KX2

T X

G2

A1

01

m1

.

-L A3~--

- ---

AZ

Due to horizontal displacement Fi~. 11.1' : Sprin~-l1Ia"

81 m3

m2

l!1odl'l for combincd

-----

T

\ 1

Due to translation and rotation horizontal

and rotational

"'1'

... 8 z

óóóùè 3

vibrations

of the deck slab

-

457

Idations of Rotary Machines

Figure 11.8 shows a typical mathematical model for a two bays frame foundation. The equivalent s ( mi) lumped over the spring i (representing frame i) is given by: m, = m ,+ m b ' + 0.33 m . + m ' ...(llAJ) I ml I Cl gl

re.

1I111/i

IIIbi

= Mass of machine resting on cross-beam of ith frame

= Mass of cross-beam of i'h frame

lIIei = Mass of columns of i'h frame l1lu",I

= Mass transfered from longitudinal girders on either side

Using Eq. (11.43), values of ml' 1112and 1".13 can 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 of fness (i.e. Kxl' K,2 and Kd) respectively. Line A1Bl shows the initial position of deck slab. The final placed position of the deck slab is represented by the line A3B3 . The deck slab rotates about the mass ltre 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 11/I ' d mL and dm 3 . e rep resents the distance between G I and G L ~

~

.

The equations of motion for the system shown in Fig. 11.8 will be: (Iml).1+ (I/1/i 1ere,

dl~lI)

\it +

I[Kti

I[ K'i

...(11.44)

(x+dmi 'V)] = F, sin cD(

...(IIA5)

(x + d,lli'V) dllli] = M. sin cD(

Fx = Horizontal unbalanced force ivL = Unbalanced moment

Denoting lI/

= 'Ill/,I = Total mass

...(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 ...( 11.48)

= x Kx + Kx . e 'V = Kx (x + e 'V) K, represents the total lateral stiffness.

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 )

= K, (x+e\jl) e+ KIjI. \jI

\jf

...( 11.49)

11

458

Soil Dynamics & Machine Foundations K", represents the equivalent torsional spring stiffness for the frame columns and is given by ., ...(11.50) K", = IK.d dki

It may be noted that IK

XI .

dk I = 0 '

Making the substitutions from Eqs, (11.46) to (11.50) in Eqs. (11.44) and 911,45), we get m .~ + K .\,\ + K X . e \jI = F .r sin W{

= Mz sin

Mmz\jl + K\ex + (K, e2 + KII')\jI

..,(11.51) ...( 11.52)

W{

Equations (11.51) and (11.52) are similar to the Eqs. (9.51) and (9.52). Proceeding exactly in the same way as discussed earlier, the solutions of Eqs. (11.51) and (11.52) can be obtained from the following equation: 4 CDn -

where

( a

22 CDnx

+ CD/I IjI

)

222

CDn + CDnx

...(11.53)

. CD nljl = 0

fK. Wnx= v-; W nII'

=

...(11.54)

~

...(11.55)

e r

...(11.56)

II'

Mm= 2

a=l+Z

,..(11.57)

m r = ~Mm= The amplitude of vibration in translation and rotation are given by e22

-

A = [

2

+

,2 CD/IX ]

2 - CD F\ ~

CD1lII'

x

]

Mc - CD2 . --=---

m

n\

Mm=

...( 11.58)

L1(CD2) 2

e

2

F X

?CD/I\r. m

-

2

-

2

CD

(conx

M~

)---=M IIIz ...( 11.59)

AIjI

where

2

=

( ) = (J)

L\ co

L1(co2)

4

(

2

2

2

2

2

- a (J)n x + co n 'v ) CD + co n x con IV

...( 11.60)

The net amplitude Air is given by \vhere,

...(11.61) Air = At + y AIjI v '= Distance of the point at which the amplitude is being calculated from the centre 01 gravity of the system

..,

Juniations of Rotary Machines 1.8 COMBINED 1

459

METHOD

fact, the resonance and amplitude methods are complimentary. Since the amplitude method is based

n a system of two degree of freedom, it is obviously an improvement over the resonance method. Howvel', the. fact that in under tuned foundation the increase in amplitude of vibration during acceleration nd deacceleration stages has been ignored in this method. In combined method which is also known as xtended resonance method, the possibilities of resonance and excessive amplitudes both during steady ibration 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 natural frequency 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 permissible limits. (iii) Amplitudes of vibration computed from Eqs. (11.20 a) and (11.23 a) should also be compared with permissible amplitudes to take care the possibility of excessive amplitudes during acceleration and deacceleration stages. A suitable value of damping ratio may be adopted to use in these equations.

