2. Applications

© 2009 Taylor & Francis Group, London, UK

Dynamics of Structure and Foundation – A Uniﬁed Approach 2. Applications

Indrajit Chowdhury Petrofac International Ltd Sharjah, United Arab Emirates

Shambhu P. Dasgupta Department of Civil Engineering Indian Institute of Technology Kharagpur, India

© 2009 Taylor & Francis Group, London, UK

Also available: Dynamics of Structure and Foundation – A Uniﬁed Approach 1. Fundamentals Indrajit Chowdhury & Shambhu P. Dasgupta 2009, CRC Press/Balkema ISBN: 978-0-415-47145-9 (Hbk) ISBN: 978-0-203-88527-7 (eBook)

CRC Press/Balkema is an imprint of the Taylor & Francis Group, an informa business © 2009 Taylor & Francis Group, London, UK Typeset by Vikatan Publishing Solutions (P) Ltd, Chennai, India. Printed and bound in Great Britain by Antony Rowe (a CPI Group company), Chippenham, Wiltshire. All rights reserved. No part of this publication or the information contained herein may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, by photocopying, recording or otherwise, without written prior permission from the publisher. Although all care is taken to ensure integrity and the quality of this publication and the information herein, no responsibility is assumed by the publishers nor the author for any damage to the property or persons as a result of operation or use of this publication and/or the information contained herein. Published by: CRC Press/Balkema P.O. Box 447, 2300 AK Leiden, The Netherlands e-mail: [email protected] www.crcpress.com – www.taylorandfrancis.co.uk – www.balkema.nl Library of Congress Cataloging-in-Publication Data Applied for ISBN: 978-0-415-49223-2 (Hbk) ISBN: 978-0-203-87922-1 (eBook)

© 2009 Taylor & Francis Group, London, UK

Contents

Preface

1

xiii

Dynamic soil structure interaction

1

1.1

1 1 2

1.2

1.3

Introduction 1.1.1 The marriage of soil and structure 1.1.2 What does the interaction mean? 1.1.3 It is an expensive analysis do we need to do it? 1.1.4 Different soil models and their coupling to superstructure Mathematical modeling of soil & structure 1.2.1 Lagrangian formulation for 2D frames or stick-models 1.2.2 What happens if the raft is f lexible? A generalised model for dynamic soil structure interaction 1.3.1 Dynamic response of a structure with multi degree of freedom considering the underlying soil stiffness 1.3.2 Extension of the above theory to system with multi degree of freedom 1.3.3 Estimation of damping ratio for the soil structure system 1.3.4 Formulation of damping ratio for single degree of freedom 1.3.5 Extension of the above theory to systems with multi-degree freedom 1.3.6 Some fallacies in coupling of soil and structure 1.3.7 What makes the structural response attenuate or amplify?

© 2009 Taylor & Francis Group, London, UK

4 6 6 6 14 28

28 29 30 31 32 40 41

vi Contents

1.4

The art of modelling 1.4.1 Some modelling techniques 1.4.2 To sum it up 1.5 Geotechnical considerations for dynamic soil structure interaction 1.5.1 What parameters do I look for in the soil report? 1.6 Field tests 1.6.1 Block vibration test 1.6.2 Seismic cross hole test 1.6.3 How do I co-relate dynamic shear modulus when I do not have data from the dynamic soil tests? 1.7 Theoretical co-relation from other soil parameters 1.7.1 Co-relation for sandy and gravelly soil 1.7.2 Co-relation for saturated clay 1.8 Estimation of material damping of soil 1.8.1 Whitman’s formula 1.8.2 Hardin’ formula 1.8.3 Ishibashi and Zhang’s formula 1.9 All things said and done how do we estimate the strain in soil, specially if the strain is large? 1.9.1 Estimation of strain in soil for machine foundation 1.9.2 Estimation of soil strain for earthquake analysis 1.9.3 What do we do if the soil is layered with varying soil property? 1.9.4 Checklist of parameters to be looked in the soil report 1.10 Epilogue

2

Analysis and design of machine foundations 2.1

2.2

Introduction 2.1.1 Case history #1 2.1.2 Case history #2 Different types of foundations 2.2.1 Block foundations resting on soil/piles 2.2.2 How does a block foundation supporting rotating machines differ from a normal foundation?

© 2009 Taylor & Francis Group, London, UK

42 42 46 46 47 49 49 50

51 52 52 58 61 61 62 63 65 65 70 77 79 80

83 83 83 84 85 85

86

Contents vii

2.2.3

Foundation for centrifugal or rotary type of machine: Different theoretical methods for analysis of block foundation 2.2.4 Analytical methods 2.2.5 Approximate analysis to de-couple equations with non-proportional damping 2.2.6 Alternative formulation of coupled equation of motion for sliding and rocking mode 2.3 Trick to by pass damping – Magnif ication factor, the key to the problem . . . 2.4 Effect of embedment on foundation 2.4.1 Novak and Beredugo’s model 2.4.2 Wolf’s model 2.5 Foundation supported on piles 2.5.1 Pile and soil modelled as f inite element 2.5.2 Piles modelled as beams supported on elastic springs 2.5.3 Novak’s (1974) model for equivalent spring stiffness for piles 2.5.4 Equivalent pile springs in vertical direction 2.5.5 The group effect on the vertical spring and damping value of the piles 2.5.6 Effect of pile cap on the spring and damping stiffness 2.5.7 Equivalent pile springs and damping in the horizontal direction 2.5.8 Equivalent pile springs and damping in rocking motion 2.5.9 Group effect for rotational motion 2.5.10 Model for dynamic response of pile 2.5.11 Dynamic analysis of laterally loaded piles 2.5.12 Partially embedded piles under rocking mode 2.5.13 Group effect of pile 2.5.14 Comparison of results 2.5.15 Practical aspects of design of machine foundations 2.6 Special provisions of IS-code 2.6.1 Recommendations on vibration isolation 2.6.2 Frequency separation 2.6.3 Permissible amplitudes 2.6.4 Permissible stresses 2.6.5 Concrete and its placing

© 2009 Taylor & Francis Group, London, UK

88 90 99 105 113 117 119 119 119 121 123 124 125 127 128 129 130 131 138 162 193 201 203 205 213 213 213 214 214 214

viii Contents

2.7

2.8

2.9

2.10

2.11

2.12 2.13

2.14

2.15 2.16 2.17

2.6.6 Reinforcements 2.6.7 Cover to concrete Analysis and design of machine foundation under impact loading 2.7.1 Introduction 2.7.2 Mathematical model of a hammer foundation Design of hammer foundation 2.8.1 Design criteria for hammer foundation 2.8.2 Discussion on the IS-code method of analysis 2.8.3 Check list for analysis of hammer foundation 2.8.4 Other techniques of analysis of Hammer foundation Design of eccentrically loaded hammer foundation 2.9.1 Mathematical formulation of anvil placed eccentrically on a foundation 2.9.2 Damped equation of motion with eccentric anvil Details of design 2.10.1 Reinforcement detailing 2.10.2 Construction procedure Vibration measuring instruments 2.11.1 Some background on vibration measuring instruments and their application 2.11.2 Response due to motion of the support 2.11.3 Vibration pick-ups Evaluation of friction damping from energy consideration Vibration isolation 2.13.1 Active isolation 2.13.2 Passive isolation 2.13.3 Isolation by trench Machine foundation supported on frames 2.14.1 Introduction 2.14.2 Different types of turbines and the generation process . . . 2.14.3 Layout planning 2.14.4 Vibration analysis of turbine foundations Dynamic soil-structure interaction model for vibration analysis of turbine foundation Computer analysis of turbine foundation based on multi degree of freedom Analysis of turbine foundation

© 2009 Taylor & Francis Group, London, UK

214 215 231 231 238 248 248 252 253 253 268 268 270 271 271 271 272 272 272 272 283 284 285 287 288 289 289 290 292 293 305 312 319

Contents ix

2.17.1 The analysis 2.17.2 Calculation of the eigen values 2.17.3 So the ground rule is . . . 2.17.4 Calculation of amplitude 2.17.5 Calculation of moments, shears and torsion 2.17.6 Practical aspects of design of Turbine foundation 2.18 Design of turbine foundation 2.18.1 Check list for turbine foundation design 2.18.2 Spring mounted turbine foundation

3

Analytical and design concepts for earthquake engineering 3.1

Introduction 3.1.1 Why do earthquakes happen in nature? 3.1.2 Essential difference between systems subjected to earthquake and vibration from machine 3.1.3 Some history of major earthquakes around the world 3.1.4 Intensity 3.1.5 Effect of earthquake on soil-foundation system 3.1.6 Liquefaction analysis 3.2 Earthquake analysis 3.2.1 Seismic coeff icient method 3.2.2 Response spectrum method 3.2.3 Dynamic analysis under earthquake loading 3.2.4 How do we evaluate the earthquake force? 3.2.5 Earthquake analysis of systems with multidegree of freedom 3.2.6 Modal combination of forces 3.3 Time history analysis under earthquake force 3.3.1 Earthquake analysis of tall chimneys and stack like structure 3.4 Analysis of concrete gravity dams 3.4.1 Earthquake analysis of concrete dam 3.4.2 A method for dynamic analysis of concrete dam 3.5 Analysis of earth dams and embankments 3.5.1 Dynamic earthquake analysis of earth dams

© 2009 Taylor & Francis Group, London, UK

319 320 321 321 321 322 322 322 330

389 389 390

391 392 394 395 395 412 412 417 424 425 431 444 448 456 481 481 485 519 519

x Contents

3.5.2

Mononobe’s method for analysis of earth dam 3.5.3 Gazetas’ method for earth dam analysis 3.5.4 Makadisi and Seed’s method for analysis of earth dam 3.5.5 Calculation of seismic force in dam and its stability 3.6 Analysis of earth retaining structures 3.6.1 Earthquake analysis of earth retaining structures 3.6.2 Mononobe’s method of analysis of retaining wall 3.6.3 Seed and Whitman’s method 3.6.4 Arango’s method 3.6.5 Steedman and Zeng’s method 3.6.6 Dynamic analysis of RCC retaining wall 3.6.7 Dynamic analysis of cantilever and counterfort retaining wall 3.6.8 Some discussions on the above method 3.6.9 Extension to the generic case of soil at a slope i behind the wall 3.6.10 Dynamic analysis of counterfort retaining wall 3.6.11 Soil sloped at an angle i with horizontal 3.7 Unyielding earth retaining structures 3.7.1 Earthquake Analysis of rigid walls when the soil does not yield 3.7.2 Ostadan’s method 3.8 Earthquake analysis of water tanks 3.8.1 Analysis of water tanks under earthquake force 3.8.2 Impulsive time period for non rigid walls 3.8.3 Sloshing time period of the vibrating fluid 3.8.4 Calculation of horizontal seismic force for tank resting on ground 3.8.5 Calculation of base shear for tanks resting on ground 3.8.6 Calculation of bending moment on the tank wall resting on the ground 3.8.7 Calculation of sloshing height

© 2009 Taylor & Francis Group, London, UK

519 522 523 526 526 526 527 530 530 532 533 533 544 544 547 560 571 571 575 577 577 581 583 583 584 584 585

Contents xi

3.9

3.10

Mathematical model for overhead tanks under earthquake 3.9.1 Earthquake Analysis for overhead tanks 3.9.2 Hydrodynamic pressure on tank wall and base 3.9.3 Hydrodynamic pressure for circular tank 3.9.4 Hydrodynamic pressure for rectangular tank 3.9.5 Effect of vertical ground acceleration 3.9.6 Pressure due to inertia of the wall 3.9.7 Maximum design dynamic pressure Practical aspects of earthquake engineering 3.10.1 Epilogue

References

© 2009 Taylor & Francis Group, London, UK

588 588 592 592 593 593 593 594 598 603 605

Preface

The monograph entitled “Dynamics of Structure and Foundation -– A Uniﬁed Approach” consists of two volumes. While in Volume 1 we dealt with background theories and formulations that constitute the above subject, this second volume deals with application of these theories to various aspects of civil engineering problems constituting topics related to dynamic soil-structure interaction, machine foundation and earthquake engineering. If we have managed to stir the wrath of the professionals in Volume 1 with mazes of tensors, differential and integral equations, it is our strong conviction that in this present volume we will be able to considerably appease this fraternity for it constitutes of a number of applications that are innovative, easy to apply and solutions to many practical problems that puts an engineer into considerable difﬁculty and uncertainties in a design ofﬁce. We start Volume 2 with the topic of Dynamic Soil Structure Interaction (DSSI). We believe this topic would play a key role in future and more so with the distinct possibility of construction of Nuclear power plants (especially in India) globally. A clear concept on this topic would surely be essential for designing such plants. Though we have dealt this topic only in terms of fundamental concepts, yet we feel that we have given sufﬁcient details to eradicate the misnomer from which many engineers suffer that “DSSI is nothing but adding some springs to the boundary of a structure and then doing the analysis through a computer”. The geotechnical aspects that play an extremely important role in selecting the soilspring value, (that are highly inﬂuenced by the strain range) have been dealt in quite detail. We hope that this section will do away with some of the major blunders that we make in DSSI analysis, and appreciate how the results thus obtained become unrealistic and questionable. We sincerely hope that engineers performing DSSI analysis, would start paying sufﬁcient attention to some of the key engineering parameters as furnished in the soil report – that are being habitually ignored in design ofﬁces. Second chapter consists of design and analysis of machine foundations (both block and frame type). In our collective experience as a consultant and academician we have seen signiﬁcant confusion on this topic as to who is responsible for this hapless orphan, structural or geotechnical engineers? While people from classical soil mechanics disowns it, as it involves the evaluation of eigen-values and vectors that are far away from

© 2009 Taylor & Francis Group, London, UK

xiv Preface

their traditional failure theories of foundation, structural engineers on the other hand are equally reluctant to shoulder the guardianship for their inherent apathy towards ‘what lies beneath the machine foundation’. As such, a design involving machine foundation throws the most challenging and interesting task in the domain of civil engineering that requires multi-discipline knowledge and should be equally interesting to an engineer having structural or geotechnical background. The matrix analysis concept that we have introduced herein is quite easy to follow and we hope would bridge the gap that is still prevalent in academics and practice alike. We would be looking forward to have some feedback from hardened professionals who are working in this area, as to how they feel about our representation which we believe is quite novel and has tried to answer a number of problems that often become burning issues on which they have spent signiﬁcant time on clarifying either to their Clients or Project Management Consultants. The last chapter of this volume deals with the most fearful force Mother Nature has created – “Earthquake”. Earthquake engineering as a topic is so vast, complex and diverse (and ever changing) that we concede that it did give us some uncomfortable moments as to what should justiﬁably constitute this chapter? Majority of the books that address this topic are far too focused on buildings and there are hardly any book around, that has addressed other specialized structures like chimneys, dams, retaining walls, water tanks etc (except some very specialized literature). It should be realized that some of these structures are expensive, important and cannot be ignored while building an earthquake resistant infrastructure. Buildings, we concede are the biggest casualties during an earthquake and are directly related to human life but damages to other structures as mentioned above can also create havoc especially in the post earthquake relief scenario. The major focus being still thrust on buildings, we were also quite surprised to ﬁnd that there is still much room for improvement in many of these structures, where technologies which are as old as 60 years are still in use (for instance earthquake response of retaining walls). We tried to improve upon many of them and believe that we have brought about a number of innovative solutions that can be adapted in a design ofﬁce environment and can also be used as a basis for further research. While presenting the topic no demarcation is made between geotechnical and structural earthquake engineering. For, as a seismic specialist our job is to minimize the destruction of property and save human lives. Thus doing a structural design we can perform the most sophisticated analysis and provide the most expensive detailing and our building still fails due to liquefaction killing people__“no medals for doing an excellent structural design!”, so if you do something do it in totality and not in isolation and this has been our major endeavour- that we have tried to communicate to you through this book. Indrajit Chowdhury Shambhu P. Dasgupta

© 2009 Taylor & Francis Group, London, UK

Chapter 1

Dynamic soil structure interaction

1.1 INTRODUCTION This chapter deals with some of the basic concepts of dynamic soil-structure interaction analysis. At the advent of this chapter we expect you to have some background on • • •

Static soil structure interaction Theory of Vibration/structural dynamics Basic theory of soil dynamics ∗∗

Based on the above , we build herein the basic concepts of dynamic soil structure interaction, which is slowly and surely gaining its importance in analytical procedure for important structures.

1.1.1 The marriage of soil and structure As was stated earlier in Chapter 4 (Vol. 1) even twenty years ago structures and foundations were dealt in complete isolation where the structural and geo-technical/foundation engineers hardly interacted1 . While the structural engineer was only bothered about the structural configuration of the system in hand he hardly cared to know anything more about soil other than the allowable bearing capacity and its generic nature, provided of course the foundation design is within his scope of work. On the other hand the geotechnical engineer only remained focused on the inherent soil characteristics like (c, φ, Nc , Nq , Nγ , eo , Cc, G etc.) and recommending the type of foundation (like isolated footing, raft, pile etc.) or at best sizing and designing the same. The crux of this scenario was that nobody got the overall picture, while in reality under static or dynamic loading the foundation and the structure do behave in tandem.

∗∗

For theoretical background on these topics please consider Volume 1. 1 Even today there are companies which has divisions like structural and civil engineering!! Where the responsibility of the structural division is to design the superstructure considering it as fixed base frame, furnish the results (Axial load, Moments and Shear) and the column layout drawing to the civil division who releases the foundation drawing based on this input data.

© 2009 Taylor & Francis Group, London, UK

2 Dynamics of Structure and Foundation: 2. Applications

In chapter 4 (Vol. 1), in the problem Example 1.3.1, we have shown how the soil stiffness can affect the bending moment and shear forces of a bridge girder and ignoring the same how we can arrive at a result which can be in significant variation to the reality. Drawing a similar analogy one can infer that ignoring the soil stiffness in the overall response (and treating it as a fixed base problem) the dynamic response of structure (the natural frequencies, amplitude etc.) can be in significant variation to the reality in certain cases. This aspect came to the attention of engineers while designing the reactor building of nuclear power plant for earthquake. Considering its huge mass and stiffness, the fundamental time period for the fixed base structure came around 0.15 sec while considering the soil effect the time period increased to 0.5 second giving a completely different response than the fixed base case. With the above understanding – that underlying soil signiﬁcantly affects the response of a structure, research was focused on this topic way back in 1970, and under the pioneering effort of academicians and engineers, the two diverging domain of technology was brought under a nuptial bond of “Dynamic soil structure interaction”, where soil and structure where married off to a uniﬁed integrated domain. To our knowledge the first signiﬁcant structure where the dynamic effect of soil was considered in the analysis in Industry in India was the 500 MW turbine foundations for Singrauli where the underlying soil was modeled as a frequency independent linear spring and the whole system was analyzed in SAP IV (Ghosh et al. 1984).

1.1.2 What does the interaction mean? We have seen earlier that considering the soil as a deformable elastic medium the stiffness of soil gets coupled to the stiffness of the structure and changes it elastic property. Based on this the characteristic response of the system also gets modified. This we can consider as the local effect of soil. On the other hand consider a case of a structure resting on a deep layer of soft soil underlain by rock. It will be observed that its response is completely different than the same system when it is located on soft soil which is of much shallow depth or resting directly on rock2 . Moreover the nature of foundation, (isolated pad, raft, pile), if the foundation is resting or embedded in soil, layering of soil, type of structure etc. has profound inﬂuence on the over all dynamic response of the system. We had shown for static soil-structure interaction (Chapter 4 (Vol. 1)) case that the soil can be modeled as equivalent springs or as finite elements and are coupled with the superstructure. Thus for a simple beam resting on an elastic support can be modeled as shown in Figure 1.1.1 and an equivalent mathematical model for the same is shown in Figure 1.1.2. Based on matrix analysis of structure the element stiffness for this element may be written as

2 The reason for these effects we will discuss subsequently.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 3

Node iii

Node j

Soil Spring Ki

Soil Spring Kj

Figure 1.1.1 Equivalent beam element connected to soil springs.

2 1

4 3

1

2

Figure 1.1.2 Mathematical model of the equivalent beam element.

⎡

6L

0

−12

6L

4L2

0

−6L

2L2

0

0

0

6L

IxL2 2Iz(1 + ν) 0

12

6L

2L2

0

6L

4L2

0

−IxL2 2Iz(1 + ν)

0

0

12

⎢ ⎢ 6L ⎢ ⎢ ⎢ ⎢ EIz ⎢ 0 [Kbeam ] = 3 ⎢ L ⎢ ⎢−12 ⎢ ⎢ 6L ⎢ ⎢ ⎣ 0

⎤

0

⎥ ⎥ ⎥ ⎥ 2 ⎥ −IxL ⎥ 2Iz(1 + ν)0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ 2 ⎦ IxL 0

(1.1.1)

2Iz(1 + ν)

and the displacement vector is given by {δ} = <δ1 θ1 θ2 δ2 θ3 θ4>T

(1.1.2)

When the soil springs are added to the nodes, the overall stiffness becomes ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ EIz [Kbeam ]= 3 ⎢ L ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

L3 Kii 12 + EIz 6L 0

⎤ 6L

0

−12

6L

4L2

0

−6L

2L2

0

IxL2 2Iz(1 + ν)

−12

6L

0

6L

2L2

0

0

0

−IxL2 2Iz(1 + ν)

0

L3 Kjj EIz 6L

12 +

0

0 6L 4L2 0

⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ −IxL2 ⎥ ⎥ 2Iz(1 + ν) ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ 2 ⎦ IxL 0

2Iz(1 + ν)

(1.1.3) © 2009 Taylor & Francis Group, London, UK

4 Dynamics of Structure and Foundation: 2. Applications where, [Kbeam ] = combined stiffness matrix for the beam and the spring; Kii = Kjj = spring values of soil at node i and node j of the beam respectively. The above is a very convenient way of representing the elastic interaction behavior of the underlying soil and can be very easily adapted in a commercially available finite element or structural analysis package.

1.1.3 It is an expensive analysis do we need to do it? This is a common query comes to the mind of an engineer before starting of an analysis. Based on this fact an engineer do become apprehensive if his/her analysis would suffer from a cost over run or whether he/she will be able to finish the design within the allocated time frame. If he is convinced that soil structure interaction do takes place and the structure is a crucial one3 our recommendation would be ‘its worth the effort rather than to be sorry later’. The additional engineering cost incurred is trivial compared to the risk and cost involved in case of a damage under an earthquake or a machine induced load. Now the first question is for what soil condition does dynamic soil structure interaction takes place? Veletsos and Meek (1974) suggest that chances of dynamic soil structure interaction can be significant for the expression Vs ≤ 20 fh

(1.1.4)

where Vs = shear wave velocity of the soil; f = fundamental frequency of the fixed base structure; h = height of the structure. Let us now examine what does Equation (1.1.4) signifies? Knowing the time period T = 1/f , the above expression can be rewritten as Vs T ≤ 20 h

(1.1.5)

For a normal framed building considering the fixed base time period as (0.1n), where n is the number of stories and thus, we have Vs n ≤ 200 h

(1.1.6)

For a normal building the average ratio of h/n (height : storey ratio) is about 3 to 3.3 meter. Thus considering h/n = 3, we have Vs ≤ 600 m/sec.

(1.1.7)

3 Like Power House, Turbine foundations, Nuclear reactor Building, Main process piper rack, distillation columns, bridges, high rise building catering to large number of people etc.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 5

From which we conclude that for ordinary framed structure, when shear wave velocity is less or equal to 600 meter/sec we can expect dynamic soil structure interaction between the frame and the soil. Incidentally, Vs = 600 m/sec is the shear wave velocity which is associated with rock. Thus it can be concluded that for all other type of soil, framed structures will behave differently than a fixed base problem-unless and until it rests on rock. For Cantilever structures like tall vessels, chimneys etc of uniform cross section fundamental time period T is given by

mh4 EI

T = 1.779

(1.1.8)

where, m = mass per unit length of the system; h = height of the structure; EI = flexural stiffness of the system. Substituting the above value in Equation (1.1.5) we have Vs T ≤ 20; h

Vs 1.779

or

mh4 EI

h

≤ 20;

11.24 or, Vs ≤ h

EI m

(1.1.9)

Considering, I = Ar2 and m = ρ · A, where A = area of cross section; r = radius of gyration; ρ = Mass density of the material, we have 11.24r Vs ≤ h

E ρ

(1.1.10)

Shear Wave Velocity for Soil-Structure interaction for Chimneys

1400.00

Shear Wave

velocity(m/sec)

1200.00 1000.00

Shear Wave velocity steel chimney

800.00 600.00

Shear Wave velocity concrete chimney

400.00 200.00

0 30

5 27

0 25

5 22

0 20

5

0

17

5

15

12

10

0

0.00

Slenderness Ratio

Figure 1.1.3 Chart to assess soil-structure interaction for steel and concrete chimney.

© 2009 Taylor & Francis Group, London, UK

6 Dynamics of Structure and Foundation: 2. Applications

For steel structure the above can be taken as, Vs ≤ 57580/λ where λ = h/r, the slenderness ratio of the structure. For concrete structure we have Vs ≤

123970 λ

(1.1.11)

Based on the above expressions one can very easily infer if soil structure interaction is significant or not. The chart in Figure 1.1.3 shows limiting shear wave velocity below which soilstructure interaction could be significant for a steel and concrete chimney.

1.1.4 Different soil models and their coupling to superstructure The various types of soil model that are used for comprehensive dynamic analysis are as follows: 1 2 3

Equivalent soil springs connected to foundations modeled as beams, plates, shell etc., Finite element models (mostly used in 2D problems), Mixed Finite element and Boundary element a concept which is slowly gaining popularity.

Of all the options, spring elements connected to superstructure still remain the most popular model in design practices due to its simplicity and economy in terms of analysis especially when the superstructure is modeled in 3-dimensions. It is only in exceptional or very important cases that the Finite elements and Boundary elements are put in to use and that too is mostly restricted to 2 dimensional cases.

1.2 MATHEMATICAL MODELING OF SOIL & STRUCTURE We present hereafter some techniques that are commonly adopted for coupling the soil to a structural system.

1.2.1 Lagrangian formulation for 2D frames or stick-models This formulation is one of the most powerful tool to couple the stiffness of soil to the superstructure-specially when one is using a stick model or a 2D model. For the frame shown hereafter we formulate the coupled stiffness and mass matrix for the soil structure system which can be effectively used for dynamic analysis. In the system shown in Figure 1.2.1, mf , Jθ = mass and mass moment of inertia of the foundation; m1 , J1 = mass and mass moment of inertia of the 1st story; m2 , J2 = mass and mass moment of inertia of the top story; Kx , Kθ = translational and rotational stiffness of the soil; and k1 , k2 = stiffness of the columns. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 7

y2 m2 J2 k2 y1

h2

m1, J 1 k1

Kx

K

h1

mf , J

Figure 1.2.1 2D Mathematical model for soil structure interaction.

The equation for kinetic energy of the system may be written as T=

1 1 1 1 m u˙ 2 + Jθ θ˙ 2 + m1 (u˙ + h1 θ˙ + y˙ 1 )2 + J1 θ˙ 2 2 f 2 2 2 1 1 + m2 (u˙ + (h1 + h2 )θ˙ + y˙ 2 )2 + J2 θ˙ 2 2 2

U=

1 1 1 1 Kx u2 + Kθ θ 2 + k1 y12 + k2 (y2 − y1 )2 2 2 2 2 d dt

Considering the expression4 , equation as ⎡

mf + m1 + m2

⎢ ⎢ ⎢ ⎢ ⎣

m1 h 1

∂T ∂ q˙ i

+

∂U ∂qi

m1 h 1 + m 2 H J + m1 h21 + m2 H 2

m1 m1 h1

m1 h1

m1

m2

m2 H

0

Kx

⎢0 ⎢ +⎢ ⎢0 ⎣ 0

0

0

Kθ

0

0

k1 + k2

0

−k2

⎤⎧ ⎫ u⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪ ⎬ 0 ⎥⎨θ ⎪ ⎥ =0 ⎥ y1 ⎪ −k2 ⎦ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k2 ⎩y2 ⎭

(1.2.2)

= 0, we have the free vibration

m1

⎡

(1.2.1)

⎤⎧ ⎫ u¨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪ ⎨ ⎬ m2 H ⎥ θ¨ ⎪ ⎥ y¨ 1 ⎪ 0 ⎥ ⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩ y¨ 2 ⎭ m2 m2

0

(1.2.3)

4 Refer Chapter 2 (Vol. 2) for further application of this formulation where we have derived a 2D soilstructure interaction model for a Turbine framed foundation.

© 2009 Taylor & Francis Group, London, UK

8 Dynamics of Structure and Foundation: 2. Applications

Figure 1.2.2 Typical finite element mesh with soil springs, for a ﬂexible raft.

where

J = Jθ + J1 + J2 sum of all mass moment of inertia;

H = h1 + h2 = the total height of the structure. Above formulation can very well be used in cases the foundation is significantly rigid and can be modeled as rigid lumped mass having negligible internal deformation5 . However for cases where the foundation is more flexible one usually resorts to finite element modeling of the base raft which is connected to the soil springs as shown in Figure 1.2.2. For the problem as shown above irrespective of the raft being modeled as a beam or a plate the soil stiffness is directly added to the diagonal element Kii of the global stiffness matrix to arrive at the over all stiffness matrix of the system. Before we proceed further we explain the above assembly by a conceptual problem hereafter.

Example 1.2.1 For the beam as shown in Figure 1.2.3, compute the global stiffness matrix when supported on a spring at its mid span. Take EI as the flexural stiffness of the beam. The spring support has stiffness @ K kN/m. Solution: For a beam having two degrees of freedom per node as shown in Figure 1.2.4, the element stiffness matrix is expressed as follows.

5 A classic example is a turbine frame foundation resting on a bottom raft whose thickness is usually greater than 2.0 meter.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 9

L

L

K

Figure 1.2.3 Spring supported beam.

2 4

1

3

Figure 1.2.4 Two degrees of freedom of a beam element.

The element matrix for such case is given by 1

⎡

12EI ⎢ L3 ⎢ ⎢ 6EI ⎢ ⎢ ⎢ L2 Kij = ⎢ ⎢ −12EI ⎢ ⎢ L3 ⎢ ⎣ 6EI L2

2 6EI L2 4EI L −6EI L2 2EI L

3 −12EI L3 −6EI L2 12EI L3 −6EI L2

4

⎤ 6EI L2 ⎥ ⎥ 2EI ⎥ ⎥ ⎥ L ⎥ ⎥ −6EI ⎥ ⎥ L2 ⎥ ⎥ 4EI ⎦ L

Assembling the element matrix for the two beams we have ⎡ 12EI ⎢ L3 ⎢ ⎢ 6EI ⎢ ⎢ ⎢ L2 ⎢ ⎢ −12EI ⎢ ⎢ ⎢ L3 ⎢ ⎢ 6EI ⎢ ⎢ [K]g = ⎢ L2 ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎣ 0

⎤

6EI L2

−12EI L3

6EI L2

0

0

0

0

4EI L

−6EI L2

2EI L

0

0

0

0

−6EI L2

12EI 12EI + L3 L3

−6EI 6EI + 2 L2 L

−12EI L3

6EI L2

0

0

2EI L

−6EI 6EI + 2 L2 L

4EI 4EI + L L

−6EI L2

2EI L

0

0

0

−12EI L3

−6EI L2

12EI L3

−6EI L2

0

0

0

6EI L2

2EI L

−6EI L2

4EI L

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

© 2009 Taylor & Francis Group, London, UK

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

10 Dynamics of Structure and Foundation: 2. Applications

As Left hand support is fixed hence we have to eliminate row and column 1 and 2. Similarly, as right hand support is hinged we have to eliminate row and column 5 from the above when we have ⎡ 24EI

0

⎢ ⎢ ⎢ [K]g = ⎢ ⎢ 0 ⎢ ⎣ 6EI L2 L3

8EI L 2EI L

6EI ⎤ L2 ⎥ ⎥ 2EI ⎥ ⎥ with appropriate boundary conditions. L ⎥ ⎥ ⎦ 4EI L

To use the spring support, the spring is now directly added to the diagonal element of the global matrix. Thus the combined stiffness matrix is given by ⎡ 24EI

+ Ks

⎢ L3 ⎢ ⎢ [K ]g = ⎢ 0 ⎢ ⎢ ⎣ 6EI L2

0 8EI L 2EI L

6EI ⎤ L2 ⎥ ⎥ 2EI ⎥ ⎥ L ⎥ ⎥ ⎦ 4EI L

The above is the normal practice adapted in global assemblage of soil spring in a finite element assembly. We further elaborate the phenomenon with a suitable practical numerical example.

Example 1.2.2 Shown in Figure 1.2.5 is a bridge girder across a river is resting at points A and B on rock abutments at ends, and resting on a pier at center of the girder (point C)

A

5.0 m

C

5.0 m Water Level

Figure 1.2.5 Bridge girder across abutments.

© 2009 Taylor & Francis Group, London, UK

B

Dynamic soil structure interaction 11

1

1

2

2

3

3

A

4

4

5

B

C

Figure 1.2.6 Idealisation of the bridge girder ignoring soil effect.

which is resting on the soil bed of the river. The ﬂexural stiffness of the girder is EI = 100,000 kN · m2 . Area of girder is 5.0 m2 . The dynamic shear modulus of soil is G = 2500 kN/m2 . The bridge pier foundation has plan dimension of 6 m × 6 m. Determine the natural frequencies of vibration of the girder considering with and without soil effect. Unit weight of concrete = 25 kN/m3. Mass moment of inertia per meter run = 30 kN · sec2 · m. Solution: The bridge girder can be mathematically represented by a continuous beam as shown in Figure 1.2.6. Here node 2 and 4 are at the center of beam. Thus, for beam element 1, 2, 3, and 4, we have element stiffness matrix as ⎡

12 ⎢ EI ⎢ 6L [Kij ] = 3 ⎢ L ⎣−12 6L

6L 4L2 −6L 2L2

⎤ −12 6L −6L 2L2 ⎥ ⎥ ⎥ 12 −6L⎦ −6L 4L2

The unconstrained combined stiffness matrix as [Kij ]

⎡

12

⎢ ⎢ 6L ⎢ ⎢ ⎢−12 ⎢ ⎢ 6L ⎢ ⎢ EI ⎢ ⎢ 0 = 3⎢ L ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎢ 0 ⎣ 0

6L

−12

6L

0

0

0

0

0

4L2

−6L

2L2

0

0

0

0

0

−6L

24

0

−12

6L

0

0

0

−6L

2

0

0

0

2L

2

0

8L

2

2L

0

−12

−6L

24

0

−12

6L

0

0

6L

2L2

0

8L2

−6L

2L2

0

0

0

0

−12

−6L

24

0

−12

0

0

0

6L

2L

0

0

0

0

0

0

0

0

Substituting the values we have © 2009 Taylor & Francis Group, London, UK

0

0

2

2

0

8L

−12

−6L

6L

2

2L

−6L 12 −6L

0

⎤

⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ 6L ⎥ ⎥ ⎥ 2L2 ⎥ ⎥ −6L ⎥ ⎦ 2 4L 0

12 Dynamics of Structure and Foundation: 2. Applications

[K] = 76800 96000 −76800 96000 0 0 0 0 0 0 96000 160000 −96000 80000 0 0 0 0 0 0 −76800 −96000 153600 0 −76800 96000 0 0 0 0 96000 80000 0 320000 −96000 80000 0 0 0 0 0 0 −76800 −96000 153600 0 −76800 −96000 0 0 0 0 96000 80000 0 320000 −96000 80000 0 0 0 0 0 0 −76800 −96000 153600 0 −76800 96000 0 0 0 0 96000 80000 0 320000 −96000 80000 0 0 0 0 0 0 −76800 −96000 76800 96000 0 0 0 0 0 0 96000 80000 −96000 160000

Now imposing the boundary condition that vertical displacement are zero at 1, 3, 5,6 we have [K] = −96000 153600 0 96000 0 0 0

160000 −96000 80000 0 0 0 0

80000 0 320000 80000 0 0 0

0 96000 80000 320000 −96000 80000 0

0 0 0 −96000 153600 0 96000

0 0 0 80000 0 320000 80000

0 0 0 0 96000 80000 160000

Lumped mass at each node is given by → Mii = 25 × 5 × 2.5/9.81 = 31.85 kN · sec2 /m. Mass moment of inertia at each node is given by → Jii = 30 × 1.25 = 37.5. Thus combined mass matrix is given by ⎡

37.5 ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 [M] = ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎣ 0 0

0 31.85 0 0 0 0 0 0

0 0 65 0 0 0 0 0

0 0 0 31.85 0 0 0 0

0 0 0 0 37.5 0 0 0

0 0 0 0 0 31.85 0 0

⎤ 0 0 0 0 ⎥ ⎥ 0 0 ⎥ ⎥ ⎥ 0 0 ⎥ ⎥ 0 0 ⎥ ⎥ 0 0 ⎥ ⎥ ⎥ 65 0 ⎦ 0 37.5

6 We assume that since the bridge is supported on hard rock at ends, displacement at node 1 and 5 are zero.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 13

1

3

2 A

5

4

B

C Kz

Figure 1.2.7 Idealisation of the bridge girder considering soil effect.

Considering the equation [K] − [M] ω2 = 0 we have MODE

1

2

3

4

5

6

7

Eigen value Natural frequency (rad/sec)

692 26.30

1328 36.44

2684 51.80

4897 69.97

7448 86.59926

7787 88.24996

11722 108.26855

Considering the effect of soil we can construct the model as in Figure 1.2.7.

4Gr0 Here Kz = 1−ν

where r0 =

LxB , Here L = B = 6.0 m π

Here r0 = 3.38 m and for G = 2500 kN/m2 and ν = 0.3 Kz = 48285.71 kN/m. Now imposing the boundary condition that vertical amplitude at node 1 and 5 are zero (node 3 is not zero) we have [K] = 160000 −96000 80000 0 0 0 0 0 −96000 153600 0 −76800 96000 0 0 0 80000 0 320000 −96000 80000 0 0 0 0 −76800 −96000 201959.1 0 −76800 −96000 0 0 96000 80000 0 320000 −96000 80000 0 0 0 0 −76800 −96000 153600 0 96000 0 0 0 96000 80000 0 320000 80000 0 0 0 0 0 96000 80000 160000 The Mass matrix remains same as derived earlier. Performing the eigen value solution we have Modes

1

2

3

4

5

6

7

8

Eigen-values Natural frequency (rad/sec)

75 8.660

692 26.30

2684 51.80

3045 56.18

7067 84.06

7448 86.30

9489 97.41

11722 108.27

© 2009 Taylor & Francis Group, London, UK

14 Dynamics of Structure and Foundation: 2. Applications

Having established the fact as to how soil affects the dynamic response let us see further what different type of soil model is possible. For design office practices spring values considered are usually based on Richart/Wolf’s model which are effectively combined with structure as shown above to find out the overall response of a system. The example above, though it has been worked out based on beam the theory, it is effective for any kind of structural elements like plates, shells, 8-nodded brick element etc. Thus implementing the above in a general purpose Finite element package is quite straight forward. For raft modeled as beam with underlain spring, the essence of arriving at individual springs at each node is same as shown in the case of static analysis based on influence zone7 . The only difference being that the nodal influence area is to be converted into an equivalent circular area to arrive at vertical spring values. The horizontal springs are based on the full area and are divided equally at the end.

1.2.2 What happens if the raft is f lexible? Methodology described in previous section is usually adapted when the raft is unconditionally rigid. However there could be cases where the raft could be perfectly ﬂexible or intermediate (i.e. somewhere between perfectly rigid and perfectly ﬂexible) when the calculation of spring values is different than what has been mentioned in the preceding. Before we get into this issue the obvious query would be what is the boundary condition for raft rigidity in terms of dynamic loading? Unfortunately there is none, and the condition pertaining to static load still applies8 . Thus as explained in Chapter 4 (Vol. 1), if L is the c/c distance between the columns, then for • • •

λL ≤ π4 the raft will behave as rigid raft For λL ≥ π the raft will behave as flexible raft For all values between π/4 ≤ λL ≤ π, the slab behave in between rigid/flexible

in which λ =

4

kB 4Ec I ,

k = modulus of sub-grade reaction, (in kN/m3 ); B = width

of raft in meter; Ec = modulus of elasticity of concrete (in kN/m2 ); I = moment of inertia of the raft (in m4 ). 1.2.2.1

Calculation of spring constant for rigid raft

The rigidity of raft plays a significant role in the soil spring values connected to the plate elements as mentioned above.

7 Refer Example 4.6.1 in Chapter 4 (Vol. 1) for further details. 8 This is not illogical for dynamic load can be conceived as a system under static equilibrium at a time t. Thus condition of rigidity as explained in Chapter 4 (Vol. 1) should hold good.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 15

When the raft is rigid the gross spring value is obtained based on the full raft dimension and then are broken up into discrete values kz = Kz

Ap AG

(1.2.4)

where, kz = value of discrete spring for the rigid finite element; Kz = value of gross spring considering the overall dimension of the raft; Ap = area of the finite element plate, and AG = gross area of the raft. 1.2.2.2

Calculation of spring constant for f lexible raft

When the raft is flexible an equivalent radius within which the load gets dispersed is first obtained from the formula r0 = 0.8ts

Ec Gs

1−ν 1 − νc2

1 3

(1.2.5)

The gross spring value is then obtained based on this equation. Finally the discrete spring for the finite element is obtained as kz

= Kz

Ap

π r20

(1.2.6)

where, r0 = equivalent radius within which the load gets dispersed; Ec = dynamic modulus of the concrete raft; Gs = dynamic shear modulus of the soil; ν = Poisson’s ratio of soil; νc = Poisson’s ratio of the raft, and ts = thickness of the raft. A suitable problem cited hereafter elaborates the above more clearly.

Example 1.2.3 A raft of dimension 30 m × 15 m is resting on a soil having dynamic shear modulus of 35000 kN/m2 and Poisson’s ratio of soil = 0.4. Determine the soil springs for plate elements of size 2.0 m × 2.0 m for finite element analysis considering, • •

The raft as rigid Considering the raft as flexible. The thickness of the raft is 1.8 m.

© 2009 Taylor & Francis Group, London, UK

16 Dynamics of Structure and Foundation: 2. Applications

Solution: Considering the raft as rigid:

30 × 15 = 11.968 meter; π 4 × 35000 × 11.96 4Gr0 Kz = = = 2790666.67 kN/m 1−ν 0.6 r0 =

For finite element of size 2 m × 2 m discrete spring value will be kz

= Kz

Ap AG

kz

➔

2×2 = 24806 kN/m = 2790666.67 30 × 15

Thus spring values at four nodes are 6201 kN/m i.e 1/4th of the above calculated value. When the raft is considered ﬂexible, we have: r0 = 0.8ts

Ec Gs

1−ν 1 − νc2

1/3

Here Ec = 3 × 108 kN/m2 ; νc = 0.25(say), then r0 = 0.8 × 1.8

Thus Kz =

3 × 108 35000

1 − 0.4 1 − 0.252

1/3 = 25.39 m

4Gr0 4 × 35000 × 25.39 = = 5924333.333 kN/m 1−ν 0.6

Thus for finite element of size 2 m × 2 m the discrete spring value is kz = 5924333.333

2×2 π × 25.392

= 11701 kN/m

Thus spring values at four nodes are 2925 kN/m It will be observed that the spring values vary considerably for the two different approach.

1.2.2.3

What sin thou make in treating foundation & the structure separately?

Difficult to pass a sweeping judgment for depending on the situation, the sin could be cardinal or even trivial. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 17

Based on a number of analysis carried out it can be stated that treating them in isolation can result in conservative design9 or dangerously un-conservative, thus resulting in an unsafe structure which could be a danger to human life and property. Having made the above statement a number of questions obviously come to mind10 like 1 2 3 4

5

How conservative or how susceptible the system can be ignoring the soil effect? Considering soil effect (specially for FEM analysis) makes the analysis more laborious and time consuming – thus more costly – is it worth? My boss is a traditionalist and under project time pressure – can I convince him it is worth the effort. Before doing the detailed analysis itself can I come up with a quantitative value based on which I can assess how far this effect will be (for good or worse) and thus convince my boss on the value addition to this effort? What is the risk in terms of cost and safety if I do not do this analysis?

The questions are surely pertinent and not always very easy to answer. However with a little bit of intelligent analysis it is not difficult to come up with a logical conclusion on this issue. We try to explain. . . The obvious answer is ‘it essentially could modify the natural frequency/time period of the system’11 . What needs to be evaluated is – what is the effect of this modified time period on the system compared to, if the soil is ignored (i.e. it is considered a fixed base problem). The two classes of problems under which dynamic soil structure interaction plays a significant role are • •

Systems subjected to vibration from machines like block foundations (machine foundations for pumps, compressors, gas turbines etc), frame foundations (turbine foundations, compressor foundations, boiler feed pump foundations) Structures subjected to earthquake.

For the machine foundation source of disturbance is the machine mounted on the system the dynamic waves generated are transferred from the machine – via structure to the surrounding soil-which is an infinite elastic half space. While for earthquake the source of disturbance is the ground itself where elastic waves generate within the soil mass due to the tectonic movement/rupture of the rock mass (geologically known as faults). It is obvious that soil will affect these two classes of problem in different ways. For instance a machine supported on a frame- the frame is usually made signiﬁcantly stiff to ensure stress induced in it are not signiﬁcant and are generally made

9 For big projects which could mean a cost over run. 10 Specially for freshman new to the topic who has got a lead engineer and a departmental HEAD to answer to. 11 We say the word “could” as because the extent of modification will depend upon the shear wave velocity of the soil. We had shown previously the boundary limits within which it can have a significant effect.

© 2009 Taylor & Francis Group, London, UK

18 Dynamics of Structure and Foundation: 2. Applications

over tuned for medium or low frequency machine when considered as a fixed based problem. But in reality considering the soil effect, the foundation may actually be under tuned or even hover near the resonance zone when the underlying soil participates in the vibration process. Thus the amplitude of vibration could significantly vary than the calculated one. Generically, considering the soil stiffness will make the system more flexible then a fixed base problem and it can be intuitively deduced that though the stress might remain within the acceptable level the amplitude of vibration will be more and could well exceed the acceptable limit which might have secondary damaging effect to the machine and its appurtenances. For earthquake the effect is quite different. In this case the structure resting on the site can be visualized as a body resting on an infinite elastic space (similar to a ship floating in sea). Due to rupture in the fault as waves dissipate in all direction the soil mass starts vibrating at its own fundamental frequency known as the free field time period of the site. In such case the earthquake acts as an electronic ﬁlter and tries to excite the superstructure resting on it to its own fundamental frequency and suppressing or even eliminating other modal frequencies12 . Thus if the fixed base frequency of the structure matches the fundamental frequency of the soil strata on which it is resting, they are in resonance and catastrophe could well be a reality. Before dwelling into the mathematical aspect of it we further substantiate the above statement by some real life facts and observations. Dowrick (2003) reports that in the Mexico earthquake in 1957 extensive damage occurred to the buildings that were tall and were found to be resting on alluvium soil of depth >1000 m. In 1967, the Caracas earthquake showed identical result where the tall structures underwent extensive damage and those were resting on deep alluvium soil overlying bedrock. In 1970 earthquake at Gediz in Turkey a part of a factory was demolished in a town about 140 Km from the epicenter while no other buildings in the town underwent any damage! Subsequent investigation revealed that the fundamental period of the building matched the free field time period of the site. The Caracas earthquake as cited earlier also showed a distinctive pattern where medium rise buildings (5–9 storeys) underwent extensive damage where depth to bedrock was less than 100 m, while buildings over 14 stories were damaged where the depth to bedrock was greater than 150 meters. Let us see why such thing happened and how does it substantiate the free field time period phenomenon as stated earlier. The free field time period of a site is given by the equation Tn =

4H (2n − 1)Vs

(1.2.7)

12 It can be visualized as a giant hand trying to shake a small body resting on it. Since the body is much weaker to the giant it tries to follow the same phase of vibration as the soil medium.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 19

Number of Stories

120 100 80 n for RCC frame

60

n for steel frame

40 20

5 1.

2

35 1.

1.

9

05 1.

0.

6

75 0.

0.

3

45 0.

0.

15 0.

0

0

Depth of soil/Shear wave Velocity

Figure 1.2.8 Limiting value of storeys for frames.

where, T = time period of the free field soil (i.e. without the structure); H = depth of soil over bedrock13 ; n = number of mode; and Vs = shear wave velocity of the soil. Thus based on the explanation above it can be argued that if the fixed base frequency of structure is in the close proximity of the free field time period of the site the structure may be subjected to significant excitation. The above statement can be extended to a very interesting hypothesis. If we equate the free field time period of the site to the fixed base time period of the structure we can arrive at some limiting design parameters which can result in significant dynamic amplification and which should be avoided at the very out set of planning of the structure. For instance as per IS-1893 RCC moment resisting frames with no infill brick work, the fundamental time period is given by T = 0.075h0.75

(1.2.8)

Thus equating it to fundamental free field time period of the site we have 0.075h0.75 =

4H , Vs

which gives h =

160H 3Vs

4/3 (1.2.9)

Considering 1 floor is of height 3.3 m, we can further simplify the equation to n = 0.303

160H 3Vs

4/3 (1.2.10)

13 Here bedrock is perceived as that level where the shear wave velocity of soil is greater or equal to 600 m/sec.

© 2009 Taylor & Francis Group, London, UK

20 Dynamics of Structure and Foundation: 2. Applications

The curves shown in Figure 1.2.8 give limiting stories for RCC and steel frames for which resonance can occur in a structure during an earthquake as per IS-189314 for various values of H/Vs . Let us now probe the problem a bit more based on a suitable numerical problem.

Example 1.2.4 A particular site has been found to consist of 100 m soil overlying bedrock when the shear wave velocity of the soil is 222.22 m/sec. Find the limiting number of stories of height 3.3 meter for an RCC frame for which resonance can occur. What would be resonance story if the depth of the overlying soft soil is only 30 m. Solution: Based on above data H/Vs = 100/222.2 = 0.45 when H = 100 m. As per the chart as shown above the limiting story for which resonance can occur is 18. Thus for a 18 storied building resonance can very well occur and the strategy would be to build the building at least (±)25% away i.e. either it should be 23 storied or more or 14 storied or less. 30 When the depth of soil is only 30 m, H/Vs = 222.2 = 0.135. Based on the above chart the limiting story height is roughly 4-storey only. Thus to avoid resonance the building should be either more than 5-storey or less than 3-storey.

The above problem well explains the phenomenon as to what happened in the Mexico and Turkey earthquakes and perhaps challenges the myth quite prevalent in many design offices – that for one or two storied building earthquake is not important and can well be ignored. It is evident from the above problem that the response depends on the depth of soil on which it is resting and depending on the free field time period the response can either amplify or attenuate. It can well affect even a one storied building. The chart in Figure 1.2.9 shows limiting story height of buildings with infill brick panels and all other type of frames as per IS 1893 for different width of building varying from 10 meter to 50 m15 . The above theory is though explained in terms of building, can very well be adapted for any class of structure for which it is possible to establish the fundamental time period expression.

14 In this case time period for steel frame is considered as T = 0.085(h)0.75 as per IS-1893. 15 Time period of the fixed base structure considered as T = 0.09h/(d)0.5 as per, Indian Standards Institution (1984, 2002). “Indian Standard Criteria for Earthquake Resistant Design of Structures”, IS: 1893 (Part 1), ISI, New Delhi, India.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 21

40 35 n for d=10m

Number of story

30

n for d=15m 25

n for d=20m

20

n for d=25m n for d=30m

15

n for d=35m

10

n for d=40m

5

n for d=45m n for d=50m

75 0.

68

6

0.

0.

53 0.

38

3

45 0.

0.

0.

23 0.

15

08

0.

0.

0

0

Depth of soil/Shear wave velocity

Figure 1.2.9 Limiting story for building with inﬁll brick panel.

Having assessed the resonance criteria and making sure at planning stage that the two periods do not match one would still like to quantify the combined time period of the overall soil structure system and assess whether there is any amplification or attenuation of the earthquake force. Before plunging into detailed analysis based on FEM or otherwise it would be useful to have a rough estimate as to how much the underlying soil affects the overall response. Veletsos and Meek (1974) has given a very useful expression based on which it is possible to estimate the modified time period of a structure, and is given by Kxh¯ 2 k¯ ¯ T =T 1+ 1+ (1.2.11) Kx Kθ where T¯ = modified time period of the structure due to the soil stiffness, T = time 2 period of the fixed base structure, k¯ = stiffnessof the fixed base structure @ 4πgTW 2 , ¯ Kx , K = horizontal and rotational spring constant of the soil (IS-1893), h = effective θ

height or inertial centroid of the system, and, W = total weight of the structure. Based on the above expression one can immediately arrive at a rough estimate as to how strong could be soil response at the very outset of a design. We elaborate the above based on two suitable problems hereafter.

Example 1.2.5 An RCC Chimney 150 meter in height has a uniform cross section area of Ac = 8.5 m2 and moment of inertia I = 92.5m4 . Evaluate the base moment and © 2009 Taylor & Francis Group, London, UK

22 Dynamics of Structure and Foundation: 2. Applications

shear under earthquake considering the problem as fixed base as well as the soil effect. The structure is located in zone IV as per IS 1893. The structure is supported on raft of diameter 18 meter. The soil has a dynamic shear wave velocity of 120 m/sec and unit weight of 19 kN/m3 . Consider 5% damping for the analysis.16 The grade of concrete used is M30 having dynamic Econc = 3.12 × 108 kN/m2 . Solution: Height of the structure = 150 m; Area of shell = 8.5 m2 Weight of chimney = 150 × 8.5 × 25 = 31875 kN (unit weight of conc. = 25 kN/m3 ) ! ! Radius of gyration of the chimney = I/A = 92.5/8.5 = 3.298 m Thus slenderness ratio H/r =

150 = 45.4. 3.298

As per IS 1893 CT = 82.8. As per IS 1893 time period of a fixed base chimney is given by, T = CT WH . 3.13 Ec Ac where, W = weight of chimney in N; Ec = Dynamic Young’s modulus of conc. @ 3 × 108 kN/m2

31875 × 150 82.8 Thus, T = = 1.13 sec 3.13 3.12 × 108 × 8.5 For 5% damping referring to chart in IS-1893 we have Sa/g = 0.10. Thus the horizontal seismic coefficient is given by, αh = βIFo

Sa . g

Here β (Soil foundation factor) = 1.0 for chimney resting on raft, I = 1.5 Importance factor, Fo = Zone factor @ 0.25 for zone IV. This gives αh = 1.0 × 1.5 × 0.25 × 0.10 = 0.0375 The Bending moment and shear force are given by,

1

4 5x 2 x 2 ¯ 0.6 x 2 + 0.4 x M = αh W H and V = Cv αh W − H H 3H 3 H

16 In this case it is presumed that reader has some idea of how to use the code IS-1893 or is at least familiar with it.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 23

Here, Cv = a coefficient depends on the slenderness ratio and as per the present ¯ = height of c.g. of the structure above base problem is 1.47 as per IS 1893; H @ 75 meter for the problem; x = distance from the top. Substituting the appropriate values, we have 1

M = 0.0375 × 31875 × 75 [0.6(1.0) 2 + 0.4(1.0)4 ] = 89648 kN/m V = 1.47 × 0.0375 × 31875 [(5/3) − (2/3)] = 1757 kN. Considering the soil effect we have the dynamic shear modulus of soil, G = ρvs2 Or G = (19/9.81) × 120 × 120 = 27890 kN/m2 . 8GR With radius of raft = 9.0 m, Kx = , ν = Poisson’s ratio of the soil 2−ν considered as 0.35, Kx = And Kθ =

8 × 27890 × 9 = 1217018.2 kN/m 2 − 0.35 8GR3 8 × 27890 × 93 which gives, Kθ = = 83412554 kN/m 3(1 − ν) 3(1 − 0.35)

The ﬁxed base stiffness of chimney is given by 4π 2 W 4 × π 2 × 31875 k¯ = = = 100458 kN/m gT 2 9.81 × 1.132 Substituting the above numerical values in Veletsos’ equation we have Kxh¯ 2 k¯ ¯ T =T 1+ 1+ Kx Kθ

1217018.2 × 752 100458 ¯ ➔ T = 1.13 1 + 1+ = 3.2 sec 1217018.2 83412554 As per IS 1893 for T = 3.2 sec, Sa/g = 0.05, which gives αh =

0.0375 × 0.05 = 0.01875 (By proportion) 0.10

The base moment and shear are given by M=

89648 × 0.01875 = 44824 kN · m; 0.0375

© 2009 Taylor & Francis Group, London, UK

24 Dynamics of Structure and Foundation: 2. Applications

and V =

1757 × 0.01875 = 878.5 kN 0.0375

The results are compared hereafter17 Case

Moment

Shear

Remarks

Without soil With soil

89648 44824

1757 878.5

Reduction in moment and shear by 34%

The problem shows a clear attenuation of the response.

We show another example hereafter.

Example 1.2.6 Shown in Figure 1.2.10 is a horizontal vessel having empty weight of 340 kN and operating weight of 850 kN is placed on two isolated footing of dimension 8.5 m × 3 m. The center to center distance between the two foundations is 5.5 meter. The center line of vessel is at height (H f ) of 4.5 meters from the bottom of the foundation. Thickness of the foundation slab is 0.3 meter. The RCC pedestal is of width 1.0 meter, length 6 meter having height of 3.45 meter. The shear wave velocity of the soil is 200 m/sec having Poisson’s ratio of 0.3. Allowable bearing capacity of the foundation is 150 kN/m2 . Calculate the design seismic moment considering the effect of soil and without it, if the site is in zone III as per IS-1893. Consider soil density @ 18 kN/m3 and unit weight of concrete as 25 kN/m3 ? Solution: Plan are of footing = 8.5 × 3 = 25.5 m2 Af 25.5 Equivalent circular radius = = = 2.849 m π π 1 BL3 = Moment of inertia of the foundation about X-axis 12 153.5313 m4 1 LB3 = Moment of inertia of the foundation about Y-axis 12 19.125 m4

1 3 × 8.53 = 12 1 8.5 × 33 = 12

17 Without an elaborate analysis it could be an effective calculation to convince the boss that you can save some money and the worth of a dynamic soil-structure interaction analysis.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 25

yp

Hf

yp Hp

Ds

Wp Y

Lp

Lf

X

Bf

Bf Ls

Figure 1.2.10 A horizontal vessel.

Equivalent circular radius about X axis =

1 64Ixx 0.25 = 3.739183 m 2 π

1 64Iyy 0.25 Equivalent circular radius about Y axis = = 2.221 m 2 π Mass density of soil (ρ) =

18 = 1.835 kN/m3 9.81

Dynamic shear modulus = ρvs2 = 1.835 × 2002 = 73400 kN/m2 © 2009 Taylor & Francis Group, London, UK

26 Dynamics of Structure and Foundation: 2. Applications

Lateral spring in X and Y direction =

Rocking spring about X axis =

Rocking spring about Y axis =

32Gr0 (1 − ν) = 1018306 kN/m 7 − 8ν

8Gr3x = 14627886 kN/m 3(1 − ν) 8Gr3y 3(1 − ν)

= 3063462 kN/m

Moment of Inertia of the pedestal about X axis =

1 × 1 × 63 = 18 m4 12

Moment of Inertia of the pedestal about X axis =

1 × 6 × 13 = 0.5 m4 12

3EIx 3 × 3.2 × 108 × 18 = = L3 3.453

Structural stiffness of pedestal about X axis = 3.95 × 108 kN/m Structural stiffness of pedestal about Y axis = 1.10 × 107 kN/m

3EIy 3 × 3.2 × 108 × 0.5 = = 3 L 3.453

Contributing mass for the vessel empty case =

340 = 17.33 kN-sec2 /m 2 × 9.81

Contributing mass for the vessel operating case 43.323 kN-sec2 /m

=

850 2 × 9.81

=

Contributing uniformly distribute load for the pedestal = 25.5 × 25/9.81 = 64.98 kN/m The mathematical model for the pedestal thus constitute of a beam element (pedestal) having a mass lumped at its tip (mass contribution from the vessel) is shown in Figure 1.2.11. The time period of such fixed base model is given by (Paz 1991) T = 2π

(M + 0.25mb ) K

mb

Figure 1.2.11 Mathematical model for the pedestal.

© 2009 Taylor & Francis Group, London, UK

M

Dynamic soil structure interaction 27

and the modified time period considering soil effect is given by ¯2 ¯ Kx h k T¯ = T 1 + 1+ Kx Kθ The time periods and the corresponding Sa/g values as per IS-1893 for 5% damping are as show hereafter.

Sl no

Case

Time period (vessel empty) about X direction

1 2

Without soil With soil effect

0.0018 0.0579

Time period (vessel empty) about Y direction

Time period (operating) about X direction

Time period (operating) about Y direction

0.011 0.1057

0.0024 0.0771

0.015 0.1408

Corresponding Sa/g value is given by

Sl no

Case

Sa/g (vessel empty) about X direction

1 2

Without soil With soil effect

0.1000 0.2

Sa/g (vessel empty) about Y direction

Sa/g (operating) about X direction

Sa/g (operating) about Y direction

0.120 0.2

0.1000 0.2

0.130 0.2

Base shear as per IS 1893 considering Importance factor as 1.0 for vessel empty case and 1.25 for vessel in operation case we have

Sl no

Case

Shear (vessel empty) about X direction

1 2

Without soil With soil effect

21.25 42.5

Shear (vessel empty) about Y direction

Shear (operating) about X direction

Shear (operating) about Y direction

25.5 42.5

26.5625 53.125

34.53125 53.125

The moment at the foundation level is given by

Sl no

Case

Moment (vessel empty) about X direction

1 2

Without soil With soil effect

100.9375 201.875

Moment (vessel empty) about Y direction

Moment (operating) about X direction

Moment (operating) about Y direction

121.125 201.875

126.1719 252.3438

164.0234 252.3438

This case clearly shows an amplification of force considering the soil effect.

© 2009 Taylor & Francis Group, London, UK

28 Dynamics of Structure and Foundation: 2. Applications

Having established a basis of how to evaluate the coupled soil-structure interaction under dynamic loading we now extend the above theory to system with multi degree of freedom where the theory can be very well be adapted as a powerful tool for a detailed yet economic dynamic analysis.

1.3 A GENERALISED MODEL FOR DYNAMIC SOIL STRUCTURE INTERACTION In this section we present a generalised model for dynamic soil-structure interaction. Though the model is developed based on 3D frames can also be adapted for a three dimensional Finite Element analysis.

1.3.1 Dynamic response of a structure with multi degree of freedom considering the underlying soil stiffness We had shown earlier that for a single degree of freedom system the modified time period of a structure considering the soil effect is given by ¯2 ¯ Kx h k 1+ T¯ = T 1 + Kx Kθ

(1.3.1)

Both ATC (1982) and FEMA has adapted this formula for practical design office usage (Veletsos & Meek 1974, Jennings & Bielek 1973). The nomenclatures of the formula are as explained earlier. Now squaring both sides of the above equation we have T¯ 2 = T 2

¯2 k¯ kh 1+ + Kx Kθ

(1.3.2)

Considering the expression T =

2π ω

we have

4π 2 m 4π 2 mh¯ 2 4π 2 4π 2 = 2 1+ 2 + ω¯ 2 ω T Kx T 2 Kθ

4π 2 mω2 4π 2 ω2 mh¯ 2 or 2 = 2 1 + + Kx Kθ ω¯ ω (1.3.3)

Simplifying and expanding the above we have 1 m mh¯ 2 1 = + + Kx Kθ ω¯ 2 ω2 1 1 1 1 = 2+ 2+ 2 ω¯ 2 ω ωx ωθ

which can be further modified to (1.3.4)

which gives the modified natural frequency relation for a system with single degree of freedom. This formulation has also been shown in, Kramer, S. (2004). © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 29

Now considering generically ω = m m m mh¯ 2 + = + , ke k Kx Kθ

or,

! k/m we have

1 1 1 h¯ 2 = + + ke k Kx Kθ

(1.3.5)

where ke = equivalent stiffness of the soil structure system having single degree of freedom. We shall extend the above basis to multi degree of freedom hereafter (Chowdhury and Dasgupta 2002).

1.3.2 Extension of the above theory to system with multi degree of freedom A 3-D frame shown in Figure 1.3.1, is considered for the presentation of the proposed method. The frame structure has n degrees-of freedom and subjected to soil reactions in the form of translational and rotational springs. For a system having n degrees of freedom the above equation can be written in the form " # [M]n×n h2 n×n [M]n×n [M]n×n [M]n×n = + + (1.3.6) Kx Kθ [Ke ]n×n [K]n×n

Y X

O Z

Kx

-- mass points

Kθ

Figure 1.3.1 A 3-D Frame having multi-degree-of freedom with representative foundation spring.

© 2009 Taylor & Francis Group, London, UK

30 Dynamics of Structure and Foundation: 2. Applications

Here, [Ke ] = equivalent stiffness matrix of the soil structure system of order n, [M] = a diagonal mass matrix of order n having masses lumped at the element diagonals, [h¯ 2 ] = radius vectors of the lumped masses to the center of the foundation springs of order n, Kx , Kθ = translation and rotation spring stiffness of the total foundation system represented by a unique value. Taking out the common factor [M], we have " 2# h [I] [I] [I] + = + Kθ [Ke ] [K] Kx

(1.3.7)

where, [I] = identity matrix of order n having its diagonal element as 1. or [I][Ke ]−1 = [I][K]−1 + [I/Kx ] + [h2 /Kθ ] ⇒ [Fe] = [F] + [Fx ] + [Fθ ]

(1.3.8)

where [F] = Flexibility matrix of the system with suffixes as mentioned earlier for stiffness matrices. Once the flexibility matrix of the equivalent soil structure system is known the stiffness matrix may be obtained from the expression [Ke ] = [Fe]−1

(1.3.9)

Now knowing the modified stiffness matrix the eigen solution may be done based on the usual procedure of [Ke ] [ϕ] = [λe ] [M] [ϕ].

(1.3.10)

1.3.3 Estimation of damping ratio for the soil structure system While calculating the damping ratio, the normal process is to guess a damping ratio for the structure like 2–5%, and consider the same damping ratio for all the mode and obtain the value of Sa/g value for the particular structure per mode corresponding to the time period based on the curves given in IS-1893. The basis of assuming this damping ratio is purely judgmental and is dependent on either the experience of the engineer, recommendation of codes, or based on field observations on the performance of similar structure under previous earthquakes. When the effect of soil is neglected it is possible to obtain the material damping ratio of the structure depending on what constitute the material like steel, RCC etc. However when the whole system is resting on soil an analyst is usually faced with the following stumbling blocks for which clear solution is still eluding us specially for modal analysis in time domain. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 31

The difficulties encountered can be summarised as follows • • •

The damping matrix of the coupled soil-structure system becomes nonproportional for which the damping matrix does not de-couple based on orthogonal transformation. As the damping ratio of the structure and the soil could be widely varying it becomes difficult to assess a common damping ratio which would affect the soil as well as structure. Even after elaborate FEM modelling of the soil, the damping ratio contribution per mode still remains guess estimation at the best.

We present hereafter a method by which one can estimate approximately the contribution of combined soil structure system under earthquake for various modes, without resorting to an elaborate modelling of the soil itself. We only estimate the contribution of the soil damping to the structural system whose response we are interested in. The estimation is surely approximate but at least gives a rational mathematical basis to arrive at some realistic damping value rather than guessing a damping value at the outset and presuming that it remain same for each mode, specially for coupled soil structure system where widely varying damping for the foundation and structure makes it difficult for the analyst to arrive at unified rational value applicable to the system.

1.3.4 Formulation of damping ratio for single degree of freedom Neglecting the higher order, the material damping ratio for a soil structure system having single degree of freedom is given (Kramer 2004) by ζ¯ ζ ζx ζθ = 2+ 2+ 2 2 ω¯ ω ωx ωθ

(1.3.11)

where, ζ¯ = damping ratio of the equivalent soil structure system; ζ = damping ratio √ of the fixed base structure; ζx = horizontal damping ratio of the soil, where ζx = 0.288 B x

(7−8ν)mg and Bx = 32(1−ν)ρ 3 , where m = total mass of the structure and foundation; g = s rx acceleration due to gravity; ν = Poisson’s ratio of the soil; ρs = mass density of the soil; rx = Equivalent circular radius in horizontal mode; ζθ = damping ratio of the 0.15√ θg and Bθ = 0.375(1−ν)J ; and Jθ = mass moment soil in rocking mode ζφx = (1+B ρ r5 ) B θ

θ

s θ

of inertia of the foundation and the structure. Converting the damping ratio equation to stiffness-mass basis we have mζ¯ mζ mζx mh2 ζθ ζ¯ ζ h 2 ζθ ζx = + + or = + + ; ke k Kx Kθ ke k Kx Kθ ζ ζx ζθ ➔ ζ¯ = ke + + k Kx Kθ © 2009 Taylor & Francis Group, London, UK

(1.3.12)

32 Dynamics of Structure and Foundation: 2. Applications

For very high value of Kx and Kθ ke → k when ζ¯ → ζ .

1.3.5 Extension of the above theory to systems with multi-degree freedom On extending the above to multi degree of freedom of order n, we have [ζ ][M]n×n [ζx ][M]n×n [ζθ ][M]n×n [h2 ]n×n [ζ¯ ][M]n×n = + + [Ke ]n×n [K]n×n Kx Kθ ➔ [ζ¯ ] = [Ke ]{[ζ ][F] + [ζx ][Fx ] + [ζθ ][Fθ ]}

(1.3.13)

[ζ¯ ] = Damping ratio matrix of the combined soil structure system having n number of modes. It is to be noted that [ζ¯ ] is non-proportional and not a diagonal matrix, and based on the matrix operation as shown above has off-diagonal terms. A study on the parametric effect shows that [ζ¯ ] becomes nearly a diagonal matrix (i.e. the off diagonal terms vanishes or approaches zero) when damping ratio of the structure and the soil foundation system are nearly equal. However, when the damping ratio are widely varying the off diagonal terms do not vanish however there magnitudes are relatively smaller than the diagonal terms (ζii ) which has the most dominant effect on the system. Thus if it is possible to arrive at a foundation layout where the damping ratio of the structure and foundation are closely spaced considering the diagonal terms as modal damping ratio per mode is quite correct. Even when the off diagonal term exists due to widely varying values for practical design engineering purpose considering the ζii term of damping ratio matrix is realistic for it gives a reasonably rational basis of estimation of the damping ratio per mode rather than guessing a value based on gut feeling. We explain the above theory based on suitable example hereafter

Example 1.3.1 Shown in Figure 1.3.1 is a three storied steel frame subjected to dynamic forces. The damping ratio for steel is found to vary between 2 to 5%. Determine • • • • • •

The fixed base natural frequencies of the structure. The fixed base eigen-vectors. Modified natural frequency with foundation stiffness. Modified eigen. Take K x = 35000 kN/m and K θ = 50000 kN/m for the soil-foundation. Analyse the floor shears for earthquake based on IS-1893 Zone III for ◦ ◦

Fixed base. Considering the soil effect.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 33

G

H

X3

E

F

X2

C

D

X1

A

B

3000

3000

3000

Figure 1.3.2

Here, 1 2 3

K AC = KDB = 1.5 × 103 kN/m M GH = 200 kN sec2 /m K CE = KDF = 1.0 × 103 kN/m M EF = 400 kN sec2 /m K EG = KFH = 0.75 × 103 kN/m M CD = 400 kN sec2 /m

Solution: The stiffness and mass matrix is given by ⎡

5000 [K] = ⎣−2000 0 •

⎤ −2000 0 3500 −1500⎦ −1500 1500

⎡

⎤

and [M] = ⎣

400 400

⎦ 200

Based on Figure 1.3.2, we have found earlier that √ √ ω √1 = 1.6426 = 1.281 rad/sec; ω2 = 10.00 = 3.162 rad/sec; ω3 = 17.104 = 4.135 rad/sec.

© 2009 Taylor & Francis Group, London, UK

34 Dynamics of Structure and Foundation: 2. Applications

Thus the time periods for the fixed base structure is given by18 T1 = 4.97 sec, T2 = 1.987 sec, T3 = 1.52 sec •

The mode shapes or the eigen-vectors are ⎡

1.00 1.0 0.5 [φ] = ⎣2.1715 2.7816 −1.50 •

⎤ 1.0 −0.9208⎦ 0.719

Normalised eigen vectors ⎡

0.01615 0.03244 [ϕi ] = ⎣0.0350718 0.01622 0.04493 −0.02433

⎤ 0.0344512 −0.03172 ⎦ 0.02477

Calculation for the combined soil-structure system Here stiffness matrix of the fixed base structure ⎡

5000 [K] = ⎣−2000 0

⎤ −2000 0 3500 −1500⎦ which on inversion gives −1500 1500

⎡

0.000333 [F] = ⎣0.000333 0.000333

0.000333 0.000833 0.000833

⎤ 0.000333 0.000833⎦ 0.003145

⎡

⎤ 1/35000 0 0 0 1/35000 0 ⎦ [Fx ] = ⎣ 0 0 1/3500 ⎡ ⎤ 2.85714 0 0 ⎦ × 10−5 0 2.85714 0 =⎣ 0 0 2.85714 ⎡

9 [h2 ] = ⎣0 0

0 36 0

⎤ 0 0⎦ 81

18 You can check the value by any of the method as explained in Chapter 5 (Vol. 1) for eigen value analysis.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 35

Thus ⎡

9/50000 0 [Fθ ] = ⎣ 0

0 36/50000 0

⎡

0.00018 0 0 0.00072 =⎣ 0 0

⎤ 0 ⎦ 0 81/50000 ⎤ 0 ⎦ 0 0.00162

As [Fe] = [F] + [Fx ] + [Fθ ] we have ⎡

0.000542 [Fe] = ⎣0.000333 0.000333

⎤ 0.000333 0.000333 0.001581905 0.000833⎦ 0.0008333 0.001982

which is combined flexibility matrix of the soil structure system. Inversion of the above flexibility matrix gives ⎡

2195.19 [Ke ] = ⎣−344.34 −224.42

⎤ −344.34 −224.4212 804.662 −306.223 ⎦ −306.223 671.0682

The above gives the combined stiffness matrix for structural system considering the soil compliance19 . Thus based on the above modified stiffness matrix and mass matrix as ⎡ [M] = ⎣

⎤

400

⎦

400 200

√ √ ω1 = 1.2163 = 1.10286 rad/sec; ω2 = √ We have, based on eigen solution, 3.9666 = 1.9916 rad/sec; ω3 = 5.8255 = 2.4136 rad/sec. Thus the time periods for the combined soil-structure system is given by T1 = 5.697 sec, T2 = 3.154 sec, T3 = 2.603 sec

19 Watch the numbers. . . . . it is symmetric and is completely different than when you add the springs directly to the diagonal. This matrix has no rigid body mode and can be used directly for static analysis too. Moreover if we take Kx and Kθ very high the Ke converges to the fixed base matrix K.

© 2009 Taylor & Francis Group, London, UK

36 Dynamics of Structure and Foundation: 2. Applications

Normalised modified eigen vectors considering soil stiffness is given by ⎡

⎤ 0.0479 0.00589 −0.00772 −0.0276⎦ −0.0169 0.05835

−0.013 [ϕi ] = ⎣−0.0409 −0.0360

Calculation of modal damping Considering, ζ = 5% for the structure, ζx = 10% for the soil in translation mode, ζθ = 15% for the soil in rocking mode We have, [ζ¯ ] = [Ke ]{[ζ ][F] + [ζx ][Fx ] + [ζθ ][Fθ ]} Substituting the values as mentioned and calculated above we have ⎡

0.092 [ζ¯ ] = ⎣−0.007 −0.002

−0.028 0.10908 −0.0126

⎤ −0.020 −0.028⎦ 0.1115

It will be seen that that the main diagonal terms are dominant and can be considered as the modal damping ratio contribution for each mode. Suppose we had closely spaced damping data like ζ = 5% for the structure; ζx = 6% for the soil in translation mode; ζθ = 5.5% for the soil in rocking mode, the modal damping matrix reduces to ⎡

⎤ 0.0525 −0.0015 −0.001016 0.05312 −0.00144 ⎦ [ζ¯ ] = ⎣ −0.0004 −0.00014 −0.00066 0.05315 When the matrix become practically diagonal dominant with off diagonal terms having very low values. Thus for the present problem ζ may be considered as ζ1 = 9.2% for first mode, ζ2 = 10.9% for second mode; and ζ3 = 11.1% for the third mode. Calculation of earthquake force fixed base structure m

φ1

mφ1

mφ12

φ2

mφ2

mφ22

φ3

mφ3

400 0.01615 6.46 0.104329 0.03244 12.976 0.420941 0.03445 13.7804 400 0.03507 14.028 0.491962 0.01622 6.488 0.105235 −0.01372 −5.488 200 0.04493 8.986 0.403741 −0.02433 −4.866 0.118389 0.02477 4.954 29.474 1.000032 14.598 0.644565 13.2464

© 2009 Taylor & Francis Group, London, UK

mφ32 0.47475407 0.07529536 0.12271058 0.67276001

Dynamic soil structure interaction 37

Modal mass participation factor κ1 =

29.474 = 29.47306 1.000032

for the first mode,

κ2 =

14.598 = 22.64777 0.644565

for the second mode,

κ3 =

13.2464 = 19.689 0.67276

for the third mode.

Assuming 5% damping for the structure we have, Mode

Time period (secs)

Sa (m/sec 2 )

Remarks

1

4.9

0.4905

2 3

1.98 1.52

0.6867 0.7848

Sa value obtained from the chart given in IS-1893 for 5% damping Do Do

For zone III: K = 1.0, β = 1.0, I = 1.2F0 = 0.2 as per the code Thus base shear is given by; V = 3i=1 K · I, β · F0 · κi · Sa mi φi Substituting data on the above formula we have Mode

Base shear V

Remarks

1 2 3

102 5.45 4.91

Fixed base case

Calculation for coupled soil-structure interaction. m

φ1

mφ1

mφ12

φ2

mφ2

mφ22

φ3

mφ3

mφ32

400 −0.013 −5.2 0.0676 0.0479 19.16 0.917764 0.006 2.4 0.0144 400 −0.041 −16.4 0.6724 −0.0077 −3.08 0.023716 −0.0276 −11.04 0.304704 200 −0.036 −7.2 0.2592 −0.0169 −3.38 0.057122 0.0583 11.66 0.679778 −28.8 0.9992 12.7 0.998602 3.02 0.998882

Modal mass participation factor κ1 =

−28.8 = −28.8231 0.9992

for the first mode,

κ2 =

12.7 = 12.717 0.998602

for the second mode, and

κ3 =

3.02 = 3.0233 0.9988

© 2009 Taylor & Francis Group, London, UK

for the third mode.

38 Dynamics of Structure and Foundation: 2. Applications

Modal damping for each mode, as calculated earlier. Mode

Damping

Time (sec)

Sa (m/sec 2 )

Remarks

1

9.2%

5.7

0.343

2 3

10.9% 11.15%

3.2 2.6

0.294 0.245

Calculated from curve based on interpolation corresponding to 9.2% damping Do- with 10.9% damping Do- with 11.15% damping

Calculation for Base shear Base shear for the frame with coupled soil-structure interaction is given by Mode

Base shear V

Remarks

1 2 3

68.4 11.4 0.537

Couple soil-foundation system

Calculation of storey forces The storey forces for the two cases are calculated hereafter Coupled soil structure system

Storey m

h mh2

mh2

3

i=1

1st 2nd Top

mh2

Base shear mode 1

Base shear mode 2

Fixed base Base shear mode 3

Base shear mode 1

Base shear mode 2

Base shear mode 3

400 3 3600 0.10526 7.20 ×10+00 1.20 ×10+00 5.66 ×10−02 1.08 ×10+01 5.74 ×10+00 5.17 ×10+00 400 6 14400 0.42105 2.88 ×10+01 4.80 ×10+00 2.26 ×10−01 4.31 ×10+01 2.29 ×10+01 2.07 ×10+01 200 9 16200 0.47368 3.24 ×10+01 5.40 ×10+00 2.55 ×10−01 4.84 ×10+01 2.58 ×10+01 2.33 ×10+01

Comparison of results Time period Structure type

T1

T2

T3

Fixed base structure Soil-structure interaction

4.9 5.697

1.987 3.154

1.52 2.603

The time periods are increasing with introduction of soil springs as predicted at the outset. Acceleration Structure type

Mode 1

Mode 2

Mode 3

Fixed base structure Soil-structure interaction

0.4905 0.34335

0.6867 0.2943

0.7848 0.245

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 39

The acceleration decreases with soil-structure effect in this case Damping Structure type

Mode 1

Mode 2

Mode 3

Fixed base structure Soil-structure interaction

5% 9.2%

5% 10.9%

5% 11.15%

Damping constant for all mode for fixed base case varies with mode for coupled analysis but is neither 5% min. nor 15% maximum but somewhere in-between which is quite logical. Base Shear (kN) Structure type

Mode 1

Mode 2

Mode 3

Fixed base structure Soil-structure interaction

102 68.4

54.5 11.4

49.1 0.537

➔ A signiﬁcant reduction in base shear, considering the soil effect, though conceptually it can be predicted that amplitude of vibration will increase. Shear Force per floor

Modes ➔ Storey 1 2 Top

Fixed base

Coupled with soil

Fixed base

Coupled with soil

Fixed base

Coupled with soil

1

1

2

2

3

3

10.8 43.1 48.4

7.2 28.8 32.4

5.74 22.9 25.8

1.2 4.8 5.4

5.17 20.7 23.3

0.0056 0.226 0.255

➔ Significant variation in floor shears per mode.

Based on the above example it can be concluded that •

• • • •

The major advantage with this technique is the calculation of the time period without resorting to an elaborate modelling of the soil. Two representative spring value for the foundation is capable of modifying the stiffness of the super-structure having any conceivable degree of freedom. This cuts down significantly the modelling as well as the cost of computation. No rigid body motion exists. Stiffness matrix of the soil structure system is symmetric and real. The structure can be discretized to as many degrees of freedoms one choose to select.

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40 Dynamics of Structure and Foundation: 2. Applications

• • • •

Beam, plates, shell, bricks anything can be used to model the super structure system thus do not generically violate the procedures followed for FEM analysis of the superstructure. Since the matrix has no rigid body mode may be also be used directly for calculating the static response too. No additional computational effort is required. Though approximate, furnishes a rational basis of estimating the modal damping ratio per mode for the coupled soil structure-system. The results are logical and in general satisfies the trend as observed based on more rigorous analysis based on complex damping and eigen value problem (where a matrix of order n × n gets inflated to the order 2n × 2n thus adding to the cost of computation).

1.3.6 Some fallacies in coupling of soil and structure (Chowdhury 2008) You will observe here that we had advocated two types of coupling of soil spring, one vide Equation (1.1.3) where the soil spring is directly added to the diagonal stiffness element of the structural matrix, meaning thereby that it is a parallel connection and the other by Equation (1.3.7) which shows that the spring are in series. The first method has developed from the theory of nodal compatibility and is a very popular technique in practice for the root of its development is in the realms of matrix analysis of structure and can very well be adapted in commercially available software. While the second formulation is developed in the frequency domain analysis as suggested by Veletsos for a harmonic oscillator having single degree of freedom coupled to a translational and rocking spring. The question that remains as to which one is more realistic and gives the true interaction of the soil with structure especially when we model the soil as boundary springs. One of the major flaws in parallel spring model is, as the boundary elements are discrete and not a continuum it only gives a local effect and also affects the structural node only locally. The intention here is not to challenge or shock the structural engineers who have been doing this for ages. But putting on the hat of a theoretical physicist and probing this formulation a bit more – it comes up with some very interesting result. Let us imagine that the beam in Figure 1.1.1 is made of RCC of say dimension 450 × 900 supported on a compliant foundation where the soil is modeled as a spring. Now we put a motor on the beam which gives some dynamic force Psin wm t-say. We want to find out the dynamic response of the beam. The problem shows no ambiguity for the beam along with the soil spring vibrates with natural frequencies that can be obtained based on the lumped mass matrix at node i & j and the stiffness matrix derived vide Equation (1.1.3) and then subsequent amplitude and stresses can be calculated by the usual procedure. Now let us presume 100 years down the road scientists have developed a material whose Young’s Modulus is say 9 × 1020 kN/m2 and we build this beam (of same dimension) with this material (not the spring which represents the soil) and pose the question as to does the system vibrate? Looking at Figure 1.1.1 one can intuitively say that yes it does vibrate, but the beam here being very stiff (lim k → ∞) undergoes rigid © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 41

body mode and the vibration is now guided by the stiffness of the spring only. Now the question is – does Equation (1.1.3) reflects this phenomenon – amazingly not! For putting this value of E = 9 × 1020 kN/m2 we find that [K]g becomes an infinitely stiff matrix where the poor Kii and Kjj (whose order would be of 105 to 106 ) is completely gobbled up by the stiffness values of the beam that are exponentially higher and would start giving time periods that are zero. Like patch test in FEM, it is a test we can use to check the sanctity of a stiffness formulation. We call this an RB (short of Rigid Body) test and we see it fails this test with parallel spring connection, especially when the structure has got signiﬁcant stiffness compared to soil. Now if we put Equation (1.3.7) which is the series connection, to RB test, we find that it passes the test with flying colors for as Limit of K → ∞ the first term in the right hand side of Equation (1.3.7) approaches zero and we are left with the soil springs values only based on which the body vibrates and satisfies RB test conditions posed earlier. In Equations (1.3.8) and (1.3.9) it is clearly seen that the soil flexibility gets directly added to the diagonal and then on inversion affects all the terms of the [Ke ] and gives the true interaction unlike parallel spring which affects only locally the interaction effects and does not possibly gives a true picture when the stiffness of the superstructure becomes quit high compared to that of the soil.

1.3.7 What makes the structural response attenuate or amplify? In Example 1.2.5 and 1.2.6 we had shown two opposite cases of dynamic soil structure response. While in the case of the chimney the response is attenuated, in case of the horizontal vessel the response is however significantly amplified. One would obviously be curious and wonder why does it happen? The riddle is surely not difficult to answer. Shown in Figure 1.3.3 is the generic nature of the acceleration curve used for design of structures under earthquake. The nature of the curve is almost common/similar for all the earthquake codes around the world. Based on the curve (Figure 1.3.3) it is evident that when the structure is very stiff or massive and its fixed base time period hovers around the vicinity of point A, the dynamic soil structure effect can show signiﬁcant ampliﬁcation so long as the coupled time period of the soil-structure system is within the zone C. Thus structures like massive gravity dams, nuclear reactor buildings, Massive turbine foundations20 , large vessels supported on short pedestals (which are stiff) could show signiﬁcant ampliﬁcation in response when the effect of soil is considered in the analysis. While for any structure whose fixed base time period is somewhere between point B and C, if exceeds the point C consideration of the soil effect can undergo a major attenuation. Normal buildings, RCC, Steel Chimneys, elevated water tanks etc would possibly fall in this category. Thus depending on the stiffness of the structure, its mass distribution, dynamic property of the soil one can either save some money (if there is

20 For instance the structural conﬁgurations used for old LMW type Russian turbo-generators used commonly in India for 210 MW plant.

© 2009 Taylor & Francis Group, London, UK

42 Dynamics of Structure and Foundation: 2. Applications

B

C

Sa g

A D

Time period (sec)

Figure 1.3.3 Generic response spectra curve for earthquake.

attenuation) or could result in more costly design (for amplified response) which may vary from case to case.

1.4 THE ART OF MODELLING Computer Modelling of soil & structure optimally to arrive at a meaningful solution is an art by itself, and can well be a topic of a complete book. We present hereafter some major techniques that has been found to effective & reasonable.

1.4.1 Some modelling techniques Experience shows that in many cases young engineers eager on get-going mode would start from the very outset with an elaborate model of the whole soil-structure system21 . They spend significant amount of time on data input and checking of such massive model and come up with a result whose qualitative difference with a much simpler model is only marginal. Moreover trying to handle a big data-base, an inadvertent modeling or input data error passing the scrutiny is not at all uncommon. So at the very outset our suggestion would be, start with a simple model without trying to over sophisticate the issue from the very out set22 . Start with a test case or a simplified model to check the results. For instance a simple model given in a book or the user manual of the software in use is a very good starting to have some idea what types of element to choose, what order of refinement suffice and what type of simplified idealization is acceptable.

21 Problem modeled with a minimum 1000 degrees of freedom! 22 Mentioning the fact that you have used eight nodded brick elements, or 9-noded plate elements based on iso-parametric formulation may look impressive as technical jargons in a design basis report but may not always be cost-effective solution.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 43

Super structures above ground

Ground level

Railway Carriage Underground Tunnel

Figure 1.4.1 An underground tunnel for movement of train in a metro city.

You will be amazed to find that in most of the cases, modeling the soil intelligently as linear springs (whose values are judiciously chosen) can be good enough for many major soil structure interaction analyses. Specially, when the structure is modeled in 3D, avoid using Finite elements to model soil and coupling it to the structure. Firstly, the model becomes huge resulting in more engineering time plus gives results which become difﬁcult to decipher and does not necessarily always gives a more accurate or better result compared to a relatively simpliﬁed model. Start with a simple model (preferably a stick model) and add the soil spring to get a first order feel of how much the soil affects its response23 . Get a basic feel as to how much the results vary in terms of fixed base problemif found significant one should then and only then resort to a much more detailed analysis. If the variation is say within 15%, one can well ignore the soil effect and consider the problem as a standard fixed base problem and proceed with the analysis. Keep your eyes open but do not be biased on the issue. Optimize your engineering effort to the best possible way. There are certain types of problem where resorting to FEM however would become almost essential. For instance for the problem considered in Figure 1.4.1, it would be impossible to arrive at reasonable solution without an application of FEM. Shown in Figure 1.4.1 is a sketch of an underground tunnel catering to movement of high speed trains. The movement of train generates dynamic forces which travels through the soil to the surface and could adversely affect the structures built on the surface like buildings, water tanks etc and becomes an important study for engineers undertaking such kind of projects.

23 A computer analysis is not mandatory at this stage, a simple hand calculation or an analysis in spread sheet or MATHCAD would suffice.

© 2009 Taylor & Francis Group, London, UK

44 Dynamics of Structure and Foundation: 2. Applications

It is but evident that for these cases of modeling, the soil as spring element will not work and a comprehensive finite element modeling of the soil based on plane strain element is required. Here also, while doing the modeling, our suggestion would be start with a crude model (say 20 to 30 elements) to get a fill of the first order effects and then progressively refine the model to get a more accurate result. In static loading case in Chapter 4 (Vol. 1) we had explained the principles of meshing of such plane strain problem. Under dynamic loading the principles meshing are generally done based on the following 1 2

3 4

Find the time period of the exciting frequency (Ts ) of the soil medium as 4H/vs . If vs is the shear wave velocity of the soil medium then for λ being the wavelength of the propagating waves they are related by vs = f λ. Here f is the natural frequency of the medium and f = 1/Ts . Thus obtain λ = vs · Ts . The mesh size should preferably be λ/10 to λ/4 for linear or bilinear/quadratic elements chosen.

One of the major limitations in FEM for wave propagation problem is that the boundary has to be taken to a signiﬁcant distance away from the source to ensure no waves are reflected back which would otherwise generate spurious modes. This often makes the problem expensive in terms of data input, checking and run time. Moreover, it is difﬁcult to gauge at the outset as to where can the boundary be terminated. Infinite finite element as discussed in Chapter 4 (Vol. 1) is one alternative which has been found to have a strong potential for catering to such problem. Other than this, paraxial boundaries or providing viscous dampers at the boundary of soil domain capable of absorbing the propagating waves are often used for this type of problems24 . Else boundary elements have also been used to model such infinite domains and are coupled to the superstructure (modeled by FEM) and an effective solution has been sought. Unfortunately most of the commercially available software do not have the provision of adding matrix which can be assembled to the FEM matrix and an engineer has to write his own special purpose software to cater to such problems. Finally a word on the soil. . . . . . Irrespective of whether we use springs or finite element to model the soil, the fundamental property on which the stiffness depends, are the value G (Dynamic shear modulus) and ν, the Poisson’s ratio. We had discussed in detail as to how to arrive at the appropriate design values of these two parameters in the next section. In spite of all the techniques used it should be clearly mentioned that the parameters are still marred by uncertainties and the results thus obtained should be mellowed with some judgment which comes out only of experience and sustained practice.

24 Refer Chapter 5 (Vol. 1) for detailed discussion on this issue.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 45

It is always preferable to do some parametric study by varying the design soil values by (±)15 to 20% (depending on how reliable and exhaustive has been the geotechnical investigation) and check how much these results affect the design values and preferably a conservative and safe value should be chosen (based on this variance). We mention in Table 1.4.1, some suggestive models for different classes of structures where we start with a primary model (i.e. to get a basic feel of the response) and a secondary model which is a further improvement to the primary model.

Table 1.4.1 Some suggestive models. Sl. Structure No. type 1

2

3

4

5

Primary model

Framed building

Secondary model

Remarks

Stick model with soil 2D or 3D frame system considered as two uniwith masses lumped at que spring (rotational nodes. Soil modeled as and translational) consti- springs under each tuting all the foundations individual foundation High-rise 2D frame for the beam 3D frame for the beam The horizontal slab building column system while the column system with need not be too with shear shear wall modeled as shear wall modeled as reﬁned and should walls an equivalent cantileplain stress elements. be good enough to ver with soil springs The horizontal slabs generate requisite under each column and modeled as plane stress stiffness in its own the shear wall elements. Soil modeled plane as springs below each foundation Chimneys 2D stick model with soil No further reﬁnement is For local effect model and elevaidealized as springs usually warranted unless the shell or superted water some local effect of soil structure as a stick tanks is required to be studied and the soil a axis on surrounding structure. symmetric plain strain element Frames 2D frames with soil A detailed 3D model Refer Chapter 2 (Vol. 2) supportmodeled as springs. constituting of beam on detailed modeling ing rotary Bottom raft considered elements with master and technique for these machines inﬁnitely stiff hence only slave node option. The type of foundation. lumped mass contribubottom raft discretised tion is taken. Soil modeinto beam or plate elemled as springs. 3 to 4 ents with soil modeled degrees of freedom as springs and connected usually sufﬁce. at each node of the raft elements. Dams and A simple stick model with A comprehensive 2D embanksoil modeled as springs model with the dam ments else time period may be broken up into plain found from formula strain element and soil suggested in code and modeled as springs or modiﬁed by Veletsos’s further reﬁned into 2D formula. plain strain element depending on the complexity of the soil or the importance of the dam

© 2009 Taylor & Francis Group, London, UK

46 Dynamics of Structure and Foundation: 2. Applications

1.4.2 To sum it up Dynamic soil structure interaction is still in its early days and investigators are still looking for answers to many problems which are encountered in practice. For instance soil are modeled as linear springs based on elastic half space theory, considering it as a linear isotropic medium, but in reality it is not so. Layered soil phenomenon, pore pressure dissipation under dynamic loading, liquefaction potential and its effect, infinite domain problem, non linear and inelastic behaviour, radiation and geometric damping are some of the important factors on which research is still in progress to arrive at a more realistic model amenable to design office practice. What has been presented in this chapter is only an introductory concept and what is in vogue in practice at the present. Hopefully in days to come our understanding in some of the issues mentioned above will be more profound and engineers and researchers would come up with results which would be more realistic and reliable. However a word of caution should be pertinent at this juncture. As stated earlier as the uncertainty plaguing the problem is many, one should not loose the final outcome of what we are trying to achieve i.e. a safe and sound structure which can stand the vagaries of nature. So one should not get lost in the maze of sophisticated mathematics and try to always economize on the structure based on what the computer out put reflects25 . For facilities important to society the results should always be mellowed with sound engineering practice like good detailing, robust geometric configuration, and good quality of time tested construction practice. All these aspects are equally important for a structure to survive the wrath of Mother Nature whose ways are still not very clearly known to us. 1.5 GEOTECHNICAL CONSIDERATIONS FOR DYNAMIC SOIL STRUCTURE INTERACTION In this section we deal with the geotechnical considerations which go into the process of a successful dynamic soil-structure interaction analysis. At the very outset we would request readers specially with a strong structural leaning not to ignore this section. For our experience shows that nemesis of many mistakes lies in misinterpretation of this particular topic. As such before launching yourself into linear or non-linear finite element analysis of soil-structure system, the conceptual aspect of the influencing soil parameters, its limitations and its effects should be clearly understood. As a pre-requisite, we expect that you have some background on. . . • •

Some fundamental concepts of Soil Mechanics Basic concepts in Soil Dynamics

25 The output is nothing but a reflection of man’s limited knowledge of nature and only an approximate quantification of an idealized mathematical model which could be in significant variance to reality in spite of our best effort.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 47

1.5.1 What parameters do I look for in the soil report? To start with we pose the above fundamental question. To readers having some background on this issue may find it intriguing for his obvious answer would be the dynamic shear modulus (G) and Poisson’s ratio (ν). The obvious query that subsequently comes to mind is, does it require a full section to be devoted to this issue? The answer would surely be an emphatic yes, for in our opinion the values adapted are often misunderstood/abused in many a case, and often makes the analysis questionable or unrealistic. The reasons that could be attributed to it are as follows: • • • • •

Geotechnical test (lab or field) based on which data evaluated are not understood properly. As the limitations of such data are not clearly made; often results in incorrect interpretation. Data considered are often not relevant or correct in terms of real situation in the field, specially for layered soil. Insufficient data and or lack of knowledge on the strain level to which the foundation-structure system will be subjected to – specially during earthquake. Lack of dynamic test data and improperly co-related value from static soil parameter which could be widely varying with the reality. Finally, often forgetting the bottom line that unlike man made material like concrete and steel, soil is far more heterogeneous and unpredictable; thus for a real soil structure interaction it is unfair to have an analysis on an absolute scale. It should preferably be done for a particular range of values and the best estimate is to be made out of it – and this is where engineering judgment would count to a large extent.

Having made the above statements, let us evaluate various aspects of dynamic property of soil which are important for an integrated soil-structure interaction analysis. Before even looking at soil report the analyst should be clear with himself on • • •

The type of structure he is dealing with Type of foundation that is anticipated like shallow foundation (could be isolated or combined footing), raft or piles etc. What analysis he is looking for like is it an analysis for machine induced load, earthquake, blast force etc.

Understanding of the above criteria will not only help him in understanding the data obtained from different tests but could also possibly make him realize their interpretation in a more realistic perspective. The engineering parameters we look for in the soil report for developing the soil model either for finite element or linear/non-linear spring dashpot model are © 2009 Taylor & Francis Group, London, UK

48 Dynamics of Structure and Foundation: 2. Applications

G2

G1

(shear stress) 2

1

1

2

Figure 1.5.1 Shear stress-strain curve of soil under cyclic loading.

1 2 3

Dynamic shear modulus (G) or shear wave velocity (vs )26 Poisson’s ratio (v) Damping value of soil both radiative and material.

The values are usually obtained either from field test, laboratory test or from theoretical co-relation with other engineering soil parameters. Before we step further into the topic it would possibly be worthwhile to understand how soil behaves under cyclic loading and what its characteristics are. It should be remembered that even under low strain, soil behavior is essentially non-linear though at low strain it does show some kind of linearity. Shown in Figure 1.5.1, is the shear stress-strain curve of soil under cyclic loading. It is evident from the above figure that shear strain varies with stress, and goes on increasing with number of cycles of loading. Thus before an analysis is being carried out one has to have an idea about the average strain range to which the soil will be subjected to under the induced dynamic loading. The characteristic curve which shows the variation of shear modulus with respect to shear strain is shown in Figure 1.5.1a. The curve shown above is otherwise known as Seed and Idriss’s (1970) curve which shows the variation of dynamic shear modulus of soil with shear strain.

26 Relationship being G = ρvs2 .

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 49

1.2 1

G/G0

0.8 0.6 0.4 0.2 0

0

0.01

0.1

1 Strain Ratio

10

100

1000

Figure 1.5.1a Variation of Shear Modulus with strain under cyclic loading. (Seed & Idriss 1970).

Soil subjected to stress by machine foundation are usually low strain and varies between to 10−4 to 10−3 %. However for an earthquake of even moderate magnitude this will be much higherhaving strain range varying to 10−2 to even 10−1 % for very severe earthquake. Since it is difficult to gauge at the outset of an analysis how much strain the soil will be subjected to, the correction factor to be used to modify the data as obtained in the soil report becomes difficult to quantify. On the contrary rendering no correction would result in assuming a more stiff soil and the result obtained based on this could be significantly varying from the reality. Fortunately or unfortunately most of the tests carried out in the field or in the laboratory for determination of the dynamic shear modulus is based on low strain range having values restricted to 10−4 %. Thus it should be clearly understood that the dynamic shear modulus data furnished in the soil report is only valid for LOW strain range and can be only used directly for analysis where the strain induced in the soil is significantly low like in design of machine foundations only. For earthquake analysis where the site is situated in an area of moderate to severe earthquake zone, direct use of such soil dynamic data may not be valid for design of normal structures, for the strain induced in soil is much higher. 1.6 FIELD TESTS The most common field tests that are carried out at site for evaluation of dynamic shear modulus or shear wave velocity are 1 2

Block Vibration Test Seismic cross hole

1.6.1 Block vibration test In block vibration test as shown in Figure 1.6.1, an oscillator is placed on a concrete block of size 1.5 m × 0.75 m × 0.7 m resting at foundation level and induces dynamic © 2009 Taylor & Francis Group, London, UK

50 Dynamics of Structure and Foundation: 2. Applications

Oscillator Lx

Fdn Level

Propagating waves H=0.6 to 1.2 m

Figure 1.6.1 Schematic diagram for block vibration test.

loading on the soil. Two geo-phones are placed at a distance to pick up the signal from the oscillator. Once the oscillator induces dynamic force on the soil the geo-phones pick up this signal and transfer them to an oscilloscope which shows an elliptical figure of Lissajous. The operating speed of the oscillator is varied till the time the natural frequency of the soil and the operating frequency of the oscillator matches (the Lissajous’ figure in the oscilloscope becomes a perfect circle). The shear wave velocity of the site is then given by vs = 4fLx

(1.6.1)

where vs = shear wave velocity of the soil; f = operating frequency of the oscillator in cps; Lx = distance between the two geo-phones. For arriving at meaningful results usually high frequency oscillators (>100 cps) are put to use for which the waves generated are of the order of 0.6 to 1.2 m. Thus results obtained from this test only influence soil of depth 0.6 to 1.2 m below the depth of foundation and should not be used where piles or other types of deep foundations having influence area propagating much deeper is used. Trying to induce lower frequency calls for much heavier oscillators which make the test uneconomical compared to other types of tests.

1.6.2 Seismic cross hole test As shown in the schematic sketch in Figure 1.6.2, a probe is placed in a bore hole to the desired depth and shear wave is generated in the soil by hitting it hard with a hammer. The waves are picked up by a geo-phone entrenched firmly to the casing of another bore hole located at a known distance (Lx ) from the first hole. The time taken to pick up the signal is measured by the oscilloscope. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 51

Oscilloscope Hammer

Ground level Bore Hole casing

Probe

Geo-Phone

Lx

Figure 1.6.2 Schematic diagram for seismic cross bore-hole test.

The dynamic shear modulus Gdyn is then obtained from the expression

Gdyn

ρ = g

Lx t

2 (1.6.2)

where, Gdyn = dynamic shear modulus of the soil; ρ = weight density of soil; g = acceleration due to gravity; Lx = distance between the two bore holes, and t = elapsed time. One of the major advantage with this test is that dynamic shear modulus can be measured to any desired depth and can very well be an integrated part of a SPT program. The test is very effective in case the soil is layered in nature where visual inspection of each layer is possible based on SPT test. However, the strain range for test is again restricted to 10−4 % which is normally less than the strain range experienced by machine foundations and earthquake analysis and needs to be corrected to arrive at the design value of G.

1.6.3 How do I co-relate dynamic shear modulus when I do not have data from the dynamic soil tests? A not so uncommon phenomenon, that even puts an experienced engineer under difﬁcult situation at times. In many cases it has been observed that no dynamic test has been carried out during the geo-technical investigation – especially if it is a building © 2009 Taylor & Francis Group, London, UK

52 Dynamics of Structure and Foundation: 2. Applications

project. Though not unusual, but should not happen as a rule, for this shows the lack of foresight on the part of the engineer while submitting the technical and commercial proposal for a project. Even at the proposal stage the process involved in a plant is well known to the bidder and all the concerned civil engineer has to do is to check with his process department and find out if rotating machines are part of the process or not. On the other hand knowing the location of a particular site one can easily find out from the codes how active this zone is seismically and if felt reasonable all he has to do is to include this additional cost of dynamic geotechnical investigation in his commercial bid. People suffer from misnomer that dynamic tests are expensive-which is actually not true, for an average dynamic test in international market takes roughly US$ 20,000–25,000 which would however be 0.25% of a small petrochemical refinery and possibly 0.1% of a combined cycle 350 MW power plant. Lack of these tests can land up some of the equipments operating in such projects into serious problem whose cost itself would constitute 30–40% of the whole project cost! So one has to decide on the risk involved – and come to a conclusion of its worth. Though theoretical co-relation exist for evaluation of dynamic shear modulus of soil from static soil test (which has been successfully used in project works), it is always preferable to have these dynamic tests carried out at site, for not only does it imbibe more conﬁdence in the design process but engineer should also be aware that “theoretically co-related values have also varied widely with respect to actual field data, and should be mellowed with judgment.” Considering the uncertainty prevalent in soil, is surely not an easy task to accomplish.

1.7 THEORETICAL CO-RELATION FROM OTHER SOIL PARAMETERS The most outstanding work in establishing theoretical co-relation for evaluating the dynamic property of soil has been done by Hardin, Drnevich, Richart, Seed, Idriss to name a few27 . The expressions suggested by them have been successfully used for many real projects by the engineers in the past. We are going to have a look at some of them hereafter and understand their limitations if any.

1.7.1 Co-relation for sandy and gravelly soil 1.7.1.1 Hardin and Richart’s Formula For rounded grained soil having void ratio less or equal to 0.8 the dynamic shear modulus is given by (Hardin and Richart 1963) G=

2630(2.17 − e)2 √ σ0 in psi 1+e

(1.7.1)

27 This is by no mean to ignore other researchers who have contributed significantly to this difficult study. We only name a few, which are popular in practice.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 53

For angular grained soil the dynamic shear modulus is given by G=

1230(2.97 − e)2 √ σ0 in psi. 1+e

(1.7.2)

where, G = dynamic shear modulus of the soil in psi, e = in-situ void ratio of the soil sample, σ0 = mean effective stress in psi = 0.333σv (1 + 2K0 ), σv = vertical effective stress in psi, σh = horizontal effective in psi = K0 σv , K0 = earth pressure at rest, and is a function of the plasticity index and the over-consolidation ratio. The relationship between plasticity index, over-consolidation ratio and K0 is as shown in the following figure.

Figure 1.7.1 Value of the K0 after Brooker & Ireland (1965) Reproduced by permission of the National Research Council of Canada from the candian geotechnical Journal Vol-2 (1965).

1.7.1.2

Seed and Idriss Formula

The formula for dynamic modulus in this case, Seed and Idriss (1970) have been related to relative density of sand which can usually be quantified from SPT test and is given by √ G = 83.3K2 σ0 in psi

(1.7.3)

Here K2 is a function of the relative density of the sand which can again be estimated from the SPT value. The relationship between SPT value and the relative density is as given Table 1.7.1. © 2009 Taylor & Francis Group, London, UK

54 Dynamics of Structure and Foundation: 2. Applications Table 1.7.1 Soil properties with SPT values. SPT value

Compactness

Relative density

Angle of friction

0–4 4–10 10–30 30–50 >50

Very loose Loose Medium Dense Very dense

0–15 15–35 35–65 65–80 >85

<28 28−30 30−36 36−41 >41

Table 1.7.2 Values of K2 versus relative density at strain of 10−3 % (Seed and Idriss 1970). Relative density (%)

K2

90 75 60 45 40 30

70 61 52 43 40 34

For case of computer programming K2 can also be represented by the expression K2 = 0.6Dr + 16

(1.7.3a)

It is to be noted that in this case to determine the relative density, the observed SPT value has to be corrected for the overburden pressure and dilatancy to arrive at the design SPT value before it is co-related with the above table. 1.7.1.3 Corrections to SPT value Though available in standard textbooks of Soil Mechanics and Foundation Engineering, for brevity we present the correction expressions as mentioned hereafter. For dilatancy correction if the observed SPT value (N0 ) is greater than 15 then the corrected SPT value N is given by (Terzaghi and Peck 1967). N = 15 +

1 (N0 − 15) 2

(1.7.4)

The overburden correction as per Peck et al. 1980 is given by N = 0.77N log10

2000 p

for p ≥ 25 kPa

(1.7.5)

For p ≤ 25 kPa, as per Murthy (1991) N =

4N 2 + 0.034p

© 2009 Taylor & Francis Group, London, UK

(1.7.6)

Dynamic soil structure interaction 55

in which, N = corrected SPT value for overburden, N = corrected SPT value for dilatancy, p = gross overburden pressure in kN/m2 . 1.7.1.4

Ohsaki and Iwasaki’s formula

Ohsaki and Iwasaki (1973) have given co-relation for dynamic shear modulus directly co-related to SPT value and is expressed as G = 12000 N 0.8 in kPa

(1.7.7)

Here N = design SPT value at the site after relevant corrections.