11.9 THREE DIMENSIONAL ANALYSIS For turbogenerator foundations of more than 100 MW capacity, a three-dimensional space frame model is 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 excitation. Nodes are specified to all bearing points, beam-column junctions, mid-points and quarter points of beams and columns and where the rhember cross- sections change significantly. Generally "thenumber of nodes specified on any member should be sufficient to calculate all the modes having frequencies less than or equal to the operating speed. Lumped-mass approach is used having lumped masses at the node points. The machine shall be modelled to lump its mass together with the mass of the foundation. Equivalent sectional properties of beams and columns are used. The computation of equivalent mass moment of inertia of the frame members pose some difficulty since these depends upon the deflection shape in each mode. These may be discretised in the first step and considered data in an iterative manner if desired. The columns may be assumed 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 into a skeleton system which retains the properties of the original structure. The stiffness matrix of the structure as a whole is assembled from the stiffness matrices of individual members. The resulting equations are 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

ÎÛÚÛÎÛÒÝÛÍ 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,"

VDI

VcrlJg. 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 tions.

vibra-

(ii) Machine foundation suffers excessive amplitudes due to the vibrations transmitted from the neighbouring machines (Fig. 12.1 b).

.~.

Machine

!F

=

Fo s;n

Machine

"'.

Z=ZoSin(Jt Foundation

(0)

Foundation

Excessive vibrations .. du
(b)

Excessive amplitu de due to vibrations transmitted from 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 suitably designed mounting system (Fig. 12.2) such that the transmitted force is reduced which in turn will reduce the amplitude. This type of isolation is termed as force isolation. This type of arrangement will also 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 mounting system 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.

Machi ne

.,

'I

Isolator

Foundation

=

Fig. 12.2: An isolator placed between machine and foundation

Machine

Founda'tion

block

Isolator

'i/i;

å óº

¢

ßÁòþ ùþ

'A.

A

.' 6

I" .~

'.

ôßù

t::>"-.l:i'.~~'~-'O""'I\'"

. -,

Concrete A.-.,

/},':

." slab'."

.:..s...".'.;::':

. ' .'

. -..'to.:".'

=

::. .~:~:..~

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 witlspring constant K and dashpot with damping constant C. This mathematical representation involves 0 basic assumption that the underlying soil or rock possess infinite rigidity. This system is identical to t ane 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

463

(llld Screening

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. This will make the system as a two- degree-freedom problem. the solutions of which are presented in the next section. The second problem in which the vibrations are transmitted from the neighbouring machines can be solved by controlling the vibrating energy reaching the desired location. This is referred to as Vibration screening. Effective screening of vibration may be achieved by proper interception. scattering. and diffraction of surface waves by using barriers such as trenches. sheet-pile walls, and piles. If the screening devices are provided near the source of vibration, then it is termed as active screening or active isolation. In case screening devIces are used by providing barriers at a point remote from the source of disturbance but near a site where vibration has to be reduced, it is termed as passive screening or passive isolation. 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 infinite rigidity, the foundation soil should be represented by a spring and not solely by a rigid sllpport as W:iS done in Sec. 8.2. Then the mathematical model becomes as shown in Fig. 12.5. The various ~termsused are explall1ed below: III I

= Mass of foundation block or mass of base slab

11l1 =

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 foundation block and base slab.

K I = Stiffness of the soil K1 = Stiffness of the isolator

'

111

~," .'t_~

..64

Soil Dynar.lics & Mac/line Folllltlatiol/s

T Zz

tF=FOSinGJt

T Z1

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

11/222 + K2 (22

The solution ofEqs.

-

...( 12.1 )

= Fa

21)

sin ((){

...( 12.2)

(J2.1) and (12.2) gives

(~

nIl n~ )Sinrol

21 =

.

W4 -

K2 + (Kl + K2)

[ 11~

]

1nl

(KI+K~) 2

[

z-

-0)-

~

1111n~

0)4_ Kz+(Kl+KZ) [~ 1nl

.,

. Fa

...(12.3)

w2 + Kl K2

~

nIl nIl

. S111 0) (

]

.F

0)2+KlK2

]

...(12.4) a

nIl 1112

The principal natural frequencies of the system shown in Fig. 12.5 can be obtained by solving the following frequency equation: .; c!) I;

I K~ -

If

L III~ 4(,r) /1

I and

2

(Kl + K~) 1111

-

]

K, Kz

W 1/ + Ill, 1112

CD1/-~ re p resent

...( 12.5)

=0

the roots of the above e quation, Eq . (12.3) can be written as K~ ~

2~'

l IIII III~- )

I

where

tl(0)2)

.F

tl (0/ )

2 4 Kz tl(w ) - (J) - -+ [~

or

. S1l1 0) I

= (0)2

...( 12.6)

0

(KI+Kz)

-0)~1)(0)2

1nl

-0)~2)

(J)2 +-KIKz

]

...(12.7

ll)

1nl n~

...(12.7 b)

IiJ

465

Vibration Isolation and Screening

Force transferred to the foundation block or base slab F{ -= K} Zj Kj K2 -

l

sin (I)f

11/} / 11-,

-

-

)

F

~(c!)2)

...( 12.8 j

. a

The transmissibility of the system will be T

Ft

-F -

...(12.9)

Fa sin OJf Kj K2

or,

1/1}/I~ ') Ll((!)~)

TF =

...( 12.9 a)

The transmissibility TF depends on the system parameters given by Eq. (12.9). ror il1ustration, a special C3se will be examined in which KI - K2 - --p 11/, /110

2

/1/')

Denoting.