Example 1.7.1 As shown in Figure 1.7.2 is a small site having dimensions 18 m × 6 m which would be supporting a Compressor unit and a few pumps, for which four boreholes were dug at four corners as shown. The soil was found to be cohesionless in nature and SPT values observed at the four bore holes are as tabled hereafter

Depth (meter)

BH1 (SPT value)

BH2 (SPT value)

BH3 (SPT value)

BH4 (SPT value)

2 4 6 8 10 14

4 8 12 15 20 22

6 6 9 12 18 24

4 6 11 16 24 28

3 5 8 11 16 20

18.0 BH1

BH2

6.0 BH3

Figure 1.7.2

© 2009 Taylor & Francis Group, London, UK

BH4

56 Dynamics of Structure and Foundation: 2. Applications

Based on Laboratory and ﬁeld analyses following parameters were further established: Ground water table = 1.6 m, below grade level Saturated density of soil = 22 kN/m3 Void ratio e0 = 0.58; Plasticity Index = 0.0; Poisson’s ratio = 0.32 Determine the best estimate of dynamic shear modulus (G) of soil at 10.6 meter below ground level presuming no dynamic soil test was done during geo-technical investigation. Solution: Average observed SPT value at a depth of 10.0 meter 20 + 18 + 24 + 16 = = 19.5 = 20 (say) 4 Average observed SPT value at a depth of 14.0 meter 22 + 24 + 28 + 20 = 23.5 = 24 (say) = 4 At a depth of 10.6 meter below ground level based on linear inter-polation average observed SPT Value =

24 − 20 × 0.6 + 20 = 20.6 ∼ = 21 (say) 4

The above observed SPT value has now to be corrected for dilatancy and overburden pressure 1 Correction for dilatancy As per Terzaghi, corrected SPT (N ) value is given by N = 15 + or N = 15 +

2

1 (N − 15) for N > 15; 2 0

1 (21 − 15) = 18 2

(1.7.8) (1.7.9)

Correction for overburden pressure

As per Peck N = 0.77N log10

2000 p

for p ≥ 25 kPa

(1.7.10)

N = Corrected SPT value for overburden; N = Corrected SPT value for dilatancy; p = Gross overburden pressure in kN/m2 . © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 57

Here p = 22 × 10.6 = 233.2 kN/m2 Substituting above in Peck’s formula we have, N = 0.77 N log10

2000 = 13 233.2

Thus, corrected design SPT value = 13 Referring to table 1.7.1 for N = 13, Dr (Relative density) = 39.5% Net overburden pressure at 10.6 meter level is expressed as σv = (22 − 10) × 9 + 22 × 1.6 = 131.6 kN/m2 (18.718 p.s.i) As there is no previous history of loading on the site O.C.R. = 1. Thus for P.I. = 0.0 and O.C.R = 1 as per Brooker and Ireland’s curve we have K0 = 0.48 Considering confining pressure

σo = 0.333σv (1 + 2K0 ); we have σ0 =

18.718 (1 + 2 × 0.48) = 12.22 p.s.i. 3

As per Hardin and Richart’s formula

G=

2630(2.17 − e)2 √ σ0 1+e

G=

2630(2.17 − 0.58)2 √ 12.22 = 14710.5 p.s.i. (101426 kN/m2 ) 1 + 0.58 (1.7.11)

As per Seed and Idriss formula Referring to the chart given above for Dr = 39.5% and strain in the range of 10−3 % (usually valid for machine foundation) K2 = 40. √ And as G = 83.3 K2 σ0 we have G = 83.3 × 40 ×

√

12.22 = 11647 p.s.i. (80308 kN/m2 )

Thus taking average value of G based on Hardin and Seed’s method Average G =

101426 + 80308 = 90867 kN/m2 2

As per Ohsaka and Iwaski’s formula © 2009 Taylor & Francis Group, London, UK

58 Dynamics of Structure and Foundation: 2. Applications

G = 12000 N 0.8 in kPa ➔ G = 12000 × (13)0.8 = 93397.6 kpa (93398 kN/m2 ) Thus it will be observed that variation with average G obtained based on Hardin, Seed’s and Ohsaka’s formula is not signiﬁcant and is of the order of 2.7%28 .

1.7.2 Co-relation for saturated clay 1.7.2.1 Hardin and Drnevich formula Hardin and Drnevich (1973) have given the following formula applicable to clayey soil as Gmax = 1230

(2.973 − e)2 (OCR)k (σ0 )0.5 in psi (1 + e)

(1.7.12)

where, e = void ratio; OCR = over consolidation ratio; σ0 = mean effective stress in psi = 0.333 (σv + 2σh ); σv = vertical effective stress in psi; σh = horizontal effective stress in psi = K0 σv , K0 = earth pressure at rest, and is a function of the plasticity index and the over-consolidation ratio. k = is a function of the plasticity index (PI) of the soil and is given as k = −5 × 10−8 (PI)3 −4 × 10−5 (PI)2 + 0.0092(PI) + 0.0025

(1.7.12a)

It is to be noted that Gmax as obtained above corresponds to a shear strain range of 0.25×10−4 % and needs to be modified for the appropriate strain range as appropriate for a problem in hand based on the expression G=

Gmax (1 + ψ/ψr )

(1.7.13)

Here ψ = desired strain range; ψr = reference strain range and is expressed as ψr =

τmax × 100 Gmax

and

28 The point we are trying to make here is not to go by one formula, but check with possibly all of them and comparing them to arrive at result which would possibly be best fit and hopefully be most realistic. Here again it is to be noted that we had not used the angular sand formula of Hardin, if the soil description does not reflect it or the soil has both rounded and angular grains an intermediate value has tobe chosen judiciously.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 59

$ τmax =

1 + K0 (σv − u) sin φ + c cos φ 2

%2

$

%2 0.5

1 − K0 (σv − u) − 2

(1.7.14)

in which, σv = total vertical stressing in soil; u = pore pressure; c = cohesion of soil; φ = angle of friction of soil, and K0 = coefficient of earth pressure at rest.

Example 1.7.2 It has been decided to place foundation of an industrial structure at 4.0 meter below the existing ground level. Based on laboratory and field tests it has been found that the Ground water table is at a depth of 1.0 meter below GL. Unconsolidated undrained triaxial tests reveal the sample to have the following values: • • •

Cohesion value c = 0.21 kg/cm2 Angle of resistance = 18 degrees Pore pressure = 0.0 kg/cm2

Consolidation tests reveal that it had a history of pre-consolidation pressure of 200 kN/m2 : • • •

Initial void ratio = 0.61 Plasticity limit PI = 35 Saturated unit weight of soil = 19 kN/m3

The site has a history of moderate to severe earthquake when from previous record it is observed to generate a strain range up to 0.1%. Calculate the dynamic shear modulus of soil for this predicted strain range. Solution: For foundation located at 4.0 meter below the ground level net vertical pressure σv = 19 × 1.0 + (19 − 10) × 3.0 = 46 kN/m2 (6.54 psi) 200 = 4.34, for plasticity index of 35 from Brooker and Ireland’s, 46 chart K0 = 1.1

OCR =

© 2009 Taylor & Francis Group, London, UK

60 Dynamics of Structure and Foundation: 2. Applications

Thus considering σo = 0.333σv (1 + 2K0 ), we have σo = 0.333 × 6.54(1 + 2 × 1.1) = 6.976 psi Gmax = 1230

(2.973 − e)2 (OCR)k (σ0 )0.5 in psi (1 + e)

Here k = 0.27 for PI = 35 as per Equation 1.7.12a Thus substituting the values we have Gmax = 1230

(2.973 − 0.61)2 (4.34)0.27 (6.976)0.5 (1 + 0.61)

= 16746 psi (115465 kN/m2 ) Calculation for Shear stress $ τmax = $ or, τmax =

1 + K0 (σv − u) sin φ + c cos φ 2

%2

%2 0.5

$

1 − K0 − (σv − u) 2 %2

1 + 1.1 (6.54) sin 18 + 3.0457 cos 18 2

%20.5 $ 1 − 1.1 (6.54) − 2

= 5.00 psi. ψr = Reference strain range and is expressed as ψr =

τmax 5.00 × 100 = × 100 = 0.0299% Gmax 16746

Thus for 0.1% strain. . . Gmax

G=

1+

ψ ψr

115465

➔G=

1+

0.1 0.0299

= 26577 kN/m2

It is thus observed that dynamic shear modulus is 23% of the theoretically calculated data. Based on the above example it would perhaps be not difﬁcult to realize that how important role does the strain range plays on the design value of dynamic shear modulus of soil. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 61

1.8 ESTIMATION OF MATERIAL DAMPING OF SOIL Damping plays a significant part in the overall response of soil structure system. While for structural members material damping plays a significant part (mostly considered as Rayleigh damping), for soil, two types of damping are basically involved. • •

Radiation damping Material damping

Radiation or geometric damping of a soil foundation system is a mean by which the energy is dissipated by means of radiation from the source and is a function of mass and inertia of the system29 . Material damping of the soil foundation system is a mean by which the energy is dissipated by hysteresis and is an inherent property of the constituting material of the soil. This can very well be found from resonant column test in the laboratory when after the soil has been vibrated the exciter is stopped and successive amplitudes are measured. If a1 and a2 are two successive amplitudes then

& a2 a2 2 2 Dm = ln 4π + ln a1 a1

(1.8.1)

The total damping ratio of a soil foundation system is sum of radiation and material damping. It is generally observed that material damping has a significant magnitude relative to radiation damping specially in rotational modes. In such cases total damping rather than geometric damping should be used to obtain the response of the structure foundation system. For translatory mode, on the contrary material damping plays an insignificant role and may be neglected in the analysis. Thus for tall narrow structures like chimney, Boiler structures, tall buildings where the coupled horizontal and rocking mode could play significant role it would perhaps be realistic to also consider the material damping of soil in order to have a meaningful response.

1.8.1 Whitman’s formula Whitman (1973) has suggested that total damping (geometric + material) for a machine foundation can be obtained from the expressions Horizontal Mode 0.31 Dh = M ρr30

29 We have dealt this detail in Chapter 2 (Vol. 2) – “Design of Machine foundations”.

© 2009 Taylor & Francis Group, London, UK

(1.8.2)

62 Dynamics of Structure and Foundation: 2. Applications

For vertical mode 0.49 Dv =

(1.8.3)

M ρr30

For rocking mode ⎡' Dθ = 0.05 + 0.1 ⎣

Iθ

(0.5

ρr50

' 1+

Iθ 4ρr50

( ⎤−1 ⎦

(1.8.4)

Here M = mass of foundation plus structure or machine vibrating; Iθ = mass moment of inertia of foundation plus machine/structure about a horizontal axis through the base of the foundation perpendicular to the plane of rocking; r0 = equivalent radius of footing, and ρ = mass density of soil.

1.8.2 Hardin’ formula Hardin (1965) has expressed material damping of sandy soil by the expression Dm =

0.985ψr0.2 √ σ0

(1.8.5)

Here notations are same as expressed earlier except the fact that the conﬁning pressure σ0 is expressed in kPa. The equation is valid for shear strain amplitude of 10−6 to 10−4 with a conﬁning pressure of 24 kPa to 144 kPa. For a particular strain range the value obtained above can be corrected based on the expression ψ/ψr Dc = Dm 1 + ψ/ψr

(1.8.6)

Example 1.8.1 For the example as shown in Example 1.7.2, estimate the damping ratio of the soil as per Hardin’s formula. The soil properties remain same as given in Example 1.7.2. Solution: Based on the solution furnished in Example 1.7.1 value of dynamic shear modulus is given by G = 90867 kN/m2 ; © 2009 Taylor & Francis Group, London, UK

σ0 = 12.22 p.s.i. (85.9 kN/m2 );

K0 = 0.48

Dynamic soil structure interaction 63

For design SPT value N = 13, φ = 31◦ , and σv = 131.6 kN/m2 $

Considering, τmax

τmax

%2 1 + K0 (σv − u) sin φ + c cos φ = 2 $ %2 0.5 1 − K0 − , we have (σv − u) 2 $ %2 $ %2 0.5 1 + 0.48 1 − 0.48 = 131.6 sin 31 − 131.6 2 2 = 36.67 kN/m2

As ψ =

τ 36.67 × 100, we have ψr = × 100 = 0.0404% G 90867

Considering Dm =

Dm =

0.985ψr0.2 , we have √ σ0

0.985(0.0404)0.2 = 0.056 √ 85.9

Thus material damping ratio is estimated as 5.6%.

1.8.3 Ishibashi and Zhang’s formula Ishibashi and Zhang (1993) has proposed an expression for the damping ratio of plastic and non-plastic soil and is given by 2 1 + exp(−0.0145PI1.3 ) G G ζ = 0.333 0.586 − 1.547 +1 2 Gmax Gmax

(1.8.7)

The notations for the above expression are already explained in earlier formulas. We show below variation of damping ratio with G/Gmax for different Plasticity Index based on the above formula. It will observed (Figure 1.8.1) that as G/Gmax reduces, as damping ratio goes on increasing meaning thereby that as strain increases damping ratio goes on increasing. Variation of Damping with strain vide Equation (1.8.6) is shown in Figure 1.8.1a. © 2009 Taylor & Francis Group, London, UK

64 Dynamics of Structure and Foundation: 2. Applications

Variation of damping ratio 0.35 0.3

PI = 10 PI = 20 PI = 30

Damping ratio

0.25

PI = 40

0.2

PI = 50

0.15

PI = 60 PI = 70

0.1

PI = 80 PI = 90 PI = 100

0.05

0

1 0.

2 0.

3 0.

4 0.

5 0.

6 0.

7 0.

8 0.

0.

1

9

0 G/Gmax

Figure 1.8.1 Variation of damping with plasticity index as per Ishibashi and Zhang (1993).

1.2 1

D/Dr

0.8 0.6 0.4 0.2 0

0

0.01

0.1

1 Strain Ratio

10

100

1000

Figure 1.8.1a Variation of damping ratio with strain under cyclic loading.

Example 1.8.2 For the clayey soil sample as shown in Example 1.7.2, determine the damping ratio for the strain range level of 0.1% based on Zhang’s formula. Consider all soil properties same as Example 1.7.2? Solution: Based on earlier example we have seen that plasticity index PI = 35. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 65

In Example 1.7.2 we have already calculated that for 0.1% strain G/Gmax = 0.230 Substituting the above in Ishibashi and Zhang’s formula we have ζ = 0.333

1 + exp(−1.47) [0.586(0.23)2 − 1.547 × 0.23 + 1] = 0.1382 2

Thus estimated damping ratio is 13.82%.

1.9 ALL THINGS SAID AND DONE HOW DO WE ESTIMATE THE STRAIN IN SOIL, SPECIALLY IF THE STRAIN IS LARGE? We acknowledge at the very outset that posing the question, though easy, is not very easy to answer. The uncertainties involved are so widely varying that it would be difﬁcult to give a precise answer to this issue. To the best of our knowledge there is no straight forward answer to this problem and the best one can achieve is a reasonable estimate or can possibly study a range of values and try to predict the overall behavior. For high speed centrifugal machine foundation it does not pose a serious problem for at the low strain range a few percent here and there does not contribute a significant variation to these values. But for impact type of machines (hammer foundations) and slow speed machines (coal mill foundations, reciprocating compressors) induced strain could be larger than strain developed during field test, for which the correct estimation of Gdyn and damping becomes important. For earthquake of course the strain would invariably be larger than measured during test, even for a moderate earthquake when as the strain range increases, degradation in soil stiffness becomes signiﬁcant and has a major contribution to the overall response. It is obvious that strain induced in soil will depend upon the strength of dynamic loading, the geological condition of the site, stress history of soil and a number of other factors. So the point remains that if there exists no previous records of strain from similar machine in same site or from previously occurring earthquake data how does one rationalize the strain? We discuss below some of the techniques which could be used for evaluation of the strain induced in the soil.

1.9.1 Estimation of strain in soil for machine foundation For machine foundations the present practice of arriving at strain dependent dynamic shear modulus and damping can be structured as follows: © 2009 Taylor & Francis Group, London, UK

66 Dynamics of Structure and Foundation: 2. Applications

• • •

Start with the field observed/lab obtained data for Gdyn and damping as furnished in the soil report which would usually correspond to the strain range of 10−4 to 10−3 %. Calculate natural frequency of the soil-foundation system based on free vibration analysis. For rotating mass type calculate the transmissibility factor based on the expression Tr = !

•

! r2 1 + (2ζ r)2 (1 − r2 )2 + (2ζ r)2

For constant force excitation (like in hammer foundation) calculate transmissibility factor based on expression ! Tr = !

• • •

1 + (2ζ r)2

(1 − r2 )2 + (2ζ r)2

(1.9.2)

where r = ωωmn , ωm = operating frequency of the machine; ωn = natural frequency of the foundation; ζ = damping ratio of the soil. Find out the pseudo-static force by multiplying the vertical unbalanced force of the machine by the transmissibility factor as mentioned above. Find out the dynamic stress induced in the soil by dividing the above force by the foundation plan area. The approximate shear strain in the soil is given by the expression. ψ(%) =

•

(1.9.1)

12qdyn G

(1.9.3)

Verify the strain obtained against the initial value. If they vary significantly find out the new G value based on the calculated strain and repeat the process as mentioned above till it converges.

The above method is surely non-rigorous but generates an answer which will give reasonably accurate results for practical analysis of machine foundations. For more complicated soil with varying properties a more rigorous analysis based on Finite element analysis is possible. This will be discussed later on. The above technique is now explained based on a suitable numerical example.

Example 1.9.1 A centrifugal turbine driven compressor has foundation dimension of 6 m × 3.2 m × 2.5 m. The weight of the compressor is 300 kN. The unbalanced mass © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 67

on the shaft is 3.5 kN · sec2 /m rotating at an eccentricity of 0.4 mm having operating frequency of 1800 rpm. The soil investigation has revealed the soil data as follows • • •

SPT = 13 (After dilatancy and overburden correction) Plasticity Index (PI) = 0 Poissons Ratio = 0.3

Calculate the correct value of dynamic shear modulus and damping. Solution: Based on Ohsaka’s formula G = 12000 × (13)0.8 = 93397.65 kN/m2 Assumed strain level = 1 × 10−4 % Weight of foundation = 6 × 3.2 × 2.5 × 25 = 1200 kN; Weight of machine = 300 kN, Total weight = 1500 kN Mass of foundation + machine = 1500/9.81 = 152.9052 kN · sec2 /m Equivalent radius of the foundation (r0 ) =

6 × 3.2 = 2.472155 m π

Equivalent vertical spring stiffness of soil30 . Kz =

4Gr0 4 × 93397.65 × 2.472 = = 1.32 × 106 kN/m 1−υ 0.7

ωn =

ωm =

Kz = m

1.32 × 106 = 92.89 rad/sec; 152.9

1800 × 2 × π = 188 radian/sec 60

2 = 3.5 × Pdyn = m · e · ωm

r = ωm /ωn = 2.029.

0.4 × (188)2 = 49.7428 kN. 1000

Considering transmissibility as

Tr = !

! r2 1 + (2ζ r)2 (1 − r2 )2 + (2ζ r)2

we have, Tr = 0.65078.

30 Refer to Chapter 5 (Vol. 1) – Basic Concepts of Soil Dynamics, for details of this formula.

© 2009 Taylor & Francis Group, London, UK

68 Dynamics of Structure and Foundation: 2. Applications

Equivalent static force on foundation = 12qdyn

Considering ψ(%) =

Considering G =

G

=

0.65078 × 49.7428 = 1.686 kN/m2 6 × 3.2

12 × 1.686 = 2.17 × 10−4 % 93397.65

Gmax (1 +

ψ ψr )

93397.65

we have, New G =

1+

2.17×10−4 1×10−4

= 29497.88 kN/m.

We proceed with second cycle of iteration with this new value of G. Shown below is such iteration for 14 cycles Cycles

1

2

3

4

5

6

7

Gdyn Damping Kz

93397.65 0.012987 1.32 × 10+06 92.89143 2.029203 0.65078 49.74281 32.37161 1.686022 2.17 × 10−04

29497.88 0.189764 4.17 × 10+05 52.20392 3.610755 0.298033 49.74281 14.82497 0.772134 3.14 × 10−04

22553.77 0.21998 3.19 × 10+05 45.64753 4.129371 0.255622 49.74281 12.71537 0.662259 3.52 × 10−04

20646.63 0.228656 2.92 × 10+05 43.67493 4.315875 0.243327 49.74281 12.10377 0.630405 3.66 × 10−04

20025.37 0.231517 2.83 × 10+05 43.01282 4.382311 0.239243 49.74281 11.90063 0.619824 3.71 × 10−04

19811.83 0.232505 2.80 × 10+05 42.78288 4.405864 0.23783 49.74281 11.83032 0.616163 3.73 × 10−04

19737.08 0.232851 2.79 × 10+05 42.7021 4.414199 0.237334 49.74281 11.80565 0.614878 3.74 × 10−04

Cycles

8

9

10

11

12

13

14

Gdyn Damping Kz

19710.75 0.232973 2.78 × 10+05 42.6736 4.417147 0.237159 49.74281 11.79696 0.614425 3.74 × 10−04

19701.45 0.233016 2.78 × 10+05 42.66353 4.418189 0.237097 49.74281 11.79388 0.614265 3.74 × 10−04

19698.17 0.233031 2.78 × 10+05 42.65998 4.418558 0.237075 49.74281 11.7928 0.614208 3.74 × 10−04

19697.01 0.233037 2.78 × 10+05 42.65872 4.418688 0.237068 49.74281 11.79241 0.614188 3.74 × 10−04

19696.60 0.233039 2.78 × 10+05 42.65827 4.418734 0.237065 49.74281 11.79228 0.614181 3.74× 10−04

19696.45 0.233039 2.78 × 10+05 42.65812 4.41875 0.237064 49.74281 11.79223 0.614179 3.74 × 10−04

19696.40 0.23304 2.78 × 10+05 42.65806 4.418756 0.237064 49.74281 11.79221 0.614178 3.74 × 10−04

ωn r(wm /wn ) Tr Pdyn Peq qdyn Strain

ωn r(wm /wn ) Tr Pdyn Peq qdyn Strain

We show hereafter the variation of strain, damping and Gdyn per cycle in Figs. 1.9.1 to 1.9.3.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 69

Variation of strain per cycle 4.00E-04 3.50E-04

Strain(%)

3.00E-04 2.50E-04 2.00E-04 1.50E-04 1.00E-04 5.00E-05 0.00E+00

1

2

3

4

5

6

7 8 9 10 Number of cycles

11

12

13

14

15

Figure 1.9.1 Variation of strain (%) per cycle.

Variation of damping ratio with strain 0.35

Damping ratio

0.3 0.25 0.2 0.15 0.1 0.05 0

1

2

3

4

5

6 7 8 9 10 Number of cycles

Figure 1.9.2 Variation of material damping per cycle.

© 2009 Taylor & Francis Group, London, UK

11

12

13 14

15

70 Dynamics of Structure and Foundation: 2. Applications

Variation of Gdyn with strain

Gdyn(kN/m2)

100000 80000 60000 40000 20000 0

1

2

3

4

5

6 7 8 9 10 11 12 13 14 Number of cycles

15

Figure 1.9.3 Variation of dynamic shear modulus per cycle.

From the tables and the above plots it is observed that the value becomes constant after 7th cycle of iteration based on which we conclude that design values are as follows Gdyn = 19700 kN/m; Material Damping ratio = 0.23, and Estimated strain range = 3.74 × 10−4 %.

Thus actual design of the foundation shall be carried out based on this corrected value instead of the initial values as mentioned in the soil report.

1.9.2 Estimation of soil strain for earthquake analysis For earthquake analysis things are surely more complicated for not only the forces induced in the soil is much more complex, the behavior itself is different from machine foundations. While in machine foundation the force is induced in the soil from the structure in earthquake the force is induced within the soil where the soil first start vibrating based on the waves propagating through it. Thus acceleration, it is excited to, depends on the free field vibration of the site. This acceleration induced in the soil generates shear strain on which the stiffness degradation and damping ratio would depend. Though the analysis shown hereafter is based for isotropic homogenous medium it can well be extended to layered soil having variable property based on weighted average. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 71

H

Propagation of Earthquake

Figure 1.9.4 Schematic diagram of an industrial site with propagating earthquake waves.

Shown in Figure 1.9.4 is a schematic diagram of an industrial site with propagating earthquake waves. The depth of the site (H) is considered to the bedrock level from where the waves are presumed to be propagating31 . The waves propagating at bedrock level travels upward and hits the surface (z = 0) when the site surface undergoes a motion. However as surface is free, it is free to shake as such no strain energy develops at the surface. The motion of such elastic waves propagating through an elastic medium can be defined by the partial differential equation ∂ 2u ∂ 2u = vs2 2 2 ∂t ∂z

(1.9.4)

Here u = displacement of the soil and is a function of time t and depth z, vs = Shear wave velocity of the soil. Considering u(z, t) = φ(z)ψ(t),

(1.9.5)

31 For site having no bedrock this level is usually considered at the depth where shear wave velocity of the site is greater or equal to 600 m/sec. Based on SPT value this can be considered as the depth where design SPT value is greater than 50.

© 2009 Taylor & Francis Group, London, UK

72 Dynamics of Structure and Foundation: 2. Applications

we have substituting in the equation above ¨ ¨ φ(z)ψ(t) = vs2 φ(z)ψ(t) or

¨ ¨ ψ(t) φ(z) = −p2 (say) = 2 φ(z) vs ψ(t) The above on separation gives two homogenous equations ¨ φ(z) + p2 φ(z) = 0

¨ and ψ(t) + p2 vs2 ψ(t) = 0

(1.9.6)

The above gives solution φ(z) = A cos pz + B sin pz

(1.9.7)

at z = 0 as there will be no shear strain hence du/dz = 0 ˙ φ(z) = −Ap sin pz + Bp cos pz = 0, at z = 0 The above gives the constant B = 0, from which we deduce, φ(z) = A cos pz At z = H as the soil is confined, hence we have, u(z, t) = 0 → A cos pH = 0 p=

(2n − 1) π 2H

(1.9.8)

Considering, ψ(t) = C cos λt + D sin λt

(1.9.9)

we have, λ = p2 vs2 , where λ is the eigen-value of the problem. Knowing from our fundamental knowledge of vibration that

ωn2 = λ =

(2n − 1)2 π 2 2 vs 4H 2

Considering T =

Tn =

2π ωn ,

➔ or ωn =

(2n − 1) π vs rad/sec 2H

(1.9.10)

we have

4H secs (2n − 1) vs

(1.9.11)

Here Tn is known as the free field time period of the site for n numbers of mode. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 73

The corresponding eigenvectors are given by φ(z) = cos

(2n − 1)πz 2H

(1.9.12)

The displacement vector is given by u(z) = κi φi (z)Sd

(1.9.13)

where Sd = Displacement spectrum of the site which can again be represented as u(z) = κi φi (z)

Sa ωn2

(1.9.14)

where Sa = acceleration spectrum of the site and is a function of the free field time period of the site32 . In which, m i φi κi = Modal mass participation factor = mi φi2

(1.9.15)

The modal participation factor can thus be considered as

)H )H πz πz * κi = = γ z cos γ z cos2 2H 2H mi φi2 mi φi

0

(1.9.16)

0

The above on integration by parts gives, κi =

8 π +2

(1.9.17)

Thus for the present problem u(z) =

32βSa H 2 (2n − 1)π z cos 2H (2n − 1)2 π 2 (π + 2)vs2

Here β =

ZI 2R

(1.9.18)

the IS code factor33

32 This response spectrum is usually available as site response spectra in absence of which charts furnished in National codes are usually followed. 33 Presently code does not have any guidline for R for soil. It has been observed that a value between R = 2 to 3 usually gives realistic results.

© 2009 Taylor & Francis Group, London, UK

74 Dynamics of Structure and Foundation: 2. Applications

Considering shear strain γz = γz = −

∂u ∂z

we have

16βSa H (2n − 1)π z sin 2 2H (2n − 1)π(π + 2)vs

(1.9.19)

Considering G = ρvs2 we have γz = −

16βSa Hρ (2n − 1)π z sin (2n − 1)(π + 2)πG 2H

(1.9.20)

Here G = dynamic shear modulus of the soil, ρ = mass density of the soil. For foundation at a particular depth below the free surface for which we have obtained the dynamic shear modulus based on field or lab test34 . We start initially to find out the shear strain in the soil based on this value considering a strain range of 10−3 /10−4 %. The steps that are followed subsequently to arrive at the corrected G and damping value are furnished hereafter (Chowdhury 2008). 1 2 3 4 5 6 7 8 9

Identify the bedrock level (H) of the site Find out the shear wave velocity from the expression G = ρvs2 4H Find out the free field time period of the site from the expression Tn = (2n−1)v s Based on the site response spectra/spectra given in code and damping value as obtained in soil report obtain the acceleration Sa Obtain shear strain for the soil proﬁle based on the expression γz = 16βSa Hρ − (2n−1)π(π+2)G sin (2n−1)πz 2H Check if this strain is near or equal to the initial strain(10−3 to 10−4 )%35 . Gmax If there exists a significant variation correct G based on the equation G = (1+ψ/ψ r) Find out the ratio G/Gmax Obtain new damping ratio based on Zhang’s expression 2 1 + exp(−0.0145PI1.3 ) G G ζ = 0.333 0.586 − 1.547 +1 2 Gmax Gmax

10 Repeat the steps as mentioned from 2 to 7 till the strain is same as previous cycle. The value for which the strain becomes constant is the corrected Dynamic shear modulus of the soil. The above steps will now be further elaborated by a suitable problem.

34 Or from theoretical co-relation. 35 In absence of this input in soil report consider 10−3 % for soft soil and 10−4 % for medium stiff soil.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 75

Example 1.9.2 For a particular site susceptible to earthquake it was observed based on soil investigation that bed rock exists at 20 meters below ground level. Seismic crosshole test reveals average dynamic shear modulus of the soil to be 154897 kN/m2 at a reference strain of 1 × 10−5 . Considering density of soil as 19 kN/m3, and plasticity index as 35. Calculate the corrected dynamic shear modulus of soil and damping at 2.5 meter below GL where foundation of a particular structure will be placed. Consider IS 1893 curves to evaluate the acceleration pertaining to a particular time period. Solution: Depth of soil over bedrock = 20 m, Density of soil = 19 kN/m3 , Thus mass 19 density of soil (ρ) = 9.81 = 1.936 kN · sec2 /m, Dynamic shear Modulus of soil 2 (G) = 154897 kN/m .

Shear wave velocity of soil (vs ) =

Considering Tn =

T1 =

G = 282.8 m/sec. ρ

4H we have for fundamental mode (2n − 1)vs

4 × 20 = 0.283 sec 282.8

Considering Zhang’s formula 2 1 + exp(−0.0145PI1.3 ) G G ζ = 0.333 − 1.547 +1 0.586 2 Gmax Gmax Taking G/Gmax = 1 for first cycle and PI = 35 we have, ζ = 0.798% For damping @ 0.798% and time period of 0.283 sec Sa = 5.75 m/sec2 from IS code. For this case the code factor Z is considered as 0.24, I = 1.2, R = 3. The value of R is chosen as 3 in this case because for PI = 35 it is assumed that the soil has high plasticity and thus has reasonable ductility. As a matter of fact there is no guideline at moment prevalent in the code and an engineer has to use his own judgment here. Substituting the above data in the expression

γz = −

(2n − 1)πz 2Sa Hρ sin (2n − 1)πG 2H

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76 Dynamics of Structure and Foundation: 2. Applications

We have at depth of 2.5 meter below ground level γz = 2.5145 × 10−5 Considering G =

Gmax (1 +

ψ ψr )

we have G = 44073.2 kN/m2

We proceed with next cycle of iteration with this new value of G. In table below we show how the data converges for 10 successive cycles Cycles

1

2

3

4

5

GkN/m2 Vs m/sec T sec Sa m/sec2 Strain G/Gmax Damping ratio(%)

154897 282.7998707 0.282885561 5.75 2.51454 × 10−05 1

44073.21001 150.8499063 0.53032847 1.9 2.9202 × 10−05 0.284532367

39512.54 142.8319 0.560099 1.85 3.17 × 10−05 0.255089

37131.85 138.4621 0.577775 1.85 3.37 × 10−05 0.23972

35405.98 135.206 0.59169 1.8 3.44 × 10−05 0.228578

0.798

12.425

13.167

13.562

13.852

Cycles

6

7

8

9

10

GkN/m2 Vs m/sec6 T sec Sa m/sec2 Strain G/Gmax Damping ratio(%)

34857.41 134.1545 0.596327 1.8 3.5 × 10−05 0.225036

34437.41 133.3438 0.599953 1.8 3.54 × 10−05 0.222325

34113.86 132.716 0.602791 1.8 3.57 × 10−05 0.220236

33863.42 132.2279 0.605016 1.8 3.6 × 10−05 0.218619

33668.85 131.8475 0.606762 1.8 3.62E−05 0.217363

13.945

14.016

14.071

14.114

14.147

Thus based on above calculation we may take, corrected G value = 33600 kN/m2 ; Damping ratio = 0.14. Variation of shear modulus and damping ratio (%) with number of cycles are shown in Figs. 1.9.5 and 1.9.6.

Varaition of Gdyn at foundation level

200000 150000 100000

G

50000 0 1

2

3

4

5

6

7

Iteration num ber

Figure 1.9.5 Variation of shear modulus with number of cycles.

© 2009 Taylor & Francis Group, London, UK

8

9

10

Damping Ratio(%)

Dynamic soil structure interaction 77

Varaition Damping ratio(%) at foundation level 15.000 10.000 5.000 0.000

1

2

3

4

5 6 7 Number of iterations

8

9

10

Figure 1.9.6 Variation of damping ratio (%) with number of cycles.

1.9.3 What do we do if the soil is layered with varying soil property? Till now the theories we have presented assumes soil as a homogenous isotropic medium but in reality in all possibility the soil encountered at a particular site will be layered in nature. Shown in Figure 1.9.7, is a typical stratified soil profile where the shear modulus, density of soil and Poisson’s ratio vary with depth. For most of the cases taking a weighted average is the normal practice where the average dynamic property may be taken as

Gav =

G1 · H 1 + G 2 · H 2 + G 3 · H 3 + G 4 · H 4 H1 + H 2 + H 3 + H 4

(1.9.21)

and same principle be applied for mass density and Poisson’s ratio. However for very important structures or site susceptible to major earthquakes methods based on finite element analysis may be applied to arrive at a design dynamic modulus and damping value36 . Shown in Figure 1.9.8, is the finite element model of a site having layered soil property. In this case the soil is modelled as plane strain element to the bedrock boundary and each individual layers having different properties can very easily be catered to. To start with we assume G value as obtained from soil report and consider the damping ratio based on Zhang’s formula considering G/Gmax = 1 at the strain level of 10−3 /10−4 % say. Suppose the previous earthquake history shows that shaking has taken place for duration of 3 sec maximum, we select duration of 6 sec for analysis.

36 In such cases preferably site response spectra of the particular should be used. Moreover some previous history of shaking and its duration should be available for analysis.

© 2009 Taylor & Francis Group, London, UK

78 Dynamics of Structure and Foundation: 2. Applications

Figure 1.9.7 Layered soil strata with varying soil property.

Super Structure

Layered Soil

Bed Rock

Figure 1.9.8 Finite element model of a site having layered soil property.

Next for duration of 6 sec we input the Sa/g curve for the particular damping and perform a time history analysis of the system for 6 sec. From the output for each layer we find the average shear strain. Based on the output average strain, we correct the value of Gdyn for the next cycle and also the damping ratio and do a second cycle of time history for 6 sec. We repeat this process for a couple of times till the values have stabilised with respect to the previous cycle. The value of G and damping considered in last cycle where the strain has stabilised are the dynamic shear modulus and damping of the soil. © 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 79

Calculation of shear modulus of soil based on the free field time period is an effective tool for assessing the dynamic shear modulus of soil. However, there is a possibility that the time period obtained by this method could be higher than the reality unless proper consideration are given for the confining effect of the surrounding soil and proper judgement of the depth is made. ATC (1982) has defined Hmax as the depth limited to 183 m having low strain shear wave velocity of 760 m/sec.

1.9.4 Checklist of parameters to be looked in the soil report Based on above discussion, the parameters which require particular attention in a soil report from the engineer are summarised as follows:

1.9.4.1

Field test

•

Has SPT test been carried out? The obvious intention is to find out N on which G value depends. This can also be utilised to cross check the field observed dynamic data. •

If SPT values are furnished are the observed data or corrections need to be done? A point to be checked for field observed data as shown earlier needs to be corrected. While the soil consultant will do this correction during his own calculation of bearing capacity of soil for foundation recommendations, usually furnishes observed field data while furnishing the bore log detail in the report. So for your calculation this data needs to be corrected. If you are not too sure you can back calculate it from recommended φ value. •

Has Ground Water Table been established during boring? Usually provided in a soil report but better to check for this has significant effect on the net vertical stress. •

Has any dynamic field test carried out? ◦ ◦

Block Vibration test Seismic cross hole test

One of the tests should be a part of the soil report. But do not take the values furnished sacrosanct. Back check with theoretical co-relation to establish if the order is close, if not you do have the right to ask your soil consultant why there is this discrepancy. There could be special geological condition which could result in such discrepancy and you should be clear about it. •

If the above tests are carried out, what is the strain range induced in the soil during the test? This is something usually not supplied by the soil consultant who usually would recommend a unique G value. This should not be acceptable to you. You should clear it at the very outset when providing him the specification for Geotechnical investigation that this is an input you are looking for and it should be © 2009 Taylor & Francis Group, London, UK

80 Dynamics of Structure and Foundation: 2. Applications

a part of his report. It is more realistic to start with this value rather than guessing a theoretical value of 10−3 /10−4 %. 1.9.4.2

Laboratory tests

•

Check Atterberg’s limit – gives values of liquid limit, plastic limit, plasticity index etc. Generally speaking37 as a ritual, structural engineers/analysts ignore this topic. Our suggestion would be, do not disregard this for this is the basic data which gives you the first insight into the fact as to how the soil behaves. Moreover plasticity index being an important property it is all the more important that you should pay attention to this. •

Triaxial test gives values of c, φ and pre-consolidation history Again given a back handed treatment by the analyst who without going through the test data would prefer to pick up the numbers which concerns him (ca and φ). We suggest go through the test and develop enough skill to interpret the pre-consolidation stress. Make sure to ask during enquiry to the consultant to supply this data. For Over Consolidation Ratio (OCR) plays a very important role in arriving at the correct value of Gdyn . During interpretation if need be, seek help of a geotechnical specialist to make sure what has been understood is correct – this will save a lot of headache in the long run. • •

Bulk density and void ratio of soil Grain distribution to check if the soil is gap graded, uniformly graded, or poorly graded. Relative density of cohesion less soil is highly dependent on this.

1.10 EPILOGUE The technology described in this chapter to our perception is still in its infancy and we are optimistic that with time and research that is being carried out all over the world, we shall be in a better position in future to predict more realistically the dynamic properties of soil which affect the response of structure. Whatever we have presented here is what we believe is simple to apply, provides reasonably realistic results and practical for day to day design office practice. There is hardly any comprehensive text which gives a defined picture on this issue. Most of the techniques developed herein are based on research papers (names furnished in the reference) and typical practices followed in some design offices38 . We urge the readers to go through these papers which we believe will give them further insight to the problem.

37 Exceptions are always there. . .. 38 Even consultants who require to use these type of technology is very limited.

© 2009 Taylor & Francis Group, London, UK

Dynamic soil structure interaction 81

The ideas presented in this chapter is to make the reader aware of the limitations prevalent with soil and also to caution him on the fact that without these values realistically estimated, the whole analysis related to dynamic soil structure interaction could become a questionable exercise. So be aware and use your judicious best to furnish a meaningful design. SUGGESTED READING39 1 Cohen, M. & Jennings, P., ‘Silent Boundary Methods For Transient Analysis’, Computational Method in Transient Analysis – Computational Method in Mechanics, Vol. 1, North Holland. 2 Dasgupta, S.P. & Kameswara Rao, N.S.V.K. 1976, ‘Some ﬁnite element solutions in the dynamics of circular footings’, Proc 2nd International Conference on Numerical Methods in Geomechanics, Blacksburg USA. 3 Dasgupta S.P. & Kameswara Rao, N.S.V.K. 1978, Dynamics of rectangular footings by Finite elements, Journal of GT Division ASCE, Vol. 104, No. 5. 4 Gazetas, G & Tassoulas, A.L. 1987, ‘Horizontal Stiffness of Arbitrarily shaped embedded foundation’, Journal of GT Division, ASCE, Vol. 113, No. 5. 5 Kameswar Rao, N.S.V. 1977, ‘Dynamic soil structure system – A Brief Review’, J. Struct. Engg., India, Vol. 4. 6 Lysmer, J. & Kuhlemeyer, R.L. 1969, ‘Finite dynamic model of inﬁnite media’, J.EM.Divn, ASCE, EM4. 7 Segol, G., Abel, J.F. & Lee, P.C.Y. 1975, ‘Finite element Mesh Gradation of surface waves’, J. GT Division, ASCE, Vol. 101, GT 11. 8 Wolf, J.P. 1985, Dynamic Soil Structure Interaction, Prenctice-Hall Englewood Cliffs, NJ. 9 Wolf, J.P. 1988, Dynamic Soil Structure Interaction in Time Domain, Prenctice-Hall, Englewood Cliffs, NJ. 10 Wolf, J.P. 1994, Foundation Vibration Analysis: Using Simple Physical Model, PrenticeHall, Englewood-Cliffs, NJ. 11 Whitman, R.V. 1970, Soil Structure Interaction – Seismic design for Nuclear power plants, The MIT press, Cambridge, Massachusets.

39 This topic being relatively new, there are not much reference books (other than reference 8, 9 & 10) which deal this topic comprehensively. Many literatures though have mentioned the interaction effect in their work. The references suggested are thus mostly restricted to research papers, which we would request you to get hold of and rummage through patiently.

© 2009 Taylor & Francis Group, London, UK

Chapter 2

Analysis and design of machine foundations

2.1 INTRODUCTION This chapter deals with vibration analysis and design of machine foundations subjected to dynamic load. As a pre-requisite to this chapter, you should be thoroughly familiar with concepts that are put in chapter 5 (Vol. 1) on • •

Basic concepts in structural dynamics. Basic concepts in soil dynamics.

Armed with these basics, we believe you will find this chapter interesting and find design of machine foundation a challenging and intellectually stimulating task. Machine foundations are one of the most important features of industrial development. In both developed and developing countries, growth of economy is largely attributed to development of industry and infra-structure facilities. In industrial facilities like Power Plants, Steel Plants, Petrochemical Complexes, Fertiliser Plants etc., consist of a number of centrifugal and reciprocating machines and these play an important role to ensure smooth operation of the process and that the output product is of right quality. If any of these equipments starts malfunctioning or breaks down due to excessive vibration or settlement of the foundations, cascading effect on the overall performance on engineering could be catastrophic at times. We give two case histories below making you aware of how far reaching could be the consequences.

2.1.1 Case history #1 In Middle-East, in one of the oil producing nation there was a plant which had been operating for last 25 years smoothly, sweetening the sour gas that was being pumped into the complex from a nearby gas-field. The authorities hit a new source of natural gas nearby this complex and the obvious choice was to pump gas from this new gas field to the existing plant for further processing. This called for upgrading the plant capacity. On engineering evaluation it was found that it necessitated certain changes in diameter of the pipes, re-routing some of the existing pipes with new supports and also changing the rating of the two-stroke reciprocating compressor which was existing at the plant. The company management wanted to expedite the issue for they perceived © 2009 Taylor & Francis Group, London, UK

84 Dynamics of Structure and Foundation: 2. Applications

that each day lost in production, the company stood to loose about 100,000 thousand US dollars in profit. In haste nobody thought to re-check performance of the foundation of the compressor under dynamic load, now that its rating was changed! When the plant started after this modification with additional gas being pumped from the new gas field, whole pipe rack started shacking violently and the compressor foundation started showing vibration amplitude that was well beyond acceptable limit. The vibration became so high at 80% production level (at which the plant would operate at most of the time) that the operation manager had no alternative but to stop the plant completely. Subsequent investigation revealed that with new rating of the compressor, the operating frequency now hovered very near to natural frequency of the foundation resulting in a resonant condition and also induced additional excitation to the fluid ﬂowing through the pipe resulting in a force which the piping system was not capable of taking care off, without undue distress. This resulted in complete overhauling of the compressor foundation and stiffening the pipe racks by additional bracings and all these re-engineering resulted in a delay of about 5 months for full scale production and also a total revenue loss to company in the tune of 300 million dollars1 .

2.1.2 Case history #2 In another case a medium scale factory requiring heavy duty power for its production opened a new unit in an industrial area (in India), where during peak summer season power supply was reported to be sporadic. To maintain optimum production level, the owner procured 3 numbers of standby generator sets to supply power during periods of power cuts. During soil investigation nobody thought of doing a test for dynamic property of the soil and the foundation was designed based on obtaining the dynamic properties by theoretical co-relation with other static engineering soil parameters. After generators were installed and started operating it was found that amplitude of vibration was well beyond tolerable limits resulting in tripping of the machines quite often. As the generator failed to meet the optimal power demand, production output nose dived quite substantially. On investigation of the problem by a consultant hired by the owner it was found that field observed dynamic properties varied widely with those considered from theoretical co-relation for the amplitude and resonance check. The consultant suggested that the generator foundation be modified by providing additional mass of concrete, adjusting the height of foundation and partially re-routing the cables adding to an additional cost of only 20% of original installation cost. The suggestion was vetoed by the owner arguing that company was going through difficult times financially and it was not possible to generate further funds for such additional capital expenditures. The owner hired a couple of mechanics and with some adjustments coerced the machines into operation but still the performance did not improve significantly. Within 1 This was time when oil was priced at 25 dollars per barrel. In todays index the loss would be 4 to 5 times the actual loss incurred.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 85

a period of six months the machines broke down completely. The equipment supplier refused to replace the machines (though they failed within the warranty period) arguing that conditions put in the contract in terms of amplitude and frequency restrictions were violated from the very outset and as such they were not responsible for bad performance of the machine. By this time the company was in such a poor condition financially due to failure of production target, that it could not generate fund to replace and overhaul the equipment and its foundations and had no other option but to declare it sick and close the unit. So lesson learnt from the above two cases are that if proper attention is not paid to the design of these type of foundations, consequences could be quite far reaching and serious in nature. 2.2 DIFFERENT TYPES OF FOUNDATIONS Foundations supporting machines can be classified into the following categories: • • • •

Block foundations resting on soil or piles Frame foundations Wall foundations Spring mounted machines resting on rafts/grade slabs

2.2.1 Block foundations resting on soil/piles These types of foundations usually consist of massive RCC blocks resting on soil or on piles and are as shown in Figures 2.2.1 and 2. Block type of foundation are usually used to support machines like 1 2 3

Pumps Motors Generators

Figure 2.2.1 Block foundation resting on soil.

© 2009 Taylor & Francis Group, London, UK

86 Dynamics of Structure and Foundation: 2. Applications

Figure 2.2.2 Block foundation on piles.

4 5 6 7

Coal mill foundations Gas Turbines Motor Driven Boiler feed pumps Centrifugal/Reciprocating type Compressors etc.

2.2.2 How does a block foundation supporting rotating machines differ from a normal foundation? The function of a foundation is basically to transfer the load coming on it from superstructure or items like vessels, tanks, skids etc. to the underlying soil. For a normal foundation supporting systems like building structure, vessels, tanks etc. the major load coming on it is static in nature. For geometric sizing of the foundation the stress induced on soil being less than the allowable bearing capacity of the soil sufﬁce. The dynamic force coming if any are quite rare and are mostly those due to earthquake forces and in case of very tall structures may be a bit more often due to wind induced vibration. In majority of the time for a normal foundation static load pre-dominates. While for machine foundation it is just the reverse. In most of the industrial facilities production being round the clock, the major load coming on the foundation is dynamic in nature and the foundation should be so designed that it is capable enough to sustain this dynamic loads over and above the static loads without causing any distress to underlying soil or to the machine it is supporting. So the question boils down to what are these conditions for which the foundation can safely sustain the dynamic load coming from the machines in operation? There are usually two conditions that are checked for while designing a machine foundation: © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 87

• •

Resonance check Amplitude check

2.2.2.1

Resonance check

All machines under operation usually induce a periodic dynamic load on the foundation and in most of the case can be represented by a function like P0 sin ωm t, where P0 = magnitude of the unbalanced force from the machine during its operation and ωm = operating frequency of the machine usually expressed in • •

radians/sec. Hertz or revolution per minute (rpm).

Due to this induced dynamic load from the machine the block foundation including some portion of the soil underlying the foundation is subjected to vibration and it is essential that the natural frequency (ωn ) of this vibration should be well away from the operating frequency of the machine i.e. resonance condition should not prevail. Irrespective of any code the normal practise is to design the foundation in such a way that its operating frequency is at least ±20% away from the natural frequency of the foundation. 2.2.2.2

Amplitude check

Under this condition, it is usually checked that the amplitude of vibration of the block foundation is well within the acceptable limits of engineering practise. The acceptable or the tolerable limits are usually suggested by the vendor supplying the equipment or in absence of such data are usually obtained from the codal stipulations. If the amplitude of vibration is more than this acceptable limit can mar the performance of the equipment in the following way • • • • •

Rapid deterioration of the machine due to heavy wear and tear. Excess amplitude of vibration inducing fatigue in the coupling and the connecting shafts leading to repeated breakdowns. Damage to the piping system connected rigidly to the equipment. Increase in decibel level during operation causing discomfort to the operators. Accumulation of explosive gases which at times could be dangerous to human life and property.

Based on the above discussion it is imperative that for a foundation designed for dynamic load the above two conditions are met. Now let us see how we mathematically model the soil-foundation system to theoretically check the two conditions as mentioned above. A block foundation as shown earlier constitute of a massive RCC block resting on the ground supporting the machine aligned over it. For all practical purpose the block and the machine is considered as a rigid lumped mass supported on an elastic base constituting the underlying soil/pile. © 2009 Taylor & Francis Group, London, UK

88 Dynamics of Structure and Foundation: 2. Applications

Kθ

Kh

Kv

Figure 2.2.3 Mathematical model of foundation in 2D.

2

2

3

3

1 1

Figure 2.2.4 Degrees of freedom in space.

Where, for analysis purpose the soil is modelled as equivalent linear springs. Shown in Figure 2.2.3 is the mathematical model of a machine foundation with soil modelled as linear springs based on mechanical analog of elastic half space theory in 2D, and 6 degrees of freedom it has on space (Figure 2.2.4). Before we go into further details of the state of the art theory for design of such foundations, it would possibly be worthwhile to look back at its evolution and study its subsequent metamorphosis to the various techniques used in present day design office practices.

2.2.3 Foundation for centrifugal or rotary type of machine: Different theoretical methods for analysis of block foundation 2.2.3.1 Tschebotarioff’s (1953) method This is one of the early methods used for calculating the natural frequency of a foundation. If fn is the natural frequency of the machine plus foundation in terms of contact © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 89

area then Tschebotarioff defined a term as reduced natural frequency fnr given by √ fnr = fn σ ,

where σ =

W t/ft 2 Af

(2.2.1)

where W is the weight of the machine plus foundation and Af is the base area. 2.2.3.2

Alpan’s (1961) method

Alpan made use of Tschebotarioff’s theory and developed an expression of natural frequency as α 1/4 fn = √ (Af ) W

(2.2.2)

where, fn = natural frequency of the foundation in cycles per minute; W = weight of the machine plus foundation in Kilogram; Af = contact area of the foundation in square meter, and α = a constant whose value depends on the nature of the soil as given in Table 2.2.1. 2.2.3.3

Newcomb’s (1951) method

Newcomb developed an empirical equation for natural frequency based on the static deflection of the soil data and is expressed as fn = 188

1 δst

(2.2.3)

where, fn = natural frequency in cycles per minute and, δst = static deflection in inches. The displacement parameter δst can be obtained from plate load test for any design bearing pressure. The above theories were mostly based on observation and experience and as such are empirical in nature. The theories can be put to use to check the resonant condition only no check for amplitude is possible by these methods. As such they shall only be used for preliminary design or sizing of the foundation only. Table 2.2.1 Value of constant, α. Sl. No.

Soil type

Value of α

1 2 3 4

Peat Plastic clay Sand Sand stone

3900 69000 82000 111000

© 2009 Taylor & Francis Group, London, UK

90 Dynamics of Structure and Foundation: 2. Applications

2.2.4 Analytical methods 2.2.4.1 Hsieh’s (1962) method Considering the soil as semi-infinite elastic medium, Hsieh put forward an analytical treatment for vibration of circular foundation as given hereunder For Translation:

mx¨ + r20 F2 ρGx˙ + r0 GF1 x = P0 sin ωm t

For rocking: ϕ θ¨ + r4θ F2 ρGx˙ + Gr3θ F1 θ = M0 sin ωm t

(2.2.4) (2.2.5)

where, m = mass of the foundation plus the machine; x, θ = displacement vectors in translational and rotational mode respectively; ϕ = mass moment of inertia; ρ = mass of the circular foundensity of soil; G = dynamic shear modulus of the soil; r0 = radius dation in translational mode which for rectangular foundations is LB/π ; rθ = radius 3

of the circular foundation in rocking mode which for rectangular foundations is 4 LB 3π 4 L3 B or 3π as the case may be; a0 = frequency factor and is given by ωm r0 ρ/G; ωm = the operating frequency of the machine; P0 , M0 = amplitude of exciting force in translation and rocking mode, and F1 , F2 = are functions whose values are given below: F1 is usually represented in the form, α1 − α2 a20 . Now substituting this value of F1 in the above differential equation we have, For translation mode, (m + α2 ρr30 )x¨ + r20 F2 ρGx˙ + r0 Gα1 x = P0 sin ωm t

(2.2.6)

For rocking mode, (ϕ + α2 ρr5θ )θ¨ + r4θ F2 ρGx˙ + α1 Gr3θ θ = M0 sin ωm t

(2.2.7)

Representing the above equations as mx x¨ + cx x˙ + kx x = P0 sin ωm t

and I θ¨ + cθ θ˙ + kθ θ = M0 sin ωm t

(2.2.8)

√ where, mx = (m + α2 ρr30 ); cx = r20 F2 ρG; kx = r0 Gα1 ; I = (ϕ + α2 ρr5θ ); cθ = √ r4θ F2 ρG; kθ = α1 Gr3θ . We had already seen earlier2 that the solution of such equation can be represented as

2 Refer Chapter 3 (Vol. 1) for solution of such equations having damped single degree of freedom.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 91

x¯ max =

P0 kx

sin ωm t

in which, r = ωm /ωnx and D = c/cc where, ωnx = of the system and is 2 mx kx . And for rocking mode,

θmax =

(2.2.9)

[(1 − r2 )2 + (2Dr)2 ]

M0 kθ

kx /mx ; and cc = critical damping

sin ωm t

(2.2.10)

[(1 − r2 )2 + (2Dr)2 ]

in which, r = ωm /ωnθand D = c/cc where, ωnθ = kθ /I; and cc = critical damping of the system and is 2 Ikθ . Here one point needs to be noticed is the additional term in the inertial coefficient mx = (m + α2 ρr30 )

and I = (ϕ + α2 ρr5θ )

(2.2.11)

Here the original mass and mass moment of inertia terms get added up with an additional term of α2 ρr30 and α2 ρr5θ respectively. This can be attributed as added mass of soil which starts vibrating with the foundation in same phase. This looks quite logical for it has indeed been observed during field observation that a part of soil below foundation do indeed participates in the vibration of the foundation system. Table 2.2.2 gives values of F1 and F2 for various modes given by Hsieh. For uniform and parabolic distribution of pressure, Hsieh suggest to use an effective radius αr0 where α is 0.78 and 0.59 respectively.

Table 2.2.2 Values of F1 and F2 . Mode

Poisson’s ratio

F1

F2

Vertical (0 < a0 < 1.5)

0.0

4.0 − 0.5a02

3.3 + 0.4a0

0.25

5.3 − 1.0a02

4.4 + 0.8a0

0.50

8.0 −

2.0a02

6.9 2.4 + 0.3a0

Horizontal (0 < a0 < 2.0)

Rocking (0 < a0 < 1.5) Torsion (0 < a0 < 2.0)

© 2009 Taylor & Francis Group, London, UK

0.0

4.5 −

0.25

4.8 −

0.50

5.3 −

0.2a02 0.2a02 0.1a02

0.0

2.5 −

0.4a02

0.4a0

All

5.1 −

0.3a02

0.5a0

2.5 + 0.3a0 2.8 + 0.4a0

92 Dynamics of Structure and Foundation: 2. Applications

2.2.4.2 Barkan or IS-2974 method This is by far the most popular method in the design office. Barkan (1962) developed this method way back in 60s and is still in vogue for design of machine foundations under rotating loads. In this method Barkan assumed the block foundation, shown in Figure 2.2.5, as a rigid lumped mass (i.e. he assumed the concrete block to have infinite stiffness in comparison to the soil and neglected any internal deformation of the concrete block itself) having three degrees of freedom. The soil medium he idealised as linear springs which he defined in terms of soil parameter cz , cx & cφ which are otherwise known as coefficient of elastic uniform compression, coefficient of elastic uniform shear, coefficient of elastic non-uniform compression respectively. 2.2.4.2.1

Vertical direction

Here in the vertical direction the spring constant is considered as kz = cz · Af

(2.2.12)

where, kz = equivalent spring in vertical direction; cz = coefficient of elastic uniform compression, and Af = plan area of the foundation.

C/L of machine shaft h

Pz sin

m

t

Mx sin

m

t

mx Px sin

x

m

t

H Zc mz

H x0

V

Figure 2.2.5 Mathematical model of Barkan for vertical and coupled sliding and rocking motion about the minor axes of the foundation.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 93

The natural frequency of the foundation is given by ωz =

kz m

and,

(2.2.13)

the amplitude of vertical vibration is given by δz =

Pz sin ωm t kz 1 − r2

(2.2.14)

where, r = ωm /ωn ; ωm = operating frequency of the machine. 2.2.4.2.2

For coupled horizontal and rocking mode

It has been observed by Barkan that when a foundation has horizontal force along its minor axis the foundation undergoes sliding and rocking simultaneously. When the foundation starts vibrating resistance is mobilised in the soil in terms of forces V and H as shown in Figure 2.2.5. The resistive force may thus be expressed as VR =

Cφ dA

(2.2.15)

and the resistive moment is expressed MR =

Cφ 2 φdA = Cφ IA φ

(2.2.16)

where = distance between rotation axis and the element of area dA; φ = angular rotation of the machine foundation; IA = second moment of area of the foundation contact surface with respect to the axis passing through the centroid of the area and perpendicular to the plane of the vibration. 2.2.4.2.3

For horizontal force, H

H = cτ Af x0 = cτ Af (x − Zc φ)

(2.2.17)

where, A = area of base contact; Zc , x, x0 etc. are shown in Figure 2.2.5. Now applying D’Alembert’s equation for dynamic equilibrium3 , we have mx¨ + H = Px sin ωm t

or mx¨ + cτ Af (x − Zc φ) = Px sin ωm t

where m is the mass of the machine foundation.

3 Refer Chapter 5 (Vol. 1) for definition of D’Alembert’s equation.

© 2009 Taylor & Francis Group, London, UK

(2.2.18)

94 Dynamics of Structure and Foundation: 2. Applications

Similarly for the moment equation about the minor axis of the foundation we have Jxφ φ¨ − cτ Af Zc x + φ(cφ IA − WZc + cτ AZc2 ) = Mx sin ωm t

(2.2.19)

where Jxφ = mass moment of inertia of the machine-foundation block about the minor axis of rotation. From Equations (2.2.18) and (2.2.19), we see that they contain both x and φ, so a coupled sliding and rocking motion will develop along this direction. Using the above equations and considering free vibrations, Barkan developed the following equation for calculation of the frequencies. ω4 −

J0 (ωφ2 + ωx2 )ω2 Jxφ

+

ωφ2 ωx2 J0 Jxφ

=0

(2.2.20)

c I −WZ

c A

where J0 = Jxφ + mZc2 ; ωφ2 = φ A J0 c and ωx2 = τm f . Based on the above, the two principal frequencies for the coupled vibration is given by ⎡ 2 ω1,2 =

J0 ⎣ 2 ωφ + ωx2 ± 2Jxφ

(ωx2 + ωφ2 )2 −

4Jxφ ωφ2 ωx2 J0

⎤ ⎦

(2.2.21)

Considering the forced vibration, the amplitudes Ax , Aφ may be expressed as Ax = Aφ =

2.2.4.2.4

2 )P ± c A Z M (cφ IA − WZc + cτ Af Zc2 − Jxφ ωm x τ f c x 2 )(ω2 − ω2 ) mJxφ (ω12 − ωm m 2 2 )M cτ Af Zc Px ± (cτ Af − mωm x 2 )(ω2 − ω2 ) mJxφ (ω12 − ωm m 2

sin ωm t

sin ωm t (2.2.22)

Torsional mode

For this mode, again the foundation considered is a lumped mass having single degree of freedom when the frequency and amplitude are given by ωψ =

Kψ Iψ

and ψ =

T sin ωm t kψ 1 − r2

(2.2.23)

where, Kψ = cψ Iψ ; r = ωm /ωn . This method is also recommended by IS 2974 “Code and design practices for machine foundation” and still remains the most popular method for vibration analysis of block foundations. But let us see the limitations of Barkans method with respect to the reality under field conditions. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 95

2.2.4.2.5

The Limitations

The major limitations that can be attributed to Barkan’s method are as follows: •

• • •

Barkan’s model does not take damping into consideration. It has been observed from field instrumentation data that damping plays a significant role in the overall response of the foundation especially when the operating frequency of the machine is low. It does not account for the embedment effect of the surrounding soil which could play a significant role on the magnitude of soil stiffness and damping. It does not take into cognisance the virtual mass of soil which vibrates in same phase with the machine and the foundation. Barkan suggested spring value (usually the coefﬁcient of uniform elastic compression) of the soil to be obtained from dynamic plate load test4 and may only give correct values for a shallow depth below the surface while this may not be valid for layered soil and also when the contact area of the foundation is large.

So based on the above limitations it was felt to upgrade the mathematical process in the design of machine foundation. Before we study the further enhancements it would be worth to write Barkan’s equation in a more generic form. The soil spring stiffness is described by the terms Kx = cτ Af

and Kxφ = cφ IA ,

(2.2.24)

while, the equations of equilibrium are defined as mx¨ + cτ Af (x − Zc φ) = Px sin ωm t,

and

Jxφ φ¨ − cτ Af Zc x + φ(cφ IA − WZc + cτ Af Zc2 ) = Mx sin ωm t

(2.2.25)

Substituting the values of Kx and Kφ , we have mx¨ + Kx (x − Zc φ) = Px sin ωm t

and

Jxφ φ¨ − Kx Zc x + φ(Kφ − WZc + Kx Zc2 ) = Mx sin ωm t

(2.2.26)

The above on writing in matrix form can be represented as

m 0

0 Jxφ

Kx x¨ + φ¨ −Kx Zc

−Kx Zc Kφ + Kx Zc2 − WZc

x Px = sin ωm t φ Mx

(2.2.27)

Based on the above equation we will see later how we develop further realistic model of the coupled horizontal and rocking mode5 . 4 Usually carried out with a plate of size 300 mm × 300 mm. 5 Structural Engineers be alert from this point.What we are going to apply herein subsequently are the theories of structural dynamics for system with two degrees of freedom.

© 2009 Taylor & Francis Group, London, UK

96 Dynamics of Structure and Foundation: 2. Applications

2.2.4.3 Richart and Lysmer’s model Richart et al. (1970) idealised the foundation as a lumped mass supported on soil which is idealised as frequency independent springs which he described in terms of soil parameter dynamic shear modulus or shear wave velocity of the soil for circular footing when footings having rectangular shape in plan can be converted into a footing having equivalent circular radius. Tables 2.2.3 and 4 along with Figure 2.2.6 show the different values of spring and damping value as per Richart and Lysmer. In which, G = dynamic shear modulus of the soil and is given by, G = ρs Vs2 ; ν = Poisson’s ratio of the soil; ρs = mass density of the soil; Vs = shear wave velocity of the soil obtained from soil testing; g = acceleration due to gravity; m = mass of the machine and foundation; J = mass moment of inertia of the machine and the foundation about the appropriate axes; K = equivalent spring stiffness of the soil; C = damping value of the soil; B = inertial factor contributing to the damping factor; D = damping ratio of the soil; r = equivalent radius of a circular foundation; L = length of the foundation, and, B = width of the foundation. Many engineers in design offices prefer to use Richart’s springs neglecting the damping and use Barkan’s formulation in matrix form as shown earlier to find out the natural frequency and amplitude of the foundation. But, by neglecting the damping, he could signiﬁcantly over-estimate the amplitude of vibration (specially for low tuned machines) thus adding to the cost by trying to restrict it within the acceptable limits. Let us now see as to what form the equations take when damping is introduced in the system.

Table 2.2.3 Values of soil springs as per Richart and Lysmer (1970) model. Sl. No.

Direction

Spring value

1

Vertical

Kz =

4Grz (1 − υ)

2

Horizontal

Kx =

32(1 − υ)Gr x (7 − 8υ)

3

3.1

4

Rocking

Rocking

Twisting

Kφ x =

Kφy =

Kψ =

© 2009 Taylor & Francis Group, London, UK

8Gr 3φx 3(1 − υ) 8Gr 3φy 3(1 − υ) 16Gr 3ψ 3

Equivalent radius LB rz = π rx =

LB π

rφx =

Remarks This is in vertical Z direction

This induce sliding in horizontal X or Y direction

LB3 3π

This produces rocking about Y axis

L3 B 3π

This produces rocking about X axes

L3 B + BL3 6π

This produces twisting about vertical Z axis

4

rφy =

4

rψ =

4

Analysis and design of machine foundations 97

Table 2.2.4 Values of soil damping as per Richart and Lysmer (1970) model. Damping ratio and Damping value

Sl. No.

Direction

Mass ratio (B)

1

Vertical

Bz =

0.25 m(1 − υ)g ρs rz3

0.425 ζz = √ , Bz √ Cz = 2ζ z Kz m

2

Horizontal

Bx =

(7 − 8υ)mg 32(1 − υ)ρs rx3

0.288 ζx = √ , Bx √ Cx = 2ζx Kx m

3

Rocking

Bφx =

0.375(1 − υ)Jφx g 5 ρs rφx

Rocking

0.375(1 − υ)Jφy g 5 ρs rφy

Bφy =

Twisting

Bψ =

Jψ g ρs rψ5

This damping value is in lateral X or Y direction

0.15 , (1 + Bφx ) Bφx = 2ζφx Kφx Jφx

This damping value is for rocking about Y direction

0.15 , (1 + Bφy ) Bφy = 2ζφy Kφy Jφy

This damping value is for rocking about Y axes

ζφy = Cφy

4

This is damping value is in vertical Z direction.

ζφx = Cφx

3.1

Remarks

0.5 , 1 + 2Bψ Cψ = 2ζψ Kψ Jψ

ζψ =

This damping value is valid for twisting about vertical Z axis.

Z Y L X

B

Figure 2.2.6 3D View of the block foundation.

2.2.4.3.1

Vertical motion considering damping of the soil

For vertical direction the equation becomes that of a lumped mass having single degree of freedom when m¨z + Cz z˙ + Kz z = P0 sin ωm t © 2009 Taylor & Francis Group, London, UK

(2.2.28)

98 Dynamics of Structure and Foundation: 2. Applications

solution is, ωz =

ωm ωn

where r = 2.2.4.3.2

Kz m

and

δz =

(P0 /Kz ) sin ωm t

(2.2.29)

(1 − r2 )2 + (2Dz r)2

and Dz = damping ratio.

Coupled horizontal and rocking motion considering damping soil

We have seen that based on Barkan’s formulation the equation of motion in matrix form is

m 0

0 Jxφ

Kx x¨ + φ¨ −Kx Zc

−Kx Zc Kφ + Kx Zc2 − WZc

x P0 = sin ωm t φ M0

(2.2.30)

Since the above equation is based on D’Alembert’s equation, the equation are said to be statically coupled when the stiffness matrix and damping matrix have the same matrix form (Meirovitch 1975). Thus, based on the above argument the damped equation of motion in coupled rocking and sliding mode becomes

m 0

0 Jxφ

Cx x¨ + φ¨ −Cx Zc

Kx + −Kx Zc

−Cx Zc Cφx + Cx Zc2 − WZc

−Kx Zc Kφx + Kx Zc2 − WZc

x˙ φ˙

x P0 = sin ωm t M0 φ

(2.2.31)

The above equations constitute the complete equation of motion for coupled sliding and rocking mode considering the damping effect of the soil. Actually for all practical calculations for finding out the dynamic response of the foundation the term −WZc is usually neglected, for it has been observed that unless the foundation is very massive and deep the term WZc has no significant effect on the overall response of the system. Based on the above argument the above equation reduces to

Cx −Cx Zc x¨ x˙ 0 + Jxφ φ¨ −Cx Zc Cφx + Cx Zc2 φ˙ Kx −Kx Zc x P0 + = sin ωm t M0 −Kx Zc Kφx + Kx Zc2 φ

m 0

(2.2.32)

The equation above surely looks elegant, but now comes the catch . . . , for this damping matrix of soil is not proportional to either the mass or the stiffness of the soil, moreover they are coupled by the term of Zc and W (the weight of the foundation) and © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 99

as such do not de-couple on orthogonal transformation6 . This forms a major headache to the designer as he is not in a position to guess the damping ratio. As we had already discussed in Chapter 5 (Vol. 1) that the most appropriate technique in such case is to resort to Time history analysis for the correct answer, many engineers find time history too intensive in terms of calculation7 and prefer to use modal response technique as a tool for analysis of the same. Of course the easiest way out is to neglect the damping and argue that the design is conservative! But this need not be done for we have already stated in Chapter 5 (Vol. 1) that it is possible to by pass this problem of orthogonal de-coupling, even when the damping matrix is non-proportional which though not exact would give still give a designer a reasonable value to estimate a more realistic amplitude of vibration (it is surely a better value than no damping considered). We will consider application of this technique in subsequent section (section 2.2.5). 2.2.4.3.3

Torsional mode

In this mode again the block foundation is again considered as a lumped mass having single degree of freedom, natural frequency and the torsional rotation, ψ is given by ωψ =

Kψ ; Iψ

and, ψ =

T sin ωm t Kψ

(2Dr)

2

(2.2.33)

+ (1 − r2 )2

where Kψ = 16Gr3ψ /3, D is the damping ratio in the torsion mode (Table 2.2.4) and r is the ratio between the natural frequency of the foundation in torsion mode and the operating frequency of the machine.

2.2.5 Approximate analysis to de-couple equations with non-proportional damping We have seen that the equation of motion is given by

Cx −Cx Zc x¨ x˙ 0 + 2 ¨ Jxφ φ −Cx Zc Cφx + Cx Zc − WZc φ˙ Kx −Kx Zc x P0 = sin ωm t + M0 −Kx Zc Kφx + Kx Zc2 − WZc φ

m 0

(2.2.34)

For finding the natural frequencies we perform the eigen value analysis when the un-damped equation becomes (neglecting-WZc for reasons as cited earlier) Kx − mλ −Kx Zc =0 (2.2.35) −Kx Zc Kφ + Kx Zc2 − Jφx λ

6 For further explanation on this property refer to Chapter 5 (Vol. 1), the topic of orthogonal transformation for modal response technique. 7 And the theory underlying time history is eluding many is not too an uncommon a fact. . ..

© 2009 Taylor & Francis Group, London, UK

100 Dynamics of Structure and Foundation: 2. Applications

Solving the above equation we find out the eigen value vis-a vis the natural frequency of the foundation system. Let the eigen values be λ1 and λ2 respectively. Let T corresponding eigen vectors be φxx φφx T and φφx φ φφ respectively, when the φφx φ . complete eigen vector matrix is expressed as, xx φφx φφφ Since the eigen vector is known separately for each mode we find out the damping ratio separately for each mode as follows. As a first step we perform the operation {φ}T [C]{φ} for each mode. For the first mode, we have φxx

φφx

Cx −Cx Zc

−Cx Zc Cφx + Cx Zc2

φxx φφx

which gives,

φxx

Cx φxx − Cx Zc φφx φφx −Cx Zc φxx + (Cφx + Cx Zc2 )φφx

2 2 = Cx φxx − 2Cx Zc φφx φxx + (Cφx + Cx Zc2 )φφx

(2.2.36)

It should be realised that the above is a unique value and we also know that the operation {φ}T [C]{φ} breaks up the equation to form 2Di ωi where i is the degrees of freedom of the system. Now considering, 2 2 − 2Cx Zc φφx φxx + (Cφx + Cx Zc2 )φφx , 2Di ωi = Cx φxx

for the first mode, D1 =

2 − 2C Z φ φ + (C 2 2 Cx φxx x c φx xx φx + Cx Zc )φφx

2ω1

(2.2.37)

where D1 = damping ratio for the first mode and; ω1 = first natural frequency of the foundation. Similarly, for the second mode proceeding in same manner it can be proved that D2 =

2 − 2C Z φ φ 2 2 Cφx φφx x c φx φφ + (Cφx + Cx Zc )φφφ

2ω2

(2.2.38)

Once the damping ratios are identiﬁed we assume, [C] = α[M] + β[K] and performing the operation {φ}T [C]{φ} = α{φ}T [M]{φ} + β{φ}T [K]{φ}, © 2009 Taylor & Francis Group, London, UK

(2.2.39)

Analysis and design of machine foundations 101

We have, for two degrees of freedom 2D1 ω1 = α + βω12

2D2 ω2 = α + βω22

and

(2.2.40)

Thus, we have two equation with two unknowns, α and β, and solving the above two equations we get the value of α and β. Once these values are known one can obtain an equivalent proportional soil damping from the operation [C] = α[M] + β[K] which is now quite suitable for modal response technique. We now further explain the above method based on a suitable numerical problem.

Example 2.2.1 For a block foundation supporting a centrifugal pump was observed to have the following design data M = 50 kN sec2 /m, J = 100 kN sec2 · m, Z c = 1.5 m, K x = 3000 kN/m, K φ = 5000 kN/m, Cx = 200 kN/m, Cφ 350 kN/m. Find out • • • •

The natural frequencies in coupled horizontal and rocking mode. The normalized eigen vectors. Find out the approximate damping ratios for each mode. Correct the damping matrix based on equivalent Rayleigh damping.

Solution: The complete equation of motion for the foundation under coupled rocking and sliding mode is given by m 0

0 Jxφ

Cx x¨ + φ¨ −Cx Zc

+

Kx −Kx Zc

−Kx Zc Kφx + Kx Zc2

−Cx Zc Cφx + Cx Zc2

x˙ φ˙

x P0 = sin ωm t φ M0

Based on the above the equation of motion becomes

x¨ 200 + −300 φ¨

50 0

0 100

+

3000 −4500

x P0 = sin ωm t φ M0

−4500 11750

© 2009 Taylor & Francis Group, London, UK

x˙ φ˙

−300 988

102 Dynamics of Structure and Foundation: 2. Applications

•

Calculation of the un-damped natural frequency To find out the natural frequencies we have

3000 − 50λ −4500

➔

−4500 =0 11750 − 100λ

35.25 × 106 −3 × 105 λ−5.875 × 105 λ+5000λ2 −20.25 × 106 = 0

The above equation on simplification reduces to λ2 − 177.5λ + 3000 = 0 177.5 + (177.5)2 − 4 × 1 × 3000 λ1 = = 158.5 ➔ ω = 12.58 rad/sec 2 177.5 − (177.5)2 − 4 × 1 × 3000 and λ2 = = 18.92 ➔ ω = 4.35 rad/sec 2

For the ﬁrst mode we have 3000 − 50 × 18.92 −4500 x =0 −4500 11750 − 100 × 18.92 φ Solving the above two homogeneous equations, considering x = 1.00, we have, x : φ = 1.00 : 0.45644. Similarly for the second mode we have 3000 − 50 × 158.5 −4500 x =0 −4500 11750 − 100 × 158.5 φ Solving the above two homogeneous equations, considering x = 1.00 We have, x : φ = 1.00 : −1.094 Thus, the complete eigen vector matrix becomes

x 1.00 = φ i=1,2 0.45644

1.00 −1.094

Calculation of normalized eigen vectors On operation {φ}T [M]{φ} we have For the first mode

50 0 1.00 0.45644 = 70.83 0 100 0.45644 √ ➔ Mr = 70.83 = 8.416

1.00

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 103

Thus

{φ}i=1 =

0.1188173 0.0542329

For the second mode, on operation {φ}T [M]{φ}, we have 1.00 ➔

− 1.094 Mr =

50 0 0 100

1.00 = 130.26 −1.094

√ 169.6836 = 13.026

0.07676 Thus {φ}i=2 = −0.08398 The normalized eigen vectors for the two modes are,

x 0.1188173 0.07676 = φ i=1,2 0.0542329 −0.08398 Calculation of the modal damping ratios Now if we perform the orthogonal operation with the complete normalized eigen vector matrix it will be observed that while {φ}T [M]{φ} diagonalise to → [I] and {φ}T [K]{φ} → [λ]. But as will be seen now that {φ}T [C]{φ} will NOT de-couple to the form 2Di ωi .

0.1188173 0.0542329 200 −300 0.1188173 0.07676 0.07676 −0.08398 −300 988 0.0542329 −0.08398 0.1188173 0.0542329 7.49359 40.546 = 0.07676 −0.08398 18.130205 −106.00024 1.8736 −0.9311626 = −0.9473666 12.014236

Since the above matrix has off-diagonal terms will NOT be equal to zero, hence we conclude that the matrix has not de-coupled due to orthogonal transformation. As such we treat the eigen vectors separately for each individual modes, Thus for the first mode we have 0.1188173

0.0542329

➔ 2D1 ω1 = 1.86308. or,

D1 =

1.86308 = 0.214 2 × 4.35

© 2009 Taylor & Francis Group, London, UK

200 −300 0.1188173 = 1.86308 −300 988 0.0542329

104 Dynamics of Structure and Foundation: 2. Applications

Similarly for the second mode we have

0.07676

➔ D2 =

200 − 0.08398 −300

−300 988

0.07676 = 12.0142 −0.08398

12.0142 = 0.4775. 2 × 12.58

Thus damping ratio is of the order of 21.4% for the first mode and 47.75% for the second mode. Correction to the damping matrix based on Rayleigh damping Let us assume the damping matrix is of the form [C] = α[M] + β[K], then on orthogonal transformation we have {φ}T [C]{φ} = α{φ}T [M]{φ} + β{φ}T [K]{φ}, which gives → 2D1 ω1 = α + βω12 and 2D2 ω2 = α + βω22 . On substituting the respective values, we have For the first mode, α + 18.92β = 2 × 0.214 × 4.35 → α + 18.92 = 1.8618 For the second mode, α + 158.5β = 2 × 0.4775 × 12.58 → α + 158.5β = 12.0139 Solving the above two equations we have → α = 0.48568 and β = 0.0727 As [C] = α[M] + β[K] we have

50 [C] = 0.48568 0

242 = −327.15

0 3000 + 0.0727 100 −4500

−4500 11750

−327.15 902.795

is the modified damping matrix which satisfies the orthogonal property. If we compare the value with the original damping matrix obtained from the soil property we see each of the term has got slightly modified.

For practical design office calculation this is usually deemed sufficient. It at least depicts a better result than no damping considered at all. We will subsequently see how data based on modified damping matrix compare with time history response which we had stated would be the most appropriate accurate method that could be adopted with non-proportional damping. But prior to that let us evaluate another form in which equations of motion for coupled rocking and sliding motion can be formulated too. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 105

2.2.6 Alternative formulation of coupled equation of motion for sliding and rocking mode We had already shown in Chapter 5 (Vol. 1) that other then formulating equation of motion based on D’Alembert’s equation we can also write them down based on energy concepts as put forward by Lagrange. We use here Lagrange’s formulation to derive the equation of motion for coupled rocking and sliding motion. We had shown previously in chapter 5 (Vol. 1) that considering a conservative system having kinetic energy as T and potential energy as U we have d(T + U) = 0 and

d(T + U) =

n ∂T d ∂T ∂U − dqi = 0 (2.2.41) + dt ∂ q˙ i ∂qi ∂qi i=1

For the machine foundation subjected to coupled rocking and sliding motion Kinetic energy (T) =

1 1 ˙ 2 + J φ˙ 2 m(x˙ + Zc φ) 2 2

The Potential Energy (U) =

and

1 1 Kx x2 + Kφ φ 2 2 2

Based on the above equation we have T=

1 1 ˙ 2 + J φ˙ 2 m(x˙ + Zc φ) 2 2

∂T ˙ and = m(x˙ + Zc φ) ∂ x˙

U=

We have,

Also,

d ∴ dt For

d dt

∂T ∂ x˙

1 1 K x x2 + Kφ φ 2 ; 2 2

T=

1 1 ˙ 2 + J φ˙ 2 ; m(x˙ + Zc φ) 2 2

∂T ∂ φ˙

U=

¨ = mx¨ + mZc φ¨ = m(x¨ + Zc φ)

∂U = Kx x ∂x ∂T ˙ + J φ˙ = mZc (x˙ + Zc φ) ∂ φ˙

= mZc x¨ + mZc2 φ¨ + J φ¨

1 1 Kx x2 + K φ φ 2 ; 2 2

∂U = Kφ φ ∂φ

Substituting the above values in the Lagrangian equation, we have mx¨ + mZc φ¨ + Kx x = 0; © 2009 Taylor & Francis Group, London, UK

and mZc x¨ + (mZc2 + J)φ¨ + Kφ φ = 0

(2.2.42)

106 Dynamics of Structure and Foundation: 2. Applications

Thus writing in matrix form the free vibration equation becomes

m mZc

mZc mZc2 + J

x¨ K + x 0 φ¨

0 Kφ

x =0 φ

(2.2.43)

Equation (2.2.42) is known as dynamically or inertially coupled equation. You will observe in contrary to the formulation based on D’Alembert’s equation, where the mass matrix is diagonal here the stiffness matrix is diagonal while the mass matrix is non-diagonal but symmetric. For equation coupled by inertia it has been observed that both the damping and stiffness matrix remains diagonal (Meirovitch 1975) and the complete equation of motion thus becomes

m mZc

mZc mZc2 + J

x¨ C + x 0 φ¨

0 Cφ

x˙ K + x 0 φ˙

0 Kφ

x P0 = sin ωm t M0 φ (2.2.44)

Now that we have established the equation the question that obviously crops up in mind is what is the advantage of this equation over the normal equation that was derived based on static coupling/Barkan’s equation. The first thing we will see subsequently that the eigen-values remain invariant with this formulation. Moreover it has been observed that damping ratio derived by this method are quantitatively closer to the values derived from classical analysis based on frequency domain analysis in complex domain. (Wolf 1988). The reason for the better prediction of damping ratio could be that the damping matrix derived by this formulation is in uncoupled form. We now further explain the above based on suitable numerical example.

Example 2.2.2 For a block foundation as described in Example 2.2.1, calculate the following based on Lagrange’s Formulation. • • •

The natural frequencies in coupled horizontal and rocking mode The normalized eigen vectors Approximate damping ratios for each mode. All design parameters pertaining to design remains same as in Example 2.2.1.

Solution: The complete equation of motion for the foundation under coupled rocking and sliding mode based on Lagrange’s formulation is given by © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 107

m mZc

mZc mZc2 + J

x¨ C + x 0 φ¨

0 Cφ

x˙ K + x 0 φ˙

0 Kφ

x P0 = sin ωm t M0 φ

Based on the above the equation of motion becomes

x¨ 200 + 0 φ¨

50 75

75 212.5

x˙ 3000 0 x + 0 5000 φ φ˙

0 350

P0 sin ωm t M0

=

To find out the natural frequencies we have 3000 − 50λ −75λ = 0 which on expansion gives, −75λ 5000 − 212.5λ 177.5 ± 139.66 :: λ1 = 2 18.92, ω1 = 4.35 rad/sec; λ2 = 158.5, ω2 = 12.58 rad/sec. 5000λ2 − 8.875 × 105 λ + 15 × 106 = 0 → λ =

It should be observed that the natural frequencies are identical to the one obtained in Example 2.2.1 based on static coupling. Calculation of the eigen vectors For the first mode we have 3000 − 50 × 18.92 −75 × 18.92

2054 −1419

➔

−75 × 18.92 x =0 5000 − 212.5 × 18.92 φ x =0 φ

−1419 979.5

Solving the above homogenous equations considering, x = 1.00 we have x : φ = 1.00 : 1.447 Similarly for the second mode 3000 − 50 × 158.5 −75 × 158.5

−75 × 158.5 x =0 5000 − 212.5 × 158.5 φ x =0 φ

−4925 −11887.5 ➔ −11887.5 −28681.25

Solving the above homogenous equations considering, x = 1.00 we have x : φ = 1.00 : −0.4143. © 2009 Taylor & Francis Group, London, UK

108 Dynamics of Structure and Foundation: 2. Applications

Calculation of the normalized eigen-vectors For the ﬁrst mode

T

{φ1 } [M]{φ1 } = 1.00

50 1.447 75

= 711.878 ➔

75 212.5

1.00 1.447

Mr = 26.68

0.03748 Thus, the normalized eigen vector is {φ1N } = . 0.05423 For the second mode we have {φ2 }T [M]{φ2 } = 1.00

− 0.4143

= 24.32945

➔

50 75

75 212.5

1.00 −0.4143

Mr = 4.93248

0.20273 Thus, the normalized eigen vector is {φ2N } = . −0.08399 Calculation of the damping ratios For the first mode 200 0.05423 0

T

{φ1 } [C]{φ1 } = 0.03748 = 1.1632 i.e. D1 =

0 350

0.03748 0.05423

➔ 2D1 ω1 = 1.1632

1.1632 = 0.133. 2 × 4.35

For the second mode {φ1 }T [C]{φ1 } = 0.20273 −0.08399

200 0

0 350

= 10.688 ➔ 2D2 ω2 = 10.688 i.e. D2 =

10.688 = 0.4248. 2 × 12.58

Thus • •

for the first mode, the damping ratio is 13.3% and for the second mode, the damping ratio is 42.48%.

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0.20273 −0.08399

Analysis and design of machine foundations 109

The table below gives comparative results based static and dynamic coupling formulations for the examples solved above.

Formulation basis Static coupling Dynamic coupling

Natural frequency 1st mode

Natural frequency 2nd mode

Normalized mode shapes 1st mode

Normalized mode shapes 2nd mode

Damping ratio 1st mode

Damping ratio 2nd mode

4.35

12.58

0.1181 : 0.0542

0.0767 : −0.0839

21.4%

47.75%

4.35

12.58

0.0375 : 0.0542

0.2027 : −0.0839

13.3%

42.48%

Based on the above it would possibly be worthwhile to know how the amplitudes vary based on the above two methods vis-à-vis the time history response which we advocated as the most appropriate and correct method for handling responses having non-proportional damping. This is what we are going to establish based on suitable numerical example hereafter. Example 2.2.3 For a block foundation as described in Example 2.2.1. Calculate the following amplitude of vibration based on • • • •

Static Coupling Dynamic coupling Time history response Discuss the results based on the three answers.

The unbalanced dynamic force is 200 KN having operating frequency of 750 r.p.m acting at a height 600 mm from the top of foundation. The total height of the concrete block (L) is 1.8 m. All other design parameters pertaining to design remains same as Example 2.2.1. Solution: The operating frequency of the machine is 750 r.p.m. = (750 × 2 × π )/60 = 78.53 rad/sec. Solution based on static coupling The equation of motion in matrix form is given by ¨ + [C]{X} ˙ + [K]{X} = {P} [M]{X} 200 P0 = sin 78.53t Here, the force matrix is given by 180 M0

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110 Dynamics of Structure and Foundation: 2. Applications

For static coupling, The normalized eigen-vector obtained earlier was 0.1188173 0.076746 [φ] = 0.0542333 −0.0839942 Considering the operation, [φ]T {P} we have 0.1188173 0.0542333 200 [φ]T {P} = sin 78.53t 0.076746 −0.0839942 180

33.52 = sin 78.53t 0.2293 On orthogonal transformation we have {φ}T [M]{φ}{ξ¨ } + {φ}T [C]{φ}{ξ˙ } + {φ}T [K]{φ}{ξ } = {φ}T {P} This in decoupled form reduces to {ξ¨1 } + 2D1 ω1 {ξ˙1 } + [ω12 ]{ξ1 } = p0 sin ωm t; {ξ¨2 } + 2D2 ω2 {ξ˙2 } + [ω22 ]{ξ2 } = m0 sin ωm t Substituting the different values, we have {ξ¨1 } + 2 × 0.214 × 4.35{ξ˙1 } + 18.92{ξ1 } = 33.52 sin 78.53t and {ξ¨2 } + 2 × 0.4776 × 12.58{ξ˙2 } + 158.5{ξ2 } = 0.230 sin 78.53t ∴ ξ1 =

33.53 sin 78.53t

(18.92 − 78.532 )2 − (1.862 × 78.53)2 = 5.45061918 × 10−3 sin 78.53t

ξ2 =

0.230 sin78.53t

(158.5 − 78.532 )2 − (12.016 × 78.53)2 = 3.816235259 × 10−5 sin 78.53t

Thus back transferring on global co-ordinate we have

x 0.1188173 0.076746 5.45061918 = × 10−3 sin 78.53t φ 0.0542333 −0.08399942 0.03816235259

0.6505566 = × 10−3 sin 78.53t 0.2923994 Solution based on dynamic coupling The normalized eigen-vector obtained earlier was 0.0374669 0.2027373 [φ] = 0.0542333 −0.0839942

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Analysis and design of machine foundations 111

Considering the operation [φ]T {P} we have

0.0374669 0.0542333 200 [φ] {P} = sin 75.53t 0.2027373 −0.0839942 180

17.255374 = sin 78.53t 25.428504 T

On orthogonal transformation we have [φ]T [M][φ]{ξ¨ } + [φ]T [C][φ]{ξ˙ } + [φ]T [K][φ]{ξ } = [φ]T {P} This in decoupled form reduces to {ξ¨1 } + 2D1 ω1 {ξ˙1 } + [ω12 ]{ξ1 } = p0 sin ωm t; {ξ¨2 } + 2D2 ω2 {ξ˙2 } + [ω22 ]{ξ2 } = m0 sin ωm t Substituting the different values we have {ξ¨1 } + 2 × 0.1505967 × 4.35{ξ˙1 } + 18.92{ξ1 } = 17.25537 sin 78.53t

and

{ξ¨2 } + 2 × 0.4248 × 12.58{ξ˙2 } + 158.5{ξ2 } = 25.429054 sin 78.53t ∴ ξ1 =

17.25537 sin 78.53t (18.92 − 78.532 )2 − (1.3102 × 78.53)2

= 2.806253 × 10−3 sin 78.53t ξ2 =

25.428504 sin 78.53t (158.5 − 78.532 )2 − (10.869 × 78.53)2

= 4.1900494 × 10−3 sin 78.53t Thus back transferring on global co-ordinate we have

x 0.0374669 0.2027373 2.806253 = × 10−3 sin 78.53t φ 0.0542333 −0.08399942 4.1900494

0.9546209 = × 10−3 sin 78.53t −0.1997474 Figure 2.2.7 gives a comparison of the amplitude value for the static and dynamic coupling case. It will be observed that • •

Dynamically coupled equation gives slightly higher amplitude than statically coupled equations. This is applicable for both translation and rotation.

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112 Dynamics of Structure and Foundation: 2. Applications

Amplitude of vibration

1.50E-03 1.00E-03

Translation Based on Static coupling

5.00E-04

Rotation based on static coupling

0.00E+00 1

20 39 58 77 96 11 5 13 4

-5.00E-04 -1.00E-03

Translation Based on Dynamic coupling Rotation based on dynamic coupling

-1.50E-03 Time steps

Figure 2.2.7 Comparison of response values static versus dynamic coupling.

Next we compare these results with time history response where we treat the complete mass, damping and stiffness matrix in uncoupled form and integrate directly the equation ¨ + [C]{X} ˙ + [K]{X} = {P sin ωM t}8 [M]{X} Figure 2.2.8 shows a very interesting result pertaining to time history vis-à-vis modal response technique based on dynamically coupled equation. The time history technique used has been Wilson-θ method having a time step of 0.0075 seconds and response has been calculated to 500 steps. It will be observed that in modal response technique we have ignored the transient response part and have only found out the response based on the steady state part, while, the step by step integration considers both the transient and the steady state responses. In comparison to a steady state response of 1 mm the time history starts with peak amplitude of approximately 5 mm in step 18 and slowly converges to a value near to 1 mm at about 295th step and becomes steady after that9 .

8 Refer Chapter 5 (Vol. 1) the topic of Time history/step by step integration. 9 Looking at the displacement value of 1 and 5 mm hardend professionals could well frown or smirk. But herein the example is deliberately designed like this to give beginners and especially students a quantitative feel which eludes many when working in micron level. Real life problems would come surely latter in the chapter.

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Analysis and design of machine foundations 113

It will be observed that the values are quite closely matching at the steady state position with step by step integration giving slightly higher values than dynamically coupled modal response. The initial response due to the transient part of the time history analysis is significant (about 5 mm), as this decays down quickly after some time (here about 2.0 seconds after the start) will really not have much effect on the over all behavior of the foundation as such, but for pipes and nozzles rigidly connected to the machine this initial high amplitude of 5 mm can have significant effect and if proper care is not taken may induce severe reversal of stress and may even induce failure.

6.00E-03 5.00E-03

Amplitude

4.00E-03 3.00E-03 Translation Modal response Translation based on time history

2.00E-03 1.00E-03 0.00E+00 -1.00E-03

1

52

103

154

205

256

307

358

409

460

-2.00E-03 -3.00E-03

Time steps

Figure 2.2.8 Comparison of response time history versus modal response.

2.3 TRICK TO BY PASS DAMPING – MAGNIFICATION FACTOR, THE KEY TO THE PROBLEM. . . A perfectionist may not like the methodology proposed regarding approximate estimate of the damping ratio based on individual mode. At the same time he might argue that for secondary equipment like medium or small capacity pumps doing time history analysis is too intense and not really called for. Fair enough, for the argument is not without some sanctity so how do we tackle this riddle? The most logical solution to the above problem could be that if we can create a condition where damping plays a negligible effect compared to un-damped situation then we can surely ignore damping from our basic equation and arrive at a result which is as good an answer with damping and the problem is solved. So the next obvious query will be, what is this condition which will give an invariable answer irrespective of damping taken or not? Before we describe this condition it would be worthwhile to understand what the magnification factor is.

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114 Dynamics of Structure and Foundation: 2. Applications

Consider Figure 2.3.1. We had seen earlier that equation of motion for a body having damped single degree of freedom is given by

δz =

P0 Kz

sin ωm t

(1 − r2 )2 + (2Dz r)2

(2.3.1)

where r = ωm /ωn (frequency ratio) and Dz = damping ratio In the above equation the denominator is known as the magnification factor. i.e. M.F. =

1 (1 − r2 )2

+ (2Dz r)2

(2.3.2)

Based on the above expression we define magnification factor as a factor which gives us a measure of how many times the dynamic amplitude gets magnified over the equivalent static deflection of P0 /K. Now the above is a very interesting equation for if we plot the above equation for various values of damping against the frequency ratio we get curve as shown in Fig. 2.3.1. In the below curve it will be observed that when the frequency ratio (ωm /ωn ) is about 3.0, all the lines irrespective of whatever is the damping converges nearly to a single line. Now what does this signify? It implies that when the frequency ratio is more than 3.0 irrespective of the damping ratio the magniﬁcation factor do not change and thus the damping ratio plays practically no part in the overall response of the system. The usual practice is that when the frequency ratio is more than 3.5 the damping effect of the soil is neglected. 8

Magnification factor

7

Mag. Factor for 5% damping Mag. Factor for 10% damping Mag. Factor for 15% damping Mag. Factor for 25% damping Mag. Factor for 37.5% damping

6 5 4 3 2 1

0. 55 0. 95 1. 35 1. 75 2. 15 2. 55 2. 95 3. 35 3. 75 4. 15 4.5 5 4. 95 5.3 5 5. 75

0.

15

0 Frequency Ratio

Figure 2.3.1 Magniﬁcation factor for various damping ratio.

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Analysis and design of machine foundations 115

One can immediately draw conclusion from the above fact that when a block foundation is resting on ground having soft to medium soil supporting machines having high operating speed it would possibly be quite justiﬁed to neglect the damping effect of soil. However, for low tuned machine it has been observed that it is difﬁcult to achieve this frequency separation for block foundation and for such cases damping cannot be ignored. We now further explain the above with a suitable numerical example.

Example 2.3.1 For the block foundation as described in Example 2.2.1. Calculate the following amplitude of vibration based on undamped equation of motion based on • • •

Static Coupling Dynamic coupling Discuss the results based on the answers.

The unbalanced dynamic force is 200 kN having operating frequency of 750 r.p.m acting at a height 600 mm from the top of foundation. The total height of the concrete block (L) is 1.8 m All other design parameters pertaining to design remains same as Example 2.2.1 Solution: The operating frequency of the machine is 750 r.p.m. = 78.53 rad/sec

750 ×2×π = 60

Solution based on static coupling ¨ ˙ The equation of motion in matrix formis given by [M]{X} + [C]{X} + [K]{X} = P0 200 {P}; the force matrix is = sin 78.53t M0 180 For Static coupling The normalized eigen-vector obtained earlier was 0.1188173 0.076746 [φ] = 0.0542333 −0.0839942 Considering the operation [φ]T {P}, we have

0.1188173 0.0542333 200 sin 78.53t 0.076746 −0.0839942 180

33.52 = sin 78.53t 0.2293

[φ]T {P} =

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116 Dynamics of Structure and Foundation: 2. Applications

We had seen earlier that the two natural frequencies are ω1 = 4.35 rad/sec; ω2 = 12.58 rad/sec 78.53 = 6.24 > 3.5 Thus the frequency separation is r = 12.58 Thus we can say that as the frequency ratio is greater than 3.5 damping effect of the soil can be neglected. Thus, with orthogonal transformation and neglecting the damping, we have [φ]T [M][φ]{ξ¨ } + [φ]T [K][φ]{ξ } = [φ]T {P} This in decoupled form reduces to {ξ¨1 } + [ω12 ]{ξ1 } = p0 sin ωm t;

{ξ¨2 } + [ω22 ]{ξ2 } = m0 sin ωm t

Substituting the different values we have or, {ξ¨1 } + 18.92{ξ1 } = 33.52 sin 78.53t

and

{ξ¨2 } + 158.5{ξ2 } = 0.230 sin 78.53t ∴ ξ1 = ξ2 =

33.53 sin 78.53t (18.92 − 78.532 )2 0.230 sin 78.53t

(158.5 − 78.532 )2

= 5.4537698 × 10−3 sin 78.53t = 3.8279354 × 10−5 sin 78.53t

Thus back transferring on global co-ordinate we have

x 0.1188173 0.076746 5.4537698 = × 10−3 sin 78.53t φ 0.0542333 −0.08399942 0.038273954

0.6509395 = × 10−3 sin 78.53t 0.2925838

Solution based on dynamic coupling The normalized eigen-vector obtained earlier was 0.0374669 0.2027373 [φ] = 0.0542333 −0.0839942 Considering the operation [φ]T {P} we have 0.0374669 0.0542333 200 [φ]T {P} = sin 75.53t 0.2027373 −0.0839942 180

17.255374 = sin 78.53t 25.428504

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Analysis and design of machine foundations 117

As we have seen earlier that the frequency separation is greater than 3.5, hence on orthogonal transformation without damping we have [φ]T [M][φ]{ξ¨ } + [φ]T [K][φ]{ξ } = [φ]T {P} This in decoupled form reduces to {ξ¨1 } + [ω12 ]{ξ1 } = p0 sin ωm t;

{ξ¨2 } + [ω22 ]{ξ2 } = m0 sin ωm t

Substituting the different values we have {ξ¨1 } + 18.92{ξ1 } = 17.25537 sin 78.53t

and

{ξ¨2 } + 158.5{ξ2 } = 25.429054 sin 78.53t 17.25537 sin 78.53t ∴ ξ1 = = 2.8066453 × 10−3 sin 78.53t (18.92 − 78.532 )2 25.428504 sin 78.53 t ξ2 = = 4.2321161 × 10−3 sin 78.53t (158.5 − 78.532 )2 Thus back transferring on global co-ordinate we have

x 0.0374669 0.2027373 2.8066453 = × 10−3 sin 78.53t φ 0.0542333 −0.08399942 4.2321161

0.963164 = × 10−3 sin 78.53t −0.2007444 On comparing the amplitudes for the damped and undamped case we have the following results. Damped amplitude

Undamped amplitude

Difference in (%)

Sl. No.

Formulation basis

X (10−3 )

φ (10−3 )

X (10−3 )

φ (10−3 )

X

φ

1 2

Static coupling Dynamic coupling

0.650525 0.9546

0.2924 −0.1997

0.6509 0.9631

0.29258 −0.20074

0.06 0.88

0.06 0.518

2.4 EFFECT OF EMBEDMENT ON FOUNDATION The theories described above dominated the scenario of design of machine foundation for quite sometime. But as the machines designed were progressively becoming heavier and having higher and higher operating frequencies the foundations in turn © 2009 Taylor & Francis Group, London, UK

118 Dynamics of Structure and Foundation: 2. Applications

were becoming more and more massive in nature10 , and it was realized that when a foundation is constructed below ground level the surrounding soil in which it is embedded plays a signiﬁcant role on the overall response of the foundation and needs to be carefully evaluated too. Number of theoretical formulations have been derived and field experiments (Gupta 1972, Erden & Stokoe 1975) have been conducted to study the embedment effect of soil on the overall response of the foundations, though there exists disagreements between the theories put forward however the general consensus about the embedment of soil on the foundation both from theoretical and field observations are as follows: • •

The embedment effect increases the natural frequency of the foundation It reduces overall amplitude of the foundation. It is not difficult conceive from the above statements that:

• •

Embedment effect increases the soil stiffness and Also has an incremental effect on the damping of the soil.

The most popular theory which is in practice in design office is given Tables 2.4.1 and 2 (Whitman 1972). Here, h = depth of embedment of the foundation in the surrounding soil; υ = Poisson’s ratio of the soil. Table 2.4.1 Embedment coefﬁcients for spring constants. Sl. No.

Direction

Coefﬁcient

1

Vertical

2

Horizontal

3

Rocking

ηφx = 1 + 1.2(1 − υ)

3.1

Rocking

ηφy = 1 + 1.2(1 − υ)

4

Twisting

h rz h ηx = 1 + 0.55(2 − υ) rx ηz = 1 + 0.6(1 − υ)

h rφx 3 h + 0.2(2 − υ) rφx h

rφy 3 h + 0.2(2 − υ) rφy None available

Equivalent radius LB rz = π LB rx = π 3 4 LB rφx = 3π

rϕy =

4

L3 B 3π

Remarks This is in vertical Z direction This induce sliding in horizontal x or y direction This produces rocking about Y axis

This produces rocking about X axis

This produces twisting about vertical Z axis

10 In most of the cases as the plan area of the foundation is dependent on the equipment general arrangement to increase the mass it was getting deeper and deeper.

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Analysis and design of machine foundations 119

Table 2.4.2 Embedment coefﬁcients for soil damping ratio. Sl. No.

Direction

Coefﬁcient

Equivalent radius

1

Vertical

αz =

1 + 1.9(1 − υ) rhz √ ηz

rz =

2

Horizontal

αz =

1 + 1.9(2 − υ) rhx √ ηx

rx =

3

Rocking

αφx

h 1 + 0.7(1 − υ) rφx + 0.6(2 − υ) = √ ηφx

1 + 0.7(1 − υ) 3.1

Rocking

αφy =

4

Twisting

None available

h rφx

3

3 h h + 0.6(2 − υ) rφy rφy √ ηφy

LB π

ηz is value as obtained as coefﬁcient for soil spring constant

LB π

ηx is value as obtained as coefﬁcient for soil spring constant

rϕx =

Remarks

4

LB3 3π

ηφx is value as obtained as coefﬁcient for soil spring constant

rφy =

4

L3 B 3π

ηφy is value as obtained as coefﬁcient for soil spring constant

It is suggested that if we multiply the spring constants available from Richart and Lysmer formulation vide Tables 2.2.3 and 4 by the above factors we get the modified spring constants valid for the embedded foundations. Damping ratio as obtained from Richart and Lysmer’s model when multiplied by the coefﬁcients as furnished in Table 2.4.2 gives the damping ratio considering the embedment effect of the soil.

2.4.1 Novak and Beredugo’s model Novak & Beredugo (1972a) model for embedded foundation has already been worked out in detail in the chapter 5 (Vol. 1) under soil dynamics and you may refer to the same for further details. Both vertical and lateral mode coupled with rocking (Novak & Beredugo 1972b) has been treated therein.

2.4.2 Wolf’s model Wolf (1985) has devised springs for dynamic foundation where he has considered additional soil mass vibrating with the foundation effect as shown in Table 2.4.3.

2.5 FOUNDATION SUPPORTED ON PILES In many cases due to poor soil condition machine foundations are loaded on piles and obviously other than static loads they are also subjected to vibrations and dynamic loading. Dynamic behaviour of piles is still to certain extent not very clearly understood though theoretical formulas exist to predict their behaviour under time dependent © 2009 Taylor & Francis Group, London, UK

120 Dynamics of Structure and Foundation: 2. Applications

Table 2.4.3 Soil spring constants as per Wolf (1985). Mode Vertical Horizontal Rocking Torsion

Spring stiffness 4Gr 0 1−υ 8Gr 0 1−υ 8Gr 3θ 3(1 − υ) 16Gr 3ψ 3

γ0

μ0

0.58

0.095

0.85

0.27 0.3

1+

3(1−υ)m 8r 5θ ρ

0.433 1+

2m rψ5 ρ

m rψ5 ρ

0.24 0.045

2 In which, C = Vsr kγ0 and m = Vsr kμ0 , where r = equivalent radius and shall be r0 , rθ , rψ as the case may be; G = Dynamic shear modulus of the soil; ρ = mass density of the soil; vs = shear wave velocity of the soil; M = mass of the soil participating in the vibration with the machine and the block foundation, and C = damping of the soil.

loading; they have been co-related with field observations for only a few simplified cases. As such the decision of using piles below machine foundations should be taken cautiously and not without some understanding of how it would behave under the load induced from the machine. Though there are very few reports on the field observation data on dynamic behaviour of piles under machine foundations it is however generally accepted that under time dependent loads piles, • • •

Have significant influence on the amplitude, especially near resonance, Increases the natural frequency of the system, Decrease the geometric damping of the soil foundation system.

Since in some cases particularly in lateral mode, the effect of piles could be adverse we repeat that it should be used with caution. For machine foundation on piles three mathematical models are usually in vogue. 1 2 3

Piles considered as frequency independent equivalent springs based on Novak’s (1974) formulation. Piles considered as beam elements connected to soil springs based on Richart’s formula as mentioned earlier. Considering the underlying soil and the pile as finite elements and executing a detailed analysis based on appropriate boundary conditions.

We will discuss all the above methods now in some detail but would like to emphasise at this point is that each one of them has its pros and cons and are not self-sufficient. As such which would be the most appropriate model for analysis varies from case to case and one method of analysis may have to be complimented by another model. While describing the model we start with the most exhaustive one and go the reverse © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 121

Figure 2.5.1 Machine foundation supported on piles.

way for we feel this will give you a better insight to the various problems that exists with dynamic behaviour of piles. Shown in Figure 2.5.1 is a machine foundation supported by piles.

2.5.1 Pile and soil modelled as f inite element This would obviously the most exhaustive model one could perceive and is shown in Figure 2.5.2. A representative and conceptual 2D model of the pile and soil are shown in Figure 2.5.2. Actually, the most appropriate model would be in 3D, where the piles are modelled as beam elements while the soil can be modelled as eight nodded brick element and a comprehensive dynamic analysis of the whole system could be performed. It can however be perceived that the analysis would time consuming and expensive (both in terms of man-hours and input data generation) and is usually not warranted except for large multi-shaft gas/steam turbines of high output resting on poor soil where such analysis could become essential in order to ensure the performance of such expensive machines under operating conditions. © 2009 Taylor & Francis Group, London, UK

122 Dynamics of Structure and Foundation: 2. Applications

Rigid body lumped as mass

Pile Cap modeled as beam element Master Node Slave Node

Figure 2.5.2 Finite element model of machine foundation with pile and soil.

One of the major disadvantages with this type of model is that the boundary of the soil has to be extended to substantial distance away both at the sides and from the pile tip in vertical direction enabling the model to predict correctly the response of the system. If this boundary limit is inadequate from the pile tip then waves transmitted to the soil due to the vibration of the machine will get reflected back and result in spurious responses which could make the analysis completely wrong. The question as to how far this boundary should extend, no rational basis has been derived yet and is completely up to the engineer’s judgement11 .

11 One thumb rule is to extend the boundary in vertical direction to 2.5 times the length of the pile.

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Analysis and design of machine foundations 123

Other than this there are certain practical problems encountered especially when the piles are long (say 20/30 m), geotechnical data may not be available to the depth to which an engineer might like to extend the boundary of the problem and as such if comprehensive soil data to the desired level is not available it may be difficult to model the system without the adequate data. In spite of the above problems the model is not without its advantage and may be summarised as follows: • • • • •

It comprehensively caters to the 3D effect of the pile soil and the foundation It can effectively model the soil if layered in nature where each of the layers has different material property. The group interaction effect of soil and pile is automatically catered for. Piles having variable geometry (tapered piles) can also be modelled without any problems. If battered piles are provided to counter any lateral thrust can also be modelled without any difficulty.

2.5.2 Piles modelled as beams supported on elastic springs In this method the piles are modelled as beam elements connected to springs in horizontal and vertical direction calculated out of the soil material property as shown Figs. 2.5.3 and 4. Again for a pile cap supporting machines a 3D model will be developed and dynamic analysis will be carried out. Shown in Figure 2.5.5 is a 3D model of pile cap, pile with soil springs. In this case the soil springs may be calculated based on Richart’s formulation or by multiplying the influence area of each node by the coefficient of uniform compression. The piles are modelled as beam elements having 6 degrees of freedom at each node. The pile cap is mathematically modelled as either beam or plate bending element depending upon the overall aspect ratio of the cap and other design considerations.

Figure 2.5.3 Pile in soil.

© 2009 Taylor & Francis Group, London, UK

Figure 2.5.4 Mathematical model.

124 Dynamics of Structure and Foundation: 2. Applications

Figure 2.5.5 Mathematical model of pile cap, piles with soil springs.

It is obvious that with respect to the previous model one of the major advantages is that it is a relatively less laborious model in terms of input generation and complexity and many engineers prefer to use this in lieu of a detailed finite element 3D model as shown previously. However the above model suffers from one serious lacuna for which it should be used with caution. The model in Figure 2.5.5 does not take into cognisance the effect of the soil which lies between the two piles and treats the soil as only discrete element based on springs. • •

This could significantly under rate or even over rate the dynamic response which depends on the nature of the soil It does not take in to cognisance the pile group interaction factor which has been observed to have significant effect on the dynamic response on the system specially when the pile spacing is between 2.5D to 3D, where D is the overall diameter of the pile.

It is recommended that this model may be used when the centre to centre distance between the piles are at least more than 5D.

2.5.3 Novak’s (1974) model for equivalent spring stiffness for piles This is possibly the most popular method used in the design offices to evaluate springs for piles subjected to dynamic loads and will be discussed in some detail. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 125

Though not without limitations the major advantage with this method is that • • • • •

It is simple to use. The spring stiffness and damping values are frequency independent. The group interaction effect of the piles can be to certain extent taken care of. The spring and damping values thus obtained can be very easily implemented as linear springs in commercially available finite element software. Standard Chart and coefficients exists for piles that are quite easy to use.

Novak’s method has found to be in excellent agreement with ﬁeld observations specially when the pile group arrangement is not complicated.

2.5.4 Equivalent pile springs in vertical direction For single end bearing piles undergoing vertical motion, the spring constants are given by the expression kbz =

Ep Ap r0

f18,1

(2.5.1)

where, kb z = equivalent spring constant for end bearing piles; Ep = Young’s modulus of pile material; Ap = cross sectional area of the pile; r0 = equivalent radius of the pile, and, f18,1 = a factor which depends on pile material (concrete, steel, timber etc.), ratio of embedded length l to radius (r0 ) and Vs /Vc (shear wave velocity of the soil above the tip to compression wave velocity in pile). The damping value in vertical direction is given by cbz =

Ep A p vs

f18,2

(2.5.2)

piles; vs = shear wave velocity of where, cbz = damping value of the end bearing the pile through which the soil is driven ( Gs g/γs ); f18,2 = is a factor as given in Table 2.5.1. Table 2.5.1 Values of factor f -as per Novak (1974) for stiffness and damping factor for single pile. For concrete piles (γs /γp = 0.7) having /r0 > 25. Slenderness ratio

Stiffness and damping function f for vertical bearing pile

20

f18,1 = 3.75(Vs /Vc )2 − 0.05(Vs /Vc ) + 0.0501 f18,2 = 15.345(Vs /Vc )2.0928

50

f18,1 = 6.25(Vs /Vc )2 + 0.05(Vs /Vc ) + 0.0199 f18,2 = −10(Vs /Vc )2 + 1.5(Vs /Vc ) − 0.012

100

f18,1 = −3.75(Vs /Vc )2 + 0.45(Vs /Vc ) + 0.0061 f18,2 = 1.4(Vs /Vc ) − 0.0083

© 2009 Taylor & Francis Group, London, UK

126 Dynamics of Structure and Foundation: 2. Applications

The values of f18,1 and f18,2 are meant for end bearing piles. However it has been observed that for friction piles having l/r0 greater than 60 or Vs /Vc greater than 0.03 these values are in small error pertaining to timber and concrete piles. For steel piles Novak has given a value of f18,1 = 0.030 and f18,2 = 0.045 where Vs /Vc = 0.033 and l/r0 greater than 80. For relatively short friction piles the following expression has been suggested by Novak for calculation of the stiffness and damping kz1 =

Ep A p r0

f18,1

and cz1 =

Ep Ap vs

f18,2

(2.5.3)

Table 2.5.2 Vertical stiffness coefﬁcients for ﬂoating piles as per Novak (1983). f 18,1 L/R

Ep /G = 10,000

Ep /G = 2500

Ep /G = 1000

Ep /G = 500

Ep /G = 250

10.8696 21.7391 32.6087 43.4783 46.7391 54.3478 65.2174 76.0870 86.9565 100.0000

0.0021 0.0031 0.0042 0.0042 0.0052 0.0052 0.0062 0.0062 0.0073 0.0083

0.0052 0.0083 0.0104 0.0125 0.0135 0.0145 0.0166 0.0177 0.0187 0.0197

0.0104 0.0166 0.0218 0.0260 0.0270 0.0281 0.0291 0.0301 0.0301 0.0301

0.0187 0.0301 0.0364 0.0405 0.0416 0.0416 0.0416 0.0416 0.0416 0.0416

0.0332 0.0509 0.0571 0.0582 0.0582 0.0582 0.0582 0.0582 0.0582 0.0582

Table 2.5.3 Vertical damping coefﬁcients for ﬂoating piles as per Novak (1983). f18,2

L/R

Ep /G = 10,000

Ep /G = 2500

Ep /G = 1000

Ep /G = 500

Ep /G = 250

10.8696 16.3043 21.7391 27.1739 32.6087 38.0435 43.4783 48.9130 54.3478 59.7826 65.2174 70.6522 76.0870 81.5217 86.9565 92.3913 100.0000

0.0032 0.0053 0.0074 0.0084 0.0105 0.0116 0.0137 0.0147 0.0147 0.0168 0.0168 0.0179 0.0189 0.0189 0.0200 0.0211 0.0211

0.0126 0.0179 0.0232 0.0263 0.0305 0.0326 0.0347 0.0368 0.0379 0.0379 0.0379 0.0379 0.0379 0.0368 0.0358 0.0358 0.0337

0.0295 0.0421 0.0495 0.0537 0.0568 0.0589 0.0579 0.0568 0.0558 0.0537 0.0526 0.0516 0.0516 0.0505 0.0505 0.0505 0.0495

0.0558 0.0695 0.0811 0.0832 0.0811 0.0789 0.0758 0.0737 0.0726 0.0716 0.0705 0.0695 0.0695 0.0695 0.0695 0.0705 0.0705

0.1032 0.1137 0.1126 0.1095 0.1053 0.1021 0.0989 0.0979 0.0979 0.0979 0.0979 0.0979 0.0979 0.0979 0.0979 0.0989 0.0989

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 127 where, f18,1 and f18,2 are stiffness and damping factors respectively as given Tables 2.5.2 and 2.5.3 respectively. It has been suggested that the values given in these charts are most appropriate 0 when used for the range of a0 = 0.1 to 0.8, where a0 = 2πvfr here f = the operating s frequency of the machine. It has also been suggested (Steven 1978) that these values are even valid for a0 as low as 0.05 which means that reasonably good results may be expected for even slender piles and low frequencies. The damping ratio for the pile may be calculated from the expression

Dz1 = cz1 /2 kz1 mc

(2.5.4)

where mc is the mass of the cap plus machinery or the portion of structure which is vibrating in the same phase as the cap. Part of the mass of the pile may also be included in the above equation but it has been generally found that this ratio of the pile mass to the mass of the supported weight is very small and is usually ignored.

2.5.5 The group effect on the vertical spring and damping value of the piles Piles in practise usually do not occur as a single pile and usually under a foundation block there will be a number of piles to form a foundation. For instance under a normal block foundation supporting some machinery the pile system could be something like as shown in Figure 2.5.6. We had shown in Section 2.5.4 as to how to calculate the spring stiffness for individual single piles. When we try to find out the equivalent spring stiffness for the pile group as shown below this does not constitute of the sum of the individual pile

Figure 2.5.6 Plan view of a block foundation supported over piles. © 2009 Taylor & Francis Group, London, UK

128 Dynamics of Structure and Foundation: 2. Applications

stiffness. It also depends upon the relative distance between the piles itself and the slenderness ratio of the piles carrying the loads and is expressed as g kz

=

N

kz1

1

N

αA

(2.5.5)

1

The equivalent damping for the pile group is given by g cz

=

N

cz1

1

N

αA

(2.5.6)

1

where, N = number of piles in a group; αA = displacement interaction factor (axial) for a typical reference pile in the group relative to itself and to all other piles in the group assuming the reference pile and all other pile are loaded to same magnitude, and the factor αA can be evaluated from the expression (Randolph and Poulos 1982); and also recommended by API 351R. αA =

0.5 ln(lp /s) ln(lp dρA )

for s ≤ lp

(2.5.6a)

Here lp = Pile length, s = spacing of piles, d = diameter of pile ρA = Gav /Gb ; Gav = Average Shear modulus along pile depth and Gb = Shear modulus at pile base. Alternatively the value can also be deduced from Poulos’s interaction curve for static interaction (Poulos and Davis 1980).

2.5.6 Effect of pile cap on the spring and damping stiffness If the pile cap is not in contact with the ground the above equations can be directly used in for the analysis. Pile caps embedded usually have a favourable effect on the response of the group and should be adapted wherever possible. It would be realistic to assume that the embedment effect generates only the side friction between the cap and the soil and that to only when dense granular backfill is used. For the soil below the pile cap which is likely to be of inferior quality can settle away from the cap for non-cohesive soil, similarly for cohesive soil this can shrink away from the sides of the pile cap and can become ineffective. Table 2.5.4 Values of S¯ 1 and S¯ 2 for various Poisson’s ratio. Poisson’s ratio

S¯ 1

S¯ 2

0.0 0.25 0.40

2.7 2.7 2.7

6.7 6.7 6.7

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 129

The expressions for calculating the stiffness and the damping constant for an embedded cap of embedd depth h is given by f kz = Gs hS¯ 1

f

and cz = hr0

Gs λs /g S¯ 2

(2.5.7)

Equation (2.5.7) should be added to the pile stiffness and damping of the pile group as presented in the previous section to arrive at the complete spring and damping constant of a pile group in vertical direction. Values of S¯ 1 and S¯ 2 for various Poisson’s ratio are given in Table 2.5.4.

2.5.7 Equivalent pile springs and damping in the horizontal direction For vibration in horizontal direction the expression for stiffness and damping is as shown below: kx1 =

Ep I p

r30

f11,1

for /r0 ≥ 25,

and cx1 =

Ep I p r20 vs

f11,2

for /r0 ≥ 25. (2.5.8)

Here Ip is the moment of inertia of the pile cross section about the centroidal axis perpendicular to the direction of the motion. Here x direction depicts the horizontal motion and f11,1 and f11,2 are factors for fixed headed piles. The values of f11,1 and f11,2 are furnished in Table 2.5.5. The group effect is expressed as g kx

N =

1 kx1 N 1 αL

and

g cx

N

= 1N 1

cx1 αL

(2.5.9)

Table 2.5.5 Values of factor f -as per Novak (1974) stiffness and damping factors for horizontal and rocking mode. Poisson’s ratio

Function f

0.25

f11,1 = 7.25(Vs /Vc )2 + 0.38(Vs /Vc ) − 0.0013 f11,2 = 17(Vs /Vc )2 + 0.915(Vs /Vc ) − 0.0032 f7,1 = −55(Vs /Vc )2 + 9.3(Vs /Vc ) + 0.1075 f7,2 = −38.75(Vs /Vc )2 + 6.55(Vs /Vc ) + 0.0734 f9,1 = −1.81(Vs /Vc ) f9,2 = 0.375(Vs /Vc )2 − 2.67(Vs /Vc ) + 0.0005

0.4

f11,1 = 7.875(Vs /Vc )2 + 0.43(Vs /Vc ) − 0.0015 f11,2 = 18.75(Vs /Vc )2 + 1.02(Vs /Vc ) − 0.0037 f7,1 = −57.5(Vs /Vc )2 + 9.65(Vs /Vc ) + 0.1113 f7,2 = −41.25(Vs /Vc )2 + 6.85(Vs /Vc ) + 0.0746 f9,1 = −1.94(Vs /Vc ) f9,2 = 0.75(Vs /Vc )2 − 2.87(Vs /Vc ) + 0.0006

© 2009 Taylor & Francis Group, London, UK

130 Dynamics of Structure and Foundation: 2. Applications Table 2.5.6 Values of Su1 and Su2 for various Poisson’s ratio. Poisson’s ratio

S¯ u1

S¯ u2

0.0 0.25 0.40

3.6 4.0 4.1

8.2 9.1 10.6

where, αL = a displacement factor for lateral motion defined in similar way to αA and is given by12 1

αLH

r 0

(1 + cos2 βp ) s r 1 0 = 0.4ρc [Ep Gc ] 7 (1 + cos2 βp ); s

αLf = 0.6ρc [Ep Gc ] 7

(2.5.9a) 2 3 αθH = αLH , αθM = αLH

(2.5.9b)

Here ρc = Gz /Gav where Gz = Shear Modulus at depth lc /4 2 lc = 2r0 [Ep /Gc ] 7 and is known as the critical length of the pile where Gc = Average shear modulus over the critical length of the pile. αLf = The horizontal interaction factor for fixed headed piles (no head rotation). αLH = The horizontal interaction factor due to horizontal force (rotation allowed). αθ H = Interaction factor due to horizontal force for rotation. αθ M = Interaction factor due to moment for rotation. βp = Angle subtended by a pile in pile group with respect to the reference pile. When the calculated interaction factor α exceeds 1/3, its value needs to be replaced √ by α = 1 − 2/ 27α, a correction made to avoid α approaching infinity as s tends to zero. Alternatively Poulos’s interaction curve for static load case under horizontal load may also be used. The stiffness and damping characteristics of the pile cap is expressed as kx = Gs hS¯ u1 f

f and cx = hr0 S¯ u2 Gs γs /g

(2.5.10)

The factors S¯ u1 and S¯ u2 are as given in Table 2.5.6.

2.5.8 Equivalent pile springs and damping in rocking motion The expression for spring stiffness and damping for simple pile has been expressed as kψ1 =

Ep I p r0

f7,1 ;

cψ1 =

Ep Ip vs

f7,2

12 Example 2.7.3 is a very good conceptual case study for the same.

© 2009 Taylor & Francis Group, London, UK

(2.5.11)

Analysis and design of machine foundations 131

Here I is the moment of inertia of the pile cross section about the axis of rotation and f7,1 and f7,2 are factors for rotational direction for fixed head piles, as furnished in Table 2.5.5.

2.5.9

Group effect for rotational motion

For a pile group the group stiffness, shown in Figure 2.5.7, is expressed as

g

kψ =

N

[kψ1 + kz1 Xr2 + kx1 Zc2 − 2Zc kxψ1 ] + kψf

(2.5.12)

1

Here Xr and Zc are shown in the Figure 2.5.7 and kz1 and kx1 are stiffness constant of single piles as described earlier. In addition kxψ1 =

Ep I p

f9,1

r20

and

kψf = Gs r0 hS¯ ψ1 + Gs r20 h

δ2 3

+

Zc r0

2

−δ

Zc r0

(2.5.13) S¯ u1

where δ = h/r0 , here h = embedment depth of pile cap, and Sψ1 is as given in Table 2.5.7. The damping matrix for the pile group is expressed by

g

cψ =

N

[cψ1 + cz1 Xr2 + cx1 Zc2 − 2Zc cxψ1 ] + cψf

(2.5.14)

1

where cz1 and cx1 are damping constant of single piles as described earlier. In addition cxψ1 =

Ep Ip r0 v s

f9,2

and

Table 2.5.7 Values of Sψ1 and Sψ2 for various Poisson’s ratio. Poisson’s ratio

S¯ ψ1

S¯ ψ2

0.0 0.25 0.40

2.5 2.5 2.5

1.8 1.8 1.8

© 2009 Taylor & Francis Group, London, UK

132 Dynamics of Structure and Foundation: 2. Applications

Machine block Soil line

h

Zc

Xr

Figure 2.5.7 Geometric description of the pile group.

f

cψ =

δr40

Gs γs /g Sψ2 +

δ2 3

+

Zc r0

2

−δ

Zc r0

S¯ u2

(2.5.15)

We now explain the above theory further by a suitable numerical problem.

Example 2.5.1 Find the vertical, horizontal and rocking stiffness of the pile group based on Novak’s formulation as shown in Fig. 2.5.8 and with the following soil properties: Length of the pile = 45.0 m; Diameter of the pile = 950 mm; Grade of concrete M20 having a dynamic modulus as 300 × 106 kN/m2 . Consider Poisson’s ratio of soil = 0.4. Solution: Since each of the layers has different velocity and thickness we take a weighted average of the shear wave velocity of the three soil layers as follows vs =

60 × 10 + 110 × 20 + 215 × 15 = 134 m/sec 45

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 133

600

1400

10 m

vs = 60 m/sec = 18 kN/m3

Layer#1

20 m

vs = 110 m/sec = 20 kN/m3

Layer#2

vs = 215 m/sec = 22 kN/m3

Layer#3

30 m

2000

2000

Geotechnical profile of soil in which the piles are located

3000

500 (Typ)

3000

3

6

9

2

5

8

1

4

7

2000

2000

Arrangement of the pile group in Plan

Figure 2.5.8 Machine foundation supported on piles.

Average weight density of soil is = γ s =

18 × 10 + 20 × 20 + 22 × 15 = 45

20.22 kN/m3 Thus dynamic shear modulus of the soil is taken as 20.22 Gavg = × (134)2 = 37010 kN/m2 ; g, acceleration due to gravity is 9.81 taken as 9.81 m/sec2 . L×B 7×5 Equivalent Radius of pile cap = = = 3.33 m π π

© 2009 Taylor & Francis Group, London, UK

134 Dynamics of Structure and Foundation: 2. Applications

Vertical stiffness of piles Cross-sectional area of each pile =

π × (0.95)2 = 0.7088 m2 4

Dynamic modulus of concrete = 300 × 106 kN/m2 Density of concrete (γc ) = 25/kN/m3 vc =

Ep × g γp

=

300 × 106 × 9.81 = 10849 m/sec; 25

vs 134 = = 1.235 × 10−2 ≈ 0.01 vc 10849 45 L = = 94.7 > 25. r0 0.475 For L/r0 = 95 and vs /vc = 0.01 we have from Novak’s Chart → f18,1 = 0.011

and f18,2 = 0.005.

Thus for each pile we have, → Kzi =

Ep A 300 × 106 × 0.7088 f18,1 = × 0.011 r0 0.475

= 4.924 × 106 kN/m and

Czi =

Ep A 300 × 106 × 0.7088 f18,2 → × 0.005 vs 134

= 7934 kN-sec/m. Calculation of group interaction factor for, λ(L/r0 ) = 95 Thus, for group effect 9

Kzg =

9 × 4.924 × 106 = 11480829 kN/m; 3.86

Czg =

9 × 7934 = 18498 kN-sec/m. 3.86

i=1 9 i=1

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 135

Calculation of group interaction factor (vertical) Pile No.

Spacing (S)

1

0.0

2 3 4 5 6 7 8 9

2.0 4.0 3.0 3.6 5.0 6.0 6.32 7.21

L/S 0.0 22.5 11.25 15.0 12.5 9.0 7.5 7.12 6.24 Total sum

α

Remarks

1.0

Pile # 1 is the reference pile

0.474 0.368 0.413 0.384 0.334 0.307 0.299 0.278 3.86

α obtained vide Eqn (7.5.6a)

Effect of the pile cap on overall pile stiffness Here the effect of layer#1 will be more dominant on the pile cap as such for finding out the stiffness properties in context of the pile cap, we have Gs =

18 × (60)2 = 6605 kN/m2 9.81

For embedded depth, h = 1.4 m and S1 = 2.7, we have, f ¯ 1 → Kzf = 6605 × 1.4 × 2.7 = 24969 kN/m. Kz = Gs hS

f Cz

6605 × 18 ¯ = hr0 Gs γs /gS2 = 1.4 × 3.33 × 6.7 = 3438.6 kN/m. 9.81

Thus, total vertical stiffness = 11480829 + 24969 = 11505798 kN/m Total damping for the pile and pile cap = 18498 + 3439 = 21937 kN-sec/m. Calculation of horizontal stiffness π × (0.95)4 = 0.04 m4 ; with l/r0 = 95, νs = 0.4 and vs /vc = 0.01, from 64 Novak’s chart f11,1 = 0.0036 and f11,2 = 0.0084

I=

Kx = Cx =

Ep I r30 Ep I r20 vs

× f11,1 =

300 × 106 × 0.04 × 0.0036 = 403091 kN/m; (0.475)3

× f11,2 =

300 × 106 × 0.04 × 0.0084 = 3334 kN-sec/m. (0.475)2 × 134

© 2009 Taylor & Francis Group, London, UK

136 Dynamics of Structure and Foundation: 2. Applications

Calculation of interaction factor (lateral) Pile number

Spacing

S/r0

β factor in degree

αl

Remarks

1 2 3 4 5 6 7 8 9

0.0 2.0 4.0 3.0 3.6 5.0 6.0 6.32 7.21

0.0 4.21 8.42 6.316 7.578 10.526 12.632 13.306 15.178

0 90 90 0 33.6 53.13 0 18.43 33.6 Total sum

0 0.246 0.123 0.328 0.231 0.134 0.164 0.148 0.115 1.489

Pile#1 is the reference pile αl calculated as per eqn (7.5.9a)

∼ =1.50

For the pile group we have, 9

Kx =

403091 × 9 = 2418546 kN/m; 1.5

Cx =

3334 × 9 = 20004 kN-sec/m. 1.50

i=1 9 i=1

Effect of the pile cap on overall pile stiffness Here the effect of layer#1 will be more dominant on the pile cap as such for finding out the stiffness properties in context of the pile cap we have

Gs =

18 × (60)2 = 6605 kN/m2 9.81

For embedded depth, h = 1.4 m and Su1 = 4.1 and Su2 = 10.6, ¯ u1 = 6605 × 1.4 × 4.1 = 37913 kN/m; Kx = Gs hS f

f Cx

6605 × 18 ¯ = hr0 Gs γs /gSu2 = 1.4 × 3.33 × 10.6 = 5440 kN/m. 9.81

Thus, the total lateral stiffness = 2418546 + 37913 = 2.456 × 106 kN/m. Total damping for the pile and pile cap in lateral direction = 20004 + 5440 = 25444 kN-sec/m.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 137

Calculation of rocking stiffness and damping The rocking stiffness of individual pile, with f7,1 = 0.202, is given by kψ1 =

Ep Ip r0

f7,1 =

300 × 106 × 0.04 0.475

× 0.202 = 5103158 kN/m

The coupled sliding and rocking stiffness with, f9,1 = −0.0194, is given by kxψ1 =

Ep I p

r20

f9,1 =

300 × 106 × 0.04 (0.475)2

× (−0.0194)

= −1031801 kN/m The pile cap stiffness is given by kψf = Gs r0 hS¯ ψ1 + Gs r20 h

δ2 3

+

Zc r0

2

Zc −δ r0

S¯ u1

Substituting the values, Gs = 6605 kN/m2 ; h = 1.4 m; r0 = 3.33 m; S¯ ψ1 = 2.5; δ = rh0 = 0.42; Zc = 1.5, Xr = 2.0 m and S¯ u1 = 4.1, we have, kψf = 84417 kN/m. Thus the total stiffness of the pile group is given by g

kψ =

N 1

[kψ1 + kz1 Xr2 + kx1 Zc2 − 2Zc kxψ1 ] + kψf

g

kψ = 9 × [5103158 + 4.924 × 106 × 4 + 403091 × 2.25 + 2 × 1.5 × 103801] + 84417 g

kψ = 2.342 × 108 kN/m Calculation of damping value in rocking mode The damping value with, f7,2 = 0.139, is given by cψ1 =

Ep Ip vs

f7,2 =

© 2009 Taylor & Francis Group, London, UK

300 × 106 × 0.04 × 0.139 = 12448 kN-sec/m. 134

138 Dynamics of Structure and Foundation: 2. Applications

The coupled sliding and rocking mode damping, with f9,2 = −0.0280, is given by cxψ1 =

Ep Ip r0 v s

f9,2 =

300 × 106 × 0.04 × (−0.0280) = −5273 kN-sec/m 0.475 × 134

The damping contributed by pile cap is given by f cψ

2 2 δ Z Z c c S¯ u2 = δr40 Gs γs /g Sψ2 + −δ + 3 r0 r0

where, Gs = 6605 kN/m2 ; h = 1.4 m; r0 = 3.33 m; S¯ ψ2 = 1.8; δ = h/r0 = 0.42; Zc = 1.5 m; S¯ u2 = 10.6; γs = 18 kN/m3 ; g = 9.81 m/sec2 , Substituting the above values we have, cψf = 14604 kN-sec/m. The damping value of the pile group is given by g

cψ =

N

[cψ1 + cz1 Xr2 + cx1 Zc2 − 2Zc cxψ1 ] + cψf

1 g

cψ = 9 × [12448 + 7934 × 4 + 3334 × 2.25 + 2 × 1.5 × 5273] + 14604 = 622145 kN-sec/m.

2.5.10 Model for dynamic response of pile In previous section we had presented the dynamic stiffness of piles as proposed by Novak. This is a very popular model for dynamic response of machine foundations in the design offices. Based on the analytical solution of Baranov (1967), Novak (1974) proposed the method for evaluating the vertical response of piles under dynamic loading. Many researchers like, Wolf and Von Arx (1978), Waas (1981), Kaynia and Kausel (1982), Banerjee and Sen (1987) have advanced solutions to this problem, yet Novak’s method remains the most popular due to its sheer simplicity in application. Solution of pile and pile-group based on the method proposed by Banerjee and Sen, which is based on Boundary Element Method gives quite accurate results but it is computationally too exhaustive to find applications in a day-to-day design office work. Applying Finite Element Method, where the pile is modeled as beam elements and the soil as Winkler springs, has yielded good results. But, they are found to be valid only when piles are single or when the distance between piles is signiﬁcant (≥5d, d being the diameter of the pile), when the pile-soil interaction can be neglected. Novak’s solution is mostly based on charts, and it furnishes stiffness and damping of a pile and the solution is addressed to the fundamental degree of freedom. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 139

In spite of its immense popularity, the model do have a few limitations as summarized below: a The solution does not take into account inertial effect of the pile; b Extrapolation is required when design data are out of range of the chart; c Charts are available only for RCC or timber piles, whether these charts are applicable to cases of steel piles13 , there is no clear-cut guideline; d The charts do not address to the case where a pile is partially embedded; e The formulation do not cater to dynamic axial load, moments or shears induced in pile due to dynamic loads. In certain cases when piles are supporting reciprocating compressors, it becomes essential to check the design for higher frequencies of the foundations to ensure that they are not matching with the second or third harmonics, when higher forces may be induced by the machine at harmonic other than the first. In such cases one has no other options but to resort to an elaborate and expensive three-dimensional Finite Element based soil-pile foundation model to arrive at an answer to this problem and in number of cases uncertainties present in such results are many. However, it was shown by Novak that results obtained for higher harmonics are not significantly different for the type of problem that is normally encountered in pile vibration studies. We present now a model (Chowdhury and Dasgupta 2006) that overcomes many of the limitations cited above. The solution is simple (yet realistic) and does not require elaborate software to be developed for the analysis. A simple spreadsheet would suffice for the problem considering the solution is basically analytic. 2.5.10.1

Proposed method

We had stated at the outset that most of the work relating to dynamic stiffness of pile is based on Baranov’s (1967) theory on the response of a soil embedded foundation. The present formulation is based on Novak and Beredugo’s (1972) approach on embedded foundation. 2.5.10.2

Vibration of friction piles

Let us consider a pile as shown in Figure 2.5.9. The pile is assumed to provide resistance both through bearing as well as friction. Let Kf represents the frictional stiffness of the pile and the pile tip bearing stiffness is taken as Kb . The longitudinal vibration of such beams having only the frictional stiffness may be represented by the expression

EA

∂ 2u ∂ 2u + K u = m(z) f ∂z2 ∂t 2

(2.5.16)

13 This is an important issue for many real life projects specially in Arctic condition (like North Siberia) or very arid region (like Sudan, Algeria) due to extreme low temperature or absence of water makes concreting hazardous and almost all the structures and foundations are built on steel piles.

© 2009 Taylor & Francis Group, London, UK

140 Dynamics of Structure and Foundation: 2. Applications

Z

Kf L

dz Kv

Kb

Figure 2.5.9 Pile embedded in ground up to a depth L and its mathematical model.

in which, E = Young’s modulus of pile; A = area of pile; Kf = dynamic frictional stiffness of soil having dimension (F/L) and u(z, t) = dynamic amplitude of pile = φ(z) q(t); and m(z) = mass of element dz. One of the solutions of equation (2.5.16) is given by q(t) = C3 sin ωt + C4 cos ωt

(2.5.17)

With the deﬁnition of u and using Equation (2.5.17), Equation (2.5.16) may be written as EA

d 2 φ(z) + Kf φ(z) = −m(z)ω2 φ(z) dz2

(2.5.18)

The above equation can further be simplified to d 2 φ(z) + p2 φ(z) = 0 dz2

(2.5.19)

where p2 = (mω2 + Kf ). If you observe Equation (2.5.19) carefully, you should realize that it suggests that the presence of frictional stiffness Kf does not affect the basic shape function of the pile and would remain same for the case had the pile would not have been embedded. However, the bearing stiffness Kb connected at the end of pile would affect the © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 141

shape function depending on the appropriate boundary condition. For computing the correct shape function of the system, one has to start with the model as shown in the Figure 2.5.9. The general solution for Equation (2.5.19) is given by Humar (1990) φ(z) = (C1 cos pz + C2 sin pz)(C3 sin ωt + C4 cos ωt)

(2.5.20)

in which, C1 , C2 , C3 and C4 are the integration constants to be determined from the appropriate boundary conditions. The pile has at the free head, z = 0, EA du = 0, which gives dz EAp[−C1 sin pz + C2 cos pz][C3 sin ωt + C4 cos ωt] = 0

➔ C2 = 0.

(2.5.21)

and at the tip, z = L, EA du = −Kb u(z)z=L , which gives dz EAp[−C1 sin pL] = −Kb C1 cos pL in which Kb = Gb r0 Cb

➔ pL tan pL = Kb L/(EA)

(2.5.22) (2.5.23)

where, Gb = dynamic shear modulus of the soil at pile tip; r0 = radius of the pile; Cb = a frequency independent dimensionless constant as suggested by Novak and Beredugo (1972) and is given in Table 2.5.8. Combining Eqns. (7.5.22) and (7.5.23), one can have pL tan pL =

Gb C b L Eπr0

(2.5.24)

It will be observed that the right hand side of Equation (2.5.24) is a dimensionless quantity. Gb Cb b Cb L If η = GEπ = r0 E π λ; where λ = slenderness ratio (L/r0 ) of the pile, Equation (2.5.24) can be represented as pL tan pL − η = 0.

(2.5.25)

Equation (2.5.25) is a transcendental equation in pL and can be solved numerically. The values of pL for various values of η for the first mode are shown in Table 2.5.9. Writing, pL = β, the arbitrary shape function of the problem is given by φ(z) = cos β

z L

(2.5.26)

The potential energy d of an element of depth dz, shown in Figure 2.5.9, is given by Shames and Dym (1995) d =

EA du 2 Kv 2 + u 2 dz 2

© 2009 Taylor & Francis Group, London, UK

(2.5.27)

142 Dynamics of Structure and Foundation: 2. Applications

where, E = Young’s modulus of pile; A = area of pile; Kv = dynamic stiffness of soil having dimension kN/m; w = displacement of pile in the z direction and may be written as φ(z)q(t). In Equation (2.5.27), Kv consists of two parts, namely, 1 2

the bearing stiffness at pile tip, and the friction stiffness along the shaft.

For a rigid circular embedded footing with embedment Df , the stiffness of the footing may be expressed as per Novak & Beredugo as Kv = Gb r0 Cb + GDf S1

(2.5.28)

where, Kv = foundation stiffness in the vertical direction; G = dynamic shear modulus of the soil along the embedment length; Gb = dynamic shear modulus of the soil at the base; r0 = radius of the foundation; Cb and S1 = dimensionless constant which are basically frequency dependent. However, it has been shown by Novak and Beredugo that considering Cb and S1 as frequency independent, no accuracy is lost for practical design problems and the analysis becomes quite simplified for rigid circular embedded footing. The frequency independent values of Cb and S1 are as given below in Table 2.5.8. However, it should be remembered that an embedded circular footing is usually considered to be rigid having infinite structural stiffness. On the contrary, a pile will be far more flexible member whose structural stiffness will be much lower, thus the above recommended value may be valid for certain pile geometry but may not be valid for others. Comparing the stiffness data of piles obtained by Novak, Dobry and Gazetas (1988) it is proposed that following value of S1 be used for dynamic analysis of piles in vertical direction. S1 =

9.553(1 + ν) λ0.333

(2.5.29)

where ν = Poisson’s ratio of the soil; and λ = slenderness ratio of the pile. This value of S1 is derived based on similar technique used earlier by Lysmer and Richart (1966) for deriving equivalent stiffness and damping of circular footings for Lysmer’s analog from the solutions of a similar elasto-dynamic analysis as proposed by Bycroft (1956). The value Cb may be taken as suggested in Table 2.5.8 for it has no bearing on the flexibility of pile and is a function of the base area only. Considering pile base area is much smaller in comparison to a footing, its contribution is only marginal. Table 2.5.8 Suggested frequency independent values suggested by Novak and Beredugo (1972) for embedded footing. Poisson’s ratio

Cb

S1

0.0 0.25 0.5

3.9 5.2 7.5

2.7 2.7 2.7

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 143

Table 2.5.9 Roots of equation pL tan(pL)-η = 0 for the ﬁrst or fundamental mode. η

pL

η

pL

η

pL

η

pL

0 0.1 0.2 0.3 0.4 0.5 0.6

0.02 0.322 0.433 0.522 0.593 0.653 0.705

0.7 0.8 0.9 1 1.25 1.5 1.75

0.75 0.791 0.828 0.86 0.931 0.988 1.036

2 2.5 3 3.5 4 5 10

1.077 1.142 1.192 1.232 1.265 1.314 1.429

15 20 25 30 35 40 50

1.473 1.496 1.51 1.52 1.527 1.533 1.54

Moreover in most of the practical cases its effect does not come into consideration (as will be shown subsequently) for analysis of such piles are either considered as bearing pile i.e. having infinite base stiffness or floating having no base effects. The first term in Equation (2.5.28) represents the contribution of base resistance, while the second term, the embedment effect of the foundation. Substituting Equation (2.5.28) in Equation (2.5.27) for an element dz, d may be written as EA du 2 Gb r0 Cb 2 GS1 dz 2 d = + u + u 2 dz 2 2

(2.5.30)

and the total potential energy over the total length of the pile (L) is given by EA = 2

L

du dz

2

GS1 dz + 2

0

L u2 dz +

G b r0 C b 2 u 2

(2.5.31)

0

Considering u(z, t) = φ(z)q(t), it can be proved (Hurty and Rubenstein 1967), that L Kij = EA

φi (z)φj (z)dz

0

L + GS1

φi (z)φj (z)dz + Gb r0 Cb φi (L)φj (L)

(2.5.32)

0

where the shape function of the problem is given by Equation (2.5.26). The first derivative of the above with respect to z is given by φ (z) = −

β z sin β L L

(2.5.33)

Using z/L = ξ implying dz = Ldξ , and converting the shape function as furnished in Equation (2.5.26) from local to generalized co-ordinates, the limits of the problem get converted to 1 to zero. © 2009 Taylor & Francis Group, London, UK

144 Dynamics of Structure and Foundation: 2. Applications

Now, if one assumes F(ξ ) = cos βξ

(2.5.34)

β Fi (ξ ) = − (sin βξ ), and (2.5.35) L +1 1 EAβi βj Kij = Fi (ξ )Fj (ξ )dξ + GS1 L Fi (ξ )Fj (ξ )dξ + Gb r0 Cb Fi (1)Fj (1). L 0

0

(2.5.36) For the fundamental mode i = j = 1 and Equation (2.5.36) reduces to EAβ 2 K1 = L

+1 1 2 F1 (ξ ) dξ + GS1 L F1 (ξ )2 dξ + Gb r0 Cb F1 (1)2 0

(2.5.37)

0

Equation (2.5.37) can be rewritten as EAβ 2 K1 = L

+1 1 2 (sin βξ ) dξ + GS1 L (cos βξ )2 dξ + Gb r0 Cb (cos β)2 0

(2.5.38)

0

Equation (2.5.38) on integration and after some simplification may be expressed as K1 = I1 + I2 + I3 in which,

1 sin 2β − ; 2 4β

I1 =

EAβ 2 L

I3 =

Gb r 0 C b (1 + cos 2β) 2

I2 = GS1 L

1 sin 2β + ; 2 4β (2.5.39)

Finally, K1 can be written as K1 =

EAβ 2 GS1 L Gb r0 Cb + + 2L 2 2

+

GS1 L EAβ − 4β 4L

sin 2β +

Gb r 0 Cb cos 2β 2 (2.5.40)

which may be further simplified to K1 = X1 + X2 sin 2β + X3 cos 2β

(2.5.41)

in which X1 =

EAβ 2 GS1 L Gb r0 Cb + + ; 2L 2 2

© 2009 Taylor & Francis Group, London, UK

X2 =

GS1 L EAβ − ; 4β 4L

X3 =

Gb r0 Cb 2 (2.5.42)

Analysis and design of machine foundations 145

Equation (2.5.42) gives the stiffness of the pile for the vertical mode, without any limitation to slenderness ratio, E/G or the material type. 2.5.10.2.1

Mass of the pile

For a conservative system, if T is the kinetic energy of the system then at any time t, the energy equations may be written as

1 T(t) = 2

H m(z) 0

Using, u(z, t) =

∂u(z, t) ∂t

n

2 dz

(2.5.43)

φi (z)qi (t)

(2.5.44)

i=1

where u(z, t) = displacement function; φi (z) = shape function; qi (t) = generalized co-ordinate; m(z) = mass of element dz and substituting Equation (2.5.44) in Equation (2.5.43), the energy equation may be written as

1 T (t) = 2

H

⎡ m(z)⎣

1 2

⎤⎡ φi (z)q˙ i (t)⎦ ⎣

j=1

0

=

n

n n

n

⎤ φj (z)q˙ j (t)⎦dz

j=1

⎡H ⎤ q˙ i (t)q˙ j (t)⎣ m(z)φi (z)φj (z)dz⎦

i=1 j=1

(2.5.45)

0

from which the mass matrix may be written as ⎡H ⎤ mij = ⎣ m(z)φi (z)φj (z)dz⎦

for i, j = 1, 2, 3 . . . n

(2.5.46)

0

Similarly the stiffness value with transformation from local to natural co-ordinate, the mass contribution of the pile may be obtained as γp AL mij = g

1 Fi (ξ )Fj (ξ )dξ

(2.5.47)

0

were γp = bulk density of pile material; A = area of pile cross section; L = pile length embedded in soil, and g = acceleration due to gravity. © 2009 Taylor & Francis Group, London, UK

146 Dynamics of Structure and Foundation: 2. Applications

For the fundamental mode, i, j = 1, and one can have γp AL m1 = g

1 F1 (ξ )2 dξ

(2.5.48)

0

The above on expansion results in γp AL m1 = g

1 (cos βξ )2 dξ

(2.5.49)

0

Integration of Equation (2.5.49) gives γp AL sin 2β m1 = 1+ 2g 2β

(2.5.50)

which is the contributory mass of the pile for the fundamental mode in the vertical direction. 2.5.10.2.2

Damping of the pile

The damping of the pile embedded in soil will constitute of two parts: • •

Material damping of the pile itself; Radiation damping of the soil-pile system.

It is obvious that the material damping of the pile will be much lower than that of the soil radiation damping. As the first step for calculating the soil damping one may ignore the material damping of the pile for the time being. Material damping of soil also is part of the system vibration. However, it has been found that for translational vibration their effect is insigniﬁcant and may be neglected without any signiﬁcant effect. Else, if one wishes, their values may be obtained from resonant column test from the laboratory when damping may be obtained from ratio of successive amplitudes. For a rigid footing embedded in soil for a depth Df , Novak and Beredugo has proposed an expression ¯ b + r0 ρG S¯ 2 Df Cz = r0 ρb Gb C

(2.5.51)

where, r0 = radius of the foundation; Gb = dynamic shear modulus at foundation base; G = dynamic shear modulus of soil in which the foundation is embedded; Df = ¯ b and S¯ 2 = frequency independent constants as deﬁned by depth of embedment; C Novak and furnished in Table 2.5.10. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 147 Table 2.5.10 Values of damping coefﬁcients based on Novak and Beredugo (1972). Poisson’s ratio

C¯ b

S¯ 2

0.0

3.438a0 + 0.5742a02 − 1.154a03 + 0.7433a04

6.059a0 +

0.25 0.5

5.06a0 7.414a0 − 2.986a02 + 4.324a03 − 1.782a04

Do Do

0.7022a0 a0 + 0.01616

where a0 = ωr/vs in which, ω operating frequency of the system in rad/sec; r = radius of the pile; vs = shear wave velocity of the soil.

With reference to Figure 2.5.1 for a pile element of length dz, embedded in the soil, the above equation may expressed as ¯ b + r0 ρG S¯ 2 dz (2.5.52) Cz = r0 ρb Gb C For systems having continuous function, the damping is usually expressed as (Mario 1987): (2.5.53) Cz = c(z) φi (z)φj (z)dz For the present case, Equation (2.5.53) can be expressed as

Cz = r0 ρG S¯ 2

L

¯ b φi (L)φj (L) φi (z)φj (z)dz + r0 ρb Gb C

(2.5.54)

0

Considering φ(z) = cos β Lz , for the fundamental mode, one can have

Cz = r0 ρGS¯ 2

L

z ¯ b cos2 β cos2 β dz + r0 ρb Gb C L

(2.5.55)

0

and hence Cz = r0

ρGS¯ 2 L

1

¯ b cos2 β cos2 βξ dξ + r0 ρb Gb C

(2.5.56)

0

Equation (2.5.56), on integration simplifies to √ ¯b ¯ 2L ρb G b C r ρG S 1 1 r 0 0 ¯b + sin 2β + cos 2β Cz = r0 ρG S¯ 2 L + r0 ρb Gb C 2 2 4β 2 (2.5.57) Equation (2.5.57) expresses the soil damping for a single pile under vertical mode ¯ b are damping coefficients which are frequency of vibration. Here the Factor S¯ 2 and C © 2009 Taylor & Francis Group, London, UK

148 Dynamics of Structure and Foundation: 2. Applications

dependent. Fortunately the damping factor is required for calculation of the amplitude when the eigen solution of the problem is already done vis-a-vis, the dimensionless ¯ b are frequency number a0 = ωr/vs term is known14 . Polynomial ﬁt curve for S¯ 2 and C available in terms of a0 which can be used to arrive at these parameters. The damping constants are given in Table 2.5.10. 2.5.10.2.3

Consideration of material damping of pile

The structural stiffness contribution of the pile is given by Equation (2.5.40), while that of the mass is given in Equation (2.5.50). Thus, if Cc be the critical damping of the pile then it can be expressed as Cc = 2 Kmp , where K (equals I1 in Equation 2.5.39) and mp are the stiffness and mass matrices of the pile. Depending on the material used for pile like (RCC, steel etc.) a suitable damping ratio (D) can be assumed. The damping (C) for the pile can expressed as Cp = DCc

(2.5.58)

This, when added to the radiation damping, calculated earlier, gives the complete damping quantity for the soil-pile system. It should be noted that for perfectly floating piles structural contribution of pile vanishes, and the material damping of the pile mentioned in the preceding need not be considered. 2.5.10.3 Vibration of bearing piles The expressions derived so far give a general case when the load is transferred from the pile to the soil, through both friction and bearing. There will be cases when the pile is pre-dominantly bearing in load transfer. Using the above formulation when lim η → ∞ (i.e. Gb is very large compared to E), pL(tan pL) = ∞, when β → π/2, the pile reduces to a perfectly bearing pile (i.e. ﬁxed at the base), however for practical case when η → 50, it will not be too erroneous to assume β → π/2, when the stiffness of the pile reduces to K1 =

EAπ 2 GS1 L + 8L 2

and

(2.5.59)

the damping may be expressed as C1 =

1 r0 ρGS¯ 2 L 2

(2.5.60)

and the mass is m1 =

γp AL . 2g

14 For calculation of damping it is considered ω = ωn for it is most critical at resonance.

© 2009 Taylor & Francis Group, London, UK

(2.5.61)

Analysis and design of machine foundations 149

2.5.10.4

Vibration of friction piles

When Gb is very small the load is transferred mainly through pile friction. In the above formulation when lim η → 0, pL tan pL = 0, when, β → 0, the pile becomes a perfectly friction pile. Thus, for β → 0, the stiffness of pile is given by K1 = (GS1 L)/2

(2.5.62)

The damping matrix may be expressed as

C1 =

1 ¯b r0 ρG S2 L + r0 ρb Gb C 2

(2.5.63)

From Equation (2.5.63) it should be noted that for a friction pile, the damping factor increases, while the stiffness term in Equation (2.5.62) is less than the bearing case in Equation (2.5.59). A similar observation has also been made by Novak (1974) in his investigation. For very poor soil, the term Gb in Equation (2.5.63) may be ignored. However for cases when piles located in medium to stiff homogenous clayey soil where G = Gb and yet the load is basically transferred through friction, the last term cannot be ignored and would further enhance the radiation damping. The mass matrix shall be same as stated in Equation (2.5.50). 2.5.10.5

Vertical vibration of partially embedded piles

In many instances, especially in the arctic condition, due to environmental reasons, the steel piles are driven into the ground when they protrude about 2 to 3 m above the ground over which the pile cap and vibrating equipments as placed (Figure 2.5.10). In such cases Novak’s (1976, 1983) chart cannot be used readily.

Rotating Machine Pile Cap Partially embedded piles G.L. L L1

Figure 2.5.10 Schematic diagram of partially embedded piles.

© 2009 Taylor & Francis Group, London, UK

150 Dynamics of Structure and Foundation: 2. Applications

To evaluate the pile stiffness for such cases, the stiffness Equation (2.5.36) is to be modified as L Kij = EA

φi (z)φj (z)dz

0

L1 + GS1

φi (z)φj (z)dz + Gb r0 Cb φi (L)φj (L)

(2.5.64)

0

where L1 = partial depth of embedment of pile and L = total length of pile. It is apparent from Equation (2.5.64) that the first and last term remains unchanged and the second term based on depth of embedment gets modiﬁed, where the integration limits changes to (L1 − 0) and the stiffness expression for the fundamental mode reduces to EAβ 2 GS1 L1 G b r0 C b EAβ G r0 C b GS1 L1 K1 = + + + − sin 2β + b cos 2β 2L 2 2 4β 4L 2 (2.5.65) The damping of the pile-soil system is given by Cz =

√ ¯ 1 1 ¯ b + r0 ρGS2 L1 sin 2β + r0 ρb Gb Cb cos 2β r0 ρGS¯ 2 L1 + r0 ρb Gb C 2 2 4β 2 (2.5.66)

The mass matrix remains the same as stated in Equation (2.5.50). It should be noted that for this case while calculating the value of S1 [Equation (2.5.29)], the slenderness ratio is to be calculated based on the embedded length of the pile. 2.5.10.5.1

Stiffness of the pile for soils with varying elastic property

In the previous section, the calculation of stiffness as well as the damping of soil was based on the dynamic shear modulus of soil invariant with depth. While this could be possible for clayey soils, there are many cases when the dynamic shear modulus of the soil has been found to vary with depth. Generically this can be expressed as G = G (z/H)α

(2.5.67)

where α = a number varying from 0–2 [considered 0 when G is constant with depth, assumed 1 for linear variation and 2 for parabolic distribution]. For instance for the soil with variable elastic property, Equation (2.5.67) may be modified to G = Gξ α where ξ = z/H. © 2009 Taylor & Francis Group, London, UK

(2.5.68)

Analysis and design of machine foundations 151

For the cases mentioned above, Novak’s (1976) chart is possibly not valid. To accommodate the above variation, the stiffness equation can be modified to

Kij =

EAβ 2 L2

+1 1 Fi (ξ )Fj (ξ )dξ + GS1 L ξ α Fi (ξ )Fj (ξ )dξ + Gb r0 Cb Fi (L)Fj (L) 0

0

(2.5.69) 2.5.10.5.2

Shear modulus having a linear variation

When the soil has linear distribution with depth, the stiffness Equation (2.5.69) may be expressed as EAβ 2 K1 = L

+1 1 2 (sin βξ ) dξ + GS1 L ξ (cos βξ )2 dξ + Gb r0 Cb (cos β)2 (2.5.70) 0

0

which, on integration and subsequent simplification, gives rise to G b r0 C b GS1 L 1 EAβ 2 1 GS1 L EAβ + 1− 2 + + − sin 2β L 4 2 2 β L β GS1 L Gb r0 Cb + + cos 2β (2.5.71) 2 4β 2

1 K1 = 2

It may be noted that while for bearing pile β = π /2, for friction pile (unlike constant G case), β = 0 is an inadmissable function in this case. For the fundamental mode the admissible function is β = π , which is the next higher mode. This is logical also for the soil having stiffness increasing with depth and the pile will have a natural tendency to wobble about its centre rather than moving en-mass. The damping matrix in this case can be expressed as

Cz = r0 ρGS¯ 2 L

1

¯ b cos2 β ξ cos2 βξ dξ + r0 ρb Gb C

(2.5.72)

0

The integration of the first term in Equation (2.5.72) being cyclic in nature and can be solved approximately by expanding the cosine function in series. On integration, Equation (2.5.72) reduces to

Cz = r0

2 ρGS¯ 2 L − 2β 2 3

© 2009 Taylor & Francis Group, London, UK

1 β2 2 4 − + β 7 33 675

¯ b cos2 β (2.5.73) + r 0 ρb G b C

152 Dynamics of Structure and Foundation: 2. Applications

2.5.10.5.3

Shear modulus having a parabolic variation

When the soil modulus has a parabolic distribution with depth, the stiffness equation may be expressed as EAβ 2 K1 = L

+1 1 2 (sin βξ ) dξ + GS1 L ξ 2 (cos βξ )2 dξ + Gb r0 Cb (cos β)2 0

0

(2.5.74) which on integration and subsequent simplification reduces to K1 =

EAβ 2 GS1 L Gb r0 Cb + + 2L 6 2

+

GS1 L + 2

1 1 + 2β 3β 3

GS1 L Gb r0 Cb + cos 2β β 2

EAβ − sin 2β 4L

(2.5.75)

In this case, the first admissible function will be β = π for a friction pile and β = π/2 for a bearing pile. The mass matrix for both the cases remains same as stated in Equation (2.5.50) while the damping matrix can be obtained from the expression

Cz = r0 ρGS¯ 2 L

1

¯ b cos2 β ξ cos2 βξ dξ + r0 ρb Gb C

(2.5.76)

0

which on integration and simplification reduces to √ √ r0 ρb Gb Cb r0 ρGS2 L r0 ρGS2 L Cz = − + sin 2β 2 4 8β √ r0 ρGS2 L r0 ρb Gb Cb + + cos 2β. 4β 2

2.5.10.6

(2.5.77)

Group effect of pile

This has already been explained in detail in section 2.5.5 and may be used for the present case also. 2.5.10.7

Effect of pile cap on pile stiffness

The pile cap has been found to affect the response of footing signiﬁcantly. Before considering its effect within the proposed framework, it would be worthwhile to recapitulate the practice in vogue. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 153

Df

Figure 2.5.11 Schematic diagram of pile and pile-cap with embedment.

The sketch given in Figure 2.5.11 can represent the pile group with pile cap. In such case usually the embedment stiffness GSf Df is directly added to the pile group stiffness and the system is considered as a lumped mass single degree freedom system where ω=

Kgroup + Gf Sf Df

(2.5.78)

M

where Gf = dynamic shear modulus of the soil surrounding the pile cap; Df = depth of embedment; Sf = constant as suggested by Novak furnished in Table 2.5.4 (as S1 ); M = mass of pile cap and machine placed on it. It may be noted that contributing effect of the pile mass is ignored in the above which could be significant for a pile group having large number of piles. To overcome the above limitation and also to derive a better response we propose a two mass lumped model has been proposed and shown in Figure 2.5.12. The mass and stiffness matrices for the above model may be written as

[K] =

Kgroup + Gf Sf Df −Gf Sf Df

© 2009 Taylor & Francis Group, London, UK

−Gf Sf Df G f Sf D f

(2.5.79)

154 Dynamics of Structure and Foundation: 2. Applications

M = Mass of pile cap + machine

K2 = GfSfDf

mp = Mass of piles in group

K1 = Kgroup

Figure 2.5.12 Proposed two mass lumped model for the pile and pile cap.

⎡ and

[M] =

nγ p AL ⎣ 2g

1+

sin 2β 2β

0

⎤ 0

⎦

(2.5.80)

M

where n = number of piles in the pile group. Since Equation (2.5.79) is statically coupled, the damping matrix is given by Cgroup + Cf C= −Cf

−Cf Cf

where Cz = r0 ρGS¯ 2 Df

(2.5.81) and Df is the embedment depth of pile cap.

(2.5.82)

Once the stiffness, mass and damping matrices are established, the natural frequency of the system may be obtained from the standard expression [K] − [M] ω2 = 0

(2.5.83)

leading to

λ1,2 =

(mp B + MA) ±

in which mp =

[(mp B + MA)2 − 4mp MAB] 2mp M

nγp AL 2g

1+

© 2009 Taylor & Francis Group, London, UK

sin 2β 2β

!

and A = Kgroup + Gf Sf Df ; B = Gf Sf Df .

(2.5.84)

Analysis and design of machine foundations 155

√ √ where, ω1 = λ1 and ω2 = λ2 here ω1 and ω2 are the natural frequency of the structure. The damping matrix generated here is non-classical in nature and will not be de-coupled on orthogonal transformation. However, since the degrees of freedom considered here is two, the same can also be converted into an equivalent Rayleigh damping (refer section 2.2.5 in this chapter where we have solved this) when the matrix will decouple and standard modal solution can be applied. 2.5.10.8

Solutions for higher modes

This case is usually not considered in design office practices and neither any guidelines presently exists for the same except treating the pile as a beam and the soil as Winkler springs and solving the same based on finite element method. Using the proposed methodology, the stiffness, damping and mass matrices can be computed for the higher modes. Referring to Equation (2.5.37), the stiffness matrix can be stated as EA [Kij] = L ⎡ ⎤ #1 #1 #1 #1 2 2 F (ξ ) β β F (ξ )F (ξ ) β β F (ξ )F (ξ ) ......... β β F (ξ )F (ξ ) β 1 1 2 1 2 1 3 1 3 1 n 1 n ⎢ ⎥ 1 ⎢ ⎥ 0 0 0 0 ⎢ ⎥ 1 1 1 # # # ⎢ ⎥ 2 2 ⎢β2 β1 F2 (ξ )F1 (ξ ) β2 F2 (ξ ) ......... ........ β2 βn F2 (ξ )Fn (ξ )⎥ ⎢ ⎥ 0 0 0 ⎢ ⎥ ⎢ ⎥ 1 1 1 1 # # # # ×⎢ ⎥ 2 2 β β F (ξ )F (ξ ) β β F (ξ )F (ξ ) β F (ξ ) .......... β β F (ξ )F (ξ ) 3 1 3 2 3 2 3 3 n 3 n ⎢ 3 1 ⎥ 3 ⎢ ⎥ 0 0 0 0 ⎢ ⎥ ............. ........ ........ ....... ........... ⎢ ⎥ ⎢ ⎥ #1 #1 ⎣ ⎦ 2 2 βn β1 Fn (ξ )F1 (ξ ) βn Fn (ξ ) 0

0

⎡

#1

F1 (ξ ) ⎢ ⎢ 0 ⎢#1 ⎢ ⎢ F2 (ξ )F1 (ξ ) ⎢0 GS1 L ⎢ ⎢#1 × dξ + 2 ⎢ ⎢ F3 (ξ )F1 (ξ ) ⎢0 ⎢ ⎢ .............. ⎢1 ⎣# Fn (ξ )F1 (ξ ) 2

#1 0

#1

F1 (ξ )F2 (ξ ) #1

#1

F1 (ξ )F3 (ξ )

.........

0

F2 (ξ )2

0

F3 (ξ )F2 (ξ )

0

......... #1

F3 (ξ )2

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

0

........

........

0

.......

#1

⎤

F1 (ξ )Fn (ξ )⎥ ⎥ ⎥ ⎥ F2 (ξ )Fn (ξ )⎥ ⎥ 0 ⎥ ⎥ #1 F3 (ξ )Fn (ξ )⎥ ⎥ ⎥ 0 ⎥ ............. ⎥ ⎥ #1 ⎦ Fn (ξ )2 0 #1

0

× dξ + Gbr0 Cb F1 (0)Fj (0) etc.

(2.5.85)

For first three modes this can simply be presented as ⎡

[K]i=1,3 j=1,3

K11 = ⎣K21 K31

© 2009 Taylor & Francis Group, London, UK

K12 K22 K32

⎤ K13 K23 ⎦ K33

(2.5.86)

156 Dynamics of Structure and Foundation: 2. Applications

where for i = j Kii =

EAβi2 GS1 L EAβi G r0 C b GS1 L − + + G b r0 C b + sin 2βi + b cos 2βi 2L 2 4βi 4L 2 (2.5.87)

For i = j we have

EAβi βj EAβi βj GS1 L sin(βi − βj ) GS1 L sin(βi + βj ) Kij = + − − 2L 2 βi − β j 2L 2 βi + β j + Gb r0 Cb cos βi βj

(2.5.88)

It should be noted at this point that there are no suggestive values available for S1 and Cb for higher modes either by Novak or any other research. However, it may be reasonably stated that for higher modes the dimensionless frequency a0 would be ≥1.0 (or near 1.0 at worse) when the curve for S1 becomes almost constant (Novak 1974) and the values furnished in Table 2.5.6 may be used without much error. The value of β for the fundamental mode is already furnished in Table 2.5.9 for the next two modes the values of β are furnished in Table 2.5.11 and Table 2.5.12. Table 2.5.11 Roots of equation pL tan(pL) − η = 0 for second mode. η

pL

η

pL

η

pL

η

pL

0 0.1 0.2 0.3 0.4 0.5 0.6

3.141 3.173 3.204 3.234 3.264 3.292 3.320

0.7 0.8 0.9 1 1.25 1.5 1.75

3.348 3.374 3.4 3.426 3.486 3.542 3.595

2 2.25 2.5 3.0 3.5 4 5

3.644 3.689 3.732 3.809 3.876 3.935 4.034

10 20 25 30 35 40 50

4.425 4.491 4.533 4.561 4.582 4.598 ∼ =3π/2

Table 2.5.12 Roots of equation pL tan(pL) − η = 0 for third mode. η

pL

η

pL

η

pL

η

pL

0 0.1 0.2 0.3 0.4 0.5 0.6

6.28 6.299 6.315 6.331 6.346 6.362 6.377

0.7 0.8 0.9 1 1.25 1.5 1.75

6.392 6.407 6.422 6.437 6.474 6.510 6.544

2 2.25 2.5 3.0 3.5 4 5

6.578 6.611 6.643 6.704 6.761 6.814 6.910

15 20 25 30 35 40 50

7.316 7.495 7.56 7.606 7.639 7.665 ∼ =5π/2

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 157

Mass matrix is similarly given by For i = j mii =

γp AL sin 2βi + M cos2 βi 1+ 2g 2βi

(2.5.89)

where M = Mass of machine plus pile cap For i = j γp AL sin(βi + βj ) sin(βi − βj ) M + − cos(βi + βj ) − cos(βi − βj ) mij = 2g βi + β j βi − β j 2 (2.5.90) The damping matrix can be obtained as For i = j Cii =

√ ¯ 1 1 ¯ b + r0 ρGS2 L sin 2βi + r0 ρb Gb Cb cos 2βi r0 ρGS¯ 2 L + r0 ρb Gb C 2 2 4β 2 (2.5.91)

For i = j sin(βi + βj ) sin(βi − βj ) − Cij = r0 ρGS¯ 2 L βi + β j βi − β j ¯b r2 ρGb C + 0 cos(βi + βj ) − cos(βi − βj ) 2

(2.5.92)

It is apparent that the dynamic analyses of piles with pile cap are standard and the validity of the same would depend on how correctly the pile stiffness values have been obtained. For this, it would worthwhile to evaluate how the present formulation matches with other established methods. To this end, the pile stiffness as obtained by Equations (2.5.59) and (2.5.62) has been compared with Novak’s chart (1974) and equation based on rigorous analysis as proposed by Dobry and Gazetas (1988). It should be noted that their expression is valid for floating piles of length say, L and embedded in an elastic half space of length 2L. The results have been compared for a single pile of various slenderness ratio λ varying from 20 to 100 and Ep /Gs value of soil varying from 250 to 10,000 for an RCC pile of diameter of 600 mm and having Ep = 30 GPa. Poisson’s ratio value for soil considered is 0.4. Here Ep = Young’s modulus of pile material; Gs = dynamic shear modulus of soil. The results for Kpile(bearing) and Kpile(friction) are shown in Figure 2.5.13 through 20 for various slenderness ratios. Finally, the natural frequency of a real life centrifugal compressor foundation supported on 9 RCC piles, 45 meter long having diameter of 950 mm, have been compared. The piles are spaced at 3.0 m c/c. The size of pile cap is 7 m × 5 m × 2.0 m, embedded to depth of 1.4 meter. The weight of the generator supported on it weighs 400 kN. The frequencies are again compared for a range of Ep /Gs varying from 250 to 10000. © 2009 Taylor & Francis Group, London, UK

158 Dynamics of Structure and Foundation: 2. Applications

Pile stiffness L/r=20

Pile stiffness (kN/m)

4.00E+06 3.50E+06 3.00E+06

Kpile(bearing) Novak(bearing) Gazetas

2.50E+06 2.00E+06 1.50E+06 1.00E+06 5.00E+05 0.00E+00

250

500

1000 Ep/Gs

2500

10000

Figure 2.5.13 Comparison of bearing pile stiffness for slenderness ratio = 20.

Stiffness (kN/m)

Stiffness of pile for L/R=20 2.00E+06 1.50E+06

Kpile(friction) Novak(friction) Gazetas

1.00E+06 5.00E+05 0.00E+00

250

500

1000 Ep/Gs

2500

10000

Vertical stiffness for L/r=40 5.0000E+06 4.0000E+06 3.0000E+06 2.0000E+06 1.0000E+06 0.0000E+00

Kpile(bearing) Novak(bearing)

0

00

00 10

Ep/Gs

50

00 10

0 50

0

Gazetas

25

Stiffness (kN/m)

Figure 2.5.14 Comparison of friction pile stiffness for slenderness ratio = 20.

Figure 2.5.15 Comparison of bearing pile stiffness for slenderness ratio = 40.

The results based on Kp(bearing) and Kp(friction) has been compared to Dobry and Gazetas’ results and presented in Table 2.5.13. The results have not been compared with Novak in this case for the charts are too crude especially in the range when the ratio of Ep /Gs = 2500–10000 and significant variation can occur based on eye estimate of stiffness function. Results have been found to be excellently matching particularly for friction piles. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 159

Stiffness of pile for L/r=40

Stiffness (kN/m)

3.00E+06 2.50E+06 2.00E+06

Kpile(friction)

1.50E+06

Novak(friction) Gazetas

1.00E+06 5.00E+05 0.00E+00

250

500

1000 Ep/Gs

5000

10000

Figure 2.5.16 Comparison of friction pile stiffness for slenderness ratio = 40.

Pile stiffness L/r=80 Pile stiffness (kN/m)

6.0000E+06 5.0000E+06 4.0000E+06

Kpile(bearing) Novak(bearing) Gazetas

3.0000E+06 2.0000E+06 1.0000E+06 0.0000E+00

250

500

1000 Ep/Gs

5000

10000

Figure 2.5.17 Comparison of bearing pile stiffness for slenderness ratio = 80.

Pile stiffness for L/r=80 Stiffness (kN/m)

5.00E+06 4.00E+06 3.00E+06 2.00E+06

Kpile(friction) Novak(friction)

1.00E+06

Gazetas

0.00E+00

250

500

1000 Ep/Gs

5000

10000

Figure 2.5.18 Comparison of friction pile stiffness for slenderness ratio = 80.

As stated earlier, the results from Equation (2.5.41) (with appropriate boundary condition for bearing and friction) have been compared with Novak’s chart and Dobry and Gazetas’ expression. The results have been studied against both the bearing and friction pile coefﬁcients as suggested by Novak and El-Sharnouby (1983). It will be observed in Figures 2.5.13 through 2.5.18 that the frictional stiffness values obtained are very © 2009 Taylor & Francis Group, London, UK

Pile stiffness (kN/m)

160 Dynamics of Structure and Foundation: 2. Applications

6.0000E+06 5.0000E+06 4.0000E+06 3.0000E+06 2.0000E+06 1.0000E+06 0.0000E+00

Pile stiffness L/r=100

Kpile(bearing) Novak(bearing) Gazetas 250

500

1000 Ep/Gs

5000

10000

Pile stiffness (kN/m)

Figure 2.5.19 Comparison of bearing pile stiffness for slenderness ratio = 100.

Pile stiffness for L/r=100

6.00E+06 5.00E+06 4.00E+06

Kpile(friction) Novak(friction) Gazetas

3.00E+06 2.00E+06 1.00E+06 0.00E+00

250

500

1000 Ep/Gs

5000

10000

Figure 2.5.20 Comparison of friction pile stiffness for slenderness ratio = 100.

Table 2.5.13 Variation of vertical frequency for compressor foundation. Sl. No.

Ep /Gs

Freq (rad/sec) for Kpile (bearing)

Freq (rad/sec) for Kpile (friction)

As per Dobry and Gazetas (rad/sec)

1 2 3 4 5 6 7

250 500 1000 2500 5000 7500 10000

196 139 99 64 47 39 35

195 138 98 62 44 36 31

197 139 98 62 44 36 31

close to Dobry and Gazetas’ results in all the cases for various L/r and E/Gs values. For the bearing piles, the values obtained are slightly higher than Dobry and Gazetas’ values but matching very closely to Novak’s data from Ep /Gs = 500 onwards. This is expected. It was pointed out by Novak and others that bearing stiffness for a pile is slightly more than that of friction stiffness. At L/r = 20 the bearing values obtained are higher than that of Dobry and Gazetas (which is logical considering his case is that of a floating pile) as well as from Novak but the difference reduces considerably from Ep /Gs = 1000 onwards, and this is © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 161

the range in which piles are commonly used in practice. The values, where Ep /Gs is ≤1000 are actually far too stiff for any piles to be bored or driven. Moreover, a pile with L/r = 20 is actually a fictitious values. For instance a standard pile of length 30 meter, the diameter becomes 3.0 m, which is actually a cassion and not a pile. It is possibly in such cases, the axial stiffness is far too high and this shows a significant higher stiffness in bearing compared to friction piles for such an unrealistic L/r ratio. For real life problems, the values of L/r is around 50–100 and Ep /Gs > 1000. It will be observed that the values obtained by the proposed method are quite close to the reported results useful for practical ranges of application. As for the frequencies obtained for various Ep /Gs values the results in Table 2.5.10 are extremely encouraging. 2.5.10.9

What is the major advantage of this model?

The major advantage with the proposed method is that instead of solving the differential equation (especially when the boundary condition gets complicated with cases like partial embedment or variable soil) the stiffness, damping and mass matrices are directly derived from energy principles and the subsequent derivation gets quite simpliﬁed. Finally, the formulation have been derived for a general case when pile can act both as bearing and friction pile for which no direct solutions are available-and this could be the reality in many cases when the pile is neither in full bearing or full ﬂoating. Comparing the results it can be well inferred that the method can be used for practical design office work without the limitations as stated at the outset. 2.5.10.10

Design steps

Based on the derivations presented, the design steps may be summarized as follows: • • • • • • • • • • • • • • •

Determine the soil properties like G, Gb , Gf and ν (Poisson’s ratio of the soil); Determine the pile properties like Length of pile L and diameter of pile (2r0 ) and also the Young’s Modulus E of the pile material; Determine the pile cap property like its mass and depth of embedment Df ; Determine the weight of machine supported on the pile cap; Obtain Novak’s stiffness and damping coefficients Cb , S1 , C b , S2 from Table 2.5.9 and Table 2.5.10, Equation (2.5.29) etc.; Establish the dimensionless parameter η = (Gb /E) (Cb /π ) λ; For the given value η determine the value of pL from Table 2.5.9; If the pile is bearing (known priori) β = π /2; Consider β = pL; Determine K1 and mp from Equations (2.5.40) and (2.5.50) respectively; Determine the embedment stiffness matrix from the Equation (2.5.79); Form the mass, stiffness; Perform eigen solution; Find the value of the frequency and obtain the dimensionless frequency number a0 ; Find the value of S2 from Beredugo’s expression as given in Table 2.5.10;

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162 Dynamics of Structure and Foundation: 2. Applications

• •

Determine the damping of the system based on Equations (2.5.57), (2.5.60) & (2.5.63); Perform Modal or time history analysis to obtain the amplitude of vibration.

2.5.11 Dynamic analysis of laterally loaded piles 2.5.11.1

Piles under dynamic lateral loading

Having presented the mathematical model of vibration in vertical direction, we now present a model of vibration of piles under lateral or horizontal load. This is an important study for the pile supporting rotating machines under centrifugal and reciprocating loads. In majority of cases it has been found that of all modes (like vertical, rocking, yawning, twisting etc.); lateral vibration (coupled with rocking) is most critical and often governs the design. Thus a study of such motion is of paramount importance for piles supporting important installations and also for facilities, which are valuable to the community under earthquake threats. Recall Novak’s method described earlier for lateral pile you will realize on retrospection that the model has got the following limitations • • • • • •

The method is coefficient based [function of the ratio of Young’s modulus of pile (Ep ), and dynamic shear modulus of soil (Gs ), as such for intermediate values one has to interpolate which may not be always very accurate. The values are given for Poisson’s ratio of 0.25 and 0.40 only. Thus for any value between 0.25 and 0.4, or beyond 0.4 another set of linear interpolation/ extrapolation is necessary. Novak and El Sharnouby (1983) has given stiffness and damping coefficients for soil having parabolic profile but in many cases the variation is linear and no coefficients are available for this case. The method does not have a solution for partially embedded piles, which is of great practical importance for piles driven in arctic condition (especially in Northern Siberia which constitute of a large number of Oil and Gas facilities). The dynamic bending moment and shear force induced on pile cannot be evaluated. Finally the formulation is valid for long piles (i.e. the failure takes place in the pile body before soil yields) and do not cater to piles, which are short.

The simplified formulas given by Dobry and Gazetas (1988) is based on more rigorous analysis, however it also does not address the issues of partial embedment, dynamic bending moment and shear, or the issue- if the pile is short (i.e. L/r < 25) etc. We now present herein (Chowdhury and Dasgupta 2008) a mathematical model for analysis of such piles under lateral load that overcomes many of the bottle necks cited above. Similar to the vertical vibration model presented earlier the present formulation is based Novak and Beredugo’s (1972) formulation for a rigid cylinder embedded in elastic half space. Shown in Figure 2.5.21 is a pile embedded in homogeneous elastic medium and considered under plane strain condition. The pile is considered long and slender, to start with. Under static conditions, the equation of equilibrium in the

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 163

M P X

Z

dz

Soil Stiffness = GSx1

Figure 2.5.21 Conceptual model of pile under lateral load.

x-direction [similar to beams on elastic foundation] is given by Timoshenko (1956) as

E p Ip

d4x = −ks x dz4

(2.5.93)

where, Ep = Young’s modulus of the pile; Ip = moment of inertia of the pile cross section; ks = elastic stiffness of the soil and is expressed as GSx1 ; G = dynamic shear modulus of the soil; Sx1 = Beredugo’s constant which are basically frequency dependent. However, it has been shown by Novak and Beredugo (1972) that considering this term frequency independent, no accuracy is lost for practical design problems and the analysis becomes quite simplified for rigid circular embedded footing. Elaboration about this parameter, in terms of piles, will be made later. The general solution of Equation (2.5.93) may be written as x = e−pz (C0 cos pz + C1 sin pz) + epz (C2 cos pz + C3 sin pz) where p =

4

(2.5.94)

GSx1 . E p Ip

For long piles under load or moment at its head, it is reasonable to assume that at significant distance from the pile head (where the load is applied), the curvature vanishes. This condition can only be satisfied when C2 and C3 in Equation (2.5.94) is

© 2009 Taylor & Francis Group, London, UK

164 Dynamics of Structure and Foundation: 2. Applications

considered insignificant. Hence, the deflection equation can be taken as x = e−pz (C0 cos pz + C1 sin pz)

(2.5.95)

Considering the pile head undergoing specified deflection and rotation as well as its head is fixed to the pile cap (same boundary condition as considered by Novak (1974)), one can have [Figure 2.5.21], At z = 0, let x = x0 ⇒ C0 = x0 , which gives x = e−pz (x0 cos pz + C1 sin pz)

(2.5.96)

= θ0 one can have Again, at z = 0, dx dz C1 = x0 +

θ0 p

(2.5.97)

Thus Equation (2.5.98) can now be represented as θ0 −pz x0 cos pz + x0 + sin pz x=e p For magnitude of rotation being small θ0 ∼ = x0 /L, x may be written as x0 x = e−pz x0 cos pz + x0 + sin pz pL 1 x −pz cos pz + 1 + =e sin pz x0 pL

(2.5.98)

(2.5.99)

(2.5.100)

Now considering β = pl and using Equation (2.5.100), for any arbitrary loading, the generic shape function in dimensionless form can be represented as −βz βz 1 βz + 1+ sin (2.5.101) φ(z) = e L cos L β L in which β=

4

GSx1 L4 , Ep Ip

L being the length of the pile.

(2.5.102)

Equation (2.5.101) can be further reduced to φ(z) = e

−βz L

βz βz cos + η sin L L

where η = 1 +

1 β

© 2009 Taylor & Francis Group, London, UK

(2.5.103) (2.5.104)

Analysis and design of machine foundations 165

The generic shape function of the pile for the fundamental mode as in Equation (2.5.103) is shown in Figure 2.5.22 for Ep /G = 5000. The potential energy d of an element of depth dz, shown in Figure 2.5.21 is then given (Shames and Dym 1995) by

Ep Ip d = 2

d2v dz2

2 +

Kh 2 v 2

(2.5.105)

where, Ep = Young’s modulus of pile; Ip = moment of inertia of pile; Kh = lateral dynamic stiffness of soil; v = displacement of the pile in the x direction and may be written as [φ(z)q(t)]. For a rigid circular embedded footing of embedment Df , the stiffness of the footing may be expressed (Beredugo and Novak (1972)) as Kh = Gb r0 Cb + GDf Sx1

(2.5.106)

where, Kh = foundation stiffness in horizontal direction; G = dynamic shear modulus of the soil along foundation surface; Gb = dynamic shear modulus of soil at the foundation base; r0 = radius of the foundation; Cb and Sx1 = constants which are basically frequency dependent. Ignoring the first term in Equation (2.5.106), which represents the contribution of base resistance, and substituting the same in Equation (2.5.105), for a cylindrical element of depth dz, embedded in soil, the potential energy d may be expressed as

Ep Ip d = 2

d2v dz2

2 +

GSx1 dz 2 v . 2

(2.5.107)

1.2 Shape function

1 0.8 0.6 0.4 0.2

9 0.

75 0.

6 0.

0. 45

3 0.

0.

0

-0.2

15

0

z/L

Figure 2.5.22 Generic shape function long pile in the horizontal mode for Ep /G = 5000.

© 2009 Taylor & Francis Group, London, UK

166 Dynamics of Structure and Foundation: 2. Applications

The total potential energy over the length of the pile (L) is then given by L

Ep Ip = 2

d2v dz2

2

GSx1 dz + 2

0

L v2 dz

(2.5.108)

0

Considering v (z, t) = φ(z)q(t), it can be proved (Hurty and Rubenstein 1967) that L Kij = Ep Ip

φi (z)φj (z)dz

L + GSx1

0

φi (z)φj (z)dz

(2.5.109)

0

where the shape function of the problem is given by Equation (2.5.103). For the fundamental mode, stiffness of the pile is then given by L K = Ep Ip

L

φi (z)2 dz

φi (z)2 dz

+ GSx1

0

(2.5.110)

0

On double differentiation, Equation (2.5.103) reduces to φ (z) =

2β 2 − βz e L L2

sin

4β 4 2βz φ (z) = 4 e− L L

2

βz βz − η cos L L

and

X Y 2βz 2βz − cos − η sin 2 2 L L

(2.5.111) (2.5.112)

where, X = 1 + η2 ; Y = 1 − η2 and η is given in Equation (2.5.104). Again from Equation (2.5.103) 2

φ(z) = e

− 2βz L

X Y 2βz 2βz + cos + η sin 2 2 L L

(2.5.113)

Substituting Equations (2.5.112) and (2.5.113) in Equation (2.5.110), the stiffness reduces to 4Ep Ip β 4 K= L4

L

e−

2βz L

X Y 2βz 2βz − cos − η sin dz 2 2 L L

0

L + GSx1

e−

2βz L

0

© 2009 Taylor & Francis Group, London, UK

X Y 2βz 2βz + cos + η sin dz 2 2 L L

(2.5.114)

Analysis and design of machine foundations 167

Equation (2.5.114) on integration by parts and on simplification may be expressed as K=

4Ep Ip β 4 L4

4Ep Ip β 4 − L4 + GSx1 + GSx1

! X L Y L −2β (1 − e−2β ) − e (sin 2β − cos 2β) + 1 2 2β 2 4β

ηL −2β (sin 2β + cos 2β)) (1 − e 4β

X L Y L −2β (1 − e−2β ) + (sin 2β − cos 2β) + 1) (e 2 2β 2 4β ηL (1 − e−2β (sin 2β + cos 2β)) 4β

(2.5.115)

In Equation (2.5.115), e−2β (sin 2β + cos 2β) and e−2β (sin 2β − cos 2β) may be ignored as their values are exceedingly small (highest is of the order 10−3 and the lowest is 10−30 for Ep /G value varying from 250 to 10,000) and has practically no effect on the stiffness value and this also considerably simplifies the expression. Based on the above simpliﬁcation, Equation (2.5.115) may be rewritten as 4Ep Ip β 4 K= L4

+ GSx1

➔ K=

X L Y L ηL (1 − e−2β ) − − 2 2β 2 4β 4β

X L Y L ηL (1 − e−2β ) + + 2 2β 2 4β 4β

(2.5.116)

Ep Ip β 3 Y −2β X(1 − e − η ) − 2 L3 +

GSx1 L Y X(1 − e−2β ) + + η 4β 2

(2.5.117)

Taking Ep Ip β 3 /L3 as common in Equation (2.5.117) and substituting the value of β from Equation (2.5.102a), Equation (2.5.117) reduces to Ep Ip β 3 K= L3 Ep Ip K= 3 L

5X 3Y 3η (1 − e−2β ) − − 4 8 4

5X −2β ) − 3Y 4 (1 − e 8 3

(η − 1)

© 2009 Taylor & Francis Group, London, UK

− 34 η

which can be further simplified to

(2.5.118)

168 Dynamics of Structure and Foundation: 2. Applications

The accuracy of Equation (2.5.118) will be dependent on the correct selection of Sx1 . For instance for rigid circular footing Beredugo and Novak (1972) has furnished a frequency independent value of Sx1 = 4.0 to 4.1 (depending on Poisson’s ratio). This has been found to give adequate accuracy for practical engineering design. Comparing the stiffness data with Novak (1974), Dobry and Gazetas (1988), it is proposed that the following values of Sx1 as furnished in Tables 2.5.14 to 16 be used for the calculation of dynamic response of the pile in the lateral direction. For a particular pile having specific slenderness ratio and Poisson’s ratio of the soil, the value of Sx1 can be selected from Tables 2.5.14, 2.5.15 and 2.5.16 and on substitution of the same in Equation (2.5.102), Equation (2.5.118), gives the solution of pile stiffness in the lateral direction. Table 2.5.14 Suggested value of Sx1 for Poisson’s ratio of soil = 0.25. Poisson’s ratio 0.25

L/r0 (Slenderness ratio)

Sx1 (250)

Sx1 (500)

Sx1 (1000)

Sx1 (2500)

Sx1 (5000)

Sx1 (10000)

25 40 60 80 100

2.00 2.19 2.30 2.36 2.39

1.83 2.05 2.17 2.24 2.28

1.66 1.90 2.05 2.12 2.17

1.43 1.70 1.87 1.96 2.01

1.25 1.55 1.74 1.84 1.90

1.07 1.39 1.60 1.71 1.78

Note: The value in parenthesis after Sx1 indicates Ep /Gs value of the soil.

Table 2.5.15 Suggested value of Sx1 for Poisson’s ratio of soil = 0.40. Poisson’s ratio 0.40

L/r0 (Slenderness ratio)

Sx1 (250)

Sx1 (500)

Sx1 (1000)

Sx1 (2500)

Sx1 (5000)

Sx1 (10000)

25 40 60 80 100

2.27 2.48 2.60 2.66 2.70

2.08 2.32 2.46 2.53 2.57

1.89 2.16 2.31 2.40 2.45

1.63 1.94 2.12 2.22 2.28

1.43 1.76 1.97 2.08 2.15

1.23 1.59 1.82 1.94 2.02

Note: The value in parenthesis after Sx1 indicates Ep /Gs value of the soil.

Table 2.5.16 Suggested value of Sx1 for Poisson’s ratio of soil = 0.50. Poisson’s ratio 0.50

L/r0 (Slenderness ratio)

Sx1 (250)

Sx1 (500)

Sx1 (1000)

Sx1 (2500)

Sx1 (5000)

Sx1 (10000)

25 40 60 80 100

2.45 2.67 2.80 2.87 2.91

2.25 2.50 2.65 2.72 2.77

2.05 2.33 2.50 2.58 2.63

1.77 2.09 2.29 2.39 2.45

1.55 1.91 2.13 2.24 2.32

1.34 1.72 1.96 2.10 2.18

Note: The value in parenthesis after Sx1 indicates Ep /Gs value of the soil.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 169

2.5.11.1.1

Estimation of contribution of pile mass

The mass matrix of the pile may be expressed as (Meirovitch 1967) Mx = m(x)

φi (z)φj (z)dz

(2.5.119)

For the present case of the pile of length L, Equation (2.5.118), can be expressed as

γp Ap Mx = g

L φ(z)2 dz

(2.5.120)

0

where, γp = unit weight of the pile material; Ap = cross sectional area of the pile; g = acceleration due to gravity.

or,

γp A p Mx = g

L e

− 2βz L

X Y 2βz 2βz + cos + η sin 2 2 L L

(2.5.121)

0

Equation (2.5.121) on integration and after simplification gives γp Ap L Mx = 4g

X(1 − e−2β ) + β

Y 2

+η

(2.5.122)

Equation (2.5.122) is the inertial contribution of the pile material for the fundamental mode. Incidentally, the inertial effect is usually ignored in design but could have signiﬁcant effect if the number of piles is large in a pile group. 2.5.11.2 Radiation damping for pile under lateral load For a rigid footing embedded in soil for a depth Df , Beredugo and Novak (1972) have proposed the expression ¯ b + r0 ρG S¯ 2 Df Cz = r0 ρb Gb C

(2.5.123)

where, r0 = radius of the foundation; Gb = dynamic shear modulus at the foundation base; G = dynamic shear modulus of the soil in which the foundation is embedded; ¯ b and S¯ 2 = frequency independent constants as defined Df = depth of embedment; C by Novak and Beredugo (1972). © 2009 Taylor & Francis Group, London, UK

170 Dynamics of Structure and Foundation: 2. Applications

With reference to Figure 2.5.21 for a pile element of length dz embedded in the soil, and ignoring the bearing effect, Equation (2.5.123) may be expressed as c(x) = r0 ρGSx2 dz

(2.5.124)

For systems having continuous response function, the damping may be expressed as Cx = c(x)

φi (z)φj (z)dz

(2.5.125)

For the pile of length L, Equation (2.5.125) may be expressed as L

φ(z)2 dz

Cx = r0 ρGSx2

(2.5.126)

0

or,

L

Cx = r0 ρGSx2

e

− 2βz L

X Y 2βz 2βz + cos + η sin 2 2 L L

(2.5.127)

0

On integration and after simplification Equation (2.5.127) reduces to Cx = r0

X(1 − e−2β ) + ρGSx2 L 4β

Y 2

+η

(2.5.128)

Equation (2.5.128) expresses the soil damping for a single pile under horizontal mode of vibration. The factor Sx2 is a frequency dependent damping coefficient. The damping factor is required for calculating the amplitude only after the eigen solution of the problem is already done vis-a-vis, the dimensionless frequency number a0 = ωr0 /vs term is known a priori. Polynomial fit curve for Sx2 are available in terms of a0 which can be used directly to obtain these parameters. Sx2 for different Poisson’s ratios are given in Table 2.5.17. Table 2.5.17 Values of Sx2 (Beredugo & Novak 1972). Poisson’s ratio 0.0 0.25 0.5

© 2009 Taylor & Francis Group, London, UK

Sx2 0.8652a0 a0 + 0.00874 41.59a0 0.83a0 + a0 + 3.90 56.559a0 0.96a0 + a0 + 4.68 7.334a0 +

Analysis and design of machine foundations 171

2.5.11.3

Material damping of pile

The structural stiffness contribution of the pile is given in the first part of Equation (2.5.117), while that of the mass is given in Equation (2.5.122). Thus, if Cc be the critical damping of the pile then it can be expressed as Cc = 2 Kmp ; K (the ﬁrst term in Equation (2.5.117)) and mp being the stiffness and mass of the pile. Depending on the material used for pile (like RCC, steel etc.) a suitable damping ratio (D) can be assumed. The damping (Cp ) for the pile can then be expressed as Cp = DCc

(2.5.129)

This, when added to the radiation damping, calculated through Equation (2.5.128) gives the complete damping quantity for the soil-pile system. 2.5.11.4

Piles with other boundary conditions

Having established the stiffness, mass and damping of the pile in lateral direction based on minimization of the potential energy of the system, the above method can be extended for the piles with other boundary conditions for which there are no standard solutions available. 2.5.11.5

Partially embedded piles

In Arctic and North Siberian condition, due to environmental reasons, the steel piles are driven into the ground when they protrude about 2–3 m above the ground over which the pile cap and vibrating equipments are placed. Piling configuration has already been shown earlier while explaining the vertical vibration of pile. In such cases the existing solutions cannot be used. However, a solution of the same is proposed hereunder. Let L be the full length of the pile and the length of the embedment in soil be L1 (refer Figure 2.5.10). For this case, one may write 4 4 GSx1 L1 βe = (2.5.130) E p Ip Here subscript “e” represents embedment of the pile. The shape function can thus be represented by φ(z) = e

and

− βLe z

φ (z) =

1

2βe2 L21

βe z βe z cos + η sin L1 L1 e

− βLe z

and hence φ (z)2 = where,

1

sin

4βe4 L41

e

βe z βe z − ηe cos L1 L1

− 2βLe z 1

(2.5.131)

Xe 2βe z Ye 2βe z − ηe sin − cos 2 2 L1 L1

Xe = 1 + ηe2 ; Ye = 1 − ηe2 and ηe = 1 +

© 2009 Taylor & Francis Group, London, UK

(2.5.132)

1 . βe

(2.5.133)

172 Dynamics of Structure and Foundation: 2. Applications

Now, considering the fact that the embedment of a pile does not have any effect on the shape function of the system, the stiffness of the pile for the fundamental mode may be written as L L1 2 K = Ep Ip φi (z) dz + GSx1 φi (z)2 dz 0

(2.5.134)

0

% Considering, α = L L1 , Equation (2.5.134) may be rewritten as

K=

4Ep Ip βe4 L41

α L1 Xe Ye 2βe z 2βe z − 2βLe z 1 dz e − cos − ηe sin 2 2 L1 L1 0

L1 + GSx1

e

− 2βLe z

0

Xe 2βe z Ye 2βe z dz + ηe sin + cos 2 2 L1 L1

(2.5.135)

Equation (2.5.135) on integration by parts and after simplification, may be expressed K=

Ep Ip βe3 L31

1 1 α 1 −2βe α −2βe (1−α) Xe +α +Ye − +ηe − α −Xe e +1 e 4 8 2 4 4 (2.5.136)

this can be further simplified to

K=

Ep Ip

Xe

1 4

! + α + Ye 18 − α2 + ηe 14 − α − Xe e−2βe α4 e−2βe (1−α) + 1

L31

(ηe − 1)3

(2.5.137)

Equation (2.5.137) gives the solution for stiffness of a partially embedded pile in the ground. The correctness of the equation can be back checked by the fact that when the pile becomes fully embedded i.e. for L1 = L α → 1, βe = β, Xe = X etc., when Equation (2.5.137) degenerates to Equation (2.5.118). Proceeding in an identical manner as done before, the mass and damping terms may be computed as γp Ap L1 Mx = 4g

Xe α(1 − e−2βe ) + Y2e α + ηe α 1/(ηe − 1)

Cx = r0 ρGSx2 L1

Xe (1 − e−2βe ) + Y2e + ηe 4/(ηe − 1)

© 2009 Taylor & Francis Group, London, UK

(2.5.138) (2.5.139)

Analysis and design of machine foundations 173

2.5.11.6

Pile embedded in soils with varying elastic property

We present now the effect of variation of shear modulus with respect to depth. In the previous section, the calculation of stiffness as well as the damping of soil was based on constant dynamic shear modulus of the soil. For varying shear modulus, the variation with depth can be expressed as G = G(z/L)m

(2.5.140)

where m = a number varying from 0–2 [considered 0 when G is constant with depth, assumed 1 for linear variation and 2 for parabolic distribution]. For a linearly varying soil the stiffness matrix can be written as

K=

4Ep Ip β 4 L4

L

e−

2βz L

X Y 2βz 2βz − cos − η sin dz 2 2 L L

0

L z − 2βz X Y 2βz 2βz + GSx1 e L + cos + η sin dz L 2 2 L L

(2.5.141)

0

Integration of above and ignoring the terms containing the factor, βe−2β · cos 2β, β · e−2βsin 2β etc., having extremely small contributions, Equation (2.5.141) reduces to K=

Ep Ip β 3 GSx1 L η Y 3Y −2β −2β X(1−e X[1 − e −η + + )− (1 + β)] + 2 4 2 L3 4β 2 (2.5.142)

and can be further simpliﬁed to

K=

Ep Ip β 3 1 1 1 3 1 −2β X 1 − e 1 + + − Y − β − η 1 − 4 4β 2 16 8β L3 (2.5.143)

The damping matrix for this case, proceeding in same manner as outlined earlier, can be represented by η r0 ρG Sx2 L 3Y −2β X[1 − e + (1 + β)] + Cx = 4 2 4β 2

(2.5.144)

The mass coefficient remains the same as expressed in Equation (2.5.122). © 2009 Taylor & Francis Group, London, UK

174 Dynamics of Structure and Foundation: 2. Applications

When the dynamic shear modulus variation is parabolic with depth, the stiffness equation of the pile can be expressed as 4Ep Ip β 4 K= L4

L e

− 2βz L

X Y 2βz 2βz − cos − η sin dz 2 2 L L

0

L + GSx1

z 2 L

e−

2βz L

2βz 2βz X Y + cos + η sin dz 2 2 L L

(2.5.145)

0

Equation (2.5.145) on integration and on subsequent simplification reduces to Ep Ip β 3 Y −2β K= X(1 − e )− −η 2 L3 GSx1 L 1 1 2 −2β + 2 + X − e − 4β β 4β 2 β2

(2.5.146)

which can be further simplified to K=

Ep Ip β 3 1 Y 1 1 −2β 3 X 1 + − e − + − − η 2 4β 2 L3 16β 2 8β 2

(2.5.147)

Equation (2.5.147) gives the stiffness expression of pile under parabolic variation of G along the length of pile. Proceeding in same manner as stated above the damping matrix may be expressed as √ 1 r0 ρGSx2 L 2 1 −2β Cx = 2+ − 2 −e X 4β β 4β 2 β

(2.5.148)

The mass coefficient remains the same as expressed in Equation (2.5.122). 2.5.11.7

Computation of bending moment and shear force

For machine foundation subjected to a lateral load of P0 sin ωm t, the amplitude of vibration is given by v(t) =

P0 K

sin ωm t

(2.5.149)

(1 − r2 )2 + (2Dr)2

where, ωm = operating frequency of the machine; P0 = unbalanced dynamic force; r = ωm /ωn = the ratio of operating and natural frequency; D = damping ratio of the system. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 175

Thus the peak amplitude is given by v(t) =

P0 K 2 (1 − r )2 + (2Dr)2

(2.5.150)

The complete displacement function is then given by v(z, t) =

or

v(z, t) =

P0 K φ(z) (1 − r2 )2 + (2Dr)2

P0 βz K e− L 2 2 2 (1 − r ) + (2Dr)

(2.5.151)

βz βz cos + η sin L L

(2.5.152)

The bending moment is given by Ep Ip v = −M(z) = −

Ep Ip P0 K 2 (1 − r )2 + (2Dr)2

2β 2 − βz e L L2

sin

βz βz − η cos L L

(2.5.153) Mz =

E p I p P0 K (1 − r2 )2 + (2Dr)2

2β 2 − βz e L L2

sin

βz βz − η cos L L

(2.5.154)

The maximum moment will be at the head i.e. at z = 0, and it can be expressed as Mmax =

2Ep Ip P0 K

(1 − r2 )2 + (2Dr)2

β(β + 1) L2

(2.5.155)

The shear force is given by Ep Ip v = − V(z) =

V(z) = −

2.5.11.8

Ep Ip P0 K

(1 − r2 )2 + (2Dr)

Ep Ip P0 K

(1 − r2 )2 + (2Dr)

2β 3 βz βz − 1) sin + + 1) cos or (η (η 3 L L 2 L

2β 3 βz βz + (η + 1) cos (η − 1) sin 3 L L 2 L

(2.5.156)

Dynamic response of short piles in the horizontal mode

There are no solutions till date for this type of piles. Existing solutions are based on long piles with the implicit assumption that under ultimate load piles fail before the © 2009 Taylor & Francis Group, London, UK

176 Dynamics of Structure and Foundation: 2. Applications

soil. However there are number of areas (e.g. Bonny River Delta in Nigeria, where the topsoil constitute of very weak clay underlain by dense sand) where the soil will yield much before the pile. Broms (1965) has shown that the displacement curvatures for such piles are completely different than that of long piles. While a long pile embedded in soil behaves as a semi-infinite beam on elastic foundation, a short pile behaves as a beam of finite length on elastic foundation. Bojtsov et al. (1982) has given solution to the generic displacement curves of such short beams on elastic foundation that is given by x = C0 cos hpz cos pz + C1 cos hpz sin pz + C2 sin hpz sin pz + C3 sin hpz cos pz (2.5.157) where p is same as expressed in Equation (2.5.96). Expressing in terms of Puzrevsky function (Karnovsky and Lebed 2001), Equation (2.5.157) can be expressed as x = C0 V0 (pz) + C1 V1 (pz) + C2 V2 (pz) + C3 V3 (pz) where, V0 (pz) = cosh pz cos pz

(2.5.158) (2.5.159)

1 V1 (pz) = √ (cosh pz sin pz + sinh pz cos pz) 2

(2.5.160)

V2 (pz) = sinh pz sin pz

(2.5.161)

1 V3 (pz) = √ (cosh pz sin pz − sinh pz cos pz) 2

(2.5.162)

Puzrevsky’s functions, defined below, have some unique functional properties, which will be used for subsequent analysis for derivation of the stiffness, damping and mass of the piles. (2.5.163)

V1 (0) = 0;

V0 (0) = 0; V0 (0) = 0; V0 (0) = 0 √ V1 (0) = p 2, V1 (0) = 0; V1 (0) = 0

V2 (0) = 0;

V2 (0) = 0;

(2.5.165)

V0 (0) = 1;

V2 (0) = 2p2 ,

V2 (0) = 0 √ V3 (0) = 2 2p3

V3 (0) = 0; V3 (0) = 0; V3 (0) = 0; √ √ V3 (pz) = p 2V2 (pz); V2 (pz) = p 2V1 (pz) √ √ V1 (pz) = p 2V0 (pz); V0 (pz) = p 2V3 (pz)

(2.5.164)

(2.5.166) (2.5.167) (2.5.168)

For a solution of the short pile one may use the model shown in Figure 2.5.23. For the analysis (similar to long piles) the pile may be assumed as fixed at base and can undergo deflection and rotation at the pile head. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 177

M P X

Z dz Soil Stiffness=GSx1

Figure 2.5.23 Conceptual model of short pile under lateral load.

Considering base of pile at z = 0, shown in Figure 2.5.23, one may write At z = 0, At z = 0,

x = 0 ⇒ C0 = 0 x = 0 ⇒ C1 = 0

which gives, x = C2 V2 (pz) + C3 V3 (pz)

(2.5.169)

At the pile head, i.e. at z = L x = 1 yielding, C2 V2 (pL) + C3 V3 (pL) = 1

(2.5.170)

Again at z = L x = 1/L which gives, C2 V2 (pL) + C3 V3 (pL) = 1/L.

(2.5.171)

Using Equations (2.5.167) and (2.5.170), one may write C2 V1 (pL) + C3 V2 (pL) =

1 √ pL 2

(2.5.172)

The above may be expressed in matrix form as & ' [V] {C} = p

(2.5.173)

which can be further reduced to {C} = [V]−1 {p} © 2009 Taylor & Francis Group, London, UK

(2.5.174)

178 Dynamics of Structure and Foundation: 2. Applications

Performing the above operation gives

1 V2 (pL) −V3 (pL) 1√ C2 = C3 1/pL 2 −V1 (pL) V2 (pL)

(2.5.175)

where = V22 (pL) − V1 (pL)V3 (pL) which implies 1 V3 (pL) V2 (pL) − C2 = √ pL 2

and

1 C3 =

V2 (pL) √ − V1 (pL) pL 2

(2.5.176)

Thus, the displacement for the given boundary condition is then expressed as 1 V3 (pL) 1 V2 (pL) x= V2 (pL) − V2 (pz) + √ − V1 (pL) V3 (pz) (2.5.177) √ pL 2 pL 2 Based on above, the generic shape function in dimensionless form is given by βz βz 1 V2 (β) 1 V3 (β) V2 (β) − √ + φ(z) = V2 √ − V1 (β) V3 L L β 2 β 2 (2.5.178) where the determinant gets modified to = V22 (β) − V1 (β)V3 (β). Considering A = C2 / and B = C3 / the shape function can now be expressed as φ(z) = AV2

βz L

+ BV3

βz L

(2.5.179)

A typical shape function for the short piles Ep /Gs = 2500 is shown in Figure 2.5.24. 0.2

Shape Function

-0.2 -0.4 -0.6 -0.8 -1

z/L

Figure 2.5.24 Generic shape function of short pile for Ep /G = 2500.

© 2009 Taylor & Francis Group, London, UK

1

9 0.

8 0.

7 0.

6 0.

5 0.

4 0.

3 0.

2 0.

1 0.

0

0

Analysis and design of machine foundations 179

Differentiating Equation (2.5.179) and using properties mentioned earlier one could have φ (z) =

βz βz 2β 2 AV + BV 0 1 2 L L L

(2.5.180)

Substituting the above functions in Equation (2.5.110), the stiffness can be expressed as 4Ep Ip β 4 K= L4

L

AV 0

0

L + GSx1

AV 2

βz L

βz L

+ BV 1

+ BV 3

βz L

βz L

2 dz

2 dz

(2.5.181)

0

Equation (2.5.181) is too complicated to solve in closed form and a numerical quadrature scheme may be used to obtain K. Considering ξ = z/L we have L · dξ = dz and as z → L; ξ → 1; as z → 0 ξ → 0; which gives 1 L 4Ep Ip β 4 2 K= [AV0 (βξ ) + BV1 (βξ )] Ldξ + GSx1 [AV2 (βξ ) + BV3 (βξ )]2 Ldξ L4 0

0

(2.5.182) Substituting the value of β [Equation (2.5.102)] in Equation (2.5.182), the stiffness may be written as ⎡ K = GSx1 L ⎣4

1

1 2

[AV0 (βξ ) + BV1 (βξ )] dξ + 0

⎤ [AV2 (βξ ) + BV3 (βξ )] dξ ⎦ 2

0

(2.5.183) ➔ K = GSx1 L[4I1 + I2 ]

(2.5.184)

1

1 2

where I1 =

[AV0 (βξ ) + BV1 (βξ )] dξ 0

and

[AV2 (βξ ) + BV3 (βξ )]2 dξ

I2 = 0

(2.5.185) The integrals I1 and I2 can very easily be solved by using Simpson’s 1/3rd rule between limits 0–1 and can be back substituted in Equation (2.5.184) to compute the stiffness for the short pile. © 2009 Taylor & Francis Group, London, UK

180 Dynamics of Structure and Foundation: 2. Applications

However, one should note that there is no theoretical or experimental benchmarking against which the stiffness values can be checked or compared. So, use of the expression must always be backed up by dynamic field test of the piles to adjust the data (especially Sx1 or Ep /G) to match with the field observed values. In absence of comparative benchmarks the design may be initiated with the suggestive values of Sx1 for various Ep /Gs given in Table 2.5.18. The values mentioned in Table 2.5.18, are based on the formulation for long pile (with L/r < 25) but may be used as a starting point for the iteration based on field observed data. The mass of pile for the fundamental mode is given by γp Ap Mx = g

L φ(z)2 dz 0

or

γp Ap L Mx = g

1 [AV2 (βξ ) + BV3 (βξ )]2 dξ

(2.5.186)

0

Mx =

γp Ap L I2 g

(2.5.187)

To start the design a value of Sx1 is selected for speciﬁc Ep /Gs from Table 2.5.18 and find out the value of the frequency ( K/M) based on Equations (2.5.184) and (2.5.187). Let this be defined as ωc where the subscript c stands for the word “computed”. Let the field-tested natural frequency of the pile be ωf , where, ωf = ωc . In most of cases it has been seen (Jadi 1999) that the field observed frequency value deviates from the computed ones and usually varies by about 30–40%. This is logical, for when the pile is bored or driven the soil gets displaced and clayey soil may loose a part of its shear strength thus resulting in reduced dynamic shear modulus compared to the value observed during geo-technical investigation. There could be cases where the field observed values might be more than the computed ones, especially in sandy soil where the soil gets densified due to pile driving. The bottom line is that in rare cases the computed and observed values would match. Table 2.5.18 Suggested for Sx1 for short piles (L/r ≤ 20) for ﬁeld data iteration. Ep /Gs

Sx1 (ν = 0.25)

Sx1 (ν = 0.4)

Sx1 (ν = 0.5)

250 500 1000 2500 5000 10000

1.53 1.35 1.17 0.95 0.95 0.95

1.75 1.54 1.34 1.09 1.09 1.09

1.89 1.68 1.46 1.46 1.46 1.46

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 181

Based on the above argument the error (ε) in the analysis is then given by ε = ωc − ωf For ε → 0 we have ωc = ωf → ωc2 = ωf2 . Considering ωc2 =

K Mx

and using Equations (2.5.184) and (2.5.187), one can have

I1 GSx1 g 4 + 1 − ωf2 = 0 γp A p I2

(2.5.188)

It will be observed that all the factors β, I 1 , I2 in Equation (2.5.188) is a function of Ep /G. The difference (= the error ε) can now be set to zero or minimum by varying the value of Ep /G for which lim ε → 0. This can very easily be done by using the standard solver or goal-seek in a spreadsheet with boundary constraint that Sx1 > 0. The solver basically uses an algorithm called generalized reduced gradient technique (GRG2) used for constrained optimization (Lasdon et al. 1978). The procedure begins with the nonlinear optimization technique with equality constraints. The necessary slack and surplus variables are added as xs or x2s to any inequality constraints, and the problem is to Optimize: y(x) Subject to: fi (x) = 0: for i = 1, 2, . . . , j where j is the number of constrained equations and n is the number of independent variables where n > m. This is a very standard technique used in all nonlinear programming and is used routinely as a mathematical tool in many standard commercially available software like MS excel, MATLAB etc. having varied applications in engineering, science and economics modeling. Use of the above will automatically revise the value of Ep /G and upgrade the values of I2 and I1 (dimensionless but a function of Ep /G), which may then be used to calculate the revised and exact stiffness and mass contribution of the pile which would closely simulate the field condition. The steps are furnished in detail in Figure 2.5.27 as to how the data are updated and corrected for the example cited in example mentioned below. Having established the mass and stiffness coefficients of the pile correctly based on field data the damping may now be established as Cx = r0 ρGSx2 LI2

(2.5.189)

where I2 is the corrected upgraded value and Sx2 is as obtained from Table 2.5.17. 2.5.11.8.1

Comparison of results

A comparison of results against established methods to ensure that the method is not an utopian exercise with differential equations and it does have applications. © 2009 Taylor & Francis Group, London, UK

182 Dynamics of Structure and Foundation: 2. Applications

Stiffness (kN/m)

6.00E+05 5.00E+05 4.00E+05

Kxx

3.00E+05

Novak

2.00E+05

Gazetas

1.00E+05

25 0 50 0 10 00 25 00 50 00 10 00 0

0.00E+00 Ep/Gs

1.20E+06 1.00E+06 8.00E+05 6.00E+05 4.00E+05 2.00E+05 0.00E+00

Kxx Novak Gazetas

25

0 50 0 10 00 25 00 50 00 10 00 0

Stiffness (kN/m)

Figure 2.5.25 Comparison of stiffness values for, r = 0.3 m and length = 30 m.

Ep/Gs

Figure 2.5.26 Comparison of stiffness values for, r = 0.6 m and length = 30 m.

For this two RCC piles of radius 0.3 m, 0.6 m of length 30 m has been has been checked with the reported results for comparison. The values Kxx [Equation (2.5.106)] is shown in Figures 2.5.25 and 2.5.26 for comparison. Next, the results of uncoupled horizontal frequency of a real time compressor foundation weighing 400 kN supported on 9 RCC piles of length 36 m and diameter 1.8 m. The pile cap size is 7 m × 5 m × 2 m. The piles are spaced at distance of 3.0 m. The natural frequencies of the foundation are compared for Ep /G value varying from 250–10,000. Weight of the compressor is 400 kN. Table 2.5.19 clearly shows that the values are in very good agreement for the base case and thus can well be used for other cases as mentioned above for which there are no direct solutions. Finally, the stiffness of a short pile has been computed. This is based on the field observed data having the following properties: Length of pile = 10 m, diameter of pile = 1.2 m. Material of pile = RCC. • • • • • •

method of installation-bored pile. based on soil test, observed Ep /G = 5000. Ep considered = 3 × 107 kN/m2 . unit weight of pile material = 25 kN/m3 . field observed natural frequency of the pile is = 58 rad/sec (9 Hz). Poisson’s ratio of soil considered = 0.4.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 183

Table 2.5.19 Comparison of frequency for a compressor foundation proposed versus Novak and Gazzetas.

Ep/G

Frequency (rad/sec) with Kproposed

Frequency (rad/sec) with KNovak

Frequency (rad/sec) with KGazzetas

250 500 1000 2500 5000 10000

252.64 192.10 146.14 101.79 77.30 58.87

251.44 194.85 150.76 107.21 94.66∗∗ 63.98

252.57 192.07 146.07 101.71 77.35 58.82

∗∗ The

stiffness value was linearly interpolated from Novak (1983) Table for Ep/G = 500.

For the above conditions: Selected value of Sx1 from Table 6 = 1.09 Ep /Gs = 5000 (given), β = 2.1512 Equation (2.5.102) A = 0.50135; B = 0.02705 Equation (2.5.179) I1 = 0.23802, I2 = 0.9035 Equation (2.5.185) Computed natural frequency Kp /Mp = 68.26 rad/sec (11 Hz) ➔ Error (ε) = 10.26 Setting the error (ε) = 0 and running the solver function in a spread sheet for changing Ep /Gs for boundary constraint Sx1 > 0, the following upgraded data have been obtained: Sx1 = 1.09; Ep /Gs = 7246; β =1.96064; A = 0.65984; B = −0.04832; I1 = 0.27266 and I2 = 0.949504. Computed natural frequency based on above data = 58 rad/sec (9 Hz). Revised error (ε) = −2.79 × 10−7 . Thus based on the above data as per Equation (2.5.184), the correct stiffness of the pile is deduced as Kpile = 9.206 × 104 kN/m. It is to be noted here that the Ep /G value has increased from 5000 to 7246 meaning thereby that the soil had lost some of its initial strength due to boring of the pile-which is quite logical. 2.5.11.8.2

Computer run steps for short pile based on f ield observed data

The following section shows the computer run for evaluation of the stiffness of the pile in lateral direction in three steps. 1 2 3

Stiffness and frequency calculation of pile based on theoretical data and calculating the error based on field observed data. The data screen just prior to run of the solver with command to change Ep /G value keeping the Sx value > 0. Final value of the stiffness and frequency of the pile after solver has optimized the data.

© 2009 Taylor & Francis Group, London, UK

184 Dynamics of Structure and Foundation: 2. Applications

Step-1: Shows the initial calculation of frequency and stiffness of pile including the error with respect to field observed frequency.

Figure 2.5.27 Steps of computation.

Steps of calculations are given in Figure 2.5.27. A comprehensive analytical solution for dynamic analysis of long piles has been presented and is in good agreement with the existing solution. Based on this, piles with boundary conditions like partial embedment and soils with varying G can also be analyzed. Considering the fact that the dynamic bending moment and shear force can also be obtained by this method, the standard practice of restricting the pile capacity to 50% of its capacity will not be necessary. It will be observed from Equations (2.5.153) and (2.5.156) that the moment and shear takes care of the dynamic magnification factor of the load at the same time gives a complete distribution of its magnitudes along the depth of the pile. This when combined with static load would give the design moment for the pile. Considering that there is no uncertainty with this formulation, one can perhaps restricts the pile load limit to 80% of its capacity in lieu of 50% as in vogue presently and this would bring significant economy in design and for large project savings could quite significant. Short piles, for which no established method exists, also can be solved by the present method. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 185

Step-2: Showing solver on the verge of optimizing by changing Ep/G value by setting the error to zero.

Figure 2.5.27 (continued).

2.5.11.9

Dynamic analysis of piles under rocking or rotational mode

We present herein the mathematical model for rocking or rotational mode. This mode generally comes coupled with translational mode. Shown in Figure 2.5.21 is a pile embedded in ground considered in a homogeneous elastic medium under plane strain condition. The pile is considered to be long and slender. Under static loading, the equation of equilibrium in the x-direction for such beam on elastic foundation is given by Timoshenko (1956)

E p Ip

d4x = −ks x dz4

(2.5.190)

where Ep = Young’s modulus of the pile; Ip = moment of inertia of the pile cross section; ks = elastic stiffness of the soil and is expressed as GSθ1 ; Gs = dynamic shear modulus of the soil; Sθ1 = Berdugo’s rotational constant which are basically frequency dependent, © 2009 Taylor & Francis Group, London, UK

186 Dynamics of Structure and Foundation: 2. Applications

Step-3: Final value of stiffness of piles after the solver has optimized the error.