...(12.10)

.

...( 12.11)

-=- = mass ratIO = 11 11/,

The natural frequency ignored, is given by

of machine

11/

foundation

system ((Oil::)in which the isolating K)

-

')

-

KI

spring (k2) IS

')

~=~

(J)~:: -= 1/11+ /I~

1+ n~ rn,j

...(12.12)

1+llm

Using Eqs. (12.10), (12.11) and (12.12), the Eq. (12.9) can be written as 1 ...(1~2.13 a)

T, = l ~~- {(I+~ )+(~'" +~f f'r II ~-{(I+~ [

01'."

, -= rI .

1 ;')

.,

., I.-

(j)-.

"(1+~;,

1 11m. ó¥

(

11~I

õî Ø¢³õó¼

Figure 12.6 shows the plots ofT I" with ~

)-(~'" +~},f't}

"

~

l

(!) 11::'

)

r

., (0':'

} 1 l(1+~m)W;

{( I+~;' H~",+~I'

J"}

1

ratio for two values of mass ratio 11 . It is evident from m

this figure that the frequency ratio ((0 I (Oil::)must exceed a particular value (depending on the magnitude of 1111/) before the transmissibility fal1s below unity. The particular values of ((0I (OJ,)for which the transmissibility equals unity are given in Fig. 12.7.

..

.

~

466

Soil Dynamics & Machille 'Foulldatiolls

10 8 6 4

l1..

.

>-

2

'-+'--

.D ,11\

,-11\ E 11\

c

1.0

0.8

0 ....

+-

0.6

c:.I u

.... 0 l1..

).Jm=1.0

0-4

,.urn= 0.5 -

--

}J.m = 0.1 .------Z

0.2

{,Jnz = kIf (rnl +rnZ) kllrnl

=

\

kZ Irn 2

\

0.1 0

0-..5

1-.0

Fr~quency

1.5

\

\

3.0

2.0

ratio,

GJ/GJnz

Fig. t 2.6: T F versus ro/ronzfor different values of mass ratio /Jm

A 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 Fa A. = ...(12.14) 2 ) (/1/ 1 + Ill)) CD . - (I)/1-

(

)

)

In most of the machines. written as

the force Fa is frequency

dependent

(2meeun.

The Eq. (12.14) can be

)

2 me emA; =

(

2

(ml + 11/2) m n;

or where,

-

A. -

2 me e

2 11/1(1+~II/) (al -1) CD

-

a I =-2b... (J)

2

- m

...(12.15

a)

...(12.15

h)

)

...(12.16)

467

VibratiOll lsolatio" a"d Scree"i"g

2.0 I

1.5

W =j<2+)Jm )(l+}Jm) GJnz

I

E =<. 1.0

0.5

0

3.0

1.0

0

4.0

GJ G.)nz

Fig. 12.7 : Mass ratio versus m/mOl

If CDlla represents the natural frequency of mass m2 resting on isolating spring, then CD

-

~ -

2

...(12.17)

IIlz

IlU

The values of maximum amplitudes of two degree freedom system obtained from Eqs. (12.3) and (12.4) for frequency dependent force will be CD~a(2 me e (i)

A=1 = ml

[

2

4 CD -(I+l1m)

2

2

(CDlla-CDII= )CD +(I+l1m)CDlla

2

2

]

...(12.18 a)

CDII=

2

or

A =1 =

(2 me e) . (/2

/Ill

where

Similarly.

[

J 2 J J 1- ( 1+ I1m) (/j + (/2 - (/j (/2

(

)]

UJ/la (/1

=

...( 12.19)

(I)

A=2 = [(l+I1I11)(/~+l1m{/i-l](2mee) m2

~

...(12.18b)

[

l-(I+l1m)

~.-- -~. ,-

2

2

((/( +(/2-(/1

2

...(12.20)

2

(/2

)]

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 isolator spring ("'2) and stiffness of isolator spring (K2)' The efficiency of isolation system is defined as A:I

11 =- A-

...(12.21)

Value of A: can be- ~omputed from Eq. (12.15a). The value of A:I may be taken equal to the permissible amplitude. The value of 11 can also expressed as given below by dividing Eq. (12.18 b) by Eq. (12.15 b). Thus 11 =

ai (1 + ~m) (a~ -1) 2

[l-(1+~m)