Figure 2.5.27 (continued).

The general solution of Equation (2.5.190) is given by x = e−qz (C0 cos qz + C1 sin qz) + eqz (C2 cos qz + C3 sin qz) GSθ1 where q = 4 Ep Ip

(2.5.191) (2.5.192)

Similar to lateral load case the deflection equation can be considered as x = e−qz (C0 cos qz + C1 sin qz)

(2.5.193)

Considering the pile head undergoing specified deflection and rotation as well as it’s head is fixed to the pile cap (same boundary condition as considered by Novak (1974)), we have At z = 0, let x = x0 ⇒ C0 = x0 , which gives x = e−qz (x0 cos qz + C1 sin qz) © 2009 Taylor & Francis Group, London, UK

(2.5.194)

Analysis and design of machine foundations 187

Again considering at z = 0,

dx = θ0 , we have dz

θ0 q

C1 = x0 +

(2.5.195)

Thus Equation (2.5.194) can now be represented as x=e

−qz

θ0 x0 cos qz + x0 + q

sin qz

(2.5.196)

Dividing each of the above term by L we have x = e−qz L

x0 cos qz + L

x0 θ0 + sin qz L qL

(2.5.197)

x0 x For magnitude of rotation being small θ0 ∼ and θz ∼ = = when we have L L θz = θ0 e

−qz

1 cos qz + 1 + sin qz qL

(2.5.197a)

Now considering β = qL and looking at Equation (2.5.197) we can say that for any arbitrary loading, the generic shape function in dimensionless form can be represented as ϕ(z) = e

−βz L

cos

βz 1 βz + 1+ sin L β L

(2.5.198)

where β=

4

GSθ 1 L4 ; Ep Ip

L = Length of the pile.

(2.5.199)

Equation (2.5.198) can thus be written as ϕ(z) = e

−βz L

βz βz cos + η sin L L

(2.5.200)

Thus it is observed that shape function for rotational mode remains invariant with respect to the lateral motion of pile for the given boundary condition. © 2009 Taylor & Francis Group, London, UK

188 Dynamics of Structure and Foundation: 2. Applications

1 0.5

F(z)

0. 9

0. 75

0. 6

0. 3

0. 45

-0.5

0. 15

0

0

Shape function

Generic Shape Function of Pile in Rotational Mode 1.5

z/L

Figure 2.5.28 Generic shape function of pile for Ep /G = 5000.

Differentiating Equation (2.5.200) we have ϕ (z) =

β −βz e L L

βz βz − (1 + η) sin L L

(η − 1) cos

1 β

when, η = 1 +

(2.5.201)

(2.5.202)

The generic shape function of the pile in fundamental mode as per Equation (2.5.200) is as shown in Figure 2.5.28 for Ep /G = 5000. The potential energy d of an element of depth dz, shown in Figure 2.5.25, under rotational mode is then given by (Craig 1981) Ep Ip d = 2

dθ dz

2 +

Kθ 2 θ 2

(2.5.203)

where, Ep = Young’s modulus of pile; Ip = moment of inertia of pile; Kθ = rotational stiffness of soil having dimension kN/m; θ = rotational displacement of pile in x direction and may be written as (z)q(t). For a rigid circular embedded footing with embedment Df , the stiffness of the footing in rotational mode may be expressed as Kθ =

Gb r30

Gs Df Cθ1 + G b r0

Sθ1 +

D2f 3r20

Sx1

(2.5.204)

where, Kθ = foundation stiffness in horizontal direction; Gs = dynamic shear modulus of the soil along foundation surface; Gb = dynamic shear modulus of the soil at foundation base; r0 = Radius of the foundation; Cθ1 and Sθ1 Sx1 = Beredugo’s Constant which are basically frequency dependent. Ignoring the first term within bracket in Equation (2.5.204) which represents the contribution of base resistance, and substituting the same in Equations (2.5.203) for a © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 189

cylindrical element of depth dz, embedded in soil, and also ignoring the term containing dz2 which is exceedingly small the potential energy d may be written as E p Ip d = 2

dθ dz

2 +

Gr20 Sθ1 dz 2 θ 2

(2.5.205)

the total potential energy over the whole length of the pile (L) is then given by Ep Ip = 2

L

dθ dz

2

Gr20 Sθ1 dz + 2

0

L θ 2 dz

(2.5.206)

0

Considering v(z, t) = (z)q(t), it can be proved that L Kij = Ep Ip

ϕi (z)ϕj (z)dz

L + Gr20 Sθ1

0

ϕi (z)ϕj (z)dz

(2.5.207)

0

where the shape function of the problem is given by Equation (2.5.200). Thus for fundamental mode the rotational stiffness of the pile is then given by L Kθ = Ep Ip

L 2

ϕ (z) dz 0

+ Gr20 Sθ1

ϕ(z)2 dz

(2.5.208)

0

where, (z) is as expressed in Equation (2.5.201). Substituting the value of φ(z) and φ (z) in Equation 2.5.208, and carrying out integration by parts and some simpliﬁcation, we ﬁnally get the rotational stiffness as ⎡ −2β ) + Y 1 + Ep Ip X (1 + ψ) (1 − e 2 ⎣ Kθ = L 2 (η − 1) 2

ψ 4

−η 1−

ψ 2

⎤ ⎦

(2.5.209)

Sθ 1 and λ = L/r0 the slenderness ratio of the pile. It is to be noted that where ψ = 4Gλ π Ep β 2 ψ is a dimensionless quantity, X, Y, η etc. are same as derived for lateral stiffness case. The accuracy of Equation (2.5.209) will be dependent on the correct selection of Sθ1 . For instance for rigid circular footing Novak and Beredugo (1972) has furnished a frequency independent value of Sθ1 = 2.5 (for any value Poisson’s ratio) which has been found to give adequate accuracy for practical engineering design. Comparing the stiffness data with Novak (1974) and Gazetas (1988) data it is proposed that the following values [Tables 2.5.20 to 22] of Sθ1 be used for the calculation of dynamic response of pile under rocking mode.

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190 Dynamics of Structure and Foundation: 2. Applications

Table 2.5.20 Suggested value of Sθ 1 for Poisson’s ratio of soil = 0.25. Poisson’s ratio 0.25

L/r0 (slenderness ratio)

Sθ1 (250)

Sθ1 (500)

Sθ1 (1000)

Sθ1 (2500)

Sθ1 (5000)

Sθ1 (10000)

25 40 60 80 100

16.968 17.358 17.567 17.674 17.736

23.089 23.656 23.961 24.110 24.199

30.776 31.586 32.016 32.225 32.348

43.412 44.678 45.333 45.648 45.833

54.647 56.390 57.272 57.688 57.930

66.877 69.253 70.418 70.958 71.267

Note: The value in Parenthesis after Sθ1 depicts the value of Ep /Gs value of the soil.

Table 2.5.21 Suggested value of Sθ 1 for Poisson’s ratio of soil = 0.40. Poisson’s ratio 0.40

L/r0 (slenderness ratio)

Sθ1 (250)

Sθ1 (500)

Sθ1 (1000)

Sθ1 (2500)

Sθ1 (5000)

Sθ1 (10000)

25 40 60 80 100

18.037 18.448 18.671 18.781 18.847

24.623 25.221 25.543 25.702 25.795

32.937 33.794 34.249 34.471 34.603

46.707 48.05 48.748 49.084 49.281

59.054 60.909 61.851 62.298 62.557

72.614 75.145 76.393 76.974 77.307

Note: The value in Parenthesis after Sθ1 depicts the value of Ep /Gs value of the soil.

Table 2.5.22 Suggested value of Sθ 1 for Poisson’s ratio of soil = 0.50. Poisson’s ratio 0.50

L/r0 (slenderness ratio)

Sθ 1 (250)

Sθ 1 (500)

Sθ 1 (1000)

Sθ1 (2500)

Sθ1 (5000)

Sθ1 (10000)

25 40 60 80 100

18.717 19.141 19.37 19.484 19.552

25.599 26.217 26.55 26.714 26.811

34.316 35.202 35.674 35.905 36.041

48.817 50.21 50.936 51.285 51.49

61.888 63.813 64.794 65.259 65.531

76.316 78.946 80.247 80.853 81.203

Note: The value in Parenthesis after Sθ1 depicts the value of Ep /Gs value of the soil.

For a particular pile having specific slenderness ratio and Poisson’s ratio of the soil we select the value of Sθ1 from the above table and on substitution of the same in Equation (2.5.199) and Equation (2.5.209) gives the solution of pile stiffness in rocking mode. 2.5.11.9.1

Estimation of mass contribution of pile

The mass matrix of the pile may be expressed as Mx = m(z) φi (z)φj (z)dz

© 2009 Taylor & Francis Group, London, UK

(2.5.210)

Analysis and design of machine foundations 191

For the present case of pile of length L, mass moment of inertia Jx is represented by

Mx Jx = L

L

r20 dz + z2 dz 4

(2.5.211)

0

Substituting Equation (2.5.210), we may now write γp Ap r20 Jx = 4g

L

γ p A p L2 ϕ(z) dz + g 2

0

L 2 z ϕ(z)2 dz L

(2.5.212)

0

where γp = weight density of the pile material; Ap = cross sectional area of pile; g = acceleration due to gravity. Equation (2.5.212) on integration by parts and simplification ﬁnally gives γp Ap r20 L Y Jx = XF(λ) + + η 16βg 2 where F(λ)

1−e

−2β

(2.5.213)

1 λ2 2 2 −2β 2+ − 2 + 2 − 4λ e β β β

and λ = L/r0 the slenderness ratio of the pile. Equation (2.5.213) gives the inertial contribution of pile in the fundamental mode. Incidentally the effect of this is usually ignored in design but could have significant effect if the number of piles is large in a pile group. 2.5.11.9.2

Radiation damping factor for pile under rocking mode

For a rigid footing embedded in soil for a depth Df , Novak and Beredugo (1972) has proposed an expression D2f D G f s Cθ = r40 ρG Cθ2 + Sθ2 + 2 Sx2 G r0 3r0

(2.5.214)

where, r0 = radius of the foundation; G = dynamic shear modulus at foundation base; Gs = dynamic shear modulus of soil in which the foundation is embedded; Df = depth of embedment; Cθ2 , Sθ2 and Sx2 = frequency independent constants as defined by Novak and Beredugo (1972). Ignoring the first term in Equation (2.5.214) which represents the contribution of base damping for a cylindrical element of depth dz, embedded in soil, and ignoring © 2009 Taylor & Francis Group, London, UK

192 Dynamics of Structure and Foundation: 2. Applications

the term, containing dz2 which is again exceedingly small, we have c(θ ) = r30 ρGs Sθ2 dz

(2.5.215)

For systems having continuous function, the damping is usually expressed as Cθ = c(θ ) φi (z)φj (z)dz

(2.5.216)

For the present case of pile of length L, Equation (2.5.216) can be expressed as

Cθ =

r30

L φ(z)2 dz

ρGs Sθ2

(2.5.217)

0

Equation (2.5.217) on integration by parts and simplification, we have Cθ =

r30

X(1 − e−2β ) + ρGSθ2 L 4β

Y 2

+η

(2.5.218)

Equation (2.5.218) expresses the soil damping for a single pile under horizontal mode of vibration. Here the Factor Sθ2 is damping coefﬁcient which is frequency dependent. Fortunately the damping factor is required for calculation of the amplitude when the eigen solution of the problem is already done vis a vis, the dimensionless frequency number a0 = ωr0 /vs term is known. Polynomial fit curve for Sθ2 are available in terms of a0 which can be used directly to arrive at these parameters. The value of Sθ2 is as given hereafter as per Novak and Beredugo (1972) Sθ 2 = 0.0144a0 + 5.263a20 − 4.177a30 + 1.643a40 − 0.2542a50

(2.5.218a)

This value unlike other Beredugo’s constant is independent of Poisson’s ratio. 2.5.11.9.3

Consideration of material damping of pile

The structural stiffness contribution of the pile is given in the first part of Equation (2.5.208), while that of the mass moment of inertia is given in Equation (2.5.212). Thus, if Cc is the critical damping of the pile then it can be expressed as Cc = 2 KJx , where K and Jx are the stiffness and mass moment of inertia of the pile. Depending on the material used for pile like (RCC, steel etc.) a suitable damping ratio (D) can be assumed. The damping (Cp ) for the pile can be expressed as Cp = D Cc

(2.5.219)

This, when added to the radiation damping, calculated in Equation (2.5.218) gives the complete damping quantity for the soil-pile system. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 193

2.5.11.9.4

Piles with other boundary conditions

Having established the stiffness, inertial and damping contribution of the pile in rocking mode based on minimization of the potential energy of the system we extend the above method for piles with other boundary conditions for which there are no standard solutions.

2.5.12 Partially embedded piles under rocking mode As stated earlier, this is a very common practice in Arctic and North Siberian condition, where due to environmental reasons; the steel piles are driven into the ground when they protrude about 2–3 m above the ground over which the pile cap and vibrating equipments are placed. In such cases Novak’s (1974, 1983) chart cannot be used, nor is Gazetas’ formulation valid. We provide the solution of the same as hereafter. Let L be the full length of the pile and let the length of the embedment is soil be L1 . For this case we have βe =

4

GSθ 1 L41 E p Ip

(2.5.220)

Here subscript “e” represents embedment of the pile. The shape function can thus be represented by φ(z) = e

− βLe z 1

βe z βe z cos + η sin L1 L1

(2.5.221)

The stiffness function can thus be represented as ϕ (z) =

βe −βL e z e 1 L1

(ηe − 1) cos

βe z βe z − (1 + ηe ) sin L1 L1

(2.5.222)

Square of the above is given by φ (z)2 =

βe2

e− 2

L1

2βe z L

Xe 2βe z 2βe z + Ye sin − 2ηe cos 2 L1 L1

(2.5.223)

Here X = 1 + ηe2 ; Y = 1 − ηe2 and ηe = 1 + β1e . Now considering the fact that embedment of a beam does not have any effect on the shape function of the system, the stiffness of the pile is expressed as L L1 2 Kθ = Ep Ip φi (z) dz + GSθ 1 [φi (z)]2 dz 0

© 2009 Taylor & Francis Group, London, UK

0

(2.5.224)

194 Dynamics of Structure and Foundation: 2. Applications

Equation (2.5.224) on integration by parts and simplification may be expressed as Ep Ip βe Kθ = Xe (α + ψ) − Xe {αe−2βe α + ψe−2βe } 2L1 α ψ ψ + Ye + − ηe α − 2 4 2

(2.5.225)

which can further be expressed as (here α = L/L1 ) −2βe α − ψe−2βe ) + Y e Ep Ip Xe (α + ψ − αe Kθ = L1 2 (ηe − 1)

α 2

+

ψ 4

− ηe α −

ψ 2

!

(2.5.226) Equation (2.5.226) gives the solution for stiffness of partially embedded piles in the ground. The correctness of the equation can be back checked by the fact that when the pile becomes fully embedded i.e. L1 = L we have α → 1, βe = β, Xe = X etc. when Equation (2.5.226) degenerates to Equation (2.5.209), the stiffness for fully embedded pile. Proceeding in identical manner as done before, the mass and damping terms can be obtained as given earlier. The mass moment of inertia of pile remains same as stated in Equation (2.5.213). The damping matrix is given by the expression Cθ = 2.5.12.1

r30

ρGSθ2 L1

Xe (1 − e−2βe ) + Y2e + ηe 4/(ηe − 1)

(2.5.227)

Stiffness of the pile for soils with varying elastic property

Considering the variation of shear modulus with depth as G = G(z/L)m

(2.5.228)

where m = a number varying from 0–2 [considered 0 when G is constant with depth, assumed 1 for linear variation and 2 for parabolic distribution] we derive the pile stiffness and other parameters as hereafter. Thus for linearly varying soil the stiffness matrix can be written as Ep Ip β 2 Kθ = L2

L

e−

2βz L

X − 2η cos

2βz 2βz − Y sin dz L L

0

+ Gr20 Sθ 1

L 2βz 2βz z − 2βz X Y e L + cos + η sin dz L 2 2 L L 0

© 2009 Taylor & Francis Group, London, UK

(2.5.229)

Analysis and design of machine foundations 195

Equation (2.5.229) can be further simplified to Kθ =

Ep Ip β ψ ψ X 1+ − e−2β 1 + (1 + β) 2L 2β 2β 1 ψ 3ψ +Y + −η 1− 8β 2 4β

(2.5.230)

The damping matrix for this case can thus be represented by √ r30 ρGSθ2 L 3Y η −2β Cx = X[1 − e (1 + β)] + + 4 2 4β 2

(2.5.231)

The mass coefficient remains same as expressed in Equation (2.5.213). When the dynamic shear modulus variation is parabolic with depth the stiffness equation of the pile is expressed as Ep Ip β 2 Kθ = L2

L 2βz 2βz − 2βz L X − 2η cos e + Y sin dz L L 0

+ Gr20 Sθ 1

L 2 z X Y 2βz 2βz − 2βz L e + cos + η sin dz L 2 2 L L

(2.5.232)

0

which can be further simplified and expressed as Kθ =

Ep Ip β ψ Y ψ 1 2 −2β − e 1 + + X 1+ 2 + − − η 2L 2 β 2 8β 2 β2 (2.5.233)

Equation (2.5.233) gives the stiffness expression of pile under parabolic variation of G along the length of pile. Proceeding in same manner as stated above the damping matrix is expressed as √ r30 ρGSθ2 L 1 1 2 −2β Cθ = 2+ − 2 X −e 4β β 4β 2 β

(2.5.234)

The mass coefficient remains same as expressed in Equation (2.5.213). 2.5.12.1.1

Calculation of dynamic bending moment and shear force in pile

Neither Novak nor Gazetas’ method can be used for this purpose. For machine foundation subjected to a dynamic moment of M0 sin ωm t, the amplitude of vibration is © 2009 Taylor & Francis Group, London, UK

196 Dynamics of Structure and Foundation: 2. Applications

given by θ (t) =

(M0 /Kθ ) sin ωm t

(2.5.235)

(1 − r2 )2 + (2Dr)2

where, ωm = operating frequency of the machine; M0 = unbalanced dynamic moment; ρ = ωm /ωn, the ratio of operating and natural frequency; D = damping ratio of the system. Thus the peak amplitude is given by θ (t) =

(M0 /Kθ )

(2.5.236)

(1 − r2 )2 + (2Dr)2

The complete displacement function is then given by θ (z, t) =

(M0 /Kθ )

βz

(1 − r2 )2 + (2Dr)2

e− L

cos

βz βz + η sin L L

(2.5.237)

Thus bending moment is given by Ep Ip θ = −M(x) or,

(Ep Ip M0 )/Kθ

β − βz M(x) = e L (1 − r2 )2 + (2Dr)2 L

βz βz (1 + η) sin − (η − 1) cos L L

(2.5.238)

The dynamic shear force is given by (Ep Ip M0 )/Kθ

2β 2 − βz Ep Ip θ = −V(z) = − e L (1 − r2 )2 + (2Dr)2 L2

βz βz sin − η cos L L

(2.5.239)

➔ 2.5.12.2

(Ep Ip M0 )/Kθ

2β 2 − βz V(z) = e L (1 − r2 )2 + (2Dr)2 L2

βz βz sin − η cos . L L

(2.5.240)

Dynamic response of short piles under rotational mode

As mentioned earlier no solution exists till date for this type of piles. Bojtsov (1982) has given solution to the generic displacement curvature of such short beams on elastic foundation which is given by x = C0 cos h pz cos pz + C1 cos h pz sin pz + C2 sin h pz sin pz + C3 sin h pz cos pz (2.5.241) where p = q is same as expressed in Equation (2.5.192). © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 197

Expressing the above in terms of Puzrevsky function (Karnovsky 2001) Equation (2.5.257) can be expressed as x = C0 V0 (pz) + C1 V1 (pz) + C2 V2 (pz) + C3 V3 (pz)

(2.5.242)

where, V0 (pz) = cos h pz cos pz

(2.5.243)

1 V1 (pz) = √ (cos h pz sin pz + sin h pz cos pz) 2

(2.5.244)

V2 (pz) = sin h pz sin pz

(2.5.245)

1 V3 (pz) = √ (cos h pz sin pz − sin h pz cos pz) 2

(2.5.246)

For analysis similar to previous case we assume the pile as fixed at base and is fixed also at pile cap level and can undergo deflection and rotation at pile head. Considering base of pile as z = 0 and applying the Puzrevsky’s functional properties as elaborated in case of piles under lateral load we have At z = 0, x = 0 ⇒ C0 = 0 At z = 0, x = 0 ⇒ C1 = 0 which gives x = C2 V2 (pz) + C3 V3 (pz)

(2.5.247)

At the pile head we have at z = L x = 1 which gives C2 V2 ( pL) + C3 V3 ( pL) = 1

(2.5.248)

Again at z = L x = 1/L which gives C2 V2 (pL) + C3 V3 (pL) = 1/L

(2.5.249)

Using the derivative properties as shown above we have C2 V1 (pL) + C3 V2 (pL) =

1 √ pL 2

(2.5.250)

Expressing the above in matrix form we have [V] {C} = {p}

(2.5.251)

which can be further expressed as {C} = [V]−1 {p} © 2009 Taylor & Francis Group, London, UK

(2.5.252)

198 Dynamics of Structure and Foundation: 2. Applications

Performing the above operation gives

⎧ 1 V2 (pL) −V3 (pL) ⎨ C2 = C3 −V1 (pL) V2 (pL) ⎩

⎫ 1 ⎬ 1 √ ⎭ pL 2

(2.5.253)

where = V22 (pL) − V1 (pL)V3 (pL) which gives 1 V3 (pL) V2 (pL) − C2 = √ pL 2

and

1 C3 =

V2 (pL) √ − V1 (pL) pL 2

(2.5.254)

Thus the displacement for the given boundary condition is then expressed as 1 V3 (pL) 1 V2 (pL) x= V2 (pL) − V2 (pz) + √ √ − V1 (pL) V3 (pz) (2.5.255) pL 2 pL 2 Considering the fact that for long piles the shape function remains invariant for rocking mode with respect to lateral motion, for same boundary condition it may be concluded that for short piles also the same condition would hold good thus the generic shape function in dimensionless form in rocking mode is given by 1 βz βz V3 (β) 1 V2 (β) φ(z) = V2 (β) − √ + V2 √ − V1 (β) V3 L L β 2 β 2 (2.5.256) where the determinant gets modified to = V22 (β) − V1 (β)V3 (β). Considering A = C2 / and B = C3 / the shape can now be expressed as φ(z) = AV2

βz L

+ BV3

βz L

(2.5.257)

Typical generic shape function for the short piles Ep /Gs = 2500 is as shown in Figure 2.5.29. Differentiation of above and using the differential properties as mentioned earlier we have √ βz βz β 2 φ (z) = AV1 + BV2 L L L

© 2009 Taylor & Francis Group, London, UK

(2.5.258)

Analysis and design of machine foundations 199

0.2

1

0. 9

0. 8

0. 7

0. 6

0. 5

0. 4

0. 3

-0.2

0. 2

Shape Function

0

0. 1

0

-0.4 -0.6 -0.8 -1

z/L

Figure 2.5.29 Generic shape function of short pile for Ep /G = 2500.

Substituting the above functions, we have L

2Ep Ip β 2 K= L2

AV1

βz L

+ BV2

βz L

2

0

L + Gr20 Sθ1

AV2

βz L

+ BV3

βz L

2 (2.5.259)

0

The above is too complicated to solve in closed form as such numerical integration may be used to arrive at the stiffness value. Considering ξ = Lz we have L · dξ = dz and as z → L; ξ → 1; as z → 0 ξ → 0; which gives 1

2Ep Ip β 2 K= L2

[AV1 (βξ ) + BV2 (βξ )]2 Ldξ 0

L + Gr20 Sθ1

[AV2 (βξ ) + BV3 (βξ )]2 Ldξ

(2.5.260)

0

Substituting the value of (from Equation 2.5.209) in Equation (2.5.260), we have ⎡

⎤ 1 1 2 K = Gr20 Sθ 1 L ⎣ [AV1 (βξ ) + BV2 (βξ )]2 dξ + [AV2 (βξ ) + BV3 (βξ )]2 dξ ⎦ ψ 0

K = Gr20 Sθ 1 L

2 I1 + I 2 ψ

© 2009 Taylor & Francis Group, London, UK

0

(2.5.261) (2.5.262)

200 Dynamics of Structure and Foundation: 2. Applications Table 2.5.23 Suggested for Sθ 1 for short piles (L/r ≤ 20) for ﬁeld data iteration. Ep /Gs

Sθ 1 (ν = 0.25)

Sθ1 (ν = 0.4)

Sθ1 (ν = 0.5)

250 500 1000 2500 5000 10000

15.563 21.046 27.873 39.05 49.07 60.187

16.561 22.468 29.860 42.041 53.014 65.311

17.197 23.372 31.135 43.976 55.576 68.598

Here 1 I1 = [AV1 (βξ ) + BV2 (βξ )]2 dξ

and

0

1 I2 = [AV2 (βξ ) + BV3 (βξ )]2 dξ 0

(2.5.263) The integrals I1 and I2 can very easily be solved by using Simpson’s 1/3rd rule between limits 0–1 and can be back substituted in Equation (2.5.261) to arrive at the stiffness for the short pile. As there is no theoretical or experimental benchmarking against which the stiffness values can be checked or compared. So use of this expression must always be backed up by dynamic field test of the piles to adjust the data (especially Sθ1 or Ep /G) to match the field observed value. In the absence of comparative benchmarks we may start the design with the following suggestive values of Sθ1 for various Ep /Gs values given in Table 2.5.23. These values as mentioned above, is based on formulation for long pile (with L/r < 25) but may be used as a starting point for the iteration based on field observed data. The mass moment of inertia of the pile for fundamental mode is given by γp Ap r20 Jx = 4g

L L γ p A p L 2 z 2 2 ϕ(z) dz + ϕ(z)2 dz g L 0

Jx =

→

Here

γp Ap r20 L

(2.5.264)

0

I1 +

γ p Ap g

L3

I3

(2.5.265)

1 I3 = ξ 2 [AV2 (ξ ) + BV3 (ξ )]dξ

(2.5.266)

4g

0

or

Jx =

where,

Mp r20 I 1 + M p L 2 I3 4 Mp =

γp Ap L g

© 2009 Taylor & Francis Group, London, UK

➔ Jx =

(2.5.267) Mp r20 [I1 + 4λ2 I2 ] 4

(2.5.268)

Analysis and design of machine foundations 201

To start the design we select a value of Sθ1 for a specified Ep /Gs from Table 2.5.23 and find out the value of the frequency based on Equation (2.5.262) and (2.5.267). Let this be defined as ωc where the subscript c stands for the word “computed”. Let the field tested natural frequency of the pile be ωf where ωf = ωc . Based on the above argument the error(ε) in the analysis is then given by ε = ωc − ωf For

ε → 0 we have, ωc = ωf or ωc2 = ωf2

Considering

ωc2

K = , Mx

4GSθ1 L we have Mp

2 ψ I1

+ I2

I1 + 4λ2 I2

− ωf2 = 0

(2.5.269)

It will be observed that all the factors β, I1 , I2 in Equation (2.5.269) is a function of Ep /Gs . The difference (which is the error ε) can now be set to zero or minimum by varying the value of Ep /Gs for which, lim ε →0. This can very easily be done by using the standard solver or goal seek in a spread sheet with boundary constraint that Sθ1 > 0. The above will automatically revise the value of E/Gs and upgrade the values of I3 , I2 and I1 (which are dimensionless functions), which may then be used to calculate the revised and exact stiffness and mass contribution of the pile which would closely simulate the field condition. Having established the mass and stiffness coefficients of the pile correctly based on field data the damping may now be established as Cθ = r30 ρGSθ2 LI2

(2.5.270)

2.5.13 Group effect of pile Refer Section 2.5.7 where this has been dealt in detail and may well be used for this case too. 2.5.13.1

Effect of pile cap on pile stiffness

The sketch given in Figure 2.5.30 represent the pile group with pile cap. In such case usually the embedment stiffness GSf Df is added to the pile group stiffness and the system is considered as a lumped mass single degree freedom system, the details of which are furnished in Novak (1974) and Prakash and Puri (1988). In conventional formulation as the stiffness matrix is statically coupled another set of stiffness Kxθ needs to be derived in addition to what has been derived above. To circumvent this issue we propose to use the following model as shown in Figure 2.5.31. © 2009 Taylor & Francis Group, London, UK

202 Dynamics of Structure and Foundation: 2. Applications

To derive the equations we use the Lagrange’s equation from the energy principle as derived earlier when we finally get the stiffness and mass matrix as ⎡

Mf ⎣ Mf Mf Zc

Mf Mf + M x Mf Zc

⎤⎧ ⎫ ⎡ M f Zc Kf ⎨x¨⎬ Mf Zc ⎦ u¨ + ⎣ 0 ⎩⎭ 0 Jx + Mf Zc2 θ¨

⎤⎧ ⎫ 0 ⎨x⎬ 0⎦ u = 0 ⎩⎭ Kθ θ

0 Kx 0

(2.5.271)

The above gives the complete free vibration equation of motion for pile plus pile cap with machine considering pile springs in translation and rocking mode. Considering the equation to be dynamically coupled the damping matrix can now be expressed as ⎡

Cf ⎣ [C] = 0 0

0 Cx 0

⎤ 0 0⎦ Cθ

(2.5.272)

Df

Zc

Figure 2.5.30 Schematic diagram of pile and pile-cap with embedment. x M = Mass of (Pile cap + Machine) Zc

Kf = Embedded stiffness of soil @ GSfDf mp = Mass of pile group K

Kx

u

Jx = Moment of inertia of Pile group

Figure 2.5.31 Mathematical model of pile group and pile cap under coupled sliding and rocking mode.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 203

where, Mf = mass of pile cap plus mass of machine; Mx = mass of pile group; Jx = mass moment of inertia of piles; Zc = center of mass of foundation plus machine along vertical axes; Kf = lateral embedded stiffness of pile cap @ G Sfx Df ; G = dynamic shear modulus of soil; Sfx = Berdugo’s constant @ 3.6, 4, 4.1 for ν = 0.0, 0.25, 0.4 respectively; Df = depth of embedment; Kx = Group lateral stiffness of pile group based on Equation (2.5.118) where Kθ = rotational stiffness of pile group; U = potential energy of the system, and T = kinetic energy of the system. It is to be noted that for pile group for calculation of mass and mass moment of inertia the mass and inertia of single pile has to be multiplied by the number of piles in the group. While for stiffness and damping the group stiffness and damping has to be derived according to Equation in section 2.5.7.

2.5.14 Comparison of results The method proposed herein is now compared withy Novak and Gazetas’ values to check their accuracy. For this two RCC piles of radius 0.4 m, 1.0 m of length 40 m has been has been checked with the reported results for comparison. The values Kθ [Equation (2.5.209)] is shown in Figures 2.5.32 and 33 for comparison. The results clearly shows that the values are in very good agreement for the base case and thus can well be used for other cases as mentioned above for which there are no direct solution. We finally calculate the stiffness of a short pile based on field observed data having the following properties. Length of pile = 10 m, Diameter of pile = 1.2 meter. Material of pile RCC. • • • • • •

Method of installation-bored pile. Based on soil test, observed Ep /G =5000. Ep considered @ 3 × 107 kN/m2 . Density of pile material = 25 kN/m3 . Field observed natural frequency of the pile is = 28 rad/sec (4 Hz). Poisson’s ratio of soil considered = 0.4. For the above conditions

Selected value of Sθ1 from Table 2.5.23 = 53.014. Ep /Gs = 5000 (given); β = 5.681 : vide Equation (2.5.198); A = −0.000912 : vide Equation (2.5.271); B = −0.003447 do; I1 = 0.0277902, I2 = 0.201259,I3 = 0.16886 : vide Equation (2.5.263). 5 Computed Stiffness = 7.78 × 10 kN/m. Computed natural frequency ( Kθ /Jx ) = 39.98 rad/sec (6 Hz). ➔ Error (ε) = 11.98 Setting the error (ε) = 0 and running the goal seek function in a spread sheet for changing Ep /Gs for boundary constraint Sθ1 > 0, we have the following upgraded data:

© 2009 Taylor & Francis Group, London, UK

204 Dynamics of Structure and Foundation: 2. Applications

Stiffness (kN/m)

1.00E+06 8.00E+05 Kxx Novak

6.00E+05 4.00E+05

Gazetas

2.00E+05

Ep/Gs

10 00 0

50 00

25 00

10 00

50 0

25 0

0.00E+00

0 00

00

Kxx Novak Gazetas

10

Ep/Gs

50

00 25

00 10

50

25

0

Comparison of Rocking Stiffness for piles 2.00E+06 1.50E+06 1.00E+06 5.00E+05 0.00E+00

0

Stiffness (kN/m)

Figure 2.5.32 Comparison of stiffness values for r = 0.4 m and length = 40 m.

Figure 2.5.33 Comparison of stiffness values for r = 1.0 m and length = 40 m.

Sx1 = 53.014; Ep /Gs = 3354; β = 6.2777; A = −0.000889; B = −0.00528; I1 = 0.00111, I2 = 0.138856, I3 = 0.11672. Computed natural frequency based on above data = 28 rad/sec (4 Hz). ➔ Revised error (ε) = −0.00092 Thus based on the above data as per Equation (2.5.261) the correct stiffness of the pile is given by Kpile = 2.64 × 105 kN/m. In case the above correction is already done for lateral pile stiffness and E/G value has been already modified to suite the field observed data, the same can directly be used without carrying out the above mentioned modification again. Referring to Figures 2.5.32 and 2.5.33, it is observed that the results are in excellent agreement with both Novak (1983) and Gazetas (1988) stiffness. Considering the base case being in such agreement formulations for other cases like partial embedment, varying shear modulus etc., can now be very easily adapted for which there are no standard solutions. The short pile case is basically a theoretical solution and needs significant field test and lab testing to arrive at a predefined Sθ1 values which would make the method more powerful. However in absence of such data the present algorithm as mentioned herein could become a very powerful tool for dynamic analysis of such piles for which no solution is available till date and yet remains a serious practical problem. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 205

Y

B X

Z

X

L Y Foundation resting on ground

Foundation on piles/springs

Figure 2.5.34

2.5.15 Practical aspects of design of machine foundations Enough gazing at the moon and theoretical contemplation, from the hallowed domain of academics let us now digress into the real world of a professional engineer and see based on the above theories how he goes about to design the foundation • •

What are the input data he looks for? What are the assumptions he considers in his process of design. There are three aspects to be considered at the start of the design

1 Environmental and economic impact. 2 Machine data. 3 Soil data. 2.5.15.1

Environmental and economic impact

This is the first point that a designer should assess, but unfortunately the effect of environmental impact on the machine foundation is often overlooked. There could be a situation, where other than the vibration of the machine itself there are external source of vibration affecting the foundation and this could be in the form of • • •

Blasting in the vicinity of the foundation Pile driving Waves transmitted by other machines operating in the vicinity of the machine foundation in question.

Our experience shows that young engineers while doing their design of machine foundations are more focussed on the quantitative magnitude of the natural frequency and the amplitude and often overlooks this point. © 2009 Taylor & Francis Group, London, UK

206 Dynamics of Structure and Foundation: 2. Applications

To asses the effect of environmental impact on the foundation, if need be, seek help of a vibration specialist and try to assess what could be the cascading effect of this secondary source of disturbance. If it is felt that this may possibly have some effect on the foundation isolate the foundation by providing pockets/cut outs all-round the foundation and leaving this space void or feeling it up with suitable dampers like cork boards, felt sheets etc. Next try to assess how important role does the machine play in the overall process system. In other words, “What would be the economic impact of the machine on the overall process vis-à-vis its performance”? For instance if a minor chemical pump stops during an engineering process the overall cost impact on the process could vary from a few hundreds of dollars to thousand dollar. While for a major generator or a compressor foundation if the performance is not up to the acceptable standard the client could stand to loose millions of dollars in terms of production output and man-hours lost. If required talk to your process engineering or mechanical engineering colleagues to asses the criticality of the machine. More important is the machine be more conservative in your design approach. Do not try to economise on the material. The money that could be saved by cutting down on a few cubic meter of concrete or hundred Kilogram of reinforcement, could be well be offset by manifolds if your company stands to pay liquidated damages due to malfunctioning of the foundation15 . For machine foundations economy lies more on the smooth performance of the machine rather than any other factors. 2.5.15.2 •

Machine data

Once you have assessed the above aspect, as a next step, you should have the machine data at your disposal in the form of a General Arrangement Drawing of the machine. On study of the drawing see if the following check list is satisfied as a minimum 1 Do the drawing furnishes the overall dimension of the machine/skid on which it is mounted? 2 Are the anchor bolt locations, size of the bolts (both diameter and length) and details of how it should be anchored to the foundation furnished by the vendor? 3 Do the drawing supply the height at which the centre line of the shaft of the machine is located from the bottom of the machine frame (which will be the top of concrete or top of grout for you)? 4 Is it clear to you what type of machine it is i.e. if it is centrifugal or reciprocating in nature?

15 And this we are sure will not have a very positive outcome on your annual performance appraisal. . . . . .

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 207

5 Does the drawing supply you with the operating speed of the machine or the range which should be cleared during the design of the machine foundation? 6 Do the foundations need to support any pipes or valves on it other than the machine itself? 7 If so, are all the loads and locations of these valves and pipes are mentioned in the drawing? 8 Does the drawing clearly mention the unbalanced mass, eccentricity or the dynamic loads generated during the operation of the machine including any speciﬁc direction? 9 Is it clear to you what would be the level of the top of concrete of the foundation? This is very important for the top of foundation usually fixed from the process engineering group and if there is any mismatch in the level in the field could create problems in terms of alignment of pipe flanges or variation in the net positive suction head (NPSH) for the pump. 10 Is the location of the equipment in terms of co-ordinates with respect to the overall plant available with you? 11 Finally has the equipment supplier defined any performance criterion which needs to be met in terms of amplitude, frequency etc. The above are very vital points both from performance and contractual point of view. For if the equipment supplier has furnished this information then it should be strictly adhered to, for once this is complied with the supplier alone stands guarantee for the performance of the machine. On the contrary if this is violated, even if the equipment supplied is faulty, the vendor can always wriggle out of the situation by saying that his specifications were violated and as such he cannot stand guarantee for the performance of the machine16 . If the vendor has not specified such conditions the usual de-fault is the local code stipulation. But do not presume this, ask him specifically to define his performance criteria and if he is unable to do so, make it clear to him (in writing) as to what performance criterion you are using based on which code (could be IS, DIN, BS, ASTM special publications etc). If possible seek his written compliance that the code-norms that is being followed by you is acceptable to him. Remember for important machines you are fiddling with millions of dollars so play safe. Guard yourself both technically as well as contractually.

2.5.15.3

My Machine is perfectly balanced. . . you don’t need to worry about the dynamic force!

A standard sales talk you will hear time and again from the equipment sales engineer. Rookie engineers often get carried away by this and fall prey to this over sales strategy.

16 Refer to case history 2 at the outset of this chapter and retrospect a bit.

© 2009 Taylor & Francis Group, London, UK

208 Dynamics of Structure and Foundation: 2. Applications

Many equipment suppliers do not supply any unbalanced dynamic load claiming their machines to be perfectly balanced! This often leaves an inexperienced engineer with the option of doing only a resonance check and leaves it at that for he has no other data as a guideline to perform any further check. What should be realised at this point is that it is possible perhaps to achieve a perfect balance in the manufacturing unit under a controlled condition at the outset. But when such machines are performing under a much gruelling conditions of operating day in day out and often left exposed to the vagaries of nature, due to normal wear and tear some imbalance will invariably be generated in the system which will induce dynamic loads on the foundation. So do not get carried away by the claims of the vendor, for you as designer alone remain responsible for the performance of the foundation. In absence of such data from the vendor you may use the following guidelines (Arya et al. 1979). 2.5.15.3.1

Design eccentricities of centrifugal machines

Ecentricities of machines under varying speeds are given in the table below.

Sl. No.

Operating speed

Eccentricity in double amplitude(inch)

1 2 3

750 1500 3000

0.014–0.032 0.008 0.002

Here unbalanced dynamic force for centrifugal machine is given by 2 Fdyn = meωm

where, m = mass of the rotating shaft; e = eccentricity developed in the shaft, and ωm = operating speed of the machine. 2.5.15.3.2 For Centrifugal compressors e(mil) = α

12,000 ≤ 1.0(mil) r.p.m.

where, α = 0.5 at installation time = 1.0 after several years of operation r.p.m. = Operating frequency of the machine. 1 mil = 0.001 inch. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 209

2.5.15.3.3

For electrical motors

Sl. No.

Motor type

Speed (RPM)

Peak to peak displacement amplitude (inch)

1

Integral horsepower electric motor

3000–4000 1500–2999 1000–1499 999 and below

0.0010 0.0015 0.0020 0.0025

2

Large induction motor

3000 and above 1500–2999 1000–1499 999 and below

0.0010 0.0020 0.0025 0.0030

2.5.15.4

Soil data

These data are furnished in the geo-technical report and should supply you with the following parameters17 • • • • • • •

Ground water table prevalent at the site Atterberg’s limits Poisson’s ratio of the soil, ν Unit weight of the soil, γ Dynamic shear modulus of the soil, G The foundation depth and bearing capacity of the soil at which the above parameters are valid All other information, regarding the static design of the foundation.

The knowledge of ground water table is essential for all block foundations and should preferably have the bottom of foundation above the ground water table for waves passing through water attenuates the dynamic response. A check on the Atterberg’s limit can give a very good indication qualitatively about the fundamental property of the soil as to how it will behave. But unfortunately very little attention is paid to this aspect in design offices. The various Atterberg’s limits like liquid limit, plastic limit etc not only give a clear indication of how the soil would behave but also holds key to the fact that if the soil is sensitive to shocks induced by vibration or not. We do not discuss the details of Atterberg’s limit and its interpretations but make you aware of one criterion which is quite important in context of machine foundation design. Generically when the natural moisture content of the soil is closer to the liquid limit the soil is deemed soft and when the natural moisture content is close to the plastic

17 Here we assume the reader has some knowledge about the static design procedure of a foundation.

© 2009 Taylor & Francis Group, London, UK

210 Dynamics of Structure and Foundation: 2. Applications

limit it is considered as stiff. However there are certain types of soils whose natural moisture content is greater than the liquid limit. If you ever encounter such case you should immediately be on the alert. For such soils generally belong to the montmorillonite group and constitutes a brittle structure. This type of soil, when disturbed by vibration, flows like a liquid. If this soil is allowed to remain in place it can be very dangerous for the foundation which may undergo sudden settlement without any notice. The liquidity index values of such soils are greater than unity. If such of soils are encountered at a level where foundation would be resting, the complete layer should be replaced by PCC or removed and back-filled with hydraulically compacted sand fill compacted to a Procter Density as specified by the soil consultant. If this strata is quite deep possibilities to be investigated to provide piles (driven/ bored) to a substantial depth below this strata and ignoring the stiffness effect of this montmorillonite clay strata while calculating the equivalent springs for the piles. The Poisson’s ratio of the soil is usually supplied in the soil report. This is required for calculation of the soil springs used for dynamic analysis of the foundation. In absence of such data υ = 0.4 would suffice for most of the cases. The weight density of soil is usually furnished in the soil report this needs to be divided by acceleration due to gravity (g) to arrive at the mass density. or, ρ = γ /g here, ρ = mass density of soil; γ =unit weight of the soil, and g = acceleration due to gravity @ 9.81m/sec2 or 32.2 ft/sec2 . The Dynamic shear modulus plays a key role in evaluation of the spring data. Though co-relation exists for theoretical evaluation of G from other engineering parameters of the soil18 for important foundations we still advocate that you insist on field test to get the field observed value of G. Try to convince the client19 , it is worth spending a few thousand dollars now rather than to pay through your nose in terms of performance compensations and could lead to a classic case of being penny wise and pound foolish. Designing a foundation with improper G value will completely waste the design effort for the said foundation. 2.5.15.5

Trial sizing of the block foundation

Based on the above input data the next step for the designer is to do a trial sizing of the block foundation with which he starts his first check for resonance and amplitude. The basic guideline for the same could be summarised as follows: •

The rigid type block foundation should be so proportioned that it should have following mass ratio with respect to the machine ◦ ◦

For centrifugal machine it should be 2 to 3 times the weight of the machine. For Reciprocating type it should be 3 to 5 times the weight of the machine.

18 Refer Chapter 1 (Vol. 2) for these theoretical co-relations. 19 Even your boss at times. . . . . .

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 211

•

•

The top of foundation is usually kept about 300 mm above the finished grade elevation to prevent damage due to surface water run-off. However this should be back checked with process department to ensure that NPSH of the pump or piping connections will not be affected adversely. The vertical thickness of the foundation should be selected based on maximum value of the following: ◦ ◦ ◦ ◦

•

The width of the foundation is selected based on the maximum value of the following: ◦ ◦ ◦ ◦

• • •

Maximum embedded length of the anchor bolts plus 250 mm One fifth the width (least dimension) of the foundation in plan One tenth of largest dimension in plan A depth of 600 mm.

Centre to centre distance of the anchor bolts plus 150 mm on both the side of the foundation Length to the edge of the machine plus 300 mm at the both the ends of the foundation 1 to 1.5 times the vertical distance from the bottom of foundation to the machine centre line Once the width and height of foundation is selected the length can be calculated based on the mass criteria as stated above.

The plan dimension of the machine should be so adjusted that c.g. of the machine assembly matches with c.g. of the foundation. For foundation resting on soil, eccentricity in c.g. of the machine and the foundation shall not be more than 5%. For large reciprocating machines the embedded depth to be so adjusted that at least 60 to 80% of the depth of the foundation is embedded in the soil. This will increase the lateral restraint and damping ratio for modes of vibration.

We now give below some useful data and mathematical expressions which could effective in day to day design office practise for design of block foundations. 2.5.15.6

Centre of gravity of the machine foundation

Here the whole machine foundation is broken into different segments having mass as mi having co-ordinates as xi , yi , zi respectively then the c.g. of the foundation is given by mi x i x¯ = i , i mi

m i yi y¯ = i i mi ,

m i zi and z = i i mi

Second moment of inertia of standard geometric shapes (Refer to Figures 2.5.35 and 36). © 2009 Taylor & Francis Group, London, UK

212 Dynamics of Structure and Foundation: 2. Applications

For foundations resting on soil the moment of inertia is calculated by

Ixx =

1 LB3 ; 12

Iyy =

1 BL3 , 12

Izz = Ixx + Iyy

and

For foundations resting on piles or springs moment of inertia is calculated by Ixx =

yi2 ;

Iyy =

i

x2i ,

and

Izz =

i

x2i + yi2 e

i

Z lx

ly

lz

X

Y

Figure 2.5.35 A solid rectangular prism.

Z

X

Y

Figure 2.5.36 A solid circular cylinders.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 213

2.5.15.7

Mass of inertia of geometrical shapes

For solid rectangular prisms (Refer to Figure 2.5.36) Jx = m/12(ly2 + lz2 ) Jy = m/12(lx2 + lz2 ) Jz = m/12(lx2 + ly2 ) For solid circular cylinders The second moment of inertia is given by m Jx = 12

3 2 2 D +l ; 4

m Jy = D2 ; 8

m Jz = 12

3 2 D + l2 4

Here D = diameter of the cylinder; L = length of the cylinder.

2.6 SPECIAL PROVISIONS OF IS-CODE We now give below some salient provisions and recommendations of IS-2974 for rotary and reciprocating types of machines that constitute the normal design office practice in India.

2.6.1 Recommendations on vibration isolation To avoid transmission of vibration to adjoining parts of the buildings or other foundations, it is necessary to provide a suitable isolation between the equipment foundation and the adjoining structures. This may be achieved by providing sand trench around the foundation block, the thickness and depth of which shall be determined for each individual case. As a rule the equipment foundation shall not be allowed to serve as a support for other structures or for machines not related to the particular equipment. In case it becomes necessary to support unimportant parts of other structures on the machine foundation itself, measure shall be taken to make the connections resilient by introducing gaskets made of rubber, cork, felt or other resilient materials.

2.6.2 Frequency separation The natural frequency of the foundation system shall be such as to avoid resonance with operating frequency of the machine and the amplitudes be kept below the permissible limit. Foundations for low frequency machine shall preferably be designed as such that the natural frequency of the foundation is higher then the operating frequency of the machine. The natural frequency of any foundation should not preferably be within 20% of the operating speed of the machine. © 2009 Taylor & Francis Group, London, UK

214 Dynamics of Structure and Foundation: 2. Applications

2.6.3 Permissible amplitudes Normally the recommendation of the vendor supplying the equipment shall guide the design, however in absence of such data the code recommends that if no resonance is to occur in adjoining structure the amplitude of vibrations of a foundation at the upper edge shall not exceed 0.20 mm in both directions. When several foundations for similar machines are erected on a common mat the computation for vibration shall proceed assuming that each machine foundation is independent of others by breaking up the raft into sections corresponding to separate foundations. The design value for the permissible amplitude of vibrations may be increased by 30%.

2.6.4 Permissible stresses Concrete of grade M15 or higher shall be used for foundations. Concrete and steel stresses are as specified in IS: 456-2000 shall be used for considering the dynamic loads separately in detailed design. The following elastic moduli of concrete may be used in design.

Grade of concrete

Edyn (kN/m2 )

M15 M20 M25 M30

250 × 106 300 × 106 340 × 106 370 × 106

2.6.5 Concrete and its placing The concrete used shall be controlled concrete conforming to design requirements. The grade of concrete should generally be M15 to M20 for block foundation and M20 for frame foundation. The concrete shall be placed and designed in accordance with IS: 456-2000. The concrete used shall be of plastic consistency having an allowable slump, which may vary between 50 to 80 mm. The water cement ratio shall not exceed 0.45. The same consistency shall be maintained throughout the foundation.

2.6.6 Reinforcements All foundation units of foundation shall be provided with top and bottom reinforcement in two directions. Reinforcement shall be provided along the surface only in case of block foundation. The reinforcement in block foundation shall not be less than 25 kg/m3 . The minimum diameter of bars shall be 12 mm with a maximum spacing of 200 mm in order to care of the shrinkage. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 215

16100 500

1

4

6

15

1442.5

9

1442.5

2

A

7

12

B

D 11

1442.5 1442.5 500

10

O

3

5

385 3285

17

6770

13

C

E 14

8

16

2400 800 800 4000 2405 425 800 PLAN VIEW OF THE BLOCK FOUNDATION

18 800

C/L of Shaft of the machine 800

800

2000

1600 (typ.) 600 3600

ELEVATION OF THE BLOCK FOUNDATION

Figure 2.6.1 Plan and elevation of a gas turbine foundation.

2.6.7 Cover to concrete For block foundation the concrete cover for protection of reinforcement shall be 75 mm at the bottom, 50 mm on both sides and 40 mm at top. We now solve a practical design problem for a Gas Turbine resting on a block foundation for your perusal and we hope that this will give you a better insight to the aspect of how to apply the previously mentioned theories to the day-to-day design office work of design of machine foundation (Figure 2.6.1).

Example 2.6.1 Design the gas turbine foundation shown in Fig. 2.6.1. Design data 1 2 3 4 5 6

Bearing capacity of soil = 200 kN/m2 Shear wave velocity of soil = 125 m/sec Density of soil = 20 kN/m3 Poisson’s ratio of soil = 0.25 C/L of shaft of the machine = 2.0 above T.O.C. Operating frequency of turbine = 2250 r.p.m

© 2009 Taylor & Francis Group, London, UK

216 Dynamics of Structure and Foundation: 2. Applications

7 8 9 10

Grade of concrete = M25 Grade of steel = Fe415 Allowable amplitude = 0.2 mm Load at various anchor locations are as shown in the table hereafter.

Equipment load data at various Anchor Bolt locations Static load

Dynamic load

Anchor bolt #

Vertical load (kN)

Vertical load (kN)

Horizontal load (kN)

1 2 3 4 5 6 7 8 9 10

−311 −42 −311 −517 −517 −311 −50 −311 −200 −200

±7

±7

±6.76

±6.76

±51

±51

11 12 13 14 15 16 17 18

−200 −350 −350 −350 −185 −185 −185 −185

±23

±23

Total

−4760

Remarks

All horizontal force is along global Y axes

Anchor Bolt for generator Anchor Bolt for generator Anchor Bolt for generator Anchor Bolt for generator

Calculate the natural frequency and amplitude based on Figure 2.6.1 and using • • • •

Richart and Lysmer model Richart and Lysmer model with embedment Wolf’s model Time history analysis based on Newmark-Beta method.

Solution: Geometric property of the foundation Area of foundation = 16.1 × 6.77 = 108.9 m2 1 1 Second Moment of Inertia = LB3 = 16.1 × 6.773 = 416.30 m4 12 12

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 217

Equivalent radius in vertical and horizontal mode: r0 =

108.9 = 5.89 m π

Equivalent radius in rocking mode about the X-axis rθ =

4

4 × 416.30 = 4.80 m π

Calculation of soil springs Shear Wave velocity = 125 m/sec; Poisson’s ratio = 0.25. Unit weight of soil = 20 kN/m3 20 Dynamic shear modulus (G) = ρVs2 = × (125)2 = 31855.25 kN/m2 9.81 Springs based on Richart’s model Kz =

4 × 31855.25 × 5.89 4Gr0 = = 1000679.6 kN/m (1 − ν) (1 − 0.25)

Ky =

32Gr0 (1 − υ) 32(1 − 0.25) × 31855.25 × 5.89 = = 900611.63 kN/m (7 − 8υ) (7 − 8 × 0.25)

Kφy =

8Gr3θ 8 × 31855.25 × (4.8)3 = = 12511907.52 kN/m. 3 (1 − υ) 3(1 − 0.25)

Table for calculation of c.g. and second moment of inertia of m/c & fdn. (explained in next page) Centre of gravity 11291.04 4997.63 = 7.65 m from the point O; y¯ = = 3.39 m from the 1476.4 1476.4 3722.95 point O; and z¯ = = 2.52 m from the bottom of the foundation. 1476.4 8.05 − 7.65 × 100 = 2.5% < 5% hence OK. Eccentricity in x direction = 16.1 3.399 − 3.385 Eccentricity in y direction = × 100 = 0.07% < 5% hence OK. 6.77

x¯ =

Total mass moment of inertia =

m 2 2 2 ly + lz2 + m(yoi + zoi ) + m¯z2 12

= 4703.00 + 4509.65 + 1476.4 × (2.52)2 = 18588.4 kN-m-sec2 .

© 2009 Taylor & Francis Group, London, UK

Lx (m)

Ly (m)

Lx (m)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 A B C D

6.5 4 5.6 0.8

6.77 6.77 6.77 5

3.6 3 3.6 1.6

E

0.8

5

1.6

Weigt ht(kN) 311 42 311 517 517 311 50 311 200 200 200 350 350 350 185 185 185 185 3960.5 2031 3412.1 160 160 14484

© 2009 Taylor & Francis Group, London, UK

mi yi

mi zi

yoi

zoi

−2.89 0 2.885 −2.89 2.885 −2.89 0 2.885 −1.44 0 1.443 −1.44 0 1.443 −2.89 2.885 −2.89 2.885 0 0 0 0

−1.078 −1.078 −1.078 −1.078 −1.078 −1.078 −1.078 −1.078 −2.078 −2.078 −2.078 −2.078 −2.078 −2.078 −1.078 −1.078 −1.078 −1.078 0.722 0.922 0.722 −1.278

300.73 4.98 300.73 499.93 499.93 300.73 5.93 300.73 130.49 88.07 130.49 228.35 154.11 228.35 178.89 178.89 178.89 178.89 210.24 175.86 181.13 26.65

−1.278

26.65 4509.65

Xi

Yi

Zi

31.70 4.28 31.70 52.70 52.70 31.70 5.10 31.70 20.39 20.39 20.39 35.68 35.68 35.68 18.86 18.86 18.86 18.86 403.72 207.03 347.82 16.31

0.385 0.385 0.385 3.67 3.65 6.075 6.075 6.075 7.3 7.3 7.3 9.7 9.7 9.7 11.3 11.3 15.3 15.3 3.25 8.5 13.3 7.3

6.27 3.385 0.5 6.27 0.5 6.27 3.385 0.5 4.828 3.385 1.943 4.828 3.385 1.943 6.27 0.5 6.27 0.5 3.385 3.385 3.385 3.385

3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 4.6 4.6 4.6 4.6 4.6 4.6 3.6 3.6 3.6 3.6 1.8 1.6 1.8 3.8

12.21 1.65 12.21 193.41 192.36 192.59 30.96 192.59 148.83 148.83 148.83 346.08 346.08 346.08 213.10 213.10 288.53 288.53 1312.08 1759.79 4625.96 119.06

198.77 14.49 15.85 330.44 26.35 198.77 17.25 15.85 98.42 69.01 39.60 172.23 120.77 69.30 118.24 9.43 118.24 9.43 1366.58 700.81 1177.36 55.21

114.13 15.41 114.13 189.72 189.72 114.13 18.35 114.13 93.78 93.78 93.78 164.12 164.12 164.12 67.89 67.89 67.89 67.89 726.69 331.25 626.07 61.98

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1977.97 946.02 1704.10 37.46

16.31 1476.40

9.7

3.385

3.8

158.21 11291.04

55.21 4997.63

61.98 3722.95

37.46 4703

Mass

mi xi

m/l2(Ly 2 + Lz2 )

0

m(yoi2 + zoi )

218 Dynamics of Structure and Foundation: 2. Applications

Loading location

Analysis and design of machine foundations 219

Calculation for damping Based on Richart’s Formula we have In the vertical direction Bz =

0.25 m(1 − υ)g 0.25 × 1476.4 × 0.75 = = 0.667 ρs r3z 2.03 × (5.89)3

0.425 0.425 →Dz = =√ = 0.52 Bz 0.667 √ Cz = 2Dz Kz m = 2 × 0.52 1000679.6 × 1476.4 = 39974 kN · sec/m In the horizontal direction 1476.4 5 (7 − 8ν) mg × = = 0.738 3 32 × 0.75 2.04 × (5.89)3 32 (1 − ν) ρs ry

By =

0.288 0.288 Dy = =√ = 0.3352 and By 0.738 √ Cy = 2Dy Ky m = 2 × 0.3352 900611.63 × 1476.4 = 24446 kN · sec/m For the rocking mode 0.375(1 − ν)Jφy g

Bφy =

ρs r5φy

=

0.375 × 0.75 × 18588.4 2.03 × (4.798)5

= 1.013

0.15 0.15 = = 0.074 √ (1 + Bφy ) Bφy 2.013 × 1.013 √ = 2Dφy Kφy m = 2 × 0.074 12511907.52 × 18588.4 = 71375 kN · sec/m

Dφy = Cφy

Calculation of natural frequencies In the vertical direction ωz =

Kz = m

100679.6 = 26 rad/sec (249 r.p.m) 1476.4

In the horizontal direction the equation of motion for free vibration is given by m 0

Cx 0 y¨ + Jyφ φ¨ −Cy Zc

Ky y˙ × ˙ + φ −Ky Zc

© 2009 Taylor & Francis Group, London, UK

−Cy Zc Cφy + Cy Zc2 − WZc

−Ky Zc y 0 = 0 Kφy + Ky Zc2 − WZc φ

220 Dynamics of Structure and Foundation: 2. Applications

For eigen value analysis we have

Ky − mλ −Ky Zc

−Ky Zc =0 Kφy + Ky Zc2 − WZc − Jφy λ

Here m = 1476.4; Jφ y = 18588.4; Zc = 2.52 m; W = 14484 kN; Ky = 900611.63 kN/m; Kφy = 12511907.52 kN/m

900611 − 1476.4λ −2269540 =0 −2269540 18194648 − 18588.4λ

or,

The above matrix on expansion and simplification reduces to λ2 − 1589λ + 409398 = 0; 1589 ± (1589)2 − 4 × 1 × 409398 λ= = 323.5, 1265 2 ➔ ω2 = 17.98 rad/sec (172 r.p.m.); and ω3 = 35.56 rad/sec (340 r.p.m.) Calculation of eigen vectors For first mode, for ω = 17.98 rad/sec, we have

900611 − 1476.4 × 323.5 −2269540 φ11 =0 −2269540 18194648 − 18588.4 × 323.5 φ12

The above on expansion gives the following two equations 422995.6φ11 − 2269540φ12 = 0

and

−2269540φ11 − 12181301φ12 = 0

Considering φ11 = 1.00 and solving the above homogenous equation we have, φ12 = 0.186379442

Thus,

φ11 1.00 = 0.186379442 φ12

For the Second mode we have 900611 − 1476.4 × 1265 −2269540 φ21 =0 −2269540 18194648 − 18588.4 × 1265 φ22 The above on expansion gives the following two equations: −967035φ21 − 2269540φ22 = 0

© 2009 Taylor & Francis Group, London, UK

and −2269540φ21 − 5319678φ22 = 0

Analysis and design of machine foundations 221

Considering φ21 = 1.00 and solving the above homogenous equation we have φ22 = −0.426092952 Thus the complete eigen vector matrix is given by 1.00 1.00 [ϕ] = 0.186379442 −0.426092952 Calculation of normalised eigen vectors For the first mode

T

{φ} [M] {φ} = 1.00

1476.4 0.186379442 0.0

The above on simplification gives,

√

Mr =

0.0 18588.4

1.00 0.186379442

√ 2122.110 = 46.06

φ11 φ12

Thus dividing each term of the eigen vector by above we have, 0.021707808 . 4.045889184 × 10−3

N =

For the second mode we have {φ}T [M] {φ} = 1.00 −0.426092952 The above on simplification gives,

√

1476.4 0.0

Mr =

0.0 18588.4

1.00 −0.426092952

√ 4851.22 = 69.65.

Thus dividing each term of the eigen vector by above we have,

0.014357356 6.11756837 × 10−3 Thus the complete normalised eigen vector matrix is

φ21 φ22

N =

[ϕ] =

21.707808 14.357356 × 10−3 4.045889184 −6.11756837

Correction of damping matrix based on Rayleigh coeff icient for modal analysis We had already stated that damping matrix obtained from soil property is non proportional and when considered in the analysis will not de-couple under orthogonal transformation as such we correct the matrix enabling us to de-couple the same20 . 20 Based on the theory of magnification factor damping may be ignored for this case for the ratio of the fundamental frequencies of the foundation to the operating frequency of the machine is more than 3.5. However for sake of clarification of the problem we continue to consider it in our analysis.

© 2009 Taylor & Francis Group, London, UK

222 Dynamics of Structure and Foundation: 2. Applications

Here,

−Cy Zc Cφy + Cy Zc2 − WZc

Cx [C] = −Cy Zc

−61604 190117.2

24446 = −61604

For the First mode we have T

{φ} [C] {φ} = 21.707808 4.045889184 × 10

−3

24446 61604

61604 190117.2

21.707808 × 10−3 4.045889184

×

The above on simpliﬁcation gives, {φ}T [C] {φ} = 3.8106 ➔ 2ζ1 ω1 = 3.8106 or ζ1 = 0.105. For the Second mode, we have −3

T

{φ} [C] {φ} = 14.357356 −6.11756837 ×10

24446 61604

61604 190117.2

14.357356 × × 10−3 −6.11756837 The above on simplification gives {φ}T [C] {φ} = 22.97

➔

2D2 ω2 = 22.97

or D2 = 0.323.

Now considering the design damping as proportional Rayleigh damping, we have [C] = α[M] + β [K]

or [φ]T [C] [φ] = α [φ]T [M] [φ] + β [φ]T [K] [φ]

and we have, 2D1 ω1 = α + βω12 and 2D2 ω2 = α + βω22 . i.e. α + 323β = 3.8106 and α + 1265β = 22.97 Solving the above two simultaneous equations, we have: α = −2.7589 and β = 0.0203 Substituting the above value of α and β we have, [C] = −2.7589

1476.4 0

14222 → [C] = −46071

© 2009 Taylor & Francis Group, London, UK

0 900611 −2269540 + 0.0203 18588.4 −2269540 18194648 −46071 . 318233

Analysis and design of machine foundations 223

Calculation of amplitude in the vertical mode Vertical force = (7 + 6.76 + 51 + 23) = 87.76 sin 236t We had calculated earlier that Dz = 0.52 and hence, δz = ➔ δz =

P0 sin ωm t/Kz (1 − r2 )2 + (2ζ r)2

,

with, r =

236 = 9.0. 26

87.76 sin 236t 1000679.6 (1 − 81)2 + (2 × 0.52 × 9)2

= 1.08822 × 10−6 sin 236t m. For, coupled sliding and rocking mode, we have Lever arm = 3.6 + 2.0 − 2.52 = 3.08 m; Horizontal force = 87.76 sin 236t Thus moment about the vertical centroid, Zc = 270 sin 236t

87.76 The force matrix can be represented as, {P} = sin 236t. 270 The equation of motion can be written as & ' & ' [M] Y¨ + [C] Y˙ + [K] {Y} = {P} With orthogonal transformation, we can write & ' & ' [φ]T [M] [φ] y¨ + [φ]T [C] [φ] y˙ + [φ]T [K] [φ] {Y} = [φ]T {P} which gives the following two equations y¨ + 2D1 ω1 y˙ + ω12 y = p sin ωm t

and θ¨ + 2D2 ω2 θ˙ + ω22 θ = m sin ωm t

T

Here [φ] {P} =

21.707808 4.045889184 14.357356 −6.11756837

=

88 p sin 236t → 270 m

3.0 sin 236t −0.388

i.e. y¨ + 2 × 0.105 × 17.98˙y + 324y = 3.0 sin 236t → y¨ + 3.776˙y + 324y = 3.0 sin 236t and θ¨ + 2 × 0.323 × 35.56θ˙ + 1265θ = −0.388 sin 236t → θ¨ + 22.97θ˙ + 1265θ = −0.388 sin 236t Thus for the horizontal translation, we have y¨ + 3.776˙y + 324y = 3.0 sin 236t © 2009 Taylor & Francis Group, London, UK

224 Dynamics of Structure and Foundation: 2. Applications

236 3.776 where r = √ = 13.11 and ζ = √ = 0.104. 324 2 324 Hence, δy =

3 sin 236t 324 (1 − 171.8)2 + (2 × 0.104 × 13.11)2

= 5.4204 × 10−5 sin 236t For the rocking mode θ¨ + 22.97θ˙ + 1265θ = −0.388 sin 236t 236 22.97 where r = √ = 6.63 and ζ = √ = 0.322 1265 2 1265 θ=

−0.388 sin 236t

1265 (1 − 43.95)2 + (2 × 0.322 × 6.63)2

= −7.1063 × 10−6 sin 236t

Thus in global co-ordinate, we have

Y 21.707808 14.357356 5.4204 = × 10−8 4.045889184 −6.11756837 −0.7106

107.462 = × 10−8 sin 236t 26.27 Net horizontal amplitude at top of foundation Y= 107.463 × 10−8 + (3.6 − 2.52) × 26.27 × 10−8 = 1.358 × 10−6 m < 0.2 mm OK Net horizontal amplitude at base of the foundation Y = 107.463 × 10−8 − 2.52 × 26.27 × 10−8 = 41.263 × 10−8 m < 0.2 mm OK We make here a very interesting comparison, shown in Figure 2.6.2, is the time history response of the block foundation with non-proportional soil damping and corrected proportional Rayleigh damping, we have obtained earlier. It will be observed that values are quite closely matching and for practical engineering work this is deemed sufficient. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 225

Comparison of amplitude based on time history 0.000015

Amplitude

0.00001 0.000005 0 1

23 45 67 89 111 133 155 177 199 221 243 265

-0.000005

Displacement with non proportional damping Displacement with corrected proportional damping

-0.00001 -0.000015

Time steps

Figure 2.6.2

Check of local vibration of the pedestals Width of the pedestal = 800 mm; Depth of the pedestal = 5000 mm; Height of pedestal = 1600 mm. I=

BD3 0.8 × 125 = = 8.33 m4 12 12

Considering the pedestal as cantilever beam, Kh =

3EI 3 × 300 × 106 × 8.33 = = 1.83 × 109 kN/m L3 (1.6)3

Self weight of the pedestal = 0.8 × 5.0 × 1.6 × 25 = 160 kN; Weight from machine = 3 × 200 + 3 × 350 = 1650 kN → Total weight = 1810 kN

2

Thus total mass(m) = W/g = 184.5 kN-sec /m and hence ω = 1.83 × 109 = 3149 rad/sec 184.5 And,

r=

ωm 236 = = 0.07. ωn 3149

As the frequency ratio is very low we neglect the damping we have, y=

P0 /k 51 sin 236t = = 2.80 × 10−8 m < 0.2 mm. 2 (1 − r ) 1.83 × 109 (1 − 0.072 )

© 2009 Taylor & Francis Group, London, UK

Kh = m

226 Dynamics of Structure and Foundation: 2. Applications

Calculation based on embedment effect We had already calculated before that based on property of soil individual soil stiffness in various modes as Kz = 1000679.6 kN/m,Cz = 39974kN.sec/m; Ky = 900611.63kN/m, and Cy = 24446kN/m, Kφy = 12511907.52 kN/m, Cϕy = 71375 kN/m. Embedment factor based on Richart’s table for soil stiffness h ηz = 1 + 0.6(1 − υ) , here h = 3.0 m as per the problem, and rz =5.89 m rz which gives, ηz = 1.23 similarly, ηy = 1 + 0.55(2 − ν)

ηφy

h and this gives ηy = 1.497 and ry

h h 3 = 1 + 1.2(1 − ν) + 0.2(2 − ν) and this gives ηφy = 1.5625. rφy rφy

Embedment factor based on Richart’s table for soil damping 1 + 1.9(1 − ν) rhz αz = = 1.556; √ ηz

αφy

1 + 1.9(2 − ν) rhy αy = = 2.20 and √ ηy

1 + 0.7(1 − ν) rhφy + 0.6(2 − ν) = √ ηφy

h rφy

3 = 1.267.

Thus considering the embedment factor the stiffness and damping value gets modified to: Kze = 1230836 kN/m,

Cze = 62204 kN · sec/m

Kye = 1348214 kN/m,

Cye = 53812 kN/m

e = 19549856 kN/m, C e = 90432 kN/m and Kφy ϕy For the vertical direction, we have

ωz =

Kze = m

1230836 = 29 rad/sec (277 r.p.m); 1476.4

Vertical force = (7 + 6.76 + 51 + 23) = 87.76 sin 236t © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 227

We had calculated earlier that Dz = 0.73; δz =

Thus

➔δz =

P0 sin ωm t/Kz (1 − r2 )2

+ (2ζ r)2

,

where r =

236 = 8.13 29

87.76 sin 236t

1230836 (1 − 66)2 + (2 × 0.52 × 8.13)2

= 1.07754 × 10−6 sin 236t m In horizontal direction the equation of motion for free vibration is given by:

Cx −Cy Zc y¨ y˙ 0 + Jyφ φ¨ −Cy Zc Cφy + Cy Zc2 − WZc φ˙

Ky −Ky Zc y 0 + = φ 0 −Ky Zc Kφy + Ky Zc2 − WZc

m 0

For eigen value analysis we have

Kye − mλ −Kye Zc

−Kye Zc =0 Kφy + Kye Zc2 − WZc − Jφy λ

Here m = 1476.4; Jφ y = 18588.4; Zc = 2.52 m; W = 14484 kN; Kye = e 1348214 kN/m; Kφy = 19549856 kN/m. →

1348214 − 1476.4λ −3397499 =0 −3397499 28075055 − 18588.4λ

The above matrix on expansion and simplification reduces to λ2 − 2423λ + 958897 = 0 → λ = 498, 1925 ➔

ω2 = 22.3 rad/sec (213 r.p.m.), and ω3 = 43.87 rad/sec (419 r.p.m.).

Now proceeding in the exact manner as explained in the previous case, we arrive at the result: Net horizontal amplitude at top of foundation Y = 107.14 × 10−8 + (3.6 − 2.52) × 26.09 × 10−8 = 1.353 × 10−6 m < 0.2 mm OK Net horizontal amplitude at the base of foundation Y = 107.14 × 10−8 − 2.52 × 26.09 × 10−8 = 41.4 × 10−8 m < 0.2 mm OK © 2009 Taylor & Francis Group, London, UK

228 Dynamics of Structure and Foundation: 2. Applications

Calculation based on Wolf’s Model Based on formulation proposed by Wolf 4Gr0 4 × 31855.25 × 5.89 = = 1000679.6 (1 − υ) (1 − 0.25) rz 5.89 × 1000679.6 × 0.58 Cz = K z γ0 = = 27348 k · sec/m Vs 125 r 2 5.89 2 z mz = Kz μ 0 = × 1000679.6 × 0.095 = 211 k-sec2 /m Vs 125

Kz =

In the horizontal direction we have 8Gry 8 × 31855.25 × 5.89 = = 2001359 k/m 1−υ 0.75 ry 5.89 × 2001359 × 0.85 Cy = K y γ0 = = 80158 k · sec/m Vs 125 r 2 5.89 2 y my = Ky μ 0 = × 2001359 × 0.27 = 1200 k-sec2 /m Vs 125

Ky =

In rocking mode, we have 8Gr3θ 8 × 31855.25 × (4.8)3 = = 12511907.52 k/m 3 (1 − υ) 3(1 − 0.25) r 2 4.8 2 y = Kφy μ0 = × 12511907 × 0.24 = 4428 k-sec2 /m Vs 125

Kφ y = Jφy

γ0 =

0.3 1+

3(1−υ)m 8r5θ ρ

Thus, Cφy =

=

0.3 1+

3 × 0.75 × 4428 8 × (4.8)5 × 2.04

= 0.242

rφy 4.8 × 12511907 × 0.242 Kφy γ0 = = 116271 k · sec/m Vs 125

For vertical direction we have ωz =

Kz = m

1000679 = 24.35 rad/sec (277 r.p.m) 1687

Vertical force = (7 + 6.76 + 51 + 23) = 87.76 sin 236t Damping Ratio Dz =

Thus

δz =

√

27348

2 1000679 × 1687

P0 sin ωm t/Kz (1 − r2 )2

© 2009 Taylor & Francis Group, London, UK

+ (2Dr)2

= 0.332

where r =

236 = 9.7 24.34

Analysis and design of machine foundations 229

➔ δz =

87.76 sin 236t

1000679 (1 − 94)2 + (2 × 0.332 × 9.7)2

= 0.94076 × 10−6 sin 236t m For coupled sliding and rocking mode Cx −Cy Zc y¨ y˙ 0 + Jyφ φ¨ −Cy Zc Cφy + Cy Zc2 − WZc φ˙ Ky −Ky Zc y 0 + = 2 φ 0 −Ky Zc Kφy + Ky Zc − WZc

m 0

For eigen value analysis we

Kye − mλ −Kye Zc

−Kye Zc =0 Kφy + Kye Zc2 − WZc − Jφy λ

Here m = 2676; Jφy = 23016; Zc = 2.52 m; W = 14484 kN; Kye = 2001359 e = 12511907 kN/m. kN/m; Kφy It is to be noted that here mass and moment of inertia is the mass/inertia of machine and foundation plus the mass/inertia of soil participating in the vibration. Thus

2001359 − 2676λ −5043425 =0 −5043425 25184838 − 23016λ

The above matrix on expansion and simplification reduces to λ2 − 1842λ + 405381 = 0

➔ λ = 255, 1587

Hence, ω2 = 15.96 rad/sec (152 r.p.m.) and ω3 = 39.83 rad/sec (380 r.p.m.) Now proceeding in the manner as explained in the case of Richart’s model we arrive at the result; Net horizontal amplitude at top of the foundation Y = 63 × 10−8 + (3.6 − 2.52) × 37.646 × 10−8 = 1.0365 × 10−6 m < 0.2 mm OK Net horizontal amplitude at the base of foundation Y = 63 × 10−8 − 2.52 × 37.646 × 10−8 = −31.86 × 10−8 m < 0.2 mm OK © 2009 Taylor & Francis Group, London, UK

230 Dynamics of Structure and Foundation: 2. Applications

The values obtained by the three methods are summarised hereafter: Comparison of natural frequencies Based on the method

ωz (r.p.m)

ωy (r.p.m)

ωθy (r.p.m)

Richart’s formula Richart with embeddment Wolf’s method

249 277 277

172 213 152

340 419 380

Comparison of amplitude Based on the method

δz (mm)

δy (mm)

δθy

Richart’s formula Richart with embeddment Wolf’s method

1.0882 × 10−3 1.077 × 10−3 0.94076 × 10−3

1.358 × 10−3 1.353 × 10−3 1.0365 × 10−3

41.263 × 10−8 41.4 × 10−8 −31.86 × 10−8

•

Based on Time History Analysis

AMPLITUDE

We perform time history analysis for springs based on Richart’s method and Wolf’s Method. Here time history response has been done for 215 steps with complete soil damping into consideration and shown in Figures 2.6.3 and 4. Time History based on Newmark Beta Method with Richarts Spring 0.000008 0.000006 0.000004 Displacement in Y direction 0.000002 Angular Rotation 0 -0.000002 1 17 33 49 65 81 97 113 129 145 161 177 193 209 -0.000004 -0.000006 Time Steps

Figure 2.6.3

AMPLITUDE

Time History based on Newmark Beta Method with Wolf's Spring Spring 0.000003 0.000002 0.000001 0 -0.000001 1 20 39 58 77 96 115 134 153 172 191 210 -0.000002 -0.000003

Figure 2.6.4

© 2009 Taylor & Francis Group, London, UK

Time Steps

Displacement in Y direction Angular Rotation

Analysis and design of machine foundations 231

2.7 ANALYSIS AND DESIGN OF MACHINE FOUNDATION UNDER IMPACT LOADING

2.7.1 Introduction In this section we will deal with foundations subjected to impact loading. These type of foundations usually constitute of hammer foundations used for forging or hydraulic stamps used to flatten steel billets to make plates out of them. The arrangement of the hammer foundation is usually as shown Figure 2.7.1. The hammer foundation, consists of a hammer or a tup which falls repeatedly on an anvil. The anvil in turn is placed on an elastic pad resting on a massive RCC block. The elastic pad is used to isolate the foundation from the surrounding and minimize the harmful effect of the vibration induced by the hammer dropping on the anvil. The elastic pad also acts as damper to reduce the net amplitude of vibration of the anvil and the foundation. Depending upon the functionality, the frame of the hammer may either rest on the foundation block as shown above or may even rest on a separate foundation. While planning the foundation it is usually ensured that the center line of the anvil is concentric with the center of gravity of the base of the foundation. This ensures that the amplitude of vibration is restricted to vertical translation only and does not give rise to any coupled motion including rocking21 . At times when the hammer is very heavy the foundation is further isolated by providing elastic pad/springs along with dampers below the RCC block too. Shown in Figure 2.7.2 is a hammer foundation where other than the anvil the RCC block is also mounted on springs and dampers to isolate the transmittal of vibration to the surrounding. The springs or the elastic pad which are placed below the RCC block is usually an expensive item and care should be taken to protect them from exposure to water, chemicals, oils etc which could otherwise damage their properties. This is usually done by providing a protective RCC trough all round the foundation and sealing the same at the top of the foundation level. The elastic material used under the anvil or the RCC block could be of cork, timber or even specialized mechanical springs and dampers supplied by vendors having technological expertise in isolation techniques of these type of foundations. 2.7.1.1

How does the behavior of a mechanical system under impact differ from externally applied harmonic loads?

We do not tender any apologies for posing so fundamental a question, for in our experience in teaching this subject, as well as interacting with professionals in the industry for over two decades, we have been somewhat startled to find that though people can arrive at the design values for various type of machine foundation quite accurately by following the code stipulation blindly, but how does the characteristics 21 Hammer foundation having eccentric anvil is though uncommon but surely not rare. We will deal with this particular case separately later.

© 2009 Taylor & Francis Group, London, UK

Frame Hammer/Tup RCC Fdn. Anvil Elastic Pad F.G.L.

Figure 2.7.1 General arrangement of a Typical Hammer Foundation.

Frame Hammer/Tup RCC Fdn. RCC Trough Anvil Elastic Pads/Springs

Figure 2.7.2 General arrangement of a Typical Hammer Foundation mounted on spring with R.C.C. Trough. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 233

of the behavior differs in the above two cases. . . the picture has remained gray to many. So before we take a plunge into the mathematical aspect of the design it would be worthwhile to understand the conceptual aspect of the problem and reflect a bit on how the two cases differ in transmitting the vibration to the system. We hypothesize two pictures for this from our day to day life. 1 Imagine a boy continuously jumping on a plank supported at two ends for some time. 2 A Karatika giving a vicious chop to the same plank at some point on it22 . 2.7.1.2

How do the planks behave under these two conditions?

For the first case if we have a stop watch we can measure the time taken by the boy when he is at rest on the plank to the time he jumps (presuming with same monotony) and again comes to rest on the plank. If we take this as his time period of vibration T it is possible to find out the frequency of his motion from the relation, ω = 1/T. If we now measure the weight of the boy we can say that the plank is subjected to a continuous external force of W sin ωt, where W is the weight of the boy and ω is the frequency with which he is jumping on the plank. Thus the plank is under a forced harmonic load and will also produce amplitudes which will be a function of the external force expressed as W sin ωt. This is called harmonic force23 . While in the second case when the Karatika executes the chop he is transferring his potential energy into a kinetic energy and is transferring this energy to the plank in a very short period of time (may be some small fraction of a second) and then it ceases to exist. This is quite unlike the earlier case when the external force continues to excite the plank till the boy continues to jump on it. These type of forces when induced on a body where it is subjected to force for a very short time is known as an impact load or in technical term we call it a transient. We consciously or otherwise often observe this phenomenon quit often in our day to day life like • • •

A hammer used to put a nail in place A mallet used to hit a golf ball A ship hitting the jetty fenders when coming to rest on a port

These are all cases of impact forces acting on a system. For instance, if we take the case of a mallet hitting the golf ball what we do is take a swing up when we concentrate our potential energy and with the down swing of the club we transfer the potential energy to kinetic energy which is then transferred to the ball at the instant of impact. Now suppose we connect the ball to a spring, we will observe that the ball starts vibrating to and fro with respect to its mean position.

22 The two cases are mutually exclusive. 23 Block Foundation dealt earlier is a typical example of this when the unbalanced mass induces an external force that is harmonic in nature.

© 2009 Taylor & Francis Group, London, UK

234 Dynamics of Structure and Foundation: 2. Applications

To understand the effect of impact further we a take a step backward and formulate a problem from our days of engineering mechanics/School Physics as hereafter.

Example 2.7.1 Shown in Figure 2.7.3 is a metal block of weight 100 kN suspended from a point O by a mass less inextensible string having a length of 2.5m. It is released from rest from a position 90 degree to vertical position of rest as shown below. The block of 100 kN hits another metal block of weight 500 KN connected to a spring of stiffness 2500 kN/m at point X. Considering the collision to be perfectly elastic find out the amplitude of vibration of the body considering friction less surface having

O

W1 = 100 kN

2500 mm

X

W2 = 500 kN K = 2500 kN/m

Figure 2.7.3 Conceptual diagram of the system.

• •

Un-damped motion. Damped Motion having a damper connected to W2 of magnitude 125 kN · sec/m

Solution: When the body is released form its position of rest it takes a swing and hits the 500 kN body at point X. The potential energy of the 100 kN body at its initial position = W1 h 1 W1 Kinetic energy of the 100 kN body at the point of impact = u21 2 g Applying the law of conservation of energy i.e. KE = PE,

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 235

we have, u1 = Now let,

2gh m/sec: For h = 2.5 m → u1 = 7m/ sec

u1 = Initial velocity of the body W1 @ 7.0 m/sec before collision; u2 = Initial velocity of the body W2 @ 0.0 m/sec before collision; v1 = Final velocity of the body W1 after collision, and v2 = Final velocity of the body W2 after collision. Then based on conservation of momentum and the collision being elastic we have, W1 W2 W1 W2 u1 + u2 = v1 + v2 g g g g or, m1 u1 + m2 u2 = m1 v1 + m2 v2

and

1 1 1 1 m1 u21 + m2 u22 = m1 v12 + m2 v22 2 2 2 2

Based on the above boundary conditions, we have m1 u1 = m1 v1 + m2 v2

and m1 u21 = m1 v12 + m2 v22

Substituting the numerical data mentioned in the problem, one can have 100v1 + 500v2 = 700

and 100v12 + 500v22 = 4900

Solving the above two equations we have ; v1 = −3.90 m/ sec and v2 = 2.18 m / sec. Here the negative value for v1 means that the 100 kN body will rebound back with a velocity of 3.9 m/sec. Now applying D’Alembert’s equation to the body connected to the spring we have mx¨ + Kx = 0

where K = spring stiffness.

→ m

dv + Kx = 0, dt

i.e. m

dv dx dx + Kx = 0, as = v we have, mvdv + Kxdx = 0 dx dt dt

where v = velocity vector of the body

The above differential equation has boundary condition as at v = v0 , x = 0 and at v = 0, x = δ 0 →

vdv +

m v0

δ Kxdx = 0, this on simplification gives, δ = v 0

© 2009 Taylor & Francis Group, London, UK

m . K

236 Dynamics of Structure and Foundation: 2. Applications

Also, δ =

v ω

where ω = the natural frequency of the structure.

ω=

Thus,

K = m

The amplitude, δ =

2500 × 9.81 = 7 rad/ sec 500 2.18 v = = 0.3114 m. ω 7

It is to be noted that from the above calculation we have managed to find out only the magnitude of the maximum amplitude. It does not tell us how the body will vibrate under this impact force. To get this history let us consider the differential equation mx¨ + Kx = 0; and let, x = C1 sin ωt + C2 cos ωt be the solution to the above. Applying the boundary condition at t = 0 x = 0 and v = 2.18 m/sec, we have At t = 0 x = 0 → C2 = 0 or x = C1 sin ωt and x˙ = C1 ω cos ωt at t = 0; → C1 = ωx˙ = vω0 from which we deduce, x=

v0 sin ωt → x = 0.3114 sin 7t ω

The above when plotted at time step of 0.05 seconds shows a curve as furnished in Figure 2.7.4. With damped vibration for single degree of freedom, we have seen earlier in Chapter 3 (Vol. 1) that amplitude of vibration is given by x = e−Dωn t [C1 cos ωd t + C2 sin ωd t] √ where, ωd = ωn (1 − D2 ) and D = c/cc and cc = 2 km.

Displacement history under initial velocity v0

0.4 0.2

-0.2 -0.3 -0.4

Time Steps

Figure 2.7.4 Displacement history undamped case.

© 2009 Taylor & Francis Group, London, UK

66

61

56

51

46

41

36

31

26

21

16

11

0 -0.1

6

0.1 1

Amplitude (meter)

0.3

Amplitude (meter)

Analysis and design of machine foundations 237

Now, for t = 0 when x = 0 → C1 = 0 which reduces the above equation to x = C2 e−Dωn t sin ωd t Again for t = 0, x˙ = v0 we have, C2 = v0 /ωd which results in the equation, x=

v0

ωn (1 − D2 )

e−Dωn t sin ωd t

Here we have, v0 = 2.18 m/ sec, ωn = 7 rad/ sec, D = 0.175. Substituting the above values, we have → x = 0.316e−1.225t sin 6.892t. Plotting the above values at time step of 0.05 sec we see the time history curve is as given in Figure 2.7.5.

Damped displacement under initial velocity v0

0.2 0.1 Amplitude (m) 65

61

57

53

49

45

41

37

33

29

25

21

17

13

9

5

0 1

Amplitude (meter)

0.3

-0.1

-0.2

Time Steps

Figure 2.7.5 Displacement history damped case.

The above response shows some very interesting results. While for un-damped motion the curve follows a sinusoidal pattern, for damped case it initially starts with peak amplitude and quickly dies down due to the inherent damping in the system in contrary to the harmonic loading, where the body continues to vibrate under the application of the externally applied force. So far so good, we have managed to arrive at the behavior pattern of a system having a single degree of freedom subjected to impact load albeit some idealization such as • • •

The string is inextensible and mass less The collision is perfectly elastic24 The spring and damper is mass less having identified definite values.

24 There is no collision in nature that is perfectly elastic for some energy is always dissipated out in form of heat or sound thus we usually use a term co-efficient of restitution. We will learn more about it subsequently.

© 2009 Taylor & Francis Group, London, UK

238 Dynamics of Structure and Foundation: 2. Applications

Now the question boils down to how does the above problem relates to a hammer foundation which we are supposed to discuss herein? To explain this, we need to clarify how does a hammer foundation work? Based on the General arrangement of hammer foundation shown earlier, the hammer or the tup either undergoes a free fall on the anvil or falls under a certain pressure (for double acting hammers). It either ﬂattens or forge the metal on the anvil to a desired shape or may even crush it to lower particle size (in case of a crusher) depending upon for what purpose the machine is being put to use. Irrespective of its function, the basic point that remains unaltered is the following: In contrary to the foundation supporting centrifugal or reciprocating type of machines where the foundation is subjected to a constant external harmonic force the hammer foundation induces a transient force at the point of collision and then ceases to exist till the next blow is induced25 . Thus based on the above statement we can postulate that for design of machine foundations of this type we need to analyze the system subject to transient shocks. Hence as a first step let us see what type of mathematical model is in vogue for analysis of these types of foundations.

2.7.2 Mathematical model of a hammer foundation For foundations resting on ground supporting anvils mounted on elastic base we usually consider a system having two degrees of freedom as shown hereafter. Shown in Figure 2.7.6 is the mathematical model of a hammer foundation resting on soil where, m = mass of the hammer or the tup; m1 = mass of the foundation block plus frame resting on it if any; m2 = mass of the anvil resting on elastic pad; k1 = soil spring to be calculated from Barkan or Richart’s formula26 ; k2 = spring value for the elastic pad, this value is normally furnished by the vendor supplying these pads; c1 = damping of the soil to be obtained from Richart’s formula; c2 = damping of the elastic pad again furnished by the vendor; H = height of the free fall of the hammer; x2 = amplitude vector of the anvil; and x1 = amplitude vector of the Foundation. We had already seen in earlier27 that for bodies having two degrees of freedom the free equation of vibration is given by

m1 0

0 m2

x¨ 1 k + k2 + 1 x¨ 2 −k2

−k2 k2

x1 = 0, x2

25 We hope by now the reader can smell the congruence with the worked out example 2.7.1. 26 Refer to section of block foundation for the formula of the springs and dampers. 27 Refer Chapter 5 (Vol. 1) on basic concepts in Structural Dynamics.

© 2009 Taylor & Francis Group, London, UK

(2.7.1)

Analysis and design of machine foundations 239

m

H

m2 x2 c2

k2

m1 x1

c1 k1

Figure 2.7.6 Mathematical model of hammer foundation resting on soil directly.

and since this is a statically coupled equation the damped free vibration of motion is given by (Meirovitch 1975)

m1 0

2.7.2.1

0 m2

x¨ 1 c + c2 + 1 x¨ 2 −c2

−c2 c2

x˙ 1 k + k2 + 1 x˙ 2 −k2

−k2 k2

x1 =0 x2

(2.7.2)

Mathematical model of foundation resting inside a trough

In this case the mathematical model for analysis of the system is as shown in Figure 2.7.7. In the above mathematical model, m = mass of the hammer or the tup; m1 = mass of the trough resting on soil; m2 = mass of the foundation plus hammer frame resting on it; if any, © 2009 Taylor & Francis Group, London, UK

240 Dynamics of Structure and Foundation: 2. Applications

m

H

m3 x3

k3

c3

m2 x2

k2

c2

x1 m1

k1

c1

Figure 2.7.7 Mathematical model of hammer foundation resting inside a trough.

m3 = mass of the anvil resting on elastic pad supplied by vendor; k1 = soil spring to deduced from either Barkan, Richarts formula; k2 = spring value for the elastic pad/spring on which the foundation is resting whose value is normally furnished by the vendor supplying these pads; k3 = spring value for the elastic pad supporting the anvil whose value is normally furnished by the vendor supplying these pads; © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 241

c1 = damping of the soil to be obtained from Richart’s formula; c2 = damping of the elastic pad /spring supporting the foundation on the trough; c3 = damping of the elastic pad supporting the anvil; H = height of the free fall of the hammer; x1 = amplitude of the trough resting on soil; x2 = amplitude of the RCC block resting on springs; and x3 = amplitude of the anvil resting on the elastic pad. The free vibration of motion for the above is given by ⎡

m1 ⎣0 0

0 m2 0

⎤⎧ ⎫ ⎡ 0 ⎨x¨ 1 ⎬ k 1 + k2 0 ⎦ x¨ 2 + ⎣ −k2 ⎩ ⎭ m3 x¨ 3 0

−k2 k2 + k 3 −k3

⎤⎧ ⎫ 0 ⎨x1 ⎬ −k3 ⎦ x2 = 0 ⎩ ⎭ k3 x3

and

the damped equation of motion is given by ⎡

m1 ⎣0 0

⎤⎧ ⎫ ⎡ c 1 + c2 0 0 ⎨x¨ 1 ⎬ −c2 m2 0 ⎦ x¨ 2 + ⎣ −c2 c2 + c 3 ⎩ ⎭ x¨ 3 0 m3 0 −c3 ⎤⎧ ⎫ ⎡ k1 + k2 −k2 0 ⎨x1 ⎬ ⎦ x2 = 0 + ⎣ −k2 k+ k −k 3 3 2 ⎩ ⎭ x3 0 −k3 k3

⎤⎧ ⎫ 0 ⎨x˙ 1 ⎬ −c3 ⎦ x˙ 2 ⎩ ⎭ c3 x˙ 3 (2.7.3)

Based on Equation (2.7.3), we have managed to formulate the equation of motion for the hammer foundation, but how it will behave under transient load is yet to be ascertained, though. Based on Example 2.7.1, we have shown the vibration characteristics of a body having single degree of freedom under impact loading, we will now extend this theory to multi-degrees of freedom. 2.7.2.2

Un-damped Response of a system under impact loading having multi-degree of freedom

We write equation of motion in matrix notation as & ' ¨ + [K] {X} = 0 [M] X

(2.7.4)

Here [M] = A square mass matrix of the order n × n; [K] = A square stiffness matrix of the order n × n; {X} = A Column deflection matrix of order n × 1 (which means n rows and 1 column). Based on the orthogonal property of the matrix28 we have ¨ + [ϕ]T [K][ϕ]{X} = 0 [ϕ]T [M][ϕ]{X}

which de-couples to

28 Refer Chapter 5 (Vol. 1) for the orthogonal property of the Matrix.

© 2009 Taylor & Francis Group, London, UK

(2.7.5)

242 Dynamics of Structure and Foundation: 2. Applications N

{{ξ¨i } + ωi2 {ξi }} = 0 where {X} =

i=1

N

[φ]{ξi }

(2.7.6)

i=1

Now let {ξi } = {Ai sin ωi t + Bi cos ωi t}. Since, {X} =

N

(2.7.7)

[φ] {ξi } , we have, {X} =

N

i=1

[φ] {Ai sin ωi t + Bi sin ωi t}

i=1

Multiplying both sides of the above expression by the term [φ]T [M] we have

[φ]T [M] {X} =

N

[φ]T [M] [φ] {Ai sin ωi t + Bi sin ωi t} which reduces to

i=1

[φ]T [M] {X} =

N

{Ai sin ωi t + Bi sin ωi t}

i=1

Now imposing the boundary condition at t = 0

{0} =

N

{X} = {0}, we have

{Ai sin ωi t + Bi cos ωi t}, which implies

i=1

{Bi } = {0}, thus we have, [φ]T [M]{X} =

N

{Ai sin ωi t}

i=1

˙ = {V0 } we have Again imposing the boundary condition at t = 0, {X}

[φ]T [M] {V0 } =

{X} =

N i=1

N

{Ai ωi }

i=1

⇒ {Ai } =

[φ] [M] {V0 } [φ] {sin ωi t} ωi

[φ] [M] {V0 } , which gives ωi

T

(2.7.8)

Based on the above we can clearly infer that for {sin ωi t} = {1}, we have the maximum value of the amplitude vector. We now explain further, the phenomenon based on a suitable numerical example. For the numerical worked out problem below, we have deliberately used a theoretical data with an objective that you can follow the process clearly. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 243

Example 2.7.2 For a system having the following data find out the amplitude of vibration when the mass m2 is subjected to an initial velocity of 0.5 m/sec.

30 −10 , [K] = −10 30

1 [M] = 0

0 , 1

0.0 and {V} = 0.5

Solution: The free vibration of motion for the body is given by m1 0

0 m2

x¨ 1 k11 + x¨ 2 −k21

−k12 k22

x1 =0 x2

The eigen value solution of the problem is expressed as

30 − λ −10

−10 =0 30 − λ

and this on simplification reduces to (λ − 20) (λ − 40) = 0 which gives λ = 20 or ⇒ ω1 = 4.35 rad/sec; and λ = 40 : ⇒ ω2 = 6.32 rad/sec. Calculation of the eigen-vectors For the first mode having λ = 20 10φ1 − 10φ2 = 0

and − 10φ1 + 10φ2 = 0

Thus for φ1 = 1, we have φ2 = 1 For the second mode having λ = 40 −10φ1 − 10φ2 = 0

and −10φ1 − 10φ2 = 0

Thus for φ1 = 1 we have φ2 = −1

This the complete eigen vector matrix as

1 [ϕ] = 1

1 . −1

The normalized eigen vector, based on orthogonal transformation theory is given by √ 1/√2 [ϕ]n = 1/ 2

√ 1/ √2 −1/ 2

© 2009 Taylor & Francis Group, London, UK

244 Dynamics of Structure and Foundation: 2. Applications

We had deduced earlier that {X}

=

N . [φ]T [M]{V0 } / ωi

i=1

[φ] {sin ωi t} ,

where {X} = [φ] {ξi } , thus based on mathematical symmetry we have {ξi } = N . [φ]T [M]{V0 } / {sin ωi t} . i=1 ωi Hence for the first mode

1 {ξ1 } = √ 2

1 √ 2

1 0 0 × sin 4.5t × 1 0.5 4.5

1 0

the above on simplification reduces to [ξ1 ] = 0.07856 sin 4.5t For the second mode we have 1 {ξ2 } = √ 2

1 −√ 2

1 0

1 0 0 × sin 6.32t × 1 0.5 6.32

the above on simplification reduces to {ξ2 } = −0.056 sin 6.32t √ 0.07856 sin 4.5t 1/ √2 −1/ 2 −0.056 sin 6.32t

0.0555 sin 4.5t − 0.0396 sin 6.32t ➔ {X} = 0.0555 sin 4.5t + 0.03965 sin 6.32t √ 1/√2 1/ 2

As {X} = [φ] {ξi } , we have, {X} =

The above when plotted at a time step of 0.05 sec gives plots as shown in Figure 2.7.8. Undamped motion of the system

0.12 0.1 0.08

Amplitude

0.06 0.04 x1

0.02

x2

0 -0.02

1

5

9

13

17

21

25

29

33

37

41

45

49

53

57

61

65

-0.04 -0.06 -0.08 -0.1

Time Steps

Figure 2.7.8 Undamped response of a system under impact loading having multi-degree of freedom.

The above plot shows how the two degree system body vibrates under a transient initial velocity of 0.5 m/sec.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 245

2.7.2.3

Damped response of a system under impact loading having multi-degree of freedom

We had already discussed in our earlier chapters that undamped motion of vibration is an idealization which is practically rare in nature and as such it would now be worthwhile to asses (Chowdhury et al. 2002) how damping affect the above phenomenon. We write equation of motion in matrix notation as ¨ + [C]{X} ˙ + [K]{X} = 0 [M]{X}

(2.7.9)

where [M] = a square mass matrix of the order n × n; [K] = a square stiffness matrix of the order n × n; [C] = a square damping matrix of order n × n; {X} = a column deflection matrix of order n × 1 (which means n rows and 1 column) Based on the orthogonal property of the matrix we have ¨ + [ϕ]T[C][ϕ]{X} ˙ + [ϕ]T [K][ϕ]{X} = 0 [ϕ]T [M][ϕ]{X}

(2.7.10)

which de-couples to N

{{ξ¨i } + 2Di ωi {ξ˙i } + ωi2 {ξi }} = 0

where {X} =

i=1

N

[φ]{ξi }

(2.7.11)

i=1

We have already proved earlier that for body having single degree of freedom the free damped equation of motion is given by x = e−Dωn t [C1 cos ωd t + C2 sin ωd t]

(2.7.12)

where, ωd = ωn [1 − D2 ]. Thus in transformed co ordinate when the equations get de-coupled we can write Let {ξi } = e−Di ωni t {Ai sin ωdi t + Bi cos ωdi t} , and since {X} =

N

(2.7.13)

[φ] {ξi } , we have,

i=1

{X} =

N

[φ] e−Di ωni t {Ai sin ωdi t + Bi cos ωdi t}

i=1

© 2009 Taylor & Francis Group, London, UK

(2.7.14)

246 Dynamics of Structure and Foundation: 2. Applications

Multiplying both sides of the above expression by the term [ϕ]T [M] we have

[φ]T [M]{X}=

N

[φ]T [M][φ]e−Di ωni t {Ai sin ωdi t + Bi cos ωdi t} which reduces to

i=1 T

[φ] [M] {X} =

N

e−Di ωni t {Ai sin ωdi t + Bi cos ωdi t}

i=1

Now imposing the boundary condition at t = 0

{0} =

N

{X} = {0}, we have,

e−Di ωni t {Ai sin ωdi t + Bi cos ωdi t} which implies,

i=1 T

[Bi ] = [0] , thus we have [φ] [M] [X] =

N

e−Di ωni t {Ai sin ωdi t}

i=1

Again imposing the boundary condition at t = 0

[φ]T [M] {V0 } =

N

{Ai ωdi } ⇒ {Ai } =

i=1

˙ = {V0 } we have, {X}

[φ]T [M] {V0 } , which gives ωi

N [ϕ]T [M]{V0 } {X} = [ϕ]e−Di ωni t {sin ωi t} ωdi

(2.7.15)

i=1

It will be interesting to note that in this case the maximum amplitude does not occur at ωi t = π/2 as in the case of undamped vibration. To get the maximum amplitude we need to plot the complete time history which was not required for the undamped case. We now further explain the phenomenon based on a suitable numerical example.

Example 2.7.3 Repeat the problem worked out in Example 2.7.1 with following damping ratio D1 = 0.15 and D2 = 0.20. All other parameters remain the same as in Example 2.7.1. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 247

Solution: The free damped equation of motion for the problem is given by m1 0

0 m2

x¨ 1 c + 11 x¨ 2 −c21

−c12 c22

x˙ 1 k11 + x˙ 2 −k21

−k12 k22

x1 =0 x2

We had already proved earlier that based on eigen solution ⇒ ω1 = 4.35 rad/sec and ⇒ ω2 = 6.32 rad/sec. The normalised eigen vector matrix for this case was calculated in the previous example as √ 1/ √2 −1/ 2

√ 1/√2 1/ 2

[ϕ]n =

N [ϕ]T [M] {V0 } Since {X} = [ϕ] e−Di ωni t {sin ωi t} and ωdi i=1

{X} =

N

[ϕ] {ξi } we have

i=1

N [ϕ]T [M] {V0 } −Di ωni t {ξi } = {sin ωi t} e ωdi i=1

Thus for first mode we have 1 {ξ1 } = √ 2

1 √ 2

1 0

0 1

1 0 × √ 0.5 4.5 (1−0.0225)

× e−0.675t × sin 4.449t or {ξ1 } = 0.07946 · e−0.675t · sin 4.449t For second mode we have 1 {ξ2 } = √ 2

1 −√ 2

1 0

0 0 × e−1.264t × sin 6.19t 1 0.5

or {ξ2 } = −0.0571168 e−1.264t sin 6.19t

© 2009 Taylor & Francis Group, London, UK

248 Dynamics of Structure and Foundation: 2. Applications

Now since √ 1/√2 {X} = [ϕ] {ξi } we have, {X} = 1/ 2 i=1

0.07946e−0.675t sin 4.49t × −0.0571168e−1264t sin 6.19t N

√ 1/ √2 −1/ 2

0.0562e−0.675t sin 4.49t − 0.0404e−1.264t sin 6.19t or {X} = 0.0562e−0.562t sin 4.49t + 0.0404e−1.264t sin 6.19t

The above equations when plotted at a time step of 0.05 sec, shows the history as given in Figure 2.7.9. Damped response of motion

0.08

Amplitude

0.06 0.04 x1 0.02

x2

0 1

4

7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67

-0.02 -0.04

Time steps

Figure 2.7.9 Damped response for the two man system.

Observations: • •

On studying the above mentioned plot you will see that the characteristic of the curve is very similar to that plotted for the single degree of freedom. Also observe that the response is largely reduced compared to Example 2.7.2 due to the consideration of damping in this case.

2.8 DESIGN OF HAMMER FOUNDATION

2.8.1 Design criteria for hammer foundation Its time we emerge form the exotic world of Newtonian Mechanics and start looking into the design parameters pertaining to hammer foundations. In the design criteria we use the following nomenclatures • • •

Wh = weight of Hammer in kN; Wa = weight of the anvil in kN; Wfr = weight of hammer resting on foundation in kN;

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 249

• • • • • • • • • • • • • • • • • •

Wf = weight of the RCC foundation in kN; La × Ba = contact Area of the anvil in m2 ; L × B = contact are of the foundation in m2; Ea = Young’s Modulus of the elastic pad below the anvil in kN/m2 ; ta = thickness of the elastic pad in m; G = Dynamic shear modulus of the soil in kN/m2 ; g = acceleration due to gravity @ 9.81 m/sec2 ; Da = critical damping ratio of the elastic pad; Ds = critical damping ratio of the soil; p = pressure on piston of double acting hammer kN/m2 ; Ap = area of piston in m2 ; L = length of the stroke in m; α = correction factor @ 1.0 for well adjusted hammer and varies between 0.5 to 0.8 for double acting hammers. Usual design value taken is 0.65; k = coefficient of impact @ 0.5 for stamping hammers and 0.25 for forging hammers; v = velocity of the hammer at the point of impact in m/sec; Va = velocity of anvil after the impact in m/sec; σs = allowable static bearing capacity of the soil in kN/m2 ; and σp = allowable stress of the elastic pad in kN/m2 .

2.8.1.1

Maximum permissible amplitudes for foundation

The maximum permissible amplitude of vibration for a hammer foundation shall not exceed 1.2 mm. 2.8.1.2

Maximum permissible amplitudes for the anvil

The maximum permissible amplitude is usually dependent on the weight of the hammer and is also at times prescribed by the vendor supplying the elastic pad who limits the deflection based on the allowable stress of the elastic pad. Following Table 2.8.1 furnishes the allowable amplitude of the anvil based on hammer weight. 2.8.1.3

Minimum weight of foundation

The minimum weight of foundation is given by Wa + Wfr Wmin = Wh 8(1 + k)v − Wh Table 2.8.1 Sl. No.

Weight of hammer (KN)

Max. permissible amplitude (mm)

1 2 3

30 20 Upto 10

3–4 2 1

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250 Dynamics of Structure and Foundation: 2. Applications

2.8.1.4

Minimum base area of the foundation

The minimum base area of the foundation may be obtained from the following expression Amin =

20(1 + k) v · Wh σs

2.8.1.5 Minimum thickness of foundation (Major 1980) This is given in Table 2.8.2. 2.8.1.6 Velocity, V, of hammer at point of impact •

For free falling hammer V = α 2gH If energy of impact Ei is given by the manufacturer then H=

•

Ei Wh

and V =

2gH

For double acting hammer V = 0.65

2.8.1.7

2g(Wh + pAp )l Wh

Velocity of anvil after of impact (1 + k)

V=

1+

Wa Wh

v

Table 2.8.2 Sl. No.

Weight of hammer (kN)

Thickness of foundation (mm)

1 2 3 4 5

>60 60 40 20 10

>2250 2250 1750 1250 1000

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Analysis and design of machine foundations 251

2.8.1.8 • • •

Calculation of natural frequency based on IS-2974 ka Calculate ωa = ma kz Calculate ωz = ma + mf + mfr Calculate ωn1 and ωn2 from the equation

ωn4 − (ωa2 + ωz2 ) (1 + α) ωn2 + (1 + α)ωa2 ωz2 = 0;

4Gr0 where, ka = EataAa and kz = (1−υ) in which, r0 = modulus of the soil. When supported on short bearing piles

kz =

α=

ma mf + mfr

L×B π

and, G = dynamic shear

kp · kz kp + kz

where, kp = vertical stiffness of pile which may be obtained from formulas derived W W earlier and ma = Wg a , mf = g f and mfr = gfr . 2.8.1.9 •

The amplitude of the RCC foundation (x1 ) is given by x1 =

•

Amplitude of vibration based on IS-2974

2 )(ω2 − ω2 ) −(ωa2 − ωn2 a n1 2 − ω2 )f ωa2 (ωn1 n2 n2

sin ωn1 t sin ωn2 t V − ωn1 ωn2

The amplitude of vibration of the anvil (x2 ) is given by x2 =

2.8.1.10

V 2 − ω2 ) (ωn1 n2

2 ) sin ω t 2 ) sin ω t (ω2 − ωn1 (ωa2 − ωn2 n1 n2 − a ωn1 ωn2

Stability of pad between anvil and block

Total deflection of the pad under impact is given by δtot = δst + δdyn where δst =

Wa + Wfr Ka

and

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δdyn =

V ωna

252 Dynamics of Structure and Foundation: 2. Applications

2.8.1.11

The loading intensity on the pad

The loading intensity on the pad is given by σp =

2.8.1.12

Ka δtot Aa Stress in soil below foundation

The stress in soil below foundation is given by σs =

Wa + Wf + Wfr + Kz ψ

in which, ψ =

Af Vˆ , 2π × fz

1+k where Vˆ = V. W 1 + Waf

2.8.2 Discussion on the IS-code method of analysis The discussion should not be deemed as a criticism of the method, for normal engineering design of standard hammer foundations the method advocated by the code is adequate. But when a foundation subjected to loading of large magnitude (heavy hammer >40 kN), the method proposed by the code may lead to conservative and expensive design. Moreover if the environmental criteria calls for more stringent restriction of amplitude or transmittal of waves, a designer may find it difficult to meet the requirements based on IS-code. The reasons attributable for the same may be summarized as follows: •

• • •

IS-code method does not take damping of the pad or that of the soil into consideration. It has been observed that damping plays a very signiﬁcant role in minimizing the amplitude of vibration for such hammer foundation (Novak and El Hifnawy 1983). It also does not take into consideration the embedment effect which could play a very significant role for heavy hammer foundation when the depth of the block could be quite large. The IS-code formula of kz = 7.6Gr0 apparently looks overestimated29 . The dynamic displacement (δdyn ) is based on uncoupled form when the actual response should based be coupled response. This could either under-estimate or could also over estimate the stress induced in the foundation.

Based on the above the design procedure suggested herein may be structured as follows.

29 For if we equate 4Gr0 /(1 − v) = 7.6Gr0 . We get v = 0.473 => 0.5. Poisson’s ratio @ 0.5 depicts perfectly plastic clay which is rarely obtained. Value of v is usually taken as 0.4.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 253

2.8.3 Check list for analysis of hammer foundation Check if the following data regarding the machine and the foundation are available with you • • • • • • • • • • • • • •

Weight of the hammer. Type of hammer (free falling or double acting etc.). Sufficient data to calculate the velocity of hammer at point of collision. (like energy of impact, height of free fall etc.). Geometric dimension and weight of anvil. Geometric dimension, stiffness and damping property of the elastic pad supporting the anvil. Anchoring detail of the elastic pad. Stiffness and damping data if mechanical springs and dampers are used in lieu of elastic pads. Mechanical detail of the springs and dampers including their ﬁxing detail as suggested by the supplier. Anchoring detail of the frame supporting the hammer on the foundation. Top elevation of the anvil based on the mechanical process. Allowable bearing capacity of the soil. Dynamic shear modulus of the soil. Grade of concrete.

2.8.4

Other techniques of analysis of Hammer foundation

For most of the cases analysis as mentioned above suffice. However there are cases where due to the massiveness of the foundation and hammer more detailed analysis is envisaged where the effect of generated shock to its surrounding could be significant30 . In such cases the best way to analyze such problem would be to resort to FEM analysis of such foundations. Shown in Figure 2.8.1 is a conceptual Finite element mathematical model of hammer foundation housed inside a structural building. If the hammer foundation is very heavy, though the hammer foundation itself may be within the acceptable limit of codal stipulation can yet have adverse effect on the building in which it is located. In such cases studying the problem based on FEM could be quit advantageous. 2.8.4.1 • •

Selection of elements

Here the anvil is modeled as a beam element having thee degrees of freedom (horizontal, vertical and rotational degrees). The elastic pad below the anvil which is usually modeled as a spring can be converted to equivalent truss element having stiffness @ AE/L for computer implementation.

30 It has not been uncommon that the shocks generated by hammer foundation has done secondary damages to the building in which it is placed or have rendered crane girders unserviceable at the Gantry level due to distortion.

© 2009 Taylor & Francis Group, London, UK

254 Dynamics of Structure and Foundation: 2. Applications

Anvil modeled as beam element Lumped mass(Typ)

Pad modeled as truss element

Figure 2.8.1 Finite element model of hammer foundation.

• • •

The block foundation itself can be modeled as a 2D plane stress element having incompatible modes. The soil medium can be modeled as 2D plane strain element again having incompatible modes. Finally the building and the foundation may be modeled as a plane frame constituting of beam elements connected to the soil elements.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 255

2.8.4.2

Boundary conditions

This is surely a major problem for failing to take a correct judgment can render the analysis useless for waves generated due to the transient shock from the hammer may get reflected from the boundary and generate spurious response of the foundation as well as the building rendering the analysis useless. We had already discussed earlier that for this type of infinite domain problem the boundary should be extended far enough ensuring that no reflection of the waves take place. This on the other hand makes the analysis voluminous and also expensive in terms of man-hours. Moreover from practical point of view geo-technical data may not be available to the depth required to ensure that no reflection takes place. This can however be done by following techniques: To provide spring and dash-pots having high damping value at the boundaries which ensures that it absorbs all energy transmitted to it 31 . Provide paraxial or quite boundaries to suppress the spurious modes32 . Providing boundaries at a distance at least 2.5 times the length of the Rayleigh waves to ensure that radiation damping is good enough to dissipate away the energy. 2.8.4.3

Material input

Following, material input shall be provided for the analysis: • • • • • •

Dynamic Modulus of concrete for the RCC block. Dynamic modulus of the elastic pad. Dynamic shear modulus of the soil33 . Poisson’s ratio of the soil. Damping property of elastic pad below anvil (in this case the truss elements). Dynamic elastic modulus of the structure enabling the computer to generate the stiffness matrix.

2.8.4.4

Input loading

The velocity of the anvil after impact shall be directly provided as input velocity for the nodes of the beam element where masses are lumped. 2.8.4.5

Method of analysis

Due to the heterogeneous properties of the material we advocate here a time history analysis having a time step of 0.1Tn where Tn is the least period of the system.

31 Unfortunately most of the commercially available FEM software does not have this feature of directly inputting dash pots except ANSYS. 32 Refer to the Chapter 5 (Vol. 1) for further explanation. 33 If the soil is layered then for each layer a separate value of shear modulus has to be provided enabling the computer to develop the material stiffness matrix for the plane strain element.

© 2009 Taylor & Francis Group, London, UK

256 Dynamics of Structure and Foundation: 2. Applications

If the building system is found to have unacceptable amplitude ways and means have to be sought to reduce the vibration transmitted to it by either providing air gaps around the foundation or by providing suitable dampers around the foundation to absorb this energy. Example 2.8.1 A hammer foundation (Figure 2.8.2) having the following data has to be designed for a particular site.

Frame Column(Typ.)

1400

2000

1400

1150

1250

1250

1150

1290

2150

Figure 2.8.2 General arrangement of the hammer foundation.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 257

• • • •

Calculate the natural frequency of vibration The damped amplitude of vibration Compare the same with time history response Re-analyse the foundation based on IS-Code or Barkan’s mathod and compare the results.

Design input data 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Weight of hammer = 20 kN Weight of frame = 85 kN Plan area of anvil = 2.0 m × 2.5 m Thickness of oak pad = 0.4 m Height of free fall = 1.8 m Unit weight of soil = 10 kN/m3 Dynamic shear modulus of soil = 120,0000 kN/m2 Poisson’s ratio = 0.40 Dynamic modulus of oak wood = 500,0000 kN/m2 Damping ratio for oak = 0.10 Plan area of foundation = 5 m × 5 m Co-efficient of restitution = 0.65 Allowable bearing capacity of soil = 180 kN/m2 Air gap between anvil and foundation = 100 mm

Let height of the R.C.C. anvil be = 1250 mm Weight of anvil (Wa ) = 2.5 × 2.0 × 1.25 × 25 = 156.25 kN Weight of hammer = 20 kN √ Velocity of hammer at point of collision = 2gH = v = 2 × 9.81 × 1.80 = 5.94 m/sec Velocity of Anvil after the impact (1 + k)

V=

1+

Wa Wh

v =

1.65 1+

156.25 20

5.94 = 1.112 m/sec

Minimum weight of foundation Wa + Wfr Wmin = Wh 8(1 + k)v − Wh 156.25 + 85 = 20 8 × 1.65 × 5.94 − = 1327.16 kN 20 Thus minimum depth required =

1327.16 = 2.12 m ∼ = 2.15 m(say) 5 × 5 × 25

Thus based on the sketch furnished earlier, Weight of foundation = (5 × 5 × 3.44 − 2.7 × 2.2 × 1.29) × 25 =1958 kN.

© 2009 Taylor & Francis Group, London, UK

258 Dynamics of Structure and Foundation: 2. Applications

Weight of the oak pad = 2.5 × 2.0 × 0.4 × 20 = 40 kN Minimum base area required = Amin =

20(1 + k) 20 × 1.65 × 5.94 × 20 = 21.78 > 25 m2 OK v · Wh = σs 180

Stiffness properties For Pad stiffness (k2 ) =

EAp 500 × 104 × 5 = = 625 × 104 kN/m t 0.4

For Soil Equivalent Radius r0 =

kz =

25 = 2.82 m π

4Gr0 4 × 120 × 104 × 2.82 = = 22560000 kN/m (1 − υ) 0.6

Damping properties 40 = 4.0 kN sec2 /m 9.81 √ Cc = 2 km = 2 625 × 105 × 4 = 31623 kN · sec/m

Mass of oak pad =

√ C = D × 2 km = 0.1 × 31623 = 3162 kN · sec/m Mass of foundation and machine = Bz =

0.25 (1 − υ) mg ρs r30

=

1958 + 156 + 85 = 224 kN · sec2 /m 9.81

0.25 × 0.6 × 224 × 9.81 19 × (2.82)3

= 0.773

0.425 = 0.4832 D= Bz Thus damping of the soil is given by √ √ Cs = D × 2 km = 2 × 0.4832 × 22560000 × 224 = 68699 kN · sec/m The mathematical model The mathematical model perceived is given in Figure 2.8.3.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 259

m2 x2 k2

c2

m1

k1

c1

Figure 2.8.3 Mathematical model of the foundation.

The equation of motion for the above is given by

m1 0

0 m2

x¨ 1 C + C2 + 1 x¨ 2 −C2

k1 + k2 −k2

+

−k2 k2

−C2 C2

x˙ 1 x˙ 2

x1 =0 x2

For free vibration we have

m1 0

0 m2

➔

x¨ 1 k + k2 + 1 x¨ 2 −k2

−k2 k2

x1 =0 x2

208 0 0 16

x¨ 1 850.6 −625 x1 + × 105 = 0 −625 625 x¨ 2 x2

For eigen solution we have

85060000 − 208λ −62500000 x1 =0 −62500000 62500000 − 16λ x2

© 2009 Taylor & Francis Group, London, UK

x1

260 Dynamics of Structure and Foundation: 2. Applications

The above on expansion and simplification reduces to λ2 − 4315192λ + 4.2367788 × 1011 = 0; → λ1 = 100525 λ2 = 4214667

⇒

ω1 = 317 rad/sec

and

ω2 = 2052 rad/sec

and

Calculation of eigen vectors For the first mode

85060000 − 20909200 −62500000 φ11 = 0; −62500000 62496672 φ12

for φ11 = 1.0

we have φ12 = 1.0264

For the second mode 85060000 − 8.7665074 × 108 −62500000

φ21 = 0; for φ21 = 1.0 φ22

−62500000 −4934672

we have φ22 = −12.665 Thus the complete eigen vector is

1.0 [ϕ] = 1.0264

1.0 −12.665

Normalization of the eigen vectors For the first mode

0 1.00 = 224.856 16 1.0264

√ 0.066688 Mr = 224.856 = 14.495 → {ϕ1 }N = 0.06844

{φ}T [M] {φ} = 1.00

1.0264

208 0

and

For the second mode

208 0 1.00 {φ} [M] {φ} = 1.00 −12.665 = 2774 0 16 −12.665

√ 0.01898 Mr = 2774 = 52.67 → {ϕ1 }N = −0.240 T

Thus the complete normalized eigen vector is [ϕ]N =

© 2009 Taylor & Francis Group, London, UK

0.0667 0.0684

and

0.01898 . −0.240

Analysis and design of machine foundations 261

Damping matrix and its correction for de-coupling

−c2 71861 −3162 = −3162 3162 c2

c1 + c2 −c2

The damping matrix is given by [C] =

The above matrix on orthogonal transformation does not de-couple as such we need to correct it. For the first mode {φ}T [C] {φ} = 0.0667

0.0684

71861 −3162

−3162 3162

0.0667 = 305 0.0684

→ 2D1 ω1 = 305 ➔ D1 = 0.480. For the second mode T

{φ} [C] {φ} =0.01898

71861 −3162 0.01898 0.0240 = 208 −3162 3162 0.240

→ 2D2 ω2 = 208 ➔ D2 = 0.050. Based on Rayleigh damping we know that on orthogonal transformation the expression should reduce to 2D1 ω1 = α + βω12

and

or α + 100525β = 305

2D2 ω2 = α + βω22 and

α + 4214667β = 208

the equation on solving gives → α = 302.63 and β = 2.3577 × 10−5 Thus, based on Rayleigh damping [C] = α [M] + β [K] → [C] = 302.63 ×

850.6 −625

208 0 + 2.3577 × 10−5 0 16

−625 64952 × 105 = 625 1474

Calculation of amplitude The equation for damped amplitude is given by N [ϕ]T [M]{V0 } {X} = [ϕ]e−Di ωnit {sin ωi t} ωdi i=1

© 2009 Taylor & Francis Group, London, UK

1474 6316

262 Dynamics of Structure and Foundation: 2. Applications

Thus for the first mode we have N [ϕ]T [M]{V0 } {ξi } = [ϕ]e−Di ωtni {sin ωi t} ωdi i=1

1 208 0 0 ξ1 = 0.0667 0.0684 × √ 0 16 1.112 317 (1 − 0.2304) √ −0.48×317t ×e × sin(317 1 − 0.2304)t

or

→ ξ1 = 4.3776e−152.16t × sin 278t × 10−3 Similarly for the second mode we have ξ2 = 0.01898 × →

0.240

208 0

0 16

0 1.112

√ 1 × e−102.6t × sin(2052 1 − 0.0025)t 2052 (1 − 0.0025) √

ξ2 = −2.08398e−102.6t × sin 2049t × 10−3

Since {X} = [ϕ] {ξ }, hence

0.0667 0.01898 4.3776e−152.16t sin 278t {X} = × 10−3 0.0684 −0.240 −2.08398e−102.6t sin 2049t 0.29198e−152.16t sin 278t − 0.03955e−102.6t sin 2049t or {X} = ×10−3 m. 0.299428e−152.16t sin 2049t + 0.50015e−102.6t sin 2049t

Amplitude (mm)

The above when plotted at a time step of 0.0005 seconds shows displacement plots as depicted in Figure 2.8.4.

0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1 1 -0.2 -0.3

Displacement History of Hammer Foundation with time step of 0.0005 seconds

Displacement amplitude of foundation (mm) Displacement amplitude of anvil (mm) 14

27

40

53

66

79

92 105 118 131 144

Time steps

Figure 2.8.4 Displacement history of foundation and the anvil.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 263

Time History analysis based on Newmark-beta method Here we do time history analysis of the equation

208 0 x¨ 1 71861 −3162 x˙ 1 + 0 16 x¨ 2 −3162 3162 x˙ 2 85060000 −62500000 x1 + =0 −62500000 62600000 x2

having boundary condition at t = 0 v = 1.112 m/sec. Solution is given in Figure 2.8.5.

Time history plot of Hammer Foundation (Newmark-beta Method)

Amplitude(meter)

0.0006 0.0004 0.0002

Amplitude of fdn. Amplitude of anvil

0 1

14

27

40

53

66

79

92 105 118 131 144

-0.0002 -0.0004

Time steps

Figure 2.8.5 Time history response of amplitude for foundation and the anvil.

It is to be noted that we use here the original non proportional damping matrix and not the corrected one used above. we give the following results for 98 steps (explanation in next page). Next we compare the response of the foundation and the anvil separately to see what is the variation in the results. The results are as plotted in Figures 2.8.6 to 7. Comparison of amplitude of foundation based on Time History and Closed form

Amplitude(mm)

0.2 0.15

Amplitude of foundation based on Newmark Method Displacement amplitude of foundation(mm)

0.1 0.05 0

1

15

29

43

-0.05

57

71

85

99 113 127 141

Time steps

Figure 2.8.6 Comparison of response, time history versus approximate damping.

© 2009 Taylor & Francis Group, London, UK

264 Dynamics of Structure and Foundation: 2. Applications Sl. Time No. step

x1 (disp)

x1 (vel)

x1 (acc)

x2 (disp)

x2 (vel)

x2 (acc)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47

0 8.03×10−06 3.90 × 10−05 9.10 × 10−05 1.40 × 10−04 1.62 × 10−04 1.53 × 10−04 1.29 × 10−04 1.15 × 10−04 1.20 × 10−04 1.38 × 10−04 1.47 × 10−04 1.36 × 10−04 1.07 × 10−04 7.67 × 10−05 5.95 × 10−05 5.87 × 10−05 6.37 × 10−05 6.04 × 10−05 4.30 × 10−05 1.79 × 10−05 −2.27 × 10−06 −9.23 × 10−06 −5.11 × 10−06 1.84 × 10−07 −2.46 × 10−06 −1.42 × 10−05 −2.79 × 10−05 −3.50 × 10−05 −3.19 × 10−05 −2.31 × 10−05 −1.64 × 10−05 −1.68 × 10−05 −2.29 × 10−05 −2.83 × 10−05 −2.76 × 10−05 −2.04 × 10−05 −1.15 × 10−05 −6.40 × 10−06 −7.06 × 10−06 −1.07 × 10−05 −1.25 × 10−05 −9.58 × 10−06 −3.15 × 10−06 2.72 × 10−06 4.66 × 10−06 2.58 × 10−06 −5.62 × 10−07

0 3.21 × 10−02 9.18 × 10−02 1.16 × 10−01 7.99 × 10−02 8.62 × 10−03 −4.58 × 10−02 −4.89 × 10−02 −9.27 × 10−03 3.21 × 10−02 3.73 × 10−02 1.53 × 10−03 −4.63 × 10−02 −6.94 × 10−02 −5.33 × 10−02 −1.53 × 10−02 1.20 × 10−02 7.88 × 10−03 −2.10 × 10−02 −4.85 × 10−02 −5.18 × 10−02 −2.89 × 10−02 1.09 × 10−03 1.54 × 10−02 5.76 × 10−03 −1.63 × 10−02 −3.05 × 10−02 −2.46 × 10−02 −3.58 × 10−03 1.58 × 10−02 1.96 × 10−02 7.21 × 10−03 −9.05 × 10−03 −1.51 × 10−02 −6.44 × 10−03 9.28 × 10−03 1.93 × 10−02 1.63 × 10−02 4.13 × 10−03 −6.79 × 10−03 −7.89 × 10−03 6.95 × 10−04 1.11 × 10−02 1.46 × 10−02 8.86 × 10−03 −1.10 × 10−03 −7.24 × 10−03 −5.33 × 10−03

0 1.28 × 10+02 1.10 × 10+02 −1.27 × 10+01 −1.32 × 10+02 −1.53 × 10+02 −6.51 × 10+01 5.27 × 10+01 1.06 × 10+02 5.96 × 10+01 −3.87 × 10+01 −1.05 × 10+02 −8.68×10+01 −5.62 × 10+00 7.01 × 10+01 8.19 × 10+01 2.72 × 10+01 −4.36 × 10+01 −7.19 × 10+01 −3.83 × 10+01 2.53 × 10+01 6.61 × 10+01 5.39 × 10+01 3.34 × 10+00 −4.19 × 10+01 −4.64 × 10+01 −1.02 × 10+01 3.39 × 10+01 5.00 × 10+01 2.75 × 10+01 −1.25 × 10+01 −3.70 × 10+01 −2.81 × 10+01 3.79 × 10+00 3.09 × 10+01 3.19 × 10+01 8.03 × 10+00 −1.97 × 10+01 −2.91 × 10+01 −1.45 × 10+01 1.01 × 10+01 2.42 × 10+01 1.74 × 10+01 −3.19 × 10+00 −1.99 × 10+01 −1.99 × 10+01 −4.67 × 10+00 1.23 × 10+01

0.00 × 10+00 4.42 × 10−04 5.39 × 10−04 2.58 × 10−04 −1.24 × 10−04 −2.80 × 10−04 −9.60 × 10−05 2.58 × 10−04 4.88 × 10−04 4.22 × 10−04 1.36 × 10−04 −1.27 × 10−04 −1.67×10−04 2.01 × 10−05 2.56 × 10−04 3.39 × 10−04 2.07 × 10−04 −2.55 × 10−05 −1.75 × 10−04 −1.39 × 10−04 2.89 × 10−05 1.74 × 10−04 1.77 × 10−04 4.07 × 10−05 −1.19 × 10−04 −1.78 × 10−04 −1.02 × 10−04 3.30 × 10−05 1.12 × 10−04 7.62 × 10−05 −3.73 × 10−05 −1.31 × 10−04 −1.32 × 10−04 −4.75 × 10−05 4.91 × 10−05 8.09 × 10−05 2.99 × 10−05 −5.43 × 10−05 −9.99 × 10−05 −7.24 × 10−05 1.62 × 10−06 6.08 × 10−05 6.10 × 10−05 8.95 × 10−06 −4.79 × 10−05 −6.27 × 10−05 −2.62 × 10−05 2.88 × 10−05

1.112 0.65707 −0.2687 −0.855 −0.6752 0.05021 0.68719 0.72961 0.19049 −0.4548 −0.6883 −0.365 0.20483 0.54351 0.40073 −0.0677 −0.4603 −0.4714 −0.1257 0.27068 0.39924 0.18314 −0.1732 −0.372 −0.2675 0.03175 0.27195 0.26895 0.04846 −0.1932 −0.2607 −0.1148 0.11024 0.22911 0.1574 −0.0303 −0.1738 −0.1628 −0.0197 0.12982 0.1662 0.07049 −0.0695 −0.1388 −0.0887 0.02976 0.11601 0.10398

0 −1819.7 −1883.2 −462.04 1181.16 1720.53 827.405 −657.74 −1498.7 −1082.6 148.758 1144.36 1135 219.711 −790.8 −1082.9 −487.52 443.13 939.469 646.244 −132.01 −732.36 −692.94 −102.31 520.295 676.689 284.117 −296.12 −585.85 −380.92 111.02 472.61 427.577 47.8826 −334.73 −415.92 −158.38 202.42 369.994 228.176 −82.648 −300.2 −259.89 −16.976 217.091 256.868 88.1138 −136.24 (Continued)

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006 0.0065 0.007 0.0075 0.008 0.0085 0.009 0.0095 0.01 0.0105 0.011 0.0115 0.012 0.0125 0.013 0.0135 0.014 0.0145 0.015 0.0155 0.016 0.0165 0.017 0.0175 0.018 0.0185 0.019 0.0195 0.02 0.0205 0.021 0.0215 0.022 0.0225 0.023 0.0235

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 265 Sl. No.

Time step

x1 (disp)

x1 (vel)

x1 (acc)

x2 (disp)

x2 (vel)

x2 (acc)

48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98

0.024 0.0245 0.025 0.0255 0.026 0.0265 0.027 0.0275 0.028 0.0285 0.029 0.0295 0.03 0.0305 0.031 0.0315 0.032 0.0325 0.033 0.0335 0.034 0.0345 0.035 0.0355 0.036 0.0365 0.037 0.0375 0.038 0.0385 0.039 0.0395 0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045 0.0455 0.046 0.0465 0.047 0.0475 0.048 0.0485 0.049

−1.37 × 10−06 1.23 × 10−06 5.37 × 10−06 7.99 × 10−06 7.37 × 10−06 4.44 × 10−06 1.76 × 10−06 1.38 × 10−06 3.28 × 10−06 5.54 × 10−06 6.11 × 10−06 4.41 × 10−06 1.71 × 10−06 −6.34 × 10−08 1.10 × 10−07 1.61 × 10−06 2.83 × 10−06 2.54 × 10−06 8.48 × 10−07 −1.01 × 10−06 −1.76 × 10−06 −1.09 × 10−06 1.90 × 10−07 8.65 × 10−07 3.52 × 10−07 −9.15 × 10−07 −1.92 × 10−06 −1.92 × 10−06 −1.02 × 10−06 −1.79 × 10−08 2.96 × 10−07 −2.38 × 10−07 −1.08 × 10−06 −1.49 × 10−06 −1.13 × 10−06 −3.02 × 10−07 3.49 × 10−07 3.75 × 10−07 −1.34 × 10−07 −6.73 × 10−07 −7.62 × 10−07 −3.25 × 10−07 2.84 × 10−07 6.01 × 10−07 4.31 × 10−07 −1.97 × 10−08 −3.47 × 10−07 −2.84 × 10−07 9.65 × 10−08 4.72 × 10−07 5.49 × 10−07

2.09 × 10−03 8.32 × 10−03 8.24 × 10−03 2.22 × 10−03 −4.66 × 10−03 −7.06 × 10−03 −3.66 × 10−03 2.16 × 10−03 5.42 × 10−03 3.65 × 10−03 −1.38 × 10−03 −5.42 × 10−03 −5.39 × 10−03 −1.70 × 10−03 2.39 × 10−03 3.61 × 10−03 1.26 × 10−03 −2.42 × 10−03 −4.35 × 10−03 −3.10 × 10−03 1.08 × 10−04 2.60 × 10−03 2.51 × 10−03 1.95 × 10−04 −2.25 × 10−03 −2.82 × 10−03 −1.19 × 10−03 1.19 × 10−03 2.40 × 10−03 1.60 × 10−03 −3.50 × 10−04 −1.79 × 10−03 −1.59 × 10−03 −5.06 × 10−05 1.49 × 10−03 1.82 × 10−03 7.82 × 10−04 −6.77 × 10−04 −1.36 × 10−03 −7.98 × 10−04 4.42 × 10−04 1.31 × 10−03 1.13 × 10−03 1.39 × 10−04 −8.16 × 10−04 −9.89 × 10−04 −3.20 × 10−04 5.72 × 10−04 9.51 × 10−04 5.52 × 10−04 −2.43 × 10−04

1.74 × 10+01 7.58 × 10+00 −7.90 × 10+00 −1.62 × 10+01 −1.13 × 10+01 1.72 × 10+00 1.19 × 10+01 1.14 × 10+01 1.64 × 10+00 −8.73 × 10+00 −1.14 × 10+01 −4.82 × 10+00 4.92 × 10+00 9.86 × 10+00 6.52 × 10+00 −1.66 × 10+00 −7.72 × 10+00 −7.02 × 10+00 −7.01 × 10−01 5.71 × 10+00 7.11 × 10+00 2.84 × 10+00 −3.20 × 10+00 −6.05 × 10+00 −3.73 × 10+00 1.44 × 10+00 5.09 × 10+00 4.44 × 10+00 3.82 × 10−01 −3.56 × 10+00 −4.26 × 10+00 −1.48 × 10+00 2.26 × 10+00 3.89 × 10+00 2.29 × 10+00 −9.82 × 10−01 −3.18 × 10+00 −2.66 × 10+00 −7.52 × 10−02 2.32 × 10+00 2.64 × 10+00 8.13 × 10−01 −1.52 × 10+00 −2.45 × 10+00 −1.37 × 10+00 6.82 × 10−01 1.99 × 10+00 1.58 × 10+00 −6.63 × 10−02 −1.53 × 10+00 −1.66 × 10+00

5.77 × 10−05 4.05 × 10−05 −4.45 × 10−06 −3.86 × 10−05 −3.55 × 10−05 −8.53 × 10−07 3.50 × 10−05 4.33 × 10−05 1.99 × 10−05 −1.41 × 10−05 −3.09 × 10−05 −1.89 × 10−05 9.41 × 10−06 2.97 × 10−05 2.63 × 10−05 3.69 × 10−06 −1.87 × 10−05 −2.34 × 10−05 −8.48 × 10−06 1.22 × 10−05 2.14 × 10−05 1.27 × 10−05 −5.56 × 10−06 −1.81 × 10−05 −1.56 × 10−05 −1.45 × 10−06 1.20 × 10−05 1.41 × 10−05 4.12 × 10−06 −8.86 × 10−06 −1.43 × 10−05 −8.42 × 10−06 3.04 × 10−06 1.05 × 10−05 8.53 × 10−06 −4.86 × 10−07 −8.65 × 10−06 −9.49 × 10−06 −2.94 × 10−06 5.19 × 10−06 8.33 × 10−06 4.48 × 10−06 −2.63 × 10−06 −7.01 × 10−06 −5.40 × 10−06 4.16 × 10−07 5.45 × 10−06 5.79 × 10−06 1.59 × 10−06 −3.39 × 10−06 −5.11 × 10−06

0.01174 −0.0804 −0.0995 −0.0371 0.04953 0.08917 0.05426 −0.0209 −0.073 −0.0629 −0.0041 0.05205 0.06115 0.02016 −0.0338 −0.0567 −0.033 0.01427 0.04544 0.03721 −0.0003 −0.0345 −0.0385 −0.0118 0.02179 0.03487 0.01895 −0.0107 −0.0291 −0.0228 0.0012 0.02214 0.02367 0.00634 −0.0144 −0.0217 −0.011 0.00762 0.01859 0.0139 −0.0013 −0.0141 −0.0144 −0.0032 0.00962 0.01365 0.0065 −0.0052 −0.0116 −0.0083 0.00138

−232.72 −136.01 59.672 189.843 156.874 1.66764 −141.3 −159.34 −49.244 89.9718 145.004 79.6655 −43.237 −120.74 −95.228 3.79594 90.9929 98.0753 26.603 −59.519 −90.463 −46.558 30.5594 76.5631 57.6405 −5.3432 −58.344 −60.105 −13.827 39.3595 56.5166 27.2158 −21.079 −48.257 −34.597 5.36525 37.4775 36.8928 7.00279 −25.792 −35.146 −15.722 14.4488 30.4011 20.7196 −4.5961 −24.005 −22.597 −3.3553 16.8157 21.7888

© 2009 Taylor & Francis Group, London, UK

Amplitude(mm)

266 Dynamics of Structure and Foundation: 2. Applications

0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1 1 -0.2 -0.3 -0.4

Comparison of amplitude of the anvil based on Time History and Closed form

16

31

46

61

76

91

106

121

136

Amplitude of anvil based on Newmark Method Displacement amplitude of anvil(mm)

Time steps

Figure 2.8.7 Comparison of response time history versus approximate damping.

The above displacements plot show some very interesting results • •

The closed form solution and the time history results are very closely matching Since the time history is done with the original soil damping matrix and compared with modal response having damping matrix corrected to Rayleigh format (and yet it gives reasonably good results), it may be concluded that the technique of separating the damping for each mode and correcting the damping matrix based on Rayleigh damping may well be adapted without any signiﬁcant error in cases where the damping matrix is non-proportional. Analysis based on Code ka = ma

ωa = ωz = α=

62500000 = 1979 rad/sec 16

kz = ma + mf + mfr

2560000 × 9.81 = 317 rad/sec 156 + 1958 + 85

ma 156 = = 0.076 mf + mfr 1958 + 85

Equation for natural frequency is given by ωn4 − (ωa2 + ωz2 )(1 + α)ωn2 + (1 + α)ωa2 ωz2 = 0 Substituting the numerical values calculated above, we have ωn4 − 4322216.7ωn2 + 4.2346974 × 1011 = 0.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 267 2 = 100302.9 → ω The above on solving gives, ωn1 n1 = 316.7 rad/sec and = 4221913.9 → ωn2 = 2054 rad/sec. The amplitude of vibration is given by the formula

2 ωn2

2 )(ω2 − ω2 ) −(ωa2 − ωn2 a n1

x1 =

sin ωn1 t sin ωn2 t V − for the foundation and, ωn1 ωn2

2 − ω2 )f ωa2 (ωn1 n2 n2 2 ) sin ω t 2 ) sin ω t (ωa2 − ωn2 (ωa2 − ωn1 V n1 n2 − for the anvil. x2 = 2 2 ) ωn1 ωn2 (ωn1 − ωn2

Undamped response of hammer foundation

0.4 0.3 0.2

145

133

121

97

Amplitude of foundation 109

85

73

61

49

37

-0.1

25

0

1

0.1 13

Amplitude of foundation block

Substituting the numerical values as calculated above and plotting @ 0.0005 sec time steps we time history plot for this case as shown in Figures 2.8.8 and 9.

-0.2 -0.3 -0.4

Time Steps

141

131

121

111

101

91

81

71

61

51

41

31

21

Undamped response of anvil

11

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1

1

Amplitude of anvil(mm)

Figure 2.8.8 Amplitude of foundation as per IS-code.

Amplitude of anvil

Time steps

Figure 2.8.9 Amplitude of anvil as per IS-code.

The maximum amplitude of vibration is 0.284 mm in lieu of 0.156 mm for damped case. This proves that without considering the effect of damping the designer may have to make the foundation more heavy (thus expensive) to reduce the amplitude if the value exceeds the acceptable limit as shown in Figure 2.8.9. In this case the undamped amplitude of the anvil is 0.754 mm in lieu of 0.471 mm when damping is considered.

© 2009 Taylor & Francis Group, London, UK

268 Dynamics of Structure and Foundation: 2. Applications

We give below a compare the various data as calculated above. Comparison of results of the hammer foundation by various techniques: Close formed (undamped solution)

Close formed (damped Solution)

Time history response of the damped equation Remarks

316.7

317

317

Practically no difference

2054

2052

2052

Practically no difference

0.754

0.471

0.539

Amplitude of 0.286 foundation (mm)

0.156

0.1622

Significant difference between undamped and damped amplitude but marginal difference with time history Comments same as above

Engineering parameter 1st Natural frequency 2nd Natural frequency Amplitude of anvil (mm)

2.9 DESIGN OF ECCENTRICALLY LOADED HAMMER FOUNDATION

2.9.1 Mathematical formulation of anvil placed eccentrically on a foundation Sometimes due to the functional requirements anvils may be placed eccentric to the RCC block as shown in Figure 2.9.1. In this case the hammer hits the anvil concentrically, but as the anvil is placed at an eccentricity e mm, say, the foundation block other than vertical mode also gets subjected to a coupled horizontal and rocking mode. The mathematical model used for such case is as shown in Figure 2.9.2. Here additional sliding (x) and rocking mode (θ ) is also simulated due to the eccentric impact.

Hammer(Wh)

e

Anvil(Wa)

Foundation(Wf)

Figure 2.9.1 Eccentrically loaded anvil.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 269

m2 z2

k2

c2

Cx X Cθ

m1

kθ

z1

kx

k1

c1

Figure 2.9.2 Mathematical model of eccentrically loaded anvil.

The nomenclatures used here are: m1 = mass of the foundation block plus frame resting on it if any; m2 = mass of the anvil resting on elastic pad; k1 = soil spring to be calculated from Barkan or Richart’s formula34 ; k2 = spring value for the elastic pad, this value is normally furnished by the vendor supplying these pads; kx = soil spring to be calculated from Barkan or Richart’s formula in x direction; kθ = soil spring to be calculated from Barkan or Richart’s formula in θ direction; c1 = damping of the soil to be obtained from Richart’s formula; c2 = damping of the elastic pad again furnished by the vendor; cx = damping of the soil to be obtained from Richart’s formula in x direction; cθ = damping of the soil to be obtained from Richart’s formula in θ direction; z2 = amplitude vector of the anvil in vertical direction; z1 = amplitude vector of the Foundation in vertical direction; x = amplitude vector of the Foundation in horizontal direction; θ = rotational amplitude of the foundation, and e = eccentricity of the anvil with respect to the c.g. of the foundation.

34 Refer to section on block foundation for the formula of the springs and dampers.

© 2009 Taylor & Francis Group, London, UK

270 Dynamics of Structure and Foundation: 2. Applications

Based on d’Alembert’s equation and the free body diagram the equation of motion for vibration are as follows: m2 z¨ 2 + k2 (z2 − z1 ) = 0 m1 z¨ 2 + k1 (z1 − θe) + k2 (z1 − z2 ) = 0 m1 x¨ + kx (x − Zc θ ) = 0 J θ¨ − kx Zc x + [kθ − Wf Zc + k1 e2 + kx Zc2 ]θ − k1 ez1 = 0

(2.9.1)

The above when written in matrix form gives the equation ⎡

m1 ⎢0 ⎢ ⎣0 0

0 m1 0 0

0 0 m2 0

⎡

k1 + k2 ⎢ 0 +⎢ ⎣ −k2 −k1 e

⎤⎧ ⎫ z¨ 1 ⎪ 0 ⎪ ⎪ ⎨ x¨ ⎪ ⎬ 0⎥ ⎥ z¨ 2 ⎪ 0⎦ ⎪ ⎪ ⎩ θ¨ ⎪ ⎭ J 0 kx 0 −kx Zc

⎤⎧ ⎫ −k1 e ⎪ ⎪z1 ⎪ ⎪ ⎥⎨x⎬ −kx Zc ⎥ ⎦ ⎪z2 ⎪ = {0} 0 ⎪ ⎩ ⎪ ⎭ 2 2 θ (kθ − Wf Zc + k1 e + kx Zc ) (2.9.2)

−k2 0 k2 0

2.9.2 Damped equation of motion with eccentric anvil The damped equation of motion is given by ¨ + [C]{X} ˙ + [K]{X} = 0 [M]{X}

(2.9.3)

Here, [M] = a square mass matrix of the order 4 × 4 as shown above [K] = a square stiffness matrix of the order 4 × 4 as shown above {X} = a Column deflection matrix of order 4 × 1 (which means 4 rows and 1 column) [C] = a square damping matrix of order 4 × 4. The equation being statically coupled, the damping matrix is given by ⎡

c1 + c2 ⎢ 0 [C] = ⎢ ⎣ −c2 −c1 e

0 cx 0 −cx Zc

−c2 0 c2 0

⎤ −c1 e ⎥ −cx Zc ⎥ ⎦ 0 (cθ − Wf Zc + c1 e2 + cx Zc2 )

(2.9.4)

Based on the above matrices one can now do the analysis in identical fashion as shown earlier35 once the initial velocity of the anvil after the impact is known. 35 Here the order of matrices being 4 × 4 eigen solution may be done by Bairstow’s method or one can directly solve for them in solution tools in computer like MATHCAD or MATLAB etc.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 271

Figure 2.10.1 Typical reinforcement detail for hammer foundation.

2.10 DETAILS OF DESIGN

2.10.1 Reinforcement detailing Usually for this type of foundation moments induced and reinforcement required for the same are nominal and nominal reinforcement is only provided. Usually 16 mm diameter bar is placed @ 150/200 mm c/c subject to the minimum condition that the volume of reinforcement shall not be less than 25 Kg/m3 . In this type of foundations, the reinforcement shall be placed along the three axis and also diagonally at the corner to prevent shear failure. Additional reinforcement shall be provided at the top side of the foundation block than at other sides. Topmost layer of reinforcement shall be provided with a cover not less than 50 mm. The sketch in Figure 2.10.1 shows a typical detail of re-bars for hammer foundations.

2.10.2 Construction procedure The foundation block should preferably be cast in one go. If this is not possible and a construction joint is unavoidable then such joints shall be horizontal in orientation and measures shall be taken to provide a proper joint by providing dowels of 12 mm or 16 mm dia bars embedded at 60 mm center to center to depth of at least 300 mm at both sides of the joint. Before placing the next layer of concrete the previously laid layer of concrete should be roughened, cleaned thoroughly and washed by water jet and then covered by a layer of rich 1:2 cement grout (1 cement, 2 sand) at least 20 mm thick. Concrete should be placed not later than 2 hours after the grout is laid. © 2009 Taylor & Francis Group, London, UK

272 Dynamics of Structure and Foundation: 2. Applications

2.11 VIBRATION MEASURING INSTRUMENTS

2.11.1 Some background on vibration measuring instruments and their application Design of machine foundation is not only a case of design of foundation and restricting the amplitude of vibration within the acceptable limit. In many cases if there are sensitive instruments around the foundation a foundation specialist is required to measure the vibration amplitude of the surrounding instruments (even though the foundation itself is safe) to ensure their safe operation. There could be properties or heritage buildings around the foundations that would require protection from the foundation vibration. In such cases an important function is to measure the vibration induced by the machine and its effect on the surrounding. This is usually done by instruments which are generally termed as vibration pick up instruments. In this section we will see what the theoretical background for development of these machines is and how they are used to measure such vibrations.

2.11.2 Response due to motion of the support In many situations vibration of a system is not due to forces acting directly on the mass but resulting from the motion of the base. Consider the situations shown in Figure 2.11.1. Figure 2.11.1(a) shows the basics of a ground-measuring device. A vibration meter shown therein measures only a relative displacement of the ground. The relative motion is usually converted to an electric voltage by making the seismic mass a magnet moving relative to coils fixed in the case shown in Figures 2.11.1(b) and (c). The electric voltage produced is a measure of the ground displacement. These types of instruments are called velocity meters. The voltage generated is proportional to the rate of cutting of the magnetic field and the output of the system will be proportional to the velocity of the vibrating body. If a rotating drum is fixed and a needle is moving on a drum, the relative motion of the instrument-soil system will be recorded on the drum [relative displacement = [z2 (t) −z1 (t)], z2 (t) being the displacement of the mass]. A basic description of a seismic pick up is shown in Figure 2.11.1(c). The relative displacement is the e. m. f. produced in the coil is the electrical signal from it mechanical counterpart. A hypothetical pick up is shown in Figure 2.11.1(d) wherein a magnetic material moves up and down in a electro-magnetic field and the electrical signal produced is a measure of the actual ground displacement.

2.11.3 Vibration pick-ups The vibration-displacement amplitudes are most often measured in soils and foundations ranging from millionths to thousandths of a centimetre and occur at frequencies ranging from less than 10 Hz to more than 100 Hz. The instruments required to measure motions of this magnitude are designed on the basis of a single-degree-of-freedom system. Instruments based on this design have two distinct advantages. First, in the S.D.O.F. system, the suspended mass is used as a reference from which vibrations are measured because in cases such as ground-motion measurements no reference is available. The second, some electrical phenomena are readily adapted to measuring © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 273

Rotating drum Vibration meter

z2(t) m

z2(t)

m

z2-z1

z2-z1

z1(t)

z1(t)

(a). General set up

(b). General set up of vibration pick up Electromagnetic field

Rotating drum z2(t) z2-z1

m

Displacement here is same as that of the ground; i.e. the actual displacement is measured

z1(t)

(c). Seismic pick up

(d). Hypothetical ground displacement measuring device

Figure 2.11.1 Seismic pick up.

z2(t)

m

k

c z1(t)

Figure 2.11.2 Displacement pick up.

the response of the system by producing an electrical signal that can be observed with an oscilloscope or recorded for subsequent analysis. An instrument that converts mechanical motion into an electrical signal is called a transducer. For vibration measurements there are three general types of transducers, namely, velocity, acceleration and displacement transducers. 2.11.3.1

Displacement transducer

In Figure 2.11.2 a schematic sketch of a displacement transducer is shown. Here we record Zr (t) = z2 (t) − z1 (t). Now if it so happens that Zr (t) ⇒1 z1 (t)

and

z2 (t) 1 z1 (t)

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274 Dynamics of Structure and Foundation: 2. Applications

then whatever Zr we measure will be the z1 (t). If z˙ 2 (t) and z¨ 2 (t) are velocities and accelerations of the mass, one can write the equation of motion of the system as m z¨ 2 + c(˙z2 − z˙ 1 ) + k(z2 − z1 ) = 0

(2.11.1)

Again, let Z(t) = z2 (t) − z1 (t), ¨ and Z(t) = z¨ 2 (t) − z¨ 1 (t),

so

˙ Z(t) = z˙ 2 (t) − z˙ 1 (t),

¨ also, z¨ 2 (t) = Z(t) + z¨ 1 (t).

Substituting above, one can write: ¨ ˙ mZ(t) + cZ(t) + kZ(t) = −m¨z1 (t)

(2.11.2)

If we assume a ground motion of the type: z1 (t) = A sin ωt ⇒ m¨z1 (t) = −mA ω2 sin ωt: Substituting the above, Equation (2.11.2) reduces to: ¨ ˙ mZ(t) + cZ(t) + kZ(t) = mA ω2 sin ωt

(2.11.3)

Solution of Equation (2.11.3) is Z=

mAω2 /k (1 − r2 )2 + (2Dr)2

and tan φ =

2Dr 1 − r2

sin (ωt − φ) = r2 κA sin(ωt − φ) = X sin(ωt − φ)

where κ =

1 (1 − r2 )2

+ (2Dr)2

(2.11.4)

Solution of Equation (2.10.4) is shown in Figure 2.11.3. It shows that when r is very high, Zmax /A approaches one. That is if we choose pick-ups having a very low ωn (natural frequency of m-k system), r will be very high. This will result in Z ≡ A. Also the phase angle φ, between the exciting force (ground motion) and the instrument should be nearly zero or 180◦ . Figure 2.11.3 shows that φ is nearly 180◦ for large values of r. Regarding D of the transducer, we see that the curve with D = 0.6 is better suited as the amplitude is not amplified near the natural frequency and secondly Zmax /A reaches unity faster. 2.11.3.2 Instrument with low natural frequency Instruments with low natural frequency, the r-values approach a large value and the relative displacement Z approaches A, regardless of the damping, D [Figure 2.11.3]. The mass here remains stationary while the supporting case moves with the vibrating body. The instrument just described is the basis of what is known as seismometer. These instruments are of large size as the relative motion of the seismic mass must be of the same order of magnitude as that of the vibration to be measured. Since the © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 275

D = 0.05 D=0 8

180 Phase angle,

D = 0.0

Zmax/A

6

4

90 0.707

40

accelerometer

D = 0.4

140

0 0

0.5

1

1.5

2

2.5

3

Frequency ratio, r D = 0.6

D = 0.707

1 seismometer 1

0

2

4

0.2

6

8

Frequency ratio, r

Figure 2.11.3 Response curves.

seismic mass (m as shown in Figure 2.11.2) is a magnet moving relative to coils fixed in the case, the voltage generated is proportional to the velocity of the vibrating body. The displacement and acceleration can be obtained from this velocity type transducer through integrator and differentiator provided in most signal-recording units. 2.11.3.3

Instrument with high natural frequency

Instruments with high natural frequency, the r-values approach a very low value and the relative displacement Z approaches A, regardless of the damping, D [Figure 2.11.3]. Again, the denominator of Equation (2.11.4), i.e.

(1 − r2 )2 + (2Dr)2 approaches unity as r → 0.

Under this condition, Equation (2.10.4) ➔ Zmax = Ar2 =

ω2 A Acceleration = , 2 ωn ωn2

implying that Zmax is now proportional to the acceleration of the motion, 1/ωn2 being the constant of proportionality. Range of accelerometer can be seen from a magni1 ﬁed plot of √ for various values of D. For D = 0.7, the useful range is 2 2 2 (1−r ) +(2Dr)

© 2009 Taylor & Francis Group, London, UK

276 Dynamics of Structure and Foundation: 2. Applications

1.04 D=0 1.02

D = 0.6

1 1 (1- r2)2 + (2Dr)2 D = 0.65

0.98 D = 0.75 0.96

D = 0.7 0

0.2

0.4 0.6 Frequency ratio, r

0.8

1

Figure 2.11.4 Acceleration error with varying frequency for various D.

0 ≤ D ≤ 0.2 with a maximum error of 0.01 percent (Figure 2.11.4). Electromagnetic type of accelerometers generally utilizes a damping around D = 0.7, which not only extends the useful frequency range but also prevents phase distortion for complex waves. 2.11.3.4

Velocity transducers

The displacement transducer described in the preceding may also be used as a velocity transducer. The relation between the relative velocity amplitude and the ground-velocity amplitude is identical to Zmax /A = r2 / (1 − r2 )2 + (2Dr)2 , since multiplication of Zmax and A by ω leaves the equation unchanged. Output from a velocity transducer is generated by coil moving through a magnetic field as mentioned earlier (Figure 2.11.5). Since voltage induced in the coil is directly proportional to the relative velocity between the coil and the magnetic field, either the coil or the magnet is made part of the mass and the other component is attached to the frame. Phase angles : tan φ1 =

2Di ω ωni

1−

ω ωni

2 : tan φ =

2Dω ωn 2 1 − ωωn

(2.11.5)

in which ω = operating frequency, frequency of the forcing function (say 50 Hz); ωni = natural frequency of the instrument (say 4.75 Hz); ωn = natural frequency of (soil mass + footing); Di = Damping coefficient of the instrument; D = damping coefficient of (soil mass + footing). For a soil-footing vibration system: Phase angles for the above system (Figure 2.11.6) may be written as tan φ1 =

2Di ω/ωni 2 ; 1 − ωωn i

© 2009 Taylor & Francis Group, London, UK

tan φ =

2Dω /ωn 2 1 − ωωn

(2.11.6)

Analysis and design of machine foundations 277

m

z 2 (t)

F0 sin t k

c z1(t) = A sin( t-

Z(t) = A sin ( t- 1= relative displacement (as Z ≈ A)

Figure 2.11.5 Ground displacement measuring instrument.

Pick up

F(t), A(t) F0 sin t

F = F0 sin t

footing

A0 sin t t

soil

+

Figure 2.11.6 Footing-soil system.

where ωn is the natural frequency of soil + footing system; other terms are as defined in the preceding. To obtain D one may use (φ1 + φ) and (φ − φ1 ), if possible. 2.11.3.5 Acceleration transducers There is no magnetic field here. The Figure 2.11.7 shows a schematic sketch of an accelerometer. The principle is “when there is a pressure difference between the faces it produces a voltage difference”, which is a measure of the force acting and hence the acceleration of the mass of the crystal. For the mass m of the crystal: m¨z2 + c(˙z2 − z˙ 1 ) + k(z2 − z1 ) = 0;

or m¨z2 + c˙z2 + kz2 = c˙z1 + kz1

(2.11.7)

Assuming, z1 = A1 sin ωt, z˙ 1 = A1 ω cos ωt, hence c˙z1 + kz1 = cA1 ω cos ωt + kA1 sin ωt Thus√the right hand side of Equation (2.11.7) reduces to F sin (ω + φ1 ): in which F = A1 c2 ω2 + k2 and tan φ1 = cω = 2D ωωn . The angle φ1 is the angle between force k (F) and the displacement of the ground z1 . © 2009 Taylor & Francis Group, London, UK

278 Dynamics of Structure and Foundation: 2. Applications

Piezo-electric crystal

O m

Accln. measured z2(t)

k

O

c

z1(t)

Figure 2.11.7 Accelerometers.

Solution of Equation (2.11.7) is given by √ (A1 c2 ω2 + k2 )/k z2 = sin (ωt + φ1 − φ2 ) = A2 sin (ωt + φ1 − φ2 ) (1 − r2 )2 + (2Dr)2

(2.11.8)

A2 1 + (2Dr)2 z2 max = = A1 A1 (1 − r2 )2 + (2Dr)2

(2.11.9)

To have A2 /A1 ⇒ 1, r-value should very low. For a typical instrument where fn = 3200 Hz (say) which much higher than the operating frequency normally encountered in practice. Within a range of r ≤ 0.2, such a situation is encountered. These are shown in Figures 2.11.8 to 10. 2.11.3.5.1

Phase angles

Between z1 and generating force: φ1 → tan φ1 =

2Dω ; ωn

Between generating force and z2 : φ2 → tan φ2 =

2Dω/ωn 2 1 − ωωn

(2.11.10)

Between z2 and z1 ➔ φ = (φ2 − φ1 ) = tan

−1

2Dr 1 − r2

− tan

−1

(2Dr) = tan

−1

2Dr3 1 − r2 (1 − 4D2 )

(2.11.11) An ideal accelerometer is the one in which instrument mass is servo-controlled to have zero relative displacement; the force necessary to accomplish this becomes a measure of the acceleration. © 2009 Taylor & Francis Group, London, UK

D=0 180 Phase angle,

A2/A1

140

Range for making A2/A1 = 1

90 0.707

40 0 0

1

0.5

1

1.5

2

2.5

Frequency ratio, r 0

2

1 √2

6

4

r

Figure 2.11.8 Accelerometer response.

4

in radian

D=.05

0.15

0.01

2

0.5

1

0 2

0

4 r

Figure 2.11.9 Phase angles.

φ1

z1 ( φ2-φ1)

cωz1

F

z2

kz1 cωz1 kz2

φ2

Reference mω2z2

ωt

Figure 2.11.10 Vector diagram for phase angles.

© 2009 Taylor & Francis Group, London, UK

6

3

280 Dynamics of Structure and Foundation: 2. Applications

The piezoelectric crystals are mounted in such a manner that under acceleration they are either compressed or bent to produce an e.m.f. which is ultimately converted to electrical signals. The natural frequencies of such accelerometers can be made very large, say in the range of 50 000 Hz. The piezoelectric crystal mounted accelerometers can be made very small in size, may be of the range of 10 mm in diameter and height and are very rugged to withstand a shock as high as 10 000 g acceleration. A typical instrument may have fni → 1 Hz to 5 Hz useful frequency, foperating → 10 Hz to 2000 Hz which means r is more than 10. Sensitivity of such instruments may be in the range 20 mV/(cm/sec) to 350 mV/(cm/sec) with maximum displacement = 0.5 cm (double amplitude) [Note Arms = 0.707 A]. Sensitivity of crystal-type accelerometer is denoted either in terms of charge, i.e. picocoulombs = pC = 10−12 coulombs per g or in terms of voltage, i.e. millivolts = mV = 10−3 V per g. Sensitivity of a crystal-type accelerometer can be established from: say a typical crystal accelerometer is 25 pC/g with crystal capacitance equal to 500 pF (picofarads). Voltage from the classical equation E = Q/C, gives the sensitivity = 25/500 = 0.05 V/g or 50 mV/g as sensitivity in terms of voltage. Again, if the accelerometer is connected to a vacuum tube voltmeter through a 3 m long cable of capacitance 300 pF, the open circuit output voltage of the accelerometer is reduced to (50)(500)/(500 + 300) = 31.3 mV/g. This loss can be avoided by using a charge ampliﬁer, in which case the capacitance of the cable has no effect. 2.11.3.6

Amplitude distortion

Normally the measured vibrations consist of a number of harmonic motions of various frequencies. Amplitude distortion occurs in an accelerometer if the acceleration of one harmonic is amplified more than another. From a harmonic solution, the amplitude of acceleration can be written as ω2 A. For an equal amplification to acceleration, it is desirable to have κ/ωn2 nearly same for all frequencies. For r = 0, κ = 1. Thus, the amplitude distortion can be defined as the change in κ/ωn2 with respect to r = 0. The percent amplitude distortion is

100 ×

κ ωn2

− 1 ωn2

1 ωn2

= 100 × (κ − 1)

It can be observed from Figure 2.11.11 that accelerometer should be built with D lying between 0.6 and 0.7 to minimize the amplitude distortion. 2.11.3.7

Phase distortion

This distortion occurs if the relative phase of the harmonics recorded is different from that of the vibration to be measured. For zero distortion, the shift φ should increase linearly with frequency of the harmonic motion. The phase shift at r = 1 is always π /2. For zero distortion, the phase shift for 0 < r < 1 should be 90r degree. Hence phase distortion in an accelerometer can be defined as: Phase distortion = (φ − 90r) degree. It can be noticed in Figure 2.11.11 that appropriate damping in an accelerometer is necessary for minimizing the phase distortion. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 281

To record any complex wave without changing its shape, the phase of all harmonics must remain unchanged with respect to the fundamental. This requires that the phase angle be zero or that all the harmonic components must be shifted equally. The first case of zero phase shift corresponds to D = 0 for r < 1. The second case of equal timewise shift of all harmonics is nearly satisfied for d = 0.7 for r < 1 (Figure 2.11.12) and when D = 0.70, the phase for r < 1 can be expressed by φ ≈ π r/2. Thus for d = 0 or 0.70, the phase distortion is completely eliminated. 10 Amplitude distortion in %

D=0 D = 0.6 + 0 D = 0.65 D = 0.75 D = 0.7

10 0

0.2

0.4 0.6 Frequency ratio, r

0.8

1

Figure 2.11.11 Amplitude distortion in accelerometer.

Phase distortion

2

0

0.2

0.4

0

D = 0.75

-2

0.70

-4

Frequency ratio, r 0.6 0.8

1

1.2

0.6

-6 -8

0.5

-10 -12

Figure 2.11.12 Phase distortion in accelerometer.

Example 2.11.1 1

A manufacturer of vibration measuring instruments gives the following specif ication for one of its vibration pick ups; Frequency range: velocity response flat from 10 Hz to 1000 Hz. Sensitivity: 0.096 V/cm/sec, both volts and velocity in rms values.

© 2009 Taylor & Francis Group, London, UK

282 Dynamics of Structure and Foundation: 2. Applications

Amplitude range: Almost no lower limit to maximum stroke between stops of 0.60 in. a

b

This instrument was used to measure the vibration of a machine with a known frequency of 30 Hz. If a reading of 0.024 V is indicated, determine the rms amplitude. Could this instrument be used to measure the vibration of a machine with known frequency of 12 Hz and double amplitude of 0.80 cm? Give reasons.

Solution: a Voltage = 0.024 V : Sensitivity = 0.096 V/(cm/sec) : Velocity = 0.024/0.096 cm/sec = 0.25 cm/sec. f = 30 Hz : ω = 2π × 30 rad/sec : Amplitude = velocity/ω = 0.132 × 10−2 cm = 0.0133 mm. f = 12 Hz : ω = 2π × 12 rad/sec : Amplitude = 0.40 cm. Velocity = 2π × 12 × 0.40 = 30.159 cm/sec b Now for f = 10 Hz : ω = 2π× 10, amplitude = 0.3 × 2.54 = 0.762 cm: velocity = 0.762 × 20π = 47.88 cm/sec. For f = 1000 Hz, velocity = 0.762 × 2000π = 478877.8 cm/sec Velocity required is 30.159 cm/sec. → So the instrument cannot be used. 2

The sensitivity of a certain crystal accelerometer is given as 18 pC/g, with its capacitance equal to 450 pF. It is used with a vacuum tube voltmeter and its cable is 5 m long with a capacitance of 50 pF/m. Determine its voltage output per g. Ans. E = 25.7 mV/g.

Solution: Sensitivity = 18 pC/g, crystal capacitance = 450 pF. Sensitivity in terms of voltage = 18/450 0.04 V/g [E = Q/C]. Total cable capacitance = 50 × 5 = 250 pF Output voltage = 40 × 450/(450 + 250) = 25.7 mV/g 3

A vibration pickup has a sensitivity of 40 mV/(cm/sec) between f = 10 Hz to 2000 Hz. If 1 g acceleration is maintained over this frequency range, what will be the output voltage at (a) 10 Hz and (b) at 2000 Hz. Ans. (a) 624.5 mV, (b) 3.123 mV.

Solution: Sensitivity = 40 mV/(cm/sec) = 0.04 V/(cm/sec) a

F = 10 Hz → ω = 2π × 10 = 62.83 rad/sec

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 283

b

Acceleration: 1 g = 981 cm/sec2 → velocity = 981/62.83 = 15.61356 cm/sec. Voltage = (40 × 981)/62.83 mV = 624.5 mV f = 2000 Hz → ω = 2π× 2000 rad/sec Velocity = 981/(2π× 2000) Voltage = (40 × 981)/(2π× 2000) = 3.123 mV.

2.12 EVALUATION OF FRICTION DAMPING FROM ENERGY CONSIDERATION E = F0 Aπ sin φ, phase angle φ at resonance is 90◦ , and the energy input is F0 Aπ . Energy dissipation, using friction concept is 4fA. Now, if 4f /π F0 < 1, energy input exceeds the energy dissipation, and the excess energy accumulated over the cycles builds up the amplitude of oscillation. Consider an embedded footing, Figure 2.12.1. Governing equation is m¨z + c˙z + Ceq z˙ + kz = F0 sin ωt

(2.12.1)

Steady state solution is A=

F0 /k sin (ωt − φ) = A0 sin (ωt − φ) 2 (1 − r2 )2 + Ceq ωk + (2Dr)2

(2.12.2)

A0 can be obtained from A0 =

F0 k

1−

4f 2 πF0

.

(2.12.3)

(1 − r2 )2 + (2Dr)2

If we have two observations with forcing functions: F1 (t) = F01 sin ωt and F2 (t) = F02 sin ωt, there will be two responses namely, A01 sin (ωt − φ) and A02 sin (ωt − φ). F = F0 sin t

f/2

f/2

m

⇒ k

Figure 2.12.1 Vibration of a footing with side friction.

© 2009 Taylor & Francis Group, London, UK

c

284 Dynamics of Structure and Foundation: 2. Applications

A0

r r=1

Figure 2.12.2

Now using Equation (2.12.3) we can write A01 F01 = A02 F02

1− 1−

4f 2 πF01

4f 2 πF02

(2.12.4)

A01 , A02 , F01 and F02 are known and hence we can obtain a representative value of f , the friction force. Now, for D = 0 Equation (2.12.2) gives 2 1 − 4f πF F0 A0 = k 1 − r2 and the response is given in Figure 2.12.2. But 4f < πF0 has to be satisfied for a real system, i.e. 4fA0 sin φ < π F0 A0 sin φ. Hence, work done by the friction force is less than the work done by exciting force. This implies building up of energy and hence a resonant situation will arise. If f is large we have to use more exact analysis for solution as the motion cannot be assumed to be harmonic. 2.13 VIBRATION ISOLATION Vibratory forces generated by machines and engines are often unavoidable; however, their effect on a dynamical system can be reduced substantially by properly designed springs, which are referred to as isolators. Protection of the base against the action of driving forces is called active isolation and protection against kinematic disturbances is called passive isolation. Thus, when the noise-making source itself is isolated from other structures, the isolation is an active isolation whereas when other structures are isolated from the noise making sources, the isolation is a passive one. In active isolation the basic problem is that of determining the force transmitted to the base; in the theory of passive isolation, it is the problem of finding the amplitude of the vibration the object is to be protected is forced into. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 285

F = F0 sin ωt

m

c

k

Figure 2.13.1 Vibration isolation (active). mω2x

FTr

F

cωx φ x

kx

Figure 2.13.2 Vector diagram for the system shown in Figure 2.13.1.

2.13.1 Active isolation Let us consider a system shown in Figure 2.13.1. Let F = F0 sin ωt acting on a SDOF system. The force transmitted to the ground through springs and damper: FTr

cω 2 2 2 = (kx) + (cωx) = kx 1 + k

(2.13.1)

The vector diagram can be shown as in Figure 2.13.2. Solution of the problem can be written as x=

F0 k (1 − r2 )2 + (2Dr)2

sin (ωt − φ) =

F0 κ sin (ωt − φ) k

(2.13.2)

Thus from Eqns. (7.13.1) and (7.13.2) one can write 1 + (2Dr)2 FTr = F0 (1 − r2 )2 + (2Dr)2 This is identical to the one developed for accelerometers. © 2009 Taylor & Francis Group, London, UK

(2.13.3)

286 Dynamics of Structure and Foundation: 2. Applications

Thus the problem of isolating a mass from motion of the support point is identical to that of isolating the disturbing forces. The ratio, FTr /F0 is called the transmissibility. Without the loss of generality, setting D = 0, Equation (2.13.3) can be written as FTr = κF0

(2.13.4)

Hence, the efﬁciency of active isolation depends on the magnitude of the magniﬁcation factor κ to be used. This requires accordingly a low value of the natural frequency ωn , which may be achieved by reducing the stiffness of the mounting of the machine or by increasing the vibrating mass. √ Transmissibility is less than unity only for r > 2. → Isolation is possible only for √ ω/ωn > 2 (refer to Figure 2.11.3). An un-damped spring is superior to a damped system in reducing transmissibility. However, to reduce amplitude near resonance some damping is desirable. It should be noted that vibration isolation of slow-speed machines (when the frequency of the disturbance is not high) may require a very low natural frequency and accordingly impractically great ﬂexibility of vibration absorbers. To overcome this difﬁculty the vibrating mass is artiﬁcially increased in such cases. This serves a twofold objective; first, the natural frequency is reduced and, second, sufﬁcient stiffness of the system is preserved. It is possible to reduce the amplitude of vibration by supporting the machine on a large mass or by other means is shown in Figure 2.13.3. Again a set of elastic constraints (vibration isolators), in the form of steel springs or rubber elements are introduced under the frame of the machine to be isolated.

m

Machine

M

(a)

(b)

Machine

(c)

Figure 2.13.3 Active isolation.

© 2009 Taylor & Francis Group, London, UK

(d)

Analysis and design of machine foundations 287

Two equally efficient types of mounting are in use. These are supporting type when vibration isolators are placed under the base of the machine (Figure 2.13.3a) and suspension type when vibration isolators are placed above the bottom of the base in the latter case the vibration isolators may be either in compression (Figure 2.13.3b) or in tension (Figure 2.13.3c). If horizontal vibration prevails in the machine to be isolated, a pendulum type suspension may be used to advantage (Figure 2.13.3d). To keep transmissibility same, k must be increased in the same ratio so that (m + M)/k remains the same. Say for some transmissibility, if we increase the mass m to m + M

1 + (2Dr)2 FTr = F0 (1 − r2 )2 + (2Dr)2

(2.13.5)

k must be increased in the same proportion so that (m + M)/k remains the same. Thus as k → increases x=

F0 k (1 − r2 )2 + (2Dr)2

sin (ωt − φ)

(2.13.6)

→ x will be reduced. When damping is negligible Transmissibility =

FTr 1 = 2 ω F0 −1

(2.13.7)

ωn

where it is understood that ω/ωn to be used is always greater than

√

2.

2.13.2 Passive isolation Consider the case in which, there is a vibration of the ground in Figure 2.11.7 instead of the force, F. Following Equation (2.11.9), we can write mx¨ 2 + cx˙ 2 + kx2 = cx˙ 1 + kx1

(2.13.8)

If x1 (t) = X1 sin ωt, Equation (2.13.8) reduces to mx¨ 2 + cx˙ 2 + kx2 = X1 [k sin ωt + cω cos ωt] = X1 k2 + c2 ω2 sin (ωt + φ) = Fx sin (ωt + φ) where

tan φ =

cω = 2Dr. k

© 2009 Taylor & Francis Group, London, UK

(2.13.9) (2.13.10)

288 Dynamics of Structure and Foundation: 2. Applications

(a)

(b)

Figure 2.13.4 Passive isolation.

The steady state solution is x2 = X1

1 + (2Dr)2 κ sin (ωt − ψ + φ)

(2.13.11)

in which tan ψ =

2Dr 1 − r2

(2.13.12)

The passive isolation is used to protect instruments and precision machines against vibrations transmitted from the supporting structure. The amplitude of vibration of an isolated object is expressed in terms of the amplitude of vibration of the base by Equation (2.13.10). Thus a passive isolation should use the same idea of making the mounting soft, as in the case of an active isolation. It is generally required that the natural frequency of the isolated object shall not exceed one-fourth of the frequency of vibration of the base. If the frequency of the disturbance is not known, it is necessary to introduce in elastic pads in the mounting system. Thus the irregularities of a road may have the shape of a sine curve with the wave length varying over a wide range. Therefore, there is a real danger that the body of a moving vehicle may be in a state of resonance; to limit resonant amplitudes the vehicle suspension is always provided with hydraulic shock absorber which dissipate a considerable amount of energy during vibration [Figure 2.13.4(a)]. This absorber system has a disadvantage: it does not afford sufficient comfort of passengers when subjected to shocks which are transmitted to the automobile body with almost no relief. To obtain the necessary softness of the suspension it may be provided with additional flexible elastic damper shown in Figure 2.13.4(b).

2.13.3 Isolation by trench An exhaustive field study was carried out by Woods to examine the effectiveness of open trenches as barrier for vibration isolation. Lamb analysed the problem of the propagation of tremors over the surface of an elastic solid. He solved the problem of spreading out of a symmetrical annular wave disturbance around a point source. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 289

These waves consist of body waves, namely longitudinal wave (Primary wave or P-wave), transverse wave (Shear wave or S-wave) and surface waves, namely, Rayleigh surface waves (R-wave). At any point on the surface P-wave arrives ﬁrst and it undergoes an oscillatory displacement. It is followed by a relatively quiet period till another oscillatory displacement owing to the arrival of the S-wave. Lamb termed this phase of motion as minor tremor. A much larger oscillatory movement is followed subsequently due to the arrival of Rayleigh waves termed as major tremor. P-wave travels faster then S-wave and R-wave is slightly slower than the S-wave. As to the nature of wave propagation, a compressional wave (i.e. P-wave) propagates radially outward from the source as hemispherical wave front whereas Rayleigh waves propagate radially outward in a cylindrical wave front. As waves travel outward the energy density decreases with the distance from the source of disturbance. This is known as radiation or geometric damping. The amplitudes of compressional wave attenuates approximately in proportion to 1/r, r is the distance of the source of disturbance. Along the surface of an elastic halfspace, the attenuation is proportional to 1/r2 . For the Rayleigh √ waves, the decrease in amplitude is proportional to 1/ r. Again, about two-third of the total energy of vibration is normally carried through the Rayleigh wave and its smaller decay with the distance in comparison to other waves. Thus, the Rayleigh wave is more important for structures near the surface, particularly in the event of earthquakes, blasts and other dynamic operations. 2.14 MACHINE FOUNDATION SUPPORTED ON FRAMES

2.14.1 Introduction In this section we will deal with machines supported on frames. These are also sometimes termed as frame foundations. These type of foundations usually support equipment like steam turbines (ST), boiler feed pumps (BFP), in power plants, compressors in petroleum reﬁneries, air blowers in automobile industry etc. Though the basic analytical principle remains the same, for the present chapter however, we will restrict our discussion mainly to foundations related to steam turbines and Boiler feed pump only. A pre-requisite to this section is again Chapter 5 (Vol. 1) (basic concepts in structural and soil dynamics) and you should have gone through the previous section on analysis and design of block foundation. We also envisage that you have some basic concepts on Matrix Analysis of Structure whose concepts we are going to use quite in detail. Turbines and Boiler feed pumps form the heart of any power plant. Thus for any developed and developing nation, capacity of supplying unhindered energy not only ensures a steady industrial growth, but also goes on to improve the quality of life in a long way. The main source of this energy is obviously electricity and this is what a turbo-generator generates, based on the electro-mechanical process. Thus if the foundation which supports these critical machines misbehave and the machine trips during operation, the cascading effect on the end users and the industry dependent on the power generated could suffer severe losses. If the shortage is severe in nature, this could even have a very adverse effect on the economic growth to a complete part of a country. © 2009 Taylor & Francis Group, London, UK

290 Dynamics of Structure and Foundation: 2. Applications

Thus for successful operation two aspects become critical for these machines • •

The machine itself should run smoothly (round the clock). The foundation supporting the equipment is capable of sustaining the various loads coming from the turbine under operation as well as those that could develop due to the vagaries of nature or otherwise like earthquake, thermal, electrical faults, short circuits etc.

In factory, quality and performance of the machine is controllable since materials used are all man made and all requisite appurtenances are manufactured under a careful controlled condition. Thus, it is not difﬁcult to arrive at a condition in the manufacturing process where quality of output for two machines coming out of a factory could be stated to have identical mechanical characteristics. However, for a civil engineer designing its foundation the situation is completely different. He neither has control on the subsoil on which it is being built nor he has any control on the vagaries of nature like earthquake, wind etc. In addition to this he has to cater to a number of uncertain loads at the start of his design like piping loads, stator loads, and electrical fault loads etc and still make sure that the foundation functions within acceptable limits of engineering norm. Considering the difﬁcult natural parameters, enormity of the machines and risk involved in terms of public outcry, turbine foundations still remain one of the most difﬁcult and challenging task in civil engineering profession. The engineer not only needs a very specialized knowledge in various aspects of civil engineering like structural mechanics, dynamic theories related to structures and soil, he should also have some interdisciplinary appreciation about mechanical and electrical aspect of the machine itself. Though advent of digital computer has made the life much simpler in terms of accurate calculations and analyzing the output results visually, for turbine foundations this should be supplemented with some engineering judgment and experience. For this is a case where the computer output numbers only, do not reflect the actual picture. The engineer has to carefully weigh the effect of the idealization in his mathematical modeling that has created these numbers and take design decision using his engineering judgment. So before we get into the main topic itself. . . We plead with our readers to be cautious with this type of foundations and not hesitate to take help of engineers who are experienced and also the construction people who has constructed and commissioned such turbines and monitored their performance36 .

2.14.2 Different types of turbines and the generation process. . . Before we go into the analysis and design aspect of such a turbine foundations it would be useful to know something about the machine itself, how it behaves and why we

36 He may be an old man not so expert with computers as our modern day engineers but remember that his experience is worth more than a million dollar software you may write for he has a feel of this giant who if starts misbehaving can have a very serious consequence.

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Analysis and design of machine foundations 291

take the trouble of mounting it on a frame when putting it on a massive block resting on ground would have made our life much easier. Irrespective of the nature of fuel like fossil fuel (coal ﬁred plant), LPG/ Naptha (like in Combined cycle or open cycle plant), processed uranium rods (in nuclear power plant) basic principle of operation of turbine remains the same. The fuel is used to generate steam to a pre-designed temperature in boiler and is allowed to expand within a turbine under pressure. This generates a mechanical energy which makes the turbine rotate. The turbine shaft in turn is connected through a coupling or a synchronous clutch to a generator rotor, that is rotated by the turbine and generates electrical forces due to mechanical movement of the generator shaft in a magnetic ﬁeld. The electric power thus generated is transferred through bus duct connections to a primary transformer where after stepping up the voltage it is supplied to power grid through a switch yard. This in essence is the simpliﬁed process of electricity generation. The machine itself is a centrifugal machine and are usually of two types • •

Gas driven Steam driven

The gas driven one basically uses Naptha or natural gas as the base fuel and even at exhaust, it contains substantial thermal energy. This is usually recycled through a heat recovery system to further heat water into steam and is passed off to a steam driven turbine to generate further electricity. While gas driven turbine does not require any condenser at the gas exhaust, steam turbines will always have a condenser connected to the steam exhaust to condense off steam coming out of the turbine. This is collected in a hot-well from where it is

Generator on bearing

Coupling

Turbine on bearing

Shaft

Condenser (spring mounted)

Figure 2.14.1 Longitudinal proﬁle of a turbine foundation with the equipment.

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292 Dynamics of Structure and Foundation: 2. Applications

further recycled back to boiler through a condensate extraction pump. For steam to condense, usually advantage of gravity force is taken and also from plant layout and pipe routing consideration, the condenser should preferably be positioned at a level which is lower than the turbine operating floor level. Based on this concept the best location for placing the condenser is usually at a location directly below the turbine. It is for this steam driven turbines are usually mounted on frames to take advantage of the space beneath it, while for gas turbines, as no such requirements are essential, are usually mounted on block foundations. Besides this, the frame mounted machines also provides easy access to electrical connections to generator and main steam pipes. Connecting the steam pipe from the bottom is preferable for it avoids dismantling of pipe work during maintenance; this also prevents pipe work draining into the turbine. A typical schematic sketch of a turbine foundation is as shown in Figure 2.14.1.

2.14.3 Layout planning For turbines placed in a power house typical layout which is most common is as shown in Figure 2.14.2.

Spring mounted Boiler feed pump (Turbine driven)

Condenser Spring mounted

Figure 2.14.2 Typical cross section of turbine pedestal and power house.

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Analysis and design of machine foundations 293

For a typical 900 MW power plant this is usually about 16/17.00 m above the power house floor level with condenser mounted on springs. The 17.00 m level is also known as the operating floor level of the power house. In such case, the boiler feed pump (BFP) which feeds the water to the boiler is driven by turbine itself and usually rest on RCC block foundation mounted on springs over steel structure. For plants of lower capacity when the top deck height is much lower, BFP is also sometimes mounted on RCC frames similar to the turbine.

2.14.4 Vibration analysis of turbine foundations We stated earlier that analysis of turbine foundation is a complicated job and requires a lot of ingenuity and deep insight to the problem of dynamics for these are machines which are massive and rotates at a very high speed37 . We present here the following methods of analysis of the framed foundation: • • • • •

Resonance or Rausch’s (1959) method Amplitude or Barkan’s (1962) method Combined or Major’s (1980) method A 2D soil structure interaction model of framed foundation that can take into cognizance the effect of underlying soil/pile as frequency independent springs (Chowdhury 1984). The 3D ﬁnite element model of the foundation considering the underlying soil into cognizance. This is analyzed through a computer.

2.14.4.1

Rausch’s method

Rausch proposed a method where the basic criteria that needs to be satisﬁed is that the fundamental natural frequency of the foundation should be out of tune to the operating frequency of machine by about (±) 20%. He suggested a mathematical model where for natural frequency in vertical direction for the individual cross frames self weight and superimposed load on longitudinal girders and the load coming from the machine is considered as lumped mass over the columns having single degree of freedom (Figure 2.14.3)38 . For horizontal direction he assumed the bottom raft to be inﬁnitely rigid and again proposed a mathematical model having single degree of freedom. He also assumed that in vertical direction the average of natural frequency of the frames is the natural frequency of the system in that direction.

37 For 50 Hz power grids the typical RPM of machines are 3000 RPM. For 60 Hz grids the speed is about 3600 RPM. For Nuclear power plants these are about 1500–1800 RPM. 50 and 60 Hz are standard Power grid cycles available globally. 38 This is surely an over simpliﬁcation of the problem.

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294 Dynamics of Structure and Foundation: 2. Applications

N

P

N Q m=(P+Q+2N)/g

KV

Typical cross frame

Equivalent mathematical model

Figure 2.14.3 Idealization in Rausch method.

1

Frequency in vertical direction

As per Rausch if there is n number of frames in the foundation and if fv is the natural frequency of the structure in the vertical direction, then fv =

n

fi /n

(2.14.1)

i=1

Knowing, ωn = Kv /m rad/sec we have, ωn = (Kv g)/W where, g = acceleration due to gravity; W = weight acting in the vertical direction. If δst is the static deflection of the frame then, δst = W/Kv i.e. ωn = g/δst rad/sec. Using T = 2π/ωn ,

δst 1 secs; f = we have, T = 2π g 2π

√ g 60 × 9.81 ∼ 30 cps, ➔f = cycles/min √ =√ δst 2π δst δst (2.14.2)

The vertical frequency of the of individual frame in vertical direction is thus given by fv = 30/ δv cpm

(2.14.3)

where δv = the total vertical deflection at mid-point of the cross beam in meters. Hence, for different types of loading as shown above, δv = δ1 + δ2 + δ3 + δ4

© 2009 Taylor & Francis Group, London, UK

(2.14.4)

Analysis and design of machine foundations 295

PL3 96EIb

where, δ1 =

2ψ + 1 ψ +2

is deflection due to concentrated load;

QL3 5ψ + 2 δ2 = is the deflection due to uniformly distributed load; 384EIb ψ + 2 Q 3 L δ3 = P+ is the deflection due to shear; 5 EAb 2

(2.14.5)

P+Q h N+ is the axial deflection of column due to the EAc 2

δ4 =

concentrated load transferred from the longitudinal girder (N) in which, P = concentrated load from the machine; Q = UDL of the cross beam (qL); q = self weight per unit length of the cross beam; N = concentrated load on the column; Ab = area of cross section of the beam; Ac = area of cross section of the column; Ib = moment of inertia of the beam; Ic = moment of Inertia for the column; E = dynamic modulus of elasticity of the frame; h = effective height of the column; L = effective length of the cross beam, and ψ = (Ib h)/(Ic L). 2

Frequency in horizontal direction

Again considering single degree of freedom the natural frequency fh is given by fh = 30

Kh1 + Kh2 + · · · · · · · · · + Khn W

where, W = total load of machine plus the top deck and Khi =

(2.14.6) 12EIc h3

6ψ + 1 . 3ψ + 2

This method does not have any provision of calculation of amplitude and suffers from following drawbacks: • •

•

Over simpliﬁcation of the mathematical model based on single degree of freedom. A resonance check does not necessarily ensure that the design is safe and the amplitudes are within acceptable limits, especially for low tuned foundation which has been observed to undergo signiﬁcant displacement when the machine speed passes through the natural frequency value during start and stopping of the machine. It considers the bottom raft as stiff and ﬁnds frequency in translational mode only, no rocking mode frequency has been calculated, and this could have signiﬁcant contribution to the overall dynamic response (which of course depends on the geometry of the foundation system).

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296 Dynamics of Structure and Foundation: 2. Applications

2.14.4.2

Amplitude or Barkan’s method

Barkan improved upon Rausch’s method by taking into consideration the following steps for the analysis: • • •

In vertical direction he considered a two mass lumped model for analysis. He derived the translational amplitude by taking coupled rotation of the top deck plain considering the top deck as rigid mass supported on a series of leaf springs which represented lateral stiffness of the columns. However, like Rausch he also assumed the frames to be supported on slab that is inﬁnitely stiff and thus ignoring the effect of elastic base (soil) supporting the bottom raft.

1 Calculation in the vertical mode Barkan argued that under vertical mode, the transverse frame will take the deformed shape as shown in Figures 2.14.4 and the mathematical idealization may be showed as given in Figure 2.14.5.

m1/2

m1/2 m2, k2 k1/2

Figure 2.14.4 Transverse frame of the foundation.

m2

z2 k2

m1 z1 k1

Fixed at base

Figure 2.14.5 Mathematical model in vertical direction.

© 2009 Taylor & Francis Group, London, UK

k1/2

Analysis and design of machine foundations 297

Since the columns are stiff and have similar inertia they would deform uniformly under axial compression while the longitudinal girders will try to resist the flexural deformation of the transverse beam based on their torsional stiffness. As torsional stiffness of the longitudinal girder is much less than axial stiffness of the columns or flexural stiffness of the transverse girder, its effect on overall dynamic response of the system is marginal and can be neglected. Similar to the proposition of Rausch he also suggested that the transverse frames can be treated independent of each other in the vertical direction39 . Based on the above he deﬁned the various analytical parameters for each transverse frame as follows: k1 =

2EAc ; h

δv =

L3 (1 + 2ψ) 3L ; + 96EIb (2 + ψ) 8GAb

k2 =

1 δv

(2.14.7)

where, G = dynamic shear modulus of concrete @ 0.5E. Calculation of mass m m2 = m0 + 0.45mb where, m0 = P/g is the concentrated mass of the machine carried by the beam and mb = the mass of the transverse girder and m1 = mL + 0.255mb + 0.35mc . in which, mL = mass from longitudinal girder transferred to the frame; mc = mass of the column. The natural frequency of each frame is then obtained from the equation m1 0 z¨ 1 k1 + k2 −k2 z1 + =0 (2.14.8) 0 m2 z¨ 2 −k2 k2 z2 Similarly the amplitude of each frame can be obtained based on the method we have explained earlier40 . For amplitude calculation, the vertical dynamic load was assumed as Pv = Ci sin ωm t

(2.14.9)

2 , in which, R = weight of the rotor; e = eccentricity of the where, Ci = (R/g)eωm rotor, and ωm = operating frequency of the machine.

39 For a modern day engineer this might appear as Barkan was trying to simplify the case but what we should realize was that he did not had a desk top computer readily available on his desk nor were computers so easily available. It was an era when most of the calculations were done manually. What is most appreciable was that he idealized and modeled an extremely complex problem to a level which was amenable to manual calculation and in-spite of the simpliﬁcation gave results which were very reasonable. 40 We have explained the method of calculation of natural frequency and amplitude of vibration for harmonic load for system with two degrees of freedom quite in detail in Chapter 5 (Vol. 1).

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298 Dynamics of Structure and Foundation: 2. Applications

2 Frequency in horizontal direction In this case it was assumed that the top deck and bottom slab as inﬁnitely rigid in its own plane and the columns act as leaf springs where the stiffness of the springs tantamount to transverse stiffness of the individual columns. Figure 2.14.6 shows the mathematical model perceived by Barkan. Mass on each horizontal spring is given by mi = m0i + mbi + 0.3mci + mLi

(2.14.10)

Here the horizontal displacement is given by δhi =

h3 (2 + 3ψ) 12EIc (1 + 6ψ)

Khi =

and

1 δhi

(2.14.11)

Here the term i represents the ith cross frame of the system.

Wb

Wa

Xgb

Xgc

Xga

Wd

Wc Xgd

H

G H

C/L axis

G

Kb

Ka Xha

A Wa

Ka

Xhb

Kc Xhc

B

Wd

Wc

Kc

Figure 2.14.6 Mathematical model in horizontal vibration.

© 2009 Taylor & Francis Group, London, UK

D

C

Wb

Kb

Kd

Kd

Analysis and design of machine foundations 299

It was argued that due to difference in geometry and shape, there will be some difference between the center of gravity (G) andcenter of stiffness (H). N While the resultant of all the masses i=1 mi will pass through the point G, there exists another point known as thecenter of elasticity (H) through which resultant of N all the column stiffness i=1 Khi will pass. As these two points do not coincide, as such other than translation the top deck will also undergo a rotation in the horizontal plane (φ) which will be coupled with the translation (x). Taking the center of gravity as the reference co-ordinate point he obtained the following differential equation of motion Mx¨ + Kh x + Kh eφ = Ph cos ωm t and Jφ φ¨ + Kh ex + (Kh e2 + γ )φ = Mh cos ωm t (2.14.12)

N for N number of frames; Kh = i=1 Khi for N number of N 2 2 frames; e = distance between the points H and G; Jφ = N i=1 mi Xgi ; γ = i=1 Khi Xhi ; N N Phi = Ci cos ωm t; Ph = i=1 Phi and Mh = i=1 Phi Xgi . Writing the above equation in the matrix form, we have in which, M =

M 0

0 Jφ

N i=1 mi

Kh x¨ + φ¨ Kh e

Kh e Kh e 2 + γ

x Ph cos ωm t = Mh cos ωm t φ

(2.14.13)

The coupled natural frequency of the system can be obtained from the equations f (ωλ2 ) = ωn4 − (αωx2 + ωϕ2 )ωn2 + ωx2 ωϕ2 = 0 Here ωx =

(2.14.14)

N 2 2 2 2 41 Kh /M , ωφ = i=1 Khi Xh /Jφ and α = (1+e )/r where r = Jφ /M i

3 Amplitude of vibration The amplitude of vibration is obtained from the expression x=

e2 r2

2 + ωx2 + ωϕ2 − ωm

f (ωλ2 )

Ph M

h − eωx2 M Jφ

;

φ=

e2 2 ω r2 x

Ph M

2 ) Mh − (ωx2 − ωm Jϕ

f (ωλ2 )

(2.14.15)

The net amplitude of horizontal vibration is given by xnet = x + X φ

(2.14.16)

41 Alternatively this can also be calculated based on the eigen value technique for two degree of freedom showed in Chapter 5 (Vol. 1).

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300 Dynamics of Structure and Foundation: 2. Applications

where, X = is the farthest point from the center of gravity point G. We can also apply the modal technique shown earlier in the chapter 5 (Vol. 1) and directly ﬁnd out the amplitude of vibration in matrix notation too. 2.14.4.3 Combined or Major’s method of analysis Major actually combined the above two methods (Barkan & Rausch) to arrive at a method which is usually known as the combined method of analysis. He realized that the resonance method of Rausch and amplitude method of Barkan are actually mutually complementary and blended the better of the two approaches42 . Thus it would be worth knowing as to what improvements Major did with respect to the previous two models. The improvements may be summarized as follows: •

•

Both Rausch and Barkan neglected the effect of underlying soil from their calculation43 , Major did try to cater for the effect of soil at least in vertical mode of vibration by adding the soil deformation to elastic deformation of the frame. As stated earlier that resonance check does not always prove to be an adequate design especially for under-tuned foundation which are found to show signiﬁcant vibration during start and stop of the machine, Major did devise a model where the foundation behavior under this transient can also be checked.

These, in essence, are the two signiﬁcant contribution of Major in his combined method. The methodology applied in this method is explained hereafter44 . 1

Frequency in vertical direction

For vertical frequency analysis Major followed in essence the method proposed by Rausch except that he took Barkan’s two-mass model as shown in Figure 2.14.5. Here, m2 = mass of the (upper slab + machine) + 0.5 times the mass of the column; m1 = mass of the bottom slab + mass of the condenser + 0.5 times the mass of the column; k2 = equivalent spring constants for the columns, and k1 = equivalent spring constants of the soil.

42 IS 2974 also recommends Major’s method for design of the Turbo-generator foundations. 43 Though Barkan acknowledged that this might affect the response but conceded that the analysis was too complex to be done manually and for very thick bottom raft, the effect of soil was negligible. 44 We apologize, for there would be some repetition with respect to earlier method of Rausch and Barkan. But we would still like to repeat it for ﬁrstly- the clarity and secondly to highlight what is the difference in approach with respect to the previous two methods.

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Analysis and design of machine foundations 301

He proposed that total vertical deflection is given by45 δv = δ1 + δ2 + δ3 + δ4 + δs

(2.14.17)

Here, PL3 δ1 = 96EIb

QL3 δ2 = 384EIb δ3 =

2ψ + 1 ψ +2

is deflection due to concentrated load;

5ψ + 2 ψ +2

is the deflection due to uniformly distributed load;

Q 3 L P+ is the deflection due to shear; 5 EAb 2

δ4 =

h EAc

N+

P+Q 2

!

(2.14.18)

is the axial deflection of column due to the concentrated

load transferred from the longitudinal girder (N). δs =

(P + Q + 2N) + Wf L f B f cu

(2.14.19)

is the elastic deformation of soil in vertical mode. Here, P = concentrated load from the machine; Q = UDL of the cross beam (qL); q = self weight per unit length of the cross beam; N = concentrated load on the column; Wf = weight of bottom slab + half the weight of the columns; Ab = area of cross section of the beam; Ac = area of cross section of the column; Ib = moment of inertia of the beam; Ic = moment of Inertia for the column; E = dynamic modulus of elasticity of the frame; h = effective height of the column; L = effective length of the cross beam; Lf = length of the foundation; Bf = width of the foundation and cu = co-efﬁcient of elastic compression of the soilψ = Ib h/(Ic L). The fundamental frequency in vertical direction is then given by fv = 30/ δv cpm. 2

(2.14.20)

Frequency in horizontal direction

Considering n number of cross frames, in horizontal direction, Major followed the same procedure of Barkan, as shown in Figure 2.14.6 like idealizing the top deck as

45 This is a very interesting proposition of adding elastic deformation of the soil directly to the structure just note it for the time being we will discuss more about it later at appropriate time.

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302 Dynamics of Structure and Foundation: 2. Applications

rigid in its own plane and considering an eccentricity e between the center of mass and center of stiffness he arrived at an expression 1 2 n 2 Khi Ih 3 fn h = 30 α0 ± α02 − i=1 cpm n i=1 Wi Jφ

(2.14.21)

h (2+3ψ) where, Khi = lateral stiffness of the ith frame i; and Khi = δ1hi where δhi = 12EI ; c (1+6ψ) Wi = total weight of the ith frame plus weight of the machine plus weight of the beam and the longitudinal beams; Jφ = mass moment of inertia ∼ = n transverse 2 ; X = distance of weight W from the resultant center of mass point G46 ; W X g i i=1 gi 2 ; X = distance of each frame from the center of rigidity H, Ih = ni=1 Khi Xhi h 3

and 3

n 1 2 ni=1 Khi Ih i=1 Khi α0 = . e + n + 2 Jϕ Jφ i=1 Wi

(2.14.22)

Calculation of amplitude

We had seen earlier in Chapter 5 (Vol. 1) that under harmonic load the amplitude of vibration is given by the expression P0 k

x¯ max =

sin ωm t

(1 − r2 )2 + (2Dr)2

(2.14.23)

where r = ωm /ωn and D = c/cc with, cc = Critical damping of the system and is √ 2 mk. For sin ωm t = 1, we have P0 k

x¯ max =

(1 − r2 )2 + (2Dr)2

➔

δst

x¯ max =

(1 − r2 )2 + (2Dr)2

(2.14.24)

Here, we introduce a term called logarithmic decrement given by ∇ = 2πD, where ∇ = Logarithmic decrement; D = damping ratio. Major replaced the 2D by ∇π and deﬁned amplitude of vibration as δst

➔ x¯ max =

(1 − r2 )2

46 Refer to Figure 2.14.6.

© 2009 Taylor & Francis Group, London, UK

+

∇ 2 π

(r)2

.

(2.14.25)

Analysis and design of machine foundations 303

4 Under-tuned foundation For under tuned foundation when ωn < ωm during starting and stopping of machine, there will be a case, when for a fleeting moment ωn = ωm and as such the frequency ratio (r) will be equal to 1.0 for that instant. During this point considering r = 1 the amplitude of vibration reduces to x¯ max = π δst /∇

(2.14.26)

Major has suggested that the logarithmic decrement (∇) be taken for concrete as 0.4 when the maximum amplitude becomes, x¯ max = 7.85 δst . 5

Over tuned foundation

For over tuned foundation when ωn > ωm the maximum amplitude can be found out from the expression δst

x¯ max =

(1 − r2 )2 +

∇ 2 π

(r)2

(2.14.27)

where δst = δv or δh as the case may be. 6

Calculation of unbalanced centrifugal force

For under tuned foundation (ωn < ωm ) the centrifugal force Ci is given by Ci = αR

ωn ωm

2 (2.14.28)

For over tuned case (ωn > ωm ) the centrifugal force Ci is given by Ci = αR

(2.14.29)

where the value of α is as given in Table 2.14.1. While in the vertical direction, Major considered the deﬂection of individual frame which when multiplied by the above factors gives the dynamic amplitude under transient condition. In horizontal direction a stick model has been considered, where the stiffness of all frames are clubbed together to arrive at a unique value of amplitude. Table 2.14.1 Values of α. Sl. No.

α

rpm rating of machine

1 2 3

0.2 0.16 0.1

≥3000 1500 750

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304 Dynamics of Structure and Foundation: 2. Applications

Since individual amplitude of each frame is also necessary he approached the problem in the following manner. If C = N i=1 Ci denotes the total centrifugal force in horizontal direction then the centrifugal force on each individual frame is distributed in terms of their individual stiffness CK X K Ci = C N hi + e N hi hi 2 i=1 Khi i=1 Khi Xhi

(2.14.30)

Here e is the distance between center of rigidity Xhi and center of the resultant of the horizontal dynamic forces, C = N i=1 Ci . Once Ci is obtained the deflection of the ith frame is obtained from the expression δhi =

Ci Khi

(2.14.31)

with the value of δhi , the amplitude of vibration in horizontal direction is obtained from the expression δhi

ahi =

(1 − r2 )2 +

∇ 2 π

(r)2

(2.14.32)

Major states that since the structure is usually more flexible in transverse direction and considering the high speed of the machine is usually under tuned in this direction and as such it is a common practice to consider for horizontal mode ahi = 7.85δhi . 7

(2.14.33)

Dynamic forces

The dynamic forces to be accounted for in structural design of the frame have been expressed by Major as follows: To account for idealization made in calculation of natural frequency it is suggested to correct the calculated natural frequency by a term, fn = fn (1 ± α), where α is a correction factor and may be considered as 0.2. For under-tuned foundation (fn < fm ) plus signed should be considered while for over tuned foundation minus sign to be considered47 . fm fm When fn lies between 1+α and 1−α , then fn = fm . 47 This actually means Major is assuming that the frequency calculation could be out from actual by (±) 20% and based on the correction factor is actually trying to develop a conservative estimate of the dynamic force.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 305 Table 2.14.2 Values of dynamic force under various conditions. Operating frequency of machine (rpm)

Case fn < fm

3000 1500 750

fn

> fm

3000

Dynamic force F = 16R

fn f

Fmax = 16R

fn fm

Fmax = 12R

m 2

F = 12R 2 f F = 8R fmn F=

2Fmax 2 2 2 fm 1− 2 + ∇ π f n

fm 1−α

< fn <

fm 1+α

Remarks

2

1500 750

Do Do

3000 1500 750

F = 16R F = 12R F = 8R

Fmax = 8R 2 fm f 2 n

Fmax = 1.0R Fmax = 0.8R Fmax = 0.5R

R = rotating weight on the frame.

Based on the above, Major suggested Table 2.14.2 for calculating the dynamic forces. For vertical dynamic force that acts on the center of the transverse beam the rotating weight on the beam only should considered as the expression R. For calculation of the horizontal dynamic force in transverse direction total rotating weight on the transverse beam plus rotating weight on the longitudinal girder transferred to the column shall also be considered while calculating the term R.

2.15 DYNAMIC SOIL-STRUCTURE INTERACTION MODEL FOR VIBRATION ANALYSIS OF TURBINE FOUNDATION We present here a further modiﬁcation of Major’s method considering the effect of underlying soil on the vibration analysis of turbine foundation. The history behind its evolution is quite interesting and would not be possibly out of context to share the background with you. Till 1980’s Indian power industry was mostly restricted to Turbine units having capacity up to 210 MW. These turbines were all supplied by BHEL48 and were prototype of LMW models used in the USSR. While the machines were quite massive and sturdy, the foundation system for these types of machines was usually wall mounted and not frame type. In reality they were actually massive RCC blocks having cutouts in it for laying the piping and ﬁxing other sundry ﬁxtures. They were generically short in height and because of their massiveness and immense rigidity these foundations were mostly over tuned. Thus conventional theories as proposed by Barkan/Major justiﬁed their analysis quite well.

48 Bharat Heavy Electrical Limited, they are the premier Turbine manufacturing company in India.

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306 Dynamics of Structure and Foundation: 2. Applications

The ﬁrst 500 MW turbines supplied in India were from Siemens KWU49 and had a complete different structural conﬁguration from the erstwhile models that were in vogue in the industry. The top deck level was much taller (about 16.0 m); they were much slicker and called for much more space below the turbine and truly represented a framed structure having under tuned characteristics. In the meantime a number of turbines in operation in various parts of India were monitored for vibration and it was observed that some of them which were designed as over-tuned system still showed transient excitation during start and stop of the machine (meaning thereby that they were showing under tuned characteristics). The question was why it was happening so? It was realized that it was possibly the soil below the bottom mat which was participating in the vibration and changing the characteristic behavior of the foundation. Wedpathak, Pandit and Guha (1977) conducted vibration monitoring on various TG foundations at different power plant in India and showed that there existed a considerable variation in amplitudes observed in the ﬁeld and those calculated theoretically. The above discrepancy suggested that there was deﬁnitely a necessity to arrive at a more realistic mathematical model to predict the response of the turbine foundations. It also proved that the assumption made in conventional analysis by Barkan and Major, that making the bottom raft thick- nulliﬁes any participation of the underlying soil in the vibration may not be true in all cases. Especially for 500 MW class of turbine where to suppress the vibration of the underlying soil the thickness of the bottom, mat would have to be so thick that the foundation could become prohibitively expensive. Moreover, due to their height and slenderness in transverse direction it was realized that translation in this direction will also induce a coupled rocking mode in the transverse plane which was not accounted for in the conventional method. Considering the inadequacy in the conventional method in the context of present day class of turbines, we started our investigation into this problem to arrive at a more rational model where the contribution of the soil in vibration of such frame foundations can be catered for. While it was always possible to solve this problem based on FEM50 , we realized that prior to that one should have the feel as to how the system is behaving and moreover considering the expense incurred for doing a major FEM analysis in terms of man hour spent in data generation, data input, checking the output and result interpretations, was there an alternative model which would give reasonable results if needed to be done manually or use computer to a minimum? That was the philosophy based on which we started our quest for a solution and the outcome is what we would like to share with you. 1 Frequency in vertical direction Unlike Major’s model we consider here a three-mass lumped system as shown in Figure 2.15.1. We use here a judicious mixture of Barkan and Major’s method and couple the soil springs based on Richart or Wolf’s formulation.

49 The ﬁrst Siemens machine of 500 MW was supplied to Trombay (Tata Electric) and the second to Singrauli NTPC. 50 This we had tackled too and will be presented at a later stage.

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Analysis and design of machine foundations 307

m3

z3 k3

m2

z2 k2

m1

k z1

Figure 2.15.1 2D-Mathematical model for soil-structure interaction (Vert. mode).

Here for n number of cross frames, we have • • •

m3 = ni=1 (Concentrated mass of the machine carried by the transverse girder +0.45 times self weight of the transverse girder) m2 = ni=1 (0.25 times the Mass of the transverse girder + mass from the longitudinal girder including machine weight if any transferred to the cross frame + 0.30 times the mass of the column) m1 = mass of the bottom slab + mass of the condenser + ni=1 (0.3 times the mass of the column) + mass of the soil participating in the vibration. For spring k3 , for the beams we have n i=1

δv =

L3 (1 + 2ψ) 3L + 96EIb (2 + ψ) 8GAb

and

k3 =

n 1 δv

(2.15.1)

i=1

n 2Ac Ec ; k1 = where, k2 = equivalent spring constants for the columns @ i=1 h equivalent spring for the soil obtained from Richart or Wolf’s formulation51 and G = dynamic shear modulus of concrete @ 0.5E. Applying D’Alembert’s equation free vibration of the system can deduced as ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ k 1 + k2 0 ⎨x¨ 1 ⎬ −k2 0 m1 0 ⎨x1 ⎬ ⎣ 0 m2 0 ⎦ x¨ 2 + ⎣ −k2 k2 + k3 −k3 ⎦ x2 = 0 (2.15.2) ⎩ ⎭ ⎩ ⎭ x¨ 3 0 −k3 k3 x3 0 0 m3

51 Refer section on block foundation for the values of the soil springs.

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308 Dynamics of Structure and Foundation: 2. Applications

Calculation is now quite straight forward for frequency analysis based on eigen value solution. Instead of soil, if the foundation is resting on piles then we can straight use pile springs based on Novak’s formulation or other methods as cited previously and use this spring as the spring k1 . The above matrix on expansion will give an equation of third degree whose characteristics roots will give the eigen values of the above problem. 2 Calculation of horizontal frequency Based on the discussion in the preceding page, it was highlighted that major lacunae lies in this mode. While conventional analysis considers translation and rotation in plan it does not consider the rotation in elevation which will also get coupled when the height of turbine deck is high. We consider all these aspects in our formulation and develop a matrix of order 4 × 4 that we feel takes into cognizance all the short comings of the conventional method we discussed. We show hereafter an analytical model conceived to cater to all the above aspects. While the conventional analysis considers lateral translation x and rotation φ in plan it considers the bottom raft to be completely rigid and the soil has no effect on the vibration. Since the major horizontal motion of the machine is in the transverse direction we have added additional degrees of freedom • •

u which is the translational displacement of the foundation. For turbines of capacity 500 MW and above as the height h of the column is quite large this will also induce a rocking of the foundation (in transverse plane) and assigned a value θ .

Thus while the conventional analysis has two degrees of freedom x and φ, in our model shown in Figure 2.15.2, we have four degrees of freedom, namely x, φ, u, θ. Here, Kx = translation spring value of soil; Kθ = rocking spring value of soil; m0 , Jφ = mass and mass moment of inertia of top deck + Machine; mf and Jθ = mass and mass moment of inertia of the bottom raft. To arrive at the equation of motion based on D’Alembert’s principle will be quite difﬁcult as the coupled motion is quite complicated. So to derive the equations we use the famous Lagrange’s equation from the energy principle when n d ∂T ∂T ∂U − dqi = 0 + d(T + U) = dt ∂ q˙ i ∂qi ∂qi

(2.15.3)

i=1

T = f (q1 , q2 , q3 . . . . . . .

qn ;

q˙ 1 , q˙ 2, q˙ 3 , . . . . . . . . . .

q˙ n )

U = f (q1 , q2 , q3 , . . . . . . . . . , qn )

and (2.15.4)

The kinetic energy, T for the system is given by T=

1 1 1 1 ˙ 2 + Jφ φ˙ 2 m u˙ 2 + Jθ θ˙ 2 + m0 (u˙ + x˙ + hθ˙ + eφ) 2 f 2 2 2

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(2.15.5)

Analysis and design of machine foundations 309

m0, Jφ

Kh h

Kx

K u mf, J

Figure 2.15.2 2D-Mathematical model for soil-structure interacton (Horz. mode).

The potential energy, U is given by U=

1 1 1 1 K x u 2 + K θ θ 2 + K h x 2 + Iφ φ 2 2 2 2 2

(2.15.6)

Differentiating, ∂T ˙ = mf u˙ + m0 (u˙ + x˙ + hθ˙ + eφ) and ∂ u˙ d ∂T ¨ = mf u¨ + m0 (u¨ + x¨ + hθ¨ + eφ) dt ∂ u˙ ∂T ˙ = Jθ θ˙ + m0 h(u˙ + x˙ + hθ˙ + eφ) and ∂ θ˙ d ∂T ¨ = Jθ θ¨ + m0 h(u¨ + x¨ + hθ¨ + eφ) dt ∂ θ˙

(2.15.7)

Similarly d dt d dt

∂T ∂ x˙

∂T ∂ φ˙

¨ = m0 (u¨ + x¨ + hθ¨ + eφ)

and (2.15.8)

¨ = Jφ φ¨ + m0 e(u¨ + x¨ + hθ¨ + eφ)

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310 Dynamics of Structure and Foundation: 2. Applications

For potential energy, we have ∂U = Kx U; ∂u

∂U = Kθ θ ; ∂θ

∂U = Kh x ∂x

and

∂U = Kh e2 φ + Iφ φ ∂φ

Substituting the above values in the equation n d ∂T ∂T ∂ d(T + U) = − dqi = 0 + dt ∂ q˙ i ∂qi ∂qi i=1

and writing in matrix form we have ⎡

m0 ⎢ m0 e ⎢ ⎣ m0 m0 h ⎡

m0 e Jφ + m0 e2 m0 e m0 eh

Kh ⎢0 +⎢ ⎣0 0

m0 m0 e m 0 + mf m0 h

0 Kh e2 + Iφ 0 0

0 0 Kx 0

⎤⎧ ⎫ x¨ ⎪ m0 h ⎪ ⎪ ⎨ ¨⎪ ⎬ φ m0 eh ⎥ ⎥ ¨⎪ m0 h ⎦ ⎪ U ⎪ ⎩ ¨⎪ ⎭ J θ + m 0 h2 θ

⎤⎧ ⎫ 0 ⎪ x⎪ ⎪ ⎨ ⎪ ⎬ ϕ 0⎥ ⎥ =0 ⎦ U⎪ 0 ⎪ ⎪ ⎩ ⎪ ⎭ θ Kθ

(2.15.9)

Equation (2.15.9) gives the complete free vibration equation of motion for the turbine foundation system considering the soil springs the translation and rocking modes. Here, Kh =

N

Khi =

i=1

Jφ =

N

N 1 δhi

where,δhi =

i=1

2 mi Xgi

i=1

and

Iφ =

N

h3 (2 + 3ψ) ; 12EIc (1 + 6ψ)

2 Ki Xhi

(2.15.10)

i=1

Before we go further a few things needs to be noticed • • •

The matrix is real and symmetric. The equations are dynamically coupled thus the reference co-ordinate is the center of rigidity and not center of mass as is the case with D’Alembert’s equation. Due to dynamic coupling, the mass matrix is a full matrix while stiffness and the damping matrix would remain in uncoupled form.

Expansion of the eigen value matrix will give a fourth order polynomial whose roots can be found based on Bairstow’s method or else can be very easily solved based on software tools like MATH CAD/ MATLAB etc. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 311

3 Calculation of amplitude of vibration We will use here the generic modal response technique to obtain the amplitude of vibration using the orthogonal property of the matrix. Moreover since the Turbine once started will continue to operate for a long time as such the steady state response is critical and we shall ignore the transient part. Thus in the vertical /horizontal direction, we have ¨ + [C]{X} ˙ + [K]{X} = {P(t)} [M]{X}

(2.15.11)

where, [M] = mass matrix of the system; [C] = damping matrix of the system; [K] = stiffness matrix of the system, and {P(t)} = P sin ωm t/P cos ωm t the dynamic force with sine or cosine function for the vertical or horizontal case respectively. Now considering the operation, ¨ + [φ]T [C][φ]{X} ˙ + [φ]T [K][φ]{X} = [φ]T {P(t)}[φ] [φ]T [M][φ]{X}

(2.15.12)

If the total numbers of degrees of freedom is j say then we have j numbers of uncoupled equation depicted by j i=1

when

ξ¨i + 2Di ωi ξ˙i + ωi 2ξi = p0i (t)

j=3 i=1 j=4 i=1

ξi =

p0i sin ωm t (1 − r2 )2 + (2Dr)2

ξi =

p0i cos ωm t (1 − r2 )2 + (2Dr)2

(2.15.13)

in the vertical direction. And (2.15.14) in the horizontal direction.

Once the displacement in uncoupled form are known the global amplitude is found out based on the expression, {X} = [φ]{ξ }. The net amplitude at the top deck, is given by the expression xinet = xi + Ui + Xhi ϕ + hθ 52

(2.15.15)

It would be worth now to objectively evaluate the advantage of this method. Some of the advantages that can be attributed to this model are: •

It takes all the fundamental degrees of freedom considered by the conventional method and also takes into consideration the effect of the soil in vibration analysis.

52 We are not trying to take a short cut. We will further elaborate the whole technique including the complete design based on a suitable problem hereafter.

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312 Dynamics of Structure and Foundation: 2. Applications

• • • •

It can take both soil or pile springs (with and without embedded effect) as an input to the overall matrix. The calculation though more intense than conventional method it is yet amenable to manual computation and gives the engineer a ﬁrst order feel as to how the coupled soil-structure is behaving under dynamic loading. It will surely give quantitatively a clear idea as to how much is the effect of soil on the overall vibration vis-a-vis ﬁxed base frequency when the effect of soil is neglected53 . It will also help in taking a better decision if further elaborate analysis based on 3D space frame model is envisaged or not.

To people of orthodox school as well as the computer buffs54 we can assure that this technique works quit ﬁne. This technique has been put into practice for a boiler feed pump framed foundation for a power plant in India and we are happy to inform that it has been operating smoothly without any problem for more than 15 years (Chowdhury and Som 1993).

2.16 COMPUTER ANALYSIS OF TURBINE FOUNDATION BASED ON MULTI DEGREE OF FREEDOM In this section we discuss the method of analysis and design of turbine foundations considering it as a frame having multi degree of freedom through computer55 . In this case the steps followed for analysis of the frame foundation is as follows: The system is broken up into three parts as shown • • •

The super structure The raft The soil

We basically use here the concept of ﬁnite element to solve the above problem. Though application of ﬁnite element is more appropriate for continuum, however basic principle of its application is well valid for this case also. Shown in Figure 2.16.1 is a typical conceptual model of a turbine foundation resting on a bottom raft which in turn is resting on soil.

53 If you are solving the problem in MATHCAD/MATLAB just put Kx = 1020 and Kθ = 1020 this will effectively make the problem a ﬁxed base one. Else delete the rows and column in the matrix pertaining to the soil degrees of freedom and reduce it to a 2X2 matrix having x and φ as the active degrees of freedom. 54 Whose staple diet is a problem having 1000 degrees of freedom. Anything less than that is surely crude! 55 It is not that we would like to continue our designs based on a paper, pencil and a calculator at best. At the door of the 21st century we do not want to carry the stigma of being Rip Van Winkle though we conﬁrm that we discourage the use of software as a black box.

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Analysis and design of machine foundations 313

Figure 2.16.1 3D computor model of a turbine frame with bottom raft and soil spring.