2

(a) +a2 -a]

...(

2"

12.22)

a; )]

Solving Eqs. (12.21) and (12.22) one can obtain the value of a2' For this a2' appropriate values oflll2 and K2 are selected. It will ensnre the amplitude of vibration to be within pf'mlissible limit ...( 12.23)

Total force on the isolator, Fa = Kz . A:2 Fa should be less than the allowable capacity of the Isolator in compression. 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 placed bet\\een the masses Ill, and 1112' in order to minimise the transmission of vibrations from the ground to the equipment. If the ground is subjected to a periodic displacement given by Zo sin w l. the equations of motIOn \vill be: nzl ZI + KI Z\

-

K2 (Z2 - ZI)

11/2Z2 +K2(l2-Z,)

= KI

...( 12.24)

Zo sin W f

...(12.25)

=0

The vafues of maximum amplitudes of motion are given by 2

~-~ ml m2

A_I = KI lo -

004 -

[

m1

...(12.26)

(KI + K2) +- K2 W2 +- Kl K2 m, 1112] 11I,n/2

~ 111) Ill.,

A :2 = K 1 Z 0 W

-

K I+J K )

4

- [(

Ill,

-

KJ

2)J

K K

...( 12.27)

+ III~] W + Ill, 111;

The displacement transmissibility of the machine (Tn) is defined as the ratio of displacement amplitude of mass 1f/2to the displacement amplitude of the rigid support. Then KIK2 A:2

111,1112

T --0- Zo - w4_

(KI+K2)+K2

[

~

111)

...( 12.28) ())2+K)K2

1112Ã

111)11/2

I

469

Vibration Isolation alld Screenillg

The Eq. (12.28) is identical to Eq. (12.9), and therefore the results shown in Figs. 12.6 and 12.7 hold good in this case also.. Equation (12.28) can also be written as: A :2 l -

To -

0

2 2 a( a2 (1 + Ilm) l-(l+llnr)

If A-1 -- is taken as permissible amplitude. and l

0

A:2i lQ is known. For this value. Cl2can be determined spring i.e. K2'

12.4 SCREENING

OF VIBRATIONS

(

2

2

al +°2 -al

2

...( 12.29)

.,

a2

)

is the applied dynamic displacement. then the ratio which in turn will give the stiffness of the isolator

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 foundation that 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 dimensions are sufficiently large compared with the wave length of the surface waves generated by the source of dIsturbance. Dolling (1966) studied the effect of size and shape of the trench on its ability to screen the vibratIOns.

F (t)

Amplitude of s u da c e displaclZmlZnt

Footing L'

Circular open t re n c h ot radius R an d depth H

I

I

~

H

1

R-1

Fig. 12.8: Vibration screening using a circular trench surrounding

the source

of vibration-Active screening (Woods. t 968)

The first comprehensive study of screening vibratIOns by use of open trenches was made by Woods and Richart ( 1967) and \Voods ( 1968). They conducted field tests by creating vertical vibrallons with a small \"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 from 150 mm to 300 mm, and the angular dimension e was varied from 900 to 3600 around the source of vibration. 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 of the 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

1.2m

uniform

Wn

= 7

silty 0/0,

Vc

3.1 m

~

~

Wn =

=

fine

sand

e = 0.61) 287 rnls

td

sandy

silt

230/0)

Cl = 0.68)

(SM) = 16.4.6 k N /

m3

(ML) '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 of ARF contour diagrams. The dimensions of the trench are expressed in non-dimensional forms by dividing Hand R by the wave length ARof Rayleigh waves. ARis obtained by determining the number of waves (/1L occuring at

in Table 12.1.

distance x from the source (AR= x/1I).Wavelengths ARfor different frequenciesare given

Table 12.1: Wavelength and wave velocity for the Rayleigh Wave at the test site (Woods, 1968) Frequency Hz

AR mm

m/s

lOO

687

137

250

513

128

300

421

126

350

336

117

VR

'".:,.

-~.,,:i,:>},\

.

(,"

.;'if'..":.

.~.

471

Vibratio" Isolatio" a"d Scree"i"g

Boundary

of screened zone

/..

-

1,25

~>1.25

D

H/"'R 1.452

1.25-0.50

RO/AR 0.726

D

1.25-0,50

f:::~::JO,50-0'25

F::,::':':lo,50-0,25

1::::~::)lo,25-0,125

1(':,\/:1 0,25 -0.125

:-,:,':,-:- < 0 ' 125 t~:;:.~

,',', '," p',-::-:,:,) < 0 ' 125

H/AR 0,596

Ro/AR 0,596

(b)

(a)

Fig, 12,10: Amplitude reduction factor contour diagrams for active screening (Woods, 1968)

The field tests of Woods (1968) thus correspond to

A

r-R

H

= 0.222 - 0.910 and -;= 0,222 - 1.82 AR'

For satisfactory screening of vibrations, Woods (1968) recommended that ARF shouid be less than or equal 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 zone screened in this case extended to a distance of atleast 10 wavelengths (I OAR)from the source of disturbance (ii) For partial circle trenches (900 < 8 < 360°), the screened zone was defined as an area outside the trench extending to at least 10 wave lengths (10 AR)from the source and bounded on the sides by

...