We discuss below step by step the concepts underlying the development of its mathematical model for analysis in computer. 1

The super structure

What element do I choose and how many nodes do I consider? The intuitive choice for the super-structure is obviously to model it as a space frame where the beams and columns are idealized as beam elements having six degrees of freedom at each node. But for modeling a turbine foundation frame there is a difference with normal building frames. Mathematical model for the beam and column are usually taken at the center line of the element as shown in Figure 2.16.2. Based on Figure 2.16.2, during computer analysis, the moment and shear output will be given at the chosen nodes. For normal building frame this does not digress from the reality much for the dimension of the columns are small. However for turbine foundation the columns are of large dimension (usually they are about 1500/2000 mm). During design of beams since we know that the design bending moment at support is to be taken at the face of the column, the large dimension of the column makes a signiﬁcant reduction in the design moment of the beam at the support. The major advantage is that it helps in reduction of congestion of reinforcement at the beam column junction. As such to correctly predict this phenomenon the model should consist of three nodes instead of one connected by rigid links as shown Figure 2.16.3. © 2009 Taylor & Francis Group, London, UK

314 Dynamics of Structure and Foundation: 2. Applications

Node (Typ.)

Actual Frame Idealized Model

Figure 2.16.2 Idealized model of a normal frame.

Moment based on one node at C/L of beam column

Design moment at column face Based on three node concept

Column Node

Beam Node

Rigid Link Bending Moment profile

Figure 2.16.3 Typical connection of beam column junction with rigid link. Note: In some software packages this may also be input as master and slave option where the beam node is usually taken as the master and the column node as the slave node.

For the beam elements as the span by depth ratio is signiﬁcant it is preferable to consider the shear deformation of the girder during the analysis. The loads that are induced by the machine to the deck are mostly transferred through the bearing/sole plate. The sole plates are not necessarily always co-aligned with beam center line. Thus to simulate this situation two of the following techniques could be used. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 315

F

F Sole plate

T=FxD

d

Link C/L Beam

Figure 2.16.4 Vertical load acting on sole plate eccentric to the center line of the beam.

• •

2

By deﬁning the load with additional torsion about the center line of the beam based on the eccentricity between the bearing plate and center line axis of the beam Providing node at the point of incident of the load and connecting this point to the mathematical model by a rigid link as shown in Figure 2.16.4. How many nodes do I consider?

Intuitively the primary choice of nodes will be the beam column junctions. On identifying these nodes we further break it up into two additional nodes based on the concepts as mentioned above. Other than this points at which direct load is transferred to the girders nodal points are to be considered also. For members under complex loading in span the number of nodes to be provided for each beam member should be sufﬁcient to plot the bending moment and the shear force diagram. For dynamic analysis enough nodes should be considered along the length of the beam and column so that all the modes having a natural frequency less than or equal to the operating frequency of the machine are simulated. The lower rigid body mode of the top deck as a unit is not affected signiﬁcantly by the number of nodes along the length of the beams. However higher modes simulating the differential deflection of the top deck are affected by the distribution of nodes. If not modeled with enough nodes these modes may be entirely missed leading to an incorrect result. The suggested number of nodes n, to be placed along the length of the span is given by the larger of the following two values56 :

n≥

L √ m 14 1 Lωm m 12 ωm :n≥ + π EI 2 π EA

(2.16.1)

56 The expressions are derived from frequencies of a simple supported beam in flexural and axial mode. The basis of this expression is that if the nth natural frequency of the beam is at or below the operating frequency of the machine then at least n mid-span nodes will be required to calculate the n modes using the discrete model.

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316 Dynamics of Structure and Foundation: 2. Applications

where, L = span of the member; E = modulus of elasticity; m = mass per unit length; ωm = operating frequency of the machine in rad/sec; I = moment of inertia about the beam about its weaker axis of bending, and A = cross sectional area of the beam. The nodal mass may be calculated either based on lumped mass approach or consistent mass approach (Archer 1963). The consistent mass approach accounts for the distributed mass and variation of deﬂection along the length of the beam. However, one major disadvantage with the consistent mass matrix is that it is a full matrix in contrary to lumped mass which is a diagonal matrix and thus calls for more computational effort. It has been observed that the natural frequency obtained by consistent mass approach is more accurate than lumped mass approach though the difference may be small for most of the practical problems. For practical analysis of Turbine foundation considering masses lumped at the nodes is the common industrial practice. Once the beam elements and the nodes are chosen and their properties like moment of inertia and sectional area etc are provided as input, the computer generates the local stiffness matrix of each beam (of size 12 × 12) and then based on their direction cosine transfers the local stiffness matrix into the global axis and assembles them to form the global stiffness matrix of the superstructure. 3

The foundation raft

The foundation raft usually consists of a slab resting on soil or pile which is about 2000/2500 mm thick. What element to use which would be optimal as well as provide the best result is still a debate among the ﬁnite element analysts. Some literature (Design Criteria for Turbine Generator Pedestal, 1970) recommend to model the raft as plate bending elements while the others (Arya et al. 1979) insist on to model it as beam elements supported on soil springs. While some advocate to use even 8 nodded brick element to model the raft. With such controversies prevailing on this issue it would possibly be worthwhile to evaluate the pros and cons of each of these elements. Plate elements Plate elements apparently look to be a good choice for physically, it best reflects the continuum. But as far as mathematical formulation of plates based on Finite Element formulation is concerned the best available element for plate bending considering its numerical convergence is the Discrete Kirchoff Triangular (DKT) plate element. The stiffness matrix formulation of DKT plate element is based on the thin plate theory having three (two translation and one rotation) degrees of freedom per node. The basic idealization is that the thickness of the plate is negligible in comparison to its plan dimension and as such the effect of transverse shear acting along the edge of the plate is neglected. For the turbine raft having thickness of 2000/2500 mm it is evident that the thickness of the raft is quite large and as such it would not be perhaps prudent to neglect the thickness vis-à-vis the effect of shear strain energy contribution of the overall system. Which catapults the problem from Kirchoff-type of thin plate to Mindlin-Reissner type of thick plate where solution is sought taking into consideration the shear deformation along the edge of the plate. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 317

Though many researcher have tried to formulate these type of plate based on FEM most of them suffers from one technical snag or other, namely • • •

Failing to pass important patch test Spurious zero energy mode Being sensitive to geometric distortion and meshing.

These can lead to poor solutions and even results which are unacceptable at times (Kardestuncer 1987). Hence, without a proper mathematical formulation of the thick plate in hand, specially the numerical problems it can create while seeking solution to the problem, we would suggest not to use such elements in modeling this problem. Brick elements Brick elements could also become a plausible choice for modeling the turbine raft. From convergence point of view brick elements are stable and have been successfully adapted to solve different class of problems in fracture, rock and fluid mechanics. However it has been observed that the eight nodded brick element usually have poor approximation capability and higher order elements having 16 or 24 nodes are usually used for efﬁcient solution. But use of such higher order elements calls for a much more expensive analysis in terms of computer time, data preparation, input, output etc and is usually not essential. Besides this brick element suffers from one serious lacunae in terms of design. Brick elements in most of the commercially available software give output in terms of normal and shear stress parameters. While this is ﬁne in terms rock or fracture mechanics problem where design check is done against allowable stresses, for the turbine raft design we are basically looking for output in terms of moment, shear and torsion. To back calculate these parameters from the computer out put and subsequent interpolation to arrive at the design moments, shears etc can be extremely tedious and chances are very high that the engineer assigned to perform this task gets lost in a maze of numbers and gets totally confused. For eigen-solution though use of brick element is OK we would however suggest users the use of brick elements for design purpose with caution for the enormous difﬁculty one could face in back calculating the stress output in terms of moment, shear and torsion. Beam elements This brings us to the last of element in use, the beam element to model the turbine raft. From convergence and correctness of results we had already discussed in quite detail in Chapter 2 (Vol. 1) that if properly modeled beam elements gives results which is very close to plate elements in simulating a raft problem57 . Moreover for derivation of stiffness matrix irrespective of the methodology used like moment area theorem, strain energy method or numerical methods like ﬁnite element, the results converge to an exact solution.

57 Refer Chapter 4 (Vol. 1) on Static soil structure Interaction where we have discussed in detail the use of beam vis-a-vis plate bending elements for simulation of rafts resting on soil.

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318 Dynamics of Structure and Foundation: 2. Applications

Even when the stiffness formulation takes into consideration transverse shear deformation unlike thick plate element the formulation is consistent and conforming. Finally computer output is in terms of moment shear and torsion directly and may be directly used for re-bar calculation without having recourse to deriving them from stress output unlike brick elements. Moreover if we take the elements with reasonable mesh reﬁnements including the transverse shear deformation into cognizance58 , we can approach a state where the energy compatibility in terms of external work done and consequent strain energy induced can be well satisﬁed. Thus in terms of ease of use as well as convergence of results beam elements do make a very attractive choice59 . For the raft, as the thickness is signiﬁcant considering the shear deformation characteristics is a must for maintaining the strain energy compatibility. 4

The soil

The basic soil parameter which needs to be known to mathematically model the soil is dynamic shear modulus (G)60 . The soil being a continuum itself can either be modeled based on FEM as 3D brick elements61 , 2D plane strain elements or discrete springs. For modeling the soil, the choice is again multiple. However as soil itself is an inﬁnite domain successful application of FEM has been mostly in cases where the problem could be simulated by a two dimensional model where the soil itself has been modeled as plane strain elements or inﬁnite ﬁnite elements to arrive at a meaningful result. Rarely, we have come across cases where in practical problems pertaining to soil has been modeled in 3D elements for the effort and cost in terms of man-hour and output interpretation can make the analysis prohibitively expensive. For the particular case of turbine foundation analysis as we are interested to know more about the behavior of the frame and the bottom raft rather then the intricate behavior of the soil itself, the common practice is to model the soil as frequency independent linear springs based on Richart or Wolf’s springs as described in section of block foundation. For practical application this has been found to be quite adequate. More sophisticated model based on frequency dependent complex stiffness is usually not warranted in this case. Depending on the soil stiffness and the stiffness of the raft a correction to the spring needs to be done for correct evaluation of the response62 . Once the spring values are evaluated they are connected to the node of the raft element based on usual ﬁnite element procedure to arrive at the complete stiffness matrix

58 Whose contribution becomes signiﬁcant as the ratio of span by depth reduces. 59 For protagonists of classical school this is to inform that many Turbines raft has been modeled as beam elements which have been analyzed, designed and constructed and has successfully stood the vagaries of nature and the test of time. 60 We have dealt this topic in detail in the Chapter 1 (Vol. 2). 61 Refer Chapter 4 (Vol. 1) where we have discussed such problems in detail in terms of static loading. 62 This we have dealt in detail in the chapter 1 (Vol. 2).

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Analysis and design of machine foundations 319

of soil foundation system. A typical example is shown in Chapter 2 (Vol. 1) where we have solved a table top centrifugal compressor foundation based on this method. 5

The machine

Do we model the machine resting on the top deck also in our analysis? A debate which has been in the profession for quite some time and we do not want to pass a sacrosanct sermon on this issue. However our objective analysis of this Shakespearean dilemma63 is as follows: For Turbines of low capacity (<350 MW) the foundations are usually designed having over tuned characteristics. Moreover as the overall dimension of the machine is also relatively smaller, as such it is reasonable to consider the whole turbine and the generator as a rigid mass whose inertial contribution as a lumped mass is taken into cognizance in the analysis only. However with increasing demand for energy, power manufactures are coming out with Turbines having higher and higher capacities. This has made the overall dimension of the turbine larger and the foundation size have also increased and have made it ﬂexible and more susceptible to dynamic excitation. For the equipment, the main shaft which connects the turbine and the generator has become longer, thus ﬂexible, and with increase in the operating speed a slight imbalance in the rotating mass can induce signiﬁcant dynamic load on the shaft and also the over all deformation of the soil, raft and the frame (specially in the ﬂexural mode) can generate a phenomenon which is know as the bowing of the turbine shaft. Bowing or bending of the shaft about its center line axis can create damage to the machine components, induce large forces at the bearing and can also reduce the operating efﬁciency of the turbine. Thus for larger turbines (>500 MW) it would be possibly justiﬁed to consider the machine as an integral part of the analysis too. For such consideration an elaborate Finite Element modeling of the turbine and the generator is usually not warranted a simple mathematical model consisting of masses lumped at strategic nodes connected by beams, springs, rigid links etc would usually sufﬁce64 .

2.17 ANALYSIS OF TURBINE FOUNDATION

2.17.1 The analysis The analysis is usually done in the computer in four steps: •

Dynamic analysis to calculate the natural frequencies of the system to ensure that it is out of tune to the operating frequency of the machine by ±20%.

63 To be or not to be . . . . 64 At this point we would strongly recommend you to take help of your equipment specialist while modeling the equipment connected to the super-structure.

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320 Dynamics of Structure and Foundation: 2. Applications

• • •

Calculation of the dynamic amplitude to check that the same are within the acceptable limits as prescribed in the code or as pre-deﬁned by the equipment supplier. Earthquake analysis if the same is perceived critical for the foundation. A pseudo–static analysis to obtain the design Moment, Shear and Torsion induced in the members check the stresses induced in the different structural elements like beam column and slabs.

2.17.2 Calculation of the eigen values For calculation of the natural frequencies or the eigen values the ﬁrst choice the user has to make as to how many modes do I consider for the analysis? First three modes, six modes or twenty modes . . . We have heard variety of such numbers65 . Unfortunately, none of the answers are universally correct, for how many modes are signiﬁcant for the analysis varies from case to case and it also depends on what we are looking for in terms of cases like checking the resonance, checking the transient response or checking the response against earthquake. The most rational basis of choice of modes would be based on modal mass participation factor66 which should always be the basis of arriving at the number of signiﬁcant modes to be considered for dynamic analysis when we are doing a resonance check. As a ﬁrst step start with say ﬁve or six signiﬁcant modes check the frequency with the operating speed of the machine and also at the same time check the modal mass participation factor for these modes67 . If the mass participation is of the order of say 50 or 60% it is evident more number of modes need to considered. Number of modes that excite at least 95 to 99% of the mass should be the basis of number of signiﬁcant modes to be considered in the analysis. The reason is as explained hereafter. Suppose for the ﬁrst six modes we ﬁnd the natural frequency of the system is below the operating speed of the machine by 20% but it has only excited say 60% of the mass while higher modes which are in the vicinity of the operating frequency has excited say 89% of the mass (say the 9th or the 10th mode) it is obvious that these modes will excite the structure much more and this we will completely miss if we restrict our analysis to a preconceived six-mode analysis only. The other advantage is that as the eigen values go on increasing with each mode there will always be some value which would match or be very near to the operating frequency of the machine. But, if nearly 100% of the mass has already participated in the vibration in the earlier modes this will have no effect on the response of the structure even though the frequency is in the vicinity of the resonance range. However, this can only be predicted conﬁdently provided you know exactly how much mass has already participated in the vibration.

65 With comments such as “From my experience”, “Normal engineering practice”, and ﬁnally “From previous experience” – from an engineer with 2 years of experience(!!!) etc. to name a few. 66 For details of modal mass participation refer to Chapter 3 (Vol. 2). 67 Most of the commercially available FEM and dynamic analysis software have this option as an output for the user to check the mass participation in the X,Y and Z direction.

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Analysis and design of machine foundations 321

2.17.3 So the ground rule is. . . Do not guess or start with the pre-conceived notion that “n” number of modes would sufﬁce check the modal mass participation factor and then decide. To arrive at the eigen value vis a vis the natural frequency, though the basic equation for the solution remains same i.e. [K]{x} = ω2 [M]{x}

(2.17.1)

computation of ω2 is surely not done in the way we have described in our earlier chapter68 . For solution of eigen-values having large degrees of freedom special numerical techniques are usually used. When earthquake analysis is also critical, number of modes signiﬁcant enough to simulate the natural frequency to 33 Hz should be considered for the analysis.

2.17.4 Calculation of amplitude Once the resonance condition is checked the next step is to ensure that the amplitude of vibrations is restricted within the acceptable limits. The techniques explained earlier based on modal analysis and orthogonal transformation69 is usually used to obtain the amplitude of vibration under operating conditions. It has been mostly seen that the response of the turbine foundation, especially considering the soil effect is usually not critical under the normal operating condition. It is only during the start and stop of the machine when the system goes on transient resonance that it shows signiﬁcant excitation. As explained and shown earlier, in the previous example of the 2D soil-structure interaction model, the best technique to ﬁnd such responses would be based on time history analysis where both the transient and steady state response needs to be checked, to ensure that such ﬂeeting response are also within the acceptable limits as prescribed by the manufacturer or the code of practice.

2.17.5 Calculation of moments, shears and torsion If earthquake load is not a governing case usually an equivalent static analysis will sufﬁce where an equivalent static load for the induced dynamic loads is obtained, based on magniﬁcation factors as suggested in the code. The table suggesting such factors has already been shown earlier while describing Major’s combined method. IS 2974 usually recommends the use of this table to obtain an equivalent static force for the rotating mass and advocates to add these loads to other loads for an equivalent static analysis and structural design of the members.

68 Different techniques used for calculation of eigen values of the system having large degrees of freedom has been dealt in detail in Chapter 5 (Vol. 1) and may please be referred to. 69 Refer the calculations for 2D model we have derived earlier or Chapter 5 (Vol. 1) for the details of such analysis.

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322 Dynamics of Structure and Foundation: 2. Applications

2.17.6 Practical aspects of design of Turbine foundation Here we digress from the theoretical contemplation and enter the real world of practicing engineer to evaluate further what other parameters and decisions form the back bone of a successful design of a Turbine foundation. As a ﬁrst step we start with a check list to see what inputs we require to start a design. 2.18 DESIGN OF TURBINE FOUNDATION

2.18.1 Check list for turbine foundation design 1 Does the drawing furnish the overall dimension of the machine? 2 Are the anchor bolt locations, size of the bolts (both diameter and length) and details of how it should be anchored to the foundation furnished by the vendor? 3 Does the drawing supply the height at which the centre line of the shaft of the machine is located from the bottom of the machine frame (which will be the top of concrete or top of grout for you). 4 Does the drawing supply you with the operating speed of the machine or the range which should be cleared during the design of the foundation? 5 Does the top deck need to support any pipes or valves on it other than the machine itself? 6 If so are all the loads and locations of these valves and pipes are mentioned in the drawing? 7 Does the drawing clearly mention the unbalanced mass, eccentricity or the dynamic loads generated during the operation of the machine? 8 Are all the cut outs in the top deck including its size and location has been made clear in the drawing? 9 Is the location of all embedded part on the top deck including their size, location and thickness has been made clear in the drawing? 10 Is the location of the condenser support including the load coming from it is available to you? 11 Is the Plan area of the working platform for accessing valves and for maintenance is made clear? 12 Different load combinations for which the turbine foundation has to be designed specially from mechanical considerations like short circuit moments, breaking of impeller, Thermal differential etc has been furnished? 13 Finally has the equipment supplier deﬁned any performance criterion which needs to be met in terms of amplitude, frequency etc. The importance of this has already been made clear previously in the chapter of block foundations. 14 Allowable bearing capacity of the soil. 15 Dynamic shear modulus of the soil. 16 Grade of concrete to be used. Once the above check list is satisﬁed the engineer starts his analysis with the tentative sizing of the geometry of the super-structure. The guideline furnished below, are suggestive as a ﬁrst trial and the adequacy of the same shall be checked against a thorough dynamic analysis. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 323

•

•

•

• • • • • • • •

The designer should give enough thought to the sizing of the equipment, its size and clearance requirements in terms of maintenance and access during operation. The size of such access corridor should be clearly discussed with the equipment vendor and also with the plant operation people to ﬁnalize the overall dimension of the top deck. All columns should be sized in such a way that they are almost equally stressed under vertical loads (i.e. σ = P/A shall be constant for all the columns as far as possible). As a rule of thumb, the columns shall have load carrying capacity of about six times the vertical load and shall be placed not less than 3.6 meters center to center. The depth of the longitudinal and the transverse beam shall be one ﬁfth the clear span with the width equal to the width of the column. Care should also be taken that if some anchor bolts are embedded in the beam the depth of the beam is adequate for generating the full strength of the anchor bolts. The deflection of the beam under static load shall be restricted to 0.5 mm. The turbine frame should in principle act as a rigid shear frame as such the flexural stiffness of the top deck beams shall be two times the flexural stiffness of the columns. The bottom of the raft shall not be placed above the level as suggested by the geo-technical consultant where the thickness (t) of the slab shall not be less than, t = 0.07L4/3 , where L is the average distance between columns. The mass of the top deck plus mass of half the length of the column shall not be less than the mass of the supported turbine and its auxiliaries on the top deck. The total mass of the frame plus the raft shall not be less than three times the mass of the machine. The stress induced in soil shall not exceed 50% of the allowable bearing capacity of the soil. For foundations supported on piles the most heavily loaded pile shall not carry 50% of its allowable load. The center of resistance for the pile group or the soil shall not be more than 300 mm from c.g. of the superimposed loads. The center of rigidity of the columns shall coincide with the c.g. of the equipment plus the top half of the structural loads both in the transverse and longitudinal direction. This shall be done based on the equations:

x¯ =

n i=1

•

x i Ix i

n 4

Ixi ,

i=1

z¯ =

n i=1

zi I z i

n 4

I zi

(2.18.1)

i=1

where, x¯ = co-ordinate of the center of rigidity in longitudinal direction; z¯ = co-ordinate of the center of rigidity in transverse direction; Ix and Iz = moment of inertia of the columns, and n = number of columns. All columns should deflect equally in vertical, transverse and longitudinal directions as far as possible when subjected to equivalent static load with a limit on deﬂection for all cases as 0.5 mm.

© 2009 Taylor & Francis Group, London, UK

324 Dynamics of Structure and Foundation: 2. Applications

•

Intermediate platforms are some times provided below the turbine deck for access from bottom and maintenance. These platforms should preferably be placed below the high pressure turbine and should be of RCC. The beams are usually of depth varying from 0.9 to 1.2 meter with a slab thickness not less than 300 mm. During computer analysis stiffening effect of such platforms on the superstructure shall be considered in the analysis and it should also be ensured that the platform itself is not in resonance with the operating speed of the machine.

1

Loads and load combinations for analysis

This we are going to deal in some detail. For unlike normal civil engineering structure the turbine foundation is a very specialized structure where different types of loading arise from the mechanical and electrical aspects of the machine. If the engineer analyzing the foundation does not have a clear idea about these loads he may land up with an analysis which could be deemed useless. Irrespective of how sophisticated FEM package you use or use the most comprehensive mathematical model if the loading input is not correct the result output is always useless. The different loads which come on the turbine are as discussed hereafter. While civil engineers are quite comfortable with loading like Dead Load (DL), Live Load (LL), Seismic load (SL) etc., our observation is that many of them are not very clear about the typical loads which come on a turbine foundation like condenser vacuum loading (CVL), normal torque loading (NTL) etc and how they could effect the behavior of the foundation. We break up the loading in three different categories: Civil Loads; Mechanical loads, and, Electrical loads. a

Civil Loads

This is constitutes of the following: 1 Dead Load (DL) As the name suggests this combines the self weight of all the frame members and weight of the foundation. 2 Live Load (LL) The live load includes those loads that vary in its magnitude and occurrence. The normal practice is to consider a Live Load @10 kN/m2 , on the top deck for the analysis and design. If based on the maintenance concept it is expected that maintenance load and lay down load shall also come on the top deck then they shall be considered as live load in the design. 3 Wind Load (WL) This is usually not considered in the analysis of Turbine foundations for in most of the cases the TG foundation is placed inside a building (the power house) where all © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 325

the wind load is transferred to the ground through the power house structure itself. There are exceptional cases only when it needs to be taken into cognizance70 . 4 Earthquake/Seismic Load (SL) This could be the major design load if the power plant is being built in area prone to major earthquakes. Earthquake is itself a major topic of study and we will not go in detail here on this issue for we have dealt this issue separately71 . 5 Loading due to Creep and Shrinkage (SCL) This usually applies to RCC frames where after initial deﬂection the structure undergoes deformation under sustained loading. This time dependent deflection at the bearing location can be two or three times more than the short term elastic deflection. However this phenomenon was not considered earlier for design due to the following reason. For a typical coal ﬁred power plant the initial machine alignment use to occur at about 24 to 36 month after the foundation has been constructed. By this time most of the deformation due to shrinkage and creep would have taken place thus further deformations were negligible and had practically no effect on the shaft alignment. However under present scenario with demand in power on the rise globally the turnkey contractors are expected to ﬁnish and hand over one whole plant in 20 to 29 months only. As such it is obvious now the loading on the turbine frame would come much earlier when the secondary deformation effect of creep and shrinkage could be signiﬁcant at the bearing level and should be carefully evaluated. b

Mechanical Loads

1

Machine Dead Load (MDL)

This constitutes of the weight of the various turbine components and is usually termed as the machine dead load. The turbine manufacturer in their equipment layout drawing usually supplies these loadings and their locations. 2

Condenser Dead Load (CDL)

We had already explained earlier that the condenser is normally mounted below the turbine top deck. Depending upon the supporting system used for installing the condenser the loading induced on the foundation varies. The common practice for installing the condenser is either of the two systems as discussed hereafter: The condenser is spring mounted on the bottom raft while the top neck is rigidly connected to the turbine exhaust nozzle. The springs are of adjustable type enabling them to transfer speciﬁed loads to the turbine exhaust nozzle. They are also sometimes

70 There are cases where the turbine deck is spring mounted and rests on steel columns, which in turn is connected to the power house structure. In such cases WL load has to be taken in consideration in the analysis specially the load combinations. In such case usually a combined power house and TG frame analysis is carried out. 71 For more detail on this issue refer to Chapter 3 (Vol. 2).

© 2009 Taylor & Francis Group, London, UK

326 Dynamics of Structure and Foundation: 2. Applications

used to balance the loading eccentricity that can develop due to circulating water pressure loads. The condenser bottom is mounted on a rigid frame and an expansion joint is provided between the condenser and the turbine exhaust nozzle to relieve the thermal expansion and variations in the condenser loads. For condenser mounted on rigid frame the total weight of the condenser is transferred to the bottom raft. For spring-mounted condensers, it is mostly welded to the turbine exhaust nozzle when the proportion of load that will be distributed between the top deck and the bottom raft depends upon the stiffness of the spring and their alignment. The equipment supplier usually supplies this loading. 3

Condenser Vacuum Load (CVL)

For condensers mounted on rigid frame we had already mentioned that an expansion joint is provided between it and the turbine exhaust nozzle, for this the difference between the atmospheric pressure on the casing of the turbine and the vacuum pressure inside the condenser develops a force on the turbine. This load can be several times in magnitude to the weight of the condenser itself and is transmitted to the foundation through the turbine soleplates. The turbine manufacturer provides the distribution of this loading. For spring mounted condensers when the condenser is rigidly connected to the turbine exhaust nozzle no vacuum load is transmitted to the turbine top deck. 4 Normal Torque Load (NTL) The steam expanding within the turbine imposes a torque on the stationary casing in the opposite direction of the rotation of the rotor. The magnitude of the torque depends on the angular speed and the power output of the turbine. The equipment vendor usually supplies this load in the vendor drawing as equivalent vertical loads on the sole plate. 5 Other Equipment Loads (OEL) Other than the turbine itself the foundation may support other equipment such as turbine stops, control valves, interceptor valves, main steam pipeline hangers etc. Thus additional dead loads from these, which are not included under the heading MDL, shall also be considered in the design. 6 Thermal Load (ThL) During operation of the turbine, temperature change of the turbine and the generator causes expansion and contraction to take place resulting in various parts to slide. As the progressive heating of the machine take place the turbine shaft expands, however the expansion does not induce any loading on the foundation for the shaft is ﬁxed longitudinally by single thrust bearing when the shaft slides freely across the journal bearings which are adequately lubricated. Unlike the shaft during the heat build up in the system during operation the turbine casing also gets heated and imposes thermal loading on the foundation. The transverse beams usually support the sole plates supporting the high pressure and the intermediate pressure turbine casing. The low-pressure turbine casings, the generator casing and the exciter are supported on the sole plates of the longitudinal and the transverse beams. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 327

During the heat built up the casing expands from their anchor points thus producing a friction load. Though exact calculation of such forces is very difﬁcult for it depends on a number of factors however the common practice is to use the following simplifying analysis in lieu of an exact analysis has been found to be adequate. The total thermal loading in longitudinal or transverse direction is considered as the vector sum of the forces acting on that direction. The magnitude of the force on any sole plate is calculated as: Force = μx (Here x is sum of machine dead load, condenser load, normal torque load and piping load if any), where, μ = coefﬁcient of friction which varies from 0.2 to 0.5. The value has to be conﬁrmed with the turbine manufacturer. At the startup condition the expansive load shall be taken as acting away from the center line of the turbine while during shutdown it will considered acting towards the center line of the turbine. In case of the longitudinal expansion an approximation is made to the direction of the force and the unbalanced force between two anchorage points, which prevent the movement of the turbine, is applied as the concentrated load at the anchorage points. 7 Turbine Casing Pipe Load (TCPL) The pipes connected to the turbine casing also induce loads to the foundation. The turbine generator manufacturer to prevent distortion or overturning of the turbine components speciﬁes maximum loads. The turbine casing may be assumed to be rigid and the forces are then calculated at the support points on the foundation. The types of piping that generate most of the loads are: main steam inlet piping; reheat steam piping, and extraction steam piping. 8

Piping Load from Equipment Attached to the Foundation (PEL)

As we had stated earlier that various auxiliary equipment are also supported on the turbine deck. Positioning and aligning piping for this equipment creates erection forces. Turbine piping is assembled and welded to these equipment and is anchored to the foundation. The remainder of the steam inlet pipes is then welded to the assembly inlet connections. Different forces are created due to thermal expansion during operation. Erection forces, static and dynamic forces should be evaluated to check if they have any signiﬁcant contribution or not. For instance a rapid closing of the steam stop valve attached to the foundation can induce a major loading. 9 Load due to Machine Unbalance (MUL) Irrespective of however care is taken in balancing the turbine generator rotor it practically impossible to do away with some imbalance in force which it will generate during its rotation. The magnitude of this imbalance depends on a number of factors like design considerations, installation and maintenance procedures. The factors which usually contribute to such imbalanced dynamic load can be summarised as follows: i Axis of rotation eccentric to the center of mass of the rotor; ii Deflection of the shaft due to gravity load; © 2009 Taylor & Francis Group, London, UK

328 Dynamics of Structure and Foundation: 2. Applications

iii iv v vi

Uneven thermal expansion; Misalignment during installation; Normal wear and tear during operation and, Corrosion.

The combined or few of the reasons as mentioned above contribute to the dynamic imbalance in the in the rotating shaft which is synchronous with the shaft rotational speed. These forces are transmitted through the bearing shaft to the foundation. The dynamic load is deﬁned by Pdyn = m · e · ω2

(2.18.2)

Here, m = unbalanced mass of the rotor; e = eccentricity of the rotor shaft, and ω = operating frequency of the machine. 10 Load due to Bowed Rotor (BRL) A bowed rotor can impose large dynamic loads on turbine generators foundation. The bowed condition of the rotor will create unbalance force which are transmitted through the machine bearings to the sole plates. The magnitude of the force will vary with the unbalanced dynamic force as mentioned above. The phenomenon can happen due to: i ii iii iv

Failure to put the rotor on turning gear when the machine is shut down; Deflection of the raft, soil and the frame in flexural mode; Water Induction and Very severe packing rub.

The largest bowed rotor response occurs at the ﬁrst critical speed for the rotor. The time taken by the turbine rotor to pass through the critical speed is shorter when going on-line. However it takes much longer time when it goes off-line and the rotor coasts through the resonant speed. Since this is a condition that usually requires turbine generator shut down it will exist only for the time required for the rotor to coast down to rest. Thus it is sufﬁcient to ensure that the foundation stresses are low enough to eliminate the chance of any permanent damage to the structure during the shut down period. The magnitude and the location of the bowed rotor is usually supplied by the manufacturer of the turbine in question and is dependent on the speciﬁc assumption made by the vendor. The force due to bowed rotor is function of the unbalanced dynamic force Pdyn = m · e · ω2

(2.18.3)

The loading is normally provided in the form of a sinusoidal function for the dynamic analysis or an equivalent static load for simpliﬁed analysis. It is to be noted that, some turbine manufacturer may not supply this load for depending on their own design some consider bowed rotor as worst case of accidental © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 329

loading for the high pressure and intermediate pressure turbine while others consider loss of turbine blade as worst case of accidental load that can come on the foundation as an emergency load. 11 Load due to Missing Rotor Blade (MRBL) A turbine rotor must be balanced dynamically to ensure satisfactory operation and cause no adverse effects on the turbine equipment or the foundation. However it has been observed from previous operational experience that in some cases (though very rare) that the last row of blade in a low pressure rotor breaks loose from the rotor and causes a severe imbalance in the system. The lose of blade which can vary in size from 500 mm to 1000 mm in length can cause substantial force on the rotor, the bearing and the foundation system. The magnitude of this unbalance is a function of the rotor blade weight, its center of gravity with respect to the rotor and rotational speed of the rotor. As this can happen with any of the several rows of last row blades a separate analysis should be made with a single unbalance equivalent to the loss of one last row blade applied to the mass point corresponding to each of the last row blade in each low pressure turbine. Since this is an emergency situation and will require the turbine to be shut down it will only exist only for the time period required for turbine to come down to a stop. Thus it is sufﬁcient to ensure that the stresses in the foundation are low enough to preclude any permanent damage during the coast down period. The loading is normally provided by the vendor in the form of a sinusoidal function for the dynamic analysis or an equivalent static load for simpliﬁed analysis. 12

Electrical Loads

1 Generator Emergency Torque (GET) Of all the loads that can occur a line-to-line short circuit at the generator terminal causes the most severe loading of the turbine generator loading. Such a fault occurs when any two of the three generators phase are shorted. The calculation of the maximum generator air gap torque during symmetrical three phase and unsymmetrical line to line or line to ground terminal short circuits is normally performed assuming no electrical damping in order to obtain greatest possible forces that can be transmitted to the foundation under different fault condition. Experience and previous data shows that the maximum torque resulting from a line to line short circuit is about 25% greater than that caused by a single terminal to ground fault and roughly 30% more than that with a symmetrical 3 phase fault at the terminal of the generator. The vendor in the form of a forcing function or an equivalent static force normally provides the loading due to generator short circuit. The use of equivalent static force for the maximum short circuit torque assume that the foundation is inﬁnitely rigid and thus must directly absorb the full impact of the severe shock forces. Since this assumption may result in over designing the foundation the more realistic approach of a dynamic analysis is on the basis of the short circuit moment as a time dependent function is usually preferred. © 2009 Taylor & Francis Group, London, UK

330 Dynamics of Structure and Foundation: 2. Applications

In view of the very severe transient nature of the maximum short circuit loading the foundation in the vicinity of the generator, the designer should perform an appropriate dynamic analysis of this abnormal load case. 2 Load combination for design The following load combination is generally considered for the design as per American practice72 . i Operating conditions The loading condition for which the foundation has to checked for and designed is = 1.4(DL+MDL+OEL+PEL +CDL+CVL)+1.7(LL+NTL+ThL+TCPL+MUL) ii

Accident Conditions

Generator Emergency − = DL+MDL+OEL+PEL+CDL+CVL+LL+GET+ThL+TCPL+MUL Bowed Rotor case − = DL+MDL+OEL+PEL+CDL+CVL+LL+NTL+BRL+ThL+TCPL Missing Rotor Blade − = DL+MDL+OEL+PEL+CDL+CVL+MRBL+NTL+ThL+TCPL Seismic load − = 0.75 [1.4(DL+MDL+OEL+PEL+CDL+CVL)+1.7(LL+NTL+ThL+TCPL +MUL+1.1 SL) It is to be noted that 1.4 and 1.7 are load factors for design of concrete section based on ACI-318. For design of sections based on other codes like IS or BS appropriate load factors in place 1.4 and 1.7 has to be taken.

2.18.2 Spring mounted turbine foundation In this section we discuss the method of analysis and design of turbine foundations mounted on springs. This is a practice which is quite common in European countries and is being put to increasing use in this part of the world now a days specially for foundations supporting Turbines of high capacity. In this case the top deck is usually mounted on springs of pre-designed speciﬁcation and is supported in turn on a frame as shown in Figure 2.18.1. From the conceptual Figure 2.18.1, the obvious question that comes to mind is why do we do such a thing and what advantage we gain from it? To understand this we take up hereafter a concept which is otherwise known as vibration isolation.

72 IS 2974 Part III though discusses the vibration analysis in detail it is silent on how and what load combinations should be considered for design.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 331

Top Deck C.G of top deck

Mechanical Springs

Frame supporting Top Deck

Bottom Raft

Figure 2.18.1 Spring mounted turbine foundation.

2.18.2.1

Theory of vibration isolation

We had seen in the section for analysis of block foundation that under dynamic load the amplitude of vibration is expressed by the formula

δz =

P0 Kz

sin ωm t

(2.18.4)

(1 − r2 )2 + (2Dz r)2

where r = ωωmn ; Dz = damping ratio, and Kz = equivalent spring of the soil. Now instead of soil spring if we support the block on some mechanical springs (Ks ) only the amplitude of vibration of the spring can be expressed as

δs =

Now,

P0 Ks

sin ωm t

1 − r2 Ks δs =

where r =

ωm . ωn

P0 sin ωm t P0 sinωm t = Ps = 1 − r2 1 − r2

© 2009 Taylor & Francis Group, London, UK

(2.18.5)

where Ps = Ks δs

(2.18.6)

332 Dynamics of Structure and Foundation: 2. Applications

Considering, TF = Ps /P0 = transmissibility factor, we have

TF =

1 1 − r2

(2.18.7)

The transmissibility factor is thus a measure of how much of the dynamic force is transmitted to the supporting springs. For transmissibility in the range less than unity the above equation is written in the form

TF =

1 r2 − 1

(2.18.8)

√ Considering the limiting case of TF = 1 we have, r2 − 1 = 1 → i.e. r = 2. Thus it is seen that√the transmissibility factor T F shall have a value less than unity for all values of r ≥ 2. For damping prevalent in the system the transmissibility factor is given by expression TF =

1 + (2Dz r)2

(1 − r2 )2 + (2Dz r)2

(2.18.9)

ωm and Dz = damping ratio. ωn If we plot the above equations for different values of frequency ratio and TF we have curves as shown in Figure 2.18.2. Observing the curves, it will be seen that even with√ damping existing in the system TF value is less than 1 when the frequency ratio r ≥ 2 i.e. the force transmitted to the support is less than the induced dynamic force. To get a further insight into how the frequency ratio affects transmissibility factor we study an expression called isolation efﬁciency expressed as where r =

I = rr2 −2 × 100 in % where r = ωωmn and is the measure of the reduction of Trans−1 missibility factor of the system (Crede 1951). We plot a curve, shown in Figure 2.18.3, based on the above √ expression. Based on this ﬁgure we ﬁnd that when frequency ratio is 2 the isolation efﬁciency is 0% i.e. 100% of the dynamic load gets transmitted to the support. However when r = 2.45 the reduction efﬁciency increases to 80% i.e. a signiﬁcant amount of reduction of force transmittal to the support system is obtained. 2

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 333

Transmissibility Factor

12 10 Damping ratio @ 5% Damping ratio @ 10% Damping ratio @ 15% Damping ratio @ 20% Damping ratio @ 25% Damping ratio @ 30%

8 6 4 2

3 3. 25

2 2. 25 2. 5 2. 75

1 1. 25 1. 5 1. 75

0. 5 0. 75

0 0. 25

0 Frequency Ratio

Figure 2.18.2 Variation in transmissibility factor.

Isolation Efficiency(%) 100 80 Isolation Efficiency(%)

60 40

4.9

4.55

4.2

3.85

3.5

3.15

2.8

1.75

1.41

0

2.45

20 2.1

Isolation efficiency(%)

120

Frequency Ratio

Figure 2.18.3 Isolation efﬁciency (%).

We give below some data showing variation of Isolation efﬁciency with respect to the frequency ratio Frequency ratio

1.414

Isolation efﬁciency(%)

0.0

1.5 20

2.0

2.5

3.0

3.5

4.0

5.0

66.66

80.95

87.5

91.11

93.33

95.8

It will be observed both from the above ﬁgures as well as from Figure 2.18.3 that up to a frequency ratio of 3.0, the reduction in transmitted force to the support is signiﬁcant but beyond that as the curve flattens asymptotically not much reduction in transmissibility is obtained. For instance if we increase the frequency ratio from 3 to 5 say the variation in isolation frequency is only 8.6% however the manufacturing cost for such mechanical springs as per some vendors nearly gets doubled. © 2009 Taylor & Francis Group, London, UK

334 Dynamics of Structure and Foundation: 2. Applications

Thus the common practise is to restrict the frequency ratio to maximum between 3 and 4 in practical engineering design. Hence it is seen that if we can provide elastic supports like springs below a foundation and can maintain a separation ratio of 3 to 4 with respect to the operating frequency of the machine following advantages may be obtained • • •

The dynamic force transmitted to the supporting system for the springs could be signiﬁcantly reduced. Based on the reduced dynamic force it is possible to restrict the amplitude of vibration to manageable limits. The foundation remains isolated/de-coupled to the surrounding and does not transfer any dynamic load.

The above points are in a nutshell major advantage gained by providing springs for vibration isolation. Moreover as the springs are man made (unlike soil where we do not have any control on its property) under a careful controlled condition, it is possible to design these springs in such a way that they do have a frequency ratio between 3 to 4 with the operating frequency of the machine. 2.18.2.2

Effect of damping on the transmissibility factor

Since any physical system in this world has some amount of damping (even air) it would be worthwhile to evaluate how damping affects the transmissibility coefﬁcient. Freq. ratio

Damping ratio

0.0

0.05

0.10

0.15

0.20

0.25

0.3

2.0 3.0 4.0

TF TF TF

0.333 0.125 0.066

0.339 0.130 0.072

0.356 0.145 0.085

0.381 0.167 0.103

0.412 0.193 0.125

0.447 0.221 0.147

0.483 0.251 0.171

Studying the above table it will be observed that having high damping value in the system is counter productive to transmissibility. On the contrary a little amount of damping in the system is advantageous in terms of transmittal of dynamic forces to the foundation. On the other hand, as we know that amplitude gets reduced due to the effect of damping in the resonant zone the most ideal damper that can be introduced in a system should thus have the following properties: • •

High damping value when the frequency of the machine is passing through the resonant range. Nominal damping value when the machine is operating at its normal speed.

Vendors specialising in supplying such viscous-dampers have their patented products which exhibits such property as discussed above thus suppressing the dynamic effect of the machine on the foundation and to its surrounding considerably. © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 335

2.18.2.3

How springs affect turbine foundation?

Based on the above theoretical discussion the query is but obvious. For Turbine foundation as shown in the Figure 2.18.1, the top deck is usually mounted on the spring and whole spring mounted assembly is then supported on a frame which could be made of either RCC or steel. The springs are supported in such a manner that the support points match with the c.g. of the top deck and machine. This helps in suppressing the coupled translation and rocking mode of the top deck. The technique of providing spring mounted Turbine top deck is mostly in vogue in Europe where the Germans pioneered this technique about 40 years ago. Surprisingly in spite of certain advantage it provides (specially for turbines operating in Nuclear power plants) in terms of cost, plant layout etc it has not been a popular concept in USA where engineers still opt for conventional framed foundations. Conventional turbine frame foundations, usually calls for columns of heavy section and also a huge base mat to suppress the dynamic effect. When the top deck is mounted on springs the major advantage is that the dynamic effect of the machine is restricted up to the spring part only and the rest of the foundation needs to be only designed for static loading. The obvious advantage is that it calls for much slicker frame resulting in considerable saving in material cost and as far as analysis is concerned, uncertainties prevalent with a comprehensive dynamic soil structure interaction analysis (specially if resting on piles) for such complex system is not required. For nuclear power plants, the operating frequency of turbine is usually around 1500 rpm unlike conventional power plants (where it is about 3000 to 3600 rpm) thus while designing the pedestals for these foundations engineers faced difﬁculties to keep them signiﬁcantly away from the operating frequency of the machine as they were becoming far too flexible to their discomfort. The obvious choice was then to mount them on springs and isolate the rest of the foundation from the dynamic effect. Though the above was a starting point of such concepts, spring mounted turbine foundations are now quite common in conventional fossil fuel power plant also. The major advantages gained in this case can be summarised as given hereunder: • • • •

The top deck remains dynamically uncoupled with respect to the supporting frame, thus the supporting frame is only subjected to static load and needs to be designed accordingly. This makes the supporting framed structure slicker and also does away with the necessity of providing a heavy bottom mat which is otherwise essential for a conventional frame foundation. The springs are capable to certain extent adjust themselves to cater to the differential settlement, if any. Even due to the overall settlement of the foundation which can cause additional stress to the critical pipe connection, adjustment can be directly made using the springs to adjust the levels and that too without interrupting the operation of the machine.

© 2009 Taylor & Francis Group, London, UK

336 Dynamics of Structure and Foundation: 2. Applications

• • • • • • •

More space is usually available below the foundation thus maintenance and laying of piping and cables become more accessible and easy. Substantial gain in material and cost is evident. Some vendors claim that with spring mounted turbine foundations saving in cost could be as high as 45% when compared to conventional frame foundation. Cost of piling is reduced as there is a signiﬁcant reduction in weight. No dynamic loads need to be considered for the piles. The structural uncoupling of the top deck allows for the use of even steel structures for the supporting frames. Use of steel structures gives additional advantage in terms of construction sequence for they can be installed parallel to the power house structure which gives a signiﬁcant saving in construction time. Differential settlements can be easily measured based on the variation of spring heights. Instrumentation techniques are available which monitors these spring heights and when it exceeds preset-values automatically give visual signals or sends alarms.

The advantages as mentioned above are making this concept progressively popular in the Industry. In many projects in India also this concept has been put to practice and the turbines are found to be operating quite smoothly without any hindrance. 2.18.2.4

Mathematical modelling of spring mounted turbines

The intuitive computer model that could be conceived for this case is to conceive spring elements connected to the top deck directly supported in the bottom frame [Figure 2.18.4]. However other than ANSYS most of the normally available structural engineering package do not have the provision of adding springs directly between two members (the basic pre-condition is one end of the spring should be ﬁxed and not an active node).

Top deck Spring Elements

Supporting frame

Figure 2.18.4 Actual model of the top deck mounted on spring.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 337

In such cases the most effective way to model the spring would be to represent it by equivalent truss element having stiffness as AE/L, where AE/L shall have a magnitude equal to the individual spring stiffness as considered in the vendor’s catalogue. 2.18.2.5

Turbine foundations concrete versus steel structure

In USA as well as India conventional Turbine Foundation design is still dominated by RCC structure. However in many countries in Europe (especially Germany, France, Hungary etc.) and Canada, Turbine foundation made of steel has been successfully implemented. One of the major advantages with the RCC is its high damping property and not requiring a very sophisticated construction technology to construct it. At the early stage of advent of turbine foundations thus RCC made a very attractive choice. However with turbine capacities increasing progressively the size of the turbine foundations are also getting bigger and the construction technology is getting more and more complex. One of the major requirements of casting of RCC Turbine Foundation is that it should be preferably cast in one go. Else additional steel has to be provided at such cold joints and additional care has to be taken during construction to ensure its monolithic property. While for a foundation having concrete volume of 500 m3 this was not a very difﬁcult task, but when foundation capacity gets increased to 1500 m3 or more it surely becomes a different ball game. Firstly one needs a complete batching plant to be erected at the turbine foundation site capable of supplying continuously concrete of same quality. This requires a very elaborate arrangement to be made by the contractor at site including a building of a make shift testing laboratory, where samples are collected and tested from different batches continuously to ensure that the concrete is of the desired speciﬁed quality. The next major difﬁculty encountered while casting is the heat of hydration which RCC generates during hardening. When the volume of concrete is large the heat of hydration can be substantial to create cracks at the surfaces and needs to be carefully controlled at site (usually controlled by using pre ﬁxed quantity of Ice blocks in lieu of water) to nullify its effects. The time consumed for laying reinforcement is substantial and needs to be thoroughly checked with respect to drawing. For large turbine foundations (>500 MW) at times client also insists in his contract that the contractor to ensure based on non-destructive test that there are no voids or honeycombing within the concrete. This calls for expensive ultra-sonic taste of the foundation which is not only an expensive exercise but time consuming too. As far as design aspect is concerned one of the major difﬁculties encountered is the laying of the embedded plates and hangers in top deck for pipe supports which are usually large in numbers. Firstly when the turbine foundation is getting designed the critical steam piping design is yet not ﬁnalised and thus the location of embedded plates and supports furnished by the piping engineer is only tentative and could be subjected to change. This surely makes the foundation designers task a difﬁcult one. © 2009 Taylor & Francis Group, London, UK

338 Dynamics of Structure and Foundation: 2. Applications

For once the turbine deck is cast no perforation or anchoring to the deck is allowed which could be detrimental to the concrete strength73 . Thus the engineer has to be doubly careful with this drawing and check it very thoroughly to ensure that not a single plate is missed, thus time taken for engineering is more. A general tendency is thus to provide more number of embedded plates or plates of bigger size to cover the eventualities. Since this is completely dependent on the engineers personal judgement it has not been uncommon that at times the tonnage has become as high as 30% extra then the estimated value and incurred unnecessary wastage. Similarly for any valves or other sundry equipment resting on the turbine top deck their location and anchor bolt details etc needs to be ﬁnalised during drawing preparation stage. If the procurement department has not ﬁnalised with these equipment or the vendor drawings are not available – the design engineer could be in a lot of difﬁculty. With steel structure on the contrary most of the difﬁculties as discussed above is not encountered. In spite of the fact that steel structure provides low damping, for large turbine foundations steel as a construction material do have some distinct advantage over RCC. Firstly every thing need not be erected at the site; the top deck, which generally consists of a rigid grillage, can be constructed at the shop under a careful controlled condition and be carried to the site and erected over the columns. As welds are susceptible to rapid fatigue failure under dynamic loads due to reversible of stress the connections are usually bolted (bolted connections also provide good damping and is more advantageous in such cases) and providing site connected bolting is not a problem at all. The major advantage in terms of construction is that the elaborate arrangement one requires for RCC structures in terms of inspection and checking of laying of rebars, controlling the concrete quality and large amount of human resources one has to deploy at the site is not required at all. In fact the fabrication of the top deck at shop can start much ahead of the erection of powerhouse it self and can be erected at site simultaneously. This signiﬁcantly saves construction cost as well as time too. From design engineering point of view one need not worry about the location of embedded plates and hangars, even with very late information welding locally steel to steel is never a problem unlike anchoring plates on concrete top deck. It can be logically perceived that steel foundation would be relatively high tuned one compared to RCC foundations due to its lower mass. However they can be suitably designed and adjusted to have the requisite frequency separation of 20%. As we had stated earlier that turbine foundation usually does not become critical during its normal operation but shows signiﬁcant excitation during the start and stop of the machine (mostly due to the soil participation) if the amplitude of vibration can limited within the acceptable limit steel structures do have a very high potential as a construction material for such type of structures.

73 Though technology exists where embedded plates can be anchored to concrete slabs after it is cast but considering the critical nature of the turbine foundation such processes are usually not allowed for Turbine top deck by the client and is not a good engineering practice too.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 339

One of the major constraints encountered for turbines mounted on steel structures is however the limitation in the available of ready made rolled sections. For large turbines composite columns made out of industrially available rolled section could become inadequate in terms of strength. This calls for usage of plate girders in lieu of composite section. Due to inherent weakness of welds under dynamic loading continuous butt welds are usually preferred instead of ﬁllet welds. Continuous butt welds specially at the flange and web junction calls for rigorous quality assurance in terms of Radiographic test or dye test to ensure 100 % weld penetration and could make the fabrication expensive. However if the steel industry in India agrees to manufacture rolled sections of higher sizes (beyond ISMB 600) steel structure can become a very strong competitor to RCC foundations. In Europe since rolled steel sections having much higher moment of inertia are available, use of steel structure as an alternate to RCC has become a viable solution there. 2.18.2.6

Design of RCC sections

The structural members are usually designed by using IS-456(2000) or the local code of the country in which it is being constructed or as speciﬁed in the contract like ACI318, BS8110, DIN etc. In most of the case the geometric sizing is decided by the equipment supplier, based on which the stress induced in the members itself are normally low and to our knowledge there has been no such cases where Turbine foundation members have misbehaved or failed due to strength failure. Most of the cases where members have misbehaved can be attributed to improper detailing or faulty construction for which cracks have been observed to develop. Based on above, proper detailing of the members are of primary importance. Some good detailing practices are mentioned hereafter which could be followed while detailing such foundations. • • • •

The vertical reinforcing bars of the column shall have sufﬁcient embedment in the base slab to develop the required stresses. Reinforcement in beams and columns shall be provided in all four sides irrespective of they are required or not. If design requirements do not guide the percentage of steel, the re-bars shall be placed symmetrically on all four sides. The minimum Steel provided in different parts of the members are mentioned hereunder

Sl. No.

Structural member type

1 2 3

Base slab 40 kg/m3 Columns 70 kg/m3 Top deck (beam and slab) 90 kg/m3

© 2009 Taylor & Francis Group, London, UK

Steel quantity

340 Dynamics of Structure and Foundation: 2. Applications

• • • • •

Shear stirrups to be provided to account for the total shear in the foundation element. Splicing in columns if any shall always be done at the mid-height. The diameter of bar in beams and columns should be so chosen that the maximum spacing of the bars are not more than 150 mm. Try to use lower diameter bar as far as practicable. For with lower diameter bars, number of bars is more and distribution of stress and transfer of load between concrete and steel is more uniform. Unless speciﬁed by the contract the cover to reinforcement is usually taken as follows: Base Slab 100 mm on top, bottom and sides. Columns and Pedestals 50 mm on sides Beams 40 mm on all sides

• •

Minimum development length for all bars irrespective of requirement shall not be less than 50 times the diameter of the bar. Beam column junction should be provided with additional steel to ensure that cracks do not develop due to continuous reversal of stresses due to the application of cyclic loads.

Example 2.18.1 Shown in Figure 2.18.5 is the layout plan of a Boiler feed pump framed foundation with location of equipment loads as shown. The dynamic loads under various operating conditions are as shown hereafter.

4590

Pump Side

4590

Motor Side

C/L Coupling

1315 Y

16.65

+3.5m(TOC)

350 3.0m(TOC)

195kN

22.1kN

35

5

508

54

775 +3.5m(T.O.C.)

+ 4.0m(T.O.C.)

X 1765

16.65

4192 =

1585 =

Figure 2.18.5 Plan view of top deck with location of equipment load.

© 2009 Taylor & Francis Group, London, UK

1200 1580 100kN 2238

Analysis and design of machine foundations 341

Table for dynamic loads on boiler feed pump top deck

Load condition

Remarks

Px

Py

Pz

Short circuit moment Total force at top deck 0.0 226 268 Operating load (1) End frame (pump side) 58 44 Middle frame 80 0 End frame (motor side) 100 8.3 Operating load (2) Load per long beam (pump side) 37 0.0 130 Operating load (2) Load per long beam (at coupling) 0 0 25

• • • • • • • • • •

Mx My Mz 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0

Operating Frequency of the machine = 5100 rpm Center line axis of shaft = 1.1 m above the top deck Bearing capacity of soil 150 kN/m2 Shear wave velocity of the soil = 115 m/sec Poisson’s Ratio of soil = 0.30 Live load of top deck slab during operation = 5 kN/m2 Unit Weight of soil = 19 kN/m3 All columns = 600 mm × 600 mm All beams = 600 mm × 900 mm Grade of concrete M25

Analyze the frame by • • • •

Rausch’s method By Barkan’s method By Major’s method By 2D soil-structure interaction model.

Compare the results of the analysis based on the above methods with time history Do detailed design of the frame. Solution: We start the problem sequentially. Here the top deck consists of a flat slab 900 mm thick supported on columns (600 mm × 600 mm) and practically does not have a framing system. Here for analysis and design we perceive a frame having edge beams in both transverse and longitudinal direction having depth of 600 mm × 900 mm as shown in Figure 2.18.5 and 6. The load from the slab is transferred to the idealized frame (as shown by the dotted lines, Figure 2.18.7) and the frame is analyzed for vibration in vertical and horizontal mode.

© 2009 Taylor & Francis Group, London, UK

342 Dynamics of Structure and Foundation: 2. Applications

3200 900

3600

(+/-)0.0

1500

Figure 2.18.6 Elevation of the frame in transverse direction.

4590

4590 A

1 2600

22.5

2 195

16.7

16.7 508 54 B

690

2571

1

1329

1417

965

2

Figure 2.18.7

Calculation of UDL load transferred to frame Thickness of slab = 900 mm Self weight of slab = 0.9 × 25 = 22.5 kN/m2

© 2009 Taylor & Francis Group, London, UK

1308

900

3

Analysis and design of machine foundations 343

Live load = 5.0 kN/m2 : Thus, DL + LL = 27.5 kN/m2 wlx 27.5 × 2.6 = = 3 3

Equivalent UDL transferred to frame 1 and 3 =

23.83 kN/m Equivalent UDL on frame 2 = 23.83 × 2 = 47.66 kN/m From Figure 2.18.7,

2 wlx 27.5 × 2.6 lx Equivalent UDL on frame A and B = = 3− × 6 ly 6 2.6 2 3− = 31.92 kN/m 4.59 Load on longitudinal beam from the area of hydraulic coupling having local projection of 1.4 m, of width 1585 mm 0.5 × 2.6 × 25 2 = 16.21 kN/m

w=

Thus the total superimposed UDL coming on the frame is as shown in Figure 2.18.8

48 KN/m 24 KN/m

32 KN/m

48 KN/m

48 KN/m

24 KN/m 32 KN/m

Figure 2.18.8 Frame with uniformly distributed load from top deck slab.

© 2009 Taylor & Francis Group, London, UK

344 Dynamics of Structure and Foundation: 2. Applications

Calculation of Concentrated load For slab panel 1 along center line axis the distribution of load is as shown in Figure 2.18.9. R=217.5 22.5

Ly R=217.5

195 Lx

690

2571

1329

2995

Figure 2.18.9 Load distribution slab panel-1.

Here we first out the point through which the resultant of this two concentrated force acts yc =

22.5 × 690 + 195 × 3261 = 2995 mm from frame 1 22.5 + 195

Now the slab being restrained at all sides subjected to a load of 217.5 at distance of 2995 mm from frame 1 it is evident that displacement at point O shall be same for long and short span. Thus considering the middle strip as a beam fixed at both ends in long direction δl =

Py a 3 b 3 3EIL3y

Here Py = the net concentrated load acting in long direction; a = 2995 mm; b = 1595 mm; Ly = 4590 mm, and Lx = 2600 mm Px L3x 192EI Here Px = The load transferred to short span. Since here due to displacement compatibility, δl = δs , we have Displacement in short span is given by, δs =

P y a3 b3 Px L3x = 192EI 3EIL3y

a3 b3 ➔ Px = 64Py 3 Ly Lx

Since by law of static V = 0 we have, Px + Py = 217.5 5 3 ab 64 +1 ➔ Py = 218 (Ly /Lx )

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 345

Now substituting the values a, b, Ly and Lx , we have → Py = 19.27 kN and Px = 198.73 kN. Since in short direction the load is symmetrical load on long beams along 198.73 Row A and B = ≈ 99.4 kN 2 In long direction, the position of the load is as shown in Figure 2.18.10. 19.3

R1

2995

R2

1595

Figure 2.18.10

19.3 × 1595 = 6.7 kN and R2 = 12.6 kN. 4590 Concentrated load transferred to edge beams from the slab panel 1 is shown in Figure 2.18.11. Thus

R1 =

99.4 12.6

6.7

99.4

Figure 2.18.11 Concentrated load transferred to edge beams from the slab panel-1.

For slab panel 2 the loading arrangement is as shown Figure 2.18.12. Ly = 4500

16.7

16.7

Lx = 2600 508

54

1417

Figure 2.18.12 Loading on slab panel-2.

© 2009 Taylor & Francis Group, London, UK

965

1308

900

346 Dynamics of Structure and Foundation: 2. Applications

Ly = 4500

Lx = 2600 194

87.4

3006

1584

Figure 2.18.13 Resultant load on slab panel-2.

Total concentrated load acting on the slab = 16.7 × 2 + 54 = 87.4 kN The c.g. of the loads in y direction (Figure 2.18.13) is given by yc =

16.7 × 1308 + 16.7 × 2273 = 684 mm 87.4

xc =

33.4 × 508 = 194 mm 87.4

and

Here Py a3y b3y

Px a3x b3x , when based on displacement compatibility 3EIL3y 3EIL3x 3 Py a3y b3y Ly a x bx 3 Px a3x b3x δl = δs when we have = ➔ P = P × y x ay by Lx 3EIL3y 3EIL3x

δl =

and δs =

Here ax = 1494 mm; bx = 1106 mm; ay = 3006 mm; by = 1584 mm; Ly = 4590 mm and Lx = 2600 mm. Substituting the values in the above equation we have, Py = 0.2299Px For V = 0 we have, Px + Py = 87.4 kN Thus, we have Px = 71.06 kN and Py = 16.34 kN. Now, proceeding in the similar manner as explained for slab panel 1 we find out the load transferred on the frame beams shown in Figure 2.18.14.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 347

Concentrated load transferred to edge beams from the slab panel 2 in Figure 2.18.14.

30.22 5.64

10.7

140.83

1494

3006

Figure 2.18.14 Concentrated load transferred to edge beams from slab panel-2.

The net distribution of concentrated force on the frame is as shown in Figure 2.18.15.

30.2 10.7 99.4 18.2 141 6.7 1584 99.4

Figure 2.18.15 Frame with concentrated load from top deck slab.

© 2009 Taylor & Francis Group, London, UK

348 Dynamics of Structure and Foundation: 2. Applications

1

Rausch’s method

Calculation of geometric properties of the frame Area of transverse beam Ab = (600 × 900) = 0.6 × 0.9 = 0.54 m2 Area of column Ac = (600 × 600) = 0.6 × 0.6 = 0.36 m2 Ib =

1 × 0.6 × 0.93 = 0.03645 m4 ; 12

Ic =

1 × 0.6 × 0.63 = 0.0108 m4 ; 12

ψ=

0.03645 × 4.05 Ib h = = 5.257 Ic L 0.0108 × 2.6

and

Calculation of load transverse frame 1 Self weight of beam = 0.6×0.9×25 = 13.5 kN/m; UDL from slab = 24 kN/m Total, UDL(q) = 24 + 13.5 = 37.5 kN/m Self weight of long beam = 13.5 kN/m; UDL on long beam = 32 kN/m Total UDL on long beam = 32 + 13.5 = 45.5 kN/m 4.59 + 0.5 × 99.4 = 154.15 kN Load from long beam = 45.5 × 2 Load from column = 0.6 × 0.6 × 1.8 × 25 = 16.2 kN Total load transferred to column (N) = 16.2 + 154.15 = 170.4 kN Load on beam from machine (P) = 6.7 kN; Modulus of concrete Ec = 3× 108 kN/m2 ➔ Ec Ib = 3 × 108 × 0.03645 = 1.0935 × 107 kN m2 ; Ec Ab = 3 × 108 × 0.54 = 1.62 × 108 kN Ec Ac = 3 × 108 × 0.36 = 1.08 × 108 kN Calculation of displacements transverse frame 1 PL3 δ1 = 96EIb

2ψ + 1 2 × 5.257 + 1 6.7 × (2.6)3 = ψ +2 5.257 + 2 96 × 1.0935 × 107 = 1.769 × 10−7 m

δ2 =

QL3 384EIb

5ψ + 2 37.5 × 2.6 × (2.6)3 5 × 5.257 + 2 = ψ +2 5.257 + 2 384 × 1.0935 × 107

= 1.5893 × 10−6 m Q 98 3 L 0.6 × 2.6 δ3 = P+ 6.7 + = = 5.364 × 10−7 m 5 EAb 2 2 1.62 × 108

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 349

δ4 =

h P+Q 6.7 + 98 4.05 N+ 170.4 + = EAc 2 2 1.08 × 108

= 8.353 × 10−6 m ➔

4

δi = 1.06557 × 10−5 m

i=1

Thus,

30 30 fv = √ = √ = 9190 rpm δv 1.06557 × 10−5

Calculation of load transverse frame 2 Self weight of beam = 0.6×0.9×25 = 13.5 kN/m; UDL from slab = 48 kN/m Total UDL(q) = 48 + 13.5 = 61.5 kN/m Self weight of long beam = 13.5 kN/m; UDL on long beam = 32 kN/m Load from machine (P) = 18.2 kN Total UDL on long beam = 32 + 13.5 = 45.5 kN/m Load from long beam = 45.5 × 4.59 = 208.85 kN; Load from column = 0.6 × 0.6 × 1.8 × 25 = 16.2 kN Shown in Figure 2.18.16.

99.4

30.2 for Row A 141 for Row B 16 kN/m

1584

1585 4590 2730 2

1

3

Figure 2.18.16 Load distribution on longitudinal girder.

Load on beam from machine on row A. Load on row A =

99.4 1584 2730 + 30.2 × + 16 × 1.585 × = 75.2 kN 2 4590 4590

Load on row B =

99.4 1584 2730 + 141 × + 16 × 1.585 × = 113.4 kN 2 4590 4590

Load on column along row A = 208.85 + 75.2 + 16.2 = 300 kN Load on column along row B = 208.85 + 113.4 + 16.2 = 338.45 kN Average load = 319.225 kN Thus loading on frame 2 is as shown in Figure 2.18.17.

© 2009 Taylor & Francis Group, London, UK

350 Dynamics of Structure and Foundation: 2. Applications

319 (kN)

18.2 (kN)

319 (kN)

62 kN/m (q)

Figure 2.18.17 Load on transverse frame-2.

Calculation of displacements for transverse frame 2

δ1 =

PL3 96EIb

2ψ + 1 2 × 5.257 + 1 18.2 × (2.6)3 = ψ +2 5.257 + 2 96 × 1.0935 × 107 = 4.832 × 10−7 m

δ2 =

QL3 384EIb

5ψ + 2 5 × 5.257 + 2 62 × 2.6 × (2.6)3 = ψ +2 5.257 + 2 384 × 1.0935 × 107 = 2.614 × 10−6 m

Q 161.2 3 L 0.6 × 2.6 δ3 = P+ 18.2 + = 5 EAb 2 2 1.62 × 108 = 9.514 × 10−7 m δ4 =

P+Q 18.2 + 161.2 h 4.05 N+ 319 + = EAc 2 2 1.08 × 108 = 1.53263 × 10−5 m

4

30 30 δi = 1.9375 × 10−5 m ➔ fv = √ = √ = 6815 rpm. δ 1.9375 × 10−5 v i=1

Calculation of load transverse frame 3 Self weight of beam = 0.6×0.9×25 = 13.5 kN/m; UDL from slab = 24 kN/m Total UDL(q) = 24 + 13.5 = 37.5 kN/m Self weight of long beam = 13.5 kN/m; UDL on long beam = 32 kN/m; © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 351

Load from machine (P) = 10.7 kN Total UDL on long beam = 32 + 13.5 = 45.5 kN/m; Load from long beam = 4.59 45.5 × = 113 kN 2 Load from column = 0.6 × 0.6 × 1.8 × 25 = 16.2 kN As shown in Figure 2.18.18, Load due to concentrated load on long beam on Row A and B 1860 3006 + 16 × 1.585 × Load on row A = 30.2 × = 30.6 kN 4590 4590 3006 1860 Load on row B = 141 × + 16 × 1.585 × = 102.62 kN 4590 4590 Average load = 66.61 kN Total load on column (N) = 66.61 + 113 ∼ = 180 kN 30.1 for Row A 141 for Row B

16 kN/m

1584 2730

2

3

Figure 2.18.18 Load on longitudinal girder.

Calculation of displacements for transverse frame 3 δ1 =

PL3 96EI b

2ψ + 1 2 × 5.257 + 1 10.7 × (2.6)3 = ψ +2 5.257 + 2 96 × 1.0935 × 107 = 2.825 × 10−7 m

δ2 =

QL3 384EI b

5ψ + 2 37.5 × 2.6 × (2.6)3 5 × 5.257 + 2 = ψ +2 5.257 + 2 384 × 1.0935 × 107 = 1.5893 × 10−6 m

δ3 =

Q 98 3 L 0.6 × 2.6 P+ 10.7 + = = 5.75 × 10−7 m 5 EAb 2 2 1.62 × 108

δ4 =

P+Q h 10.7 + 98 4.05 N+ 180 + = EAc 2 2 1.08 × 108 = 8.788 × 10−6 m

© 2009 Taylor & Francis Group, London, UK

352 Dynamics of Structure and Foundation: 2. Applications 4

δi = 1.1234 × 10−5 m;

i=1

30 30 Thus, fv = √ = √ = 8950 rpm. δv 1.1234 × 10−5

Thus average vertical frequency of the frame = 8318 > 5100 rpm

9190 + 6815 + 8950 = 3

Frequency in horizontal direction Weight of top deck = 9.78 × 3.2 × 0.9 × 25 = 704 kN Weight from machine = 22.1 + 195 + 16.65 × 2 + 54 + 100 = 404 Total weight = 704 + 404 = 1108 kN Khi =

12EIc h3

6ψ + 1 6 × 5.257 + 1 12 × 3 × 108 × 0.0108 × = 3ψ + 2 3 × 5.257 + 2 (4.05)3 = 1071751 kN/m

Kh1 + Kh2 + · · · · · · + KhN 3 × 1071751 fh = 30 = 30 W 1108 = 1616 < 5100 rpm. The method does not have any provision of amplitude check and only check for resonance with the operating frequency of the machine. 2

Barkan’s method

Calculation of stiffness for transverse frame 1 in vertical direction 2EAc 2 × 3 × 108 × 0.6 × 0.6 is the stiffness of the columns = = h 4.05 5.333 × 107 kN/m For transverse beam

k1 =

δv = =

L3 (1 + 2ψ) 3L + 96EI b (2 + ψ) 8GAb 3 × 2.6 (2.6)3 (1 + 2 × 5.257) × + 7.257 96 × 1.0935 × 107 8 × 1.5 × 108 × 0.6 × 0.9

= 3.86014 × 10−8 m k2 =

1 → k2 = 2.6 × 107 kN/m δv

The stiffness matrix thus becomes

k + k2 [K] = 1 −k2

−k2 7.93 −2.6 = × 107 kN/m −2.6 2.6 k2

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 353

m1 = mL + 0.255mb + 0.35mc (24 + 13.5) × 2.6 mb = Mass of cross girder = = 9.938 kN · sec2 /m 9.81 mL = Mass transferred from long girder 104.45 × 2 + 99.4 = 31.43 kN · sec2 /m 9.81 0.6 × 0.6 × 3.6 × 25 × 2 mc = Mass of column = = 6.605 KN/sec2 /m 9.81 m1 = 0.255 × 9.938 + 0.35 × 6.605 + 31.43 = 36.27 kN · sec2 /m m2 = m0 + 0.45mb =

m0 = Load from machine on transverse girder; 6.7 + 0.45 × 9.938 = 5.15kN · sec2 /m m2 = 9.81 36.27 0 Thus, [M] = 0 5.15 Thus based on eigen value solution =

−2.6 × 107 7.93 × 107 − 36.27λ =0 −2.6 × 107 2.6 × 107 − 5.15

The above on solution gives74 λ1 = 1.2369 × 106 λ2 = 5.998 × 106

➔ ➔

ω1 = 1112 rad/sec(10168 rpm) > 5100 rpm ω2 = 2449 rad/sec(23386 rpm) > 5100 rpm

The normalized eigen vector is given by 0.14837 [ϕ]n = 0.19654

−0.07413 0.39428

Based on loading table for dynamic load, Pv = 44 + 130 = 174 KN. Thus equation of motion becomes

36.27 0

z¨ 1 7.93 −2.6 z 0.0 + × 107 1 = sin 534t −2.6 2.6 174 z¨ 2 z2

0 5.15

74 We will not solve this equation directly. Please refer to Chapter 5 (Vol. 1) on structural dynamics where we have solved in detail such eigen value problem.

© 2009 Taylor & Francis Group, London, UK

354 Dynamics of Structure and Foundation: 2. Applications

Calculation of amplitude for transverse frame 1 in vertical direction 0.14837 0.19654 34.19 0 Thus [ϕ]T = Considering 5% n {P} = −0.07413 0.39428 174 68.60 damping for concrete on orthogonal transformation of the above equation of motion we have ξ¨1 + 111ξ˙1 + 1.2369 × 106 ξ1 = 34.19 sin 534t; ξ¨2 + 245ξ˙2 + 5.998 × 106 ξ2 = 68.60 sin 534t Here, r =

ωm 534 = 0.48 for mode 1 and r = = 0.21, ωn 2449

➔ ξ1 =

andξ2 =

34.19 sin 534t 1.2369×106 2 (1−(0.48) )2 −(2×0.05×0.48)2

68.60 sin 534t 5.998×106 2 (1−(0.21) )2 −(2×0.05×0.21)2

= 3.584×10−5 sin 534t m

= 11.92×10−6 sin 534t m

0.14837 z1 = We have then, {Z} = [ϕ]{ξ } = 0.19654 z2

4.38 × 10−6 sin 534t = sin 534t × 10−6 m 11.657

−0.07413 0.39428

35.48 11.92

Calculation of stiffness for transverse frame 2 in vertical direction 2 × 3 × 108 × 0.6 × 0.6 2EAc is the stiffness of the columns = h 4.05 5.333 × 107 kN/m

k1 =

=

For transverse beam δv =

=

3L L3 (1 + 2ψ) + 96EI b (2 + ψ) 8GAb 3 × 2.6 (2.6)3 (1 + 2 × 5.257) × + 7 7.257 96 × 1.0935 × 10 8 × 1.5 × 108 × 0.6 × 0.9

= 3.86014 × 10−8 m ➔

k2 =

1 = 2.6 × 107 kN/m δv

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 355

The stiffness matrix thus becomes

−k2 7.93 −2.6 = × 107 kN/m −2.6 2.6 k2

k + k2 [K] = 1 −k2

m1 = mL + 0.255mb + 0.35mc (48 + 13.5) × 2.6 = 16.29 kN · sec2 /m 9.81 mL = Mass transferred from long girder

mb = Mass of cross girder =

=

(32 + 13.5) × 4.95 75.2 113.4 + + = 42.20 kN sec2 /m. 9.81 9.81 9.81

mc = Mass of column =

0.6 × 0.6 × 3.6 × 25 × 2 = 6.605 kNsec2 /m 9.81

m1 = 0.255 × 16.29 + 0.35 × 6.605 + 42.20 = 48.65 kN × sec2 /m m2 = m0 + 0.45mb m0 = Load from machine on transverse girder 18.2 + 0.45 × 16.29 = 9.185 kN · sec2 /m. 9.81 48.65 0 Thus [M] = 0 9.185 7.93 × 107 − 48.65λ −2.6 × 107 Thus the eigen value solution = = 0 −2.6 × 107 2.6 × 107 − 9.185λ

m2 =

The above on solution gives75 λ1 = 861700

➔

λ2 = 3.599 × 10

6

ω1 = 928 rad/sec(8861 rpm) > 5100 rpm ➔

ω2 = 1897 rad/sec(18115 rpm) > 5100 rpm

0.1215 The normalized eigen vector is given by, [ϕ]n = 0.1748

−0.0754 0.27985

Based on loading table for dynamic load Pv = 25 + 130 = 155 kN.

75 We will not solve this equation directly. Please refer to Chapter 5 (Vol. 1) where we have solved in detail such eigen value problem.

© 2009 Taylor & Francis Group, London, UK

356 Dynamics of Structure and Foundation: 2. Applications

Thus equation of motion becomes

48.65 0

z¨ 1 7.93 + z¨ 2 −2.6

0 9.185

−2.6 0.0 7 z1 × 10 = sin 534t z2 2.6 155

Calculation of amplitude for transverse frame 2 in vertical direction [ϕ]T n {P} =

0.1215 −0.0759

0.1748 0 27.094 = 0.27985 155 43.3768

Considering 5% damping for concrete on orthogonal transformation of the above equation of motion we have ξ¨1 + 92.8ξ˙1 + 861700ξ1 = 27.094 sin 534t; ξ¨2 + 189.7ξ˙2 + 3.599 × 106 ξ2 = 43.3768 sin 534t

Here r =

ωm 5100 = 0.575 for mode 1 and r = = 0.28 ωn 18115

➔ ξ1 =

and

27.094 sin 534t 861700 (1 − (0.575)2 )2 − (2 × 0.05 × 0.575)2

ξ2 =

43.3768 sin 534t 3.599×106 2 (1 − (0.28) )2 − (2 × 0.05 × 0.28)2

= 4.68 × 10−5 sin 534t m

= 13.07 × 10−6 sin 534t m

Since {Z} = [ϕ] {ξ },

z1 0.1215 −0.0754 46.8 = sin 534t × 10−6 0.1748 0.27985 13.07 z2

4.70 = sin 534t × 10−6 m. 11.838

Calculation of stiffness for transverse frame 3 in vertical direction Referring to the previous calculation the stiffness matrix is

k + k2 [K] = 1 −k2

−k2 7.93 −2.6 = × 107 kN/m −2.6 2.6 k2

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 357

m1 = mL + 0.255mb + 0.35mc mb = Mass of cross girder =

(24 + 13.5) × 2.6 = 9.93 kN · sec2 /m 9.81

mL = Mass transferred from long girder (32 + 13.5) × 4.95 30.6 102.62 = + + = 25.06 kN · sec2 /m. 9.81 × 2 9.81 9.81 mc = Mass of column =

0.6 × 0.6 × 3.6 × 25 × 2 = 6.605 kN · sec2 /m 9.81

m1 = 0.255 × 9.93 + 0.35 × 6.605 + 25.06 = 29.90 kN · sec2 /m m2 = m0 + 0.45mb m0 = Load from machine on transverse girder 10.7 + 0.45 × 9.93 = 5.56 kN · sec2 /m 9.81 30 0 Thus [M] = 0 5.56

m2 =

Thus based on eigen value solution 7.93 × 107 − 30λ −2.6 × 107 = =0 −2.6 × 107 2.6 × 107 − 5.56λ The above on solution gives76 λ1 = 1404600

➔

ω1 = 1185 rad/sec(11316 rpm) > 5100 rpm

λ2 = 5915000

➔

ω2 = 2432 rad/sec(23224 rpm) > 5100 rpm

The normalized eigen vector is given by 0.1555 [ϕ]n = 0.2194

−0.0956 0.3612

Based on loading table for dynamic load Pv = 8.3 + 25 = 33.3 kN.

76 We will not solve this equation directly. Please refer to Chapter 5 (Vol. 1) where we have solved in detail such eigen value problem.

© 2009 Taylor & Francis Group, London, UK

358 Dynamics of Structure and Foundation: 2. Applications

Thus equation of motion becomes

z¨ 1

30

0

0

5.56

+

z¨ 2

7.93

−2.6

−2.6

× 10

2.6

7

z1 z2

=

0.0

33.3

sin 534t

Calculation of amplitude for transverse frame 3 in vertical direction Thus

[ϕ]T n {P}

=

0.1555

0.2194

0

−0.0956

0.3612

33.3

=

7.30 12.03

Considering 5% damping for concrete on orthogonal transformation of the above equation of motion we have ξ¨1 + 118.5ξ˙1 + 1404600ξ1 = 7.30 sin 534t; ξ¨2 + 243.2ξ˙2 + 5915000ξ2 = 12.03 sin 534t. ωm = 0.45 for mode 1 and r = 0.22 ωn

Here r =

ξ1 =

7.30 sin 534t 1404600 2 2 (1 − (0.45) ) − (2 × 0.05 × 0.45)2

= 6.5 × 10−6 sin 534t

ξ2 =

12.03 sin 534t 5.915×106 2 (1 − (0.22) )2 − (2 × 0.05 × 0.22)2

= 2.136 × 10−6 sin 534t

and

Since {Z} = [ϕ]{ξ } Hence

z1 0.1555 −0.0956 6.5 = sin 534t × 10−6 0.2194 0.3612 2.136 z2

=

0.822 sin 534t × 10−6 m 2.219

Calculation of horizontal frequency Table for calculation of mass and stiffness Frame

m0

mb

mc

ml

mi

δhi

K hi

1 2 3

0.683 1.855 1.0907

9.938 16.29 9.93

6.605 6.605 6.605

31.43 42.2 25.06

44.030 62.326 38.060 144.416

9.333E-07 9.333E-07 9.333E-07

1071467 1071467 1071467 3214401

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 359

Here mi = m0i + mbi + 0.3mci + mLi and δhi =

h3 (2 + 3ψ) 12EIc (1 + 6ψ)

Table for calculation of second, mass moment and stiffness inertia Frame mi 1 2 3

Khi

di

midi

44.030 1071467 0 0 62.326 1071467 4.59 286.08 38.060 1071467 9.18 349.39 635.47

Xg = Thus

xgi (Xg − di ) Xhi

Khi dI

0 4.40 4918032.8 −0.190 9836065.6 −4.78 14754098

mi xgi2

4.59 852.5194 22573770 0 2.243947 0 −4.59 869.5175 22573770 1724.281 45147541

635.476 14754098 = 4.40 m; Xh = = 4.59 m 144.416 3214401 e = 4.59 − 4.40 = 0.190 m

Dynamic loads Phi = 58 + 80 + 100 = 238 kN Mh =

N

and

Phi Xgi = 238 × 4.40 = 1047.2

i=1

Thus equation of motion becomes Kh Kh e x¨ M 0 x Ph cos ωm t + = Mh cos ωm t 0 Jϕ ϕ¨ Kh e Kh e2 + γ ϕ 144.416 0.0 x¨ 3.215 × 106 0.6095 × 106 x + ϕ 0.0 1724.3 ϕ¨ 0.6095 × 106 4.53 × 106

238 cos 534t = 1047.2 cos 534t

Thus for natural frequency we have

3.215 × 106 − 144.416λ 0.6095 × 106 =0 0.6095 × 106 4.53 × 106 − 1724.3λ

The above on expansion and solution gives λ1 = 21919

➔ ω1 = 148 rad/sec(1413 rpm) < 5100 rpm

λ2 = 26614

➔ ω2 = 163 rad/sec(1557 rpm) < 5100 rpm

© 2009 Taylor & Francis Group, London, UK

Khi Xhi2

360 Dynamics of Structure and Foundation: 2. Applications

The normalized eigen vector is given by, 0.08011 0.022485 −6.5109 × 10−3 0.023186 238 cos 534t 0.08011 −6.5109 × 10−3 T [ϕ]n {P} = 1047.2 cos 534t 0.022485 0.023186

12.25 cos 534t = 248.15 cos 534t

[ϕ]n =

Considering 5% damping for concrete on orthogonal transformation of the above equation of motion we have ξ¨1 + 14.8ξ˙1 + 21919ξ1 = 12.25 cos 534t; ξ¨2 + 16.3ξ˙2 + 26614ξ2 = 248.15 cos 534t. Here r =

i.e.

ωm = 3.609 for mode 1 and r = 3.27 ωn

ξ1 =

12.25 cos 534t 21919 (1− (3.6)2 )2 −(2 × 0.05 × 0.3.6)2

and ξ2 =

248.1552 cos 534t 26614 2 2 (1−(3.27) ) −(2×0.05×3.27)2

= 4.6707 × 10−5 cos 534t = 9.61416 × 10−4 cos 534t

Since {X} = [ϕ]{ξ }, hence ➔

0.08011 −0.022485 x 4.6707 = cos 534t × 10−5 ϕ 96.1416 −6.5109 × 10−3 0.023186

2.5359 = cos 534t × 10−5 m 2.1987

Thus, displacement of frame 3 This is generically given by xnet = x + X ϕ where X = is the farthest point form the center of gravity point G x3 = 2.5359 × 10−5 + 4.78 × 2.1987 × 10−5 = 1.30456 × 10−4 Displacement of frame 2 x2 = 2.5359 × 10−5 + 0.190 × 2.1987 × 10−5 = 2.953 × 10−5 Displacement of frame 3 x1 = 2.5359 × 10−5 − 4.40 × 2.1987 × 10−5 = −7.13828 × 10−5

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 361

3

Major’s combined method of analysis

In this case the natural frequency in vertical direction is same as shown in Rausch’s method earlier except for the case of soil when Major considers the total frame as stick model and combines with soil displacement. Thus for transverse frame 1, we have 4

δi = 1.06557 × 10−5 m

i=1

For calculation of

4

i=1 δi

refer to previous calculation by Rausch’s method

30 30 and fv = √ = √ = 9190 r.p.m δv 1.06557 × 10−5 While Major calculated the dynamic force based on Rotor weights, here dynamic force coming on the frame has directly been given, thus Load on transverse beam (Cb ) = 44 kN Load on column from Long beams(Cc ) = 130 kN

Thus

C L3 δ1 = b 96EIb

2ψ + 1 ψ +2

3 L h + C + 5 EAb b EAc

Cb + C c 2

44 × (2.6)3 11.514 0.6 × 2.6 × 44 × + 7 7.257 96 × 1.0935 × 10 1.62 × 108

=

4.05 + 1.08 × 108

44 + 130 2

= 4.855 × 10−6 m As ωn > ωm hence corrected value of operating frequency is ωn = 0.8 × 9190 = 7352 r.p.m 1

M.F =

(1 − r2 )2 + ( ∇π )2 (r)2

,

here r =

5100 = 0.69 7352

Substituting the above values, we have M.F. = 1.900 Thus δv1 = 4.855 × 10−6 × 1.9 = 9.2245 × 10−6 m

© 2009 Taylor & Francis Group, London, UK

and = 0.4

362 Dynamics of Structure and Foundation: 2. Applications

For transverse frame 2, we have 4

δi = 1.9375 × 10−5 m

i=1

For calculation of and

4

i=1 δi

refer to previous calculation by Rausch’s method

30 30 fv = √ = √ = 6815 rpm δv 1.9375 × 10−5

While Major calculated the dynamic force based on Rotor weights, here dynamic force coming on the frame has directly been given, thus Load on transverse beam (Cb ) = 0.0 kN Load on column from Long beams(Cc ) = 130 + 25 = 155 kN C L3 Thus δ1 = b 96EIb

2ψ + 1 ψ +2

= 0.0 + 0.0 +

h 3 L Cb + + 5 EAb EAc

4.05 1.08 × 108

155 2

Cb + C c 2

= 2.9063 × 10−6 m

As ωn > ωm hence corrected value of operating frequency is ωn = 0.8×6815 = 4948 rpm which is less than the operating speed of 5100 r.p.m. At transient resonant condition as per Major M.F = 7.85, here r = 5100 = 0.69 7352 Substituting the above values we have Thus, δv2 = 2.9063 × 10−6 × 7.85 = 2.281 × 10−5 m For transverse frame 3 we have 4

δi = 1.1234 × 10−5 m

i=1

For calculation of and

4

i=1 δi

refer to previous calculation by Rausch’s method

30 30 fv = √ = √ = 8950 rpm δv 1.1234 × 10−5

While Major calculated the dynamic force based on Rotor weights, here dynamic force coming on the frame has directly been given, thus Load on transverse beam (Cb ) = 8.30 kN

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 363

Load on column from Long beams(Cc ) = 25 kN Thus

δ1 = =

Cb L3 96EIb

2ψ + 1 ψ +2

+

h 3 L C + 5 EAb b EAc

Cb + C c 2

8.3 × (2.6)3 11.514 0.6 × 2.6 × 8.3 × + 7 7.257 96 × 1.0935 × 10 1.62 × 108 8.3 + 25 4.05 = 9.2478 × 10−7 m + 2 1.08 × 108

As ωn > ωm hence corrected value of operating frequency is ωn = 0.8×8950 = 7160 rpm which is greater than the operating speed of 5100 rpm. Here M.F =

1 [(1 − r2 )2 + ( ∇π )2 (r)2 ]

,

r=

5100 = 0.69 7352

and

= 0.4

which gives M.F. = 1.995 Substituting the above values we have δv3 = 9.2478 × 10−7 × 1.995 = 1.8446 × 10−6 m Calculation of vertical frequency including the soil effect In this case we consider the total top deck including the column as a stick model 4 −5 Thus, for frame 1, For frame 2, i=1 δi = 1.06557 × 10 ; 4 −5 × 10 m; i=1 δi = 1.9375 For frame 3, 4i=1 δi = 1.1234 × 10−5 m; Thus, δav = 1.37549 × 10−5 m. Shear wave velocity = 115 m/sec; unit weight of soil = 19 kN/m3 ; mass density of soil = 1.936 kN sec2 /m4 . Dynamic shear modulus, G = 1.936 × 115 × 115 = 25614 kN/m2 . 9.78 × 3.2 Base Area = 9.78 m × 3.2 m; equivalent radius r0 = = 3.156 m π 4Gr0 4 × 25614 × 3.156 Thus vertical spring stiffness of soil = = = (1 − ν) 0.7 461930 kN/m Weight of top deck = 9.78 × 3.2 × 0.9 × 25 = 704.16 kN Weight of column = 6 × 0.36 × 3.625 = 194.4 kN Weight of machine = 22.1 + 195 + 2 × 16.65 + 54 + 100 = 404 kN Weight of base mat = 9.78 × 3.2 × 1.5 × 25 = 1173.6 kN

© 2009 Taylor & Francis Group, London, UK

364 Dynamics of Structure and Foundation: 2. Applications

Total vertical Load = 704.16 + 194.4 + 404 + 1173.6 = 2476.16 kN δs =

Pv 2476.16 = = 5.36046 × 10−3 m Kv 461930

Total displacement δv = δ1 + δ2 + δ3 + δ4 + δs = 1.37549 × 10−5 + 5.36046 × 10−3 = 5.37422 × 10−3 m 30 30 Knowing fv = √ cpm; we have fs = √ = 409 rpm δv 5.37422 × 10−3 Calculation of horizontal frequency For horizontal frequency we know that 1 1 2 2 N 2 2 2 Khi Ih 2 3 3 (fn )h = 30 α0 ± α0 s − i=1 c.p.m N J i=1 Wi ϕ Here, Khi = Lateral stiffness of the ith frame i and Khi =

1 where δhi = δhi

h3 (2 + 3ψ) 12EIc (1 + 6ψ) Wi = total weight of the ith frame plus weight of the machine plus weight of the transverse beam and the longitudinal beams, Jφ = Mass moment of inertia ∼ = N 2 i=1 Wi Xgi ; Xg = distance of weight W from the resultant center of mass point 2 G; Ih = N i=1 Khi Xhi ; Xh = distance of each frame from the centre of rigidity H,

and

N 1 2 N I K K hi hi h i=1 α0 = + i=1 + e N 2 Jϕ Jϕ i=1 Wi

3 × 1.0717 × 106 1 3 × 1.0717 × 106 Here, α0 = 0.1902 + 2 1724.3 × 9.81 144.416 × 9.81 2 × 2.257 × 107 + 1724.3 × 9.81

= 2472.42

Here all the data within the parenthesis were calculated while doing the calculation based on Barkan’s method. 1 2 N 2 Khi Ih 3 2 Thus (fn )1 = 30 α0 − α0 − i=1 N J i=1 Wi ϕ

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 365

or,

1 2 2 3

(fn )1 = 30 2472.4 −

6112853 −

3 × 1.0717 × 106 2 × 2.257 × 107 × 144.416 × 9.81 1724.3 × 9.81

√ = 30 2472.4 − 56797 ➔

√ (fn )1 = 30 2472.4 − 238 = 1418 rpm(148 rad/sec);

√ (fn )2 = 30 2472.4 + 238 = 1562 rpm(164 rpm/sec).

Calculation of Horizontal amplitude as per Figure 2.18.19

4590

4590

58 kN

80 kN

100 kN

Figure 2.18.19 Top deck slab with transverse load.

The resultant of the Horizontal dynamic load acts at x¯ =

80 × 4.59 + 58 × 4.59 × 2 = 3.78 m 100 + 58 + 80

Thus eccentricity between center of rigidity and x¯ is 0.81 m is CK X K Ci = C N hi + e N hi hi hence for frame 1 we have 2 i=1 Khi i=1 Khi Xhi

C1 = 238

1.0716×106 238×1.0716×106 ×4.59 + 0.81 = 100.33 kN 3×1.0716× 106 2×2.257×107

© 2009 Taylor & Francis Group, London, UK

366 Dynamics of Structure and Foundation: 2. Applications

Ci 100.33 we have, δh1 = = 9.36264 × 10−5 Khi 1.0717 × 106 As the foundation is under tuned thus maximum amplitude at transient condition is given by Knowing δhi =

δh1 = 9.36264 × 10−5 × 7.85 = 7.34967 × 10−4 m. For frame 2, we have

C2 = 238

1.0716 × 106 238 × 1.0716 × 106 × 0.0 + 0.81 = 79.33 kN 3 × 1.0716 × 106 2 × 2.257 × 107

Knowing δhi =

Ci 79.33 we have, δh2 = = 7.4277 × 10−5 Khi 1.0717 × 106

As the foundation is under tuned thus maximum amplitude at transient condition is given by δh2 = 7.4277 × 10−5 × 7.85 = 5.8307 × 10−4 .

For frame 3 we have

C1 = 238

1.0716 × 106 238 × 1.0716 × 106 × −4.59 + 0.81 = 58.33 kN 6 3 × 1.0716 × 10 2 × 2.257 × 107

Knowing δhi =

Ci 58.33 we have, δh3 = = 5.44306 × 10−5 Khi 1.0717 × 106

As the foundation is under tuned thus maximum amplitude at transient condition is given by δh3 = 5.44306 × 10−5 × 7.85 = 4.273 × 10−4 m

4

Based on 2D soil structure interaction model

Calculation in vertical direction The mathematical model for this case is as shown in Figure 2.18.20. Mathematical model of the turbine foundation with soil spring in vertical direction.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 367

The equation of motion is given by ⎡

m1 ⎢ ⎣0 0

0 m2 0

⎤⎧ ⎫ ⎡ ⎪z¨ 1 ⎪ c 1 + c2 0 ⎨ ⎬ ⎢ ⎥ 0 ⎦ z¨ 2 + ⎣ −c2 ⎪ ⎪ m3 ⎩z¨ 3 ⎭ 0

⎡

k1 + k2 + ⎣ −k2 0

−k2 k2 + k 3 −k3

−c2 c2 + c 3 −c3

⎤⎧ ⎫ 0 ⎪ ⎬ ⎨z˙ 1 ⎪ ⎥ −c3 ⎦ z˙ 2 ⎪ ⎪ c3 ⎩z˙ 3 ⎭

⎤⎧ ⎫ ⎧ ⎫ 0 ⎨z1 ⎬ ⎨ 0 ⎬ −k3 ⎦ z2 = 0 sin ωm t ⎩ ⎭ ⎩ ⎭ Pv k3 z3

m3

k3

m2 k2 m1

k1

Figure 2.18.20 Mathematical model of the turbine foundation with soil spring in vertical direction.

Here, mass of column

mc =

0.36 × 3.6 × 6 × 25 = 19.816 9.81

Here m1 = Mass of the bottom raft

m1 =

9.78 × 3.2 × 1.5 × 25 + 5.944 = 125.57 ≡ 126 kN · sec2 /m 9.81

© 2009 Taylor & Francis Group, London, UK

368 Dynamics of Structure and Foundation: 2. Applications

m2 = 0.25 times the weight of the transverse girder + weight of machine from longitudinal girder + self weight from longitudinal girder + 0.3 times the weight of column. m2 =

24 + 13.5 48 + 13.5 × 2.6 × 2 + × 2.6 × 0.25 9.81 9.81

+

4.59 × 2 × (32 + 13.5) 16 × 1.585 × 2 + 9.81 9.81

+

99.4 × 2 + 30.2 + 141 0.3 × 6 × 0.36 × 3.6 × 25 + 9.81 9.81

= 100.451 ≡ 100 kN · sec2 /m m3 = Concentrated mass on transverse girder + 0.45 times the self weight For frame 1 we have m31 =

6.7 (24 + 13.5) + × 2.6 × 0.45 = 5.15 kN · sec2 /m 9.81 9.81

For frame 2 we have m32 =

18.2 (48 + 13.5) + × 2.6 × 0.45 = 9.19 kN · sec2 /m 9.81 9.81

For frame 3 we have m33 =

10.7 (24 + 13.5) + × 2.6 × 0.45 = 5.56 kN · sec2 /m 9.81 9.81

Thus for the complete frame we have, m3 = 5.15 + 9.19 + 5.56 = 19.9 ≡ 20 The mass matrix is thus given by ⎡

126 [M] = ⎣ 0 0

⎤ 0 0 100 0 ⎦ 0 20

Calculation of stiffness matrix In vertical direction the displacement of the transverse girder is given by δv =

L3 (1 + 2ψ) 3L + 96EIb (2 + ψ) 8GAb

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 369

δv =

3 × 2.6 2.63 × (1 + 2 × 5.257) + 7 96 × 1.0935 × 10 × (2 + 5.257) 8 × 1.5 × 108 × 0.54

= 3.86014 × 10−8 3 = 7.772 × 107 3.86014 × 10−8 2 × 3 × 108 × 0.36 2EAc = × 3 = 18.0 × 107 For columns we have, k2 = h 3.6 For the soil the equivalent spring stiffness is given by Thus for three frames we have, k3 =

4Gr0 = 461930 1−ν ⎡ ⎤ k1 + k 2 −k2 0 k2 + k3 −k3 ⎦ [K] = ⎣ −k2 0 −k3 k3 ⎡ ⎤ 18.0461930 −18.0 0 −18 25.772 −7.772⎦ × 107 =⎣ 0 −7.772 7.772 k1 =

Thus for eigen value solution we have77 ⎡

⎤ 180461930 − 126λ −180000000 0 ⎣ −180000000 257720000 − 100λ −77720000 ⎦ = 0 0 −77720000 77720000 − 20λ This gives the eigen values and the corresponding three natural frequencies as ⎡

0.002 × 106 ⎣ [λ] = 0 0

0 2.5641 × 106 0

⎤ 0 ⎦ 0 6 5.3294 × 10

The corresponding eigen vector are given by ⎡

⎤ 0.5768 −0.3675 0.1266 [ϕ] = ⎣0.5775 0.2983 −0.3454⎦ ; 0.5778 0.8771 0.9299 ⎡ ⎤ 44.721 0 0 1601 0 ⎦ rad/sec [ω] = ⎣ 0 0 0 2308

77 We have solved the eigen problem in Math-Cad directly.

© 2009 Taylor & Francis Group, London, UK

370 Dynamics of Structure and Foundation: 2. Applications

Based on orthogonal transformation [ϕ]T [M][ϕ] ⎡

⎤

81.9436

[ϕ]T [M][ϕ] = ⎣

⎦

42.1484 31.2429

Thus the scale factors are given by √ √ 81.9436 = 9.052; Mr2 = 42.1484 = 6.492; √ = 31.2429 = 5.5895.

Mr1 = Mr3

Thus the normalised eigen vector is given by ⎡

0.06372 −0.05661 [ϕ]n = ⎣0.06379 0.04595 0.06383 0.13510

⎤ 0.02265 −0.06179⎦ 0.16636

Calculation of damping matrix √ Critical damping, Cc = 2 km Let the damping ratio for RCC structure be 5%. Thus critical damping Cc is given by c3 = 0.05 × (2 7.772 × 107 × 20) = 3942.6; c2 = 0.05 × (2 18.00 × 107 × 100) = 13416 The soil spring is calculated based on Richart’s formula as shown hereafter78 Bz =

(1 − ν)Wf 4ρs r30

=

0.7 × 1173.6 = 0.3438; 4 × 19 × (3.156)3

√ c1 = 0.7247 × (2 461930 × 126) = 11057.6

0.425 Dz = = 0.7247 Bz

Thus the damping matrix is given by ⎡

c1 + c2 [C] = ⎣ −c2 0

−c2 c2 + c 3 −c3

⎤ ⎡ ⎤ 0 24474 −13416 0 −c3 ⎦ = ⎣−13416 17359 −3943⎦ 0 −3943 3943 c3

78 For further details refer to the section of Design and Analysis of Block Foundation.

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 371

Thus on orthogonal transformation for each individual mode we have ⎡

⎤

44.8982

[ϕ]T [C][ϕ] = ⎣

⎦

207.9 306.57

→ 2D1 ω1 = 44.8902

⇒ D1 = 0.502

Calculation for load vectors The total vertical dynamic force is given by, Pv = 44 + 8.3 + 130 × 2 + 25 × 2 = 362.3 kN Performing the operation ⎫ ⎤⎧ 0.06372 0.06379 0.06383 ⎨ 0 ⎬ ⎣ 0.04595 0.13510⎦ 0 [ϕ]T sin 534t n {P} = −0.05661 ⎩ ⎭ 0.02265 −0.06179 0.16636 362.3 ⎡

⎧ ⎫ ⎨23.1256⎬ = 48.9467 sin 534t ⎩ ⎭ 60.2772 Thus the three uncoupled equation of motion is given by ξ¨1 + 44.89ξ˙1 + 2000ξ1 = 23.1256 sin 534t ξ¨2 + 207.89ξ˙2 + 2564100ξ2 = 48.9467 sin 534t ξ¨3 + 207.89ξ˙3 + 2564100ξ3 = 60.2772 sin 534t ξ1 = 0.502,

ξ2 = 0.0649

and ξ3 = 0.0664

Based on the above the displacement vector is given by

ξi =

3 i=1

pi sin ωm t (1 − r2i )2 + (2Di ri )2

Once we know the displacement vectors in un-coupled state the displacement in the global structural co-ordinate is given by {Z} = [ϕ]{ξ } The results are shown hereafter in tabular form

© 2009 Taylor & Francis Group, London, UK

© 2009 Taylor & Francis Group, London, UK

53.664 1280.8 1846.4

44.72 1601 2308

Amplitude

z1 z2 z3

0.06372 0.06379 0.06383

Eigen vector 1st mode

Eigen vector 3rd mode 0.2265 0.06179 0.16636

Eigen vector 2nd mode −0.05661 0.04595 0.1351 23.13 48.95 60.28

P

Natural frequency

44.72 1601 2308

Z

Z1 Z2 Z3

0.06372 0.06379 0.06383

Eigen vector 1st mode

Eigen vector 3rd mode 0.2265 0.06179 0.16636

Eigen vector 2nd mode −0.05661 0.04595 0.1351 23.1256 48.9467 60.2772

P 0.502 0.065 0.066

Damping ratio

Transient frequency during start and stop of machine = 44.72 rad/sec

Case 2

Corrected frequency

Natural frequency

Operating frequency of the machine = 534 rad/sec

Case 1

1 0.027933 0.019376

r

0.502 0.065 0.066

Damping ratio

0.996016 1.000774 1.000372

M.F.

9.950 0.416 0.289

r

7.3537 × 10−04 7.3627 × 10−04 7.3774 ×10−04

1.1517 × 10−02 1.9111 × 10−05 1.1320 × 10−05

8.96 × 10−06 9.30 × 10−06 1.26 × 10−05

1.17 × 10−04 2.304 × 10−05 1.235 × 10−05

Z

Z

Disp (uncoupled)

Disp (uncoupled)

0.01014 1.20780 1.09033

M.F.

372 Dynamics of Structure and Foundation: 2. Applications

Analysis and design of machine foundations 373

The results are compared hereafter by bar chart shown in Figure 2.18.21. Vertical amplitude columns

0.8

Amplitude (mm)

0.7 0.6 0.5 Vertical amplitude columns

0.4 0.3 0.2 0.1 0

0.8

Operating Ist Transient Operating Case

Vertical amplitude Transverse girder

Amplitude (mm)

0.7 0.6 0.5 Vertical amplitude Transverse girder

0.4 0.3 0.2 0.1 0

Operating Ist Transient Operating Case

0.8

Vertical amplitude Bottom Raft

Amplitude (mm)

0.7 0.6 0.5 Vertical amplitude Bottom Raft

0.4 0.3 0.2 0.1 0

Operating Ist Transient Operating Case

Figure 2.18.21 Comparison of transient and operating response.

© 2009 Taylor & Francis Group, London, UK

374 Dynamics of Structure and Foundation: 2. Applications

Analysis in coupled horizontal and rocking mode The mathematical model for the turbine foundation for this mode is perceived as Figure 2.18.22.

m0, Jφ

y

Kh

Kx

h

Kθ θ

u

Figure 2.18.22 2D mathematical model for coupled translation and rocking including soil springs.

The un-damped equation of motion for free vibration in coupled horizontal and rocking motion is given by ⎤⎧ ⎫ m0 m0 e m0 m0 h y¨ ⎪ ⎪ ⎪ ⎨ ϕ¨ ⎪ ⎬ ⎢ m0 e Jϕ + m0 e2 ⎥ m e m eh 0 0 ⎢ ⎥ ¨⎪ ⎣ m0 m0 e m0 + m f m0 h ⎦ ⎪ U ⎪ ⎩ ¨⎪ ⎭ 2 θ m0 eh m0 h Jθ + m0 h m0 h ⎡ ⎤⎧ ⎫ Kh 0 0 0 ⎪ ⎪y⎪ ⎪ ⎢ 0 Kh e2 + Iϕ 0 ⎥ ⎨ϕ ⎬ 0 ⎥ +⎢ =0 ⎣0 U⎪ 0 Ky 0 ⎦ ⎪ ⎪ ⎩ ⎪ ⎭ θ 0 0 0 Kθ ⎡

Calculation of mass matrix Here m0 = mass of the top deck + weight of the machine + 0.3 times the weight of the column e = distance between the centre of gravity and centre of rigidity © 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 375

h = height between centre of the top deck mass to the centre of the bottom raft mass mf = mass of the bottom raft. Jφ = mass moment of inertia of top deck in plan for the lumped masses @ 2 Jφ = N i=1 mi Xgi Jθ = mass moment of inertia of bottom raft in transverse plane Referring to calculation done in Barkan’s method earlier we have m0 = 144.416 kN · sec2 /m;

Jφ =

N

2 mi Xgi = 1724.3;

i=1

e = 4.59 − 4.40 = 0.19 m 44.03 × 0.45 + 62.326 × 0.45 + 38.06 × 0.45 = 0.45 (refer to calcu144.416 lations based on Barkan’s method for individual mass data) Thus, h = 3.6 + (0.9 − 0.45) + 1.5/2 = 4.8 m z¯ =

9.78 × 3.2 × 1.5 × 25 = 120 kN · sec2 /m; 9.81 m 2 120 Jθ = (ly + lz2 ) = (3.22 + 1.52 ) = 125 12 12

mf =

Thus ⎡

m0 m0 e ⎢ m0 e Jϕ + m0 e2 [M] = ⎢ ⎣ m0 m0 e m0 eh m0 h ⎡

144.416 ⎢ 27.44 =⎢ ⎣144.416 693

m0 m0 e m0 + m f m0 h

⎤ m0 h m0 eh ⎥ ⎥ m0 h ⎦ J θ + m 0 h2

⎤ 144.416 693 27.44 132 ⎥ ⎥ kN · sec2 /m 264 693 ⎦ 693 3452

27.44 1730 27.44 132

Calculation of stiffness matrix The stiffness matrix of the frame including the soil spring is given by ⎡

Kh ⎢0 [K] = ⎢ ⎣0 0

0 Kh e2 + Iϕ 0 0

0 0 Ky 0

⎤ 0 0⎥ ⎥ 0⎦ Kθ

Here shear wave velocity of the soil = 115 m/sec © 2009 Taylor & Francis Group, London, UK

376 Dynamics of Structure and Foundation: 2. Applications

Unit weight of soil γs = 19 kN/m3 ; Poisson’s ratio of soil νs = 0.3 19 Thus dynamic shear modulus of soil G = (115)2 = 25614 kN/m2 9.81 3 9.78 × 3.2 4 9.78 × (3.2) = 3.156 m; rθ = ry = = 2.414 m π 3π Thus, Ky = 105 kN/m Kθ =

32 (1 − νs ) Gry 32 × 0.7 × 25614 × 3.186 = = 3.93645 × 7 − 8νs 7 − 2.4

8Gr3θ 8 × 25614 × (2.414)3 = = 13.73 × 105 kN/m 3(1 − νs ) 2.1

Referring to the table for calculation of stiffness in Barkan’s method, done earlier 3

Khi = 3 × 1.072 × 106 = 3.216 × 106 kN/m;

i=1

Iφ =

N

2 Ki Xhi = 2 × 2.257 × 107 = 4.514 × 107

i=1

We had calculated earlier Thus Kh e2 + Iφ = 3.216 × 107 × 0.192 + 4.514 × 107 = 4.5256 × 107 kN/m Thus the stiffness matrix can be represented by ⎡

3.216 × 106 ⎢ 0 [K] = ⎢ ⎣ 0 0

0 4.5256 × 107 0 0

0 0 3.94 × 105 0

⎤ 0 ⎥ 0 ⎥ ⎦ 0 6 1.373 × 10

Calculation for eigen value For eigen value analysis we have ⎡

3.216 × 106 − 144.416λ −27.44 | 7 − 1730λ | ⎢ −27.44 4.5256 × 10 ⎢ ⎣ −144.416 −27.44 | −693 −132 | ⎤ | −144.416 −693 ⎥ | −27.44 −132 ⎥=0 5 ⎦ | 3.94 × 10 − 264 −693 | −693 1.373 × 106 − 3452

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 377

Solving the above we have79 ⎡

⎤

644590

⎢ [λ] = ⎢ ⎣

⎥ ⎥ ⎦

340 3650 26260

which gives the natural frequency as ⎡ ⎤ 803 ⎢ ⎥ 18.43 ⎥ rad/ sec80 [ω] = ⎢ ⎣ ⎦ 60.41 157 The corresponding eigen vectors are given by ⎡ ⎤ 0.9812 0.0675 −0.0128 −0.0043 ⎢−0.0005 0.0009 −0.0002 −0.9992⎥ ⎥ [ϕ] = ⎢ ⎣−0.0411 0.6156 0.9761 0.0051 ⎦ −0.1888 0.7851 −0.2170 0.0386 Now performing the operation [ϕ]T [M][ϕ]⎤we have, ⎡ 4.9 ⎢ ⎥ 2984.4 ⎥ [ϕ]T [M][ϕ] = ⎢ ⎣ ⎦ 120.8 1722.3 Thus the scaled factors are given by √ √ 4.9 = 2.213; Mr2 = 2984.4 = 54.63; √ √ = 120.8 = 10.99, and Mr4 = 1722.3 = 41.5.

Mr1 = Mr3

Thus the normalized eigen vector is given by ⎡

0.4438 ⎢−2.2593764 × 10−4 [ϕ]n = ⎢ ⎣ −0.018798 −0.085314

1.2392 × 10−3 1.6474 × 10−5 0.0112685 0.0142961

| | | |

79 We have solved the eigen value problem directly in Math CAD. 80 Reader to check that in case we ignore the effect of soil, the mass matrix reduces to M = 144.416 27.44 3.216 0 and the stiffness matrix K = × 106 and the above on 27.44 1730 0 45.256 eigen value solution gives natural frequencies as 148 rad/sec and 163 rad/sec which is exactly same as what we have obtained based on Barkan or Major’s method.

© 2009 Taylor & Francis Group, London, UK

378 Dynamics of Structure and Foundation: 2. Applications

| −1.1646952 × 10−3 | −1.8198362 × 10−5 | 0.08882 | −0.0197452

⎤ −1.0361446 × 10−4 ⎥ −0.0240771 ⎥ −4 1.2289157 × 10 ⎦ 9.30212048 × 10−4

Calculation of damping matrix Let the damping ratio for the RCC frame be 5% then we have Frame No.

Mass

Kh

Xh

√ Cc = 2 km

1 2 3

44.03 62.326 38.06 144.416

1.0717 × 106 1.0717 × 106 1.0717 × 106

4.59 0.0 −4.59

13738 16346 12773

Ch (5% of Cc )

Ch · xh2

687 817 639 2143.05

14474 0.0 13463 27937

Calculation for soil damping By =

(7 − 8ν) Wf 32 (1 − ν) γs r3x

=

(7 − 2.4) × 1177.2 32 × 19 × 0.7 × (3.156)3

= 0.404

and

0.288 0.288 Dy = √ =√ = 0.452 Bx 0.404 Cy = 2Dy Ky m = 2 × 0.452 3.94 × 105 × 120 = 6216 kN · sec /m Bφ = Dφ =

0.375(1 − υ)Jφ g γs

r5

=

0.375 × 0.7 × 125 × 9.81 19 × (2.414)5

φ

= 0.2066

0.15 0.15 = 0.274 = 1.2066 × 0.454 (1 + Bφ ) Bφ

√ Thus Cφ = 2Dφ Kφ Jφ = 2 × 0.274 × 1.373 × 106 × 125 = 7179 KN · sec/m. The equation being dynamically coupled, the damping matrix is given by ⎡

Ch ⎢0 [C] = ⎢ ⎣0 0

0 Ch e2 + Ch Xh2 0 0

0 0 Cy 0

⎤ 0 0⎥ ⎥ 0⎦ Cθ

Substituting the values calculated above we have ⎡ ⎤ 2143 0 0 0 ⎢ 0 28014 0 0 ⎥ ⎥ [C] = ⎢ ⎣ 0 0 6216 0 ⎦ 0 0 0 7179

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 379

Now performing the operation [ϕ]T [C][ϕ] and for each mode separately we have ⎡ ⎤ 476.532 0 0 0 ⎢ ⎥ 0 2.26 0 0 ⎥ [2Di ωi ] = ⎢ ⎣ ⎦ 0 0 51.84 0 0 0 0 16.2462 Thus dividing each term of the above matrix by 2ωi , we have ⎡ ⎤ 0.296 0 0 0 ⎢ 0 0.0605 0 0 ⎥ ⎥ [Di ] = ⎢ ⎣ 0 0 0.429 0 ⎦ 0 0 0 0.052 Calculation of load vector

⎧ ⎫ 238 ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ 1092.42 Here the load vector is given by {Ph } = , here Mφ = N i=1 Phi Xhi 0.0 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1511.3 N and Mθ = i=1 Phi h and h = 3.6 + 0.9 + 1.1 + 0.75 = 6.35 m, this is from the center line of shaft to the c.g of the bottom raft. ⎧ ⎫ ⎪−23.5575⎪ ⎪ ⎪ ⎨ ⎬ 21.9186 T T Performing the operation [ϕ] {P} we have, [ϕ] {P} = ⎪ −30.138 ⎪ ⎪ ⎪ ⎩ ⎭ −24.9211 Thus the four uncoupled equation of motion is given by ξ¨1 + 476.532ξ˙1 + 644590ξ1 = −23.5575 sin 534t ξ¨2 + 2.26ξ˙2 + 340ξ2 = 21.9186 sin 534t ξ¨3 + 51.849ξ˙3 + 3650ξ3 = −30.138 sin 534t ξ¨4 + 16.2462ξ˙4 + 24670ξ4 = −24.9211 sin 534t Once we know the displacement vectors in un-coupled state the displacement in the global structural co-ordinate is given by {Y} = [ϕ]{ξ } → ξi =

4 i=1

pi sin ωm t (1 − r2i )2 + (2Di ri )2

The results are tabulated as follows: Here we have calculated the response for four cases 1 2

One during normal operation For three transient case during start and stop of the machine

© 2009 Taylor & Francis Group, London, UK

Operating frequency = 534 rad/sec

Displacement

Frequency

Corrected frequency

Y φ U θ

802 18 60 157

642 22 72 188

Eigen Vectors 1st mode

Eigen Vectors 2nd mode

Eigen Vectors 3rd mode

Eigen Vectors 4th mode

Uncoupled force vector

Damping Ratio

Frequency Ratio

M.F. -

Uncoupled amplitude

Net displacement

0.443 −2.26 ×10−04 −0.0188 −0.0853

1.24 ×10−03 1.65 ×10−05 0.11268 0.01429

−1.16 ×10−03 −1.82 ×10−05 0.08882 0.01975

−1.04 ×10−04 −2.41×10−02 1.23 ×10−04 9.30 ×10−04

−23.6 21.9 −30.13 −24.9

0.30 0.06 0.43 0.05

0.83 24.15 7.37 2.83

1.72 0.00 0.02 0.14

−6.32 ×10−05 1.11 ×10−04 0.000154 −0.00014

−2.77 ×10−05 −3.48 ×10−06 −1.17 ×10−08 9.88 ×10−06

Displacement under 1st transient Transient frequency = 18.43 rad/sec

Displacement

Frequency

Eigen Vectors 1st mode

Y φ U θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314

© 2009 Taylor & Francis Group, London, UK

Eigen Vectors 2nd mode

Eigen Vectors 3rd mode

Eigen Vectors 4th mode

Uncoupled Damping vector

Ratio force

Frequency

M.F.

Uncoupled amplitude

Net displacement

1.24 1.65×10−05 0.11268 0.01429

−1.16×10−03 −1.82×10−05 0.08882 0.019745

−1.04×10−04 −2.41×10−05 1.23×10−04 9.30×10−04

−23.55 21.91 −30.13 −24.9

0.296 0.061 0.429 0.052

0.02 1.00 0.31 0.12

1.00 8.26 1.06 1.01

−3.672×10−05 0.5333 −0.00875 −0.001025

6.55×10−04 3.36×10−05 5.93×10−02 7.80×10−03

380 Dynamics of Structure and Foundation: 2. Applications

Displacement under normal operation

Displacement under 2nd transient Transient frequency = 60.41 rad/sec

Displacement

Frequency

Eigen Vectors 1st mode

Y φ ϒ θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314

Eigen Vectors 2nd mode

Eigen Vectors 3rd mode

Eigen Vectors 4th mode

Uncoupled force vector

Damping Ratio

Frequency Ratio

M.F. -

Uncoupled amplitude

Net displacement

1.24×10−03 1.65×10−05 0.112685 0.014296

−1.16×10−03 −1.82 ×10−05 0.08882 0.019745

−1.04×10−04 −2.41×10−02 1.23×10−04 9.30×10−04

−23.55 21.91 −30.13 −24.92

0.296 0.0605 0.429 0.052

0.08 3.28 1.00 0.38

1.00 0.10 1.17 1.17

−3.687×10−05 0.006617 −0.009625 −0.00118543

−3.18×10−05 2.88×10−05 −1.09×10−03 2.87 ×10−04

Displacement under 3rd transient

Displacement

Frequency

Eigen Vectors 1st mode

Y φ U θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314

© 2009 Taylor & Francis Group, London, UK

Eigen Vectors 2nd mode

Eigen Vectors 3rd mode

Eigen Vectors 4th mode

Uncoupled force vector

Damping Ratio

Frequency Ratio

M.F. -

Uncoupled amplitude

Net displacement

1.24 ×10−03 1.65 ×10−05 0.112685 0.014296

−1.16 ×10−03 −1.82×10−05 0.08882 0.019745

−1.04 ×10−04 −2.41 ×10−02 1.23×10−04 9.30 ×10−04

−23.55 21.91 −30.13 −24.92

0.296 0.0605 0.429 0.052

0.20 8.52 2.60 1.00

1.03 0.01 0.16 9.62

−3.789 ×10 − 05 0.00090156 −0.001334 −0.009722

−1.31 ×10−05 2.34 ×10−04 1.77×10−05 3.35×10−05

Analysis and design of machine foundations 381

Transient frequency = 157 rad/sec

382 Dynamics of Structure and Foundation: 2. Applications

Net amplitude of the frames The net amplitude of the frame is given by the expression yinet = yi + Ui + Xhi ϕ + hθ where h = 4.8 m and Xhi = +4.59 m, 0.0, −4.59 m respectively for Frame 1, 2 and 3. The results are as shown hereafter Operation condition

Frame 1

Frame 2

Frame 3

Normal operating case 1st Transient (18.43 rad/sec) 2nd Transient (60 rad/sec) 3rd Transient (157 rad/sec)

3.57 × 10−05 9.76 × 10−02 1.4 × 10−03 1.20 × 10−03

1.97 × 10−05 9.74 × 10−02 1.27 × 10−03 1.30 × 10−04

3.78 × 10−05 9.73 × 10−03 1.14 × 10−03 −9.45 × 10−04

The results are shown based on bar chart as given in Figure 2.18.23. Amplitude of the frames

Displacement of top deck

1.20E-01 1.00E-01 8.00E-02

Normal operating force 6.00E-02

1st transient

4.00E-02

2nd transient

2.00E-02

3rd transient

0.00E+00 -2.00E-02

Frame1

Frame2

Frame3

Frame Numbers

Figure 2.18.23 Comparison of displacement operating vs transient.

The results of natural frequencies are compared based on various methods. Frequency in vertical direction Mode number

Rausch

Barkan

Major

2D soil structure interaction

1st Mode 2nd Mode 3rd Mode

– 8318 –

– 10265 21575

409 8318 –

427 15288 22040

Frequency in horizontal direction Mode number

Rausch

Barkan

Major

2D soil structure interaction

1st Mode 2nd Mode 3rd Mode 4th Mode

– 1616 – –

– 1413 1557 –

– 1418 1562 –

176 577 1547 7667

© 2009 Taylor & Francis Group, London, UK

Analysis and design of machine foundations 383

Comparison of results with time history response Here we compare the results obtained above with time history response which we have stated earlier as the most appropriate method of analysis for such coupled soil-structure interaction analysis We show below the plots (Figures 2.18.24 and 25) for the amplitude of vibration and displacement of the frames under operating frequency of 534 radians/sec.

Amplitude based on Newmark -

method

0.00004 0.00003 Y

0.00001

0.

5

25

09

14

0.

0.

-0.00002

0.

0 75

-0.00001

0. 19 23 75 0. 28 0. 5 33 25 0. 3 0. 8 42 75 0. 47 0. 5 52 25 0. 5 0. 7 61 75

0 04

Amplitude

0.00002

-0.00003 -0.00004 Time steps in seconds

Figure 2.18.24 Amplitudes under operating condition having frequency @ 534 rad/sec.

Amplitude of frames at top deck

1.50E-04 1.00E-04 Y1 Y2 Y3

5.00E-05 0.62

0.55

0.58

0.52

0.49

0.46

0.42

0.36

0.39

0.33

0.26

0.29

0.23

0.2

0.16

0.13

0.1

-5.00E-05

0.07

0

0.00E+00 0.03

Amplitude (meter)

2.00E-04

-1.00E-04 -1.50E-04

Time steps (sec)

Figure 2.18.25 Amplitude of displacement of the top deck under operating frequency of 534 rad/sec.

Under transient load like Major we assumed the operating frequency in resonance with the natural frequency instantaneously and considering the function, sin ωm t = 1. For time history response we consider the operating frequency equal to first, second and third transient frequency respectively and find out the transient response (peak amplitude). The results are plotted graphically in Figures 2.18.26 through 31.

© 2009 Taylor & Francis Group, London, UK

384 Dynamics of Structure and Foundation: 2. Applications

Amplitude based on Newmark- method

0.006 0.004

5.85

5.4

4.95

4.5

4.05

3.6

3.15

2.7

2.25

1.8

1.35

0.9

Y

0.45

0.002 0 -0.002

0

Amplitude

0.01 0.008

U

-0.004 -0.006 -0.008 -0.01

Time steps in seconds

Y1 Y2 Y3

6

5.1 5.4 5.7

3.9 4.2 4.5 4.8

3 3.3 3.6

1.8 2.1 2.4 2.7

0.6 0.9 1.2 1.5

Amplitude of frames at top deck

5.00E-02 4.00E-02 3.00E-02 2.00E-02 1.00E-02 0.00E+00 -1.00E-02 -2.00E-02 -3.00E-02 -4.00E-02 -5.00E-02

0 0.3

Amplitude (meter)

Figure 2.18.26 Amplitudes at the ﬁrst transient frequency @ 18.43 rad/sec.

Time steps (sec)

Figure 2.18.27 Amplitude of top deck at the ﬁrst transient frequency @ 18.43 rad/sec.

Amplitude based on Newmark - method 0.0015

Y

0.0005 3

2.8

2.6

2.4

2.2

2

1.8

1.6

1.4

1.2

1

0.8

0.6

0.4

-0.0005

0.2

0 0

Amplitude

0.001

-0.001 -0.0015 Time steps in seconds

Figure 2.18.28 Amplitudes at the second transient frequency @ 60 rad/sec.

© 2009 Taylor & Francis Group, London, UK

U

Analysis and design of machine foundations 385

Amplitude of frames at top deck

2.50E-03 1.50E-03 1.00E-03

Y1 Y2 Y3

3

2.7

2.85

2.4

2.55

2.1

2.25

1.8

1.95

1.5

1.65

1.2

1.35

0.9

1.05

0.6

0.75

0.3

-5.00E-04

0.45

0

5.00E-04 0.00E+00 0.15

Amplitude (meter)

2.00E-03

-1.00E-03 -1.50E-03 -2.00E-03 -2.50E-03

Time steps (sec)

Figure 2.18.29 Amplitudes of top deck at the second transient frequency @ 60 rad/sec.

Amplitude based on Newmark - method 0.0003 0.0002 Y

Amplitude

0.0001

0.58

0.54

0.5

0.45

0.41

0.36

0.32

0.27

0.23

0.14

0.18

0.09

-0.0001

0.05

0

0

U

-0.0002 -0.0003 Time steps in seconds

Figure 2.18.30 Amplitudes at third transient frequency @ 157 rad/sec. 2.00E-03

Amplitude of frames at top deck

1.00E-03 5.00E-04 0.00E+00

0 0. 03 0. 0 0. 5 08 0. 1 0. 13 0. 1 0. 5 18 0. 2 0. 23 0. 2 0. 5 28 0. 3 0. 33 0. 3 0. 5 38 0. 4 0. 43 0. 4 0. 5 48 0. 5 0. 5 0. 3 55 0. 57 0. 6

Amplitude (meter)

1.50E-03

-5.00E-04 -1.00E-03 -1.50E-03 -2.00E-03

Time steps (sec)

Figure 2.18.31 Amplitude of top deck at third transient frequency @ 157 rad/sec.

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Y1 Y2 Y3

386 Dynamics of Structure and Foundation: 2. Applications

Discussions of the results The basic amplitudes as obtained by the two methods are shown hereafter Modal Sl. No. Displacements response

Time history

Critical frequency Remarks

1 2

Y φ

2.77 × 10−05 2.617 × 10−05 534 2.34 × 10−04 2.09 × 10−04 157

3

ϒ

1.09 × 10−04 8.33 × 10−04

60.41

4

θ

7.80 × 10−03 7.38 × 10−03

18.43

This is under operating case Operating frequency in resonance with this mode Operating frequency in resonance with this mode Operating frequency in resonance with this mode

The displacement of the top deck is as shown hereafter

Sl. No.

Displacements

Modal response

Time history

Critical frequency

Remarks

1 2 3 4

Y Y Y Y

3.57 × 10−05 1.20 × 10−03 1.4 × 10−03 9.76 × 10−02

1.6 × 10−04 1.31 × 10−03 2.21 × 10−03 4.12 × 10−02

534 157 60 18.43

This is under operating case Amplitude at third transient Amplitude at second transient Amplitude at first transient

Discussion on the results It is pretty obvious from the results that time history gives a clearer picture so far as the amplitude is concerned. While by modal technique, taking sin ωm t = 1, we calculate the maximum amplitude at various resonance points but actually at the time when the operating frequency passes this particular frequency sin ωm t = 1 but something lesser than that. Since we do not know this data as a conservative value this is usually taken as unity. Thus it is evident from the result that while by time history we get a value of deflection of top deck as 41 mm this gives about 98 mm by modal technique. While in all other cases though the order of displacement does remain same they do show discrepancy in top deck amplitude. Specially under operating condition while modal response gives amplitude of deﬂection of top deck as 3.57E-05 m time history gives an amplitude of 1.6E-04 (about 5 times). This can be attributed to the fact that while by modal response we only find the steady state part, by time history the transient part of the vibration is also considered in the response and this possibly gives an initial higher response at operating frequency. Thus for resonance check we can follow the modal response technique to find out the eigen values, however for amplitude check carrying out time history response specially to check the transient response is much more sensible.

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Analysis and design of machine foundations 387

Based on the above calculation following points can be concluded • • • •

•

Soil has significant influence on the natural frequency of the system. Major’s hypothesis of transient response is correct, and there too soil could play a signiﬁcant role (Though Major’s method of calculating the response is conservative). Increasing the thickness of bottom raft does not necessarily help. It could reduce the peak operating frequency response but will not have much effect on the transient response of the system during start and stop of the machine. Transient response can be signiﬁcant, and all the signiﬁcant transients to be checked for the frame. Though this may not structurally effect the frame (for soil transients would mostly be rigid body displacements) its effect on the stress level of the connected piping and load induced at nozzles could be significant. For critical structures amplitude should preferably be checked based on time history response to ensure that amplitudes are within acceptable limit. This should be done specially under operating frequency where modal technique usually ignores the transient part and only gives the maximum value for the steady state part.

SUGGESTED READING 1 2 3 4 5 6

Barkan, D.D. 1962, Dynamics of Bases and Foundations, McGraw-Hill Book Co. NY. Srinivasalu, P. & Vadiyanathan, C.V. 1977, Handbook of Machine Foundations, Tata Mcgraw-Hill, New Delhi. Major, A. 1980, Dynamics in Civil Engineering – Analysis and Design, Vols. I–IV, Akademia Kiado, Budaapest and Collets Holding London. Arya, S.C., O’Neill, M.W. & Pincus, G. 1979, Design of Structures and Foundations for Vibrating Machines, Gulf Publishing Co., Houston, Texas. Kameswara Rao, N.S.V. 1998, Vibration Analysis and Foundation Dynamics, Wheeler Publishing, New Delhi. Verma, C.V.J. & Lal, P.K. Ed., Treatise on the design, analysis and testing of High capacity Turbo Generator foundation, Central Board of Irrigation and Power Publication #262.

We also furnish some selected papers which we feel would further ameliorate your insight to the problem 1 2 3

4 5

Almuti, A.M. 1976, ‘Large ﬂexible Turbine foundation’, Methods of Structural Analysis, ASCE, NY, pp. 707–719. Aneja, I. 1975, ‘Dynamic Response of Systems – Turbine generators on Various Foundations’, Proceedings of the American Power conference, Vol. 37, pp. 528–540. Arya, A.S. & Drewer, R. 1997, ‘Mathematical modeling and computer simulation of elevated foundations supporting vibrating Machinery’, Transaction of IMACS, Vol. XIX, No. 4, Dec. Design Criteria for Turbine Generator Pedestal 1970, Journal of Power Division, ASCE, Vol. 96, Jan, pp. 1–22. Kasten, H.L. & Kirkland, W.D. 1970, ‘Spring mounted Turbine Generator Spins Quitely, Efﬁciently’, Electric Light and Power, E/D Edition, Nov., pp. 38–40.

© 2009 Taylor & Francis Group, London, UK

388 Dynamics of Structure and Foundation: 2. Applications 6 7 8 9 10 11

Shen, G.T. & Stone, N.E. 1975, ‘Natural frequencies of turbine foundation’, Structural Design of Nuclear Plant Facilities, Vol. II, ASCE, NY, pp. 302–330. Srinivasulu, P. & Lakshmannan, N. 1978, ‘Dynamic response of turbo-generator pedestal’, ASCE, Spring convention, Pittsburgh, Pensylvania, April, pp. 24–28. Chowdhury, I. & Som, P.K. 1993, ‘Dynamic Pile structure interaction of Boiler Feed Pump Frame Foundation’, Indian Geotechnical Conference, Vol. 1, pp. 411–414. Task Committee on Turbine Foundations 1987, Design of large steam turbine-generator foundations, ASCE, NY. Rausch, E. 1959, ‘Maschinen Fundamente und andere dynamisch beanspruchte Baukonstructionen’, VDI Verlag, Dusseldorf. Wedpathak, A.V., Pandit, V.K. Guha, S.K. 1977, ‘Soil-Foundation interaction under sinusoidal and impact type dynamic loads’, Int. Symp. on Soil-Structure Interaction, University of Roorkee, Roorkee.

© 2009 Taylor & Francis Group, London, UK

Chapter 3

Analytical and design concepts for earthquake engineering

3.1 INTRODUCTION In this chapter we will deal with some of the fundamental concepts pertaining to earthquake engineering. On completion of this chapter you should have an understanding of • • • •

Why earthquake happens in nature. Essential engineering parameters, which affect the geo-technical and structural aspect of a system under earthquake. Basic concepts of dynamic analysis as applied to Earthquake engineering pertaining to buildings, and different types of industrial and infra-structural systems like chimney, retaining wall, water tank, RCC and earth dams etc. Have an understanding of different provisions of IS 1893 (2002) code.

Before reading this chapter we however feel that you should have following background as a pre-requisite. 1 2

Basic concepts in structural and soil dynamics as furnished in Chapter 5 (Vol. 1). Also have some fundamental awareness of how earthquake can affect a structurefoundation system.

Earthquake is perhaps the most complex natural phenomenon which human being is trying to understand, combat and harness, from the early history of mankind. In spite of scientific study of the subject for the last 100 years or so, it is felt that we are still in the infancy of our knowledge on the subject. The parameters affecting this phenomenon are so large and varying and also covering different branches of science, we can at best arrive at a simplified model of the problem amenable to human perception and try to arrive at a solution which would in all probability survive this nature’s assault with some limited damage, if ever the structure faces such vagary. The basic objective of an earthquake resistant design is not to make the structure fool proof but to limit its damage to the extent of minimizing the loss of human life and property. Though earthquake is a global phenomenon, yet there are some countries in the world like USA, Japan, Turkey, India, Iran, Newzealand etc that are severely affected by earthquakes leading to signiﬁcant loss of human life and properties, while © 2009 Taylor & Francis Group, London, UK

390 Dynamics of Structure and Foundation: 2. Applications

there are others whose geological characteristics are considered seismically inert like United Kingdom, Gulf countries like Oman, Kuwait, UAE, Qatar etc. which have no signiﬁcant history of earthquakes. Based on the above, it is evident that there are countries where significant research and investigation have been carried out to develop procedures for earthquake resistant design of structures. Countries like USA, Japan, India, Mexico etc have contributed significantly on this issue.

3.1.1 Why do earthquakes happen in nature? The topic itself can be subject matter of a complete book. A detailed discussion on this is beyond the scope of this work, however as civil engineers to design structures, which can withstand such calamity- some fundamental understanding on this issue is essential. As shown in Figure 3.1.1 the earth constitute of a central core, consisting of molten magma which is undergoing continuous upheaval. While the outer core, which has solidified in million of years forms the outer earth crust. The inner magma (the molten core) is continuously creating a pressure on the outer core and trying to come out by seeking some weaknesses in earth crust. Whenever it can come out it generates what is known as a volcanic eruption. When it cannot, it tries to push the crust upward thus creating folds and faults resulting in a source which stores a significant amount of potential/strain energy. As by law of nature, all systems in course of time try to achieve minimum state of energy; these storehouses of potential energy keep on releasing their stored strain energy as kinetic energy generating waves on surface of the outer crust which is commonly known as earthquake. It is said that Himalayan mountain range is one such formation due to pressure of the inner magma. Deformation which the earth crust underwent, due to formation of the mountain range, is still being adjusted naturally.

Earth Crust

Earth Core

Figure 3.1.1 Earth with its central core.

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Molten Magma

Analytical and design concepts for earthquake engineering 391

It is for this reason, areas in its close proximity like Assam, Nepal and portions of south China is often subjected to severe earthquakes. There is also a phenomenon called seismotectonic movement otherwise known as continental drift that generates earthquake at certain location of the earth. According to this theory, the outer crust of earth is made up of undistorted plates of lithosphere. These plates are in differential motion, and at places they move away from each other where new plates are added from the interior of the earth while in places they collide with each other. All major earthquakes which mark the active zones of the earth closely follows the plate boundaries and has been found to be a function of the movements of these plates (Stevens 1980). Human interference can also sometimes modify stresses on the earth surface to trigger minor or even moderate earthquakes. In many mining areas tremors and shocks results due to underground explosion in mines, causing damages to structures on ground. One of the classic cases of man made earthquake was Koyna Dam incident in 1967 in India, when pounding of large amount of water behind the dam resulted in an earthquake causing extensive damage to surrounding (Chopra & Chakrabarti 1973).

3.1.2 Essential difference between systems subjected to earthquake and vibration from machine In Chapter 5 (Vol. 1) and Chapter 2 (Vol. 2) we had discussed in detail response of machine foundation under dynamic loading. In machine foundations, unbalanced force from the machine gets transmitted to ground via structure/foundation to the soil media. In such cases normally a limited part of the soil is affected significantly. Moreover the strain range induced in soil is usually limited to low strain range (usually 10−3 %). However in case of earthquake the phenomenon is quite different. In this case when an earthquake shock is generated due to rupture of a fault within the earth surface it generates waves within soil that induces a much larger strain (10−2 to 10−1 %) for a major earthquake. Shown in Figure 3.1.2, is a typical propagation of waves through the soil medium and is usually a combination of four types of waves namely, 1 2 3 4

P-waves (body waves) S-waves (body waves) Rayleigh Waves (surface waves) Love waves (surface waves)

The primary or P-waves are the fastest traveling of all waves and generally produce longitudinal compression and extension within a soil medium. This wave can travel both through soil and water and is the first one to arrive at a site. However soil being relatively more resistant to compression and dilation, effect of its impact on ground distortion is minimal. The S-waves, also otherwise known as secondary or shear waves usually cause shear deformation in the medium through which they propagate. The S-waves can usually © 2009 Taylor & Francis Group, London, UK

392 Dynamics of Structure and Foundation: 2. Applications

Time History response Velocity

0.1

19 . 4

18 . 4

17 . 3

16 . 2

15 . 1

14

13

11 . 9

10 . 8

9 .7 2

8 .6 4

7.56

6.48

4.32 5.4

3.42

2.16

-0.1

1.08

0 0

Velocity v(m/sec)

0.2

-0.2 -0.3

Time steps

Figure 3.1.2 Typical propagation of earthquake waves through surface.

propagate through soil only1 . It travels at a much slower speed through the ground than primary waves and soil being weak in resisting shear deformation; it is found to cause maximum damage to ground surface. Rayleigh waves are surface waves which are found to produce ripples on surface of the ground2 . These waves produce both horizontal and vertical movement of earth surface as the waves travel away from the source. Love waves are similar to S-waves and produces transverse shear deformation to the ground. These entire waves combine together to produce shock waves from which an engineer extracts value of the maximum ground acceleration (amax ) which is the major parameter that governs his design. Based on above it is apparent that mechanics of earthquake is opposite to dynamics of machine foundation in the sense that here forces are transmitted from soil to the structure. It is the shock within the ground which excites the structure and induces inertial force in the system.

3.1.3 Some history of major earthquakes around the world A number of major earthquakes have been recorded that resulted in massive losses of human lives and destructions of thousands of buildings and structures. Some of them are cited in Table 3.1.1.

1 Since liquid have no shear resistance it cannot travel through water. 2 This is very much similar to ripples produced by a pebble dropped in a pond.

© 2009 Taylor & Francis Group, London, UK

Table 3.1.1 Various Earthquakes & their casualities. Name, Location & Year

Casualty

Calcutta, 1737

Destroyed 300 000 lives.

Portugal, Spain & Northern Morocco, subjected to three strong shocks in the afternoon of Nov. 1, 1775

Devastated Lisbon, loss of life was heavy. The disaster was more detrimental since the first shock was followed by a colossal whirling wall of water sweeping everybody and everything it met on its path.

The Alma-Ata earthquake, 1910

Demonstrated continuous vibrations, that lasted for 5 minutes.

Tokyo and Yokohama, Sept. 1, 1923 11 000 buildings were ruined, 59 000 houses burned in Yokohama. Entire area of Tokyo was affected. Death toll was 100 000, while 43 000 missing. 300 000 houses were damaged. Himalayan earthquake, 1950

One of the severest events recorded instrumentally. Equivalent to the explosion of 100 000 A-bombs.

Mongolian earthquake, December 4, 1956

Vast devastation, A mountain peak was split into two parts. Part of mountain, 400 m high, collapsed and fell down a precipice. A depression, up to 18 km in length and 800 m in width, originated. Broad fissures, up to 20 m in width appeared on the ground surface. One of these broke for length of 250 km. The intensity of the earthquake approached a force 11.

Alaskan earthquake, 1964

The most severe of all known seismic events. Intensity was over 11.

Chile, 1960

The most violent earthquake of the 20th century. It affected an area over 200 000 km2 and caused numerous landslides.

Latur and Osmanabad, September 1993

Magnitude of earthquake was 6.3. Total of 7601 people lost their lives in Latur and Osmanabad. Number of houses destroyed in the earthquake was about 30 000.

Bhuj earthquake, 26 January 2001

Some historical structures that survived the 1819 (M = 7.7) earthquake have been destroyed in the 2001 earthquake. The death toll was 19 727 and the number of injured 166 000. Indications are that 600 000 people were left homeless, with 348 000 houses destroyed and an additional 844 000 damaged. Magnitude was 7.6. Liquefaction – On 4 February liquefaction phenomenon were reported by hydrologists and by local villagers, with an indication that the flow was sufficient in some cases to activate desert rivers that have been dry for more than a century. Widespread liquefaction was confirmed by SPOT imagery and by field observation (5 Feb.). Many mudvolcanoes in the Rann of Kachchh have dimensions of hundreds of meters: one covers a 5 km diameter stretch of the southern Rann with dark sand and mud. Numerous ancient river channels have been illuminated by a pock mark pattern of sand vents, and some have clearly flowed, and breached their old channels.

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394 Dynamics of Structure and Foundation: 2. Applications

3.1.4 Intensity The severity of shaking of an earthquake as felt through damage is described as intensity at a certain place on an arbitrary scale. One such scale is Modiﬁed Mercalli Scale (MMS). This is shown in Table 3.1.2.

Table 3.1.2 Modified Mercalli Intensity Scale. Class of earthquakes

Description

I

Not felt except by a few under specially favourable circumstances.

II

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

III

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 heavy truck 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 pop.

VI

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

VII

Everybody runs outdoors, damage negligible 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 in specially designed structures; considerable in ordinary substantial buildings with partial collapse; very heavy in poorly built structures; panel walls thrown out of framed structures; 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; damage very heavy in substantial buildings with partial collapse; buildings shifted off foundations; ground cracked conspicuously; and underground pipes broken.

X

Some well built wooden structures 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 rails bent greatly.

XII

Total damage; waves seen on ground surface; lines of sight and level distorted; and objects thrown upward into the air.

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Analytical and design concepts for earthquake engineering 395

3.1.5 Effect of earthquake on soil-foundation system Having explained the primary source of disturbance is in the soil itself, it is important to assess and know what could be the effects of an earthquake on the soil on which a structure is built. For it should be understood that irrespective of how well an earthquake resistant design is carried out for a structure if the ground supporting it fails, the structure will invariably undergo significant damage and which at times could even be catastrophic3 . The major effect on soil affected by an earthquake can be classified as follows: 1 2 3 4 5

Liquefaction of soil Settlement of foundation due to deep seated liquefaction failure Reduction of bearing capacity Ground Subsidence Land Slides

Of all the phenomena defined above, liquefaction is perhaps the most important factor that has caused major damage in many previous earthquakes, and unfortunately gets very little attention from structural engineers in a design office4 . Thus it is important to understand what the phenomenon is and what are the methods available to assess and mitigate it?

3.1.6 Liquefaction analysis 3.1.6.1

What is liquefaction?

Conceptually speaking liquefaction is very much akin to giving a rapid squeeze to a sponge ball saturated with water. When the squeeze is applied, we observe that water stored inside the sponge comes out and the sponge feels lighter as the water comes out. For soil sample (especially when it is cohesionless) shear strength is given by the expression s = (σ − u) tan ϕ

(3.1.1)

where, s = shear strength of the soil; σ = overburden pressure of the soil sample; u = in-situ pore pressure within the soil sample, and φ = angle of internal friction of the soil sample. When earthquake force acts on the soil sample it produces a rapid shock or a squeeze on the soil body, by virtue of which there is a sudden increase in pore pressure. But unlike the sponge ball the pore pressure cannot dissipate readily. When force due to earthquake is significantly high (M ≥ 6.5) which also results in ground shaking for a good amount of time the pore pressure increment becomes sufficiently high such that it equals the overburden pressure and the soil looses its shear

3 Nigaata Earthquake (1964) in Japan was one of the primary example where a number of structures underwent significant damages due to ground subsidence and liquefaction of soil. 4 Especially in India where in previous earthquakes a significant damage has been recorded due to this phenomenon.

© 2009 Taylor & Francis Group, London, UK

396 Dynamics of Structure and Foundation: 2. Applications

strength altogether (i.e. s = 0) and starts flowing like a liquid. This phenomenon is otherwise known as liquefaction of soil. When such phenomenon is observed during an earthquake soil collapses completely and sand boils are observed in the ground. Even c-φ soils losses signiﬁcant part of its strength resulting in bearing capacity failures of foundation and or signiﬁcant settlement. Liquefaction of soil has been observed in a number of earthquakes throughout the world like Nigaata in Japan (1964), Kobe in Japan (1995), Dhubri and Koyna (1967) earthquakes in India. From the above discussion it is obvious that non-plastic cohesionless soils under saturated condition are most susceptible to liquefaction during an earthquake. As SPT value has been extensively used to define the static engineering strength of cohesionless soil consistently it was but natural that researchers tried to co-relate SPT values of cohesion less sandy soil to liquefaction potential of soil samples due to earthquake shocks. Pioneering research in this area was done by Seed et al. (1984) who correlated the observed SPT values to cyclic resistance ratio which is one of the major parameters used to define the liquefaction potential of a soil sample. We will talk more about this later; first let us see how liquefaction is measured for a particular soil sample. The susceptibility of a soil sample undergoing liquefaction is measured by a term called liquefaction potential, which is measured as a Factor of Safety (FS) against Cyclic Resistance Ratio (CRR) to Cyclic Stress Ratio (CSR). It is defined as FS =

CRR ≥ 1.0 CSR

(3.1.2)

In other words (based on Equation (3.1.2)), if the factor of safety is less than or equal to 1.0, the soil has very good possibility of undergoing liquefaction under an earthquake, however if the value is greater than 1.0, the possibility of soil failure due to liquefaction is remote. Thus it is obvious that we need to first understand what does CSR and CRR stand for. During earthquake soil under the influence of an earthquake will be subjected to repetitive shear stress (known as cyclic shear stress) and it is estimated by the expression CSR =

τav amax σv r = 0.65 σv g σv d

(3.1.3)

where, amax = maximum acceleration at the ground surface; σv = total overburden pressure at the design depth; σv = effective overburden pressure at the design depth; g = acceleration due to gravity, and, rd = stress reduction factor which varies with depth and is given by rd = 1.0 − 0.000765z

for z ≤ 9.15 m

(3.1.4a)

rd = 1.174 − 0.0267z

for 9.15 m ≤ z ≤ 23 m

(3.1.4b)

rd = 0.744 − 0.008z rd = 0.5

for 23 m ≤ z ≤ 30 m

for z ≥ 30 m

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(3.1.4c) (3.1.4d)

Analytical and design concepts for earthquake engineering 397

For ease of electronic computation rd may also be expressed by the expression (Blake et al. 2002)5 rd =

1.000 − 0.4113z0.5 + 0.04052z + 0.001753z1.5

1.000 − 0.4177z0.5 + 0.5729z − 0.006205z1.5 + 0.001210z2

(3.1.5)

The maximum acceleration of the ground (amax ) is another factor, which needs careful evaluation. For practical design office purpose one of the expressions used to evaluate amax is amax = 0.184 × 100.320M (D)−0.8 g

(3.1.6)

where amax = maximum ground acceleration; M = expected moment magnitude of earthquake, D = maximum epicenter distance in km, and g = acceleration due to gravity @ 9.81 m/sec2 . It may be noted that if more reliable observed earthquake data is available for the site (predicting ground acceleration more accurately) it may well be used in lieu of the above formula. Having calculated the cyclic stress ratio based on the above expressions it is essential to evaluate the cyclic resistance ratio (CRR) of the in-situ soil. It is evident that the CRR value of the soil sample will depend on its in-situ strength. Since Laboratory testing can be carried out under a better controlled environment, one of the plausible methods which have been tried is to collect in-situ undisturbed soil sample for evaluation of the parameter CRR in the laboratory. However, one of the major difﬁculties encountered in this case is that generally the in-situ stress state cannot be established in the laboratory, and specimens of granular soil retrieved with typical drilling techniques are far too disturbed to yield any meaningful results. Only through very specialized sampling techniques such as ground freezing, sufﬁciently undisturbed sample can be obtained, which again becomes prohibitively expensive for all but most critical projects. It is for this reason, co-relating the CRR value with field observed test data is still the state of the art practice. 3.1.6.2

Co-Relation between CRR and SPT value

For calculation of CRR based on observed SPT value (No ), as a first step, the observed SPT value is subjected to certain corrections is as expressed by (N1 )60 = No (CN )(CE )(CB )(CR )(CS )

(3.1.7)

in which, No = measured SPT value at the site; CN = a correction factor for overburden pressure; CE = a correction factor for hammer energy ratio; CB = a correction 5 This formula was proposed as guidelines for analyzing and mitigating landslide hazards in California, Southern California Earthquake Center, Univ. of Southern California, Los Angeles.

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398 Dynamics of Structure and Foundation: 2. Applications Table 3.1.3 Correction factors to observed SPT values. Factor

Equipment parameter

Term

Overburden pressure Energy ratio

Independent of Equipment Safety Hammer Doughnut Hammer 3 to 4 m 4 to 6 m 6 to 10 m 10 to 30 m >30 m 65 to 115 mm 150 mm 200 mm Standard Sampler Sampler without Liners

CN CE

Rod length

Bore Hole Diameter Sampling Method

CR

CB CS

Correction factor Pa σ v

0.6 to 1.17 0.45 to 1.0 0.75 0.85 0.95 1.0 >1.0 1.0 1.05 1.15 1.0 1.2

In the table, Pa = atmospheric pressure or 100 kPa (100 kN/m2 ); σv = effective overburden pressure at depth of the standard penetration sample.

factor for borehole diameter; CR = a correction factor for rod length; CS = a correction factor for sampler with or without liners, and, (N1 )60 = corrected SPT value with 60% hammer efﬁciency. The correction factors for various equipment parameters are as shown in Table 3.1.3. Having established the design SPT value (N1 )60 the cyclic resistance ratio (CRR) is given by the expression for clean sands (i.e. <5% contents) as

CRR =

a + by + cy2 + dy3 1 + ey + fy2 + gy3 + hy4

(3.1.8)

where, a = 0.048, b = 0.004721, c = 0.0006136, d = −1.673 × 10−5 , e = −0.1248, f = 0.009578, g = −0.0003285, h = 3.714 × 10−6 , and y = (N1 )60 . Equation (3.1.8) is valid for (N1 )60 less than 30. For clean granular soil having N > 30 are far too dense to liquefy and are generally classed as non-liquefiable. Another expression, which is used for clean sand base for computation of CRR is6

CRR7.5 =

1 (N1 )60 1 50 + − + 34 − (N1 )60 135 200 [10 × (N1 )60 + 45]2

where, CRR7.5 = the cyclic resistance ratio at earthquake magnitude of 7.5.

6 After Alan. F. Rauch at the University of Texas, 1998.

© 2009 Taylor & Francis Group, London, UK

(3.1.9)

Analytical and design concepts for earthquake engineering 399

3.1.6.3

Inf luence of f ine contents on CRR value

While developing the original expression Seed et al. (1984) noted an apparent increase of CRR value with an increase in fine contents. Whether this can be attributed to an increase in resistance or decrease in penetration resistance is not clear. However to cater to this, correction has been recommended to SPT values for the influence of fine contents. Other grain characteristics like Plasticity index (PI) may also affect the liquefaction resistance as well, however is not so well defined till date. Hence, corrections based solely on fine contents are used and should be mellowed with judgment and caution. Seed et al. (1983) proposed corrections of (N1 )60 to an equivalent clean sand value (N1 )60CS given by (N1 )60CS = α + β(N1 )60

(3.1.10)

where α and β are determined from the following relationships as shown in Table 3.1.4. Table 3.1.4 Modification factor to SPT value based on fine contents. Sl. No.

Values of α and β

Fine content

1

α = 0

For FC ≤ 5%

α α β β β

2 3 4 5 6

1.76−

190 FC2

=e = 5.0 = 1.0 1.5

= 0.99 + FC 1000 = 1.2

5% ≤ FC ≤ 35% FC ≥ 35% For FC ≤ 5% 5% ≤ FC ≤ 35% FC ≥ 35%

The above equations can now be used for routine liquefaction resistance calculation for soil subjected to SPT at field. 3.1.6.4

Effect of earthquake magnitude on liquefaction resistance

The original study of the liquefaction potential was based on an earthquake magnitude of 7.5. To evaluate the potential of earthquake at other magnitudes, correction factors were proposed that allows induced stress ratios for other magnitudes be adjusted to a magnitude of 7.5 by dividing the stress ratios by the factors as shown in Table 3.1.5. The magnitude scaling factor (MSF) as proposed in Table 3.1.5 – based on recent research is now believed to be very conservative for moderate size earthquake. A new set of MSF has now been proposed by Idriss where the MSF is defined as function of Moment Magnitude and is given by MSF =

102.24 M2.56

© 2009 Taylor & Francis Group, London, UK

(3.1.11)

400 Dynamics of Structure and Foundation: 2. Applications Table 3.1.5 Magnitude scaling factor as proposed by Seed and Idriss (1970). Sl. No.

Earthquake magnitude

Magnitude scaling factor

1 2 3 4 5

5.25 6 6.75 7.5 8.5

1.5 1.32 1.13 1.0 0.89

Table 3.1.6 Magnitude scaling factor as proposed by various investigators**.

Sl. No.

Magnitude

Seed and Idriss (original) (1970)

1 2 3 4 5 6 7

5.5 6 6.5 7 7.5 8 8.5

1.43 1.32 1.19 1.08 1 0.94 0.89

Idriss (1999) 2.2 1.76 1.44 1.19 1 0.84 0.72

Arango 3 2 1.6 1.25 1 0.75

2.2 1.65 1.4 1.1 1 0.85

Ambreseys (1995)

Andrus & Stokoe

2.86 2.2 1.69 1.3 1 0.67 0.44

2.8 2.1 1.6 1.25 1 0.8 0.65

** American Practice.

We furnish in Table 3.1.6, the data furnished by other researchers on the MSF value varying with earthquake magnitude. The factor of safety against Liquefaction can now be expressed as FS =

CRR7.5 CSR

MSF

(3.1.12)

where CRR7.5 = Cyclic resistance ratio for an earthquake magnitude 7.5. Whatever has been discussed previously will now be further clarified by a suitable problem, which covers the whole gamut of the above conditions.

Example 3.1.1 As shown in Figure 3.1.3, is a site soil profile which consists of 3.0 m of silty clay underlain by 6 m of sand whose average SPT value is 13. The ground water table is observed to be at a level of 1.0 meter below ground level. The dry density of the silty clay is 18 kN/m3 , while that in saturated condition is 20 KN/m3 . The saturated density of sand is 19.6 kN/m3 . Sieve analysis shows the sand to have Fines content as 15%. Find the liquefaction potential when the site is considered to be 150 km away from the epicentre having an earthquake moment magnitude of 6.5? The SPT test was carried out by standard sampler with safety hammer & having rod length of 6.0 m. The diameter of the bore hole was 150 mm. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 401

Density of soil =18/m3

1.0m GWL

2.0 m

Density of soil =20kN/m3 Average SPT Value=13 Saturated Density of soil=19.6 kN/m3

6.0 m

Figure 3.1.3 Soil Profile of a site with typical soil properties.

Solution: Considering amax = 0.184 × 100.320M (D)−0.8 g. Here M = 6.5 and D = 150 km which gives, amax = 0.184 × 100.320 × 6.5 (150)−0.8 g = 0.4017 g. Effective vertical stress at center of the sand layer is, σv = 18 × 1.0 + 10 × 2 + 9.6 × 3 = 66.8 kN/m2 . The gross vertical pressure at center of the sand layer, σv = 18 × 1.0+20 × 2+ 19.6 × 3 = 116.8 kN/m2 . The depth below ground, where liquefaction potential is calculated is 1 + 2 + 3 = 6 m. Thus z = 6.0 m < 9.15 m, which gives, rd = 1.0 − 0.000765z → rd = 1.0 − 0.000765 × 6 = 0.9954. amax σv r , we have, CSR = 0.65 Considering, CSR = 0.65 g σ v d 0.4017g 116.8 × 0.9954 = 0.4544 g 66.8 The corrected SPT value is given by (N1 )60 = No (CN )(CE )(CB )(CR )(CS )

100 Here, No = 13, CN = , CE = 1.0, CB = 0.85, CR = 1.05, CS = 1.0 66.8 100 Thus (N1 )60 = 13 × × 1.0 × 0.85 × 1.05 × 1.0 = 14.2 66.8 For FC = 15% we have

α=e

1.76−

190 FC 2

=e

© 2009 Taylor & Francis Group, London, UK

1.76−

190 152

= 2.498;

and

402 Dynamics of Structure and Foundation: 2. Applications

β = 0.99 +

FC1.5 1000

= 0.99 +

151.5 1000

= 1.048.

Thus corrected SPT value is given by (N1 )60CS = α + β(N1 )60 i.e. (N1 )60CS = 2.498 + 1.048 × 14.2 = 17.38 ≡ 17 (say) 1 (N1 )60 50 1 Considering CRR7.5 = + + − 34 − (N1 )60 135 [10 × (N1 )60 + 45]2 200 1 17 50 1 We have, CRR7.5 = + + − = 0.1808 34 − 17 135 [10 × 17 + 45]2 200 The Magnitude scaling factor is given by MSF =

102.24 102.24 = = 1.44; M2.56 6.52.56

Thus, FS =

CRR7.5 CSR

MSF =

0.1808 0.4544

× 1.44 = 0.572 < 1.0.

Hence, as the factor of safety being less than 1.0, the soil has a high chance of liquefaction during the earthquake.

3.1.6.5 Correlation between CRR and CPT value Other than SPT, cone penetration test (CPT) is also used in field for evaluation of geotechnical engineering parameters. As such investigators have also tried to co-relate the CPT value with CRR for evaluation of liquefaction potential. One of the advantages with CPT being that since it is a continuous process; thin layers of soil that one can miss by SPT will not be missed in this case. As stated earlier, Equation (3.1.3) is used to determine the CSR value. The CRR value is indirectly co-related to CPT by developing relationship between CPT and SPT value. As per Seed and Idriss qc = 4 to 5 N

for clean sand, and

qc = 3.5 to 4.5 N

for silty sand

(3.1.13)

where, qc = the observed CPT value in MPa. Once an equivalent SPT value is obtained from the observed qc , rest of the procedure remains same as stated earlier. Murthy et al. (1991) has given the relationship which can also be used to obtain equivalent SPT values from the observed cone penetration values. This is given in Table 3.1.7. Schmertmann (1978) presented a relationship between SPT and CPT values for various types of soil [Table 3.1.8] which are also extensively used in the design offices to determine equivalent SPT values from the observed CPT values. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 403 Table 3.1.7 Relationship between relative density of fine sand, SPT, cone resistance and angle of friction. State of sand

Dr

N

qc (MPa)

φ

Very loose Loose Medium dense Dense Very dense

<0.2 0.2–0.4 0.4–0.6 0.6–0.8 0.8–1.0

<4 4–10 10–30 30–50 >50

<2.0 2–4 4–12 12–20 >20

<30 30–35 35–40 40–45 >45

Table 3.1.8 Relationship between SPT, CPT values for different types of soil after Schmertmann (1978). Type of soil

qc /N

Sand and gravel mixture Sand Sandy silt Clay-silt sand mixture Insensitive clay

6 4 3 2 1.5

3.1.6.6

Liquefaction of clay

Normally clay is deemed non liquefiable. However based on experience of earthquake in China it is now established that there are certain types of clay, which under shaking do undergo liquefaction. As a rule of thumb, a clay sample will be deemed liquefiable provided all of the following criteria as mentioned below are complied with, • • •

Weight of soil particles finer than 0.005 mm is less than 15% of the dry weight of the soil. The liquid limit (LL) of the soil is less than 35%. The moisture content of the soil is less than 0.9 times the liquid limit of soil.

Clayey soil meeting not all of the above criteria are usually considered non liqueﬁable. 3.1.6.7

Settlement of foundation due to liquefaction failure

We had stated in our earlier section of liquefaction that during an earthquake, due to shock there is a sudden increase in pore pressure that cannot dissipate immediately resulting in lose of shear strength of soil. However, in course of time, this pore pressure dissipates away towards the surface resulting in volumetric deformation of the ground. Considering the above phenomenon and heterogeneous nature of soil, the soil may undergo differential settlement which could be critical for building foundations and underground lifelines. © 2009 Taylor & Francis Group, London, UK

404 Dynamics of Structure and Foundation: 2. Applications

A technique to estimate the ground settlement has been proposed by Ishihara and Yoshimine (1992) wherein they developed a chart based on which the post liquefaction volumetric strain is co-related to the FS value (CRR/CSR) and the SPT value as shown in Fig. 3.1.4.

Figure 3.1.4 Curves for volumetric strain versus FS after Yoshimine (1992).

Based on above, once we know FS and SPT value, the volumetric strain is read off from the curve and the settlement is obtained by multiplying this strain with the depth of the soil. The above is now further elaborated by a problem shown below. Example 3.1.2 For the soil sample as described in Example 3.1.1 estimate the settlement of the sandy layer considering all other boundary conditions remaining identical. Solution: From previous example we have seen FS =

0.1808 0.4544

× 1.44 = 0.572 < 1.0,

which shows that the soil can undergo liquefaction. We has also seen that the corrected SPT value of the soil is N = 17.38 say 17. Referring to Ishihara & Yoshimine’s chart we find volumetric strain = 2.0%. 2.0 Thus settlement of the sand layer of 6 m is = 100 × 6000 = 120 mm.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 405

3.1.6.8

Reduction of bearing capacity of soil

Normally it is believed that earthquake has marginal effect on bearing capacity of soil. As a matter of fact it is often a common practice and advised in many codes to increase the allowable bearing capacity by 25%. The reason for this can be attributed to the fact that normally when we find bearing capacity of soil, we find out ultimate bearing capacity of soil. Dividing it by a factor of safety we arrive at the allowable bearing capacity of soil. This is mostly as per the general shear failure theory of soil, where Terzaghi, Meyerhof or Brinch Hansen’s formula is used. However in many cases and especially for cohesive soil, it is the settlement – that governs the design bearing capacity of soil. Thus during an earthquake which is considered once in a lifetime phenomenon on the structure, a lowering of the factor of safety on the bearing capacity is usually deemed acceptable, and hence allowable bearing capacity is increased. However it should be made clear that such increment is valid for a case when the foundation is resting on • • •

Crystalline rocks having no horizontal fragments or laminations; Dense compacted sand having SPT value >30; Stiff to very stiff clay with nominal plastic flow.

If the soil is otherwise made of fragmented rock, loose sand or soft plastic clay sensitive to vibration this increased bearing capacity value should not be used7 . In such cases there could be significant reduction in strength when the foundation can undergo either a local shear failure (when the foundation punches through overlying soil due to liquefaction of bottom layer) or undergo a general shear failure when there is a significant change in soil property for which bearing capacity factors Nc , Nq , and Nγ undergo reduction resulting in a reduced bearing capacity. 3.1.6.9

Punching shear failure of soil

To understand how local shear failure can occur, let us consider the soil profile as shown in Figure 3.1.5. Let us consider the case of a foundation resting on the top layer of shallow clayey soil which is non liquefiable, underlain by a layer of loose sand susceptible to liquefaction. It is apparent from the figure that depth of the layer below the footing to the top of liqueﬁable sand layer is quite less and it might so happen that if bottom layer looses its strength and the foundation is subjected to heavy load from superstructure, the foundation may punch through this thin layer of soil and collapse, causing serious damage to the super-structure. Similar to a column punching through a RCC footing here the whole foundation punches through the soil along the vertical dotted line to collapse.

7 Unfortunately many design engineers hardly give consideration to this and believes that this increase of bearing capacity of foundation almost a sacrosanct issue.

© 2009 Taylor & Francis Group, London, UK

406 Dynamics of Structure and Foundation: 2. Applications

P

Figure 3.1.5 Soil Profile of a site with foundation resting on top layer on non-liquefiable soil.

To prevent this happening we calculate a factor of safety (FS) expressed as FS = FS =

2(B + L)Zτ f P 2Zτ f P

for isolated footing, and,

for strip footing.

(3.1.14)

where, B = width of foundation in meter; L = length of foundation in meter; Z = depth of soil layer from bottom of footing to the top of liquefiable soil, and τf = shear strength of un-liquefiable layer of soil in kN/m2 . If the top layer of non liquefiable soil is cohesive in nature (clay) then the shear strength is given by τf = Su ,

where Su = undrained shear strength of the soil.

(3.1.15a)

For c − ϕ soil (undrained shear strength parameters) the shear strength is given by τf = c + σh tan ϕ

(3.1.15b)

where σh = horizontal total stress in kN/m2 ; for cohesive soil this is often assumed as 0.5σv . For a non-liquefiable soil layer of cohesionless soil, the shear strength is given by τf = σh tan ϕ = k0 σv tan ϕ σh

(3.1.15c)

where = effective horizontal stress in kN/m2 and is equal to the coefficient of passive pressure at rest times the vertical effective stress σv , and, φ = effective angle of friction of the cohesionless soil. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 407

We now show the application of the above based on a suitable problem as shown in Example 3.1.3.

Example 3.1.3 Shown in Figure 3.1.5, is a footing of size 3 m × 2 m placed on a stiff clayey–silt layer of undrained shear strength Su = 50 kN/m2 and φ = 10◦ . The footing has maximum load of 650 kN on it (including its own weight). The clay layer (3.0 m deep) is underlain by a layer of loose sand 9.0 meter deep which is susceptible to liquefaction. Find the factor of safety of the foundation under punching shear failure. The foundation is resting at depth of 1.5 m below ground level. Unit weight of soil of the top layer is 20 kN/m3 . Solution: As per the problem Z = 3.0 − 1.5 = 1.5 m; σv = 20 × 1.5 = 30 kN/m2 . Thus, σh = 0.5 × 30 = 15 kN/m2 and τf = 50 + 15 tan 10 = 52.64 kN/m2 . The resistive force = 2(B + L) Zτf = 2(3 + 2) × 1.5 × 52.64 = 789.6 kN. And, FS = 789.6/650 = 1.214. Considering the uncertainty in soil, FS = 1.2 could be a low value.

3.1.6.10

General shear failure capacity reduction due to liquefaction

This phenomenon is generally observed in case of the soil supporting the foundation is a stiff clay layer underlain by sandy layer susceptible to liquefaction. The ultimate bearing capacity of foundation based on general shear failure theory is given by Terzaghi’s equation as qult = cNc + qNq +

1 γs BNγ 2

(3.1.16)

The first term cNc gives the strength of the soil due to its cohesive property. The second term depicts the effect of overburden soil which goes on to increase the bearing capacity of the soil and the last term 12 γs BNγ gives the frictional strength of the soil where the term Nγ is a function of the friction angle φ. For clayey soil, as φ = 0, it gives Nγ = 0 and Nq = 1; For spread footing, considering the aspect ratio (B/L) correction, we have

qult qult

B = cNc 1 + 0.3 + γ Df , further modified to L B + γ Df . = Su Nc 1 + 0.3 L

© 2009 Taylor & Francis Group, London, UK

(3.1.17)

408 Dynamics of Structure and Foundation: 2. Applications

For shallow foundation near the ground as the second term has minimal effect, for all practical purpose we can consider the equation to be B qult = Su Nc 1 + 0.3 L

(3.1.18)

For the bottom layer of liquefiable soil there is obviously a reduction in value of Nc and this is usually function of the ratio of Z/B as given in Table 3.1.9. Table 3.1.9 Reduction in value of Nc for Z/B ratio. Z/B

Nc

0 0.25 0.5 1.0 1.5 ∞

0 0.7 1.3 2.5 3.8 5.5

where B = width of the foundation; Z = height of soil from bottom of foundation to the top of liquefiable soil.

Example 3.1.4 For the example problem cited in Example 3.1.3, find the reduced bearing capacity of the foundation considering the top layer of soil as stiff clay of undrained shear strength of 50 kN/m2 . All other parameters remain the same as the earlier problem. Solution: Under unliquefied state the ultimate bearing capacity is given by B qult = Su Nc 1 + 0.3 + γ Df L For φ = 0 Nc = 5.5, qult = 50 × 5.5 1 + 0.3 23 + 20 × 1.5 = 357.3 kN/m2 . Considering foundation size as 2 m × 3 m we have, Qult = 357.3 × 2 × 3 = 2143.5 kN ➔

FS =

2143.5 = 3.3 650

When the bottom soil is liqueﬁed considering, Z/B = 1.5/2.0 = 0.75, referring to Table 3.1.8, reduced Nc value = 1.9. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 409

Thus, qult = 50 × 1.9 1 + 0.3 × 23 + 20 × 1.5 = 144 kN/m2 ➔ Qult = 144 × 2 × 3 = 864 kN. Thus, FS = 864 650 = 1.3, which is low and should preferably be about 1.5.

3.1.6.11

Ground subsidence due to earthquake

During an earthquake of major magnitude there are many cases of ground subsidence and land slides which have wrecked havoc on many structures and especially underground services which may get severely damaged due to this. In the San Francisco Bay Earthquake (1906), the major source of damage was the fire which broke out as an aftermath of the earthquake and could not be contained as most of the underground water pipe lines were severely damaged due to ground subsidence and became non-functional. The major reason for this subsidence is again deep seated liquefaction for which the soil starts to flow and due to differential or non uniform flow can split apart a structure built on it. Roads and pavements were observed to undergo extensive damage due to subsidence and similar was a major observation in Chi–Chi earthquake in Taiwan (1999). When the slope of the ground is less or equal to 6% the flow of soil is generally defined as a lateral displacement of soil. When this slope is more than 6% the same is know as a land slide. A number of researches have been carried out to develop a mathematical model, which would effectively predict the subsidence of the ground during a major earthquake. However, parametric functions being so many in numbers and uncertain that there is yet a model which can be stated as unconditionally applicable. The most used mathematical model for practical engineering purpose is one empirical model by Bartlet & Youd (1992) developed based on historical data collected from the six earthquakes in USA and two in Japan. They proposed two expressions one for sites near steep banks with a free face, the other with sites having gently sloped terrain. For free faced condition log DH = −16.3658 + 1.1782M − 0.9275 log R − 0.0133R + 0.6572 log W + 0.3483 log T15 + 4.5270 log(100 − F15 ) − 0.9224D5015

(3.1.19)

For sloped terrain condition log DH = −15.7870 + 1.1782M − 0.9275 log R − 0.0133R + 0.4293 log S + 0.3483 log T15 + 4.5270 log(100 − F15 ) − 0.9224D5015

(3.1.20)

in which, DH = estimated average ground displacement in meters; D5015 = average mean grain size of the liqueﬁable layers included in T15 in mm; M = moment © 2009 Taylor & Francis Group, London, UK

410 Dynamics of Structure and Foundation: 2. Applications

magnitude of the earthquake; R = epicentral distance in kM; F15 = average fine content (passing ASTM 200 sieve) for the liqueﬁable layer in% included in T15 ; T15 = the cumulative thickness (in meter) of the saturated granular layer having blow count <15; S = ground slope in percent, and, W = ratio of height (H) of the free face to the distance (L) from the base of the free pace to point in question percent. Example 3.1.5 Shown in Figure 3.1.6 is a site soil proﬁle which consists of 3.0 m of clay underlain by 6 m of sand whose average SPT value is 13 which is susceptible to earthquakes. The site consists of a canal ﬂowing across as shown in the figure shown below. The unit weight of the clay is 20 kN/m3 . The saturated unit weight of sand is 19.6 kN/m3 . Sieve analysis shows the sand to have fines content as 15%. The average grain size diameter of the sand layer is 0.032. A power house is to be in built on this site located at distance of 30 meter from the canal bank. The site is considered to be 50 km away from the epicentre having an earthquake Moment magnitude of 6.75. Find the estimated movement of soil with this free face condition.

30m

Unit weight of soil =20kN/m

3

3.0m

Average SPT Value=13 Saturated unit weight of soil=19.6 kN/m3

6.0m

Figure 3.1.6 Soil Profile of a site with typical soil properties.

Solution: Here, R = 50 km; M = 6.75; W = H/L = 3/30 = 0.1 = 10%; T = 6 m, and, D50 = 0.32. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 411

Considering, log DH = −16.3658 + 1.1782M − 0.9275 log R − 0.0133R + 0.6572 log W + 0.3483 log T15 + 4.5270 log(100 − F15 ) − 0.9224D5015 ,

we have

log DH = −16.3658 + 1.1782 × 6.75 − 0.9275 log 50 − 0.0133 × 50 + 0.6572 × log 10 + 0.3483 log 6 + 4.5270 log(100 − 10) − 0.9224 × 0.032 → log DH = −16.3658 + 7.95285 − 1.576 − 0.665 + 0.6572 + 0.271 + 8.846 − 0.295 = −1.1751

which gives DH = 0.068 m. ➔ Considering uncertainties this value can vary from half to double thus estimated value is 0.034 m to 0.136 m.

3.1.6.12

Effect of earthquake on structures

From above discussion it is obvious that earthquake has a profound influence on soil, and since a structure is built on this soil – it do also affects its response. Potential energy stored in earth faults are released due to its rupture and generates kinetic energy in form of stress waves in soil which propagates as P- and S-waves on the surface of earth and induces acceleration on structures and foundations built on the surface of the earth. Thus, as per Newton’s law of motion the structure is subjected to force as a result of its inertial mass, which it has to resist depending on its stiffness and ensure that stresses and deformations induced in the structure and foundation are within safe limits. Above in essence is the basic philosophy of earthquake resistant design. The analytical methods adapted for earthquake analysis for different class of structures and foundations may be classiﬁed into following category: • • •

Seismic coefficient method or equivalent static method Response spectrum method or psuedo static analysis Dynamic analysis which is further subdivided into: ◦ ◦

Modal analysis Time history analysis.

We, as a first step, would study in general the basic principles underlying the above methods and finally see their application to different class of structures and foundations like buildings, tall chimneys, elevated water tank, retaining walls, earth dams etc. © 2009 Taylor & Francis Group, London, UK

412 Dynamics of Structure and Foundation: 2. Applications

3.2 EARTHQUAKE ANALYSIS

3.2.1 Seismic coeff icient method This is an approach where the earthquake force is treated as an equivalent static force based on the zonal classification of a country8 . Though earthquake force in essence is dynamic in nature based on the potential occurrence of earthquakes in a particular zone, the soil condition, the type of foundation, code recommends a certain percentage of weight of the structure which it is expected to resist as lateral force. It should be noted that this method is now obsolete in terms of latest code – IS-1893 2002 and may only be used with caution just to get an idea about the extent of force it may generate in a particular zone for a particular type of structure, and that too only for cases where large number of human life is not endangered – either due to direct or indirect effect of earthquake. Based on the seismic zoning, soil foundation system, importance factor etc we derive a factor, αh , which is given by αh = βIα0

(3.2.1)

where, β = a coefﬁcient depending on the soil foundation system as given in Table 3.2.1, I = importance factor as furnished in Table 3.2.2, α0 = basic horizontal seismic coefﬁcient as given in Table 3.2.3. Based on above having derived, the value of αh , the base shear acting at the soil foundation level, is given by V = KCαh W, V = αh W,

for multistoried frames or buildings and

for all other type of structures

(3.2.2)

where, V = base shear on the structure due to a given earthquake; K = a factor known as the performance factor of the frame; C = a coefficient defining flexibility of the structure with increase in number of storey, depending on fundamental time period. The value of flexibility factor C versus time is as given in Figure 3.2.1. The value of performance factor K for different type of framing is as given in Table 3.2.4. For calculation of time period (T), code has furnished some empirical formulas from which T may be found out as follows: •

For moment resisting frame without bracings or shear walls resisting lateral loads T = 0.1n

(3.2.3)

here n = number of storey including basement. 8 It is presumed the reader has a copy of the earthquake code like IS-1893 (1984 and 2002) for cross reference.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 413 Table 3.2.1 Soil foundation factor β for various soil foundation system as per IS-1893, 1984. Type of soil constituting the foundation

Pile passing through any soil but resting on rock

Piles on any other soil

Rock or hard soil Medium soil Soft soil

1.0 1.0 1.0

Not applicable 1.0 1.2

Raft foundations

Combined or Isolated RCC foundation with tie beams

Isolated Fdn without tie beams

Well foundations

1.0 1.0 1.0

1.0 1.0 1.2

1.0 1.2 1.5

1.0 1.2 1.5

Table 3.2.2 Value of importance factor I as per IS-1893, 1984. Type of structure

Importance factor (I)

Dams (all types) Containers of inflammable or poisonous gases or liquids Important service and community structures such as hospitals, water towers and tanks, schools important bridges, important power houses, monumental structures, emergency buildings like telephone exchange fire bridge, large assembly buildings like structures like cinemas, assembly halls and subway stations All others

3.0 2.0

1.5 1.0

Table 3.2.3 Basic seismic coefficient α0 as per IS-1893, 1984.

•

Zone classification

Seismic coefficient (α0 )

V IV III II I

0.08 0.05 0.04 0.02 0.01

For all others T=

0.09H √ d

(3.2.4)

where, H = total height of the main structure in meters and, d = maximum base dimension of building in meters in direction parallel to the applied seismic force. The above formulations are valid only for buildings which are regular in shape and have regular distribution of mass or stiffness both in horizontal and/or vertical plane. The value of α0 @ 0.08 (Table 3.2.3) has been obtained for zone V based on observations of earthquake occurrence in that zone however the values for other has been reduced proportionally, the basis of this reduction has never been very explicit. Though the above method has now been made obsolete in the recent code (IS-18932002) but it still remains in practice in design offices to estimate preliminarily the magnitude of earthquake force before a more detailed analysis is carried out. © 2009 Taylor & Francis Group, London, UK

414 Dynamics of Structure and Foundation: 2. Applications

Figure 3.2.1 Value of flexibility factor C as per IS-1893, 1984. Table 3.2.4 Value of performance factor I as per IS-1893, 1984. Structural framing system Moment resistance (MR) frame with appropriate ductility details as given in IS-4326 Frame as above with RC shear walls or steel bracing members designed for ductility Frame with either steel bracing members or plain or nominally reinforced concrete infill panels MR Frame as above in combination with masonry infill Reinforced concrete framed buildings (Not covered by 1 or 2 above)

Value of performance factor K 1.0 1.0 1.3 1.6 1.6

Example 3.2.1 An RCC building having frame layout is as shown in Figure 3.2.2. The transverse cross section of the frame is also shown in the figure. Given the following loading and geometric dimensions of the various structural members calculate the base shear on the building as per seismic coefﬁcient method IS-1893 (1984) considering zone IV. Consider soil foundation system as of medium stiffness. Loadings • • • • • •

Live load on roof = 2 kN/m2 Live load on other floors = 4 kN/m2 Parapet wall on roof = 1.5 m all round Internal Partition walls = 1 kN/m2 Floor finish = 1.5 kN/m2 Cement plaster on ceiling = 50 mm.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 415

4

4

6.0

6.0

6.0

6.0

Plan view of the frame EL 116.4

EL 112.8

EL 109.2

EL 105.6

EL102.0

Tie beam all round

EL 100.0 4.0

4.0

Transverse elevation of the frame

Figure 3.2.2 A four storied RCC frame.

Geometric properties (Dimensions in mm) • • • • •

Column size = 300 × 600 Beam size in transverse direction = 300 × 450 Beam size in longitudinal direction = 300 × 600 Average thickness of water proofing on roof = 75 mm All external walls 250 mm thick.

Material properties • • • •

Unit weight of concrete = 25 kN/m3 Unit weight of brick = 20 kN/m3 Unit weight of cement plaster = 24 kN/m3 Grade of concrete = M25.

© 2009 Taylor & Francis Group, London, UK

416 Dynamics of Structure and Foundation: 2. Applications

Seismic zone properties • • •

Seismic zone = Zone IV Soil type Medium stiff Foundation type = Isolated footings with tie beams at 1.0 m below Ground level.

Consider no live load on roof and 50% reduction in live load for other floors during earthquake. Solution: Calculation of roof load (El 116.4) Assume slab thickness = 125 mm; Wt of slab = 0.125 × 24 × 8 × 25 = 600 kN; Live Load on roof = 2.0 × 24 × 8 = 384 kN; Parapet wall (1.5 m high) = 1.5 × 0.25 × 2 (24 + 8) × 20 = 480 kN; Water proofing on roof = 0.075 × 24 × 8 × 24 = 345.6 kN; Cement plaster on ceiling = 0.05 × 24 × 8 × 24 = 230.4 kN; Wt of long beam = 0.3 × (0.6 − 0.125) × 24 × 3 × 25 = 256.5 kN; Wt of short beam = 0.3 × (0.450 − 0.125) × 8 × 5 × 25 = 97.5 kN; Wt of columns = 0.3 × 0.6 × 1.8 × 15 × 25 = 121.5 kN, and, Total load on roof = 600 + 384 + 480 + 346 + 230 + 257 + 98 + 122 = 2517 kN. Calculation of load on other floors (El 112.8 109.2 and 105.6) Wt of slab = 0.125 × 24 × 8 × 25 = 600 kN; Live Load on floor = 4.0 × 24 × 8 = 768 kN Wt of partition wall = 1.0 × 24 × 8 = 192 kN; Load from external brick wall = (3.6 − 0.475) × 0.25 × 48 × 20 + (3.6 − 0.325) × 0.25 × 16 × 20 = 1012 kN; Cement plaster on ceiling = 0.05 × 24 × 8 × 24 = 230.4 kN; Flooring on slab = 1.5 × 24 × 8 = 288 kN; Wt of long beam = 0.3 × (0.6 − 0.125) × 24 × 3 × 25 = 256.5 kN; Wt of short beam = 0.3 × (0.450 − 0.125) × 8 × 5 × 25 = 97.5 kN; Wt of columns = 0.3 × 0.6 × 3.6 × 15 × 25 = 243 kN, and, Total load on each floor = 600 + 768 + 192 + 1012 + 230 + 288 + 257 + 98 + 243 = 3688 kN. Calculation of load on ground floor (El 102.0) Load from external brick wall = (3.6 − 0.475) × 0.25 × 48 × 20 + (3.6 − 0.325) × 0.25 × 16 × 20 = 1012 kN;

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 417

Wt of long beam = 0.3 × (0.6 − 0.125) × 24 × 3 × 25 = 256.5 kN; Wt of short beam = 0.3 × (0.450 − 0.125) × 8 × 5 × 25 = 97.5 kN; Wt of columns = 0.3 × 0.6 × 2.8 × 15 × 25 = 189 kN, and, Total load on ground floor = 1016 + 258 + 98 + 189 = 1561 kN. Total load to be considered for earthquake Load at roof level = 2517 − 384 = 2133 kN (Considering no live load on roof during earthquake); Load at EL 112.8 = 3688 − 768 + 0.5 × 768 = 3304 kN (Considering 50% live load on each floor during earthquake); Load at El 109.2 = 3304 kN (Same as other floor); Load at El 104.6 = 1561 kN, and, Total Weight = 2133 + 3 × 3304 + 1561 = 13606 kN. Calculation of seismic coefficient As stated in the theory above, ah = βIα0 For Seismic zone IV, α0 = 0.05 For medium stiff soil with isolated foundations connected by tie beam, β = 1.0 For normal residential building importance factor I = 1.0 Thus, ah = 1.0 × 1.0 × 0.05 = 0.05 Considering T = 0.1n where n = number of storey, we have, T = 0.5 secs, based on which as IS-1893 1984 flexibility factor C = 0.75. Considering Moment resistant frame with ductile detailing, K = 1.0 And Vb = KCαh W = 1.0 × 0.75 × 0.05 × 13606 = 510.225 ∼ = 510 kN Thus, the total base shear acting on building for an earthquake force acting in either transverse or longitudinal direction is = 510 KN9 .

3.2.2 Response spectrum method This method has undergone almost a radical change compared to what is furnished in IS-1893 2002 and that what was furnished in IS-1893 1984. In the previous code (1984 version) it was observed that base shear developed based on seismic coefficient method and that by response spectrum method were almost matching or were very close for 5% damping in the system. However with the present version (2002) this force is almost double to the previous version. This we believe would significantly enhance the project cost of all projects to come in near future.

9 This is strictly not correct for we will see later that time period will vary in both direction based on its stiffness and mass thus earthquake force will also vary accordingly. Moreover the force calculated herein is the total force acting on the building considered as stick model.

© 2009 Taylor & Francis Group, London, UK

418 Dynamics of Structure and Foundation: 2. Applications

Figure 3.2.3 Response Spectrum Curve Sa /g as per 1893 (1984).

3.2.2.1 Response spectrum method as per 1984 version Though the 1984 version has been made obsolete however for historical reason and also for comparison with the present code we present below the steps followed in this method. The 1984 version gave a set of curves representing the values Sa /g versus different time period in seconds for different level of damping. The sets of curves are as shown in Figure 3.2.3. The curve as shown in Figure 3.2.3 is actually based on the curves generated by Housner based on his observations and average spectrum obtained using four earthquake time histories. Based on the response spectra curve as furnished in Figure 3.2.3 for a particular time period of a structure, the corresponding Sa/g is obtained for a particular damping ratio. Based on the zonal demarcation like I, II, III, IV etc. code gives a values of response spectrum factor F0 10 based on which the coefﬁcient of horizontal seismic force is given by αh = βIF0

Sa g

(3.2.5)

Here β and I are as already defined factors in the seismic coefficient method and factor F0 is as defined in Table 3.2.5. 10 This is exactly 5 times the value of α0 as given for seismic coefficient method.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 419 Table 3.2.5 Value of Seismic zone factor F0 as per IS-1893, 1984. Zone classification

Seismic zone factor (F0 )

V IV III II I

0.40 0.25 0.20 0.10 0.05

Once the value of αh is known the rest of the procedure remains same as that for seismic coefficient method. It may be noted that here that the time period may either be obtained based on formulations as given in code or may be found out based on a detailed dynamic analysis and forces are then obtained based on modal response technique11 . We now explain the above procedure based on a suitable numerical problem.

Example 3.2.2 For the building cited in Example 3.2.1, find the base shear as per response spectrum technique based on IS-1893, 1984. Consider the site to be zone 4 with medium stiff soil. Consider 5% damping ratio for the structure. Solution: Referring to Example 3.2.1 the time period of the building is given by T = 0.1 n = 0.5 sec. For 0.51 sec and 5% damping the Sa /g obtained from the curve as shown in Fig. 8.2.4 is Sa /g = 0.16 As stated previously in Example 3.2.1, β = 1.0 and I = 1.0 and F0 = 0.25 as per Table 3.1.6. Thus

αh = βIF0

Sa g

Or αh = 1.0 × 1.0 × 0.25 × 0.16 = 0.04 As shown in Example 3.2.1, total weight of the structure W = 10310 kN For T = 0.5 sec C = 0.75 K = 1.0.

11 This we are going to study in detail subsequently.

© 2009 Taylor & Francis Group, London, UK

420 Dynamics of Structure and Foundation: 2. Applications

Thus considering V = KCαh W, we have, V = 1.0 × 0.75 × 0.04 × 13606 = 408.18 kN = 408 kN.

3.2.2.2 Response spectrum method as per IS-1893 2002 As stated at the outset, the method has undergone a drastic modification with respect to the present code. In lieu of the soil foundation factor (β) considered in the earlier code, the latest version now defines the Sa /g curve for different type of soil starting with rock to soft soil. Sa /g curve for various type of soil as per IS-1893 (2003) is shown in Figure 3.2.4 for 5% damping.

Spectral Response as per IS -1893 2002

Spectral Acceleration Coefficient (Sa/g)

3 2.5 2

Sa/g(Hard soil/Rock)

1.5

Sa/g(Medium soil)

1 0.5

Sa/g(Soft soil) 3.74

3.4

3.06

2.72

2.38

2.04

1.7

1.36

1.02

0.68

0

0.34

0

Time Period (secs)

Figure 3.2.4 Response Spectrum Curve Sa /g as per IS-1893 (2002).

Moreover as computer analysis has almost become a daily routine work in day to day design office practice-where it is preferable to have digitised data of Sa /g for computer input, the code now defines the Sa /g curve by direct formulas enabling one to furnish numerical input for earthquake analysis by computer. The formulas suggested by code for various types of soil as per Clause 6.4.4 of the code for 5% damping ratio are as shown in Table 3.2.6: The code has given factors based on which the values of Sa /g obtained above may be modified for different damping ratio. Typical Sa /g curve for soft soil with different damping ratio are shown in Figure 3.2.5 while multiplication factors to be considered for different damping ratios are furnished in Table 3.2.7. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 421 Table 3.2.6 Expressions for Sa /g for different types of soil as per IS-1893 2002. Type of soil

Value of Sa /g

Range

Rock or hard soil

1 + 15T 2.5 1.00/T 1 + 15T 2.5 1.36/T 1 + 15T 2.5 1.67/T

0.0 < T < 0.1 0.1 < T < 0.4 0.4 < T < 4.0 0.0 < T < 0.1 0.1 < T < 0.55 0.55 < T < 4.0 0.0 < T < 0.1 0.1 < T < 0.67 0.67 < T < 4.0

Medium soil Soft soil

Spectral Acceleration Soft soil

Spectral acceleration coefficients (Sa/g)

3 Sa/g(5%)

2 .5

Sa/g(7%)

2

Sa/g(10%)

1 .5

Sa/g(15%)

1

Sa/g(20%) Sa/g(25%)

0 .5

Sa/g(30%)

3 .92

3 .64

3 .36

3 .08

2 .8

2 .52

2 .24

1 .96

1 .4

1 .68

1 .12

0 .84

0 .56

0 .28

0

0

Time period(secs)

Figure 3.2.5 Response Spectrum Curve Sa /g for soft soil as per IS-1893 (2002). Table 3.2.7 Multiplying factors for obtaining values for other damping as per IS-1893 (2002). Damping ratio (%) Factors

0 3.2

2 1.4

5 1.0

7 0.9

10 0.8

15 0.7

20 0.6

25 0.55

30 0.5

Table 3.2.8 Seismic zone factor as per IS-1893 (2002). Seismic zone Seismic intensity Z

II Low 0.1

III Moderate 0.16

IV Severe 0.24

V Very severe 0.36

The country unlike previously that was classified into 5 zones (zone I to V) in the present code zone I has now been merged with zone II and the zones now constitute of zone II to V only. The zone factors to be considered as per the present code are as presented in Table 3.2.8. © 2009 Taylor & Francis Group, London, UK

422 Dynamics of Structure and Foundation: 2. Applications Table 3.2.9 Response reduction factor R as per IS-1893 (2002). Sl. No.

Lateral load resistant system

R

1 2

Ordinary moment resistant frame Special Moment resisting frame specially detailed to provide ductile behaviour Steel Frame with Concentric Bracing Eccentric Bracing Special moment resistant frame with ductile detailing Buildings with shear walls Load bearing Masonry wall buildings Un-reinforced Reinforced with horizontal RC band Reinforced with horizontal RC band and vertical bars at corners of rooms and jamb openings Ordinary reinforced concrete shear walls Ductile shear walls Buildings with Dual systems Ordinary shear wall with OMRF Ordinary shear wall with SMRF Ductile shear wall with OMRF Ductile shear wall with SMRF

3.0

3 3a 3b 4 5 5a 5b 5c 6 7

5.0 4.0 5.0 5.0

1.5 2.5 3.0 3.0 4.0 3.0 4.0 4.5 5.0

The importance factor, I has remained unchanged and as such the factors furnished earlier in Table 3.2.2 still holds good. To bring it in line with international practice followed by other countries12 , the code has now introduced a new factor R which is known as the response reduction factor and also called the ductility factor in many literatures. This is the property of a body to dissipate energy by means of its ductile behaviour and may be generated by means of special detailing13 . The R factor for buildings is basically a function of the structural configuration of the building like whether it is a Ordinary Moment Resistant frame (OMRF), special moment resistant frame (SMRF) or has shear wall etc. The value of the response reduction factor R for different types of structural system as defined in IS-1893 2002 is furnished in Table 3.2.9. Based on the above data the design horizontal seismic coefficient Ah for a structure is determined by the expression

Ah =

ZISa 2Rg

12 Especially UBC 1997, and NEHRP as followed in USA. 13 We will discuss more about this later.

© 2009 Taylor & Francis Group, London, UK

(3.2.6a)

Analytical and design concepts for earthquake engineering 423

and the base shear is furnished by the expression V = Ah W

(3.2.6b)

The empirical relation furnished by time period has also undergone some modiﬁcations. As per the latest code the approximate fundamental time period in seconds for a moment resistant frame without brick infill panels may be estimated by the empirical expression Ta = 0.075h0.75 Ta = 0.085h

0.75

for RCC frame building for steel buildings

(3.2.7)

where, h = height of the building. For all other buildings including moment resistant frame buildings with brick infill panels is estimated from the formula as furnished in Equation (3.2.4). Based on above we now solve a numerical problem to illustrate how base shear is obtained as per latest IS-1893. Example 3.2.3 For the building cited in Example 3.1.1, find the base shear as per response spectrum technique based on IS-1893, 2002. Consider the site to be zone 4 with medium stiff soil. Consider 5% damping ratio for the structure. Solution: Referring to Example 3.1.1, the time period of the building is given by 0.09h T= √ d Here h = 16.4 meter and d = 8 m in transverse direction and d = 24 m in long direction thus 0.09 × 16.4 = 0.5218 sec in short direction √ 8 0.09 × 16.4 = 0.3012 sec in long direction. T= √ 24

T=

and

Thus based on the response spectrum curve Sa /g = 2.50 for both short and long direction As per IS-1893 2002 for Zone IV Z = 0.24 © 2009 Taylor & Francis Group, London, UK

424 Dynamics of Structure and Foundation: 2. Applications

Considering SMRF with ductile detailing as per Table 3.2.9, R = 5.0. Thus Ah =

0.24 × 1 × 2.5 ZISa = = 0.06 2Rg 2×5

As shown in Example 10.1 total weight of the structure W = 13606 kN Thus considering, V = Ah W we have, V = 0.06 × 13606 = 816.36 kN. Thus based on the above three examples if we compare the base shear, for the given building we have as follows: Sl. No.

Code

Method

Base shear (kN)

1

IS-1893-1984

510

2

IS-1893-1984

Seismic coefficient method Response spectrum Method

3

IS-1893-2002

do

816

408

Remarks

Shear force value very near to seismic coefficient method Shear force almost double the value of base as calculated by IS-1893-1984

Comparison of base shear as per IS-1893 (1984) and IS-1893 2002.

3.2.3 Dynamic analysis under earthquake loading To understand the basic concept, we start with system having single degree of freedom and subsequently extend this to system having multi-degree of freedom. As shown in Figure 3.2.6, a single bay portal is subjected to an earthquake force for which the body moves through a distance ug at base and undergoes additional deformation of ut at top. We had shown earlier that under time dependent force the equation of motion is given by mu¨ + cu˙ + ku = 0

(3.2.8)

where m = mass of the system; c = damping of the system (usually represented by a dash pot); k = stiffness of the system; u, ¨ u, ˙ u = acceleration, velocity and displacement vectors, respectively. As during the motion the body undergoes a rigid body motion in terms of ug , it does not affect the stiffness and damping of the system, which are affected by ut only. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 425

Y

ut

X ug

Figure 3.2.6 Single bay portal subjected to earthquake force.

Thus the above equation may represented as m(u¨ g + ut ) + cu˙ t + kut = 0

(3.2.9)

From which we arrive at the expression mu¨ t + cu˙ t + kut = −mu¨ g

or mu¨ t + cu˙ t + kut = Fe

(3.2.10)

where Fe = the earthquake force induced on the system and is equal to the mass of the body times ground acceleration due to earthquake.

3.2.4 How do we evaluate the earthquake force? Before we proceed further to analyse the above equation of equilibrium, it is essential to understand the nature and characteristics of earthquake force and how do we evaluate it. The earthquake force in essence is a transient force and acts on a body for a small instant of time. In terms of Newtonian mechanics this can also be termed as an impulsive force acting on a body. According to the basic law of physics an impulse force is expressed as, Fˆ = F(t)dt. This expression means a force F which is function of time is acting upon a body for a very small duration of time dt and is normally defined as an impulse. As,

F=m

dv dt

we can write this as, Fdt = mdv.

ˆ is acting on a body, it will result in a sudden change in Thus if an impulse force, F, its velocity without significant change in its displacement. © 2009 Taylor & Francis Group, London, UK

426 Dynamics of Structure and Foundation: 2. Applications

For spring mass system under free vibration we had seen earlier that the displacement is given by x = A sin ωn t + B cos ωn t,

(3.2.11)

where A and B are integration constants and their magnitudes depend on the boundary condition. For boundary conditions at t = 0, velocity = v0 and displacement x = x0 , the above expression can be written as v0 x= sin ωn t + x0 cos ωn t ωn

where ωn =

k m

(3.2.12)

Thus for the spring mass initially at rest and acted upon by an impulse force is given by x=

Fˆ sin ωn t mωn

(3.2.13)

When considering damping for the system the free vibration equation is written as x = Ae−Dωn t sin

1 − D 2 ωn t + φ

(3.2.14)

Considering the impulse load the above can modified to x=

Fˆ e−Dωn t sin 1 − D2 ωn t √ mωn 1 − D2

(3.2.15)

where D is the damping ratio of the system While considering earthquake the above expression can be further reduced to the expression x=

√

x¨

ωn 1 − D 2

e−Dωn t sin 1 − D2 ωn t

(3.2.16)

Under earthquake the shock induced on the ground is generally represented by response spectra or velocity spectra. Moreover, as we are interested in the peak value (or maximum force in the system) the above integral can effectively used to obtain the peak velocity from which maximum displacement and acceleration are obtained subsequently a shown here after. We had seen earlier that equation of motion for the portal structure under earthquake is given by the expression mu¨ t + cu˙ t + kut = Fe © 2009 Taylor & Francis Group, London, UK

(3.2.17)

Analytical and design concepts for earthquake engineering 427

Dividing each term by m we have u¨ t +

c k Fe u˙ t + ut = m m m

or u¨ t + 2Dn ωn u˙ t + ωn2 ut = u¨ g

(3.2.18)

Since the force is impulsive in nature acting for duration of time ξ (say), the displacement ut can be represented by

ut =

ωn

√

1 1 − D2

t

u¨ g (ξ )e−Dωn (t−ξ ) sin

1 − D2 ωn (t − ξ )dξ

(3.2.19)

0

Differentiating the above we have t −1 u˙ t = u¨ g (ξ )e−Dωn (t−ξ ) − Dωn sin 1 − D2 ωn (t − ξ )dξ √ ωn 1 − D 2 0 2 + ωn 1 − D cos 1 − D2 ωn (t − ξ ) dξ

(3.2.20)

Considering, t C1 =

u¨ g (ξ )−Dωn t cos 1 − D2 ωn ξ dξ

and

0

t C2 =

u¨ g (ξ )e−Dωn t sin 1 − D2 ωn ξ dξ ,

the velocity can be expressed as

0

e−Dωn t u˙ t = C1 D − C2 1 − D2 sin 1 − D2 ωn t 1 − ζ2

+ C1 1 − D2 + C2 D cos 1 − D2 ωn t →

e−Dωn t u˙ t = √ C12 + C22 sin 1 − D 2 ωn t − φ 1 − D2

(3.2.21)

The velocity spectrum or the peak velocity is given by the maximum value of the above e−Dωn t 2 2 C1 + C 2 i.e. Sv = u˙ g = √ 1 − D2

max

when the maximum displacement is given by © 2009 Taylor & Francis Group, London, UK

428 Dynamics of Structure and Foundation: 2. Applications

Sd =

Sv ωn

and Sd =

Sa ωn2

(3.2.22)

where Sa is the acceleration spectrum. Thus the maximum force the system may experience is given by Fmax = mω2n Sd

(3.2.23)

It is obvious that that for response spectrum analysis the value Sa is function of the time period or natural frequency of the system which is given by the expression ω=

k m

and T =

2π . ω

(3.2.24)

Certain type of structures can very well be modelled as systems with single degree of freedom and the base force can be found out as follows: Example 3.2.4 Shown in Figure 3.2.7 is an air cooler of weight 450 kN is supported on a structure as shown. Determine the force on the system calculating time period based on dynamic analysis. Consider the soil is medium stiff and the site is in zone III. Consider 5% damping for the structure. For beams and columns section properties are as follows I xx = 1268.6 cm4 , I yy = 568 cm4 and A = 78 cm2 , Area of the bracing members = 12 cm2 , Esteel = 2 × 108 kN/m2 . Unit weight of column material = 78.5 kN/m3 What will be the force on the frame based formulation as given in the code?

6500

6000

Figure 3.2.7 Structure supporting an air cooler.

© 2009 Taylor & Francis Group, London, UK

3000

Analytical and design concepts for earthquake engineering 429

Solution: For earthquake force in transverse direction Stiffness of each column is given by K = 12EI/L3 Here I = 1268.6 cm4 = 1.2686 × 10−5 m4 E = 2 × 108 kN/m2 L = 6.5 m, 12 × 2 × 108 × 1.2686 × 10−5 = 110.86 kN/m (4.5)3 4 Considering four columns Ki = 4 × 110.86 = 443.46 kN/m

Thus,

K=

i=1

Weight of the air cooler = 450 kN 450 Mass of the air cooler = = 45.87 kN-sec2 /m 9.81 7.8 × 10−3 × 78.5 Mass of each column = = 0.0624 kN-sec2 /m 9.81 Considering 1/3rd weight of column contributing to top mass of 4 columns 4

mi =

i=1

0.0624 × 4 = 0.0832 kN − sec2 /m 3

Weight of top beam = (6 + 3) × 2 × 78.5 × 7.8 × 10−3 = 11 kN 11 Mass of beam = = 1.123 kN-sec2 /m 9.81 Thus total mass = 45.87 + 0.0832 + 1.123= 47.07 kN-sec2 /m m 47.07 Considering T = 2π we have, T = 2π = 2.04 sec. K 443.46 for which as per IS-1893(2002), Sa /g = 0.666 ZISa Considering Ah = , 2Rg where Z = 0.16 for zone III, I-1.0; Sa /g = 0.666 and R = 3.0, 0.16 × 1.0 × 0.666 we have, Ah = = 0.0177 2×3 Thus Vh = 0.0177 × 47.07 × 9.81 = 8.20 kN. For earthquake in longitudinal direction (i.e. in the direction of the braced bay) Stiffness of per column (considered hinged at base, Figure 3.2.8) 3EI 3 × 2.0 × 108 × 5.68 × 10−6 = = 12.41 KN/m L3 (6.5)3 6.5 θ = tan−1 = 65.22◦ 3.0 =

© 2009 Taylor & Francis Group, London, UK

430 Dynamics of Structure and Foundation: 2. Applications

6500

6000

Figure 3.2.8 Frame in longitudinal direction

Stiffness of each bracing

AE L

cos2 θ

=

1.2×10−3 ×2.0×108 6.5

cos2 65.22 =

6486 kN/m. Thus total stiffness of the frame in longitudinal direction = 4 × 12.4 + 6486 × 4 = 25993.6 kN/m. m Considering T = 2π we have K T = 2π

47.07 = 0.267 25993.6

for which as per IS-1893(2002), Sa /g = 2.5.

ZISa , here Z = 0.16 for zone III, I-1.0 and R = 4.0 (for 2Rg 0.16 × 1.0 × 2.5 concentric bracing) we have, Ah = = 0.05. 2×4 Considering Ah =

Thus Vh = 0.05 × 48.2 × 9.81 = 23.6 kN in longitudinal direction. As per code for steel frame (vide Equation 3.2.9), Ta = 0.085 h0.75 ➔ Ta = 0.085 × 6.50.75 = 0.346 sec for which the value Sa /g = 2.5. ZISa here Z = 0.16 for zone III, I-1.0 and R = 4.0 2Rg 0.16 × 1.0 × 2.5 we have, Ah = = 0.05. 2×4 Thus considering Ah =

Thus, the maximum force on the frame = 23.1 kN, this is same as we obtained using the dynamic analysis.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 431

3.2.5 Earthquake analysis of systems with multi-degree of freedom Before we delve into the detailed dynamic analysis of systems with multi-degree of freedom under earthquake force (based on modal analysis or time history response), we deal with a particular technique often used in practical engineering design where for many buildings effect of fundamental time period is most pre-dominant. In such cases higher mode participation vis-a-vis its effect being insignificant are ignored without causing any significant errors. 3.2.5.1 Analysis based on assumed shape function This is a technique in which a multi-degree freedom system is converted into an equivalent system having mass and stiffness of that of a single degree of freedom based on an assumed shape function to find out the time period of a system. To start with let us consider a stick model of a system having multi-degree of freedom as shown in Figure 3.2.9. The kinetic energy of the system is given by T(t) =

n 1 ∂y(z, t) 2 mi 2 ∂t

(3.2.25)

i=1

We consider here, y(z, t) = ϕ(z)ξ(t)

(3.2.26)

where, ϕ(z) = admissible shape function which satisfies the boundary condition of the system; ξ(t) = generalized co-ordinate.

Mn Kn Displaced Shape(1st Mode)

M3 K3 M2 K2 M1

K1

Figure 3.2.9 A stick model having multi-degree of freedom.

© 2009 Taylor & Francis Group, London, UK

432 Dynamics of Structure and Foundation: 2. Applications

Thus, T(t) =

n 1

2

1 → T(t) = 2

⎡ ⎤ n n mi ⎣ ϕj (z)ξ˙j (t) ϕk (z)ξ˙ k (t)⎦

i=1

j=1

n n

ξ˙j (t)ξ˙ k (t) mi

j=1 k=1

k=1 n

ϕj (z)ϕ k (z)

(3.2.27)

i=1

from which we conclude that the generalized mass of the system is given by, ∗

M = mi

n

ϕj (z)ϕ k (z)

(3.2.28)

i=1

Thus for fundamental mode for j = k we have ∗

M =

n

mi ϕi2 (z)

(3.2.29)

i=1

Similarly potential energy is given by 1 ki [y(z, t)]2 2 n

V(t) =

(3.2.30)

i=1

where, = difference in the displacement between two adjacent level. Hence, ⎡ ⎤ n n n 1 ⎣ ki ϕ j (z)ξj (t) ϕ k (z)ξ k (t)⎦ V (t) = 2 i=1 j=1 k=1

n n n 1 → V(t) = ξj (t)ξ k (t) ki φ j (z)φ k (z) 2 j=1 k=1

(3.2.31)

i=1

Thus for the fundamental mode, j = k, we have ∗

K =

n

ki ϕ 2i (z)

(3.2.32)

i=1

Now knowing, T = 2π ∗

T = 2π

M∗ K∗

© 2009 Taylor & Francis Group, London, UK

m K,

we have for this generalized case

(3.2.33)

Analytical and design concepts for earthquake engineering 433

From the above mathematical derivation it is obvious that if we know what could be the assumed shape function correctly it is possible to arrive at the fundamental time period of the system. Based on the aspect ratio (H/D), Naeem (1989) has proposed the following shape functions which may be considered for buildings modeled as stick having multi-degrees of freedom. Sl. No.

H/D

Shape function

1

H/D < 1.5

2

1.5 < H/D < 3

3

H/D > 3.0

πx sin 2H x H πx 1 − cos 2H

where, H = height of the building; D = width of building in direction of the earthquake force considered. We will now solve the previous building problem (vide Example 3.2.1) to see how base shear results differ with what we have calculated earlier.

Example 3.2.5 Refer the problem as shown in Example 3.2.1 calculate the time period of the building based on assumed shape function method and calculate the base shear in both transverse and longitudinal direction and find out the base shear based on IS-1893-2002. Consider all other boundary conditions remains same as was defined in the previous problem (Figure 3.2.10). EL 116.4

EL 112.8

EL 109.2

EL 105.6

EL102.0 Tie beam all round EL 100.0 4.0

4.0

Figure 3.2.10 Transverse elevation of frame.

Solution: Considering the frame as a stick model in transverse direction we have the model as shown in Figure 3.2.11. © 2009 Taylor & Francis Group, London, UK

434 Dynamics of Structure and Foundation: 2. Applications

El-116.4 K1 EL-112.8 K2 El-109.2 K3 El-105.6 K4 EL-102.00 K5 EL-100.0

Figure 3.2.11 Stick model of the frame.

Dimension of column = 300 × 600 Moment of inertia of the column = 0.0054 m4 Stiffness of column = 12EI H3 Here, Econc = 2.85 × 107 kN/m2 ∴ Ki =

1 × 300 × 6003 = 5400000000 mm4 = 12

12 × 2.85 × 107 × 0.0054 = 39583.33 KN/m (3.6)3

For ﬁfteen column per level total stiffness Ki = 15×39583.33 = 593750 kN/m Thus, K1 = K2 = K3 = K4 = 593750 kN/m 12 × 2.85 × 107 × 0.0054 And K5 = 15 × = 3462750 (2)3 16.4 Since H/D in transverse direction is = = 2.05 < 3.0 thus shape function 8 considered is x/H Level

Weight

Mass

5

2133

217.4

Stiffness

φi

φi

1.00 593750

4

3304

337.0

3

3304

337.0

2

3304

337.0

1

1561

159.12

593750

28737.5 205.03

0.219 0.561

593750

28476.84 106.06

0.22 0.341

593750

28737.5 39.18

0.22 0.121

28737.5 2.33

0.121 567.67

ki φi2

217.4 0.22

0.780

3462750

mi φi2

50698.12 165387.46

M∗ 567.67 ∗ Considering = 2π we have, T = 2π = 0.368 sec K∗ 165387.46 Based on response spectrum curve, Sa /g = 2.5 T∗

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 435

Considering all other parameters remaining constant, Ah =

ZISa 2Rg

=

0.24 × 1.0 × 2.5 = 0.06. 2×5 Base shear = 0.06 × 13606 = 816 kN, which is same as we got earlier, based on method as suggested by the code. For longitudinal direction, we have Dimension of column = 300 × 600 1 Moment of inertia of the column = × 600 × 3003 = 1350000000 mm4 = 12 0.00135 m4 12EI Stiffness of column = H3 Here, Econc = 2.85 × 107 kN/m2 → Ki =

12 × 2.85 × 107 × 0.00135 = 9895.833 kN/m (3.6)3

For ﬁfteen column per level total stiffness Ki = 15 × 9895.833 = 148437.5 kN/m Thus K1 = K2 = K3 = K4 = 148437.5 kN/m And K5 = 15 ×

12 × 2.85 × 107 × 0.00135 = 865687.5 kN/m (2)3

Since H/D in transverse direction is = πx function is sin 2h Level

Weight

Mass

Stiffness

5

2133

217.4

16.4 = 0.683 < 1.5 thus shape 24

φi 1.00

148438 4

3304

337

3

3304

337

2

3304

337

1

1561

159.12

φi

mi φi2 217.4

0.059 0.941

148438

516.71 298.4

0.179 0.771

148438

4756.1 200.32

0.260 0.511

148438

10034.4 87.99

0.321 0.190

865688

ki φi2

15295.2 5.744

0.190 809.85

31251.33 61853.74

M∗ 809.85 ∗ Considering = 2π we have, T = 2π = 0.7189 sec ∗ K 61853.74 Based on response spectrum curve, Sa /g = 1.39 Considering all other parameters remaining constant

T∗

Ah =

ZISa 0.24 × 1.0 × 1.39 = = 0.0334 2Rg 2×5

© 2009 Taylor & Francis Group, London, UK

436 Dynamics of Structure and Foundation: 2. Applications

Thus, the base shear = 0.0334 × 13606 = 454 kN. We see that when actual stiffness and mass distribution of the system is considered for calculation of the time period the base shear is in signiﬁcant variation to that as to what has been considered in the code.

3.2.5.2 Dynamic analysis of systems having multi-degree of freedom under earthquake forces In this section we discuss the time history and modal analysis technique as applied to earthquake. We had already discussed in detail the basic concepts underlying the same in Chapter 5 (Vol. 1) (basic concepts of structural dynamics) as applied to harmonic forces. The fundamental steps for earthquake analysis, essentially remains the same as that of harmonic force except the fact that calculation of amplitude and interpretation of forces in the system is different. For a structural system having N degrees of freedom we have seen earlier in Chapter 5 (Vol. 1) that the equation of motion is expressed as ¨ + [C]{X} ˙ + [K]{X} = {P(t)} [M]{X}

(3.2.34)

were [M] = mass matrix of the system of size N × N [C] = damping matrix of size N × N [K] = stiffness matrix of size N × N ¨ {X}, ˙ {X} = acceleration, velocity, displacement vector of the system {X}, Considering the displacement vector as {X} = [ϕ(x) ]{ξ(t) } the eigen value of the problem is given by [K] − [M]ω2 ][ϕ] = 0,

(3.2.35)

from which we find out the time period of the system for m number of significant modes. The different techniques to find out the eigenvalues for the above equation have already been discussed in Chapter 5 (Vol. 1). The equation of motion can now be expressed as [M][ϕ]{ξ¨ } + [C][ϕ]{ξ˙ } + [K][ϕ]{ξ } = {P(t)}

(3.2.36)

Pre-multiplying the above by [ϕ]T we have [ϕ]T [M][ϕ]{ξ¨ } + [ϕ]T [C][ϕ]{ξ˙ } + [ϕ]T [K][ϕ]{ξ } = [ϕ]T {P(t)}

(3.2.37)

Based on orthogonal property we had seen earlier that the above de-couples into N number of equations expressed by {ξ¨n } + 2Dω{ξ˙n } + ω2 {ξn } =

© 2009 Taylor & Francis Group, London, UK

[ϕn ]T {P(t)} [ϕn ]T [M][ϕ]

(3.2.38)

Analytical and design concepts for earthquake engineering 437

For earthquake as the force induced in the system can be expressed as {P(t)} = [M]{u¨ g }, the above general equation can be modified into {ξ¨n } + 2Dω{ξ˙n } + ω2 {ξn } =

Ln {u¨ g } [ϕn ]T [M][ϕ]

(3.2.39)

where, Ln = [ϕ]T [M][I] here [I] is a unit column vector of dimension, N. The solution of the above equation for nth mode at any time, t, is given by the expression14 1 Ln ξn (t) = T φn [M]φn ωn

t

u¨ g (τ )e−Dn ωn (t−τ ) sin ωn (t − τ )dτ

(3.2.40)

0

The displacement for each mass i at time t is then obtained by superimposition of all modes calculated at this time t and is given by xi =

N

φin ξn (t)

(3.2.41)

n=1

The earthquake force on the structure is then expressed in terms of the effective acceleration ξ¨neff (t) = ωn2 ξn (t)

(3.2.42)

Considering f = kδ, we have, the earthquake force at any floor i at time is t is given by fin (t) = kin xin

or fin (t) = kin φin ξn (t)

(3.2.43)

Since based on the eigen value expression, we have kφ = mω2 φ

(3.2.44)

Substituting the value of k in terms of inertial force, we have fin (t) = mi ωn2 φin ξn (t)

(3.2.45)

Superimposing all modal contribution the earthquake force on the structure is expressed as f (t) = [M]ωn2 [ϕ]ξ(t)

14 This, we explained in the case of systems having single degree of freedom.

© 2009 Taylor & Francis Group, London, UK

(3.2.46)

438 Dynamics of Structure and Foundation: 2. Applications

where [M] = mass matrix of the system of order N × N; [ϕ] = relative amplitude distribution of order N × N; [ω2 ] = diagonal matrix of order N × N having eigenvalues in the diagonal term. Based on the above theory the entire history of displacement and force response can be defined for any multi-degree of freedom system having calculated the modal response amplitudes. When the above theory is applied to response spectrum, as discussed earlier with single degree of freedom the maximum response for the each mode is considered. If the maximum value of ξn max of the Duhamel integral is considered, the maximum displacement in that mode is given by xn max = φn ξn max = φn

Ln T φn [M]φn

Svn ωn

(3.2.47)

The maximum earthquake force in the structure is then given by fn max = [M]φn

Ln San φn2 [M]φn

(3.2.48)

where San , Svn are spectral velocity as furnished in the code. The base shear which is the algebraic sum of all the force is given by

V0 (t) =

N

fn max (t) =

i=1

N L2n San φn M n

(3.2.49)

i=1

L2

The expression Mnn is usually called the effective modal mass of the system and when divided by the total mass (represented in percentage), reflects the percentage of modal mass responding to the earthquake force in each mode. We now further illustrate the above theory by a suitable numerical problem. Example 3.2.6 Shown in Figure 3.2.12 is a three storied RCC frame subjected to earthquake in zone IV having medium soil condition. The damping ratio for RCC considered is 5%. Determine • • • • • • •

The natural frequencies of the structure. The eigen-vectors. The acceleration, velocity and displacement as per IS-1893 2002 based on response spectrum method. Effective Modal mass participation for each mode. The nodal displacement per mode. The nodal shear force per mode. Base shear for the three modes.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 439

G

H

X3

E

F

X2

C

D

X1

A

B

Figure 3.2.12 A three story RCC Frame.

Here 1 K AC = K DB = 15000 kN/m M GH = 20 kN sec2 /m 2 K CE = K DF = 10000 kN/m M EF = 40 kN sec2 /m 3 K EG = K FH = 5000 kN/m M CD = 40 kN sec2 /m Solution: The free body diagram of the structure is as shown below in Figure 3.2.13:

k2(x2-x1)

k3(x3-x2) m3x3

k3(x3-x2)

m1x1

m2x2

k2(x2-x1)

k1x1

Figure 3.2.13 Free body diagram at each floor.

Based on the free-body-diagram, for free vibration we have, m3 x¨ 3 + k3 (x3 − x2 ) = 0 m2 x¨ 2 + k2 (x2 − x1 ) − k3 (x3 − x2 ) = 0 m1 x¨ 1 + k1 x1 − k2 (x2 − x1 ) = 0

© 2009 Taylor & Francis Group, London, UK

440 Dynamics of Structure and Foundation: 2. Applications

The above on simplification while writing in the matrix form gives ⎡

m1 ⎣0 0

0 m2 0

⎤⎧ ⎫ ⎡ k1 + k2 −k2 0 ⎨x¨ 1 ⎬ k2 + k3 0 ⎦ x¨ 2 + ⎣ −k2 ⎩ ⎭ 0 −k3 x¨ 3 m3

⎤⎧ ⎫ 0 ⎨x1 ⎬ −k3⎦ x2 = 0 ⎩ ⎭ k3 x3

The above on substituting the values gives the following matrices. ⎡

50000 −20000 [K] = ⎣−20000 30000 0 −10000 •

⎤ 0 −10000⎦ 10000

⎡

⎤

40

and [M] = ⎣

40

⎦ 20

Natural frequencies and modal values of the system

Applying any one of the methods for determination of eigen-values and eigenvectors as cited in Chapter 5 (Vol. 1) we have λ1 = 156.93, Thus,

•

λ2 = 750,

√ ω1 = √156.93 = 12.527 rad/sec, which implies T1 = 0.502 sec ω2 = √750 = 27.386 rad/sec, which implies T2 = 0.229 sec ω3 = 1593 = 39.913 rad/sec, which implies T3 = 0.157 sec

The corresponding eigenvectors are given as For the first mode, For second mode, For third mode,

•

λ3 = 1593

{ϕ}T 1 = −0.314 −0.686 −1 {ϕ}T 2 = −0.5 −0.5 1 {ϕ}T 3 = 1 −0.686 0.314

The Matrix M n and Ln

The identity matrix is given by [I]T = 1

1

1

For first mode Mn1 = {ϕ}T 1 [M]{ϕ}1 = −0.314 −0.686

⎡ 40 −1 ⎣

⎧ ⎫ ⎨−0.314⎬ × −0.686 = 42.772 ⎩ ⎭ −1

© 2009 Taylor & Francis Group, London, UK

⎤ 40

⎦ 20

Analytical and design concepts for earthquake engineering 441

Ln1 = {ϕ}T 1 [M]{I} = −0.314 ⎡

40

×⎣

−0.686

−1

⎤⎧ ⎫ ⎨1⎬ ⎦ 1 = −60 40 ⎩ ⎭ 20 1

For second mode

Mn2 = {ϕ}T 2 [M]{ϕ}2 = −0.5

⎡ 40 1 ⎣

−0.5

⎫ ⎤⎧ ⎨−0.5⎬ ⎦ −0.5 = 40 40 ⎩ ⎭ 20 1

⎡ Ln2 = {ϕ}T 2 [M]{I} = −0.5 −0.5

40

1 ⎣

40

⎤⎧ ⎫ ⎨1⎬ ⎦ 1 = −20 ⎩ ⎭ 20 1

Similarly for third mode Mn3 = {ϕ}T 3 [M]{ϕ}3 = 60.802;

Ln3 = {ϕ}T 3 [M]{I} = 18.832

⎧ ⎫ ⎨84.167⎬ 10 Thus for three modes are given = this, when divided by ⎩ ⎭ Mn Mn 5.833 the total mass of the system (i.e. 40 + 40 + 20 = 100 kN), and multiplied by 100 we have L2n

L2n

⎧ ⎫ ⎨84.167⎬ 10 ℵ= % ⎩ ⎭ 5.833

which represents the percentage mass participating in each mode. •

Calculation of Acceleration and velocity based on IS 1893 (2002)

For the first mode, considering T1 = 0.502 sec Sa = 2.5 g

or Sa = 2.5 × 9.81 = 24.525 m/sec2

© 2009 Taylor & Francis Group, London, UK

442 Dynamics of Structure and Foundation: 2. Applications

Considering zone IV severe earthquake condition Z = 0.24 and considering ductility factor R = 3.0 (for ordinary moment resisting frame) 0.24 × 24.525 = 0.981 m/sec2 ; 2×3 Sa 0.981 Sv (design) = = = 0.078 m/sec. ω 12.527

Sa (design) =

For the second mode, considering T2 = 0.229 sec Sa = 2.5 g

→ Sa = 2.5 × 9.81 = 24.525 m/sec2

Considering Z = 0.24 as before ductility factor R = 3.0 (for ordinary moment resisting frame) Sa (design) =

0.24 × 24.525 = 0.981 m/sec2 ; 2×3

Sv (design) =

Sa 0.981 = = 0.036 m/sec. ω 27.386

For third mode considering, T3 = 0.157 sec Sa = 2.5 g

→ Sa = 2.5 × 9.81 = 24.525 m/sec2 .