,~

"""

.t72


Fouudations

radial lines from the centre of source through points 45° from ends of trench. In this case also. a minimum 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 open trenches in passive screening (Fig. 12.11), A typical layout of these tests consisting of two vibration e:\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 mm deep, Frequencies of excitation varied from 200 to 350 Hz.

Amp! itude 0 f s u rfa c e vibration

Source of disturbance .'

. .

..

,

Amplitude of surface vibration ..

,

Equipmrznt to be protected

.

~

.' ,

,

, ,

. '.

~

..

.

"

-.

. .' t.

r

, . :'

-.

R

.

"

::.

,. . Open t rrznch Fi~. 1:!.I] : "ihration

~ ';

H

"

.'

1

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 these tests that the zones screened by the trench would be symmetrical about the 0° line. Figure 12.13 shows the ARf contour diagram for one of these tests. For satisfactory screening. Woods (1968) recommended that the ARF should be less than or equal to 0,25 in a semi-CIrcular zone of radius (1/2) L behind the trench. The conclusions made on the hasis of this 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 ,~ I'R

R -=2 'i at -=7 .) 'R

AI -, ,/'R

R = 7.0 .C6.0 at /' 'R

-.

0

..

m 473

Vibratioll Isolatioll alld Screenillg

(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 are the most effective. However the open (unfilled) trenches may present instability problems necessitating trenches backfilled with sawdust, sand or bentonite slurry. The performance of open trench \vith sawdust was found better as compared with sand or bentonite slurry.

320 240

/

/

0

N

0

0

/

~

",

0

0

0,

/

I

0 ""'

0

0

O

0

I

0

0 0 L

~'

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

-', ;{~:', ,

0

0

0

,/

.

"

\,

" /

.\,

0

0

l.Sm

./

/'t 0

75 pi c k u P benches

0

r>.

..

0 0

0

0

',0 0 ',('", ", , ' " "

0

/

0

~

0

<, " '<:(

0

0

0

0

O

/".

/

0

0

OOA

/

0

0

/0

0

0

0

0

0

0

0

0

/

0

80

0

6.1 m /

0

0

/

[°

0

0

0

16

v' \,

'."

K

Tr en' ch

barriers

,

,,"

'>

/

"v :?s B

//

0 0

/ l.Sm

/

vibration

exciter

footings

Fig. 12.12 : Plan \iew of the field site layout for passive screening (Woods. 1968)

474

Soil Dynamics & Mac/,ine Forllldatioll.'i

Centqr tin q

lliII1J ARF>1.25

0

ARF 1.25-0.5

[tJ

+

AA F 0.5-0.25

n

ARF

ó¬ó´ó

+

+

. Q

4.6m

Trench

0.25-0.125

~ ARF<0.125

m HI AR

=

2.38

LIAR

=

4.76

=

5.96 0.17

1.5

R I AA W/'AR

=

Fig. 12.\3: Amplitude reduction factor contour diagrams for !Jassive screening

12.5 PASSIVE SCREE1\I;\C

BY USE OF PILE BARRIERS

There may be situations in which Rayleigh wavelengths may be in the range of 40 to SOm. For such a casc. 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 piles as 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-space to evaluate the effect of void cylinderical obstacles on reduction of vibration amplitudes. A box of size 1000 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 two consecutive void holes through which energy can pass through the barrier. The numerical evaluation of barrier effectiveness was made by obtaining the average of the values of ARF obtained on several lines beyond the barrier in a section ::I:15° of both sides of an axis through the source of disturbance and perpendicular \0 the barrier. In all tests. H/AR and LIARwere kept as 1.4 and 2.5 respectively. The Isolation effectiveness is defined as

Effectiveness = I

-

ARF

...(12.31)

;"'ii[;

Vibration

Isolation

475

and Screeni1lg

F(!)

Fig. 12.14 : Definition of parameters

t

for cylindrical hole barriers (Woods et aI., 1974)

1,0

LL

a::

« I

11

0,5

."

111

"'-

111 t>I

C

I

t>I > .;: u

\\\ \ \

I

'O"

"-

,

.:

"',

~',

,

0:

"

,

.

"

"

'

0,:

"

':""

J

. '. . . . '0.: . '0:~, :', ': :' "'"

0.4

'0,. 0 ""

' :'0

J:-

'

,

0,15

t>I

....

....... W

. . l

" , 0,.

I

,,\\

"-

\\0.125

\\

0.21

"-

"-

\ 0.075

"

\

0 0

0.05

0,10

0.15

0.20

0.25

Snf"r Fig. 12.15: Isolation effectiveness as a function of hole diameter and spacing (Woods et al.. 1974)

0,30

""1iItJ'i

",","',,'

">,,',,'

-

,

.P6


L Slllg the data of different tests, a non-dimensional plot of the isolation effectiveness versus S/A.R ratio for different vaues ofD/A.R was plotted as shown in Fig. 12.15. Woods et al. (1974) recommended that J row of void cylinderical holes may act as an isolation barrier if D 1

-

->-6

AR

I

SI! -
:.1nJ

...( 12.32)

...( 12.33)

4

Soil

type

Barrier

materials

107

f 10

Infinitely rigid pile

6

VI

-0 C

Steel

0

ZI

VI

N

10

.

5

E

Concrete

QJ

u c

0 -0

10

4

Gravel

Tim ber

0.

'-E > 0

0 ens e san d Hard clay

103 -

Si!t Loose sand Very soft clay

..c

crI

-

>0 Cl:

10

2

Plasti c foam

1 10 e--

.

Void boreho le

0 10 '--- ! 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 effect of dlal1l~.t..:r.spacing. and material properties of the soil pile system on the isolation effectiveness They ~'\

'~\'II1~'iLld~d

that:

!Z.'...,'~J

477

Vibration Isolation and Screelling

(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 is impedance ratio which is defined as IR = where.

Pp

Pp VRP Ps VRS

...( 12.34)

= Density of pile material

Ps = 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 V R) for various soils and pile materials. .

(iv) Two rows of barriers are more effective than single row barriers,

jILLUSTRATIVE

EXAMPLES'

Example 12.1 Determine the stiffness of the isolator to be kept between a reciprocating machine and the foundation shown in Fig. 12.17 to bring the vibration amplitude to less than 0.02 mm. The weight of the machine is 18 kN, and it produces an unbalanced force of 4.0 kN when operated at a speed of 600 rpm. The dynamic shear modulus and Poisson's ratio of the soil are 2 x 104 kN/m2 and '0.35 respectively, Solution: 1.

Mass of the foundation

block =

24x4.0x3.0x1.5

9.81

')

= 44 kNs-/m

Mass of machine = 9~~1 = 1.83 kNs2/m Total mass of machine and foundation = 44 + 1.83 = 55.83 kNs2/111 Stiffness of the soil,

K==

r0 =

4 G ro 1- ~l

f;fA = ~4 1tx 3 = 1.95 m 4

K = 4x2xl0 =

1-0.35

x 1.95 =24

Natural frequency of the whole system without any isolation will be:

x

104 kN/m

478

Soil Dynamics & Machine Foundation

Machine Weight

18kN

=

T 1.5m

1

blo c k

Foundation

///,\,\

r

/\\

~

4.0 m Section

3.0m

~

~

4.0 m Plan Fig. 12.17: Machine-foundation

CD n;

~

-

--

m

system (Example 12.1)

4 24 X 10 1 55.83

= 65.0

rad/s

2 Ttx 550 CD= 60 = 57.6 rad!s (A ) ; max -

Fa 2

2 -

m(oo//;-oo)

4.0

2 ., 55.83(65.0 -57.6- )

=-07.9¨ 10-501=0.0790101

479

Vibratioll Isolatioll alld Screellillg

The amplitude is greater than the permissible amplitude i.e. 0.02 mm. Hece isolator is required between machine and foundation. 2. Let the isolator be having stiffness Kz. Adopting the two degrees freedom system as shown in Fig. 12.5. 2

11/2 =

1.83kNs Im 2 11/1 = 44 kNs Im 4 Kl = K: = 24 x 10 kN/m - = 65.0 rad/s 1/-

(0

0.02 = 0.253

AZI -

11= Az - 0.079

= 65.0 = 1.13

Cl)n:

a I = ---;;'nz

57.0

- 1.83 = 0.0416

/-1./11 = 1111-

44

From Eq. (12.22) 11 =

or

'"' -0.k53

or

a~

a~ (l+/-I.m)(a~-l) 2

[l-(1-llm)

2

(al +a2 -al

2

2

a2 )J

2 ., a2(1+0.0416)(1.13--1) 2

2

[1- (1 + 0.0416) (1.13 + a2 -1.13

2

2

a2 )J

= 0.2310 i.e. a2 = 0.481

(jJ

11i!...= 0.481. It gives (011(1 = 0.481 x 57.6 = 27.7 rad's

(j)

.. 3.