Considering Z = 0.24 and R = 3.0, we have Sa (design) =

0.24 × 24.525 = 0.981 m/sec2 ; 2×3

Sv (design) =

Sa 0.981 = = 0.025 m/sec. ω 39.913

Thus based on above for the three modes, we have ⎧ ⎫ ⎨0.981⎬ {Sa } = 0.981 m/sec2 ⎩ ⎭ 0.981

© 2009 Taylor & Francis Group, London, UK

and

⎧ ⎫ ⎨0.078⎬ {Sv } = 0.036 m/sec ⎩ ⎭ 0.025

Analytical and design concepts for earthquake engineering 443

•

Calculation of displacement

The displacement in the first mode is given by (Equation 3.2.49) ⎫ ⎧ ⎧ ⎫ ⎨2.752 × 10−3 ⎬ ⎨−0.314⎬ 0.078 −60 Ln1 Sv1 × = 6.017 × 10−3 m = −0.686 × = [ϕ] ⎭ ⎩ ⎭ 42.772 12.527 ⎩ Mn1 ω1 −1 8.769 × 10−3 The displacement in the second mode is given by ⎫ ⎧ ⎫ ⎧ ⎨−0.5⎬ −20 ⎨ 3.27 × 10−4 ⎬ 0.036 Ln2 Sv2 = −0.5 × × = 3.27 × 10−4 m = [ϕ] ⎭ ⎩ ⎭ ⎩ Mn2 ω2 40 27.386 1 −6.54 × 10−4 The displacement in the third mode is given by ⎫ ⎧ ⎫ ⎧ ⎨ 1.00 ⎬ 18.832 ⎨ 1.907 × 10−4 ⎬ 0.025 Ln3 Sv3 = −0.686 × × = −1.309 × 10−4 m = [ϕ] ⎭ ⎩ ⎭ 60.802 39.913 ⎩ Mn3 ω3 0.314 5.986 × 10−5 •

The shear force per floor is given by

[V]i=n = [M]φn

Ln ωn Svn Mn

Thus for the first mode we have ⎡

[V]i=1

40 =⎣0 0

0 40 0

⎫ ⎤⎧ 0 ⎨−0.314⎬ −60 0 ⎦ −0.686 ⎩ ⎭ 42.772 20 −1 n=1

⎧ ⎫ ⎨220.618⎬ × 12.527 × 0.078 = 482.302 kN ⎩ ⎭ 351.46 For the second mode we have ⎡

[V]i=2

40 =⎣0 0

0 40 0

© 2009 Taylor & Francis Group, London, UK

⎧ ⎫ ⎫ ⎤⎧ 9.81 ⎬ 0 ⎨−0.5⎬ ⎨ −20 0 ⎦ −0.5 ×27.386×0.036 = 9.81 kN ⎩ ⎭ ⎩ ⎭ −9.81 20 1.0 n=2 40

444 Dynamics of Structure and Foundation: 2. Applications

Thus for the third mode we have ⎡

[V]i=3

40 =⎣0 0

0 40 0

⎫ ⎤⎧ 0 ⎨ 1.00 ⎬ 18.832 0 ⎦ −0.686 ⎩ ⎭ 20 0.314 n=3 60.802

⎧ ⎫ ⎨ 12.153 ⎬ × 39.913 × 0.025 = −8.339 kN ⎩ ⎭ 1.907 •

Base shear per mode

The base shear per mode is given by For the first mode, Vb = 221 + 482 + 351 = 1054 kN; the second Mode, Vb = 9.8 + 9.8 − 9.8 = 9.8 kN; and for the third mode, Vb = 12 − 8 + 2 = 6 kN.

Thus,

⎧ ⎫ ⎨1054⎬ Vb = 9.8 kN ⎩ ⎭ 6

3.2.6 Modal combination of forces Once the maximum response for each mode is obtained as described above, it is essential to obtain the combined response of all modes. As the modal maxima may or may not occur at the same time and nor have the same sign they cannot be combined to give accurate total maximum response. The most convenient way to represent this is to combine them based on probability basis. Three techniques often used for modal combination of forces are • • •

Absolute Sum Method (ABSSUM) Square root of Sum Square (SRSS) Complete Quadratic Combination (CQC)

3.2.6.1

The absolute sum method (ABSSUM)

As the name suggests by this method the modal combination of all responses are obtained by summing up the absolute values of the response without considering their algebraic signs. Thus, based on above λn =

n

|λi |

i=1

© 2009 Taylor & Francis Group, London, UK

(3.2.50)

Analytical and design concepts for earthquake engineering 445

where, |λi | represents the absolute value of the responses, without consideration of their algebraic sign. This method though still in practice sometimes has been observed to give results which are too conservative and is now a days only used in case of non-critical structure. Use of this method for important and critical structures has almost been abolished. 3.2.6.2

The square root of sum square method (SRSS)

In this method the modal response are obtained by summing up the square of the responses and taking its root and has been found to give a much better result (Rosenblueth 1951). This method is however valid only when the frequencies of the structure are widely spaced. For structures having repeated roots or closely spaced roots, CQC is found to be superior, however when eigen values are widely spaced SRSS and CQC converges to almost identical results. Thus based on the above we have, $ % n % λn = & λ2i

(3.2.51)

i=1

3.2.6.3

The complete quadratic combination method (CQC)

In this method (Der Kiureghian 1981) the response of the system is obtained by the expression $ % n % n λn = & λi ρij λj

(3.2.52)

i=1 j=1

in which, n = number of modes being considered; λi = response quantity in mode i; λj = response quantity in mode j, and, ρij = Cross modal coefficient and is given by

ρij =

3 8 Di Dj (Di + βij Dj )βij2

(1 − βij2 )2 + 4Di Dj βij (1 + βij )2

(3.2.53)

where, Di = Modal damping ratio for mode i; Dj = modal damping ratio for mode j, and, βij = frequency ratio (ωi /ωj ). For normal seismic dynamic analysis the damping ratio is usually considered constant for all modes when the above equation reduces to 3

ρij =

8D2 (1 + βij )βij2 (1 − βij2 )2 + 4D2 βij (1 + βij )2

© 2009 Taylor & Francis Group, London, UK

(3.2.54)

446 Dynamics of Structure and Foundation: 2. Applications

Cross modal coefficient

Cross modal coefficient

1.2 2% DR

1

5% DR 0.8

7% DR

0.6

10% DR 15% DR

0.4

20%DR 0.2

25% DR

1. 94

1. 76

1. 58

1. 4

1. 22

1. 04

0. 86

0. 5 0. 68

0 Frequency Ratio

Figure 3.2.14 Variation of cross modal frequency for different frequency ratios.

The variation of the cross modal response with frequency ratio for various damping ratio is as shown in Figure 3.2.14. From the curve we make a very interesting observation. The cross modal ratio plays a significant part in the magnitude when the frequency ratio varies between 0.88 to 1.14. For other frequencies (which are widely apart) they diminish rapidly and their contribution is insignificant. In other words for widely space frequencies the CQC method in effect converges to the SRSS method. We now further elaborate the above theories based on a suitable example

Example 3.2.7 For a typical three storied frame, the natural frequencies calculated are 4.257, 8.66 and 14.382 ⎧ ⎫ rad sec respectively. The corresponding base shear estimated ⎨330⎬ are V b = 75 kN, find out the combined maximum base shear based on ⎭ ⎩ 33 • • •

Absolute sum method. Square root of sum square method. Complete quadratic combination method. Consider 5% damping in all modes.

Solution: •

Absolute sum method

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Analytical and design concepts for earthquake engineering 447

As per this the base shear is given by Vb = 330 + 75 + 33 = 438 kN. •

Square root of sum square method

As per this the base shear is given by Vb = •

3302 + 752 + 332 = 340 kN

Complete quadratic combination method

As a first step we find out the values βij = (ωi /ωj ) for the three modes which is as given below Mode

1

2

3

1 2 3

1 2.034296 3.378436

0.49157 1 1.660739

0.295995 0.602142 1

Considering 5% damping as constant for all mode we have the cross modal values as 3 8D2 1 + βij βij2 ρij = 2 (1 − βij2 )2 + 4D2 βij 1 + βij Mode

1

2

3

1 2 3

1 0.025022 0.009162

0.0123 1 0.045746

0.002712 0.027546 1

Now considering Vb n

i=1

n

j=1 λi ρij λj ,

⎧ ⎫ ⎨330⎬ 75 kN. and applying the equation, λn = = ⎩ ⎭ 33

we have base shear based on CQC expression as

Mode

1

2

3

1 2 3

108900 619.3049 99.77034

304.432 5625 113.2222

29.53152 2475 1089

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448 Dynamics of Structure and Foundation: 2. Applications

Adding all the nine terms in the above table and taking square root we have Vb =

√ 119255.3 = 345.33 kN.

Thus it will be observed that based on CQC method base shear is 345 kN in lieu of 340 kN based on SRSS method. Since the frequencies are widely spaced the variation is only marginal – about 1.56% only.

3.3 TIME HISTORY ANALYSIS UNDER EARTHQUAKE FORCE Time history analysis under earthquake force is possibly the most comprehensive analysis one can undertake. However in spite of its rigorous mathematical basis15 , modal response technique has still remained a more popular method in day to day design office practice. The reason underlying the same can be attributed primarily to lack of site accelerograms which is the basic input for such an analysis. Previously site speciﬁc ground acceleration data available was few and far for which engineers always preferred to use the modal response technique using the response spectrum curve which is available in all codes of all countries having a speciﬁc earthquake code. However in last thirty years there has been a significant technological advancement based on which earthquake accelerograms are now almost globally available for all major earthquakes. All major and minor tremors occurring around the World are now being manned constantly. This has signiﬁcantly enhanced our data base and in years to come for important structures time history analysis would hopefully become a routine affair16 . We show hereafter a typical acceleration spectrum for the famous El-Centro Earthquake in Figure 3.3.1. When an earthquake occurs anywhere in the world the seismic monitoring station picks up the tremor signals and based on such data ground acceleration/velocity at different time steps are obtained. This data is further used as input ground acceleration for time history analysis of structure to be build at that site or at its close proximity. The theory underlying the method remains the same as shown in Chapter 5 (Vol. 1) except the fact that we had earlier solved the problem with the forcing function as harmonic force which in case of earthquake is the ground acceleration, {¨ag }17 . Thus, the basic equation of motion is ¨ t+t } + [C]{X ˙ t+t } + [K]{Xt+t } = {Rt+t } [M]{X

(3.3.1)

15 Refer Chapter 5 (Vol. 1) where we have discussed the various techniques of time history analysis. 16 For Nuclear power plants time history response analysis is now mandatory for all class 1 type structures like turbine building, reactor building, spent fuel chamber etc. 17 This is usually obtained as an input from the site based on observed data like the one as shown for the El-Centro Earthquake.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 449

0.4

0.3 0.2

0.1

Acceleration, g

0.0 0.0

5.0

10.0

15

20

25

30

35

-0.1

-0.2

Time in seconds

-0.3

-0.4

Figure 3.3.1 Accelerogram of EL centro earthquake of May 18, 1940 NS component.

The term Rt+t is obtained by multiplying the ground acceleration data by the mass, [M]. In other words, here Rt+t = [M]{¨ag }t+t at every time step, t + t. Thus once the force Rt+t is known, rest of the procedure remains the same as what has been described earlier in Chapter 5 (Vol. 1). For instance the steps of Newmark-β method gets slightly modified for earthquake case as follows:

Steps for Newmark-β method for earthquake analysis • • • •

Assemble the mass matrix, [M], damping matrix [C] and stiffness matrix [K]. Evaluate {X¨ 0 } (This will be obtained from the accelerometer data of the site). Select time step size t and parameters δ and β where δ ≥ 0.50 and β = 0.25(0.5 + δ)2 . Calculate integration constant 1 δ 1 1 , α2 = , α3 = − 1, , α1 = βt βt 2β βt 2 t δ α5 = − 2 , α6 = t(1 − δ), α7 = t δ. 2 β

α0 =

•

α4 =

δ − 1, β (3.3.2)

Form the modified stiffness matrix as ˆ = [K] + α0 [M] + α1 [C] [K]

© 2009 Taylor & Francis Group, London, UK

(3.3.3)

450 Dynamics of Structure and Foundation: 2. Applications

•

Calculate modified load at time t + t ˙ t + α3 X ¨ t} ˆ t+t } = [M]{¨ag }t+t + [M]{α0 Xt + α2 X {R ˙ t + α5 X ¨ t} + [C]{α1 Xt + α4 X

•

(3.3.4)

Solve for displacement vector ˆ ˆ [K]{X t+t } = {Rt+t }

•

(3.3.5)

Calculate the acceleration and velocity at time t + t ˙ t } − α3 {X ¨ t} ¨ t+t } = α0 {Xt+t − Xt } − α2 {X {X

(3.3.6)

˙ t+t } = {X ˙ t } + α6 {X ¨ t } + α7 {X ¨ t+t } {X

(3.3.7)

For the sake of brevity we now explain the above through a suitable numerical problem.

Example 3.3.1 A frame foundation supporting a compressor is subjected to El-Centro accelerogram as shown in Figure 3.3.1. The stiffness, mass and damping (non-proportional) matrix are given. Determine the response of the machine foundation based on the time history response.

200 [M] = 0

0 1000

and

7000 [C] = −2800

3000 −1200 [K] = −1200 51000

−2800 12300

Solution: The displacement history is shown in tabular form for the first 10 steps at time step of 0.02 seconds and the results of displacement and acceleration for node 2 and node 1 are finally shown graphically for 1566 steps in Figures 3.3.2 and 3. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 451

Sl. No.

Displacement at node 1

Velocity at node 1

Acceleration at node 1

Displacement at node 2

Velocity at node 2

Acceleration at node 2

1 2 3 4 5 6 7 8 9 10

0 2.98 × 10−06 1.18 × 10−05 2.59 × 10−05 4.98 × 10−05 9.13 × 10−05 1.50 × 10−04 2.13 × 10−04 2.66 × 10−04 3.07 × 10−04

0 2.98 × 10−04 5.82 × 10−04 8.35 × 10−04 1.55 × 10−03 2.59 × 10−03 3.25 × 10−03 3.06 × 10−03 2.27 × 10−03 1.80 × 10−03

0 2.98 × 10−02 −1.36 × 10−03 2.66 × 10−02 4.53 × 10−02 5.83 × 10−02 7.55 × 10−03 −2.68 × 10−02 −5.15 × 10−02 4.09 × 10−03

0.00 × 10+00 3.24 × 10−06 1.32 × 10−05 2.99 × 10−05 5.83 × 10−05 1.07 × 10−04 1.78 × 10−04 2.57 × 10−04 3.26 × 10−04 3.80 × 10−04

0 0.000324 0.000672 0.001001 0.001838 0.003077 0.003967 0.003899 0.003049 0.002377

0 0.032396 0.002417 0.030511 0.053184 0.07067 0.018316 −0.02506 −0.06 −0.00713

Displacement(d2)

2.00E-02 1.00E-02

19.6

18.2

16.8

15.4

14

12.6

11.2

9.8

8.4

7

5.6

4.2

1.4

-1.00E-02

2.8

0.00E+00 0

Displacement(m)

3.00E-02

-2.00E-02 Time step(sec)

Figure 3.3.2 Displacement History at node 2.

4 2

-4 Time steps

Figure 3.3.3 Acceleration response at node 1.

© 2009 Taylor & Francis Group, London, UK

19.8

18.5

17.2

15.8

14.5

13.2

11.9

10.6

9.24

7.92

5.28

3.96

2.64

1.32

-2

6.6

0 0

Acceleration at node 1 (m/sec2)

Acceleration(node 1)

452 Dynamics of Structure and Foundation: 2. Applications

This method as discussed earlier in Chapter 5 (Vol. 1) is applicable on full matrix when the mass, stiffness and damping matrix are all known. The technique is particularly suitable for cases, which has non-classical damping (where the matrix on orthogonalization does not de-couple). However for systems with large degree of freedom we rarely know the complete damping matrix and we normally deal with the modal damping ratio usually defined as a constant value for each mode for normal structural analysis. For instance in case of analysis of 3D framed building structure we do not (or cannot) define the damping matrix and the usual input is the modal damping ratio assumed constant for all modes. In such cases we can either form the Rayliegh damping coefficient α and β adapting the method as stated in Chapter 5 (Vol. 1) or proceed as mentioned hereafter. • •

As a first step we perform the usual eigen-value analysis and obtain the frequencies and the eigen vectors. Now knowing the modal damping ratio D (which is usually pre-defined) we decouple the equation into n number of equations (here n is the total numbers of degree of freedom of the system) of the form 2 {ξ¨i=1,n } + 2Di=1,n ωi = 1, n{ξ˙i=1,n } + ωi=1,n {ξi=1,n } = {u¨ g }

(3.3.8)

For a given time history the above can be expressed as 2 {ξ¨i=1,n } + 2Di=1,n ωi=1,n {ξ˙i=1,n } + ωi=1,n {ξi=1,n }

= √

−1 1 − D2

t

u¨ g (τ )e−Dωn (t−τ ) − Dωn sin 1 − D2 ωn (t − τ )dτ

0

+ ωn 1 − D2 cos 1 − D2 ωn (t − τ ) dτ (3.3.9)

•

•

For each of this equation we perform the time history response either by integration of the Duhamel Integral or by numerical integration based on any one of the methods as explained in Chapter 5 (Vol. 1) and find out the values of the displacement, velocity and acceleration and finally do a modal combination to obtain the response for the different mode. n In ¨t} = 1 [ϕn ]{ξt } and acceleration {u nsuch case the displacement {ut } = ¨ [ϕ ]{ ξ }. t 1 n The corresponding effective earthquake force is given by {Vn (t)} = [M] {u¨ t } .

© 2009 Taylor & Francis Group, London, UK

(3.3.10)

Analytical and design concepts for earthquake engineering 453

The total modal force for all modes are given by {Vn (t)} = [M] [ϕ] ωn2 {ξt }

(3.3.11)

For many large complex structures or finite element system with many degrees of freedom even the above process could be time consuming and very laborious, fortunately for many such systems, it is the first few modes which contribute signiﬁcantly to the inertial forces when the subsequent higher modes can be neglected without any appreciable error. In such case if for a system N × N if J number of modes (J << N) are deemed to be significant (which can very well be estimated from the modal mass participation). Then the mass matrix [M]N×N , stiffness matrix [K]N×N and the damping matrix [C]N×N can well be crunched down to a matrix of order J × J by the following operations ˆ J×J = [ϕ]J×N [M]N×N [ϕ]N×J [M] ˆ J×J = [ϕ]J×N [K]N×N [ϕ]N×J [K]

similarly, and

ˆ J×J = [ϕ]J×N [C]N×N [ϕ]N×J . [C]

(3.3.12)

ˆ J×J = modified stiffness matrix ˆ J×J = modified mass matrix of size J × J; [K] were, [M] of size J × J; [C]J×J = modified damping matrix of size J × J, and, [φ]J×N = the eigen vector for the first J modes of the structure of size N × N. Once the modified matrix is known we can very well undertake a time history analysis of this modified matrix and greatly reduce our computation time. We now explain the above theory by a suitable numerical problem.

Example 3.3.2 Shown in Figure 3.3.4 is a three-storied frame subjected to dynamic forces based on EL-Centro Earthquake as shown in Figure 3.3.1. The damping ratio for the structure is considered as 5%. Determine G

H

X3

E

F

X2

C

D

X1

3000

3000

3000 A

B

( All dimensions are in mm )

Figure 3.3.4 Sketch diagram of three-storied space frame.

© 2009 Taylor & Francis Group, London, UK

454 Dynamics of Structure and Foundation: 2. Applications

• • •

The fixed base natural frequencies of the structure. The fixed base eigen vectors. Displacement and shear force. Let us take, KAC = KDB = 1.5 × 103 kN/m;

KCE = KDF = 1.0 × 103 kN/m;

KEG = KFH = 0.75 × 103 kN/m; MEF = 400 KN sec2 /m;

MGH = 200 kN sec2 /m;

MCD = 400 KN sec2 /m.

Solution: The stiffness and mass matrix is given by ⎡

5000 [K] = ⎣−2000 0

⎤ −2000 0 3500 −1500⎦ −1500 1500

⎡

⎤

400

and

[M] = ⎣

400

⎦ 200

Considering, [[K] − [M][ω2 ]][ϕ] = [0] ω1 = 1.281 rad/sec; ω2 = 3.162 rad/s ec;

ω3 = 4.135 rad/sec.

Thus the time periods for the fixed base structure is given by T1 = 4.97 sec,

T2 = 1.987 sec,

T3 = 1.52 sec .

The mode shapes or the eigen vectors and normalised eigen vectors are ⎡

⎤ 1.00 1.0 1.0 0.5 −0.9208⎦ ; [φ] = ⎣2.1715 2.7816 −1.50 0.719 ⎡ ⎤ 0.01615 0.03244 0.0344512 −0.03172 ⎦ [ϕi ] = ⎣0.0350718 0.01622 0.04493 −0.02433 0.02477 Considering the orthogonal equation {ξ¨i } + 2Di ωi {ξ˙i } + ω2 {ξi } = {¨ag } we have the three equations as: ξ¨1 + 0.281ξ˙1 + 1.640961ξ1 = a¨ g ; ξ¨3 + 0.4135ξ˙3 + 17.098ξ3 = a¨ g .

© 2009 Taylor & Francis Group, London, UK

ξ¨2 + 0.3162ξ˙2 + 9.99824ξ2 = 0;

Analytical and design concepts for earthquake engineering 455

Performing the time history analysis based on Wilson-θ method for input accelerogram of El-Centro earthquake and combining the response based on the equation, {X} = [ϕ]{ξ }. We plot below the displacement and force history in Figures 3.3.5 and 6.

Displacement History 0.05 0.04 0.03

Modal disp1

0.01

Modal disp2 28.9

31.1

26.6

24.4

20

Modal disp3 22.2

15.5

17.8

13.3

8.88

11.1

4.44

-0.01

6.66

0

0 2.22

Displacement(meter)

0.02

-0.02 -0.03 -0.04 Time steps(sec)

Figure 3.3.5 Displacement history of the frame for the three modes.

Modal shear response(kN)

400

Shear1 shear 2 31.1

28.9

26.6

24.4

22.2

20

17.8

15. 5

13. 3

11. 1

8.88

6.66

4.44

2.22

0 0

Force(kN)

200

shear 3

-200 -400 Time step(sec)

Figure 3.3.6 Modal shear history of the frame for the three modes.

It will be observed that the major contribution is from the fundamental mode, the higher mode contribution is practically insignificant.

What has been explained above is the generic theory pertaining to earthquake dynamic and pseudo-static analysis. Though the above has been explained with respect to frames (or buildings) can be very easily be extended to a generic finite element model with the underlying principle remaining the same be the analysis is done based on response spectrum method or step by step integration. © 2009 Taylor & Francis Group, London, UK

456 Dynamics of Structure and Foundation: 2. Applications

We now show application of the above theories as applied to some special structures which are important to society and industry, have got some unique features and require some special analytical techniques.

3.3.1 Earthquake analysis of tall chimneys and stack like structure Tall chimneys and vertical self-supporting vessels are an important feature of power and petrochemical industry. Damage to them during an earthquake can have a severe consequence both in terms of economy and loss of human life. While it is expected that a power plant remains functional after an earthquake, which is essential to fight the aftermath of the disaster, leakage or damage of vertical vessels in refinery or chemical plants containing flammable or toxic liquid can create havoc to the environment and surrounding life18 . Though the analysis herein is discussed in terms of tall chimneys can well be applied for vertical self-supporting vessels also. With ever growing demand for power, engineering industry is churning out power plants of progressively higher capacity. To maintain the ecological balance as well as limit the environmental pollution, chimneys emitting the spent flue gas are also getting higher and higher everyday. In India it is now mandatory that for all fossil fuel power plants the height of chimneys be minimum 220 meter for 210 MW unit and 275 meter for 500 MW unit. While this though reduces the ground pollution concentration signiﬁcantly has posed new challenges to the structural engineers to come up with a safe design of these tall chimneys especially under wind and earthquake, which affects its behaviour signiﬁcantly. Unlike other tall structures the most dangerous thing about chimney is that these structures are basically a cantilever structure having one line of defence (the structures itself) and has practically no redundancy built in it. Thus during an earthquake if any portion of it develops a hinge would invariably make the system a mechanism with collapse being imminent. It is for this knowing the dynamic behaviour of the same under an earthquake loading is of primary importance. Fortunately as these structures have uniform distribution of mass and stiffness are more amenable to classical mathematical treatment; however one of the major controversies that remain with its behaviour is the level of damping to be considered in the analysis. While one school of thought prefers to use the standard damping ratio as used for RCC (5–7%), the other school of thought is that since of its huge mass (due to its self weight and lining) a major portion of the chimney remains under compression even under wind and seismic loading and thus remains un-cracked. Since propagation of cracking enhances the damping property of the system and does not occur in the major portion of the chimney, a much lower damping ratio of say 2% should be a

18 Though the reason was different some of the readers may remember the Bhopal gas tragedy in 1980’s in India where huge number of people perished and got disabled for life due to leakage of toxic gas from vessels in the plant of a multi-national Company.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 457

Plan View EI = constant

Figure 3.3.7 Typical multi- flue-chimney with its mathematical model.

more reasonable value. Unfortunately very little field observed instrumented data are available to come to any decisive conclusion on this issue. Chimneys, shown in Figure 3.3.7, are usually of two types • •

Multi-flue chimneys (used to cater to more than two power units at a time) having uniform cross section. Single flues (used to cater one or two units) usually having a tapered profile.

3.3.1.1

Analysis as proposed in IS-code

Before we start with the dynamic analysis of such tall structures we present herein the method as proposed in IS Code: As per IS code the time period of such chimneys considered as shown in Figure 3.3.4 are fixed at base and is given by ' T = CT

WH EAg

(3.3.13)

where, W = weight of chimney plus lining and all other accessories; H = height of chimney above the base; E = modulus of elasticity of the structural shell; A = area of cross section of the base; g = acceleration due to gravity; CT = constant which is a © 2009 Taylor & Francis Group, London, UK

458 Dynamics of Structure and Foundation: 2. Applications Table 3.3.1 Slenderness ratio

CT

Cv

5 10 15 20 25 30 35 40 45 50 or more

14.4 21.2 29.6 38.4 47.2 56 65 73.8 82.8 1.8 × (H/r)

1.02 1.12 1.19 1.25 1.30 1.35 1.39 1.43 1.47 1.50

function of the slenderness ratio; For circular section A = 2π rt; r = mean radius of the shell and t = thickness of the shell. The design base shear and moment for ﬁxed base is given by

z 0.5 z z 4 V = Cv Ah W 1.1 ; + 0.75 + 0.9 H H H 0.5 z 4 ¯ 0.4 z M = Ah W H + 0.6 H H

(3.3.14)

¯ = height of where, Cv = a coefficient which is a function of slenderness ratio; H ZI centre of gravity of the structure above base, and Ah = 2R , the seismic coefficient as per code. The values of Cv and CT are as furnished in Table 3.3.1. IS code does not furnish any expression for the tip deflection. 3.3.1.2

Dynamic analysis of tall chimneys

We start with the analysis of a multi flue chimney of uniform cross section based on Rayleigh Ritz technique to arrive at a closed form solution before extending the same to a numerical solution for a tapered cantilever. Since the outer core of a multi-flue chimney (usually termed as the wind shield) is of uniform cross section we consider it as a cantilever beam fixed at base. The free vibration equation of such beam is given by the expression (Hurty and Rubenstein 1967), EI

∂ 4w ∂z4

2 ∂ w + ρA =0 ∂t 2

(3.3.15)

here, E = elastic modulus of the beam material; I = moment of inertia of the beam; ρ = mass density of the beam material; A = area of cross section of the beam, and, © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 459

w = displacement of the beam and is a function of time and geometry and is depicted as w(z, t) = Y(z) · q(t)

(3.3.16)

Based on separation of variable technique the above partial differential equation can be separated into two linear differential equation and one of which is ( EI

)

d4Y

− λ4 Y = 0

dz4

where λ4 =

ρAω2 EI

(3.3.17)

The generic solution to this equation is given by (Murray 1967) Y = C1 sin λz + C2 cos λz + C3 sin hλz + C4 cos hλz

(3.3.18)

Imposing the four boundary conditions: 1

Y=0

2

dY =0 dz

3 4

d3Y dz

3

d2 Y dz2

at z = 0; at z = 0;

=0

at z = L;

=0

at z = L.

(3.3.19)

We have the shape function solution as Ym = sin

μm z μm z μm z μm z − sin h − αm cos − cos h H H H H

Here m = number of modes 1, 2, 3, . . . . . . . . . . μm = 1.875, 4.694, 7.855,

2m − 1 π. 2

For m = 1, 2, 3 . . . . . . m, etc. αm =

sin μm + sin hμm cos μm + cos hμm

We apply here the Rayliegh Ritz technique as described below. © 2009 Taylor & Francis Group, London, UK

(3.3.20)

460 Dynamics of Structure and Foundation: 2. Applications

For a conservative system if T is kinetic energy and V is the Potential energy of the system then at any time t the energy equations may be written in the form 1 T(t) = 2

H m(z) 0

here

y(z, t) =

n

∂y(z, t) ∂t

2 dz

(3.3.21)

ϕi (z)qi (t)

(3.3.22)

i=1

where y (z, t) = displacement function; ϕi (z) = admissible function, and, qi (t) = generalized co-ordinate. Substituting the above in the energy equation we have

T(t) =

1 2

H m(z)

n

⎤

⎡ n ϕi (z)q˙ i (t) ⎣ ϕj (z)q˙ j (t)⎦dz

i=1

j=1

0

⎡H ⎤ n n 1 = q˙ i (t)q˙ j (t) ⎣ m(z)ϕ i (z)ϕ j (z)dz⎦ 2 i=1 j=1

(3.3.23)

0

from which we conclude that the mass coefficient has the form ⎡H ⎤ mij = ⎣ m(z)ϕ i (z)ϕ j (z)dz⎦

for i, j = 1, 2, 3 . . . . . . n

(3.3.24)

0

For potential energy V we have 1 V(t) = 2

H 0

∂ 2 y(z, t) EI(z) ∂z2

2 dz

⎤ ⎡H n n 2 ϕ (z) d 2 ϕ (z) 1 d j i ⎦ dz, = q˙ i (t)q˙ j (t) ⎣ EI(z) 2 dz2 dz2 i=1 j=1

(3.3.25)

0

from which we conclude that stiffness has the form H kij =

EI(z) 0

d 2 ϕi (z) d 2 ϕj (z) dz2

© 2009 Taylor & Francis Group, London, UK

dz2

dz

for i, j = 1, 2, 3. . . . . . n

(3.3.26)

Analytical and design concepts for earthquake engineering 461

Since a multi-flue stack is considered to have a constant EI the stiffness and mass expression is given as H kij = EI

d 2 ϕi (z) d 2 φj (z) dz2

0

dz2

⎡ γA and mij = ⎣ g

dz

H

⎤ ϕi (z)ϕ j (z)dz⎦

(3.3.27)

0

Considering the shape function as μi z μi z μi z μi z − sin h − αi cos − cos h H H H H μj z μj z μj z μj z ϕj = sin − sin h − αj cos − cos h H H H H

ϕi = sin

and (3.3.28)

The double derivative of the above is given by ϕi = ϕj =

μ2i μi z μi z μi z μi z

−sin cos − sin h + α + cos h i H H H H H2 μ2j H2

−sin

and (3.3.29)

μ z

μj z μj z μj z j − sin h + αj cos + cos h H H H H

Before performing the integration we change the above to generalized co-ordinate dz z when dξ = by considering, ξ = and as z → 0, ξ → 0 and as z → H, ξ → 1 H H based on above we can now express the double derivative as f (ξ )i = f (ξ )j =

μ2i −sin μi ξ − sin hμi ξ + αi (cos μi ξ + cos hμi ξ ) , 2 H

and (3.3.30)

μ2j H2

−sin μj ξ − sin hμj ξ + αj (cos μj ξ + cos hμj ξ )

Thus stiffness of the system can now be expressed as kij =

1 EIμ2i μ2j

H3

f (ξ )i f (ξ )j dξ

(3.3.31)

0

and mass of the system is given by γ AH mij = g

1 f (ξ )i f (ξ )j dξ ,

where i = j = 1, 2, 3, . . . . . . m,

(3.3.32)

0

For most of the chimneys it is found that first three modes are sufficient to predict the dynamic response, as modal mass participation is almost 100% by this. © 2009 Taylor & Francis Group, London, UK

462 Dynamics of Structure and Foundation: 2. Applications

Thus for the first three modes the stiffness matrix19 is given by ⎡ μ41

1

⎤ f (ξ )21 dξ

⎢ ⎢ 0 ⎢ 1 EI ⎢ 2 2 ⎢ [K]ij = 3 ⎢μ2 μ1 f (ξ )2 f (ξ )1 dξ H ⎢ 0 ⎢ ⎣ 2 2 1 μ3 μ1 f (ξ )3 f (ξ )1 dξ

μ42 μ23 μ22

0

1

1 0

f2 (ξ )2 dξ

f (ξ )3 f (ξ )2 dξ

0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1 ⎦ μ43 f (ξ )23 dξ 0

(3.3.33) and the mass matrix is given by ⎡

1

⎤ (ξ )21 dξ

f ⎢ ⎢ 0 ⎢1 γ AH ⎢ ⎢ f (ξ )2 f (ξ )1 dξ [M]ij = ⎢ g ⎢0 ⎢1 ⎣ f (ξ )3 f (ξ )1 dξ

1 1

0

f2 (ξ )2 dξ

0

f (ξ )3 f (ξ )2 dξ

0

1 0

f (ξ )23 dξ

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.3.34)

The above integrals can very easily be solved based on Simpson’s 1/3rd rule between the limits 1 to 0 when we have

[K]3×3

⎡ 22.936 EI ⎣ −0.002 = 3 H 0.006

−0.002 468.044 −0.11

⎤ 0.006 −0.11 ⎦ 3812.81

and the mass matrix is given by, ⎡ ⎤ 1.855 0 0 W⎣ 0 0.964 0 ⎦ [M]3×3 = g 0 0 1.002

(3.3.35)

(3.3.36)

where, W = total weight of the shell + internal slabs + brick linings. Converting the above into standard eigen-value form of Aϕ = λϕ and applying the generalized Jacobi technique20 we have ⎡ 12.364 0 EIg ⎣ 0 485.523 [λ] = WH 3 0 0

⎤ 0 0 ⎦ and 3805

19 The stiffness and mass matrix is symmetric about is diagonal. 20 The technique has been worked in detail in Chapter 5 (Vol. 1).

© 2009 Taylor & Francis Group, London, UK

(3.3.37)

Analytical and design concepts for earthquake engineering 463

Eigen vectors for the first three modes 3 1

f1(x)

0

f2(x) 1

0. 9

0. 8

0. 7

0. 6

0. 5

0. 4

0. 3

0. 1

-2

0. 2

-1 0

Eigenvectors

2

f3(x)

-3 Z/H

Figure 3.3.8 Eigen vectors for the first three modes.

the eigen vectors are given as ⎡

−1.0

⎢ −6 [ϕ] = ⎢ ⎣4.384 × 10 1.579 × 10

−6

2.278 × 10−6 −1 −3.307 × 10−5

⎤⎡

⎤ f1 (ξ ) ⎥⎢ ⎥ −3.437 × 10−5 ⎥ ⎦ ⎣f2 (ξ )⎦ f3 (ξ ) 1 8.528 × 10−7

(3.3.38)

The eigen vector plots for the first three modes are as shown in Figure 3.3.8. since [λ] = ω2 and T = 2π ω we have ⎡

1.787 [T] = ⎣ 0 0

0 0.285

⎤' 0 WH 3 0 ⎦ EIg 0.102

Thus for the first three modes we have Mode number 1

2

3

© 2009 Taylor & Francis Group, London, UK

Time period (secs) ' WH3 T1 = 1.787 EIg ' WH3 T2 = 0.285 EIg ' WH3 T2 = 0.102 EIg

(3.3.39)

464 Dynamics of Structure and Foundation: 2. Applications

3.3.1.3

Transformation to the format of IS-code

We have shown above that for fundamental mode the time period is given by ' T1 = 1.787

WH 3 , EIg

(3.3.40)

Now, considering I = Ar2 , where A = area of the stack at the base and r = radius of gyration, the equation can be written in the format of ' T1 = 1.787ψ

WH EAg

(3.3.41)

where ψ = slenderness ratio of the stack @ H/r. Considering, CT = 1.7873ψ, we have ' T1 = CT

WH EAg

(3.3.42)

which is the same format as presented in the code. If we compare the values of CT as furnished in code and as derived here it will be observed that code gives a higher value of time period vis-a-vis what is presented here. Since the accuracy of Rayleigh Ritz Method is dependent on the choice of the assumed shape function it is evident that code had used a different shape function then what has been presented herein21 . The various values of CT as proposed by the present method and what has been proposed in the code are as mentioned hereunder22 . Slenderness ratio (H/r)

CT (as per IS code)

CT (1st Mode)

CT (2nd Mode)

CT (3rd Mode)

5 10 15 20 25 30 35 40 45 50

14.4 21.2 29.6 38.4 47.2 56 65 73.8 82.8 1.8 × (H/r)

8.935 17.87 26.81 35.74 44.675 53.61 62.54 71.5 80.41 89.35

1.425 2.85 4.275 5.70 7.125 8.55 9.975 11.4 12.83 14.25

0.51 1.02 1.53 2.04 2.55 3.06 3.57 4.08 4.59 5.10

21 Present analysis would give slightly different (higher) values of moments and shears then what has been proposed in the code. 22 IS-1893 does not propose any CT values for 2nd or 3rd mode.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 465

Calculation of amplitude In terms of response spectrum analysis displacement Sd is given by, Sd = Sa /ω2 . Expressing it in terms of codal formulation, we may express it as Sd = κi

ZI Sa 2R ω2

(3.3.43)

where, κi = modal participation factor and is given by

n i=1

mi ϕ i /

n i=1

mi ϕi2 .

For an element of length dz the above can be expressed as κi =

γA H g 0 ϕi dz γA H 2 g 0 ϕi dz

1 = 10

fi (ξ ) dξ

0 fi

(ξ )2 dξ

(3.3.44)

in which, Z = zone coefficient; I = importance factor, and, R = ductility factor. Integration of the mass participation factor within limits 1 to 0 for the first three modes gives Mode number

Mass participation factor (κi )

1 2 3

0.575 0.442 0.254

ZI Now considering, β = 2R , an IS code factor, we can write the time dependent function of displacement as

Sd = κi β

Sa ω2

(3.3.45)

Thus for the first mode, we have Sd = 0.575β

Sa1 WH 3 Sa1 WH 3 = 0.0465β 12.364EIg EIg

(3.3.46)

Let the complete function is given by, w(z, t) = φ(z) · q(t), thus for this case w(z, t) = 0.0465β

Sa1 WH 3 [f1 (ξ ) + 4.384 × 10−6 f2 (ξ ) + 1.579 × 10−6 f3 (ξ )], EIg

neglecting the influence of the second and third mode whose influence are negligible we have w(z, t) = −0.0465β

Sa1 WH 3 [f 1 (ξ )] EIg

© 2009 Taylor & Francis Group, London, UK

(3.3.47)

466 Dynamics of Structure and Foundation: 2. Applications 2

For calculation of moment and shear we know that, EI ddzw2 = −Mz , and hence, 1 Sa1 Mz = −0.0465β(WH ) [−μ21 f (ξ )] g H2 Sa1 = 0.163β (WH) [f1 (ξ )] g

3

Again considering, Vz = Vz = 0.306βW

Sa1 g

dMz , dz

(3.3.48)

we have

[f1 (ξ )]

(3.3.49)

Proceeding in identical manner for the second mode, we have

w(z, t) = 9.103 × 10−4 β

Sa2 WH 3 × [− f2 (ξ )], EIg

(3.3.50)

ignoring the influence of mode one and three as their influence are very small.

Mz = −9.103 × 10

= 2.005 × 10

−4

−2

βWH

βWH

Vz = 9.415 × 10−2 βW

Sa2 g

Sa2 g

Sa2 g

[−μ22 f2 (ξ )]

[−f2 (ξ )]

and

[f2 (ξ )]

(3.3.51)

Similarly for the third mode, we have

w(z, t) = 6.675 × 10−5 β

Sa3 WH 3 [f 3 (ξ )] EIg

Mz = 4.12 × 10−3 βWH Vz = 0.0323βW

© 2009 Taylor & Francis Group, London, UK

Sa2 g

Sa3 g

[f3 (ξ )],

[f3 (ξ )]

and

(3.3.52)

Analytical and design concepts for earthquake engineering 467

The above can thus be generalized along the height of the chimney as Sai WH 3 , EIg Sai Mξ = (Coeff m)β · W · H , g

wi (ξ , t) = (Coeff d)β

and

Vξ = (Coeff v)βW

Sai g

.

(3.3.53)

Here ξ = z/H, the height ratio and i = number of mode. It will be observed that once we know the values of coefficients within parenthesis for i = 1, 2, 3, we can immediately find out the dynamic amplitude, shear and moments without going through the elaborate process of dynamic analysis. The coefficients for dynamic amplitude, moment and shears are as stated hereafter ξ = z/H Coeff d1 Coeff d2

Coeff d3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0.00000 0.44401 0.00001 0.41345 0.00003 0.38291 0.00006 0.35244 0.00008 0.32211 0.00010 0.29203 0.00010 0.26234 0.00009 0.23318 0.00007 0.20472 0.00004 0.17717 0.00000 0.15072 −0.00003 0.12559 −0.00006 0.10203 −0.00008 0.08026 −0.00009 0.06054 −0.00007 0.04313 −0.00005 0.02829 −0.00001 0.01628 0.00004 0.00736 0.00009 0.00181 0.00015 −0.00011

0.00000 0.00054 0.00212 0.00466 0.00809 0.01232 0.01729 0.02291 0.02911 0.03584 0.04300 0.05054 0.05840 0.06652 0.07483 0.08330 0.09188 0.10053 0.10922 0.11793 0.12664

0.00000 0.00005 0.00017 0.00034 0.00054 0.00075 0.00094 0.00110 0.00122 0.00128 0.00128 0.00120 0.00105 0.00084 0.00057 0.00024 −0.00012 −0.00052 −0.00093 −0.00136 −0.00178

Coeff m1

Coeff m2

Coeff m3

Coeff v1

Coeff v2

Coeff v3

0.03938 0.02997 0.02063 0.01149 0.00276 −0.00531 −0.01248 −0.01851 −0.02320 −0.02642 −0.02809 −0.02821 −0.02689 −0.02429 −0.02068 −0.01639 −0.01180 −0.00736 −0.00356 −0.00088 0.00015

0.00825 0.00502 0.00189 −0.00095 −0.00325 −0.00479 −0.00542 −0.00509 −0.00390 −0.00204 0.00019 0.00244 0.00439 0.00575 0.00635 0.00614 0.00523 0.00387 0.00242 0.00136 0.00119

−0.61200 −0.61189 −0.61112 −0.60907 −0.60518 −0.59892 −0.58982 −0.57746 −0.56144 −0.54143 −0.51713 −0.48829 −0.45470 −0.41618 −0.37259 −0.32382 −0.26979 −0.21044 −0.14574 −0.07568 −0.00024

−0.18830 −0.18793 −0.18549 −0.17947 −0.16887 −0.15328 −0.13276 −0.10786 −0.07951 −0.04899 −0.01780 0.01241 0.03993 0.06308 0.08032 0.09027 0.09181 0.08409 0.06652 0.03877 0.00071

−0.06460 −0.06401 −0.06040 −0.05212 −0.03896 −0.02192 −0.00290 0.01567 0.03129 0.04183 0.04589 0.04301 0.03382 0.01995 0.00380 −0.01175 −0.02373 −0.02945 −0.02674 −0.01412 0.00933

We now elaborate the theory with a suitable numerical solution. Example 3.3.3 A multi-Flue chimney has height of 220 m. Its estimated weight including lining and internal slab is 175,000 kN. The diameter of the chimney at is 22.0 m having average shell thickness of 650 mm. The chimney is situated in a place depicted by zone IV as per IS-code resting on medium soil. Find the deflection, moment and shear for first three modes and the maximum design moments and shears. Consider grade of concrete used as M30 and damping ratio as 5% for the three modes. © 2009 Taylor & Francis Group, London, UK

468 Dynamics of Structure and Foundation: 2. Applications

Solution: Outside diameter of chimney = 22 m; Shell thickness = 650 mm, Inside diameter of chimney = 20.7 m. √ Young’s Modulus of concrete = 5700 30 × 103 = 31220186 kN/m2 . Moment of Inertia at base = Area of chimney at base =

π 4 22 − 20.74 = 2486.39 m4 64

π 2 22 − 20.72 = 43.59745 m2 4

I 220 = 29.18139. = 7.552 m; Slenderness Ratio = 7.552 A ' WH Considering Time period = T1 = 1.787ψ for first mode, we have EAg

Radius of gyration =

T1 = 1.787 × 29.18139

175000 × 220 = 2.8 sec . 31220186 × 43.59745 × 9.81

Similarly for the mode 2, we have T2 = 0.285 × 29.18139

175000 × 220 31220186 × 43.59745 × 9.81

= 0.446 sec, givesSa /g = 2.5. And for the mode 3, we have T2 = 0.102 × 29.18139

175000 × 220 31220186 × 43.59745 × 9.81

= 0.1594 sec s, gives Sa /g = 2.5. For Zone IV medium stiff soil as per IS-1893 Z = 0.24 I = 1.5 R = 2.0 ➔ β =

0.24 × 1.5 = 0.09 2×2

Substituting the values of W, H, E, I, Sa /g and β, and multiplying it by the coefficients as furnished earlier we have

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 469

ξ

1 (m)

2 (m)

3 (m)

M1 (kN · m)

M2 (kN · m)

M3 (kN · m)

V1 (kN)

V2 (kN)

V3 (kN)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0 −4.9E-06 −7.5E-05 −0.00036 −0.00109 −0.00253 −0.00498 −0.00875 −0.01413 −0.0214 −0.03082 −0.04258 −0.05685 −0.07374 −0.09334 −0.11566 −0.14071 −0.16845 −0.19883 −0.23182 −0.26735

0 −1.2E-05 −0.00016 −0.00068 −0.00173 −0.00333 −0.00529 −0.00729 −0.00893 −0.00983 −0.00974 −0.00861 −0.00664 −0.00423 −0.00192 −0.00035 −9.4E-05 −0.00163 −0.00524 −0.01107 −0.01911

0 7.97E-06 9.23E-05 0.00032 0.000648 0.000931 0.001015 0.00084 0.000491 0.000151 6.82E-07 0.00011 0.000398 0.000678 0.000767 0.0006 0.000276 2.37E-05 9.28E-05 0.000657 0.001775

327879 284303.7 243858.8 206595.1 172575.8 141858.9 114482.5 90452.24 69730.34 52228.61 37803.75 26256.23 17332.5 10730.79 6110.51 3105.154 1338.709 445.3734 92.42883 6.071341 0

12543.67 7265.948 3441.049 1067.111 61.55624 228.489 1260.757 2773.24 4358.467 5651.426 6388.65 6447.994 5859.733 4785.996 3472.652 2184.266 1137.157 447.2201 107.6952 8.071409 0

865.5315 320.4488 45.18539 11.61869 134.9782 292.6448 374.0941 330.6002 194.2332 53.75978 0.338409 73.71389 239.4717 409.8315 495.0058 454.2715 316.2084 156.1138 44.97968 3.881338 0

2823.571 2822.533 2815.416 2796.584 2760.962 2704.16 2622.622 2513.781 2376.22 2209.821 2015.894 1797.282 1558.446 1305.506 1046.263 790.1897 548.3924 333.5583 159.8887 43.02797 5.67E-05

1303.114 1297.935 1264.562 1183.756 1048.157 863.554 647.893 427.7108 232.5303 88.35721 11.71511 5.593723 58.30672 145.6543 236.0995 298.0526 307.9525 257.7331 160.4891 53.6873 1.04E-05

242.3699 237.9722 211.8813 157.8184 88.21899 27.95214 0.499664 14.16074 56.56307 101.0252 121.2799 106.0343 64.83385 21.74064 0.493604 9.901529 38.26835 60.66253 55.77621 24.17968 1.43E-05

The design values are obtained by the SRSS values of the three modes and are as given here under

ξ

D(comb) m

Mcomb kN · m

Vcomb kN

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0 1.55E-05 0.000203 0.000836 0.002147 0.004283 0.007337 0.011419 0.01672 0.023551 0.032319 0.043441 0.057237 0.073869 0.093362 0.115663 0.140709 0.168457 0.198904 0.23208 0.268037

328120 284396.7 243883 206597.9 172575.9 141859.4 114490.1 90495.35 69866.69 52533.5 38339.78 27036.49 18297.8 11756.85 7045.756 3823.528 1784.729 650.1806 148.8775 10.82006 0

3119.199 3115.759 3093.635 3040.899 2954.543 2838.836 2701.465 2549.947 2388.24 2213.893 2019.572 1800.416 1560.883 1313.786 1072.572 844.5906 630.1059 425.8726 233.3069 72.92729 5.94E-05

© 2009 Taylor & Francis Group, London, UK

470 Dynamics of Structure and Foundation: 2. Applications

Bending Moment under earthquake

300000

M1 M2

200000

M3 Mcomb

100000

9 0.

6

75

0.

0.

45

0.

0.

0

0.

3

0 -100000

15

Momnet(kN.M)

400000

Z/H

Figure 3.3.9 Bending Moment diagram of the chimney.

Shear force(kN)

Shear force diagram for three modes 4000 V1

3000

V2

2000

V3

1000

Vcomb 9 0.

6

75 0.

0.

3

45 0.

0.

0.

15

0

0

Z/H

Figure 3.3.10 Shear force diagram of the chimney.

We show in Figs. 3.3.9 and 10, the Modal moments shear and SRSS values.

3.3.1.4 Analysis of single f lue tapered chimney For single flue chimney due to thermodynamic reason and to enhance the exit velocity of the flue gas23 , are usually provided with a tapered section as shown in Figure 3.3.11. Besides the reason cited, the choice is also structurally reasonable for the moment and shear increases from zero at top to maximum at base. Based on this the obvious economic design be that which has a section minimum at top and maximum at base. As the section has a varying profile (generally linear) the mathematical treatment as shown for the multi-flue chimney with constant EI becomes complex for a closed form solution except for the fundamental mode.

23 The flue gas needs sufficient exit velocity to reduce the ground level pollution concentration.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 471

Variable EI

Figure 3.3.11 Single Flue Tapered Chimney and its mathematical model.

However this can very easily be solved by applying numerical techniques and arrive at an accurate answer. In case of tapered chimneys the numerical solution is preferable because though in most of the cases the profile is linear however from stress point of view and also to diminish the amplitude at the top, the profile usually has a number of transition zones (i.e. the slope often changes at two or three positions thus have varying integral functions with different limits). Secondly the brick liner inside the chimney shell which reduces the temperature differential across the chimney shell also undergoes change in thickness after a certain level thus making the mass function discontinuous which surely makes the choice of a numerical solution more attractive. However one additional step on has to do in this case is to perform the eigen value analysis which was already implicit in the calculation for chimneys with constant sections. The theory presented earlier can be modified for numerical analysis as follows: As the moment of inertia of the section is varying the stiffness equation can be expressed as

kij =

1 Eμ2i μ2j

H4

Iz ϕ (z)i ϕ (z)j dz

(3.3.54)

0

In which, Iz is moment of inertia considered to be varying at different height z. where, μ2i μi z μi z μi z μi z

−sin cos − sin h + α + cos h i H H H H H2 2 2 μ μ μi z μi z − sin h ➔ ϕi = i2 F (z) and F (z) = i2 −sin H H H H μi z μi z

+ cos h + αi cos H H

ϕi =

© 2009 Taylor & Francis Group, London, UK

(3.3.55)

472 Dynamics of Structure and Foundation: 2. Applications

Thus the stiffness matrix can now be written as [K]3×3 ⎡ H 2 4 ⎢ μ1 Iz F1 (z)dz ⎢ 0 ⎢ H H E ⎢ 2 2 ⎢ = 4 ⎢μ2 μ1 Iz F2 (z)F1 (z)dz μ42 Iz F2 2 (z)dz H ⎢ 0 0 ⎢ H ⎣ 2 2 H 2 2 μ3 μ1 Iz F3 (z)F1 (z)dz μ3 μ2 Iz F3 (z)F2 (z)dz 0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ H ⎦ μ43 Iz F3 2 (z)dz 0

(3.3.56) Similarly Mass equation can now be written as ⎡

⎤

H (γc Ac + γb Ab ) ϕ12 (z)

⎢ ⎢ 0 H 1⎢ ⎢ [M]ij = ⎢(γc Ac + γb Ab ) ϕ2 (z)ϕ1 (z)1 g⎢ 0 ⎢ ⎣ H (γc Ac + γb Ab ) ϕ3 (z)ϕ1 (z)

H

(γc Ac + γb Ab ) ϕ22 (z) H

0

(γc Ac + γb Ab ) ϕ3 (z)ϕ2 (z)

0

0

H (γc Ac + γb Ab ) ϕ32 (z)

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0

(3.3.57) where, γc = unit weight of concrete; γb = unit weight of brick lining; Ac = area of RCC shell at any height z, and Ab = area of Brick lining at any height z. Each of the above term of the stiffness and mass matrix are to be obtained by numerical integration between the limits 0-H. For numerical solution for μ and α following values are to be adopted. Mode

1

2

3

μ α

1.875 1.362221

4.694 0.981868

7.855 1.000776105

After the stiffness and mass matrix are formed, an eigen values analysis needs to performed based on the equation [K] − [M]ω2 = 0

(3.3.58)

Once the eigen values vis-á-vis time periods are known, the Sa /g values can be obtained from the response curve as furnished in the codes. The displacement amplitude is thus furnished by the equation ZI Sa (3.3.59) Sd = κi 2R ω2 and the complete solution is given by the expression24 24 Here κi remains same as the case with constant EI as show earlier.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 473

wi (z, t) = κi

ZI Sa [ϕii ] Fi (z). 2R ω2

(3.3.60)

Thus w1 (z, t) = κ1

ZI Sa1 [ϕ11 F1 (z) + ϕ12 F2 (z) + ϕ13 F3 (z)] 2R ω12

w2 (z, t) = κ2

ZI Sa2 [ϕ21 F1 (z) + ϕ22 F2 (z) + ϕ23 F3 (z)] 2R ω22

w3 (z, t) = κ3

ZI Sa3 [ϕ31 F1 (z) + ϕ32 F2 (z) + ϕ33 F3 (z)] 2R ω32

and

(3.3.61)

where, [ϕii ] = eigen vector value for the mode I. The moment and shear are then obtained from the equation EI

d2w = −Mz dz2

and EI Sa1 EI

M1 (z, t) = −κ1 β where, β =

d3w = −Vz , dz3

which gives

[ϕ11 μ21 F1 (z) + ϕ12 μ22 F2 (z) + ϕ13 μ23 F3 (z)]

ω12 H 2

(3.3.62)

ZI 2R .

M2 (z, t) = −κ2 β M3 (z, t) = −κ3 β

Sa2 EI ω22 H 2 Sa3 EI ω32 H 2

[ϕ21 μ21 F1 (z) + ϕ22 μ22 F2 (z) + ϕ23 μ23 F3 (z)]

and

[ϕ31 μ21 F1 (z) + ϕ32 μ22 F2 (z) + ϕ33 μ23 F3 (z)]

(3.3.63)

3

Considering EI ddzw3 = −Vz , V1 (z, t) = −κ1 β V2 (z, t) = −κ2 β V3 (z, t) = −κ3 β

Sa1 EI ω12 H 3 Sa2 EI ω22 H 3 Sa3 EI ω32 H 3

[ϕ11 μ31 F1 (z) + ϕ12 μ32 F2 (z) + ϕ13 μ33 F3 (z)] [ϕ21 μ31 F1 (z) + φ22 μ32 F2 (z) + ϕ23 μ33 F3 (z)] [ϕ31 μ31 F1 (z) + φ32 μ32 F2 (z) + ϕ33 μ33 F3 (z)]

and (3.3.64)

where, μi z μi z μi z μi z

Fi (z) = −cos − cos h + αi −sin + sin h H H H H We explain the above with a suitable numerical example. © 2009 Taylor & Francis Group, London, UK

(3.3.65)

474 Dynamics of Structure and Foundation: 2. Applications

Example 3.3.4 A 220 m tall RCC chimney has properties as shown hereafter. Calculate the first three fundamental time period and seismic response for seismic zone IV, with site having medium soil. The data for the chimney are given below: Height of chimney = 220 m; Diameter of shell at bottom = 22 m; Shell thickness at bottom = 650 mm; Diameter of shell at top = 5.0; Shell thickness at top = 250 mm; Air gap between shell and lining = 100 mm althrough; Thickness of brick lining = 150 mm from 220 to 150 m; Thickness of brick lining = 230 mm from 150 to 25 m; unit weight of concrete = 25 kN/m3 ; Unit weight of brick = 22 kN/m3 ; Grade of concrete = M35; Zone coefficient = 0.24; Importance factor = 1.5; R(Ductility factor) = 2.0. Solution: For the problem the earthquake factor =

ZI 0.24 × 1.5 = = 0.09 2R 4

Econc = 31220185.78 kN/m2

z

Outside diameter

Inside diameter

Outside dia lining

Inside dia Lining

Area of lining

Area of concrete

Moment of inertia

0 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

22 21.15 20.3 19.45 18.6 17.75 16.9 16.05 15.2 14.35 13.5 12.65 11.8 10.95 10.1 9.25 8.4 7.55 6.7 5.85 5

20.7 19.8925 19.085 18.2775 17.47 16.6625 15.855 15.0475 14.24 13.4325 12.625 11.8175 11.01 10.2025 9.395 8.5875 7.78 6.9725 6.165 5.3575 4.55

0 0 0 18.0775 17.27 16.4625 15.655 14.8475 14.04 13.2325 12.425 11.6175 10.81 10.0025 9.195 8.3875 7.58 6.7725 5.965 5.1575 4.35

0 0 0 17.6175 16.81 16.0025 15.195 14.3875 13.58 12.7725 11.965 11.1575 10.35 9.5425 8.895 8.0875 7.28 6.4725 5.665 4.8575 4.05

0 0 0 12.89600222 12.31252993 11.72905763 11.14558534 10.56211304 9.978640746 9.39516845 8.811696154 8.228223859 7.644751563 7.061279268 4.262355833 3.881830423 3.501305012 3.120779602 2.740254192 2.359728782 1.979203372

43.5974521 40.5351404 37.5834816 34.7424755 32.0121223 29.3924218 26.8833741 24.4849792 22.1972371 20.0201477 17.9537111 15.9979274 14.1527964 12.4183182 10.7944927 9.28132008 7.87880022 6.58693314 5.40571884 4.33515733 3.37524861

2486.389939 2135.781491 1823.567498 1546.84251 1302.815147 1088.808108 902.2581633 740.7161567 601.8470066 483.4297052 383.3573185 299.6369863 230.3899223 173.851414 128.3708229 92.41158428 64.55120714 43.48127451 28.00744321 17.04944393 9.641081217

Next we define the function fi (ξ ) for the first three modes and then multiplying H and integrating the expression m1 = (γc Ac + γb Ab ) 0 φ12 (z)dz etc. we obtain mass matrix as shown hereafter. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 475

z

f1 (ξ ) f2 (ξ )

f3 (ξ )

f1 (x) · f1 (x) f2 (x) · f1 (x) f2 (x) · f2 (x) f3 (x) · f1 (x) f3 (x) · f2 (x) f3 (x) · f3 (x)

0 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

0.00 0.01 0.05 0.10 0.17 0.27 0.37 0.49 0.63 0.77 0.92 1.09 1.26 1.43 1.61 1.79 1.98 2.16 2.35 2.54 2.72

0.00 0.13 0.46 0.85 1.21 1.45 1.51 1.38 1.05 0.58 0.04 −0.50 −0.95 −1.24 −1.32 −1.16 −0.79 −0.23 0.46 1.22 2.00

0.00 0.14 1.96 11.60 32.43 69.73 126.80 205.03 303.73 420.17 549.82 686.68 823.77 953.56 942.19 1019.18 1070.23 1091.06 1078.68 1031.49 949.38

0.00 0.05 0.18 0.37 0.59 0.82 1.03 1.21 1.34 1.41 1.40 1.32 1.16 0.92 0.62 0.27 −0.14 −0.57 −1.03 −1.49 −1.96

0.00 0.59 7.81 42.86 110.18 215.60 352.36 504.71 650.94 767.47 833.09 832.57 759.11 615.44 364.48 150.83 −74.41 −288.90 −472.15 −607.65 −684.29

114105.17 43303.49

0.00 2.51 31.09 158.35 374.36 666.56 979.13 1242.41 1395.06 1401.84 1262.32 1009.45 699.53 397.22 141.00 22.32 5.17 76.50 206.67 357.97 493.21

0.00 1.59 19.60 98.31 225.52 381.57 516.23 573.05 510.50 318.23 23.32 −315.27 −621.92 −824.51 −770.21 −661.89 −427.87 −116.81 210.18 495.41 697.70

0.00 6.77 78.02 363.21 766.23 1179.70 1434.49 1410.63 1094.07 581.27 35.34 −382.25 −573.11 −532.15 −297.95 −97.96 29.75 30.93 −92.00 −291.85 −502.88

112513.56 4217.34

54348.47

0.00 18.25 195.82 833.10 1568.28 2087.90 2101.62 1601.62 858.02 241.03 0.99 144.75 469.53 712.92 629.62 429.85 171.06 12.51 40.95 237.94 512.74 136827.21

Integrating each of the above term by Simpson’s 1/3rd rule25 and dividing each of the above terms by g = 9.81 we have the mass matrix as ⎡

11631.5 [M] = ⎣ 4414.2 429.997

4414.2 11469.27 5540

⎤ 429.997 5540 ⎦ kN-sec2 /m 13948

Again for stiffness matrix we show the functions fi (ξ ) as hereafter and Eμ2 μ2 1 applying the expression kij = Hi 4 j 0 Izφ (z)i φ (z)j dz we have f1 (x) f2 (x) f3 (x) f1 (x) · f1 (x) f2 (x) · f1 (x) f2 (x) · f2 (x) f3 (x) · f1 (x) f3 (x) · f2 (x) f3 (x) · f3 (x)

z 0 11 22 33 44 55

2.72 1.96 2.00 3283.58 2.54 1.49 1.22 2445.70 2.35 1.03 0.46 1791.12 2.16 0.57 −0.23 1287.15 1.98 0.14 −0.79 905.58 1.79 −0.27 −1.16 622.12

14833.25 9030.09 4913.99 2136.53 395.01 −576.65

67007.83 33341.16 13481.69 3546.40 172.30 534.50

42337.64 20605.66 6118.56 −2422.39 −6355.70 −7091.03

191256.31 76080.79 16786.46 −4020.89 −2772.32 6572.75

545891.08 173607.84 20901.32 4558.87 44606.72 80825.06 (continued)

25 I = h/3[(y0 + 4(y1 + y3 + y5) + · · · · · · · · · + yn − 1 + 2)(y2 + y4 + · · · · · · · · · + yn − 2) + yn]

© 2009 Taylor & Francis Group, London, UK

476 Dynamics of Structure and Foundation: 2. Applications

z

f1 (x) f2 (x) f3 (x) f1 (x) · f1 (x) f2 (x) · f1 (x) f2 (x) · f2 (x) f3 (x) · f1 (x) f3 (x) · f2 (x) f3 (x) · f3 (x)

66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

1.61 1.43 1.26 1.09 0.93 0.77 0.63 0.49 0.37 0.27 0.17 0.10 0.05 0.01 0.00

−0.62 −0.92 −1.16 −1.32 −1.40 −1.41 −1.34 −1.21 −1.03 −0.82 −0.59 −0.37 −0.18 −0.05 0.00

−1.32 −1.24 −0.95 −0.50 0.04 0.58 1.05 1.38 1.51 1.45 1.21 0.85 0.46 0.13 0.00

−1008.36 −1091.32 −976.03 −772.62 −554.21 −362.68 −215.99 −115.90 55.01 −22.39 −7.41 −1.81 −0.26 0.01 0.00

2443.96 4413.37 5635.75 5869.80 5261.92 4150.99 2900.50 1787.65 957.77 433.67 157.71 41.78 6.48 0.30 0.00

−5968.29 −4094.23 −2238.81 −818.80 43.83 421.35 474.44 368.52 225.67 110.94 42.45 11.59 1.83 0.08 0.00

14465.37 16557.31 12927.29 6220.62 −416.12 −4822.53 −6371.21 −5684.07 −3929.12 −2148.97 −903.64 −268.21 −45.51 −2.23 0.00

85618.10 62116.76 29652.64 6592.42 32.91 5602.72 13994.95 18073.30 16118.72 10648.68 5177.64 1721.86 319.55 16.79 0.00

1.1961× 2.3089× 10+05 10+05

1.3445× 10+06

2.4450× 10+05

2.3870× 10+06

8.9746× 10+06

416.04 269.86 169.03 101.70 58.37 31.69 16.08 7.51 3.16 1.16 0.35 0.08 0.01 0.00 0.00

Integrating each of the above term numerically by Simpson’s 1/3rd rule we have stiffness matrix as ⎡

1.1961 × 105

⎢ ⎢ [K] = ⎢2.3089 × 105 ⎣ 2.445 × 105

2.3089 × 105 1.3445 × 106 2.387 × 106

2.445 × 105

⎤

⎥ ⎥ 2.387 × 106 ⎥ ⎦ 8.9746 × 106

Performing the eigen value analysis by any of the methods as shown in Chapter 5 (Vol. 1) we have λ

ω(rad/sec)

T(sec)

6.382 76.771 664.385

2.53 8.76 25.77

2.48 0.717 0.244

Thus for T = 2.48 sec we have Sa /g = 0.548, for T = 0.717; Sa /g = 1.896 and for T = 0.244 Sa /g = 2.5. The corresponding eigen vectors are given by 0.967 −0.25 0.052

0.165 −0.958 0.236

0.061 −0.227 0.972

ZI Sa [ϕii ]Fi (z) for the first three 2R ω2 modes and performing an SRSS we have the deflection as Now applying the expression wi (z, t) = κi

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 477

Z

d 1 (meter)

d 2 (meter)

d 3 (meter)

d (comb)

0 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

0 0.000251703 0.000966707 0.002091173 0.003590126 0.005461638 0.007748398 0.010542958 0.013984374 0.018245647 0.023513101 0.029960377 0.037720885 0.046863104 0.057373044 0.069147478 0.082000282 0.095682647 0.109916343 0.124437733 0.139049403

0 −0.00013216 −0.0005533 −0.00129389 −0.00236493 −0.00374318 −0.00535884 −0.00708952 −0.00876281 −0.01016819 −0.01107722 −0.01126932 −0.01055956 −0.00882398 −0.00601826 −0.00218613 0.00254487 0.00798104 0.01389279 0.02005545 0.02629749

0 0.00012267 0.00041479 0.00076611 0.00107735 0.00126895 0.00128912 0.0011191 0.00077447 0.00030182 −0.00022885 −0.00073501 −0.00113413 −0.00135609 −0.00135336 −0.00110751 −0.00063079 3.7726E-05 0.00084109 0.00171779 0.00261821

0 0.000309628 0.001188575 0.002575668 0.004431993 0.006741745 0.00950877 0.012754123 0.016521178 0.020889875 0.02599275 0.03201815 0.039187443 0.047705891 0.057703702 0.069190892 0.082042187 0.096014933 0.110794041 0.126055232 0.141538509

Again applying the expressions

Miz = κi βEIz

Sai

3

ωi2 H 2

i=1

ϕi μ2i Fi (z)

z

M1 (kN · m)

M2 (kN · m)

M3 (kN · m)

M(comb)

0 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

3.63×10+05 2.75×10+05 2.07×10+05 1.60×10+05 1.34×10+05 1.25×10+05 1.27×10+05 1.33×10+05 1.37×10+05 1.36×10+05 1.26×10+05 1.10×10+05 8.92×10+04 6.66×10+04 4.55×10+04 2.79×10+04 1.50×10+04 6.67×10+03 2.20×10+03 3.81×10+02 0.00×10+00

−1.74×10+05 −1.72×10+05 −1.63×10+05 −1.44×10+05 −1.13×10+05 −7.34×10+04 −3.00×10+04 1.12×10+04 4.47×10+04 6.70×10+04 7.69×10+04 7.57×10+04 6.63×10+04 5.22×10+04 3.70×10+04 2.33×10+04 1.27×10+04 5.76×10+03 1.92×10+03 3.35×10+02 0.00×10+00

1.96×10+05 1.01×10+05 2.96×10+04 −1.80×10+04 −4.43×10+04 −5.29×10+04 −4.86×10+04 −3.66×10+04 −2.18×10+04 −7.96×10+03 2.62×10+03 9.03×10+03 1.15×10+04 1.10×10+04 8.79×10+03 6.01×10+03 3.48×10+03 1.64×10+03 5.65×10+02 1.01×10+02 0.00×10+00

448077.79 340170.59 265471.40 215984.91 180701.73 154065.57 138948.07 138323.60 146038.98 151530.17 147996.14 133959.24 111706.12 85366.09 59278.77 36847.91 19963.65 8961.31 2979.07 517.31 0.00

© 2009 Taylor & Francis Group, London, UK

478 Dynamics of Structure and Foundation: 2. Applications

Considering

Viz = κi βEIz

Sai

3

ωi2 H 3

i=1

φi μ3i Fi (z)

z

V 1 (kN)

V 2 (kN)

V 3 (kN)

V (comb)

0 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

3878.895 3227.414 2217.082 859.165 −597.419 −1856.457 −2693.050 −3008.120 −2831.386 −2289.459 −1557.132 −808.201 −177.805 257.491 481.345 525.028 447.041 311.860 173.245 65.076 −0.008

2420.965 1972.271 1129.664 −97.013 −41453.868 −2636.536 −3411.105 −3670.259 −3437.311 −2834.863 −2036.228 −1216.009 −511.899 −4.045 287.389 390.325 360.932 262.670 149.620 57.113 −0.011

7221.983 6144.799 4938.797 3591.885 2226.129 997.189 30.551 −609.604 −925.257 −971.298 −832.931 −601.699 −356.071 −149.873 −9.173 63.902 83.191 69.421 42.549 16.980 −0.007

8.5477×10+03 7.2156×10+03 5.5302×10+03 3.6945×10+03 2.7251×10+03 3.3752×10+03 4.3462×10+03 4.7845×10+03 4.5484×10+03 3.7711×10+03 2.6953×10+03 1.5792×10+03 6.4842×10+02 2.9796×10+02 5.6069×10+02 6.5734×10+02 5.8055×10+02 4.1361×10+02 2.3283×10+02 8.8233×10+01 1.4923×10−02

The Bending moment diagram is shown Figure 3.3.12.

Figure 3.3.12 Bending Moment diagram of tapered chimney.

© 2009 Taylor & Francis Group, London, UK

8

0 22

6

4

Height z(m)

19

17

0

2

15

13

11

88

66

44

M1 M2 M3 M(comb) 22

5.00E+05 4.00E+05 3.00E+05 2.00E+05 1.00E+05 0.00E+00 -1.00E+05 -2.00E+05 -3.00E+05

0

Bending moment(kN-m)

Bending Moment Diagram

Analytical and design concepts for earthquake engineering 479

3.3.1.5

Computer analysis of tall chimneys

Many corporate houses have developed in-house computer program for analysis and design of tall chimneys based on IS-4998 or ACI 307, CICIND etc. The seismic analysis part however can also be done in generic finite element commercially available package like GTSTRUDL, SAP 2000, STAAD PRO etc. For dynamic analysis of tall chimneys normally a stick model with masses lumped at convenient nodes sufﬁce. The structural element constitutes of beam elements with the mass of the shell and brick lining lumped at each end nodes i and j. The computer assembles the stiffness matrix based on the principle of finite element [K] = [B]T D[B]dz and forms the lumped mass matrix [M] which is diagonal in nature. On formation of these elements it performs the eigen value analysis based on the expression [K][ϕ] = [M][ϕ]ω2 = 0 and then perform the modal analysis based on response spectrum method as explained earlier.

3.3.1.6

Discussion on factors affecting the dynamic analysis of tall chimneys

The major factors which affect the dynamic response of tall chimneys under earthquake are the code factors • • •

Z (zone factor) I (Importance factor) R (Ductility factor)

While IS code recommends the value of Z for different zones, the importance factor for chimney considering its slenderness requires special consideration while the usual practice is to apply a factor of 1.5, however for zones which a more susceptible to earthquake (like zone IV and V) it is recommended (Wilson 2003) that importance factor considered be 2.0. For structures of category 2(RCC Chimney) IS code recommend an Importance factor of 1.75. CICIND recommends ductility factor R = 1 for non ductile detailing and R = 2 when ductile detailing is to be adopted. IS-code recommends a value of R = 3.0 for RCC Chimneys.

3.3.1.7

Do we consider soil structure interaction for dynamic analysis of chimney?

This is a question which has plagued many engineers undertaking the task of design of tall chimneys. While research papers (Ghosh and Batavyal 1985, Navarro 1992, Luco 1986) and code do recommend considering this, some assessments need to be made whether it has any value addition in undertaking this complicated task. Most of the tall chimneys are structurally flexible in nature and in all possibility have its fixed base time period which would induce a base acceleration of the type k/T where k varies with the nature of soil. Considering the soil as equivalent springs based on Richart/Wolf’s formulation and correcting the time for soil-structure interaction based © 2009 Taylor & Francis Group, London, UK

480 Dynamics of Structure and Foundation: 2. Applications

on the expression $ ( ) % % ¯2 ¯ Kx h k 1+ T¯ = T &1 + Kx Kθ

(3.3.66)

where, T¯ = modiﬁed time period of the structure due to the soil stiffness; T = time period of the fixed base structure; k¯ = stiffness of the fixed base structure @ 4π 2 W/(gT 2 ); Kx , Kθ = horizontal and rotational spring constant of the soil26 ; h¯ = effective height or inertial centroid of the system, and W = total weight of the structure. It will be observed that in most of the cases the time period will further prolong and which would reduce the value of Sa/g as given in code. While one may feel happy that it would give a more economic design considering the attenuation of response, however is not true in all cases specially for chimney like structures. Firstly for such flexible structures vibrating during earthquake the acceleration at the top portion of the chimney will be subjected to much higher acceleration then the ground acceleration input we furnish in the analysis. Only if we do a time history analysis it will be observed that the acceleration at top is indeed much more than the input base acceleration. Thus forces in reality could be more at top portion then what we observe considering soil-structure interaction and further reducing the design moments and shears may not always be a safe decision, even with the soil damping attenuating the responses further at the higher mode. UBC 97 tries to cater to this phenomenon by a provision of a ﬁctitious force Ve = 0.07 V.T for time period greater than 0.7 seconds. As IS code does not have this provision, considering soil compliance may under rate the response on the top portion of the structure. Moreover as the ductility-design for these types of chimneys is still not well defined, it would perhaps be preferable to design it as a fixed base structure and render a conservative design. Unless the structure is itself so rigid that one is reasonably sure that considering soil structure interaction can amplify the response instead of attenuation. It has however been observed that soil-structure interaction analysis plays a critical role in aerodynamic response of such chimneys especially the along wind response which shows ampliﬁcation while considering the foundation compliance effect (Sadhegpour & Chowdhury 2008). 3.3.1.8 • • •

Thus to sum it up. . . .

A chimney should preferably be designed as a fixed base cantilever. Minimum three modes should be considered for dynamic analysis since the higher mode at times can give higher response in terms of shears and moments. If by three modes at least 90% mass participation is not there higher mode participation needs to be considered.

26 Refer Chapter 2 (Vol. 2) for the expressions of Kx and Kθ .

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 481

• • • •

Consider damping ratio somewhere between 2–5%. As the frequencies in most cases are widely spaced SRSS method of modal combination would sufﬁce. If ground acceleration spectra for the particular site is available on may undertake a time history analysis, when soil structure interaction can be considered. In case codal spectrum is used for design, fixed base analysis is always preferable.

3.4 ANALYSIS OF CONCRETE GRAVITY DAMS

3.4.1 Earthquake analysis of concrete dam The topic of engineering of concrete dam can itself be a subject of a book. For dams only of very large magnitude are built with concrete and obviously requires a detailed analysis as the failure risk is far too catastrophic in nature. However based on the state of art as present till date no concrete dam in the world has yet undergone a failure (except some few cracks) due to earthquake though many of them has faced seismic force as high as 1.0 g. While we are writing this section, IS-1893 (2002) code is yet to come up with procedural practice for earthquake analysis of concrete dams. As such, our discussion and benchmarking herein will be based on IS-1893 (1984) which has been in practice for last 20 years and based on which a number of dams have been built. 3.4.1.1

The IS-code method

Shown in Figure 3.4.1 is a typical concrete dam with water upstream. During an earthquake it is usually assumed that the dam vibrates on its own thus generating a force (W/g) a¨ due to its inertia. It is also subjected to a hydrodynamic force due to propagation of waves through the liquid see figure 3.4.1 that produces additional force on the dam face over and above the hydrostatic force it sustains. Water level

z

W a g Hydrodynamic pressure W

Figure 3.4.1 Typical concrete dam with water in upstream.

© 2009 Taylor & Francis Group, London, UK

482 Dynamics of Structure and Foundation: 2. Applications

The earthquake dynamic analysis for a dam is based on certain simplifying assumptions as stated hereunder • • •

The water the dam supports is incompressible. The dam is rigid and has the same motion throughout its body. The dam and the reservoir motion is uncoupled and the interaction between them is negligible for all practical purpose27 .

The code has also recommended to adapt seismic coefficient method for dam up to 100 meter high while response spectrum method with dynamic analysis (i.e. calculation of time period and taking its effect) for dams that are greater than 100 m. The seismic coefficient method is quite straightforward and does not require any elaboration as the steps are same for building analysis except the fact that importance factor to be considered should be as recommended by the code (for dams I = 3.0). For response spectrum analysis IS-1893 (1984) code recommends to derive the fundamental time period of concrete dam as

T = 5.55

H2 B

γc gEs

(3.4.1)

where, B = width of the dam base in m; H = height of the dam in m; γc = unit weight of material of dam in kg/m3 ; g = acceleration due to gravity @ 9.81 m/sec2 , and, Es = modulus of elasticity of the material of the dam in kg/m2 . However, the basis of derivation of this formula has not been elaborated either in the code or in its explanatory manual28 . Looking at the formula it appears that Eqn. (3.4.1) is derived by applying the Rayleigh Ritz method to some assumed shape functions and considering the dam as a cantilever beam having varying cross section. The code also does not give any value of time periods for higher modes whose effects are perhaps considered as non-critical. The base shear and moment for the dam is given by the expression VB = 0.6α h W; where αh = βIF 0

MB = 0.9α h W h¯

Sa g

(3.4.2) (3.4.3)

the notations in Eqn. (3.4.3) are as explained earlier in the section where we have discussed on the code IS-1893 (1984).

27 This may not be true in all cases and could have significant effect. We will study this later on. 28 Except for note that this has been developed based on some research work carried out at University of Roorkee India.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 483

As per latest version of the code the seismic design coefficient can be taken as αh =

ZI Sa 2R g

(3.4.4)

where, W = weight of the concrete dam; h¯ = height of the center of gravity of the dam, above its base. The value Sa /g is read from the chart and is a function of the time period of the dam as mentioned in Equation (3.4.1). For any horizontal section at a depth z below the top of dam the shear force Vz and bending moment Mz may be obtained from the expression Vz = Cv VB

and Mz = CM MB

(3.4.5)

where the values of Coefficients CV and CM are as furnished in figure below.

Figure 3.4.2 Values of Cv and Cm along the height of dam.

Considering a concrete dam as massive where the weight plays a significant part in its stability, it is evident that unlike other structures, the vertical mode of earthquake acceleration plays a significant part in its stability and cannot be ignored. As per IS code based on response spectrum the force due to vertical acceleration is considered as 0.75 times the value of αh at the top of the dam and reducing linearly to zero at base. 3.4.1.2

Hydrodynamic pressure on dam from the reservoir

As wave propagates through the ground there is an instantaneous hydrodynamic pressure (or suction) exerted on the dam over and above the hydrostatic pressure it sustains. © 2009 Taylor & Francis Group, London, UK

1

9

6 0.

0.

5 0.

8

4 0.

0.

3 0.

7

2 0.

0.

1

0

1.2 1 0.8 0.6 0.4 0.2 0 0.

Value of Cs

484 Dynamics of Structure and Foundation: 2. Applications

z/H

Figure 3.4.3 Value of Cs with depth.

Based on the assumption that the water is incompressible the hydrodynamic pressure at any depth z below the reservoir is determined by the expression p = Cs αh γw h

(3.4.6)

where, p = hydrodynamic pressure of water; Cs = a coefficient which varies with shape and depth; αh = seismic design coefficient as explained in Equation 3.4.4. γw = unit weight of water, and h = height of water in the reservoir. The variation of Cs with depth is given by the expression Cm Cs = 2

z z z z 2− + 2− h h h h

(3.4.7)

where Cm value varies (almost) linearly from a value of 0.735, when the vertical upstream angle of the dam face varies from 0 degree (i.e. perfectly vertical), to a value of 0.0, when this angle is 90 degree. The variation of Cs with depth is shown in Figure 3.4.3. The values in graph multiplied by the value Cm will give the value Cs . The approximate values of Shear and Moment at depth z below the free surface is given by the expression Vz = 0.726 pz

and Mz = 0.299 pz2 .

(3.4.8)

where, p = hydrodynamic pressure is given in Equation (3.4.6). 3.4.1.3

Some comments and review of the IS-code method

The above method though has been in practice for quite some time, yet there are some approximation and one assumption that is perhaps not conceptually correct.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 485

Firstly, the value of time period as proposed in Equation (3.4.1) over estimates the time period of a dam by about 25% and is not unique. The top width at crest has a significant effect on the time period29 . Secondly, while calculating the hydrodynamic pressure vide Equation (3.4.6) code uses the value of αh as obtained based on Equation (3.4.1). This is in violation to the basic assumption made at the outset that the dam face is rigid and the vibration of the two systems (dam and the fluid in the reservoir) is uncoupled/independent. In this case as per the above mentioned assumption the free field time period of the fluid in the reservoir (assumed tending to infinity) in horizontal direction should govern the value of αh g rather than the time period of the dam whose stiffness is considered as infinite in comparison to the fluid. The present assumption could lead to a significant variation in end results in some cases.

3.4.2 A method for dynamic analysis of concrete dam We present (Chowdhury & Dasgupta 2008) here a method wherein we have tried to overcome a number of deficiencies as mentioned in the above IS-code method. The salient feature of the method are summarized as hereafter 1 2

3

4

Calculation of time period of the dam having varying cross section for three modes and studying the effect of the varying section on the time period. Calculating the free field time period of the fluid in the reservoir and estimating the hydrodynamic pressure assuming the dam wall to be perfectly rigid when the two systems are not coupled. A practical simpliﬁed approach considering fluid-structure interaction (for fundamental mode analysis) wherein we study the effect of hydrodynamic pressure on the wall vis-á-vis the response of the dam when the dam is considered to have a finite stiffness and the fluid is considered compressible. How to model the fluid stiffness and mass when we carry out a finite element analysis of the dam section considering fluid structure interaction.

3.4.2.1

Calculation of time period of the dam having variable cross section

As shown in Figure 3.4.4, we show a dam of height H having base width B and top width as Bt supporting water of height Hw . The dam is assumed to be resting on firm ground (usually rock) and is considered to be fixed at base. Thus for mathematical analysis the dam is considered as a cantilever flexural member (fixed at base free at top) having varying width Bz where this variation is considered as linear with respect to H.

29 This we are going to study in some detail subsequently.

© 2009 Taylor & Francis Group, London, UK

486 Dynamics of Structure and Foundation: 2. Applications

Bt

a=∞

H

Hw

Z

B

X

Figure 3.4.4 A dam supporting water in reservoir.

As derived in the case of the chimney, solving the fourth order differential equation and differentiating the potential and kinetic energy of the system, the stiffness of the dam can be expressed as H kij =

EI(z)

d 2 ϕi (z) d 2 ϕj (z) dz2

0

dz2

dz

(3.4.9)

for i, j = 1, 2, 3 . . . . . . n and the mass matrix is expressed as ⎡ γc Az mij = ⎣ g

H

⎤ ϕi (z)ϕ j (z)dz⎦ .

(3.4.10)

0

For a cantilever beam the fourth order differential equation is EI

∂ 4w ∂z4

2 ∂ w + ρA =0 ∂t 2

(3.4.11)

where, w(z) = ϕi (z)q(t), and ϕi = sin

μi z μi z μi z μi z − sin h − αi cos − cos h H H H H

(3.4.12)

in which, μi = 1.875, 4.694, 7.855, 2m−1 2 π for i = 1, 2, 3 . . . . . . m, and αi =

sin μi + sin hμi cos μi + cos hμi

© 2009 Taylor & Francis Group, London, UK

(3.4.13)

Analytical and design concepts for earthquake engineering 487

Since the Moment of inertia and the area of the dam vary with depth. At any height z from the bottom the moment of inertia at any height z is expressed as z 3 Iz = I0 1 + ψ H

(3.4.14)

where, I0 = moment of inertia of the dam at base, and Bt − B . B

ψ=

(3.4.15)

Similarly area at any height z is given by z Az = A0 1 + ψ H

(3.4.16)

Based on above the stiffness and mass equation gets modified to

kij = EI 0 ⎡ and

H

1+ψ

0

γc A0 mij = ⎣ g

z 3 d 2 ϕi (z) d 2 ϕj (z) dz H dz2 dz2

(3.4.17)

⎤ z 1+ψ ϕi (z)ϕ j (z)dz⎦ H

H

(3.4.18)

0

The double derivative of Equation (3.4.12) is given by φi =

μ2i μi z μi z μi z μi z

−sin − sin h cos + α + cos h i H H H H H2

and

(3.4.19)

To evaluate the stiffness and mass matrix in generic form by integration we change the above to generalized co-ordinate by considering; ξ=

z H

when dξ =

dz and as z → 0, ξ → 0 and as z → H, ξ → 1 H

based on above we can now express the double derivative as μ2i − sin μi ξ − sin hμi ξ + αi (cos μi ξ + cos hμi ξ ) 2 H μ2 = i2 f (ξ )i (say) H

F (ξ )i =

© 2009 Taylor & Francis Group, London, UK

(3.4.20) (3.4.21)

488 Dynamics of Structure and Foundation: 2. Applications

Thus stiffness of the system can now be expressed as 1 EI0 μ2i μ2j

kij =

H3

(1 + ψξ )3 f (ξ )i f (ξ )j dξ

(3.4.22)

0

and mass of the system is given by γ A0 H mij = g

1 (1 + ψξ ) f (ξ )i f (ξ )j dξ ;

where, i = j = 1, 2, 3, . . . . . . . . . , m

0

(3.4.23) Thus, for the first three modes, the stiffness matrix is given by [K]ij = ⎡

EI H3

⎤ 1 μ41 (1 + ψξ )3 f (ξ )21 dξ Symmetrical ⎢ ⎥ 0 ⎢ ⎥ ⎢ ⎥ 1 1 ⎢ ⎥ × ⎢μ22 μ21 (1 + ψξ )3 f (ξ )2 f (ξ )1 dξ μ42 (1 + ψξ )3 f2 (ξ )2 dξ ⎥ ⎢ ⎥ 0 0 ⎢ ⎥ ⎣ ⎦ 1 1 1 3 3 3 2 2 2 2 2 4 μ3 μ1 (1 + ψξ ) f (ξ )3 f (ξ )1 dξ μ3 μ2 (1 + ψξ ) f (ξ )3 f (ξ )2 dξ μ3 (1 + ψξ ) f (ξ )3 dξ 0

0

0

(3.4.24) and the mass matrix30 is given by ⎡

1 (1 + ψξ ) f (ξ )21 dξ ⎢ ⎢ 0 γ AH ⎢ ⎢1 [M]ij = ⎢ (1 + ψξ ) f (ξ )2 f (ξ )1 dξ g ⎢0 ⎢ ⎣1 (1 + ψξ ) f (ξ )3 f (ξ )1 dξ 0

⎤ 1 1 0

(1 + ψξ ) f2 (ξ )2 dξ

0

(1 + ψξ ) f (ξ )3 f (ξ )2 dξ

1 0

(1 + ψξ ) f (ξ )23 dξ

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.4.25) It is apparent that the values of the stiffness and mass matrix are dependent on the parameter ψ, which would surely influence these values. As an example we solve the above for Bt /B = 0.1 (which is the most standard for concrete dams of about 100 m high) i.e. the top width is 10% of base width for some space is normally kept at the top of the dam for motor and pedestrian access, maintenance and inspection. Thus, considering ψ = −0.9, we have based on numerical integration between 1 to 0. 30 The matrix is symmetric about is diagonal.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 489

The stiffness matrix is

[K]3×3

⎡ 13.965 EI0 ⎣ = 3 24.495 H 23.911

24.495 165.575 289.015

⎤ 23.911 289.015 ⎦ 1.167 × 103

The mass matrix is given by

[M]3×3

⎡ 0.509 γc A0 H ⎣ 0.185 = g −0.025

⎤ 0.185 −0.025 0.449 0.169 ⎦ 0.169 0.522

Converting the above into standard eigen-value form of Aφ = λφ and applying the generalized Jacobi technique, we have ⎡ ⎤ 21.62866 0 0 EI0 g ⎣ ⎦ and the corresponding eigen0 255.8406 0 [λ] = γc A 0 H 4 0 0 2288.56 vectors are given as ⎡

0.986 [ϕ] = ⎣−0.167 0.020 have, ⎡

1.3509 [T] = ⎣ 0 0

⎫ ⎤T ⎧ −0.242 0.127 ⎨f1 (ξ )⎬ 0.945 −0.169⎦ f2 (ξ ) , since [λ] = ω2 and T = ⎩ ⎭ −0.218 0.977 f3 (ξ )

0 0.3928

2π ω ,

we

⎤' 0 4 ⎦ γc A 0 H 0 EI 0 g 0.13135

Now, considering width of the dam as 1 m in the Y direction (perpendicular to the plane of the paper),

A0 = B × 1m2

and I0 = 1 ×

B3 4 m . 12

Substituting the above, we have for the first three modes: Mode number 1 2 3

© 2009 Taylor & Francis Group, London, UK

Time period (secs) 2 γc T1 = 4.68 HB Eg 2 γc T2 = 1.36 HB Eg 2 γc T3 = 0.455 HB Eg

490 Dynamics of Structure and Foundation: 2. Applications 2

The above can be generalized to an expression, T1 = CT HB

γc . Eg

We present here the values of CT for the first three modes for various values of Bt /B Bt /B 0.05

0.1

0.15

0.2

0.25

0.3

Eigen value

Natural frequency (rad/sec)

Time period factor (CT ) in form as presented in code

24.306 261.863 2348 21.62866 255.84 2288 19.807 257.77 2268 18.485 264.047 2270 17.477 273.01 2289 16.678 283.7957 2322

4.93011156 16.1821816 48.4561658 4.65066232 15.9949992 47.833043 4.45050559 16.0552172 47.6235236 4.29941857 16.2495231 47.644517 4.1805502 16.5230143 47.8434949 4.08387071 16.846237 48.1871352

4.41482756 1.34503449 0.44918107 4.68010595 1.36077483 0.45503257 4.89058871 1.355671 0.45703448 5.0624502 1.33946038 0.4568331 5.20639421 1.31728945 0.45493316 5.32964775 1.29201509 0.45168886

To substantiate further, we have also compared the fundamental natural frequency of a tapered cantilever beam of various width ratio (Bt /B) derived by exact solution (Karnovsky & Lebed 2001) with the method as proposed above and compared hereafter. Bt /B

Proposed analysis

Exact solution

0 0.1 0.2 0.3

5.348832 4.650806 4.299419 4.083871

5.3151 4.6307 4.2925 4.0817

3.4.2.1.1

Comparison of natural frequency (rad/sec) proposed and exact analysis

It is evident from above that the value of time period as given in code is over estimated31 . For even with Bt /B as 0.3 (which is very high) the time period coefﬁcient is 5.33 in lieu of 5.55. While for a 10% crest width the error is of the order of (+)19%.

31 Though we agree that this may not have much effect on Sa /g finally in some cases.

© 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 491

3.4.2.2

Fundamental uncoupled amplitudes of vibration of the dam

The maximum amplitude of vibration as per modal response analysis is expressed as Sd = Sa /ω2 , thus based on provisions of code it may be further written as Sd = κi β where κi =

Sa ω2

(3.4.26)

n

i=1 mi φi /

n

2 i=1 mi φi

, 1

1 κi =

for the present problem this is expressed as

(1 + ψξ ) fi (ξ )2 dξ

(1 + ψξ )fi (ξ )dξ 0

0

For a typical value of Bt /B = 0.1, the κi values for the three modes are given as: ZI κ1 = 0.726, κ2 = 0.771 and κ2 = 0.432, β = 2R , and the code factors remain the same as explained earlier. The displacement along the height z is thus expressed as w(z) = κi β

Sa [φ11 f1 (z) + φ12 f2 (z) + φ13 f3 (z)] ω2

(3.4.27)

where, fi (z) = sin μHi z − sin h μHi z − αi cos μHi z − cosh μHi z . 2 γc in which, CT = 4.68, 1.36, 0.455 etc. and also Considering, T1 = CT HB Eg considering, ω = 2π/T, we have finally, w(z) =

κi βCT2 Sa γc H 4 [φ11 f1 (z) + φ12 f2 (z) + φ13 f3 (z)] 4π 2 B2 E g

(3.4.28)

where for Bt /B = 0.1, the displacement for fundamental mode may be expressed as w(z) =

3.4.2.3

0.403κi β Sa γc H 4 [0.986f1 (z) − 0.167f2 (z) + 0.020f3 (z)]. B2 E g

(3.4.29)

Dynamic Bending Moment and Shear force for the fundamental mode

For a beam element the bending moment is given by M(z) = −EIz

d2w , dz2

© 2009 Taylor & Francis Group, London, UK

(3.4.30)

492 Dynamics of Structure and Foundation: 2. Applications

for the present case, this may be expressed as z 3 κi βC 2T H 4 γc Sa d 2 M(z) = −EI 0 1 + ψ (φ11 f1 (z) + φ12 f2 (z) + φ13 f3 (z)) H g dz2 4π 2 B2 E κi βCT2 Sa z 3 γc BH 2 1 + ψ =− (φ11 μ21 f1 (z) + φ12 μ22 f2 (z) + φ13 μ23 f3 (z)) 2 g H 48π (3.4.31) For Bt /B = 0.1, Equation (3.4.30) is expressed as M(z) = −0.0335β

Sa g

z 3 γc BH 2 1 + ψ H

× (3.466f1 (z) − 3.679f2 (z) + 1.234f3 (z))

(3.4.32)

Similarly, the shear force is written as V(z) =

dM(z) dz

=

κi βCT2 48π 2

Sa z 3 γc BH 1 + ψ (φ11 μ31 f1 (z) + φ12 μ32 f2 (z) + φ13 μ33 f3 (z)) g H

➔ V(z) = −0.0335

κi βC 2T 48π

2

Sa g

z 3 γc BH 1 + ψ H

× (6.5f1 (z) − 17.27φ12 f2 (z) + 9.693f3 (z))

(3.4.33)

The values of f (z) and f (z) are as given hereafter Z/H

f1 (z)

f2 (z)

f3 (z)

f1 (Z)

f2 (Z)

f3 (Z)

0 0.2 0.4 0.6 0.8 1

2.72444111 1.9765624 1.2564103 0.62639921 0.17408595 0

1.96373508 0.13756457 −1.1575438 −1.3421761 −0.5912631 0

2.001552 −0.790419 −0.948172 1.052816 1.209796 0

−2 −1.977702 −1.834738 −1.485855 −0.881407 −0.000284

−2 −1.793709 −0.844848 0.423056 0.972256 0.000179

−2 −1.20662 0.966178 1.034407 −0.79471 0.000486

3.4.2.4 Free f ield time period of the reservoir and the hydrodynamic pressure from reservoir In this section we determine the free field time period of the water extending to infinity in horizontal direction and having a depth Hw as shown in Figure 3.4.4. As assumed by the code we presume the dam wall is acting rigidly with respect to the fluid and the fluid vibration remains uncoupled with respect to the vibration of the dam. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 493

Since the dam profile is considered in two dimensions we start with two dimensional propagation of wave due to earthquake through the water. This is expressed as 1 ∂ 2 w(x, z, t) ∂ 2 w(x, z, t) ∂ 2 w(x, z, t) + = 2 2 2 ∂x ∂z c ∂t 2

(3.4.34)

in which, c = velocity of sound in water and is expressed c = Bm /ρw , normally taken as 1439 m/sec; where, Bm = bulk modulus of water (usually considered as 2.11 × 106 kN/m2 ), and ρw = mass density of water and w = displacement of fluid medium. We will not solve this problem in detail here since we have already solved an identical problem having same boundary condition for estimation of dynamic pressure due to earthquake on a rigid wall later under the topic of earth retaining structures (Section 3.7.1). Based on this analysis the fundamental frequency of the reservoir can be expressed as ω1 = cπ/(2Hw )(Hw = height of water in the reservoir). Considering, T = 2π/ω, we have, T = 4Hw /c and the eigenvector is expressed as , where n = 1, 2, 3. . .; the number of modes. φ(z) = cos (2n−1)πz 2H Based on modal response technique, the maximum amplitude function can be defined as Sd =

Sa ω2

(3.4.35)

where Sd = maximum displacement; Sa = acceleration which is the function of time period 4Hw /c, and can be read off from normalized response given in the code. Considering β = ZI/2R a code factor depending on earthquake zone (low, moderate, severe etc.), Importance factor, ductility factor etc. (here R may be considered as 1.5 for un-reinforced concrete), we can write w(z) = κi β

Sa ϕ(z) ω2

(3.4.36)

Now substituting the value of ω and using Bm = ρw c2 , we have w(z) = =

2 Sa Hw πz 4 κ β cos i 2H π2 c2 2 S a γw H w 4 πz κ β cos i 2 Bm g 2H π

where γw = unit weight of water, κi = modal mass participation factor = g = acceleration due to gravity. © 2009 Taylor & Francis Group, London, UK

(3.4.37) mϕ i 2i , mi ϕi

and,

494 Dynamics of Structure and Foundation: 2. Applications

Thus, the modal participation factor can be considered as

κi =

mi ϕi /

⎛H ⎞1 ⎛ H ⎞ πz ⎠ ⎝ γw z cos2 π z dz⎠ mi ϕi2 = ⎝ γw z cos dz 2H 2H 0

0

The above on integration by parts gives, κi =

Thus,

w(z) =

8 π+2

32 S a γw H 2 πz β cos 2 B 2H g π (π + 2) m

(3.4.39)

The deformation in fluid in z direction is given by [as εz = εz =

(3.4.38)

∂w ∂z ]:

16 S a γw H w πz β cos π (π + 2) Bm g 2H

(3.4.40)

Thus dynamic pressure is given by (considering 15% damping for fluid) pdyn = Bm

∂u 12 S a γw H w πz =− β sin , ∂z π (π + 2) g 2H

(3.4.41)

the negative sign indicates that the pressure is acting on the dam. Here the factor 16 in Equation (3.4.40) is multiplied by a factor 0.75 as per IS-1893 (2002) to cater to 15% damping ratio. The variation of pressure is as shown in Figure 3.4.5. Based on above the pressure can be expressed as pdyn = −Coeff

βSa γw Hw g

(3.4.42)

z/H

Figure 3.4.5 Variation of hydrodynamic pressure with depth.

© 2009 Taylor & Francis Group, London, UK

1

0. 8

0. 6

0. 4

0. 2

1.2 1 0.8 0.6 0.4 0.2 0 0

pressure coefficient

where the coefficients can be read from the graph in Figure 3.4.5.

Analytical and design concepts for earthquake engineering 495

Z

dy

dz dx Y X

Figure 3.4.6 An elastic elemental body in space.

Considering Vz =

Hw 0

pz dz, the shear force on the dam face is

Vdyn = −0.645pz × z,

and the moment is given by

Mdyn = −0.4014pz × z2

(3.4.43) (3.4.44)

where, pz = pressure at any point z from top of the dam, and z = distance from top of the dam. 3.4.2.5

A f luid-structure interaction model for earthquake analysis of dam

In previous section we had proposed a technique for dynamic analysis where the vibration is uncoupled. By this we mean the two vibrations are independent of each other. The stiffness of one system does not affect the stiffness of the other vis-a-vis the frequency of each of the system is also mutually exclusive. In reality this is only an idealization and the fluid and structure do have finite stiffness and vibrate in coupled mode. We present a here two practical easy to apply model based on which such fluid structure interaction can be carried out. There are off course sophisticated finite element models available where the fluid and the structure could both be modeled in 2D for a comprehensive analysis but is not without its lacunae. Firstly such analysis is quite expensive and requires a significant large model for problems such as dams. Moreover fluid elements having no shear strength and being almost an incompressible medium often gives rise to numerical difficulty in arriving at a meaningful solution. In the present model as proposed herein we use a simplified lumped mass–spring based model. 3.4.2.6 Determination of the f luid spring and mass We show in Figure 3.4.6, an elemental body in space of length dx, dy and dz. © 2009 Taylor & Francis Group, London, UK

496 Dynamics of Structure and Foundation: 2. Applications

For such a body under load, the strain energy equation as per theory of elasticity in three dimensions is given by the expression V=

G λe2 2 2 2 + G εx2 + εy2 + εz2 + γxy + γyz + γxz 2 2

(3.4.45)

If the body is a fluid medium then as G = 0, we have, V = (Bm e2 )/2 where, Bm = bulk modulus of water, and e = εx + εy + εz, thus V=

Bm (εx + εy + εz )2 2

(3.4.46)

For two dimensional case for a reservoir as x → ∞ in horizontal direction, we have εx = 0, and hence, V=

Bm (εy + εz )2 2

(3.4.47)

Similarly, since we are only considering unit width of the dam in the direction perpendicular to the paper, the displacement is invariant in this direction which gives εy = 0 and we are finally left with V=

Bm εz2 Bm = 2 2

∂w ∂z

2 (3.4.48)

Considering, w(z) = φ(z), q(t) one can have ∂V ∂w ∂ = Bm ∂qr ∂z ∂qr ➔

∂w ∂z

∂V ∂ϕi ∂ϕr qi qr = Bm ∂qr ∂z ∂z

mf =

Kf =

Hw2 2g

8

Figure 3.4.7 Equivalent stiffness and lumped mass of the fluid element.

© 2009 Taylor & Francis Group, London, UK

(3.4.49)

Analytical and design concepts for earthquake engineering 497

where, φ(z) = generalized shape function with respect to the x and z co-ordinates, and q(t) = displacement function with respect to time in the generalized coordinate. From which it can be proved that the stiffness and mass matrix can be written as Hw ∂ϕi ∂ϕr Kir = Bm dz ∂z ∂z

γw and Mir = g

0

Hw ϕi ϕr dz

(3.4.50)

0

where K = stiffness matrix of the fluid medium; M = mass matrix of the fluid medium; i and r are different modes 1, 2, 3. . . K and M for the fundamental mode are given by Hw K11 = Bm 0

∂ϕ ∂z

2 · dz

and M11

γw = g

Hw (ϕ)2 dz

(3.4.51)

0

and substituting it in Considering the shape function as given in ϕ(z) = cos (2n−1)πz 2H w Equation (3.4.50) and by integrating, we have

K11 =

π 2 Bm 8H w

and M11 =

γw Hw 2g

It may be observed that the unit of the stiffness derived here is kN/m2 , which means that the expression gives stiffness per unit area. Thus to determine the total stiffness of the water in contact with dame face one has to multiply this by the contact area which in this case is Hw x1. 2 This gives K11 = π 8Bm kN/m per meter width. γ H2

Similarly the effective mass is given by M11 = s2gw kN · sec2 /m per meter width. Considering, T = 2π M/K and substituting it in the above expression, one can arrive at the same expression, T = 4Hw /c, as was derived earlier. This shows that the stiffness and mass matrix formulation as represented here is dimensionally correct. Based on above as shown in Figure 3.4.7, we have managed to derive an equivalent stiffness and lumped mass of the fluid contained by the dam whose fundamental frequency matches with the free field time period of the fluid continuum. 3.4.2.7

A semi-analytic model for f luid structure interaction

The spring element derived above is obtained per unit area this means this spring is distributed on the surface of the dam. © 2009 Taylor & Francis Group, London, UK

498 Dynamics of Structure and Foundation: 2. Applications

The potential energy d of an element of depth dz, of the dam section and the fluid shown in Figure 3.4.3, is then given by

E c Iz d = 2

d2w dz

2

2 +

Kf 2

w2

(3.4.52)

where, Ec = Young’s modulus of the concrete dam; Iz = moment of inertia of dam B3 expressed 12 (1 + ψ Hz )3 Kf = fluid stiffness; w = displacement of the fluid – dam system and may be written as [φ(z)q(t)]. The total potential energy over the height (H) of the dam is then given by

2 Hw π 2 Bm z 3 d 2 v dz + v2 dz 1+ψ H 16H w dz2

H

Ec I0 = 2

0

(3.4.53)

0

Considering w(z, t) = φ(z)q(t), the stiffness expression becomes

Kij = Ec Io

H

Hw φi (z)φj (z)dz

0

0

π 2 Bm z 3 ϕi (z)φ j (z)dz + 1+ψ H 8H w

(3.4.54)

Thus, for the fundamental mode, Equation (3.4.54) may be written as

K = Ec Io

H

Hw ϕ(z)2 dz

0

0

π 2 Bm z 3 2 ϕ (z) dz + 1+ψ H 8H w

(3.4.55)

It has been shown by Timoshenko that when a beam is supported by distributed spring it only affects the natural frequency while the mode shape remains unaltered with respect to that the original beam (i.e. in air), as such like in the case of the uncoupled dynamic analysis carried out previously (where effect of water was ignored) we consider ϕi = sin

μi z μi z μi z μi z − sin − αi cos − cos h H H H H

where, μi = 1.875, 4.694, 7.855, 2m−1 2 π; for i = 1, 2, 3. . .. . . m. © 2009 Taylor & Francis Group, London, UK

(3.4.56)

Analytical and design concepts for earthquake engineering 499

Thus, in natural co-ordinates, the stiffness expression can be expressed as Ec I0 μ41 K= H3

1

π 2 Bm (1 + ψξ ) f (ξ ) dξ + 8 3

1

2

0

f (ξ )2 dξ

(3.4.57)

0

The above can be simplified to ⎡ 1 ⎤ Ec I0 μ41 ⎣ (1 + ψξ )3 f (ξ )2 dξ + χs f (ξ )2 dξ ⎦ K= H3

(3.4.58)

0

where, χs = 1.2

Bm Ec

H B

, a constant number.

Thus, K may be written as

K=

Ec I0 μ41 [I1 + χs I2 ] H3

(3.4.59)

1 1 where, I1 = 0 (1 + ψξ )f (ξ )2 dξ and I2 = 0 f (ξ )2 dξ . Similarly the mass coefficient based on kinetic energy principle can be as expressed as

γc A0 M= g

H

Hw ϕ(z)2 dz

0

0

z γ w Hw 1+ψ ϕ(z)2 dz + H 2g

(3.4.60)

The above in natural co-ordinate can now be expressed as γc A0 H M= g

1

2 γ w Hw (1 + ψξ ) f (ξ ) dξ + 2g

1

2

0

f (ξ )2 dξ

(3.4.61)

0

This can be further expressed as ⎤ ⎡ 1 1 γc A0 H ⎣ M= (1 + ψξ ) f (ξ )2 dξ + χm f (ξ )2 dξ ⎦ g 0

where, χm =

1 5

Hw H

0

2 H B

© 2009 Taylor & Francis Group, London, UK

, a constant number.

(3.4.62)

500 Dynamics of Structure and Foundation: 2. Applications

Thus, M may be expressed as γc A 0 H M= [I3 + χm I2 ] ; g

1 where, I3 =

(1 + ψξ )f (ξ )2 dξ

(3.4.63)

0

Now considering the expression, T = 2π M K and substituting it in the above expressions and after