Ko = 1112 A

Z2

=

(jJ

2

1

Il{/

= 1.83 x 27.7- = 1.40

x

[(l+llm)a~+Il",a;-1][2mee(O2] 2 2 1 1 2 1112(0[ 1-(l+1l",) al +az -aj ai

(

=

3

10 kN/m

)

[(1+0.0416)x1.l32+0.0416xO.2310-1.-(4.0) ~ 1 1.83 x 57.6-

= - 8.5

1-(1+0.0416)

(1.13- +0.2310-1.13-

~ A

0.2310)

-~

x

Force in the isolator = Ko' An "'1.18kt'

10 m

= lAD x

103 x 8.5 x 10-~kN

110

480

Soil Dynamics

& Machine

Foundations

A suitable isolation system may be selected which has total stiffness of 1.40 x 103kN/m and allowable compressive 'load more than 1.18 kN.

Example 12.2 Determine the stiffness of the isolation system if it is placed between the foundation block and base slab as 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 Isolators are placed between this block and base slab as shown in Fig. 12.18. Then

with machine.

Machine

T

Foundation

1.0 rn

block

i Isolator

T

O.Srn

Base

-L

~

slab

~

4.0 rn Fi~. 12.18: :\-Iachine-foundation

1//2

==

1.83 +

isolator system (Example 12.2)

-}x 44

= 31.2 kNs2/m

I 111( == 3"

J

x

44 = 14.7 kNs-/m 4

2

= 24 x 10 kNs fm (j)n:: = 65.0 rad/s Kl

11 = 0.253

From eq. (12.22),

al = 1.13 31.2 2 12 1111/ = t4.7 = . 2 2 a2 (1+2.12)(1.13 -1) 0.253 = 2 2 221.13 '+a2 -1.13 ([2)] [ 1-(1+2.12)

(

.-

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

bratioll Isolation

,"'ti'~:,,<.

,."

';",-,:,.

"~,<~Atj;!~";,;

"'.,:,' ">:!',., E

481

alld Screellillg

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/s 0) 2

4

Ka = 11120)2 na =31.2 x 48 = 7.2 x 10 kN/m (ii)

[0+2.12)

- An =

.,

x 1.132 +2.12 x 0.697 -1]4.0

r

31.2 x 57.6- Ll-(1+2.12)

2

(1.13

2'

+0.697~1.13

x 0.697) ]

17.85

= 103514[-2.38] = - 7.24 x 10--5m Force in the isolator = Ka A.,=- ==7.2 x 104 x 7.24 x 10- 5 = 5.2 kN On comparing with the results of example 12.1, it can be concluded that the stiffness of the isolator and force on it depends significantly on the location of the isolator.

Example 12.3 It is planned to install a compressor having operating speed of 1000 rpm at a distance of 50m from a precision machine. Suggest a suitable open trench barrier to provide effective vibration isolation. The velocity of shear waves at the site was found as 140 mIs. Solution: Active screening 1000 f = 60 = 16.7 Hz

.

OperatIng frequency,

Rayleigh wave velocity VR may be taken approximately equal to shear wave velocity i.e. 140 m/so V Therefore,

Wave

length

AI{ ==

;

140 == 16.7 ==8.4 m

Depth of the trench for active screening is given by

H

==0.6 AI{

= 0.6 x 8.4 ==5.04 m

A partial ci~cle trench with e = 120° may be located at 4.0m distance from the source (Fig. 12.19 a), Passive screening Depth 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). -AR 84. R = 4.52, AT = 25 4.52-2.0(6.0-2.5)=4.26 ')..2 . + 7.0-2.0 AR R

"1I

Soil Dynamics & Machille Foulldations 4.26 x 8.4 Length of trench =

2

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.

Precision machine

Precision

machine

T 12n

Tren ch

)

I 27m

{

I 50m I

38m. 0

120

Tre

../

,/

eññ£³

.

Co m pre ssor

Compre ssor (a)

Active

isolation

12.19.-Layout

-'-

(b)

Passive

isolation

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", 1nL Conf. 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.

483

'atio" Iso/atio" a"d Scree"i"g

(1967), "Screening of elastic waves by trenches", Proc. Int. Symp. Wave Propag. Dyn. Prop. Earth Mater, Albuquerque., NM, 275-284.

)ds, R. D., and Richart, F. E. Jr.

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 system for 'force isolation'. Represent it by a mathematical model and then give the procedure of getting the 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 procedure of 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 sinusoidally varying unbalanced force of 4kN in the vertical direction. The operating speed of the machine is

800 rpm. The dynamic shear modulus 'of the soil is 2.5 x 104kN/m2. Assume suitably any data not given. 6. A compressor having an operating speed of 1300 rpm was installed in an industrial unit. Later on it was planned to place a precision machine at a distance of 60 m from it. It was felt necessary to protect this precision machine from any damaging vibration caused by the compressor. Design 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 F f

A

Culmann's construction:

f

modified, 191

f

Amplitude of motion, 15 Amplitude method, 454

Cyclic: mobility, 279

Antiliquefaction measures, 324

simple shear test, 118, 128, 286, 296 torsional shear test, 118, 131, 286

B

triaxial compression test, 118, 133

Bandwidth method, 57 Bearing capacity of footings, 238 factors, 241

D Damping:

critical, factor, ratio,

Blasting tests, field, 314 Block foundations: degree of freedom. 352 effect of shape on response, 394 embedded. 394 method of analysis, 353 modes of vibration, 352 rocking, vibrations, 354, 360, 383, 396 rocking and sliding vibrations, 354, 363, 389, 397 sliding vibrations, 354, 359, 379, 395 vertical vibrations, 354, 358, 370, 394 yawing vibrations, 354. 362, 386 Block-resonance test: horizontal, 154

Degrees of freedom, 14 Displacement analysis of retaining wall, 20 I in pure rotation, 210 in pure translation, 205 - using Richard Elms model, 201 using Saran et al model, 214 Dynamic bearing capacity, 238 dynamic analysis, 252 pseudo-static analysis, 238 Dynamic earth pressure, 187 effect of saturation, 187 effect of submergence, 190 effect of uniform surcharge, 189 for c- soils, 193 Mononobe-Okabe's theory. 200

vertical,151

c CodTicienr of clastic:

point of application, 200 pseudo-static methods, 238

nonllniform col11pn.:ssion,356 nonuniform shear, 357 uniform compreSSion,354 uniform shear. 356 CrItical distance, 96,98, 100

'

E Earthquake: epit:enter, 3 equivalent dynamic load, 6 focus, 3

J

'ct Index

485

M

:ensity, 4 19nitude, 5

Machines

constants, 68

types, 340

constants, 68 half-space method, 370 ded block foundation, 394

reciprocating, 340 impact, 344 rotary, 345 Machine foundations categories of, 340 criteria for satisfactory action, 347

sS :

permissible amplitudes of, 348

:aring capacity, 238 :neralised bearing capacity equation, 240 )rizontal displacement, 249 ttlement, 249 t, 249

Mononobe-Okabe's theory, 187 Motion amplitude of, 15. harmonic, 15 periodic, 15 isolation, 35

ncy: Ircing, 15 ltural, 15

N Natural frequency,

ysical prospecting, 93

15

p t machine foundations,

421

Periodic motion, 15

~sign procedure,

432

Permissible amplitudes, 348

v'namic analysis,

426

'Pt time, %,97,

100

Phase Jag, 16 Phase lead, 16

R spring method, 354

Resonance method, 450

'action, 279

Resonant column test, 118, 119

Ictors affecting, 323 eld tests for, 314

Rotating mass type excitation, 30

.om standard penetration tests, 319 litial, 279 Iboratory studies, 283 lechanism, 281 :andard curves and correlatIOns, 30 I .:rr11lnology,279 one of, 306

Rotary machines design criteria, 445 loads on, 446 three dimensional analysis of, 460 resonance method of analysis, 450 amplitude method of analysis, 454 combined method of analysis, 459

486


s

fiee, 19,22,40,43

Sc\eening of waves, 469 Seismic:

isolation, 32 force, 32 motion, 35

coefficients, 9 cross-borehole survey, 147, 148 down-hole survey, 147, ISO forces, 9 zones, 9 refraction, 94, 147, 151 up-hole survey, 147,150

measuring instruments, 36 acceleratioI" pickup, 38,39 displacement pickup, 38 velocity pick up, 39 Vibrations of rod end conditions, 74 fixed-fixed, 75. 79 fixed-free, 78 free-free, 74, 76

Shake table tests, 286, 309 Shear modulus, 163,165,171

of finite length, 76 longitudinal, 76 torsional, 80

Soil mass participation in vibration, 400 Standard penetration test, 159

of infinite length, 70 longitudinal, 70 torsional, 72

Strain, 67

T Theory of vibrations,

w

13

Wave propagation

Time period, IS Transient tests

in elastic half space. 86 in elastic rods. 70, 72, 76, 80 in infinite medium, 81

u

Waves

Ultrasonic pulse test, 118,126

compressIon, 83, 108 head, 95

Lndamped free vibration, 19

Rayleigh, 9\ shear. 84, 108

v Vibration absorber, 48

z

forced, 25,42,47

Zone of liquefaction, 306

ODD

if-:"

. _'M-

_.-

~

,

'----""~"'-:--~"'--:-'""~'l ",

.\. ,- ',.

~ . :,

1

., ~

\

~

I

k~,

" '-

J

~L.._":~",-"

-,.~.~

:.c

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

Soil Dynamics and Machine Foundations Swami Saran_2

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